Calculo 1, Solucionario (solo Pares) - Ron Larson - 9na Edición
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C H A P T E R 1 Limits and Their Properties Section 1.1
A Preview of Calculus . . . . . . . . . . . . . . . . . . . 305
Section 1.2
Finding Limits Graphically and Numerically . . . . . . . 305
Section 1.3
Evaluating Limits Analytically . . . . . . . . . . . . . . . 309
Section 1.4
Continuity and One-Sided Limits
Section 1.5
Infinite Limits
Review Exercises
. . . . . . . . . . . . . 315
. . . . . . . . . . . . . . . . . . . . . . . 320
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327
C H A P T E R 1 Limits and Their Properties Section 1.1
A Preview of Calculus
Solutions to Even-Numbered Exercises 4. Precalculus: rate of change slope 0.08
2. Calculus: velocity is not constant Distance 20 ftsec15 seconds 300 feet 6. Precalculus: Area 2
8. Precalculus: Volume 326 54
2
2 5 5 5 10.417 2 3 4 5 5 5 5 5 5 5 1 9.145 Area 5 2 1.5 2 2.5 3 3.5 4 4.5
10. (a) Area 5
(b) You could improve the approximation by using more rectangles.
Section 1.2 2.
x
1.9
1.99
1.999
2.001
2.01
2.1
f x
0.2564
0.2506
0.2501
0.2499
0.2494
0.2439
lim
x→2
4.
x2 0.25 x2 4 3.1
3.01
3.001
2.999
2.99
2.9
f x
0.2485
0.2498
0.2500
0.2500
0.2502
0.2516
lim
1 x 2
x3
Actual limit is 14 .
0.25
x
3.9
3.99
3.999
4.001
4.01
4.1
f x
0.0408
0.0401
0.0400
0.0400
0.0399
0.0392
lim
x→4
8.
Actual limit is 14 .
x
x→3
6.
Finding Limits Graphically and Numerically
xx 1 45 0.04 x4 0.1
x f x lim
x→0
0.0500
0.01 0.0050
cos x 1 0.0000 x
Actual limit is 251 .
0.001 0.0005
0.001
0.01
0.0005
0.0050
0.1 0.0500
(Actual limit is 0.) (Make sure you use radian mode.)
305
306
Chapter 1
Limits and Their Properties
10. lim x2 2 3
12. lim f x lim x2 2 3
x→1
x→1
1 does not exist since the x3 function increases and decreases without bound as x approaches 3.
x→1
16. lim sec x 1
14. lim
x→3
18. lim sinx 0
x→0
x→1
20. Ct 0.35 0.12 t 1 (a)
1
0
5 0
(b)
t Ct
3
3.3
3.4
3.5
3.6
3.7
4
0.59
0.71
0.71
0.71
0.71
0.71
0.71
lim Ct 0.71
t→3.5
(c)
3
2.5
2.9
3
3.1
3.5
4
0.47
0.59
0.59
0.59
0.71
0.71
0.71
t Ct
lim Ct does not exist. The values of C jump from 0.59 to 0.71 at t 3.
t→3.5
22. You need to find such that 0 < x 2 < implies f x 3 x2 1 3 x2 4 < 0.2. That is,
0.2 4 0.2 3.8 3.8 3.8 2
< x2 4 < x2 < x2 < x < x2
< < < < <
0.2 4 0.2 4.2 4.2 4.2 2
So take 4.2 2 0.0494.
Then 0 < x 2 < implies 4.2 2 < x 2 < 4.2 2 3.8 2 < x 2 < 4.2 2.
Using the first series of equivalent inequalities, you obtain
f x 3 x2 4 < 0.2.
24. lim 4 x→4
x 2 2
4
x 2 < 0.01 2 2
x < 0.01 2
1 x 4 < 0.01 2
Hence, if 0 < x 4 < 0.02, you have 0 < x 4 < 0.02
1 x 4 < 0.01 2 2
4
x < 0.01 2
x 2 < 0.01 2
f x L < 0.01
Section 1.2 26. lim x2 4 29
28. lim 2x 5 1
x→5 2
x→3
x 4 29 < 0.01
Finding Limits Graphically and Numerically
Given > 0:
x2 25 < 0.01
2x 5 1 <
2x 6 < 2 x 3 <
x 5x 5 < 0.01 0.01
x 5 < x 5
x 3
If we assume 4 < x < 6, then 0.0111 0.0009.
0.01 , you have 11
x5 <
0.01 1 < 0.01 11 x5
Hence, if 0 < x 5 <
x 5
x 5 < 0.01
x2 25 < 0.01
x2 4 29 < 0.01
f x L < 0.01
Hence, if 0 < x 3 < , you have 2
9
2 3
x 1
2 3x
29 3
2 3
2x 6 <
2x 5 1 <
f x L < 32. lim 1 1 x→2
Given > 0:
0, you have
<
1 1 <
f x L <
< 32
Hence, let 32.
x 3 < 2
x→1
2 3x
2
Hence, let 2.
2 2 29 30. lim 3 x 9 3 1 9 3
Given > 0:
<
3 Hence, if 0 < x 1 < 2, you have
2 3x
x 1 < 32
2 3x
9
2 3
29 3
< <
f x L <
34. lim x 4 2 x→4
Given > 0:
x 2
x 2 <
x 4 <
x→3
x 2 <
x 2
x 3 0
x 3
0 < x 4 < 3 ⇒ x 4 < x 2
Given > 0:
x 2
Assuming 1 < x < 9, you can choose 3. Then,
36. lim x 3 0
⇒ x 2 < .
< <
Hence, let .
Hence for 0 < x 3 < , you have
x 3 <
x 3 0 <
f x L <
307
308
Chapter 1
Limits and Their Properties
40. f x
38. lim x2 3x 0 x→3
Given > 0:
x2 3x 0
xx 3
x2
x3 4x 3 −3
1 2
lim f x
<
4
x→3
5
−4
<
The domain is all x 1, 3. The graphing utility does not show the hole at 3, 12 .
x 3 < x If we assume 4 < x < 2, then 4.
Hence for 0 < x 3 < , you have 4
1
1
x 3 < 4 < x
xx 3 <
x 3x 0 <
f x L < 2
42. f x
x3 x2 9
lim f x
x→3
1 6
44. (a) No. The fact that f 2 4 has no bearing on the existence of the limit of f x as x approaches 2.
3
−9
(b) No. The fact that lim f x 4 has no bearing on the x→2 value of f at 2.
3
−3
The domain is all x ± 3. The graphing utility does not show the hole at 3, 16 . 46. Let px be the atmospheric pressure in a plane at altitude x (in feet).
48.
0.002
(1.999, 0.001) (2.001, 0.001)
lim px 14.7 lbin2
x→0
1.998 0
2.002
Using the zoom and trace feature, 0.001. That is, for
0 < x 2 < 0.001,
50. True
x2 4 4 < 0.001. x2
52. False; let f x
x10, 4x, 2
x4 x4
.
lim f x lim x2 4x 0 and f 4 10 0
x→4
x2 x 12 7 x→4 x4
54. lim
n
4 0.1 n
f 4 0.1 n
n
4 0.1 n
x→4
f 4 0.1 n
1
4.1
7.1
1
3.9
6.9
2
4.01
7.01
2
3.99
6.99
3
4.001
7.001
3
3.999
6.999
4
4.0001
7.0001
4
3.9999
6.9999
Section 1.3
56. f x mx b, m 0. Let > 0 be given. Take
If 0 < x c <
. m
xc
, then m
That is, 12L < gx L < 12L 1 2L
gx
<
3
< 2L
Hence for x in the interval c , c , x c, 1 gx > 2L > 0.
x→c
Evaluating Limits Analytically (a) lim gx 2.4
10
x→4
(b) lim gx 4 x→0
0
4.
(a) lim f t 0
10
t→4
(b) lim f t 5
−5
t→1
10
10
−5
− 10
gx
12 x 3 x9
8. lim 3x 2 33 2 7 x→3
x→2
10. lim x2 1 12 1 0 x→1
14. lim
x→3
18. lim
x→3
2 2 2 x 2 3 2
x 1
x4
3 1
34
f t t t 4
6. lim x3 23 8
2
22. lim 2x 13 20 13 1 x→0
12. lim 3x3 2x2 4 313 212 4 5 x→1
16. lim
x→3
2x 3 23 3 3 x5 35 8
3 3 x 4 442 20. lim x→4
24. (a) lim f x 3 7 4 x→3
(b) lim gx 42 16 x→4
(c) lim g f x g4 16 x→3
26. (a) lim f x 242 34 1 21 x→4
3 21 6 3 (b) lim gx
28. lim tan x tan 0 x→
x→21
(c) lim g f x g21 3 x→4
30. lim sin x→1
34.
1 such that 0 < x 0 < implies gx L < 2L.
which shows that lim mx b mc b.
2.
x sin 1 2 2
lim cos x cos
x→53
5 1 3 2
309
1 58. lim gx L, L > 0. Let 2L. There exists > 0
mx c < mx mc < mx b mc b <
Section 1.3
Evaluating Limits Analytically
32. lim cos 3x cos 3 1 x→
36. lim sec x→7
6x sec 76 23
3
310
Chapter 1
Limits and Their Properties
38. (a) lim 4f x 4 lim f x 4 x→c
x→c
32 6
3 3 3 lim f x 27 3 f x 40. (a) lim x→c
(b) lim f x gx lim f x lim gx x→c
x→c
x→c
(c) lim f xgx lim f x lim gx x→c
x→c
x→c
x→c
lim f x x→c f x 27 3 (b) lim x→c 18 lim 18 18 2
3 1 2 2 2
x→c
(c) lim f x lim f x2 272 729
32 12 43
2
x→c
lim f x 32 f x x→c 3 x→c gx lim gx 12
x→c
(d) lim f x23 lim f x23 2723 9
(d) lim
x→c
x→c
x→c
42. f x x 3 and hx
x2 3x agree except at x 0. x
44. gx
1 x and f x 2 agree except at x 0. x1 x x
(a) lim hx lim f x 5
(a) lim f x does not exist.
(b) lim hx lim f x 3
(b) lim f x 1
x→2
x→1
x→2
x→0
x→0
x→0
2x2 x 3 and gx 2x 3 agree except at x1 x 1.
x3 1 and gx x2 x 1 agree except at x1 x 1.
46. f x
48. f x
lim f x lim gx 5
x→1
lim f x lim gx 3
x→1
x→1
x→1
7
4
−8
4
−4
50. lim
x→2
2x x 2 lim x2 4 x→2 x 2x 2 lim
x→2
54. lim
x
lim
x→0
x 1 2
x→3
x3
x→4
x2 5x 4 x 4x 1 lim x2 2x 8 x→4 x 4x 2
lim
x→3
lim
x→4
2 x 2
x
x→0
lim
56. lim
52. lim
1 1 x2 4
2 x 2
x→0
4 −1
−8
2x2
2 x 2 2 x 2
2 x 2 x
x 1 2
x3
x 1 3 1 x 2 6 2
lim
x→0
1 2 x 2
x 1 2 x 1 2
lim
x→3
2 1 4 22
x3 1 1 lim x 3x 1 2 x→3 x 1 2 4
1 1 4 x 4 x4 4 4x 4 1 1 lim lim 58. lim x→0 x→0 x→0 4x 4 x x 16
60. lim
x→0
x x2 x2 x2 2x x x2 x2
x2x x lim lim lim 2x x 2x
x→0
x→0
x→0
x
x
x
Section 1.3
62. lim
x→0
Evaluating Limits Analytically
311
x x3 x3 x3 3x2 x 3x x2 x3 x3 lim
x→0
x
x lim
x→0
64. f x
x3x2 3x x x2 lim 3x2 3x x x2 3x2
x→0
x
4 x x 16
1
0
x
15.9
15.99
15.999
16
16.001
16.01
16.1
f x
.1252
.125
.125
?
.125
.125
.1248
20
−1
It appears that the limit is 0.125.
4 x 4 x lim x→16 x 16 x→16 x 4 x 4
Analytically, lim
lim
x→16
1 x 4
1 . 8
x5 32 80 x→2 x 2
100
66. lim
x f x
1.9
1.99
72.39
79.20
1.999 79.92
1.9999
2.0
79.99
?
2.0001 80.01
2.001 80.08
2.01
2.1
80.80
88.41
x5 32 x 2x4 2x3 4x2 8x 16 lim x→2 x 2 x→2 x2
Analytically, lim
lim x4 2x3 4x2 8x 16 80. x→2
(Hint: Use long division to factor x5 32.)
68. lim
x→0
31 cos x 1 cos x lim 3 x→0 x x
30 0
sin x tan2 x sin2 x lim lim 2 x→0 x→0 x cos x x→0 x x
72. lim
cos2 x sin x
10 0
76. lim
x→ 4
1 tan x cos x sin x lim sin x cos x x→4 sin x cos x cos2 x sin x cos x lim x→ 4 cos xsin x cos x 1 lim x→ 4 cos x lim sec x x→ 4
2
78. lim
x→0
sin 2x sin 2x lim 2 sin 3x x→0 2x
13 sin3x3x 21 13 1 32
70. lim
→0
cos tan sin lim 1 →0
74. lim sec 1 →
−4
3 −25
312
Chapter 1
Limits and Their Properties
80. f h 1 cos 2h 0.1
h f h
4
0.01
1.98
0.001
1.9998
2
0
0.001
0.01
0.1
?
2
1.9998
1.98
−5
5
−4
Analytically, lim 1 cos 2h 1 cos0 1 1 2.
The limit appear to equal 2.
h→0
82. f x
sin x 3 x
2
−3
0.1
x f x
0.01
0.215
0.0464
0.001
0
0.001
0.01
0.1
0.01
?
0.01
0.0464
0.215
−2
The limit appear to equal 0.
sin x 3 2 sin x lim x 01 0. 3 x→0 x→0 x x
Analytically, lim
84. lim
h→0
x h x x h x f x h f x lim lim h→0 h→0 h h h
lim
h→0
3
x h x x h x
xhx 1 1 lim h→0 h x h x 2x x h x
f x h f x x2 2xh h2 4x 4h x2 4x x h2 4x h x2 4x lim lim h→0 h→0 h→0 h h h
86. lim
lim
h→0
h2x h 4 lim 2x h 4 2x 4 h→0 h
88. lim b x a ≤ lim f x ≤ lim b x a x→a
x→a
x→a
90. f x x sin x 6
b ≤ lim f x ≤ b x→a
Therefore, lim f x b. x→a
− 2
2 −2
lim x sin x 0
x→0
92. f x x cos x
94. hx x cos
6
− 2
0.5
2
− 0.5
0.5
−6
− 0.5
lim x cos x 0
x→0
1 x
lim x cos
x→0
1 0 x
Section 1.3 x2 1 and gx x 1 agree at all points x1 except x 1.
96. f x
Evaluating Limits Analytically
98. If a function f is squeezed between two functions h and g, hx ≤ f x ≤ gx, and h and g have the same limit L as x → c, then lim f x exists and equals L. x→c
100. f x x, gx sin2 x, hx
sin2 x x When you are “close to” 0 the magnitude of g is “smaller” than the magnitude of f and the magnitude of g is approaching zero “faster” than the magnitude of f. Thus, g f 0 when x is “close to” 0
2
g −3
3
h f −2
102. st 16t2 1000 0 when t s lim
t→5102
5 210 st
5 10 seconds 1000 16 2
510 2
t
104. 4.9t2 150 0 when t
lim
t→5102
0 16t2 1000 510 t 2
16 t2 lim
t→5102
lim
t→5102
125 2
510 t 2
16 t
t 5 210
t 5 10
2
16 t lim
t→5102
510 2
510 8010 ftsec 253 ftsec 2
1500
5.53 seconds. 150 4.9 49
The velocity at time t a is sa st 4.9a2 150 4.9t2 150 4.9a ta t lim lim t→a t→a t→a at at at
lim
lim 4.9a t 2a4.9 9.8a msec. t→a
Hence, if a 150049, the velocity is 9.8150049 54.2 msec. 106. Suppose, on the contrary, that lim gx exists. Then, since lim f x exists, so would lim f x gx, which is a x→c
x→c
contradiction. Hence, lim gx does not exist.
x→c
x→c
108. Given f x x n, n is a positive integer, then lim x n lim x x n1 lim xlim x n1
x→c
x→c
x→c
x→c
c lim x x n2 clim xlim x n2 x→c
x→c
cclim x→c
x→c
. . . c n.
x x n3
110. Given lim f x 0: x→c
For every > 0, there exists > 0 such that f x 0 < whenever 0 < x c < .
Now f x 0 f x
313
f x 0 < for x c < . Therefore, lim f x 0. x→c
314
Chapter 1
Limits and Their Properties
f x ≤ f x ≤ f x lim f x ≤ lim f x ≤ lim f x x→c x→c x→c
112. (a) If lim f x 0, then lim f x 0. x→c
x→c
0 ≤ lim f x ≤ 0 x→c
Therefore, lim f x 0. x→c
(b) Given lim f x L: x→c
For every > 0, there exists > 0 such that f x L < whenever 0 < x c < . Since f x L ≤ f x L < for x c < , then lim f x L .
x→c
116. False. Let f x
114. True. lim x3 03 0 x→0
3x xx 11
,c1
Then lim f x 1 but f 1 1. x→1
1 118. False. Let f x 2 x2 and gx x2. Then f x < gx for all x 0. But lim f x lim gx 0. x→0
120. lim
x→0
1 cos x 1 cos x lim x→0 x x
x→0
1 cos x
1 cos x
1 cos2 x sin2 x lim x→0 x1 cos x x →0 x1 cos x
lim lim
x→0
sin x x
lim
x→0
sin x
1 cos x
sin x x
lim 1 sincosx x x→0
10 0
122. f x
sec x 1 x2
(a) The domain of f is all x 0, 2 n. (b)
x→0
2
(d) − 3 2
3 2
1 2
(c) lim f x
sec x 1 sec x 1 x2 x2
−2
The domain is not obvious. The hole at x 0 is not apparent.
Hence, lim
x→0
sec x 1
1 sin2 x 1 tan2 x x2sec x 1 cos2 x x2 sec x 1
sec x 1 1 sin2 x 1 lim 2 x→0 x cos2 x x2 sec x 1
11
124. The calculator was set in degree mode, instead of radian mode.
sec2 x 1
sec x 1 x2sec x 1
12 21.
Section 1.4
Section 1.4 2. (a) (b)
Continuity and One-Sided Limits
315
Continuity and One-Sided Limits
lim f x 2
4. (a)
lim f x 2
(b)
x→2
x→2
lim f x 2
6. (a)
lim f x 2
(b)
x→2
x→2
lim f x 0
x→1
lim f x 2
x→1
(c) lim f x 2
(c) lim f x 2
(c) lim f x does not exist.
The function is continuous at x 2.
The function is NOT continuous at x 2.
The function is NOT continuous at x 1.
x→2
8. lim x→2
x→2
2x 1 1 lim x2 4 x→2 x 2 4
x→1
10. lim x→4
x 2
x4
lim x→4
lim x→4
lim x→4
12. lim x→2
x 2
14. lim x→0
x2
lim
x→2
x 2
x4
x 2 x 2
x4 x 4 x 2 1 x 2
1 4
x2 1 x2
x x2 x x x2 x x2 2xx x2 x x x2 x lim x→0 x x lim x→0
2xx x2 x x
lim 2x x 1 x→0
2x 0 1 2x 1 16. lim f x lim x2 4x 2 2 x→2
x→2
lim f x lim
x→2
x→2
x2
18. lim f x lim 1 x 0 x→1
x→1
4x 6 2
lim f x 2
x→2
20. lim sec x does not exist since x→ 2
lim
x→ 2
sec x and
lim
x→ 2
24. lim 1 x→1
x→2
sec x do not exist.
2x 1 1 2
22. lim 2x x 22 2 2
26. f x
x2 1 x1
has a discontinuity at x 1 since f 1 is not defined.
x, 28. f x 2, 2x 1,
x < 1 x 1 has discontinuity at x 1 since f 1 2 lim f x 1. x→1 x > 1
30. f t 3 9 t2 is continuous on 3, 3.
32. g2 is not defined. g is continuous on 1, 2.
316
Chapter 1
Limits and Their Properties
x is continuous for all real x. 2
34. f x
1 is continuous for all real x. x2 1
38. f x
x has nonremovable discontinuities at x 1 and x 1 since lim f x and lim f x do not exist. x→1 x→1 x2 1
36. f x cos
x3 has a nonremovable discontinuity at x 3 since lim f x does not exist, and has a removable discontinuity x→3 x2 9 at x 3 since
40. f x
lim f x lim
x→3
42. f x
x→3
1 1 . x3 6
x1 x 2x 1
44. f x
has a nonremovable discontinuity at x 2 since lim f x does not exist, and has a removable discontinux→2 ity at x 1 since lim f x lim
x→1
46. f x
x→1
x 3
x3 has a nonremovable discontinuity at x 3 since lim f x x→3 does not exist.
1 1 . x2 3
3,
2x x,
x < 1 x ≥ 1
2
has a possible discontinuity at x 1. 1. f 1 12 1 2.
lim f x lim 2x 3 1
x→1
x→1
x→1
lim f x 1 x→1
lim f x lim x2 1 x→1
3. f 1 lim f x x→1
f is continuous at x 1, therefore, f is continuous for all real x.
48. f x
2x, x 4x 1, 2
x ≤ 2 has a possible discontinuity at x 2. x > 2
1. f 2 22 4 2.
lim f x lim 2x 4
x→2
x→2
x→2
x→2
lim f x does not exist. x→2
lim f x lim x2 4x 1 3
Therefore, f has a nonremovable discontinuity at x 2.
50. f x
csc x , 6 2,
1. f 1 csc
2 6
2. lim f x 2 x→1
3. f 1 lim f x x→1
x 3 x 3
≤ 2 > 2
csc x , 6 2,
f 5 csc
1 ≤ x ≤ 5 x < 1 or x > 5
has possible discontinuities at x 1, x 5.
5 2 6
lim f x 2
x→5
f 5 lim f x x→5
f is continuous at x 1 and x 5, therefore, f is continuous for all real x.
Section 1.4
x has nonremovable discontinuities at each 2 2k 1, k is an integer.
58. lim g(x lim
20
x→0
x→0
lim fx 0
x→0
4 sin x 4 x
lim gx lim a 2x a
x→0
f is not continuous at x 4
x→0
−8
x→0
8
Let a 4. −10
x2 a2 x→a x a
60. lim gx lim x→a
lim x a 2a x→a
Find a such that 2a 8 ⇒ a 4.
62. f gx
1 x 1
Nonremovable discontinuity at x 1. Continuous for all x > 1. Because f g is not defined for x < 1, it is better to say that f g is discontinuous from the right at x 1. 66. hx
64. f gx sin x2 Continuous for all real x
1 x 1x 2
Nonremovable discontinuity at x 1 and x 2. 2
−3
4
−2
68. f x
cos x 1 , x < 0
5x, x
lim f x lim
x→0
x→0
3
x ≥ 0 −7
f 0 50 0
cos x 1 0 x
2
−3
lim f x lim 5x 0
x→0
x→0
Therefore, lim f x 0 f 0 and f is continuous on the entire real line. (x 0 was the only possible discontinuity.) x→0
70. f x xx 3 Continuous on 3,
317
54. f x 3 x has nonremovable discontinuities at each integer k.
52. f x tan
56. lim f x 0
Continuity and One-Sided Limits
72. f x
x1 x
Continuous on 0,
318
Chapter 1
74. f x
Limits and Their Properties
x3 8 x2
76. f x x3 3x 2 is continuous on 0, 1. f 0 2 and f 1 2
14
By the Intermediate Value Theorem, f x 0 for at least one value of c between 0 and 1. −4
4 0
The graph appears to be continuous on the interval 4, 4. Since f 2 is not defined, we know that f has a discontinuity at x 2. This discontinuity is removable so it does not show up on the graph.
78. f x
4 x tan is continuous on 1, 3. x 8
f 1 4 tan
3 4 < 0 and f 3 tan > 0. 8 3 8
By the Intermediate Value Theorem, f 1 0 for at least one value of c between 1 and 3.
82. h 1 3 tan
80. f x x3 3x 2 f x is continuous on 0, 1. f 0 2 and f 1 2 By the Intermediate Value Theorem, f x 0 for at least one value of c between 0 and 1. Using a graphing utility, we find that x 0.5961. 84. f x x2 6x 8
h is continuous on 0, 1.
f is continuous on 0, 3.
h0 1 > 0 and h1 2.67 < 0.
f 0 8 and f 3 1
By the Intermediate Value Theorem, h 0 for at least one value between 0 and 1. Using a graphing utility, we find that 0.4503.
1 < 0 < 8 The Intermediate Value Theorem applies. x2 6x 8 0
x 2x 4 0 x 2 or x 4 c 2 (x 4 is not in the interval.) Thus, f 2 0.
86. f x
x2 x x1
The Intermediate Value Theorem applies.
f is continuous on 2 , 4. The nonremovable discontinuity, x 1, lies outside the interval. 5
f
5 35 20 and f 4 2 6 3
20 35 < 6 < 6 3
x2 x 6 x1 x2 x 6x 6 x2 5x 6 0
x 2x 3 0 x 2 or x 3 c 3 (x 2 is not in the interval.) Thus, f 3 6.
Section 1.4 88. A discontinuity at x c is removable if you can define (or redefine) the function at x c in such a way that the new function is continuous at x c. Answers will vary. (a) f x
x 2
Continuity and One-Sided Limits
1, 0, (c) f x 1, 0,
x2 sinx 2 (b) f x x2
319
if x ≥ 2 if 2 < x < 2 if x 2 if x < 2
y 3 2 1 −3
−2
−1
x −1
1
2
3
−2 −3
90. If f and g are continuous for all real x, then so is f g (Theorem 1.11, part 2). However, fg might not be continuous if gx 0. For example, let f x x and gx x2 1. Then f and g are continuous for all real x, but fg is not continuous at x ± 1.
1.04, 92. C 1.04 0.36t 1, 1.04 0.36t 2,
0 < t ≤ 2 t > 2, t is not an integer t > 2, t is an integer
You can also write C as C
Nonremovable discontinuity at each integer greater than 2.
1.04, 1.04 0.362 t,
0 < t ≤ 2 . t > 2
C 4 3 2 1 t 1
2
3
4
94. Let st be the position function for the run up to the campsite. s0 0 (t 0 corresponds to 8:00 A.M., s20 k (distance to campsite)). Let rt be the position function for the run back down the mountain: r0 k, r10 0. Let f t st rt. When t 0 (8:00 A.M.),
f 0 s0 r0 0 k < 0.
When t 10 (8:10 A.M.), f 10 s10 r10 > 0. Since f 0 < 0 and f 10 > 0, then there must be a value t in the interval 0, 10 such that f t 0. If f t 0, then st rt 0, which gives us st rt. Therefore, at some time t, where 0 ≤ t ≤ 10, the position functions for the run up and the run down are equal. 96. Suppose there exists x1 in a, b such that f x1 > 0 and there exists x2 in a, b such that f x2 < 0. Then by the Intermediate Value Theorem, f x must equal zero for some value of x in x1, x2 (or x2, x1 if x2 < x1). Thus, f would have a zero in a, b, which is a contradiction. Therefore, f x > 0 for all x in a, b or f x < 0 for all x in a, b. 98. If x 0, then f 0 0 and lim f x 0. Hence, f is x→0 continuous at x 0. If x 0, then lim f t 0 for x rational, whereas t→x
lim f t lim kt kx 0 for x irrational. Hence, f is not
t →x
t →x
continuous for all x 0.
100. True 1. f c L is defined. 2. lim f x L exists. x→c
3. f c lim f x x→c
All of the conditions for continuity are met.
320
Chapter 1
Limits and Their Properties
102. False; a rational function can be written as PxQx where P and Q are polynomials of degree m and n, respectively. It can have, at most, n discontinuities.
104. (a)
S 60 50 40 30 20 10 t 5
10
15 20 25 30
(b) There appears to be a limiting speed and a possible cause is air resistance. 106. Let y be a real number. If y 0, then x 0. If y > 0, then let 0 < x0 < 2 such that M tan x0 > y (this is possible since the tangent function increases without bound on 0, 2). By the Intermediate Value Theorem, f x tan x is continuous on 0, x0 and 0 < y < M, which implies that there exists x between 0 and x0 such that tan x y. The argument is similar if y < 0. 108. 1. f c is defined. 2. lim f x lim f c x f c exists. x→0
x→c
Let x c x. As x → c, x → 0 3. lim f x f c. x→c
Therefore, f is continuous at x c. 110. Define f x f2x f1x. Since f1 and f2 are continuous on a, b, so is f. f a f2a f1a > 0 and
f b f2b f1b < 0.
By the Intermediate Value Theorem, there exists c in a, b such that f c 0. f c f2c f1c 0 ⇒ f1c f2c
Section 1.5 2.
Infinite Limits
1 x2 1 lim x→2 x 2 lim
4.
x→2
6. f x x
x 4
lim sec
x 4
x→2
x x2 9 3.5
f x 1.077
3.1
3.01
3.001
2.999
2.99
2.9
2.5
5.082 50.08
500.1
499.9
49.92
4.915
0.9091
lim f x
x→3
lim f x
x→3
lim sec
x→2
Section 1.5
8. f x sec
321
x 6
3.5
x
Infinite Limits
f x 3.864
3.1
3.01
19.11
191.0
3.001 2.999 2.99 2.9
2.5
1910
3.864
1910
191.0
19.11
lim f x
x→3
lim f x
x→3
10. lim x→2
lim
x→2
4 x 23
12. lim x→0
2x 2x lim 2 x→0 x 1 x x 1 x 2
Therefore, x 0 is a vertical asymptote.
4 x 23
lim
2x x21 x
lim
2x x21 x
x→1
Therefore, x 2 is a vertical asymptote.
x→1
Therefore, x 1 is a vertical asymptote.
14. No vertical asymptote since the denominator is never zero.
16.
lim hs and lim hs .
s→5
s→5
Therefore, s 5 is a vertical asymptote. lim hs and lim hs .
s→5
s→5
Therefore, s 5 is a vertical asymptote.
18. f x sec x x
1 has vertical asymptotes at cos x
2n 1 , n any integer. 2
20. gx
12x3 x2 4x 1 xx2 2x 8 3x2 6x 24 6 x2 2x 8
1 x, 6 x 2, 4 No vertical asymptotes. The graph has holes at x 2 and x 4.
22. f x
x
x3
4x2 x 6 4x 3x 2 4 , x 3, 2 2x2 9x 18 xx 2x2 9 xx 3
Vertical asymptotes at x 0 and x 3. The graph has holes at x 3 and x 2.
24. hx
x 2x 2 x2 4 x3 2x2 x 2 x 2x2 1
has no vertical asymptote since lim hx lim
x→2
x→2
4 x2 . x2 1 5
26. ht
tt 2 t ,t2 t 2t 2t 2 4 t 2t 2 4
Vertical asymptote at t 2. The graph has a hole at t 2.
322
Chapter 1
28. g
Limits and Their Properties
tan sin has vertical asymptotes at cos
x2 6x 7 lim x 7 8 x→1 x→1 x1
30. lim
2n 1 n, n any integer. 2 2
2 −3
3
There is no vertical asymptote at 0 since lim
→0
tan 1.
−12
Removable discontinuity at x 1
sinx 1 1 x1
32. lim
x→1
2
34. lim x→1
Removable discontinuity at x 1
−3
2x 1x
3
−2
36. lim x→4
40. lim
x→3
44.
48.
x2
x2 1 16 2
38.
x2 1 x2 9
lim
x→ 2
lim
x→ 12
lim
x→12
6x2 x 1 3x 1 5 lim 4x2 4x 3 x→12 2x 3 8
42. lim x2 x→0
2 cos x
46. lim
x→0
x2 tan x and
lim
x→ 12
x2 tan x .
Therefore, lim x2 tan x does not exist. x→ 12
1 x
x 2 lim x 2tan x 0 x→0 cot x
50. f x
x2
x3 1 x1
lim f x lim x 1 0
x→1
x→1
4
−8
8
−4
52. f x sec
x 6
54. The line x c is a vertical asymptote if the graph of f approaches ± as x approaches c.
lim f x
x→3
6
−9
9
−6
56. No. For example, f x
1 has no x2 1
58. P
vertical asymptote. lim
V→0
k V k k (In this case we know that k > 0.) V
Section 1.5
200 ftsec 6 3 (b) r 50 sec2 200 ftsec 3
60. (a) r 50 sec2
(c)
lim
→ 2
62. m
Total distance Total time
50
2d dx dy
50
2xy yx
m0
lim m lim
v→c
(b)
v→c
m0 1 v2c2
x
30
40
50
60
y
150
66.667
50
42.857
(c) lim x→25
25x x 25
As x gets close to 25 mph, y becomes larger and larger.
50y 50x 2xy 50x 2xy 50y 50x 2yx 25 25x y x 25 Domain: x > 25 1 1 1 1 66. (a) A bh r2 1010 tan 102 2 2 2 2
(b)
(c)
f
50 tan 50 Domain:
0.3
0.6
0.9
1.2
1.5
0.47
4.21
18.0
68.6
630.1
0, 2 (d)
100
0
lim A
→ 2
1.5 0
68. False; for instance, let f x
70. True
x2 1 . x1
The graph of f has a hole at 1, 2, not a vertical asymptote.
72. Let f x
1 1 and gx 4, and c 0. x2 x
1 1 2 and lim 4 , but x→0 x x→0 x lim
x1 x1 lim x x 1 0. 2
lim
x→0
2
4
x→0
4
323
1 v2c2
50 sec2
64. (a) Average speed
Infinite Limits
74. Given lim f x , let g x 1. then lim x →c
by Theorem 1.15.
x →c
gx 0 fx
324
Chapter 1
Limits and Their Properties
Review Exercises for Chapter 1 2. Precalculus. L 9 12 3 12 8.25 4.
0.1
x f x
0.01
0.358
0.5
0.001
0.001
0.01
0.1
0.354
0.354
0.353
0.349
0.354
−1
1
lim f x 0.2
x→0
6. gx
−0.5
3x x2
(b) lim gx 0
(a) lim gx does not exist.
x→0
x→2
Assuming 4 < x < 16, you can choose 5.
8. lim x 9 3. x→9
Let > 0 be given. We need
x→5
x 9 < x 3
x 9 < 5 < x 3
x 3 < ⇒ x 3 x 3 < x 3
10. lim 9 9. Let > 0 be given. can be any positive
Hence, for 0 < x 9 < 5, you have
x 3 <
f x L <
12. lim 3 y 1 3 4 1 9 y→4
number. Hence, for 0 < x 5 < , you have
9 9 < f x L < 14. lim t→3
t2 9 lim t 3 6 t→3 t3
16. lim
4 x 2
x→0
x
lim lim
x→0
18. lim
s→0
x 1 4 x 2
11 s 1 lim 11 s 1 11 s 1 s→0
s s 11 s 1 lim
s→0
20. lim
4 x 2
x→0
x→2
11 s 1 1 1 lim s 11 s 1 s→0 1 s 11 s 1 2
x2 4 x 2x 2 lim x3 8 x→2 x 2x2 2x 4 lim
x→2
x2 x2 2x 4
4 1 12 3
22.
lim
x→ 4
4x 44 tan x 1
4 x 2 4 x 2
1 4
Review Exercises for Chapter 1
24. lim
x→0
cos x 1 cos cos x sin sin x 1 lim x→0 x x lim
x→0
cos x x 1
lim sin
x→0
sin x x
0 01 0 3 2 7 26. lim f x 2gx 4 23 12 x→c
28. f x (a)
3 x 1 x1
x f x
lim
x→1
(c) lim x→1
1.1
1.01
1.001
0.3228
0.3322
0.3332
3 x 1 0.333 x1
0.3333
3
−3
3 x 1 3 x 3 3 x 2 1 x
1x 3 3 x 11 x x2
lim
1 3 x 3 x2 1
−3
2
lim
x→1
2
Actual limit is 13 .
3 x 3 x 1 1 lim x→1 x1 x1
x→1
(b)
1.0001
1 3
30. st 0 ⇒ 4.9t2 200 0 ⇒ t2 40.816 ⇒ t 6.39 sec When t 6.39, the velocity is approximately lim t→a
sa st lim 4.9a t t→a at lim 4.96.39 6.39 62.6 msec. t→6.39
32. lim x 1 does not exist. The graph jumps from 2 to 3 x→4
34. lim gx 1 1 2. x→1
at x 4.
36. lim f s 2 s→2
38. f x
3x2 x 2 , x1 0,
lim f x lim
x→1
x→1
x 1 x1
3x x 2 x1 2
lim 3x 2 5 0 x→1
Removable discontinuity at x 1 Continuous on , 1 1,
325
326
Chapter 1
52xx,3,
40. f x
Limits and Their Properties
x x 1 1 1x 1 lim 1
x
x ≤ 2 x > 2
42. f x
lim 5 x 3
x→2
x→0
lim 2x 3 1
Domain: , 1 , 0,
x→2
Nonremovable discontinuity at x 2 Continuous on , 2 2,
Nonremovable discontinuity at x 0 Continuous on , 1 0,
x1 2x 2
44. f x
46. f x tan 2x Nonremovable discontinuities when
x1 1 x→1 2x 1 2 lim
x
Removable discontinuity at x 1 Continuous on , 1 1,
2n 1 4
Continuous on
2n 4 1, 2n 4 1 for all integers n. 48. lim x 1 2 x→1
lim x 1 4
x→3
Find b and c so that lim x2 bx c 2 and lim x2 bx c 4. x→1
x→3
1bc2
Consequently we get Solving simultaneously,
b
and 9 3b c 4.
3 and
50. C 9.80 2.50 x 1 , x > 0
c 4. 52. f x x 1x (a) Domain: , 0 1,
9.80 2.50 x 1
(b) lim f x 0
C has a nonremovable discontinuity at each integer.
x→0
30
(c) lim f x 0 x→1
0
5 0
54. hx
4x 4 x2
56. f x csc x Vertical asymptote at every integer k
Vertical asymptotes at x 2 and x 2
58.
62.
lim
x→ 12
lim
x→1
66. lim x→0
x 2x 1
x2 2x 1
x1
sec x
x
60.
lim
x→1
64. lim x→2
68. lim x→0
x1 1 1 lim x4 1 x→1 x2 1x 1 4 1
3 2 x 4
cos2 x
x
Problem Solving for Chapter 1 tan 2x x
70. f x (a)
327
0.1
x f x lim
x→0
0.01
2.0271
0.001
2.0003
2.0000
0.001
0.01
0.1
2.0000
2.0003
2.0271
tan 2x 2 x
(b) Yes, define f x
2, x
tan 2x ,
x0 x0
.
Now f x is continuous at x 0.
Problem Solving for Chapter 1 1 1 x 2. (a) Area PAO bh 1x 2 2 2
4. (a) Slope
1 1 y x2 Area PBO bh 1y 2 2 2 2 (b) ax
40 4 30 3
(b) Slope
3 Tangent line: y 4 x 3 4
3 4
x22 Area PBO x Area PAO x2
3 25 y x 4 4
x
4
2
1
0.1
0.01
Area PAO
2
1
12
120
1200
Area PBO
8
2
12
1200
120,000
ax
4
2
1
110
(c) Let Q x, y x, 25 x2 mx
x3
(d) lim mx lim x→3
1100
x→3
25 x2 4
x3
25 x2 4 25 x2 4
25 x2 16 x→3 x 3 25 x2 4
lim
(c) lim ax lim x 0 x→0
25 x2 4
x→0
lim
x→3
lim
x→3
3 x3 x x 325 x2 4 3 x 25 x2 4
3 6 44 4
This is the slope of the tangent line at P.
6.
a bx 3
x
a bx 3
x
a bx 3 a bx 3
x
x→0
Setting
b 3 3
b 3 bx 3
3, you obtain b 6.
Thus, a 3 and b 6.
ax tan x a because lim 1 x→0 tan x x
a2 2 a
bx lim x→0 x 3 bx 3 lim
x→0
Thus,
Thus, lim
x→0
x→0
Letting a 3 simplifies the numerator.
x→0
x→0
lim f x lim
a bx 3 xa bx 3
3 bx 3
8. lim f x lim a2 2 a2 2
.
a2 a 2 0
a 2a 1 0 a 1, 2
328
Chapter 1
Limits and Their Properties 1 (a) f 4 4 4
y
10.
f 3
1 3
3 2
0
f 1 1 1
1 x
−1
1
−1
(b) lim f x 1
(c) f is continuous for all real numbers except
x→1
lim f x 0
x→1
x 0, ± 1, ± 12, ± 13, . . .
lim f x
x→0
lim f x
x→0
−2
v2
12. (a)
192,000 v02 48 r
192,000 v2 v02 48 r r lim r
v→0
192,000 v v02 48 192,000 48 v02
Let v0 48 43 feetsec. (b)
v2
1920 v02 2.17 r
1920 v2 v02 2.17 r r lim r
v→0
1920 v2 v02 2.17 1920 2.17 v02
Let v0 2.17 misec 1.47 misec. r
(c)
lim r
v→0
10,600 v2 v02 6.99 10,600 6.99 v02
Let v0 6.99 2.64 misec. Since this is smaller than the escape velocity for earth, the mass is less.
14. Let a 0 and let > 0 be given. There exists 1 > 0 such that if 0 < x 0 < , then f x L < . Let 1 a . Then for 0 < x 0 < 1 a , you have
x < a1
ax < 1 f ax L < .
As a counterexample, let f x Then lim f x 1 L, x→0
but lim f ax lim f 0 2. x→0
x→0
12
x0 . x0
C H A P T E R 1 Limits and Their Properties Section 1.1
A Preview of Calculus . . . . . . . . . . . . . . . . . . . . 27
Section 1.2
Finding Limits Graphically and Numerically . . . . . . . . 27
Section 1.3
Evaluating Limits Analytically
Section 1.4
Continuity and One-Sided Limits . . . . . . . . . . . . . . 37
Section 1.5
Infinite Limits . . . . . . . . . . . . . . . . . . . . . . . . 42
. . . . . . . . . . . . . . . 31
Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
C H A P T E R 1 Limits and Their Properties Section 1.1
A Preview of Calculus
Solutions to Odd-Numbered Exercises 1. Precalculus: 20 ftsec15 seconds 300 feet
3. Calculus required: slope of tangent line at x 2 is rate of change, and equals about 0.16.
5. Precalculus: Area 12 bh 12 53 15 2 sq. units
7. Precalculus: Volume 243 24 cubic units
9. (a)
6
(1, 3) −4
8 −2
(b) The graphs of y2 are approximations to the tangent line to y1 at x 1. (c) The slope is approximately 2. For a better approximation make the list numbers smaller:
0.2, 0.1, 0.01, 0.001 11. (a) D1 5 12 1 52 16 16 5.66 5 5 5 5 5 5 (b) D2 1 2 1 2 3 1 3 4 1 4 1 2
2
2
2
2.693 1.302 1.083 1.031 6.11 (c) Increase the number of line segments.
Section 1.2 1.
x
1.9
1.99
1.999
2.001
2.01
2.1
f x
0.3448
0.3344
0.3334
0.3332
0.3322
0.3226
lim
x→2
3.
Finding Limits Graphically and Numerically
x2 0.3333 x2 x 2
0.1
x f x lim
x→0
0.01
0.2911
0.001
0.2889
x 3 3
x
Actual limit is 13 .
0.2887
0.2887
0.001
0.01
0.1
0.2887
0.2884
0.2863
Actual limit is 1 23.
27
28
Chapter 1
5.
x
2.9
f x
0.0641
lim
x→3
7.
Limits and Their Properties
2.999
3.001
3.01
3.1
0.0627
0.0625
0.0625
0.0623
0.0610
1x 1 14 0.0625 x3
0.1
x
2.99
f x
0.01
0.9983
lim
x→0
0.001
0.99998
sin x 1.0000 x
Actual limit is 161 .
1.0000
0.001
0.01
0.1
1.0000
0.99998
0.9983
(Actual limit is 1.) (Make sure you use radian mode.)
11. lim f x lim 4 x 2
9. lim 4 x 1
x→2
x→3
13. lim
x→2
x 5 does not exist. For values of x to the left of 5, x 5 x 5 equals 1,
x5 whereas for values of x to the right of 5, x 5 x 5 equals 1. x→5
15. lim tan x does not exist since the function increases and x→ 2
17. lim cos1x does not exist since the function oscillates x→0
between 1 and 1 as x approaches 0.
decreases without bound as x approaches 2. 19. Ct 0.75 0.50 t 1 (a)
(b)
3
t
3
3.3
3.4
3.5
3.6
3.7
4
C 1.75
2.25
2.25
2.25
2.25
2.25
2.25
2.5
2.9
3
3.1
3.5
4
1.75
1.75
1.75
2.25
2.25
2.25
lim Ct 2.25
t→3.5
(c)
5
0 0
t
2
C 1.25
lim Ct does not exist. The values of C jump from 1.75 to 2.25 at t 3. t→3
21. You need to find such that 0 < x 1 < implies 1 f x 1 1 < 0.1. That is, x
0.1 <
1 1 < 0.1 x
1 0.1 <
1 x
< 1 0.1
9 < 10
1 x
11 < 10
10 > 9
x
>
10 11
10 10 1 > x 1 > 1 9 11 1 1 > x 1 > . 9 11
So take
1 . Then 0 < x 1 < implies 11
1 1 < x1 < 11 11
1 1 < x1 < . 11 9
Using the first series of equivalent inequalities, you obtain
f x 1
1 1 < < 0.1. x
Section 1.2 23. lim 3x 2 8 L x→2
3x 2 8
3x 6 3 x 2
< 0.01 < 0.01 < 0.01
0.01 0.0033 3 0.01 Hence, if 0 < x 2 < , you have 3
0 < x2 <
3x 6 < 0.01
3x 2 8 < 0.01
f x L < 0.01 3 x 2 < 0.01
27. lim x 3 5
Finding Limits Graphically and Numerically
25. lim x2 3 1 L x→2
x2 3 1
x2 4
x 2x 2
x 2 x 2
< 0.01 < 0.01
If we assume 1 < x < 3, then 0.015 0.002.
Hence, if 0 < x 2 < 0.002, you have 1
x 2
x 2 < 0.01
x2 4 < 0.01
x2 3 1 < 0.01
f x L < 0.01
12 x 1 12 4 1 3
x 1 3 <
x 2 <
<
1 2
<
Hence, let .
1 2
1 2
Hence, if 0 < x 2 < , you have
x2 <
x 3 5 <
f x L <
x 4
x 4
< < 2
Hence, let 2.
Hence, if 0 < x 4 < 2, you have
x 4 < 2
x 2
x 1 3 1 2
1 2
< <
f x L <
31. lim 3 3
3 33. lim x0
x→6
x→0
Given > 0:
x <
3 x 0 < Given > 0:
33 <
3
x
0 <
< 3
Hence, any > 0 will work.
Hence, let 3.
Hence, for any > 0, you have
Hence for 0 < x 0 < 3, you have
1
x 2 < 0.002 50.01 < x 2 0.01
Given > 0:
x 3 5
x 2
< 0.01
0.01
x→4
Given > 0:
< 0.01
x 2 < x 2
29. lim
x→2
f x L < 33 <
29
x <
0:
Given > 0:
x 2 4
x 2 4
x 2 x 2 x 2
x2 1 2
x2 1
x 1x 1
<
x 2 < 0
< <
Hence, .
x 2 <
x 2 4 <
x 2 4 <
f x L <
If we assume 0 < x < 2, then 3.
Hence for 0 < x 1 < , you have 3
1
1
x2 1 <
x2 1 2 <
f x 2 < x9
0.5
41. f x
x4 1 6
x 1 < 3 < x 1
(because x 20)
x 5 3
lim f x
<
x2 <
x→4
<
x 1 < x 1
Hence for 0 < x 2 < , you have
39. f x
<
−6
10
x 3
lim f x 6
6
x→9
−0.1667
0
10 0
The domain is 5, 4 4, . 1 The graphing utility does not show the hole at 4, 6 .
The domain is all x ≥ 0 except x 9. The graphing utility does not show the hole at 9, 6.
43. lim f x 25 means that the values of f approach 25 as x gets closer and closer to 8. x→8
45. (i) The values of f approach different numbers as x approaches c from different sides of c:
(ii) The values of f increase without bound as x approaches c:
(iii) The values of f oscillate between two fixed numbers as x approaches c:
y
y
y
6
4
5
3
4
2
3
4 3
2
1
1
x
−4 −3 −2 −1 −1
1
2
3
4
−3 −2 −1 −1
3
4
−3
−4
−4
47. f x 1 x1x lim 1 x1x e 2.71828
x→0
y 7
3
(0, 2.7183)
2 1 −3 −2 −1 −1
x 1
2
2
5
−2
−3
x
−4 −3 −2
x 2
3
4
5
x
f x
x
f x
0.1
2.867972
0.1
2.593742
0.01
2.731999
0.01
2.704814
0.001
2.719642
0.001
2.716942
0.0001
2.718418
0.0001
2.718146
0.00001
2.718295
0.00001
2.718268
0.000001
2.718283
0.000001
2.718280
3
4
Section 1.3 49. False; f x sin xx is undefined when x 0. From Exercise 7, we have lim
x→0
51. False; let f x
sin x 1. x
Evaluating Limits Analytically
31
53. Answers will vary.
x10, 4x, 2
x4 . x4
f 4 10 lim f x lim x2 4x 0 10
x→4
x→4
55. If lim f x L1 and lim f x L2, then for every > 0, there exists 1 > 0 and 2 > 0 such that x c < 1 ⇒ f x L1 < and x→c
x→c
x c < 2 ⇒ f x L2 < . Let equal the smaller of 1 and 2. Then for x c L1 L2 L1 f x f x L2 ≤ L1 f x f x L2 < . Therefore, L1 L2 < 2. Since > 0 is arbitrary, it follows that L1 L2.
< , we have
57. lim f x L 0 means that for every > 0 there exists > 0 such that if x→c
0 < x c < , then
f x L 0
< .
This means the same as f x L < when 0 < x c < .
Thus, lim f x L. x→c
Section 1.3 1.
Evaluating Limits Analytically (a) lim hx 0
7
x→5
−8
3.
(b) lim hx 6
x→0
π
−π
x→1
13
(a) lim f x 0
4
−7
(b) lim f x 0.524
−4
hx x2 5x
f x x cos x 7. lim 2x 1 20 1 1
5. lim x4 24 16 x→2
x→0
9. lim x2 3x 32 33 9 9 0 x→3
11. lim 2x2 4x 1 232 43 1 18 12 1 7 x→3
13. lim
x→2
17. lim
x→7
1 1 x 2 5x x 2
15. lim
x→1
57 7 2
35 9
35 3
x3 13 2 2 x2 4 12 4 5 5
19. lim x 1 3 1 2 x→3
x→ 3
6
32
Chapter 1
Limits and Their Properties
21. lim x 32 4 32 1
23. (a) lim f x 5 1 4
x→4
x→1
(b) lim gx 43 64 x→4
(c) lim g f x g f 1 g4 64 x→1
25. (a) lim f x 4 1 3
27. lim sin x sin x→ 2
x→1
(b) lim gx 3 1 2
1 2
x→3
(c) lim g f x g3 2 x→1
29. lim cos x→2
33.
x 2 1 cos 3 3 2
lim sin x sin
x→56
31. lim sec 2x sec 0 1 x→0
5 1 6 2
35. lim tan x→3
37. (a) lim 5gx 5 lim gx 53 15 x→c
x→c
(b) lim f x gx lim f x lim gx 2 3 5 x→c
x→c
x→c
(c) lim f xgx lim f x lim gx 23 6 x→c
x→c
x→c
lim f x
f x 2 x→c (d) lim x→c gx lim gx 3
39. (a) lim f x3 lim f x3 43 64 x→c
x→c
(b) lim f x lim f x 4 2 x→c
x→c
(c) lim 3 f x 3 lim f x 34 12 x→c
x→c
(d) lim f x32 lim f x32 432 8 x→c
x→c
41. f x 2x 1 and gx x 0.
4x tan 34 1
2x2 x agree except at x
x→c
43. f x xx 1 and gx
x3 x agree except at x 1. x1
(a) lim gx lim f x 1
(a) lim gx lim f x 2
(b) lim gx lim f x 3
(b) lim gx lim f x 0
x→0
x→1
45. f x
x→1
x→0
x→1
x→1
x2 1 and gx x 1 agree except at x 1. x1
lim f x lim gx 2
x→1
x→1
47. f x x 2.
x→1
x→1
x3 8 and gx x2 2x 4 agree except at x2
lim f x lim gx 12
x→2
3
x→2
12 −3
4
−4
−9
9 0
49. lim
x→5
x5 x5 lim x2 25 x→5 x 5x 5 lim
x→5
1 1 x 5 10
51. lim
x→3
x2 x 6 x 3x 2 lim x→3 x 3x 3 x2 9 lim
x→3
x 2 5 5 x 3 6 6
Section 1.3
53. lim
x 5 5
lim
x
x→0
lim
x→0
55. lim
x 5 3
x4
x→4
x 5 5
lim
x→4
x 5 5 x 5 5
5 x 5 5 1 1 lim x x 5 5 x→0 x 5 5 2 5 10
x 5 3
x4
x→4
lim
x
x→0
Evaluating Limits Analytically
x 5 3 x 5 3
1 x 5 9 1 1 lim x 4 x 5 3 x→4 x 5 3 9 3 6
1 1 2 2 x 1 1 2x 2 22 x 57. lim lim lim x→0 x→0 x→0 22 x x x 4
59. lim
x→0
2x x 2x 2x 2 x 2x lim lim 2 2
x→0
x→0
x
x
x x2 2x x 1 x2 2x 1 x2 2x x x2 2x 2 x 1 x2 2x 1 lim
x→0
x→0
x
x
61. lim
lim 2x x 2 2x 2
x→0
63. lim
x 2 2
x→0
x
x
0.1
f x
0.354 0.01
0.358
2
0.001
0
0.001
0.01
0.1
0.345
?
0.354
0.353
0.349
0.354
−3
3
−2
Analytically, lim
x 2 2
x
x→0
lim
x 2 2
x
x→0
lim
x→0
x 2 2
x 2 2
x22
x x 2 2
lim
x→0
1 x 2 2
1 2 2
2
4
1 1 2x 2 1 65. lim x→0 x 4
0.354
3
−5
x
0.1
0.01
0.001
0
0.001
f x
0.263
0.251
0.250
?
0.250
1 1 2x 2 2 2 x Analytically, lim lim x→0 x→0 x 22 x
1
0.01
0.1
0.249
0.238
x
1
1
−2
1
1
lim . x x→0 lim 22 x x x→0 22 x 4
33
34
Chapter 1
67. lim
x→0
Limits and Their Properties
sin x lim x→0 5x
sin x x
15 1 15 51
69. lim
x→0
1 sin x1 cos x lim x→0 2 2x2
sin x x
1 cos x x
1 10 0 2
1 cos h2 1 cos h 1 cos h lim h→0 h→0 h h
sin2 x sin x lim sin x 1 sin 0 0 x→0 x→0 x x
71. lim
73. lim
00 0
75. lim
x→ 2
cos x lim sin x 1 x→ 2 cot x
79. f t
f t
t→0
sin 3t sin 3t lim t→0 2t 3t
4
0.01
2.96
0.001
2.9996
3
0
0.001
0.01
0.1
?
3
2.9996
2.96
− 2
2 −1
The limit appear to equal 3.
sin 3t sin 3t lim 3 31 3. t→0 t→0 t 3t
Analytically, lim
81. f x
32 1 32 23
sin 3t t 0.1
t
77. lim
sin x2 x
1
− 2
x
0.1
0.01 0.001 0 0.001 0.01 0.1
f x
0.099998 0.01 0.001 ? 0.001 0.01 0.099998
2
−1
sin x2 sin x2 lim x 01 0. x→0 x→0 x x2
Analytically, lim
83. lim
h→0
f x h f x 2x h 3 2x 3 2x 2h 3 2x 3 2h lim lim lim 2 h→0 h→0 h→0 h h h h
4 4 4 4 f x h f x 4x 4x h xh x 85. lim lim lim lim 2 h→0 x hx h→0 h→0 h→0 h h x hxh x 87. lim 4 x2 ≤ lim f x ≤ lim 4 x2 x→0
x→0
x→0
89. f x x cos x 4
4 ≤ lim f x ≤ 4 x→0
Therefore, lim f x 4. x→0
− 3 2
3 2
−4
lim x cos x 0
x→0
Section 1.3
91. f x x sin x
93. f x x sin
6
2
1 x
−0.5
0.5
−6
−0.5
lim x sin x 0
lim x sin
x→0
95. We say that two functions f and g agree at all but one point (on an open interval) if f x gx for all x in the interval except for x c, where c is in the interval.
1 0 x
97. An indeterminant form is obtained when evaluating a limit using direct substitution produces a meaningless fractional expression such as 00. That is, lim
x→c
f x gx
for which lim f x lim gx 0 x→c
99. f x x, gx sin x, hx
When you are “close to” 0 the magnitude of f is approximately equal to the magnitude of g. Thus, g f 1 when x is “close to” 0.
f h
−5
x→c
sin x x
3
g
5
−3
101. st 16t2 1000 lim t→5
s5 st 600 16t2 1000 16t 5t 5 lim lim lim 16t 5 160 ftsec. t→5 t→5 t→5 5t 5t t 5
Speed 160 ftsec 103. st 4.9t2 150 s3 st 4.932 150 4.9t2 150 4.99 t2 lim lim t→3 t→3 t→3 3t 3t 3t
lim
lim
x→3
4.93 t3 t lim 4.93 t 29.4 msec x→3 3t
105. Let f x 1x and gx 1x. lim f x and lim gx do not exist. x→0
x→0
lim 0 0
1 1 lim f x gx lim x→0 x→0 x x
x→0
107. Given f x b, show that for every > 0 there exists a > 0 such that f x b < whenever x c < . Since f x b b b 0 < for any > 0, then any value of > 0 will work.
35
0.5
− 2
x→0
Evaluating Limits Analytically
109. If b 0, then the property is true because both sides are equal to 0. If b 0, let > 0 be given. Since lim f x L,
x→c
there exists > 0 such that f x L < b whenever 0 < x c < . Hence, wherever 0 < x c < , we have
b f x L
<
or
bf x bL
which implies that lim bf x bL. x→c
<
36
Chapter 1
Limits and Their Properties
M f x ≤ f xgx ≤ M f x
111.
113. False. As x approaches 0 from the left,
lim M f x ≤ lim f xgx ≤ lim M f x
x→c
x→c
x→c
x 1. x
2
M0 ≤ lim f xgx ≤ M0 x→c
−3
0 ≤ lim f xgx ≤ 0
3
x→c
Therefore, lim f xgx 0.
−2
x →c
115. True.
117. False. The limit does not exist. 4
−3
6
−2
119. Let f x
4,4,
if x ≥ 0 if x < 0
lim f x lim 4 4.
x→0
x→0
lim f x does not exist since for x < 0, f x 4 and for x ≥ 0, f x 4.
x→0
rational 0,1, ifif xx isis irrational 0, if x is rational g x x, if x is irrational
121. f x
lim f x does not exist.
x→0
No matter how “close to” 0 x is, there are still an infinite number of rational and irrational numbers so that lim f x does not x→0 exist. lim gx 0.
x→0
When x is “close to” 0, both parts of the function are “close to” 0.
123. (a) lim
x→0
1 cos x 1 cos x lim x→0 x2 x2 lim
x→0
1 cos2 x x 1 cos x 2
sin2 x x→0 x2
lim 1
1 cos x
1 cos x
1
1 cos x
12 21
(b) Thus,
1 cos x 1 1 ⇒ 1 cos x x2 x2 2 2 1 ⇒ cos x 1 x2 for x 0. 2
1 (c) cos0.1 1 0.12 0.995 2 (d) cos0.1 0.9950, which agrees with part (c).
Section 1.4
Section 1.4
3. (a) lim f x 0
(b) lim f x 1
(b) lim f x 0
x→3
x→4
x→3
The function is continuous at x 3.
x→0
x→4
(c) lim f x does not exist
(c) lim f x 0
x→3
x
lim
x→0
9.
does not exist because
x x2 9
grows
x
1
1
lim x x→0 xx x x
x→0
x→3
x x2 9
x 1. x
lim
x→3
lim
x→3
without bound as x → 3 .
1 1 x x x x x x lim 13. lim x→0 x→0 x xx x
15. lim f x lim
The function is NOT continuous at x 4.
The function is NOT continuous at x 3.
x5 1 1 lim x2 25 x→5 x 5 10
x
lim f x 2
x→4
(b) lim f x 2
x→3
(c) lim f x 1
11. lim
5. (a)
x→3
x→3
x→5
37
Continuity and One-Sided Limits
1. (a) lim f x 1
7. lim
Continuity and One-Sided Limits
1 xx x
1 1 2 xx 0 x
x2 5 2 2
17. lim f x lim x 1 2 x→1
x→1
lim f x lim x3 1 2
x→1
x→1
lim f x 2
x→1
21. lim 3x 5 33 5 4
19. lim cot x does not exist since x→
x→4
x 3 for 3 < x < 4
lim cot x and lim cot x do not exist.
x→
x→
23. lim 2 x does not exist x→3
because lim2 x 2 3 5
x→3
and
25. f x
1 x2 4
27. f x
has discontinuities at x 2 and x 2 since f 2 and f 2 are not defined.
x x 2
has discontinuities at each integer k since lim f x lim f x. x→k
x→k
lim 2 x 2 4 6.
x→3
29. gx 25 x2 is continuous on 5, 5 .
31. lim f x 3 lim f x. x→0
x→0
f is continuous on 1, 4 .
33. f x x2 2x 1 is continuous for all real x.
38
Chapter 1
Limits and Their Properties
35. f x 3x cos x is continuous for all real x.
37. f x
x is not continuous at x 0, 1. Since x2 x
x 1 for x 0, x 0 is a removable x2 x x 1 discontinuity, whereas x 1 is a nonremovable discontinuity.
39. f x
x is continuous for all real x. x2 1
41. f x
x2 x 2x 5
has a nonremovable discontinuity at x 5 since lim f x x→5 does not exist, and has a removable discontinuity at x 2 since lim f x lim
x→2
43. f x
x 2 has a nonremovable discontinuity at x 2 since
45. f x
x,x ,
x2
x→2
1 1 . x5 7
lim f x does not exist.
x→2
x ≤ 1 x > 1
2
has a possible discontinuity at x 1. 1. f 1 1 2.
lim f x lim x 1
x→1
x→1
x→1
x→1
lim f x 1
lim f x lim x2 1
x→1
3. f 1 lim f x x→1
f is continuous at x 1, therefore, f is continuous for all real x. x 1, 47. f x 2 3 x,
x ≤ 2
1. f 2
x > 2
2 12 2
lim f x lim
x→2
2.
has a possible discontinuity at x 2.
x→2
2x 1 2
lim f x lim 3 x 1
x→2
x→2
lim f x does not exist.
x→2
Therefore, f has a nonremovable discontinuity at x 2.
49. f x
x tan 4 , x,
x x
1. f 1 1 2. lim f x 1 x→1
3. f 1 lim f x x→1
x < 1 tan 4 , ≥ 1 x,
1 < x < 1 has possible discontinuities at x 1, x 1. x ≤ 1 or x ≥ 1
f 1 1 lim f x 1
x→1
f 1 lim f x x→1
f is continuous at x ± 1, therefore, f is continuous for all real x.
Section 1.4
Continuity and One-Sided Limits
39
51. f x csc 2x has nonremovable discontinuities at integer multiples of 2.
53. f x x 1 has nonremovable discontinuities at each integer k.
55. lim f x 0
57. f 2 8
50
x→0
lim f x 0
Find a so that lim ax2 8 ⇒ a
x→0
x→2
f is not continuous at x 2.
−8
8 2. 22
8 −10
59. Find a and b such that lim ax b a b 2 and lim ax b 3a b 2. x→1
x→3
a b 2
3a b 2 4
4a
a 1 b
2, f x x 1, 2,
x ≤ 1 1 < x < 3 x ≥ 3
2 1 1
61. f gx x 12
63. f gx
Continuous for all real x.
Nonremovable discontinuities at x ± 1
67. f x
65. y x x Nonremovable discontinuity at each integer 0.5
1 1 x2 5 6 x2 1
2xx 2x,4, 2
x ≤ 3 x > 3
Nonremovable discontinuity at x 3 5
−3
3
−5
7
−1.5 −5
69. f x
x x2 1
71. f x sec
Continuous on ,
73. f x
sin x x
Continuous on: . . . , 6, 2, 2, 2, 2, 6, 6, 10, . . . 1 75. f x 16x4 x3 3 is continuous on 1, 2 .
f 1 33 16 and f 2 4. By the Intermediate Value Theorem, f c 0 for at least one value of c between 1 and 2.
3
−4
x 4
4
−2
The graph appears to be continuous on the interval 4, 4 . Since f 0 is not defined, we know that f has a discontinuity at x 0. This discontinuity is removable so it does not show up on the graph.
40
Chapter 1
Limits and Their Properties
77. f x x2 2 cos x is continuous on 0, . f 0 3 and f 2 1 > 0. By the Intermediate Value Theorem, f c 0 for the least one value of c between 0 and .
81. gt 2 cos t 3t
79. f x x3 x 1 f x is continuous on 0, 1 . f 0 1 and f 1 1 By the Intermediate Value Theorem, f x 0 for at least one value of c between 0 and 1. Using a graphing utility, we find that x 0.6823. 83. f x x2 x 1
g is continuous on 0, 1 .
f is continuous on 0, 5 .
g0 2 > 0 and g1 1.9 < 0.
f 0 1 and f 5 29
By the Intermediate Value Theorem, gt 0 for at least one value c between 0 and 1. Using a graphing utility, we find that t 0.5636.
1 < 11 < 29 The Intermediate Value Theorem applies. x2 x 1 11 x2 x 12 0
x 4x 3 0 x 4 or x 3 c 3 (x 4 is not in the interval.) Thus, f 3 11. 85. f x x3 x2 x 2
87. (a) The limit does not exist at x c.
f is continuous on 0, 3 .
(b) The function is not defined at x c.
f 0 2 and f 3 19
(c) The limit exists at x c, but it is not equal to the value of the function at x c.
2 < 4 < 19 The Intermediate Value Theorem applies.
(d) The limit does not exist at x c.
x3 x2 x 2 4 x3 x2 x 6 0
x 2x2 x 3 0 x2 (x2
x 3 has no real solution.) c2
Thus, f 2 4. 89.
91. The functions agree for integer values of x:
y 5 4 3 2 1 −2 −1
gx 3 x 3 x 3 x f x 3 x 3 x x 1
3 4 5 6 7
−2 −3
However, for non-integer values of x, the functions differ by 1. f x 3 x gx 1 2 x.
The function is not continuous at x 3 because lim f x 1 0 lim f x.
x→3
for x an integer
x→3
1 1 For example, f 2 3 0 3, g2 3 1 4.
Section 1.4
t 2 2 t
Continuity and One-Sided Limits
N
93. Nt 25 2 t
0
1
1.8
2
3
3.8
Nt
50
25
5
50
25
5
Number of units
50 40 30 20 10
t 2
Discontinuous at every positive even integer.The company replenishes its inventory every two months.
4
6
8
10 12
Time (in months)
95. Let V 43 r 3 be the volume of a sphere of radius r. V1 43 4.19 V5 3 53 523.6 4
Since 4.19 < 275 < 523.6, the Intermediate Value Theorem implies that there is at least one value r between 1 and 5 such that Vr 275. (In fact, r 4.0341.) 97. Let c be any real number. Then lim f x does not exist since there are both rational and x→c
irrational numbers arbitrarily close to c. Therefore, f is not continuous at c. y
1, if x < 0 0, if x 0 99. sgnx 1, if x > 0
4 3 2 1
(a) lim sgnx 1
−4 −3 −2 −1
x→0
x 1
2
3
4
−2
(b) lim sgnx 1
−3
x→0
−4
(c) lim sgnx does not exist. x→0
101. True; if f x gx, x c, then lim f x lim gx and x→c
x→c
103. False; f 1 is not defined and lim f x does not exist. x→1
at least one of these limits (if they exist) does not equal the corresponding function at x c.
105. (a) f x
b
0 ≤ x < b b < x ≤ 2b
0
(b) gx
y
2b
x 2
0 ≤ x ≤ b
b
x 2
b < x ≤ 2b
y
2b
b
x b
b
2b
NOT continuous at x b.
x b
2b
Continuous on 0, 2b .
41
42
Chapter 1
107. f x
Limits and Their Properties
x c2 c
x
, c > 0
Domain: x c2 ≥ 0 ⇒ x ≥ c2 and x 0, c2, 0 0, x c2 c
lim
x
x→0
lim
x c2 c
x→0
x
x c2 c x c2 c
x c2 c2 1 1 lim x→0 xx c2 c x→0 x c2 c 2c
lim
Define f 0 12c to make f continuous at x 0. 109. hx xx
15
h has nonremovable discontinuities at x ± 1, ± 2, ± 3, . . . . −3
3 −3
Section 1.5 1.
Infinite Limits
lim 2
x x2 4
lim 2
x x2 4
x→2
x→2
3.
lim tan
x 4
lim tan
x 4
x→2
x→2
5. f x
1 x2 9
x
3.5
3.1
3.01
3.001
2.999
2.99
2.9
2.5
f x
0.308
1.639
16.64
166.6
166.7
16.69
1.695
0.364
lim f x
x→3
lim f x
x→3
7. f x
x2 x2 9
x
3.5
3.1
3.01
3.001
2.999
2.99
2.9
2.5
f x
3.769
15.75
150.8
1501
1499
149.3
14.25
2.273
lim f x
x→3
lim f x
x→3
Section 1.5
9. lim x→0
1 1 lim x2 x→0 x2
Therefore, x 0 is a vertical asymptote.
Infinite Limits
43
x2 2 x 2x 1
11. lim x→2
x2 2 x 2x 1
lim
x→2
Therefore, x 2 is a vertical asymptote. lim
x2 2 x 2x 1
lim
x2 2 x 2x 1
x→1
x→1
Therefore, x 1 is a vertical asymptote.
13.
lim
x→2
x2
x2 x2 and lim 2 x→2 x 4 4
15. No vertical asymptote since the denominator is never zero.
Therefore, x 2 is a vertical asymptote. lim
x→2
x2 x2 and lim 2 x→2 x 4 x2 4
Therefore, x 2 is a vertical asymptote.
17. f x tan 2x x
21.
lim
x→2
lim
x→2
sin 2x has vertical asymptotes at cos 2x
2n 1 n , n any integer. 4 4 2 x
x 2x 1
x x 2x 1
Therefore, x 2 is a vertical asymptote. lim
x x 2x 1
lim
x x 2x 1
x→1
x→1
19. lim 1 t→0
4 4 lim 1 2 t→0 t2 t
Therefore, t 0 is a vertical asymptote.
x3 1 x 1x2 x 1 x1 x1
23. f x
has no vertical asymptote since lim f x lim x2 x 1 3
x→1
x→1
Therefore, x 1 is a vertical asymptote.
25. f x
x 5x 3 x3 ,x5 x 5x2 1 x2 1
No vertical asymptotes. The graph has a hole at x 5.
27. st
t has vertical asymptotes at t n, n sin t
a nonzero integer. There is no vertical asymptote at t 0 since lim t→0
t 1. sin t
44
Chapter 1
Limits and Their Properties
x2 1 lim x 1 2 x→1 x 1 x→1
29. lim
31.
lim
x2 1 x1
lim
x2 1 x1
x→1
2
−3
x→1
3
8
−3
3
Vertical asymptote at x 1
−8
−5
Removable discontinuity at x 1
33. lim x→2
37.
x3 x2
lim
x→3
45. lim
x→
x→3
x2 2x 3 x1 4 lim x→3 x 2 x2 x 6 5
41. lim 1 x→0
35. lim
1 x
x2 x x 1 lim 2 x→1 x 1x 1 x→1 x 1 2
39. lim
43. lim x→0
x lim x sin x 0 csc x x→
x2 x 3x 3
47.
2
2 sin x
lim
x→ 12
x sec x and
lim
x→ 12
x sec x .
Therefore, lim x sec x does not exist. x→ 12
49. f x
x2 x 1 x3 1
lim f x lim
x→1
x→1
51. f x
1 x1
1 x2 25
lim f x
x→5
0.3
3
−8
−4
8
5
−0.3
−3
53. A limit in which f x increases or decreases without bound as x approaches c is called an infinite limit. is not a number. Rather, the symbol
55. One answer is f x
x3 x3 . x 6x 2 x2 4x 12
lim f x
x→c
says how the limit fails to exist. 57.
k , 0 < r < 1. Assume k 0. 1r k lim S lim (or if k < 0) r→1 r→1 1 r
59. S
y 3 2 1 −2
x
−1
1 −1 −2
3
Section 1.5
61. C
528x , 0 ≤ x < 100 100 x
63. (a) r
(a) C25 $176 million
(b) r
(b) C50 $528 million (c) C75 $1584 million
x
x→25
f x
1
0.5
0.2
0.1
0.01
0.001
0.0001
0.1585
0.0411
0.0067
0.0017
0
0
0
0.5
lim
x→0 − 1.5
x sin x 0 x
1.5
− 0.25
(b)
x f x
1
0.5
0.2
0.1
0.01
0.001
0.0001
0.1585
0.0823
0.0333
0.0167
0.0017
0
0
0.001
0.0001
0.1667 0.1667
0.1667
0.25
− 1.5
lim
1.5
x→0
x sin x 0 x2
− 0.25
(c)
x f x
1
0.5
0.2
0.1
0.1585
0.1646
0.1663
0.1666
0.01
0.25
− 1.5
lim
1.5
x→0
x sin x 0.1167 16 x3
− 0.25
(d)
x f x
1
0.5
0.2
0.1
0.1585
0.3292
0.8317
1.6658
0.01
0.001
0.0001
16.67
166.7
1667.0
1.5
− 1.5
1.5
− 1.5
For n ≥ 3, lim x→0
x sin x . xn
lim
x→0
x sin x x4
7 ftsec 12
215 3 ftsec 2 625 225
(c) lim
528 Thus, it is not possible. (d) lim x→100 100 x 65. (a)
27 625 49
2x 625 x2
Infinite Limits
45
46
Chapter 1
Limits and Their Properties (b) The direction of rotation is reversed.
67. (a) Because the circumference of the motor is half that of the saw arbor, the saw makes 17002 850 revolutions per minute.
(d)
(c) 220 cot 210 cot : straight sections. The angle subtended in each circle is
2 2 2
(e)
2 .
0.3
0.6
0.9
1.2
1.5
L
306.2
217.9
195.9
189.6
188.5
450
Thus, the length of the belt around the pulleys is 20 2 10 2 30 2 .
0
Total length 60 cot 30 2
(f)
0, 2
Domain:
2
0
lim
→ 2
L 60 188.5
(All the belts are around pulleys.) (g) lim L →0
71. False; let
69. False; for instance, let f x
1, f x x 3,
x2 1 or x1
x0 x 0.
The graph of f has a vertical asymptote at x 0, but f 0 3.
x gx 2 . x 1 73. Given lim f x and lim gx L: x→c
x→c
(2) Product:
If L > 0, then for L2 > 0 there exists 1 > 0 such that gx L < L2 whenever 0 < x c < 1. Thus, L2 < gx < 3L2. Since lim f x then for M > 0, there exists 2 > 0 such that f x > M2L whenever
x c
x→c
< 2. Let be the smaller of 1 and 2. Then for 0 < x c < , we have f xgx > M2LL2 M.
Therefore lim f xgx . The proof is similar for L < 0. x→c
(3) Quotient: Let > 0 be given.
There exists 1 > 0 such that f x > 3L2 whenever 0 < x c < 1 and there exists 2 > 0 such that gx L <
L2 whenever 0 < x c < 2. This inequality gives us L2 < gx < 3L2. Let be the smaller of 1 and 2. Then for 0 <
x c
gx f x
<
3L2 . 3L2
Therefore, lim
x→c
75. Given lim
x→c
< , we have
gx 0. f x
1 0. f x
Suppose lim f x exists and equals L. Then, x→c
lim 1 1 1 x→c 0. x→c f x lim f x L lim
x→c
This is not possible. Thus, lim f x does not exist. x→c
Review Exercises for Chapter 1
Review Exercises for Chapter 1 1. Calculus required. Using a graphing utility, you can estimate the length to be 8.3. Or, the length is slightly longer than the distance between the two points, 8.25. 3.
x
0.1
0.01 0.001
f x
0.26
0.25
0.250
1
0.001
0.01
0.1
0.2499
0.249
0.24
−1
lim f x 0.25
x→0
5. hx
1
−1
x2 2x x
7. lim 3 x 3 1 2
(a) lim hx 2
x→1
x→0
Let > 0 be given. Choose . Then for
(b) lim hx 3
0 < x 1 < , you have
x→1
x 1 < 1 x < 3 x 2 < f x L < 9. lim x2 3 1 x→2
1
x 2.
Let > 0 be given. We need x2 3 1 < ⇒ x2 4 x 2x 2 < ⇒ x 2 < Assuming, 1 < x < 3, you can choose 5. Hence, for 0 < x 2 < 5 you have
1
x 2 < 5 < x 2
x 2x 2 < x2 4 < x2 3 1 < f x L <
11. lim t 2 4 2 6 2.45
13. lim
t→2
t→4
15. lim
x→4
x 2
x4
lim
x→4
lim
x→4
x 2
17. lim
x 2x 2 1 x 2
1 4 2
x→0
1 4
x3 125 x 5x2 5x 25 lim x→5 x 5 x→5 x5
19. lim
lim x2 5x 25 x→5
75
21. lim
x→0
1 t2 1 lim t2 4 t→2 t 2 4
1x 1 1 1 x 1 lim x→0 x xx 1 1 1 lim x→0 x 1
1 cos x x lim x→0 sin x sin x
1 xcos x 10 0
47
48
Chapter 1
23. lim
x→0
Limits and Their Properties
sin 6 x 12 sin 6 cos x cos 6 sin x 12 lim x→0 x x 1 2
lim
x→0
0
3 sin x cos x 1 lim x x→0 2 x
3
2
1
3
2
25. lim f x gx 4 3 2 3
2
1
x→c
27. f x (a)
2x 1 3
x1
x
1.1
1.01
1.001
1.0001
f x
0.5680
0.5764
0.5773
0.5773
lim
2x 1 3
x1
x→1
(c) lim
2x 1 3
x→1
x1
Actual limit is 33.
0.577 lim
2x 1 3
x1
x→1
lim
2 2x 1 3
2 23
sa st 4.942 200 4.9t2 200 lim t→a t→4 at 4t t→4
2 0
1 3 3 3
29. lim
lim
−1
2x 1 3
2x 1 3 x 1 2x 1 3
x→1
2
2x 1 3
lim x→1
(b)
31. lim x→3
x 3 x3
lim
x→3
x 3 1 x3
4.9t 4t 4 4t
lim 4.9t 4 39.2 msec t→4
33. lim f x 0 x→2
35. lim ht does not exist because lim ht 1 1 2 and t→1
t→1
lim ht 121 1 1.
t→1
37. f x x 3 lim x 3 k 3 where k is an integer.
x→k
lim x 3 k 2 where k is an integer.
x→k
Nonremovable discontinuity at each integer k Continuous on k, k 1 for all integers k 41. f x lim
x→2
1 x 22
1 x 22
Nonremovable discontinuity at x 2 Continuous on , 2 2,
39. f x
3x2 x 2 3x 2x 1 x1 x1
lim f x lim 3x 2 5
x→1
x→1
Removable discontinuity at x 1 Continuous on , 1 1, 43. f x
3 x1
lim f x
x→1
lim f x
x→1
Nonremovable discontinuity at x 1 Continuous on , 1 1,
Problem Solving for Chapter 1 45. f x csc
x 2
49
47. f 2 5 Find c so that lim cx 6 5.
Nonremovable discontinuities at each even integer. Continuous on
x→2
c2 6 5
2k, 2k 2
2c 1
for all integers k. c
x2 4 x2 x 2 x2 x2
51. f x
49. f is continuous on 1, 2. f 1 1 < 0 and f 2 13 > 0. Therefore by the Intermediate Value Theorem, there is at least one value c in 1, 2 such that 2c3 3 0.
1 2
(a) lim f x 4 x→2
(b) lim f x 4 x→2
(c) lim f x does not exist. x→2
53. gx 1
2 x
55. f x
Vertical asymptote at x 10
Vertical asymptote at x 0
57.
lim
x→2
2x2 x 1 x2
61. lim
x2 2x 1 x1
65. lim
sin 4x 4 sin 4x lim x→0 5x 5 4x
69. C
80,000p , 0 ≤ 0 < 100 100 p
x→1
x→0
8 x 102
59.
lim
x→1
x1 1 1 lim x3 1 x→1 x2 x 1 3
63. lim x x→0
54
67. lim x→0
(a) C15 $14,117.65
(b) C50 $80.000
(c) C90 $720,000
(d)
lim
p→100
1 x3
csc 2x 1 lim x →0 x sin 2x x
80,000p 100 p
Problem Solving for Chapter 1 1. (a) Perimeter PAO x2 y 12 x2 y2 1
x2
x2
1
Perimeter PBO x 1 2
2
y2
x2
x2
x4
y2
1
1
x 12 x4 x2 x4 1
(c) lim r x x→0
101 2 1 101 2
(b) rx
x2 x2 12 x2 x4 1 x 12 x4 x2 x4 1
x
4
2
1
0.1
0.01
Perimeter PAO
33.02
9.08
3.41
2.10
2.01
Perimeter PBO
33.77
9.60
3.41
2.00
2.00
rx
0.98
0.95
1
1.05
1.005
50
Chapter 1
Limits and Their Properties
3. (a) There are 6 triangles, each with a central angle of 60 3. Hence,
5. (a) Slope
12bh 6 121 sin 3
Area hexagon 6
(b) Slope of tangent line is
3 3 2.598. 2
y 12 h = sin θ
h = sin 60°
y
1
1
mx Error:
3 3 0.5435. 2
12bh n 121 sin 2n n sin 22 n.
An
12
2.598
3
24
48
3.106
An
12 169 x2 12 169 x2
lim
x2 25 x 512 169 x2
lim
x 5 12 169 x2
3.139
Letting x 2 n,
144 169 x2 x 512 169 x2
x→5
(d) As n gets larger and larger, 2 n approaches 0.
12 169 x2 x5
lim
x→5
96
3.133
x→5
x→5
An n
6
5 169 x Tangent line 12 12
(d) lim mx lim
(b) There are n triangles, each with central angle of 2 n. Hence,
n
5 x 5 12
169 x2 12 x5
x→5
(c)
5 . 12
(c) Q x, y x, 169 x2
θ
60°
12 5
5 10 12 12 12
This is the same slope as part (b).
sin2 n sin2 n sin x 2 n 2 n x
which approaches 1 .
7. (a) 3 x1 3 ≥ 0
(b)
(c)
0.5
x1 3 ≥ 3
x ≥ 27
− 30
Domain: x ≥ 27, x 1 (d) lim f x lim x→1
12 − 0.1
3 x1 3 2
x1
x→1
3 x1 3 2 3 x1 3 2
3 x1 3 4 x→1 x 1 3 x1 3 2
lim lim
x→1 x1 3
lim
x→1
1
x2 3
x1 3 1 x1 3 1 3 x1 3 2
1 x2 3 x1 3 1 3 x1 3 2
1 1 1 1 12 2 12
9. (a) lim f x 3: g1, g4 x→2
(b) f continuous at 2: g1 (c) lim f x 3: g1, g3, g4 x→2
lim f x
x→27
3 271 3 2
27 1
2 1 0.0714 28 14
Problem Solving for Chapter 1 11.
y
13. (a)
y
4 3
2
2 1 −4 −3 −2 −1
x 1
2
3
4
1
−2 −3
x a
−4
(a)
f 1 1 1 1 1 0
(b) (i) lim Pa, bx 1 x→a
f 0 0
(ii) lim Pa, bx 0
f 12 0 1 1
(iii) lim Pa, bx 0
f 2.7 3 2 1 (b)
b
lim f x 1
x→1
lim f x 1
x→1
lim f x 1
x→1 2
(c) f is continuous for all real numbers except x 0, ± 1, ± 2, ± 3, . . .
x→a
x→b
(iv) lim Pa, bx 1 x→b
(c) Pa, b is continuous for all positive real numbers except x a, b. (d) The area under the graph of u, and above the x-axis, is 1.
51
C H A P T E R 2 Differentiation Section 2.1
The Derivative and the Tangent Line Problem . . . 53
Section 2.2
Basic Differentiation Rules and Rates of Change . 60
Section 2.3
The Product and Quotient Rules and Higher-Order Derivatives . . . . . . . . . . . . . . 67
Section 2.4
The Chain Rule . . . . . . . . . . . . . . . . . . . 73
Section 2.5
Implicit Differentiation . . . . . . . . . . . . . . . 79
Section 2.6
Related Rates . . . . . . . . . . . . . . . . . . . . 85
Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . 92
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . 98
C H A P T E R Differentiation Section 2.1
2
The Derivative and the Tangent Line Problem
Solutions to Odd-Numbered Exercises
1. (a) m 0
(c) y
3. (a), (b)
(b) m 3 y
f )4) 4
f )1) )x 1
f )1)
1)
x
1
y
f 4 f 1 x 1 f 1) 41 3 x 1 2 3
1x 1 2
6
f )4)
5
x1
5 )4, 5)
4
f )4)
f )1)
3
3 2
f )1) )1, 2)
2
1
x 1
5. f x 3 2x is a line. Slope 2
2
3
4
5
6
7. Slope at 1, 3 lim
x→0
g1 x g1 x
1 x2 4 3 x→0 x
lim
lim
x→0
1 2x x2 1 x
lim 2 2x 2 x→0
9. Slope at 0, 0 lim
t→0
f 0 t f 0 t
3t t2 0 t→0 t
lim
lim 3 t 3 t→0
11. f x 3 f x lim
x→0
lim
x→0
f x x f x x 33 x
lim 0 0 x→0
13. f x 5x fx lim
x→0
lim
x→0
15. hs 3 f x x f x x 5x x 5x x
lim 5 5 x→0
hs lim
2 s 3
s→0
hs s hs s
2 2 3 s s 3 s 3 3 lim s→0 s
2 s 3 2 lim s→0 s 3
53
54
Chapter 2
Differentiation
17. f x 2x2 x 1 f x x f x x
fx lim
x→0
lim
x→0
2x x2 x x 1 2x2 x 1 x
2x2 4xx 2x2 x x 1 2x2 x 1 x→0 x
lim
4xx 2x2 x lim 4x 2x 1 4x 1 x→0 x→0 x
lim
19. f x x3 12x f x x f x x
fx lim
x→0
lim
x→0
x x3 12x x x3 12x x
x3 3x2x 3xx2 x3 12x 12x x3 12x x→0 x
lim
3x2x 3xx2 x3 12x x→0 x
lim
lim 3x2 3xx x2 12 3x2 12 x→0
21. f x
1 x1
fx lim
x→0
f x x f x x
1 1 x x 1 x 1 lim x→0 x lim
x 1 x x 1 xx x 1x 1
lim
x xx x 1x 1
lim
1 x x 1x 1
x→0
x→0
x→0
1 x 12
23. f x x 1 fx lim
x→0
lim
x→0
lim
x→0
lim
x→0
f x x f x x x x 1
x
x 1
x x 1 x 1
xx x 1 x 1 1 x x 1 x 1
1 1 x 1 x 1 2x 1
x x 1 x 1 x x 1 x 1
Section 2.1
25. (a) f x x2 1 fx lim
x→0
The Derivative and the Tangent Line Problem
17. (b)
8
f x x f x x
(2, 5) −5
x x2 1 x2 1 lim x→0 x
5 −2
2xx x2 x→0 x
lim
lim 2x x 2x x→0
At 2, 5, the slope of the tangent line is m 22 4. The equation of the tangent line is y 5 4x 2 y 5 4x 8 y 4x 3. 27. (a) f x x3 fx lim
x→0
18. (b)
10
(2, 8)
f x x f x x
−5
5
x x3 x3 lim x→0 x
−4
3x2x 3xx2 x3 x→0 x
lim
lim 3x2 3xx x2 3x2 x→0
At 2, 8, the slope of the tangent is m 322 12. The equation of the tangent line is y 8 12x 2 y 12x 16. 29. (a) f x x fx lim
x→0
lim
18. (b) f x x f x x x x x
x
x→0
lim
x→0
lim
x→0
(1, 1)
x x x x x x
x x x x x x x 1 x x x
1 2x
At 1, 1, the slope of the tangent line is m
1 1 . 21 2
The equation of the tangent line is y1 y
1 x 1 2 1 1 x . 2 2
3
−1
5 −1
55
56
Chapter 2
31. (a) f x
Differentiation
(b)
4 x
(4, 5)
f x x f x fx lim x→0 x
lim
x x
x→0
− 12
4 4 x x x x x
xx xx x 4x x 2x x 4x x xxx x
lim
x3 2x 2x xx2 x3 x 2x 4x xxx x
x→0
12
−6
lim
x→0
10
x2x xx2 4x x→0 xxx x
lim
x2 xx 4 x→0 xx x
lim
x2 4 4 1 2 x2 x
At 4, 5, the slope of the tangent line is m1
4 3 16 4
The equation of the tangent line is 3 y 5 x 4 4 3 y x2 4 33. From Exercise 27 we know that fx 3x2. Since the slope of the given line is 3, we have
fx
3x2 3 x ± 1.
1 . 2xx
Since the slope of the given line is 12 , we have
Therefore, at the points 1, 1 and 1, 1 the tangent lines are parallel to 3x y 1 0. These lines have equations y 1 3x 1
35. Using the limit definition of derivative,
and
y 3x 2
y 1 3x 1 y 3x 2.
1 1 2xx 2 x 1.
Therefore, at the point 1, 1 the tangent line is parallel to x 2y 6 0. The equation of this line is 1 y 1 x 1 2 1 1 y1 x 2 2 1 3 y x . 2 2
37. g5 2 because the tangent line passes through 5, 2 g5
20 2 1 5 9 4 2
39. f x x ⇒ fx 1
b
Section 2.1
The Derivative and the Tangent Line Problem
43.
41. f x x ⇒ fx matches (a)
57
y 4
decreasing slope as x →
3 2 1 −4 −3 −2 −1 −1
x 1
2
3
4
−2 −3 −4
Answers will vary. Sample answer: y x 45. (a) If fc 3 and f is odd, then fc fc 3 (b) If fc 3 and f is even, then fc f c 3 47. Let x0, y0 be a point of tangency on the graph of f. By the limit definition for the derivative, fx 4 2x. The slope of the y line through 2, 5 and x0, y0 equals the derivative of f at x0: 7
5 y0 4 2x0 2 x0
6
(2, 5)
5 4
5 y0 2 x04 2x0
3 2
5 4x0 x02 8 8x0 2x02
(3, 3) (1, 3)
1 x
−2
0 x0 4x0 3 2
1
2
3
6
0 x0 1x0 3 ⇒ x0 1, 3 Therefore, the points of tangency are 1, 3 and 3, 3, and the corresponding slopes are 2 and 2. The equations of the tangent lines are y 5 2x 2
y 5 2x 2
y 2x 1
y 2x 9
49. (a) g0 3 (b) g3 0 (c) Because g1 3 , g is decreasing (falling) at x 1. 8
(d) Because g4 3 , g is increasing (rising) at x 4. 7
(e) Because g4 and g6 are both positive, g6 is greater than g4, and g6 g4 > 0. (f) No, it is not possible. All you can say is that g is decreasing (falling) at x 2. 1 51. f x 4 x3
2
By the limit definition of the derivative we have fx
3 2 4x . −2
2
1.5
1
0.5
0
0.5
1
1.5
2
f x
2
27 32
14
1 32
0
1 32
1 4
27 32
2
fx
3
27 16
3 4
3 16
0
3 16
3 4
27 16
3
x
2
−2
58
Chapter 2
Differentiation
55. f 2 24 2 4, f 2.1 2.14 2.1 3.99
f x 0.01 f x 0.01
53. gx
2x 0.01 x 0.012 2x x2 100
f2
3.99 4 0.1 Exact: f2 0 2.1 2
3
g f −2
4 −1
The graph of gx is approximately the graph of fx. 57. f x
1 x
and fx
1 . 2x3 2
5
As x → , f is nearly horizontal and thus f 0.
f −2
5
f′ −5
59. f x 4 x 32 Sx x
f 2 x f 2 x 2 f 2 x 4 2 x 32 3 1 x 12 x 2 3 x 2 3 x 2x 2 3 x x 1: Sx x 2 3 x 1
(a) x
5
S 0.1
3 3 x 2 3 x x 0.5: Sx 2 2 x 0.1: Sx
19 19 4 x 2 3 x 10 10 5
S1 −2
7
S 0.5 −1
(b) As x → 0, the line approaches the tangent line to f at 2, 3. 61. f x x2 1, c 2 f x f 2 x2 1 3 x 2x 2 lim lim lim x 2 4 x→2 x →2 x→2 x2 x2 x2
f2 lim
x→2
63. f x x3 2x2 1, c 2 f x f 2 x2x 2 x3 2x2 1 1 lim lim lim x2 4 x →2 x→2 x→2 x2 x2 x2
f2 lim
x→2
67. f x x 62 3, c 6
65. gx x , c 0 g0 lim
x→0
As x → 0 , As x → 0 ,
x gx g0 lim . Does not exist. x→0 x0 x
f6 lim
x→6
f x f 6 x6
1 →
lim
x 62 3 0 x6
lim
1 x 61 3
x
x
x
x
1
x
x
→
x→6
x→6
f
Does not exist.
Section 2.1
69. h x x 5, c 5 h5 lim
x→5
x 5 0
lim
x 5
x→5
x→5
71. f x is differentiable everywhere except at x 3. (Sharp turn in the graph.)
hx h5 x 5
lim
The Derivative and the Tangent Line Problem
x5
x5
Does not exist. 73. f x is differentiable everywhere except at x 1. (Discontinuity)
75. f x is differentiable everywhere except at x 3. (Sharp turn in the graph)
77. f x is differentiable on the interval 1, . (At x 1 the tangent line is vertical)
79. f x is differentiable everywhere except at x 0. (Discontinuity)
83. f x
81. f x x 1
The derivative from the left is lim
x→1
xx 11 ,, xx >≤ 11 3 2
The derivative from the left is
f x f 1 x1 0 lim 1. x→1 x1 x1
lim
x→1
f x f 1 x 13 0 lim x→1 x1 x1 lim x 12 0.
The derivative from the right is lim
x→1
x→1
f x f 1 x1 0 lim 1. x→1 x1 x1
The one-sided limits are not equal. Therefore, f is not differentiable at x 1.
The derivative from the right is lim
x→1
f x f 1 x 12 0 lim x→1 x1 x1 lim x 1 0. x→1
These one-sided limits are equal. Therefore, f is differentiable at x 1. f1 0 85. Note that f is continuous at x 2. f x
4xx 1,3, xx >≤ 22 2
f x f 2 x2 1 5 lim lim x 2 4. x→2 x→2 x2 x2
The derivative from the left is lim x→2
The derivative from the right is lim x→2
f x f 2 4x 3 5 lim lim 4 4. x→2 x→2 x2 x2
The one-sided limits are equal. Therefore, f is differentiable at x 2. f2 4 87. (a) The distance from 3, 1 to the line mx y 4 0 is d
y
Ax1 By1 C A2 B2
3
m3 11 4 3m 3 . m2 1
m2 1
2
1 x
(b)
1
5
2
3
4
The function d is not differentiable at m 1. This corresponds to the line y x 4, which passes through the point 3, 1.
−4
4 −1
59
60
Chapter 2
Differentiation
89. False. the slope is lim
x→0
f 2 x f 2 . x
91. False. If the derivative from the left of a point does not equal the derivative from the right of a point, then the derivative does not exist at that point. For example, if f x x , then the derivative from the left at x 0 is 1 and the derivative from the right at x 0 is 1. At x 0, the derivative does not exist.
93. f x
0,x sin1x,
x0 x0
Using the Squeeze Theorem, we have x ≤ x sin1x ≤ x , x 0. Thus, lim x sin1x 0 f 0 and f is continuous at x→0 x 0. Using the alternative form of the derivative we have lim
x→0
f x f 0 x sin1x 0 1 lim lim sin . x→0 x→0 x0 x0 x
Since this limit does not exist (it oscillates between 1 and 1), the function is not differentiable at x 0. gx
x0, sin1x, xx 00 2
Using the Squeeze Theorem again we have x2 ≤ x2 sin1x ≤ x2, x 0. Thus, lim x2 sin1x 0 f 0 and f is continux→0 ous at x 0. Using the alternative form of the derivative again we have lim
x→0
f x f 0 x2 sin1x 0 1 lim lim x sin 0. x→0 x→0 x0 x0 x
Therefore, g is differentiable at x 0, g0 0.
Section 2.2
Basic Differentiation Rules and Rates of Change
y x12
1. (a)
y 12 x12 y1 12 3. y 8 y 0
11. f x x 1 fx 1
19. y y
sin cos 2 cos sin 2
(b)
y x32
y x2
(c)
y 32 x12
(d)
y 2x
y1 32
y 3x2
y1 2
5. y x6
7. y
y 6x5
y1 3
1 x7 x7
y 7x8
5 x x15 9. y
1 1 y x45 45 5 5x
7 x8
15. gx x 2 4x3
13. f t 2t2 3t 6
17. st t3 2t 4
gx 2x 12x 2
fx 4t 3
21. y x2
1 cos x 2
y 2x
1 sin x 2
y x3
st 3t2 2
23. y
1 3 sin x x
y
1 3 cos x x2
Section 2.2
Function
Rewrite
Derivative
Simplify
5 y x2 2
y 5x3
y
5 x3
3 2x3
3 y x3 8
y
y
9 8x4
x
y x12
1 y x32 2
5 25. y 2 2x 27. y 29. y
Basic Differentiation Rules and Rates of Change
x
31. f x
3 3x2, 1, 3 x2
fx
6x3
9 4 x 8
y
1 7 33. f x x3, 2 5
6 3 x
1 2x32
0, 21
4x2 4x 1
21 fx x2 5
y 8x 4 y0 4
f0 0
f1 6
2 2 39. f x x 5 3x
37. f 4 sin , 0, 0
41. gt t2
fx 2x 6x3 2x
f 4 cos 1
6 x3
4 t2 4t3 t3
gt 2t 12t4 2t
f0 41 1 3 43. f x
y 2x 12, 0, 1
35.
x3 3x2 4 x 3 4x2 x2
45. y xx2 1 x3 x y 3x2 1
8 x3 8 fx 1 3 x x3 3 x x12 6x13 47. f x x 6
1 1 2 fx x12 2x23 2 2 x x23
49. hs s45 s23 4 2 4 2 h(s s45 s13 15 13 5 3 5s 3s
51. f x 6 x 5 cos x 6x12 5 cos x fx 3x12 5 sin x
3 x
5 sin x
55. (a) f x
53. (a) y x4 3x2 2 y 4x3 6x
fx
At 1, 0: y 413 61 2. y 0 2x 1
Tangent line:
3
−2
2x34
3 74 3 x 74 2 2x
At 1, 2, f1
2x y 2 0 (b)
2 4 3 x
Tangent line:
3 2 3 y 2 x 1 2 3 7 y x 2 2
2
(1, 0)
3x 2y 7 0
−1
(b)
5
(1, 2) −2
7 −1
12 t4
61
62
Chapter 2
Differentiation
59. y
57. y x4 8x2 2 y 4x3 16x
1 x2 x2
y 2x3
4xx2 4 4xx 2x 2
2 cannot equal zero. x3
Therefore, there are no horizontal tangents.
y 0 ⇒ x 0, ± 2 Horizontal tangents: 0, 2, 2, 14, 2, 14 61. y x sin x, 0 ≤ x < 2
63. x 2 kx 4x 9
y 1 cos x 0
2x k 4
cos x 1 ⇒ x
x 2 2x 4x 4x 9 ⇒ x 2 9 ⇒ x ± 3.
Horizontal tangent: ,
3 k x3 x 4
k 3 x2 4
Equate derivatives
Hence, k 2x 4 and
At x , y .
65.
Equate functions
For x 3, k 2 and for x 3, k 10.
Equate functions
Equate derivatives
3 2 x 3 2 3 3 4 3 3 x 3 ⇒ x x 3 ⇒ x 3 ⇒ x 2 ⇒ k 3. Hence, k x and 4 x 4 4 4 2 67. (a) The slope appears to be steepest between A and B.
(c)
y
(b) The average rate of change between A and B is greater than the instantaneous rate of change at B.
f B C A
D
E x
69. gx f x 6 ⇒ gx fx
y
71. 3
f f
1
x 3
2
1
1
2
3
2
If f is linear then its derivative is a constant function. f x ax b fx a
Section 2.2
Basic Differentiation Rules and Rates of Change
73. Let x1, y1 and x2, y2 be the points of tangency on y x2 and y x2 6x 5, respectively. The derivatives of these functions are y 2x ⇒ m 2x1
and
y 2x 6 ⇒ m 2x2 6.
m 2x1 2x2 6 x1 x2 3 Since y1 x12 and y2 x22 6x2 5,
y
5
m
y2 y1 x22 6x2 5 x12 2x2 6. x2 x1 x2 x1
4 3
)2, 3)
2
)1, 1)
1
x22 6x2 5 x2 32 2x2 6 x2 x2 3
x 2
3
)1, 0) 2
3
−1
x22 6x2 5 x22 6x2 9 2x2 62x2 3 2x22 12x2 14 4x22 18x2 18 2x22 6x2 4 0
y
5
2x2 2x2 1 0
4
)2, 4)
3
x2 1 or 2
2
x2 1 ⇒ y2 0, x1 2 and y1 4
1
x −1
Thus, the tangent line through 1, 0 and 2, 4 is
−2
40 y0 x 1 ⇒ y 4x 4. 21
x2 2 ⇒ y2 3, x1 1 and y1 1 Thus, the tangent line through 2, 3 and 1, 1 is y1
32 11x 1 ⇒ y 2x 1.
75. f x x, 4, 0
77. f1 1
1 1 fx x12 2 2 x 1 2 x
3.64
0y 4 x 0.77 3.33
4 x 2 x y 4 x 2 x x 4 x 2x x 4, y 2 The point 4, 2 is on the graph of f. Tangent line:
y2
02 x 4 4 4
4y 8 x 4 0 x 4y 4
1.24
63
64
Chapter 2
Differentiation
79. (a) One possible secant is between 3.9, 7.7019 and 4, 8: y8
20
8 7.7019 x 4 4 3.9
(4, 8) −2
y 8 2.981x 4
12 −2
y Sx 2.981x3.924 3 3 (b) fx x12 ⇒ f4 2 3 2 2 Tx 3x 4 8 3x 4 Sx is an approximation of the tangent line Tx. (c) As you move further away from 4, 8, the accuracy of the approximation T gets worse. 20
f T
−2
12
−2
(d)
x
3
f 4 x
1
T4 x
1
2
1
0.5
2.828
5.196
6.548
2
5
6.5
81. False. Let f x x2 and gx x2 4. Then fx gx 2x, but f x gx.
0.1
0
0.1
0.5
7.702
8
8.302
9.546
11.180
14.697
18.520
7.7
8
8.3
9.5
11
14
17
1
2
83. False. If y 2, then dydx 0. 2 is a constant.
85. True. If gx 3f x, then gx 3fx.
87. f t 2t 7, 1, 2 ft 2 Instantaneous rate of change is the constant 2. Average rate of change: f 2 f 1 22 7 21 7 2 21 1 (These are the same because f is a line of slope 2.)
3
1 89. f x , 1, 2 x fx
1 x2
Instantaneous rate of change:
1, 1 ⇒ f1 1
2, 21 ⇒ f2 41 Average rate of change: f 2 f 1 12 1 1 21 21 2
Section 2.2
Basic Differentiation Rules and Rates of Change
91. (a) st 16t2 1362
st 4.9t2 v0t s0
93.
4.9t2 120t
vt 32t (b)
65
vt 9.8t 120
s2 s1 1298 1346 48 ftsec 21
v5 9.85 120 71 msec
(c) vt st 32t
v10 9.810 120 22 msec
When t 1: v1 32 ftsec. When t 2: v2 64 ftsec. (d) 16t2 1362 0 t2 (e) v
1362 1362 9.226 sec ⇒ t 16 4
1362
4
32
1362
4
8 1362 295.242 ftsec 2 97. v 40 mph 3 mimin
v
23 mimin6 min 4 mi
Velocity (in mph)
60 50 40
v 0 mph 0 mimin
30
0 mimin2 min 0 mi
20 10
v 60 mph 1mimin
t 2
4
6
8
s
Distance (in miles)
95.
10
Time (in minutes)
10 8
(10, 6) 6
(6, 4) 4
(8, 4) 2
(0, 0)
t 2
1 mimin2 min 2 mi
4
6
8
(The velocity has been converted to miles per hour) (b) Using a graphing utility, you obtain
99. (a) Using a graphing utility, you obtain R 0.167v 0.02.
B 0.00586v2 0.0239v 0.46.
(c) T R B 0.00586v2 0.1431v 0.44
(d)
60
T
dT 0.01172v 0.1431 (e) dv
B R
For v 40, T40 0.612.
0
For v 80, T80 1.081.
(f) For increasing speeds, the total stopping distance increases.
For v 100, T100 1.315.
101. A s2,
dA 2s ds
100
0
103.
When s 4 m, dA 8 square meters per meter change in s. ds
C
1,008,000 6.3Q Q
dC 1,008,000 6.3 dQ Q2 C351 C350 5083.095 5085 $1.91 When Q 350,
dC $1.93. dQ
105. (a) f1.47 is the rate of change of the amount of gasoline sold when the price is $1.47 per gallon. (b) f1.47 is usually negative. As prices go up, sales go down.
10
Time (in minutes)
66
Chapter 2
Differentiation
107. y ax2 bx c Since the parabola passes through 0, 1 and 1, 0, we have
0, 1: 1 a02 b0 c ⇒ c 1 1, 0: 0 a12 b1 1 ⇒ b a 1. Thus, y ax2 a 1x 1. From the tangent line y x 1, we know that the derivative is 1 at the point 1, 0. y 2ax a 1 1 2a1 a 1 1a1 a2 b a 1 3 Therefore, y 2x2 3x 1. 109. y x3 9x y 3x2 9 Tangent lines through 1, 9: y 9 3x2 9x 1
x3 9x 9 3x3 3x2 9x 9 0 2x3 3x2 x22x 3 x 0 or x 32 3 81 3 9 The points of tangency are 0, 0 and 32 , 81 8 . At 0, 0 the slope is y0 9. At 2 , 8 the slope is y 2 4 .
Tangent lines: y 0 9x 0
9 3 y 81 8 4 x 2
and
y 94 x 27 4
y 9x 9x y 0
111. f x
9x 4y 27 0
xax , b, 3
2
x ≤ 2 x > 2
f must be continuous at x 2 to be differentiable at x 2. lim f x lim ax3 8a
x→2
x→2
lim f x lim x2 b 4 b
x→2
x→2
fx
3ax , 2x, 2
8a 4 b 8a 4 b
x < 2 x > 2
For f to be differentiable at x 2, the left derivative must equal the right derivative. 3a22 22 12a 4 a 13 4 b 8a 4 3
Section 2.3
The Product and Quotient Rules and Higher-Order Derivatives
67
113. Let f x cos x. f x x f x x
fx lim
x→0
lim
cos x cos x sin x sin x cos x x
lim
cos xcos x 1 sin x lim sin x x→0 x x
x→0
x→0
0 sin x1 sin x
Section 2.3
The Product and Quotient Rules and Higher-Order Derivatives
1. gx x 2 1x 2 2x gx x 2 12x 2 x 2 2x2x 2x3 2x 2 2x 2 2x3 4x 2 4x3 6x 2 2x 2
3 tt2 4 t13t2 4 3. ht
1 ht t132t t 2 4 t 23 3 2t 43 7t2 4 3t23
5. f x x 3 cos x
x x2 1
7. f x
fx x 3sin x cos x3x 2 3x cos x 2
9. hx
x3
sin x
3 x x13 3 x3 1 x 1
1 x 3 1 x23 x133x 2 3 hx x 3 12
1 x2 x 2 11 x2x 2 2 2 x 1 x 12
fx
11. gx
t2 4 3t23
sin x x2
gx
x 2cos x sin x2x x cos x 2 sin x x 22 x3
x3 1 x9x 2 3x23x 3 12 1 8x 3 x 3 12
3x23
13. f x x3 3x2x2 3x 5 fx x3 3x4x 3 2x2 3x 53x2 3 10x4 12x3 3x2 18x 15
15. f x fx
x2 4 x3
x 32x x 2 41 2x 2 6x x 2 4 x 32 x 32
f0 15
f1
f x x cos x
17.
fx xsin x cos x1 cos x x sin x f
4
2
2
2 2 4 4 2 8
164 1 1 32 4
x 2 6x 4 x 32
68
Chapter 2
Differentiation
Function
Rewrite
Derivative
Simplify 2x 2 3
19. y
x 2 2x 3
1 2 y x2 x 3 3
2 2 y x 3 3
y
21. y
7 3x3
y
7 3 x 3
y 7x4
y
23. y
4x32 x
y 4x, x > 0
y 2x12
y
25. f x fx
7 x4
2 x
3 2x x2 x2 1
x2 12 2x 3 2x x22x x2 12
2x2 4x 2 2x 12 2 x2 12 x 12
2 ,x1 x 12
27. f x x 1 fx 1
4 4x x x3 x3
x 34 4x1 x 2 6x 9 12 x 32 x 32
29. f x
2x 5 2x12 5x1 2 x
5 5 fx x12 x32 x32 x 2 2
x 2 6x 3 x 32
31. hs s3 22 s6 4s3 4 hs 6s5 12s2 6s2s3 2 1 2x 1 x 2x 1 33. f x x3 xx 3 x 2 3x 2
fx
x 2 3x2 2x 12x 3 2x 2 6x 4x 2 8x 3 x 2 3x2 x 2 3x2 2x 2 2x 3 2x 2 2x 3 2 x 2 3x2 x x 32
35. f x 3x3 4xx 5x 1 fx 9x2 4x 5x 1 3x3 4x1x 1 3x3 4xx 51 9x2 4x2 4x 5 3x 4 3x 3 4x 2 4x 3x 4 15x3 4x2 20x 9x 4 36x3 41x 2 16x 20 6x 4 12x 3 8x 2 16x 15x 4 48x 3 33x 2 32x 20
37. f x fx
x2 c2 x2 c2
x2 c22x x2 c22x x2 c22 4xc x2 c22 2
39. f x t2 sin t ft t2 cos t 2t sin t tt cos t 2 sin t
2x 5 2x 5 2x32 2xx
Section 2.3
41. f t ft
The Product and Quotient Rules and Higher-Order Derivatives
cos t t
43. f x x tan x
t sin t cos t t sin t cos t t2 t2
4 t 8 sec t t14 8 sec t 45. gt
1 1 gt t34 8 sec t tan t 34 8 sec t tan t 4 4t
fx 1 sec2 x tan2 x
47. y
31 sin x 3 sec x tan x 2 cos x 2
3 3 y sec x tan x sec2 x sec xtan x sec x 2 2 3 sec x tan x tan2 x 1 2 51. f x x2 tan x
49. y csc x sin x
fx x2 sec2 x 2x tan x
y csc x cot x cos x
xx sec2 x 2 tan x
cos x cos x sin2 x
cos xcsc2 x 1 cos x cot2 x 53. y 2x sin x x 2 cos x
55. gx
y 2x cos x 2 sin x x 2sin x 2x cos x 4x cos x 2 sin x x 2 sin x 57. g g
1 sin cos
sin 12
y y
61.
(form of answer may vary)
1 csc x 1 csc x
1 csc xcsc x cot x 1 csc xcsc x cot x 2 csc x cot x 1 csc x2 1 csc x2
6 212 2 3 43
2
ht ht h
2x2 8x 1 x 22
1 sin
y
59.
gx
xx 122x 5
sec t t tsec t tan t sec t1 t2 sec tt tan t 1 t2 sec tan 1 1 2 2
(form of answer may vary)
69
70
Chapter 2
Differentiation
63. (a) f x x3 3x 1x 2, 1, 3
53. (b)
10
fx x3 3x 11 x 23x2 3 − 10
4x3 6x2 6x 5
10
(1, − 3)
f1 1 slope at 1, 3.
− 10
Tangent line: y 3 1x 1 ⇒ y x 2 65. (a) f x
x , 2, 2 x1
51. (b)
fx
1 x 11 x1 x 12 x 12
f2
1 1 slope at 2, 2. 2 12
6
(2, 2) −3
6
−3
Tangent line: y 2 1x 2 ⇒ y x 4
4 , 1
f x tan x,
67. (a)
fx sec2 x f
55. (b)
4
( ( π ,1 4
−
4 2 slope at 4 , 1.
−4
Tangent line:
y12 x y 1 2x
4
2
4x 2y 2 0
69. f x fx
x2 x1
6 x 23 3x1 x 22 x 22
71. fx
x 12x x21 x 12
gx
x2 2x xx 2 x 12 x 12
gx
fx 0 when x 0 or x 2.
6 x 25 5x 41 x 22 x 22
5x 4 3x 2x 4 f x 2 x 2 x 2 x 2
f and g differ by a constant.
Horizontal tangents are at 0, 0 and 2, 4. 73. f x x n sin x fx x n cos x nx n1 sin x x
n1
x cos x n sin x
When n 1: fx x cos x sin x.
75. Area At 2t 1t 2t32 t12 At 2
When n 4: fx x3x cos x 4 sin x. For general n, fx x n1 x cos x n sin x.
12
3t12
When n 2: fx xx cos x 2 sin x. When n 3: fx x2x cos x 3 sin x.
32t 21t
12
1 12 t 2
6t 1 2 cm sec 2t
Section 2.3
77.
C 100
x , 200 x x 30 2
The Product and Quotient Rules and Higher-Order Derivatives
1 ≤ x
dC 400 30 100 3 dx x x 302
79.
Pt 500 1
4t 50 t2
50 t24 4t2t 50 t22
Pt 500
200 4t 2
50 t
(a) When x 10:
dC $38.13. dx
500
(b) When x 15:
dC $10.37. dx
2000
(c) When x 20:
dC $3.80. dx
2 2
50 t2
50 t 2 2
P2 31.55 bacteria per hour
As the order size increases, the cost per item decreases. 1 cos x
sec x
81. (a)
d d 1 cos x0 1sin x sin x 1 sec x dx dx cos x cos x2 cos x cos x cos x
d d 1 sin x0 1cos x cos x 1 csc x dx dx sin x sin x2 sin x sin x sin x
cot x
(c)
sin x
cos x sec x tan x
1 sin x
csc x
(b)
cos x csc x cot x sin x
cos x sin x
d d cos x sin xsin x cos xcos x sin2 x cos2 x 1 2 csc2 x cot x 2 dx dx sin x sin x sin2 x sin x
85. f x
83. f x 4x32 fx 6x12 f x 3x12
3
fx
1 x 11 x1 x 12 x 12
f x
2 x 13
x
87. f x 3 sin x
89. fx x2
fx 3 cos x
f x 2x
f x 3 sin x
93.
95. f x 2gx hx
y 4
fx 2gx hx
3
f2 2g2 h2
2
22 4
1 x 1
2
3
4
f 2 0 One such function is f x x 22.
0
x x1
91. f x 2x f 4x
97. f x
1 1 2x12 2 x
gx hx
fx
hxgx gxhx hx 2
f2
h2g2 g2h2 h2 2
12 34 12
10
71
72
Chapter 2
Differentiation
101. vt 36 t2, 0 ≤ t ≤ 6
y
99. f′
at 2t
2
f
1
v3 27 msec x
2
1
1
a3 6 msec
2
The speed of the object is decreasing. f″
It appears that f is cubic; so f would be quadratic and f would be linear. 103. vt at
100t 2t 15
(a) a5
2t 15100 100t2 2t 152
(b) a10
1500
1.2 ftsec2 210 15 2
1500 2t 152
(c) a20
1500
0.5 ftsec2 220 15 2
1500 2.4 ftsec2 25 15 2
105. f x gxhx fx gxhx hxgx
(a)
f x gxh x gxhx hxg x hxgx gxh x 2gxhx hxg x f x gxh x gxh x 2gxh x 2g xhx hxg x hxg x gxh x 3gx h x 3g xhx g xhx f
4x
gxh4x gxh x 3gxh x 3g xh x 3g xh x 3g xhx g xhx g4xhx gxh4x 4gxh x 6g xh x 4g xhx g4xhx
(b) f nx gxhnx
nn 1n 2 . . . 21 nn 1n 2 . . . 21 gxhn1x g xhn2x 1 n 1n 2 . . . 21 21 n 2n 3 . . . 21
nn 1n 2 . . . 21 g xhn3x . . . 321 n 3n 4 . . . 21
nn 1n 2 . . . 21 n1 g xhx gnxhx n 1n 2 . . . 21 1
gxhnx
n! n! gxhn1x g xhn2x . . . 1!n 1! 2!n 2! n! gn1xhx gnxhx n 1!1!
Note: n! nn 1 . . . 3
2 1 (read “n factorial.”)
Section 2.4
f
3 cos 3 12
fx sin x
f
3 sin 3 23
f x cos x
f
3 cos 3 21
107. f x cos x
(a) P1x fax a f a
3
2
x 3 21
(b)
1 x 4 3
2
3
2
2
P2
1 P2x f ax a2 f ax a f a 2
The Chain Rule
− 2
x 3 21
−2
(c) P2 is a better approximation.
109. False. If y f xgx, then
P1
f
(d) The accuracy worsens as you move farther away from x a 3. 111. True
dy f xgx gxfx. dx
113. True
hc f cgc gcfc f c0 gc0 0
115. f x x x
x , if x ≥ 0 x , if x < 0 2
2
2x, if x ≥ 0 2 x
2x, if x < 0 2, if x > 0 f x 2, if x < 0 fx
f 0 does not exist since the left and right derivatives are not equal.
Section 2.4
The Chain Rule
y f gx
u gx
y f u
1. y 6x 54
u 6x 5
y u4
3. y x2 1
u x2 1
y u
5. y csc3 x
u csc x
y u3
7. y 2x 73 y 32x 722 62x 72 11. f x 9 x223 2 4x fx 9 x2132x 3 39 x213
9. gx 34 9x4 gx 124 9x39 1084 9x3 13. f t 1 t12 1 1 ft 1 t121 2 21 t
73
74
Chapter 2
Differentiation
17. y 24 x 214
15. y 9x2 413 1 6x y 9x2 42318x 3 9x2 423
y 2
2 34
2x
x
4 4 x 23
21. f t t 32
19. y x 21 y 12 x21 23.
144 x
1 x 22
ft 2t 33
2 t 33
25. f x x2x 24
y x 212
fx x2 4x 231 x 242x
dy 1 1 x 232 dx 2 2x 232
2xx 23 2x x 2 2xx 233x 2
27. y x1 x2 x1 x212
121 x
y x
2 12
29. y
2x 1 x2121
x21 x212 1 x212 1 x
2
xx 52
x2 132 x2 x2 1
35. y y
1 x2 132
2
33. f v
2
gx 2
x5 x2 2
x 2 2 x 52x x 2 22
x2x2 132 x2 112
2
1 2x2 1 x2 31. gx
xx2 112
1 y x x2 1322x x2 1121 2
x 1 x
2 12
x x2 1
112vv
fv 3
2x 52 10x x 2 x 2 23
3
1 2v 1v
1 2v 1 v2 1 v 2
2
91 2v2 1 v4
3t 2 37. gt 2 t 2t 1
x 1
x 1 2
1 3x2 4x32 2xx2 12
gt
The zero of y corresponds to the point on the graph of y where the tangent line is horizontal.
3tt2 3t 2 t2 2t 132
The zeros of g correspond to the points on the graph of g where the tangent lines are horizontal. 24
2
y −1
g′
5
y′ −2
g
−5
3 −2
Section 2.4
x x 1
39. y
y
41. st
x 1x
st
2xx 1
The Chain Rule
22 t1 t 3 t 1 t
The zero of st corresponds to the point on the graph of st where the tangent line is horizontal.
y has no zeros. 4
3
y s′
−5
4
−3
6
y′ s −2 −3
43.
y
cos x 1 x
3
y
dy x sin x cos x 1 dx x2
−5
5
y′
x sin x cos x 1 x2
−3
The zeros of y correspond to the points on the graph of y where the tangent lines are horizontal. 45. (a)
(b) y sin 2x
y sin x y cos x
y 2 cos 2x
y0 1
y0 2
1 cycle in 0, 2
2 cycles in 0, 2 The slope of sin ax at the origin is a.
47.
49. gx 3 tan 4x
y cos 3x
gx 12 sec2 4x
dy 3 sin 3x dx 51. y sin x2 sin 2 x 2 y cos x2 2 2x 2 2x cos 2x 2
55. f x
53. hx sin 2x cos 2x hx sin 2x2 sin 2x cos 2x2 cos 2x 2
cos2
75
2x 2
sin2
2x
2 cos 4x. Alternate solution:
fx
hx
1 sin 4x 2
hx
1 cos 4x4 2 cos 4x 2
cos x cot x 2 sin x sin x sin2 xsin x cos x2 sin x cos x sin4 x sin2 x 2 cos2 x 1 cos2 x sin3 x sin3 x
76
Chapter 2
Differentiation
57. y 4 sec2 x y 8 sec x sec x tan x 8 sec2 x tan x
59. f 14 sin2 2 14 sin 2 2 f 2 14 sin 2 cos 2 2 sin 2 cos 2 12 sin 4
61. f x 3 sec2 t 1
63.
ft 6 sec t 1 sec t 1 tan t 1 6 sin t 1 6 sec t 1 tan t 1 cos3 t 1
y x
1 sin2x 2 4
x
1 sin4x2 4
2
dy 1 12 1 x cos4x28x dx 2 4 1 2x cos2x2 2x
65.
y sincos x
67.
dy coscos x sin x dx
st t 2 2t 812, 2, 4 st
sin x coscos x
s2
69.
f x
3 3x3 41, x3 4
fx 3x3 423x2 f1
1, 53 9x2 x3 42
71. f t ft
1 2 t 2t 8122t 2 2 t1 t 2 2t 8
3 4 3t 2 , 0, 2 t1 5 t 13 3t 21 t 12 t 12
f0 5
9 25
y 37 sec32x, 0, 36
73.
y 3 sec22x 2 sec(2x tan2x 6 sec32x tan2x y0 0 75. (a) f x 3x2 2 , 3, 5
77. (a) f x sin 2x, , 0 fx 2 cos 2x
1 fx 3x2 2126x 2
f 2
3x 3x2 2 f3
Tangent line: y 2x ⇒ 2x y 2 0
9 5
63. (b)
2
Tangent line: 9 y 5 x 3 ⇒ 9x 5y 2 0 5 (b)
−2
7
(3, 5) −5
5
−3
0
(π , 0)
2
Section 2.4
79. f x 2x2 13
The Chain Rule
77
81. f x sin x 2
fx 6x2 122x
fx 2x cos x 2
12xx 4 2x2 1
f x 2x 2xsin x2 2 cos x2
12x5 24x3 12x
2 cos x2 2x2 sin x2
f x 60x 4 72x2 12 125x2 1x2 1
83.
y
85.
y
f′
3
3
2
2
f
1
x
x 2
2
3
3
1
2
f 2
f′
3
The zeros of f correspond to the points where the graph of f has horizontal tangents.
The zeros of f correspond to the points where the graph of f has horizontal tangents.
87. gx f 3x gx f3x3 ⇒ gx 3 f3x 89. (a) f x gxhx
73. (b) f x ghx
fx gxhx gxhx
fx ghxhx
f5 32 63 24
f5 g32 2g3 Need g3 to find f5.
73. (c) f x
gx hx
fx
hxgx gxhx hx 2
f5
36 32 12 4 32 9 3
73. (d) f x gx 3 f x 3 gx 2gx f5 3326 162
(b) f 132,400331 v1
91. (a) f 132,400331 v1
f 1132,400331 v21
f 1132,400331 v21
132,400 331 v 2
When v 30, f 1.016.
When v 30, f 1.461. 93. 0.2 cos 8t The maximum angular displacement is 0.2 (since 1 ≤ cos 8t ≤ 1. d
0.2 8 sin 8t 1.6 sin 8t dt When t 3, d dt 1.6 sin 24 1.4489 radians per second.
132,400 331 v 2
95.
S CR 2 r 2
dR dr dS C 2R 2r dt dt dt
Since r is constant, we have drdt 0 and dS 1.76 10521.2 102105 dt 4.224 102 0.04224.
78
Chapter 2
Differentiation
97. (a) x 1.6372t3 19.3120t2 0.5082t 0.6161 (b) C 60x 1350 601.6372t3 19.3120t2 0.5082t 0.6161 1350 dC 604.9116t2 38.624t 0.5082 dt 294.696t2 2317.44t 30.492 The function
dC is quadratic, not linear. The cost function levels off at the end of the day, perhaps due to fatigue. dt
99. f x sin x
101. (a) rx fgxgx r1 fg1g1
(a) fx cos x f x 2 sin x
Note that g1 4 and f4
f x 3 cos x
Also, g1 0. Thus, r1 0
f 4 4 sin x (b) f x 2 f x 2 sin x 2sin x 0 f 2kx 1k 2k sin x
(c)
f 2k1x 1k1 2k1 cos x
(b) sx gf xf x s4 gf 4f 4 5 64 1 5 and Note that f 4 , g 2 2 62 2 5 f4 . 4
Thus, s4
103.
g xx n
2x n 2x2 nx
2x n2 xx n
x x n 2 xx n
a g
107. hx x cos x
2x 3 , gx 2 2x 3
dg 1 2 x nx122x n dx 2
hx x sin x
1 5 5 . 2 4 8
105. gx 2x 3
nx
x2
x
x
50 5 62 4
cos x,
x 0
x
3 2
Section 2.5
109. (a) f x tan
x 4
f 1 1
f x
x sec2 4 4
f x
x sec2 2 4
tan
x 4 4
P1x f1x 1 f 1 P2x (b)
Implicit Differentiation
f1
2 4 2
f 1
21 8 4
x 1 1. 2
1 x 12 f1x 1 f 1 x 12 x 1 1 2 4 8 2
2
P1 P2 f 0
3 0
(c) P2 is a better approximation than P1 (d) The accuracy worsens as you move away from x c 1. 111. False. If y 1 x12, then y 12 1 x121.
Section 2.5 1.
113. True
Implicit Differentiation
x2 y2 36
3.
2x 2yy 0 y
1 12 1 12 x y y 0 2 2
x y
y
x3 xy y2 4
5.
7.
3x2 xy y 2yy 0
y
3x 2 3x 2y 6xy 4xyy 2y2 0
4xy 3x 2y 6xy 3x 2 2y 2 y1
6xy 3x 4xy 3x 2
2y 2
yx
x3y3 y x 0
3x3y2 1y 1 3x2y3
y 3x2 2y x
9. x3 3x 2 2xy 2 12
x12 y12
3x3y2y 3x2y3 y 1 0
2y xy y 3x2
2
x12 y12 9
y
11.
1 3x2y3 3x3y2 1
sin x 2cos 2y 1 cos x 4sin 2yy 0 y
cos x 4 sin 2y
79
80
Chapter 2
Differentiation
13. sin x x1 tan y cos x x
sec2
y
y sinxy
15.
yy 1 tan y1
y xy y cosxy y x cosxyy y cosxy
cos x tan y 1 x sec2 y
y 17. (a) x2 y2 16 y2
16
(b)
y
x2
6
y ± 16 x2
(c) Explicitly:
6
x
y=−
16 − x 2
2x 2yy 0 y
x x x 2 y 16 x2 ± 16 x 19. (a) 16y2 144 9x2
(b)
x y
y 6
1 9 144 9x2 16 x2 16 16
4
y = 34
16 − x2
2
3 y ± 16 x2 4
−6
−2
x 2
6
−4 −6
y = − 43
16 − x2
(d) Implicitly:
(c) Explicitly:
18x 32yy 0
dy 3 ± 16 x2122x dx 8
y
3x 3x 9x 416 x2 443y 16y
xy 4
23.
xy y1 0 x y y y x
At 4, 1: y
2
(d) Implicitly:
dy 1 ± 16 x2122x dx 2
y
16 − x 2
−2
−6
21.
y=
2 −6
y2
y cosxy 1 x cosxy
1 4
y2
9x 16y
x2 4 x2 4
2yy
x 2 42x x 2 42x x 2 42
2yy
16x x 2 42
y
8x yx 2 42
At 2, 0, y is undefined.
Section 2.5
25.
x23 y23 5
1 y sec2x y 1 1 sec2x y y sec2x y tan2x y sin2x y tan2x y 1 x2 2 x 1
yx
x13 y13
81
tanx y x
27.
2 13 2 13 x y y 0 3 3 y
Implicit Differentiation
3
1 At 8, 1: y . 2
At 0, 0: y 0.
29.
x2 4y 8
x2 y22 4x2y
31.
2x2 y22x 2yy 4x2y y8x
x2 4y y2x 0 y
4x3 4x2yy 4xy2 4y3y 4x2y 8xy
2xy x2 4
4x2yy 4y3y 4x2y 8xy 4x3 4xy2
2x8x2 4 x2 4
4yx2y y3 x2 42xy x3 xy2 y
16x 2 x 42 At 2, 1: y
At 1, 1: y 0.
32 1 64 2
Or, you could just solve for y: y x
2
33.
8 4
tan y x
35.
ysec2 y 1
x 2 y 2 36 2x 2yy 0
1 y cos2 y, < y < sec2 y 2 2
y
sec2 y 1 tan2 y 1 x2
x2 y2 16 2x 2yy 0 y
x y
x yy 0 1 yy y2 0
xy
39.
y
3x2 3x2 2y 2y
y
2x3y 3y2 4x2
y2 y3y x2
y
y2 x2 16 3 y3 y
y2
y 2 x 2 36 3 y3 y
2yy 3x2
2
yx
y x
y2 x3
0
1 yy
x y
y1 xy y y2
1 y 1 x2 37.
2xy x3 xy2 x2y y3 x2
2x3
xy
3y
3y2x 6y 4x2
3y 3x 4x2 4y
x3
3y
xy 2x y2 2x
82
Chapter 2
Differentiation
41. x y 4
9
1 12 1 12 x y y 0 2 2 y
(9, 1)
y x
−1
14 −1
At 9, 1, y
1 3
1 Tangent line: y 1 x 9 3 1 y x4 3 x 3y 12 0 43. x2 y2 25 y
x y
At 4, 3:
6
Tangent line: y 3
4 x 4 ⇒ 4x 3y 25 0 3
(4, 3) −9
9
3 Normal line: y 3 x 4 ⇒ 3x 4y 0. 4
−6
At 3, 4:
6
3 Tangent line: y 4 x 3 ⇒ 3x 4y 25 0 4 Normal line: y 4
45.
4 x 3 ⇒ 4x 3y 0. 3
(−3, 4) −9
9
−6
x2 y2 r 2 2x 2yy 0 y
x slope of tangent line y
y slope of normal line x Let x0, y0 be a point on the circle. If x0 0, then the tangent line is horizontal, the normal line is vertical and, hence, passes through the origin. If x0 0, then the equation of the normal line is y y0 y
y0 x x0 x0 y0 x x0
which passes through the origin.
Section 2.5
47. 25x2 16y2 200x 160y 400 0
Implicit Differentiation
y
(− 4, 10)
50x 32yy 200 160y 0 y
200 50x 160 32y
2516
6
(− 8, 5)
(0, 5) 4
(− 4, 0)
Horizontal tangents occur when x 4: 16y2
10
x
−10 − 8 − 6 − 4
2004 160y 400 0
−2
2
y y 10 0 ⇒ y 0, 10 Horizontal tangents: 4, 0, 4, 10. Vertical tangents occur when y 5: 25x2 400 200x 800 400 0 25xx 8 0 ⇒ x 0, 8 Vertical tangents: 0, 5, 8, 5. 49. Find the points of intersection by letting y2 4x in the equation 2x2 y2 6. 2x2 4x 6
x 3x 1 0
and
The curves intersect at 1, ± 2.
y 2 = 4x
Parabola:
Ellipse: 4x 2yy 0 y
4
(1, 2) −6
2yy 4
2x y
y
6
(1, − 2) 2x 2 + y 2 = 6
2 y
−4
At 1, 2, the slopes are: y 1
y 1.
At 1, 2, the slopes are: y 1
y 1.
Tangents are perpendicular. 51. y x and x sin y
4
x = sin y
Point of intersection: 0, 0
−6
y x:
x sin y:
y 1
1 y cos y
6
(0, 0)
−4
x+y=0
y sec y At 0, 0, the slopes are: y 1
y 1.
Tangents are perpendicular. 53.
xy C
x2 y2 K
xy y 0
2x 2yy 0
y
y x
y
x y
At any point of intersection x, y the product of the slopes is yxxy 1. The curves are orthogonal.
2
2
C=4 −3
3
C=1 K = −1 −2
−3
3
K=2 −2
83
84
Chapter 2
Differentiation
55. 2y2 3x4 0 (a) 4yy 12x3 0
(b) 4y
4yy 12x3
dx dy 12x3 0 dt dt y
12x3 3x3 y 4y y
dx dy 3x3 dt dt
57. cos y 3 sin x 1 (b) sin y
(a) sin yy 3 cos x 0 y
3 cos x sin y
dy dx 3 cos x 0 dt dt dy dx sin y 3 cos x dt dt
59. A function is in explicit form if y is written as a function of x: y f x. For example, y x3. An implicit equation is not in the form y f x. For example, x 2 y 2 5. 61. (a) x4 44x2 y2
10
4y2 16x2 x4
− 10
10
1 y2 4x2 x4 4 y±
(b)
− 10
1 4x2 x4 4
1 y 3 ⇒ 9 4x 2 x4 4 36 16x2 x4 x4 16x2 36 0 x2
16 ± 256 144 8 ± 28 2
Note that x2 8 ± 28 8 ± 27 1 ± 7 2. Hence, there are four values of x: 1 7, 1 7, 17, 1 7 To find the slope, 2yy 8x x3 ⇒ y
x8 x2 . 23
1 For x 1 7, y 3 7 7, and the line is 1 1 y1 37 7x 1 7 3 37 7x 87 23 . 1 For x 1 7, y 3 7 7, and the line is
y2 137 7x 1 7 3 137 7x 23 87 . 1 For x 1 7, y 3 7 7, and the line is
y3 137 7x 1 7 3 137 7x 23 87 . 1 For x 1 7, y 3 7 7, and the line is 1 1 y4 37 7x 1 7 3 37 7x 87 23 .
—CONTINUED—
10
− 10
10
y1
y3
y2 − 10
y4
Section 2.6
Related Rates
61. —CONTINUED— (c) Equating y3 and y4,
1 7 7 x 1 7 3 13 7 7 x 1 7 3 3
7 7 x 1 7 7 7 x 1 7 7x 7 7 7x 7 77 7x 7 7 7x 7 77
167 14x x If x
87 7
87 87 , then y 5 and the lines intersect at ,5 . 7 7
63. Let f x xn xpq, where p and q are nonzero integers and q > 0. First consider the case where p 1. The derivative of f x x1q is given by d 1q f x x f x f t f x x lim lim x→0 t→x dx x tx where t x x. Observe that f t f x t1q x1q t1q x1q 1q q tx tx t x1qq
t1q
x1q
t11q
t1q x1q . . . t1qx12q x11q
t12qx1q
1 . t11q t12qx1q . . . t1qx12q x11q
As t → x, the denominator approaches qx11q. That is, d 1q 1 1 x 11q x1q1. dx qx q Now consider f x xpq xp1q. From the Chain Rule,
1 1 d p p p fx xp1q1 xp xp1q 1pxp1 xpqp p1 xpq 1 nxn1 n . q dx q q q q
Section 2.6 1.
Related Rates
y x
xy 4
3.
1 dx dy dt 2x dt
x
dx dy y 0 dt dt
dx dy 2x dt dt
y dx dy dt x dt
(a) When x 4 and dxdt 3,
dx x dy dt y dt
dy 1 3 3 . dt 24 4 (b) When x 25 and dydt 2, dx 225 2 20. dt
(a) When x 8, y 12, and dxdt 10, dy 12 5 10 . dt 8 8 (b) When x 1, y 4, and dydt 6, 1 3 dx 6 . dt 4 2
85
86
5.
Chapter 2
Differentiation
y x2 1
7.
y tan x
dx 2 dt
dx 2 dt
dy dx 2x dt dt
dy dx sec2 x dt dt
(a) When x 1,
(a) When x 3,
dy 212 4 cmsec. dt (b) When x 0,
dy 222 8 cmsec. dt (b) When x 4,
dy 202 0 cmsec. dt (c) When x 1, dy 212 4 cmsec. dt 9. (a)
dy dx negative ⇒ positive dt dt
(b)
dy dx positive ⇒ negative dt dt
13.
dy 2 22 4 cmsec. dt (c) When x 0, dy 122 2 cmsec. dt dy dx a . 11. Yes, y changes at a constant rate: dt dt No, the rate
dy dx is a multiple of . dt dt
D x2 y2 x2 x2 12 x4 3x2 1 dx 2 dt dx 2x3 3x dx 4x3 6x dD 1 4 x 3x2 1 124x3 6x 4 2 dt 2 dt x 3x 1 dt x4 3x2 1
15.
A r2
17. (a) sin
dr 3 dt
cos
12b ⇒ b 2s sin 2 s 2 h ⇒ h s cos 2 s 2 A
dA dr 2 r dt dt (a) When r 6,
1 1 bh 2s sin 2 2 2
s cos 2
s2 s2 2 sin cos sin 2 2 2 2
dA 2 63 36 cm2min. dt θ
(b) When r 24,
s
s h
dA 2 243 144 cm2min. dt
b
(b)
dA s2 d d 1 cos where radmin. dt 2 dt dt 2 When
12
dA s2 3 , 6 dt 2 2
3s 2
8
dA s2 1 1 s2 , 3 dt 2 2 2 8 (c) If d dt is constant, dAdt is proportional to cos . When
Section 2.6
dV 4 800 V r 3, 3 dt
19.
21.
ds dx 12x dt dt
dr 1 dV 1 800 dt 4 r 2 dt 4 r 2 (a) When r 30,
dr 1 2 800 cmmin. dt 4 302 9
(b) When r 60,
dr 1 1 800 cmmin. dt 4 60 2 18
1 1 9 V r 2h h2 h 3 3 4
s 6x2 dx 3 dt
dV dr 4 r 2 dt dt
23.
(a) When x 1, ds 1213 36 cm2sec. dt (b) When x 10, ds 12103 360 cm2sec. dt
since 2r 3h
3 3 h 4
h
dV 10 dt
r
dh 4dVdt dV 9 2 dh h ⇒ dt 4 dt dt 9h2 When h 15, 25.
8 410 dh ftmin. dt 9 152 405 12
6
Related Rates
1
3 1
(a) Total volume of pool
1 2126 1612 144 m3 2
Volume of 1m. of water
1 166 18 m3 2
2 h=1
(see similar triangle diagram) 18 % pool filled 144 100% 12.5%
(b) Since for 0 ≤ h ≤ 2, b 6h, you have 1 V bh6 3bh 36hh 18h2 2 dV dh 1 dh 1 1 1 36h ⇒ mmin. dt dt 4 dt 144h 1441 144
12 b=6
87
88
Chapter 2
Differentiation
x2 y2 252
27. 2x
dx dy 2y 0 dt dt dy x dt y
dx dx 2x since 2. dt y dt
25
y
x
(a) When x 7, y 576 24,
dy 27 7 ftsec. dt 24 12
When x 15, y 400 20, When x 24, y 7,
dy 215 3 ftsec. dt 20 2
dy 224 48 ftsec. dt 7 7
(b)
1 A xy 2
dx dA 1 dy x y dt 2 dt dt
From part (a) we have x 7, y 24, and
dx 2, dt
dy 7 . dt 12
7 dA 1 Thus, dt 2 7 12 242 527 21.96 ft2sec. 24 tan
(c)
sec2
x y
d 1 dt y
dx x 2 dt y
d 1 cos2 dt y
Using x 7, y 24,
dy dt
dx x 2 dt y
θ
dy dt
x
d dx dy 7 24 24 2, and cos , we have dt dt 12 25 dt 25
241 2 247 127 121 rad sec.
29. When y 6, x 122 62 63, and 12 − y
s x2 12 y2
2x
2
2
s x
108 36 12. x2
25
y
( x, y )
y 12
12 y 2
s2
dy ds dx 212 y1 2s dt dt dt x
dx dy ds y 12 s dt dt dt
Also, x2 y2 122 2x
dy dy x dx dx 2y 0⇒ . dt dt dt y dt
Thus, x
x dx ds dx y 12 s dt y dt dt
12x ds dx sy dx xx s ⇒ dt y dt dt 12x dy x dx 63 dt y dt 6
ds 126 1 3 0.2 msec (horizontal) dt 15 12 63 53
3 1 msec (vertical). 15
5
Section 2.6
s2 x2 y2
31. (a)
dx 450 dt
)
les
in e(
dy 600 dt
mi
y
c 200 tan
s
Di
s
100
x 200
100
dx dy ds 2x 2y 2s dt dt dt
Distance (in miles)
ds xdxdt ydydt dt s When x 150 and y 200, s 250 and ds 150450 200600 750 mph. dt 250 (b) t
250 1 hr 20 min 750 3
s2 902 x2
33.
2nd
x 30 30 ft
dx 28 dt 2s
3rd
x
1st s
ds ds x dx 2x ⇒ dt dt dt s
dx dt
90 ft Home
When x 30, s 902 302 3010 ds 28 30 28 8.85 ftsec. dt 10 3010 35. (a)
15 y ⇒ 15y 15x 6y 6 yx y
5 x 3
15
dx 5 dt dy 5 dt 3 (b)
6
dx 5 25 5 ftsec dt 3 3
10 d y x dy dx 25 5 ftsec dt dt dt 3 3
x y
Related Rates
89
90
Chapter 2
37. xt
Differentiation
39. Since the evaporation rate is proportional to the surface area, dVdt k4 r 2. However, since V 43 r 3, we have
1 t sin , x 2 y 2 1 2 6
(a) Period:
2 12 seconds 6
1 (b) When x , y 2 Lowest point:
1 12 2
dr dV 4 r 2 . dt dt 2
3
2
m.
0, 23
1 (c) When x , y 4
Therefore, k4 r 2 4 r 2
1 14
2
15
4
dr dr ⇒k . dt dt
and t 1
dx 1 t t cos cos dt 2 6 6 12 6
x2 y2 1 2x
dx dy dy x dx . 2y 0⇒ dt dt dt y dt
Thus, dy 14 dt 154 Speed
12 cos 6
1 23 24 15 1205.
15 12
5 5 msec 120 120
pV1.3 k
41. 1.3 pV 0.3
dV dp V1.3 0 dt dt
dV dp V 0 dt dt
V 0.3 1.3p
1.3p tan
43.
dV dp V dt dt
y 30
y
dy 3 msec. dt sec2
d 1 dy dt 30 dt 1 d cos2 dt 30
y
θ
dy dt
When y 30, 4 and cos 22. Thus, d 1 1 1 3 radsec. dt 30 2 20
x
30
Section 2.6
y tan , y 5 x
45.
L
dx 600 mihr dt d 5 sec2 2 dt x
y=5
θ x
dx dt
d 5 dx x2 5 dx cos2 2 2 2 dt x dt L x dt
L5 15dxdt sin 15600 120 sin
47.
2
2
2
2
(a) When 30 ,
d 120 1 30 radhr radmin. dt 4 2
(b) When 60 ,
d 3 3 120 90 radhr radmin. dt 4 2
(c) When 75 ,
d 120 sin2 75 111.96 radhr 1.87 radmin. dt
d 10 revsec2 radrev 20 radsec dt x (a) cos 30
P 30
θ x
d 1 dx sin dt 30 dt
x
dx d 30 sin 30 sin 20 600 sin dt dt (b)
2000
4π
0
− 2000
is least when dxdt n or n 180 .
(c) dxdt 600 sin is greatest when sin 1 ⇒ 2 n or 90 n
(d) For 30 , For 60 ,
49. tan
dx 1 600 sin30 600 300 cmsec. dt 2 3 dx 600 sin60 600 3003 cmsec dt 2
x ⇒ x 50 tan 50 d dx 50 sec2 dt dt 2 50 sec2
d dt
d 1 cos2 , ≤ ≤ dt 25 4 4
180
Related Rates
91
92
Chapter 2
Differentiation
51. x2 y2 25; acceleration of the top of the ladder
First derivative: 2x
d 2y dt 2
dy dx 2y 0 dt dt dy dx y 0 dt dt
x Second derivative: x
d 2x dx dt 2 dt
d 2y dy dx y 2 dt dt dt
dy 0 dt
xddt x dxdt dydt
1 d 2y dt 2 y When x 7, y 24,
2
2
2
2
dy 7 dx dx d 2x , and 2 (see Exercise 27). Since is constant, 2 0. dt 12 dt dt dt
d 2y 1 7 70 22 dt 2 24 12
49 1 625 241 4 144 24 144 0.1808 ft sec 2
2
53. (a) Using a graphing utility, you obtain ms 0.881s2 29.10s 206.2 (b)
dm dm ds ds 1.762s 29.10 dt ds dt dt
(c) If t s 1995, then s 15.5 and Thus,
ds 1.2. dt
dm 1.76215.5 29.101.2 2.15 million. dt
Review Exercises for Chapter 2 1. f x x2 2x 3 fx lim
x→0
f x x f x x
x x2 2x x 3 x2 2x 3 x→0 x
lim
x2 2xx x2 2x 2x 3 x2 2x 3 x→0 x
lim
2xx x2 2x lim 2x x 2 2x 2 x→0 x→0 x
lim
5. f is differentiable for all x 1.
3. f x x 1 fx lim
x→0
lim
x→0
lim
x→0
lim
x→0
lim
x→0
f x x f x x
x x 1 x 1 x x x x
x
x x x x x x
x x x x x x x 1 x x x
1 2 x
Review Exercises for Chapter 2
7. f x 4 x 2
1 4 9. Using the limit definition, you obtain gx x . 3 6
(a) Continuous at x 2. (b) Not differentiable at x 2 because of the sharp turn in the graph.
4 1 3 At x 1, g1 3 6 2
y 7 6 5 4 3 2 x
−1
1 2 3 4 5 6
−2 −3
11. (a) Using the limit defintion, fx 3x 2.
13. g2 lim
x→2
At x 1, f1 3. The tangent line is y 3x 1
x3 x 2 4 x→2 x2
lim
0
−4
x 2x 1 4 x→2 x2
lim
y 2 3x 1
(b)
gx g2 x2
2
lim
(−1, −2)
x→2
x 2x 2 x 2 x2
lim x 2 x 2 8 x→2
−4
15.
19. f x x8
17. y 25
y
f′
f
fx 8x7
y 0
2
1
x −1
1
21. ht 3t 4
23. f x x3 3x2
ht 12t 3
fx 3x2 6x 3xx 2
3 x 6x1 2 3x1 3 25. hx 6 x 3
hx 3x1 2 x2 3 29. f 2 3 sin f 2 3 cos
3 x
2 27. gt t2 3
1 3 x2
gx
4 3 4 t 3 3 3t
31. f 3 cos
sin 4
f 3 sin
cos 4
93
94
Chapter 2
Differentiation
F 200 T
33.
Ft
35.
st 16t2 s0 s9.2 169.22 s0 0
100 T
s0 1354.24
(a) When T 4, F4 50 vibrations/sec/lb.
The building is approximately 1354 feet high (or 415 m).
(b) When T 9, F9 3313 vibrations/sec/lb. 37. (a)
(c) Ball reaches maximum height when x 25.
y
y x 0.02x2
(d)
15
y 1 0.04x
10
y0 1
5
y10 0.6
x 20
40
60
Total horizontal distance: 50 (b) 0 x 0.02x2
y25 0 y30 0.2 y50 1
x 0x 1 implies x 50. 50 39. xt t2 3t 2 t 2t 1 (a) vt xt 2t 3
(e) y25 0
3 (b) vt < 0 for t < 2 .
(d) xt 0 for t 1, 2.
at vt 2 3 (c) vt 0 for t 2 . 3 3 1 1 1 x 2 2 2 1 2 2 4
v1 21 3 1
v2 22 3 1 The speed is 1 when the position is 0. 43. hx x sin x x1 2 sin x
41. f x 3x2 7x2 2x 3 fx 3x2 72x 2 x2 2x 36x 2
6x3
9x2
16x 7
1 sin x x cos x 2 x
x2 x 1 x2 1 x2 12x 1 x2 x 12x fx x2 12 2 x 1 2 x 12
45. f x 2x x2
47. f x
fx 2 2x3 2 1
hx
1 x3
2x3 1 x3
51. y
49. f x 4 3x21 fx 4 3x226x 53. y 3x 2 sec x y 3x 2 sec x tan x 6x sec x
6x 4 3x22
y
x2 cos x cos x 2x x 2sin x 2x cos x x 2 sin x cos2 x cos2 x
55. y x tan x y x sec2 x tan x
Review Exercises for Chapter 2
59. gt t3 3t 2
57. y x cos x sin x
gt 3t2 3
y x sin x cos x cos x x sin x
g t 6t 61. f 3 tan
y 2 sin x 3 cos x
63.
f 3 sec2
y 2 cos x 3 sin x
f 6 sec sec tan 6 sec tan
y 2 sin x 3 cos x
2
y y 2 sin x 3 cos x 2 sin x 3 cos x 0 65. f x 1 x31 2 1 fx 1 x31 23x2 2
3x2
2 1 x3
s2
1
s
s2
1
3s2
2
xx 31x
2
2
11 x 32x x 2 12
2x 3x 2 6x 1 x 2 13
71. y 3 cos3x 1
5 1 2s
5 2
x3 x2 1
hx 2
69. f s s2 15 2s3 5 fs
67. hx
s3
5 2
s2
3 2
y 9 sin3x 1
3s 1 5 5
3 2
s2
s3
ss2 13 28s3 3s 25
73. y
1 csc 2x 2
1 y csc 2x cot 2x2 2
y
csc 2x cot 2x
77. y
x sin 2x 2 4
75. y
1 1 cos 2x2 2 4
1 1 cos 2x sin2 x 2
2 3 2 2 sin x sin7 2x 3 7
y sin1 2 x cos x sin5 2 x cos x
79. y y
cos x sin x1 sin2 x
sin x x2
x 2 cos x sin x x 22
cos3 x sin x 81. f t t2t 15
83. gx 2xx 11 2
ft tt 147t 2 The zeros of f correspond to the points on the graph of f where the tangent line is horizontal. 0.1
gx
g does not equal zero for any value of x in the domain. The graph of g has no horizontal tangent lines.
f′
4
−0.1
1.3
g′
f −0.1
x2 x 13 2
−2
7
g −2
95
96
Chapter 2
Differentiation
85. f t t 11 2t 11 3 t 15 6 ft
87. y tan 1 x
5 6t 11 6
y
f does not equal zero for any x in the domain. The graph of f has no horizontal tangent lines.
sec2 1 x 2 1 x
y does not equal zero for any x in the domain. The graph has no horizontal tangent lines.
5
5
y
f − 20
f′ −2
7
2
y′
−1
−4
91. f x cot x
89. y 2x2 sin 2x
93. f t
t 1 t2
ft
t1 1 t3
f t
2t 2 1 t4
fx csc2 x
y 4x 2 cos 2x
f 2 csc xcsc x cot x
y 4 4 sin 2x
2 csc2 x cot x
95. g tan 3 sin 1 g 3 sec2 3 cos 1 g 18 sec2 3 tan 3 sin 1 97. T 700t2 4t 101 T
1400t 2 t2 4t 102
(a) When t 1, T
(b) When t 3,
14001 2 18.667 deg hr. 1 4 102
T
(d) When t 10,
(c) When t 5, T
14003 2 7.284 deg hr. 9 12 102
14005 2 3.240 deg hr. 25 30 102
T
140010 2 0.747 deg hr. 100 40 102
x2 3xy y3 10
99.
2x 3xy 3y 3y2y 0 3x y2y 2x 3y 2x 3y 3x y2
y
y x x y 16
101. y
12x x 1 2
y x
1 2
12y
1 2
y y1 2 0
x 2 x yy y 2 y x 2 xy x 2 xy y y 2 y 2 x y
2 xy y 2 x
2 y 2 xy x
2y x y y 2x y x x
Review Exercises for Chapter 2
x sin y y cos x
103.
105.
x cos yy sin y y sin x y cos x
6
(2, 4)
2x 2yy 0 −9
yx cos y cos x y sin x sin y y
x2 y2 20
97
9
x y y
y sin x sin y cos x x cos y
−6
1 At 2, 4: y 2 1 Tangent line: y 4 x 2 2 x 2y 10 0 Normal line: y 4 2x 2 2x y 0
107.
y x dy 2 units sec dt dx dy 1 dx dy ⇒ 2 x 4 x dt dt dt 2 x dt 1 dx 2 2 units/sec. (a) When x , 2 dt
109.
(b) When x 1,
dx 4 units/sec. dt
(c) When x 4,
dx 8 units/sec. dt
s 1 2 h 2
111. st 60 4.9t2 st 9.8t
1 s h 4
s 35 60 4.9t2 4.9t2
dV 1 dt
w 2 2s 2 2
14h 4 2 h
tan 30
dV 5 dh 4 h dt 2 dt 2dV dt dh dt 54 h When h 1,
dh 2 m min. dt 25
1 2
1 3
5 4.9 st xt
xt 3 st
4h 5 5 2 h 8 hh 2 2 4
25
t
Width of water at depth h:
V
s (t)
ds 5 dx 3 39.8 dt dt 4.9 1 2
2
38.34 m sec
s 2 h 2
30˚ x(t )
C H A P T E R 2 Differentiation Section 2.1
The Derivative and the Tangent Line Problem . . 330
Section 2.2
Basic Differentiation Rules and Rates of Change 338
Section 2.3
The Product and Quotient Rules and Higher-Order Derivatives . . . . . . . . . . . . . 344
Section 2.4
The Chain Rule . . . . . . . . . . . . . . . . . . 350
Section 2.5
Implicit Differentiation . . . . . . . . . . . . . . 356
Section 2.6
Related Rates . . . . . . . . . . . . . . . . . . . 361
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 367 Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . 373
C H A P T E R Differentiation Section 2.1
2
The Derivative and the Tangent Line Problem
Solutions to Even-Numbered Exercises
2. (a) m
1 4
4. (a)
f 4 f 1 5 2 1 41 3 f 4 f 3 5 4.75 0.25 43 1
(b) m 1
Thus,
f 4 f 1 f 4 f 3 > 41 43
(b) The slope of the tangent line at 1, 2 equals f1. This slope is steeper than the slope of the line through 1, 2 and 4, 5. Thus, f 4 f 1 < f1. 41 6. g x
3 3 x 1 is a line. Slope 2 2
8. Slope at 2, 1 lim
x→0
g2 x g2 x
5 2 x2 1 x→0 x
lim
5 4 4x x2 1 x→0 x
lim
lim 4 x 4 x→0
10. Slope at 2, 7 lim
t→0
h2 t h2 t
2 t2 3 7 t→0 t
lim
4 4t t2 4 t→0 t
lim
lim 4 t 4 t→0
14. f x 3x 2 fx lim
x→0
lim
x→0
lim
x→0
f x x f x x
330
gx lim
x→0
gx x gx x
lim
5 5 x
lim
0 0 x
x→0
x→0
1 16. f x 9 2x fx lim
x→0
f x x f x x
3x x 2 3x 2 x
lim
9 1 2x x 9 1 2x x
3x x
lim
21 21
lim 3 3 x→0
12. gx 5
x→0
x→0
Section 2.1
The Derivative and the Tangent Line Problem
331
18. f x 1 x2 fx lim
x→0
f x x f x x
1 x x2 1 x2 x→0 x
lim lim
x→0
1 x2 2xx x2 1 x2 x
2xx x2 lim 2x x 2x x→0 x→0 x
lim
20. f x x3 x2 fx lim
x→0
f x x f x x
x x3 x x2 x3 x2 x→0 x
lim
x3 3x2x 3xx2 x3 x2 2xx x2 x3 x2 x→0 x
lim lim
x→0
3x2x 3xx2 x3 2xx x2 x
lim 3x2 3xx x2 2x x 3x2 2x x→0
22. f x
1 x2
24. f x
fx lim
x→0
f x x f x x
1 1 x x2 x2 lim x→0 x x x x xx x2x2 2
fx lim
x→0
f x x f x x 4
lim
x→0
2
x x
4
x
x
lim
4 x 4 x x x x x x
2xx x2 2 2 x→0 xx x x
lim
4x 4x x x x x x x x x
2x x x x2x2
lim
4
x x x x x x
lim
x→0
lim lim
x→0
4
x
2x x4
2 x3
x→0
x→0
x→0
4 2
x x x x x x
x x x
x x x
332
Chapter 2
Differentiation
26. (a) f x x2 2x 1 fx lim
x→0
18. (b)
5
f x x f x x
(−3, 4)
−6
x x2 2x x 1 x2 2x 1 x→0 x
3
lim
lim
x→0
−1
2xx x2 2x x
lim 2x x 2 2x 2 x→0
At 3, 4, the slope of the tangent line is m 23 2 4. The equation of the tangent line is y 4 4x 3 y 4x 8. 28. (a) f x x3 1 f x x f x fx lim x→0 x
18. (b)
4
(1, 2) −6
6
x x3 1 x3 1 lim x→0 x lim
x3
−4
x 3xx x 1 x
3x 2
x→0
2
3
x3
1
lim 3x 2 3xx x2 3x 2 x→0
At 1, 2, the slope of the tangent line is m 312 3. The equation of the tangent line is y 2 3x 1 y 3x 1. (b)
30. (a) f x x 1
(5, 2)
f x x f x fx lim x→0 x lim
x x 1 x 1
x
x→0
lim
x→0
lim
x→0
−2
x x 1 x 1
x x 1 x 1
x x 1 x 1 x x x 1 x 1 1
x x 1 x 1
1 2 x 1
At 5, 2, the slope of the tangent line is m
1 1 2 5 1 4
The equation of the tangent line is 1 y 2 x 5 4 1 3 y x 4 4
4
10
−4
Section 2.1
The Derivative and the Tangent Line Problem
1 x1
32. (a) f x
(b)
f x x f x x→0 x 1 1 x x 1 x 1 lim x→0 x x→0
−6
3
−3
x 1 x x 1 xx x 1x 1
lim x→0
3
(0, 1)
fx lim
lim
333
1 x x 1x 1
1 x 12
At 0, 1, the slope of the tangent line is m
1 1. 0 12
The equation of the tangent line is y x 1. 34. Using the limit definition of derivative, fx 3x 2. Since the slope of the given line is 3, we have
36. Using the limit definition of derivative, fx
1 Since the slope of the given line is , we have 2
3x 2 3 x 2 1 ⇒ x ± 1.
1 1 2x 132 2
Therefore, at the points 1, 3 and 1, 1 the tangent lines are parallel to 3x y 4 0. These lines have equations
1 x 132 1x1⇒x2
y 3 3x 1 and y 1 3x 1 y 3x
1 . 2x 13 2
At the point 2, 1, the tangent line is parallel to x 2y 7 0. The equation of the tangent line is
y 3x 4
1 y 1 x 2 2 1 y x2 2 38. h1 4 because the tangent line passes through 1, 4 h1
40. f x x 2 ⇒ fx 2x
(d)
64 2 1 3 1 4 2
42. f does not exist at x 0. Matches (c)
44.
Answers will vary.
y 4
Sample answer: y x
3 2 1 x
−4 −3 −2
1 −2 −3 −4
2
3
4
334
Chapter 2
46. (a) Yes. lim
x→0
Differentiation
f x 2x f x f x x f x lim fx x→0 2Dx x
(b) No. The numerator does not approach zero. (c) Yes. lim
x→0
f x x f x x f x x f x f x x f x lim x→0 2x 2x lim
x→0
(d) Yes. lim
x→0
f x f x x f x f x x 2x 2x
1 1 fx fx fx 2 2
f x x f x fx x
48. Let x0, y0 be a point of tangency on the graph of f. By the limit definition for the derivative, fx 2x. The slope of the line y through 1, 3 and x0, y0 equals the derivative of f at x0: 3 y0 2x0 1 x0
10
(3, 9)
8 6
3 y0 1 x02x0
4
(−1, 1)
3 x02 2x0 2x02
−8 −6 −4 −2 −2
x02 2x0 3 0
−4
x 2
4
6
(1, −3)
x0 3x0 1 0 ⇒ x0 3, 1 Therefore, the points of tangency are 3, 9 and (1, 1, and the corresponding slopes are 6 and 2. The equations of the tangent lines are y 3 6x 1
y 3 2x 1
y 6x 9
y 2x 1
50. (a) f x x2 fx lim
x→0
(b) gx lim
x→0
f x x f x x
g x x g x x
x x3 x3 x→0 x
lim
x x2 x2 x→0 x
lim
x2 2xx x2 x2 x→0 x
x3 3x2x 3xx2 x3 x3 x
lim
x3x2 3xx x2 x
lim
x→0
lim
lim
x→0
x2x x x
x→0
lim 3x2 3xx x2 3x2 x→0
lim 2x x 2x
At x 1, g1 3 and the tangent line is
x→0
At x 1, f1 2 and the tangent line is y 1 2x 1
or
y 2x 1.
At x 0, f0 0 and the tangent line is y 0. At x 1, f1 2 and the tangent line is y 2x 1.
y 1 3x 1
y 3x 2.
At x 1, g1 3 and the tangent line is y 1 3x 1 or y 3x 2. 2
2
−3
or
At x 0, g0 0 and the tangent line is y 0.
3
−3
For this function, the slopes of the tangent lines are always distinct for different values of x.
−3
3
−2
For this function, the slopes of the tangent lines are sometimes the same.
Section 2.1
The Derivative and the Tangent Line Problem
1 52. f x 2 x2
3
By the limit definition of the derivative we have fx x. 2
x f x
2
fx
2
1.5
1
1.125 1.5
0.5
0.5 1
0.125 0.5
0
0.5
1
1.5
2
0
0.125
0.5
1.125
2
0
0.5
1
1.5
2
f x 0.01 f x 0.01
54. gx
−2
2 −1
1 56. f 2 23 2, f 2.1 2.31525 4
3 x 0.01 3 3 100
f2
8
2.31525 2 3.1525 Exact: f2 3 2.1 2
f
g −1
8 −1
The graph of gx is approximately the graph of fx. 58. f x
x3 3 3x and fx x2 3 4 4 6
f′ −9
9
f −6
60. f x x
1 x
f 2 x f 2 x 2 f 2 Sx x x
1 5 2 x 2 5 x 2 x 2
2 x
5 2x 3 5 22 x2 2 52 x x 2 x 2 22 x x 2 22 x 2
(a) x
1: Sx
5 5 5 5 x 2 x 6 2 6 6
4 5 4 9 x 0.5: Sx x 2 x 5 2 5 10 x 0.1: Sx
16 5 16 41 x 2 x 21 2 21 42
(b) As x → 0, the line approaches the tangent line to f at 2, 52 . 62. g x xx 1 x 2 x, c 1 gx g1 x2 x 0 xx 1 lim lim x→1 x→1 x→1 x 1 x1 x1
g1 lim
lim x 1 x→1
4
−6
6
S0.1 S0.5 S1
f −4
335
336
Chapter 2
Differentiation
64. f x x3 2x, c 1
x 1x2 x 3 f x f 1 x3 2x 3 lim lim lim x2 x 3 5 x→1 x →1 x→1 x1 x1 x1
f1 lim
x→1
1 66. f x , c 3 x f x f 3 1 x 1 3 3x lim lim x→3 x→3 x3 x3 3x
f3 lim
x→3
lim x 3 x→3 3x 9 1
1
1
68. g x x 31 3, c 3 gx g3 1 x 31 3 0 lim lim x→3 x→3 x→3 x 32 3 x 3 x3
g3 lim
Does not exist.
70. f x x 4 , c 4 f4 lim
x→4
x4 0 x4 f x f 4 lim lim x→4 x→4 x 4 x4 x4
Does not exist. 72. f x is differentiable everywhere except at x ± 3. (Sharp turns in the graph.) 74. f x is differentiable everywhere except at x 1. (Discontinuity) 76. f x is differentiable everywhere except at x 0. (Sharp turn in the graph) 78. f x is differentiable everywhere except at x ± 2. (Discontinuities) 80. f x is differentiable everywhere except at x 1. (Discontinuity) 82. f x 1 x2 The derivative from the left does not exist because lim
x→1
1 x2 0
1 x2 f x f 1 lim lim x→1 x→1 x1 x1 x1
1 x2
1 x2
lim x→1
1x
1 x2
. (Vertical tangent)
The limit from the right does not exist since f is undefined for x > 1. Therefore, f is not differentiable at x 1. 84. f x
xx,, xx >≤ 11 2
The derivative from the left is lim
x→1
f x f 1 x1 lim lim 1 1. x→1 x 1 x→1 x1
The derivative from the right is lim
x→1
f x f 1 x2 1 lim lim x 1 2. x→1 x 1 x→1 x1
These one-sided limits are not equal. Therefore, f is not differentiable at x 1.
Section 2.1
86. Note that f is continuous at x 2. f x
1 2x
The Derivative and the Tangent Line Problem
337
1, x < 2 x ≥ 2
2x,
The derivative from the left is lim
x→2
f x f 2 1 x 1 2 lim 12 x 2 1. lim 2 x→2 x→2 x 2 x2 x2 2
The derivative from the right is lim
x→2
2x 2 f x f 2 lim x→2 x2 x2
lim x→2
2x 2
2x 2
2x 4 2x 2 2 1 lim lim x 2 2x 2 x→2 x 2 2x 2 x→2 2x 2 2.
The one-sided limits are equal. Therefore, f is differentiable at x 2. f2 12 88. (a) f x x2 and fx 2x
72. (b) gx x3 and gx 3x2
y
y
5
3
4
f
g′
3
2
2
−4 −3 −2 −1
f'
g
1
1 x 1
2
3
4
−2
x
−1
1
2
−1 −3
72. (c) The derivative is a polynomial of degree 1 less than the original function. If hx x n, then hx nx n1. 72. (d) If f x x4, then fx lim
x→0
f x x f x x
lim
x x4 x4 x
lim
x4 4x3x 6x2x2 4xx3 x4 x4 x
lim
x4x3 6x2x 4xx2 x3 x
x→0
x→0
x→0
lim 4x3 6x2x 4xx2 x3 4x3 x→0
Hence, if f x x , then fx 4x3 which is consistent with the conjecture. However, this is not a proof, since you must verify the conjecture for all integer values of n, n ≥ 2. 4
90. False. y x 2 is continuous at x 2, but is not differentiable at x 2. (Sharp turn in the graph) 92. True—see Theorem 2.1
94.
3
−3
3 −1
As you zoom in, the graph of y1 x2 1 appears to be locally the graph of a horizontal line, whereas the graph of y2 x 1 always has a sharp corner at 0, 1. y2 is not differentiable at 0, 1.
338
Chapter 2
Differentiation
Section 2.2
Basic Differentiation Rules and Rates of Change
y x12
2. (a)
y x1
(b)
y x2
y 12 x32
5 15 y 3x4 2 sin x 2 sin x 8 8x4
4 y x3 3
y
4 3x3
y
2 9x3
28. y
3x2
y
30. y
4 x3
y 4x3
2 x 9
35, 2
y
2 3 x 9
y 12x2
y 12x2 34.
y 3x3 6, 2, 18
36. f x 35 x2, 5, 0
y 9x2
3 ft 2 5t
38.
5 5 2 cos x x3 2 cos x 2x3 8
Simplify
2 y x2 3
f
fx 6x2 2x 3
Derivative
2 3x2
3 , 5t
18. f x 2x3 x2 3x
Rewrite
26. y
f t 3
24. y
y cos x
Function
1 1 y x34 34 4 4x
8 x9
y 3x 2
22. y 5 sin x
gt sin t
4 x x14 10. y
16. y 8 x3
y 2t 2
20. gt cos t
y1 2
1 x8 x8
y 8x9
14. y t2 2t 3
gx 3
32.
8. y
y 8x7
f x 0
3x2 30x 75
y2 36
fx 6x 30 f5 0
35 35 gt 2 3 cos t, , 1
40. f x x2 3x 3x2
gt 3 sin t g 0
44. hx
y 2x3
y1 32
6. y x8
4. f x 2
y x2
(d)
y 32 x52
y1 1
y1 12
12. gx 3x 1
y x32
(c)
2x2 3x 1 2x 3 x1 x
1 2x2 1 hx 2 2 x x2 3 x 5 x x13 x15 48. f x
1 1 1 1 fx x23 x45 23 45 3 5 3x 5x
42. f x x x2
fx 2x 3 6x3 2x 3
fx 1 2x3
6 x3
1
2 x3
46. y 3x6x 5x 2 18x 2 15x3 y 36x 45x 2
50. f t t23 t13 4 2 1 2 1 ft t13 t23 13 23 3 3 3t 3t
Section 2.2
52. f x fx
2 3 x
Basic Differentiation Rules and Rates of Change
339
3 cos x 2x13 3 cos x
2 43 2 x 3 sin x 43 3 sin x 3 3x
54. (a) y x3 x
(b)
y 3x2 1
5
−5
5
At 1, 2: y 312 1 4. y 2 4x 1
Tangent line:
−7
4x y 2 0 56. (a) y x2 2xx 1
(b)
12
x3 3x2 2x y 3x2 6x 2 −3
At 1, 6: y 312 61 2 11.
3 −2
Tangent line: y 6 11x 1 0 11x y 5 60. y x2 1
58. y x3 x y 3x2 1 > 0 for all x.
y 2x 0 ⇒ x 0
Therefore, there are no horizontal tangents.
At x 0, y 1. Horizontal tangent: 0, 1 64. k x 2 4x 7
62. y 3x 2 cos x, 0 ≤ x < 2 y 3 2 sin x 0 sin x
3
2
⇒ x
2x 4
2 or 3 3
At x
3 3 ,y . 3 3
At x
2 23 3 ,y . 3 3
Horizontal tangents:
3 ,
3 3
3
66. kx x 4
Equate functions
k 1 2x
Equate derivatives
Hence, k 2x and
Equate functions Equate derivatives
Hence, x 2 and k 4 8 7 ⇒ k 3
, 23, 2
3 3
3
2x x x 4 ⇒ 2x x 4 ⇒ x 4 ⇒ k 4
68. The graph of a function f such that f > 0 for all x and the rate of change the function is decreasing i.e. f < 0 would, in general, look like the graph at the right.
y
x
340
Chapter 2
Differentiation
72.
70. gx 5f x ⇒ gx 5fx
y 2
f
1 −2
x
−1
1
3
4
f′ −3 −4
If f is quadratic, then its derivative is a linear function. f x ax2 bx c fx 2ax b 74. m1 is the slope of the line tangent to y x. m2 is the slope of the line tangent to y 1x. Since y x ⇒ y 1 ⇒ m1 1 and y
1 1 1 ⇒ y 2 ⇒ m2 2 . x x x
The points of intersection of y x and y 1x are 1 ⇒ x2 1 ⇒ x ± 1. x
x
At x ± 1, m2 1. Since m2 1m1, these tangent lines are perpendicular at the points intersection. 2 76. f x , 5, 0 x fx
78. f4 1 16
2 x2
2 0y x2 5 x
−10
10 2x x2y 10 2x x2
2x
10 2x 2x 4x 10 5 4 x ,y 2 5 The point 52 , 45 is on the graph of f. The slope of the
8 tangent line is f 52 25 .
Tangent line:
19 −1
y
8 5 4 x 5 25 2
25y 20 8x 20 8x 25y 40 0
Section 2.2
Basic Differentiation Rules and Rates of Change
(b) fx 3x2
80. (a) Nearby point: 1.0073138, 1.0221024 Secant line: y 1
Tx 3x 1 1 3x 2
1.0221024 1 x 1 1.0073138 1
(c) The accuracy worsens at you move away from 1, 1.
y 3.022x 1 1 (Answers will vary.)
341
2
2
(1, 1) −3
(1, 1) −3
3
f
T
3 −2 −2
(d)
x
3
2
1
f x
8
1
0
Tx
8
5
2
0.5 0.125 0.5
0.1
0
0.1
0.5
1
2
3
0.729
1
1.331
3.375
8
27
64
0.7
1
1.3
2.5
4
7
10
The accuracy decreases more rapidly than in Exercise 59 because y x3 is less “linear” than y x32. 82. True. If f x gx c, then fx gx 0 gx. 86. False. If f x
1 n xn, then fx nxn1 n1 xn x
88. f t t2 3, 2, 2.1 ft 2t
2, 1 ⇒ f2 22 4 2.1, 1.41 ⇒ f2.1 4.2 Average rate of change: f 2.1 f 2 1.41 1 4.1 2.1 2 0.1
st 16t2 22t 220 vt 32t 22 v3 118 ftsec st 16t2 22t 220 112 height after falling 108 ft 16t2
6
90. f x sin x, 0, fx cos x
Instantaneous rate of change:
92.
84. True. If y x 1 x, then dydx 11 1.
22t 108 0
2t 28t 27 0 t2 v2 322 22 86 ftsec
Instantaneous rate of change:
0, 0 ⇒ f0 1
6 , 12 ⇒ f6
3
2
0.866
Average rate of change: f 6 f 0 12 0 3 0.955 6 0 6 0 94. st 4.9t2 v0t s0 4.9t2 s0 0 when t 6.8. s0 4.9t2 4.96.82 226.6 m
342
Chapter 2
96.
Differentiation
98. This graph corresponds with Exercise 75.
v
s 10
40
Distance (in miles)
Velocity (in mph)
50
30 20 10 t 2
4
6
8
8
4 2
10
Time (in minutes)
(10, 6)
6
(0, 0)
(4, 2)
(6, 2) t
2
4
6
8
10
Time (in minutes)
(The velocity has been converted to miles per hour) 1 100. st at2 c and st at. 2 Average velocity:
st0 t st0 t 12at0 t2 c 12at0 t2 c t0 t t0 t 2t
12at02 2t0t t2 12at02 2t0t t2 2t
2at0t 2t
at0 st0 Instantaneous velocity at t t0 102. V s3,
dV 3s2 ds
When s 4 cm, 104.
dV 48 cm2. ds
C gallons of fuel usedcost per gallon
18,750 1.25 15,000 x x
dC 18,750 dx x2 x
10
C
1875
dC dx
187.5
15 1250 83.333
20 537.5 46.875
25 750 30
30 625 20.833
35
40
535.71
468.75
15.306
11.719
The driver who gets 15 miles per gallon would benefit more from a 1 mile per gallon increase in fuel efficiency. The rate of change is larger when x 15. 106.
dT KT Ta dt
Section 2.2
Basic Differentiation Rules and Rates of Change
1 108. y , x > 0 x y
y
1 x2
2
( )
(a, b) = a, a1
At a, b, the equation of the tangent line is
1
x 1 1 2 2x a or y 2 . a a a a
y
343
x 1
2
3
The x-intercept is 2a, 0.
2a.
The y-intercept is 0,
1 2 1 2. The area of the triangle is A bh 2a 2 2 a 110. y x2 y 2x (a) Tangent lines through 0, a: y a 2xx 0 x2 a 2x2 a x2 ± a x
The points of tangency are ± a, a. At a, a the slope is y a 2a. At a, a the slope is y a 2a. Tangent lines: y a 2a x a and y a 2a x a y 2a x a
y 2a x a
Restriction: a must be negative. (b) Tangent lines through a, 0: y 0 2xx a x2 2x2 2ax 0 x2 2ax xx 2a The points of tangency are 0, 0 and 2a, 4a2. At 0, 0 the slope is y0 0. At 2a, 4a2 the slope is y2a 4a. Tangent lines: y 0 0x 0 and y 4a2 4ax 2a y0
y 4ax 4a2
Restriction: None, a can be any real number.
f2x sin x is differentiable for all x 0.
112. f1x sin x is differentiable for all x n, n an integer.
You can verify this by graphing f1 and f2 and observing the locations of the sharp turns.
344
Chapter 2
Differentiation
Section 2.3
The Product and Quotient Rules and Higher-Order Derivatives
3 2. f x 6x 5x 2
4. gs s4 s2 s124 s2
fx 6x 53x 2 x3 26
4 s2 1 gs s122s 4 s 2 s12 2s 32 2 2 s12
18x3 15x 2 6x3 12 24x3 15x2 12
6. gx x sin x
8. gt
gx x cos x sin x
10. hs
hs
21 x x cos x 21 x sin x
gt
s s 1
12. f t
s 11 s
12s
ft
12
s 12 1 2 s 12
s 1 s
16. f x
fx x2 2x 13x2 x3 12x 2
x 1 2x 1 2
x 1 2
5x2
t2 2 2t 7
2t 72t t2 22 2t2 14t 4 2t 72 2t 72 cos t t3 t 3sin t cos t3t 2 t sin t 3 cos t t 32 t4
s 2 2 2s 1
14. f x x2 2x 1x3 1
3x2
4 5s 2 2s12
2
x2
fx
x 1
2x 2
f1 0
x1 x1
x 11 x 11 x 12 x1x1 x 12
f2 18.
f x
sin x x
fx
xcos x sin x1 x2
f
2 x 12 2 2 2 12
x cos x sin x x2
6 6 3236 12
2
33 18 2
3 3 6 2
Function 20. y
5x 2 3 4
Rewrite
Derivative
3 5 y x2 4 4
y
10 x 4
Simplify y
5x 2
Section 2.3
Function
The Product and Quotient Rules and Higher-Order Derivatives
Rewrite
Derivative
Simplify
22. y
4 5x2
4 y x2 5
8 y x3 5
y
24. y
3x2 5 7
5 3 y x2 7 7
y
6x 7
6 y x 7
26. f x fx
2 4 4 x 1 28. f x x 1 x 1 x x 1
x3 3x 2 x2 1
x2 13x2 3 x3 3x 22x x2 12 x4 6x2 4x 3 x2 12
12x x 12
5 16 x x23 6
5 1 6x16 x23
12
fx x4
3
13x 23
x 1x 1x 1 xx 11 4x
2x3
3 x x 3 x13x12 3 30. f x
fx x13
8 5x3
3
2
2xxx1 2 2
2
32. hx x2 12 x4 2x2 1 hx 4x3 4x 4xx2 1
Alternate solution: 3 x x 3 f x
x56 3x13 5 fx x16 x23 6
5 1 6x16 x23
34. gx x 2
2x x 1 1 2x x x 1
gx 2
2
x 12x x 21 2x 2 2x 1 x 2 2x x 2 2x 2 x 12 x 12 x 12
36. f x x2 xx2 1x2 x 1 fx 2x 1x2 1x2 x 1 x2 x2xx2 x 1 x2 xx2 12x 1 2x 1x 4 x3 2x2 x 1 x2 x2x3 2x2 2x x2 x2x3 x2 2x 1 2x5 x 4 3x3 x 1 2x5 2x2 2x5 x 4 x3 x2 x 6x5 4x3 3x2 1
38. f x
c2 x2 c2 x2
c2 x22x c2 x22x fx c2 x22
4xc2 c2 x22
40. f 1 cos f 1sin cos 1 cos 1 sin
345
346
Chapter 2
42. f x fx 46. hs
Differentiation
44. y x cot x
sin x x
y 1 csc2 x cot2 x
x cos x sin x x2 1 10 csc s s
hs
48. y
1 10 csc s cot s s2
y
sec x x x sec x tan x sec x x2 sec xx tan x 1 x2
52. f x sin x cos x
50. y x sin x cos x
fx sin xsin x cos xcos x
y x cos x sin x sin x x cos x
cos 2x 54. h 5 sec tan h 5 sec tan 5 sec
56. f x sec2
tan
x x x 1 3x
fx 2 58. f f
sin 1 cos
2
2
2
x 1
x5 2x3 2x2 2 (form of answer may vary) x2 12
60. f x tan x cot x 1
1 cos 1 cos 1 1 cos 2
fx 0 f1 0
(form of answer may vary) f x sin xsin x cos x
62.
fx sin xcos x sin x sin x cos xcos x sin x cos x sin2 x sin x cos x cos2 x sin 2x cos 2x f
4 sin 2 cos 2 1
64. (a) f x x 1x2 2, 0, 2
51. (b)
4
fx x 12x x2 21 3x2 2x 2 −4
f0 2 slope at 0, 2.
4
Tangent line: y 2 2x ⇒ y 2x 2 66. (a) f x
x1 , x1
2, 13
fx
x 11 x 11 2 x 12 x 12
f2
2 1 slope at 2, . 9 3
Tangent line: y
1 2 1 2 x 2 ⇒ y x 3 9 9 9
−4
54. (b)
4
−3
6
−4
Section 2.3
f x sec x,
68. (a)
3 , 2
The Product and Quotient Rules and Higher-Order Derivatives
. (b)
347
6
fx sec x tan x f
3 23 slope at 3 , 2.
−
−2
Tangent line:
y 2 23 x
3
63x 3y 6 23 0 70. f x fx
x2 x2 1
72. f x
xcos x 3 sin x 3x1 x cos x sin x x2 x2
gx
xcos x 2 sin x 2x1 x cos x sin x x2 x2
x2 12x x22x 2x 2 x2 12 x 12
fx 0 when x 0.
sin x 2x sin x 3x 5x f x 5 x x
gx
Horizontal tangent is at 0, 0.
f and g differ by a constant.
74. f x
cos x xn cos x xn
76. V r 2h t 2
fx xn sin x nxn1 cos x
xn1
1 t32 2t12 2
x sin x n cos x
Vt
x sin x n cos x xn1
When n 1: fx
x sin x cos x . x2
When n 2: fx
x sin x 2 cos x . x3
When n 3: fx
x sin x 3 cos x . x4
When n 4: fx
x sin x 4 cos x . x5
For general n, fx
x sin x n cos x . x n1
78.
P
fx gx 1 sin x sec x tan x cos x cos x 1 ⇒ sec x tan x csc x cot x ⇒ 1 ⇒ csc x cot x 1 cos x sin x sin x sin3 x 3 1 ⇒ tan x 1 ⇒ tan x 1 cos3 x 3 7 , 4 4
dP k 2 dV V
gx csc x, 0, 2
x
1 3 12 3t 2 cubic inchessec t t12 2 2 4t12
k V
f x sec x
80.
12t
348
Chapter 2
Differentiation
82. (a) nt 9.6643t2 90.7414t 77.5029 vt 276.4643t2 2987.6929t 1809.9714 (b) A
vt 276.46t2 2987.69t 1809.97 nt 9.66t2 90.74t 77.50
86. f x
fx 1
64 x3
192 x4
40.46x 2 2.09x 17.83 x 2 9.39x 8.022
1 x2 2x 1 x2 x x
fx 1 f x
32 x2
f x
A represents the average retail value (in millions of dollars) per 1000 motor homes. (c) At
84. f x x
88. f x sec x fx sec x tan x
1 x2
f x sec xsec2 x tan xsec x tan x sec xsec2 x tan2 x
2 x3 92. f 4x 2x 1
90. f x 2 2x1 f x 2x2
94. The graph of a differentiable function f such that f > 0 and f < 0 for all real numbers x would in general look like the graph below.
f 5x 2
2 x2
f 6x 0
y
f x
98. f x gxhx
96. f x 4 hx
100.
fx hx
fx gxhx hxgx
f2 h2 4
f2 g2h2 h2g2
y 3
f′
f f ′′
34 12
−2
14
−1
x −1
2
3
4
−2
It appears that f is quadratic; so f would be linear and f would be constant. 102. st 8.25t2 66t
Average velocity on:
vt 16.50t 66 at 16.50 tsec st ft
0
1
2
3
0
57.75
99
123.75
vt st ftsec
66
49.5
33
16.5
at vt ftsec2
16.5
16.5
16.5
16.5
4
0, 1 is
57.75 0 57.75. 10
1, 2 is
99 57.75 41.25. 21
2, 3 is
123.75 99 24.75. 32
3, 4 is
132 123.75 8.25. 43
132 0 16.5
Section 2.3
104. (a)
The Product and Quotient Rules and Higher-Order Derivatives
f x x n
f x
86. (b)
f n x nn 1n 2 . . . 21 n! f nx Note: n! nn 1 . . . 3
1 x
1nnn 1n 2 . . . 21 x n1 1nn! x n1
2 1 (read “n factorial.”)
106. xf x xfx f x
xf x xf x f x f x xf x 2f x
xf x xf x f x 2f x xf x 3f x In general, xf x n xf nx nf n1x. f
2 1
fx cos x
f
2 0
f x sin x
f
2 1
(a) P1x fax a f a 0 x P2x
11 2
(b)
1 1 f ax a2 fax a f a 1 x 2 2 2
1
1 x 2 2
2
2
1
2
(π2 , 1
(
108. f x sin x
− 2
3 2
−2
(c) P2 is a better approximation than P1. (d) The accuracy worsens as you move farther away from x a 110. True. y is a fourth-degree polynomial.
112. True
114. True. If vt c then at vt 0.
dny 0 when n > 4. dx n 116. (a) fg fg fg fg fg f g fg f g
True
(b) fg fg fg fg fg fg f g fg 2f g f g
fg f g
. 2
False
349
350
Chapter 2
Differentiation
Section 2.4
The Chain Rule
y f gx 2. y
1 x 1
4. y 3 tan x 2 6. y cos
3x 2
u gx
y f u
ux1
y u12
u x2
y 3 tan u
u
3x 2
y cos u
8. y 2x3 12
10. y 34 x 25
y 22x3 16x 2 12x22x3 1
y 154 x 22x 30x4 x 2 14. gx 5 3x 5 3x12
12. f t 9t 223
1 3 gx 5 3x123 2 25 3x
2 6 ft 9t 2139 3 3 9t 2
16. gx x2 2x 1 x 12 x 1 gx
3 27 fx 2 9x349 4 42 9x34
1,1, xx >< 11
20. st t 2 3t 11
22. y 5t 33
st 1t 2 3t 122t 3
18. f x 32 9x14
y 15t 34
2t 3 t2 3t 12
24. gt t2 212
15 t 34
26. f x x3x 93
1 t gt t2 2322t 2 2 t 232
fx x33x 923 3x 931 3x 929x 3x 9 27x 324x 3
1 28. y x 216 x 2 2
30. y
1 1 y x 2 16 x 2122x x16 x 212 2 2
x3 x16 x 2 216 x 2
x3x 2 32 216 x 2
32. ht
ht 2
t2 t3 2
t
3
t2 2
t
3
22t t23t2 t3 22
2t24t t4 2t34 t3 3 t3 23 t 23
1 x 4 4121 x x 4 4124x 3 2 y x4 4
2
x x4 4
x 4 4 2x 4 4 x4 4 32 x 4 x 4 432
Section 2.4
34. gx
gx 3 36. y y
3x 2 2 2x 3
2
2
2
22
33x 2 226x 2 18x 4 63x 2 223x 2 9x 2 2x 34 2x 34
x 2x 1
38. f x x2 x2 fx
1 2xx 132
x 25x 2 2x
The zeros of f correspond to the points on the graph of f where the tangent lines are horizontal.
y has no zeros. 7
4
y f′
y′ −6
6
f
−3
6
−1 −2
40. y t 2 9t 2 y
42. gx x 1 x 1
5t2 8t 9 2t 2
gx
The zero of y corresponds to the point on the graph of y where the tangent line is horizontal.
1 1 2x 1 2x 1
g has no zeros. 6
15
g
y′
g′
y −3
6
−2
10 −2
− 15
44.
y x2 tan
1 x
6
y
1 1 dy 2x tan sec2 dx x x
−4
y sin 3x
5
y′
The zeros of y correspond to the points on the graph of y where the tangent lines are horizontal.
46. (a)
351
3
3x2x 32 2x 36x2x 33x 2
The Chain Rule
−6
(b)
y sin
y 3 cos 3x y0 3 3 cycles in 0, 2
y y0
2x
12 cos 2x 1 2
Half cycle in 0, 2 The slope of sin ax at the origin is a.
352
48.
Chapter 2
Differentiation
y sin x
50. hx secx2 hx 2x secx2 tanx2
dy cos x dx 52. y cos 1 2x2 cos 1 2x2 y sin 1 2x221 2x2 41 2x sin1 2x2 54. g sec g sec
12 tan 12
12 sec 12 12 tan 12 sec 12 tan 12 12 2
sec 12 tan 12
1 1 sec 2 2
2
2
cos v cos v sin v csc v
56. gv
gv cos vcos v sin vsin v cos2 v sin2 v cos 2v 60. gt 5 cos 2 t 5cos t2
58. y 2 tan3 x y 6 tan2 x sec2 x
gt 10 cos tsin t 10 sin tcos t 5 sin 2 t
62. ht 2 cot2 t 2
64.
ht 4 cot t 2csc2 t 2
y 3x 5 cos x 2 3x 5 cos 2 x 2
4 cot t 2 csc2 t 2
dy 3 5 sin 2 x 22 2x dx 3 10 2x sinx2
66. y sin x13 sin x13 y cos x13
68.
13x 31sin x 23
cos x 1 cos x13 3 x23 sin x23
23
cos x
y 3x3 4x15, 2, 2 1 y 3x3 4x459x2 4 5 y2
70. f x
x2
1 x2 3x2, 3x2
fx 2x2 3x32x 3 f4 74. y
y
22x 3 x2 3x3
72. f x fx
1 2 x1 , 2, 3 2x 3
2x 31 x 12 5 2x 32 2x 32
f2 5
5 32
1 cos x, x
4, 161
9x2 4 53x3 4x45
2 , 2
1 sin x x2 2cos x
y2 is undefined.
Section 2.4
1 76. (a) f x xx2 5 , 2, 2 3
f x tan2 x,
78. (a)
f
x2 1 x2 5 3x2 5 3
62. (b)
4 , 1
4 212 4
Tangent line:
4 1 13 f2 3 33 3 9
y2
353
fx 2 tan x sec2 x
1 1 1 fx x x2 5122x x2 512 3 2 3
y14 x 64. (b)
Tangent line:
The Chain Rule
13 x 2 ⇒ 13x 9y 8 0 9
⇒ 4x y 1 0 4
4
−
6 −4 −9
9
−6
80. f x x 21 fx x 22 f x 2x 23
1 x 22
2 x 23
82. f x sec2 x fx 2 sec x sec x tan x 2 sec2 x tan x f x 2 sec2 xsec2 x 2 tan x2 sec2 x tan x 2 2 sec4 x 4 2 sec2 x tan2 x 2 2 sec2 xsec2 x 2 tan2 x 2 2 sec2 x3 sec2 x 2 84.
y
f
86.
3
f′
y
4
f
3
2
2
1 −3
f′
−2
−1
x −1
1
2
3
−2 −3
f is decreasing on , 1 so f must be negative there. f is increasing on 1, so f must be postive there. 88. gx f x 2 gx fx 22x ⇒ gx 2x fx 2
f x
−1 −2
4
f′
−3 −4
The zeros of f correspond to the points where the graph of f has horizontal tangents.
354
Chapter 2
Differentiation
90. (a) gx sin2 x cos 2 x 1 ⇒ gx 0
92. y 13 cos 12t 14 sin 12t v y 13 12 sin 12t 14 12 cos 12t
gx 2 sin x cos x 2 cos xsin x 0
4 sin 12t 3 cos 12t
(b) tan2 x 1 sec2 x
When t 8, y 0.25 feet and v 4 feet per second.
gx 1 f x Taking derivatives of both sides, gx fx. Equivalently, fx 2 sec x sec x tan x and gx 2 tan x sec2 x, which are the same. 94. y A cos t (a) Amplitude: A
3.5 1.75 2
(b) v y 1.75
y 1.75 cos t Period: 10 ⇒
0.35 sin
2 10 5
y 1.75 cos
t 5
t 5
96. (a) Using a graphing utility, or by trial and error, you obtain a model of the form
t 1 T t 64.18 22.15 sin 6
(b)
t sin 5 5
(c) Tt 22.15 cos
6t 1 6
11.60 cos
6t 1
20
100 0
13
−20 0
13 0
(d) The temperature changes most rapidly when t 4.1 (April) and t 10.1 (October). The temperature changes most slowly Tt 0 when t 1.1 (January) and t 7.1 (July). 98. (a) gx f x2 ⇒ gx fx (b) hx 2 f x ⇒ hx 2 fx (c) rx f 3x ⇒ rx f3x3 3 f3x Hence, you need to know f3x. 1 r0 3 f0 3 3 1
r1 3 f3 34 12 (d) sx f x 2 ⇒ sx fx 2 Hence, you need to know fx 2. s2 f0 13 , etc.
x fx gx hx
2
1
0
1
2
3
4
2 3
13
1
2
4
4
2 3
13
1
2
4
8
4 3
23
2
4
8
12
1
1
2
rx sx
1
3
4
Section 2.4
100. f x p f x for all x.
The Chain Rule
102. If f x f x, then
(a) Yes, fx p fx, which shows that f is periodic as well.
d d f x f x dx x fx1 fx
(b) Yes, let gx f 2x, so gx 2 f2x. Since f is periodic, so is g.
fx fx. Thus, fx is even.
104.
u u2 d 1 uu d u u u 2 u2122uu u , u 0 dx dx 2 u2 u
fx 2x
xx
2 2
108. f x sin x
106. f x x2 4
4 , x ±2 4
fx cos x
110. (a) f x sec2x
(b)
x , x k
sin sin x
6
P2
fx 2sec 2xtan 2x
P1
f x 22sec 2xtan 2x tan 2x 2sec 2xsec2 2x2 4sec 2xtan2 2x sec3 2x f
6 sec 3 2
f
6 2 sec 3 tan 3 43
f
6 423 2 56
f 0.78
0 0
3
P1x 43 x P2x
2 6
1 56 x 2 6
28 x
6
2
2
43 x
43 x
2 6
2 6
(c) P2 is a better approximation than P1. 112. False. If f x sin2 2x, then fx 2sin 2x2 cos 2x. 114. False. First apply the Product Rule.
(d) The accuracy worsens as you move away from x 6.
355
356
Chapter 2
Differentiation
Section 2.5 2.
Implicit Differentiations
x2 y2 16 2x 2yy 0 y
x3 y3 8
4.
3x2 3y2y 0
x y
y x2y y2x 2
6.
x2 y2
xy12 x 2y 0
8.
1 xy12xy y 1 2y 0 2
x2y 2xy y2 2yxy 0
x2 2xyy y2 2xy
x y y 1 2y 0 2xy 2xy
yy 2x y xx 2y
xy y 2xy 4xy y 0 y
2 sin x cos y 1
10.
2sin xsin yy cos ycos x 0 y
2xy y 4xy x
sin x cos y2 2
12.
2sin x cos y cos x sin yy 0
cos x sin yy 0
cos x cos y sin x sin y
y
cot x cot y 14.
cot y x y
16. x sec
csc2 yy 1 y y
1
1 1 csc2 y
y
1 tan2 y cot 2 y 18. (a) x2 4x 4 y2 6y 9 9 4 9
x 22 y 32 4 (Circle) y 32 4 x 22 y 3± 4 x 22 (c) Explicitly:
1 y
y 1 1 sec tan y2 y y
y2 1 1 y2 cos cot sec1y tan1y y y
(b)
y x 1
2
3
−1
4
5 2
y = −3 + 4 − (x − 2)
−2 −3 −4 −5
dy 1 ± 4 x 22122x 2 dx 2
x 2
4 x 22
x 2 ± 4 x 22
x 2 3 ± 4 x 22 3
x 2 y3
y = −3 −
4 − (x − 2)2
(d) Implicitly: 2x 2yy 4 6y 0
2y 6y 2x 2 y
x 2 y3
cos x sin y
Section 2.5
20. (a) 9y 2 x 2 9 y2 y
x2 9
(b) x2
1
9 9
4
−6
6
± x 2 9
−4
3
dy (c) Explicitly: dx (d) Implicitly:
9y 2
Implicit Differentiation
1 2
± x 2 9122x
3
±x
3x 9 2
x ±x 3± 3y 9y
x 9 2
18yy 2x 0 18yy 2x y
2x x 18y 9y
x2 y3 0
22.
x y3 x3 y3
24.
2x 3y2y 0
x3 3x2y 3xy2 y3 x3 y3
2x 3y2
3x2y 3xy2 0
2 At 1, 1: y . 3
x 2y 2xy 2xyy y 2 0
y
x 2 y xy 2 0
x2 2xyy y2 2xy y
y y 2x xx 2y
At 1, 1: y 1. x3 y3 4xy 1
26.
3x 2 3y 2y 4xy 4y
3y 2
4xy 4y y
At 2, 1, y
30.
x cos y 1
28.
xy sin y cos y 0
3x 2
y
4y 3x 2 3y 2 4x
4 12 8 38 5
1 cot y cot y x x
3 : y 21 3.
At 2,
4 xy2 x3 4 x2yy y21 3x2 y
cos y x sin y
x3 y3 6xy 0
32.
3x2 3y2y 6xy 6y 0
3x2 y2 2y4 x
y3y2 6x 6y 3x2 y
At 2, 2: y 2. At
6y 3x2 2y x2 3y2 6x y2 2x
169 32 4 . 43, 83 : y 163 649 83 40 5
357
358
Chapter 2
Differentiation
cos y x
34.
sin y y 1 y
1 ,0 < y < sin y
sin2 y cos2 y 1 sin2 y 1 cos2 y sin y 1 cos2 y 1 x 2 y
1 1 x 2
, 1 < x < 1 x2y2 2x 3
36.
2x2yy 2xy2 2 0 x2yy xy2 1 0 y
1 xy2 x2y
2xyy x2 y2 x2yy 2xyy y2 0 4xyy x2y2 x2yy y2 0 4 4xy2 1 xy22 x2yy y2 0 x x2y2 4xy2 4x2y4 1 2xy2 x2y4 x 4y3y x2y4 0 x 4y3y 2x2y4 2xy2 1 y 38. 1 xy x y
2x2y4 2xy2 1 x4y3
y xy x 1 y
y2 4x
40.
2yy 4
x1 1 1x
y
y 0
y 2y2y
y 0 42.
y2 2yy y
At 2,
2 y
x1 x2 1
x2 11 x 12x x2 12 x2 1 2x2 2x x2 12 1 2x x2 2yx2 12 5
5
: y 2 15544 4 1 10
Tangent line: y
1
5
5
2
1 x 2 105
105y 10 x 2 x 105y 8 0
5
.
1
(2, ) 5 5
−1
5
−1
2 4 y
2 y y 2
3
Section 2.5
Implicit Differentiation
44. x2 y2 9 y
x y
4
(0, 3)
At 0, 3:
−6
6
Tangent line: y 3 Normal line: x 0.
−4
At 2, 5 :
4
Tangent line: y 5 Normal line: y 5 46.
2 x 2 ⇒ 2x 5y 9 0 5 5
2
(2, 5 ) −6
6
x 2 ⇒ 5x 2y 0.
−4
y2 4x 2yy 4 y
2 1 at 1, 2 y
Equation of normal at 1, 2 is y 2 1x 1, y 3 x. The centers of the circles must be on the normal and at a distance of 4 units from 1, 2. Therefore,
x 12 3 x 22 16 2x 12 16 x 1 ± 22 . Centers of the circles: 1 22, 2 22 and 1 22, 2 22 Equations: x 1 22 2 y 2 22 2 16
x 1 22 2 y 2 22 2 16 48. 4x2 y2 8x 4y 4 0 8x 2yy 8 4y 0 y
8 8x 4 4x 2y 4 y2
Horizontal tangents occur when x 1: 41 2
y2
81 4y 4 0
y2 4y y y 4 0 ⇒ y 0, 4 Horizontal tangents: 1, 0, 1, 4. Vertical tangents occur when y 2: 4x2 22 8x 42 4 0 4x2 8x 4xx 2 0 ⇒ x 0, 2 Vertical tangents: 0, 2, 2, 2.
y
(1, 0) −1
1
x 2
3
−1
(2, − 2)
(0, − 2) −3 −4 −5
(1, − 4)
4
359
360
Chapter 2
Differentiation
50. Find the points of intersection by letting y2 x3 in the equation 2x2 3y2 5. 2x2 3x3 5
and
3x3 2x2 5 0
2
2x2 + 3y2 = 5
Intersect when x 1.
(1, 1) −2
Points of intersection: 1, ± 1 y2 x3:
2x 2 3y 2 5:
2yy 3x2
4x 6yy 0
y
4
(1, − 1)
3x2 2y
y
−2
y 2= x 3
2x 3y
At 1, 1, the slopes are: y
3 2
2 y . 3
At 1, 1, the slopes are: y
3 2
2 y . 3
Tangents are perpendicular. 52. Rewriting each equation and differentiating, x3 3y 1 y
x3y 29 3
3
x 1 3
y
y x2
x (3y − 29) = 3 15
1 3 29 3 x
y
x 3 = 3y − 3
1 . x2
−15
12 −3
For each value of x, the derivatives are negative reciprocals of each other. Thus, the tangent lines are orthogonal at both points of intersection. 54.
x2 y2 C2 2x 2yy 0
y Kx
2
2
K=1
y K −3
x y y
K = −1
C=1
−3
3
3
C=2
At the point of intersection x, y the product of the slopes is xyK xKxK 1. The curves are orthogonal.
−2
−2
56. x2 3xy2 y3 10 (a) 2x 3y2 6xyy 3y2y 0
(b) 2x
dx dy dy dx 3y 2 6xy 3y2 0 dt dt dt dt
6xy 3y2y 3y2 2x
2x 3y2
3y2 2x y 2 3y 6xy 58. (a) 4 sin x cos y 1
(b) 4 sin xsin y
4 sin xsin yy 4 cos x cos y 0 y
cos x cos y sin x sin y
cos x cos y
dx dy 6xy 3y2 dt dt
dy dx 4 cos x cos y 0 dt dt
dx dy sin x sin y dt dt
Section 2.6
60. Given an implicit equation, first differentiate both sides with respect to x. Collect all terms involving y on the left, and all other terms to the right. Factor out y on the left side. Finally, divide both sides by the left-hand factor that does not contain y.
Related Rates
62.
18
00
1671
B 1994
A 18
00
Use starting point B.
x y c
64. 1 2x
1
dy
2y dx
0
y dy dx x
Tangent line at x0, y0: y y0
y0
x x0
x0
x-intercept: x0 x0y0, 0 y-intercept: 0, y0 x0y0 Sum of intercepts:
x0 x0y0 y0 x0y0 x0 2x0y0 y0 x0 y0 2 c 2 c.
Section 2.6 2.
Related Rates
y 2x 2 3x dy dx 4x 6 dt dt
2x
dy dx 2y 0 dt dt
x dx dy dt y dt
dx 1 dy dt 4x 6 dt (a) When x 3 and
x2 y2 25
4.
dx dy 2, 43 62 12 dt dt
(b) When x 1 and
dy dx 1 5 5, 5 dt dt 41 6 2
dx y dy dt x dt (a) When x 3, y 4, and dxdt 8, dy 3 8 6 dt 4 (b) When x 4, y 3, and dydt 2, 3 3 dx 2 . dt 4 2
361
362
6.
Chapter 2
y
Differentiation
8.
1 1 x2
dx 2 dt
dx 2 dt
dx dy cos x dt dt
dy 2x dx dt 1 x22 dt
y sin x
(a) When x 6,
(a) When x 2,
dy cos 2 3 cmsec. dt 6
dy 222 8 cmsec. dt 25 25
(b) When x 4,
(b) When x 0,
dy cos 2 2 cmsec. dt 4
dy 0 cmsec. dt
dy cos 2 1 cmsec. dt 3
dy 222 8 cmsec. dt 25 25
(b) 14.
dx dy negative ⇒ negative dt dt
(c) When x 3,
(c) When x 2,
10. (a)
12. Answers will vary. See page 145.
dx dy positive ⇒ positive dt dt
D x2 y2 x2 sin2 x dx 2 dt dx x sin x cos x dx 2 2 sin x cos x dD 1 2 x sin2 x122x 2 sin x cos x dt 2 dt x2 sin2 x dt x2 sin2x
16.
A r2 dA dr 2 r dt dt If drdt is constant, dAdt is not constant. dA dr depends on r and . dt dt
18.
4 V r3 3 dr 2 dt dr dV 4 r 2 dt dt (a) When r 6,
dV 4 622 288 in3min. dt
When r 24,
dV 4 2422 4608 in3min. dt
(b) If drdt is constant, dVdt is proportional to r2.
20.
V x3 dx 3 dt dV dx 3x 2 dt dt (a) When x 1, dV 3123 9 cm3sec. dt
(b) When x 10, dV 310 2 3 900 cm3sec. dt
Section 2.6
22.
1 1 V r 2h r 23r r 3 3 3
Related Rates
1 1 25 3 25 3 V r 2h h h 3 3 144 3144
24.
dr 2 dt
By similar triangles, 5r 12h ⇒ r 125 h.
dV dr 3 r 2 dt dt
dV 10 dt
(a) When r 6,
dV 25 2 dh dh 144 dV h ⇒ dt 144 dt dt 25 h 2 dt
dV 3 622 216 in3min. dt
363
When h 8,
(b) When r 24,
dh 144 9 ftmin. 10 dt 2564 10 5
dV 3 2422 3456 in3min. dt r 12
h
1 26. V bh12 6bh 6h2 since b h 2 (a)
dV dh dh 1 dV 12h ⇒ dt dt dt 12h dt When h 1 and
(b) If
x2 y2 25
28. 2x
dy dx 2y 0 dt dt dx y dt x
dV dh 1 1 2, 2 ftmin dt dt 121 6
dh 3 dV 3 and h 2, then 122 9 ft3min. dt 8 dt 8
0.15y dy dy since 0.15. dt x dt
When x 2.5, y 18.75,
18.75 dx 0.15 0.26 msec dt 2.5
12 ft 3 ft
5
y
h ft
3 ft
x
30. Let L be the length of the rope. (a)
L2 144 x2 dL dx 2L 2x dt dt dx L dL 4L dL since 4 ftsec. dt x dt x dt When L 13, x L2 144 169 144 5 413 52 dx 10.4 ftsec. dt 5 5 Speed of the boat increases as it approaches the dock.
(b) If
dx 4, and L 13, dt
dL x dx dt L dt
5 4 13
20 ftsec 13
4 ft/sec 13 ft 12 ft
As L → 0,
dL increases. dt
364
Chapter 2
32.
x 2 y 2 s2 2x
Differentiation
34. s2 902 x2
since dydt 0
dx ds 0 2s dt dt dx s ds dt x dt
When s 10, x 100 25 75 53 dx 10 480 240 1603 277.13 mph. dt 3 53
2nd
30 ft
x 60
x
dx 28 dt ds x dt s
3rd
1st
s 90 ft
dx dt
Home
When x 60, s 902 602 3013 ds 60 56 28 15.53 ftsec. dt 3013 13
y
x
s
5 mi
x
20 y 6 yx
36. (a)
20y 20x 6y 20
14y 20x 10 x y 7
6 x y
dx 5 dt 50 dy 10 dx 10 5 ftsec dt 7 dt 7 7 (b)
50 35 15 d y x dy dx 50 5 ftsec dt dt dt 7 7 7 7
38. xt
3 sin t, x2 y2 1 5
(a) Period:
(c) When x
2 2 seconds
3 (b) When x , y 5 Lowest point:
3 1 5
2
3 ,y 10
1 14
2
15
4
and
3 1 1 3 sin t ⇒ sin t ⇒ t 10 5 2 6
4 m. 5
dx 3 cos t dt 5
4 0, 5
x2 y2 1 2x Thus,
dx dy dy x dx 2y 0⇒ . dt dt dt y dt 310 dy dt 154
5 cos 6 3
95 9 . 125 255
Speed
95 0.5058 msec 125
Section 2.6
1 1 1 R R1 R2
40.
42.
rg tan v2
32r sec2
dR2 1.5 dt 1 dR 2 dt R1
365
32r tan v2, r is a constant.
dR1 1 dt
1 R2
Related Rates
d dv 2v dt dt d dv 16r sec2 dt v dt
dR1 1 2 dt R2
dR2 dt
Likewise,
d v dv cos2 . dt 16r dt
When R1 50 and R2 75, R 30
1 1 dR 302 1 1.5 dt 502 752 0.6 ohmssec. sin
44.
10 x
dx 1ftsec dt cos
x 10
dx dt ddt 10 x
θ
2
d 10 dx 2 sec dt x dt
10 10 1 2 221 25 1 0.017 radsec 252 102 252 25 521 2521 525 Police
x 50
tan
46.
d 302 60 radmin radsec dt sec2
θ 50 ft
ddt 501 dxdt dx d 50 sec2 dt dt
(a) When 30 ,
dx 200 ftsec. dt 3
(c) When 70 ,
dx 427.43 ftsec. dt
48. sin 22
x y
(b) When 60 ,
dx 200 ftsec. dt
50. (a) dydt 3dxdt means that y changes three times as fast as x changes.
0 dx x dt y
x
x y2
dy 1 dt y
dx dt
dy sin 22 240 89.9056 mihr dt
y x 22˚
(b) y changes slowly when x 0 or x L. y changes more rapidly when x is near the middle of the interval.
366
Chapter 2
Differentiation
52. L2 144 x2; acceleration of the boat
First derivative: 2L L
d 2x . dt 2
dx dL 2x dt dt dx dL x dt dt
Second derivative: L
d 2L dL dt 2 dt
dL d 2x dx x 2 dt dt dt
dx dt
L ddtL dLdt dxdt
1 d 2x dt 2 x When L 13, x 5,
2
2
2
2
dx dL d 2L dL is constant, 2 0. 10.4, and 4 (see Exercise 30). Since dt dt dt dt
d 2x 1 130 42 10.42 dt 2 5 1 1 16 108.16 92.16 18.432 ftsec2 5 5 54.
yt 4.9t2 20
y
dy 9.8t dt y1 4.9 20 15.1 y1 9.8
20
y 20 By similar triangles, x x 12 20x 240 xy. When y 15.1, 20x 240 x15.1
20 15.1x 240 x
240 . 4.9
20x 240 xy 20
dx dy dx x y dt dt dt dx x dy dt 20 y dt
At t 1,
dx 2404.9 9.8 97.96 msec. dt 20 15.1
y 12
(0, 0)
x
x
92
Chapter 2
Differentiation
51. x2 y2 25; acceleration of the top of the ladder
First derivative: 2x
d 2y dt 2
dy dx 2y 0 dt dt dy dx y 0 dt dt
x Second derivative: x
d 2x dx dt 2 dt
d 2y dy dx y 2 dt dt dt
dy 0 dt
xddt x dxdt dydt
1 d 2y dt 2 y When x 7, y 24,
2
2
2
2
dy 7 dx dx d 2x , and 2 (see Exercise 27). Since is constant, 2 0. dt 12 dt dt dt
d 2y 1 7 70 22 dt 2 24 12
49 1 625 241 4 144 24 144 0.1808 ft sec 2
2
53. (a) Using a graphing utility, you obtain ms 0.881s2 29.10s 206.2 (b)
dm dm ds ds 1.762s 29.10 dt ds dt dt
(c) If t s 1995, then s 15.5 and Thus,
ds 1.2. dt
dm 1.76215.5 29.101.2 2.15 million. dt
Review Exercises for Chapter 2 1. f x x2 2x 3 fx lim
x→0
f x x f x x
x x2 2x x 3 x2 2x 3 x→0 x
lim
x2 2xx x2 2x 2x 3 x2 2x 3 x→0 x
lim
2xx x2 2x lim 2x x 2 2x 2 x→0 x→0 x
lim
5. f is differentiable for all x 1.
3. f x x 1 fx lim
x→0
lim
x→0
lim
x→0
lim
x→0
lim
x→0
f x x f x x
x x 1 x 1 x x x x
x
x x x x x x
x x x x x x x 1 x x x
1 2 x
Review Exercises for Chapter 2
7. f x 4 x 2
1 4 9. Using the limit definition, you obtain gx x . 3 6
(a) Continuous at x 2. (b) Not differentiable at x 2 because of the sharp turn in the graph.
4 1 3 At x 1, g1 3 6 2
y 7 6 5 4 3 2 x
−1
1 2 3 4 5 6
−2 −3
11. (a) Using the limit defintion, fx 3x 2.
13. g2 lim
x→2
At x 1, f1 3. The tangent line is y 3x 1
x3 x 2 4 x→2 x2
lim
0
−4
x 2x 1 4 x→2 x2
lim
y 2 3x 1
(b)
gx g2 x2
2
lim
(−1, −2)
x→2
x 2x 2 x 2 x2
lim x 2 x 2 8 x→2
−4
15.
19. f x x8
17. y 25
y
f′
f
fx 8x7
y 0
2
1
x −1
1
21. ht 3t 4
23. f x x3 3x2
ht 12t 3
fx 3x2 6x 3xx 2
3 x 6x1 2 3x1 3 25. hx 6 x 3
hx 3x1 2 x2 3 29. f 2 3 sin f 2 3 cos
3 x
2 27. gt t2 3
1 3 x2
gx
4 3 4 t 3 3 3t
31. f 3 cos
sin 4
f 3 sin
cos 4
93
94
Chapter 2
Differentiation
F 200 T
33.
Ft
35.
st 16t2 s0 s9.2 169.22 s0 0
100 T
s0 1354.24
(a) When T 4, F4 50 vibrations/sec/lb.
The building is approximately 1354 feet high (or 415 m).
(b) When T 9, F9 3313 vibrations/sec/lb. 37. (a)
(c) Ball reaches maximum height when x 25.
y
y x 0.02x2
(d)
15
y 1 0.04x
10
y0 1
5
y10 0.6
x 20
40
60
Total horizontal distance: 50 (b) 0 x 0.02x2
y25 0 y30 0.2 y50 1
x 0x 1 implies x 50. 50 39. xt t2 3t 2 t 2t 1 (a) vt xt 2t 3
(e) y25 0
3 (b) vt < 0 for t < 2 .
(d) xt 0 for t 1, 2.
at vt 2 3 (c) vt 0 for t 2 . 3 3 1 1 1 x 2 2 2 1 2 2 4
v1 21 3 1
v2 22 3 1 The speed is 1 when the position is 0. 43. hx x sin x x1 2 sin x
41. f x 3x2 7x2 2x 3 fx 3x2 72x 2 x2 2x 36x 2
6x3
9x2
16x 7
1 sin x x cos x 2 x
x2 x 1 x2 1 x2 12x 1 x2 x 12x fx x2 12 2 x 1 2 x 12
45. f x 2x x2
47. f x
fx 2 2x3 2 1
hx
1 x3
2x3 1 x3
51. y
49. f x 4 3x21 fx 4 3x226x 53. y 3x 2 sec x y 3x 2 sec x tan x 6x sec x
6x 4 3x22
y
x2 cos x cos x 2x x 2sin x 2x cos x x 2 sin x cos2 x cos2 x
55. y x tan x y x sec2 x tan x
Review Exercises for Chapter 2
59. gt t3 3t 2
57. y x cos x sin x
gt 3t2 3
y x sin x cos x cos x x sin x
g t 6t 61. f 3 tan
y 2 sin x 3 cos x
63.
f 3 sec2
y 2 cos x 3 sin x
f 6 sec sec tan 6 sec tan
y 2 sin x 3 cos x
2
y y 2 sin x 3 cos x 2 sin x 3 cos x 0 65. f x 1 x31 2 1 fx 1 x31 23x2 2
3x2
2 1 x3
s2
1
s
s2
1
3s2
2
xx 31x
2
2
11 x 32x x 2 12
2x 3x 2 6x 1 x 2 13
71. y 3 cos3x 1
5 1 2s
5 2
x3 x2 1
hx 2
69. f s s2 15 2s3 5 fs
67. hx
s3
5 2
s2
3 2
y 9 sin3x 1
3s 1 5 5
3 2
s2
s3
ss2 13 28s3 3s 25
73. y
1 csc 2x 2
1 y csc 2x cot 2x2 2
y
csc 2x cot 2x
77. y
x sin 2x 2 4
75. y
1 1 cos 2x2 2 4
1 1 cos 2x sin2 x 2
2 3 2 2 sin x sin7 2x 3 7
y sin1 2 x cos x sin5 2 x cos x
79. y y
cos x sin x1 sin2 x
sin x x2
x 2 cos x sin x x 22
cos3 x sin x 81. f t t2t 15
83. gx 2xx 11 2
ft tt 147t 2 The zeros of f correspond to the points on the graph of f where the tangent line is horizontal. 0.1
gx
g does not equal zero for any value of x in the domain. The graph of g has no horizontal tangent lines.
f′
4
−0.1
1.3
g′
f −0.1
x2 x 13 2
−2
7
g −2
95
96
Chapter 2
Differentiation
85. f t t 11 2t 11 3 t 15 6 ft
87. y tan 1 x
5 6t 11 6
y
f does not equal zero for any x in the domain. The graph of f has no horizontal tangent lines.
sec2 1 x 2 1 x
y does not equal zero for any x in the domain. The graph has no horizontal tangent lines.
5
5
y
f − 20
f′ −2
7
2
y′
−1
−4
91. f x cot x
89. y 2x2 sin 2x
93. f t
t 1 t2
ft
t1 1 t3
f t
2t 2 1 t4
fx csc2 x
y 4x 2 cos 2x
f 2 csc xcsc x cot x
y 4 4 sin 2x
2 csc2 x cot x
95. g tan 3 sin 1 g 3 sec2 3 cos 1 g 18 sec2 3 tan 3 sin 1 97. T 700t2 4t 101 T
1400t 2 t2 4t 102
(a) When t 1, T
(b) When t 3,
14001 2 18.667 deg hr. 1 4 102
T
(d) When t 10,
(c) When t 5, T
14003 2 7.284 deg hr. 9 12 102
14005 2 3.240 deg hr. 25 30 102
T
140010 2 0.747 deg hr. 100 40 102
x2 3xy y3 10
99.
2x 3xy 3y 3y2y 0 3x y2y 2x 3y 2x 3y 3x y2
y
y x x y 16
101. y
12x x 1 2
y x
1 2
12y
1 2
y y1 2 0
x 2 x yy y 2 y x 2 xy x 2 xy y y 2 y 2 x y
2 xy y 2 x
2 y 2 xy x
2y x y y 2x y x x
Review Exercises for Chapter 2
x sin y y cos x
103.
105.
x cos yy sin y y sin x y cos x
6
(2, 4)
2x 2yy 0 −9
yx cos y cos x y sin x sin y y
x2 y2 20
97
9
x y y
y sin x sin y cos x x cos y
−6
1 At 2, 4: y 2 1 Tangent line: y 4 x 2 2 x 2y 10 0 Normal line: y 4 2x 2 2x y 0
107.
y x dy 2 units sec dt dx dy 1 dx dy ⇒ 2 x 4 x dt dt dt 2 x dt 1 dx 2 2 units/sec. (a) When x , 2 dt
109.
(b) When x 1,
dx 4 units/sec. dt
(c) When x 4,
dx 8 units/sec. dt
s 1 2 h 2
111. st 60 4.9t2 st 9.8t
1 s h 4
s 35 60 4.9t2 4.9t2
dV 1 dt
w 2 2s 2 2
14h 4 2 h
tan 30
dV 5 dh 4 h dt 2 dt 2dV dt dh dt 54 h When h 1,
dh 2 m min. dt 25
1 2
1 3
5 4.9 st xt
xt 3 st
4h 5 5 2 h 8 hh 2 2 4
25
t
Width of water at depth h:
V
s (t)
ds 5 dx 3 39.8 dt dt 4.9 1 2
2
38.34 m sec
s 2 h 2
30˚ x(t )
98
Chapter 2
Differentiation
Problem Solving for Chapter 2 1. (a) x 2 y r2 r2 Circle
3
x 2 y Parabola Substituting,
−3
y r2 r2 y
3 −1
y 2 2r y r 2 r 2 y y2 2r y y 0 y y 2r 1 0 Since you want only one solution, let 1 2r 0 ⇒ r 12 Graph y x 2 and x 2 y 12 2 14
(b) Let x, y be a point of tangency: x 2 y b2 1 ⇒ 2x 2 y by 0 ⇒ y
x circle. by
y x 2 ⇒ y 2x (parabola). Equating, 2x
x by
3
2b y 1 by
−3
1 1 ⇒by 2 2
3 −1
Also, x 2 y b2 1 and y x 2 imply
y y b2 1 ⇒ y y y
1 2
1 ⇒ y 21 1 ⇒ y 43 and b 45.
54
Center: 0,
Graph y x 2 and x 2 y
3. (a)
5 4
2
1
f x cos x
P2x a0 a1x a2x 2
P10 a0 ⇒ a0 1
f 0 1
P20 a0 ⇒ a0 1
P10 a1 ⇒ a1 0
f0 0
P20 a1 ⇒ a1 0
f 0 1
1 P20 2a2 ⇒ a2 2
f x cos x
P1x a0 a1x
f 0 1 f0 0
(b)
P1x 1
1 P2x 1 2x 2
(c)
x
1.0
0.1
0.001
0
0.001
0.1
1.0
cos x
0.5403
0.9950
1
1
1
0.9950
0.5403
P2x
0.5
0.9950
1
1
1
0.9950
0.5
P2x is a good approximation of f x cos x when x is near 0. (d)
f x sin x
P3x a0 a1x a2 x 2 a3x3
f 0 0
P30 a0 ⇒ a0 0
f0 1
P30 a1 ⇒ a1 1
f 0 0
P30 2a2 ⇒ a2 0
f 0 1
P30 6a3 ⇒ a3 16
P3x x 6 x3 1
Problem Solving for Chapter 2 5. Let px Ax3 Bx 2 Cx D px 3Ax 2 2Bx C At 1, 1: A B C D 1
Equation 1
3A 2B C
Equation 2
14
At 1, 3: A B C D 3 3A 2B C
Equation 3
2
Equation 4
Adding Equations 1 and 3: 2B 2D 2 Subtracting Equations 1 and 3: 2A 2C 4 Adding Equations 2 and 4: 6A 2C 12 Subtracting Equations 2 and 4: 4B 16 1 Hence, B 4 and D 22 2B 5 1 Subtracting 2A 2C 4 and 6A 2C 12, you obtain 4A 8 ⇒ A 2. Finally, C 24 2A 0
Thus, px 2x3 4x 2 5. x4 a2 x 2 a 2 y 2
7. (a)
a2 y 2 a2x 2 x 4 y
± a2x 2 x 4
a
Graph: y1 (b)
a2x 2 x4
a
and y2
a2x 2 x4
a
2
a = 12 −3
3
a=2 a=1 −2
± a, 0 are the x-intercepts, along with 0, 0. (c) Differentiating implicitly, 4x 3 2a2 x 2a2 y y y
a2 2
±a 2a2x 4x3 xa2 2x 2 0 ⇒ 2x 2 a2 ⇒ x . 2 2a2y a2y
2
a2
a2 a y 2
2 2
a4 a4 a2y 2 4 2 a2y 2
a4 4
y2
a2 4
y±
a 2
Four points:
a2, a2, a2, 2a, a2, a2, a2, 2a
99
100
Chapter 2
9. (a)
Differentiation Line determined by 0, 30 and 90, 6:
y
(0, 30)
30
y 30
(90, 6) (100, 3) x 90
100
30 6 24 4 4 x 0 x x ⇒ y x 30 0 90 90 15 15
When x 100, y
4 10 100 30 > 3 ⇒ Shadow determined by man. 15 3
Not drawn to scale
(b)
Line determined by 0, 30 and 60, 6:
y
30
(0, 30)
y 30
(60, 6) (70, 3) x 60
70
Not drawn to scale
30 6 2 2 x 0 x ⇒ y x 30 0 60 5 5
When x 70, y
2 70 30 2 < 3 ⇒ Shadow determined by child. 5
(c) Need 0, 30, d, 6, d 10, 3 collinear. 30 6 63 24 3 ⇒ ⇒ d 80 feet 0d d d 10 d 10 (d) Let y be the length of the street light to the tip of the shadow. We know that For x > 80, the shadow is determined by the man. dy 5 dx 25 5 y yx ⇒ y x and . 30 6 4 dt 4 dt 4 For x < 80, the shadow is determined by the child. y y x 10 10 100 dy 10 dx 50 . ⇒y x and 30 3 9 9 dt 9 dt 9 Therefore,
25 dy 4 dt 50 9
x > 80 0 < x < 80
dy is not continuous at x 80. dt
11. Lx lim
x→0
Lx x Lx x
lim
Lx Lx Lx x
lim
Lx x
x→0
x→0
Also, L0 lim
x→0
Lx L0 x
But, L0 0 because L0 L0 0 L0 L0 ⇒ L0 0. Thus, Lx L0, for all x. The graph of L is a line through the origin of slope L0.
dx 5. dt
Problem Solving for Chapter 2 13. (a)
z (degrees)
0.1
0.01
0.0001
sin z z
0.0174524
0.0174533
0.0174533
(b) lim
z→0
sin z 0.0174533 z
In fact, lim
z→0
(c)
sin z z 180
d sin z z sin z sin z lim z →0 dz z lim
sin z cos z sin z cos z sin z z
lim
sin z
z →0
z →0
cos z 1 z
sin z0 cos z (d) S90 sin
lim
z →0
cos z
sin z z
cos z 180 180
90 sin 1; C180 cos 180 1 180 2 180
d d Sz sincz c coscz Cz dz dz 180 (e) The formulas for the derivatives are more complicated in degrees. 15. jt at (a) jt is the rate of change of the acceleration. (b) From Exercise 102 in Section 2.3, st 8.25t 2 66t vt 16.5t 66 at 16.5 at jt 0
101
Review Exercises for Chapter 2
Review Exercises for Chapter 2
2. f x
x1 x1
x x 1 x 1 f x x f x x x 1 x 1 fx lim lim x→0 x→0 x x
x x 1x 1 x x 1x 1 xx x 1x 1
lim
x→0
x2 xx x x x 1 x2 xx x x x 1 x→0 xx x 1x 1
lim
2x 2 2 lim xx x 1x 1 x→0 x x 1x 1 x 12
lim
x→0
2 4. f x x
6. f is differentiable for all x 3.
fx lim
x→0
f x x f x x
2 2 x x x lim x→0 x lim
2x 2x 2x xx xx
lim
2x xx xx
lim
2 2 2 x xx x
x→0
x→0
x→0
8. f x
x1 4x4xx2,,
if x < 2 if x ≥ 2
2
2
10. Using the limit defintion, you obtain hx
(a) Nonremovable discontinuity at x 2.
At x 2, h2
(b) Not differentiable at x 2 because the function is discontinuous there. y 5 4
1 −5 −4
−2 −1 −1
x 1
2
−2
12. (a) Using the limit definition, fx
2 . x 12
At x 0, f0 2. The tangent line is y 2 2x 0 y 2x 2
(b)
4
(0, 2) −6
6
−4
3 67 42 . 8 8
3 4x. 8
367
368
Chapter 2
14. f2 lim
x→2
Differentiation
f x f 2 x2
16.
y
1
1 1 x1 3 lim x→2 x2 lim
3x1 x 2x 13
lim
1 1 x 13 9
x→2
x→2
18. y 12
−
π 2
x
π 2
−1
f′
22. f t 8t 5
20. gx x12
y 0
f
ft 40t 4
gx 12x11
24. gs 4s4 5s2
2 28. hx x2 9
26. f x x12 x12
gs 16s3 10s
1 1 x1 fx x12 x32 32 2 2 2x
30. g 4 cos 6
32. g
g 4 sin g
hx
4 3 4 x 3 9 9x
5 sin 2 3 5 cos 2 3
34. s 16t2 s0 First ball: 16t2 100 0 t
10 2.5 seconds to hit ground 100 16 4
Second ball: 16t2 75 0 t2
7516 5 4 3 2.165 seconds to hit ground
Since the second ball was released one second after the first ball, the first ball will hit the ground first. The second ball will hit the ground 3.165 2.5 0.665 second later. 36. st 16t2 14,400 0 16t2 14,400 t 30 sec 1 1 Since 600 mph 6 mi/sec, in 30 seconds the bomb will move horizontally 6 30 5 miles.
Review Exercises for Chapter 2
38.
y
v02 64
(
v02 v02 , 64 128
) 2 0
( v32 , 0 ) x v02 128
(a) y x
32 32 2 x x 1 2x v02 v0
0 if x 0 or x
(b) y 1
v02 . 32
64 x v02
When x
v02 64 v02 , y 1 2 0. 64 v0 64
Projectile strikes the ground when x v0232. Projectile reaches its maximum height at x v0264. (one-half the distance) (c) y x
32 2 32 x x 1 2x 0 v02 v0
(d) v0 70 ftsec
when x 0 and x x0232. Therefore, the range is x v0232. When the initial velocity is doubled the range is x
Range: x
Maximum height: y
2v0 4v0 32 32 2
v02 702 153.125 ft 32 32
2
v02 702 38.28 ft 128 128
50
or four times the initial range. From part (a), the maximum height occurs when x v0264. The maximum height is 0
v02 v02 32 v02 2 y 64 64 v0 64
2
v02 v02 v02 . 64 128 128
0
160
If the initial velocity is doubled, the maximum height is
y
v02 2v02 2v02 4 64 128 128
or four times the original maximum height. 40. (a) y 0.14x2 4.43x 58.4
(b)
320
0
60 0
(c)
(d) If x 65, y 362 feet.
12
0
60 0
(e) As the speed increases, the stopping distance increases at an increasing rate.
369
370
Chapter 2
Differentiation
42. gx x3 3xx 2
44. f t t 3 cos t
gx x3 3x1 x 23x 2 3
ft t3sin t cos t3t2
x3 3x 3x3 6x 2 3x 6
t3 sin t 3t2 cos t
4x3 6x 2 6x 6 46. f x fx
x1 x1
48. f x
x 11 x 11 x 12
fx
2 x 12
50. f x 93x2 2x1 fx 93x2 2x26x 2 54. y 2x x 2 tan x
52. y 181 3x 3x2 2x2
y 56. y
y 2 x 2 sec2 x 2x tan x y 58. vt 36 t2, 0 ≤ t ≤ 6
sin x x2
x2 cos x sin x2x x cos x 2 sin x x4 x3 1 sin x 1 sin x
1 sin x cos x 1 sin xcos x 1 sin x2 2 cos x 1 sin x2
f x
a4 8 msec
h t 4 sin t 5 cos t
23 5x 3x2 x2 12
fx 3x34
v4 36 16 20 msec
ht 4 cos t 5 sin t
x2 16 6x 52x x2 12
60. f x 12x14
at vt 2t
62. ht 4 sin t 5 cos t
6x 5 x2 1
9 74 9 x 74 4 4x
y
64.
10 cos x x
xy cos x 10 xy y sin x 0 xy sin x y xy y sin x y y sin x
66. f x x2 113 1 fx x2 1232x 3 2x 3x2 123
68. f x x2
5
1 x
2x x1
1 x
fx 5 x2
4
2
Review Exercises for Chapter 2
70. h
371
72. y 1 cos 2x 2 cos2 x
1 3
y 2 sin 2x 4 cos x sin x
1 3 31 21 h 1 6
2 2 sin x cos x 4 sin x cos x 0
1 21 3 2 1 1 6 1 4 74. y csc 3x cot 3x
76. y
y 3 csc 3x cot 3x 3 csc2 3x 3 csc 3xcot 3x csc 3x
sec7 x sec5 x 7 5
y sec6 xsec x tan x sec4 xsec x tan x sec5 x tan xsec2 x 1 sec5 x tan3 x
3x
78. f x
80. y
x 2 1
1 3x 2 112 3x x 2 1122x 2 fx x2 1
y
x 1 sinx 1 cosx 11 x 12
3x 2 1 3x 2 x 2 132
cosx 1 x1
1 x 1 sinx 1 cosx 1 x 12
3 x 2 132
82. f x x 2x 4 2 x2 2x 82
100
f′
fx 4x3 3x2 6x 8 4x 2x 1x 4
−7
The zeros of f correspond to the points on the graph of f where the tangent line is horizontal.
gx
2x2 1 x2 1
y
g does not equal zero for any value of x. The graph of g has no horizontal tangent lines.
y does not equal zero for any x in the domain. The graph has no horizontal tangent lines. 75
y′
y
g′ 6
−3
3
g −3
− 60
3x 227x 2 23x
5
−6
5
86. y 3xx 23
84. gx xx2 112
− 25
f
372
Chapter 2
Differentiation
88. y 2 csc3 x y
3 x
10
csc3 x cot x
y
The zero of y corresponds to the point on the graph of y where the tangent line is horizontal.
−4
2 sec xsec x tan x
y 2 cos 2x
2 2 sec2 x tan x x3
96. hx xx2 1
x2x2 3 x2 132
x2 9y2 4x 3y 0
23x2 5x 3 x2 12
g x
26x3 15x2 18x 5 x2 13
(a) When h 9,
4 dv ftsec. dh 3
(b) When h 4,
dv 2 ftsec. dh
y2 x3 x2y xy y2
102.
2x 18yy 4 3y 0
0 x3 x2y xy 2y2
36y 1y 4 2x y
0 3x2 x2y 2xy xy y 4yy
x2 x 4yy 3x2 2xy y
4 2x 36y 1
cosx y x
104.
gx
4 dv dh h
x2 1
h x
6x 5 x2 1
v 2gh 232h 8h
98.
2x2 1
hx
100.
94. gx
y 2 sin x cos x sin 2x
y x2 sec2 x y 2x
8
92. y sin2 x
90. y x1 tan x 3
y′
−1
106.
1 y sinx y 1
3x2 2xy y x2 x 4y
x2 y2 16
10
2x 2yy 0
y sinx y 1 sinx y y
y
1 sinx y sinx y
cscx 1 1
y
−10
x y
At 5, 3: y
5 3
5 Tangent line: y 3 x 5 3 5x 3y 16 0 3 Normal line:y 3 x 5 5 3x 5y 30 0 108. Surface area A 6x 2, x length of edge. dx 5 dt da dx 12x 124.55 270 cm2sec dt dt
10
−10
Problem Solving for Chapter 2 tan x
110.
d 32 radmin dt sec2
1
θ
ddt dxdt
x
dx tan2 16 6x2 1 dt 1 dx 1 15 When x , 6 1 kmmin 450 kmhr. 2 dt 4 2
Problem Solving for Chapter 2 2.
Let a, a2 and b, b2 2b 5 be the points of tangency.
y 10 8 6 4 −8 −6 −4 −2
For y x 2, y 2x and for y x 2 2x 5, y 2x 2. Thus, 2a 2b 2 ⇒ a b 1, or a 1 b. Furthermore, the slope of the common tangent line is
x 2 4 6 8 10
−4 −6
a2 b2 2b 5 1 b2 b2 2b 5 2b 2 ab 1 b b ⇒
1 2b b2 b2 2b 5 2b 2 1 2b
⇒ 2b2 4b 6 4b2 6b 2 ⇒ 2b2 2b 4 0 ⇒ b2 b 2 0 ⇒ b 2b 1 0 b 2, 1 For b 2, a 1 b 1 and the points of tangency are 1, 1, 2, 5. The tangent line has slope 2: y 1 2x 1 ⇒ y 2x 1 For b 1, a 1 b 2 and the points of tangency are 2, 4 and 1, 8. The tangent line has slope 4: y 4 4x 2 ⇒ y 4x 4. 4. (a) y x 2, y 2x. Slope 4 at 2, 4. Tangent line: y 4 4x 2 y 4x 4 1
(b) Slope of normal line: 4. 1 Normal line: y 4 4x 2
y 14x 92 y 14x 92 x 2 ⇒ 4x 2 x 18 0 ⇒ 4x 9x 2 0 x 2, 94. Second intersection point: 94, 81 16 (c) Tangent line: y 0 Normal line: x 0 —CONTINUED—
373
374
Chapter 2
Differentiation
4. —CONTINUED— (d) Let a, a2, a 0, be a point on the parabola y x 2. Tangent line at a, a2 is y 2ax a a2. Normal line at a, a2 is 1 y x a a2. To find points of intersection, solve 2a x2 x2 x2
1 x a a2 2a
1 1 x a2 2a 2
1 1 1 1 x a2 2a 16a2 2 16a2
x 4a1 a 4a1 2
x
2
1 1 ± a 4a 4a
x
1 1 a ⇒ x a Point of tangency 4a 4a
x
1 1 1 2a2 1 a ⇒ x a 4a 4a 2a 2a
The normal line intersects a second time at x
2a2 1 . 2a
6. f x a b cos cx fx bc sin cx At 0, 1: a b 1 At
Equation 1
4 , 32: a b cosc4 23
Equation 2
c4 1
Equation 3
bc sin
From Equation 1, a 1 b. Equation 2 becomes 1 b b cos From Equation 3, b
1 cos
c4 23 ⇒ b b cos c4 12
1 1 1 1 c . Thus cos c c c 4 2 c sin c sin c sin 4 4 4
c4 21 c sinc4
Graphing the equation gc
1 c c c sin cos 1, you see that many values of c will work. 2 4 4
1 3 3 1 One answer: c 2, b , a ⇒ f x cos 2x 2 2 2 2
Problem Solving for Chapter 2 8. (a) b2y 2 x3a x; a, b > 0 y2
(b) a determines the x-intercept on the right: a, 0.
x a x b2 3
b affects the height.
x3a x
Graph y1
b
and y2
x3a x
b
(c) Differentiating implicitly. 2b 2 y y 3x 2a x x 3 3ax 2 4x 3 y
3ax 2 4x3 0 2b2y
⇒ 3ax 2 4x3 3a 4x x b2y 2 y2
3a . 4
1 a 3a4 a 3a4 27a 64 4 3
3
27a 4 33a2 ⇒y± 256b 2 16b
Two points:
10. (a) y x13 ⇒
3a4 , 3 16b3a , 3a4 , 316b3a
2
2
dy 1 23 dx x dt 3 dt
1 dx 1 823 3 dt dx 12 cmsec dt (b) D x 2 y 2 ⇒
dD 1 2 dx dy x y 2 2x 2y dt 2 dt dt
dx dy y dt dt x 2 y 2 x
y (c) tan ⇒ sec2 x
812 21 98 49 cmsec. 64 4 68 17
dy dx x y d dt dt dt x2
68 2
θ 8
From the triangle, sec
68
8
. Hence
d 81 212 16 4 radsec dt 68 68 17 64 64
375
376
Chapter 2
Differentiation
12. Ex lim
Ex x Ex
x
lim
ExE x Ex
x
x→0
x→0
lim Ex
E x x 1
Ex lim
E x 1
x
x→0
x→0
But, E0 lim
x→0
E x E0 E x 1 lim 1.
x→0
x
x
Thus, Ex ExE0 Ex exists for all x. For example: Ex e x.
14. (a) vt
27 t 27 ftsec 5
at (b) vt
27 ftsec2 5 27 27 t 27 0 ⇒ t 27 ⇒ t 5 seconds 5 5
S5
27 2 5 275 6 73.5 feet 10
(c) The acceleration due to gravity on Earth is greater in magnitude than that on the moon.
C H A P T E R 3 Applications of Differentiation Section 3.1
Extrema on an Interval
. . . . . . . . . . . . . . 103
Section 3.2
Rolle’s Theorem and the Mean Value Theorem
Section 3.3
Increasing and Decreasing Functions and the First Derivative Test . . . . . . . . . . . . . . 113
Section 3.4
Concavity and the Second Derivative Test . . . . 121
Section 3.5
Limits at Infinity
Section 3.6
A Summary of Curve Sketching
Section 3.7
Optimization Problems . . . . . . . . . . . . . . 145
Section 3.8
Newton’s Method . . . . . . . . . . . . . . . . . 155
Section 3.9
Differentials . . . . . . . . . . . . . . . . . . . . 160
. 107
. . . . . . . . . . . . . . . . . 129 . . . . . . . . . 136
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 163 Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . 172
C H A P T E R 3 Applications of Differentiation Section 3.1
Extrema on an Interval
Solutions to Odd-Numbered Exercises
1. f x fx
x2 x2 4
3. f x x
x2 42x x22x 8x 2 x2 42 x 42
f0 0
5.
fx 1 27x3 1 f3 1
f x x 223
27 27 x x2 2x2 x 27 x3
27 110 33
7. Critical numbers: x 2 x 2: absolute maximum
2 fx x 213 3 f2 is undefined. 9. Critical numbers: x 1, 2, 3
11. f x x2x 3 x3 3x2
x 1, 3: absolute maximum
fx 3x2 6x 3xx 2
x 2: absolute minimum
Critical numbers: x 0, x 2 15. hx sin2 x cos x, 0 < x < 2
13. gt t4 t, t < 3
124 t
gt t
1 4 t12
12
1 4 t12t 24 t 2
hx 2 sin x cos x sin x sin x2 cos x 1 On 0, 2, critical numbers: x
5 , x , x 3 3
8 3t 24 t
8 Critical number is t . 3 17. f x 23 x, 1, 2
19. f x x2 3x, 0, 3
fx 2 ⇒ No critical numbers
fx 2x 3
Left endpoint: 1, 8 Maximum
Left endpoint: 0, 0 Minimum
Right endpoint: 2, 2 Minimum
Critical number:
32 , 94 Maximum
Right endpoint: 3, 0 Minimum
103
104
Chapter 3
Applications of Differentiation
3 21. f x x3 x2, 1, 2 2 fx 3x2 3x 3xx 1 Left endpoint:
1, 25 Minimum
23. f x 3x23 2x, 1, 1 fx 2x13 2
3 x 2 1 3 x
Left endpoint: 1, 5 Maximum
Right endpoint: 2, 2 Maximum
Critical number: 0, 0 Minimum
Critical number: 0, 0
Right endpoint: 1, 1
1, 21
Critical number:
25. gt gt
t2
t2 , 1, 1 3
27. hs
6t t2 32
Left endpoint:
hs
1, 41 Maximum
Critical number: 0, 0 Minimum Right endpoint:
1 s 22
Left endpoint:
0, 21 Maximum
Right endpoint: 1, 1 Minimum
1, 14 Maximum 6
29. f x cos x, 0,
1
31. y
fx sin x Left endpoint: 0, 1 Maximum Right endpoint:
1 , 0, 1 s2
1 3 Minimum , 6 2
y
4 x tan , 1, 2 x 8 4 x 0 sec2 x2 8 8
x 4 sec2 2 8 8 x On the interval 1, 2, this equation has no solutions. Thus, there are no critical numbers. Left endpoint: 1, 2 3 1, 4.4142 Maximum Right endpoint: 2, 3 Minimum
33. (a) Minimum: 0, 3 Maximum: 2, 1 (b) Minimum: 0, 3
35. f x x2 2x (a) Minimum: 1, 1 Maximum: 1, 3
(c) Maximum: 2, 1
(b) Maximum: 3, 3
(d) No extrema
(c) Minimum: 1, 1 (d) Minimum: 1, 1
Section 3.1
37. f x
2x4x , 2, 2
0 ≤ x ≤ 1 1 < x ≤ 3
Extrema on an Interval
105
3 , 1, 4 x1 Right endpoint: 4, 1 Minimum
39. f x
Left endpoint: 0, 2 Minimum
4
Right endpoint: 3, 36 Maximum 36 −1
4 −1
−1
3 −4
41. (a)
f x 3.2x5 5x3 3.5x, 0, 1
(b)
5
(1, 4.7)
0
fx 16x4 15x2 3.5 16x4 15x2 3.5 0
1
(0.4398, − 1.0613)
x2
−2
Maximum: 1, 4.7 (endpoint)
Minimum: 0.4398, 1.0613
x
15 ± 152 4163.5 216 15 ± 449 32
15 32
449
0.4398
f 0 0 f 1 4.7 Maximum (endpoint) f
15 32
449
1.0613
Minimum: 0.4398, 1.0613 43. f x 1 x312, 0, 2 3 fx x21 x312 2 3 f x x4 4x1 x332 4 3 fx x6 20x3 81 x352 8 Setting f 0, we have x6 20x3 8 0. x3 x
20 ± 400 418 2 3 10
± 108 3 1
In the interval 0, 2, choose x
3 10
± 108 3 1 0.732.
3 10 108 f 1.47 is the maximum value.
45.
f x x 123, 0, 2 fx
2 x 113 3
2 f x x 143 9 8 fx x 173 27 f 4x f 5x
56 x 1103 81
560 x 1133 243 56
f 40 81 is the maximum value.
106
Chapter 3
Applications of Differentiation
47. f x tan x f is continuous on 0, 4 but not on 0, . y
49.
51. (a) Yes
53. (a) No
(b) No
(b) Yes
5 4 3
lim tan x .
x → 2
f
2 1
x
−2 −1
1
3
4
5
6
−2 −3
55. P VI RI 2 12I 0.5I 2, 0 ≤ I ≤ 15
S 6hs
57.
P 0 when I 0.
3s2 3 cos , ≤ ≤ 2 sin 6 2
dS 3s2 3csc cot csc2 d 2
P 67.5 when I 15. P 12 I 0
Critical number: I 12 amps
3s 2 csc 3cot csc 0 2
csc 3cot
When I 12 amps, P 72, the maximum output.
sec 3
No, a 20-amp fuse would not increase the power output. P is decreasing for I > 12.
arcsec3 0.9553 radians S
6 6hs 3s2 3
S
2 6hs 3s2 3
2
2
S arcsec3 6hs
3s 2 2 2
S is minimum when arcsec3 0.9553 radians. 59. (a) y ax2 bx c
y
A
y 2ax b
B
The coordinates of B are 500, 30, and those of A are 500, 45. From the slopes at A and B,
9% −500
6%
x 500
1000a b 0.09 1000a b 0.06. Solving these two equations, you obtain a 340000 and b 3200. From the points 500, 30 and 500, 45, you obtain 30
3 3 c 5002 500 40000 200
45
3 3 c. 5002 500 40000 200
In both cases, c 18.75 y
75 . Thus, 4
3 75 3 x2 x . 40000 200 4
—CONTINUED—
Section 3.2
Rolle’s Theorem and the Mean Value Theorem
107
59. —CONTINUED— (b)
x
500
400
300
200
100
000
100
200
300
400
500
d
0
.75
3
6.75
12
18.75
12
6.75
3
.75
0
For 500 ≤ x ≤ 0, d ax2 bx c 0.09x. For 0 ≤ x ≤ 500, d ax2 bx c 0.06x. (c) The lowest point on the highway is 100, 18, which is not directly over the point where the two hillsides come together. 61. True. See Exercise 25.
Section 3.2
63. True.
Rolle’s Theorem and the Mean Value Theorem
1. Rolle’s Theorem does not apply to f x 1 x 1 over 0, 2 since f is not differentiable at x 1.
3. f x x2 x 2 x 2x 1 x-intercepts: 1, 0, 2, 0 1 fx 2x 1 0 at x . 2
5. f x x x 4
7. f x x2 2x, 0, 2
x-intercepts: 4, 0, 0, 0
f 0 f 2 0
1 fx x x 41 2 x 41 2 2
f is continuous on 0, 2. f is differentiable on 0, 2. Rolle’s Theorem applies.
x 41 2
2x x 4
3 8 fx x 4 x 41 2 0 at x 2 3 9. f x x 1x 2x 3, 1, 3
fx 2x 2 2x 2 0 ⇒ x 1 c value: 1 11. f x x2 3 1, 8, 8
f 1 f 3 0
f 8 f 8 3
f is continuous on 1, 3. f is differentiable on 1, 3. Rolle’s Theorem applies.
f is continuous on 8, 8. f is not differentiable on 8, 8 since f0 does not exist. Rolle’s Theorem does not apply.
f x x3 6x2 11x 6 fx 3x2 12x 11 3x2 12x 11 0 ⇒ x c
6 3 6 3 ,c 3 3
6 ± 3 3
108
Chapter 3
13. f x
Applications of Differentiation
x2 2x 3 , 1, 3 x2
f 1 f 3 0 f is continuous on 1, 3. (Note: The discontinuity, x 2, is not in the interval.) f is differentiable on (1, 3. Rolle’s Theorem applies. fx
x 22x 2 x2 2x 31 0 x 22 x2 4x 1 0 x 22 x
4 ± 2 5 2 ± 5 2
c value: 2 5
15. f x sin x, 0, 2
17. f x
f 0 f 2 0
f 0 f
f is continuous on 0, 2. f is differentiable on 0, 2. Rolle’s Theorem applies.
0, 6
6 0
f is continuous on 0, 6. f is differentiable on 0, 6. Rolle’s Theorem applies.
fx cos x c values:
6x 4 sin2 x,
3 , 2 2
fx
6 8 sin x cos x 0
6 8 sin x cos x 1 3 sin 2x 4 2 3 sin 2x 2
3 1 arcsin x 2 2 x 0.2489 c value: 0.2489 19. f x tan x, 0,
f x x 1, 1, 1
21.
f 0 f 0
f 1 f 1 0
f is not continuous on 0, since f 2 does not exist. Rolle’s Theorem does not apply.
f is continuous on 1, 1. f is not differentiable on 1, 1 since f0 does not exist. Rolle’s Theorem does not apply. 1
−1
1
−1
Section 3.2
23.
4, 4 1 1
f x 4x tan x,
25. f t 16t2 48t 32
(b) v ft must be 0 at some time in 1, 2.
f is continuous on 1 4, 1 4. f is differentiable on 1 4, 1 4. Rolle’s Theorem applies.
ft 32t 48 0 t
fx 4 sec2 x 0 sec2 x
109
(a) f 1 f 2 64
1 1 f 0 4 4
f
Rolle’s Theorem and the Mean Value Theorem
3 seconds 2
4
sec x ± x±
2
1 2 1
arcsec ± arccos
2
± 0.1533 radian c values: ± 0.1533 radian 0.5
− 0.25
0.25
− 0.5
27.
29. f x
y
tangent line
1 , 0, 6 x3
f has a discontinuity at x 3.
(c2, f(c2)) f
(a, f(a)) (b, f(b)) (c1, f(c1)) a tangent line
b secant line
x
31. f x x2 is continuous on 2, 1 and differentiable on 2, 1. f 1 f 2 1 4 1 1 2 3 1 fx 2x 1 when x . Therefore, 2 1 c . 2
33. f x x2 3 is continuous on 0, 1 and differentiable on 0, 1. f 1 f 0 1 10 2 fx x1 3 1 3 x
23
c
8 27
3
8 27
110
Chapter 3
Applications of Differentiation
35. f x 2 x is continuous on 7, 2 and differentiable on 7, 2. f 2 f 7 0 3 1 2 7 9 3 fx
37. f x sin x is continuous on 0, and differentiable on 0, . f f 0 0 0 0 0
1 1 3 2 2 x
fx cos x 0 c
2 2 x 3
39. f x (a)
2 x
3 2
2x
9 4
x
1 4
c
1 4
2
x 1 on , 2 . x1 2 (c) fx
1
f
tangent − 0.5
2
secant
1 2 x 12 3
x 12
−1
3 2
x 1 ±
(b) Secant line: f 2 f 1 2 2 3 1 2 slope 2 1 2 5 2 3 y
2 2 x 2 3 3
6
2
In the interval 1 2, 2, c 1 6 2. f c
1 6 2
1 6 2 1
Tangent line: y 1
3y 2 2x 4 3y 2x 2 0
32 1 ±
y1
2 2 6 1
6
6
6 2 2 1 x 3 2
6
6
3
6 2 2 x 3 3 3
3y 2x 5 2 6 0
Section 3.2
Rolle’s Theorem and the Mean Value Theorem
111
41. f x x, 1, 9
1, 1, 9, 3 m (a)
31 1 91 4 (c)
3
tangent
1 2 x
f 9 f 1 1 91 4
secant
f
fx
1
9 1
1 (b) Secant line: y 1 x 1 4 y
3 1 x 4 4
1 1 2 c 4
c 2
c4
c, f c 4, 2
0 x 4y 3
m f4
1 4
1 Tangent line: y 2 x 4 4 y
1 x1 4
0 x 4y 4 43. st 4.9t 2 500 (a) Vavg
45. No. Let f x x2 on 1, 2.
s3 s0 455.9 500 14.7 m sec 30 3
(b) st is continuous on 0, 3 and differentiable on 0, 3. Therefore, the Mean Value Theorem applies.
fx 2x f0 0 and zero is in the interval (1, 2 but f 1 f 2.
vt st 9.8t 14.7 m sec t
14.7 1.5 seconds 9.8
47. Let St be the position function of the plane. If t 0 corresponds to 2 P.M., S0 0, S5.5 2500 and the Mean Value Theorem says that there exists a time t0, 0 < t0 < 5.5, such that St0 vt0
2500 0 454.54. 5.5 0
Applying the Intermediate Value Theorem to the velocity function on the intervals 0, t0 and t0, 5.5, you see that there are at least two times during the flight when the speed was 400 miles per hour. 0 < 400 < 454.54
112
Chapter 3
Applications of Differentiation
49. (a) f is continuous on 10, 4 and changes sign, f 8 > 0, f 3 < 0. By the Intermediate Value Theorem, there exists at least one value of x in 10, 4 satisfying f x 0.
(b) There exist real numbers a and b such that 10 < a < b < 4 and f a f b 2. Therefore, by Rolle’s Theorem there exists at least one number c in 10, 4 such that fc 0. This is called a critical number.
y
(c)
y
(d)
8
8
4
4 x
−8
−4
x −8
4
−4
4
−4
−4
−8
−8
(e) No, f did not have to be continuous on 10, 4. 51. f is continuous on 5, 5 and does not satisfy the conditions of the Mean Value Theorem. ⇒ f is not differentiable on 5, 5. Example: f x x
53. False. f x 1 x has a discontinuity at x 0.
y
8
f )x)
x
6
)5, 5)
) 5, 5) 4 2
x 4
2
2
4
2
55. True. A polynomial is continuous and differentiable everywhere. 57. Suppose that px x2n1 ax b has two real roots x1 and x2. Then by Rolle’s Theorem, since px1 px2 0, there exists c in x1, x2 such that pc 0. But px 2n 1x2n a 0, since n > 0, a > 0. Therefore, px cannot have two real roots. 59. If px Ax2 Bx C, then px 2Ax B
f b f a Ab2 Bb C Aa2 Ba C ba ba
Ab2 a2 Bb a ba
b aAb a B ba
Ab a B. Thus, 2Ax Ab a and x b a 2 which is the midpoint of a, b. 61. f x 12 cos x differentiable on , . fx 12 sin x 12 ≤ fx ≤
1 2
⇒ fx < 1 for all real numbers.
Thus, from Exercise 60, f has, at most, one fixed point. x 0.4502
Section 3.3
Section 3.3
Increasing and Decreasing Functions and the First Derivative Test
Increasing and Decreasing Functions and the First Derivative Test
1. f x x2 6x 8
3. y
Increasing on: 3,
Decreasing on: 2, 2 2 7. gx x 2x 8
5. f x 1 x2 x2
gx 2x 2
2 x3
Critical number: x 1
Discontinuity: x 0 Test intervals: Sign of fx:
< x < 0
Conclusion:
0 < x <
f > 0
f < 0
Increasing
Decreasing
Test intervals:
< x < 1
Sign of gx:
g < 0
g > 0
Decreasing
Increasing
Conclusion:
1 < x <
Increasing on: 1,
Increasing on , 0
Decreasing on: , 1
Decreasing on 0, Domain: 4, 4
9. y x16 x2 y
x3 3x 4
Increasing on: , 2, 2,
Decreasing on: , 3
fx
113
2 8 2 x 22x 22 16 x2 16 x2 x2
Critical numbers: x ± 22 Test intervals:
4 < x < 22
22 < x < 22
22 < x < 4
y < 0
y > 0
y < 0
Decreasing
Increasing
Decreasing
Sign of y: Conclusion:
Increasing on 22, 22 Decreasing on 4, 22, 22, 4 11. f x x2 6x
13. f x 2x2 4x 3
fx 2x 6 0
fx 4x 4 0
Critical number: x 3
Critical number: x 1
Test intervals: Sign of fx:
< x < 3
Conclusion:
3 < x <
Test intervals:
f < 0
f > 0
Sign of fx:
Decreasing
Increasing
Conclusion:
< x < 1
1 < x <
f > 0
f < 0
Increasing
Decreasing
Increasing on: 3,
Increasing on: , 1
Decreasing on: , 3
Decreasing on: 1,
Relative minimum: 3, 9
Relative maximum: 1, 5
114
Chapter 3
Applications of Differentiation
15. f x 2x3 3x2 12x fx 6x2 6x 12 6x 2x 1 0 Critical numbers: x 2, 1 Test intervals: Sign of fx:
< x < 2
2 < x < 1
f > 0
f < 0
f > 0
Increasing
Decreasing
Increasing
Conclusion:
1 < x <
Increasing on: , 2, 1, Decreasing on: 2, 1 Relative maximum: 2, 20 Relative minimum: 1, 7 17. f x x23 x 3x2 x3 fx 6x 3x2 3x2 x Critical numbers: x 0, 2 Test intervals:
< x < 0
0 < x < 2
Sign of fx:
f < 0
f > 0
f < 0
Conclusion:
Decreasing
Increasing
Decreasing
2 < x <
Increasing on: 0, 2 Decreasing on: , 0, 2, Relative maximum: 2, 4 Relative minimum: 0, 0
19. f x
x5 5x 5
fx x4 1 Critical numbers: x 1, 1 Test intervals: Sign of fx: Conclusion:
< x < 1
1 < x < 1
f > 0
f < 0
f > 0
Increasing
Decreasing
Increasing
Increasing on: , 1, 1, Decreasing on: 1, 1 Relative maximum: 1, 45 Relative minimum: 1, 45
1 < x <
Section 3.3
Increasing and Decreasing Functions and the First Derivative Test
21. f x x13 1
23. f x x 123
1 1 fx x23 23 3 3x
fx
Critical number: x 0
Critical number: x 1
< x < 0
Test intervals: Sign of fx: Conclusion:
0 < x <
2 3x 113
< x < 1
Test intervals:
f > 0
f > 0
Sign of fx:
Increasing
Increasing
Conclusion:
Decreasing
Increasing
No relative extrema
Decreasing on: , 1 Relative minimum: 1, 0
1, x < 5 x5 x5 1, x > 5
Critical number: x 5 Test intervals: Sign of fx:
< x < 5
Conclusion:
5 < x <
f > 0
f < 0
Increasing
Decreasing
Increasing on: , 5 Decreasing on: 5, Relative maximum: 5, 5 27. f x x fx 1
1 x 1 x2 1 2 x x2
Critical numbers: x 1, 1 Discontinuity: x 0 Test intervals: Sign of fx: Conclusion:
< x < 1
1 < x < 0
0 < x < 1
f > 0
f < 0
f < 0
f > 0
Increasing
Decreasing
Decreasing
Increasing
Increasing on: , 1, 1, Decreasing on: 1, 0, 0, 1 Relative maximum: 1, 2 Relative minimum: 1, 2
1 < x <
f > 0
Increasing on: 1,
25. f x 5 x 5
1 < x <
f < 0
Increasing on: ,
fx
115
116
Chapter 3
29. f x fx
Applications of Differentiation
x2 x2 9 18x x2 92x x22x 2 x2 92 x 92
Critical number: x 0 Discontinuities: x 3, 3 Test intervals: Sign of fx:
< x < 3
3 < x < 0
0 < x < 3
f > 0
f > 0
f < 0
f < 0
Increasing
Increasing
Decreasing
Decreasing
Conclusion:
3 < x <
Increasing on: , 3, 3, 0 Decreasing on: 0, 3, 3, Relative maximum: 0, 0 31. f x fx
x2 2x 1 x1
x 12x 2 x2 2x 11 x2 2x 3 x 3x 1 x 12 x 12 x 12
Critical numbers: x 3, 1 Discontinuity: x 1 Test intervals: Sign of fx:
< x < 3
3 < x < 1
1 < x < 1
f > 0
f < 0
f < 0
f > 0
Increasing
Decreasing
Decreasing
Increasing
Conclusion:
1 < x <
Increasing on: , 3, 1, Decreasing on: 3, 1, 1, 1 Relative maximum: 3, 8 Relative minimum: 1, 0 33. f x fx
x cos x, 0 < x < 2 2 1 sin x 0 2
Critical numbers: x
Test intervals:
5 , 6 6
0 < x <
6
5 < x < 6 6
5 < x < 2 6
Sign of fx:
f > 0
f < 0
f > 0
Conclusion:
Increasing
Decreasing
Increasing
Increasing on: Decreasing on:
0, 6 , 56, 2
6 , 56
6 , 126 3 5 5 6 3 Relative minimum: , 6 12 Relative maximum:
Section 3.3
Increasing and Decreasing Functions and the First Derivative Test
117
35. f x sin2 x sin x, 0 < x < 2 fx 2 sin x cos x cos x cos x2 sin x 1 0
7 3 11 , , , 2 6 2 6
Critical numbers: x
0 < x <
Test intervals:
2
7 < x < 2 6
7 3 < x < 6 2
11 3 < x < 2 6
11 < x < 2 6
Sign of fx:
f > 0
f < 0
f > 0
f < 0
f > 0
Conclusion:
Increasing
Decreasing
Increasing
Decreasing
Increasing
0, 2 , 76, 32 , 116, 2
Increasing on: Decreasing on:
2 , 76 , 32, 116
Relative minima:
76, 41 , 116, 14
Relative maxima:
2 , 2 , 32, 0
37. f x 2x9 x2, 3, 3 (a) fx
29 2x2 9 x2
(c)
y
(b) f′
Critical numbers: x ±
f
10 8
3 2
±
32 2
(d) Intervals:
4 2
3, 3 2 2 3 2 2, 3 2 2 3 2 2, 3
x 1
29 2x2 0 9 x2
1
2
8 10
fx < 0
fx > 0
fx < 0
Decreasing
Increasing
Decreasing
f is increasing when f is positive and decreasing when f is negative. 39. f t t 2 sin t, 0, 2 (a) ft t2 cos t 2t sin t tt cos t 2 sin t (b)
(c) tt cos t 2 sin t 0 t 0 or t 2 tan t t cot t 2
y 40
t 2.2889, 5.0870 (graphing utility)
f′
30
Critical numbers: t 2.2889, t 5.0870
20 10
−10 −20
t
π 2
2π
f
(d) Intervals:
0, 2.2889
2.2889, 5.0870
5.0870, 2
ft > 0
ft < 0
ft > 0
Increasing
Decreasing
Increasing
f is increasing when f is positive and decreasing when f is negative.
118
41.
Chapter 3
f x
Applications of Differentiation
x5 4x3 3x x2 1x3 3x x3 3x, x ± 1 x2 1 x2 1
y
f x gx x3 3x for all x ± 1.
(− 1, 2)
fx 3x2 3 3x2 1, x ± 1
fx 0 −4 −3
f symmetric about origin
x
−1
1 2 3 4 5
−2 −3 −4 −5
zeros of f: 0, 0, ± 3, 0
(1, − 2)
Holes at 1, 2 and 1, 2
No relative extrema 43. f x c is constant ⇒ fx 0
45. f is quadratic ⇒ f is a line.
y
y
4
4
2
f′
2
f′ −4
5 4 3
−2
2
x
x −4
4
−2
2
−2
−2
−4
−4
4
47. f has positive, but decreasing slope y
4 2
f′ x −4
−2
2
4
−2 −4
In Exercises 49–53, f x > 0 on , 4, f x < 0 on 4, 6 and f x > 0 on 6, . 49. gx f x 5
51.
gx fx
gx f x
53. gx f x 10
gx fx
gx fx 10 g0 f10 > 0
g6 f6 < 0
g0 f0 < 0
> 0, x < 4 ⇒ f is increasing on , 4. 55. fx undefined, x 4 < 0, x > 4 ⇒ f is decreasing on 4, . Two possibilities for f x are given below. (a)
y
(b)
y
6
2
4
1
x 1
2
1 x 2 −2
6
8
3
3
4
5
Section 3.3
Increasing and Decreasing Functions and the First Derivative Test
57. The critical numbers are in intervals 0.50, 0.25 and 0.25, 0.50 since the sign of f changes in these intervals. f is decreasing on approximately 1, 0.40, 0.48, 1, and increasing on 0.40, 0.48.
y
1
x
−1
Relative minimum when x 0.40.
1
Relative maximum when x 0.48.
−1
59. f x x, gx sin x, 0 < x < (a)
0.5
x
1
1.5
2
2.5
3
f x
0.5
1
1.5
2
2.5
3
gx
0.479
0.841
0.997
0.909
0.598
0.141
f x seems greater than gx on 0, . (b)
(c) Let hx f x gx x sin x
5
hx 1 cos x > 0 on 0, . Therefore, hx is increasing on 0, . Since h0 0, hx > 0 on 0, . Thus,
0 −2
x sin x > 0
x > sin x on 0,
x > sin x f x > gx on 0, .
61. v kR rr2 kRr2 r3
63.
v k2Rr 3r2
P
dP R R22vR1 vR1R22R1 R21 1 dR2 R1 R24
kr2R 3r 0 2
r 0 or 3 R Maximum when r
vR1R2 , v and R1 are constant R1 R22
2 3 R.
vR1R1 R2 0 ⇒ R2 R1 R1 R23
Maximum when R1 R2 . 65. (a) B 0.1198t4 4.4879t3 56.9909t2 223.0222t 579.9541 (b)
1500
0
20 0
(c) B 0 for t 2.78, or 1983, (311.1 thousand bankruptcies) Actual minimum: 1984 (344.3 thousand bankruptcies) 3 1 (c) The solution is a0 a1 0, a2 , a3 : 2 2
67. (a) Use a cubic polynomial f x a3 x 3 a 2 x 2 a1x a0.
1 3 f x x 3 x 2. 2 2
(b) fx 3a 3 x 2 2a 2 x a1.
0, 0:
2, 2:
0 a0
f 0 0
0 a1
f0 0
2 8a 3 4a 2
f 2 2
0 12a3 4a 2
f2 0
(d)
4
(2, 2) −2
(0, 0)
−4
4
119
120
Chapter 3
Applications of Differentiation
69. (a) Use a fourth degree polynomial f x a4 x 4 a 3 x 3 a 2 x 2 a1 x a0. (b) fx 4a4x3 3a3x2 2a2x a1 (0, 0:
4, 0:
2, 4:
0 a0
f 0 0
0 a1
f0 0
0 256a4 64a3 16a2
f 4 0
0 256a4 48a3 8a2
f4 0
4 16a4 8a3 4a2
f 2 4
0 32a4 12a3 4a2
f2 0
1 (c) The solution is a0 a1 0, a2 4, a3 2, a4 . 4 1 f x x4 2x3 4x2 4 (d)
5
(2, 4)
−2
(0, 0)
(4, 0)
5
−1
71. True
73. False
Let hx f x gx where f and g are increasing. Then hx fx gx > 0 since fx > 0 and gx > 0.
Let f x x3, then fx 3x2 and f only has one critical number. Or, let f x x3 3x 1, then fx 3x2 1 has no critical numbers.
75. False. For example, f x x3 does not have a relative extrema at the critical number x 0. 77. Assume that fx < 0 for all x in the interval a, b and let x1 < x2 be any two points in the interval. By the Mean Value Theorem, we know there exists a number c such that x1 < c < x2, and fc
f x2 f x1 . x2 x1
Since fc < 0 and x2 x1 > 0, then f x2 f x1 < 0, which implies that f x2 < f x1. Thus, f is decreasing on the interval. 79. Let f x 1 xn nx 1. Then fx n1 xn1 n n1 xn1 1 > 0 since x > 0 and n > 1. Thus, f x is increasing on 0, . Since f 0 0 ⇒ f x > 0 on 0,
1 xn nx 1 > 0 ⇒ 1 xn > 1 nx.
Section 3.4
Section 3.4
Concavity and the Second Derivative Test
Concavity and the Second Derivative Test 3. f x
1. y x2 x 2, y 2 Concave upward: ,
24 1444 x2 , y x 12 x2 123 2
Concave upward: , 2, 2, Concave downward: 2, 2
5. f x
x2 1 43x2 1 , y 2 2 x 1 x 13
7. f x 3x2 x3 fx 6x 3x2
Concave upward: , 1, 1,
f x 6 6x
Concave downward: 1, 1
Concave upward: , 1 Concave downward: 1,
9. y 2x tan x,
2 , 2
11. f x x3 6x2 12x fx 3x2 12x 12
y 2 sec2 x
f x 6x 2 0 when x 2.
y 2 sec2 x tan x Concave upward:
,0 2
Concave upward: 2,
2
0,
Concave downward:
13.
The concavity changes at x 2. 2, 8 is a point of inflection. Concave downward: , 2
1 f x x4 2x2 4 fx x3 4x f x 3x2 4 f x 3x2 4 0 when x ±
Test interval:
< x <
Sign of f x: Conclusion: Points of inflection:
2 3
2 3
.
2 3
< x <
2
2
3
3
< x <
f x > 0
f x < 0
f x > 0
Concave upward
Concave downward
Concave upward
± 23, 209
121
122
Chapter 3
Applications of Differentiation
15. f x xx 43 fx x3x 4 2 x 43 x 424x 4 f x 4x 12x 4 4x 42 4x 42x 1 x 4 4x 43x 6 12x 4x 2 f x 12x 4x 2 0 when x 2, 4. < x < 2
2 < x < 4
f x > 0
f x < 0
f x > 0
Concave upward
Concave downward
Concave upward
Test interval: Sign of f x: Conclusion:
4 < x <
Points of inflection: 2, 16, 4, 0 17. f x xx 3, Domain: 3, fx x f x
12x 3
1 2
x 3
3x 2 2x 3
6x 3 3x 2x 31 2 3x 4 4x 3 4x 33 2
f x > 0 on the entire domain of f (except for x 3, for which f x is undefined). There are no points of inflection. Concave upward on 3, 19. f x
x x2 1
fx
1 x2 x2 12
f x
2xx2 3 0 when x 0, ± 3 x2 13
Test intervals:
< x < 3
3 < x < 0
0 < x < 3
Sign of fx:
f < 0
f > 0
f < 0
f > 0
Conclusion:
Concave downward
Concave upward
Concave downward
Concave upward
Test interval:
0 < x < 2
Points of inflection:
21. f x sin
3,
2x , 0 ≤ x ≤ 4
1 x fx cos 2 2
1 x f x sin 4 2
f x 0 when x 0, 2, 4. Point of inflection: 2, 0
3
4
, 0, 0,
3,
3
4
3 < x <
Sign of f x: Conclusion:
2 < x < 4
f < 0
f > 0
Concave downward
Concave upward
Section 3.4
, 0 < x < 4 2
tan x 2 2
23. f x sec x fx sec x
Concavity and the Second Derivative Test
f x sec3 x
sec x tan2 x 0 for any x in the domain of f. 2 2 2
Concave upward: 0, , 2, 3 Concave downward: , 2, 3, 4 No points of inflection 25. f x 2 sin x sin 2x, 0 ≤ x ≤ 2 f x 2 cos x 2 cos 2x f x 2 sin x 4 sin 2x 2 sin x1 4 cos x f x 0 when x 0, 1.823, , 4.460. Test interval:
0 < x < 1.823
1.823 < x <
< x < 4.460
4.460 < x < 2
Sign of f x:
f < 0
f > 0
f < 0
f > 0
Concave downward
Concave upward
Concave downward
Concave upward
Conclusion:
Points of inflection: 1.823, 1.452, , 0, 4.46, 1.452 27. f x x4 4x3 2
29. f x x 52
fx 4x3 12x2 4x2x 3
fx 2x 5
f x 12x 2 24x 12xx 2
f x 2
Critical numbers: x 0, x 3
Critical number: x 5
However, f 0 0, so we must use the First Derivative Test. fx < 0 on the intervals , 0 and 0, 3; hence, 0, 2 is not an extremum. f 3 > 0 so 3, 25 is a relative minimum. 31. f x x3 3x2 3
f 5 > 0 Therefore, 5, 0 is a relative minimum.
33. gx x26 x3
fx 3x2 6x 3xx 2
gx xx 6212 5x
f x 6x 6 6x 1
g x 46 x5x2 24x 18
Critical numbers: x 0, x 2
12 Critical numbers: x 0, 5 , 6
f 0 6 < 0 Therefore, 0, 3 is a relative maximum. f 2 6 > 0 Therefore, 2, 1 is a relative minimum.
g 0 432 > 0 Therefore, 0, 0 is a relative minimum. g 12 5 155.52 < 0 Therefore, 5 , 268.7 is a relative minimum. 12
g 6 0 Test fails by the First Derivative Test, 6, 0 is not an extremum.
123
124
Chapter 3
Applications of Differentiation 4 x
37. f x x
35. f x x2 3 3 fx
2 3x1 3
fx 1
f x
2 9x4 3
f x
Critical number: x 0
4 x2 4 2 x x2
8 x3
Critical numbers: x ± 2
However, f 0 is undefined, so we must use the First Derivative Test. Since fx < 0 on , 0 and fx > 0 on 0, , 0, 3 is a relative minimum.
f 2 < 0 Therefore, 2, 4 is a relative maximum. f 2 > 0 Therefore, 2, 4 is a relative minimum.
39. f x cos x x, 0 ≤ x ≤ 4 fx sin x 1 ≤ 0 Therefore, f is non-increasing and there are no relative extrema. 41. f x 0.2x2x 33, 1, 4 (a) fx 0.2x5x 6x 32
(c)
y
f x x 34x 9.6x 3.6 2
0.4x 310x 24x 9
2 1
(b) f 0 < 0 ⇒ 0, 0 is a relative maximum. f
6 5
x 2
> 0 ⇒ 1.2, 1.6796 is a relative minimum.
4
f is increasing when f > 0 and decreasing when f < 0. f is concave upward when f > 0 and concave downward when f < 0.
3, 0, 0.4652, 0.7049, 1.9348, 0.9049
1 1 sin 3x sin 5x, 0, 3 5
(a) fx cos x cos 3x cos 5x fx 0 when x
f x 0 when x
5 ,x ,x . 6 2 6 5 ,x , x 1.1731, x 1.9685 6 6
6 , 0.2667, 1.1731, 0.9638, 1.9685, 0.9637,
4
56, 0.2667
Note: 0, 0 and , 0 are not points of inflection since they are endpoints.
f
2
−2
2 < 0 ⇒ 2 , 1.53333 is a relative maximum.
Points of inflection:
y
(c)
f x sin x 3 sin 3x 5 sin 5x
(b) f
1
f
Points of inflection:
43. f x sin x
f ′′
f′
2
π 4
π 2
f′
π
x
−4 −6 −8
f ′′
The graph of f is increasing when f > 0 and decreasing when f < 0. f is concave upward when f > 0 and concave downward when f < 0.
Section 3.4 f < 0 means f decreasing
y
45. (a)
Concavity and the Second Derivative Test
f increasing means concave upward
4 3
f > 0 means f increasing
y
(b)
f increasing means concave upward
4 3
2
2
1
1 x 1
3
2
x 1
4
47. Let f x x4.
49.
3
2
4
y
f
f x 12x2
2
f′
f 0 0, but 0, 0 is not a point of inflection.
f ′′
y
x
−2
6
1 −1
5 4 3 2 1 −3
−2
x
−1
51.
1
2
3
53.
y
f ′′
f′
y
f
4
4
2
x 2
2
(2, 0) (4, 0) x
2
2
4
55.
4
6
2
y
57.
y
3 2
f
1
(2, 0)
(4, 0) x
1 1 2
2
3
4
5
x 4
8 8
125
12
f ′′
f is linear. f is quadratic. f is cubic. f concave upwards on , 3, downward on 3, .
126
Chapter 3
Applications of Differentiation
59. (a) n 1:
n 2:
n 3:
n 4:
f x x 2
f x x 22
f x x 23
f x x 24
fx 1
fx 2x 2
fx 3x 22
fx 4x 23
f x 0
f x 2
f x 6x 2
fx 12x 22
No inflection points
No inflection points
Inflection point: 2, 0
No inflection points:
Relative minimum: 2, 0
6
−9
9
Relative minimum: 2, 0
6
−9
6
9
6
Point of inflection −6
−9
9
−6
−6
−9
9
−6
Conclusion: If n ≥ 3 and n is odd, then 2, 0 is an inflection point. If n ≥ 2 and n is even, then 2, 0 is a relative minimum. (b) Let f x x 2n, fx nx 2n1, f x nn 1x 2n2. For n ≥ 3 and odd, n 2 is also odd and the concavity changes at x 2. For n ≥ 4 and even, n 2 is also even and the concavity does not change at x 2. Thus, x 2 is an inflection point if and only if n ≥ 3 is odd. 61. f x ax3 bx 2 cx d Relative maximum: 3, 3 Relative minimum: 5, 1 Point of inflection: 4, 2 fx 3ax 2 2bx c, f x 6ax 2b
f 3 27a 9b 3c d 3 98a 16b 2c 2 ⇒ 49a 8b c 1 f 5 125a 25b 5c d 1 f3 27a 6b c 0, f 4 24a 2b 0 49a 8b c 1
24a 2b
27a 6b c
22a 2b 1
22a 2b a
1 2,
0
1
b 6, c
45 2 ,
2a
d 24
1 45 f x 2 x3 6x 2 2 x 24
0
1
Section 3.4
Concavity and the Second Derivative Test
127
63. f x ax3 bx2 cx d Maximum: 4, 1 Minimum: 0, 0 (a) fx 3ax2 2bx c,
f x 6ax 2b
f 0 0 ⇒ d 0
(b) The plane would be descending at the greatest rate at the point of inflection.
f 4 1 ⇒ 64a 16b 4c 1 f4 0 ⇒
48a 8b c 0
f0 0 ⇒
c0
f x 6ax 2b
3 3 x 0 ⇒ x 2. 16 8
Two miles from touchdown.
1 3 Solving this system yields a 32 and b 6a 16 . 1 3 3 2 f x 32 x 16 x
65. D 2 x4 5L x3 3L 2x 2
C 0.5x 2 15x 5000
67.
D 8x3 15L x 2 6L 2x x8x 2 15L x 6L 2 0 x 0 or x
15 ± 33 15L ± 33L L 16 16
x
15 16 33 L 0.578L.
S
5000t2 8 t2
dC 5000 0.5 2 0 when x 100 dx x
St
80,000t 8 t22
S t
80,0008 3t2 8 t23
C 5000 0.5x 15 x x
C average cost per unit
By the Second Derivative Test, the deflection is maximum when
69.
C
By the First Derivative Test, C is minimized when x 100 units.
S t 0 for t 8 3 1.633. Sales are increasing at the greatest rate at t 1.633 years. 71.
4 22
f x 2sin x cos x,
f
fx 2cos x sin x,
f
f x 2sin x cos x,
P1
0 4
−2
22 4
1 22 x 4 2 4
2
f
22 f 4
P1x 22 0 x
4
P2 −4
P1x 0 P2x 22 0 x
P2x 22 x
4
2
22 2 x
4
2
P2x 22 The values of f, P1, P2, and their first derivatives are equal at x 4. The values of the second derivatives of f and P2 are equal at x 4. The approximations worsen as you move away from x 4.
128 73.
Chapter 3
Applications of Differentiation
f x 1 x, fx f x
f 0 1
1 , 21 x
f0
1 , 41 x3 2
f 0
5
P1
1 2
f −8
4
P2
1 4
−3
21x 0 1 2x
P1x 1 P1x
1 2
21x 0 21 41x 0
P2x 1
2
1
x x2 2 8
1 x P2x 2 4 P2x
1 4
The values of f, P1, P2, and their first derivatives are equal at x 0. The values of the second derivatives of f and P2 are equal at x 0. The approximations worsen as you move away from x 0. 75. f x x sin
fx x f x x
1x
1
sin1x 1x cos1x sin1x
1 1 cos x2 x
−1
x1 cos1x x1 cos1x x1 sin1x 0
1 1 1 sin x x2 x
2
2
1
( π1 , 0) −1
3
1
Point of inflection:
1 , 0
When x > 1 , f < 0, so the graph is concave downward. 77. Assume the zeros of f are all real. Then express the function as f x ax r1x r2x r3 where r1, r2, and r3 are the distinct zeros of f. From the Product Rule for a function involving three factors, we have fx ax r1x r2 x r1x r3 x r2x r3 f x ax r1 x r2 x r1 x r3 x r2 x r3 a6x 2r1 r2 r3. Consequently, f x 0 if x
2r1 r2 r3 r1 r2 r3 Average of r1, r2, and r3. 6 3
79. True. Let y ax3 bx2 cx d, a 0. Then y 6ax 2b 0 when x b 3a, and the concavity changes at this point.
Section 3.5
Limits at Infinity
83. False. Concavity is determined by f .
81. False. f x 3 sin x 2 cos x fx 3 cos x 2 sin x 3 cos x 2 sin x 0 3 cos x 2 sin x 3 2
tan x
Critical number: x tan132 f tan1 32 3.60555 is the maximum value of y.
Section 3.5 1. f x
Limits at Infinity
3x2 x2 2
x x2 2
3. f x
5. f x
4sin x x2 1
No vertical asymptotes
No vertical asymptotes
No vertical asymptotes
Horizontal asymptote: y 3
Horizontal asymptote: y 0
Horizontal asymptotes: y 0
Matches (f)
Matches (d)
Matches (b)
7. f x
4x 3 2x 1
x
100
101
102
103
104
7
2.26
2.025
2.0025
2.0003
f x
10
105 106 2
− 10
10
2 − 10
lim f x 2
x→
9. f x
6x
10
4x2 5
x
100
101
102
103
104
105
106
f x
2
2.98
2.9998
3
3
3
3
− 10
10
− 10
lim f x 3
x→
11. f x 5
1 x2 1
6
x
100
101
102
103
104
105
106
f x
4.5
4.99
4.9999
4.999999
5
5
5
lim f x 5
x→
−1
8 0
129
130
Chapter 3
Applications of Differentiation
f x 5x3 3x2 10 10 5x 3 2 x2 x2 x
13. (a) hx
lim hx
x→
x2 2 0 x→ x3 1
15. (a) lim
(Limit does not exist)
x2 2 1 x2 1
(c) lim
x2 2 x1
x→
f x 5x3 3x2 10 3 10 5 3 x3 x3 x x
(b) hx
(b) lim
x→
lim hx 5
x→
(Limit does not exist)
f x 5x3 3x2 10 5 10 3 2 4 x4 x4 x x x
(c) hx
lim hx 0
x→
5 2x32 0 x→ 3x2 4
19. lim
17. (a) lim
x→
2x 1 2 1x 2 0 2 lim 3x 2 x→ 3 2x 3 0 3
2 5 2x32 x→ 3x32 4 3
(b) lim (c) lim
x→
21. lim
x →
5 2x32 3x 4
(Limit does not exist)
x 1x 0 lim 0 x2 1 x → 1 1x2 1
23.
lim
x →
5x 2 5x lim x 3 x → 1 3x
Limit does not exist.
25.
lim
x →
x x2
x
1 , x x2
lim
lim
1
x →
1 1x
1 2 2x 1 x 27. lim lim x→ x2 x x→ x2 x x2
lim
x→
for x
x2
x →
2 1x x (1x
< 0 we have x x2
1
for x < 0, x x2
2
29. Since 1x ≤ sin2xx ≤ 1x for all x 0, we have by the Squeeze Theorem, lim
x →
1 sin2x 1 ≤ lim ≤ lim x → x → x x x
0 ≤ lim
x →
sin2x ≤ 0. x
Therefore, lim
x →
sin2x 0. x
31. lim
x →
1 0 2x sin x
Section 3.5
33. (a) f x
x x1
4
lim
y=1
y = −1
x 1 lim x → x 1 x →
Limits at Infinity
−6
6
x 1 x1
−4
Therefore, y 1 and y 1 are both horizontal asymptotes.
35. lim x sin x →
1 sin t lim 1 x t→0 t
Let x 1t.
37.
x x 3 lim x x2 3 x x2 3 x → x → 2 lim
39. lim x x2 x lim x →
x →
lim
x →
41.
x f x
x
x
x2 x
x
2
x x
lim
x →
2
101
102
103
104
105
106
1
0.513
0.501
0.500
0.500
0.500
0.500
x x2 x x → 1 x →
lim
x →
3 0 x x2 3
x 1 1 lim 2 x x2 x x → 1 1 1x
lim
f x
100
x →
x
x 3
x x2 x
lim x xx 1 lim
43.
2
x x2 x x x2 x
8
−2
x x x2 x 1 1 1 1x
1 2
100
101
102
103
104
105
106
0.479
0.500
0.500
0.500
0.500
0.500
0.500
Let x 1t. sint2 1 sint2 1 1 lim x sin lim lim x → t →0 t →0 2 2x t t2 2
−1
1
−2
2
−1
131
132
Chapter 3
45. (a)
Applications of Differentiation
47. Yes. For example, let f x
y 4 3
y
f′
2
1
8 x
−4
1
2
3
4 4
−3
2
−4
x
(b) lim f x 3
−4
lim fx 0
x →
−2
2
4
6
−2
x →
(c) Since lim f x 3, the graph approaches that of a x →
horizontal line, lim f x 0. x →
49. y
2x 1x
y 3
Intercepts: 2, 0, 0, 2
2
−3 −2 −1
Symmetry: none Horizontal asymptote: y 1 since 2x 2x 1 lim lim . x → 1 x x → 1 x
x 1
2
3
4
5
−2 −3 −4 −5
Discontinuity: x 1 (Vertical asymptote) 51. y
x2
x 4
53. y
x2
x2 9
Intercept: 0, 0
Intercept: 0, 0
Symmetry: origin
Symmetry: y-axis
Horizontal asymptote: y 0
Horizontal asymptote: y 1 since
Vertical asymptote: x ± 2
x →
y
x2
x2 x2 1 lim 2 . x → x 9 9
Relative minimum: 0, 0
5 4 3 2 1 −1
lim
y 4
x 2 3 4 5
3
−2 −3 −4 −5
2 1 −3 −2 −1
x −1 −2
1
2
3
6x2 . x 22 1
Section 3.5
55. y
2x2 4
Limits at Infinity
57. xy2 4
x2
Domain: x > 0
Intercept: 0, 0
Intercepts: none
Symmetry: y-axis
Symmetry: x-axis
Horizontal asymptote: y 2
Horizontal asymptote: y 0 since
Vertical asymptote: x ± 2
2
lim
y
x →
8
x
0 lim x →
2 x
.
Discontinuity: x 0 (Vertical asymptote)
6 4
y
2 −4
4
x
−2
4
2
6
3 2 1 −1 −1
x 1
2
3
4
5
−2 −3 −4
59. y
2x 1x
61. y 2
3 x2
Intercept: 0, 0
Intercepts: ± 32, 0
Symmetry: none
Symmetry: y-axis
Horizontal asymptote: y 2 since
Horizontal asymptote: y 2 since
lim
x →
2x 2x 2 lim . x → 1 x 1x
lim
x →
Discontinuity: x 1 (Vertical asymptote)
2
lim
x →
y
y 4
1
3 x 1
2
3
4
2
5
1
−2
−4 −3 −2
−3
x 2
3
4
−4 −5 −6
63. y 3
2 x
y
2 2 2 2 Intercept: y 0 3 ⇒ 3 ⇒ x , 0 x x 3 3
8 7 6 5 4 3 2 1
Symmetry: none Horizontal asymptote: y 3 Vertical asymptote: x 0
2 x3 . 2
Discontinuity: x 0 (Vertical asymptote)
2
−3 −2 −1
2 x3 2
−4 −3 −2 −1
x 1 2 3 4 5
133
134
Chapter 3
65. y
Applications of Differentiation
x3 x2 4
67. f x 5
1 5x2 1 2 x x2
Domain: , 2, 2,
Domain: , 0, 0,
Intercepts: none
fx
2 ⇒ No relative extrema x3
Symmetry: origin 6 ⇒ No points of inflection x4
Horizontal asymptote: none
f x
Vertical asymptotes: x ± 2 (discontinuities)
Vertical asymptote: x 0
y
Horizontal asymptote: y 5
20 16 12 8 4 −5 −4 −3 −2 −1
7
y=5 1 2 3 4 5
−8 − 12 − 16 − 20
69. f x fx f x
x=0
x
−6
6 −1
x x2 4
x2 4 x2x x2 42
3
−4
x2 4 0 for any x in the domain of f . x2 42
5
x = − 2 −3
x=2
x2 422x x2 42x2 42x x2 42 2xx2 12 0 when x 0. x2 43
Since f x > 0 on 2, 0 and f x < 0 on 0, 2, then 0, 0 is a point of inflection. Vertical asymptotes: x ± 2 Horizontal asymptote: y 0 x2 x2 71. f x x2 4x 3 x 1x 3
x2 4x 3 x 22x 4 x2 4x 5 fx 2 0 x2 4x 32 x 4x 32 f x
x2 4x 322x 4 x2 4x 52x2 4x 32x 4 x2 4x 34 2x3 6x2 15x 14 0 when x 2. x2 4x 33
Since f x > 0 on 1, 2 and f x < 0 on 2, 3, then 2, 0 is a point of inflection. Vertical asymptote: x 1, x 3 Horizontal asymptote: y 0
2
x=3 −1
5
y=0 x=1 −2
Section 3.5 3x 73. f x 2 4x 1 fx f x
Limits at Infinity
2
y= 3
3 ⇒ No relative extrema 4x2 132
−3
2
3
y= −3 2
36x 0 when x 0. 4x2 152
−2
Point of inflection: 0, 0 Horizontal asymptotes: y ±
3 2
No vertical asymptotes
75. gx sin
gx
x x 2, 3 < x <
2 cos
1.2
x x 2
y = sin(1)
x 22
3
Horizontal asymptote: y 1 Relative maximum:
( π2−π 2 , 1)
12 0
2
x ⇒ x 5.5039 x2 2
2
No vertical asymptotes 77. f x
x3 3x2 2 2 , gx x xx 3 xx 3
(a)
(c)
8
− 80
f=g −4
(b) f x
− 70
x3 3x2 2 xx 3 x2x 3 2 xx 3 xx 3
x
2 gx xx 3
79. C 0.5x 500 C x
C 0.5 lim
x →
80
8 −2
C
70
500 x
0.5 500x 0.5
The graph appears as the slant asymptote y x.
135
136
Chapter 3
Applications of Differentiation
83. (a) T1t 0.003t 2 0.677t 26.564
81. line: mx y 4 0 y
(b)
90
5
T1 y = mx + 4
3 − 10
2
130 − 10
(3, 1)
1
x −2 −1 −1
1
2
3
4
(c)
90
T2
(a) d
Ax1 By1 C m3 11 4 A2 B2
m2 1
− 10
3m 3
m2 1
(b)
120 − 10
T2
7
1451 86t 58 t
(d) T10 26.6 −6
T20 25.0
6 −1
(c) lim dm 3 lim dm m →
t→
m →
The line approaches the vertical line x 0. Hence, the distance approaches 3.
85. Answers will vary. See page 195.
Section 3.6
(e) lim T2
86 86 1
(f) The limiting temperature is 86. T1 has no horizontal asymptote.
87. False. Let f x
2x x2 2
. (See Exercise 2.)
A Summary of Curve Sketching
1. f has constant negative slope. Matches (D) 5. (a) fx 0 for x 2 and x 2
3. The slope is periodic, and zero at x 0. Matches (A) (c) f is increasing on 0, .
f > 0
f is negative for 2 < x < 2 (decreasing function). f is positive for x > 2 and x < 2 (increasing function). (b) f x 0 at x 0 (Inflection point). f is positive for x > 0 (Concave upwards). f is negative for x < 0 (Concave downward).
(d) fx is minimum at x 0. The rate of change of f at x 0 is less than the rate of change of f for all other values of x.
Section 3.6
7. y y
x2
x2 3
A Summary of Curve Sketching
y
y=1
6x 0 when x 0. x2 32
1
181 x2 0 when x ± 1. y 2 x 33
1,
1 4
1 4
1,,
x 4
2
(0, 0))
4
Horizontal asymptote: y 1 y < x < 1 1 4
x 1 1 < x < 0 x0
0
0 < x < 1 1 4
x1
1 < x <
9. y
y
Conclusion
Decreasing, concave down
0
Point of inflection
Decreasing, concave up
0
Relative minimum
Increasing, concave up
0
Point of inflection
Increasing, concave down
1 3 x2
y y
y
11. y
1 < 0 when x 2. x 22
2 x 23
No relative extrema, no points of inflection
73, 0, 0, 27
Intercepts:
Vertical asymptote: x 2 Horizontal asymptote: y 3 y
x
2x x2 1
y
2x2 1 < 0 if x ± 1. x2 12
y
4xx2 3 0 if x 0. x2 13
Inflection point: 0, 0 Intercept: 0, 0 Vertical asymptote: x ± 1 Horizontal asymptote: y 0 Symmetry with respect to the origin
2
x
1
7 , 0 3
y
x
1
4
x 4
y
0
2
x 2
4
0,,
7 2
y
3
(0, 0)
4
137
138
Chapter 3
13. gx x gx 1 g x
Applications of Differentiation
4 x2 1 8x x4 2x2 8x 1 0 when x 0.1292, 1.6085 2 2 x 1 x2 12
3 8 1 0 when x ± x2 13 3
3x2
4
3 , 2.423 3 )1.6085, 2.724) ) 1.3788, 0) x 3
2
1
g 0.1292 < 0, therefore, 0.1292, 4.064 is relative maximum.
2
3
2
y
g 1.6085 > 0, therefore, 1.6085, 2.724 is a relative minimum. Points of inflection:
)0.1292, 4.064) 3 , 3.577 3
y
)0, 4)
x
33, 2.423, 33, 3.577
Intercepts: 0, 4, 1.3788, 0 Slant asymptote: y x
15. f x
x2 1 1 x x x
fx 1
y 4
1 0 when x ± 1. x2
2 f x 3 0 x
y=x
2
(1, 2) −4
x
−2
2
x=0
−4
Relative maximum: 1, 2
4
(−1, −2)
Relative minimum: 1, 2 Vertical asymptote: x 0 Slant asymptote: y x
17. y
x2 6x 12 4 x2 x4 x4
y 1
4 x 42
y
x
4
8 6
(6, 6)
4
y
x 2x 6 0 when x 2, 6. x 42 8 x 43
y < 0 when x 2. Therefore, 2, 2 is a relative maximum. y > 0 when x 6. Therefore, 6, 6 is a relative minimum. Vertical asymptote: x 4 Slant asymptote: y x 2
(0, −3)
x
y
2
2 x
6
8
(2, −2)
10
Section 3.6
A Summary of Curve Sketching
19. y xx 4,
y
Domain: , 4
y y
( 83 ,
4
16 3 3
2
8 3x 8 0 when x and undefined when x 4. 3 24 x
(0, 0)
(4, 0) x
−2
2
4
3x 16 16 0 when x and undefined when x 4. 44 x3 2 3
Note: x
16 3
is not in the domain. y
y
Conclusion
Increasing, concave down
0
Relative maximum
Decreasing, concave down
Undefined
Undefined
y < x < x
8 3
8 3
16 33
8 < x < 4 3 x4
0
21. hx x9 x2
Endpoint
Domain: 3 ≤ x ≤ 3
y
9 2x2 3 32 hx 0 when x ± ± 9 x2 2 2
−5 −4
(
5 4 3 2 1
(− 3, 0)
x2x2 27 h x 0 when x 0 9 x23 2
(0, 0)
)
(3, 0)
1 2 3 4 5
(
−
32 9
3 2, 9 2 2
x
−2 −1
2 , 2 3 2 9 Relative minimum: , 2 2 Relative maximum:
3 2, 9 − 2 2
)
−5
Intercepts: 0, 0, ± 3, 0 Symmetric with respect to the origin Point of inflection: 0, 0 23. y 3x2 3 2x y 2x1 3 2
y
21 x1 3 x1 3
5
0 when x 1 and undefined when x 0. y
(1, 1)
2 < 0 when x 0. 3x4 3 y
< x < 0 x0
0
0 < x < 1 x1 1 < x <
1
(
( 278 , 0 ) x
(0, 0) 2
y
y
Undefined
Undefined
Increasing, concave down
0
Relative maximum
Decreasing, concave down
Conclusion Decreasing, concave down Relative minimum
1
2
3
5
139
140
Chapter 3
Applications of Differentiation
25. y x3 3x2 3
y
y 3x2 6x 3xx 2 0 when x 0, x 2
4
(−0.879, 0)
(0, 3)
y 6x 6 6x 1 0 when x 1 (1, 1)
y < x < 0 x0
3
0 < x < 1 x1
1
1 < x < 2 x2 2 < x <
1
y
y
Conclusion
Increasing, concave down
0
Relative maximum
Decreasing, concave down
0
Point of inflection
Decreasing, concave up
0
Relative minimum
Increasing, concave up
(2.532, 0) x
2
4
(2, (1.347, 0)
2
27. y 2 x x3
1)
y
5
y 1 3x2
4
No critical numbers (0, 2)
y 6x 0 when x 0.
1
(1, 0) x
y < x < 0 x0 0< x <
2
y
y
3
2
1
2
3
Conclusion Decreasing, concave up
0
Point of inflection
Decreasing, concave down
29. f x 3x3 9x 1
y
(−1.785, 0) 8 ( 1, 7)
fx 9x2 9 9x2 1 0 when x ± 1 f x 18x 0 when x 0 f x < x < 1 x 1
7
1 < x < 0 x0
1
0 < x < 1 x1 1 < x <
5
fx
f x
(0, 1) (1.674, 0)
Conclusion
Increasing, concave down
0
Relative maximum
Decreasing, concave down
0
Point of inflection
Decreasing, concave up
0
Relative minimum
Increasing, concave up
x 3
1
1
2
3
2 4 6
(1, 5) (0.112, 0)
Section 3.6
A Summary of Curve Sketching
31. y 3x4 4x3
y
y 12x3 12x2 12x2x 1 0 when x 0, x 1. y 36x2 24x 12x3x 2 0 when x 0, x 23 . y < x < 1 x 1
1
1 < x <
23
x 23
16
27
23 < x < 0 x0
0
0 < x <
y
y
Decreasing, concave up
0
Relative minimum
Increasing, concave up
0
Point of inflection
Increasing, concave down
0
0
Point of inflection
Increasing, concave up
2
1
(− 43 , 0)
Conclusion
(0, 0) 1
fx f x
12x 2
12x 2
y
16 4x 1x 2 0 when x 1, x 2.
15
24x 12xx 2 0 when x 0, x 2. fx
f x
Decreasing, concave up
0
Relative minimum
Increasing, concave up
0
Point of inflection
Increasing, concave down
0
0
Point of inflection
Increasing, concave up
< x < 1 x 1
11
1 < x < 0 x0
0
0 < x < 2 x2 2 < x <
20
2
f x
(− 23 , − 1627 (
(−1, −1)
33. f x x4 4x3 16x 4x3
16
Conclusion
35. y x5 5x
(2, 16)
10 5
(0, 0)
−3
y 4
1 < x < 0 0
0 < x < 1 1 < x <
4
3
4
y
< x < 1
x1
2
(−1.679, 0)
)
4 5, 0
6
( 1, 4)
4
y 20x3 0 when x 0.
x0
x
1
(−1, −11)
(
y 5x4 5 5x4 1 0 when x ± 1.
x 1
x
−2
(0, 0) 2
1
1
y
y
Conclusion
2
Increasing, concave down
4
0
Relative maximum
Decreasing, concave down
6
0
Point of inflection
Decreasing, concave up
0
Relative minimum
Increasing, concave up
( 4 5, 0 )
(1,
2
4)
x
141
142
Chapter 3
Applications of Differentiation
37. y 2x 3 y
y
22x 3 3 undefined at x . 2x 3 2
4
(0, 3)
3
y 0
2 1
y
y < x <
3 2
x 32 3 2
0
< x <
Undefined
Conclusion
x
Decreasing
Increasing
1 sin 3x, 0 ≤ x ≤ 2 18 1 3 y cos x cos 3x 0 when x , . 6 2 2
y 2 1
1 5 7 11 . y sin x sin 3x 0 when x 0, , , , , 2 6 6 6 6
19 , 2 18
Relative minimum:
32 , 19 18
Inflection points:
−1
π 2
π
x 3π 2
−2
4
Relative minimum
39. y sin x
Relative maximum:
3
3 , 0 2
6 , 94, 56 , 49, , 0, 76 , 49, 116 , 94
41. y 2x tan x,
< x < 2 2
43. y 2csc x sec x, 0 < x <
2
y 2 sec2 x 0 when x ± . 4
y 2sec x tan x csc x cot x 0 ⇒ x 4
y 2sec2 x tan x 0 when x 0.
Relative minimum:
Relative maximum:
4 , 2 1
Relative minimum:
4 , 1 2
Inflection point: 0, 0
Vertical asymptotes: x ± 2
Vertical asymptotes: x 0, x y 16 12 8 4
y −4
2 1
− π 2
−1 −2
π 4
π 2
x
4 , 42
4
2
x
2
Section 3.6
45. gx x tan x,
3 3 < x < 2 2
gx
x sin x cos x 0 when x 0 cos2 x
g x
2cos x x sin x cos3 x
Vertical asymptotes: x
47. f x
A Summary of Curve Sketching
20x 1 19x2 1 x 1 x xx2 1 2
10
− 15
15
− 10
3 3 , , , 2 2 2 2
x 0 vertical asymptote
Intercepts: , 0, 0, 0, , 0
y 0 horizontal asymptote
Symmetric with respect to y-axis.
Minimum: 1.10, 9.05 Maximum: 1.10, 9.05
2 and 2 , 32
Increasing on 0,
Points of inflection: 1.84, 7.86, 1.84, 7.86
Points of inflection: ± 2.80, 0 y 10 8 6 4 2 −π
49. y
π 4
−2 −4 −6 −8 − 10
π
3π 2
x
x
51. f is cubic.
x2 7
f is quadratic.
2
f is linear. −4
4
y
f ′′
f −2
0, 0 point of inflection
x 2
2 1
y ± 1 horizontal asymptotes
53.
f′
y
2
y
4
f ′′
4
f 2 x −4
−2
2
4
x −4
−2
2
−2 −4
(any vertical translate of f will do)
−4
4
143
144
Chapter 3
Applications of Differentiation
y
55.
y
4 2
4
f
2
f ′′
x −4
x −8
8
−4
4
−2
−2
−4
−4
8
(any vertical translate of f will do)
57. Since the slope is negative, the function is decreasing on 2, 8, and hence f 3 > f 5.
59. f x
4x 12 4x 5
x2
Vertical asymptote: none Horizontal asymptote: y 4 9
−6
9 −1
The graph crosses the horizontal asymptote y 4. If a function has a vertical asymptote at x c, the graph would not cross it since f c is undefined. 61. hx
6 2x 3x
63. f x
23 x 2, if x 3 3x Undefined, if x 3
The rational function is not reduced to lowest terms.
x2 3x 1 3 x 1 x2 x2
3
−3
6
3 −3
−2
The graph appears to approach the slant asymptote y x 1.
4 −1
hole at 3, 2
65. f x (a)
cos2 x , 0, 4 x2 1 (b) fx
1.5
0
4
− 0.5
On 0, 4 there seem to be 7 critical numbers: 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5
cos xx cos x 2 x2 1sin x 0 x2 13 2
1 3 5 7 Critical numbers , 0.97, , 1.98, , 2.98, . 2 2 2 2 The critical numbers where maxima occur appear to be integers in part (a), but approximating them using f shows that they are not integers.
Section 3.7 67. Vertical asymptote: x 5
71. f x
Slant asymptote: y 3x 2
1 x5
y 3x 2
(b) As b varies, the position of the vertical asymptote changes: x b. Also, the coordinates of the minimum a > 0 or maximum a < 0 are changed.
3xn x4 1
(a) For n even, f is symmetric about the y-axis. For n odd, f is symmetric about the origin. (b) The x-axis will be the horizontal asymptote if the degree of the numerator is less than 4. That is, n 0, 1, 2, 3.
(d) There is a slant asymptote y 3x if n 5: 3x 3x5 3x 4 . x4 1 x 1 (e)
(c) n 4 gives y 3 as the horizontal asymptote.
75. (a)
2750
1
8 0
(b) When t 10, N10 2434 bacteria. (c) N is a maximum when t 7.2 (seventh day). (d) Nt 0 for t 3.2 (e) lim Nt t→
Section 3.7 1. (a)
1 3x2 13x 9 x5 x5
ax x b2
(a) The graph has a vertical asymptote at x b. If a > 0, the graph approaches as x → b. If a < 0, the graph approaches as x → b. The graph approaches its vertical asymptote faster as a → 0.
73. f x
145
69. Vertical asymptote: x 5
Horizontal asymptote: y 0 y
Optimization Problems
13,250 1892.86 7
Optimization Problems
First Number, x
Second Number
Product, P
10
110 10
10110 10 1000
20
110 20
20110 20 1800
30
110 30
30110 30 2400
40
110 40
40110 40 2800
50
110 50
50110 50 3000
60
110 60
60110 60 3000
—CONTINUED—
n
0
1
2
3
4
5
M
1
2
3
2
1
0
N
2
3
4
5
2
3
146
Chapter 3
Applications of Differentiation
1. —CONTINUED— (b)
First Number, x
Second Number
Product, P
10
110 10
10110 10 1000
20
110 20
20110 20 1800
30
110 30
30110 30 2400
40
110 40
40110 40 2800
50
110 50
50110 50 3000
60
110 60
60110 60 3000
70
110 70
70110 70 2800
80
110 80
80110 80 2400
90
110 90
90110 90 1800
100
110 100
100110 100 1000
The maximum is attained near x 50 and 60. (c) P x110 x 110x x2 (d)
(e)
3500
(55, 3025)
dP 110 2x 0 when x 55. dx d 2P 2 < 0 dx 2
0
120 0
The solution appears to be x 55. 3. Let x and y be two positive numbers such that xy 192. Sxyx
192 x
192 dS 1 2 0 when x 192. dx x d 2S 384 3 > 0 when x 192. dx2 x S is a minimum when x y 192. 7. Let x be the length and y the width of the rectangle. 2x 2y 100 y 50 x A xy x50 x dA 50 2x 0 when x 25. dx d 2A 2 < 0 when x 25. dx2 A is maximum when x y 25 meters.
P is a maximum when x 110 x 55. The two numbers are 55 and 55.
5. Let x be a positive number. Sx
1 x
dS 1 1 2 0 when x 1. dx x 2 d 2S 3 > 0 when x 1. dx2 x The sum is a minimum when x 1 and 1x 1.
9. Let x be the length and y the width of the rectangle. xy 64 y
64 x
P 2x 2y 2x 2
64x 2x 128x
dP 128 2 2 0 when x 8. dx x d 2P 256 3 > 0 when x 8. dx2 x P is minimum when x y 8 feet.
Section 3.7 11. d x 42 x 02
x4 4x 174
Since d is smallest when the expression inside the radical is smallest, you need only find the critical numbers of
Since d is smallest when the expression inside the radical is smallest, you need only find the critical numbers of
f x x2 7x 16.
f x x4 4x 17 4 .
fx 2x 7 0
fx 4x3 4 0
x 72
x1
By the First Derivative Test, the point nearest to 4, 0 is 72, 72 .
By the First Derivative Test, the point nearest to 2, 12 is 1, 1. y
y 4
4
3
3
( x, x )
2
2
3
(2, 12(
d
x
x 2
( x, x 2 )
1
d
1
1
147
13. d x 22 x2 12 2
x2 7x 16
15.
Optimization Problems
−2
(4, 0)
dQ kxQ0 x kQ0x kx2 dx
−1
1
2
17. xy 180,000 (see figure)
S x 2y x
d 2Q kQ0 2kx dx2
360,000 where S is the length x
of fence needed.
kQ0 2x 0 when x
Q0 . 2
360,000 dS 1 0 when x 600. dx x2
d 3Q Q 2k < 0 when x 0. dx3 2
d 2S 720,000 > 0 when x 600. dx2 x3
dQdx is maximum when x Q02.
S is a minimum when x 600 meters and y 300 meters.
y
x
19. (a) A 4area of side 2area of Top
(b) V lengthwidthheight
(a) A 4311 233 150 square inches
(a) V 3311 99 cubic inches
(b) A 455 255 150 square inches
(b) V 555 125 cubic inches
(c) A 43.256 266 150 square inches
(c) V 663.25 117 cubic inches
(c) S 4xy 2x 2 150 ⇒ y V x 2y x 2
150 2x 2 4x
x x
150 4x 2x 752 x 21 x 2
75 3 2 V x 0 ⇒ x ±5 2 2
y
3
x x
By the First Derivative Test, x 5 yields the maximum volume. Dimensions: 5 5 5. (A cube!)
148
Chapter 3
Applications of Differentiation s 2
V xs 2x2, 0 < x <
21. (a)
dV 2xs 2x2 s 2x2 dx s s s 2xs 6x 0 when x , s2 is not in the domain. 2 6 d 2V 24x 8s dx2 s d 2V < 0 when x . dx2 6 V
5 2s 3 is maximum when x . 27 6
2 2 (b) If the length is doubled, V 27 2s3 8 27 s3. Volume is increased by a factor of 8.
23.
16 2y x
2x
32 4y 2x x y
32 2x x 4
A xy
x 2 2
x 2
32 2x4 x x 2
y
x2 8
x
1 8x x 2 x 2 x 2 2 4 8 dA 8x x x8x 1 dx 2 4 4
8 32 . 1 4 4
0 when x d 2A 1 dx 2 4
y
< 0 when x 4 32
32 2 324 324 16 4 4
The area is maximum when y
25. (a)
32 16 feet and x feet. 4 4
02 y2 01 x1 y2
(b)
2 x1
L x 2 y 2
10
x
2
—CONTINUED—
4
(2.587, 4.162)
x 2 x 2 1 2
2
8 4 , x > 1 x 1 x 12
0
10 0
L is minimum when x 2.587 and L 4.162.
Section 3.7
Optimization Problems
25. —CONTINUED—
1 1 2 x (c) Area Ax xy x 2 x 2 2 x1 x1 Ax 1
x 1 x 1 1 0 x 12 x 12
x 12 1 x 1 ±1 x 0, 2 (select x 2) Then y 4 and A 4. Vertices: 0, 0, 2, 0, 0, 4 27.
see figure
A 2xy 2x25 x2
y
252x x 225 x
1 dA 2x dx 2 2
8
2
6
2
25252xx 0 when x y 5 2 2 3.54.
±
Width:
(
x
2
−6 −4 −2 −2
2
4
6
−4
52 52 52 ,0 , ± , . 2 2 2 52 ; Length: 52 2
29. xy 30 ⇒ y
30 x
x+2 x
30 2 A x 2 x
see figure
dA 30 30 2x2 30 x 2 2 0 when x 30. 2 dx x x x2
y
25 − x 2
2
By the First Derivative Test, the inscribed rectangle of maximum area has vertices
( x,
30 30
y
y+2
30
By the First Derivative Test, the dimensions x 2 by y 2 are 2 30 by 2 30 (approximately 7.477 by 7.477). These dimensions yield a minimum area. 31. V r 2h 22 cubic inches or h (a)
22 r2
Radius, r
Height
0.2
22 0.22
2 0.2 0.2
0.4
22 0.42
2 0.4 0.4
0.6
22 0.62
2 0.6 0.6
0.8
22 0.82
2 0.8 0.8
—CONTINUED—
Surface Area
22 220.3 0.22
22 111.0 0.42
22 75.6 0.62
22 59.0 0.82
149
150
Chapter 3
Applications of Differentiation
31. —CONTINUED— (c) S 2 r 2 2 rh
(b) Radius, r
Height
0.2
22 0.22
2 0.2 0.2
0.4
22 0.42
2 0.4 0.4
Surface Area
2 rr h 2 r r
22 220.3 0.22
22 111.0 0.42
(d)
22 44 2 r 2 r2 r
100
(1.52, 43.46) −1
4 −10
0.6
22 0.62
2 0.6 0.6
0.8
22 0.82
22 2 0.8 0.8 59.0 0.82
22 50.3 1.0 2
22 45.7 1.22
22 43.7 1.42
22 43.6 1.62
22 44.8 1.82
22 47.1 2.02
22 1.02
2 1.0 1.0
1.2
22 1.22
2 1.2 1.2
1.4
22 1.42
2 1.4 1.4
1.6
22 1.62
2 1.6 1.6
1.8
22 1.82
2 1.8 1.8
2.0
22 2.02
2 2.0 2.0
1.0
22 75.6 0.62
The minimum seems to be 43.46 for r 1.52. (e)
44 dS 3 4 r 2 0 when r 11 1.52 in. dr r 22 3.04 in. r2 Note: Notice that 22 1113 22 2 13 2r. h 2 23 r 11 h
The minimum seems to be about 43.6 for r 1.6. 33. Let x be the sides of the square ends and y the length of the package. P 4x y 108 ⇒ y 108 4x V x 2y x 2108 4x 108x 2 4x3 dV 216x 12x2 dx 12x18 x 0 when x 18. d 2V 216 24x 216 < 0 when x 18. dx2 The volume is maximum when x 18 inches and y 108 418 36 inches.
Section 3.7
35.
Optimization Problems
151
1 1 V x 2h x 2 r r 2 x2 see figure 3 3 x3 x dV 1 2x r r2 x2 2r2 2rr2 x2 3x2 0 dx 3 r2 x2 3r2 x2
2r 2 2rr 2 x2 3x2 0
(0, r)
2rr 2 x2 3x2 2r 2 h
4r 2r 2 x2 9x4 12x 2 r 2 4r4
x 0,
r
x
0 9x4 8x2r 2 x29x2 8r 2 22r 3
(x, −
r 2 − x2
(
By the First Derivative Test, the volume is a maximum when x
4r 22r and h r r 2 x 2 . 3 3
Thus, the maximum volume is
4r3 3281 r
1 8r 2 V 3 9
3
cubic units.
37. No, there is no minimum area. If the sides are x and y, then 2x 2y 20 ⇒ y 10 x. The area is Ax x10 x 10x x2. This can be made arbitrarily small by selecting x 0. 4 V 12 r 3 r 2h 3
39.
h
41. Let x be the length of a side of the square and y the length of a side of the triangle.
12 43 r 3 12 4 r r2 r2 3
4x 3y 10
12 4 r S 4 r 2 2 rh 4 r 2 2 r r 2 3
1 3 A x2 y y 2 2
24 8 2 4 2 24 r r 4 r 2 r 3 3 r 24 dS 8 3 9 1.42 cm. r 2 0 when r dr 3 r 48 d 2S 8 3 9 cm. 3 > 0 when r dr2 3 r
10 3y2 3 2 y 16 4
3 dA 1 10 3y3 y0 dy 8 2
30 9y 43y 0 y
3 9
The surface area is minimum when r cm and h 0. The resulting solid is a sphere of radius r 1.42 cm.
30 9 43
d 2A 9 43 > 0 dy 2 8 A is minimum when
r h
y
30 103 . and x 9 43 9 43
152
Chapter 3
Applications of Differentiation
43. Let S be the strength and k the constant of proportionality. Given h2 w2 242, h2 242 w 2,
45.
R
v02 sin 2
g
3 dR 2v02 cos 2 0 when , . d
g 4 4
S kwh2 S kw576 w2 k576w w3 dS k576 3w3 0 when w 83, h 86. dw d 2S 6k w < 0 when w 83. dw 2
d 2R 4v 2 0 sin 2 < 0 when . d 2 g 4 By the Second Derivative Test, R is maximum when 4.
These values yield a maximum.
47. sin tan I
h h ⇒s , 0 < < s sin 2 h 2 tan ⇒ h 2 tan ⇒ s 2 sec 2 sin
k sin k sin k sin cos2 s2 4 sec2 4
α
α
dI k sin 2 sin cos cos2 cos d 4
k cos cos 2 2 sin2 4
k cos 1 3 sin2 4
0 when
s
h
4 ft
3 1 , , or when sin ± . 2 2 3
Since is acute, we have sin
1 3
⇒ h 2 tan 2
12 2 feet.
Since d 2Id 2 k4 sin 9 sin2 7 < 0 when sin 13, this yields a maximum.
49.
S x2 4, L 1 3 x2 Time T
x2 4
2
x2 6x 10
4
dT x x3 0 dx 2x2 4 4x2 6x 10
S=
2 x
x2 + 4 3−x 1
L=
1 + (3 − x 2(
Q
x2 9 6x x2 x2 4 4x2 6x 10 x4 6x3 9x2 8x 12 0 You need to find the roots of this equation in the interval 0, 3 . By using a computer or graphics calculator, you can determine that this equation has only one root in this interval x 1. Testing at this value and at the endpoints, you see that x 1 yields the minimum time. Thus, the man should row to a point 1 mile from the nearest point on the coast.
Section 3.7
51.
T
x2 4
v1
Optimization Problems
x2 6x 10
v2
dT x x3 0 dx v1x2 4 v2x2 6x 10
θ1
2
3−x
x
1
Since
θ2
x x2 4
sin 1 and
x3 x2 6x 10
Q
sin 2
we have sin 1 sin 2 sin 1 sin 2 0⇒ . v1 v2 v1 v2 Since d 2T 4 1 > 0 dx2 v1x2 432 v2x2 6x 1032 this condition yields a minimum time. 53. f x 2 2 sin x
(a) Distance from origin to y-intercept is 2. Distance from origin to x-intercept is 2 1.57.
y
(b) d x2 y2 x2 2 2 sin x2
3
3 2 1
−
π 4
−1
π 4
π 2
x
(0.7967, 0.9795)
− 4
2
−1
Minimum distance 0.9795 at x 0.7967. (c) Let f x d 2x x 2 2 2 sin x2. fx 2x 22 2 sin x2 cos x Setting fx 0, you obtain x 0.7967, which corresponds to d 0.9795. 55. F cos kW F sin
2 k +1
kW F cos k sin
dF kWk cos sin 0 d
cos k sin 2 k cos sin ⇒ k tan ⇒ arctan k Since cos k sin
1 k2 k2 1, 1 k2 1
k2
the minimum force is F
kW kW . cos k sin k2 1
1
θ
k
153
154
Chapter 3
57. (a)
Applications of Differentiation (b)
Base 1
Base 2
Altitude
Area
22.1
8
8 16 cos 10
8 sin 10
22.1
8 sin 20
42.5
8
8 16 cos 20
8 sin 20
42.5
8 16 cos 30
8 sin 30
59.7
8
8 16 cos 30
8 sin 30
59.7
8
8 16 cos 40
8 sin 40
72.7
8
8 16 cos 40
8 sin 40
72.7
8
8 16 cos 50
8 sin 50
80.5
8
8 16 cos 50
8 sin 50
80.5
8
8 16 cos 60
8 sin 60
83.1
8
8 16 cos 60
8 sin 60
83.1
8
8 16 cos 70
8 sin 70
80.7
8
8 16 cos 80
8 sin 80
74.0
8
8 16 cos 90
8 sin 90
64.0
Base 1
Base 2
Altitude
Area
8
8 16 cos 10
8 sin 10
8
8 16 cos 20
8
The maximum cross-sectional area is approximately 83.1 square feet. (c) A a b
h 2
(d)
8 8 16 cos
8 sin
2
64cos cos2 sin2 642 cos2 cos 1
641 cos sin , 0 < < 90 (e)
dA 641 cos cos 64 sin sin
d
642 cos 1cos 1
100
(60°, 83.1)
0 when 60 , 180 , 300 . The maximum occurs when 60 .
0
90 0
59. C 100
x ,1 ≤ x 200 x x 30 2
C 100
400 30 x3 x 302
Approximation: x 40.45 units, or 4045 units 61.
S1 4m 12 5m 62 10m 32 64 dS1 24m 14 25m 65 210m 310 282m 128 0 when m . dm 141 Line: y
S 4
63. S3
64 x 141
64 64 64 1 5 6 10 3 141 141 141
256 320 640 858 1 6 3 6.1 mi 141 141 141 141
4m 1 5m 6 10m 3
m2
1
m2
1
m2
S3
1
30
Using a graphing utility, you can see that the minimum occurs when x 0.3. Line: y 0.3x
20
10
40.3 1 50.3 6 100.3 3 S3 4.5 mi. 0.32 1
(0.3, 4.5) m 1
2
3
Section 3.8
Section 3.8
Newton’s Method
1. f x x2 3 fx 2x x1 1.7
n
xn
f xn
fxn
f xn fxn
1
1.7000
0.1100
3.4000
0.0324
1.7324
2
1.7324
0.0012
3.4648
0.0003
1.7321
3. f x sin x fx cos x x1 3
n
xn
1
3.0000
2
3.1425
5. f x x3 x 1 fx 3x2 1 Approximation of the zero of f is 0.682.
xn
f xn fxn
fxn
f xn fxn
0.1411
0.9900
0.1425
3.1425
0.0009
1.0000
0.0009
3.1416
f xn
xn
f xn fxn
n
xn
f xn
fxn
f xn fxn
1
0.5000
0.3750
1.7500
0.2143
0.7143
2
0.7143
0.0788
2.5307
0.0311
0.6832
3
0.6832
0.0021
2.4003
0.0009
0.6823
n
xn
f xn
fxn
f xn fxn
1
1.2000
0.1416
2.3541
0.0602
1.1398
2
1.1398
0.0181
3.0118
0.0060
1.1458
3
1.1458
0.0003
2.9284
0.0001
1.1459
n
xn
f xn
fxn
f xn fxn
1
1.5000
0.3750
6.7500
0.0556
7. f x 3x 1 x fx
Newton’s Method
3 1 2x 1
Approximation of the zero of f is 1.146.
xn
xn
f xn fxn
f xn fxn
Similarly, the other zero is approximately 7.854. 9. f x x3 3 fx
3x2
Approximation of the zero of f is 1.442.
1.4444
1.4444
0.0134
6.2589
0.0021
1.4423
3
1.4423
0.0003
6.2407
0.0001
1.4422
fx 3x2 7.8x 4.79
n
xn
f xn
fxn
f xn fxn
1
0.5000
0.3360
1.6400
0.2049
0.7049
2
0.7049
0.0921
0.7824
0.1177
0.8226
3
0.8226
0.0231
0.4037
0.0573
0.8799
4
0.8799
0.0045
0.2495
0.0181
0.8980
5
0.8980
0.0004
0.2048
0.0020
0.9000
6
0.9000
0.0000
0.2000
0.0000
0.9000
—CONTINUED—
f xn fxn
2
11. f x x3 3.9x2 4.79x 1.881
Approximation of the zero of f is 0.900.
xn
xn
f xn fxn
155
156
Chapter 3
Applications of Differentiation
11. —CONTINUED—
n
xn
f xn
fxn
f xn fxn
1
1.1
0.0000
0.1600
0.0000
xn
f xn fxn
1.1000
Approximation of the zero of f is 1.100. n
xn
f xn
fxn
f xn fxn
1
1.9
0.0000
0.8000
0.0000
xn
f xn fxn
1.9000
Approximation of the zero of f is 1.900. 13. f x x sinx 1 fx 1 cosx 1 Approximation of the zero of f is 0.489.
15. hx f x gx 2x 1 x 4
n
xn
f xn
fxn
f xn fxn
1
0.5000
0.0206
1.8776
0.0110
0.4890
2
0.4890
0.0000
1.8723
0.0000
0.4890
hxn
h xn hxn
0.6000
0.0552
1.7669
0.0313
0.5687
0.5687
0.0001
1.7661
0.0000
0.5687
n
xn
h xn
hxn
h xn hxn
1
4.5000
0.1373
21.5048
0.0064
4.4936
2
4.4936
0.0039
20.2271
0.0002
4.4934
xn
1 2
hx 1 sec2 x Point of intersection of the graphs of f and g occurs when x 4.493.
Point of intersection of the graphs of f and g occurs when x 0.569. 17. hx f x gx x tan x
19. f x x2 a 0
21. xi1
fx 2x xi2 a xi1 xi 2xi
23. xi1
2xi 2 xi2 a xi2 a xi a 2xi 2xi 2 2xi 3xi4 6 4xi3
i
1
2
3
4
xi
1.5000
1.5694
1.5651
1.5651
4 6 1.565
f xn fxn
h xn
n
1 hx 2 2x 4
xn
xn
xn
h xn hxn
h xn hxn
xi2 7 2xi
i
1
2
3
4
5
xi
2.0000
2.7500
2.6477
2.6458
2.6458
7 2.646
Section 3.8 25. f x 1 cos x
Newton’s Method
157
fx sin x
n
xn
f xn
fxn
f xn fxn
Approximation of the zero: 3.141
1
3.0000
0.0100
0.1411
0.0709
3.0709
2
3.0709
0.0025
0.0706
0.0354
3.1063
3
3.1063
0.0006
0.0353
0.0176
3.1239
4
3.1239
0.0002
0.0177
0.0088
3.1327
5
3.1327
0.0000
0.0089
0.0044
3.1371
6
3.1371
0.0000
0.0045
0.0022
3.1393
7
3.1393
0.0000
0.0023
0.0011
3.1404
8
3.1404
0.0000
0.0012
0.0006
3.1410
27. y 2x3 6x2 6x 1 f x
29. y x3 6x2 10x 6 f x
y 6x2 12x 6 fx
y 3x2 12x 10 fx
x1 1
x1 2
fx 0; therefore, the method fails.
x2 1
n
xn
f xn
fxn
1
1
1
0
xn
f xn fxn
x3 2 x4 1 and so on. Fails to converge y
31. Answers will vary. See page 222. Newton’s Method uses tangent lines to approximate c such that f c 0.
1
First, estimate an initial x1 close to c (see graph). Then determine x2 by x2 x1
−1
f x1 . fx1
Calculate a third estimate by x3 x2
x a 3
f(x)
x1
x2 c
2
b
x
−1 −2
f x2 . fx2
Continue this process until xn xn1 is within the desired accuracy. Let xn1 be the final approximation of c.
33. Let gx f x x cos x x gx sin x 1. The fixed point is approximately 0.74.
n
xn
g xn
gxn
g xn gxn
1
1.0000
0.4597
1.8415
0.2496
2
0.7504
0.0190
1.6819
0.0113
0.7391
3
0.7391
0.0000
1.6736
0.0000
0.7391
xn
g xn gxn
0.7504
158
Chapter 3
Applications of Differentiation
35. f x x3 3x2 3, fx 3x2 6x (a)
(b) x1 1
4
f x1 1.333 fx1
x2 x1 −4
5
Continuing, the zero is 1.347. −2
(d)
y
3x
1 (c) x1 4
4
y
f
x2 x1
f x1 2.405 fx1
3
Continuing, the zero is 2.532.
x 2
(e) If the initial guess x1 is not “close to” the desired zero of the function, the x-intercept of the tangent line may approximate another zero of the function.
1
y
4
1.313x
5
3.156
The x-intercepts correspond to the values resulting from the first iteration of Newton’s Method.
37. f x
1 a0 x
fx
1 x2
xn1 xn
1xn a 1 xn xn2 a xn xn xn2a 2xn xn2a xn2 axn 1xn2 xn
39. f x x cos x, 0,
y
fx x sin x cos x 0
1
)0.860, 0.561)
Letting F x fx, we can use Newton’s Method as follows.
x 2
Fx 2 sin x x cos x
1
n
xn
F xn
Fxn
F xn Fxn
1
0.9000
0.0834
2.1261
0.0392
0.8608
2
0.8608
0.0010
2.0778
0.0005
0.8603
Approximation to the critical number: 0.860
xn
F xn Fxn
2 3
Section 3.8
Newton’s Method
41. y f x 4 x2, 1, 0
y
d x 12 y 02 x 12 4 x22 x4 7x2 2x 17
5
d is minimized when D x4 7x2 2x 17 is a minimum.
2 1
gx 12x2 14 g xn gxn
2.0000
34.0000
0.0588
1.9412
1.9412
0.0830
31.2191
0.0027
1.9385
1.9385
0.0012
31.0934
0.0000
1.9385
1
2.0000
2 3
(1, 0) x
−3
gxn
xn
(1.939, 0.240)
3
gx D 4x3 14x 2
n
159
g xn
xn
−1 −1
1
3
g xn gxn
x 1.939 Point closest to 1, 0 is 1.939, 0.240. Minimize: T
43.
T T
Distance rowed Distance walked Rate rowed Rate walked x2 4
3
x 3x2 4
x2 6x 10
4 x3 4x2 6x 10
0
4xx2 6x 10 3x 3x2 4 16x2x2 6x 10 9x 32x2 4 7x4 42x3 43x2 216x 324 0 Let f x 7x4 42x3 43x2 216x 324 and fx 28x3 126x2 86x 216. Since f 1 100 and f 2 56, the solution is in the interval 1, 2. f xn fxn
f xn fxn
f xn
fxn
1.7000
19.5887
135.6240
0.1444
1.5556
1.5556
1.0480
150.2780
0.0070
1.5626
1.5626
0.0014
49.5591
0.0000
1.5626
n
xn
1 2 3
xn
Approximation: x 1.563 miles 2,500,000 76x3 4830x2 320,000
45.
76x3 4830x2 2,820,000 0 Let f x 76x3 4830x2 2,820,000 fx 228x2 9660x. From the graph, choose x1 40. f xn fxn
f xn fxn
n
xn
f xn
fxn
1
40.0000
44000.0000
21600.0000
2.0370
37.9630
2
37.9630
17157.6209
38131.4039
0.4500
38.4130
3
38.4130
780.0914
34642.2263
0.0225
38.4355
4
38.4355
2.6308
34465.3435
0.0001
38.4356
xn
The zero occurs when x 38.4356 which corresponds to $384,356.
160
Chapter 3
Applications of Differentiation
47. False. Let f x x2 1x 1. x 1 is a discontinuity. It is not a zero of f x. This statement would be true if f x pxqx is given in reduced form. 49. True 51. f x 14 x3 3x2 34 x 2
y
3 3 fx 4 x2 6x 4
60 40
Let x1 12.
20 x
n
xn
f xn
fxn
f xn fxn
1
12.0000
7.0000
36.7500
0.1905
11.8095
2
11.8095
0.2151
34.4912
0.0062
11.8033
3
11.8033
0.0015
34.4186
0.0000
11.8033
xn
−10 −5
f xn fxn
5
15
20
Approximation: x 11.803
Section 3.9
Differentials
1. f x x2
x f x
fx 2x Tangent line at 2, 4: y f 2 f2x 2
x2
T x 4x 4
1.9
1.99
2
3.6100
3.9601
3.6000
3.9600
2.01
2.1
4
4.0401
4.4100
4
4.0400
4.4000
y 4 4x 2 y 4x 4 3. f x x5
1.99
2
2.01
fx 5x4
x f x x5
24.7610
1.9
31.2080
32
32.8080
40.8410
2.1
Tangent line at 2, 32: y f 2 f2x 2
T x 80x 128
24.0000
31.2000
32
32.8000
40.0000
y 32 80x 2 y 80x 128 5. f x sin x fx cos x Tangent line at 2, sin 2: y f 2 f2x 2
x
1.9
1.99
2
2.01
2.1
f x sin x
0.9463
0.9134
0.9093
0.9051
0.8632
T x cos 2x 2 sin 2
0.9509
0.9135
0.9093
0.9051
0.8677
y sin 2 cos 2x 2 y cos 2x 2 sin 2 1 3 7. y f x 2 x3, fx 2 x2, x 2, x dx 0.1
y f x x f x f 2.1 f 2 0.6305
dy fxdx f20.1 60.1 0.6
Section 3.9
Differentials
9. y f x x4 1, fx 4x3, x 1, x dx 0.01 y f x x f x
11.
dy fx dx
f 0.99 f 1
f10.01
0.994 1 14 1 0.0394
40.01 0.04
13. y
y 3x2 4 dy 6x dx
dy
15. y x 1 x2
dy x
19.
y
17.
dy sin
3 dx 2x 12
y 2x cot2 x dy 2 2 cot x csc2 xdx
x 1 2x2 1 x2 dx dx 2 1 x 1 x2
6x 1 1 cos 3 2
x1 2x 1
2 2 cot x 2 cot3 xdx
6x2 1 dx
21. (a) f 1.9 f 2 0.1 f 2 f20.1
23. (a) f 1.9 f 2 0.1 f 2 f20.1 1 12 0.1 1.05
1 10.1 0.9 (b) f 2.04 f 2 0.04 f 2 f20.04
(b) f 2.04 f 2 0.04 f 2 f20.04 1 2 0.04 0.98
1 10.04 1.04 25. (a) g2.93 g3 0.07 g3 g30.07 8
12
1
27. (a) g2.93 g3 0.07 g3 g30.07
0.07 8.035
8 00.07 8 (b) g3.1 g3 0.1 g3 g30.1
(b) g3.1 g3 0.1 g3 g30.1 8 29.
A x2
8 00.1 8 A r 2 r 14
x dx
1 ± 64
r dr ± 14
dA 2x dx A dA 212
0.1 7.95 31.
x 12
3 ±8
12
1 ± 64
square inches
A dA 2 r dr 28 ± 14 ± 7 square inches
161
162
Chapter 3
33. (a)
Applications of Differentiation
x 15 centimeter
35.
x dx ± 0.05 centimeters A
r dr ± 0.02 inches
x2
4 (a) V r 3 3
dA 2x dx 215± 0.05
dV 4 r 2 dr 462± 0.02 ± 2.88 cubic inches
± 1.5 square centimeters
(b) S 4 r 2
Percentage error:
dS 8r dr 86± 0.02 ± 0.96 square inches
± 1.5 2 dA 0.00666. . . % A 152 3
(b)
r 6 inches
4 r 2 dr 3dr dV (c) Relative error: V 43 r 3 r
dA 2x dx 2 dx 2 ≤ 0.025 A x x
0.025 dx ≤ 0.0125 1.25% x 2
3 0.02 0.01 1% 6
dS 8 r dr 2dr Relative error: S 4 r 2 r
20.02 2 0.000666 . . . % 6 3
37. V r 2h 40 r 2, r 5 cm, h 40 cm, dr 0.2 cm V dV 80 r dr 8050.2 80 cm3 (b) 0.0025360024 216 seconds
39. (a) T 2 Lg dT
3.6 minutes
dL g Lg
Relative error: dT dL g Lg T 2 Lg
dL 2L
1 relative error in L 2
1 0.005 0.0025 2
Percentage error:
1 dT 100 0.25% % T 4
41. 2645 26.75 h
d ± 15 ± 0.25 (a)
h 9.5 csc
b
dh 9.5 csc cot d dh cot d h
9.5
θ
dh cot 26.750.25 h
Converting to radians, cot 0.46690.0044 0.0087 0.87% in radians.
(b)
dh cot d ≤ 0.02 h 0.02 0.02 tan d ≤ cot d 0.02 tan 26.75 0.02 tan 0.4669 ≤ 26.75 0.4669 0.0216 2.16% in radians
Review Exercises for Chapter 3
43. r
v02 sin 2 32
45. Let f x x, x 100, dx 0.6. f x x f x f x dx
v0 2200 ftsec
changes from 10 to 11 dr
x
22002 cos 2 d 16
10
f x x 99.4
180
d 11 10
1 dx 2x
100
1 0.6 9.97 2100
Using a calculator: 99.4 9.96995
180
r dr
22002 20 cos 16 180
180 4961 feet
4961 feet 49. Let f x x, x 4, dx 0.02, f x 1 2x .
4 x, x 625, dx 1. 47. Let f x 4 x f x x f x f x d x
Then
1 dx 44x3
f 4.02 f 4 f 4 dx
1 4 624 4 625 f x x 1 4 625 3 4 5
4.02 4
1 1 0.02 2 0.02. 4 24
1 4.998 500
4 624 4.9980. Using a calculator,
51. In general, when x → 0, dy approaches y. 53. True
55. True
Review Exercises for Chapter 3 1. A number c in the domain of f is a critical number if f c 0 or f is undefined at c.
y 4
f ′(c) is 3 undefined.
f ′(c) = 0
x −4 −3
−1 −2 −3 −4
3. gx 2x 5 cos x, 0, 2
18
(6.28, 17.57)
g x 2 5 sin x 2
0 when sin x 5 . Critical numbers: x 0.41, x 2.73 Left endpoint: 0, 5 Critical number: 0.41, 5.41 Critical number: 2.73, 0.88 Minimum Right endpoint: 2, 17.57 Maximum
(2.73, 0.88) − 4
2 −1
1
2
4
163
164
Chapter 3
Applications of Differentiation
5. Yes. f 3 f 2 0. f is continuous on 3, 2 , differentiable on 3, 2.
7. f x 3 x 4 y
(a)
f x x 33x 1 0 for x 13.
6 4
c 13 satisfies f c 0.
2 x
−2
2
4
6
10
−4 −6
f 1 f 7 0 (b) f is not differentiable at x 4.
9.
f x x23, 1 ≤ x ≤ 8
f x x cos x,
11.
2 f x x13 3
f x 1 sin x f b f a 2 2 1 ba 2 2
f b f a 4 1 3 ba 81 7
f c 1 sin c 1
2 3 f c c13 3 7 c
13.
149
3
≤ x ≤ 2 2
c0
2744 3.764 729
f x Ax2 Bx C f x 2Ax B f x2 f x1 Ax22 x12 Bx2 x1 x2 x1 x2 x1 Ax1 x2 B f c 2Ac B Ax1 x2 B 2Ac Ax1 x2 c
x1 x2 Midpoint of x1, x2
2
15. f x x 12x 3 f x x 1 1 x 32x 1 2
x 13x 7 7 Critical numbers: x 1 and x 3
17. hx xx 3 x32 3x12 Domain: 0, 3 3 h x x12 x12 2 2 3 3x 1 x12x 1 2 2x Critical number: x 1
Interval:
< x < 1
1 < x <
7 3
7 3
< x <
Sign of f x:
f x > 0
f x < 0
f x > 0
Conclusion:
Increasing
Decreasing
Increasing
Interval:
0 < x < 1
Sign of h x:
h x < 0
h x > 0
Conclusion:
Decreasing
Increasing
1 < x <
Review Exercises for Chapter 3 19. ht 14t 4 8t
Test Interval: < t < 2
h t t 3 8 0 when t 2. Relative minimum: 2, 12
165
2 < t <
Sign of h t:
h t < 0
h t > 0
Conclusion:
Decreasing
Increasing
1 1 cos12t sin12t 3 4
21. y
v y 4 sin12t 3 cos12t (a) When t
1 , y inch and v y 4 inches/second. 8 4
(b) y 4 sin12t 3 cos12t 0 when
sin12t 3 3 ⇒ tan12t . cos12t 4 4
3 4 Therefore, sin12t and cos12t . The maximum displacement is 5 5 y (c) Period:
1345 41 53 125 inch. 2 12 6 1 6 6
Frequency:
23. f x x cos x, 0 ≤ x ≤ 2 f x 1 sin x f x cos x 0 when x Points of inflection:
3 , . 2 2
Test Interval: Sign of f x: Conclusion:
3 < x < 2 2
3 < x < 2 2
f x < 0
f x > 0
f x < 0
Concave downward
Concave upward
Concave downward
0 < x <
2
2 , 2 , 32, 32
25. gx 2x21 x2
y
g x 4x2x2 1 Critical numbers: x 0, ±
(−
1 2
1, 1 2 2
)
1
−2
g x 4 24x
(
1, 1 2 2
)
(0, 0)
2
x
−1
2
−2
g 0 4 > 0
Relative minimum at 0, 0
1 1 Relative maximums at ± , 2 2
y
27. 6
(5, f(5))
5 4
(3, f(3))
2 1 −1
29. The first derivative is positive and the second derivative is negative. The graph is increasing and is concave down.
7
3
−3
1 g ± 8 < 0 2
(6, 0) (0, 0) 2 3 4 5
x 7
166
Chapter 3
Applications of Differentiation
31. (a) D 0.0034t4 0.2352t3 4.9423t2 20.8641t 94.4025 (b)
369
0
29 0
(c) Maximum at 21.9, 319.5 1992 Minimum at 2.6, 69.6 1972 (d) Outlays increasing at greatest rate at the point of inflection 9.8, 173.7 1979
33. lim
x →
2x2 2 2 lim 5 x → 3 5x2 3
35. lim
2x 3 x4
39. f x
3x2
37. hx
Discontinuity: x 4 lim
x →
x →
5 cos x 0, since 5 cos x ≤ 5. x
3 2 x
Discontinuity: x 0
2x 3 2 3x lim 2 x → 1 4x x4
lim
x →
3x 2 2
Vertical asymptote: x 4
Vertical asymptote: x 0
Horizontal asymptote: y 2
Horizontal asymptote: y 2
41. f x x3
243 x
43. f x
x1 1 3x2
Relative minimum: 3, 108
Relative minimum: 0.155, 1.077
Relative maximum: 3, 108
Relative maximum: 2.155, 0.077 0.2
200
−2
−5
5
5
− 1.4
− 200
Vertical asymptote: x 0
Horizontal asymptote: y 0
45. f x 4x x2 x4 x Domain: , ; Range: , 4 f x 4 2x 0 when x 2. f x 2 Therefore, 2, 4 is a relative maximum. Intercepts: 0, 0, 4, 0
y
5
)2, 4)
4 3 2 1
x 1
2
3
5
Review Exercises for Chapter 3 47. f x x16 x2, Domain: 4, 4 , Range: 8, 8
y
2
2, 8
8
Domain: 4, 4 ; Range: 8, 8
6 4
16 2x2 f x 0 when x ± 22 and undefined when x ± 4. 16 x2 f x
2xx2 24 16 x232
2
(− 4, 0)
(4, 0) x
8
6
2
2
4
6
8
(0, 0)
8
2
2,
8
f 22 > 0
Therefore, 22, 8 is a relative minimum. f 22 < 0
Therefore, 22, 8 is a relative maximum. Point of inflection: 0, 0 Intercepts: 4, 0, 0, 0, 4, 0 Symmetry with respect to origin 49. f x x 13x 32
y
Domain: , ; Range: ,
4
f x x 12x 35x 11 0 when x 1,
11 , 3. 5
f x 4x 15x2 22x 23 0 when x 1,
( 115 , 1.11(
(1, 0)
(2.69, 0.46) (3, 0)
2
x
−2
4 −2
11± 6 . 5
6
(1.71, 0.60)
−4
f 3 > 0 Therefore, 3, 0 is a relative minimum. f
115 < 0
Therefore,
is a relative maximum. 115, 3456 3125
Points of inflection: 1, 0,
11 5
6
11 5
, 0.60 ,
6
, 0.46
Intercepts: 0, 9, 1, 0, 3, 0 51. f x x13x 323
y
Domain: , ; Range: ,
4 3
x1 f x 0 when x 1 and undefined when x 3, 0. x 313x23 2 f x 53 is undefined when x 0, 3. x x 343 3 4 is By the First Derivative Test 3, 0 is a relative maximum and 1, a relative minimum. 0, 0 is a point of inflection.
Intercepts: 3, 0, 0, 0
2 1
) 3, 0)
)0, 0) x
5
4
2
) 1,
1
1
1.59) 3
2
167
168
Chapter 3
Applications of Differentiation
x1 x1
53. f x
x
1
y
Domain: , 1, 1, ; Range: , 1, 1, f x
2 < 0 if x 1. x 12
f x
4 x 13
4
y
1 2
x 2
2
4
2
Horizontal asymptote: y 1 Vertical asymptote: x 1 Intercepts: 1, 0, 0, 1 55. f x
4 1 x2
y 5
Domain: , ; Range: 0, 4
8x 0 when x 0. f x 1 x22 3 81 3x2 . 0 when x ± f x 1 x23 3
(0, 4)
4
(−
3,3 3
(
(
3,3 3
1
2
(
2 1 −3
−2 −1
x −1
3
f 0 < 0 Therefore, 0, 4 is a relative maximum. Points of inflection: ± 33, 3 Intercept: 0, 4 Symmetric to the y-axis Horizontal asymptote: y 0
57. f x x3 x
y
4 x
10
Domain: , 0, 0, ; Range: , 6 , 6, f x 3x2 1 f x 6x
4 x2
3x4
4 0 when x ± 1. x2 x2
8 6x4 8 0 x3 x3
f 1 < 0 Therefore, 1, 6 is a relative maximum. f 1 > 0 Therefore, 1, 6 is a relative minimum. Vertical asymptote: x 0 Symmetric with respect to origin
5
(1, 6) x
2
1
(−1, −6) −5
1
x
2
0
Review Exercises for Chapter 3
59. f x x2 9
y
Domain: , ; Range: 0, f x
2xx2 9 0 when x 0 and is undefined when x ± 3. x2 9
10
5
2x2 9 is undefined at x ± 3. f x 2 x 9
)0, 9)
) 3, 0)
)3, 0) x
4
f 0 < 0
2
2
4
Therefore, 0, 9 is a relative maximum. Relative minima: ± 3, 0 Points of inflection: ± 3, 0 Intercepts: ± 3, 0, 0, 9 Symmetric to the y-axis 61. f x x cos x
y
)2 , 2
Domain: 0, 2 ; Range: 1, 1 2
3 3 , 2 2
f x 1 sin x ≥ 0, f is increasing. f x cos x 0 when x Points of inflection:
1)
2
3 , . 2 2
3 3 , , , 2 2 2 2
)0, 1)
, 2 2 x
2
Intercept: 0, 1 63. x2 4y2 2x 16y 13 0 (a) x 2 2x 1 4y 2 4y 4 13 1 16
y
x 1 4y 2 4 x 12 y 22 1 4 1 The graph is an ellipse: 2
2
4
(1, 3) 3 2 1
Maximum: 1, 3
(1, 1) x −1
Minimum: 1, 1
1
2
3
(b) x2 4y2 2x 16y 13 0 2x 8y
dy dy 2 16 0 dx dx dy 8y 16 2 2x dx dy 2 2x 1x dx 8y 16 4y 8
The critical numbers are x 1 and y 2. These correspond to the points 1, 1, 1, 3, 2, 1, and 2, 3. Hence, the maximum is 1, 3 and the minimum is 1, 1.
169
170
Chapter 3
Applications of Differentiation
65. Let t 0 at noon.
(100 − 12t, 0) (0, 0)
L d 2 100 12t2 10t2 10,000 2400t 244t 2
A
(100, 0)
d
300 dL 2400 488t 0 when t 4.92 hr. dt 61
B (0, −10t)
Ship A at 40.98, 0; Ship B at 0, 49.18 d 2 10,000 2400t 244t 2 4098.36 when t 4.92 4:55 P.M.. d 64 km 67. We have points 0, y, x, 0, and 1, 8. Thus,
y
08 8x y8 or y . m 01 x1 x1
(0, y)
10
(1, 8)
8 6
Let f x L 2 x 2
x 8x 1 . 2
4 2
x f x 2x 128 x1 x
(x, 0)
x 1 x 0 x 12
x 2
4
6
8
10
64x 0 x 13
x x 13 64 0 when x 0, 5 minimum. Vertices of triangle: 0, 0, 5, 0, 0, 10 69.
A Average of basesHeight
x 2 s
3s2 2sx x2
2
s
see figure
s
dA 1 s xs x 3s2 2sx x2 dx 4 3s2 2sx x2
x−s 2
22s xs x 0 when x 2s. 43s2 2sx x2 A is a maximum when x 2s. 71. You can form a right triangle with vertices 0, 0, x, 0 and 0, y. Assume that the hypotenuse of length L passes through 4, 6. 60 6x y6 or y 04 4x x4
Let f x L2 x2 y2 x 2 f x 2x 72
x 6x 4 . 2
0 x x 4 x 4 4 2
3 x x 43 144 0 when x 0 or x 4 144.
L 14.05 feet
s
x−s 2 x
m
3s 2 + 2sx − x 2 2
Review Exercises for Chapter 3 csc
73. csc
L1 or L1 6 csc 6
2 9 or L L2
2
9 csc
see figure
L1 θ
2
L2
θ 9
L L1 L2 6 csc 9 csc
171
6
(π2 − θ(
2 6 csc 9 sec
dL 6 csc cot 9 sec tan 0 d tan3
3 2 2 ⇒ tan 3 3 3
sec 1 tan2 csc L6
1 23
23
323 223
313
sec 323 223 tan 213
323 22312 323 22312 9 3323 22332 ft 21.07 ft Compare to Exercise 72 using a 9 and b 6. 13 2 313
75. Total cost Cost per hourNumber of hours T
v 110 11v 550 5 600 v 60 v 2
dT 11 550 11v 2 33,000 2 dv 60 v 60v 2 0 when v 3000 1030 54.8 mph. d 2T 1100 3 > 0 when v 1030 so this value yields a minimum. dv 2 v 77. f x x3 3x 1 From the graph you can see that f x has three real zeros. f x 3x2 3 f xn
f xn
f xn f xn
1.5000
0.1250
3.7500
0.0333
1.5333
2
1.5333
0.0049
4.0530
0.0012
1.5321
n
xn
f xn
f xn
f xn f xn
1
0.5000
0.3750
2.2500
0.1667
0.3333
2
0.3333
0.0371
2.6667
0.0139
0.3472
3
0.3472
0.0003
2.6384
0.0001
0.3473
n
xn
f xn
f xn
f xn f xn
1
1.9000
0.1590
7.8300
0.0203
1.8797
2
1.8797
0.0024
7.5998
0.0003
1.8794
n
xn
1
xn
f xn f xn
xn
xn
f xn f xn
f xn f xn
The three real zeros of f x are x 1.532, x 0.347, and x 1.879.
C H A P T E R 3 Applications of Differentiation Section 3.1
Extrema on an Interval
. . . . . . . . . . . . . . 378
Section 3.2
Rolle’s Theorem and the Mean Value Theorem
Section 3.3
Increasing and Decreasing Functions and the First Derivative Test . . . . . . . . . . . . . . 387
Section 3.4
Concavity and the Second Derivative Test . . . . 394
Section 3.5
Limits at Infinity
Section 3.6
A Summary of Curve Sketching
Section 3.7
Optimization Problems . . . . . . . . . . . . . . 419
Section 3.8
Newton’s Method . . . . . . . . . . . . . . . . . 429
Section 3.9
Differentials . . . . . . . . . . . . . . . . . . . . 434
. 381
. . . . . . . . . . . . . . . . . 402 . . . . . . . . . 410
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 437 Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . 445
C H A P T E R 3 Applications of Differentiation Section 3.1
Extrema on an Interval
Solutions to Even-Numbered Exercises
2. f x cos
x 2
4.
f x 3x x 1
12 x 1
fx 3x
x fx sin 2 2
12
x 1 3
3 x 112x 2x 1 2
f0 0 f2 0
3 x 1123x 2 2
32 0
f
6. Using the limit definition of the derivative,
lim
f x f 0 4 x 4 lim 1 x→0 x0 x
lim
f x f 0 4 x 4 lim 1 x→0 x0 x0
x→0
x→0
8. Critical number: x 0. x 0: neither
f0 does not exist, since the one-sided derivatives are not equal. 10. Critical numbers: x 2, 5
12. gx x2x2 4 x4 4x2
x 2: neither
gx 4x3 8x 4xx2 2
x 5: absolute maximum
Critical numbers: x 0, x ± 2
14. f x fx
4x x2 1
x2 14 4x2x 41 x2 2 x2 12 x 12
Critical numbers: x ± 1
16. f 2 sec tan , 0 < < 2 f 2 sec tan sec2 sec 2 tan sec sin
cos cos
sec 2
1
sec2 2 sin 1 On 0, 2, critical numbers:
378
7 11 , 6 6
Section 3.1
18. f x fx
2x 5 , 0, 5 3
20. f x x2 2x 4, 1, 1 fx 2x 2 2x 1
2 ⇒ No critical numbers 3
Left endpoint:
Extrema on an Interval
Left endpoint: (1, 5 Minimum Right endpoint: 1, 1 Maximum
0, 53 Minimum
Right endpoint: 5, 5 Maximum 22. f x x3 12x, 0, 4
3 24. gx x, 1, 1
fx 3x2 12 3x2 4
gx
Left endpoint: 0, 0
Left endpoint: 1, 1 Minimum
Critical number: 2, 16 Minimum
Critical number: 0, 0
Right endpoint: 4, 16 Maximum
Right endpoint: 1, 1 Maximum
Note: x 2 is not in the interval.
26. y 3 t 3 , 1, 5
28. ht
From the graph, you see that t 3 is a critical number. 4
−1
1 3x23
ht
t , 3, 5 t2 2 t 22
Left endpoint: 3, 3 Maximum
5
Right endpoint: −4
5, 53 Minimum
Left endpoint: 1, 1 Minimum Right endpoint: 5, 1 Critical number: 3, 3 Maximum
30. g x sec x,
6 , 3
32. y x2 2 cos x, 1, 3 y 2x sin x
gx sec x tan x Left endpoint:
Right endpoint:
2 ,
, 1.1547 6 3 6
3 , 2 Maximum
Left endpoint: 1, 1.5403 Right endpoint: 3, 7.99 Maximum Critical number: 0, 3 Minimum
Critical number: 0, 1 Minimum 34. (a) Minimum: 4, 1 Maximum: 1, 4
36. (a) Minima: 2, 0 and 2, 0 Maximum: 0, 2
(b) Maximum: 1, 4
(b) Minimum: 2, 0
(c) Minimum: 4, 1
(c) Maximum: 0, 2
(d) No extrema
(d) Maximum: 1, 3
379
380
Chapter 3
38. f x
Applications of Differentiation
22 3x,x ,
1 ≤ x < 3 3 ≤ x ≤ 5
2
2 , 0, 2 2x Left endpoint: 0, 1 Minimum
40. f x
Left endpoint: 1, 1 Maximum
3
Right endpoint: 5, 13 Minimum 3
(1, 1)
−3
12
(0, 1)
−1
(3, − 7)
5
−1
(5, − 13) −15
42. (a)
Maximum:
3
(2, ) 8 3
0
2, 83
Minimum: 0, 0, 3, 0 3
0
4 (b) f x x 3 x, 0, 3 3 fx
4 1 x 3 x121 3 x121 3 2
4 1 3 x12 x 23 x 3 2
44. f x
1 1 , ,3 x2 1 2
fx
2x x2 12
f x
21 3x2 x2 13
fx
24x 24x3 x2 14
Setting f 0, we have x 0, ± 1. 1
f 1 2 is the maximum value.
26 3x 62 x 22 x 3 x 3 3 x 3 3 x
Critical number: x 2 f 0 0 Minimum f 3 0 Minimum f 2
8 3
Maximum:
46.
f x f x f 4x f 5x
2, 38
1 , 1, 1 x2 1 24x 24x3 See Exercise 44. x2 14
48. Let f x 1x. f is continuous on 0, 1 but does not have a maximum. f is also continuous on 1, 0 but does not have a minimum. This can occur if one of the endpoints is an infinite discontinuity.
245x4 10x2 1 x2 15 240x
x2 16 3x4
10x2
y 2
3
f 40 24 is the maximum value.
1
−2
x
−1
1 −1
2
Section 3.2 y
50. 5 4
Rolle’s Theorem and the Mean Value Theorem
52. (a) No
54. (a) No
(b) Yes
(b) Yes
381
f
3 2
x
−2 −1
1
2
3
4
5
6
−2 −3
56. x
v 2 sin 2 3 , ≤ ≤ 32 4 4
C 2x
58.
C1 300,002
d is constant. dt
C300 1600
dx d dx by the Chain Rule dt d dt
300,000 , 1 ≤ x ≤ 300 x
C 2
v 2 cos 2 d 16 dt
300,000 0 x2
2x2 300,000 x2 150,000
In the interval 4, 34, 4, 34 indicate minimums for dxdt and 2 indicates a maximum for dxdt. This implies that the sprinkler waters longest when 4 and 34. Thus, the lawn farthest from the spinkler gets the most water.
x 10015 387 > 300 outside of interval C is minimized when x 300 units. Yes, if 1 ≤ x ≤ 400, then x 387 would minimize C.
60. f x x
y 3
The derivative of f is undefined at every integer and is zero at any noninteger real number. All real numbers are critical numbers.
2 1 −2
x
−1
1
2
3
−2
64. False. Let f x x2. x 0 is a critical number of f.
62. True. This is stated in the Extreme Value Theorem.
gx f x k x k2 x k is a critical number of g.
Section 3.2
Rolle’s Theorem and the Mean Value Theorem 4. f x xx 3
2. Rolle’s Theorem does not apply to f x cotx2 over , 3 since f is not continuous at x 2.
x-intercepts: 0, 0, 3, 0 3 fx 2x 3 0 at x . 2
6. f x 3xx 1 x-intercepts: 1, 0, 0, 0
1 x fx 3x x 112 3x 112 3x 112 x 1 2 2 fx 3x 112
32x 1 0 at x 32.
382
Chapter 3
Applications of Differentiation
8. f x x2 5x 4, 1, 4
f x x 3x 12, 1, 3
10.
f 1 f 4 0
f 1 f 3 0
f is continuous on 1, 4. f is differentiable on 1, 4.
f is continuous on 1, 3. f is differentiable on 1, 3. Rolle’s Theorem applies.
Rolle’s Theorem applies.
fx x 32x 1 x 12
fx 2x 5 2x 5 0 ⇒ x c value:
x 12x 6 x 1
5 2
x 13x 5
5 2
c value:
12. f x 3 x 3 , 0, 6
5 3
f x
14.
x2 1 , 1, 1 x
f 0 f 6 0
f 1 f 1 0
f is continuous on 0, 6. f is not differentiable on 0, 6 since f3 does not exist. Rolle’s Theorem does not apply.
f is not continuous on 1, 1 since f 0 does not exist. Rolle’s Theorem does not apply.
16. f x cos x, 0, 2 f 0 f 2 1 f is continuous on 0, 2. f is differentiable on 0, 2. Rolle’s Theorem applies.
f x cos 2x,
18.
f f
fx sin x
c value:
12 , 6
3 12 2
6 21
f
f 12 0
Rolle’s Theorem does not apply.
20.
f x sec x,
4 , 4
f f 2 4 4
f is continuous on 4, 4. f is differentiable on 4, 4. Rolle’s Theorem applies.
22. f x x x13, 0, 1 f 0 f 1 0 f is continuous on 0, 1. f is differentiable on 0, 1. (Note: f is not differentiable at x 0.) Rolle’s Theorem applies. fx 1
fx sec x tan x sec x tan x 0
1
x0
3 x2
c value: 0
x2 x c value:
1 3 x2 3
1 3 1 27
271
3
9
1 0 3 x2 3
3
9
0.1925
1
0
−1
1
Section 3.2
f x
24.
x x sin , 1, 0 2 6
f 1 f 0 0 f is continuous on 1, 0. f is differentiable on 1, 0. Rolle’s Theorem applies. 1 x fx cos 0 2 6 6
x 3 cos 6 x
Rolle’s Theorem and the Mean Value Theorem
26. Cx 10
1x x x 3
(a) C3 C6
25 3
Cx 10
(b)
1 3 0 x2 x 32
1 3 x2 6x 9 x2 2x2 6x 9 0
6 3 arccos Value needed in 1, 0.
x
0.5756 radian
c value: 0.5756
6 ± 63 3 ± 33 4 2
In the interval 3, 6: c
0.02
−1
6 ± 108 4
3 33 4.098. 2
0
−0.01
28.
30. f x x 3 , 0, 6
y
f is not differentiable at x 3.
f x
a
b
32. f x xx2 x 2 is continuous on 1, 1 and differentiable on 1, 1. f 1 f 1 1 1 1 fx 3x2 2x 2 1
3x 1x 1 0 1 c 3
34. f x x 1x is continuous on 12, 2 and differentiable on 12, 2. f 2 f 12 32 3 1 2 12 32 fx
1 1 x2
x2 1 c1
383
384
Chapter 3
Applications of Differentiation
36. f x x3 is continuous on 0, 1 and differentiable on 0, 1.
38. f x 2 sin x sin 2x ferentiable on 0, .
f 1 f 0 1 0 1 10 1 fx 3x2 1 x±
f f 0 0 0 0 0 fx 2 cos x 2 cos 2x 0 2cos x 2 cos2 x 1 0
3
3
In the interval 0, 1: c
is continuous on 0, and dif-
22 cos x 1cos x 1 0 3
3
.
1 2
cos x
cos x 1
5 , , 3 3
x In the interval 0, : c
. 3
40. f x x 2 sin x on , (c) fx 1 2 cos x 1
2
(a) tangent − 2
secant 2 tangent
f
− 2
c± , 2
f
2 2 2
2 2 2
f
(b) Secant line: slope
cos x 0
f f 1 2
y 1x
Tangent lines: y
2 2 1 x 2 yx2
y
yx
2 1 x 2 2
yx2 42. f x x4 4x3 8x2 5, 0, 5, 5, 80 m
80 5 15 50
(a)
150
(c) First tangent line: y f c mx c y 9.59 15x 0.67
tangent f
secant
0 15x y 0.46
tangent 5
0 0
(b)
Secant line: y 5 15x 0 0 15x y 5 fx 4x3 12x2 16x f 5 f 1 15 51 4c3 12c2 16c 15 0 4c3 12c2 16c 15 c 0.67 or c 3.79
Second tangent line: y f c mx c y 131.35 15x 3.79 0 15x y 74.5
Section 3.2
44. St 200 5 (a)
9 2t
Rolle’s Theorem and the Mean Value Theorem
S12 S0 2005 914 2005 92 450 12 0 12 7 St 200
(b)
2 9 t 450 7 2
1 1 2 t2 28 2 t 27 t 27 2 3.2915 months St is equal to the average value in April. 46. f a f b and fc 0 where c is in the interval a, b. (a) gx f x k
gx f x k
(b)
(c)
ga k gb k f a
ga gb f a k gx fx ⇒ gc 0
gx fx k
gx f k x g
ak g bk f a
gx k fk x
Interval: a, b
gc k fc 0
Critical number of g: c
Interval: a k, b k Critical number of g: c k
g
ck k fc 0
Interval:
ak, bk
Critical number of g:
c k
48. Let Tt be the temperature of the object. Then T0 1500 and T5 390 . The average temperature over the interval 0, 5 is 390 1500 222 Fhr. 50 By the Mean Value Theorem, there exists a time to, 0 < t0 < 5, such that Tt0 222. 2 x 50. f x 3 cos 2 ,
fx 6 cos
2x sin 2x 2
3 cos (a)
2x sin 2x (b) f and f are both continuous on the entire real line.
7
−2
2
−7
(c) Since f 1 f 1 0, Rolle’s Theorem applies on 1, 1. Since f 1 0 and f 2 3, Rolle’s Theorem does not apply on 1, 2.
(d) lim fx 0 x→3
lim fx 0
x→3
385
386
Chapter 3
Applications of Differentiation
52. f is not continuous on 5, 5. Example: f x
0,1x,
54. False. f must also be continuous and differentiable on each interval. Let
x0 x0
f x
y
x3 4x . x2 1
f (x) = 1x
4 2
(5, 15) x 2
(− 5, − 15)
4
−5
56. True 58. Suppose f x is not constant on a, b. Then there exists x1 and x2 in a, b such that f x1 f x2. Then by the Mean Value Theorem, there exists c in a, b such that fc
f x2 f x1 0. x2 x1
This contradicts the fact that fx 0 for all x in a, b. 60. Suppose f x has two fixed points c1 and c2. Then, by the Mean Value Theorem, there exists c such that fc
f c2 f c1 c2 c1 1. c2 c1 c2 c1
This contradicts the fact that fx < 1 for all x. 62. Let f x cos x. f is continuous and differentiable for all real numbers. By the Mean Value Theorem, for any interval a, b, there exists c in a, b such that f b f a fc ba cos b cos a sin c ba cos b cos a sin cb a
cos b cos a sin c
b a
cos b cos a ≤ b a since sin c
≤ 1.
Section 3.3
Section 3.3
Increasing and Decreasing Functions and the First Derivative Test
Increasing and Decreasing Functions and the First Derivative Test
2. y x 12
4. f x x4 2x2
Increasing on: , 1
Increasing on: 1, 0, 1,
Decreasing on: 1,
Decreasing on: , 1, 0, 1
6. y y
x2 x1 xx 2 x 12
Critical numbers: x 0, 2
Discontinuity: x 1
Test intervals: < x < 2
2 < x < 1
1 < x < 0
0 < x <
Sign of fx:
y > 0
y < 0
y < 0
y > 0
Conclusion:
Increasing
Decreasing
Decreasing
Increasing
Increasing on , 2, 0, Decreasing on 2, 1, 1, 0 8. hx 27x x3 hx 27 3x2 33 x3 x hx 0 Critical numbers: x ± 3 Test intervals:
< x < 3
3 < x < 3
Sign of hx:
h < 0
h > 0
h < 0
Decreasing
Increasing
Decreasing
Conclusion:
3 < x <
Increasing on 3, 3 Decreasing on , 3, 3, 10. y x y
387
4 x
x 2x 2 x2
Critical numbers: x ± 2
Discontinuity: 0
Test intervals: < x < 2
2 < x < 0
0 < x < 2
2 < x <
Sign of y:
y > 0
y < 0
y < 0
y > 0
Conclusion:
Increasing
Decreasing
Decreasing
Increasing
Increasing: , 2, 2, Decreasing: 2, 0, 0, 2
388
Chapter 3
Applications of Differentiation
12. f x x2 8x 10
14. f x x2 8x 12
fx 2x 8 0
fx 2x 8 0
Critical number: x 4
Critical number: x 4
Test intervals: Sign of fx:
< x < 4 4 < x <
Conclusion:
f < 0
f > 0
Decreasing
Increasing
Increasing on: 4,
fx 3x2 12x 3xx 4 Critical numbers: x 0, 4 Test intervals:
< x < 0
0 < x < 4
Sign of fx:
f > 0
f < 0
f > 0
Conclusion:
Increasing
Decreasing
Increasing
4 < x <
Increasing on , 0, 4, Decreasing on 0, 4 Relative maximum: 0, 15 Relative minimum: 4, 17 18. f x x 22x 1 fx 3xx 2 Critical numbers: x 2, 0 Test intervals:
< x < 2
2 < x < 0
Sign of fx:
f > 0
f < 0
f > 0
Conclusion:
Increasing
Decreasing
Increasing
Relative minimum: 0, 4
f > 0
f < 0
Conclusion:
Increasing
Decreasing
Relative maximum: 4, 4
16. f x x3 6x2 15
Relative maximum: 2, 0
Sign of fx:
Decreasing on: 4,
Relative minimum: 4, 6
Decreasing on: 2, 0
< x < 4
Increasing on: , 4
Decreasing on: , 4
Increasing on: , 2, 0,
Test intervals:
0 < x <
4 < x <
Section 3.3
Increasing and Decreasing Functions and the First Derivative Test 22. f x x23 4
20. f x x4 32x 4 fx 4x3 32 4x3 8
2 2 fx x13 13 3 3x
Critical number: x 2 Test intervals: Sign of fx: Conclusion:
389
Critical number: x 0
< x < 2
2 < x <
f < 0
f > 0
Test intervals:
< x < 0
Decreasing
Increasing
Sign of fx:
f < 0
f > 0
Conclusion:
Decreasing
Increasing
Increasing on: 2, Decreasing on: , 2
Increasing on: 0,
Relative minimum: 2, 44
Decreasing on: , 0
0 < x <
Relative minimum: 0, 4
24. f x x 113
26. f x x 3 1
1 fx 3x 123
fx
Critical number: x 1
Critical number: x 3
Test intervals: Sign of fx: Conclusion:
< x < 1
1 < x <
1, x > 3 x3 x3 1, x < 3
Test intervals:
f > 0
f > 0
Sign of fx:
Increasing
Increasing
Conclusion:
< x < 3 Decreasing
Increasing
Increasing on: 3,
No relative extrema
Decreasing on: , 3 Relative minimum: 3, 1
fx
x x1
x 11 x1 1 x 12 x 12
Discontinuity: x 1 Test intervals: Sign of fx: Conclusion:
< x < 1
f > 0
f > 0
Increasing
Increasing
Increasing on: , 1, 1, No relative extrema
1 < x <
f > 0
Increasing on: ,
28. f x
3 < x <
f < 0
390
Chapter 3
30. f x
Applications of Differentiation
x3 1 3 2 x2 x x
fx
1 6 x 6 x2 x3 x3
Critical number: x 6 Discontinuity: x 0 Test intervals: Sign of fx:
< x < 6
6 < x < 0
f < 0
f > 0
f < 0
Decreasing
Increasing
Decreasing
Conclusion:
0 < x <
Increasing on: 6, 0 Decreasing on: , 6, 0, Relative minimum:
32. f x fx
6, 121
x2 3x 4 x2
x 22x 3 x2 3x 41 x2 4x 10 x 22 x 22
Discontinuity: x 2 < x < 2
Test intervals: Sign of fx: Conclusion:
2 < x <
f > 0
f > 0
Increasing
Increasing
Increasing on: , 2, 2, No relative extrema 34. f x sin x cos x
1 sin 2x, 0 < x < 2 2
fx cos 2x 0 Critical numbers: x
3 5 7 , , , 4 4 4 4
0 < x <
Test intervals:
4
3 < x < 4 4
3 5 < x < 4 4
5 7 < x < 4 4
7 < x < 2 4
Sign of fx:
f > 0
f < 0
f > 0
f < 0
f > 0
Conclusion:
Increasing
Decreasing
Increasing
Decreasing
Increasing
Increasing on: Decreasing on:
0, 4 , 34, 54, 74, 2 4 , 34, 54, 74
Relative maxima:
4 , 21, 54, 21
Relative minima:
34, 21, 74, 12
Section 3.3
36. f x fx
Increasing and Decreasing Functions and the First Derivative Test
sin x , 0 < x < 2 1 cos2x
Test intervals:
0 < x <
cos x2 sin2x 0 1 cos2x2
Sign of fx: Conclusion:
Critical numbers: x
3 , 2 2
0, 2 , 32, 2
Increasing on: Decreasing on:
2 , 32
3 < x < 2 2
3 < x < 2 2
f > 0
f < 0
f > 0
Increasing
Decreasing
Increasing
2
Relative maximum:
2 , 1
Relative minimum:
32, 1
38. f x 10 5 x2 3x 16 , 0, 5 (a) fx
52x 3 x2 3x 16
(b)
y 15
f
12
6
f′
3 −1
(c)
52x 3 0 x2 3x 16
x 3
1
−3
4
(d) Intervals:
3 Critical number: x 2
0, 23
32, 5
fx > 0
fx < 0
Increasing
Decreasing
f is increasing when f is positive and decreasing when f is negative.
40. f x
x x cos , 0, 4 2 2
(a) fx
1 1 x sin 2 2 2
(b)
y 8 6
f 4 2
f′ π
(c)
x 1 1 sin 0 2 2 2 sin
x 1 2 x 2 2
Critical number: x
2π
3π
4π
x
(d) Intervals:
0,
, 4
fx > 0
fx > 0
Increasing
Increasing
f is increasing when f is positive.
391
392
Chapter 3
Applications of Differentiation 44. f x is a line of slope 2 ⇒ fx 2.
42. f t cos2t sin2t 2 sin2t gt, 2 < t < 2 ft 4 sin t cos t 2 sin 2t
6
f symmetric with respect to y-axis zeros of f: ±
4
−6
6 −2
Relative maximum: 0, 1
2 , 1, 2 , 1
Relative minimum: 2
−3
3
−2
46. f is a 4th degree polynomial ⇒ f is a cubic polynomial.
48. f has positive slope y
y 6
4
f′
3 2
−6 −4 −2
x 2
4
f′
6
x
−3 −2 −1
1
2
3
−2
In Exercises 50–54, f x > 0 on , 4, f x < 0 on 4, 6 and f x > 0 on 6, . gx 3f x 3
50.
52. gx f x
54. gx f x 10
gx fx
gx fx 10
g0 f0 > 0
g8 f2 < 0
gx 3fx g5 3f5 > 0
58. st 4.9sin t 2
56. Critical number: x 5 f4 2.5 ⇒ f is decreasing at x 4.
(a) vt 9.8sin t
(b) If 2, the speed is maximum,
f6 3 ⇒ f is increasing at x 6.
5, f 5 is a relative minimum. 60. C
vt 9.8 t.
3t ,t ≥ 0 27 t3
(a)
t
0
0.5
1
1.5
2
2.5
3
Ct
0
0.055
0.107
0.148
0.171
0.176
0.167
The concentration seems greater near t 2.5 hours. (b)
(c) C
0.25
0
3 0
The concentration is greatest when t 2.38 hours.
speed 9.8 sin t
27 t 33 3t3t 2 27 t 32 327 2t 3 27 t 32
3 C 0 when t 3 2 2.38 hours.
By the First Derivative Test, this is a maximum.
Section 3.3 62. P 2.44x P 2.44
Increasing and Decreasing Functions and the First Derivative Test
x2 5000, 0 ≤ x ≤ 35,000 20,000
Test intervals: Sign of P:
0 < x < 24,400
24,400 < x < 35,000
P > 0
P < 0
x 0 10,000
x 24,400 Increasing when 0 < x < 24,400 hamburgers. Decreasing when 24,400 < x < 35,000 hamburgers. 64. R 0.001T 4 4T 100 (a) R
0.004T 3 4 0 2 0.001T 4 4T 100
(b)
125
T 10 , R 8.3666
−100
100 −25
The minimum resistance is approximately R 8.37 at T 10 . 66. f x 2 sin3x 4 cos3x 6
−
−6
The maximum value is approximately 4.472. You could use calculus by finding fx and then observing that the maximum value of f occurs at a point where fx 0. For instance, f0.154 0, and f 0.154 4.472. 68. (a) Use a cubic polynomial f x a3x3 a2x2 a1x a0. (b) fx 3a3 x 2 2a 2 x a1
0, 0:
4, 1000:
0 a0
f 0 0
0 a1
f0 0
1000 64a3 16a2
f 4 1000
0 48a3 8a2
f4 0
(c) The solution is a0 a1 0, a2 f x (d)
125 3 375 2 x x. 4 2
1200
(4, 1000)
−3
(0, 0) −400
8
375 125 , a3 2 4
393
394
Chapter 3
Applications of Differentiation
70. (a) Use a fourth degree polynomial f x a4 x4 a3 x3 a 2 x 2 a1x a0. (b) fx 4a4x3 3a3x2 2a2x a1
1, 2:
1, 4:
3, 4:
2 a4 a3 a2 a1 a0
f 1 2
0 4a4 3a3 2a2 a1
f1 0
4 a4 a3 a2 a1 a0
f 1 4
0 4a4 3a3 2a2 a1
f1 0
4 81a4 27a3 9a2 3a1 a0
f 3 4
0 108a4 27a3 6a2 a1
f3 0
(c) The solution is a0
23 8 ,
a1
32 ,
a2
1 4,
a3
1 2,
a4 18
f x 18 x 4 12 x3 14 x2 32 x 23 8 . (d)
6
(−1 , 4)
(3, 4)
(1, 2)
−4
6 −2
74. True
72. False Let hx f xgx where f x gx x. Then hx x2 is decreasing on , 0.
If f x is an nth-degree polynomial, then the degree of fx is n 1.
76. False. The function might not be continuous. 78. Suppose fx changes from positive to negative at c. Then there exists a and b in I such that fx > 0 for all x in a, c and fx < 0 for all x in c, b. By Theorem 3.5, f is increasing on a, c and decreasing on c, b. Therefore, f c is a maximum of f on a, b and thus, a relative maximum of f.
Section 3.4
Concavity and the Second Derivative Test
2. y x3 3x2 2, y 6x 6 Concave upward: , 1 Concave downward: 1,
6. y
1 2 3x5 40x3 135x, y xx 2x 2 270 9
Concave upward: , 2, 0, 2 Concave downward: 2, 0, 2,
4. f x
x2 1 6 , y 2x 1 2x 13
Concave upward: , 12 Concave downward: 12 , 8. hx x5 5x 2 hx 5x4 5 hx 20x3 Concave upward: 0, Concave downward: , 0
Section 3.4 10. y x 2 csc x, ,
fx 6x2 6x 12 f x 12x 6
2 cot xcsc x cot x
csc2 x
2
csc3 x
csc x
cot2 x
f x 12x 6 0 when x 12.
Concave upward: 0,
< x <
Test interval
Concave downward: , 0
Sign of f x Conclusion Point of inflection:
14.
395
12. f x 2x3 3x2 12x 5
y 1 2 csc x cot x y 2 csc x
Concavity and the Second Derivative Test
1 2
1 2
< x <
f x < 0
f x > 0
Concave downward
Concave upward
12 , 132
f x 2x4 8x 3 fx 8x3 8 f x 24x 2 0 when x 0. However, 0, 3 is not a point of inflection since f x ≥ 0 for all x. Concave upward on ,
16. f x x3x 4 fx x3 3x2x 4 x2x 3x 4 4x2x 3 f x 4x 2 8xx 3 4xx 2x 3 12xx 2 0 f x 12xx 2 0 when x 0, 2. Test interval Sign of f x Conclusion
< x < 0
0 < x < 2
f x > 0
f x < 0
f x > 0
Concave upward
Concave downward
Concave upward
2 < x <
Points of inflection: 0, 0, 2, 16 18. f x xx 1, Domain: 1, 3x 2 1 fx x x 112 x 1 2 2x 1 6x 1 3x 2x 112 3x 4 f x 4x 1 4x 132 f x > 0 on the entire domain of f (except for x 1, for which f x is undefined). There are no points of inflection. Concave upward on 1, 20. f x
x1 x
Domain: x > 0
x1 fx 32 2x 3x f x 52 4x Point of inflection:
3, 43 3, 4 3 3
3 < x <
Test intervals
0 < x < 3
Sign of f x
f > 0
f < 0
Conclusion
Concave upward
Concave downward
396
Chapter 3
22. f x 2 csc
3x , 0 < x < 2 2
fx 3 csc f x
Applications of Differentiation
3x 3x cot 2 2
9 3x 3x 3x csc3 0 for any x in the domain of f. csc cot2 2 2 2 2
Concave upward:
0, 23 , 43, 2 23, 43
Concave downward:
No points of inflection 24. f x sin x cos x, 0 ≤ x ≤ 2 fx cos x sin x f x sin x cos x f x 0 when x
3 7 , . 4 4 0 < x <
Test interval: Sign of f x: Conclusion:
3 4
3 7 < x < 4 4
7 < x < 2 4
f x < 0
f x > 0
f x < 0
Concave downward
Concave upward
Concave downward
3 < x < 2 2
3 3 < x < 2 2
Points of inflection:
34, 0 , 74, 0
26. f x x 2 cos x, 0, 2 fx 1 2 sin x f x 2 cos x f x 0 when x
3 , 2 2 2
Test intervals:
0 < x <
Sign of f x:
f < 0
f > 0
f < 0
Conclusion:
Concave downward
Concave upward
Concave downward
Points of inflection:
2 , 2 , 32, 32
28. f x x2 3x 8
30. f x x 52
fx 2x 3
fx 2x 5
f x 2
f x 2
3 Critical number: x 2
Critical number: x 5
f 32 > 0
f 5 < 0
Therefore, 2 , 4 is a relative minimum. 3
41
Therefore, 5, 0 is a relative maximum.
Section 3.4 32. f x x3 9x2 27x fx 3x 2 18x 27 3x 32 f x 6x 3 Critical number: x 3 However, f 3 0, so we must use the First Derivative Test. fx ≥ 0 for all x and, therefore, there are no relative extrema.
Concavity and the Second Derivative Test
1 34. gx x 22x 42 8 gx
x 4x 1x 2 2
3 g x 3 3x x2 2 Critical numbers: x 2, 1, 4 g 2 9 < 0
2, 0 is a relative maximum. g 1 92 > 0
1, 10.125 is a relative minimum. g 4 9 < 0
4, 0 is a relative maximum. 36. f x x2 1 fx
38. f x
x
fx
x2 1
Critical number: x 0 1 f x 2 x 132
x x1 1 x 12
There are no critical numbers and x 1 is not in the domain. There are no relative extrema.
f 0 1 > 0 Therefore, 0, 1 is a relative minimum. 40. f x 2 sin x cos 2x, 0 ≤ x ≤ 2 fx 2 cos x 2 sin 2x 2 cos x 4 sin x cos x 2 cos x1 2 sin x 0 when x f x 2 sin x 4 cos 2x f
6 < 0
f
2 > 0
f
56 < 0
f
32 > 0
Relative maxima:
6 , 23 , 56, 23
Relative minima:
2 , 1 , 32, 3
5 3 , , , . 6 2 6 2
397
398
Chapter 3
Applications of Differentiation
42. f x x26 x2, 6, 6 (a) fx
3x4 x2 6 x2
(c)
y
f
6
fx 0 when x 0, x ± 2. f x
6x4 9x2 12 6 x232
f x 0 when x ±
x
−3
9 2
33
.
3
f'
f ''
−6
(b) f 0 > 0 ⇒ 0, 0 is a relative minimum. The graph of f is increasing when f > 0 and decreasing when f < 0. f is concave upward when f > 0 and concave downward when f < 0.
f ± 2 < 0 ⇒ ± 2, 42 are relative maxima. Points of inflection: ± 1.2758, 3.4035 44. f x 2x sin x, 0, 2 (a) fx 2x cos x
sin x 2x
(c)
y 4
Critical numbers: x 1.84, 4.82
f f'
2
cos x cos x sin x f x 2x sin x 2x 2x 2x2x
2
f ''
−2
2cos x 4x2 1sin x 2x 2x2x
x
π
−4
4x cos x 4x2 1sin x 2x2x
f is increasing when f > 0 and decreasing when f < 0. f is concave upward when f > 0 and concave downward when f < 0.
(b) Relative maximum: 1.84, 1.85 Relative minimum: 4.82, 3.09 Points of inflection: 0.75, 0.83, 3.42, 0.72 46. (a)
f < 0 means f decreasing
y 4
(b) 4
f decreasing means concave downward
3 2
f > 0 means f increasing
y
f decreasing means concave downward
3 2
1
1 x 1
2
3
x
4
48. (a) The rate of change of sales is increasing. S > 0
1
50.
2
4
3
y
f
3
f'
f ''
(b) The rate of change of sales is decreasing. S > 0, S < 0 (c) The rate of change of sales is constant. S C, S 0 (d) Sales are steady. S C, S 0, S 0 (e) Sales are declining, but at a lower rate. S < 0, S > 0 (f) Sales have bottomed out and have started to rise. S > 0
−2
x
−1
3 −1
Section 3.4 52.
Concavity and the Second Derivative Test
54.
y
399
y
3
2
f ''
2
1
f
1
(0, 0) x
−1
1
(2, 0) x
−1
3
1
3
−1 −2
f'
−3
56.
58. (a)
y
d 12
3 2
(0, 0)
(2, 0) x
−1
1
3
−1
t 10
(b) Since the depth d is always increasing, there are no relative extrema. fx > 0 (c) The rate of change of d is decreasing until you reach the widest point of the jug, then the rate increases until you reach the narrowest part of the jug’s neck, then the rate decreases until you reach the top of the jug. 3 x 60. (a) f x
y 3
fx 13 x23
2
f x 29 x53
1
(0, 0)
Inflection point: 0, 0
−6
(b) f x does not exist at x 0.
−4
x
−2
2
4
6
−2 −3
62. f x ax3 bx 2 cx d Relative maximum: 2, 4 Relative minimum: 4, 2 Point of inflection: 3, 3 fx 3ax 2 2bx c, f x 6ax 2b
f 2 8a 4b 2c d 4 56a 12b 2c 2 ⇒ 28a 6b c 1 f 4 64a 16b 4c d 2 f2 12a 4b c 0, f4 48a 8b c 0, f 3 18a 2b 0 28a 6b c 1
18a 2b
12a 4b c
16a 2b 1
16a 2b 1 2,
b
f x
1 3 2x
a
1
92 ,
0
2a
c 12, d 6
9 2 2x
12x 6
0
1
400
Chapter 3
Applications of Differentiation
64. (a) line OA: y 0.06x
slope: 0.06
line CB: y 0.04x 50
y
slope: 0.04
150
f x ax3 bx 2 cx d (−1000, 60) A
fx 3ax2 2bx c
1000, 60:
C (0, 50)
60 10003a 10002b 1000c d 0.06 10002 3a 2000b c
1000, 90:
(1000, 90) B
100
−1000
O
x 1000
90 10003a 10002b 1000c d 0.04 10002 3a 2000b c
The solution to this system of 4 equations is a 1.25 108, b 0.000025, c 0.0275, and d 50. (b) y 1.25 108x3 0.000025x2 0.0275x 50
(c)
0.1
100 −1100
1100
− 0.1 −1100
1100 −10
(d) The steepest part of the road is 6% at the point A. 66. S
5.755T 3 8.521T 2 0.654T 0.99987, 0 < T < 25 108 106 104
(a) The maximum occurs when T 4 and S 0.999999. (b)
(c) S20 0.9982
S
1.001 1.000 0.999 0.998 0.997 0.996 T
5
68. C 2x C 2
10
15
20
25
30
300,000 x
300,000 0 when x 10015 387 x2
By the First Derivative Test, C is minimized when x 387 units.
70. S (a)
100t2 , t > 0 65 t2 100
0
35 0
(b) St St
13,000t 65 t22 13,00065 3t2 0 ⇒ t 4.65 65 t23
S is concave upwards on 0, 4.65, concave downwards on 4.65, 30. (c) St > 0 for t > 0. As t increases, the speed increases, but at a slower rate.
Section 3.4 72.
f x 2sin x cos x,
Concavity and the Second Derivative Test
f 0 2
401
4
P2
f
fx 2cos x sin x,
f0 2
f x 2sin x cos x,
f 0 2
−6
6
P1
P1x 2 2x 0 21 x
−4
P1x 2 1 P2x 2 2x 0 2 2x 0 2 2 2x x 2
P2x 2 2x P2x 2 The values of f, P1, P2, and their first derivatives are equal at x 0. The values of the second derivatives of f and P2 are equal at x 0. The approximations worsen as you move away from x 0. 74.
f x
x , x1
f 2 2
3
x 1 fx , 2xx 12
32 f2 4 22
3x2 6x 1 f x 32 , 4x x 13
23 232 f 2 16 82
P2x
5 −1
32 32 52 x 2 x 4 4 2
32 1 232 x 2 x 22 2 32 x 2 232 x 22 4 2 16 4 32
32 4
P2x 2 P2x
P2 f
−1
P1x 2 P1x
P1
3
32 232 x 2 4 16
232 16
The values of f, P1, P2 and their first derivatives are equal at x 2. The values of the second derivatives of f and P2 are equal at x 2. The approximations worsen as you move away from x 2. 76. f x xx 62 x3 12x2 36x fx 3x2 24x 36 3x 2x 6 0 f x 6x 24 6x 4 0 Relative extrema: 2, 32 and 6, 0 Point of inflection 4, 16 is midway between the relative extrema of f.
402
Chapter 3
Applications of Differentiation
px ax3 bx2 cx d
78.
px 3ax2 2bx c px 6ax 2b 6ax 2b 0 x
b 3a
The sign of px changes at x b3a. Therefore, b3a, pb3a is a point of inflection.
p
b b3 b2 b 2b3 bc d a b c d 3 2 3a 27a 9a 3a 27a 2 3a
When px x3 3x2 2, a 1, b 3, c 0, and d 2. x0
3 1 31
y0
233 30 2 2 0 2 0 2712 31
The point of inflection of px x3 3x2 2 is x0, y0 1, 0. 80. False. f x 1x has a discontinuity at x 0. 82. True y sinbx Slope: y b cosbx b ≤ y ≤ b
Assume b > 0
84. False. For example, let f x x 24.
Section 3.5 2. f x
Limits at Infinity
2x x2 2
4. f x 2
x4
x2 1
6. f x
2x2 3x 5 x2 1
No vertical asymptotes
No vertical asymptotes
No vertical asymptotes
Horizontal asymptotes: y ± 2
Horizontal asymptote: y 2
Horizontal asymptote: y 2
Matches (c)
Matches (a)
Matches (e)
8. f x x f x
2x2 x1 100
101
102
1
18.18
198.02
lim f x
x→
20
103 1998.02
(Limit does not exist.)
104 19,998
105 199,998
106 1,999,998
0 −2
10
Section 3.5
10. f x
8x
Limits at Infinity
10
x2 3
x
101
102
103
104
105
106
107
f x
8.12
8.001
8
8
8
8
8
0
15 0
lim f x 8
x→
12. f x 4 x f x
3 x2 2
10
100
101
102
103
104
105
106
5
4.03
4.0003
4.0
4.0
4
4
0
15 0
lim f x 4
x→
14. (a) hx
f x 5x2 3x 7 7 5x 3 x x x
lim hx
x→
(b) hx
f x x2
(Limit does not exist) 5x2
16. (a) lim
x→
(b) lim
3x 7 3 7 5 2 x2 x x
x→
(c) hx
3 2x 2 3x 1 3
3 2x2 x→ 3x 1
(c) lim
lim hx 5
x→
3 2x 0 3x3 1
(Limit does not exist)
f x 5x2 3x 7 5 7 3 2 3 x3 x3 x x x
lim hx 0
x→
18. (a) lim
x→
(b) lim
x→
5x32 0 4x2 1
5x32 x→ 4x 1
22. lim 4
26. lim
x→
3x3 2 3 1 9x3 2x2 7 9 3
24. lim
12x x4
x→
5x32 5 4x 32 1 4
(c) lim
x→
20. lim
(Limit does not exist)
3 404 x
x x2 1
lim
x→
lim
x→
x→
1
x2 1 x2
1 x (1x
for x < 1
0, x x2
2
(Limit does not exist)
403
404
28.
Chapter 3
lim
x →
Applications of Differentiation
3x 1 3 1x , for x < 0 we have x2 x lim x2 x x → x2 x x2 lim
3 1x
x →
30. lim
x →
1 1x
x cos x cos x lim 1 x → x x
3
32. lim cos x→
1x cos 0 1
101 Note: lim
x →
cos x 0 by the Squeeze Theorem since x
cos x 1 1 ≤ ≤ . x x x
34. f x
3x
36. lim x tan
6
x2 2
lim f x 3
x →
−9
x→
1 tan t sin t lim lim t →0 x t →0 t t
cos t 1
11 1
9
lim f x 3
Let x 1t.
x→
−6
Therefore, y 3 and y 3 are both horizontal asymptotes. 38. lim 2x 4x2 1 lim x →
40.
x →
2x
lim 4x 1
2x 4x2 1
4x2 1
2x
x →
3x 9x x lim 3x 9x2 x 3x 9x2 x x → x → 2 lim
lim
x →
3x
2
9x x
2x 4x2 1
0
x 3x 9x2 x 1
lim
x →
1
2
3
9x2 x
for x
< 0 we have x x2
x2 1 1 lim x → 3 9 1x 6 42.
x
100
101
102
103
104
105
f x
1.0
5.1
50.1
500.1
5000.1
50,000.1
lim
x →
x2 xx2 x 1
Limit does not exist.
30
106 500,000.1
x2 xx2 x x3 lim x2 xx2 x x → x2 xx2 x
0
50 0
Section 3.5 44.
f x lim
101
102
103
104
105
106
2.000
0.348
0.101
0.032
0.010
0.003
0.001
x1
x →
3
100
x
xx
Limits at Infinity
0
0
25
−1
46. x 2 is a critical number.
48. (a) The function is even:
fx < 0 for x < 2.
(b) The function is odd:
fx > 0 for x > 2.
lim f x 5
x→
lim f x 5
x→
lim f x lim f x 6
x →
x →
For example, let f x
6 6. 0.1x 22 1
y
8
4
x 2
2
4
6
x3 x2
50. y
2x 9 x2
52. y
Intercept: 0, 0
23
Intercepts: 3, 0, 0,
Symmetry: origin
Symmetry: none
Horizontal asymptote: y 0
Horizontal asymptote: y 1 since
Vertical asymptote: x ± 3
x3 x3 . 1 lim lim x → x 2 x → x 2
y 6 5 4 3 2 1
Discontinuity: x 2 (Vertical asymptote) y 5 4 3 2 −4 −3 −2 −1
−5−4
x 1
3 4 5 6
−1 −2 −3 −4 −5 −6
−2 −3 −4 −5
54. y
x2
x2 9
y 5 4 3 2
Intercept: 0, 0 Symmetry: y-axis Horizontal asymptote: y 1 since lim
x →
x2
x2 x2 1 lim 2 . x → x 9 9
Discontinuities: x ± 3 (Vertical asymptotes) Relative maximum: 0, 0
−5 −4
−1 −2 −3 −4 −5
x 1
4 5
x 1 2
6
405
406
Chapter 3
56. y
Applications of Differentiation
2x2 4
58. x2y 4
x2
Intercept: 0, 0
Intercepts: none
Symmetry: y-axis
Symmetry: y-axis
Horizontal asymptote: y 2
Horizontal asymptote: y 0 since
Relative minimum: 0, 0 y
lim
x →
4 4 0 lim 2 . x → x x2
Discontinuity: x 0 (Vertical asymptote)
5 4 3
y
1 x
−5 −4 −3 −2 −1
1 2 3 4 5
−2 −3 −4 −5
4 3 2 1 x
−5 −4 −3 −2 −1
60. y
2x 1 x2
1 2 3 4 5
1 x
62. y 1
Intercept: 0, 0
Intercept: 1, 0
Symmetry: origin
Symmetry: none
Horizontal asymptote: y 0 since
Horizontal asymptote: y 1 since
lim
x →
2x 2x 0 lim . x → 1 x2 1 x2
Discontinuities: x ± 1 (Vertical asymptotes)
lim
x →
1 1x 1
lim
x →
1 1x .
Discontinuity: x 0 (Vertical asymptote) y
y 5 4 3 2 1
5 4 3 2 x
−5 −4 −3 −2
2
x
−5 −4 −3 −2 −1
1 2 3 4 5
−3 −4 −5
64. y 4 1
1 x2
66. y
x x2 4
Intercepts: ± 1, 0
Domain: , 2, 2,
Symmetry: y-axis
Intercepts: none
Horizontal asymptote: y 4
Symmetry: origin
Vertical asymptote: x 0
Horizontal asymptotes: y ± 1 since
y
lim
x2 4
1, lim
x →
x x2 4
1.
Vertical asymptotes: x ± 2 (discontinuities)
3 2 1 −5 −4 −3 −2
x
x →
5
y
x 2 3 4 5
5 4 3 2 −5 −4 −3
−1 −2 −3 −4 −5
x 1
3 4 5
Section 3.5 68. f x fx f x
x2
x2 1
x = −1
2x x2 12x x22x 2 0 when x 0. x2 12 x 12
Limits at Infinity
4
x=1
−3
(0, 0)
y=1
x2 122 2x2x2 12x 23x2 1 2 x2 14 x 13
407
3
−4
Since f 0 < 0, then 0, 0 is a relative maximum. Since f x 0, nor is it undefined in the domain of f, there are no points of inflection. Vertical asymptotes: x ± 1 Horizontal asymptote: y 1 70. f x fx f x
1 1 x2 x 2 x 1x 2
x = −1
2x 1 1 0 when x . x2 x 22 2
x2
( 12 , − 49( −3
x 2 2 2x 12 x2 x 24 2
x=2
2
x2
3
y=0
x 22x 1
−2
6 x 1 x2 x 23 x2
Since f 12 < 0, then 12 , 49 is a relative maximum. Since f x 0, nor is it undefined in the domain of f, there are no points of inflection. Vertical asymptotes: x 1, x 2 Horizontal asymptote: y 0 72. f x fx f x
x1 x2 x 1 xx 2 0 when x 0, 2. x2 x 12 2x3 3x2 1 0 when x 0.5321, 0.6527, 2.8794. x2 x 13
f 0 < 0 Therefore, 0, 1 is a relative maximum. f 2 > 0 Therefore,
2, 31 is a relative minimum. Points of inflection: 0.5321, 0.8440, 0.6527, 0.4491 and 2.8794, 0.2931 Horizontal asymptote: y 0
(− 0.6527, 0.4491) 2
(0.5321, 0.8440)
−3
3
(
−2, − 1 3
( −2
(0, 1)
(− 2.8794, − 0.2931)
408
Chapter 3
74. gx gx
Applications of Differentiation
2x
4
3x2 1
y= 2
3
2 3x2 132
−6
6
y=− 2
18x gx 3x2 152
3 −4
No relative extrema. Point of inflection: 0, 0. Horizontal asymptotes: y ±
2 3
No vertical asymptotes
76. f x fx
2 sin 2x Hole at 0, 4 x
6
4x cos 2x 2 sin 2x x2
−
2
There are an infinite number of relative extrema. In the interval 2 , 2 , you obtain the following.
5
2 −2
Relative minima: ± 2.25, 0.869, ± 5.45, 0.365 Relative maxima: ± 3.87, 0.513 Horizontal asymptote: y 0 No vertical asymptotes
78. f x (a)
1 x3 2x2 2 1 , gx x 1 2 2x2 2 x f=g
(c)
4
−6
70
−80
6
80
−70
−4
(b) f x
x3 2x2 2 2x2
2xx
3 2
2x2 2 2x2 2x2
The graph appears as the slant asymptote y 12 x 1.
1 1 x 1 2 gx 2 x
80.
lim
v1v2 →
100 1
1 100 1 0 100% v1v2c
82. y
3.351t2 42.461t 543.730 t2
(a)
5
20
100 0
(b) Yes. lim y 3.351 t→
Section 3.5
Limits at Infinity
409
100t2 , t > 0 65 t2
84. S (a)
120
5
30 0
100 100 1
(b) Yes. lim S t→
px a xn . . . a1x a0 lim n m . . . x → qx x → bm x b1x b0
86. lim
Divide px and qx by xm. an a1 a0 . . . m1 m px x mn x x 0. . .00 0 lim Case 1: If n < m: lim . . .00b 0 x qx x→ b b b 1 m m bm . . . m1 m0 x x a1 a0 an . . . m1 m an . . . 0 0 an px x x lim Case 2: If m n: lim . . .00b . b b x → qx x → b 1 0 m m bm . . . m1 m x x px lim Case 3: If n > m: lim x → qx x →
a1 a0 an x nm . . . m1 m x x ± . . .0 . . . ± . b b b 0 1 m bm . . . m1 m0 x x
88. False. Let y1 x 1, then y10 1. Thus, y1 1 2x 1 and y10 12. Finally, 1 1 and y10 . 4x 132 4
y1
1 1 1 1 Let p ax2 bx 1, then p0 1. Thus, p 2ax b and p0 2 ⇒ b 2 . Finally, p 2a and p0 4 ⇒ a 8 . Therefore,
18x2 12x 1,
x < 0
x 1,
x ≥ 0
f x
fx
12
f x
14x 1
12 14 x,
and f 0 1,
x < 0
1 and f0 , and x > 0 2
x 1 ,
14, ,
32
x < 0
1 and f 0 . x > 0 4
f x < 0 for all real x, but f x increases without bound.
410
Chapter 3
Section 3.6
Applications of Differentiation
A Summary of Curve Sketching
2. The slope of f approaches as x → 0, and approaches as x → 0. Matches (C)
4. The slope is positive up to approximately x 1.5. Matches (B)
6. (a) x0, x2, x4
(b) x2, x3 (d) x1
(c) x1 (e) x2, x3
8. y y
x x2 1
y 1
1 x2 1 xx 1 0 when x ± 1. x2 12 x2 12
x 3
3
4
3 < x < 1 x 1
1 2
1 < x < 0 x0
0
0 < x < 1 1 2
x1 1 < x < 3
3
x 3 3 < x <
4
3, 3 4
)
(−1, − 12 ) 1
(0, 0)
(−
Horizontal asymptote: y 0
< x < 3
(
x
2x3 x2 y 2 0 when x 0, ± 3. x 13
y
(1, 12 )
y
y
Conclusion
Decreasing, concave down
0
Point of inflection
Decreasing, concave up
0
Relative minimum
Increasing, concave up
0
Point of inflection
Increasing, concave down
0
Relative maximum
Decreasing, concave down
0
Point of inflection
Decreasing, concave up
3, − 3 4
)
2
Section 3.6
10. y
x2 1 x2 9
12. f x
A Summary of Curve Sketching
x2 2 1 x x
y
20x 0 when x 0 x2 92
fx
2 < 0 when x 0. x2
y
60x2 3 < 0 when x 0 x2 93
f x
4 0 x3
Therefore, 0, Intercept:
1 is a relative maximum. 9
1 0, 9
Intercept: 2, 0 Vertical asymptote: x 0 Horizontal asymptote: y 1 y
Vertical asymptotes: x ± 3
5
Horizontal asymptote: y 1
y
x 1
2
3
4
x=0
y=1 x
−1 −2 −3 −4 −5
14. f x x
4 5 1
0, − 9
(
)
x=3
32 x2
16. f x
64 x 4x2 4x 16 0 when x 4. x3 x3
fx 1 f x
2
(−2, 0)
5 4 3 2
x = −3
3
y=1
Symmetric about y-axis
−5 − 4
4
192 > 0 if x 0. x4
x3 4x x 2 x2 4 x 4
fx
x2x2 12 0 x2 42
f x
8xx2 12 0 when x 0 x2 43
when x 0, ± 23
Therefore, 4, 6 is a relative minimum.
Intercept: 0, 0
3 4, 0 Intercept; 2
Relative maximum: 23, 33
Vertical asymptote: x 0
Relative minimum: 23, 33
Slant asymptote: y x
Inflection point: 0, 0 Vertical asymptotes: x ± 2
y
(−2 4, 0) 3
Slant asymptote: y x
(4, 6)
8 6
y
4
y=x
2
8 6 4
x
−8 −6
2 −4 −6
4
6
8
x=0 −8 − 6 − 4
x 4 6 8 10
411
412
Chapter 3
18. y
3 2x2 5x 5 2x 1 x2 x2
y 2 y
Applications of Differentiation
y
(4 +2 6, 7.899 (
12
3 2x2 8x 5 4 ± 6 . 0 when x 2 x 2 x 22 2
8
(
6 0 x 23
4+ 6 , − 1.899 2
(
−8
y = 2x − 1 x
−4
4
8
12
x=2
4 2 4 Relative minimum: 2
6
Relative maximum:
6
−8
, 1.8990
, 7.8990
Intercept: 0, 52 Vertical asymptote: x 2 Slant asymptote: y 2x 1 20. gx x9 x Domain: x ≤ 9
Domain: 4 ≤ x ≤ 4
22. y x16 x2
gx
36 x 0 when x 6 29 x
y
g x
3x 12 < 0 when x 6 49 x32
y
28
x2
0 when x ± 22
16 x2
2xx2 24 0 when x 0 16 x232
Relative maximum: 6, 63
Relative maximum: 22, 8
Intercepts: 0, 0, 9, 0
Relative minimum: 22, 8
Concave downward on , 9
Intercepts: 0, 0, ± 4, 0
y
(0,
10 8 6 4 0) 2
−8 − 6 − 4 −2
(6, 6
Symmetric with respect to the origin
3)
Point of inflection: 0, 0 y
(9, 0)
x
2 4 6 8 10
(−4, 0) −5
(−2
10 8 6 4 2
−3 −2 −1
2 , −8)
(2
2 , 8)
(0, 0)
(4, 0)
x
1 2 3 4 5
−6 −8 −10
24. y 3x 123 x 12 y
2 2 2x 143 2x 1 0 when x 0, 2 13 x 1 x 113
y undefined for x 1 y
y 5 4 3
2 2 < 0 for all x 1 3x 143
Concave downward on , 1 and 1, Relative maximum: 0, 2, 2, 2 Relative minimum: 1, 0 Intercepts: 0, 2, 1, 0, 1.280, 0, 3.280, 0
(2, 0)
(2, 2) x
−5 − 4 −3
−1 −2 −3 −4 −5
1 2 3 4 5
(1, 0)
Section 3.6
A Summary of Curve Sketching
26. y 13 x3 3x 2
y
y x 1 0 when x ± 1 2
2
y 2x 0 when x 0
1
(−2, 0)
(1, 0) x
y < x < 1 4
x 1
3
1 < x < 0 2
x0
3
0 < x < 1 x1
0
1 < x <
y
−1
y
Conclusion
Decreasing, concave up
0
Relative minimum
Increasing, concave up
0
Point of inflection
Increasing, concave down
0
Relative maximum
Decreasing, concave down
(
0,
(−1, − 43 (
28. f x 13 x 13 2
1 −2 3
2
(
−2
y
fx x 12 0 when x 1.
4
f x 2x 1 0 when x 1.
3
(1, 2) 2
f x < x < 1 x1
2
1< x <
fx
f x
Conclusion
Increasing, concave down
0
0
Point of inflection
Increasing, concave up
( 5) 0, 3
(−0.817, 0) x 1
30. f x x 1x 2x 5
3
3
y
fx x 1x 2 x 1x 5 x 2x 5 x2
2
12
(2 −
3, 6 3 ) (2, 0)
4x 1 0 when x 2 ± 3 . −12 −9 −6 −3
f x 6x 2 0 when x 2. f x < x < 2 3 x 2 3
63
2 3 < x < 2 x2
0
2 < x < 2 3 x 2 3 2 3 < x <
63
fx
f x
Conclusion
Increasing, concave down
0
Relative maximum
Decreasing, concave down
0
Point of inflection
Decreasing, concave up
0
Relative minimum
Increasing, concave up
Intercepts: 0, 10, 1, 0, 2, 0, 5, 0
x 3
(2 +
6
9 12
3, −6 3)
413
414
Chapter 3
Applications of Differentiation
32. y 3x4 6x2
5 3
y
y 12x3 12x 12xx2 1 0 when x 0, x ± 1. y 36x2 12 123x2 1 0 when x ± y < x < 1 x 1
43
1 < x < x
3
3
3
3
0
3 < x < 0
x0
53
0 < x < x 3
3
3
3
3
3
0
3 < x < 1
x1 1 < x <
43
3
3
(− 33 , 0(
.
y
y
Conclusion
Decreasing, concave up
0
Relative minimum
Increasing, concave up
0
Point of inflection
Increasing, concave down
0
Relative maximum
Decreasing, concave down
0
Point of inflection
Decreasing, concave up
0
Relative minimum
Increasing, concave up
−5 − 4 −3 −2 4 −1, − 3
(
34. f x x4 8x3 18x2 16x 5
f x 12x 2 48x 36 12x 3x 1 0 when x 3, x 1.
< x < 1 x1 x3
4 < x <
Conclusion
Decreasing, concave up
0
0
Point of inflection
Decreasing, concave down
16
0
Point of inflection
Decreasing, concave up
0
Relative minimum
Increasing, concave up
27
f x
3 < x < 4 x4
fx
0
1 < x < 3
(0, 53 ) ( 33 , 0( x 2 3 4 5 4 1, − 3
)−3 (
)
y
fx 4x3 24x2 36x 16 4x 4x 12 0 when x 1, x 4.
f x
7 6 5 4 3 2
4 −4
(0, 5) (1, 0) 1
(5, 0) 3
4
6
−12 −20 −28
(3, −16) (4, −27)
x 7
Section 3.6
A Summary of Curve Sketching
36. y x 15
y
y 5x 14 0 when x 1.
2
y 20x 13 0 when x 1.
1
y < x < 1 x1
0
1 < x <
(1, 0)
y
y
Conclusion
Increasing, concave down
0
0
Point of inflection
Increasing, concave up
2
3
y
2x 3x2 6x 5 2x 3x 5x 1 x2 6x 5 x 5x 1
1 −1
38. y x2 6x 5 y
x
−1
6
5
(3, 4)
4
0 when x 3 and undefined when x 1, x 5.
3 2
2x2 6x 5 2x 5x 1 y 2 undefined when x 1, x 5. x 6x 5 x 5x 1
y
y
0
Undefined
Undefined
40. y cos x
Increasing, concave down Relative maximum
Decreasing, concave down
0
Undefined
Undefined
5
x
6
Increasing, concave up
5 . y sin x sin 2x sin x1 2 cos x 0 when x 0, , , 3 3 y cos x 2 cos 2x cos x 22 cos2 x 1 1 ± 33 0.8431, 0.5931. 4 cos x cos x 2 0 when cos x 8
y 2 1
π −1 −2
Therefore, x 0.5678 or 5.7154, x 2.2057 or 4.0775.
3 , 43, 53 , 43
Relative minimum:
4
Relative minimum, point of inflection
1 cos 2x, 0 ≤ x ≤ 2 2
Relative maxima:
(5, 0) 3
Relative minimum, point of inflection
2
2
Decreasing, concave up
0
5 < x <
1
Conclusion
3 < x < 5 x5
y
4
1 < x < 3 x3
(1, 0)
< x < 1 x1
1
, 23
Inflection points: 0.5678, 0.6323, 2.2057, 0.4449, 5.7154, 0.6323, 4.0775, 0.4449
2π
x
415
416
Chapter 3
Applications of Differentiation
42. y 2x 2 cot x, 0 < x < y 2 csc2 x 0 when x
y 5 4 3 2 1
3 , 4 4
y 2 csc2 x cot x 0 when x
2
Relative maximum:
34 , 32 5
Relative minimum:
4 , 2 3
Point of inflection:
2 , 4
x
π
−1 −2 −3 −4 −5
Vertical asymptotes: x 0, 44. y sec2
8x 2 tan 8x 1, 3 < x < 3
y 2 sec2
y 5 4 3 2
8x tan 8x 8 2 sec 8x 8 0 ⇒ x 2 2
x
Relative minimum: 2, 1
−5 − 4 −3 −2 −1
3 4 5
46. gx x cot x, 2 < x < 2 gx
y 4
sin x cos x x sin2x
3 2
x 1. x→0 tan x
−2π
g0 does not exist. But lim x cot x lim x→0
π
−1
2π
x
−3 −4
3 3 ,0 , ,0 , ,0 , ,0 2 2 2 2
−π
−2
Vertical asymptotes: x ± 2 , ± Intercepts:
(2, −1)
−2 −3 −4 −5
Symmetric with respect to y-axis. Decreasing on 0, and , 2
48. f x 5
x 1 4 x 1 2
50. f x
6
4x
52. f is constant.
x2 15
f is linear.
6
f is quadratic. −9
9
−8
8
y
f −6
x 2, 4 vertical asymptote y 0 horizontal asymptote
f ''
−6
y ± 4 horizontal asymptotes
x
0, 0 point of inflection f'
Section 3.6 54.
y
A Summary of Curve Sketching
y
120 100
10
f
80
8
60
f ′′
6 4 2 x
−6 −3
3
6
9
x −4 −2 −2
12 15
2
4
6
8
10
(any vertical translate of f will do) 56.
y
y
2 1
1 x
−2
1
x −2
2
−1
1
−1
−1
−2
−2
2
(any vertical translate of the 3 segments of f will do) 58. If fx 2 in 5, 5 , then f x 2x 3 and f 2 7 is the least possible value of f 2. If fx 4 in 5, 5 , then f x 4x 3 and f 2 11 is the greatest possible value of f 2.
60. gx
3x4 5x 3 x4 1
62. gx
Vertical asymptote: none
Horizontal asymptote: y 3
x2 x 2 x1
x 2x 1 x 2, if x 1 x1 Undefined, if x 1
The rational function is not reduced to lowest terms.
7
4
−8
−6
4
6 −1 −4
The graph crosses the horizontal asymptote y 3. If a function has a vertical asymptote at x c, the graph would not cross it since f c is undefined.
64. gx
2x2 8x 15 5 2x 2 x5 x5 18
−10
20 −2
The graph appears to approach the slant asymptote y 2x 2.
hole at 1, 3
417
418
Chapter 3
Applications of Differentiation
66. f x tansin x (a)
(b) f x tansin x tansin x
3
−2
tansin x f x
2
Symmetry with repect to the origin (d) On 1, 1, there is a relative maximum at 2, tan 1 and a relative minimum at 12, tan 1. 1
−3
(c) Periodic with period 2 (e) On 0, 1, the graph of f is concave downward. 68. Vertical asymptote: x 3
70. Vertical asymptote: x 0
Horizontal asymptote: none
Slant asymptote: y x
y
x2 x3
y x
1 1 x2 x x
1 1 72. f x ax2 ax axax 2, a 0 2 2 1 fx a2x a aax 1 0 when x . a f x a2 > 0 for all x. (a) Intercepts: 0, 0,
2a, 0
(b)
y
a=2
a = −2
5
Relative minimum:
1a, 21
4
a=1
Points of inflection: none
x
−3
a = −1
2
−1
3
74. Tangent line at P: y y0 fx0x x0 (a) Let y 0: y0 fx0x x0
(b) Let x 0: y y0 fx0x0
fx0x x0 fx0 y0 x x0 x-intercept:
y0 f x0 x0 fx0 fx0
f x0 x0 ,0 fx0
(c) Normal line: y y0 Let y 0: y0
y y0 x0 fx0 y f x0 x0 fx0 y-intercept: 0, f x0 x0 fx0
(d) Let x 0: y y0
1 x x0 fx0
1 x x0 fx0
y-intercept:
x x0 y0 fx0 x0 f x0 fx0
0
(f) PC 2 y02
f x0 f x0 x0 fx0 fx0
0, y
x0 fx0
ffxx f x ffxx f x 0
0
0
f x01 fx0 2 PC 2 fx0
x-intercept: x0 f x0 fx0, 0
x0 fx0
y y0
y0 fx0 x x0
(e) BC x0
1 x0 fx0
(g) AB x0 x0 f x0 fx0 f x0 fx0
x0 y0 AP f x01 fx0 2
(h) AP f x0 2
2 f
2
2
2
0
2
0
0
2
2
Section 3.7
Section 3.7
Optimization Problems
Optimization Problems
2. Let x and y be two positive numbers such that x y S. P xy xS x Sx x2
4. Let x and y be two positive numbers such that xy 192. S x 3y
S dP S 2x 0 when x . dx 2
192 3y y
192 dS 3 2 0 when y 8. dy y
S d 2P 2 < 0 when x . dx 2 2
d 2S 384 3 > 0 when y 8. dy 2 y
P is a maximum when x y S2.
S is minimum when y 8 and x 24.
6. Let x and y be two positive numbers such that x 2y 100. P xy y100 2y 100y 2y2 dP 100 4y 0 when y 25. dy d 2P 4 < 0 when y 25. dy 2 P is a maximum when x 50 and y 25. 8. Let x be the length and y the width of the rectangle. 2x 2y P y
P 2x P x 2 2
A xy x
P2 x P2 x x
y
2
x
P dA P 2x 0 when x . dx 2 4 d 2A P 2 < 0 when x . dx2 4 A is maximum when x y P4 units. (A square!) 10. Let x be the length and y the width of the rectangle. xy A y
A x
P 2x 2y 2x 2
y
Ax 2x 2Ax
2A dP 2 2 0 when x A. dx x d 2P 4A 3 > 0 when x A . dx2 x P is minimum when x y A centimeters. (A square!)
419
x
420
Chapter 3
Applications of Differentiation 14. f x x 12, 5, 3
12. f x x 8, 2, 0
d x 52 x 12 3 2
From the graph, it is clear that 8, 0 is the closest point on the graph of f to 2, 0.
x2 10x 25 x2 2x 22
y
x2 10x 25 x4 4x3 8x 4 x4 4x3 x2 18x 29
4
Since d is smallest when the expression inside the radical is smallest, you need to find the critical numbers of
2
gx x4 4x3 x2 18x 29
x 2
4
6
8
10
12
gx 4x3 12x2 2x 18 2x 12x2 8x 9 By the First Derivative Test, x 1 yields a minimum. Hence, 1, 4 is closest to 5, 3.
16.
F
v 22 0.02v 2
18. 4x 3y 200 is the perimeter. (see figure)
dF 22 0.02v 2 dv 22 0.02v 2 2
A 2xy 2x
200 3 4x 38 50x x 2
dA 8 50 2x 0 when x 25. dx 3
0 when v 1100 33.166. By the First Derivative Test, the flow rate on the road is maximized when v 33 mph.
d 2A 16 < 0 when x 25. dx2 3 A is a maximum when x 25 feet and y
100 3
y x
20. (a)
(c)
Height, x
Length & Width
Volume
1
24 21
124 21 2 484
2
24 22
224 22 2 800
3
24 23
324 23 2 972
4
24 24
424 24 2 1024
5
24 25
524 25 2 980
6
24 26
624 26 2 864
x
(b) V x24 2x2, 0 < x < 12 (d)
1200
0
12 0
The maximum volume seems to be 1024.
dV 2x24 2x2 24 2x2 24 2x24 6x dx 1212 x4 x 0 when x 12, 4 12 is not in the domain. 2V
d 122x 16 dx2 d 2V < 0 when x 4. dx2 When x 4, V 1024 is maximum.
feet.
Section 3.7 22. (a) P 2x 2 r 2x 2
y 2
2y
Optimization Problems
y
x
2x y 200 ⇒y
200 2x 2 100 x
(b) Length, x
Width, y
Area, xy
10
2 100 10
2 10 100 10 573
20
2 100 20
20
2 100 20 1019
30
2 100 30
30
2 100 30 1337
40
2 100 40
40
2 100 40 1528
50
2 100 50
50
2 100 50 1592
60
2 100 60
60
2 100 60 1528
The maximum area of the rectangle is approximately 1592 m2. (c) A xy x (e)
2 2 100 x 100x x2
2000
0
100 0
Maximum area is approximately 1591.55 m2 x 50 m.
24. You can see from the figure that A xy and y
6x . 2
y 6
6x 1 6x x2. Ax 2 2
5 4
y=
6−x 2
3
dA 1 6 2x 0 when x 3. dx 2 d 2A 1 < 0 when x 3. dx2 A is a maximum when x 3 and y 32.
2
( x, y )
1 x 1
2
3
4
5
6
421
422
Chapter 3
26. (a)
A
Applications of Differentiation
1 base height 2 1 216 h2 4 h 2
4 4
h
16 h24 h
16 − h 2
dA 1 16 h2122h4 h 16 h212 dh 2 16 h212h4 h 16 h2
2h2 2h 8 2h 4h 2 16 h2 16 h2
dA 0 when h 2, which is a maximum by the First Derivative Test. dh Hence, the sides are 216 h2 43, an equilateral triangle. Area 123 sq. units. (b) cos tan
4h 84 h
4 h 8
16 h2
Area 2
8
α
4+h
4
4h
4
h
1216 h 4 h 2
16 − h 2
4 h2 tan 64 cos4 tan A 64cos4 sec2 4 cos3 sin tan 0 ⇒ cos4 sec2 4 cos3 sin tan 1 4 cos sin tan 1 sin2 4 sin
1 ⇒ 30 and A 123. 2
(c) Equilateral triangle
28. A 2xy 2xr 2 x 2 see figure 2r dA 2r2 2x2 0 when x . dx 2 r2 x2
By the First Derivative Test, A is maximum when the rectangle has dimensions 2r by 2r2.
y
(−x,
r 2 − x2
(−r, 0)
( ( x,
r 2 − x2
( x
(r, 0)
Section 3.7 30. xy 36 ⇒ y
Optimization Problems
36 x
x+3 x
36 A x 3y 3 x 3 3 x
y
y+3
108 36 3x 9 x dA 108 3 0 ⇒ 3x2 108 ⇒ x 6, y 6 dx x2 Dimensions: 9 9
32. V r 2h V0 cubic units or h
S 2 r 2 2 rh 2 r 2
V0 r
V dS 2 2r 20 0 when r dr r h
V0 r2
2V units. 0
3
V0 V0223 2V013 3 V 2 2 V 23 213 2r 0 0
By the First Derivative Test, this will yield the minimum surface area. V r 2x
34.
x 2r 108 ⇒ x 108 2 r V
see figure
108 2 r 108r 2 2r3
r 2
r
x
dV 216r 6 r 2 6 r36 r dr 0 when r
36 and x 36.
36 d 2V 216 12 r < 0 when r . dr 2 Volume is maximum when x 36 inches and r 36 11.459 inches. 36.
V x 2h x 2 2r2 x2 2 x 2r 2 x2
see figure
dV 1 2 2 x2 r x 2122x 2xr 2 x2 dx 2
2 x r2 x2
6r 2r 2 ⇒x . 3 3
By the First Derivative Test, the volume is a maximum when 6r
3
and h
2r 3
.
Thus, the maximum volume is V
23 r 2r3 43r3 . 3
2
r2 − x2
(
(x, −
r2 − x2
(
x r
h
2r 2 3x2
0 when x 0 and x 2
x
(x,
423
424
Chapter 3
Applications of Differentiation
38. No. The volume will change because the shape of the container changes when squeezed. 4 40. V 3000 r 3 r 2h 3 3000 4 r r2 3
h
Let k cost per square foot of the surface area of the sides, then 2k cost per square foot of the hemispherical ends.
C 2k4 r 2 k2 rh k 8 r 2 2 r
6000 32 dC k r 2 0 when r dr 3 r
16 4 r k r 3000 r 3 3
2
2
6000 r
5.636 feet and h 22.545 feet. 1125 2 3
By the Second Derivative Test, we have
d 2C 12,000 32 k dr 2 3 r3
. > 0 when r 1125 2 3
Therefore, these dimensions will produce a minimum cost. 42. (a) Let x be the side of the triangle and y the side of the square. 3 4 A cot x 2 cot y 2 where 3x 4y 20 4 3 4 4
A
x
3
4
43 0
3 x2 5 x 2 4
A
5 4 cot x 2 cot y 2 where 4x 5y 20 4 4 4 5
4 2 x2 1.7204774 4 x , 0 ≤ x ≤ 5. 5
3 20 x2 5 x , 0 ≤ x ≤ . 4 3
3
(b) Let x be the side of the square and y the side of the pentagon.
2
A 2x 2.75276384 4
4 x 0 5
x 2.62 When x 0, A 27.528, when x 2.62, A 13.102, and when x 5, A 25. Area is maximum when all 20 feet are used on the pentagon.
60 43 9
When x 0, A 25, when x 60 43 9, A 10.847, and when x 203, A 19.245. Area is maximum when all 20 feet are used on the square. (c) Let x be the side of the pentagon and y the side of the hexagon. A
6 5 cot x 2 cot y 2 where 5x 6y 20 4 5 4 6
2 3 20 5x 2 5 cot x 3 , 0 ≤ x ≤ 4. 4 5 2 6
5 5 A cot x 33 2 5 6
20 5x 0 6
x 2.0475 When x 0, A 28.868, when x 2.0475, A 14.091, and when x 4, A 27.528. Area is maximum when all 20 feet are used on the hexagon
(d) Let x be the side of the hexagon and r the radius of the circle. A
2 6 cot x r 2 where 6x 2 r 20 4 6
10 3x 2 10 33 2 x ,0 ≤ x ≤ . 2 3
A 33 6
10 3x 0
x 1.748 When x 0, A 31.831, when x 1.748, A 15.138, and when x 103, A 28.868. Area is maximum when all 20 feet are used on the circle. In general, using all of the wire for the figure with more sides will enclose the most area.
Section 3.7 44. Let A be the amount of the power line.
Optimization Problems
425
y
A h y 2x2 y2
(0, h) h−y
x 2y dA 0 when y . 1 dy x2 y2 3
y
2x2 x d 2A 2 > 0 for y . 2 dy x y232 3
x
(−x, 0)
(x, 0)
The amount of power line is minimum when y x3 . 46. f x 12 x2 (a)
1 4 gx 16 x 12 x2 on 0, 4
(c) fx x, Tangent line at 22, 4 is
9
y 4 22 x 22
f g −1
y 22x 4.
4
gx 4 x3 x, Tangent line at 22, 0 is 1
−3
1 y 0 4 22 3 22 x 22
1 4 1 4 (b) dx f x gx 12 x2 16 x 12 x2 x2 16 x
y 22x 8.
dx 2x 14 x3 0 ⇒ 8x x3 ⇒ x 0, 22 in 0, 4 The maximum distance is d 4 when x 22.
The tangent lines are parallel and 4 vertical units apart. (d) The tangent lines will be parallel. If dx f x gx, then dx 0 fx gx implies that fx gx at the point x where the distance is maximum.
48. Let F be the illumination at point P which is x units from source 1. F
kI2 kI1 x2 d x2
2kI2 2kI 2kI2 dF 2kI1 0 when 3 1 . dx x3 d x3 x d x3 3 I 1 3 I 2
x dx
3 I x 3 I d x 1 2 3 I x 3 I d 3 I1 1 2
x
3 I d 1 3 I 2
3 I 1
3 6kI2 d I1 d 2F 6kI1 4 > 0 when x 3 . 2 4 3 I dx x d x I1 2
This is the minimum point.
50. (a) T θ
x2 4
2
3 x 4
x
x2 + 4
x2 4
2 x
dT x 1 0 dx 2x2 4 4
(b)
3−x Q
1 2
2x2 x2 4 x2 4 x2 T2 2
—CONTINUED—
1 hours 4
426
Chapter 3
Applications of Differentiation
50. —CONTINUED— T
(c)
x2 4
v1
3 x v2
(d) Cost x2 4 C1 3 xC2
dT 1 x 0 dx v1x2 4 v2
T
3 x 1C2 1C1 C2 1C2 C1
x2 d12
v1
v1 only. v2
d22 a x2
54. Cx 2kx2 4 k4 x
v2
Cx
xa dT x 0 dx v1x2 d12 v2d22 a x2
3x2 4 x
we have sin 1 sin 2 sin 1 sin 2 0⇒ . v1 v2 v1 v2
2 3
Or, use Exercise 50(d): sin
Since 2
Thus, x
2
d1 d2 d > 0 d x 2 v1x2 d1232 v2d22 a x2 32
θ
1 1 V r 2h r 2144 r 2 3 3
dV 1 1 r2 144 r 2122r 2r144 r 2 dr 3 2
1 288r 3r3 r96 r 2 0 when r 0, 46. 3 144 r 2 144 r 2
By the First Derivative Test, V is maximum when r 46 and h 43. Area of circle: A 122 144 Lateral surface area of cone: S 46 46 2 43 2 486 Area of sector: 144 486
.
x2 + 4 x
2 3
2
this condition yields a minimum time.
k0
4x2 x2 4
x xa sin 1 and sin 2 2 2 2 x d1 d2 a x2
2T
2xk x2 4
2x x2 4
Since
56.
v1 v2
depends on
52.
1C1
From above, sin
v x 1 x2 4 v2 sin
x2 4
1 2
r 72
2
144 486 2 3 6 1.153 radians or 66 72 3
4−x
C2 1 ⇒ 30. C1 2
Section 3.7
Optimization Problems
58. Let d be the amount deposited in the bank, i be the interest rate paid by the bank, and P be the profit. P 0.12 d id d ki 2 since d is proportional to i 2 P 0.12ki 2 iki 2 k0.12i 2 i3 0.24 dP k0.24i 3i 2 0 when i 0.08. di 3 d 2P k0.24 6i < 0 when i 0.08 Note: k > 0. di 2 The profit is a maximum when i 8%. P
60.
(a)
1 3 s 6s 2 400 10
3 3 dP s 2 12s ss 40 0 when x 0, s 40. ds 10 10 3 d 2P s 12 ds 2 5 d 2P 0 > 0 ⇒ s 0 yields a minimum. ds 2 d 2P 40 < 0 ⇒ s 40 yields a maximum. ds2 The maximum profit occurs when s 40, which corresponds to $40,000 P $3,600,000.
(b)
3 d 2P s 12 0 when s 20. ds 2 5 The point of diminishing returns occurs when s 20, which corresonds to $20,000 being spent on advertising.
62. S2 4m 1 5m 6 10m 3
Using a graphing utility, you can see that the minimum occurs when m 0.3. Line y 0.3x
S2 40.3 1 50.3 6 100.3 3 4.7 mi. S2
30
20
10
(0.3, 4.5) m 1
2
3
427
428
Chapter 3
Applications of Differentiation
h
x and cos . The area A of the 2 r 2 r cross equals the sum of two large rectangles minus
64. (a) Label the figure so that r2 x2 h2. Then, the area A is 8 times the area of the region given by OPQR:
(b) Note that sin
the common square in the middle.
h
A 22x2h 4h2 8xh 4h2
θ 2
R x
O
8r2 sin
h Q
P r
cos 4r2 sin2 2 2 2
4r2 sin sin2
12h
A8
2
cos sin
x2 x r2 x2r2 x2
2
A 4r2 cos sin
x hh
12r
8
2
cos 0 2 2
cos 2 sin
2 2
tan 2
8xr2 x2 4x2 4r2 Ax 8r2 x2
arctan2 1.10715 or 63.4
8x2 8x 0 r2 x2
8x2 8x 8r2 x2 r2 x2 x2 xr2 x2 r2 x2 2x2 r2 xr2 x2 4x4 4x2r2 r4 x2r2 x2 5x4 5x2r2 r4 0 Quadratic in x2. 5r2 ± 25r4 20r4 r2 5 ± 5 . 10 10 Take positive value. x2
xr
5 10
5
(c) Note that x2
0.85065r
Critical number
r2 r2 5 5 and r2 x2 5 5. 10 10
Ax 8xr2 x2 4x2 4r2
10r 5 5 10r 5 5
8
2
10r 20
8
4
12
2
12
4
r2 5 5 4r2 10
2 2r2 5r2 4r2 5
8 25 2 r25 2r2 r 5 5
5 4 2r2 5 1 2r25 1 5 5
Using the angle approach, note that tan 2, sin
Thus, A 4r2 sin sin2
2
2 1 1 4r2 1 5 2 5
4r25 1 2r25 1 2
2 5
and sin2
2 211 cos 211 15.
Section 3.8
Section 3.8
Newton’s Method
Newton’s Method
2. f x 2x2 3 fx 4x x1 1
f xn fxn
n
xn
f xn
fxn
1
1
1
4
0.125
5.0
0.025
xn
1 4
f xn fxn 5 4
2
5 1.25 4
fx sec2 x
n
xn
f xn
fxn
f xn fxn
x1 0.1
1
0.1000
0.1003
1.0101
0.0993
0.0007
2
0.0007
0.0007
1.0000
0.0007
0.0000
4. f x tan x
6. f x x5 x 1 fx 5x 1 4
Approximation of the zero of f is 0.755.
Approximation of the zero of f is 4.8284.
xn
xn
f xn
fxn
f xn fxn
1
0.5000
0.4688
1.3125
0.3571
0.8571
2
0.8571
0.3196
3.6983
0.0864
0.7707
3
0.7707
0.0426
2.7641
0.0154
0.7553
4
0.7553
0.0011
2.6272
0.0004
0.7549
Approximation of the zero of f is 0.7937.
xn
n
xn
f xn
fxn
f xn fxn
1
5
0.1010
0.5918
0.1707
4.8293
2
4.8293
0.0005
0.5858
.00085
4.8284
n
xn
f xn
fxn
f xn fxn
1
1
1
6
0.1667
0.8333
2
0.8333
0.1573
4.1663
0.0378
0.7955
3
0.7955
0.0068
3.7969
0.0018
0.7937
4
0.7937
0.0000
3.7798
0.0000
0.7937
10. f x 1 2x3 fx 6x2
f xn fxn
n
8. f x x 2x 1 1 fx 1 x 1
1.225
xn
f xn fxn
xn
f xn fxn
f xn fxn
429
430
Chapter 3
Applications of Differentiation
1 12. f x x 4 3x 3 2 fx 2x3 3 Approximation of the zero of f is 0.8937.
Approximation of the zero of f is 2.0720.
14. f x x3 cos x fx 3x2 sin x Approximation of the zero of f is 0.866.
16. hx f x gx 3 x hx 1
n
xn
f xn
fxn
f xn fxn
1
1
0.5
5
0.1
0.9
2
0.9
0.0281
4.458
0.0063
0.8937
3
0.8937
0.0001
4.4276
0.0000
0.8937
xn
f xn fxn
n
xn
f xn
fxn
f xn fxn
1
2
1
13
0.0769
2.0769
2
2.0769
0.0725
14.9175
0.0049
2.0720
3
2.0720
0.0003
14.7910
0.0000
2.0720
xn
f xn fxn
n
xn
f xn
fxn
f xn fxn
1
0.9000
0.1074
3.2133
0.0334
0.8666
2
0.8666
0.0034
3.0151
0.0011
0.8655
3
0.8655
0.0001
3.0087
0.0000
0.8655
1 x2 1
2x x2 12
xn
f xn fxn
n
xn
h xn
hxn
h xn hxn
1
2.9000
0.0063
0.9345
0.0067
2.8933
2
2.8933
0.0000
0.9341
0.0000
2.8933
n
xn
h xn
hxn
h xn hxn
1
0.8000
0.0567
2.3174
0.0245
0.8245
2
0.8245
0.0009
2.3832
0.0004
0.8241
xn
h xn hxn
Point of intersection of the graphs of f and g occurs when x 2.893. 18. hx f x gx x2 cos x hx 2x sin x One point of intersection of the graphs of f and g occurs when x 0.824. Since f x x2 and gx cos x are both symmetric with respect to the y-axis, the other point of intersection occurs when x 0.824. 20. f x xn a 0 fx nxn1 xi1 xi
xin a nxin1
nxin xin a n 1xin a nxin1 nxin1
xn
h xn hxn
Section 3.8
22. xi1
xi2 5 2xi
24. xi1
Newton’s Method
431
2xi3 15 3xi2
i
1
2
3
4
i
1
2
3
4
xi
2.0000
2.2500
2.2361
2.2361
xi
2.5000
2.4667
2.4662
2.4662
3 15 2.466
5 2.236
26. f x tan x fx sec2 x
n
xn
f xn
fxn
f xn fxn
Approximation of the zero: 3.142
1
3.0000
0.1425
1.0203
0.1397
3.1397
2
3.1397
0.0019
1.0000
0.0019
3.1416
3
3.1416
0.0000
1.0000
0.0000
3.1416
y x1
3 2
f xn fxn
30. f x 2 sin x cos 2x
28. y 4x3 12x2 12x 3 f x 12x2
xn
fx 2 cos x 2 sin 2x
24x 12 fx
x1
Fails because fx1 0.
fx2 0; therefore, the method fails. n
xn
f xn
fxn
f xn fxn
1
3 2
3 2
3
2
1
1
0
3 2
f xn fxn
n
xn
f xn
fxn
1 2
1
1
3 2
3
0
—
—
g xn gxn
xn
32. Newton’s Method could fail if fc 0, or if the initial value x1 is far from c. 34. Let gx f x x cot x x gx csc2 x 1. The fixed point is approximately 0.86.
g xn gxn
n
xn
g xn
gxn
1
1.0000
0.3579
2.4123
0.1484
0.8516
2
0.8516
0.0240
2.7668
0.0087
0.8603
3
0.8603
0.0001
2.7403
0.0000
0.8603
xn
36. f x sin x, fx cos x (a)
(d)
2
y 2
−
−2
(b) x1 1.8 x2 x1
(3, 0.141) (6.086, 0)
1
π 2 −1
π
x
(3.143, 0)
−2
f x1 6.086 fx1
(c) x1 3 x2 x1
(1.8, 0.974)
f x1 3.143 fx1
The x-intercepts correspond to the values resulting from the first iteration of Newton’s Method. (e) If the initial guess x1 is not “close to” the desired zero of the function, the x-intercept of the tangent line may approximate another zero of the function.
432
Chapter 3
Applications of Differentiation (b) xn1 xn2 11xn
38. (a) xn1 xn2 3xn
1 3
i
1
2
3
4
i
1
2
3
4
xi
0.3000
0.3300
0.3333
0.3333
xi
0.1000
0.0900
0.0909
0.0909
1 11
0.333
0.091
40. f x x sin x, 0,
y
fx x cos x sin x 0
4 3
Letting Fx fx, we can use Newton’s Method as follows.
Fx 2 cos x x sin x n
xn
1
2.0000
2
2.0290
1
Fxn
F xn Fxn
0.0770
2.6509
0.0290
2.0290
0.0007
2.7044
0.0002
2.0288
F xn
(2.029, 1.820)
2
xn
F xn Fxn
π 4
π 2
x
3π 4
Approximation to the critical number: 2.029 42. y f x x2, 4, 3
y
d x 42 y 32 x 42 x2 32 x4 7x2 8x 25
3 2
d is minimum when D x4 7x2 8x 25 is minimum.
(0.529, 0.280)
1
x
gx D 4x3 14x 8
−2 −1 −1
gx 12x2 14
1
2
3
4
−2 −3
n
xn
g xn
gxn
g xn gxn
1
0.5000
0.5000
17.0000
0.0294
0.5294
2
0.5294
0.0051
17.3632
0.0003
0.5291
3
0.5291
0.0001
17.3594
0.0000
0.5291
xn
(4, − 3)
g xn gxn
x 0.529 Point closest to 4, 3 is approximately 0.529, 0.280.
44. Maximize: C C
3t 2 t 50 t3 3t4 2t3 300t 50 0 50 t32
n
xn
f xn
fxn
f xn fxn
1
4.5000
12.4375
915.0000
0.0136
4.4864
2
4.4864
0.0658
904.3822
0.0001
4.4863
Let f x 3t4 2t3 300t 50 fx 12t3 6t2 300. Since f 4 354 and f 5 575, the solution is in the interval 4, 5. Approximation: t 4.486 hours
xn
f xn fxn
Section 3.8
Newton’s Method
46. 170 0.808x3 17.974x2 71.248x 110.843, 1 ≤ x ≤ 5 Let f x 0.808x3 17.974x2 71.248x 59.157 fx 2.424x2 35.948x 71.248. From the graph, choose x1 1 and x1 3.5. Apply Newton’s Method. n
xn
f xn
fxn
f xn fxn
1
1.0000
5.0750
37.7240
0.1345
1.1345
2
1.1345
0.2805
33.5849
0.0084
1.1429
3
1.1429
0.0006
33.3293
0.0000
1.1429
n
xn
f xn
fxn
f xn fxn
1
3.5000
4.6725
24.8760
0.1878
3.6878
2
3.6878
0.3286
28.3550
0.0116
3.6762
3
3.6762
0.0009
28.1450
0.0000
3.6762
xn
f xn fxn
xn
f xn fxn
The zeros occur when x 1.1429 and x 3.6762. These approximately correspond to engine speeds of 1143 revmin and 3676 revmin. 48. True
50. True
52. f x 4 x2 sinx 2 Domain: 2, 2 x 2 and x 2 are both zeros. fx 4 x2 cosx 2 Let x1 1.
x 4 x2
sinx 2
n
xn
f xn
fxn
f xn fxn
1
1.0000
0.2444
1.7962
0.1361
2
1.1361
0.0090
1.6498
0.0055
1.1416
3
1.1416
0.0000
1.6422
0.0000
1.1416
Zeros: x ± 2, x 1.142 y 1 x −2
1
−2 −3
2
xn
f xn fxn
1.1361
433
434
Chapter 3
Applications of Differentiation
Section 3.9
Differentials
2. f x 6 6x2 x2 fx 12x3
x
y
1.99
2
2.01
2.1
6 x2
1.6620
1.5151
1.5
1.4851
1.3605
9 3 Tx x 2 2
1.65
1.515
1.5
1.485
1.35
f x
12 x3
Tangent line at 2,
1.9
3 : 2
3 12 3 x 2 x 2 2 8 2 9 3 y x 2 2
4. f x x fx
1 2x
x
1.9
1.99
2
2.01
2.1
f x x
1.3784
1.4107
1.4142
1.4177
1.4491
1.3789
1.4107
1.4142
1.4177
1.4496
T x
Tangent line at 2, 2 :
x 2
1 2
y f 2 f2x 2 y 2
1 x 2 22
y
x 1 22 2
6. f x csc x fx csc x cot x Tangent line at 2, csc 2: y f 2 f2x 2 y csc 2 csc 2 cot 2x 2 y csc 2 cot 2x 2 csc 2 x
1.9
1.99
2
2.01
2.1
f x csc x
1.0567
1.0948
1.0998
1.1049
1.1585
T x csc 2 cot 2x 2 csc 2
1.0494
1.0947
1.0998
1.1048
1.1501
8. y f x 1 2x2, fx 4x, x 0, x dx 0.1 y f x x f x
dy fx dx
f 0.1 f 0
f00.1
1 20.1 1 20 0.02 2
2
00.1 0
10. y f x 2x 1, fx 2, x 2, x dx 0.01 y f x x f x
dy fx dx
f 2.01 f 2
f20.01
22.01 1 22 1 0.02
20.01 0.02
Section 3.9 12.
y 3x2 3
14. y 9 x2
dy 2x1 3dx
16.
2 dx x1 3
dy
1
y x
18.
x
dy
1
sec2 x x2 1
x
2
x 1 9 x21 22xdx dx 9 x2 2
y x sin x dy x cos x sin x dx
1 x1 dx dx dy 2x 2xx 2xx
20. y
Differentials
22. (a) f 1.9 f 2 0.1 f 2 f20.1 1 10.1 1.1
12 sec2 x tan x sec2 x2x dx x2 12
(b) f 2.04 f 2 0.04 f 2 f20.04 1 10.04 0.96
x tan x x
2 sec xx tan dx x 1 2
2
2
2
24. (a) f 1.9 f 2 0.1 f 2 f20.1
26. (a) g2.93 g3 0.07 g3 g30.07
1 00.1 1
8 30.07 7.79
(b) f 2.04 f 2 0.04 f 2 f20.04
(b) g3.1 g3 0.1 g3 g30.1
1 00.04 1 28. (a) g2.93 g3 0.07 g3 g30.07 8 50.07 7.65 (b) g3.1 g3 0.1 g3 g30.1 8 50.1 8.5
8 30.1 8.3 30.
1 A 2 bh, b 36, h 50
db dh ± 0.25 dA 12 b dh 12 h db A dA 2 36± 0.25 2 50± 0.25 1
1
± 10.75 square centimeters 32.
x 12 inches
34. (a)
x dx ± 0.03 inch
C 56 centimeters C dC ± 1.2 centimeters
(a) V x3
C 2 r ⇒ r
dV 3x2 dx 3122± 0.03
A r2
± 12.96 cubic inches (b) S 6x2
dA
dS 12x dx 1212± 0.03
C 2
2C
2
1 2 C 4
1 1 33.6 C dC 56± 1.2 2 2
33.6 dA 0.042857 4.2857% A 1 4562
± 4.32 square inches
(b)
dA 1 2C dC 2dC ≤ 0.03 A 1 4C2 C dC 0.03 ≤ 0.015 1.5% C 2
435
436
36.
Chapter 3
Applications of Differentiation
P 500x x2
12x
2
77x 3000 , x changes from 115 to 120
dP 500 2x x 77dx 577 3x dx 577 3115120 115 1160 1160 dP 100 100 2.7% P 43517.50
Approximate percentage change:
38. V 43 r3, r 100 cm, dr 0.2 cm
E IR
40.
V dV 4 r dr 4 100 0.2 8000 cm 2
2
3
R
E I
dR
E dI I2
dR E I 2 dI dI R E I I
dR dI dI R I I
1 1 A baseheight 9.5cot 9.5 45.125 cot 2 2 dA 45.125
h 50 tan
44.
42. See Exercise 41.
csc2
0.0044 sin 0.4669cos 0.4669
0.0109 1.09% in radians 3 x, x 27, dx 1 46. Let f x 3 x f x x f x fxdx
3 26 3 27
d
dh 50 1.2479 d ≤ 0.06 x 50 tan1.2479
csc2 d d dA A cot sin cos 0.25 sin 26.75cos 26.75
dh 50 sec2
d
71.5 1.2479 radians
1 dx 3 x2 3
1 1 1 3 2.9630 3 272 3 27
3 26 2.9625 Using a calculator,
48. Let f x x3, x 3, dx 0.01. f x x f x fx dx x3 3x2 dx f x x 2.993 33 3320.01 27 0.27 26.73 Using a calculator: 2.993 26.7309
sec2
9.9316 d ≤ 0.06 2.9886
d ≤ 0.018
h
θ
50 ft
Review Exercises for Chapter 3 50. Let f x tan x, x 0, dx 0.05, fx sec2 x.
52. Propagated error f x x f x,
Then
relative error
f 0.05 f 0 f0dx
437
dy dy , and the percent error 100. y y
tan 0.05 tan 0 sec2 00.05 0 10.05.
54. True,
y dy a x dx
56. False Let f x x, x 1, and x dx 3. Then y f x x f x f 4 f 1 1 and dy fx dx
1 3 3 . 21 2
Thus, dy > y in this example.
Review Exercises for Chapter 3 2. (a) f 4 f 4 3 (c)
(b) f 3 f 3 4 4
y
(d) Yes. Since f 2 f 2 1 1 and f 1 f 1 2, the Mean Value says that there exists at least one value c in 2, 1 such that
6 4
x −6 −4 −2
4
fc
6
−4
(e) No, lim f x exists because f is continuous at 0, 0.
−6
x →0
At least six critical numbers on 6, 6. 4. f x
f 1 f 2 2 1 1. 1 2 12
x x2 1
, 0, 2
(f) Yes, f is differentiable at x 2.
6. No. f is not differentiable at x 2.
1 fx x x2 1322x x2 112 2 1 x2 132 No critical numbers Left endpoint: 0, 0 Minimum Right endpoint: 2, 25 Maximum
8. No; the function is discontinuous at x 0 which is in the interval 2, 1 .
10.
1 f x , 1 ≤ x ≤ 4 x fx
1 x2
f b f a 14 1 34 1 ba 41 3 4 fc
1 1 c2 4
c2
12.
f x x 2x, 0 ≤ x ≤ 4 fx
1 2x
2
f b f a 6 0 3 ba 40 2 fc
1 3 2 2 2c
c1
438
Chapter 3
Applications of Differentiation
f x 2x2 3x 1
14.
fx 4x 3 f b f a 21 1 5 ba 40 fc 4c 3 5 c 2 Midpoint of 0, 4 16. gx x 13
Critical number: x 1
18. f x sin x cos x, 0 ≤ x ≤ 2
Critical numbers: x
gx
5 ,x 4 4
1 < x <
Sign of gx
gx > 0
gx > 0
Conclusion
Increasing
Increasing
0 < x <
Interval
fx cos x sin x
20. gx
< x < 1
Interval
gx 3x 12
4
5 < x < 4 4
5 < x < 2 4
Sign of fx
fx > 0
fx < 0
fx > 0
Conclusion
Increasing
Decreasing
Increasing
3 x sin 1 , 0, 4 2 2
Test Interval
3 x 1 cos 2 2 2
Sign of gx
gx > 0
gx < 0
gx > 0
Conclusion
Increasing
Decreasing
Increasing
0 when x 1
2 2 ,3
Relative maximum:
1 2 , 32
Relative minimum:
3 2 , 23
22. (a) y A sinkm t B coskm t
A sinkm t A ⇒ tankm t . B coskm t B
Therefore, sin km t cos km t
A A2 B2
B A2 B2
.
When v y 0, yA
A A B B A B B A
2
2
2
2
2
(b) Period:
y Akm coskm t Bkm sinkm t 0 when
0 < x < 1
2
B2.
1
2 2 < x < 3
3
2 km
Frequency:
1 1 km 2 km 2
2 < x < 4
Review Exercises for Chapter 3 24. f x x 22x 4 x3 12x 16 fx 3x2 12 f x 6x 0 when x 0.
439
Test Interval
< x < 0
Sign of f x
f x < 0
f x > 0
Concave downward
Concave upward
Conclusion
0 < x <
Point of inflection: 0, 16 26.
ht t 4t 1 Domain: 1, ht 1
2 t 1
y
28. 7
0 ⇒ t3
6 5 4
1 h t t 132 h 3
30.
C
1 > 0 8
3, 5 is a relative minimum.
Qx s 2x r
r Qs dC 2 0 dx x 2
3 2 1 x
−1
1
2
3
4
5
6
7
32. (a) S 0.1222t3 1.3655t2 0.9052t 4.8429 (b)
r Qs x2 2
25
1
14 0
2Qs x2 r x
34. lim
x →
2Qs r
2x 2x lim 0 3x2 5 x → 3 5x2
38. gx
5x2 2
x2
lim
x →
5x2 5 lim 5 2 x → 1 2x2
x2
(c) St 0 when t 3.7. This is a maximum by the First Derivative Test. (d) No, because the t3 coefficient term is negative.
36. lim
3x
x →
x2 4
40. f x
lim
x →
3 1 4x2
3
3x x2 2
lim
x →
3x x2 2
Horizontal asymptote: y 5
lim
x →
lim
x →
lim
x →
3x x2 2
3xx x2 2x2
3 1 2x2
lim
x →
lim
x →
3
3xx
x2 2 x2
3 3 1 2x2
Horizontal asymptotes: y ± 3
440
Chapter 3
Applications of Differentiation
2 4 cos x cos 2x 3
44. gx
42. f x x3 3x2 2x xx 1x 2 Relative minima: (0, 0, 1, 0, 2, 0
Relative minima: 2 k, 0.29 where k is any integer.
Relative maxima: 1.577, 0.38, 0.423, 0.38
Relative maxima: 2k 1 , 8.29 where k is any integer.
3
10
−2
4 − 5 2
−1
46. f x 4x3 x4 x34 x
y 30
Domain: , ; Range: , 27
25 20
fx 12x 4x 4x 3 x 0 when x 0, 3. 2
3
5 2
0
2
15 10
f x 24x 12x 2 12x2 x 0 when x 0, 2.
5 x
f 3 < 0
−2
1
2
3
5
Therefore, 3, 27 is a relative maximum. Points of inflection: 0, 0, 2, 16 Intercepts: 0, 0, 4, 0 48. f x x2 42
y
Domain: , ; Range: 0,
24 20
fx 4xx2 4 0 when x 0, ± 2. 23 f x 43x2 4 0 when x ± . 3 f 0 < 0
(0, 16) (−2, 0)
(2, 0) 8 4 x
−3 −2 −1
Therefore, 0, 16 is a relative maximum.
1
2
3
f ± 2 > 0 Therefore, ± 2, 0 are relative minima. Points of inflection: ± 233, 649 Intercepts: 2, 0, 0, 16, 2, 0 Symmetry with respect to y-axis 50. f x x 3x 23 Domain: , ; Range:
y
16,875 256 ,
−4
fx x 33x 22 x 23 4x 7x 22 0 when x 2,
f 74 > 0 Therefore, 74 , 16,875 256 is a relative minimum. Points of inflection: 2, 0, 12 , 625 16 Intercepts: 2, 0, 0, 24, 3, 0
−2
2
x
4
− 20
(0,−24)
7 4.
f x 4x 72x 2 x 224 62x 1x 2 0 when x 2,
(3, 0)
(−2, 0)
− 40 − 60
1 2.
( 74 , − 16.875 256 )
Review Exercises for Chapter 3 52. f x x 213x 123
y
Graph of Exercise 39 translated 2 units to the right x replaces by x 2.
3 2
1, 0 is a relative maximum.
1,
1
(−1, 0)
4 is a relative minimum.
3
1
2, 0 is a point of inflection.
−2
2x 1 x2
(1, 41/3)
y 3
Domain: , ; Range: 1, 1
2
(1, 1)
21 x1 x 0 when x ± 1. fx 1 x22 f x
x 3
−3
Intercepts: 1, 0, 2, 0 54. f x
(2, 0)
−3 −2
1 x −1
1
2
3
(−1, −1)
2x3 x2 0 when x 0, ± 3. 1 x23
−2 −3
f 1 < 0 Therefore, 1, 1 is a relative maximum. f 1 > 0 Therefore, 1, 1 is a relative minimum. Points of inflection: 3, 32, 0, 0, 3, 32 Intercept: 0, 0 Symmetric with respect to the origin Horizontal asymptote: y 0 56. f x
x2 1 x4
2
Domain: , ; Range: 0,
1
fx
1 x42x x24x3 2x1 x1 x1 x2 0 when x 0, ± 1. 1 x42 1 x42
f x
1 x422 10x4 2x 2x521 x44x3 21 12x4 3x8 0 when x ± 1 x44 1 x43
6 ± 3
33
4
f ± 1 < 0
Therefore, ± 1,
1 are relative maxima. 2
y 1
f 0 > 0 Therefore, 0, 0 is a relative minimum. Points of inflection:
±
4
6 33 , 0.29 , ± 3
Intercept: 0, 0 Symmetric to the y-axis Horizontal asymptote: y 0
3 4 1 2
(−1, ) 1 2
4
6 33 , 0.40 3
(1, 21) (0, 0) x
−3 −2 −1 − 21
1
2
3
.
441
442
Chapter 3
58. f x x2
Applications of Differentiation
1 x3 1 x x
Domain: , 0, 0, ; Range: ,
2x 4, 60. f x x 1 x 3 2, 2x 4,
1 2x3 1 1 fx 2x 2 0 when x 3 . x x2 2 2 2x3 1 0 when x 1. f x 2 3 x x3
Domain: , Range: 2,
Intercept: 0, 4 y 4
>0
1 f 3 2
(0, 4)
3 2
1 3 Therefore, 3 , 3 is a relative minimum. 2 4
1
Point of inflection: 1, 0
x 1
2
3
4
Intercept: 1, 0 Vertical asymptote: x 0 y 3 2 1
(−1, 0) −3 −2
62. f x
( 12 , 34 ) 3
3
x 1
2
3
1 2 sin x sin 2 x
Domain: 1, 1 ; Range:
y 1
32 3, 32 3
fx 2cos x cos 2 x 22 cos x 1cos x 1 0
1 2
x − 12
− 12
1 2
−1
2 Critical Numbers: x ± , 0 3
f x 2 sin x 2 sin 2 x 2 sin x1 4 cos x 0 when x 0, ± 1, ± 0.420. By the First Derivative Test:
32, 32 3 is a relative minimum. 23, 32 3 is a relative maximum.
Points of inflection: 0.420, 0.462, 0.420, 0.462, ± 1, 0, 0, 0 Intercepts: (1, 0, 0, 0, 1, 0 Symmetric with respect to the origin 64. f x xn, n is a positive integer. (a) fx nxn1 The function has a relative minimum at 0, 0 when n is even. (b) f x nn 1xn2 The function has a point of inflection at 0, 0 when n is odd and n ≥ 3.
x ≤ 1 1 < x ≤ 3 x > 3
Review Exercises for Chapter 3 x2 y2 1 1, y 144 x2 144 16 3
66. Ellipse:
A 2x
y 12
2 4 144 x2 x144 x2 3 3
144 − x 2
−12
(
12 −8 −12
4 144 2x2 0 when x 72 62. 3 144 x2
1 3
x
x2 dA 4 144 x2 dx 3 144 x2
( x,
8
2 The dimensions of the rectangle are 2x 122 by y 144 72 42. 3 68. We have points 0, y, x, 0, and 4, 5. Thus, m
50 5x y5 or y . 04 4x x4
Let f x L 2 x 2
x 5x 4
L (4, 5)
2
5 (x, 0)
x fx 2x 50 x4 x
(0, y)
x4x 0 x 42
4
100x 0 x 43
3 100 . x x 43 100 0 when x 0 or x 4
L
x
2
3 100 4 25x2 x x 42 25 10023 25 12.7 feet 2 3 100 x 4 x4
70. Label triangle with vertices 0, 0, a, 0, and b, c. The equations of the sides of the triangle are y cbx and y cb a x a. Let x, 0 be a vertex of the inscribed rectangle. The coordinates of the upper left vertex are x, cbx. The y-coordinate of the upper right vertex of the rectangle is cbx. Solving for the x-coordinate x of the rectangle’s upper right vertex, you get c c x x a b ba
y= cx b
b ax bx a
(
ba ab x xaa x. b b
(0, 0)
Finally, the lower right vertex is
a a b b x, 0 . Width of rectangle: a Height of rectangle:
ab xx b
c x see figure b
A WidthHeight a
A
ab xx b
bc x a ba x bc x
b x 0 when x . ba acb 2ac b 2
a c dA c a x x dx b b b
x, c x b
2
b2 a ba b2 bc b2 a2 2c 41 ac 21 12 ac 21 Area of triangle
(
(x, 0)
(b, c) c ( x − a) b−a
y=
(a − a b− b x, bc x( (a, 0)
(a − a b− b x, 0(
443
444
Chapter 3
Applications of Differentiation
72. You can form a right triangle with vertices 0, y, 0, 0, and x, 0. Choosing a point a, b on the hypotenuse (assuming the triangle is in the first quadrant), the slope is m
b0 bx yb . ⇒y 0a ax ax
Let f x L2 x2 y2 x2
. abx x 2
fx 2x 2
ab abx x a x 2
2x a x3 ab2 3 ab2. 0 when x 0, a a x3 3 Choosing the nonzero value, we have y b a 2b. 3 ab2 2 b L a 3 a2b 2
a2 3a43b23 3a23b43 b212 a23 b2332 meters 74. Using Exercise 73 as a guide we have L1 a csc and L2 b sec . Then dLd a csc cot b sec tan 0 when 3 ab, sec tan
a23 b23
b13
, csc
L L1 L2 a csc b sec a
a23 b23
a13
and
a23 b2312 a23 b2312 b a23 b2332. a13 b13
This matches the result of Exercise 72. 76. Total cost Cost per hourNumber of hours T
11v 825 v 110 7.50 500 v 50 v 2
dT 11 825 11v 2 41,250 2 dv 50 v 50v 2 0 when v 3750 256 61.2 mph. d 2T 1650 3 > 0 when v 256 so this value yields a minimum. dv 2 v 78. f x x3 2x 1 From the graph, you can see that f x has one real zero. fx 3x2 2 f changes sign in 1, 0 . n
xn
f xn
f xn
f xn fxn
1
0.5000
0.1250
2.7500
0.0455
0.4545
2
0.4545
0.0029
2.6197
0.0011
0.4534
On the interval 1, 0 : x 0.453.
xn
f xn fxn
Review Exercises for Chapter 3
43. r
v02 sin 2 32
45. Let f x x, x 100, dx 0.6. f x x f x f x dx
v0 2200 ftsec
changes from 10 to 11 dr
x
22002 cos 2 d 16
10
f x x 99.4
180
d 11 10
1 dx 2x
100
1 0.6 9.97 2100
Using a calculator: 99.4 9.96995
180
r dr
22002 20 cos 16 180
180 4961 feet
4961 feet 49. Let f x x, x 4, dx 0.02, f x 1 2x .
4 x, x 625, dx 1. 47. Let f x 4 x f x x f x f x d x
Then
1 dx 44x3
f 4.02 f 4 f 4 dx
1 4 624 4 625 f x x 1 4 625 3 4 5
4.02 4
1 1 0.02 2 0.02. 4 24
1 4.998 500
4 624 4.9980. Using a calculator,
51. In general, when x → 0, dy approaches y. 53. True
55. True
Review Exercises for Chapter 3 1. A number c in the domain of f is a critical number if f c 0 or f is undefined at c.
y 4
f ′(c) is 3 undefined.
f ′(c) = 0
x −4 −3
−1 −2 −3 −4
3. gx 2x 5 cos x, 0, 2
18
(6.28, 17.57)
g x 2 5 sin x 2
0 when sin x 5 . Critical numbers: x 0.41, x 2.73 Left endpoint: 0, 5 Critical number: 0.41, 5.41 Critical number: 2.73, 0.88 Minimum Right endpoint: 2, 17.57 Maximum
(2.73, 0.88) − 4
2 −1
1
2
4
163
164
Chapter 3
Applications of Differentiation
5. Yes. f 3 f 2 0. f is continuous on 3, 2 , differentiable on 3, 2.
7. f x 3 x 4 y
(a)
f x x 33x 1 0 for x 13.
6 4
c 13 satisfies f c 0.
2 x
−2
2
4
6
10
−4 −6
f 1 f 7 0 (b) f is not differentiable at x 4.
9.
f x x23, 1 ≤ x ≤ 8
f x x cos x,
11.
2 f x x13 3
f x 1 sin x f b f a 2 2 1 ba 2 2
f b f a 4 1 3 ba 81 7
f c 1 sin c 1
2 3 f c c13 3 7 c
13.
149
3
≤ x ≤ 2 2
c0
2744 3.764 729
f x Ax2 Bx C f x 2Ax B f x2 f x1 Ax22 x12 Bx2 x1 x2 x1 x2 x1 Ax1 x2 B f c 2Ac B Ax1 x2 B 2Ac Ax1 x2 c
x1 x2 Midpoint of x1, x2
2
15. f x x 12x 3 f x x 1 1 x 32x 1 2
x 13x 7 7 Critical numbers: x 1 and x 3
17. hx xx 3 x32 3x12 Domain: 0, 3 3 h x x12 x12 2 2 3 3x 1 x12x 1 2 2x Critical number: x 1
Interval:
< x < 1
1 < x <
7 3
7 3
< x <
Sign of f x:
f x > 0
f x < 0
f x > 0
Conclusion:
Increasing
Decreasing
Increasing
Interval:
0 < x < 1
Sign of h x:
h x < 0
h x > 0
Conclusion:
Decreasing
Increasing
1 < x <
Review Exercises for Chapter 3 19. ht 14t 4 8t
Test Interval: < t < 2
h t t 3 8 0 when t 2. Relative minimum: 2, 12
165
2 < t <
Sign of h t:
h t < 0
h t > 0
Conclusion:
Decreasing
Increasing
1 1 cos12t sin12t 3 4
21. y
v y 4 sin12t 3 cos12t (a) When t
1 , y inch and v y 4 inches/second. 8 4
(b) y 4 sin12t 3 cos12t 0 when
sin12t 3 3 ⇒ tan12t . cos12t 4 4
3 4 Therefore, sin12t and cos12t . The maximum displacement is 5 5 y (c) Period:
1345 41 53 125 inch. 2 12 6 1 6 6
Frequency:
23. f x x cos x, 0 ≤ x ≤ 2 f x 1 sin x f x cos x 0 when x Points of inflection:
3 , . 2 2
Test Interval: Sign of f x: Conclusion:
3 < x < 2 2
3 < x < 2 2
f x < 0
f x > 0
f x < 0
Concave downward
Concave upward
Concave downward
0 < x <
2
2 , 2 , 32, 32
25. gx 2x21 x2
y
g x 4x2x2 1 Critical numbers: x 0, ±
(−
1 2
1, 1 2 2
)
1
−2
g x 4 24x
(
1, 1 2 2
)
(0, 0)
2
x
−1
2
−2
g 0 4 > 0
Relative minimum at 0, 0
1 1 Relative maximums at ± , 2 2
y
27. 6
(5, f(5))
5 4
(3, f(3))
2 1 −1
29. The first derivative is positive and the second derivative is negative. The graph is increasing and is concave down.
7
3
−3
1 g ± 8 < 0 2
(6, 0) (0, 0) 2 3 4 5
x 7
166
Chapter 3
Applications of Differentiation
31. (a) D 0.0034t4 0.2352t3 4.9423t2 20.8641t 94.4025 (b)
369
0
29 0
(c) Maximum at 21.9, 319.5 1992 Minimum at 2.6, 69.6 1972 (d) Outlays increasing at greatest rate at the point of inflection 9.8, 173.7 1979
33. lim
x →
2x2 2 2 lim 5 x → 3 5x2 3
35. lim
2x 3 x4
39. f x
3x2
37. hx
Discontinuity: x 4 lim
x →
x →
5 cos x 0, since 5 cos x ≤ 5. x
3 2 x
Discontinuity: x 0
2x 3 2 3x lim 2 x → 1 4x x4
lim
x →
3x 2 2
Vertical asymptote: x 4
Vertical asymptote: x 0
Horizontal asymptote: y 2
Horizontal asymptote: y 2
41. f x x3
243 x
43. f x
x1 1 3x2
Relative minimum: 3, 108
Relative minimum: 0.155, 1.077
Relative maximum: 3, 108
Relative maximum: 2.155, 0.077 0.2
200
−2
−5
5
5
− 1.4
− 200
Vertical asymptote: x 0
Horizontal asymptote: y 0
45. f x 4x x2 x4 x Domain: , ; Range: , 4 f x 4 2x 0 when x 2. f x 2 Therefore, 2, 4 is a relative maximum. Intercepts: 0, 0, 4, 0
y
5
)2, 4)
4 3 2 1
x 1
2
3
5
Review Exercises for Chapter 3 47. f x x16 x2, Domain: 4, 4 , Range: 8, 8
y
2
2, 8
8
Domain: 4, 4 ; Range: 8, 8
6 4
16 2x2 f x 0 when x ± 22 and undefined when x ± 4. 16 x2 f x
2xx2 24 16 x232
2
(− 4, 0)
(4, 0) x
8
6
2
2
4
6
8
(0, 0)
8
2
2,
8
f 22 > 0
Therefore, 22, 8 is a relative minimum. f 22 < 0
Therefore, 22, 8 is a relative maximum. Point of inflection: 0, 0 Intercepts: 4, 0, 0, 0, 4, 0 Symmetry with respect to origin 49. f x x 13x 32
y
Domain: , ; Range: ,
4
f x x 12x 35x 11 0 when x 1,
11 , 3. 5
f x 4x 15x2 22x 23 0 when x 1,
( 115 , 1.11(
(1, 0)
(2.69, 0.46) (3, 0)
2
x
−2
4 −2
11± 6 . 5
6
(1.71, 0.60)
−4
f 3 > 0 Therefore, 3, 0 is a relative minimum. f
115 < 0
Therefore,
is a relative maximum. 115, 3456 3125
Points of inflection: 1, 0,
11 5
6
11 5
, 0.60 ,
6
, 0.46
Intercepts: 0, 9, 1, 0, 3, 0 51. f x x13x 323
y
Domain: , ; Range: ,
4 3
x1 f x 0 when x 1 and undefined when x 3, 0. x 313x23 2 f x 53 is undefined when x 0, 3. x x 343 3 4 is By the First Derivative Test 3, 0 is a relative maximum and 1, a relative minimum. 0, 0 is a point of inflection.
Intercepts: 3, 0, 0, 0
2 1
) 3, 0)
)0, 0) x
5
4
2
) 1,
1
1
1.59) 3
2
167
168
Chapter 3
Applications of Differentiation
x1 x1
53. f x
x
1
y
Domain: , 1, 1, ; Range: , 1, 1, f x
2 < 0 if x 1. x 12
f x
4 x 13
4
y
1 2
x 2
2
4
2
Horizontal asymptote: y 1 Vertical asymptote: x 1 Intercepts: 1, 0, 0, 1 55. f x
4 1 x2
y 5
Domain: , ; Range: 0, 4
8x 0 when x 0. f x 1 x22 3 81 3x2 . 0 when x ± f x 1 x23 3
(0, 4)
4
(−
3,3 3
(
(
3,3 3
1
2
(
2 1 −3
−2 −1
x −1
3
f 0 < 0 Therefore, 0, 4 is a relative maximum. Points of inflection: ± 33, 3 Intercept: 0, 4 Symmetric to the y-axis Horizontal asymptote: y 0
57. f x x3 x
y
4 x
10
Domain: , 0, 0, ; Range: , 6 , 6, f x 3x2 1 f x 6x
4 x2
3x4
4 0 when x ± 1. x2 x2
8 6x4 8 0 x3 x3
f 1 < 0 Therefore, 1, 6 is a relative maximum. f 1 > 0 Therefore, 1, 6 is a relative minimum. Vertical asymptote: x 0 Symmetric with respect to origin
5
(1, 6) x
2
1
(−1, −6) −5
1
x
2
0
Review Exercises for Chapter 3
59. f x x2 9
y
Domain: , ; Range: 0, f x
2xx2 9 0 when x 0 and is undefined when x ± 3. x2 9
10
5
2x2 9 is undefined at x ± 3. f x 2 x 9
)0, 9)
) 3, 0)
)3, 0) x
4
f 0 < 0
2
2
4
Therefore, 0, 9 is a relative maximum. Relative minima: ± 3, 0 Points of inflection: ± 3, 0 Intercepts: ± 3, 0, 0, 9 Symmetric to the y-axis 61. f x x cos x
y
)2 , 2
Domain: 0, 2 ; Range: 1, 1 2
3 3 , 2 2
f x 1 sin x ≥ 0, f is increasing. f x cos x 0 when x Points of inflection:
1)
2
3 , . 2 2
3 3 , , , 2 2 2 2
)0, 1)
, 2 2 x
2
Intercept: 0, 1 63. x2 4y2 2x 16y 13 0 (a) x 2 2x 1 4y 2 4y 4 13 1 16
y
x 1 4y 2 4 x 12 y 22 1 4 1 The graph is an ellipse: 2
2
4
(1, 3) 3 2 1
Maximum: 1, 3
(1, 1) x −1
Minimum: 1, 1
1
2
3
(b) x2 4y2 2x 16y 13 0 2x 8y
dy dy 2 16 0 dx dx dy 8y 16 2 2x dx dy 2 2x 1x dx 8y 16 4y 8
The critical numbers are x 1 and y 2. These correspond to the points 1, 1, 1, 3, 2, 1, and 2, 3. Hence, the maximum is 1, 3 and the minimum is 1, 1.
169
170
Chapter 3
Applications of Differentiation
65. Let t 0 at noon.
(100 − 12t, 0) (0, 0)
L d 2 100 12t2 10t2 10,000 2400t 244t 2
A
(100, 0)
d
300 dL 2400 488t 0 when t 4.92 hr. dt 61
B (0, −10t)
Ship A at 40.98, 0; Ship B at 0, 49.18 d 2 10,000 2400t 244t 2 4098.36 when t 4.92 4:55 P.M.. d 64 km 67. We have points 0, y, x, 0, and 1, 8. Thus,
y
08 8x y8 or y . m 01 x1 x1
(0, y)
10
(1, 8)
8 6
Let f x L 2 x 2
x 8x 1 . 2
4 2
x f x 2x 128 x1 x
(x, 0)
x 1 x 0 x 12
x 2
4
6
8
10
64x 0 x 13
x x 13 64 0 when x 0, 5 minimum. Vertices of triangle: 0, 0, 5, 0, 0, 10 69.
A Average of basesHeight
x 2 s
3s2 2sx x2
2
s
see figure
s
dA 1 s xs x 3s2 2sx x2 dx 4 3s2 2sx x2
x−s 2
22s xs x 0 when x 2s. 43s2 2sx x2 A is a maximum when x 2s. 71. You can form a right triangle with vertices 0, 0, x, 0 and 0, y. Assume that the hypotenuse of length L passes through 4, 6. 60 6x y6 or y 04 4x x4
Let f x L2 x2 y2 x 2 f x 2x 72
x 6x 4 . 2
0 x x 4 x 4 4 2
3 x x 43 144 0 when x 0 or x 4 144.
L 14.05 feet
s
x−s 2 x
m
3s 2 + 2sx − x 2 2
Review Exercises for Chapter 3 csc
73. csc
L1 or L1 6 csc 6
2 9 or L L2
2
9 csc
see figure
L1 θ
2
L2
θ 9
L L1 L2 6 csc 9 csc
171
6
(π2 − θ(
2 6 csc 9 sec
dL 6 csc cot 9 sec tan 0 d tan3
3 2 2 ⇒ tan 3 3 3
sec 1 tan2 csc L6
1 23
23
323 223
313
sec 323 223 tan 213
323 22312 323 22312 9 3323 22332 ft 21.07 ft Compare to Exercise 72 using a 9 and b 6. 13 2 313
75. Total cost Cost per hourNumber of hours T
v 110 11v 550 5 600 v 60 v 2
dT 11 550 11v 2 33,000 2 dv 60 v 60v 2 0 when v 3000 1030 54.8 mph. d 2T 1100 3 > 0 when v 1030 so this value yields a minimum. dv 2 v 77. f x x3 3x 1 From the graph you can see that f x has three real zeros. f x 3x2 3 f xn
f xn
f xn f xn
1.5000
0.1250
3.7500
0.0333
1.5333
2
1.5333
0.0049
4.0530
0.0012
1.5321
n
xn
f xn
f xn
f xn f xn
1
0.5000
0.3750
2.2500
0.1667
0.3333
2
0.3333
0.0371
2.6667
0.0139
0.3472
3
0.3472
0.0003
2.6384
0.0001
0.3473
n
xn
f xn
f xn
f xn f xn
1
1.9000
0.1590
7.8300
0.0203
1.8797
2
1.8797
0.0024
7.5998
0.0003
1.8794
n
xn
1
xn
f xn f xn
xn
xn
f xn f xn
f xn f xn
The three real zeros of f x are x 1.532, x 0.347, and x 1.879.
172
Chapter 3
Applications of Differentiation
79. Find the zeros of f x x4 x 3. fx 4x3 1 From the graph you can see that f x has two real zeros. f changes sign in 2, 1.
n
xn
f xn
f xn
f xn fxn
1
1.2000
0.2736
7.9120
0.0346
1.1654
2
1.1654
0.0100
7.3312
0.0014
1.1640
xn
f xn fxn
On the interval 2, 1: x 1.164. f changes sign in 1, 2. f xn
f xn fxn
0.5625
12.5000
0.0450
1.4550
1.4550
0.0268
11.3211
0.0024
1.4526
1.4526
0.0003
11.2602
0.0000
1.4526
n
xn
1
1.5000
2 3
f xn
xn
f xn fxn
On the interval 1, 2; x 1.453. 81.
y x1 cos x x x cos x dy 1 x sin x cos x dx dy 1 x sin x cos x dx
83.
S 4 r 2. dr r ± 0.025 dS 8r dr 89± 0.025 ± 1.8 square cm dS 8 r dr 2 dr 100 100 100 S 4 r 2 r
2± 0.025 100 ± 0.56% 9
4 V r3 3 dV 4 r 2 dr 492± 0.025 ± 8.1 cubic cm dV 4 r 2 dr 3 dr 100 100 100 V 43r 3 r
3± 0.025 100 ± 0.83% 9
Problem Solving for Chapter 3 1. Assume y1 < d < y2. Let gx f x dx a. g is continuous on a, b and therefore has a minimum c, gc on a, b. The point c cannot be an endpoint of a, b because ga fa d y1 d < 0 gb fb d y2 d > 0 Hence, a < c < b and gc 0 ⇒ fc d.
Problem Solving for Chapter 3 3. (a) For a 3, 2, 1, 0, p has a relative maximum at 0, 0. For a 1, 2, 3, p has a relative maximum at 0, 0 and 2 relative minima. (b) px 4ax3 12x 4xax2 3 0 ⇒ x 0, ±
3a
p x 12ax2 12 12ax2 1 For x 0, p 0 12 < 0 ⇒ p has a relative maximum at 0, 0. (c) If a > 0, x ±
3a are the remaining critical numbers.
3a 12 3a 12 24 > 0 ⇒ p has relative minima for a > 0.
p ±
(d) 0, 0 lies on y 3x2. Let x ±
3a. Then
px a
3a
2
6
3 a
2
a=1 a=3 y
2ax 2x
2x2
8 7 6 5 4 3 2
a
For a ≥ 0, there is one relative minimum at 0, 0.
(c) For a < 0, there are two relative minima at x ±
−2
2a.
(d) There are either 1 or 3 critical points. The above analysis shows that there cannot be exactly two relative extrema. c x2 x c c c 2x 0 ⇒ 2 2x ⇒ x3 ⇒ x x2 x 2
cx 3
2c 2 x3
If c 0, f x x2 has a relative minimum, but no relative maximum.
2c is a relative minimum, because f 2c > 0. c If c < 0, x is a relative minimum too. 2 If c > 0, x
3
3
Answer: all c.
a=2
a=0
a = −1 a = −2 a = −3
(b) For a < 0, there is a relative maximum at 0, 1.
f x
a = −1 a = −3
3x2 is satisfied by all the relative extrema of p.
p x 16x2 2a
fx
1
2
3
3a a9 18a a9.
5. px x 4 ax2 1
7. f x
a=1 x
−3
(a) px
a=3
2
9 Thus, y 3 ± a
4x4
y
a=2
3
x −1 −2
2
−8
a = −2 a=0
173
174
Chapter 3
9. Set
Applications of Differentiation
f b f a f ab a k. b a2
Define Fx f x f a fax a kx a2. Fa 0, Fb f b f a fab a kb a2 0 F is continuous on a, b and differentiable on a, b. There exists c1, a < c1 < b, satisfying Fc1 0. Fx fx fa 2kx a satisfies the hypothesis of Rolle’s Theorem on a, c1: Fa 0, Fc1 0. There exists c2, a < c2 < c1 satisfying F c2 0. Finally, F x f x 2k and F c2 0 implies that k
f c2 . 2
Thus, k
11. E E
f b f a f ab a f c2 1 ⇒f b f a fab a f c2b a2. b a2 2 2
tan 1 0.1 tan 10 tan tan2 0.1 tan 1 10 tan
1 10 tan 10 sec2 2 tan sec2 10 tan tan2 10 sec2 0 1 10 tan
⇒ 1 10 tan 10 sec2 2 tan sec2 10 tan tan2 10 sec2 ⇒ 10 sec2 2 tan sec2 100 tan sec2 20 tan2 sec2 100 tan sec2 10 tan2 sec2 ⇒ 10 2 tan 10 tan2 ⇒ 10 tan2 2 tan 10 0 tan
2 ± 4 400 0.90499, 1.10499 20
Using the positive value, 0.7356, or 42.14 . 13. v 2400 sin v 2400 cos 0
3 2n, 2n, n an integer 2 2
Problem Solving for Chapter 3 x y 4 1 or y x 4. 3 4 3
15. The line has equation Rectangle:
4 4 Area A xy x x 4 x2 4x. 3 3 8 8 3 Ax x 4 0 ⇒ x 4 ⇒ x 3 3 2 Dimensions:
3 2 2
Calculus was helpful.
Circle: The distance from the center r, r to the line
r
r r 1 3 4
x y 1 0 must be r: 3 4
12 7r 12 7r 12 5 12 5
1 1 9 16 5r 7r 12 ⇒ r 1 or r 6.
Clearly, r 1. x y 1 and satisfies x y r. 3 4
Semicircle: The center lies on the line
7 12 r r 1 ⇒ r 1 ⇒ r . No calculus necessary. 3 4 12 7
Thus
17. y 1 x21 y
2x 1 x22
y
3 23x2 1 1 0 ⇒ x± ± 3 x2 13 3
y′′:
+++ −−−− −−−− +++ −
3 3
0
3 3
The tangent line has greatest slope at
19. (a)
x sin x
3 3
3
,
4
and least slope at 33, 34 .
0.1
0.2
0.3
0.4
0.5
1.0
0.09983
0.19867
0.29552
0.38942
0.47943
0.84147
sin x ≤ x (b) Let f x sin x. Then fx cos x and on 0, x you have by the Mean Value Theorem, fc cosc Hence,
f x f 0 , 0 < c < x x0 sin x x
sin x cosc ≤ 1 x
⇒ sin x ≤ x ⇒ sin x ≤ x
175
Review Exercises for Chapter 3 50. Let f x tan x, x 0, dx 0.05, fx sec2 x.
52. Propagated error f x x f x,
Then
relative error
f 0.05 f 0 f0dx
437
dy dy , and the percent error 100. y y
tan 0.05 tan 0 sec2 00.05 0 10.05.
54. True,
y dy a x dx
56. False Let f x x, x 1, and x dx 3. Then y f x x f x f 4 f 1 1 and dy fx dx
1 3 3 . 21 2
Thus, dy > y in this example.
Review Exercises for Chapter 3 2. (a) f 4 f 4 3 (c)
(b) f 3 f 3 4 4
y
(d) Yes. Since f 2 f 2 1 1 and f 1 f 1 2, the Mean Value says that there exists at least one value c in 2, 1 such that
6 4
x −6 −4 −2
4
fc
6
−4
(e) No, lim f x exists because f is continuous at 0, 0.
−6
x →0
At least six critical numbers on 6, 6. 4. f x
f 1 f 2 2 1 1. 1 2 12
x x2 1
, 0, 2
(f) Yes, f is differentiable at x 2.
6. No. f is not differentiable at x 2.
1 fx x x2 1322x x2 112 2 1 x2 132 No critical numbers Left endpoint: 0, 0 Minimum Right endpoint: 2, 25 Maximum
8. No; the function is discontinuous at x 0 which is in the interval 2, 1 .
10.
1 f x , 1 ≤ x ≤ 4 x fx
1 x2
f b f a 14 1 34 1 ba 41 3 4 fc
1 1 c2 4
c2
12.
f x x 2x, 0 ≤ x ≤ 4 fx
1 2x
2
f b f a 6 0 3 ba 40 2 fc
1 3 2 2 2c
c1
438
Chapter 3
Applications of Differentiation
f x 2x2 3x 1
14.
fx 4x 3 f b f a 21 1 5 ba 40 fc 4c 3 5 c 2 Midpoint of 0, 4 16. gx x 13
Critical number: x 1
18. f x sin x cos x, 0 ≤ x ≤ 2
Critical numbers: x
gx
5 ,x 4 4
1 < x <
Sign of gx
gx > 0
gx > 0
Conclusion
Increasing
Increasing
0 < x <
Interval
fx cos x sin x
20. gx
< x < 1
Interval
gx 3x 12
4
5 < x < 4 4
5 < x < 2 4
Sign of fx
fx > 0
fx < 0
fx > 0
Conclusion
Increasing
Decreasing
Increasing
3 x sin 1 , 0, 4 2 2
Test Interval
3 x 1 cos 2 2 2
Sign of gx
gx > 0
gx < 0
gx > 0
Conclusion
Increasing
Decreasing
Increasing
0 when x 1
2 2 ,3
Relative maximum:
1 2 , 32
Relative minimum:
3 2 , 23
22. (a) y A sinkm t B coskm t
A sinkm t A ⇒ tankm t . B coskm t B
Therefore, sin km t cos km t
A A2 B2
B A2 B2
.
When v y 0, yA
A A B B A B B A
2
2
2
2
2
(b) Period:
y Akm coskm t Bkm sinkm t 0 when
0 < x < 1
2
B2.
1
2 2 < x < 3
3
2 km
Frequency:
1 1 km 2 km 2
2 < x < 4
Review Exercises for Chapter 3 24. f x x 22x 4 x3 12x 16 fx 3x2 12 f x 6x 0 when x 0.
439
Test Interval
< x < 0
Sign of f x
f x < 0
f x > 0
Concave downward
Concave upward
Conclusion
0 < x <
Point of inflection: 0, 16 26.
ht t 4t 1 Domain: 1, ht 1
2 t 1
y
28. 7
0 ⇒ t3
6 5 4
1 h t t 132 h 3
30.
C
1 > 0 8
3, 5 is a relative minimum.
Qx s 2x r
r Qs dC 2 0 dx x 2
3 2 1 x
−1
1
2
3
4
5
6
7
32. (a) S 0.1222t3 1.3655t2 0.9052t 4.8429 (b)
r Qs x2 2
25
1
14 0
2Qs x2 r x
34. lim
x →
2Qs r
2x 2x lim 0 3x2 5 x → 3 5x2
38. gx
5x2 2
x2
lim
x →
5x2 5 lim 5 2 x → 1 2x2
x2
(c) St 0 when t 3.7. This is a maximum by the First Derivative Test. (d) No, because the t3 coefficient term is negative.
36. lim
3x
x →
x2 4
40. f x
lim
x →
3 1 4x2
3
3x x2 2
lim
x →
3x x2 2
Horizontal asymptote: y 5
lim
x →
lim
x →
lim
x →
3x x2 2
3xx x2 2x2
3 1 2x2
lim
x →
lim
x →
3
3xx
x2 2 x2
3 3 1 2x2
Horizontal asymptotes: y ± 3
440
Chapter 3
Applications of Differentiation
2 4 cos x cos 2x 3
44. gx
42. f x x3 3x2 2x xx 1x 2 Relative minima: (0, 0, 1, 0, 2, 0
Relative minima: 2 k, 0.29 where k is any integer.
Relative maxima: 1.577, 0.38, 0.423, 0.38
Relative maxima: 2k 1 , 8.29 where k is any integer.
3
10
−2
4 − 5 2
−1
46. f x 4x3 x4 x34 x
y 30
Domain: , ; Range: , 27
25 20
fx 12x 4x 4x 3 x 0 when x 0, 3. 2
3
5 2
0
2
15 10
f x 24x 12x 2 12x2 x 0 when x 0, 2.
5 x
f 3 < 0
−2
1
2
3
5
Therefore, 3, 27 is a relative maximum. Points of inflection: 0, 0, 2, 16 Intercepts: 0, 0, 4, 0 48. f x x2 42
y
Domain: , ; Range: 0,
24 20
fx 4xx2 4 0 when x 0, ± 2. 23 f x 43x2 4 0 when x ± . 3 f 0 < 0
(0, 16) (−2, 0)
(2, 0) 8 4 x
−3 −2 −1
Therefore, 0, 16 is a relative maximum.
1
2
3
f ± 2 > 0 Therefore, ± 2, 0 are relative minima. Points of inflection: ± 233, 649 Intercepts: 2, 0, 0, 16, 2, 0 Symmetry with respect to y-axis 50. f x x 3x 23 Domain: , ; Range:
y
16,875 256 ,
−4
fx x 33x 22 x 23 4x 7x 22 0 when x 2,
f 74 > 0 Therefore, 74 , 16,875 256 is a relative minimum. Points of inflection: 2, 0, 12 , 625 16 Intercepts: 2, 0, 0, 24, 3, 0
−2
2
x
4
− 20
(0,−24)
7 4.
f x 4x 72x 2 x 224 62x 1x 2 0 when x 2,
(3, 0)
(−2, 0)
− 40 − 60
1 2.
( 74 , − 16.875 256 )
Review Exercises for Chapter 3 52. f x x 213x 123
y
Graph of Exercise 39 translated 2 units to the right x replaces by x 2.
3 2
1, 0 is a relative maximum.
1,
1
(−1, 0)
4 is a relative minimum.
3
1
2, 0 is a point of inflection.
−2
2x 1 x2
(1, 41/3)
y 3
Domain: , ; Range: 1, 1
2
(1, 1)
21 x1 x 0 when x ± 1. fx 1 x22 f x
x 3
−3
Intercepts: 1, 0, 2, 0 54. f x
(2, 0)
−3 −2
1 x −1
1
2
3
(−1, −1)
2x3 x2 0 when x 0, ± 3. 1 x23
−2 −3
f 1 < 0 Therefore, 1, 1 is a relative maximum. f 1 > 0 Therefore, 1, 1 is a relative minimum. Points of inflection: 3, 32, 0, 0, 3, 32 Intercept: 0, 0 Symmetric with respect to the origin Horizontal asymptote: y 0 56. f x
x2 1 x4
2
Domain: , ; Range: 0,
1
fx
1 x42x x24x3 2x1 x1 x1 x2 0 when x 0, ± 1. 1 x42 1 x42
f x
1 x422 10x4 2x 2x521 x44x3 21 12x4 3x8 0 when x ± 1 x44 1 x43
6 ± 3
33
4
f ± 1 < 0
Therefore, ± 1,
1 are relative maxima. 2
y 1
f 0 > 0 Therefore, 0, 0 is a relative minimum. Points of inflection:
±
4
6 33 , 0.29 , ± 3
Intercept: 0, 0 Symmetric to the y-axis Horizontal asymptote: y 0
3 4 1 2
(−1, ) 1 2
4
6 33 , 0.40 3
(1, 21) (0, 0) x
−3 −2 −1 − 21
1
2
3
.
441
442
Chapter 3
58. f x x2
Applications of Differentiation
1 x3 1 x x
Domain: , 0, 0, ; Range: ,
2x 4, 60. f x x 1 x 3 2, 2x 4,
1 2x3 1 1 fx 2x 2 0 when x 3 . x x2 2 2 2x3 1 0 when x 1. f x 2 3 x x3
Domain: , Range: 2,
Intercept: 0, 4 y 4
>0
1 f 3 2
(0, 4)
3 2
1 3 Therefore, 3 , 3 is a relative minimum. 2 4
1
Point of inflection: 1, 0
x 1
2
3
4
Intercept: 1, 0 Vertical asymptote: x 0 y 3 2 1
(−1, 0) −3 −2
62. f x
( 12 , 34 ) 3
3
x 1
2
3
1 2 sin x sin 2 x
Domain: 1, 1 ; Range:
y 1
32 3, 32 3
fx 2cos x cos 2 x 22 cos x 1cos x 1 0
1 2
x − 12
− 12
1 2
−1
2 Critical Numbers: x ± , 0 3
f x 2 sin x 2 sin 2 x 2 sin x1 4 cos x 0 when x 0, ± 1, ± 0.420. By the First Derivative Test:
32, 32 3 is a relative minimum. 23, 32 3 is a relative maximum.
Points of inflection: 0.420, 0.462, 0.420, 0.462, ± 1, 0, 0, 0 Intercepts: (1, 0, 0, 0, 1, 0 Symmetric with respect to the origin 64. f x xn, n is a positive integer. (a) fx nxn1 The function has a relative minimum at 0, 0 when n is even. (b) f x nn 1xn2 The function has a point of inflection at 0, 0 when n is odd and n ≥ 3.
x ≤ 1 1 < x ≤ 3 x > 3
Review Exercises for Chapter 3 x2 y2 1 1, y 144 x2 144 16 3
66. Ellipse:
A 2x
y 12
2 4 144 x2 x144 x2 3 3
144 − x 2
−12
(
12 −8 −12
4 144 2x2 0 when x 72 62. 3 144 x2
1 3
x
x2 dA 4 144 x2 dx 3 144 x2
( x,
8
2 The dimensions of the rectangle are 2x 122 by y 144 72 42. 3 68. We have points 0, y, x, 0, and 4, 5. Thus, m
50 5x y5 or y . 04 4x x4
Let f x L 2 x 2
x 5x 4
L (4, 5)
2
5 (x, 0)
x fx 2x 50 x4 x
(0, y)
x4x 0 x 42
4
100x 0 x 43
3 100 . x x 43 100 0 when x 0 or x 4
L
x
2
3 100 4 25x2 x x 42 25 10023 25 12.7 feet 2 3 100 x 4 x4
70. Label triangle with vertices 0, 0, a, 0, and b, c. The equations of the sides of the triangle are y cbx and y cb a x a. Let x, 0 be a vertex of the inscribed rectangle. The coordinates of the upper left vertex are x, cbx. The y-coordinate of the upper right vertex of the rectangle is cbx. Solving for the x-coordinate x of the rectangle’s upper right vertex, you get c c x x a b ba
y= cx b
b ax bx a
(
ba ab x xaa x. b b
(0, 0)
Finally, the lower right vertex is
a a b b x, 0 . Width of rectangle: a Height of rectangle:
ab xx b
c x see figure b
A WidthHeight a
A
ab xx b
bc x a ba x bc x
b x 0 when x . ba acb 2ac b 2
a c dA c a x x dx b b b
x, c x b
2
b2 a ba b2 bc b2 a2 2c 41 ac 21 12 ac 21 Area of triangle
(
(x, 0)
(b, c) c ( x − a) b−a
y=
(a − a b− b x, bc x( (a, 0)
(a − a b− b x, 0(
443
444
Chapter 3
Applications of Differentiation
72. You can form a right triangle with vertices 0, y, 0, 0, and x, 0. Choosing a point a, b on the hypotenuse (assuming the triangle is in the first quadrant), the slope is m
b0 bx yb . ⇒y 0a ax ax
Let f x L2 x2 y2 x2
. abx x 2
fx 2x 2
ab abx x a x 2
2x a x3 ab2 3 ab2. 0 when x 0, a a x3 3 Choosing the nonzero value, we have y b a 2b. 3 ab2 2 b L a 3 a2b 2
a2 3a43b23 3a23b43 b212 a23 b2332 meters 74. Using Exercise 73 as a guide we have L1 a csc and L2 b sec . Then dLd a csc cot b sec tan 0 when 3 ab, sec tan
a23 b23
b13
, csc
L L1 L2 a csc b sec a
a23 b23
a13
and
a23 b2312 a23 b2312 b a23 b2332. a13 b13
This matches the result of Exercise 72. 76. Total cost Cost per hourNumber of hours T
11v 825 v 110 7.50 500 v 50 v 2
dT 11 825 11v 2 41,250 2 dv 50 v 50v 2 0 when v 3750 256 61.2 mph. d 2T 1650 3 > 0 when v 256 so this value yields a minimum. dv 2 v 78. f x x3 2x 1 From the graph, you can see that f x has one real zero. fx 3x2 2 f changes sign in 1, 0 . n
xn
f xn
f xn
f xn fxn
1
0.5000
0.1250
2.7500
0.0455
0.4545
2
0.4545
0.0029
2.6197
0.0011
0.4534
On the interval 1, 0 : x 0.453.
xn
f xn fxn
Problem Solving for Chapter 3 80. Find the zeros of f x sin x x 1. fx cos x 1 From the graph you can see that f x has three real zeros.
n
xn
f xn
f xn
f xn fxn
1
0.2000
0.2122
3.5416
0.0599
0.2599
2
0.2599
0.0113
3.1513
0.0036
0.2635
3
0.2635
0.0000
3.1253
0.0000
0.2635
n
xn
f xn
f xn
f xn fxn
1
1.0000
0.0000
2.1416
0.0000
n
xn
f xn
f xn
f xn fxn
1
1.8000
0.2122
3.5416
0.0599
1.7401
2
1.7401
0.0113
3.1513
0.0036
1.7365
3
1.7365
0.0000
3.1253
0.0000
1.7365
xn
xn
f xn fxn
f xn fxn
1.0000
xn
f xn fxn
The three real zeros of f x are x 0.264, x 1, and x 1.737.
82.
y 36 x2
84.
x dy 1 36 x2122x dx 2 36 x2 dy
p 75
1 x 4
p p8 p7
75
x dx 36 x2
8 7 1 75 4 4 4
1 1 1 dp dx 1 4 4 4
p dp because p is linear
Problem Solving for Chapter 3 2. (a)
dV 3x2dx 3x2 x V x x3 x3 3x2x 3xx2 x3
V dV 3xx2 x3 3xx x2x
x, where → 0 as x → 0. (b) Let
y fx. Then → 0 as x → 0. x
Furthermore, y dy y fxdx x.
445
446
Chapter 3
Applications of Differentiation
4. Let hx gx f x, which is continuous on a, b and differentiable on a, b. ha 0 and hb gb f b.
y
By the Mean Value Theorem, there exists c in a, b such that
g
f
hb ha gb f b hc . ba ba
x a
b
Since hc gc fc > 0 and b a > 0, gb f b > 0 ⇒ gb > f b. 6. (a) f 2ax b, f 2a 0. No points of inflection. (b) f 3ax2 2bx c, f 6ax 2b 0 ⇒ x
b . One point of inflection. 3a
(c) y kyL y kLy ky2 y kLy 2kyy kyL 2y L If y , then y 0 and this is a point of inflection because of the analysis below. 2 y′′:
++++++
−−−−−−
L y= 2
8.
d 5 13
θ 12
x d 132 x2, sin . d Let A be the amount of illumination at one of the corners, as indicated in the figure. Then A
kI kIx sin 132 x2 132 x232
Ax kI
x2 169321 x
32 x
2
169122x
169 x23
⇒ x2 16932 3x2x2 16912 x2 169 3x2 2x2 169 x
13 2
9.19 feet
By the First Derivative Test, this is a maximum.
0
Problem Solving for Chapter 3 10. Let T be the intersection of PQ and RS. Let MN be the perpendicular to SQ and PR passing through T. Let TM x and TN b x. SN MR bx MR ⇒ SN bx x x NQ PM bx PM ⇒ NQ bx x x SQ
bx bx d MR PM x x
1 1 bx 1 b x2 1 2x2 2bx b2 Ax Area dx d b x d x d 2 2 x 2 x 2 x
1 x4x 2b 2x2 2bx b2 Ax d 2 x2
Ax 0 ⇒ 4x2 2xb 2x2 2bx b2 2x2 b2 x
b 2
bx b b2 d d 2 1d. x b2
Hence, we have SQ
Using the Second Derivative Test, this is a minimum. There is no maximum. S
Q
N b−x
b
T x P
d
M
R
12. (a) Let M > 0 be given. Take N M. Then whenever x > N M, you have f x x2 > M. (b) Let > 0 be given. Let M
1. Then whenever x > M 1,
you have x2 >
1 1 1 ⇒ 2 < ⇒ 2 0 < . x x
(c) Let > 0 be given. There exists N > 0 such that f x L < whenever x > N. 1 1 Let Let x . N. y If 0 < y <
f x L
1 1 1 , then < ⇒ x > N and N x N
f
1 L < . y
447
448
Chapter 3
Applications of Differentiation
14. Distance 42 x2 4 x2 42 f x fx
x 42 x2
4x 4 x2 42
0
x4 x2 42 x 442 x2 x2 16 8x x2 16 x2 8x 1616 x2 32x2 8x3 x 4 x 4 8x3 32x2 128x 256 128x 256 x2 The bug should head towards the midpoint of the opposite side. Without Calculus: Imagine opening up the cube: P x
Q
The shortest distance is the line PQ, passing through the midpoint.
v 16. (a) s
km m 1000 5 hr km v sec 18 3600 hr
v
20
40
60
80
100
s
5.56
11.11
16.67
22.22
27.78
d
5.1
13.7
27.2
44.2
66.4
dt 0.071s2 0.389s 0.727 (c)
(b) The distance between the back of the first vehicle and the front of the second vehicle is dt, the safe stopping distance. The first vehicle passes the given point in 5.5s seconds, and the second vehicle takes dss more seconds. Hence, T (d)
ds 5.5 . s s
T s 0.071s 0.389
6.227 s
10
Ts 0.071
6.227 6.227 ⇒ s2 s2 0.071 ⇒ s 9.365 msec
0
30
T 9.365 1.719 seconds
0
9.365 msec
1 5.5 T 0.071s2 0.389s 0.727 s s The minimum is attained when s 9.365 msec. 18. (a)
x
0
0.5
1
2
1 x
1
1.2247
1.4142
1.7321
1 x1 2
1
1.25
1.5
2
(e) d9.365 10.597 m
(b) Let f x 1 x. Using the Mean Value Theorem on the interval 0, x, there exists c, 0 < c < x, satisfying fc
1 f x f 0 1 x 1 . 21 c x0 x
Thus 1 x
3600 3.37 kmhr 1000
x x 1 < 1 because 1 c > 1. 21 c 2
Review Exercises for Chapter 4
Review Exercises for Chapter 4 2.
y
4.
u 3x du 3 dx
f′
2
3 3x
x
dx
2 3x133 dx 3x23 C 3
f
6.
x3 2x2 1 dx x2
x 2 x2 dx
8.
5 cos x 2 sec2 x dx 5 sin x 2 tan x C
1 1 x2 2x C 2 x 12. 45 mph 66 ftsec
10. f x 6x 1 fx
6x 1 dx 3x 12 C1
Since the slope of the tangent line at 2, 1 is 3, it follows that f2 3 C1 3 when C1 0. fx 3x 12 f x
3x 12 dx x 13 C2
f 2 1 C2 1 when C2 0.
30 mph 44 ftsec at a vt at 66 since v0 66 ftsec. a st t 2 66t since s0 0. 2 Solving the system vt at 66 44 a s(t t2 66t 264 2
f x x 13
we obtain t 245 and a 5512. We now solve 5512t 66 0 and get t 725. Thus, s
72 725 5512 2 5
2
66
725 475.2 ft.
Stopping distance from 30 mph to rest is 475.2 264 211.2 ft. 14. at 9.8 msec2 vt 9.8t v0 9.8t 40 st 4.9t2 40t s0 0 (a) vt 9.8t 40 0 when t
40 4.08 sec. 9.8
(b) s4.08 81.63 m (c) vt 9.8t 40 20 when t (d) s2.04 61.2 m
20 2.04 sec. 9.8
483
484
Chapter 4
Integration
16. x1 2, x2 1, x3 5, x4 3, x5 7 (a)
1 16 1 5 x 2 1 5 3 7 5 i1 i 5 5
(b)
x 2 1 5 3 7 210
5
1
i1
1
1
5
(c)
1
1
37
i
2x x
22 22 21 12 25 52 23 32 27 72 56
2 i
i
i1 5
(d)
x x i
1 2 5 1 3 5 7 3 5
i1
i2
1 18. y 9 x2, x 1, n 4 4
9 41 4 9 41 9 9 41 16 9 41 25
S4 1
22.5
9 419 9 41 16 9 41 25 9 9
s4 1
14.5 20. y x2 3, x Area lim
12
f ci x
8 6
lim
y
right endpoints 10
n
n→ i1
n→
2 n
n
2i n
i1
2
3
2 n
4 2 x 1
2 n 4i2 lim 3 n→ n i1 n2 lim
2 4 nn 12n 1 3n n n2 6
lim
43 n 1n2n 1 6 38 6 263
n→
n→
2
1 2 22. y x3, x 4 n Area lim
y 20
n
f ci x
15
n→ i1
n
1 2i lim 2 n→ i1 4 n
10
3
2 n
5 x
1 n 24i 24i2 8i3 lim 8 2 3 n→ 2n i1 n n n
4 n 3i 3i2 i3 1 2 3 n i1 n n n
lim
4 3 nn 1 3 nn 12n 1 1 n2n 12 n 2 3 n n 2 n 6 n 4
n→
4 6 4 1 15
1
lim
n→
2
2
3
4
Review Exercises for Chapter 4
24. (a) S m
b4b4 m2b4 b4 m3b4 b4 m4b4 b4 mb16 1 2 3 4 5mb8 2
s m0 (b) Sn
y
y = mx
b4 mb4b4 m2b4 b4 m3b4 b4 mb16 1 2 3 3mb8 2
f n n n n mn i n
bi
n
b
i1
sn
2
mbi
b
b
i1
n1 b bi m n n i0
bi f n
2 n
i1
n1 i0
mb2 nn 1 mb2n 1 2 n 2 2n
x=b
mb2 n 1n mb2n 1 i 2 n 2 2n i0
b b m n n
2
2 n1
mb2n 1 mb2n 1 1 2 1 1 lim mb bmb baseheight n→ n→ 2n 2n 2 2 2
(c) Area lim
b
(d)
mx dx
0
n
26.
lim
→ i1
12mx
b
2
0
1 mb2 2
3
3ci9 ci 2 xi
3x9 x2 dx
28.
y
1
6 5 3 2 1 x − 4 −3 −2 −1
1
2
3
4
−2
4
4
6
30. (a)
0
3
(b)
3
f x dx
0
16 x2 dx
1 42 8 2
6
f x dx
f x dx 4 1 3
3
6
f x dx
6
f x dx 1 1
3
4
(c)
f x dx 0
4
6
(d)
6
10 f x dx 10
3
12 12x2 dx x3 2
x4 2x2 5 dx
1
6 x 3
3
2
1
2
36.
2
2
38.
1
1 1 dx x2 x3
4
40.
1
5 x5
34.
1
1 1 x2 x3 dx 2 x 2x 1
4
4
t3
1 1 2
2 8 1 2 8 2 1
1
1
1
1 1
2
52 15
3 2t
2
2
t2 2 dt
2x3 5x 3
325 163 10 325 163 10
1
16 6 6 , (d) 9 3
sec2 t dt tan t
4
f x dx 101 10
3
3
32.
485
1
14 3
(semicircle)
x
486
Chapter 4
Integration
2
42.
x 4 dx
0
2
x2 4x 2
0
2
10
44.
1
x2 x 2 dx
6
5 4 y
2 1
3 x
−2 −1
2
1
3
4
2
5
x
−2
1
3
−1 −2
3
1
x1 x dx x12 x32 dx
48. Area
0
sec2 x dx
0
23x
2 2 4 3 5 15
32
3
2 x52 5
1
tan x
0
0
3
y 5 4
y
3 1
2
π 6
x 1
50.
1 20
2
x3 dx
0
8 x4
2 0
y
2
8
2
6
3 2 x
4
x3
2
( 3 2 , 2) x 1
52. Fx
1 x2
56.
x
2 1
8 1 1 2 24 3 3 2
y 7
46.
x3 x2 2x 3 2
1 x
2
2
54. Fx csc2 x
dx
x2 2 x2 dx
x3 1 2x C 3 x
π 3
x
10 7 9 3 6 2
Review Exercises for Chapter 4 58. u x3 3, du 3x2 dx
x2x3 3 dx
1 2 x3 312 3x2 dx x3 332 C 3 9
60. u x2 6x 5, du 2x 6 dx
x3 1 2x 6 1 2 1 dx x 6x 51 C C x2 6x 52 2 x2 6x 52 2 2x2 6x 5
62.
64.
66.
68.
70.
x sin 3x2 dx
cos x sin x
dx
1 1 sin 3x26x dx cos 3x2 C 6 6
sin x12 cos x dx 2sin x12 C 2sin x C
1 cot4 csc2 d cot 4csc2 d cot 5 C 5
1
x2x3 13 dx
0
6
72.
1 1 sec 2x tan 2x2 dx sec 2x C 2 2
sec 2x tan 2x dx
3
x3 133x2 dx
x2 8122x dx
1 3
x 1 dx 6 3x2 8
1
1 x3 14 12
0
6
3
13x
1 0
812
2
1 5 16 1 12 4
6 3
1 27 1 3
74. u x 1, x u 1, dx du When x 1, u 0. When x 0, u 1.
0
2
1
1
x2x 1 dx 2
u 12u du
0 1
2
u52 2u32 u12 du 2
0
4
76.
27u
72
4 2 u52 u32 5 3
1 0
32 105
sin 2x dx 0 since sin 2x is an odd function.
4
78. u 1 x, x 1 u, dx du When x a, u 1 a. When x b, u 1 b.
b
Pa, b
a
1155 3 1155 x 1 x32 dx 32 32
1155 32
1b
1 u3 u32 du
1a
1b
u92 3u72 3u52 u32 du
1a
1155 2u52 105u3 385u2 495u 231 32 1155
(a) P0, 0.25
16 105u
385u2 495u 231
(b) P0.5, 1
16 105u
385u2 495u 231
u52
u52
3
3
1155 2 11 2 2 92 6 72 2 52 u u u u 32 11 3 7 5
1b 1a
16 105u u52
0.025 2.5%
0.736 73.6%
0.75 1
0 0.5
3
1b 1a
385u2 495u 231
1b 1a
487
488
Chapter 4
2
80.
1.75 sin
0
Integration
t 2 t dt 1.75 cos 2 2
2
2 7 1.751 1 2.2282 liters
Increase is 7 5.1 1.9 0.6048 liters.
1
82. Trapezoidal Rule n 4:
0
1
Simpson’s Rule n 4:
0
x32 1 21432 21232 23432 1 0 0.172 2 dx 2 2 3x 8 3 14 3 12 3 342 2
x32 1 41432 21232 43432 1 0 0.166 2 dx 2 2 3x 12 3 14 3 12 3 342 2
Graphing utility: 0.166
84. Trapezoidal Rule n 4:
1 sin2 x dx 3.820
0
Simpson’s Rule n 4: 3.820 Graphing utility: 3.820
Problem Solving for Chapter 4
x
2. (a) Fx
sin t2 dt
2
x Fx (b) Gx x Gx
0
1.0
1.5
1.9
2.0
2.1
2.5
3.0
4.0
5.0
0.8048
0.4945
0.0265
0.0611
0
0.0867
0.3743
0.0312
0.0576
0.2769
1 x2
x
sin t2 dt
2
1.9
1.95
1.99
2.01
2.05
2.1
0.6106
0.6873
0.7436
0.7697
0.8174
0.8671
lim Gx 0.75
x→2
(c) F2 lim
x→2
Fx F2 x2
1 x→2 x 2
lim
x
sin t2 dt
2
lim Gx x→2
Since Fx sin x2, F2 sin 4 lim Gx. x→2
Note: sin 4 0.7568
Problem Solving for Chapter 4 4. Let d be the distance traversed and a be the uniform acceleration. We can assume that v0 0 and s0 0. Then at a vt at 1 s t at2. 2 st d when t
2da. 2da
2ad.
1 2ad 0 2
ad2 .
The highest speed is v a The lowest speed is v 0. The mean speed is
The time necessary to traverse the distance d at the mean speed is t
d ad2
2da
which is the same as the time calculated above. 6. (a)
y 100 80 60 40 20 x 0.2
0.4
0.6
0.8
1.0
(b) v is increasing (positive acceleration) on 0, 0.4 and 0.7, 1.0. (c) Average acceleration
v0.4 v0 60 0 150 mihr2 0.4 0 0.4
(d) This integral is the total distance traveled in miles.
1
vt dt
0
1 385 38.5 miles
0 220 260 240 240 65 10 10
(e) One approximation is a0.8
v0.9 v0.8 50 40 100 mihr2 0.9 0.8 0.1
(other answers possible)
489
490
Chapter 4
Integration
x
8.
x
f tx t dt
0
d dx
Thus,
0
x
x f t dt
x
t f t dt x
0
x
0
x
f tx t dt x f x
0
x
f t dt
t f t dt
0
x
f t dt x f x
0
f t dt
0
Differentiating the other integral, d dx
x
x
0
x
f v dv dt
0
f v dv.
0
Thus, the two original integrals have equal derivatives,
x
x
f tx t dt
0
t
0
f v dv dt C
0
Letting x 0, we see that C 0.
1
10. Consider
2 3
1
x dx x32
0
0
2 . The corresponding 3
Riemann Sum using right-hand endpoints is Sn Thus, lim
1 n
1 n
2 . . . n
1
1 2 . . . n n32
1 2 . . . n
n32
n→
3
12. (a) Area
nn
2 . 3
3
3
9 x2 dx 2
9 x2 dx
y
0
2 9x
x3 3
10
3
8 7 6 5 4 3 2 1
0
2 27 9 36 (b) Base 6, height 9. Area
2 2 bh 69 36. 3 3
−4
x
−2 −1
1 2
4 5
(c) Let the parabola be given by y b2 a2x2, a, b > 0.
ba
Area 2
b2 a2x2 dx
0
2
b2x
y
a2
x3 3
ba b2
0
a a3 a
2 b2
2
b
b
3
a 3 a 34 a
2 Base
b3
1 b3
b3 −
2b , height b2 a
Archimedes’ Formula: Area
2 2b 2 4 b3 b 3 a 3 a
b a
b a
x
Problem Solving for Chapter 4 14. (a) 1 i3 1 3i 3i2 i3 ⇒ 1 i3 i3 3i2 3i 1 3i2 3i 1 i 13 i3
(b) n
3i
3i 1
2
i1
n
i 1
i3
3
i1
23 13 33 23 . . . n 13 n3 n 13 1 n
Hence, n 13
3i
2
3i 1 1
i1
(c) n 13 1
n
3i
2
3i 1
i1 n
3i
⇒
2
n
i
2
2
i1
3nn 1 n 2
3nn 1 n 2
n3 3n2 3n
i1
⇒
n
3i
2n3 6n2 6n 3n2 3n 2n 2
2n3 3n2 n 2
nn 12n 1 2
nn 12n 1 6
i1
20
16. (a) C 0.1
12 sin
8
18
(b) C 0.1
12 sin
10
t 8 14.4 t 8 dt cos 12 12
0
0
b b
0 b
0
Then,
f x dx. f x f b x
f b u du f b u f u f b u du f b u f u f b x dx f b x f x
b
2A
0
f x dx f x f b x
b
0
f b x dx f b x f x
b
1 dx b.
0
b Thus, A . 2
1
(b) b 1 ⇒
0
14.4 1 1 $9.17 18 10
3 14.4 3 10.8 6 $3.14 14.4 2 2
Let u b x, du dx. A
8
Savings 9.17 3.14 $6.03. b
t 8 14.4 t 8 6 dt cos 0.6t 12 12
18. (a) Let A
20
sin x 1 dx sin1 x sin x 2
491
C H A P T E R Integration
4
Section 4.1
Antiderivatives and Indefinite Integration . . . . . . . . . 177
Section 4.2
Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
Section 4.3
Riemann Sums and Definite Integrals . . . . . . . . . . . 188
Section 4.4
The Fundamental Theorem of Calculus . . . . . . . . . . 192
Section 4.5
Integration by Substitution . . . . . . . . . . . . . . . . . 197
Section 4.6
Numerical Integration
Review Exercises
. . . . . . . . . . . . . . . . . . . 204
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
C H A P T E R Integration Section 4.1
4
Antiderivatives and Indefinite Integration
Solutions to Odd-Numbered Exercises
1.
d 3 d 9 C 3x3 C 9x4 4 dx x3 dx x
3.
d 1 3 x 4x C x2 4 x 2x 2 dx 3
5.
dy 3t2 dt
7.
dy x3 2 dx
y t3 C Check:
x x
13.
1 dx 2x3
15.
11.
Integrate
Simplify
x 4 3
3 4 3 x C 4
4 3
x1 2 C 1 2
1 x2 C 2 2
x2 3x C 2
x
17.
21.
x2 3 dx
3 x5 3 C x5 3 C 5 3 5
d 3 5 3 3 x2 x C x2 3 dx 5
C
2x 3x 2dx x 2 x3 C
Check:
d 1 4 x 2x C x3 2 dx 4
3 x2 dx
d 2 5 2 x C x3 2 dx 5
1 C 4x2
1 x3 2 dx x 4 2x C 4
Check:
2
d x2 3x C x 3 dx 2
C
x3 2 dx
1 3 x dx 2
x 3dx
Check:
23.
x1 3 dx
dx
Check:
19.
3 x dx
1
Check:
Rewrite
9.
2 y x5 2 C 5
d 3 t C 3t2 dt
Given
2 x3 2 2x 1 dx x5 2 x2 x C 5
Check:
25.
d 2 x x3 C 2x 3x 2 dx
1 dx x3
Check:
d 2 5 2 x x2 x C x3 2 2x 1 dx 5
x3 dx
x2 1 C 2C 2 2x
d 1 1 C 3 dx 2x2 x
177
178
27.
Chapter 4
x2 x 1 dx x
Check:
29.
Integration
2 2 2 x3 2 x1 2 x1 2 dx x5 2 x3 2 2x1 2 C x1 23x2 5x 15 C 5 3 15
d 2 5 2 2 3 2 x2 x 1 x x 2x1 2 C x3 2 x1 2 x1 2 dx 5 3 x
x 13x 2 dx
3x2 x 2 dx
31.
1 x3 x2 2x C 2 Check:
y2 y dy
2 y5 2 dy y7 2 C 7
d 2 7 2 y C y5 2 y2 y dy 7
Check:
d 3 1 2 x x 2x C 3x2 x 2 dx 2 x 13x 2
33.
37.
dx
1 dx x C
Check:
d x C 1 dx
1 csc t cot t dt t csc t C
39.
d t csc t C 1 csc t cot t dt
tan2 y 1 dy
2 sin x 3 cos x dx 2 cos x 3 sin x C d 2 cos x 3 sin x C 2 sin x 3 cos x dx
Check:
Check:
41.
35.
sec2 sin d tan cos C d tan cos C sec2 sin d
Check:
43. f x cos x
sec2 y dy tan y C
y
d Check: tan y C sec2 y tan2 y 1 dy
3 2
C
3
2C
0
x 2 2
3
45. f x 2
47. fx 1 x2
f x 2x C
f x x
f )x)
2x
2
y
x3 3
f )x)
5
f )x)
f )x)
2x
x
3 2
x 3
2
1
2
3
x 3
2
3 1
Answers will vary.
f
2
Answers will vary.
2x 1 dx x2 x C
1 12 1 C ⇒ C 1
x3 3
4
3
dy 2x 1, 1, 1 dx y
x y
2
49.
x3 C 3
4
f′
C
3
y x2 x 1
Section 4.1
51.
dy cos x, 0, 4 dx y
Antiderivatives and Indefinite Integration
179
53. (a) Answers will vary. y
5
cos x dx sin x C
4 sin 0 C ⇒ C 4 y sin x 4
x
−3
5
−3
(b)
dy 1 x 1, 4, 2 dx 2 x2 xC 4
y
6
−4
42 2 4C 4
8 −2
2C x2 x2 4
y
55. fx 4x, f 0 6 f x
4x dx 2x 2 C
57. ht 8t3 5, h1 4 ht
8t3 5dt 2t4 5t C
f0 6 202 C ⇒ C 6
h1 4 2 5 C ⇒ C 11
f x 2x 2 6
ht 2t4 5t 11 61. f x x3 2
59. f x 2 f2 5
f4 2
f 2 10
f 0 0
fx
fx
2 dx 2x C1
f2 4 C1 5 ⇒ C1 1 fx 2x 1 f x
2x 1 dx x2 x C2
f 2 6 C2 10 ⇒ C2 4 f x x x 4 2
x3 2 dx 2x1 2 C1
fx f x
2 x
1.5t 5 dt 0.75t 2 5t C
h0 0 0 C 12 ⇒ C 12 ht 0.75t2 5t 12 (b) h6 0.7562 56 12 69 cm
C1
3
2x1 2 3 dx 4x1 2 3x C2
f x 4x1 2 3x 4 x 3x
x
2 f4 C1 2 ⇒ C1 3 2
f 0 0 0 C2 0 ⇒ C2 0
63. (a) ht
2
180
Chapter 4
Integration
65. f 0 4. Graph of f is given. (a) f4 1.0
(f) f is a minimum at x 3.
(b) No. The slopes of the tangent lines are greater than 2 on 0, 2. Therefore, f must increase more than 4 units on 0, 4.
(g)
y 6 4
(c) No, f 5 < f 4 because f is decreasing on 4, 5. (d) f is an maximum at x 3.5 because f3.5 0 and the first derivative test.
2 x −2
(e) f is concave upward when f is increasing on , 1 and 5, . f is concave downward on 1, 5. Points of inflection at x 1, 5. 67. at 32 ft sec2 vt
v0 st 32t v0 0 when t time to reach 32 maximum height.
32t 60dt 16t 2 60t C2 s
st 16t2 60t 6 Position function
The ball reaches its maximim height when
15 16 15 18 8
v0
2
60
32 550 v0
v02 v02 550 64 32
73. From Exercise 71, f t 4.9t2 10t 2. v t 9.8t 10 0 (Maximum height when v 0.)
9.8 dt 9.8t C1
9.8t 10
v0 v0 C1 ⇒ vt 9.8t v0
t
9.8t v0 dt 4.9t v0t C2 2
f 0 s0 C2 ⇒ f t 4.9t 2 v0t s0
f
10 9.8
10 9.8 7.1 m
a 1.6
st
v0
158 6 62.26 feet
vt
2
15 seconds 8
71. at 9.8
75.
v0
v0 187.617 ft sec
32t 60 t
32 1632 v02 35,200
vt 32t 60 0
f t
8
st 16t2 v0t
32 dt 32t C1
vt
6
−6
s0 6 C2
s
4
69. From Exercise 68, we have:
v0 60 C1 st
2
1.6 dt 1.6t v0 1.6t, since the stone was dropped, v0 0.
1.6t dt 0.8t2 s0
s20 0 ⇒ 0.8202 s0 0 s0 320 Thus, the height of the cliff is 320 meters. vt 1.6t v20 32 m sec
Section 4.1
0 ≤ t ≤ 5
77. xt t3 6t2 9t 2
Antiderivatives and Indefinite Integration
79. vt
(a) vt xt 3t 12t 9 2
xt
3t2 4t 3 3t 1t 3
t1 2
t > 0
vt dt 2t1 2 C
x1 4 21 C ⇒ C 2
at vt 6t 12 6t 2 (b) vt > 0 when 0 < t < 1 or 3 < t < 5.
xt 2t1 2 2 position function
(c) at 6t 2 0 when t 2.
1 1 at vt t3 2 3 2 acceleration 2 2t
v2 311 3 81. (a) v0 25 km hr 25
1000 250 m sec 3600 36
v13 80 km hr 80
1000 800 m sec 3600 36
83. Truck: vt 30 st 30t Let s0 0. Automobile: at 6
at a constant acceleration
vt 6t Let v0 0.
vt at C
st 3t2 Let s0 0.
v0 v13
At the point where the automobile overtakes the truck:
250 250 ⇒ vt at 36 36
30t 3t2
800 250 13a 36 36
0 3t2 30t 0 3tt 10 when t 10 sec.
550 13a 36 a st a
(b)
s13 85.
1 t
(a) s10 3102 300 ft (b) v10 610 60 ft sec 41 mph
550 275
1.175 m sec2 468 234
t 2 250 t 2 36
s0 0
275 132 250 13 189.58 m 234 2 36
1 mi hr5280 ft mi 22 ft sec 3600 sec hr 15 (a)
t
0
5
10
15
20
25
30
V1ft sec
0
3.67
10.27
23.47
42.53
66
95.33
V2ft sec
0
30.8
55.73
74.8
88
93.87
95.33
(c) S1t S2t
V1t dt
0.1068 3 0.0416 2 t t 0.3679t 3 2
V2t dt
0.1208t3 6.7991t2 0.0707t 3 2
In both cases, the constant of integration is 0 because S10 S20 0 S130 953.5 feet S230 1970.3 feet The second car was going faster than the first until the end.
(b) V1t 0.1068t2 0.0416t 0.3679 V2t 0.1208t2 6.7991t 0.0707
181
182
Chapter 4
Integration
87. at k vt kt k st t2 since v0 s0 0. 2 At the time of lift-off, kt 160 and k2t2 0.7. Since k2t2 0.7,
1.4k 1.4 1.4 160 v k k k t
1.4k 1602 ⇒ k
1602 1.4
18,285.714 mihr2 7.45 ftsec2. 89. True
91. True
93. False. For example,
95. fx
1,3x,
x
x dx
x dx
x dx because
x2 x3 C C1 3 2
x2 C 2
2
0 ≤ x < 2 2 ≤ x ≤ 5
x C1, 0 ≤ x < 2 f x 3x2 C2, 2 ≤ x ≤ 5 2 f 1 3 ⇒ 1 C1 3 ⇒ C1 2 f is continuous: Values must agree at x 2: 4 6 C2 ⇒ C2 2
0 ≤ x < 2 x 2, f x 3x 2 2, 2 ≤ x ≤ 5 2 The left and right hand derivatives at x 2 do not agree. Hence f is not differentiable at x 2.
Section 4.2 1.
5
5
5
i1
i1
i1
2i 1 2 i 1 21 2 3 4 5 5 35 4
3.
k
2
k0 9
7.
Area
1 1 1 1 1 158 1 1 2 5 10 17 85 9.
i1 20
15.
i1
58 3 8
1
3i
j
4
5.
k1
i 2 20
i1
2021 420 2
n
11.
2 n n i1
17.
i 1
j1
2i 2
c c c c c 4c 2i
20
2
3
i1
2in
13.
19
i
2
i1
192039 2470 6
3 n 3i 2 1 n i1 n
2
15
19.
ii 1
2
i1
15
i
2
3
i1
15
i
2
15
i
i1
21. sum seqx
>
Section 4.2
2 3, x, 1, 20, 1 2930 (TI-82)
i1
20
i
152162 151631 1516 2 4 6 2
2
3
i1
14,400 2,480 120
2020 1220 1 320 6
202141 60 2930 6
12,040 9 33 23. S 3 4 2 51 2 16.5
s 1 3 4 1 9 2
25 2
25. S 3 3 51 11
12.5
s 2 2 31 7
14 14 12 14 34 14 114 1 2 8 3 2 0.768 1 2 3 1 1 1 1 1 3 1 s4 0 0.518 4 4 4 2 4 4 4 8
27. S4
29. S5 1 s5
n→
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0.646 65 5 75 5 85 5 95 5 2 5 6 7 8 9 10
81n n n 4 1 814 lim n 2
31. lim
1 1 1 1 1 1 1 1 15 65 5 75 5 85 5 95 5 51 61 71 81 91 0.746
2
4
4
n→
81 81 2n3 n2 1 n4 4 4
18n nn 2 1 182 lim n n n 1821 9 2
33. lim n→
2
2i 1 1 n 1 nn 1 n2 n Sn 2 2i 1 2 2 2 n n n 2 n i1 i1 n
35.
2
n→
S10
12 1.2 10
S100 1.02 S1000 1.002 S10,000 1.0002 6kk 1 6 n 2 6 nn 12n 1 nn 1 3 k k 3 3 n n k1 n 6 2 k1 n
37.
6 2n2 3n 1 3n 3 1 22n2 2 Sn n2 6 n
S10 1.98 S100 1.9998 S1000 1.999998 S10,000 1.99999998
n n
39. lim
n→ i1
16i 2
lim
n→
16 n 16 nn 1 n2 n i lim 2 lim 8 n→ n n→ n2 i1 2 n2
Area
8 lim 1 n1 8 n→
183
184
Chapter 4
Integration
1 1 n1 1 n 1n2n 1 i 12 lim 3 i 2 lim 3 3 n→ i1 n n→ n i1 n→ n 6 n
41. lim
lim
n→
1 2n3 3n2 n 1 2 3n 1n2 lim n→ 6 6 n3 1
1 nn 1 2
1 nn 2 lim n 1 n i 2 lim n n n n
43. lim
i
45. (a)
n
1
2
n→ i1
n→
1
i1
n
1
n→
i1
y
(e)
x
31
2 lim 1 n 2n n 21 21 3 2
2
n→
5
10
50
100
3
sn
1.6
1.8
1.96
1.98
2
Sn
2.4
2.2
2.04
2.02
1
i 12n 2n lim n4 i 1 n
1
(b) x
n
(f) lim
x
n→
3
n→
i1
Endpoints:
lim
n→
2n < 22n < . . . < n 12n < n2n 2 n
f x
i n n n
2
lim
n→
(c) Since y x is increasing, f mi f xi1 on xi1, xi. sn
2
i1
x
n4 nn 2 1
lim
2n 1 2 n
n→
i1
n 2 2 i 1 n n i1
2 n
2n n 1 n4 2
4 n i n→ n2 i1 lim
i1
2i 2 f n
lim
i1
n
i1
4 nn 1 n lim 2 n→ n 2
20 2 n n
0
49. Because the curve is concave upward, the midpoint approximation will be less than the actual area: <
dx
Section 4.3
51. f x
1 x4
53.
is not integrable on the interval 3, 5 and f has a discontinuity at x 4.
Riemann Sums and Definite Integrals
y
4 3 2 1 1
2
3
x
4
a. A 5 square units 55.
y
2 3 2
1 1 2
x 1 2
1
3 2
2
1
d.
1 2 sin x dx 12 1 2
0
3
57.
x 3 x dx
0
n
4
8
12
16
20
Ln
3.6830
3.9956
4.0707
4.1016
4.1177
Mn
4.3082
4.2076
4.1838
4.1740
4.1690
Rn
3.6830
3.9956
4.0707
4.1016
4.1177
4
8
12
16
20
Ln
0.5890
0.6872
0.7199
0.7363
0.7461
Mn
0.7854
0.7854
0.7854
0.7854
0.7854
Rn
0.9817
0.8836
0.8508
0.8345
0.8247
2
59.
sin2 x dx
0
n
61. True
63. True
65. False
2
0
67. f x x2 3x, 0, 8 x0 0, x1 1, x2 3, x3 7, x4 8 x1 1, x2 2, x3 4, x4 1 c1 1, c2 2, c3 5, c4 8 4
f c x f 1 x i
1
f 2 x2 f 5 x3 f 8 x4
i1
41 102 404 881 272
x dx 2
191
192
Chapter 4
69. f x
1,0,
Differentiation
x is rational x is irrational
is not integrable on the interval 0, 1. As → 0, f ci 1 or f ci 0 in each subinterval since there are an infinite number of both rational and irrational numbers in any interval, no matter how small. 71. Let f x x2, 0 ≤ x ≤ 1, and xi 1 n. The appropriate Riemann Sum is
f c x n n
n
i
i
2
i
i1
i1
lim
n→
1 n 1 3 i 2. n n i1
1 2 1 1 22 32 . . . n2 lim 3 n→ n n3
n→
1. f x
0
n2n 1n 1 6
1 1 1 2n2 3n 1 1 lim n→ 3 6n2 2n 6n2 3
lim
Section 4.4
The Fundamental Theorem of Calculus
4 x2 1
5
5
3. f x x x2 1
2
4 dx is positive. x2 1
−5
2
5
x x2 1 dx 0
−5
−5
−2
2x dx x2
t2 2 dt
t3 2t
2t 12 dt
0
101
0
1
9.
1
1
1
4
1
u2 du
u
34t
t1 3 t2 3 dt
1
3
23.
2t
1 1
43t
3
2t2 t
34t
4 3
3 2
1 0
4 1 21 3 3
23 4
4u1 2
3 t5 3 5
23u
1 x2 2 3 2 x 3 2 3
x x1 2 dx
0
0
0
1
3 2 0
3
1
23 4 32
4 4
1 0
1 1 2 1 3 2 3 18
2x 3 dx split up the integral at the zero x
3 2
3x x2
4
27 34 53 20
3
3 2x dx
0
0
34 2 34 2 4
3 2
2x 3 dx
0
4 3
1
0
21.
1
1
1
x x 1 dx 3 3
0
0
2
2
u1 2 2u1 2 du
1
x2 2x
13 2 31 2 103
4
x 2 dx
23 2 3 1 21
3 t 2 dt
1
19.
4t2 4t 1 dt
3 3 1 dx x x2 x
1
17.
1
0
2
15.
7.
1
0
13.
1
3
1
11.
0
1
1
5.
5
x2 3x
3 3 2
3 2
92 49 0 9 9 94 29 292 49 29
12 2 52
Section 4.4
3
25.
0
2
x 2 4 dx
3
4 x 2 dx
0
8
27.
x3 3
2 0
4 sec tan d 4 sec
6
6
3
10,000t 6 dt 10,000
0
3
37. A
3
3 3
3
2
3
233
3 0
2
0
x x2 dx
0
1
35. A
$135,000
x 5 x10 2x
2 3x1 2 x3 2 dx 2x3 2 x5 2 5 0
cos x dx sin x
0
3
42 42 0
t2 6t 2
3
3 x x dx
0
39. A
1 0 1 2
0
2
sec2 x dx tan x
3
33.
3
3
3
8 8 9 12 8 3 3
6
x3 4x
1 sin x dx x cos x
6
31.
23 3
0
29.
x 2 4 dx
2
4x
The Fundamental Theorem of Calculus
3
3
0
0
x2 x3
3 1
2
12 3 5
41. Since y ≥ 0 on 0, 2,
1
2
A
2
0
x3 x dx
2
43. Since y ≥ 0 on 0, 2,
45.
x4 x2 4 2
2 0
0
4 2 6.
3x2 1 dx x3 x
0
A
1 6
0
x 2 x dx
x2 4x3
3 2 2
2
0
f c2 0
6 8 2 3
c 2 c
3 4 2 3
c 2 c 1
2 0
8 2 10.
2
8 2 3
3 4 2 1 3
c 12 6 34
2
6 34 2 64 2 c 1 ± 3
c 1 ±
2
c 0.4380 or c 1.7908
193
194
Chapter 4
4
47.
Integration
4
2 sec2 x dx 2 tan x
4
21 21 4
4
4 4 4
f c
2 sec2 c
8
sec2 c
4
sec c ±
2
c ± arcsec
2
± arccos
49.
2
1 2 2
2
4 x2 dx
Average value 4 x2
51.
1 0
2
± 0.4817
2
1 1 4x x3 4 3
2
1 4
8 38 8 38 38
sin x dx
0
1 cos x
0
3 3
, 8 3
(
5
(
2
3 3
, 8 3
(0.690, π2 ( − 2
(2.451, π2 (
−1
6
57.
0
0
6
2 f x dx
0
0
f x dx Fb Fa.
a
3 2
f x dx area of region A 1.5
6
b
then
2
2
3
53. If f is continuous on a, b and F x f x on a, b,
2
2
0
6
f x dx f x dx
f x dx 1.5 5.0 6.5
2
6
2 dx
f x dx
0
12 3.5 15.5 61. (a) F x k sec2 x
(b)
F 0 k 500
1 3 0
3
500 sec2 x dx
0
F x 500 sec2 x
3
1500 tan x
0
1500 3 0
826.99 newtons 827 newtons
63.
1 50
5
0
(
−1
2
x 0.690, 2.451
59.
2
−3
8 8 2 3 ± 1.155. when x2 4 or x ± 3 3 3
sin x
55.
(−
8 3
2 Average value
0.1729t 0.1552t2 0.0374t3 dt
1 0.08645t2 0.05073t3 0.00935t4 5
5 0
0.5318 liter
Section 4.4
65. (a)
The Fundamental Theorem of Calculus
(b)
1
0
10
24
0
−1
24 0
The average value of S appears to be g.
The area above the x-axis equals the area below the x-axis. Thus, the average value is zero. 67. (a) v 8.61 104t3 0.0782t2 0.208t 0.0952 (b)
90
− 10
70 − 10
60
(c)
vt dt
0
8.61 4 10
4t 4
x
69. Fx
t 5 dt
0
t2 5t 2
x 0
0.0782t3 0.208t 2 0.0952t 3 2
x2 5x 2
60 0
2476 meters
x
71. Fx
x
10 2 dv 1 v
1
10 1 10 10 1 x x
F2
4 52 8 2
F5
25 25 55 2 2
F2 10
12 5
F8
64 58 8 2
F5 10
45 8
F8 10
78 354
x
73. Fx
cos d sin
1
x 1
x
sin x sin 1
75. (a)
t 2 dt
0
F2 sin 2 sin 1 0.0678
10 v
10v2 dv
x
t2 2t 2
0
x 1
1 x2 2x 2
d 1 2 (b) x 2x x 2 dx 2
F5 sin 5 sin 1 1.8004 F8 sin 8 sin 1 0.1479
77. (a)
x 3 t
dt
8
(b)
4t
3
x
4 3
8
79. (a)
x 4
d 3 4 3 3 x x 12 x1 3 dx 4
2
(b)
sec2 t dt tan t
x x 4
tan x 1
d tan x 1 sec2 x dx
x
x
81. F x
x
3 3 x 4 3 16 x 4 3 12 4 4
t 2 2t dt
F x x2 2x
83. F x
1
x
t 4 1 dt
F x x4 1
85. F x
t cos t dt
0
F x x cos x
195
196
Chapter 4
Integration
x2
87. Fx
4t 1 dt
Alternate solution:
x
x2
Fx
2t t 2
x2
4t 1 dt
x
x
2x 22 x 2 2x2 x
0
x
8x 10
x2
4t 1 dt
0
x
F x 8
4t 1 dt
x2
4t 1 dt
0
4t 1 dt
0
F x 4x 1 4x 2 1 8
sin x
89. Fx
t dt
0
23t 3 2
sin x 0
2 sin x3 2 3
F x sin x1 2 cos x cos x sin x
x3
91. Fx
sin t 2 dt
0
F x sinx32
3x2 3x2 sin x 6
Alternate solution
sin x
Fx
t dt
0
F x sin x
d sin x sin xcos x dx
x
93. gx
f t dt
0
1 1 g0 0, g1 , g2 1, g3 , g4 0 2 2
45t
x
5000 25
2
(b)
f
g
2
3
x
5 4
0
12 5 4 x 1000125 12x5 4 5
C1 1000125 121 $137,000 C5 1000125 1255 4 $214,721
x 1
t1 4 dt
0
5000 25 3
y
1
95. (a) Cx 5000 25 3
C10 1000125 12105 4 $338,394
4
−1 −2
g has a relative maximum at x 2.
1
99. False;
97. True
1
0
x2 dx
1
1
x2 dx
x2 dx
0
Each of these integrals is infinite. f x x2 has a nonremovable discontinuity at x 0.
1 x
101. f x
0
1 dt t 1 2
x
1 dt 2 t 1 0
By the Second Fundamental Theorem of Calculus, we have f x
1 1 1 2 1 x2 1 x2 x 1
1 1 0. 1 x2 x2 1
Since f x 0, f x must be constant.
Section 4.5 103. xt t 3 6t 2 9t 2 xt 3t 2 12t 9 3t 2 4t 3 3t 3t 1
5
Total distance
xt dt
0
5
3 t 3t 1 dt
0
1
3
t 2 4t 3 dt 3
3
0
5
t 2 4t 3dt 3
1
t 2 4t 3dt
3
4 4 20 28 units
4
105. Total distance
xt dt
1
4
vt dt
1 4 1
1
t
dt
2t12
4 1
22 1 2 units
Section 4.5
1. 3. 5.
7.
f g x
g x dx
u g x
du g x dx
5x2 1210x dx
5x2 1
10x dx
x2 1
2x dx
tan x
sec2 x dx
x
x2 1
dx
tan2 x sec2 x dx
1 2x4 2 dx
Check:
9.
Integration by Substitution
1 2x5 C 5
d 1 2x5 C 21 2x4 dx 5
9 x2122x dx
Check:
9 x232 2 C 9 x232 C 32 3
2 d 2 9 x232 C dx 3 3
3
29 x2122x 9 x22x
Integration by Substitution
197
198
11.
Chapter 4
x3x4 32 dx
Check:
13.
17.
t2 232 1 2 1 t2 232 C C t 2122t dt 2 2 32 3
32t2 2122t d t2 232 C t2 212t dt 3 3
5 5 1 x2132x dx 2 2
d 15 15 1 x243 C dx 8 8
1 x243 15 C 1 x243 C 43 8
4
3 1 x2 31 x2132x 5x1 x213 5x
1 x 1 d C 21 x232x dx 41 x22 4 1 x23
x2 1 1 1 x31 1 dx 1 x323x2 dx C C 1 x32 3 3 1 31 x3
x
1
d 1 1 x2 C 11 x323x2 3 dx 31 x 3 1 x32
1 x2
1 1 1 x22 x 1 dx 1 x232x dx C C 2 3 1 x 2 2 2 41 x22
Check:
27.
5x1 x213 dx
Check:
25.
Check:
23.
1 1 x3 15 x3 15 x3 143x2 dx C C 3 3 5 15
d x3 15 5x3 143x2 C x2x3 14 dx 15 15
tt2 2 dt
Check:
21.
Check:
19.
x2x3 14 dx
Check:
x4 33 1 1 x4 33 C C x4 324x3 dx 4 4 3 12
d x4 33 3x4 32 3 C 4x x4 32x3 dx 12 12
Check:
15.
Integration
dx
1 1 1 x212 1 x2122x dx C 1 x2 C 2 2 12
d 1 x 1 x212 C 1 x2122x 1 x2 dx 2 1 t
t1 dt 1 1t t1 dt 1 41t 3
3
2
4
2
1 1 1 1t 4 d C 4 1 dt 4 4 t
t1 t1 1 1t 3
2
1 1 1 2x12 dx 2x12 2 dx C 2x C 2 2 12 2x
Check:
d 2x C 212x122 1 dx 2x
C
2
3
Section 4.5
29.
31.
33.
x2 3x 7 dx x
2 2 x32 3x12 7x12 dx x52 2x32 14x12 C xx2 5x 35 C 5 5
Check:
d 2 52 x2 3x 7 x 2x32 14x12 C dx 5 x
2 dt t
Check:
d 14 2 t t2 C t 3 2t t 2 t dt 4 t
t2 t
1 t3 2t dt t 4 t2 C 4
9 yy dy
35. y
Check:
9y12 y32 dy 9
23y 52y 32
52
2 C y3215 y C 5
d 2 32 d 2 6y32 y52 C 9y12 y32 9 yy y 15 y C dy 5 dy 5
4x
4x 16 x2
dx
37. y
4 x dx 2 16 x2122x dx 4
Integration by Substitution
x2 2 16 12x C 2
2 12
2x2 416 x2 C
39. (a) 3
1 x2 2x 322x 2 dx 2
1 x2 2x 31 C 2 1
1 C 2x2 2x 3
x4 x2 dx
x 2
dy x4 x2, 2, 2 dx y
−2
x1 dx x2 2x 32
(b)
y
−1
1 2
1 4 x2122x dx 2
2
1
34 x232 C 34 x232 C
1 2, 2: 2 4 2232 C ⇒ C 2 3 1 y 4 x232 2 3 2
−2
2
−1
41.
45.
sin x dx cos x C
1 1 1 1 1 cos d cos 2 d sin C 2
43.
sin 2x dx
1 1 sin 2x2x dx cos 2x C 2 2
199
200
Chapter 4
Integration
1 2
49.
1 tan5 x C tan5 x C 5 5
51.
47.
1 1 sin 2x2 1 sin 2x2 cos 2x dx C sin2 2x C 2 2 2 4
sin 2x cos 2x dx
sin 2x cos 2x dx sin 2x cos 2x dx
tan4 x sec2 x dx
1 1 cos 2x2 1 cos 2x2 sin 2x dx C1 cos2 2x C1 2 2 2 4 2 sin 2x cos 2x dx
1 2
sin 4x dx
OR
1 cos 4x C2 8
csc2 x dx cot x3csc2 x dx cot3 x
53.
OR
cot2 x dx
cot x2 1 1 1 1 C C tan2 x C sec2 x 1 C sec2 x C1 2 2 cot2 x 2 2 2
csc2 x 1 dx cot x x C
55. f x
cos
x x dx 2 sin C 2 2
Since f 0 3 2 sin 0 C, C 3. Thus, f x 2 sin 57. u x 2, x u 2, dx du
xx 2 dx
u 2u du u32 2u12 du
4 2 u52 u32 C 5 3
2u32 3u 10 C 15
2 x 232 3x 2 10 C 15
2 x 2323x 4 C 15
59. u 1 x, x 1 u, dx du
x21 x dx 1 u2u du u12 2u32 u52 du
23u
2u32 35 42u 15u2 C 105
2 1 x32 35 421 x 151 x2 C 105
2 1 x3215x2 12x 8 C 105
32
4 2 u52 u72 C 5 7
x 3. 2
Section 4.5
1 1 61. u 2x 1, x u 1, dx du 2 2
x2 1 dx 2x 1
12u 1 2 1 1 du 2 u
1 12 2 u u 2u 1 4 du 8 1 u32 2u12 3u12 du 8
1 2 52 4 32 u u 6u12 C 8 5 3
u12 2 3u 10u 45 C 60
2x 1
60
32x 12 102x 1 45 C
1 2x 112x2 8x 52 C 60
1 2x 13x2 2x 13 C 15
63. u x 1, x u 1, dx du
x dx x 1 x 1
u 1 du u u
du
u 1 u 1
u u 1
1 u12 du u 2u12 C u 2u C x 1 2x 1 C x 2x 1 1 C x 2x 1 C1 where C1 1 C. 65. Let u x2 1, du 2x dx.
1
1
xx2 13 dx
1 2
1
1
x2 132x dx
8x
67. Let u x3 1, du 3x2 dx
2
1
2x2x3 1 dx 2
1 3
2
23 x
x3 1123x2 dx
1
3
132 32
2 1
4 3 x 132 9
2 1
4 27 22 12 982 9
1
2
14
1 1
0
Integration by Substitution
201
202
Chapter 4
Integration
69. Let u 2x 1, du 2 dx.
4
0
1 1 dx 2 2x 1
71. Let u 1 x, du
9
1
4
2x 1122 dx 2x 1
0
4
9 1 2
0
1 dx. 2x
1 dx 2 2 x 1 x
9
1
1 x 2
21 x dx 1 2x
9 1
1 1 1 2 2
73. u 2 x, x 2 u, dx du When x 1, u 1. When x 2, u 0.
2
1
23x dx 32 sin23x
0
0
u32 u12 du
1
2
cos
0
2 u 1 u du
1
2
75.
0
x 12 x dx
5u 2
52
2 u32 3
0 1
5 3 154 2
2
3 3 33 2 2 4
77. u x 1, x u 1, dx du When x 0, u 1. When x 7, u 8.
7
Area
8
3 x 1 dx x
0
3 u du u 1
1 8
u43 u13 du
1
79. A
2 sin x sin 2x dx 2 cos x
0
81. Area
4
83.
0
23
2
2x dx 2
23
sec2
2
sec2
73
3 3 12 384 7 7 4
3 u43 4
0
23
2
1
2 3 1
7
85.
xx 3 dx 28.8
3
144 5
2x 12 dx
2x 12 dx
0
−1 0
1 1 4 1 2x 12 2 dx 2x 13 C1 x3 2x2 x C1 2 6 3 6 4 4x2 4x 1 dx x3 2x2 x C2 3
1 They differ by a constant: C2 C1 . 6
d 7.377 6
4
8
0
−1
cos 5
5
3
87.
15
−1
1209 28
4
2x 12 dx 2 tan2x
x 10 dx 3.333 3 2x 1
8
1 cos 2x 2
3
89.
37u
−1
Section 4.5
91. f x x2x2 1 is even.
2
93. f x xx2 13 is odd.
2
2
x2x2 1 dx 2
x4 x2 dx 2
0
325 3 8
2
2
95.
0
x2 dx
3 x3
0
2 0
2
x2 dx
x2 dx
0
2
0
4
2
2
0
0
xx2 13 dx 0
272 15
2
8 3
(b)
2
(d)
2
4
x3 6x2 2x 3 dx
2
x2 dx 2
4
4
16 3
2
3x2 dx 3
x2 dx 8
0
4
x3 2x dx
x2 dx
0
0
8 x2 dx 3
4
97.
2
2
x2 dx
(c)
x5 x3 5 3
8 ; the function x2 is an even function. 3 2
(a)
Integration by Substitution
4
6x2 3 dx 0 2
6x2 3 dx 2 2x3 3x
0
4 0
232
99. Answers will vary. See “Guidelines for Making a Change of Variables” on page 292.
2
101. f x xx2 12 is odd. Hence,
2
xx2 12 dx 0.
k dV dt t 12
103.
Vt
k k C dt t 12 t1
V0 k C 500,000 1 V1 k C 400,000 2 Solving this system yields k 200,000 and C 300,000. Thus, Vt
200,000 300,000. t1
When t 4, V4 $340,000.
105.
1 ba (a) (b) (c)
b
74.50 43.75 sin
a
262.5 t 1 74.50t cos 3 6
t 1 262.5 t dt 74.50t cos 6 ba 6
262.5 t 1 74.50t cos 3 6
3 0
1 262.5 t 74.50t cos 12 6
6 3
b a
1 262.5 223.5 102.352 thousand units 3
1 262.5 447 223.5 102.352 thousand units 3
12 0
1 262.5 262.5 894 74.5 thousand units 12
203
204
Chapter 4
107.
1 ba (a) (b)
Integration
b
1 1 1 cos60 t sin120t b a 30 120
2 sin60 t cos120 t dt
a
1240 0
30 1 2 1201 301
240
2x 12 dx
1 1 2x 12 2 dx 2x 13 C 2 6
(c)
1 1 1 cos60 t sin120 t 130 0 30 120
109. False
10
10
10
ax3 bx2 cx d dx
30 301 0 amps
30
1
10
10
ax3 cx dx
10
Odd 113. True
2 5 2 2 1.382 amps
111. True
130 0
1
60
0
1 1 1 cos60 t sin120 t 1240 0 30 120
10
bx2 d dx 0 2
bx2 d dx
0
Even
4 sin x cos x dx 2 sin 2x dx cos 2x C 115. Let u x h, then du dx. When x a, u a h. When x b, u b h. Thus,
b
bh
f x h dx
a
x2 dx
0 2
Trapezoidal:
x2 dx
0 2
Simpson’s:
x2 dx
0
2
3. Exact:
x3 dx
0 2
Trapezoidal:
x3 dx
0 2
Simpson’s:
f x dx.
ah
Numerical Integration 2
1. Exact:
bh
f u du
ah
Section 4.6
0
a
30 0 301 4 1.273 amps
160
1 1 1 cos60 t sin120 t 160 0 30 120
b
x3 dx
3x 1
2
3
0
8 2.6667 3
2
2
1 1 02 4 2 1 1 04 6 2
212 2
32
2
212 4
32
2
213 2
32
3
213 4
32
3
4
4.000
3
3
x4
2 0
1 1 02 4 2 1 1 04 6 2
11 2.7500 4
8 2.6667 3
17 4.2500 4
24 4.0000 6
22 22
23 23
Section 4.6
2
5. Exact:
x3 dx
0 2
Trapezoidal:
x3 dx
0 2
Simpson’s:
x3 dx
0
14x
2 0
x dx
x dx
4
x dx
4
2
1 2
Trapezoidal:
23x
9
32
4
1
2
1
3
24
2 3
34
4
3
34
213 2 3
54
213 4
3
2
54
3
64
2
3
64
2 3
74
4
3
74
8 4.0625
3
8 4.0000
16 38 12.6667 3 3
18
378 2 214 2 478 2 264 2 578 2 314 2 678 3
378
1 1 dx x 12 x1
2 1
478
21 4
678 3 12.6667
31 4
1 1 1 0.1667 3 2 6
578
26 4
1 1 1 1 1 1 1 2 dx 2 2 x 12 8 4 54 12 32 12 74 12 9
1 1 32 8 32 1 0.1676 8 4 81 25 121 9
1 1 1 1 1 1 1 4 dx 2 4 x 12 12 4 54 12 32 12 74 12 9
1 x3 dx
0 2
1 x3 dx
0
1 1 64 8 64 1 0.1667 12 4 81 25 121 9
2
Simpson’s:
2
5 24 24
11. Trapezoidal:
3
24
5 22 16
Simpson’s:
2
12.6640
9
9. Exact:
3
1 1 04 12 4
9
Simpson’s:
1 1 02 8 4
4
Trapezoidal:
4.0000
4
9
7. Exact:
Numerical Integration
1 1 2 1 18 2 2 2 1 278 3 3.283 4 1 1 4 1 18 2 2 4 1 278 3 3.240 6
Graphing utility: 3.241
1
13.
1
x 1 x dx
0
x1 x dx
0
x1 x dx
x1 x dx
1
Trapezoidal:
0
1
Simpson’s:
0
Graphing utility: 0.393
1 02 8
1 04 12
121 21 2 341 43 0.342
1 1 1 2 4 4
121 21 4 341 43 0.372
1 1 1 2 4 4
205
206
Chapter 4
Integration
2
15. Trapezoidal:
cosx2 dx
2
8
0
2
2
cos 0 2 cos
4
2
2
2 cos
2
2 cos
2
2
2
2
2 cos
2
4 cos
2
4
2
2
2
cos
2
0.957
2
Simpson’s:
cosx2 dx
2
12
0
cos 0 4 cos
4
2
4
2
cos
2
0.978 Graphing utility: 0.977
1.1
17. Trapezoidal:
1 sin1 2 sin1.0252 2 sin1.052 2 sin1.0752 sin1.12 0.089 80
sin x2 dx
1 1.1
Simpson’s:
1 sin1 4 sin1.0252 2 sin1.052 4 sin1.0752 sin1.12 0.089 120
sin x2 dx
1
Graphing utility: 0.089
4
19. Trapezoidal:
x tan x dx
0
4
Simpson’s:
x tan x dx
0
2 2 3 3 02 tan 2 tan 2 tan 0.194 32 16 16 16 16 16 16 4
2 2 3 3 04 tan 2 tan 4 tan 0.186 48 16 16 16 16 16 16 4
Graphing utility: 0.186
21. (a)
23.
y
f x x3 fx 3x2
y = f ( x)
f x 6x f x 6 f 4x 0 x
a
b
(a) Trapezoidal: Error ≤
The Trapezoidal Rule overestimates the area if the graph of the integrand is concave up.
f x is maximum in 0, 2 when x 2. (b) Simpson’s: Error ≤ f 4x 0.
25. f x
2 in 1, 3 . x3
(a) f x is maximum when x 1 and f 1 2. Trapezoidal: Error ≤ f 4x
23 12n2
2 < 0.00001, n2 > 133,333.33, n > 365.15; let n 366.
24 in 1, 3 x5
(b) f 4x is maximum when x 1 and when f 41 24. Simpson’s: Error ≤
25 180n4
2 03 12 0.5 since 1242
24 < 0.00001, n4 > 426,666.67, n > 25.56; let n 26.
2 05 0 0 since 18044
Section 4.6
Numerical Integration
27. f x 1 x (a) f x
1 in 0, 2 . 41 x32 1
f x is maximum when x 0 and f 0 4.
< 0.00001, n
8 1 12n2 4
Trapezoidal: Error ≤
> 16,666.67, n > 129.10; let n 130.
2
15 in 0, 2 161 x72
(b) f 4x
15
f 4x is maximum when x 0 and f 40 16. Simpson’s: Error ≤
< 0.00001, n
32 15 180n4 16
4
> 16,666.67, n > 11.36; let n 12.
29. f x tanx2 (a) f x 2 sec2x2 1 4x2 tanx2 in 0, 1 .
f x is maximum when x 1 and f 1 49.5305. 1 03 49.5305 < 0.00001, n2 > 412,754.17, n > 642.46; let n 643. 12n2
Trapezoidal: Error ≤
(b) f 4x 8 sec2x2 12x2 3 32x4 tanx2 36x2 tan2x2 48x4 tan3x2 in 0, 1
f 4x is maximum when x 1 and f 41 9184.4734. Simpson’s: Error ≤
1 05 9184.4734 < 0.00001, n4 > 5,102,485.22, n > 47.53; let n 48. 180n4
31. Let f x Ax3 Bx2 Cx D. Then f 4x 0. Simpson’s: Error ≤
b a5 0 0 180n4
Therefore, Simpson’s Rule is exact when approximating the integral of a cubic polynomial.
1
Example:
x3 dx
0
1 1 04 6 2
3
1
1 4
This is the exact value of the integral. 33. f x 2 3x2 on 0, 4 . Ln
Mn
Rn
Tn
Sn
4
12.7771
15.3965
18.4340
15.6055
15.4845
8
14.0868
15.4480
16.9152
15.5010
15.4662
10
14.3569
15.4544
16.6197
15.4883
15.4658
12
14.5386
15.4578
16.4242
15.4814
15.4657
16
14.7674
15.4613
16.1816
15.4745
15.4657
20
14.9056
15.4628
16.0370
15.4713
15.4657
n
207
208
Chapter 4
Integration
35. f x sin x on 0, 4 . Ln
Mn
Rn
Tn
Sn
4
2.8163
3.5456
3.7256
3.2709
3.3996
8
3.1809
3.5053
3.6356
3.4083
3.4541
10
3.2478
3.4990
3.6115
3.4296
3.4624
12
3.2909
3.4952
3.5940
3.4425
3.4674
16
3.3431
3.4910
3.5704
3.4568
3.4730
20
3.3734
3.4888
3.5552
3.4643
3.4759
n
37. A
2
x cos x dx
0
Simpson’s Rule: n 14
2
x cos x dx
0
28 cos 28 2 14 cos 14 4 328 cos 328 . . . 2 cos 2
0 cos 0 4 84
0.701 y
1
1 2
π
π
4
2
x
5
39. W
100x 125 x 3 dx
0
Simpson’s Rule: n 12
5
100x 125 x3 dx
0
400
12
41.
0
6
1 x2
12 0 36
15 125 15 12
12
3
200
10 125 10 12
12
3
. . . 0 10,233.58 ft lb
Simpson’s Rule, n 6
6 46.0209 26.0851 46.1968 26.3640 46.6002 6.9282
1000 125 2125 2120 2112 290 290 295 288 275 235 89,250 sq m 210
t
45.
3
1 113.098 3.1416 36
43. Area
dx
125 125
5 5 0 400 312 12
sin x dx 2, n 10
0
By trial and error, we obtain t 2.477.
C H A P T E R Integration
4
Section 4.1
Antiderivatives and Indefinite Integration . . . . . . . . . 450
Section 4.2
Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456
Section 4.3
Riemann Sums and Definite Integrals . . . . . . . . . . . 462
Section 4.4
The Fundamental Theorem of Calculus . . . . . . . . . . 466
Section 4.5
Integration by Substitution . . . . . . . . . . . . . . . . . 472
Section 4.6
Numerical Integration
Review Exercises
. . . . . . . . . . . . . . . . . . . 479
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488
C H A P T E R Integration Section 4.1
4
Antiderivatives and Indefinite Integration
Solutions to Even-Numbered Exercises
2.
1 d 4 1 x C 4x3 2 dx x x
4.
d 2x2 3 d 2 32 x 2x12 C C dx dx 3 3x
x12 x32 6.
dr d
8.
r C Check:
2x2 1 C 2 c 2 x
Check:
d 1 C 2x3 dx x 2
Given
Rewrite
Integrate
Simplify
10.
1 dx x2
x1 C 1
1 C x
12.
xx2 3 dx
x3 3x dx
x4 x2 3 C 4 2
1 4 3 2 x x C 4 2
14.
1 x1 C 9 1
1 C 9x
16.
1 dx 3x 2
1 2 x dx 9
5 xdx 5x
x2 C 2
Check:
20.
18.
x4 2x 2 2x C 4
4x3 6x 2 1dx x 4 2x 3 x C d 4 x 2x 3 x C 4x 3 6x 2 1 dx
d x4 2x 2 2x C x3 4x 2 dx 4
x
dx x 2x 1
12
1 x32 1 x12 2 x12 dx C x32 x12 C 2 32 2 12 3
d 2 32 1 1 x x12 C x12 x12 x dx 3 2 2x
Check:
x 3 4x 2dx
Check:
450
x2 d 5x C 5 x dx 2
Check:
22.
x2 dx
x2 1 x32
dy 2x3 dx y
d C d
Section 4.1
24.
4 x3 1 dx
Check:
28.
4 x34 1 dx x74 x C 7
x2 2x3 3x4dx
x1 2x2 3x3 C 1 2 3
1 1 1 2 3C x x x
36.
40.
x4 dx
d 1 1 C 4 dx 3x3 x
2t2 12 dt
30.
4t4 4t2 1 dt
4 4 t5 t3 t C 5 3 Check:
d 45 43 t t t C 4t 4 4t2 1 dt 5 3 2t2 12
x 2 2x 3 x4
1 3 t2 3t3 dt t3 t4 C 3 4
3 dt 3t C
34.
d 13 34 t t C t2 3t3 1 3tt2 dt 3 4
Check:
1 t 2 sin t dt t3 cos t C 3
1 2 sec2 d 3 tan C 3
sec ytan y sec y dy
Check:
sec y tan y sec2 y dy
42.
d 3t C 3 dt
38.
d 13 t cos t C t2 sin t dt 3
d 1 3 tan C 2 sec2 d 3
cos x dx 1 cos2 x
sec y tan y C
d Check: sec y tan y C sec y tan y sec2 y dy
Check:
sec ytan y sec y
cos x dx sin2 x
46. fx x
y 6
f x C=3
x2 C 2 y
C=0 C = −2 4
6
d 1 csc x C csc x cot x dx sin x
y
2
x
f ( x) = 2
2
f′ f (x) = − 1x + 1
f′
4
−4
−4
x
−2
4 −2
−4
−2
cos x sin x
1 x2
6
−2
cos x 1 cos2 x
2 f ( x) = x + 2
8 8
x dx cos sin x
1 f x C x
x
−2
1 sin x
48. fx
4 2
csc x cot x dx csc x C
44. f x x
451
x3 1 C 3C 3 3x
1 3tt2 dt
Check:
d 1 1 1 2 3 C x2 2x3 3x4 dx x x x
Check:
1 dx x4
Check:
32.
26.
d 4 74 4 x3 1 x x C x34 1 dx 7
x 2 2x 3 dx x4
Check:
Antiderivatives and Indefinite Integration
x 2
4
−4
f (x) = − 1x
452
50.
Chapter 4
Integration
dy 2x 1 2x 2, 3, 2 dx y
52.
2x 1 dx x2 2x C
2 32 23 C ⇒ C 1 y x2 2x 1
54. (a)
y
(b)
5
4
3
1 C ⇒ C2 1
y
1 2, x > 0 x
x2 dx
1 C x
5
(−1, 3)
y
x3 xC 3
3
13 1 C 3
−4
4
−5
1 3 1C 3
−5
C y
56. gx 6x 2, g0 1 gx
y
dy x2 1, 1, 3 dx
x
−4
1 dy 2 x2, 1,3 dx x
6x 2 dx 2x3 C
7 3 7 x3 x 3 3 58. fs 6s 8s 3, f 2 3 f s
6s 8s 3ds 3s 2 2s 4 C
g0 1 203 C ⇒ C 1
f 2 3 322 224 C 12 32 C ⇒ C 23
gx 2x3 1
f s 3s 2 2s 4 23 62. f x sin x
60. f x x2 f0 6
f0 1
f 0 3 fx
f 0 6
1 x 2 dx x3 C1 3
fx
sin x dx cos x C1
f0 0 C1 6 ⇒ C1 6
f0 1 C1 1 ⇒ C1 2
1 fx x3 6 3
fx cos x 2
f x
1 3 1 x 6 dx x 4 6x C2 3 12
f 0 0 0 C2 3 ⇒ C2 3 f x
1 4 x 6x 3 12
f x
cos x 2 dx sin x 2x C2
f 0 0 0 C2 6 ⇒ C2 6 f x sin x 2x 6
Section 4.1
64.
dP kt, 0 ≤ t ≤ 10 dt Pt
2 kt12 dt kt32 C 3
Antiderivatives and Indefinite Integration
66. Since f is negative on , 0, f is decreasing on , 0. Since f is positive on 0, , f is increasing on 0, . f has a relative minimum at 0, 0. Since f is positive on , , f is increasing on , .
P0 0 C 500 ⇒ C 500
y 3
2 P1 k 500 600 ⇒ k 150 3 2 Pt 150t32 500 100t32 500 3
f′
2
−3
x
−2
1
70. v0 16 ftsec
f 0 s0
(a)
st 16t2 16t 64 0 16t2 t 4 0
32 dt 32t C1
t
f0 0 C1 v0 ⇒ C1 v0 ft 32t v0 f t st
1 ± 17 2
Choosing the positive value,
32t v0 dt 16t2 v0t C2
f 0 0 0 C2 s0 ⇒ C2 s0
3
−3
s0 64 ft
2
−2
f
f0 v0
ft vt
f ′′
1
P7 100732 500 2352 bacteria 68. f t at 32 ftsec2
453
t
1 17
2.562 seconds. 2 vt st 32t 16
(b)
f t 16t 2 v0t s0 v
1 2 17 321 2 17 16
1617 65.970 ftsec 72. From Exercise 71, f t 4.9t2 1600. (Using the canyon floor as position 0.) f t 0 4.9t2 1600 4.9t2 1600 t2
1600 ⇒ t 326.53 18.1 sec 4.9
74. From Exercise 71, f t 4.9t2 v0t 2. If f t 200 4.9t2 v0t 2, then vt 9.8t v0 0 for this t value. Hence, t v09.8 and we solve 4.9
9.8 v0
2
v0
9.8 2 200 v0
v2 4.9 v02 0 198 2 9.8 9.8 4.9 v02 9.8 v02 9.82 198 4.9 v02 9.82 198 v02 3880.8 ⇒ v0 62.3 msec.
454
76.
Chapter 4
v dv GM
Integration
1 dy y2
1 2 GM v C 2 y
15t 9
(b) vt > 0 when 0 < t <
5 and 3 < t < 5. 3
7 (c) at 6t 14 0 when t . 3
1 GM C v02 2 R 1 2 GM 1 2 GM v v0 2 y 2 R
v
73 373 573 3 2 32 34
2GM 2GM v02 y R
v2 v02 2GM
80. (a) at cos t
f t
7t2
at vt 6t 14
1 2 GM v C 2 0 R
vt
t3
(a) vt xt 3t2 14t 15 3t 5t 3
When y R, v v0.
v2
0 ≤ t ≤ 5
78. xt t 1t 32
1y R1
at dt
cos t dt sin t C1 sin t since v0 0
vt dt
sin t dt cos t C2
f 0 3 cos0 C2 1 C2 ⇒ C2 4 f t cos t 4 (b) vt 0 sin t for t k, k 0, 1, 2, . . . v0 45 mph 66 ftsec
(a) 16.5t 66 44
30 mph 44 ftsec
t
15 mph 22 ftsec at a
s
vt at 66
(b) 16.5t 66 22 t
ec
132 when a at 16.5 vt 16.5t 66
st 8.25t 2 66t
mp
h=
66
30
33 16.5. 2
44
ft/s
h=
(c)
66 66 a
mp
2
44 16.5 117.33 ft ft/s ec mp 0m h= ph 22 f t/se c
s
66 at 66 0 when t . a
44
2.667 16.5
0
15
vt 0 after car moves 132 ft.
66 a 66 s a 2 a
22
1.333 16.5
22 16.5 73.33 ft
a st t2 66t Let s0 0. 2
45
82.
132 73.33 feet
117.33 feet
It takes 1.333 seconds to reduce the speed from 45 mph to 30 mph, 1.333 seconds to reduce the speed from 30 mph to 15 mph, and 1.333 seconds to reduce the speed from 15 mph to 0 mph. Each time, less distance is needed to reach the next speed reduction.
Section 4.1
Antiderivatives and Indefinite Integration
30
84. No, car 2 will be ahead of car 1. If v1t and v2t are the respective velocities, then
0
86. (a) v 0.6139t3 5.525t2 0.0492t 65.9881
30
v2t dt >
455
v1t dt.
0
0.6139t4 5.525t3 0.0492t2 65.9881t 4 3 2 Note: Assume s0 0 is initial position
(b) st
vtdt
s6 196.1 feet
88. Let the aircrafts be located 10 and 17 miles away from the airport, as indicated in the figure. vAt kA t 150
vB kB t 250
1 sAt kA t2 150t 10 2
Airport
A
B
0
10
17
1 sB kB t2 250t 17 2
(a) When aircraft A lands at time tA you have vAtA kA tA 150 100 ⇒ kA
50 tA
1 sAtA kA t2A 150tA 10 0 2
1 50 2 t 150tA 10 2 tA A 125tA 10 tA
kA
10 . 125
50 125 625 2 t 150t 10 50 625 ⇒ SAt S1t tA 10 2
Similarly, when aircraft B lands at time tB you have vBtB kB tB 250 115 ⇒ kB
135 tB
1 sBtB kB t2B 250tB 17 0 2
1 135 2 t 250tB 17 2 tB B 365 t 17 2 B tB
kB (b)
34 . 365
135 365 49,275 49,275 2 135 ⇒ SBt S2t t 250t 17 tB 34 34 68 (c) d sBt sAt
20
20
Yes, d < 3 for t > 0.0505. sB
d 3
sA 0.1
0
0
0
90. True
0.1
0
92. True
456
Chapter 4
Integration
94. False. f has an infinite number of antiderivatives, each differing by a constant.
96.
d sx2 cx2 2sxsx 2cxcx dx 2sxcx 2cxsx 0 Thus, sx cx2 k for some constant k. Since, 2
s0 0 and c0 1, k 1. Therefore,
sx2 cx2 1. [Note that sx sin x and cx cos x satisfy these properties.]
Section 4.2
Area 5
6
2.
kk 2 31 42 53 64 50
4.
4
i 1
2
1
1
1
47
j3
k3
6.
1
j 3 4 5 60
i 13 0 8 1 27 4 64 9 125 238
i1
10.
2
2 n 2i 1 1 n i1 n
18.
i
10
15
2i 3 2 i 315
i1
12.
j1
15
16.
j 1 4
4
5 1 i i1
2
i1
i1
10
2
1
i1
10
i
3
i1
10
i
22. sum seqx 15
i
1011 102112 3080 4 2
3
2i
15 215 1 2 1515 1 2 4 2 152162 1516 14,160 4
2 1 55 3 2 6
2 1 1 9 4.5 3 2 3 2
26. S 5 2 1 s 21
14 2 14 12 2 14 34 2 14
i1
3 2x, x, 1, 15, 1 14,160 (TI-82)
i1
s 4 4 2 01 10
28. S8
1 2
14
54 2 14 32 2 14 74 2 14
2 2
5 6 7 1 1 2 3 16 1 2 6.038 4 2 2 2 2 2 2
1 s8 0 2 4
10
24. S 5 5 4 21 16
i 1 n
1 n1 n i0
1
i1
2
14.
101121 10 375 6
15216 45 195
ii
2
i1
2
20.
10
i
1
>
15
8.
14 2 14 12 2 14 . . . 74 2 14 5.685
14
2
Section 4.2
30. S5 1 s5
15 1 15 15 1 25 15 1 35 15 1 45 15 2
2
n→
2
24 21 16 9 1 1 0.859 5 5 5 5 5
1 15 15 1 25 15 1 35 15 1 45 15 0 0.659 2
2
2
64n nn 16 2n 1 646 lim 2n
3
32. lim
2
3
n→
2
3n2 n 64 64 2 n3 6 3
n1 nn 2 1 21 lim n n n 211 21 2
34. lim n→
2
4j 3 1 n 1 4nn 1 2n 5 3n Sn 2 4j 3 2 2 n j1 n 2 n j1 n n
36.
2
n→
S10
25 2.5 10
S100 2.05 S1000 2.005 S10,000 2.0005 4 n 3 4 n2n 12 nn 12n 1 4i2i 1 4 i i2 4 4 n n n 4 6 i1 i1 n
38.
4 n3 2n2 n 2n2 3n 1 n3 4 6
1 3n3 6n2 3n 4n2 6n 2 3n3
1 3n3 2n2 3n 2 Sn 3n3
S10 1.056 S100 1.006566 S1000 1.00066567 S10,000 1.000066657
40. lim
n n
42. lim
1 n n
n
2i
2
n→ i1
n
n→ i1
2i
2
lim
n→
2
4 n 4 nn 1 4 1 i lim 2 1 2 lim n→ n n→ 2 n2i1 2 n
lim
n→
2 lim
n n
2
i1
i 4 i
n
n
i1
i1
4n
2
2 3 nn 1 4nn 12n 1 n 4n n3 2 6
n→
2 n→ n3 n→
2 n n 2i2 n3i1
lim lim
1 2 n2 34 n2 3n2
2 12
2
4 26 3 3
Area
457
458
Chapter 4
1 n n 2 lim n n 2i n
3
2i
44. lim n→
Integration
2
n
1
n→
i1
3
4
i1
1 n 3 n 6n2i 12ni 2 8i 3 n→ n4 i1
2 lim 2 lim
1 4 nn 1 nn 12n 1 n2n 12 n 6n2 12n 8 n4 2 6 4
2 lim
1 3 n3 4 n6 n2 2 n4 n2
n→
2 lim 10
46. (a)
2
n→
n→
2
4 13 2 20 n n
(b) x
y
31 2 n n
4
Endpoints: 3
1 < 1
2 1
1 16,666.67, n > 129.10; let n 130. 12n2
Trapezoidal: Error ≤ f 4 x
24 in 0, 1 1 x 5
(b) f 4 x is maximum when x 0 and f 4 0 24. Simpson’s: Error ≤
1 24 < 0.00001, n4 > 13,333.33, n > 10.75; let n 12. (In Simpson’s Rule n must be even.) 180n4
28. f x x 1 23 (a) f x
2 in 0, 2 . 9x 1 43 2
f x is maximum when x 0 and f 0 9.
< 0.00001, n
8 2 12n4 9
Trapezoidal: Error ≤ (b) f 4 x
> 14,814.81, n > 121.72; let n 122.
2
56 in 0, 2 81x 1 103 56
f 4 x is maximum when x 0 and f 4 0 81.
< 0.00001, n
32 56 Simpson’s: Error ≤ 180n4 81 be even.)
4
> 12,290.81, n > 10.53; let n 12. (In Simpson’s Rule n must
30. f x sinx2 (a) f x 2 2x2 sinx2 cosx2 in 0, 1 .
f x is maximum when x 1 and f 1 2.2853. 1 0 3 2.2853 < 0.00001, n2 > 19,044.17, n > 138.00; let n 139. 12n2 (b) f 4 x 16x4 12 sinx2 48x2 cosx2 in 0, 1 Trapezoidal: Error ≤
f 4 x is maximum when x 0.852 and f 4 0.852 28.4285. Simpson’s: Error ≤
1 0 5 28.4285 < 0.00001, n4 > 15,793.61, n > 11.21; let n 12. 180n4
32. The program will vary depending upon the computer or programmable calculator that you use. 34. f x 1 x2 on 0, 1 . Ln
Mn
Rn
Tn
Sn
4
0.8739
0.7960
0.6239
0.7489
0.7709
8
0.8350
0.7892
0.7100
0.7725
0.7803
10
0.8261
0.7881
0.7261
0.7761
0.7818
12
0.8200
0.7875
0.7367
0.7783
0.7826
16
0.8121
0.7867
0.7496
0.7808
0.7836
20
0.8071
0.7864
0.7571
0.7821
0.7841
n
482
Chapter 4
Integration
sin x on 1, 2 . x
36. f x
Ln
Mn
Rn
Tn
Sn
4
0.7070
0.6597
0.6103
0.6586
0.6593
8
0.6833
0.6594
0.6350
0.6592
0.6593
10
0.6786
0.6594
0.6399
0.6592
0.6593
12
0.6754
0.6594
0.6431
0.6593
0.6593
16
0.6714
0.6594
0.6472
0.6593
0.6593
20
0.6690
0.6593
0.6496
0.6593
0.6593
n
38. Simpson’s Rule: n 8
2
8 3
1 32 sin d 2
0
3
1 32 sin 0 4 1 32 sin 16 2 1 32 sin 8 . . . 1 32 sin 2
2
2
2
2
6 17.476
40. (a) Trapezoidal:
2
f x dx
0
2 4.32 24.36 24.58 25.79 26.14 27.25 27.64 28.08 8.14 12.518 28
Simpson’s:
2
f x dx
0
2 4.32 44.36 24.58 45.79 26.14 47.25 27.64 48.08 8.14 12.592 38
(b) Using a graphing utility, y 1.3727x3 4.0092x2 0.6202x 4.2844
2
Integrating,
y dx 12.53
0
42. Simpson’s Rule: n 6
1
4
0
4 4 2 4 2 4 1 1 dx 1 1 x2 36 1 16 2 1 26 2 1 36 2 1 4 6 2 1 56 2 2
3.14159 44. Area
120 75 281 284 276 267 268 269 272 268 256 242 223 0 212
7435 sq m 46. The quadratic polynomial px
x x2 x x3 x x1 x x3 x x1 x x2 y y y x1 x2 x1 x3 1 x2 x1 x2 x3 2 x3 x1 x3 x2 3
passes through the three points.
Review Exercises for Chapter 4
Review Exercises for Chapter 4 1.
y
3.
7.
f
2 1 2x2 x 1 dx x3 x2 x C 3 2
f
x
5.
x3 1 dx x2
x
1 1 1 dx x2 C x2 2 x
9. fx 2x, 1, 1 f x
11.
2x dx x2 C
When x 1:
4x 3 sin x dx 2x2 3 cos x C
at a vt
a dt at C1
v0 0 C1 0 when C1 0.
y 1 C 1
vt at
C2
st
y 2 x2
a at dt t2 C2 2
s0 0 C2 0 when C2 0. a st t 2 2 a s30 302 3600 or 2 a
23600 8 ftsec2. 302
v30 830 240 ftsec 10
13. at 32
15. (a)
i1
vt 32t 96
n
(b)
st 16t2 96t (b) s3 144 288 144 ft
(d) s
i
3
i1
(a) vt 32t 96 0 when t 3 sec.
(c) vt 32t 96
2i 1
96 3 when t sec. 2 2
32 1694 9632 108 ft
10
(c)
4i 2
i1
209
210
Chapter 4
17. y
Integration
1 10 , x , n 4 x2 1 2
1 10 10 10 10 2 1 122 1 12 1 322 1
Sn S4
13.0385
10 10 10 10 1 2 122 1 1 1 322 1 22 1
s n s4
9.0385 9.0385 < Area of Region < 13.0385 4 19. y 6 x, x , right endpoints n
8
n
6
f ci x
Area lim n→
y
lim
4
i1
n
n→ i1
2
4i 4 n n
6
lim
4 4 nn 1 6n n n 2
lim
24 8
n→
n→
21. y 5 x2, x Area lim
x
−2
2
3 n
y 6
f ci x
n→ i1
4 3 2
3i 5 2 n
2
3 n 12i 9i2 2 1 n→ n i1 n n
lim
1
3 n
−4 −3
3 12 nn 1 9 nn 12n 1 n 2 n n 2 n 6
lim
3 18 n n 1 92 n 1n2n 1
n→
x
−1
1
2
3
4
−2
lim
n→
8
n1 24 8 16 n
n→ i1
lim
6
n
n
4
−2
2
3 18 9 12
23. x 5y y2, 2 ≤ y ≤ 5, y Area lim
3 n
y 6
52 n 2 n n n
3i
3i
2
3
4
n→ i1
lim
n→
n
9i2
3 15i i 10 4 12 2 n i1 n n n
3 n 3i 9i2 lim 6 2 n→ n i1 n n lim
n→
3
18
9 27 9 2 2
1 x 1
3 3 nn 1 9 nn 12n 1 6n 2 n n 2 n 6
2
2
3
4
5
6
Review Exercises for Chapter 4
n
25.
2ci 3 xi
lim
→
i1
6
2x 3 dx
211
y
27.
4
12 9 6 3 x
−3
3
6
9
−3
5
0
5
5 x 5 dx
0
0
(triangle)
6
29. (a)
2
2 6
(c)
gx dx 210 33 11
2
f x dx 510 50
43
2 x dx 2x
x2 2
xx dx
4
3
0
9
4 16 8 4 1
x
9
4
0
4
2x 1 dx x2 x
1
8
52
sin d cos
3
4
3
34
3
1
9
0
8
25x
x3 2 dx
4
34
41.
2
4x
3 x 1 dx
1
39.
6
2
8
37.
f x dx 3
6
5f x dx 5
2
33.
gx dx 10 3 7
2
6
6
31.
2f x 3gx dx 2
2
(d)
6
f x dx
2
gx dx 10 3 13
2
6
f x gx dx
6
f x dx
2
6
(b)
6
f x gx dx
35.
1
2
2
11
2
2
43.
x2 9 dx
3
y
6
4
2
x3
6
643 36 9 27
64 54 10 3 3 3
6 4 2 x
2
4
6
4
x 4
0
3
y
2
1
3 9x
8 2
1
2 4
6
4t 3 2t dt t 4 t 2
2 2
3 1
2 9 5 4 5 52243 32 422 5 5
0
73 (c) 4 1
16 16 2
8
5
5 5 x dx
x dx
25 2
212
Chapter 4
Integration
1
45.
x x3 dx
0
2 4 x2
x4
1 0
9
1 1 1 2 4 4
47. Area
1
4 x
dx
12 4x12
9 1
83 1 16
y
y 10 8
1
6 4
x 1
2 −2
1
49.
1 94
9
4
1 x
5 2x
9
1
dx
4
2 2 3 2 Average value 5 5
x 2
−2
4
6
y
1
x 2
51. Fx x21 x3
x2 13 dx
x2 x3 3
dx
4
6
8
10
53. Fx x2 3x 2
x6 3x4 3x2 1 dx
57. u x3 3, du 3x2 dx
25 2 , 4 5
5 2
25 x 4
55.
10
2
2 1 5 x x
8
x3 312 x2 dx
x7 3 5 x x3 x C 7 5
1 2 x3 312 3x2 dx x3 312 C 3 3
59. u 1 3x2, du 6x dx
x1 3x24 dx
61.
63.
65.
67.
69.
sin3 x cos x dx
sin 1 cos
1 1 1 1 3x246x dx 1 3x25 C 3x2 15 C 6 30 30
1 4 sin x C 4
1 cos 12 sin d 21 cos 12 C 21 cos C
dx
tann x sec2 x dx
tann1 x C, n 1 n1
1 sec x2 sec x tan x dx 2
1
xx2 4 dx
1 2
1 1 1 sec x2 sec x tan x dx 1 sec x3 C 3
2
1
x2 42x dx
1 x2 42 2 2
2
1 9 0 9 4 4 1
Review Exercises for Chapter 4
3
71.
0
3
1 dx 1 x
1 x12 dx 21 x12
0
3 0
213
422
73. u 1 y, y 1 u, dy du When y 0, u 1. When y 1, u 0.
1
2
0
y 11 y dy 2
0
1 u 1u du
1 0
2
u32 2u12 du 2
1
75.
cos
0
x dx 2 2
cos
0
2x 12 dx 2 sin2x
0
25u
52
0
4 u32 3
1
28 15
2
77. u 1 x, x 1 u, dx du When x a, u 1 a. When x b, u 1 b.
b
Pa, b
a
15 4
15 15 x1 x dx 4 4
1 uu du
1a
u32 u12 du
1a
1b
1b
(a) P0.50, 0.75 (b) P0, b
1 x32 3x 2 2
1 x32 3x 2 2
b 0
15 2 52 2 32 u u 4 5 3 0.75 0.50
1b 1a
15 2u32 3u 5 4 15
1b 1a
1 x32 3x 2 2
b a
0.353 35.3%
1 b32 3b 2 1 0.5 2
1 b 3b 2 1 32
b 0.586 58.6% 79. p 1.20 0.04t C
15,000 M
t1
p ds
t
(b) 2005 corresponds to t 15.
(a) 2000 corresponds to t 10. C
15,000 M
11
1.20 0.04t dt
C
10
11
15,000 1.20t 0.02t2 M
2
81. Trapezoidal Rule n 4:
1
2
Simpson’s Rule n 4:
1
10
15,000 1.20t 0.02t2 M
16 15
27,300 M
24,300 M
1 1 1 2 2 2 1 dx 0.257 1 x3 8 1 13 1 1.253 1 1.53 1 1.753 1 23
1 1 1 4 2 4 1 dx 0.254 1 x3 12 1 13 1 1.253 1 1.53 1 1.753 1 23
Graphing utility: 0.254 83. Trapezoidal Rule n 4:
2
x cos x dx 0.637
85. (a) R < I < T < L
0
Simpson’s Rule n 4: 0.685
(b) S4
Graphing Utility: 0.704
40 f 0 4f 1 2f 2 4f 3 f 4 34
1 1 1 4 42 21 4 5.417 3 2 4
214
Chapter 4
Integration
Problem Solving for Chapter 4
1
1. (a) L1
1
(b) Lx
1 dt 0 t
1 by the Second Fundamental Theorem of Calculus. x
L1 1
x
(c) Lx 1
1
2.718
1 dt for x 2.718 t
1 dt 0.999896 t
1
x1
(d) We first show that
1
To see this, let u
x1
Then
1
1 dt t
1 dt t
1
1 dt. 1x1 t
1 t and du dt. x1 x1
1
1 x1 du 1x1 ux1
Lx1x2
Now,
(Note: The exact value of x is e, the base of the natural logarithm function.)
x1x2
1
1
1 du 1x1 u x2
x2
1 du 1x1 u x1
1
1 du u
1
1
1x1 t
dt.
t 1 du using u u x 1x1 1
1 dt t
1
1
x2
1
1 du u
1 du u
Lx1 Lx2 .
x
3. Sx
0
sin
2t dt 2
y
(a)
y
(b)
2
3
1 2 x 1
3
1
−1
x
−2
1
2
The zeros of y sin
3
2
5
x2 x2 0 ⇒ n ⇒ x2 2n ⇒ x 2n, n integer. 2 2
Relative maximum at x 2 1.4142 and x 6 2.4495 Relative minimum at x 2 and x 8 2.8284 (d) S x cos
2x x 0 ⇒ 2x 2
2
n ⇒ x2 1 2n ⇒ x 1 2n, n integer 2
Points of inflection at x 1, 3, 5, and 7.
72 23
x2 correspond to the relative 2
extrema of Sx. (c) Sx sin
6
Problem Solving for Chapter 4 y
5. (a) 5 4 3 2 1
(b) (6, 2)
(8, 3) f
x
0
1
Fx
0
(0, 0)
2
1 2
3
2
4
7 2
5
4
7 2
215
6
7
8
2
1 4
3
x 2
−1 −2 −3 −4 −5
4 5 6 7 8 9
(2, −2)
1, 1, (d) F x fx 1, 2
0 ≤ x < 2 x, x 4, 2 ≤ x < 6 (c) f x 1 x 1, 6 ≤ x ≤ 8 2
0 ≤ x < 2 x22, Fx f t dt x22 4x 4, 2 ≤ x < 6 0 14x2 x 5, 6 ≤ x ≤ 8
x
0 < x < 2 2 < x < 6 6 < x < 8
x 2 is a point of inflection, whereas x 6 is not. ( f is not continuous at x 6.)
Fx f x. F is decreasing on 0, 4 and increasing on 4, 8. Therefore, the minimum is 4 at x 4, and the maximum is 3 at x 8.
1
7. (a)
1
cos x dx cos
1
1
1 3
1
cos x dx sin x
1
cos 13 2 cos 13 1.6758
2 sin1 1.6829
Error. 1.6829 1.6758 0.0071
1
(b)
1 1
1 1 3 1 dx x2 1 13 1 13 2
(Note: exact answer is 2 1.5708)
(c) Let px ax3 bx2 cx d.
1
1
ax4
4
px dx
p
1 3
b
a
1 Fx dx 2
b
f x fx dx
a
b
1
2b 2d 3
11. Consider
x5 dx
0
1 Fx 2
1
p 13 b3 d b3 d 2b3 2d 1
9. Consider Fx f x2 ⇒ Fx 2f x fx. Thus,
bx3 cx2 dx 3 2
x6 6
1 0
1 . 6
The corresponding Riemann Sum using right endpoints is Sn
1 n
1n 2n 5
5
n . . . n
5
a
1 Fb Fa 2 1 f b2 f a2 2
1 5 1 25 . . . n5 n6
Thus, lim Sn lim n→
n→
15 25 . . . n5 1 . n6 6
216
Chapter 4
Integration
b
13. By Theorem 4.8, 0 < f x ≤ M ⇒
b
f x dx ≤
a
b
Similarly, m ≤ f x ⇒ mb a
b
f x dx.
m dx ≤
a
M dx Mb a.
a
a
b
Thus, mb a ≤
f x dx ≤ Mb a. On the interval 0, 1, 1 ≤ 1 x4 ≤ 2 and b a 1.
a
1
Thus, 1 ≤
1 x4 dx ≤ 2.
0
15. Since f x ≤ f x ≤ f x ,
b
a
17.
1 365
365
0
b
f x dx ≤
a
100,000 1 sin
b
f x dx ≤
1
Note:
1 x4 dx 1.0894
0
f x dx ⇒
a
b
a
f x dx ≤
b
a
f x dx.
2t 60 100,000 365 2t 60 dt t cos 365 365 2 365
365
0
100,000 lbs.
C H A P T E R 5 Logarithmic, Exponential, and Other Transcendental Functions Section 5.1
The Natural Logarithmic Function: Differentiation . . . . 493
Section 5.2
The Natural Logarithmic Function: Integration . . . . . . 498
Section 5.3
Inverse Functions . . . . . . . . . . . . . . . . . . . . . . 503
Section 5.4
Exponential Functions: Differentiation and Integration . . 509
Section 5.5
Bases Other than e and Applications . . . . . . . . . . . . 516
Section 5.6
Differential Equations: Growth and Decay . . . . . . . . . 522
Section 5.7
Differential Equations: Separation of Variables
Section 5.8
Inverse Trigonometric Functions: Differentiation . . . . . 535
Section 5.9
Inverse Trigonometric Functions: Integration
. . . . . . 527
. . . . . . . 539
Section 5.10 Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . 543 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 548 Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554
C H A P T E R 5 Logarithmic, Exponential, and Other Transcendental Functions Section 5.1
The Natural Logarithmic Function: Differentiation
Solutions to Even-Numbered Exercises 2. (a)
(b)
y
3
3 0.2
4
2 1 −4 x 1
2
3
4
The graphs are identical.
−1
4. (a) ln 8.3 2.1163
8.3
(b)
1
6. (a) ln 0.6 0.5108
0.6
1 dt 2.1163 t
(b)
1
8. f x ln x
1 dt 0.5108 t
10. f x ln x
Reflection in the x-axis
Reflection in the y-axis and the x-axis
Matches (d)
Matches (c)
14. f x ln x
12. f x 2 ln x
16. gx 2 ln x
Domain: x 0
Domain: x > 0 y
Domain: x > 0 y
y 3
2
3
2 1 1
2
x 1
2
3
4
−3
−2
x
−1
1
2
3
−1
1
−2 −2
18. (a) ln 0.25 ln 14 ln 1 2 ln 2 1.3862
x
−3
1
2
3
20. ln23 ln 232 32 ln 2
(b) ln 24 3 ln 2 ln 3 3.1779 3 12 1 2 ln 2 ln 3 0.8283 (c) ln 3
1 (d) ln 72 ln 1 3 ln 2 2 ln 3 4.2765
22. ln xyz ln x ln y ln z
1 24. lna 1 lna 112 2 lna 1
26. ln 3e2 ln 3 2 ln e 2 ln 3
28. ln
1 ln 1 ln e 1 e
493
494
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
30. 3 ln x 2 ln y 4 ln z ln x3 ln y 2 ln z4 ln
x 3y 2 z4
32. 2 ln x lnx 1 lnx 1 2 ln
34.
x x ln 2 x 1x 1 x 1
3 3 x2 1 lnx2 1 lnx 1 lnx 1 ln ln 2 2 x 1x 1
36.
2
xx
1 1
2 2
3
38. lim ln6 x
3
40. lim ln x →5
x →6
x x 4
f=g −1
5 −1
42. y ln x32 y
3 ln x 2
44. y ln x12
3 2x
y
46. hx ln2x2 1
50.
48.
1 lnx2 4 2
1 dy x ln x 1 ln x dx x
52. f x ln
dy 1 2x x 2 dx 2 x2 4 x 4 ln t t
54. ht
58. y ln y
xx 11 31 lnx 1 lnx 1
fx
56.
t1t ln t 1 ln t ht t2 t2
3
1 1 1 1 2 2 3 x1 x1 3 x2 1 3x2 1
y x ln x
1 4x 4x 2 2x2 1 2x 1
y lnx2 4
1 2x
1 At 1, 0, y 2 .
3 At 1, 0, y 2.
hx
1 ln x 2
x 2x 3 ln 2x lnx 3
1 1 3 x x 3 xx 3
y lnln x 1 dy 1x dx ln x x ln x
60. f x ln x 4 x2 fx
1 x 1 4 x2 x 4 x2 1 4 x2
ln 5 1.6094
Section 5.1
62.
x2 4 1 2 x2 4 ln 2x2 4 x
y
x2 4
2x2
The Natural Logarithmic Function: Differentiation
dy 2x2 xx2 4 4xx2 4 1 1 dx 4x 4 4 2 x2 4
1 1 ln 2 x2 4 ln x 4 4
x x 4 4x1 2
Note that: 1 1 2 x2 4 2 x2 4 Hence,
2 x2 4 2 x2 4 x2 2 x2 4
x2 4 1 1 2 x2 4 x 1 dy 3 2 dx 2xx 4 x 4 x2 4x x2 4
1 12 2 x2 4 x2 4 1 3 2 x 4x 2xx 4
x2 4 x2 4 1 x2 4 3 2 x 4x x3 4xx 4
64. y ln csc x y
66.
csc x cot x cot x csc x
dy sec x tan x sec2 x dx sec x tan x
68.
y ln1 sin2 x
y ln sec x tan x
1 ln1 sin2 x 2
sec xsec x tan x sec x sec x tan x
ln x
70. gx
t 2 3dt
l
dy 1 2 sin x cos x sin x cos x dx 2 1 sin2 x 1 sin2 x
gx ln x2 3
d ln x2 3 ln x dx x
(Second Fundamental Theorem of Calculus)
72. (a)
y 4 x2 ln
12 x 1 ,
0, 4
1 1 dy 2x dx 12x 1 2 2x When x 0,
1 x2 dy 1 . dx 2
1 Tangent line: y 4 x 0 2 1 y x4 2 (b)
8
−4
4
−4
74.
lnxy 5x 30 ln x ln y 5x 30 1 1 dy 50 x y dx 1 dy 1 5 y dx x y 5xy y dy 5y dx x x
495
496
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
y xln x 4x
76.
y x
1x ln x 4 3 ln x
x y xy x x ln x 4x x3 ln x 0 78. y x ln x
6
Domain: x > 0 y 1 y
1 0 when x 1. x
(1, 1) −1
6 −1
1 > 0 x2
Relative minimum: 1, 1 80. y
ln x x
2
(e, e−1)
3 2
3 − 23
,2e
−1
Domain: x > 0 y
(e
) 6
x1x ln x 1 ln x 0 when x e. x2 x2
−4
x21x 1 ln x2x 2ln x 3 y 0 when x e32. x4 x3 Relative maximum: e, e1 3 Point of inflection: e32, 2 e32
x 82. y x2 ln . 4 y x2
Domain x > 0
x x 1 2x ln x 1 2 ln 0 when x 4 4
x x 1 1 2 ln ⇒ ln ⇒ x 4e12 4 4 2 y 1 2 ln
f x x ln x,
84.
x 1 x 2x 3 2 ln 4 x 4
fx 1 ln x, 1 f x , x
Point of inflection: 4e32, 24e3 4
(4e−3/2, −24e−3)
−4
−4
(4e−1/2, −8e−1)
f 1 1
P2x f 1 f1x 1 x 1 P1x 1,
1 x 1 2, 2
P11 0
1 f 1x 12 2 P21 0
P11 1
P2x 1 x 1 x, P2 x x,
8
f 1 1
P1x f 1 f1x 1 x 1,
y 0 when x 4e32 Relative minimum: 4e12, 8e1
f 1 0
P21 1
P2 1 1
The values of f, P1, P2, and their first derivatives agree at x 1. The values of the second derivatives of f and P2 agree at x 1. 3
f P1
P2 −2
4 −1
Section 5.1 86. Find x such that ln x 3 x.
The Natural Logarithmic Function: Differentiation
88.
f x x ln x 3 0 fx 1 xn1
f xn 4 ln xn xn xn fxn 1 xn
n
1
xn
2
2
f xn
0.3069
y x 1x 2x 3 ln y
1 x
1 lnx 1 lnx 2 lnx 3
2
1 dy 1 1 1 1 y dx 2 x1 x2 x3
3
2.2046
2.2079
0.0049
0.0000
497
1 3x2 12x 11 2 x 1x 2x 3
dy 3x2 12x 11 dx 2y
3x2 12x 11 2x 1x 2x 3
Approximate root: x 2.208
xx
2
y
90.
2
1 1
92.
dy dx
x
x2 2
1 2x 1 x4 1
x 1x 2 x 1x 2
ln y lnx 1 lnx 2 lnx 1 lnx 2
1 ln y lnx2 1 lnx2 1
2 1 dy 1 2x 2x y dx 2 x2 1 x2 1
y
1 1 dy 1 1 1 y dx x1 x2 x1 x2
dy 2 4 6x2 12 y 2 2 y 2 dx x 1 x 4 x 1x2 4
x2 1122x 2 x 112x2 1x2 1
x 1x 2 6x2 2 x 1x 2 x 1x 1x 2x 2
2x 2 x 132x2 112
6x2 2 x 12x 22
94. The base of the natural logarithmic function is e. 96. gx ln f x, f x > 0 gx
fx f x
(a) Yes. If the graph of g is increasing, then gx > 0. Since f x > 0, you know that fx gxf x and thus, fx > 0. Therefore, the graph of f is increasing. 98. t (a)
(b) No. Let f x x2 1 (positive and concave up). gx lnx2 1 is not concave up.
5.315 , 1000 < x 6.7968 ln x (d)
50
dt 1 5.3156.7968 ln x2 dx x
4000
1000 0
(b) t1167.41 20 years T 1167.412012 $280,178.40 (c) t1068.45 30 years T 1068.453012 $384,642.00
5.315 x6.7968 ln x2
When x 1167.41, dtdx 0.0645. When x 1068.45, dtdx 0.1585. (e) There are two obvious benefits to paying a higher monthly payment: 1. The term is lower 2. The total amount paid is lower.
498
Chapter 5
100. (a)
Logarithmic, Exponential, and Other Transcendental Functions (b) T p
350
34.96 3.955 p p
(c)
30
T10 4.75 deglbin2 0
T70 0.97 deglbin2
100
0
100 0
0
lim Tp 0
p→
102. y 10 ln (a)
10
100 x2
x
100 x
2
10 ln 10 100 x2 ln x 100 x2 (c) lim
20
x→10
0
dy 0 dx
10 0
(b)
dy x 1 x 10 dx x 100 x2 10 100 x2 100 x2
x
10
10
100 x2 10 100 x2
x
100 x2 10 100 x2
10x 100x x
2
1
10 x
100 x2 10 x 2 2 x 100 x 10 100 x
x 10 x 10 100 x2
2 x 10 100 x2 10 100 x x x x2
When x 5, dydx 3. When x 9, dydx 199. 104. y ln x y
106. False
is a constant.
1 > 0 for x > 0. x
Since ln x is increasing on its entire domain 0, , it is a strictly monotonic function and therefore, is one-to-one.
Section 5.2 2.
6.
d ln 0 dx
The Natural Logarithmic Function: Integration
10 1 dx 10 dx 10 ln x C x x
1 1 1 dx 3 dx 3x 2 3 3x 2 1 ln 3x 2 C 3
4. u x 5, du dx
1 dx ln x 5 C x5
8. u 3 x3, du 3x2 dx
x2 1 1 dx 3x2 dx 3 x3 3 3 x3 1 ln 3 x3 C 3
Section 5.2
The Natural Logarithmic Function: Integration
499
10. u 9 x2, du 2x dx
12.
x 9 x2
x3
1 9 x2122x dx 9 x2 C 2
dx
1 xx 2 3x2 6x dx dx u x3 3x2 4 2 3 3x 4 3 x 3x2 4
14.
1 ln x3 3x2 4 C 3
2x2 7x 3 dx x2
2x 11
19 dx x2
16.
x3 6x 20 dx x5
x2 11x 19 ln x 2 C
18.
x3 3x2 4x 9 dx x2 3
3 x
3x
3 ln 1 x13 C
x2 5x 19
1
x dx
1 ln ln x C 3
xx 2 dx x 13
x2 2x 1 1 dx x 13
x 12 dx x 13
1 dx x1
1 dx x 13
ln x 1
26. u 1 3x, du
1 dx 1 3x
3 2 dx ⇒ dx u 1 du 23x 3
12 u 1 du u3
2 3
1
1 du u
2 u ln u C 3
2 1 3x ln1 3x C 3
2 2 3x ln1 3x C1 3 3
115 dx x5
x3 5x2 19x 115 ln x 5 C 3 2
1 1 1 dx x lnx3 3 ln x
24.
1 1 1 dx 3 dx 1 x13 1 x13 3x23
x23
20.
x2 1 lnx2 3 C 2 2
1 dx 3x23
22. u 1 x13, du
x dx x2 3
1 dx x 13
1 C 2x 12
500
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
28. u x13 1, du
3 x
x 1 3
dx
1 dx ⇒ dx 3u 12 du 3x23
u1 3u 12 du u u1 2 u 2u 1 du u
3 3
u2 3u 3
u3 3
3
x
13
3
C
3u2 3u ln u 2
13 3x13 12 3x13 1 ln x13 1 3 2
tan 5 d
C
3 ln x13 1
30.
1 du u
3x23 3x13 x C1 2
1 5 sin 5 d 5 cos 5
32.
sec
1 ln cos 5 C 5
34. u cot t, du csc2 t dt
csc2
2 ln sec
36.
x x tan C 2 2
sec t tan t dt lnsec t tan t lncos t C
t dt ln cot t C cot t
x x 1 dx 2 sec dx 2 2 2
ln
sec t tan t C cos t
ln sec tsec t tan t C
38. y
2x dx x2 9
40. r
0, 4: 4 ln0 9 C ⇒ y lnx2 9 4 ln 9 ln x2 9 C
sec2 t dt tan t 1
ln tan t 1 C C 4 ln 9
, 4: 4 ln0 1 C ⇒ C 4 r ln tan t 1 4 10
8
(0, 4)
−9
9 −8
(π , 4) −2
−4
42.
dy ln x , 1, 2 dx x (a)
8
(b)
y
y
ln x ln x2 dx C x 2
2
y1 2 ⇒ 2
1 x
4 −1 −2
Hence, y
ln 12 C ⇒ C 2 2
ln x2 2. 2
Section 5.2
1
44.
The Natural Logarithmic Function: Integration
1 1
1 dx ln x 2 1 x 2
46. u ln x, du
2
e
ln 3 ln 1 ln 3
e
1
48.
0
x1 dx x1
1
1
1 dx
0
0
2
1 dx x ln x
e
e
e2
1 1 dx ln ln x ln x x
e
ln 2
2 dx x1
0 1 2 ln 2
1
x 2 ln x 1
0.2
50.
1 dx x
0.2
csc 2 cot 2 2 d
0.1
csc2 2 2 csc 2 cot 2 cot2 2 d
0.1 0.2
2 csc2 2 2 csc 2 cot 2 1 d
0.1
cot 2 csc 2
52. ln sin x C ln
ln
0.2 0.1
0.0024
1 C ln csc x C csc x
54. ln csc x cot x C ln
56.
csc x cot xcsc x cot x csc2 x cot2 x C ln C csc x cot x csc x cot x
1 C ln csc x cot x C csc x cot x
1 x 2 dx 1 x 6 1 x 4 ln 1 x C1 1 x 4x x 4 ln 1 x C where C C1 5.
58.
60.
tan2 2x 1 dx ln sec 2x tan 2x sin 2x C sec 2x 2
4
sin2 x cos2 x dx ln sec x tan x 2 sin x cos x 4
2
ln
2 1 2 1
4
4
2 1.066
Note: In Exercises 62 and 64, you can use the Second Fundamental Theorem of Calculus or integrate the function.
62. F x
tan t dt
0
Fx tan x
x2
x
64. F x
1
1 dt t
2x 2 Fx 2 x x
501
502
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
4
66.
68. A
y
1
x4 dx x
4
x 4 ln x
2 1
1
1
4 dx x 4 1
4 4 ln 4 1
x 1
3
2
4
3 4 ln 4 8.5452
−1 −2
6
A3 Matches (a) 9
0 0
4
70.
10 ln cos0.3x 3
10 10 ln cos1.2 1 ln cos0.3 11.7686 3 3
2x tan0.3x dx x2
1
16
1
4
8
0
5
−2
72. Substitution: u x2 4 and Power Rule
74. Substitution: u tan x and Log Rule
76. Answers will vary.
78. Average value
1 42
4
2
2
4
2
4x 1 dx x2
1 1 2 dx x x
1 x
1 1 ln 2 4 2
1 1 ln 4 1.8863 4 2
2 ln x 2 ln 4 2 ln 2
82. t
84.
10 ln 2
300
250
80. Average value
300 250
St
k dt k ln t C k ln t C since t > 1. t
S2 k ln 2 C 200 S4 k ln 4 C 300 Solving this system yields k 100ln 2 and C 100. Thus, St
100 ln t ln t 100 100 1 . ln 2 ln 2
x dx 6
3 ln2 3 ln1 0
3 ln2 3
4.1504 units of time
dS k dt t
sec
10 10 4 ln 200 ln 150 ln ln 2 ln 2 3
0
12 6 ln sec 6x tan 6x
1 dT T 100
10 lnT 100 ln 2
2
4 2
1 20
2 0
Section 5.3 86. k 1: f1x x 1 k 0.5: f0.5x k 0.1: f0.1x
y
x 1
0.5
2 x 1
f1 8 6
1 x 1 10 10 0.1
10 x
f
4
f
0.1 x
2
88. False
Section 5.3
(b)
10
8
f 4
3x 3x 34 x 4 4
2 x
−2
3 3 4x g f x g3 4x x 4
2
f x 1 x3
(b) 2
g −2
x
−1
16 12
16 16 x x
f
8
g
g f x g16 x2 16 16 x2
x 8
x2 x
f x
1 , x ≥ 0 1x
gx
1x , 0 < x ≤ 1 x
20
20
f gx f 16 x 16 16 x2
16
y
(b)
gx 16 x
1 g f x g 1x
3
−2
3 x3 x 1 x3
1 x x
2 −1
x3
f x 16 x2, x ≥ 0
f gx f
8
f 3
1 1 x x 3 1
g
y
3 1 x 1 f gx f 3 1 x 3
g f x g1
4
−2
3 1 x gx
8. (a)
8
y
3x 4
f gx f
6. (a)
6
Inverse Functions
f x 3 4x
4
90. False; the integrand has a nonremovable discontinuity at x 0.
d 1 ln x dx x
4. (a)
0.5
2
k→0
gx
503
10
lim fk x ln x
2. (a)
Inverse Functions
(b)
1 1x x 1 1x 1x
y
3
g
2
1 1 x 1x 1 1 x x 1
12
1
f x 1
1x x 1
2
3
504
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
10. Matches (b)
12. Matches (d) x2 4 Not one-to-one; does not have an inverse
16. F x
14. f x 5x 3 One-to-one; has an inverse y
18. gt
x2
1 t 2 1
Not one-to-one; does not have an inverse
y −2
3
x
−1
1
2 1
−1
1 2
−3
3
x
−1
−3
−
1 2
1 2
1
1
−1
− 2
−1
20. f x 5xx 1
22. hx x 4 x 4
One-to-one; has an inverse
Not one-to-one; does not have an inverse
12
9
−9
0
9
6 0
−9
24. f x cos
3x 2
3 3x 2 4 fx sin 0 when x 0, , , . . . 2 2 3 3 f is not strictly monotonic on , . Therefore, f does not have an inverse. 28. f x lnx 3, x > 3
26. f x x3 6x2 12x fx 3x2 12x 12 3x 22 ≥ 0 for all x.
fx
f is increasing on , . Therefore, f is strictly monotonic and has an inverse.
f x 3x y
30.
32.
f is increasing on 3, . Therefore, f is strictly monotonic and has an inverse.
f x x3 1 y
y 3
3 y 1 x
x y 3
3 x 1 f1x
x
1 > 0 for x > 3. x3
y
3 x
f x x2 y, 0 ≤ x
34.
x y y x
1 f
x x
1
y
x x 3
f 1
5 4 3 2
y 3 2
−3
1
−3
2
3
f
2
−1
1
f x
−2
3
f −1 2 3 4 5
f −1
1
f
4
x
−5 − 4 −3
f
y
−4 −5
x 1
2
3
4
Section 5.3 f x x2 4 y, x ≥ 2
36.
Inverse Functions
505
5 2x 1 y f x 3
38.
y5 243 486 x5 243 y 486
x y2 4
x
y x2 4 f 1x x2 4, x ≥ 0
f 1x
y
x5 243 486
5
f
4
−1
6
f
f
3
The graphs of f and f 1 are reflections of each other across the line y x.
f −1
2
−4
8
1 x 1
2
3
4
−6
5
f x x3 5 y
40.
f x
42.
x y5 3
x2 y x
x
2 y1
y
2 x1
f 1x
2 x1
y x5 3
44.
x
0
2
4
f 1x
6
2
0
y
f 1x x5 3 2
f −1 f −3
3
8 6 4
f
−1
(4, 0) f
The graphs of f and f 1 are reflections of each other across the line y x.
−6
(2, 2)
2
4 −2
(0, 6)
2
4
6
x 8
6
−4
The graphs of f and f 1 are reflections of each other across the line y x. 46. f x k2 x x3 is one-to-one for all k 0. Since f 13 2, f 2 3 k2 2 23 12k ⇒ k 14 .
x 2 1 1 f x
48. f x x 2 on 2, x2
> 0 on 2,
f is increasing on 2, . Therefore, f is strictly monotonic and has an inverse.
50. f x cot x on 0, fx csc2 x < 0 on 0, f is decreasing on 0, . Therefore, f is strictly monotonic and has an inverse.
2
52. f x sec x on 0,
2
fx sec x tan x > 0 on 0,
f is increasing on 0, 2. Therefore, f is strictly monotonic and has an inverse.
506
54.
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
3 y on 0, 10 x2
f x 2
f −1
2x2 3 x2y
f
x22 y 3
10
0 0
3 y ± 2x 3 2y
x±
f 1x
The graphs of f and f 1 are reflections of each other across the line y x.
5
2 3 x, x < 2
56. (a), (b)
58. (a), (b)
2
f −1 f
h −1
−3
10
h
−10
3
10
−10
−2
(c) Yes, f is one-to-one and has an inverse. The inverse relation is an inverse function.
(c) h is not one-to-one and does not have an inverse. The inverse relation is not an inverse function. 60. f x 3
62. f x ax b
Not one-to-one; does not have an inverse
f is one-to-one; has an inverse ax b y x
yb a
y
xb a
f 1x
64. f x 16 x4 is one-to-one for x ≥ 0. 16
x4
x3y
x4
xy3
yx
yx3
4 16 x y
f 1
x
f
fx
16 x, x ≤ 16 70. Yes, the area function is increasing and hence one-to-one. The inverse function gives the radius r corresponding to the area A.
1 5 1 x 2x3; f 3 243 54 11 a. 27 27 1 5x4 6x2 27
f 111
x x 3, x ≥ 0
1
4
68. No, there could be two times t1 t2 for which ht1 ht2.
72. f x
66. f x x 3 is one-to-one for x ≥ 3.
y
16 y 4 16
xb ,a 0 a
1 1 27 1 f f 111 f3 534 632 17
Section 5.3
Inverse Functions
f x cos 2x, f 0 1 a
74.
fx 2 sin 2x
f 11
1 1 1 1 which is undefined. f f 11 f0 2 sin 0 0
f x x 4, f 8 2 a
76.
fx
f 12
1 2x 4 1 1 1 1 4 f f 12 f8 1 28 4 1 4
78. (a) Domain f Domain f 1 ,
(d)
f x 3 4x, 1, 1
(b) Range f Range f 1 ,
fx 4
(c)
f1 4
y
f f
x
−5 −4 −3 −2 −1 −2 −3 −4 −5
2 3
80. (a) Domain f 0, , Domain f 1 0, 4 (b) Range f 0, 4, Range f 1 0, (c)
y 4
2
1 4
f 11
1 4
f x
4 1 x2
fx
8x , f1 2 x2 12
x 1
2
4 x x 2
f 1x
f −1 f
82.
(d)
f 1x
f 1x
3
1
3x , 1, 1 4
f 1x
3 2
−1
3
4
x2
4x x
, f 12
1 2
x 2 lny2 3 12
dy 1 2y y2 3 dx
dy 16 3 13 dy y2 3 . At 0, 4, . dx 4y dx 16 16 In Exercises 84 and 86, use the following. f x 18 x 3 and g x x3 3 f1 x 8 x 3 and g1 x x
84. g1 f 13 g1 f 13 g10 0
3 4 86. g1 g14 g1g14 g1
3 3 9 4 4
507
508
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
In Exercises 88 and 90, use the following. f x x 4 and g x 2x 5 f1 x x 4 and g1 x
x5 2
88. f 1 g1x f 1g1x f 1
90. g f x g f x
x 2 5
x5 4 2
x3 2
92. The graphs of f and f 1 are mirror images with respect to the line y x.
gx 4 2x 4 5 2x 3
Note: g f 1
94. Theorem 5.9: Let f be differentiable on an interval I. If f has an inverse g, then g is differentiable at any x for which fgx 0. Moreover, gx
96. f is not one-to-one because different x-values yield the same y-value.
34 53
Example: f 3 f
x3 2 f 1 g1
Hence, g f 1x
1 , fgx 0 fgx
98. If f has an inverse, then f and f 1 are both one-to-one. Let f11x y then x f 1 y and f x y. Thus, f 11 f.
Not continuous at ± 2. 100. If f has an inverse and f x1 f x2, then f 1 f x1 f 1 f x2 ⇒ x1 x2. Therefore, f is one-to-one. If f x is one-toone, then for every value b in the range, there corresponds exactly one value a in the domain. Define gx such that the domain of g equals the range of f and gb a. By the reflexive property of inverses, g f 1. 102. True; if f has a y-intercept.
104. False Let f x x or gx 1 x .
106. From Theorem 5.9, we have: gx
1 fgx
g x
fgx0 f gxgx fgx2
f gx 1 fgx fgx2
f gx fgx3
If f is increasing and concave down, then f > 0 and f < 0 which implies that g is increasing and concave up.
Section 5.4
Section 5.4
Exponential Functions: Differentiation and Integration
e2 0.1353. . .
2.
Exponential Functions: Differentiation and Integration
4.
ln 0.1353. . . 2
ln 0.5 0.6931. . .
6. eln 2x 12 2x 12
1 e0.6931. . . 2
x6 10. 6 3ex 8
8. 4ex 83 ex
3ex 14
83 4
ex
83 x ln 3.033 4
x ln
12. 200e4x 15 e4x
14 3
143 1.540
14. ln x2 10 x2 e10
15 3 200 40
x ± e5 ± 148.4132
3 4x ln 40 x
40 1 ln 0.648 4 3 18. lnx 22 12
16. ln 4x 1
x 22 e12
4x e e x
x 2 e6
e 0.680 4
x 2 e6 405.429
1 20. y 2 ex
22. y ex2
y
y
4
4
3
3
2
2
1 x
−1
1
24. (a)
2
3
−2
(b)
10
−8
−1
10 −2
Horizontal asymptotes: y 0 and y 8
x 1
2
10
−8
10 −2
Horizontal asymptote: y 4
509
510
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
28. y
26. y Ceax
C 1 eax
Horizontal asymptote: y 0 lim
Reflection in the y-axis
x →
Matches (d)
lim
C C 1 eax
x →
C 0 1 eax
Horizontal asymptotes: y C and y 0 Matches (b) 30. f x ex3 gx ln
32. f x ex1
y
f
4
6
3
4
2
2
1 x
−2
lim
x →
y
8
−2
34. In the same way,
gx 1 ln x
3 ln x
x3
g
2
36. 1 1
4
6
x
er for r > 0.
f
g
x
−1
8
1 xr
1
3
2
4
−1
1 1 1 1 1 1 2.71825396 2 6 24 120 720 5040 e 2.718281828 e > 11
1 1 1 1 1 1 2 6 24 120 720 5040
38. (a) y e2x
(b) y e2x
y 2e2x
y 2e2x
At 0, 1, y 2.
At 0, 1, y 2.
40. f x e1x
42.
fx e1x
y ex
2
44.
y x2ex dy x2ex 2xex dx
dy 2 2xex dx
xex 2 x
46. gt e3t
2
6 gt e3t 6t 3 3 3t2 t e 2
48.
y ln
11 ee x
50.
x
ln1 ex ln1 ex dy ex ex x dx 1 e 1 ex
52.
y
ex ex 2
dy ex ex dx 2
2ex 1 e2x
y ln
x
ex 2
lnex ex ln 2 dy ex ex dx ex ex
54.
e
y xex ex ex x 1 dy ex ex x 1 xex dx
e2x 1 e2x 1
Section 5.4
Exponential Functions: Differentiation and Integration
56. f x e3 ln x
58.
e3 x
fx
y ln ex x dy 1 dx
62. gx x ex ln x
exy x2 y2 10
60.
x dxdy ye
xy
2x 2y
dy 0 dx
gx
1 ex ex ln x x 2x
dy xy xe 2y yexy 2x dx
g x
dy yexy 2x xy dx xe 2y
1 xex ex ex ex ln x 32 4x x2 x ex2x 1 1 ex ln x x2 4xx
y ex3 cos 2x 4 sin 2x
64.
y ex6 sin 2x 8 cos 2x ex3 cos 2x 4 sin 2x ex10 sin 2x 5 cos 2x 5ex2 sin 2x cos 2x y 5ex4 cos 2x 2 sin 2x 5ex2 sin 2x cos 2x 5ex5 cos 2x 25ex cos 2x y 2y 25ex cos 2x 25ex2 sin 2x cos 2x 5ex3 cos 2x 4 sin 2x 5y Therefore, y 2y 5y ⇒ y 2y 5y 0.
66. f x
ex ex 2
2
fx
ex ex > 0 2
fx
ex ex 0 when x 0. 2
−3
(0, 0)
3
−2
Point of inflection: 0, 0
68. gx gx g x
1 2
1 2
1 2
ex3 2 2
x 3ex3 2 2
(
2,
x 2x 4ex3 2
e− 0.5 2π
( (
( 4,
0
e− 0.5 2π
( 6
0
3, 0.399 1 1 e , 4, e 2, 0.242, 4, 0.242 2 2
1
2
12
12
70. f x xex
2
fx xex ex ex1 x 0 when x 1. f x ex ex1 x exx 2 0 when x 2. Relative maximum: 1, e1 Point of inflection: 2,
1 2π
2
3, Points of inflection: 2, Relative maximum:
(
3,
0.8
2e2
(1, e −1) −2
(2, e−2) 4
−2
511
512
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
72. f x 2 e3x4 2x
( 53 , 96.942) ( 70.798 )
100
fx e3x2 3e3x4 2x e3x10 6x 0 when x
5 3.
4 , 3
fx e3x6 3e3x10 6x e3x24 18x 0 when x 43 . Relative maximum: Point of inflection:
53 , 96.942 43 , 70.798
2.5 0
e
(b) Ax xf c x 10
f c f c x
74. (a)
− 0.5
c cx cx ec e (d) c
cex c x
x ex 1
2
cex c x c
x ex 1
0
10x2 x1ex (c) Ax x e e 1
4 0
lim c 1
x →0
lim c 0
6
x →
(2.118, 4.591)
9
0 0
The maximum area is 4.591 for x 2.118 and f x 2.547. 76. Let x0, y0 be the desired point on y ex.
y
y ex
3 2
Slope of tangent line
y ex
1
1 ex y
Slope of normal line
y ex0 ex0x x0 We want 0, 0 to satisfy the equation: ex0 x0ex0 1 x0e2x0 x0e2x0 1 0 Solving by Newton’s Method or using a computer, the solution is x0 0.4263.
0.4263, e0.4263
( 0.4263, e − 0.4263 ) x
−1
1 −1
2
3
x x exe 1 1
10x2 x1ex e x e 1
10cec 10c xecx
cecx c xec
x
Section 5.4
Exponential Functions: Differentiation and Integration
513
78. V 15,000e0.6286t , 0 ≤ t ≤ 10 (a)
(b)
20,000
10
0 0
dV 9429e0.6286t dt
(c)
When t 1,
dV 5028.84. dt
When t 5,
dV 406.89. dt
20,000
10
0 0
80. 1.56e0.22t cos 4.9t ≤ 0.25 (3 inches equals one-fourth foot.) Using a graphing utility or Newton’s Method, we have t ≥ 7.79 seconds.
2
10
0
−2
82. (a) V1 1686.79t 23,181.79 V2 109.52t2 3220.12t 28,110.36
(b) The slope represents the rate of decrease in value of the car.
20,000
0
10 0
(c) V3 31,450.770.8592t 31,450.77e0.1518t
(d) Horizontal asymptote: lim V3t 0 t→
As t → , the value of the car approaches 0. (e)
84.
dV3 4774.2e0.1518t dt For t 5,
dV3 2235 dollarsyear. dt
For t 9,
dV3 1218 dollarsyear. dt
f x ex 2, f 0 1 2
2
P1
fx xex 2, f0 0 2
fx
2 x2ex 2
2 ex 2
2 ex 2
x2
1, f0 1
P1x 1 0x 0 1, P10 1
f
−3
3
P2 −2
P1x 0, P10 0 P2x 1 0x 0
1 x2 x 02 1 , P20 1 2 2
P2x x, P20 0 P2x 1, P20 1 The values of f, P1, P2 and their first derivatives agree at x 0. The values of the second derivatives of f and P2 agree at x 0.
514
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions 88. Let u x4, du 4x3 dx.
86. nth term is xnn! in polynomial: y4 1 x
Conjecture: ex 1 x
4
90.
94.
4
e1 1 1
1 e
92.
2
xex 2 dx 2
0
3
x2ex 2 dx
ex 2x dx ex 2 2
0
2
2
0
1 e1
3
1 1 e2x 2e2x dx dx ln1 e2x C 1 e2x 2 1 e2x 2
e1 e
2ex 2ex dx 2 ex ex2ex ex dx ex ex2
e2x 2ex 1 dx ex
ex 2 ex dx
ex 2x ex C
108.
lne2x1 dx
2x 1 dx
106.
1 e sec 2x sec 2x tan 2x dx e sec 2x C 2
u sec 2x, du 2 sec 2x tan 2x
1 sin x e2x dx cos x e2x C1 2
f0 1
1 1 C1 ⇒ C1 1 2 2
fx cos x f x
1 2x e 1 2
cos x
1 2x e 1 dx 2
1 sin x e2x x C2 4 f 0
1 1 C2 ⇒ C2 0 4 4
f x x sin x
1 2x e 4
2 C ex ex
110. y
x2 x C
112. fx
2 x32 3x2 2 3 e dx ex 2 C 3 2 3
102. Let u ex ex, du ex ex dx.
ex ex dx lnex ex C ex ex
C
96. Let u 1 e2x, du 2e2x dx.
100. Let u ex ex, du ex exdx.
104.
4
x
4x dx e
x2 , du x dx. 2
2
4
x2 x3 ... 2! 3!
3
2
x
e
e1x 1 1 2 2 2 dx e1x 3 dx e1x C x3 2 x 2
98. Let u
e3x dx e3x
3
x2 x3 x4 2! 3! 4!
ex ex 2 dx
e2x 2 e2x dx
1 1 2x e 2x e2x C 2 2
Section 5.4
114. (a)
Exponential Functions: Differentiation and Integration
(b)
y
4
y
xe0.2x dx
x
−4
4
−4
0, 23
dy 2 xe0.2x , dx 2
1 0.4
e0.2x 0.4x dx 2
1 0.2x2 2 e C 2.5e0.2x C 0.4
0, 23: 23 2.5e C 2.5 C ⇒ C 1 0
y 2.5e0.2x 1 2
b
116.
ex dx ex
a
b a
2
ea eb
118.
0
1 e2x 2 dx e2x 2x 2
2 0
1 1 e4 4 4.491 2 2 a
4
b
−2
4 0
4
120. (a)
x
x ex dx, n 12
122.
0
0.30.3t dt
0 x
e
Midpoint Rule: 92.1898
1 2
e0.3x 1
1 2
e0.3x
1 2
0
Simpson’s Rule: 92.7385 Graphing Utility: 92.7437
1 2
0.3t
Trapezoidal Rule: 93.8371
2
(b)
2xex dx, n 12
0.3x ln
0
Midpoint Rule: 1.1906 x
Trapezoidal Rule: 1.1827
1 ln 2 2
ln 2 2.31 minutes 0.3
Simpson’s Rule: 1.1880 Graphing Utility: 1.18799 124.
t
0
1
2
3
4
R
425
240
118
71
36
6.052
5.481
4.771
4.263
3.584
ln R
515
(a) ln R 0.6155t 6.0609 R e0.6155t6.0609 428.78e0.6155t (b)
450
−1
5 0
4
(c)
0
4
Rt dt
0
428.78e0.6155t dt 637.2 liters
516
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
126. The graphs of f x ln x and gx ex are mirror images across the line y x.
ln
130.
128. (a) Log Rule: u ex 1 (b) Substitution: u x2
ea ln ea ln eb a b eb
ln eab a b Therefore, ln
ea ea ln eab and since y ln x is one-to-one, we have b eab. eb e
Section 5.5 2. y
2 1
Bases Other than e and Applications
t8
4. y
At t0 16, y
8. loga
12
168
1 4
2 1
t5
6. log27 9 log27 2723 23
At t0 2, y
1 loga 1 loga a 1 a
10. (a)
25
0.7579
12. (a) log3 19 2
2723 9 log27 9
(b)
12
32 19
2 3
4912 7
(b)
1634 8
log49 7 12
3 log16 8 4 18. y 3x
16. y 2x
14. y 3x1
2
x
1
0
1
2
3
x
2
1
0
1
2
x
0
±1
±2
y
1 9
1 3
1
3
9
y
16
2
1
2
16
y
1
1 3
1 9
y
y
12
y
8 1
10 6 8 4
6 4
−1
20. (a) log3
1
2
2 −2
x
−1
x 1
2
3
1 x 81 3x
1 81
x 4 (b) log6 36 x 6x 36 x2
4
−2
−1
−1
x 1
2
22. (a) logb 27 3 b3 27 b3 (b) logb 125 3 b3 125 b5
Section 5.5 24. (a) log3 x log3x 2 1
Bases Other than e and Applications
(b) log10x 3 log10 x 1
log3 xx 2 1 xx 2 x2
log10
31
x3 1 x x3 101 x
2x 3 0
x 1x 3 0
x 3 10x
x 1 OR x 3
3 9x
x 3 is the only solution since the domain of the logarithmic function is the set of all positive real numbers. 56x 8320
26.
x
35x1 86
28.
6x ln 5 ln 8320 x
5x1
ln 8320 0.935 6 ln 5
86 3
x 1ln 5 ln
863
ln x1
863
ln 5 ln
x1
1 0.10 365
30.
365t
ln 5
3.085
t 3 102.6
0.10 365t ln 1 ln 2 365 1 365
863
32. log10t 3 2.6
2
t
t 3 102.6 401.107 ln 2 6.932 0.10 ln 1 365
34. log5 x 4 3.2 x 4 53.2
x 4 53.22 56.4 x 4 56.4 29,748.593 36. f t 3001.007512t 735.41 Zero: t 10
38. gx 1 2 log10 xx 3 Zeros: x 0.826, 3.826 5
10
(10, 0) 0
−10
20
1 3
−5
(− 0.826, 0) 5
(3.826, 0)
−5
517
518
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
40. f x 3x
x
gx log3 x
2
1
0
1
2
1 9
1 3
1
3
9
f x
y
9
f 6
1 9
1 3
1
3
9
2
1
0
1
2
x g x
42. gx 2x
g x
−3
3
6
9
−3
46. f t
y x62x
44.
gx ln 2 2x
3
dy x 2ln 662x 62x dx
ft
62x 2xln 6 1
62x1 2x ln 6 48. g 52 sin 2
50.
1 g 52 2 cos 2 2 ln 5 52 sin 2
52. hx log3
54.
1 3x 2 ln 3 2xx 1
56. f t t 32 log2 t 1 t 32 ft
1 lnt 1 2 ln 2
y log10 2x log10 2 log10 x
y log10
x2 1 x
dy 2x 1 dx x2 1 ln 10 x ln 10
1 0 x 1 ln 3
1 1 1 ln 3 x 2x 1
32t 2t ln 3 1 t2
log10x2 1 log10 x
1 log3 x 1 log3 2 2
1 1 hx x ln 3 2
t2 ln 3 32t 32t t2
dy 1 1 0 dx x ln 10 x ln 10
x x 1 2
log3 x
32t t
58.
1 1 2x ln 10 x2 1 x
x2 1 1 ln 10 xx2 1
y xx1 ln y x 1ln x
1 1 3 t32 t12 lnt 1 2 ln 2 t1 2
1 dy 1 x 1 ln x y dx x dy x1 y ln x dx x
x x2 x 1 x ln x
60.
y 1 x1x ln y
62.
1 ln1 x x
0
64.
2
0
33 52 dx
2
27 25 dx
2 dx
2
2x
lnx 1 1 x1x 1 x x1 x
5x C ln 5
0
5x dx
1 1 1 dy 1 ln1 x 2 y dx x 1x x dy y 1 lnx 1 dx x x 1 x
0 2
4
Section 5.5
66.
2
3 x 7 3x dx
70. (a)
1 3x2 23 x 7 dx 2
68.
sin x
2
Bases Other than e and Applications
519
cos x dx, u sin x, du cos x dx
1 sin x 2 C ln 2
1 2
73x C 2 ln 7
(b)
y
dy esin x cos x dx
6
y
4 2
sin x
e
, 2 sin x
cos x dx e
C
, 2: 2 esin C 1 C ⇒ C 1 x
y esin x 1
10 −2
72. logb x
ln x log10 x ln b log10 b
74. f x log10 x (a) Domain: x > 0
(d) If f x < 0, then 0 < x < 1.
y log10 x
(e) f x 1 log10 x log10 10
(b)
log1010x
10y x
x
f 1 (c)
10x
x must have been increased by a factor of 10.
log10 1000 log10 103 3 log1010,000 log10
104
(f) log10
4
x log x1
10 x1
log10 x2
2
3n n 2n
If 1000 ≤ x ≤ 10,000, then 3 ≤ f x ≤ 4.
Thus, x1x2 102n 100n . 76. f x ax (a) f u v auv au av f u f v
78. Vt 20,000 (a)
(b) f 2x a2x ax 2 f x 2
34
t
(b)
V
34
3 dV 20,000 ln dt 4
t
(c)
V ′(x) x
20,000
dV When t 1: 4315.23 dt
16,000 12,000 8,000
When t 4:
4,000 t 2
V2 20,000
4
6
8
10
34 $11,250 2
dV 1820.49 dt
−1000
2
4
6
12
− 2000 − 3000 − 4000 − 5000 − 6000
Horizontal asymptote: v 0 As the car ages, it is worth less each year and depreciates less each year, but the value of the car will never reach $0.
520
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
80. P $2500, r 6% 0.06, t 20
A 2500 1
0.06 n
20n
n
1
2
4
12
365
Continuous
A
8017.84
8155.09
8226.66
8275.51
8299.47
8300.29
A 2500e0.0620 8300.29 82. P $5000, r 7% 0.07, t 25
0.07 A 5000 1 n
n
1
2
4
12
365
Continuous
A
27,137.16
27,924.63
28,340.78
28,627.09
28,768.19
28,773.01
t
1
10
20
30
40
50
P
94,176.45
54,881.16
30,119.42
16,529.89
9071.80
4978.71
25n
A 5000e0.0725 84. 100,000 Pe0.06t ⇒ P 100,000e0.06t
86. 100,000 P 1
0.07 365
365t
⇒ P 100,000 1
0.07 365
365t
t
1
10
20
30
40
50
P
93,240.01
49,661.86
24,663.01
12,248.11
6082.64
3020.75
88. Let P $100, 0 ≤ t ≤ 20. A 100e0.03t
(a)
A = 100e0.06t A = 100e
0.05t
(b)
A = 100e0.03t
A 100e0.05t A20 271.83
n→
400
A20 182.21 (b)
90. (a) lim
P
or 86%
0.860.25e0.25n 0.215e0.25n 0.25n 1 e 1 e0.25n
P3 0.069
20
0
0.86 0.86 1 e0.25n
P10 0.016
0
A 100e0.06t
(c)
A20 332.01
92. (a)
(d)
12,000
p t
38,000 t5 1 19et5 e 0 5 1 19et53
19et5 1 0
40 0
(b) Limiting size: 10,000 fish (c)
pt
10,000 1 19et5
pt
et5 19 10,000 1 19et52 5
38,000et5 1 19et52
p1 113.5 fishmonth p10 403.2 fishmonth
t ln 19 5 t 5 ln 19 14.72
Section 5.5 94. (a) y1 6.0536x 97.5571
521
(c) The slope of 6.0536 is the annual rate of change in the amount given to philanthropy.
y2 100.0751 17.8148 ln x
(d) For 1996, x 6 and y1 6.0536, y2 2.9691,
y3 99.45571.0506x
y3 6.6015, y4 3.2321.
y4 101.2875x0.1471 (b)
Bases Other than e and Applications
y3 is increasing at the greatest rate in 1996.
150
0
8 100
y3 seems best.
3
96. A
x
x
3 dx
0
ln 3
3
3
0
98.
26 23.666 ln 3
y
x
1
101
102
104
106
1 x1x
2
2.594
2.705
2.718
2.718
30
20
10
x 1
100.
2
3
t
0
1
2
3
4
y
600
630
661.50
694.58
729.30
y Ckt When t 0, y 600 ⇒ C 600. y 600kt 630 661.50 694.58 729.30 1.05, 1.05, 1.05, 1.05 600 630 661.50 694.58 Let k 1.05. y 6001.05t
102. True.
104. True.
f en1 f en ln en1 ln en n1n 1
n
d y Cex dxn y for n 1, 2, 3, . . .
106. True. f x gxex 0 ⇒ gx 0 since ex > 0 for all x.
522
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
108. y x sin x ln y ln xsin x sin x ln x 1 y sin x cos x ln x y x
y xsin x At
sin x cos x ln x x
2 , 2 , y
2
sin2
sin2 cos ln 2 2 2
2 0 1 2
1 x 2 2
Tangent line: y
yx
Section 5.6 2.
Differential Equations: Growth and Decay
dy 4x dx y
dy 4y dx
4.
4 x dx 4x
x2 C 2
y
6.
dy dx 4y 1 dy 4y
4 y exC1 Cex
9y2 4x32 C
y 4 Cex y x1 y
xy y 100x
10.
y 100x x y x100 y
y x 1y y dx 1y
dy 1y
x dx x dx
x2 ln1 y C1 2 1 y ex 2 C1 2
y eC1 ex 2 1
y x 100 y y dx 100 y
1 dy 100 y
ln100 y
x dx x dx
x2 C1 2
2
Cex 2 1 2
x dx
3y2 2 32 x C1 2 3
ln 4 y dy x C1
8.
3y
3yy x
3yy dx
dx
x
ln100 y
x2 C1 2
100 y ex 2 C1 2
y eC1 ex 2 100 2
y 100 Cex 2 2
Section 5.6 dP k10 t dt
12.
dP dt dt
Differential Equations: Growth and Decay dy kxL y dx
14.
1 dy kx L y dx
k10 t dt
k dP 10 t2 C 2 k P 10 t2 C 2
1 dy dx L y dx
kx dx
kx 2 1 dy C1 Ly 2
lnL y
kx 2 C1 2
L y ekx 2 C1 2
y L eC1 ekx 2 2
y L Cekx 2 2
16. (a)
y
(b)
4
dy xy, dx dy x dx y
(0, 12 ) x
−4
0, 21
4
ln y
x2 C 2
y ex 2C C1ex 2
−4
2
2
0, 21: 21 C e 1
18.
dy 3 t, 0, 10 dt 4
dy
20.
15
3 t dt 4
0
1 y t32 C 2
dy y
10
ln y
−5
3 dt 4
(0, 10) −5
3 t C1 4
5 −5
10 Ce0 ⇒ C 10 y 10e3t4
dN kN dt
24.
dP kP dt P Cekt
(Theorem 5.16)
0, 250: C 250
(Theorem 5.16)
0, 5000: C 5000
1, 400: 400 250ek ⇒ k ln
400 8 ln 250 5
When t 4, N 250e4ln85 250eln85
4
250
40
eC1 e34t Ce3t4
1 y t 32 10 2
N Cekt
1 1 2 ⇒ y ex 2 2 2
y e34tC1
1 10 032 C ⇒ C 10 2
22.
⇒ C1
dy 3 y, 0, 10 dt 4
(0, 10)
0
85
4
8192 5
1, 4750: 4750 5000ek ⇒ k ln
19 20
When t 5, P 5000eln19205 5000
19 20
5
3868.905
523
524
Chapter 5
26. y Cekt, 0, 4,
Logarithmic, Exponential, and Other Transcendental Functions
5, 12
C4
3, 12, 4, 5
1 Ce3k 2
y 4ekt
5 Ce4k
1 4e5k 2 k
y Cekt,
28.
2Ce3k
ln18 0.4159 5
1 4k Ce 5
10e3k e4k
y 4e0.4159t
10 ek k ln 10 2.3026 y Ce2.3026t 5 Ce2.30264 C 0.0005 y 0.0005e2.3026t
30. y
dy ky dt
32.
dy 1 2 xy dx 2 dy > 0 when y > 0. Quadrants I and II. dx
34. Since y Ce ln121620t , we have 1.5 Ce ln1216201000 ⇒ C 2.30 which implies that the initial quantity is 2.30 grams. When t 10,000, we have y 2.30e ln12162010,000 0.03 gram. 36. Since y Ce ln125730t, we have 2.0 Ce ln12573010,000 ⇒ C 6.70 which implies that the initial quantity is 6.70 grams. When t 1000, we have y 6.70e ln1257301000 5.94 grams. 38. Since y Ce ln125730t, we have 3.2 Ce ln1257301000 ⇒ C 3.61. Initial quantity: 3.61 grams. When t 10,000, we have y 1.08 grams. 40. Since y Ce ln1224,360t, we have 0.4 Ce ln1224,36010,000 ⇒ C 0.53 which implies that the initial quantity is 0.53 gram. When t 1000, we have y 0.53e ln1224,3601000 0.52 gram.
42. Since
dy ky, y Cekt or y y0ekt. dx
1 y y0e5730k 2 0 k
ln 2 5730
0.15y0 y0eln 25730t ln 0.15 t
ln 2t 5730 5730 ln 0.15 15,682.8 years. ln 2
44. Since A 20,000e0.055t, the time to double is given by 40,000 20,000e0.055t and we have 2 e0.055t ln 2 0.055t t
ln 2 12.6 years. 0.055
Amount after 10 years: A 20,000e0.05510 $34,665.06
Section 5.6 46. Since A 10,000ert and A 20,000 when t 5, we have the following. 20,000 10,000e5r
48. Since A 2000ert and A 5436.56 when t 10, we have the following. 5436.56 2000e10r
ln 2 0.1386 13.86% 5
r
Differential Equations: Growth and Decay
Amount after 10 years: A 10,000e ln 2510 $40,000
r
ln5436.562000 0.10 10% 10
The time to double is given by 4000 2000e0.10t t
0.06 12
50. 500,000 P 1
ln 2 6.93 years. 0.10
1240
0.09 12
52. 500,000 P 1
P 500,0001.005480 $45,631.04
1225
P 500,000 1
0.09 12
300
$53,143.92
(c) 2000 1000 1
54. (a) 2000 10001 0.6t 2 1.06t
2 1
ln 2 t ln1.06 t
ln 2 11.90 years ln 1.06
(b) 2000 1000 1
2 1
0.06 12
0.06 12
12t
1 12
0.055 12
ln 2 12t ln 1 1 t 12
12t
0.055 365
0.055 365
ln 2 365t ln 1 t
12t
1 365
365t
365t
0.055 365
ln 2 12.60 years 0.055 ln 1 365
(d) 2000 1000e0.055t
0.055 12
ln 2 12.63 years 0.055 ln 1 12
ln 2 11.55 years 0.06
ln 2 12.95 years ln 1.055
t
2 1
0.055 2 1 12
0.06 365 1 ln 2 t 11.55 years 0.06 365 ln 1 365
ln 2 t ln1.055
ln 2 365t ln 1
(c) 2000 1000 1
2 1.055t
(b) 2000 1000 1
365t
ln 2 0.06t
56. (a) 2000 10001 0.055t
t
365t
2 e0.06t
ln 2 11.58 years 0.06 ln 1 12
(d) 2000 1000e0.06t
0.06 ln 2 12t ln 1 12 t
12t
0.06 365
0.06 365
2 e0.055t ln 2 0.055t t
ln 2 12.60 years 0.055
525
526
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
58. P Cekt Ce0.031t
60. P Cekt Ce0.004t
P1 11.6 Ce0.0311 ⇒ C 11.9652
P1 3.6 Ce0.0041 ⇒ C 3.5856
P 11.9652e0.031t
P 3.5856e0.004t
P10 16.31
or
P10 3.45
16,310,000 people in 2010
62. (a) N 100.15961.2455t
C 742,000 632,000 742,000e2k
Analytically, 400 100.15961.2455t 1.2455t
k
400 3.9936 100.1596
t
When t 4, y $538,372.
ln 3.9936 6.3 hours. ln 1.2455
20 301 e30k
e0.0366t
ln13 ln 3 0.0366 30 30 e0.0366t
25 301 e0.0366t
(b)
30e30k 10
N 301
ln632742 0.0802 2
y 742,000e0.0802t
t ln 1.2455 ln 3.9936
k
3,450,000 people in 2010
y Cekt, 0, 742,000, 2, 632,000
64.
(b) N 400 when t 6.3 hours (graphing utility)
66. (a)
or
t
1 6 ln 6 49 days 0.0366
68. S 251 ekt (a) 4 251 ek1 ⇒ 1 ek
4 21 21 ⇒ ek ⇒ k ln 0.1744 25 25 25
(b) 25,000 units lim S 25 t→
(c) When t 5, S 14.545 which is 14,545 units.
(d)
25
0
8 0
70. (a) R 979.39931.0694t 979.3993e0.0671t I 0.1385t4 2.1770t3 9.9755t2 23.8513t 266.4923 (b)
2000
Rate of growth Rt 65.7e0.0671t 0
10 0
(c)
(d) Pt
500
I R
1
0
10 0
0
10 0
Section 5.7
72.
93 10 log10
I 10log10 I 16 1016
Differential Equations: Separation of Variables
6.7 log10 I ⇒ I 106.7 I 80 10 log10 16 10log10 I 16 10
1 dy y 80
k dt
When t 0, y 1500. Thus, C ln 1420.
10 106.7
10
Percentage decrease:
dy ky 80 dt
lny 80 kt C.
8 log10 I ⇒ I 108 6.7
Since
74.
8
100 95%
When t 1, y 1120. Thus, k1 ln 1420 ln1120 80 k ln 1040 ln 1420 ln Thus, y 1420eln104142 t 80. When t 5, y 379.2.
76. True
78. True
Section 5.7
Differential Equations: Separation of Variables
2. Differential equation: y
2xy x2 y2
Solution: x 2 y 2 Cy Check: 2x 2yy Cy y
2x 2y C
y
2xy 2xy 2xy 2xy 2y 2 Cy 2y 2 x 2 y 2 y 2 x 2 x 2 y 2
4. Differential equation: y 2y 2y 0 Solution: y C1ex cos x C2ex sin x y C1 C2ex sin x C1 C2ex cos x
Check:
y 2 C1ex sin x 2C2ex cos x y 2y 2y 2C1ex sin x 2C2ex cos x 2 C1 C2ex sin x C1 C2ex cos x 2C1ex cos x C2ex sin x 2C1 2C1 2C2 2C2ex sin x 2C2 2C1 2C2 2C1ex cos x 0 2 6. y 3 e2x ex
y 23 2e2x ex y 23 4e2x ex 2 2 Substituting, y 2y 3 4e2x ex 2 3 2e2x ex 2ex.
104 . 142
527
528
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
In Exercises 8 –12, the differential equation is y 4 16y 0. 8.
y 3 cos 2x
y 5 ln x
10.
y4 48 cos 2x y4 16y 48 cos 2x 48 cos 2x 0,
y4 Yes. y4 16y
12.
30 x4 30 80 ln x 0, x4
No.
y 3e2x 4 sin 2x y4 48e2x 64 sin 2x y4 16y 24e2x 64 sin 2x 163e2x 4 sin 2x 0,
Yes
In 14–18, the differential equation is xy 2y x3ex. 14. y x 2ex, y x 2ex 2xex ex x2 2x xy 2y x ex
x2
2x 2
x2ex
x3ex,
16. y sin x, y cos x xy 2y xcos x 2sin x x3ex,
Yes.
18. y x2ex 5x2, y x2ex 2xex 10x xy 2y xx2ex 2xex 10x 2x2ex 5x2 x3ex,
Yes. 22. 2x 2 y 2 C passes through 3, 4
20. y A sin t
29 16 C ⇒ C 2
d 2y A 2 sin t dt 2
Particular solution: 2x 2 y 2 2
Since d 2ydt 2 16y 0, we have A 2 sin t 16A sin t 0. Thus, 2 16 and ± 4. 24. Differential equation: yy x 0 General solution: x 2 y 2 C Particular solutions: C 0, Point C 1, C 4, Circles
26. Differential equation: 3x 2yy 0 General solution: 3x2 2y2 C 6x 4yy 0 23x 2yy 0 3x 2yy 0 Initial condition: y1 3: 312 232 3 18 21 C Particular solution: 3x2 2y2 21
y 2 1
1
2
x
No.
Section 5.7
Differential Equations: Separation of Variables Initial conditions: y2 0, y2
28. Differential equation: xy y 0 General solution: y C1 C2 ln x
0 C1 C2 ln 2
1 1 y C2 , y C2 2 x x 1 1 xy y x C2 2 C2 0 x x
1 2
y
C2 x
1 C2 ⇒ C2 1, C1 ln 2 2 2 Particular solution: y ln 2 ln x ln
x 2
30. Differential equation: 9y 12y 4y 0 General solution: y e2x3C1 C2 x
2 2 2 y e2x3C1 C2x C2e2x3 e2x3 C1 C2 C2 x 3 3 3 y
2 2 2 2x3 2 2 2 2 e C C2 C2 x e2x3 C2 e2x3 C1 2C2 C2 x 3 3 1 3 3 3 3 3
9y 12y 4y 9
23 e 23 C 2x3
1
2C2
2 2 2 C x 12e2x3 C1 C2 C2 x 4e2x3C1 C2 x 0 3 2 3 3
Initial conditions: y0 4, y3 0 0 e2C1 3C2 4 1C1 0 ⇒ C1 4 0 e24 3C2 ⇒ C2
4 3
4 Particular solution: y e2x3 4 x 3
32.
dy x3 4x dx y
36.
34.
x4 2x2 C 4
x3 4x dx
dy x cos x 2 dx y
x cosx2 dx
1 sinx2 C 2
u x 2, du 2x dx
40.
dy x 5 x. Let u 5 x, u2 5 x, dx 2u du dx y
y
38.
x 5 x dx
5 u2u2u du
10u2 2u4 du
10u3 2u5 C 3 5
10 2 5 x32 5 x52 C 3 5
dy ex dx 1 ex
ex dx ln1 ex C 1 ex
dy tan2 x sec2 x 1 dx y
sec2 x 1 dx tan x x C
529
530
42.
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
dy 5ex2 dx y
dy x 2 2 dx 3y 2
44.
21 dx
5ex2 dx 52 ex2
3y 2 dy
10ex2 C
dr 0.05s ds
46.
y3
48.
dr
dy y
r 0.025s 2 C
x 2 2dx
x3 2x C 3
xy y
0.05s ds
y
50.
dx x
dy 6 cos x dx
y dy
y2
dy 5x dx
dy
4y
54. 5x
x2 9
y 5x2 912 C 56. x y y 0
dy 3ex dx
4y dy
dx
3ex dx
2y2 3ex C 58. 2xy ln x2 0
y12 dy x12 dx
2x
dy 2 ln x dx
2 32 2 y x32 C1 3 3
dy
ln x dx x
y
ln x2 C 2
y32 x32 C Initial condition: y 1 4, 432 132 8 1 9 C
y1 2: 2 C 1 y ln x2 2 2
Particular solution: y32 x32 9
y 1 x2
60.
dy x 1 y 2 dx
1 y212 y dy
1 x212 x dx
1 y212 1 x212 C y0 1:
0 1 C ⇒ C 1
1 y 2 1 x 2 1
62.
6 cos x dx
y2 6 sin x C1 2
ln y ln x ln C ln Cx y Cx
52. x2 9
dr er2s ds
erdr
e2s ds
1 er e2s C 2 1 1 r0 0: 1 C ⇒ C 2 2 1 1 er e2s 2 2 1 1 er e2s 2 2 r ln r ln
12e
2s
1 e2s 1 ln 2 2
1 2e 2s
12 sin x C
Section 5.7
64. dT kT 70 dt 0
Differential Equations: Separation of Variables dy 2y dx 3x
66.
dT k dt T 70
lnT 70 k t C1
3 dy y
2 dx x
ln y3 ln x2 ln C
T 70 Cekt
y3 Cx2
Initial condition: T0 140; 140 70 70 Ce0 C
Initial condition: y8 2, 23 C82, C
Particular solution: T 70 70ekt, T 701 ekt
Particular solution: 8y3 x2, y
m
68.
dy y 0 y dx x 0 x
dy y
70.
1 23 x 2
f x, y x3 3x2y2 2y2 f tx, ty t3x3 3t4x2y2 2t2y2
dx x
Not homogeneous
ln y ln x C1 ln x ln C ln Cx y Cx f x, y
72.
f t x, t y
xy
x2 y2
74.
f t x, t y tantx t y tant x y
tx ty x2 t 2 y 2
t 2
f x, y tanx y
Not homogeneous
2
xy t xy t
x 2 y 2 t x 2 y 2
Homogeneous of degree 1
f x, y tan
76.
f t x, t y tan
y x ty y tan tx x
Homogeneous of degree 0
78. y
x3 y3 xy2
y
80.
x y 2 dy x3 y3 dx y vx,
dy x dv v dx
xvx x dv v dx x3 vx3 dx 2
x4 v2 dv x3 v3 dx x3 dx v3 x3 dx xv 2 dv dx
v 2 dv
vx
1 dx x
v3 ln x C 3
yx 3 lnx C 3
y3 3x3 ln x Cx3
dv x 2 v 2 x 2 dx 2x 2 v
2v dx 2x dv
x2 y2 , y vx 2xy
1 v2 dx v
2v dx dv v2 1 x
lnv 2 1 ln x ln C ln v2 1
C x
y2 C 1 x2 x y 2 x2 Cx
C x
1 8
531
532
Chapter 5
82.
y
Logarithmic, Exponential, and Other Transcendental Functions
2x 3y , y vx x
dv 2x 3vx vx 2 3v dx x dv 2 2v ⇒ dx
x
ln 1 v ln
x2
x 2 v 2 dx x 2 x 2 vv dx x dv 0
dv dx 2 1v x
ln C ln
y 2 dx xx y dy 0, y vx
84.
x2C
1v dx dv v x
v ln v ln x ln C1 ln
C1 x
v ln
C1 xv
1vxC 2
1
y x2C x
C1 ev vx
y x2C 1 x
C1 eyx y
y Cx3 x
y Ceyx Initial condition: y1 1, 1 Ce1 ⇒ C e Particular solution: y e1yx
86. 2x 2 y 2 dx xy d y 0
88.
dy x dx y
Let y vx, dy x dv v dx.
v 2x 2
2x 2 v 2
2
2v 2
2x 2
y
dx xv xx d v v d x 0
2x 2
dx
x3v
dv 0
d x x v d v
x2 C1 v 212
12
C x2 y212 x
1 Cx2 y212 x y1 0: 1 C1 0 ⇒ C 1 1 x2 y2 x 1 x x2 y2
2
4
−4
ln x2 ln1 v 212 ln C
x
−2 −2
1 ln1 v 2 C1 2
1 y2 C 1 2 x2 x
2 −4
2 v dx dv x 1 v2 2 ln x
4
y dy x dx y 2 x 2 C1 2 2 y 2 x2 C
Section 5.7 dy 0.25x4 y dx
90.
y
8
dy 0.25x dx 4y
dy y4
Differential Equations: Separation of Variables
4
0.25x dx
−4
x
−2
2
4
−2
1 x dx 4
1 ln y 4 x 2 C1 8
y 4 eC1 18x Ce18x 2
2
y 4 Ce18x
2
dy 2 y, y0 4 dx
92.
94.
dy 0.2x2 y, y0 9 dx
9
10
−5
5
−5
−1
96.
dy ky, y Cekt dt Initial conditions: y0 20, y1 16 20 Ce C 0
98.
5 0
dy kx 4 dx The direction field satisfies dydx 0 along x 4: Matches (b).
16 20ek k ln
4 5
Particular solution: y 20et ln45 When 75% has been changed: 5 20et ln45 1 et ln45 4 t
100.
ln14 6.2 hr ln45
dy ky2 dx
The direction field satisfies dydx 0 along y 0, and grows more positive as y increases. Matches (d).
102. From Exercise 101, w 1200 Cekt, k 1 w 1200 Cet w0 w0 1200 C ⇒ C 1200 w0 w 1200 1200 w0et
533
534
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
104. Let the radio receiver be located at x0, 0. The tangent line to y x x2 joins 1, 1 and x0, 0.
y
2
Transmitter (−1, 1) 1
y=x−
x2
Radio
−1
(b) Now let the transmitter be located at 1, h.
(a) If x, y is the point of tangency on the y x x2, then
1 2x
y 1 x x2 1 1 2x x1 x1
x2 2x h 1 0
2x 2 0
2 ± 4 8 x 2
1
x
3
1 3 3 5
10 1 x0 1 1 3
3 1 x0 6 3 3
(c)
3 2 h h 4
3
Then,
h0 h 3 2 h h 4 1 x0 1 1 2 h
4 3 6 1.155 6 3 3
2h 4 3 2 h 2 h
x0 1
2 h h 2h 4 3 2 h
10
x0 0.25
2
y x x2
6 3 3
6 3 3 x0 6 3 3 x0
2 ± 4 4h 1
1 2 h
y x x2 3 3 5 Then
y h x x2 h x1 x1
x 2x 2 1 2x x x 2 h
x 2x2 1 2x x x2 1 x2
x
x 1 x0
h 2 h 1 2h 4 3 2 h
3
−2
There is a vertical asymptote at h 14 , which is the height of the mountain. x2 2y2 C
106. Given family (hyperbolas):
2x 4yy 0 y y
Orthogonal trajectory:
2yy 2C
x 2y
2y x
y
−2
2x y
y dy 2x dx y2 x 2 K1 2
k x2
2x 2 y 2 K 4
2
3
y
ln y 2 ln x ln k
−3
C y2 1 y y 2x y 2x
Orthogonal trajectory (ellipse):
dy 2 dx y x
y kx2
y 2 2Cx
108. Given family (parabolas):
−6
6
−4
Section 5.8
Inverse Trigonometric Functions: Differentiation
110. Given family (exponential functions): y Cex y
Cex
y
Orthogonal trajectory (parabolas):
535
4
y −6
1 y
6
−4
y dy dx y2 x K1 2 y2 2x K 114. Two families of curves are mutually orthogonal if each curve in the first family intersects each curve in the second family at right angles.
112. The number of initial conditions matches the number of constants in the general solution.
116. True dy x 2y 1 dx 118. True x 2 y 2 2Cy
x 2 y 2 2Kx dy K x dx y
x dy dx C y x Cy
Kx Kx x2 2Kx 2x2 x2 y 2 2x2 y 2 x2 2 2 1 2 2 2 2 y Cy y 2Cy 2y x y 2y x y2
Section 5.8
Inverse Trigonometric Functions: Differentiation
2. y arccos x (a)
1
x y
0.8
3.142
(b)
0.6
2.499
y
0.4
2.214
1.982
0.2 1.772
0 1.571
0.2
0.4
0.6
0.8
1
1.369
1.159
0.927
0.634
0
(d) Intercepts:
(c)
π
No symmetry −1
1 0
x
−1
4.
1
1, 4 4
,
,
3,
3 , 6 3 6
0, 2 and 1, 0
6. arcsin 0 0
3, 3
536
Chapter 5
8. arccos 0
12. arccos
Logarithmic, Exponential, and Other Transcendental Functions
2
3
2
10. arc cot 3
56
18. (a) tan arccos
2
2
14. arcsin0.39 0.40
tan4 1
16. arctan3 1.25
(b) cos arcsin 2
5 6
5 12 13 13
13 5
θ
2
12
θ 2
53
20. (a) sec arctan
34
5
65 5 1111
(b) tan arcsin
θ
5
−3
11
θ
34 −5
6
22. y secarc tan 4x
24. y cosarccot x
1 + 16x 2
arctan 4x
θ
y sec 1 16x2
1
y cos
26. y sec arcsinx 1
arcsinx 1
x−1
θ
1 y sec 2x x2
30.
2x −
arcsin x2
xh r
x
r
θ r 2 − (x − h)2
r 2 x h2
r
32. arctan2x 5 1 2x 5 tan1
f=g −2
2
1 x tan1 5 1.721 2
−5
Asymptote: x 0 x 2
2 4 − x2
x cos 2 tan
x x2 1
xh r
y cos
5
arccos
1
θ
28. y cos arcsin 1
x2 + 1
arccot x
4x
θ x
4 x2
x
x−h
Section 5.8
Inverse Trigonometric Functions: Differentiation
537
34. arccos x arcsec x x cosarcsec x x
1 x
x
x2 − 1
θ 1
x2 1 x ±1
36. (a) arcsinx arcsin x, x ≤ 1.
(b) arccosx arccos x, x ≤ 1.
Let y arcsinx. Then,
Let y arccosx. Then,
x sin y ⇒ x sin y ⇒ x siny.
x cos y ⇒ x cos y ⇒ x cos y.
Thus, y arcsin x ⇒ y arcsin x. Therefore, arcsinx arcsin x.
38. f x arctan x
x tan y 2
2
Thus, y arccos x ⇒ y arccos x. Therefore, arccosx arccos x.
y
40. f x arccos
π
π 2
−6 −4 −2
2
4
2t
fx
1 1 x21 x 2x11 x 2
52. y lnt2 4 y
1 x x4 x2 4 arcsin 2 2
56.
1 1 1 x 4 x212 2x 4 x2 2 1 x22 2 2 1 x2 4 4 x2 4 x2 2 4 x2
4
x2
2
2x 4x2 1
hx 2x arctan x
fx 0
2
4
x
6
1
x 4x2 1
48. hx x2 arctan x
50. f x arcsin x arccos x
y
−2
44. f x arcsec 2x
1 t 4
−6 −4
Range: 0,
46. f x arctan x
54. y
π
(4, 0)
Domain: 4, 4
6
42. f t arcsin t 2
fx
(−4, π )
x 4 cos y x
f x is the graph of arctan x shifted 2 units upward.
ft
y
x cos y 4
Domain: , Range: 0,
4x
2t 1 t2 4 2
x2 1 x2
1 t arctan 2 2 1
t 1 2
2
12
1 2t 1 2t 2 t2 4 t2 4 t 4
y x arctan 2x
1 ln1 4x2 4
dy 2x 1 8x arctan2x arctan2x dx 1 4x2 4 1 4x2
538
Chapter 5
58. y 25 arcsin y 5
Logarithmic, Exponential, and Other Transcendental Functions x x25 x2 5
1 1 x22
60. y arctan
1 25 x2 x 25 x212 2x 2
y
1 1 1 x2 422x 2 1 x22 2
25 x2 25 x2 2 2 25 x 25 x 25 x2
2 x x2 4 x2 42
2x2 25 x2
2x2 8 x x2 42
y
62. f x arctan x, a 1 fx
1 1 x2
f x
2x 1 x22
P1 (x) π 2
f
π 4 −4
−2
x 2
P2 (x)
P1x f 1 f 1x 1
1 x 1 4 2
P2x f 1 f 1x 1
1 1 1 f 1x 12 x 1 x 12 2 4 2 4
64. f x arcsin x 2x
66. f x arcsin x 2 arctan x
1 1 fx 2 0 when 1 x2 or 2 2 1 x x± f x f
1 x 2 2x2 4
3
2
1 1 x2
x4 6x2 3 0 x ± 0.681
Relative minimum:
3
2
2 , 0.68 3
23, 0.68
68. arctan 0 0. is not in the range of y arctan x. x 3
3x
3 dx d 2 dt x 9 dt If x 10,
70. The derivatives are algebraic. See Theorem 5.18.
74. cos
arccot
If x 3,
By the First Derivative Test, 0.681, 0.447 is a relative maximum and 0.681, 0.447 is a relative minimum.
0
(b)
fx
d 11.001 radhr. dt
d 66.667 radhr. dt
A lower altitude results in a greater rate of change of .
750 s
s
arccos d d dt ds
750s
θ 750
ds 1 750 ds 2 dt s2 dt 1 750s
750 ds ss2 7502 dt
Section 5.9 76. (a) Let y arcsin u. Then
Inverse Trigonometric Functions: Integration (b) Let y arctan u. Then
1
sin y u y
cos y y u
1 − u2
sec2 y
u u dy . dx cos y 1 u2
u y
dy u dx
1
dy u u . dx sec2 y 1 u2
(c) Let y arcsec u. Then
(d) Let y arccos u. Then
u
sec y u
u2− 1
1
cos y u
y
dy sec y tan y u dx
1 + u2
tan y u
u
539
1
1 − u2
y
sin y
dy u u . dx sec y tan y u u2 1
dy u dx
u
u u dy . dx sin y 1 u2
Note: The absolute value sign in the formula for the derivative of arcsec u is necessary because the inverse secant function has a positive slope at every value in its domain.
(f) Let y arccsc u. Then u
csc y u
1
y
(e) Let y arccot u. Then
csc y cot y
1 + u2
cot y u
1
dy u dx
u2 − 1
dy u u dx csc y cot y u u2 1
y
csc2 y
dy u dx
u
Note: The absolute value sign in the formula for the derivative of arccsc u is necessary because the inverse cosecant function has a negative slope at every value in its domain.
u u dy . dx csc2 y 1 u2
78. f x sin x
3
gx arcsinsin x
f
(a) The range of y arcsin x is 2 ≤ y ≤ 2.
− 2
2
g
(b) Maximum: 2
−3
Minimum: 2 82. False
80. False
The range of y arcsin x is , . 2 2
Section 5.9
1
6.
4 4 3 4 dx dx arctan3x C 1 9x2 3 1 9x2 3
10.
1 1 x1 dx arctan C 4 x 12 2 2
3 4x2
2
1
Inverse Trigonometric Functions: Integration
2.
2
arcsin2 0 arccos2 0 0
dx
3
2
2 1
4x2
dx
3 2
arcsin2x C
1
4.
0
8.
12.
1 x 2 dx arcsin 2 4 x
3
1 x 1 arctan 2 dx 3 3 3 9 x x4 1 dx x2 1
1
0
3 3
6
36
1 x2 1 dx x3 x C 3
540
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
14. Let u t 2, du 2t dt.
t4
16. Let u x2, du 2x dx.
t 1 1 t2 1 dt 2t dt arctan C 2 2 2 16 2 4 t 8 4
1 1 1 dx 2x dx 2 x2x22 22 xx 4 4
18. Let u arccos x, du
12
0
1 dx. 1 x2
12
arccos x dx 1 x2
0
2
22.
1
1 dx 3 x 22
2
24.
0
20. Let u 1 x2, du 2x dx. 0
12
2
1
arccos x dx 1 x2
1 arccos2 x 2
0
3
2
0
x 1 dx 1 x2 2
32 0.925 32
1 1 x2 dx arctan 32 x 22 3 3
cos x dx arctansin x 1 sin2 x
1 x2 arcsec C 4 2
4
26.
2 1
0
3
1 2x dx 1 x2
12 ln1 x
0
2
3
ln 2
3
18
3 1 dx. u x, du dx, dx 2u du 2x1 x 2x
3 2u du du 3 3 arctan u C 2 u1 u2 1 u2 3 arctanx C
28.
1 x2
30.
x2 2x 2 1 dx dx x 12 4 2 x 12 4
4x 3
dx 2
2
2
34.
36.
x2
2 x2 4x
dx
dx 41 x2 3 arcsin x C
3 dx x 12 4
2
dx 1 x2 arctan 2 3 3 2 x 2 9
2 2
4 1 arctan 3 3
2x 2 1 dx 7 dx ln x2 2x 2 7 arctanx 1 C x2 2x 2 1 x 12
2 4 x2 4x 4
2 dx 4 x 22
2 arcsin
40.
1 1 x2
3 x1 1 lnx2 2x 5 arctan C 2 2 2
dx 4x 13
2x 5 dx x2 2x 2
dx 3
32.
2x 1 x2
1 dx x 1x2 2x
dx
38. Let u x2 2x, du 2x 2 dx.
x1 x2 2x
x2 C 2
1 dx arcsecx 1 C x 1x 12 1
dx
1 x2 2x122x 2 dx 2
x2 2x C
Section 5.9
Inverse Trigonometric Functions: Integration
42. Let u x2 4, du 2x dx.
x 9 8x2 x4
1 2x 1 x2 4 dx arcsin C 2 25 x2 42 2 5
dx
44. Let u x 2, u2 2 x, 2u du dx
x 2
x1
dx
2u2 du 3
u2
2u
46. The term is
48. (a) (b)
50. (a) (b) (c)
32
2
2u2 6 6 1 du 2 du 6 2 du u2 3 u 3
u 6 arctan C 2x 2 23 arctan 3 3
9 9 9 3 : x2 3x x2 3x x 4 4 4 2
2
9 4
(c)
x 3 2 C
2
ex dx cannot be evaluated using the basic integration rules. xex dx
2
1 x2 e C, u x2 2
1 1x 1 e dx e1x C, u x2 x
1 dx cannot be evaluated using the basic integration rules. 1 x4
x 1 2x 1 dx dx arctanx2 C, u x2 1 x4 2 1 x22 2
x3 1 4x3 1 dx dx ln1 x 4 C, u 1 x 4 4 1x 4 1 x4 4
52. (a)
54.
y
4
dy 2y , y0 2 dx 16 x2 3
x
−1.25
1.25 −3 −4
(b)
−1
dy x16 y2, 0, 2 dx dy 16 y2
x dx
4y x2 C 2
arcsin
42 C ⇒ C 6
0, 2: arcsin
4y x2 6 2
arcsin
3
y x2 sin 4 2 6
y 4 sin
x2 6
2
541
542
Chapter 5
1
56. A
0
Logarithmic, Exponential, and Other Transcendental Functions
x 1 dx arcsin 2 4 x2
1 0
1
6
58.
arcsin x dx 0.571
0
y
y
π
1
2
π 4
x
60. Fx
1 2
x
1
1 2
x2
x
1 2
1
2 dt t2 1
(a) Fx represents the average value of f x over the interval x, x 2. Maximum at x 1, since the graph is greatest on 1, 1.
(b) Fx arctan t Fx
62.
1 6x x2
x2 x
arctanx 2 arctan x
1 1 4x 1 1 x2 x2 4x 5 2 0 when x 1. 2 2 1 x 2 1x x2 1x2 4x 5 x 1x2 4x 5
dx
(a) 6x x2 9 x2 6x 9 9 x 32
1 dx 6x x2
(c)
4
y2
dx x3 arcsin C 3 9 x 32
y1 −1
(b) u x, u2 x, 2u du dx
1 6u2 u4
2u du
7
−2
2 6 u2
du 2 arcsin
u6 C 2 arcsin 6 C x
The antiderivatives differ by a constant, 2. Domain: 0, 6
64. Let f x arctan x fx
x 1 x2
1 1 x2 2x2 > 0 for x > 0. 1 x2 1 x22 1 x2
y 5
x Since f 0 0 and f is increasing for x > 0, arctan x > 0 for x > 0. Thus, 1 x2 x arctan x > . 1 x2 Let gx x arctan x gx 1
1 x2 > 0 for x > 0. 2 1x 1 x2
Since g0 0 and g is increasing for x > 0, x arctan x > 0 for x > 0. Thus, x > arctan x. Therefore, x < arctan x < x. 1 x2
y3
4 3 2
y2
1
y1 2
4
6
x 8
10
Section 5.10
Section 5.10
Hyperbolic Functions
e0 e0 1 2
2. (a) cosh0
4. (a) sinh10 0 (b) tanh10 0
2 (b) sech1 0.648 e e1
6. (a) csch12 ln (b) coth13
8.
1 2 5 0.481
1 4 ln 0.347 2 2
1 cosh 2x 1 e2x e2x2 e2x 2 e2x ex ex 2 2 4 2
10. 2 sinh x cosh x 2
e
x
ex 2
e
x
2
cosh2 x
ex e2x e2x sinh 2x 2 2
xy2
12. 2 cosh
Hyperbolic Functions
x 2 y coshx 2 y 2 e e
x
2
exy2 2
xy2
e
exy2 2
ey ey ex ex ex ey ey 4 2 2
cosh x cosh y tanh x
14.
1 2
2
Putting these in order:
1 2
sech2 x 1 ⇒ sech2 x
3
3 ⇒ sech x 4 2
23 1 cosh x 3 32 1 2 coth x 12 sinh x tanh x cosh x csch x
1 33
122 3 3
3
3
csch x 3
cosh x
23 3
sech x
3
tanh x
1 2
coth x 2
3
3
3
16. y coth3x y 3
sinh x
18. gx lncosh x
3x
csch2
20. y x cosh x sinh x y x sinh x cosh x cosh x x sinh x
gx
1 sinh x tanh x cosh x
22. ht t coth t ht 1 csch2 t coth2 t
2
543
544
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
24. gx sech2 3x
28. y sechx 1
26. f x esinh x
gx 2 sech3x sech3x tanh3x3
y sechx 1 tanhx 1
fx cosh xesinh x
6 sech 3x tanh 3x 2
30. f x x sinhx 1 coshx 1
6
fx x coshx 1 sinhx 1 sinhx 1 x coshx 1 fx 0 for x 0. By the First Derivative Test,0, cosh1 0, 1.543 is a relative minimum.
−6
6 (0, −1.543) −2
32. hx 2 tanh x x
34.
y a cosh x y a sinh x
2
y a cosh x
(0.88, 0.53) −3
3
Therefore, y y 0.
(− 0.88, − 0.53) −2
Relative maximum: 0.88, 0.53 Relative minimum: 0.88, 0.53 36. f x cosh x
f 1 cosh0 1
f x sinh x
f 1 sinh0 0
f x cosh x
f 1 cosh0 1
3
f P2 P1 −2
P1x f 0 f0x 0 1
2 0
1 P2x 1 2 x2
38. (a) y 18 25 cosh
x , 25 ≤ x ≤ 25 25
(b) At x ± 25, y 18 25 cosh1 56.577. At x 0, y 18 25 43.
80
x (c) y sinh 25. At x 25, y sinh1 1.175 −25
25 −10
40. Let u x, du
1 dx. 2x
42. Let u cosh x, du sinh x dx.
cosh x 1 dx 2 cosh x dx 2 sinhx C x 2x
44. Let u 2x 1, du 2 dx.
sech22x 1 dx
1 sech22x 12 dx 2 1 tanh2x 1 C 2
sinh dx 1 sinh2 x
sinh x 1 dx C cosh2 x cosh x
sech x C 46. Let u sech x, du sech x tanh x dx.
sech3 x tanh x dx sech2 xsech x tanh x dx 1 sech3 x C 3
Section 5.10
48.
cosh2 x dx
4
1 cosh 2x dx 2
50.
0
Hyperbolic Functions
1 x dx arcsin 2 5 25 x
4 0
1 sinh 2x x C 2 2
1 1 x sinh 2x C 2 4
52.
2 1 1 1 4x2 dx 2 2 dx 2 ln C 2 2 2x x1 4x 2x1 2x
56. y tanh1
54. Let u sinh x, du cosh x dx.
cosh x 9 sinh2 x
dx arcsin arcsin
sinh3 x C e
x
58. y sech1cos 2x, 0 < x < y
e 6
x
y
C
2x
1 1 2 1 x22 2 4 x2
4
1 2 sin 2x 2 2 sin 2x 2 sec 2x, cos 2x sin 2x cos 2x cos 2x1 cos2 2x
since sin 2x ≥ 0 for 0 < x < 4. 60. y csch1 x2 csch x x 11 x 2 x 1x
y 2 csch1 x
1
2
2
62. y x tanh1 x ln1 x2 x tanh1 x y x
1 1 x tanh
1
2
x
1 ln1 x2 2
64. See page 401, Theorem 5.22.
x tanh1 x 1 x2
66. Equation of tangent line through P x0, y0: y a sech1
y
a2 x02 x0 a2 x02 x x0 a x0
When x 0,
Q
a P
y a sech1
x0 x a2 x02 a2 x02 a sech1 0. a a
Hence, Q is the point 0, a sech1x0a.
x02 2
a
x 1 2x 1 1 3 x2 dx dx ln C 4 2 2 9x 2 9 x 2 6 3 x2
1 3 ln C 12 3 x2 x2
x
L
Distance from P to Q: d x02 a2
68.
(a, 0)
arcsin
4 5
545
546
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
3 70. Let u x32, du x dx. 2
72.
x 1 x3
dx
2 2 1 3 2 x dx sinh1x32 C ln x32 1 x3 C 3 1 x322 2 3 3
dx x 2x2 4x 8
dx x 2x 22 4
2 x 22 4 1 C ln 2 x2
74.
1 dx x 12x2 4x 8
1 dx x 12x 12 6
3 x 12 3 1 1 dx 1 ln C x1 2 x 1x 12 32 6
76. Let u 2x 1, du 2 dx. y
78. y
1 dx x 14x2 8x 1 1 2x dx 4x x2
ln 4x x2
2 2x 1 3 2x 12 2
4 2x 1 dx 3 dx 4x x2 x 22 4
3
e2x
0
2
1 2
0
e2x dx e2x
2 lne
1 1 lne4 e4 ln 2 2 2
e
ln
4
dx
1 2e2x e2x dx e2x e2x 2x
6 x2 4
6 ln x x2 4
1
5
2 2x e
82. A
tanh 2x dx
0
3 4x2 8x 1 1 ln C 2x 1 3
3 x 2 2 3 x4 ln C ln 4x x2 ln C 4 x 2 2 4 x
2
80. A
dx
e2x
2 0
5 3
6 ln 5 21 6 ln 3 5 6 ln
53 215 3.626
e4 1.654 2
84. (a) vt 32t
(b) st
vt dt
32t dt 16t2 C
s0 1602 C 400 ⇒ C 400 st 16t2 400 —CONTINUED—
Section 5.10
Hyperbolic Functions
547
84. —CONTINUED— dv 32 kv2 dt
(c)
dv kv2 32
t→
dt
st
32 k v 1 ln t C 232 32 k v
Since v0 0, C 0. ln
32 k v 32 k v
32 k v 32 k v
v
k e2
32k t
e32k t e32k t
32 e 32 k
ex ex tan y sinh x. 2
1
k 32k
ln cosh 32k t C
s2t 0 when t 8.3 seconds
e32k t
32k t
When air resistance is not neglected, it takes approximately 3.3 more seconds to reach the ground.
tanh 32k t
x
e +e
−x x
e −e
−x
y 2
Thus, y arctansinh x. Therefore, arctansinh x arcsintanh x. y sech1 x sech y x sech ytanh yy 1 1 1 1 sech ytanh y sech y1 sech2 y x1 x2
y sinh1 x sinh y x
cosh yy 1 y
32
tanh 32k t dt
s1t 0 when t 5 seconds.
86. Let y arcsintanh x. Then, ex ex and ex ex
k
s1t 16t2 400.
e 32k t 1 e
32
s2t 400 100 ln cosh0.32 t 32k t
k
When k 0.01,
32 e232k t 1
90.
.
400 ⇒ 400 1k ln cosh 32k t.
y
k
When t 0, s0 C
e232k t
v k k e232k t 32e232k t 1
88.
32
1 ln cosh 32k t C. k
232k t
32 k v e232k t 32 k v
sin y tanh x
k
tanh32k t
(e) Since tanhct dt 1c ln coshct (which can be verified by differentiation), then
Let u k v, then du k dv. 1
32
The velocity is bounded by 32k.
dv dt 32 kv2
k
(d) lim
1 1 1 cosh y sinh2 y 1 x2 1
548
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
Review Exercises for Chapter 5 2. f x lnx 3
4. lnx2 1x 1 lnx2 1 lnx 1
y 3
Horizontal shift 3 units to the right Vertical asymptote: x 3
x=3
2 1
x 1
2
4
5
6
−1 −2 −3
x 25x 1 3
6. 3ln x 2 lnx2 1 2 ln 5 3 ln x 6 lnx2 1 ln 52 ln x3 lnx2 16 ln 25 ln
8. ln x lnx 3 0
10. hx ln
ln xx 3 0
hx
xx 3 e0
2
6
xx 1 ln x lnx 1 lnx 2 x2
1 1 1 x2 4x 2 x x 1 x 2 x3 3x2 2x
x2 3x 1 0 x
3 ± 13 2
x
3 13 3 13 < 0. only since 2 2
12. f x lnxx2 22 3 ln x fx
16.
2 lnx2 2 3
1 2 2x 7x2 6 3 x 3 x2 2 3x 6x
y
a bx bx 1 2 ax2a bx x a bx
lnx x
0
tan
dx
e
1 dx x
x 2x 1 1 dx dx ln x2 1 C x2 1 2 x2 1 2
1 b a 1 b 2 ax2 a2 xa bx ax axa bx
4
24.
20. u ln x, du
18. u x2 1, du 2x dx
1 a bx a lna bx b2
1 b lna bx ln x ax a2
y
dy 1 ab x b dx b2 a bx a bx
1 b a bx ln ax a2 x
1 b dy 1 b 1 2 2 dx a x a a bx x
14.
22.
1
1 1 1 2 ln x dx ln x C 2 x 4
4
4 x dx ln cos 4 x 0 ln
0
12 2 ln 2 1
ln x dx x
e
1
ln x1
e
1 1 dx ln x2 x 2
1
1 2
Review Exercises for Chapter 5 26. (a)
f x 5x 7
(b)
y 5x 7
6
f
−10
−1 6
y7 x 5 x7 y 5 x7 f 1x 5
28. (a)
f x x3 2
f
−10
(c) f 1 f x f 15x 7 f f 1x f
(b)
x 5 7 5 x 5 7 7 x f
f
−1
−4
3 y2x
5x 7 7 x 5
3
y x3 2
5
3 x2y 3 x 2 f 1x
549
−3
3 x3 2 2 x (c) f 1 f x f 1x3 2 3 x 2 f f 1x f 3 x 2 2 x 3
30. (a)
f x x2 5, x ≥ 0
(b)
4
f −1
y x2 5
−6
6
y 5 x
f
x 5 y
f 1
x x 5
−6
(c) f 1 f x f 1x2 5 x2 5 5 x for x ≥ 0. f f 1x f x 5 x 5 5 x 2
f x xx 3
32.
f x ln x
34.
f 4 4
x ex
f 1
f 1x ex
1 fx x 3 xx 31 2 2
f 10 e0 1
f4 1 2 3
f 14
36. (a)
1 1 f4 3
f x e1x
(b)
4
y e1x ln y 1 x x 1 ln y y 1 ln x f 1x 1 ln x
f
−4
f
−1
5
−2
(c) f 1 f x f 1 e1x 1 lne1x 1 1 x x f f 1x f 1 ln x e1 1ln x eln x x
550
Chapter 5
38. y 4ex
Logarithmic, Exponential, and Other Transcendental Functions 40. gx ln
2
y
1 e e
x
x
ln ex ln1 ex x ln1 ex
5 4
ex 1 1 ex 1 ex
gx 1 1 x
−5 − 4 −3 −2 −1
1 2 3 4 5
−2 −3 −4 −5
42. hz ez 2
1 46. f esin 2 2
44. y 3e3 t
2
hz zez 2 2
y 3e3 t3t2
2x sin x2 xey
dy ey dx
dy 2x sin x2 ey dx xey 52. Let u e2x e2x, du 2e2x e2x dx.
dx
3 1
x2ex
1 x3 1 2 1 3 e 3x dx ex 1 C 3 3
1 lne2x e2x C 2
e2x 1 1 dx 2e2x dx e2x 1 2 e2x 1
e1 x 1 dx e1 x 2 dx e1 x C x2 x
54. Let u x3 1, du 3x2 dx.
e2x e2x 1 2e2x 2e2x dx dx 2x 2x e e 2 e2x e2x
56.
f cos 2 esin 2
1 1 50. Let u , du 2 dx. x x
cos x2 xey
48.
9e3 t t2
58. (a), (c)
10,000
1 lne2x 1 C 2 0
5 0
V 8000e0.6t, 0 ≤ t ≤ 5
(b)
Vt 4800e0.6t V1 2634.3 dollars year V4 435.4 dollars year
2
60. Area
2ex dx 2ex
0
62. gx 62x
2 0
2e2 2 2
2 1.729 e2 64. y log4 x2
2
y
y
5 4 3 2 1
8 6
−5 − 4 −3 −2 −1 2
−4
−2
x 2
4
−2 −3 −4 −5
x 1 2 3 4 5
Review Exercises for Chapter 5 68. y x4x
66. f x 4xex
y 4x x 4x ln 4
fx 4xex ln 44xex 4xex1 ln 4 x log5 x log5x 1 x1
70. hx log5 hx
1 1 1 1 1 ln 5 x x1 ln 5 x x 1
74. t 50 log10
72.
21 t 1 1 t dt 2 C t2 ln 2
18,000 18,000 h
t 50 log10
(c)
(a) Domain: 0 ≤ h < 18,000 (b)
551
10t 50
t
18,000 18,000 h
18,000 18,000 h
18,000 h 18,00010t 50
100
h 18,0001 10t 50
80
As h → 18,000, t → .
60 40 20 h 4000
12,000
(d) t 50 log10 18,000 50 log1018,000 h dt 50 dh ln 1018,000 h 50 d 2t dh2 ln 1018,000 h2
Vertical asymptote: h 18,000
No critical numbers As t increases, the rate of change of the altitude is increasing.
76. 2P Pe10r
y5
78.
2 e10r y600 5
ln 2 10r r
dy 0.012y, s > 50 ds
1 0.012
dy y
ds
1 ln y s C1 0.012 y Ce0.012s When s 50, y 28 Ce0.01250 ⇒ C 28e0.6 y 28e0.60.012s, s > 50 e2x dy dx 1 e2x
dy
12
t 1620
600 1620
3.868 grams
ln 2 6.93% 10
80. (a)
82.
12
e2x 1 2e2x dx dx 2x 1e 2 1 e2x
1 y ln1 e2x C 2
(b) Speed(s)
50
55
60
65
70
Miles per Gallon (y)
28
26.4
24.8
23.4
22.0
552
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
84. y ey sin x 0 dy ey sin x dx
ey dy
sin x dx
ey cos x C1 ey
1 cos x C
y ln
86.
C C1
1 ln cos x C cos x C
dy 3x y (homogeneous differential equation) dx x 3x y dx x dy 0 Let y vx, dy x dv v dx. 3x vx dx xx dv v dx 0
3x 2vx dx x2 dv 0 3 2v dx x dv
1 dx x
ln x
1 dv 3 2v
1 ln 3 2v C1 ln3 2v1 2 ln C2 2
x C23 2v1 2
x2 C3 2v C 3 2
yx
x3 C3x 2y 3Cx 2Cy y
88.
dv kv 9.8 dt (a)
dv kv 9.8
x3 3Cx 2C
(b) lim vt t→
dt
9.8 k
1 9.8 kv0 9.8ekt dt k
(c) st
1 1 9.8t kv0 9.8ekt C k k
1 9.8t 2kv0 9.8ekt C k k
1 ln kv 9.8 t C1 k ln kv 9.8 kt C2 kv 9.8
ektC2
C3
ekt
1 v 9.8 C3ekt k At t 0,
1 v0 9.8 C3 ⇒ C3 kv0 9.8 k 1 v 9.8 kv0 9.8ekt k
Note that k < 0 since the object is moving downward.
s0
1 1 kv 9.8 C ⇒ C s0 2kv0 9.8 k2 0 k
st
9.8t 1 1 2kv0 9.8ekt s0 2kv0 9.8 k k k
9.8t 1 2kv0 9.8ekt 1 s0 k k
Review Exercises for Chapter 5 90. hx 3 arcsin2x
92. (a) Let arccot 2
5
cot 2
y
1
θ
4
1 tanarccot 2 tan . 2 −π 2
553
−π 4
π 4
x
π 2
2
(b) Let arcsec 5
−2
sec 5
5 2
−4
1 . cosarcsec 5 cos 5
θ 1
94. y arctanx2 1 y
96. y
2x 2x 1 x2 12 x4 2x2 2
y
1 arctan e2x 2
1 1 e2x 2e2x 4x 2 1e 1 e4x
x 98. y x2 4 2 arcsec , 2 < x < 4 2 y
x x 4 2
x2 4 1 4 x x2 4 2 2 2 2 x x 2x 2 1 x 4 x x 4 x x 4
100. Let u 5x, du 5 dx.
1 1 dx 3 25x2 5
102.
104.
3
1 5x 1 5 dx arctan C 3 5x2 53
1 1 x dx arctan C 16 x2 4 4
4x 1 1 x dx 4 dx 4 x21 22x dx 4 arcsin 4 x2 C 2 2 2 2 4 x 4 x
106. Let u arcsin x, du
2
1 dx. 1 x2
arcsin x 1 dx arcsin x2 C 2 1 x2
y
108. π 2
y = arcsin x π 4
x 0.25
0.5
0.75
1
2
Since the area of region A is 1
0
1
the shaded area is
arcsin x dx
0
110. y x tanh1 2x y x
1 0.571. 2
112. Let u x3, du 3x2 dx.
1 2 4x tanh 2
sin y dy ,
1
2x
2x tanh1 2x 1 4x2
x2sech x32 dx
1 1 sech x323x2 dx tanh x3 C 3 3
554
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
Problem Solving for Chapter 5 2. (a)
y
(b)
y
4
2
3
1
π 2
−1
π
3π 2
2
x
2π
1
−2
π 2
sin x dx
0
2
sin x dx ⇒
2
2
sin x dx 0
0
(d) 1
π 2
1 2
1
π 4
1
1
arccos x dx 2
2
y
4. y 0.5x and y 1.2x intersect y x. y
π 2
y = 0.5 x
4 , 12. 6
2
0
y = 2x y = x
does not intersect y x. y = 1.2 x −6
Suppose y x is tangent to y ax at x, y.
6 −2
ax x ⇒ a x1x. y ax ln a 1 ⇒ x ln x1x 1 ⇒ ln x 1 ⇒ x e, a e1e For 0 < a ≤ e1e 1.445, the curve y ax intersects y x. 6. (a) y f x arcsin x sin y x Area A
Area B
126 12 0.2618
4
6
22
(b)
arcsin x dx AreaC
12
6
2
2
3
2
3 2
2
4 22 A B
2 3 2 8 2 12
—CONTINUED—
4
sin y dy cos y
82 121
2 3
2
0.1346
x
1 is symmetric with respect to the 1 tan x2
point
2x
x
sin x 2 dx 22 4
x
2π
y
π
−1
3π 2
0
y
(c)
π
0.1589
1 1 dx 1 tan x2 2 2 4
Problem Solving for Chapter 5
555
6. —CONTINUED—
y
ln 3
(c) Area A
ey dy
y = ln x
0
ln 3
ey
0
ln 3
312
ey = x
A B
3
Area B
ln x dx 3ln 3 A 3 ln 3 2 ln 27 2 1.2958
x 1
2
3
1
(d)
tan y x Area A
3
tan y dy
4 3
4
ln cos y
y
y = arctan x
2 1 1 ln ln ln2 ln 2 2 2 2
3
Area C
arctan x dx
1
A
3 3 21 ln 2 4 1
C
B
x 1
3
43 3 21 ln 2 0.6818 12 10. Let u tan x, du sec2 x dx
y ex
8.
π 3 π 4
y
ex
yb
ea
Area
x a
4
0
1 dx sin x 4 cos2 x 2
y eax aea b Tangent line
bx ab b b ea
sec2 x dx tan2 x 4
du u2 4
12 arctanu2
1 0
xa1 ca1
Thus, a c a a 1 1. dy 1 y1 y, y0 dt 4
1 1 dy y 1y
0
0
eax aea b
4
1
If y 0,
12. (a)
1 1 arctan 2 2
y
(b)
1
dt
ln y ln 1 y t C
y ln tC 1y y etC C1et 1y y C1et yC1et y y0
C1et 1 1 C1et 1 C2et
1 1 ⇒ C2 3 4 1 C2
Hence, y
1 . 1 3et
—CONTINUED—
−6
−4
−2
x 2
4
6
dy y1 y y y2 dt d2y 1 y y 2yy ⇒ y 0 for y dt 2 2 d2y 1 1 d2y > 0 if 0 < y < and 2 < 0 if < y < 1. 2 dt 2 2 dt 1 Thus, the rate of growth is maximum at y , the 2 point of inflection.
556
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
12. —CONTINUED— (c) y y1 y, y0 2 1 1 C2et
As before, y y0 2 Thus, y
1 1 ⇒ C2 1 C2 2
1 2 . 1 t 2 et 1 e 2
The graph is different: y 6 4 2 −6
−4
−2
x 2
4
6
14. (a) u 985.93 985.93
v 985.93
0.095 120,0000.095 1 12 12
0.095 120,0000.095 1 12 12
12t
(d) t 12.7 years Again, the larger part goes for interest.
v
(c) The slopes are negatives of each other. Analytically,
u15 v15 14.06.
u
12t
(b) The larger part goes for interest. The curves intersect when t 27.7 years.
du dv u 985.93 v ⇒ dt dt
1000
0
35 0
C H A P T E R 5 Logarithmic, Exponential, and Other Transcendental Functions Section 5.1
The Natural Logarithmic Function: Differentiation . . . . 218
Section 5.2
The Natural Logarithmic Function: Integration . . . . . . 223
Section 5.3
Inverse Functions . . . . . . . . . . . . . . . . . . . . . . 227
Section 5.4
Exponential Functions: Differentiation and Integration . . 233
Section 5.5
Bases Other than e and Applications . . . . . . . . . . . . 240
Section 5.6
Differential Equations: Growth and Decay . . . . . . . . . 246
Section 5.7
Differential Equations: Separation of Variables
Section 5.8
Inverse Trigonometric Functions: Differentiation . . . . . 259
Section 5.9
Inverse Trigonometric Functions: Integration
. . . . . . 251
. . . . . . . 263
Section 5.10 Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . 267 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278
C H A P T E R 5 Logarithmic, Exponential, and Other Transcendental Functions Section 5.1
The Natural Logarithmic Function: Differentiation
Solutions to Odd-Numbered Exercises 1. Simpson’s Rule: n 10 x
0.5
x
1
1 dt t
0.6932
0.5
Note:
1
1 dt t
1.5
2
2.5
3
3.5
4
0.4055
0.6932
0.9163
1.0987
1.2529
1.3865
1
1 dt 0.5 t
3. (a) ln 45 3.8067
45
(b)
1
5. (a) ln 0.8 0.2231
0.8
1 dt 3.8067 t
(b)
1
7. f x ln x 2
1 dt 0.2231 t
9. f x ln x 1
Vertical shift 2 units upward
Horizontal shift 1 unit to the right
Matches (b)
Matches (a)
11. f x 3 ln x
13. f x ln 2x
Domain: x > 0
15. f x lnx 1
Domain: x > 0
Domain: x > 1 y
y
y 3
2
2 2
1 1
1
x 3
x 1
2
3
4
1
1
2
3
1
1
2
4
x
5
2
3
17. (a) ln 6 ln 2 ln 3 1.7917
2 19. ln 3 ln 2 ln 3
(b) ln 23 ln 2 ln 3 0.4055 (c) ln 81 ln 34 4 ln 3 4.3944 (d) ln 3 ln 312 12 ln 3 0.5493
21. ln
218
xy ln x ln y ln z z
3 a2 1 lna2 113 23. ln
1 lna2 1 3
Section 5.1
x x 1 2
25. ln
3
3
The Natural Logarithmic Function: Differentiation
27. ln zz 12 ln z lnz 12
3lnx2 1 ln x3
ln z 2 lnz 1
3lnx 1 lnx 1 3 ln x
29. lnx 2 lnx 2 ln
31.
x2 x2
1 xx 32 1 2 lnx 3 ln x lnx2 1 ln 2 ln 3 3 x 1
33 . 2 ln 3
35.
xxx 31
2
3
2
1 9 lnx2 1 ln 9 lnx2 1 ln 2 x2 1 37. lim lnx 3
3
x →3
f=g 9
0
−3
39. lim lnx23 x ln 4 1.3863
41. y ln x3 3 ln x
x →2
y
3 x
At 1, 0, y 3. 45. gx ln x 2 2 ln x
43. y ln x2 2 ln x y
2 x
gx
47.
y ln x4 dy 1 4ln x3 4ln x3 dx x x
2 x
At 1, 0, y 2.
49.
y ln xx2 1 ln x
1 lnx2 1 2
51. f x ln
dy 1 1 2x 2x2 1 dx x 2 x2 1 xx2 1
ln t t2
53. gt
55.
t 21t 2t ln t 1 2 ln t gt t4 t3
57.
xx 11 21 lnx 1 lnx 1
y ln
fx
1 dy 1 1 1 dx 2 x 1 x 1 1 x2
x ln x lnx2 1 x2 1
1 2x 1 x2 2 x x 1 xx2 1
y lnln x2 1 d 2xx2 2 1 dy ln x2 2 dx ln x dx ln x2 x ln x2 x ln x
59. f x ln fx
4 x2
x
1 ln4 x2 ln x 2
x 1 4 4 x2 x xx2 4
219
220 61.
Chapter 5 y
Logarithmic, Exponential, and Other Transcendental Functions
x2 1 lnx x2 1 x
dy xxx2 1 x2 1 1 dx x2 x x2 1
63.
1 1 x 1 x x2 1
2 2
x
x2 1 x x 1 2
y ln sin x
1
x
x x2 1
1 1 1 x2 x2 1 2 2 2 x 1 x 1 x x 1 x2
2 2
65.
y ln
dy cos x cot x dx sin x
cos x cos x 1
ln cos x ln cos x 1
dy sin x sin x sin x tan x dx cos x cos x 1 cos x 1
67.
y ln
1 sin x 2 sin x
69. f x sin 2x ln x2 2 sin 2x ln x
fx 2 sin 2x
ln 1 sin x ln 2 sin x dy cos x cos x dx 1 sin x 2 sin x
3 cos x sin x 1sin x 2 (b)
4
(1, 3)
1 dy 6x dx x When x 1,
2 sin 2x 2x cos 2x ln x x
2 sin 2x x cos 2x ln x2 x
y 3x2 ln x, 1, 3
71. (a)
1x 4 cos 2x ln x
−1
dy 5. dx
2
−3
Tangent line: y 3 5x 1 y 5x 2 0 5x y 2 73.
x2 3 ln y y2 10 2x
3 dy dy 2y 0 y dx dx 2x
y
dy 3 2y dx y
x2 2x 0
2
Domain: x > 0 1 x 1x 1 0 when x 1. x x
y 1
1 > 0 x2
1 Relative minimum: 1, 2
2 x2
xy y x
x2 ln x 2
y x
2 x
y
dy 2x 2xy dx 3y 2y 3 2y 2
77. y
y 2ln x 3
75.
(1, 12 ) 0
3 0
2
Section 5.1
The Natural Logarithmic Function: Differentiation
79. y x ln x
221
2
Domain: x > 0 y x
1x ln x 1 ln x 0 when x e
1.
0
3
(e−1 , −e−1 ) −1
1 > 0 x
y
Relative minimum: e1, e1
81. y
x ln x
4
( e, e )
Domain: 0 < x < 1, x > 1
(e2, e2/2)
0
y
ln x1 x1x ln x 1 0 when x e. ln x2 ln x2
y
ln x21x ln x 12x ln x 2 ln x 0 when x e 2. ln x4 xln x3
9
−4
Relative minimum: e, e Point of inflection: e2, e22 83.
f x ln x,
f 1 0
1 fx , x
The values of f, P1, P2, and their first derivatives agree at x 1. The values of the second derivatives of f and P2 agree at x 1.
f1 1
2
1 f x 2, x
f 1 1
P1 f
P1x f 1 f1x 1 x 1, P2x f 1 f1x 1 x 1 P1x 1,
P11 0
1 x 12, 2
P21 0
P21 1
P21 1
85. Find x such that ln x x.
87.
f x ln x x 0 1 1 x
n
1
xn
0.5 0.1931
y xx2 1 ln y ln x
f xn 1 ln xn xn xn fxn 1 xn
f xn
−2
P11 1
P2x 1,
xn1
5
P2
1 f 1x 12 2
P2x 1 x 1 2 x,
fx
−1
2
1 dy 1 x 2 y dx x x 1
2x2 1 2x2 1 dy y 2 dx xx 1 x2 1
3
0.5644
0.5671
0.0076
0.0001
Approximate root: x 0.567
1 lnx2 1 2
222
Chapter 5
y
89.
Logarithmic, Exponential, and Other Transcendental Functions
x23x 2 x 12
ln y 2 ln x
y
91.
1 ln3x 2 2 lnx 1 2
xx 132 x 1
ln y ln x
3 1 lnx 1 lnx 1 2 2
1 dy 3 2 2 y dx x 23x 2 x 1
1 3 1 1 dy 1 1 y dx x 2 x1 2 x1
dy 3x2 15x 8 y dx 2x3x 2x 1
dy y 2 3 1 dx 2 x x1 x1
3x3 15x2 8x 2x 133x 2
y 4x2 4x 2 2x2 2x 1x 1 2 xx2 1 x 132
93. Answers will vary. See Theorem 5.1 and 5.2.
95. ln ex x because f x ln x and gx ex are inverse functions.
97. (a) f 1 f 3
(b) fx 1
10 log10
99.
1010
10I ln1010 ln 10I ln1010ln I 16 ln 10 160 10 log 16
16
10
I
10 10 10 ln 1010 16 ln 10 10 ln 10 16 ln 10 6 ln 10 60 decibels ln 10 ln 10 ln 10
101. (a) You get an error message because ln h does not exist for h 0. (b) Reversing the data, you obtain h 0.8627 6.4474 ln p. (c)
2 0 for x 2. x
(d) If p 0.75, h 2.72 km. (e) If h 13 km, p 0.15 atmosphere. (f)
h 0.8627 6.4474 ln p 1 6.4474
25
1 dp (implicit differentiation) p dh
p dp dh 6.4474 0
1 0
For h 5, p 0.55 and dpdh 0.0853 atmos/km. For h 20, p 0.06 and dpdh 0.00931 atmos/km. As the altitude increases, the rate of change of pressure decreases.
103. (a) f x ln x, gx x
4 x (b) f x ln x, gx 15
25
g g
f f
0
500 0
20,000
0 0
1 1 fx , gx x 2x
1 1 fx , gx 4 3 x 4x
For x > 4, gx > fx. g is increasing at a faster rate than f for “large” values of x.
For x > 256, gx > fx. g is increasing at a faster rate than f for “large” values of x. f x ln x increases very slowly for “large” values of x.
Section 5.2
The Natural Logarithmic Function: Integration
105. False ln x ln 25 ln25x lnx 25
Section 5.2
1.
The Natural Logarithmic Function: Integration
5 1 dx 5 dx 5 ln x C x x
3. u x 1, du dx
5. u 3 2x, du 2 dx
1 dx ln x 1 C x1
7. u x2 1, du 2x dx
1 1 1 dx 2 dx 3 2x 2 3 2x
x 1 1 dx 2x dx x2 1 2 x2 1
1 ln 3 2x C 2
1 lnx2 1 C 2
lnx2 1 C
9.
x2 4 dx x
x
4 dx x
11. u x3 3x2 9x, du 3x2 2x 3 dx
x2 4 ln x C 2
x2 2x 3 1 3x2 2x 3 dx dx x3 3x2 9x 3 x3 3x2 9x
13.
x2 3x 2 dx x1
17.
x4 x 4 dx x2 2
6 dx x1
x4
x2 2
1
dx
x3 3x2 5 dx x3
x dx x2 2
21. u x 1, du dx x 1
x2 4x 6 ln x 1 C 2
19. u ln x, du
x3 1 2x lnx2 2 C 3 2
15.
23.
x 11 2 dx
2x 1
1 2
C
2x 1 C
1 ln x3 3x2 9x C 3
x2
5 dx x3
x3 5 ln x 3 C 3
1 dx x
1 ln x2 3 dx ln x C x 3
2x dx x 12
2
2x 2 2 dx x 12
2x 1 1 dx 2 dx x 12 x 12 1 1 dx dx 2 x1 x 12
2 ln x 1
2 C x 1
223
224
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions 1 dx ⇒ u 1 du dx 2x
25. u 1 2x, du
1 dx 1 2x
u 1 du u
u ln u C1
1 du u
u
1 2x ln 1 2x C1 2x ln 1 2x C where C C1 1.
27. u x 3, du
x x 3
dx 2
1 dx ⇒ 2u 3 du dx 2x
2 6u 9 lnu C
2
u 32 u2 6u 9 du 2 du 2 u u
u2
1
u6
9 du u
u2 12u 18 ln u C1
x 3 12 x 3 18 ln x 3 C1 2
x 6x 18 ln x 3 C where C C1 27.
29.
cos d ln sin C sin
31.
csc 2x dx
u sin , du cos d
33.
1 ln csc 2x cot 2x C 2
cos t dt ln 1 sin t C 1 sin t
37. y
35.
3 dx 2x
3
(1, 0) −10
3 ln x 2 C
tan2 d
(0, 2)
4
−3
3
−3
1 0, 2: 2 lncos 0 C ⇒ C 2 2
1 s ln cos 2 2 2
dy 1 , 0, 1 dx x 2 (a)
y 3 ln x 2
41.
1 tan22 d 2 1 ln cos 2 C 2
−10
sec x tan x dx ln sec x 1 C sec x 1
10
1, 0: 0 3 ln1 2 C ⇒ C 0
39. s
10
1 dx x2
1 csc 2x2 dx 2
(b)
y
(0, 1)
y
1 dx ln x 2 C x2
3
3
y0 1 ⇒ 1 ln 2 C ⇒ C 1 ln 2 x
−2
4
−3
Hence, y ln x 2 1 ln 2 ln
x2 1. 2
−3
6
−3
Section 5.2
4
43.
0
47.
0
2 x
2
x1
0
1
49.
cos d ln sin sin
21
1
0 ln 3 2
ln
e
1
7 3
1 2
2 sin 2 1.929 1 sin 1
1 C ln sec x C cos x
ln
1 dx x1
225
1 dx x
1 ln x2 1 dx 1 ln x3 x 3 1 e
53. ln sec x tan x C ln
55.
x ln x 1
2
51. ln cos x C ln
45. u 1 ln x, du
5 ln 13 4.275 3
2 dx x1
2 2 x
4
0
5 5 dx ln 3x 1 3x 1 3
The Natural Logarithmic Function: Integration
sec x tan xsec x tan x sec2 x tan2 x C ln C sec x tan x sec x tan x
1 C ln sec x tan x C sec x tan x
1 dx 2 1 x 2 ln 1 x C1 1 x 2 x ln 1 x C where C C1 2.
57.
cos1 x dx sin1 x C 2
59.
4
2
csc x sin x dx lncsc x cot x cos x
4
ln2 1
2
2
0.174
Note: In Exercises 61 and 63, you can use the Second Fundamental Theorem of Calculus or integrate the function.
63. F x
Fx
1 x
Fx
3x
x
61. F x
1 dt 1 t
x
1 dt t
3x
1
1 dt t
x
1
1 dt t
65.
y
2
3 1 0 3x x
x
−1
−1 2
A 1.25 Matches (d)
67. A
4 2 x
1
4 dx x
4
1
x
4 dx x
10
2 4 ln x
15 8 ln 2 13.045 square units 2
x2
4 1
8 4 ln 4
1 2 0
6 0
1 2
1
226
Chapter 5
2
69.
2 sec
0
Logarithmic, Exponential, and Other Transcendental Functions
x 12 dx 6
2
sec
0
6x6 dx
10
12 ln sec 6x tan 6x
12 12 ln sec tan ln 1 0 3 3
12 ln 2 3 5.03041
1 42
77. Average value
4
2
8 dx 4 x2
0
4 0
4
75. Divide the polynomials: x2 1 x1 x1 x1
79. Average value
x2dx
2
4 4
0
73. Substitution: u x2 4 and Log Rule
71. Power Rule
81. Pt
2
1 x
1 e1
e
1
ln x 1 ln x2 dx x e1 2
4 2
14 21 1
e
1
1 1 e1 2
1 0.291 2e 2
3000 0.25 dt 30004 dt 12,000 ln 1 0.25t C 1 0.25t 1 0.25t
P0 12,000 ln 1 0.250 C 1000 C 1000
Pt 12,000 ln 1 0.25t 1000 1000 12 ln 1 0.25t 1 P3 1000 12ln 1.75 1 7715
83.
1 50 40
50
40
50
40 $168.27
90,000 dx 3000 ln 400 3x 400 3x
85. (a) 2x2 y2 8
(b) y2 e1 xdx eln xC eln1 xeC
y2 2x2 8 Let k 4 and graph y2
y1 2x2 8 y2
2x2
8
4 x
1 k x
yy 2 2 x x 1
2
10
10 − 10 − 10
10
10 − 10 − 10
(c) In part (a),
2x2 y2 8 4x 2yy 0 y
2x . y
In part (b),
y2 2yy y
4 4x1 x 4 x2 2 2y 2y y 2 2 . yx 2 y x 4x 2x
Using a graphing utility the graphs intersect at 2.214, 1.344. The slopes are 3.295 and 0.304 1 3.295, respectively.
Section 5.3 87. False
Inverse Functions
89. True
1 ln x lnx12 ln x12 2
1 dx ln x C1 x
ln x ln C ln Cx , C 0
Section 5.3 1. (a)
Inverse Functions
f x 5x 1 gx
(b) 3
x1 5
f 2 1
x1 x1 5 1x f gx f 5 5
g( f x g5x 1
3. (a)
y
g x
−3
1
2
3
5x 1 1 x 5
f x x3
(b)
y
3 x gx
3
f
2
3 x f gx f 3 x 3 x
g
1
x
3 3 g f x gx3 x x
−3 −2
1
2
3
−2 −3
5. (a)
f x x 4
(b)
gx x2 4, x ≥ 0
g 10
f gx f x2 4
8 6
x2 4 4 x2 x
4
g f x g x 4
f x
1 x
f
2 x
x 4 2 4 x 4 4 x
7. (a)
y 12
2
4
6
(b)
8
10
12
y 3
1 gx x f gx
1 x 1x
g f x
1 x 1x
9. Matches (c)
2
f=g
1 x −1
11. Matches (a)
1
2
3
227
228
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
13. f x 34 x 6
1 3 s2 One-to-one; has an inverse
17. hs
15. f sin Not one-to-one; does not have an inverse
One-to-one; has an inverse y
1 −4
y
8
8 1 6 4
π 2
2
−8
−6
−4
−7
θ
3π 2
x
−2
−1
19. f x ln x
21. gx x 53
One-to-one; has an inverse
One-to-one; has an inverse
2
200
−1
5 −10
2 − 50
−2
23. f x x a3 b
25. f x
fx 3x a2 ≥ 0 for all x.
x4 2x2 4
f x x3 4x 0 when x 0, 2, 2.
f is increasing on , . Therefore, f is strictly monotonic and has an inverse.
f is not strictly monotonic on , . Therefore, f does not have an inverse.
27. f x 2 x x3 fx 1 3x2 < 0 for all x. f is decreasing on , . Therefore, f is strictly monotonic and has an inverse. 29.
f x 2x 3 y x y
31.
f x x5 y
y3 2 x3 2
f x x y
33.
5 y x
x y2
5 x y
y x2 f 1x x2, x ≥ 0
5 x x15 f 1x y
x3 x 2
f 1
y
f 2
f y
3
1
f
1
1
2 4
f
x 2
1
2
1
2
f
1
x 2
2 2
f
4
2
x 1
2
3
Section 5.3 f x 4 x2 y, 0 ≤ x ≤ 2
35.
Inverse Functions
229
3 x 1 y f x
37.
x 4 y2
x y3 1
y 4 x2
y x3 1 f 1x x3 1
f 1x 4 x2, 0 ≤ x ≤ 2
2
y
f −1 f
3
−3
f
2
3
1
f
The graphs of f and f 1 are reflections of each other across the line y x.
−2
1
x 1
2
3
f x x23 y, x ≥ 0
39.
f x
41.
x y32 y x32 y
f 1x x32, x ≥ 0 4
f −1
The graphs of f and f 1 are reflections of each other across the line y x.
f
0
f 1x
x x2 7
x
7y 1 y2
7x 1 x2 7x 1 x2
, 1 < x < 1
2
f −1 f
6 −3
0
y
3
The graphs of f and f 1 are reflections of each other across the line y x.
−2
43.
x
1
2
3
4
f 1x
0
1
2
4
y
45. (a) Let x be the number of pounds of the commodity costing 1.25 per pound. Since there are 50 pounds total, the amount of the second commodity is 50 x. The total cost is y 1.25x 1.6050 x
(4, 4)
4
0.35x 80
0 ≤ x ≤ 50.
3 2
(b) We find the inverse of the original function:
(3, 2)
y 0.35x 80
(2, 1)
1
(1, 0) 1
x 2
3
4
0.35x 80 y x 100 35 80 y 20 Inverse: y 100 35 80 x 7 80 x.
x represents cost and y represents pounds. (c) Domain of inverse is 62.5 ≤ x ≤ 80. (d) If x 73 in the inverse function, 100 y 100 35 80 73 5 20 pounds.
230
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
47. f x x 42 on 4,
4 on 0, x2
49. f x
fx 2x 4 > 0 on 4,
fx
f is increasing on 4, . Therefore, f is strictly monotonic and has an inverse.
8 < 0 on 0, x3
f is decreasing on 0, . Therefore, f is strictly monotonic and has an inverse.
51. f x cos x on 0, fx sin x < 0 on 0, f is decreasing on 0, . Therefore, f is strictly monotonic and has an inverse. x y on 2, 2 x2 4
f x
53.
Domain: all x Range: 2 < y < 2
x 2 y 4y x x2 y x 4y 0 a y, b 1, c 4y
f
1 ± 1 4 y4y 1 ± 1 16y2 x 2y 2y y f 1x
1 0,
55. (a), (b)
The graphs of f and f 1 are reflections of each other across the line y x.
2
−3
3
f
−1
−2
1 16x2 2x, if x 0
if x 0 57. (a), (b)
6
4
f
g
f −1 −5
−6
10
6
g−1 −4
(c) Yes, f is one-to-one and has an inverse. The inverse relation is an inverse function. 59. f x x 2, Domain: x ≥ 2 fx
1 > 0 for x > 2. 2x 2
f is one-to-one; has an inverse x 2 y
x 2 y2 x y2 2 yx 2
−4
(c) g is not one-to-one and does not have an inverse. The inverse relation is not an inverse function.
61. f x x 2 , x ≤ 2 x 2 2x f is one-to-one; has an inverse 2xy 2yx f 1x 2 x, x ≥ 0
2
f 1
x x2 2, x ≥ 0
63. f x x 32 is one-to-one for x ≥ 3.
x 3 y 2
x 3 y x y 3 y x 3 f 1x x 3, x ≥ 0 (Answer is not unique)
65. f x x 3 is one-to-one for x ≥ 3. x3y
xy3 yx3 f 1x x 3, x ≥ 0 (Answer is not unique)
Section 5.3 67. Yes, the volume is an increasing function, and hence one-to-one. The inverse function gives the time t corresponding to the volume V.
f x x3 2x 1, f 1 2 a
71.
1 1 1 2 f f 12 f1 312 2 5
6 21 a
f x sin x, f
73.
fx cos x
1
f 1
12 f f
1 1
1 32
12
1 1 f 6 cos 6
23 3
4 f x x3 , f 2 6 a x
75.
fx 3x2
f 16
4 x2
1 1 1 1 f f 16 f2 322 422 13
77. (a) Domain f Domain f 1 ,
79. (a) Domain f 4, , Domain f 1 0,
(b) Range f Range f 1 ,
(b) Range f 0, , Range f 1 4,
(c)
(c)
y
f f −1
8 6
x 1
2
3
4
x
−3
2
f x x3,
(d)
6
8
10
12
f x x 4, 5, 1 fx
1 2x 4
12 43
f5
1 2
3 x, f 1x
12, 18
4
fx 3x2 f
f 1
f
2
−2
(d)
f −1
10
2
−3 −2
y 12
3
1
x
f 1
18, 12
1 3 2 3 x
18 34
231
69. No, Ct is not one-to-one because long distance costs are step functions. A call lasting 2.1 minutes costs the same as one lasting 2.2 minutes.
fx 3x2 2 f 1
Inverse Functions
f 1x x2 4, 1, 5
f 1x 2x f 11 2
232 81.
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
x y3 7y2 2 1 3y2
dy dy 14y dx dx
1 1 1 dy dy . At 4, 1, . dx 3y2 14y dx 3 14 11 Alternate solution: let f x x3 7x2 2. Then fx 3x2 14x and f1 11. Hence,
1 1 dy . dx 11 11
In Exercises 83 and 85, use the following. f x 18 x 3 and g x x3 3 x f1 x 8 x 3 and g1 x
83. f 1 g11 f 1g11 f 11 32
85. f 1 f 16 f 1 f 16 f 172 600
In Exercises 87 and 89, use the following. f x x 4 and g x 2x 5 f1 x x 4 and g1 x
x5 2
87. g1 f 1x g1 f 1x g1x 4
x 4 5 2
x1 2
89. f gx f gx f 2x 5 2x 5 4 2x 1 x1 2 g1 f 1
Hence, f g1x
Note: f g1 91. Answers will vary. See page 335 and Example 3.
93. y x2 on , does not have an inverse.
95. f is not one-to-one because many different x-values yield the same y-value.
97. Let f gx y then x f g1y. Also,
Example: f 0 f 0
2n 1 Not continuous at , where n is an integer 2
f gx y f gx y gx f 1y x g1 f 1y g1 f 1y Since f and g are one-to-one functions, f g1 g1 f 1.
99. Suppose gx and hx are both inverses of f x. Then the graph of f x contains the point a, b if and only if the graphs of gx and hx contain the point b, a. Since the graphs of gx and hx are the same, gx hx. Therefore, the inverse of f x is unique.
Section 5.4
Exponential Functions: Differentiation and Integration 103. True
101. False Let f x x2.
x
105. Not true
f x
107.
Let f x
1 x, x,
0 ≤ x ≤ 1 1 < x ≤ 2
2
.
fx
f is one-to-one, but not strictly monotonic.
Section 5.4
dt 1 t 4
, f 2 0
1 1 x 4
f 10
1 1 17 f2 1 17
Exponential Functions: Differentiation and Integration
1. e0 1
3.
ln 2 0.6931
5. eln x 4 x4
e0.6931. . . 2
ln1 0
9. 9 2ex 7
7. ex 12
2ex 7
x ln 12 2.485
ex 1 x0 13. ln x 2
11. 50ex 30 ex
x e2 7.3891
3 5
x ln x ln
35 53 0.511
15. lnx 3 2 x3
17. lnx 2 1 x 2 e1 e
e2
x 2 e2
x 3 e2 10.389
x e2 2 5.389 19. y ex
21. y ex
y
2
Symmetric with respect to the y-axis Horizontal asymptote: y 0
4 3
y
2
x 1
1
2
3
x 1
1
233
234
Chapter 5
23. (a)
Logarithmic, Exponential, and Other Transcendental Functions (b)
7
(c)
3
g
f
7
f
f
q
−2 −5
4
h
7 −1
−4
Horizontal shift 2 units to the right
8 −1
−3
Vertical shift 3 units upward and a reflection in the y-axis
A reflection in the x-axis and a vertical shrink 27. y C1 eax
25. y Ceax Horizontal asymptote: y 0
Vertical shift C units
Matches (c)
Reflection in both the x- and y-axes Matches (a)
29. f x e2x
31. f x ex 1
gx lnx
gx lnx 1
1 ln x 2
y y 6
f
6
f
4
4
g
2 2
g
x 2
x
−2
2
4
4
6
6
−2
33.
35.
3
1 1 1,000,000
1,000,000
g
2.718280469
e 2.718281828
f −1
4
e >
−1
1 1 1,000,000
As x → , the graph of f approaches the graph of g. 0.5 x e0.5 lim 1 x → x
37. (a) y e3x
(b) y e3x
y 3e3x
y 3e3x
At 0, 1, y 3.
At 0, 1, y 3.
39. f x e2x
41. f x e2xx
2
fx 2e2x
45. gt et et 3 gt 3et et 2et et
43.
dy ex dx 2x
dy 2 2x 1e2xx dx
47.
y ln ex x2 2
dy 2x dx
y ex
49.
y ln1 e2x 2e2x dy dx 1 e2x
1,000,000
Section 5.4
51.
y
Exponential Functions: Differentiation and Integration
2 2ex ex 1 ex ex
dy ex2x 2 exx2 2x 2 x2ex dx
dy 2ex ex 2ex ex dx
y x2ex 2xex 2ex ex x2 2x 2
53.
2ex ex ex ex 2
55. f x ex ln x fx ex
y ex sin x cos x
57.
1x e
x
ln x ex
1x ln x
dy excos x sin x sin x cos xex dx ex2 cos x 2ex cos x 61. f x 3 2xe3x
xey 10x 3y 0
59. xey
fx 3 2x3e3x 2e3x
dy dy ey 10 3 0 dx dx
7 6xe3x
dy y xe 3 10 ey dx
fx 7 6x3e3x 6e3x 36x 5e3x
dy 10 ey y dx xe 3 y ex cos2x sin2x
63.
y ex 2 sin 2x 2 cos 2x ex cos2x sin 2x ex 1 2 cos2x 1 2 sin2x
y ex 2 2sin 2x 2 2cos 2x ex 1 2 cos2x 1 2sin 2x ex 1 22 sin 2x 1 22 cos 2x
2y 3y 2ex 1 2cos 2x 1 2sin 2x 3ex cos2x sin 2x ex 1 22 cos 2x 1 22 sin 2x y
Therefore, 2y 3y y ⇒ y 2y 3y 0. 65. f x fx
ex ex 2
6
ex ex 0 when x 0. 2 (0, 1)
−3
ex ex fx > 0 2
3
0
Relative minimum: 0, 1 67. gx
1 2
ex2 2 2
2,
1 2 gx x 2ex2 2 2
(
1,
1 2 g x x 1x 3ex2 2 2
2, Points of inflection: 1, Relative maximum:
(
0.8
0
2, 0.399 1 e , 3, 12 e 1, 0.242, 3, 0.242 2
1
2
1 2
1 2
e− 0.5 2π
1 2π
( (
(
3,
e− 0.5 2π
( 4
0
235
236
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
69. f x x2ex
3
fx x2ex 2xex xex2 x 0 when x 0, 2. f x ex2x x2 ex2 2x
(2, 4 e −2 )
(0, 0)
exx2 4x 2 0 when x 2 ± 2.
−1
5 0
Relative minimum: 0, 0
(2 ±
2, (6 ± 4
2)e− (2 ± 2)
Relative maximum: 2, 4e2 x 2 ± 2 y 2 ± 22e2 ± 2 Points of inflection: 3.414, 0.384, 0.586, 0.191 71. gt 1 2 tet
5
(− 1, 1 + e)
gt 1 tet g t tet
(0, 3)
−6
6
Relative maximum: 1, 1 e 1, 3.718
−3
Point of inflection: 0, 3 73.
A baseheight 2xex
2
y
dA 2 2 4x2ex 2ex dx
3 2
2ex 1 2x2 0 when x 2
2
2
(
. −2
A 2e1 2
75. y
a aL x b e L ex b b b y 1 aex b2 1 aex b2
y
1 aex b2
e aL aLb e 21 ae b x b
2
x b
x b
1 aex b4
1 aex b
e aL 2aLb e ab e b x b
2
x b
x b
1 aex b3
Laex b aex b 1 1 aex b3 b2
y 0 if aex b 1 ⇒ yb ln a
x 1 ln ⇒ x b ln a b a
L L L 1 aeb ln a b 1 a1 a 2
Therefore, the y-coordinate of the inflection point is L 2.
ab e x b
) x
−1
1 −1
L , a > 0, b > 0, L > 0 1 aex b
2 −1 2 ,e 2
2
)
Section 5.4
Exponential Functions: Differentiation and Integration
77. ex x ⇒ f x x ex
79. (a)
237
4
fx 1 ex xn1 xn
f xn x exn xn n fxn 1 exn
−4
−2
x1 1 x2 x1
f x1 0.5379 fx1
x3 x2
f x2 0.5670 fx2
x4 x3
f x3 0.5671 fx3
(b) When x increases without bound, 1 x approaches zero, and e1 x approaches 1. Therefore, f x approaches 2 1 1 1. Thus, f x has a horizontal asymptote at y 1. As x approaches zero from the right, 1 x approaches , e1 x approaches and f x approaches zero. As x approaches zero from the left, 1 x approaches , e1 x approaches zero, and f x approaches 2. The limit does not exist since the left limit does not equal the right limit. Therefore, x 0 is a nonremovable discontinuity.
We approximate the root of f to be x 0.567.
81.
h
0
5
10
15
20
P
10,332
5,583
2,376
1,240
517
ln P
9.243
8.627
7.773
7.123
6.248
(a)
5
(b) ln P ah b
12
P eahb ebeah P Ceah, C eb −2
For our data, a 0.1499 and C e9.3018 10,957.7
22 0
P 10,957.7e0.1499h
y 0.1499h 9.3018 is the regression line for data h, ln P. (c)
(d)
12,000
dP 10,957.710.1499e0.1499h dh 1642.56e0.1499h
0
22
For h 5,
0
83.
f x ex 2, f 0 1
dP dP 776.3. For h 18, 110.6. dh dh
7
f
1 1 fx ex 2, f0 2 2 1 1 f x ex 2, f 0 4 4 P1x 1
P1
P2 −6
6 −1
1 x x 0 1, P10 1 2 2
1 1 P1x , P10 2 2 1 1 x2 x P2x 1 x 0 x 02 1, P20 1 2 8 8 2 1 1 1 P2x x , P20 4 2 2 1 1 P2x , P20 4 4 The values of f, P1, P2 and their first derivatives agree at x 0. The values of the second derivatives of f and P2 agree at x 0.
238
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
85. (a) y ex
(b) y ex
y1 1 x
y2 1 x 4
x2 2
4
y
y
y1 −2
y2
2 −2
−1
2 −1
(c) y ex y3 1 x
x2 x3 2 6
4
y y 3
−2
2 −1
87. Let u 5x, du 5 dx.
5x
89. Let u 2x, du 2 dx.
1
5x
e 5dx e C
91.
0
xex dx 2
1 x2 1 2 e 2xdx ex C 2 2
93.
1 2
1
1 e2x 2 dx e2x 2 0 1 e2 1 1 e2 2 2e2
e2x dx
e x 1 dx 2 ex dx 2ex C x 2x
95. Let u 1 ex, du ex dx.
ex ex ex dx dx ln1 ex C ln x C x lnex 1 C x x 1e 1e e 1
3 3 97. Let u , du 2 dx. x x
3 e3 x
1
x2
dx
1 3
99. Let u 1 ex, du ex dx.
e3 x
1
1 e 3 x 3
3 1
105.
2 1 e x3 2 C 3
e e2 1 3
103.
ex ex dx ln ex ex C ex ex
e sin x cos x dx
ex1 ex dx 1 ex1 2ex dx
2
101. Let u ex ex, du ex ex dx.
x3 dx
3
1 sin x e cos x dx 1 sin x e C
5 ex dx e2x
5e2x dx
ex dx
5 e2x ex C 2
107.
ex tanex dx
tanex ex dx
ln cosex C
1 0
Section 5.4
109. Let u ax2, du 2ax dx. Assume a 0
y
Exponential Functions: Differentiation and Integration
111. fx
1 x 1 e ex dx ex ex C1 2 2
xeax dx
f0 C1 0
f x
2
1 2 1 ax2 e 2ax dx eax C 2a 2a
1 x 1 e ex dx ex ex C2 2 2
f 0 1 C2 1 ⇒ C2 0 1 f x ex ex 2
113. (a)
y
dy 2ex 2, dx
(b)
5
y (0, 1)
1 2ex 2 dx 4 ex 2 dx 2
4ex 2 C
x
−2
0, 1
5
0, 1: 1 4e0 C 4 C ⇒ C 5
−2
y 4ex 2 5 6
−4
8 −2
5
115.
ex dx ex
0
5 0
6
e5 1 147.413
117.
xex 4dx 2ex 4 2
0
150
2
6
0
2e3 2
2 1.554
3
0
−4.5
6
4.5
0
−3
119. (a) f u v euv eu ev (b) f k x e
kx
x
123.
eu f u ev f v
e f x . x k
k
60
121. 0.0665
e0.0139t48 dt 2
48
Graphing Utility: 0.4772 47.72%
x
et dt ≥
0
1 dt
0
x
e t
0
≥
t
x
0
ex 1 ≥ x ⇒ ex ≥ 1 x for x ≥ 0
125. f x ex. Domain is , and range is 0, . f is continuous, increasing, one-to-one, and concave upwards on its entire domain. lim ex 0 and lim ex .
x→
x→
2
127. Yes. f x Cex, C a constant.
129. ex > 0 ⇒
0
exdx > 0.
239
240
Chapter 5
131. f x
Logarithmic, Exponential, and Other Transcendental Functions
ln x x
(a) fx
y
1 ln x 0 when x e. x2
1 2
On 0, e, fx > 0 ⇒ f is increasing. On e, , fx < 0 ⇒ f is decreasing.
x
e
2
6
4
8
− 12
(b) For e ≤ A < B, we have: ln A ln B > A B B ln A > A ln B ln AB > ln BA AB > BA. (c) Since e < , from part (b) we have e > e.
Section 5.5 1. y
2 1
Bases Other than e and Applications
t3
3. y
At t0 6, y
12
63
1 4
2 1
t7
5. log2 18 log2 23 3
At t0 10, y
7. log7 1 0
12
107
0.3715
11. (a) log10 0.01 2
23 8
9. (a)
102 0.01
log2 8 3 (b)
31
(b) log0.5 8 3
1 3
0.53 8
1 log3 1 3
15. y
13. y 3x x
2
1
0
1
2
y
1 9
1 3
1
3
9
12 3 8
13 3 x
17. hx 5x2
x
x
2
1
0
1
2
y
9
3
1
1 3
1 9
y
x
1
0
1
2
3
y
1 125
1 25
1 5
1
5
y y
4
4 4
3
3
3 2
2 1
x 2
1
1
2
x 2
1
1
2
x 1
2
3
4
Section 5.5 19. (a) log10 1000 x
Bases Other than e and Applications
21. (a) log3 x 1
10x 1000
31 x x 13
x3
(b) log2 x 4
(b) log10 0.1 x 10x 0.1
24 x
x 1
x 16
1
x2 x log5 25
23. (a)
x2
x log5
52
(b) 3x 5 log2 64 2
3x 5 log2 26 6
x2 x 2 0
3x 1
x 1x 2 0
x 13
x 1 OR x 2 32x 75
25.
3 xln 2 ln 625
2x ln 3 ln 75 x
23x 625
27.
1 ln 75 1.965 2 ln 3
3x
ln 625 ln 2
x3
1 0.09 12
29.
12t
31. log2x 1 5
3
x 1 25 32
0.09 12t ln 1 ln 3 12 t
ln 625 6.288 ln 2
1 12
x 33
ln 3 12.253 0.09 ln 1 12
33. log3 x2 4.5 x2 3 4.5 x ± 34.5 ± 11.845 37. hs 32 log10s 2 15
35. gx 621x 25
Zero: s 2.340
Zero: x 1.059
40
30
(− 1.059, 0) −4
10 −1
−30
8
(2.340, 0) −20
241
242
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
39. f x 4x
x
gx log4 x
y
2
1
0
1 2
1
1 16
1 4
1
2
4
f x
f
3 2
g
1 16
1 4
1
2
4
2
1
0
1 2
1
x gx
41. f x 4x
x
43.
fx ln 4 4x
1
2
3
1
45. gt t2 2t
y 5x2
gt t 2 ln 2 2t 2t 2t
dy ln 5 5x2 dx
t 2t t ln 2 2 2t t2 t ln 2
47. h 2 cos
49.
h 2 sin ln 22 cos
2
51. f x log2
1 dy dx x ln 3
ln 2 cos sin
x2 x1
53.
2 log2 x log2 x 1 fx
55. gt gt
y log3 x
y log5 x2 1 dy 1 dx 2
2 1 x ln 2 x 1 ln 2
1 log5 x2 1 2
2x
x
x2 1ln 5 x2 1ln 5
x2 ln 2xx 1
10 log4 t 10 ln t t ln 4 t 10 t 1t ln t ln 4 t2
y x2x
57.
ln y
2 ln x x
2 1 2 2 1 dy ln x 2 2 1 ln x y dx x x x x
10 5 1 ln t 2 1 ln t t 2 ln 4 t ln 2
dy 2y 2 1 ln x 2x2x 21 ln x dx x
59.
y x 2x1
61.
ln y x 1 lnx 2
1 dy 1 x 1 lnx 2 y dx x2 dy x1 y lnx 2 dx x2
xx 21 lnx 2
x 2x1
x
x
3 dx
3 C ln 3
Section 5.5
2
63.
x
x
1
2 dx
ln 2
2
2
65.
1
1 1 4 ln 2 2
Bases Other than e and Applications
x
2
x5
dx
7 7 2 ln 2 ln 4
67.
1 x2 5 2x dx 2 1 5x C 2 ln 5
2
1 2 5x C 2 ln 5
32x dx, u 1 32x, du 2ln 332x dx 1 32x 1 2 ln 3
69.
243
2x 1 2 ln 33 2x ln1 3 C 2x dx 13 2 ln 3
1 dy 0.4x3, 0, dx 2 y
x3
0.4
(a)
x3
dx 3 0.4
(b)
4
4
13 dx
−6
6
x
−4
3 0.4x3 C 3ln 2.50.4x3 C ln 0.4 y 3 ln 2.50.4x3
y
(0, 12 (
4
−4
−4
1 3 ln 2.5 2
31 0.4x3 1 ln 2.5 2
71. Answers will vary. Example: Growth and decay problems. 73.
y
x
1
2
8
y
0
1
3
4
(8, 3)
3
(a) y is an exponential function of x: False (b) y is a logarithmic function of x: True; y log2 x (c) x is an exponential function of y: True, 2y x
2
(d) y is a linear function of x: False
(2, 1)
1
(1, 0) x 2
4
6
8
1 x ln 2 gx xx ⇒ gx xx 1 ln x
75. f x log2 x ⇒ fx
77. Ct P1.05t (a) C10 24.951.0510
[Note: Let y gx. Then: ln y ln xx x ln x 1 1 y x ln x y x y y1 ln x y x 1 ln x gx.
x
hx x ⇒ hx 2x 2
kx 2x ⇒ kx ln 22x From greatest to smallest rate of growth: gx, kx, hx, f x
$40.64 (b)
dC Pln 1.051.05t dt When t 1:
dC 0.051P dt
When t 8:
dC 0.072P dt
dC t (c) dt ln 1.05 P1.05
ln 1.05Ct The constant of proportionality is ln 1.05.
244
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
79. P $1000, r 312 % 0.035, t 10
0.035 A 1000 1 n
n
1
2
4
12
365
Continuous
A
1410.60
1414.78
1416.91
1418.34
1419.04
1419.07
n
1
2
4
12
365
Continuous
A
4321.94
4399.79
4440.21
4467.74
4481.23
4481.69
10n
A 1000e0.03510 1419.07 81. P $1000, r 5% 0.05, t 30
0.05 A 1000 1 n
30n
A 1000e0.0530 4481.69 83. 100,000 Pe0.05t ⇒ P 100,000e0.05t
85. 100,000 P 1
0.05 12
t
1
10
20
30
40
50
P
95,122.94
60,653.07
36,787.94
22,313.02
13,583.53
8208.50
12t
⇒ P 100,000 1
0.05 12
12t
t
1
10
20
30
40
50
P
95,132.82
60,716.10
36,864.45
22,382.66
13,589.88
8251.24
87. (a) A 20,000 1
0.06 365
3658
89. (a) lim 6.7e48.1t 6.7e0 6.7 million ft3
$32,320.21
t →
(b) A $30,000
(c) A 8000 1
(b) 0.06 365
3658
20,000 1
0.06 365
3654
1 0.06 365
3658
1
0.06 365
3654
322.27 48.1t e t2
V20 0.073 million ft3yr V60 0.040 million ft3yr
$12,928.09 25,424.48 $38,352.57 (d) A 9000
V
1
$34,985.11 Take option (c).
91. y (a)
300 3 17e0.0625x (c) If y 66.67%, then x 38.8 or 38,800 egg masses.
100
(d) y 3003 17e0.0625x 1
0
y
318.75e0.0625x 3 17e0.0625x2
y
19.921875e0.0625x 17e0.0625x 3 3 17e0.0625x 3
100 0
(b) If x 2 (2000 egg masses), y 16.67 16.7%.
17e0.0625x 3 0 ⇒ x 27.8 or 27,800 egg masses.
Section 5.5 93. (a) B 4.75396.7744d 4.7539e1.9132d (b)
Bases Other than e and Applications
245
Bd 9.0952e1.9132d
(c)
B0.8 42.03 tonsinch
120
B1.5 160.38 tonsinch
0
2 0
4
95. (a)
(c) The functions appear to be equal: f t gt ht
f t dt 5.67
0
Analytically,
4
gt dt 5.67
f t 4
0 4
ht dt 5.67
2t3
8 4 4 gt
4
3
913
23 t
t
ht 4e0.653886t 4 e0.653886 t 40.52002t
0
(b)
38
6
gt 4
−1
94 40.52002 13 t
t
No. The definite integrals over a given interval may be equal when the functions are not equal.
5 −1
10
97. P
2000e0.06t dt
99.
0
2000 e 0.06
0.06t
10
t
0
1
2
3
4
y
1200
720
432
259.20
155.52
0
y Ckt
$15,039.61
When t 0, y 1200 ⇒ C 1200. y 1200kt 720 432 259.20 155.52 0.6, 0.6, 0.6, 0.6 1200 720 432 259.20 Let k 0.6. y 12000.6t
101. False. e is an irrational number.
103. True.
105. True.
f gx 2 elnx2 2x2x g f x ln2 ex 2 ln ex x
d x d e ex and ex ex dx dx ex ex when x 0.
e0e0 1
246
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
dy 8 5 y y , y0 1 dt 25 4
107.
8 4 dy dt ⇒ y54 y 25 5 ln y ln ln
1 1 dy y 54 y
8 dt ⇒ 25
54 y 52 t C
54y y 52 t C y e25tC C1e25t 54 y
y 0 1 ⇒ C1 4 ⇒ 4e25t ⇒ 4e25t
54 y y ⇒ 5e
⇒ y
5e25t 5 1.25 1 4 e0.4t 1 0.25e0.4t
Section 5.6
1.
25t
4e25t y y 4e25t 1y
4e25t
Differential Equations: Growth and Decay
dy x2 dx y
y 54 y
dy y2 dx
3.
x 2dx
x2 2x C 2
dy dx y2 1 dy y2
dx
ln y 2 x C1 y 2 exC1 Cex y Cex 2
y
5.
5x y
yy 5x
yy dx y dy
5x dx 5x dx
1 2 5 2 y x C1 2 2
y x y
7.
y x y
y dx y
x dx
dy y
x dx
ln y
2 32 x C1 3
y e23x
32
y 2 5x 2 C
C1
eC1 e23x
32
Ce23x
32
Section 5.6
9. 1 x2y 2xy 0 y
2x y y 1 x2
k dQ 2 dt t
11.
2xy 1 x2
Differential Equations: Growth and Decay
dQ dt dt
dN k250 s ds
13.
k dt t2
dN ds ds
2x dx 1 x2
dy y
2x dx 1 x2
k250 s ds
k dN 250 s 2 C 2
k dQ C t
y dx y
247
k N 250 s2 C 2
k Q C t
ln y ln1 x2 C1 ln y ln1 x 2 ln C ln y ln C1 x2 y C1 x 2 15. (a)
dy x6 y, 0, 0 dx
(b)
y 9
7
dy x y6
ln y 6 x
−5
−1
−6
x 2 C 2
6 −1
y 6 ex 2C C1ex 2 2
5
(0, 0)
2
y 6 C1ex
22
0, 0: 0 6 C1 ⇒ C1 6 ⇒ y 6 6ex 2 2
17.
dy 1 t, 0, 10 dt 2
dy
19.
16
1 t dt 2 −4
dy y
4
21.
0, 4: 4
y 10et2
23.
(Theorem 5.16)
0, 20,000: C 20,000
C
3, 10: 10 4e3k ⇒ k
1 5 ln 3 2
4, 12,500: 12,500 20,000e4k ⇒ k
When x 6, y 4e13 ln526 4eln52
2
52
dV kV dt V Cekt
(Theorem 5.16)
4
10 −1
10 Ce0 ⇒ C 10
dy ky dx
Ce0
−1
y et2 C1 eC1 et2 Cet2
1 2 t 10 4
y Cekx
(0, 10)
1 dt 2
1 ln y t C1 2
−1
1 10 02 C ⇒ C 10 4 y
16
(0, 10)
1 y t2 C 4
dy 1 y, 0, 10 dt 2
2
25
1 5 ln 4 8
When t 6, V 20,000e14 ln586 20,000eln58
32
20,000
58
32
9882.118
248
Chapter 5
25. y Cekt,
Logarithmic, Exponential, and Other Transcendental Functions
0, 12, 5, 5
27.
1 Cek
1 C 2 y
y Cekt, 1, 1, 5, 5 5 Ce5k 5Cek Ce5k
1 kt e 2
5ek e5k
1 5 e5k 2 k
ln 10 0.4605 5
y
1 0.4605t e 2
5 e4k k
ln 5 0.4024 4
y Ce0.4024t 1 Ce0.4024 C 0.6687 y 0.6687e0.4024t
29. A differential equation in x and y is an equation that involves x, y and derivatives of y.
31.
dy 1 xy dx 2 dy > 0 when xy > 0. Quadrants I and III. dx
33. Since the initial quantity is 10 grams, y 10eln121620t . When t 1000, y 10eln1216201000 6.52 grams. When t 10,000, y 10eln12162010,000 0.14 gram. 35. Since y Celn121620t, we have 0.5 Celn12162010,000 ⇒ C 36.07. Initial quantity: 36.07 grams. When t 1000, we have y Celn1216201000 23.51 grams. 37. Since the initial quantity is 5 grams, we have y 5.0eln125730t. When t 1000, y 4.43 g. When t 10,000, y 1.49 g. 39. Since y Celn1224,360t, we have 2.1 Celn1224,3601000 ⇒ C 2.16. Thus, the initial quantity is 2.16 grams. When t 10,000, y 2.16eln1224,36010,000 1.63 grams.
41. Since
dy ky, y Cekt or y y0ekt. dx
1 y y0e1620k 2 0 k
ln 2 1620
y y0eln 2t 1620. When t 100, y y0eln 216.2 y00.9581. Therefore, 95.81% of the present amount still exists.
43. Since A 1000e0.06t, the time to double is given by 2000 1000e0.06t and we have 2 e0.06t ln 2 0.06t t
ln 2 11.55 years. 0.06
Amount after 10 years: A 1000e0.0610 $1822.12
Section 5.6 45. Since A 750ert and A 1500 when t 7.75, we have the following.
ln 2 0.0894 8.94% 7.75
Amount after 10 years: A 750e0.089410 $1833.67
r
ln1292.85500 0.0950 9.50% 10
The time to double is given by 1000 500e0.0950t t
0.075 12
49. 500,000 P 1
1220
0.075 12
P 500,000 1
240
2 1
ln 2 10.24 years ln 1.07
0.07 12
0.007 12
1 t 12
t
ln 2 9.90 years 0.07
0.085 12
12t
0.085 365
0.085 365
ln 2 365t ln 1 t
12t
1 365
365t
365t
0.085 365
ln 2 8.16 years 0.085 ln 1 365
(d) 2000 1000e0.085t
0.085 12
ln 2 8.18 years 0.085 ln 1 12
0.07 365
ln 2 9.90 years 365 ln1 0.07365
(c) 2000 1000 1
ln 2 8.50 years ln 1.085
365t
ln 2 0.07t
ln 2 t ln1.085
ln 2 12t ln 1
365t
2 1
0.085 2 1 12
0.07 365
2 e0.07t
2 1.085t
420
(d) 2000 1000e0.07t
55. (a) 2000 10001 0.085t
0.07 365
ln 2 365t ln 1 t
ln 2 9.93 years 12 ln1 0.0712
(b) 2000 1000 1
12t
t
0.08 12
12t
0.07 ln 2 12t ln 1 12 t
ln 2 t ln 1.07
2 1
1235
P 500,000 1
(c) 2000 1000 1
2 1.07t
$30,688.87
53. (a) 2000 10001 0.07t
(b) 2000 1000 1
0.08 12
51. 500,000 P 1
$112,087.09
t
ln 2 7.30 years. 0.095
249
47. Since A 500ert and A 1292.85 when t 10, we have the following. 1292.85 500e10r
1500 750e7.75r r
Differential Equations: Growth and Decay
2 e0.085t ln 2 0.085t t
ln 2 8.15 years 0.085
250
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
57. P Cekt Ce0.009t
59. P Cekt Ce0.036t
P1 8.2 Ce0.0091 ⇒ C 8.1265
P1 4.6 Ce0.0361 ⇒ C 4.7686
P 8.1265e0.009t
P 4.7686e0.036t
P10 7.43
or
P10 6.83
7,430,000 people in 2010
61. If k < 0, the population decreases.
or
6,830,000 people in 2010
P Cekx, 0, 760, 1000, 672.71
63.
C 760
If k > 0, the population increases.
672.71 760e1000x x
ln672.71760 0.000122 1000
P 760e0.000122x When x 3000, P 527.06 mm Hg. 65. (a)
19 301 e20k 30e20k 11 k
25 301 e0.0502t
(b)
e0.0502t
ln1130 0.0502 20
t
N 301 e0.0502t
1 6 ln 6 36 days 0.0502
67. S Cekt (a)
S 5 when t 1
(b) When t 5, S 20.9646 which is 20,965 units.
5
(c)
Cek
30
lim Cekt C 30
t →
5 30ek 0
k ln 16 1.7918
40 0
S 30e1.7918t 69. At Vte0.10t 100,000e0.8 t e0.10t 100,000e0.8 t0.10t
0.4 dA 100,000 0.10 e0.8 t0.10t 0 when 16. dt t The timber should be harvested in the year 2014, 1998 16. Note: You could also use a graphing utility to graph At and find the maximum of At. Use the viewing rectangle 0 ≤ x ≤ 30 and 0 ≤ y ≤ 600,000. I 71. I 10 log10 , I0 1016 I0 (a) 1014 10 log10 (b) 109 10 log10
1014 20 decibels 1016
109 70 decibels 1016
106.5 (c) 106.5 10 log10 16 95 decibels 10 (d) 104 10 log10
73. R
ln I 0 , I eR ln 10 10R ln 10
(a) 8.3
ln I 0 ln 10
I 108.3 199,526,231.5 (b) 2R
ln I 0 ln 10
I e2R ln 10 e2R ln 10 eR ln 102 10R 2
104 120 decibels 1016
Increases by a factor of e2R ln 10 or 10R. (c)
1 dR dI I ln 10
Section 5.7
Differential Equations: Separation of Variables
75. False. If y Cekt, y Ckekt constant.
Section 5.7
77. True
Differential Equations: Separation of Variables
1. Differential equation: y 4y Solution: y Ce4x Check: y 4Ce4x 4y 3. Differential equation: y y 0 Solution: y C1 cos x C2 sin x y C1 sin x C2 cos x
Check:
y C1 cos x C2 sin x y y C1 cos x C2 sin x C1 cos x C2 sin x 0
5. y cos x ln sec x tan x y cos x
1 sec x tan x sec2 x sin x ln sec x tan x sec x tan x
cos x sec xtan x sec x sin x ln sec x tan x sec x tan x
1 sin x ln sec x tan x
1 y sin x sec x tan x sec2 x cos x ln sec x tan x sec x tan x
sin xsec x cos x ln sec x tan x Substituting,
y y sin xsec x cos x ln sec x tan x cos x ln sec x tan x tan x. In Exercises 7–11, the differential equation is y4 16y 0. y 3 cos x
7.
y e2x
9.
y4 3 cos x
y4 16e2x
y4 16y 45 cos x 0,
y4 16y 16e2x 16e2x 0,
No.
Yes. y C1e2x C2e2x C3 sin 2x C4 cos 2x
11.
y4 16C1e2x 16C2e2x 16C3 sin 2x 16C4 cos 2x y4 16y 0, Yes.
251
252
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
In 13–17, the differential equation is xy 2y x3ex. 15. y x 22 ex , y x 2ex 2x2 ex
13. y x 2, y 2x xy 2y x2x 2x2 0 x3ex
xy 2y xx2ex 2xex 4x 2x2ex 2x2 x3ex,
No.
Yes.
1 x 1 xy 2y x 2 ln x x3ex, x
17. y ln x, y
19.
No.
21. y 2 Cx3 passes through 4, 4
y Cekx
16 C64 ⇒ C 14
dy Ckekx dx Since dy dx 0.07y, we have Thus, k 0.07.
Ckekx
Particular solution: y2 14 x3 or 4y 2 x 3
0.07Cekx.
23. Differential equation: 4yy x 0
2
C=0
General solution: 4y 2 x 2 C
−3
Particular solutions: C 0, Two intersecting lines C ± 1, C ± 4, Hyperbolas 2
−2
2
2
C = −1
C=1 −3
3
3
−3
−2
3
−2
2
C = −4
C=4 −3
−3
3
3
−2
−2
25. Differential equation: y 2y 0 General Solution: y Ce2x y 2y C2e2x 2Ce2x 0 Initial condition: y0 3, 3 Ce0 C Particular solution: y 3e2x
27. Differential equation: y 9y 0 General solution: y C1 sin 3x C2 cos 3x y 3C1 cos 3x 3C2 sin 3x, y 9C1 sin 3x 9C2 cos 3x y 9y 9C1 sin 3x 9C2 cos 3x 9C1 sin 3x C2 cos 3x 0
Initial conditions: y 2 C1 sin
6 2, y 6 1
2 C
2
cos
2 ⇒ C
1
2
y 3C1 cos 3x 3C2 sin 3x 1 3C1 cos
2 3C sin 2
2
3C2 ⇒ C2
1 3
Particular solution: y 2 sin 3x
1 cos 3x 3
Section 5.7
Differential Equations: Separation of Variables Initial conditions: y2 0, y2 4
29. Differential equation: x 2 y 3xy 3y 0 General solution: y C1 x C2 x 3
0 2C1 8C2
y C1 3C2 x2, y 6C2 x
y C1 3C2 x 2
x2y 3x y 3y x26C2 x 3xC1 3C2 x2
4 C1 12C2 C1 4C2 0
3C1x C2 x3 0
C1 12C2 4
1 C2 , C1 2 2
1 Particular solution: y 2x x3 2
31.
dy 3x2 dx y
33.
3x 2 dx x 3 C
x dy dx 1 x2 y
1 x dx ln1 x2 C 1 x2 2
u 1 x 2, du 2x dx
35.
2 dy x 2 1 dx x x y
1
37.
y
u 2x, du 2dx
dy xx 3 dx y
xx 3 dx
u2 3u2udu
dy 2 xex dx y
u5 u C 52 x 3 5
3
5 2
2x 33 2 C
43.
xex dx 2
1 sin 2x dx cos 2x C 2
Let u x 3, then x u2 3 and dx 2u du.
2 u4 3u 2 du 2
41.
2 dx x
x 2 ln x C x ln x 2 C
39.
dy sin 2x dx
1 x2 e C 2
u x 2, du 2x dx
dy x dx y
y dy
x dx
y2 x2 C1 2 2 y2 x2 C
45.
dr 0.05r ds
dr r
0.05 ds
ln r 0.05s C1 r e0.05sC1 Ce0.05s
47. 2 xy 3y
dy y
3 dx 2x
ln y 3 ln2 x ln C ln C2 x3 y Cx 23
253
254
Chapter 5
49.
yy sin x
y dy y2 2 y2
Logarithmic, Exponential, and Other Transcendental Functions
51. 1 4x2
sin x dx
dy x dx dy
x 1 4x2
cos x C1
dy
2 cos x C
dx
x 1 4x2
1 8
dx
1 4x21 28x dx
1 y 1 4x21 2 C 4 53. y ln x x y 0
dy y
55. yy ex 0 ln x dx x
u ln x, du dxx
y dy
2 C 1
ex dx
y2 ex C1 2
1 ln y ln x 2 C1 2 y e1 2lnx
y 2 2ex C
Celnx 2 2
Initial condition: y0 4, 16 2 C, C 14 Particular solution: y 2 2ex 14 57. yx 1 y 0
59. y1 x2
dy x 1 dx y
ln y
dy x1 y 2 dx
y x dy dx 1 y2 1 x2
x 12 C1 2
1 1 ln1 y 2 ln1 x 2 C1 2 2 ln1 y 2 ln1 x 2 ln C lnC1 x 2
y Cex1 2 2
Initial condition: y2 1, 1 Ce1 2, C e1 2 Particular solution: y e1 x1 2 ex 2
2
2 2x
1 y 2 C1 x2 y0 3: 1 3 C ⇒ C 4 1 y2 41 x2 y 2 3 4x 2
61.
du uv sin v 2 dv
63. dP kP dt 0
du u
v sin v 2 dv
dP k dt P
ln P kt C1
1 ln u cos v 2 C1 2
P Cekt
u Cecos v 2 2
Initial condition: P0 P0, P0 Ce0 C
Initial condition: u0 1, C
1 e1 2 e1 2
Particular solution: u e1cosv 2 2
Particular solution: P P0 ekt
Section 5.7 dy 9x dx 16y
65.
Differential Equations: Separation of Variables
67.
0y y dy dx x 2 x 2
dy y
16y dy 9x dx 8y 2
m
9 2 x C 2
1 dx 2
1 ln y x C1 2 y Cex 2
9 25 Initial condition: y 1 1, 8 C, C 2 2 Particular solution: 8y 2
9 2 25 x , 2 2
16y 2 9x 2 25
69.
71. f x, y
f x, y x 3 4xy 2 y 3 f t x, t y t 3 x 3 4t xt 2 y 2 t3 y3
f tx, ty
t 3x3 4xy 2 y3
f x, y 2 ln x y
75.
f t x, t y 2 ln t x t y
vx x
2
xy , y vx 2x
tx x 2 ln ty y
y
79.
dv x v x dx 2x
vx
dv 1 v v dx 2
dv 1v
x y
Homogeneous degree 0
Not homogeneous y
f x, y 2 ln f t x, t y 2 ln
2 ln t 2 x y 2ln t 2 ln x y
77.
t4x2y2 x2y2 t3 2 2 2 x y2 ty
t2x2
Homogeneous of degree 3
Homogeneous of degree 3
73.
x2y2 x2 y2
ln1 v 2 ln x ln C ln Cx 1 Cx 1 v 2
dv x xv dx x xv
v dx x dv
dx x
xy , y vx xy
1v dx 1v
v1 dx dv v2 2v 1 x
C
1 Cx 1 y x2
x2 Cx x y2
x Cx y
2
C1 1 ln v 2 2v 1 ln x ln C1 ln 2 x
v 2 2v 1 x 2
y2 y C 2 1 2 x2 x x
y 2 2xy x 2 C
255
256
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
y
81.
xy , y vx x2 y 2
xv dx x dv 2xev vx dx 0
x2 v dv 2 vx dx x x2 v 2 v dx x dv
1v dv v3 2
ev dv
v dx 1 v2
2 dx x
ev ln C1 x 2
dx x
ey x ln C1 ln x 2 ey x C ln x 2
1 2 ln v ln x ln C1 ln C1 x 2v
x dy 2xey x y dx 0, y v x
83.
Initial condition: y1 0, 1 C
1 ln C1x v 2v 2
Particular solution: ey x 1 ln x 2
x 2 ln C1 y 2y 2
y Cex 2y 2
2
x sec xy y dx x dy 0, y v x
85.
87.
dy x dx
x sec v xvdx xv dx x dv 0
cos v dv
y 4
sec v v dx v dx x dv dx x
x 2
2
sin v ln x ln C1 4
x Cesin v Cesiny x Initial condition: y1 0, 1
y Ce0
C
x dx
1 2 x C 2
Particular solution: x esiny x
89.
dy 4y dx
y 8
dy 4y
dx
ln 4 y x C1 4 y ex C1
4
y 4 Cex −4 −3
91.
x 1
2
3
4
dy 0.5y, y0 6 dx
93.
dy 0.02y10 y, y0 2 dx
12
−6
12
6 −4
− 12
48 −2
Section 5.7
95.
dy ky, y Cekt dt Initial conditions:
Differential Equations: Separation of Variables
97. y0 y0
dy k y 4 dx The direction field satisfies dy dx 0 along y 4; but not along y 0. Matches (a).
y0 y1620 2 C y0 y0 y0e1620k 2 k
ln1 2 1620
Particular solution: y y0etln 2 1620 When t 25, y 0.989y0, y 98.9% of y0.
99.
dy k yy 4 dx The direction field satisfies dy dx 0 along y 0 and y 4. Matches (c). dw k1200 w dt
101.
dw 1200 w
k dt
dv Wv
k dt
lnW v k t C1
ln1200 w kt C1 1200 w
dv kW v dt
103. (a)
ektC1
v W Cekt
Cekt
w 1200 Cekt
Initial conditions:
w0 60 1200 C ⇒ C 1200 60 1140
W 20, v 0 when t 0, and
w 1200 1140e
v 5 when t 1.
kt
(a)
1400
C 20, k ln3 4
1400
Particular solution: v 201 eln3 4t 201 e0.2877t 0
10 0
0
10 0
(b) s 1400
201 e0.2877t dt
20t 3.4761e0.2877t C Since S0 0, C 69.5 and we have s 20t 69.5e0.2877t 1. 0
10 0
(b) k 0.8: t 1.31 years k 0.9: t 1.16 years k 1.0: t 1.05 years (c) Maximum weight: 1200 pounds lim w 1200 t→0
257
258
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
105. Given family (circles): x 2 y 2 C
107. Given family (parabolas): x2 Cy 2x Cy
2x 2yy 0 y Orthogonal trajectory (lines):
y
x y
y
y x
y
Orthogonal trajectory (ellipses):
dy y
2x 2x 2y 2 C x y x
dx x
x 2y
2 y dy x dx
ln y ln x ln K
y2
y Kx
x2 K1 2
x 2 2y 2 K
4 4 −6
6 −6
6
−4 −4
109. Given family: y 2 Cx3
111. A general solution of order n has n arbitrary constants while in a particular solution initial conditions are given in order to solve for all these constants.
2yy 3Cx2 y
3Cx2 3x2 y2 3y 2y 2y x3 2x y
Orthogonal trajectory (ellipses):
2x 3y
3 y dy 2 x dx 3y 2 x 2 K1 2 3y 2 2x 2 K 4
−6
6
−4
113. Mx, ydx Nx, ydy 0, where M and N are homogeneous functions of the same degree. 117. False f t x, t y t 2x 2 t 2 xy 2 t 2 f x, y
115. False. Consider Example 2. y x3 is a solution to xy 3y 0, but y x3 1 is not a solution.
Section 5.8
Section 5.8
Inverse Trigonometric Functions: Differentiation
Inverse Trigonometric Functions: Differentiation
1. y arcsin x (a)
x
1
0.8
0.6
0.4
0.2
0
0.2
0.4
0.6
0.8
1
y
1.571
0.927
0.644
0.412
0.201
0
0.201
0.412
0.644
0.927
1.571
(b)
y
(c)
π 2
(d) Symmetric about origin: arcsinx arcsin x
2
−1
Intercept: 0, 0
1
x
−1
1 −2 −π 2
3. False.
5. arcsin
arccos
1 2 6
1 2 3
since the range is 0, .
7. arccos
1 2 3
11. arccsc 2
9. arctan
4
3
3
6
13. arccos0.8 2.50
15. arcsec1.269 arccos
1 1.269
0.66
3 3 17. (a) sin arctan 4 5
21 cot 6 3
19. (a) cot arcsin 5 3
3
θ
θ 4
1
2
4 5 (b) sec arcsin 5 3
(b) csc arctan
5
4
5 12
135
12
θ
θ
3
5 13
23. y sinarcsec x
21. y cosarcsin 2x 1
arcsin 2x
2x
y cos 1 4x
2
θ
arcsec x, 0 ≤ ≤ ,
1− 4 x 2
y sin
x2 1
x
2
x x2 − 1
θ 1
The absolute value bars on x are necessary because of the restriction 0 ≤ ≤ , 2, and sin for this domain must always be nonnegative.
259
260
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
x 3
25. y tan arcsec
x
θ
x arcsec 3 y tan
27. y csc arctan
x2 − 9
arctan
3
x2 9
2x
29. sinarctan 2x
x2 + 2 x
θ 2
2 x2 2
x
31. arcsin3x 12
1 4x2
3x sin 12
2
x 13 sin 12 1.207
f=g
−2
x
y csc
3
x 2
2
−2
Asymptotes: y ± 1 arctan 2x
1 + 4 x2
tan 2x sin
2x
θ
2x 1 4x2
1
33. arcsin2x arccos x
2x sin arccos x
2x 1 x, 0 ≤ x ≤ 1
2x 1 x 1
3x 1
1−x
θ
1 x 3
x
1 35. (a) arccsc x arcsin , x ≥ 1 x
(b) arctan x arctan
Let y arccsc x. Then for
1 ,x > 0 x 2
Let y arctan x arctan1 x. Then,
≤ y < 0 and 0 < y ≤ , 2 2
tan y
csc y x ⇒ sin y 1 x. Thus, y arcsin1 x. Therefore, arccsc x arcsin1 x.
tanarctan x tanarctan1 x 1 tanarctan x tanarctan1 x
x 1 x 1 x1 x
x 1 x (which is undefined). 0
Thus, y 2. Therefore, arctan x arctan1 x 2. 37.
f x arcsinx 1
y
π
x 1 sin y Domain: 0, 2 Range:
f x is the graph of arcsin x shifted 1 unit to the right.
( ) x
−1
1 −π 2 −π
(0, − π2 )
2
f x arcsec 2x 2x sec y
2, π 2
π 2
x 1 sin y
, 2 2
39.
x
3
Domain: Range:
y
(− 12 , π (
1 sec y 2
, 21 , 12,
0, 2 , 2 ,
π 2
( 12 , 0 ( −2
−1
x
1
2
Section 5.8
41. f x 2 arcsinx 1 2 2 f x 1 x 12 2x x2
45. f x arctan f x
Inverse Trigonometric Functions: Differentiation
43. gx 3 arccos g x
x a
47. gx
1 a a 1 x2 a2 a2 x2
g x
49. ht sinarccos t 1 t 2
x 2
31 2 3 2 1 x 4 4 x2 arcsin 3x x x 3 1 9x2 arcsin 3x x2 3x 1 9x2 arcsin 3x x21 9x2
51. y x arccos x 1 x2
1 t h t 1 t 21 22t 2 1 t2
y arccos x
x 1 x2
1 1 x21 22x 2
arccos x
53.
y
1 1 x1 ln arctan x 2 2 x1
55.
x 1 dy x arcsin x arcsin x 2 dx 1 x 1 x2
1 1 lnx 1 lnx 1 arctan x 4 2
y x arcsin x 1 x2
dy 1 1 1 2 1 1 dx 4 x 1 x 1 1 x2 1 x4
57. y 8 arcsin y 2
x x16 x2 4 2
59. y arctan x
16 x2 1 x 16 x21 22x 2 1 x 4 2 4
y
x 1 x2
1 1 x2 x2x 2 1x 1 x22
16 x2 8 x2 2 16 x 2 216 x2
1 x2 1 x2 1 x22
16 16 x2 x2 x2 2 16 x2 216 x
2 1 x22
61. f x arcsin x, a
1 2
y 1.5 1.0
1 f x 1 x2
0.5 x
1 1 P1x f f 2 2 P2x f
0.5 1.0 1.5
P2
x f x 1 x23 2
P1
− 1.0
f
1 23 1 x x 2 6 3 2
12 f 12 x 21 21 f 12 x 21
2
− 1.5
23 1 23 1 x x 6 3 2 9 2
2
261
262
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
f x arcsec x x 1 f x 1 x x2 1
63.
65.
f x
1 1 0 1 x2 1 x 42
1 x2 1 x 42
0 when x x2 1 1. x2x2 1 1
0 8x 16
x4 x2 1 0 when x2 x±
f x arctan x arctanx 4
x2
1 5 or 2
By the First Derivative Test, 2, 2.214 is a relative maximum.
1 5 ± 1.272 2
Relative maximum: 1.272, 0.606 Relative minimum: 1.272, 3.747 69. y arccot x, 0 < y <
67. The trigonometric functions are not one-to-one on , , sot their domains must be restricted to intervals on which they are one-to-one.
x cot y tan y
1 x
So, graph the function y arctan
d (b) dt
x 71. (a) cot 5
arccot
x 5
ht 16t 2 256
73. (a)
16t 2 256 0 when t 4 sec. (b) tan
h 16t2 256 500 500
500t
arctan
16
2
h
θ 500
16
d 8t 125 1000t dt 1 4 125t 2 162 15,625 1616 t 22 When t 1, d dt 0.0520 rad sec. When t 2, d dt 0.1116 rad sec.
1x for x > 0 and y arctan1x for x < 0.
1 5 x 1 5
2
dx 5 dx 2 dt x 25 dt
If
d dx 400 and x 10, 16 rad hr. dt dt
If
dx d 400 and x 3, 58.824 rad hr. dt dt
Section 5.9
75. tanarctan x arctan y
Inverse Trigonometric Functions: Integration
263
tanarctan x tanarctan y xy , xy 1 1 tanarctan x tanarctan y 1 xy
Therefore, arctan x arctan y arctan
1x xyy , xy 1.
Let x 12 and y 13 . arctan
13 56 56 arctan arctan arctan 1 12 arctan13 arctan 1 12 12 13 1 16 56 4
77. f x kx sin x f x k cos x ≥ 0 for k ≥ 1 fx k cos x ≤ 0 for k ≤ 1 Therefore, f x kx sin x is strictly monotonic and has an inverse for k ≤ 1 or k ≥ 1. 81. True
79. True
d sec2 x sec2 x arctantan x 1 dx 1 tan2 x sec2 x
1 d arctan x > 0 for all x. dx 1 x2
Section 5.9 1.
5
9
x2
Inverse Trigonometric Functions: Integration
dx 5 arcsin
3 C x
3. Let u 3x, du 3 dx.
16
0
5.
1 1 dx 3
1 9x2
16
0
16
1 1 3 dx arcsin3x 3
1 3x2
0
18
7 7 x dx arctan C 16 x2 4 4
7. Let u 2x, du 2 dx.
32
0
1 1 dx 1 4x2 2
9.
1 dx x 4x2 1
11.
x2
13.
1 x 12
x3 dx 1
1
32
0
32
2 1 2x
2
dx
12 arctan2x
0
6
2 dx arcsec 2x C 2x 2x2 1
x
x dx x2 1
x dx
dx arcsinx 1 C
1 2x 1 1 dx x2 lnx2 1 C (Use long division.) 2 x2 1 2 2
15. Let u t 2, du 2t dt.
t
1 t 4
dt
1 1 1 2t dt arcsint 2 C 2 1 t 22 2
264
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions 1
17. Let u arcsin x, du
1 2
0
19. Let u 1 x2, du 2x dx.
dx.
1 x2
0
1 2
arcsin 1 dx arcsin2 x 2
1 x2
0
x 1 dx 2 2
1x 12
2 0.308 32
0
12
1 x2122x dx
1 x2
0
12
3 2
2
0.134 21. Let u e2x, du 2e2x dx.
23. Let u cos x, du sin x dx.
1 1 e2x 2e2x e2x arctan C 4x dx 2x 2 dx 4e 2 4 e 4 2
sin x dx 2 2 1 cos x
2 1
sin x dx cos2 x
25.
1
x 1 x
dx. u x, x u2, dx 2u du
27.
arctancos x
2
x3 1 2x 1 dx dx 3 2 dx x2 1 2 x2 1 x 1
1 du 2u du 2 2 arcsin u C
1 u2 u 1 u2
1 lnx2 1 3 arctan x C 2
2 arcsin x C
29.
x5
9 x 32
dx
x 3
9 x 32
dx
8
9 x 32
9 x 32 8 arcsin 6x x2 8 arcsin
3x 1 C
1 dx arctanx 1 1 x 12
2x dx x2 6x 13
2x 6 1 dx 6 2 dx x2 6x 13 x 6x 13
2
0
1
x2 4x
dx
2 0
x 2 3 C
dx arcsin
x 2 2 C
ln x2 6x 13 3 arctan
35.
x 3 3 C
1 dx x2 2x 2
0
33.
dx
2
31.
1
4 x 22
2
2x 6 1 dx 6 dx x2 6x 13 4 x 32
37. Let u x2 4x, du 2x 4 dx.
x2
x2 4x
3
39.
2
dx
2x 3 dx
4x x2
3
2
1 x2 4x122x 4 dx x2 4x C 2
2x 4 dx
4x x2
3
2
1 dx
4x x2
2 4x x2 arcsin
x 2 2
3 2
3
4x x2124 2x dx
2
4 2 3
4
3
2
1.059 6
1
4 x 22
dx
Section 5.9
Inverse Trigonometric Functions: Integration
265
41. Let u x2 1, du 2x dx.
1 x 2x dx dx 1 arctanx2 1 C x4 2x2 2 2 x2 12 1 2
43. Let u et 3. Then u2 3 et, 2u du et dt, and
et 3 dt
2u2 du u2 3
2u 2 3 arctan
2 du
6
2u du dt. u2 3
1 du u2 3
u C 2 et 3 2 3 arctan
3
e 3 3 C t
45. A perfect square trinomial is an expression in x with three terms that factor as a perfect square. Example: x2 6x 9 x 32
47. (a) (c)
49. (a)
1
1 x2
dx arcsin x C, u x
(b)
x
1 x2
dx 1 x2 C, u 1 x2
1 dx cannot be evaluated using the basic integration rules. x 1 x2
x 1 dx
2 x 132 C, u x 1 3
(b) Let u x 1. Then x u2 1 and dx 2u du.
x x 1 dx
u2 1u2u du 2 u4 u2 du 2
u5 u3 C 5
3
2 2 2 3 2 u 3u 5 C x 1323x 1 5 C x 1323x 2 C 15 15 15
(c) Let u x 1. Then x u2 1 and dx 2u du.
x dx
x 1
u2 1 u3 2 2 2u du 2 u2 1 du 2 u C uu2 3 C x 1x 2 C u 3 3 3
Note: In (b) and (c), substitution was necessary before the basic integration rules could be used. 51. (a)
y
(b)
5
dy 3 , 0, 0 dx 1 x2
y3 x
−5
5
3 2
dx 3 arctan x C 1 x2
−8
8
0, 0: 0 3 arctan0 C ⇒ C 0 −5
53.
y 3 arctan x
(0, 0)
dy 10 , y 3 0 dx x x2 1 4
−6
55. A
3
1
12
−8
− 3 2
x2
1 dx 2x 1 4
2 arctan 1
x1 2
3 1
3
1
1 dx x 12 22
1 arctan1 0.3927 2 8
266
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
57. Area 11 1
y 2
Matches (c)
3 2
−1
x −1
1 2
2
1
59. (a)
0
4 dx 4 arctan x 1 x2
1 0
1
4 arctan 1 4 arctan 0 4
4 40
(b) Let n 6.
1
4
0
4 1 3.1415918 1 1 4136 1 219 1 414 1 249 1 2536 2
4 1 dx 4 1 x2 36
(c) 3.1415927
61. (a)
1 u u u d arcsin C 2 2 2 dx a a
1 u a
a u2
Thus, (b)
du
a2 u2
arcsin
ua C.
1 1 u u ua u d 1 arctan C 2 2 2 dx a a a 1 ua2 a a u2a2 a u2
Thus,
du a2 u2
1 u u dx arctan C. a2 u2 a a
(c) Assume u > 0. 1 1 u u ua u d 1 arcsec C . The case u < 0 is handled in a simidx a a a ua ua2 1 a u u2 a2a2
u u2 a2 lar manner.
Thus,
du u u2 a2
63. (a) vt 32t 500
u u u2 a2
dx
1 u arsec C. a a
(b) st
550
vt dt
32t 500 dt
16t 2 500t C s0 160 5000 C 0 ⇒ C 0 0
20 0
st 16t 2 500t When the object reaches its maximum height, vt 0. vt 32t 500 0 32t 500 t 15.625 s15.625 1615.6252 50015.625 3906.25 ft Maximum height
—CONTINUED—
Section 5.10
Hyperbolic Functions
63. —CONTINUED—
(c)
1 dv 32 kv2
dt
32k v t C k arctan v 32k t C 32
1
arctan
32k
1
32k v tanC 32k t 32 v tan C k
32k t
When t 0, v 500, C arctan 500k 32 , and we have vt
32k tan arctan 50032k
32k t .
(d) When k 0.001, v(t 32,000 tanarctan 5000.00003125 0.032 t. 500
0
7 0
vt 0 when t0 6.86 sec.
6.86
(e) h
32,000 tan arctan 5000.00003125 0.032 t dt
0
Simpson’s Rule: n 10; h 1088 feet (f) Air resistance lowers the maximum height.
Section 5.10 1. (a) sinh 3
Hyperbolic Functions
e3 e3 10.018 2
(b) tanh2
3. (a) cschln 2
sinh2 e2 e2 2 0.964 cosh2 e e2
(b) cothln 5
2 2 4 eln 2 eln 2 2 1 2 3 coshln 5 eln 5 eln 5 ln 5 sinhln 5 e eln 5
5. (a) cosh12 ln 2 3 1.317 (b) sech1
23 ln 1
7. tanh2 x sech2 x
ee
x x
1 4 9
2 3
ex ex
0.962
e 2
x
2 ex
2
e2x 2 e2x 4 e2x 2 e2x 2x 1 ex ex2 e 2 e2x
5 1 5 13 5 1 5 12
267
268
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
9. sinh x cosh y cosh x sinh y
e
x
ex 2
e
y
ey ex ex 2 2
e
y
ey 2
1 exy exy exy exy exy exy exy exy 4 exy exy 1 sinhx y 2exy exy 4 2
11. 3 sinh x 4 sinh3 x sinh x3 4 sinh2 x
e
x
ex 2
e
x
ex 1 3 e2x 2 e2x ex exe2x e2x 1 2 2
3 4 e
x
ex 2
2
e3x e3x 1 sinh3x e3x ex ex ex e3x ex 2 2
sinh x
13.
cosh2 x
32
2
3 2
1 ⇒ cosh2 x
tanh x
3 2 13 2
13 13 ⇒ cosh x 4 2
313 13
csch x
2 1 3 2 3
sech x
213 1 13 13 2
coth x
13 1 3 3 13
15. y sinh1 x2
17. f x lnsinh x
y 2x cosh1 x2
19. y ln tanh y
x 2
fx
21. hx
x 1 2 1 sech2 tanhx 2 2 2 sinhx 2 coshx 2
23. f t arctansinh t ft
1 cosh t 1 sinh2 t
cosh t sech t cosh2 t
hx
1 cosh x coth x sinh
1 x sinh2x 4 2 1 1 cosh2x 1 cosh2x sinh2 x 2 2 2
1 csch x sinh x 25. Let y gx. y xcosh x ln y cosh x ln x
1 dy cosh x sinh x ln x y dx x dy y cosh x xsinh x ln x dx x
xcosh x cosh x xsinh x ln x x
Section 5.10
Hyperbolic Functions
27. y cosh x sinh x2 y 2cosh x sinh xsinh x cosh x 2cosh x sinh x2 2e2x 29. f x sin x sinh x cos x cosh x, 4 ≤ x ≤ 4
(−π , cosh π )
12
(π , cosh π )
fx sin x cosh x cos x sinh x cos x sinh x sin x cosh x 2 sin x cosh x 0 when x 0, ± . − 2
Relative maxima: ± , cosh Relative minimum: 0, 1
31. gx x sech x
x cosh x
33.
y a sinh x y a cosh x
1
y a sinh x
(1.20, 0.66) −
2
(0, − 1) −2
y a cosh x
Therefore, y y 0. (−1.20, − 0.66) −1
Relative maximum: 1.20, 0.66 Relative minimum: 1.20, 0.66 35. f x tanh x
f 1 tanh1 0.7616
2
P1
1 0.4200 cosh21
fx sech2 x
f1
fx 2 sech2 x tanh x
f 1 0.6397
f −3
4
P2 −2
P1x f 1 f1x 1 0.7616 0.42x 1 P2x 0.7616 0.42x 1
37. (a) y 10 15 cosh
0.6397 x 12 2
x , 15 ≤ x ≤ 15 15
(b) At x ± 15, y 10 15 cosh1 33.146. At x 0, y 10 15 cosh1 25.
y
(c) y sinh
30
x . At x 15, y sinh1 1.175 15
20 10
− 10
x 10
20
39. Let u 1 2x, du 2 dx.
sinh1 2x dx
1 sinh1 2x2 dx 2
1 cosh1 2x C 2
41. Let u coshx 1, du sinhx 1 dx.
cosh2x 1 sinhx 1 dx
1 cosh3x 1 C 3
269
270
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
45. Let u
43. Let u sinh x, du cosh x dx.
cosh x dx ln sinh x C sinh x
x csch2
1 1 47. Let u , du 2 dx. x x
0
csch2
x2 x2 x dx coth C 2 2
1 5x 1 ln dx 25 x2 10 5 x
4 0
51. Let u 2x, du 2 dx.
1 1 ln 9 ln 3 10 5
2 4
0
x 2x 1 1 dx dx arctanx2 C x4 1 2 x22 1 2
y
57. y sinh1tan x 1 sec2 x sec x y tan2 x 1
3 9x2 1
59. y coth1sin 2x
y
1 2 cos 2x 2 sec 2x 1 sin2 2x
61. y 2x sinh12x 1 4x2 y 2x
1 2 4x 2 sinh
1
2
2x
4x 1 4x2
2 sinh12x
63. See page 395.
65. y a sech1
ax a
2
x2
x a2 x x2 a2 a2 x2 1 dy 2 2 2 2 2 2 2 2 2 2 dx x a x x a1 x a a x xa x xa x
67.
1 dx 1 e2x
69. Let u x, du
71.
1 x1 x
1 dx. 2x
dx 2
1 dx 4x x2
ex 1 1 e2x dx csch1ex C ln C x 2 ex e
ex1
1
dx 2 sinh 2 x 1 x 1
1
2
x C 2 ln x 1 x C
1 1 x4 1 x 2 2 ln C dx ln x 22 4 4 x 2 2 4 x
2 4
1 2 dx arcsin2x 1 2x2
55. y cosh13x
53. Let u x2, du 2x dx.
x2 dx 2
csch1 x coth1 x 1 1 1 1 dx csch coth 2 dx csch C x2 x x x x 4
49.
x2 , du x dx. 2
0
4
Section 5.10
73.
1 dx 1 4x 2x2
1 1 dx 3 2x 12 2
77. y
1 80 8x 16x2
x3 21x dx 5 4x x2
2
2
2
2x 1 3
dx
2x 1 3 2x 1 3 1 1 ln C ln C 26 2x 1 3 26 2x 1 3
75. Let u 4x 1, du 4 dx. y
4 4x 1 1 1 dx arcsin C 4 81 4x 12 4 9
dx
x 4
20 dx 5 4x x2
x 4 dx 20
4
sech
0 4
2
0
0
2
x dx 2
81. A
0
2
dx
2 dx ex 2
ex 2 dx 1
52 ln x
5 ln 4 17 5.237 2
ex 2 2
8 arctanex 2
4 0
5 2
5x
x4 1
ex 2
4
4
0
2x x22 1 2
8 arctane 2 5.207 2
1 dx 32 x 22
x2 x2 x2 20 x 2 3 10 x 1 10 x 5 C 4x C C 4x ln ln 4x ln 2 6 x 2 3 2 3 x5 2 3 x1
79. A 2
83.
Hyperbolic Functions
3k dt 16 3kt 16
1 dx x2 12x 32
1 x 6 2 1 x8 1 dx ln C ln C x 62 4 22 x 6 2 4 x4
When x 0:
t0 1 C ln2 4
When x 1:
t 10
30k 1 7 7 1 1 ln ln2 ln 16 4 3 4 4 6 k When t 20:
7 2 ln 15 6
163 152 ln 76 20 41 ln 2xx 88 ln
76
2
ln
x8 2x 8
x8 49 36 2x 8 62x 104 x
104 52 1.677 kg 62 31
dx
x4 1
2 0
271
272
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
85. As k increases, the time required for the object to reach the ground increases. ex ex 2
87. y cosh x y
y cosh1 x
89.
cosh y x
ex ex sinh x 2
sinh y y 1 y
91. y sech x
1 1 1 sinh y cosh2 y 1 x2 1
2 ex ex
y 2ex ex2ex ex
e
x
2 ex
ee
x x
ex sech x tanh x ex
Review Exercises for Chapter 5 1. f x ln x 3
y
Vertical shift 3 units upward Vertical asymptote: x 0
5 4
x=0
3 2 1 x 1
4x4x
2
3. ln
5
2
5. ln 3
2
3
4
5
1 1 2x 12x 1 1 ln ln2x 1 ln2x 1 ln4x2 1 1 5 4x2 1 5
3 4 x2 3 1 3 4 x2 ln x ln ln4 x2 ln x ln 3 ln 3 x
9. gx lnx
7. lnx 1 2 x 1 e2
gx
x 1 e4
1 ln x 2
1 2x
x e4 1 53.598 11. f x xln x
13.
x 1 fx ln x1 2 ln x 2 x 1 2 ln x 1 ln x 2 ln x 2ln x
15.
1 a bx 1 y ln lna bx ln x a x a
dy 1 b 1 1 dx a a bx x xa bx
y
1 a lna bx b2 a bx
b x dy 1 ab dx b2 a bx a bx2 a bx2
17. u 7x 2, du 7dx
1 1 1 1 dx 7 dx ln 7x 2 C 7x 2 7 7x 2 7
Review Exercises for Chapter 5
19.
4
sin x sin x dx dx 1 cos x 1 cos x
21.
1
x1 dx x
4
1
1
4
1 3 ln 4
1 dx x ln x x
ln 1 cos x C
3
23.
sec d ln sec tan
0
3
0
ln 2 3
f x 12 x 3
25. (a)
y
1 2x
(b)
7
3
f
−1
− 11
2 y 3 x
10
f
2x 3 y f 1x 2x 6
−7
1 1 (c) f 1 f x f 1 2 x 3 2 2 x 3 6 x 1 f f 1x f 2x 6 2 2x 6 3 x
f x x 1
27. (a)
(b)
4
f −1
y x 1
f
y2 1 x
−3
x2 1 y f 1
x
x2
6
−2
1, x ≥ 0
(c) f 1 f x f 1 x 1 x2 12 1 x f f 1x f x2 1 x2 1 1 x2 x for x ≥ 0.
3 x 1 f x
29. (a)
(b)
4
f −1
3 x 1 y
f
y3 1 x x3
−4
5
1y
f 1x x3 1
−2
3 x 1 (c) f 1 f x f 1 3 x 1 1 x 3
3 x3 1 1 x f f 1x f x3 1
31.
f x x3 2
f x tan x
33.
f 1x x 21 3
f
1 f 1x x 22 3 3
1 0.160 35 3
3
3
fx sec2 x
1 1 f 11 1 22 3 3 332 3
6
f
f 1
6 34
33 f1 6 43
273
274
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
f x ln x
35. (a)
(b)
2
f −1
y lnx
f −3
ey x e2y x
3
−2
e2x y
(c) f 1 f x f 1 ln x e2 ln x e ln x x
f 1x e2x
f f 1x f e2x ln e2x ln ex x
37. y ex 2
39. f x lnex x2 2
fx 2x
y 6 4 2 x
−2
2
4
−2
41. gt t2et
43. y e2x e2x
gx t2et 2tet tett 2
45. gx
x2 ex
e2x e2x 1 y e2x e2x1 22e2x 2e2x 2 e2x e2x yln x y2 0
47.
ex 2x x2ex x2 x gx e2x ex
y
1x ln xdydx 2ydydx 0 2y ln x
dy y dx x y dy dx x2y ln x
49. Let u 3x2, du 6x dx.
xe3x dx 2
51.
e4x e2x 1 dx ex
1 3x2 1 2 e 6x dx e3x C 6 6
xe1x dx 2
1 1x2 e 2x dx 2
1 2 e1x C 2 57.
e3x ex ex dx
1 e3x ex ex C 3
53.
e4x 3e2x 3 C 3ex
55. Let u ex 1, du ex dx.
ex dx ln ex 1 C ex 1
y exa cos 3x b sin 3x y ex3a sin 3x 3b cos 3x exa cos 3x b sin 3x ex3a b sin 3x a 3b cos 3x y ex33a b cos 3x 3a 3b sin 3x ex3a b sin 3x a 3b cos 3x ex6a 8b sin 3x 8a 6b cos 3x y 2y 10y ex6a 8b 23a b 10b sin 3x 8a 6b 2a 3b 10a cos 3x 0
Review Exercises for Chapter 5
4
59. Area
0
1 2 2 xex dx ex 2
4 0
1 e16 1 0.500 2 63. y log2x 1
61. y 33 2
y
y
6 5
4 3
4 3
2 1
2
x
−1 1
2
2
3 4
5
6 7
−2 −3
x
−4 −3 −2 −1
1
3 4
−4
−2
65. f x 3x1
y x2x1
67.
fx 3x1 ln 3
ln y 2x 1 ln x y 2x 1 2 ln x y x y y
69. gx log31 x gx
1 log31 x 2
71.
2x x 1 2 ln x x 2x x 1 2 ln x 2x1
x 15x1 dx 2
1 1 x12 5 C 2 ln 5
1 1 1 2 1 xln 3 2x 1ln 3
73. (a) y x a y ax a1
(b) y ax
(c)
y xx
(d) y aa
ln y x ln x
y ln aax
y 0
1 1 y x 1 ln x y x y y1 ln x y xx1 ln x 75. 10,000 Pe0.0715 P
77.
Ph 30ekh P18,000 30e18,000k 15
10,000 $3499.38 e1.05
k
ln 2 ln1 2 18,000 18,000
Ph 30eh ln 2 18,000 P35,000 30e35,000 ln 2 18,000 7.79 inches
79.
P Ce0.015t 2C Ce0.015t 2 e0.015t ln 2 0.015t t
ln 2 46.21 years 0.015
81.
dy x2 3 dx x
dy y
x
3 dx x
x2 3 ln x C 2
275
276
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
83. y 2xy 0
dy 2xy dx 1 dy y
2x dx
ln y x2 C1 2 C 1
ex
y
y Cex
2
85.
dy x2 y2 (homogeneous differential equation) dx 2xy
x2 y2 dx 2xy dy 0 Let y vx, dy x dv v dx.
x2 v2x2 dx 2xvxx dv v dx 0 x2 v2x2 2x2v2 dx 2x3v dv 0 x2 x2v2 dx 2x3v dv 1 v2 dx 2x dv
dx x
2v dv 1 v2
ln x ln 1 v 2 C1 ln 1 v 2 ln C Cx2
x
C C 1 v 2 1 y x2 x2 y2
1
Cx x2 y2
or C1
x x2 y2
87. y C1x C2 x3 y C1 3C2 x2 y 6C2 x x2 y 3xy 3y x26C2 x 3xC1 3C2 x2 C1x C2 x3 6C2 x3 3C1x 9C2 x3 3C1x 3C2 x3 0 x 2, y 0: 0 2C1 8C2 ⇒ C1 4C2 x 2, y 4: 4 C1 12C2 1 4 4C2 12C2 8C2 ⇒ C2 , C1 2 2 1 y 2x x3 2 89. f x 2 arctanx 3
y
4
−6
−4
x
−2
2 −2 −4
Review Exercises for Chapter 5
Let arcsin
91. (a)
277
1 2 2
1 2
sin
1
θ
1 1 sin arcsin sin . 2 2 Let arcsin
(b)
3 1 cos . 2 2
cos arcsin
x
93. y tanarcsin x y
1 2
1 2
sin
3
95. y x arcsec x
1 x2
y
1 x21 2 x21 x21 2 1 x23 2 1 x2
x arcsec x x x2 1
97. y xarcsin x2 2x 21 x2 arcsin x y
1 x2 2x arcsin x 2x 2 2 2 arcsin x arcsin x arcsin x2 1 x2 1 x2 1 x2
99. Let u e2x, du 2e2x dx.
1 dx e2x e2x
1 1 e2x 1 dx 2e2x dx arctane2x C 1 e4x 2 1 e2x2 2
101. Let u x2, du 2x dx.
x 1 x4
2x , du 4 2 x
2
arctanx 2 1 dx 4 x2 2
107.
dy A2 y2
1 1 1 2x dx arcsin x2 C 2 1 x22 2
dx
105. Let u arctan
103. Let u 16 x2, du 2x dx.
y arcsin A
arctan
x 2
dx.
4 2 x dx 41arctan 2x 2
k dt m
k tC m
2
C
109. y 2x coshx y 2
Since y 0 when t 0, you have C 0. Thus,
mk t Ay
sin
mk t
y A sin 111. Let u x2, du 2x dx.
x x4 1
dx
1 1 x 1 dx 2x dx ln16 x2 C 16 x2 2 16 x2 2
1 1 1 2x dx ln x2 x4 1 C 2 x22 1 2
1 sinh x 2 sinh x 2x 2x
272
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
85. As k increases, the time required for the object to reach the ground increases. ex ex 2
87. y cosh x y
y cosh1 x
89.
cosh y x
ex ex sinh x 2
sinh y y 1 y
91. y sech x
1 1 1 sinh y cosh2 y 1 x2 1
2 ex ex
y 2ex ex2ex ex
e
x
2 ex
ee
x x
ex sech x tanh x ex
Review Exercises for Chapter 5 1. f x ln x 3
y
Vertical shift 3 units upward Vertical asymptote: x 0
5 4
x=0
3 2 1 x 1
4x4x
2
3. ln
5
2
5. ln 3
2
3
4
5
1 1 2x 12x 1 1 ln ln2x 1 ln2x 1 ln4x2 1 1 5 4x2 1 5
3 4 x2 3 1 3 4 x2 ln x ln ln4 x2 ln x ln 3 ln 3 x
9. gx lnx
7. lnx 1 2 x 1 e2
gx
x 1 e4
1 ln x 2
1 2x
x e4 1 53.598 11. f x xln x
13.
x 1 fx ln x1 2 ln x 2 x 1 2 ln x 1 ln x 2 ln x 2ln x
15.
1 a bx 1 y ln lna bx ln x a x a
dy 1 b 1 1 dx a a bx x xa bx
y
1 a lna bx b2 a bx
b x dy 1 ab dx b2 a bx a bx2 a bx2
17. u 7x 2, du 7dx
1 1 1 1 dx 7 dx ln 7x 2 C 7x 2 7 7x 2 7
Review Exercises for Chapter 5
19.
4
sin x sin x dx dx 1 cos x 1 cos x
21.
1
x1 dx x
4
1
1
4
1 3 ln 4
1 dx x ln x x
ln 1 cos x C
3
23.
sec d ln sec tan
0
3
0
ln 2 3
f x 12 x 3
25. (a)
y
1 2x
(b)
7
3
f
−1
− 11
2 y 3 x
10
f
2x 3 y f 1x 2x 6
−7
1 1 (c) f 1 f x f 1 2 x 3 2 2 x 3 6 x 1 f f 1x f 2x 6 2 2x 6 3 x
f x x 1
27. (a)
(b)
4
f −1
y x 1
f
y2 1 x
−3
x2 1 y f 1
x
x2
6
−2
1, x ≥ 0
(c) f 1 f x f 1 x 1 x2 12 1 x f f 1x f x2 1 x2 1 1 x2 x for x ≥ 0.
3 x 1 f x
29. (a)
(b)
4
f −1
3 x 1 y
f
y3 1 x x3
−4
5
1y
f 1x x3 1
−2
3 x 1 (c) f 1 f x f 1 3 x 1 1 x 3
3 x3 1 1 x f f 1x f x3 1
31.
f x x3 2
f x tan x
33.
f 1x x 21 3
f
1 f 1x x 22 3 3
1 0.160 35 3
3
3
fx sec2 x
1 1 f 11 1 22 3 3 332 3
6
f
f 1
6 34
33 f1 6 43
273
274
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
f x ln x
35. (a)
(b)
2
f −1
y lnx
f −3
ey x e2y x
3
−2
e2x y
(c) f 1 f x f 1 ln x e2 ln x e ln x x
f 1x e2x
f f 1x f e2x ln e2x ln ex x
37. y ex 2
39. f x lnex x2 2
fx 2x
y 6 4 2 x
−2
2
4
−2
41. gt t2et
43. y e2x e2x
gx t2et 2tet tett 2
45. gx
x2 ex
e2x e2x 1 y e2x e2x1 22e2x 2e2x 2 e2x e2x yln x y2 0
47.
ex 2x x2ex x2 x gx e2x ex
y
1x ln xdydx 2ydydx 0 2y ln x
dy y dx x y dy dx x2y ln x
49. Let u 3x2, du 6x dx.
xe3x dx 2
51.
e4x e2x 1 dx ex
1 3x2 1 2 e 6x dx e3x C 6 6
xe1x dx 2
1 1x2 e 2x dx 2
1 2 e1x C 2 57.
e3x ex ex dx
1 e3x ex ex C 3
53.
e4x 3e2x 3 C 3ex
55. Let u ex 1, du ex dx.
ex dx ln ex 1 C ex 1
y exa cos 3x b sin 3x y ex3a sin 3x 3b cos 3x exa cos 3x b sin 3x ex3a b sin 3x a 3b cos 3x y ex33a b cos 3x 3a 3b sin 3x ex3a b sin 3x a 3b cos 3x ex6a 8b sin 3x 8a 6b cos 3x y 2y 10y ex6a 8b 23a b 10b sin 3x 8a 6b 2a 3b 10a cos 3x 0
Review Exercises for Chapter 5
4
59. Area
0
1 2 2 xex dx ex 2
4 0
1 e16 1 0.500 2 63. y log2x 1
61. y 33 2
y
y
6 5
4 3
4 3
2 1
2
x
−1 1
2
2
3 4
5
6 7
−2 −3
x
−4 −3 −2 −1
1
3 4
−4
−2
65. f x 3x1
y x2x1
67.
fx 3x1 ln 3
ln y 2x 1 ln x y 2x 1 2 ln x y x y y
69. gx log31 x gx
1 log31 x 2
71.
2x x 1 2 ln x x 2x x 1 2 ln x 2x1
x 15x1 dx 2
1 1 x12 5 C 2 ln 5
1 1 1 2 1 xln 3 2x 1ln 3
73. (a) y x a y ax a1
(b) y ax
(c)
y xx
(d) y aa
ln y x ln x
y ln aax
y 0
1 1 y x 1 ln x y x y y1 ln x y xx1 ln x 75. 10,000 Pe0.0715 P
77.
Ph 30ekh P18,000 30e18,000k 15
10,000 $3499.38 e1.05
k
ln 2 ln1 2 18,000 18,000
Ph 30eh ln 2 18,000 P35,000 30e35,000 ln 2 18,000 7.79 inches
79.
P Ce0.015t 2C Ce0.015t 2 e0.015t ln 2 0.015t t
ln 2 46.21 years 0.015
81.
dy x2 3 dx x
dy y
x
3 dx x
x2 3 ln x C 2
275
276
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
83. y 2xy 0
dy 2xy dx 1 dy y
2x dx
ln y x2 C1 2 C 1
ex
y
y Cex
2
85.
dy x2 y2 (homogeneous differential equation) dx 2xy
x2 y2 dx 2xy dy 0 Let y vx, dy x dv v dx.
x2 v2x2 dx 2xvxx dv v dx 0 x2 v2x2 2x2v2 dx 2x3v dv 0 x2 x2v2 dx 2x3v dv 1 v2 dx 2x dv
dx x
2v dv 1 v2
ln x ln 1 v 2 C1 ln 1 v 2 ln C Cx2
x
C C 1 v 2 1 y x2 x2 y2
1
Cx x2 y2
or C1
x x2 y2
87. y C1x C2 x3 y C1 3C2 x2 y 6C2 x x2 y 3xy 3y x26C2 x 3xC1 3C2 x2 C1x C2 x3 6C2 x3 3C1x 9C2 x3 3C1x 3C2 x3 0 x 2, y 0: 0 2C1 8C2 ⇒ C1 4C2 x 2, y 4: 4 C1 12C2 1 4 4C2 12C2 8C2 ⇒ C2 , C1 2 2 1 y 2x x3 2 89. f x 2 arctanx 3
y
4
−6
−4
x
−2
2 −2 −4
Review Exercises for Chapter 5
Let arcsin
91. (a)
277
1 2 2
1 2
sin
1
θ
1 1 sin arcsin sin . 2 2 Let arcsin
(b)
3 1 cos . 2 2
cos arcsin
x
93. y tanarcsin x y
1 2
1 2
sin
3
95. y x arcsec x
1 x2
y
1 x21 2 x21 x21 2 1 x23 2 1 x2
x arcsec x x x2 1
97. y xarcsin x2 2x 21 x2 arcsin x y
1 x2 2x arcsin x 2x 2 2 2 arcsin x arcsin x arcsin x2 1 x2 1 x2 1 x2
99. Let u e2x, du 2e2x dx.
1 dx e2x e2x
1 1 e2x 1 dx 2e2x dx arctane2x C 1 e4x 2 1 e2x2 2
101. Let u x2, du 2x dx.
x 1 x4
2x , du 4 2 x
2
arctanx 2 1 dx 4 x2 2
107.
dy A2 y2
1 1 1 2x dx arcsin x2 C 2 1 x22 2
dx
105. Let u arctan
103. Let u 16 x2, du 2x dx.
y arcsin A
arctan
x 2
dx.
4 2 x dx 41arctan 2x 2
k dt m
k tC m
2
C
109. y 2x coshx y 2
Since y 0 when t 0, you have C 0. Thus,
mk t Ay
sin
mk t
y A sin 111. Let u x2, du 2x dx.
x x4 1
dx
1 1 x 1 dx 2x dx ln16 x2 C 16 x2 2 16 x2 2
1 1 1 2x dx ln x2 x4 1 C 2 x22 1 2
1 sinh x 2 sinh x 2x 2x
278
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
Problem Solving for Chapter 5 1. tan 1 3 x tan 2
6 10 x
6
Minimize 1 2:
3
f x 1 2 arctan fx
1 9 1 2 x
3 x 2
3 6 arctan x 10 x
1
θ1 0
θ
x
θ2 a
10
6 0 36 10 x2 1 10 x2 3 6 x2 9 10 x2 36
10 x2 36 2x2 9 100 20x x2 36 2x2 18 x2 20x 118 0 x a 10 218 4.7648
20 ± 202 4118 10 ± 218 2
f a 1.4153
1 2 1.7263 or 98.9 Endpoints: a 0: 1.0304 a 10: 1.2793 Maximum is 1.7263 at a 10 218 4.7648. 3. f x sinln x or 0,
(a) Domain: x > 0
(f)
2k. 2 2 2 2 . Two values are x e , e
(b) f x 1 sinln x ⇒ ln x
(c) f x 1 sinln x ⇒ ln x
3 2k. 2
1 (e) fx cosln x x
k ⇒ 2
x e2 on 1, 10
f 10 0.7440
1 lim f x seems to be . (This in incorrect.) 2
(g) For the points x e2, e32, e72, . . . we have f x 1. For the points x e2, e52, e92, . . .
fx 0 ⇒ cosln x 0 ⇒ ln x
−2
x→0
(d) Since the range of the sine function is 1, 1 , parts (b) and (c) show that the range of f is 1, 1 .
f 1 0
5
0
Two values are x e2, e32.
f e2 1
2
we have f x 1. That is, as x → 0, there is an infinite number of points where f x 1, and an infinite number where f x 1. Thus lim sinln x does not exist. x→0
Maximum is 1 at x
e2
4.8105
You can verifiy this by graphing f x on small intervals close to the origin.
Problem Solving for Chapter 5 5. (a)
279
Area sector t t t ⇒ Area sector Area circle 2 2 2
1 (b) Area AOP baseheight 2 At
1 cosh t sinh t 2
cosh t
x2 1 dx
1 cosh t
x2 1 dx
1
1 At cosh2 t sinh2 t cosh2 t 1 sinh t 2 1 cosh2 t sinh2 t sinh2 t 2 1 1 cosh2 t sinh2 t 2 2 1 At t C. But, A0 C 0 ⇒ C 0 2 1 Thus, At t or t 2 At. 2 9. Let u 1 x, x u 1, x u2 2u 1,
y ln x
7.
y
dx 2u 2du.
1 x
Area
1 y b x a a
4
1
1 dx x x
1 y x b 1 Tangent line a
3
2
If x 0, c b 1. Thus, b c b b 1 1.
u 1 du u2 u
32
2
2u 2 du u 1 u2 2u 1
32
2
u
du
2 ln u
3 2
2 ln 3 2 ln 2 2 ln
32
0.8109
11. (a)
dy y1.01 dt
y1.01 dy
(b)
y1 dy
y0.01 y0.01
y kt C y
0.01t C 1 C 0.01t
1 y C 0.01t100 y0 1: 1 Hence, y
1 ⇒ C1 C100
1 . 1 0.01t100
For T 100, lim y . t→T
k dt
y kt C1
dt
y0.01 t C1 0.01 1
1 C kt1
y0 y0
1 1 1 ⇒ C1 ⇒ C C1 y0 y0
Hence, y
1 1 kt y0
For t →
1 , y → . y0 k
1
.
280
Chapter 5
13. Since
Logarithmic, Exponential, and Other Transcendental Functions
dy k y 20, dt 1 dy y 20
k dt
ln y 20 k t C y Cekt 20. When t 0, y 72. Therefore, C 52. When t 1, y 48. Therefore, 48 52ek 20, ek 2852 713, and k ln713. Thus, y 52e ln713 t 20. When t 5, y 52e5 ln713 20 22.35.
15. (a)
dS k1SL S dt S
(b)
dS 4 ln S100 S dt 9
S dSdt 100 S dSdt
4 d 2S ln dt2 9
L is a solution because 1 Cekt
dS L1 Cekt 2C kekt dt
LC kekt 1 Cekt 2
Lk 1 LCe
1 Cekt
kt
k L L 1 Cekt
ln
49100 2S dSdt
0 when S 50 or
Choosing S 50, we have:
C Lekt
L L 1 Cekt
50
L 100. Also, S 10 when t 0 ⇒ C 9. And, S 20 when t 1 ⇒ k ln49.
100 1 9eln49t
2 1 9eln49t ln19 t ln49
k k1SL S, where k1 . L
100 Particular Solution. S 1 9eln49t
dS 0. dt
t 2.7 months (d)
S 140 120 100
100 1 9e0.8109t (c)
80 60 40 20
125
t 1
2
3
4
(e) Sales will decrease toward the line S L. 0
10 0
PA R T
I I C H A P T E R 6 Applications of Integration Section 6.1
Area of a Region Between Two Curves . . . . . . . . . . 264
Section 6.2
Volume: The Disk Method . . . . . . . . . . . . . . . . . 271
Section 6.3
Volume: The Shell Method
Section 6.4
Arc Length and Surfaces of Revolution . . . . . . . . . . 282
Section 6.5
Work
Section 6.6
Moments, Centers of Mass, and Centroids . . . . . . . . . 290
Section 6.7
Fluid Pressure and Fluid Force . . . . . . . . . . . . . . . 297
Review Exercises
. . . . . . . . . . . . . . . . 278
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 287
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305
C H A P T E R 6 Applications of Integration Section 6.1
Area of a Region Between Two Curves
Solutions to Even-Numbered Exercises
2. A
2x 5 x 2 2x 1 dx
4. A
x 2 x 3 dx
2
2
2
2
x 2 4 dx
1
1
0
0
1
3
1
8.
x 1 3 x 1 dx
6. A 2
1 x 2 x 2 1 dx
10.
2
4
x3 x x dx 3 3
12.
sec2 x cos x dx
4
y
y
y
7 2
3
(3, 6)
6
−π,2 4 2
(
5 4
)
( 4π , 2)
3 x
−2
2
2 1
(2, 23 )
(− 4π , 22 ) (π4 , 22 )
(3, 1) x
−1
−2
2
4
5
6
−π 4
7
8
1 14. f x 2 2 x
16. A
2
gx 2 x
8
A 1
2
Matches (a)
y
1 3 10 x xx 8 2 8
dx
3 2 7 x x 10 dx 8 2
x8 7x4 3
2
8
10x
2
64 112 80 1 7 20 18
4
y 3
(2, 9) (0, 2)
8
(8, 6)
6 1
(4, 0)
4 x
1
2
3
(2, 92 )
2
(8, 0) x 2
264
π 4
4
6
10
x
Section 6.1 18. The points of intersection are given by x2
Area of a Region Between Two Curves
20. The points of intersection are given by:
4x 1 x 1
x 2 4x 2 x 2
x2 3x 0
x3 x 0 when x 0, 3
x2 3x when x 0, 3
3
A
x2 4x 1 x 1 dx
A
3
3
x2 3x dx
3
2 3
x 3x 3 2
x 2 4x 2 x 2 dx
0
0 3
f x gx dx
0
0
3
x 2 3x dx
0
0
y
27 9 9 2 2
6
(3, 5)
5
y
4
6
3
5
(0, 2)
(3, 4)
4 3 2
x
−1
(0, 1)
1
2
3
5
4
x 1
5
22. A
3
2
4
5
6
1 1 0 dx x2 x
1
5 1
4 5
y
(1, 1)
1.0 0.8 0.6 0.4 0.2
(5, 0.04) x 1
2
3
4
5
24. The points of intersection are given by 3 x 1 x 1
y
x 1 x 13 x3 3x2 3x 1 3x 2x 0
x3
1
(2, 1)
2
(1, 0)
xx2 3x 2 0 xx 2x 1 0 ⇒ x 0, 1, 2
1
A2
0
3 x 1 dx x 1
2 x 4x 1
2
x2
3
1
4 3
0
12 1 0 43 21
2
x 2
(0, −1)
x3
3
3 2 x 2
3 0
9 2
265
266
Chapter 6
Applications of Integration
26. The points of intersection are given by: 2y y 2 y
y
(−3, 3)
y y 3 0 when y 0, 3
3
3
A
f y g y dy
1
0
(0, 0)
3
−3
2y y 2 y dy
x
−2
1 −1
0 3
3y y 2 dy
0
32 y
2
1 3 y 3
3 0
9 2
3
28. A
y
f y g y dy 4
0
3
0
y
16 y 2
(
3
3 ,3 7
)
2
3
1 2
16 y 21 22y dy
1
0
16 y 2
1
30. A
0 dy
4
0
3 0
4 7 1.354
x
−1
1
4 dx 2x
y
(1, 4)
0
1
4x 4 ln 2 x
3
2
3
(0, 2)
4 4 ln 2
1
1.227
x
−1
34. The points of intersection are given by:
32. The point of intersection is given by:
x 4 2x 2 2x 2
x 3 2x 1 2x
1
x 2x 2 4 0 when x 0, ± 2
x 3 1 0 when x 1
A2
f x gx dx
1
2x 2 x 4 2x 2 dx
0 2
1
2
1
A
3
1
2
x 3 2x 1 2x dx
1
x4 x x 1 dx 4 1 3
Numerical Approximation: 2.0
1 1
4x 2 x 4 dx
0
2
2
4x3
3
x5 5
2 0
128 15
Numerical Approximation: 8.533 10
y
(−2, 8)
(2, 8)
(−1, 2) −4
(1, 0) −2
x
−1
2 −1 −2
(1, −2)
4 −2
(0, 0)
Section 6.1
Area of a Region Between Two Curves
36. f x x 4 4x 2, gx x 3 4x
y
g 4
The points of intersection are given by:
3
x 4 4x 2 x 3 4x −4 −3
x 4 x 3 4x 2 4x 0 xx 1x 2x 2 0 when x 2, 0, 1, 2
0
A
2
1
3
4
−3 −4
1
x 3 4x x 4 4x 2 dx
x
−1
f
2
x 4 4x 2 x 3 4x dx
0
x3 4x x4 4x2 dx
1
248 37 53 293 30 60 60 30
Numerical Approximation: 8.267 0.617 0.883 9.767
3
38. A
0
x2
3 ln x 2 1
40. A
3
( 3, )
3
2
0
−1
5
(0, 0) −1
−1
6.908
5
−1
Numerical Approximation: 6.908
6
44 xx dx 3.434
x
0
9 5
3 ln 10
42. A
4
6x 0 dx 1
1
44. A
cos 2x sin x dx
2
0
6
43 23 0 3 4 3 1.299
1 sin 2x cos x 2
2
y
(
π, 1 6 2
2 4 x 4 sec
)
2 4
2
x2 4x
2 4
2
2
2
x x tan dx 4 4
4 x sec 4
1
0
4 4 4 2
2
4 1 2 2.1797
y
−π 2
π 6
(− π2 , −1)
4
x
3
−1
2
(1,
2)
1 x 1
46. From the graph we see that f and g intersect twice at x 0 and x 1.
1
A
48. A
2
(0, 1)
2x 1 3 x dx
2
0
1 x 3 x2 x ln 3
1 sin 2x 2
(1, 3)
1
2 cos x
3
0
2 sin x cos 2x 0 dx
0
y
gx f x dx
0
1 2 1 0.180 ln 3
−1
5 4
− 4
(0, 1)
1
x 1
2
(π, 1)
−2
0
4
267
268
Chapter 6
5
50. A
Applications of Integration
4 ln x 0 dx x
1
2ln x 2
52. (a) y x e x, y 0, x 0, x 1
1
5
(b) A
2ln 5 2 5.181
1
x e x dx.
0
No, it cannot be evaluated by hand.
2
(5, 1.29) 0
(c) 1.2556
y = xe x
4
6
(1, 0)
−2
−1
2 −1
x
54. Fx
0
12 1 t 2 dt t3 2t 2 6
x
x3 2x 6
0
(b) F4
(a) F0 0
43 56 24 6 3
(c) F6 36 12 48 y
y y 20
20 20
16
16 16
12
12 12
8
8 8
4
4 4
x
x 1
2
3
4
5
x
6 1
y
56. F y
1
4ex 2 dx 8ex 2
1
(b) F0 8 8e1 2 3.1478
30
25
25
20
20
20
15
15
15
10
10
10
5
5
2
4
2
3
2
74 x
2
6
4
7
2
y = 29 x − 12 (4, 6) y = − 25 x + 16
2
(6, 1) x 6
2
8
−2
(2, −3)
y=x−5
10
6
4
2
21x
6 4
1
2
3
4
5 x 16 x 5 dx 2
7 x 21 dx 2
4
4 x
7x
4
−4
9 x 12 x 5 dx 2
7 x 7 dx 2
4
5
6
5 x
−1
4
3
y
30
1
2
(c) F4 8e2 8e1 2 54.2602
25
x
6
1
6
30
4
y
5
y
−1
4
8e y 2 8e1 2
y
3
y
(a) F1 0
58. A
2
7 7 14
−1
x 1
2
3
4
Section 6.1
Area of a Region Between Two Curves
1 x2 1
60. f x
f x
y
(0, 1)
2x x 2 1 2
(1, 21 )
1 2
1 1 At 1, , f 1 . 2 2 Tangent line:
1 4
y=− 1x+1 2
x 1 2
1 1 1 x 1 or y x 1 2 2 2
y
1
0
1 1 x1 1 2
x2
3 2
1
1 at x 0. x2 1
The tangent line intersects f x A
1 f ( x) = 2 x +1
3 4
dx arctan x x4 x
2
1 0
3 0.0354 4
62. Answers will vary. See page 417. 64. x 3 ≥ x on 1, 0
y
x 3 ≤ x on 0, 1
(1, 1)
1
Both functions symmetric to origin
0
(0, 0)
1
1
x 3 x dx
x 3 x dx.
1
0
1
Thus,
x
−1
1
−1
x 3 x dx 0.
1
A2
x x 3 dx 2
0
(− 1, − 1)
x2 x4
4 1
2
0
1 2
66. Proposal 2 is better, since the cummulative deficit (the area under the curve) is less.
9
68. A 2
9 x dx 2 9x
0
9b
2
0
x2 2
9 0
9 b x dx
0
12 9
81 9 x b dx 2
9b
2
y
81
x2 2
9b
81 2
81 2
9 b9 b
81 2
2 9 bx
0
9b b9
9
2
9
2
2.636
6
(− (9 −b), b)
−6
(9 −b, b) x
−3
3 −3 −6
6
2
269
270
Chapter 6 n
70. lim
→0 i1
5
4i 4 and x is the same as n n
2
2
y
4 x i2 x
where xi 2
Applications of Integration
4 x 2 dx 4x
x3 3
2 2
f ( x) = 4 − x 2
3 2
32 . 3
1
(− 2, 0) −3
(2, 0) x
−1
1
3
−1
5
72.
5
7.21 0.26t 0.02t 2 7.21 0.1t 0.01t 2 dt
0
0.01t 2 0.16t dt
0
0.01t 3
29 billion $ 2.417 billion 12
3
0.16t 2 2
5 0
74. 5% : P1 893,000 e 0.05t 312 %: P2 893,000 e 0.035t Difference in profits over 5 years:
5
893,000e 0.05t 893,000e 0.035t dt 893,000
0
e0.05 e0.035
0.05t
0.035t 5 0
893,00025.6805 34.0356 20 28.5714 893,0000.2163 $ 193,156 Note: Using a graphing utility you obtain $193,183. 76. The curves intersect at the point where the slope of y2 equals that of y1, 1. y2 0.08x2 k ⇒ y2 0.16x 1 ⇒ x
1 6.25 .16
6.25
(b) Area 2
(a) The value of k is given by
y2 y1 dx
0
y1 y 2
6.25
2
6.25 0.086.25 2 k
0.08x 2 3.125 x dx
0
k 3.125.
2
0.08x 3
3
3.125x
x2 2
6.25 0
26.510417 13.02083
16 2 8 5.908
78. (a) A 6.031 2
1
2
(b) V 2A 25.908 11.816 m 3 80. True
1
2
(c) 5000V 500011.816 59,082 pounds
Section 6.2
Section 6.2
3
2
4 x 2 2 dx
0
4. V
x 4 8x 2 16 dx
0
9 x2
0
3
2 dx
9 x2 dx
5 x5
8x 3 16x 3
2
x3 3
3 0
256 15
0
V
x2
8
4
16 y 2
0
y3 16y 3
4 0
2 dy
16 y 2 dy
128 3
x4
80
2
x5
8
V
4
0
8 0
y 4 8y 3 16y 2 dy
1
y5 2y 5
4
16y 3 3
2
4x 4 dx
0
y
4 1
−4
8
8
6
6
4
4
2
−4
4
(c) Rx 8, rx 8 2x 2
2
32x 2
dx 4
4x 4
8
8x 2
x4
dx
0
0
8 3 1 5 x x 3 5
2 0
896 15
2y dy y y 4 4 dy 2 2 2
2
0
2
0
4y
y
42 3 2 y 2 y 3 4
y 8
−4
−2
128 5
4
8
2
x
−2
V
64 64 32x 2 4 x 4 dx
0
4
0
(d) R y 2 y 2, r y 0
2
5 2
2 x
−2
V
4x5
y
2
6
6
4
4
2
2 x 2
4
−4
−2
0
4
y 2 4y 2 dy
22
4482 132.69 15
459 153 15 5
V
16
12 dx
(b) Rx 2x 2, rx 0
y y2 dy 4y 2 4
2
2x3 12x 3
12. y 2x 2, y 0, x 2 (a) R y 2, r y y 2
22 dx
1282 322 242 80 3
1
0
16 2x
2
4
4
2
22
0
10. V
x2
2
x ± 22
8. y 16 x 2 ⇒ x 16 y 2
4 4
22
8 16 x2
18
22
x2 4
6. 2 4
0
9x
V
271
Volume: The Disk Method
2
2. V
Volume: The Disk Method
x 4
8
0
16 3
272
Chapter 6
Applications of Integration
14. y 6 2x x 2, y x 6 intersect at 3, 3 and 0, 6. (a) Rx 6 2x x 2, rx x 6
(b) Rx 6 2x x2 3, rx x 6 3
0
V
3
V
0
0
6 2x x 2 2 x 6 2 dx
3 2x x 2 2 x 3 2 dx
3 0
x 4 4x 3 9x 2 36x dx
3
5 x 1
5
x 4 3x 3 18x 2
0 3
243 5
x 4 4x 3 3x 2 18x dx
3
5x 1
x 4 x 3 9x 2
5
y
−6
−4
2
V
0
8
8
4
4
2
2 x
x3 , rx 0 2
dx
x6 4
x7 28
32 16
x
−2
2
3
4 2 4 sec x 2 dx
dx
3
8 sec x sec 2 x dx
0
128 144 28 7
5
3 3 2
1 x 1
2
3
1
4
−1
π 9
20. R y 6, r y 6 6 y y
4
V
x
π 3
2π 9
4π 9
5π 9
y 5
6 2 y 2 dy
4
0
36y
y3 3
4 0
368 3
0
8 ln 2 3 3 27.66 y
2
3
8 ln 2 3 3 8 ln 1 0 0
0
(2, 4)
8 ln sec x tan x tan x
2
y
−1
108 5
0
0
16x
−4
V
16 4x3 x4
18. Rx 4, rx 4 sec x
2
2
−6
2
x3 4 2
3
y
−2
16. Rx 4
0
3 2 1 x 1 −1
2
3
4
5
Section 6.2 6 22. R y 6 , r y 0 y
6
V
6
2
6
V 2
2
dy
2
2
1 y
35
rx 0
x4 x 2 dx 2
2
36 y 2 ln y
6
0
2 1 1 2 dy y y
36
36
24. Rx x4 x 2, 2
6 y
Volume: The Disk Method
2
6 2
32 2 ln 2
2 ln 6
4x 2 x 4 dx
0
4x3
3
x5 5
2 0
128 15 y
13 1 36 2 ln 1213 6 ln 3 241.59 3 3
3 2
y
1 6 −3
5
x
−1
1
2
3
4 −2 3 −3 2 1 x 1
2
3
5
3 , rx 0 x1
26. Rx
8
V
4
0
3 x1
28. Rx e x 2, rx 0
dx
4
x 1
2
dx
8 0
e x dx
0
0
1 9 x1
e x 2 2 dx
0
8
9
4
V
2
e x
8
4 0
e 4 1 168.38
y
y
4 8 3 6 2 4 1 2 x 2
4
6
8 −2
4
30. V
0
4
0
1 4 x 2
2
x
56 88 48 3 3
4 0
8
dx
x2 5x 16 dx 4
x3 5x2 16x 12 2
2
4
8
4
x
x 2
4
2 4 21 x dx
6
2
y 4
x2 5x 16 dx 4
x3 5x2 16x 12 2
2
8 4
(4, 2)
3
1 x −2
2 −1
4
6
8
10
273
274
Chapter 6
Applications of Integration
32. y 9 x 2, y 0, x 2, x 3
y 9 8 7 6 5 4 3 2 1
x 9 y
5
V
9 y
0
2 22 dy
5
5 y dy
(2, 5)
x
0
1 2 3 4 5 6 7 8 9
y2 2
25 25 2 2
5y 25
34. V
2
5 0
3
cos x 2 dx 2.4674
36. V
0
ln x 2 dx 3.2332
1
y
2
1
x 2
1
5
38. V
2 arctan 0.2x 2 dx 15.4115
0
b
3 40. A 4
42. V
y
d
or
V
a
1
Matches (b)
Ax dx
Ay dy
c
3 4 1 2 1 4
x 1 4
44. (a)
1 2
3 4
z
1
(b) 2
4
2 4 x
8
y
x
8
8
−4
h
0
r2 2 x dx h2
r3h x 2
y
r
y= r x h
(h, r)
h
3
2
2
y
8 x
a < c < b.
r 46. Rx x, rx 0 h
z 4
4
2
V
(c)
z
4
0
1 r 2 h 3 r 2h 3h 3
x h
16
y
Section 6.2
Volume: The Disk Method
4
50. (a) V
48. x r 2 y 2 , R y r 2 y 2 , r y 0
0
r
V
r 2 y 2
h
2 dy
c
r y dy 2
2
r3
r3 h3 r 2h 3 3
2r3
h3 3
3
c2
y3 3
r2y
x dx
0
h
2 d x
4
x dx
0
2x
2 4 0
8
Let 0 < c < 4 and set
r
x
275
r
x2
2
c 0
c 2 4. 2
8
c 8 22
h
r 2h
Thus, when x 22, the solid is divided into two parts of equal volume.
c
(b) Set
x dx
0
2r 3 3r 2h h 3 3
8 (one third of the volume). Then 3
43 c 2 8 2 16 4 , c , c . 2 3 3 3 3
d
y
To find the other value, set of the volume). Then
r
h
x dx
0
16 (two thirds 3
32 46 d 2 16 2 32 , d , d . 2 3 3 3 3
x
The x-values that divide the solid into three parts of equal volume are x 43 3 and x 46 3.
52. y
2.95,0.1x
11.5
V
0
3
2.2x 2 10.9x 22.2, 0 ≤ x ≤ 11.5 11.5 < x ≤ 15
0.1x 3 2.2x 2 10.9x 22.2 2 dx
0.1x 4 2.2x 3 10.9x 2 22.2x 4 3 2
11.5 0
y 8
15
6
2.95 2 dx
11.5
2.95 2x
4
15 2
11.5 x
1031.9016 cubic centimeters
4
8
12
54. (a) First find where y b intersects the parabola: b4
x2 4 z
x 2 16 4b 44 b 5
x 24 b
24b
V
0
4
4
4
0
4
0
x2 b 4
x2 b 4
2
4
dx
24b
b4
x2 4
2
dx
2
2
dx
x
x4 bx 2 2x 2 b 2 8b 16 dx 16 2
80
2x 3 bx 3 b 2x 8bx 16x 3 6
5
128 32 64 512 b 4b 2 32b 64 4b 2 b 3 3 3 15
x5
64
—CONTINUED—
4 0
2
4
y
16
276
Chapter 6
Applications of Integration
54. —CONTINUED—
(b) graph of Vb 4b 2
64 512 b 3 15
(c) Vb 8b
120
64 64 3 8 223 0 ⇒ b 3 8 3
V b 8 > 0 ⇒ b
0
8 is a relative minimum. 3
4 0
Minimum Volume is 17.87 for b 2.67
10
56. (a) V
f x 2 dx
0
Simpson’s Rule: b a 10 0 10, V
n 10
2.1 2 41.9 2 22.1 2 42.35 2 22.6 2 42.85 2 22.9 2 42.7 2 22.45 2 4 2.2 2 2.3 2 3 178.405 186.83 cm 3 3
(b) f x 0.00249x 4 0.0529x 3 0.3314x 2 0.4999x 2.112 6
0
10 0
10
(c) V
f x 2 dx 186.35 cm 3
0
1 58. V 1023 30 m3 2
60.
1 1 (b) Ax bh 24 x 2 34 x 2 2 2
y 3
3 4 x 2
1
2
V 3
x
−3
3
1 −1
2
4 x 2 dx
3 4x
−3
x3 3
2 2
Base of Cross Section 24 x 2 (a) Ax b 2 24 x 2
2 4 − x2
2
2 4 − x2
2
V
2
44 x 2 dx
4 4x
3 2
x 3
2
2 4 − x2
128 3 2 4 − x2
—CONTINUED—
2 4 − x2
323 3
Section 6.2
Volume: The Disk Method
60. —CONTINUED— 1 (d) Ax bh 2
1 (c) Ax r 2 2
2
V
4 x 2 2 2 4 x 2 2
2
2
4 x 2 dx
x3 4x 2 3
2 2
1 24 x 2 4 x 2 4 x 2 2
2
16 3
V
2
4 x 2 dx 4x
x3 3
2 2
32 3
4 − x2 2 4 − x2 2 4 − x2
62. The cross sections are squares. By symmetry, we can set up an integral for an eighth of the volume and multiply by 8. A y b 2 r 2 y 2
2 y
r
V8
x
r 2 y 2 dy
0
1 3 y 3
8 r2y
r 0
16 3 r 3
64. V
R 2 r 2
R 2 r 2 R 2 r 2
2
R 2 x 2 2 r 2 dx R
R 2 r 2 x 2 dx
0
2 R 2 r 2 x
2
R2
r 2 3 2
x3 3
r
R 2 r 2
0
R 2 r 23 2 3
R2 − r 2
R
4 R 2 r 2 3 2 3
When a 2: x2 y2 1 represents a circle. (b) y 1 xa1 a
66. (a) When a 1: x y 1 represents a square.
1
A2
1
1 xa1 a dx 4
1
1 x a1 a dx
y
1
a=1
−1
1
0
To approximate the volume of the solid, form n slices, each of whose area is approximated by the integral above. Then sum the volumes of these n slices.
a=2
−1
x
277
278
Chapter 6
Applications of Integration
Section 6.3
Volume: The Shell Method
2. px x
4. px x
hx 1 x
hx 8 x 2 4 4 x 2
1
V 2
2
x1 x dx
V 2
0
x4 x 2 dx
0
1
x x2 dx 2
2
0
x2 3 2
x3
1 0
3
2
2
4x x 3 dx
0
2 2x 2
6. px x
8. px x
6
0
0
8
hx 4 2x
2
2
V 2
1 3 x dx 2
4 6
x 4
2
10. px x
hx 4 x
1 hx x2 2 V 2
x4 4
0
V 2
0
2 0
2
8
2
4x 2x 2 dx
0
y
2 2 2x 2 x 3 3
y 18
x4 2x dx
0
1 2 2x 2 x 4 4
324
2
4x x 3 dx
2 0
16 3
y
3
15 4
2
12 9
3
1
6 3 −1 −3
−2
x 1
2
3
4
5
6
2
x
−1
1
2 1 x
−1
12. px x hx
0
2
x
2
sinx x dx
1
sin x dx 2 cos x
0
14. p y y p y ≥ 0 on 2, 0 h y 4 2 y 2 y
2
y2 y dy
2
π 2
x
π
3π 4
−1
16. p y y h y 16 y2
y 4
V 2
y16 y2 dy
0
3 2 1
4
0
2
4
π 4
4
0
2
3
3
0
V 2
2
y
sin x x
V 2
1
2y y 2 dy
y 2
y3 3
0 2
8 3
2
16y y3 dy
0
y4 2 8y2 4
4 0
2 128 64 128
x −1 −2 −3 −4
4
8
12
Section 6.3 20. px 6 x
18. px 2 x hx 4x x x 4x 2x 2
2
hx x
2
2 x4x 2x 2 dx
V 2
6 x x dx
0
0
4
2
2
4
2
V 2
Volume: The Shell Method
8x 8x 2 2x 3 dx
2
6x 1 2 x 3 2 dx
0
0
8 1 2 4x 2 x 3 x 4 3 2
2 0
16 3
2 2 4x3 2 x5 2 5
y
4 0
192 5
y 4
4
3 3 2 2
1 x
1
1
2
3
4
5
−1 x
−1
1
−2
3
22. (a) Disk
(b) Shell
5
V
y
10 , rx 0 x2
Rx
10 x2
1
8
2
dx
5
2 −1
1
x3 100 3
20
x 1
−2
2
3
4
5
1
x
1
4
x4 dx
5
V 2
6
5
100
Rx x, rx 0
10
10 dx x2
1 dx x
5
20 ln x
1
5 1
20 ln 5
100 1 496 1 3 125 15
(c) Disk 10 x2
Rx 10, rx 10
5
V
102 10
1
200 100 3x x
5
3
1
10 x2
dx 2
1904 15 y
24. (a) Disk
(b) Same as part a by symmetry
Rx a 2 3 x 2 3 3 2
y
(0, a)
rx 0
(0, a)
a
V
a
(a, 0)
a 2 3 x 2 3 3 dx
(−a, 0)
(a, 0)
x x
a
2
a 2 3a 4 3x 2 3 3a 2 3x 4 3 x 2 dx
0
9 4 3 5 3 9 2 3 7 3 1 3 a x a x x 5 7 3
9 3 9 3 1 3 32 a 3 a a a 5 7 3 105
2 a 2x 2 a 3
a 0
(0, −a)
279
280
Chapter 6
Applications of Integration
z
26. (a)
(b)
z
(c)
z
2
2
2
5
x
x
x
5
5
a < c < b
−2
4
28. 2
y
5
2
x
0
2x dx
y
30. (a) 1
represents the volume of the solid generated by revolving the region bounded by y x 2, y 0, and x 4 about the y-axis by using the Shell Method.
2
10
y
3 4 1 2 1 4
2
16 2y dy 2
0
4 2y dy 2
2
x 1 4
0
1 2
represents this same volume by using the Disk Method.
3 4
1
1
y
(b) V 2
4
x 1 x 3 dx 2.3222
0
3 2 1 x 1
2
3
4
5
−1
Disk Method
32. (a)
34. y tan x, y 0, x 0, x
y 4
4
Volume 1
3
Matches (e)
2
y
1 x 1
2
3
3
(b) V 2
1
2
4
2x dx 19.0162 1 e 1 x
1
π 4
x
π 2
36. Total volume of the hemisphere is 12 43 r 3 23 3 3 18. By the Shell Method, px x, hx 9 x 2. Find x0 such that
x0
6 2
x 9 x 2 dx
y
0
6
x0
9 x21 2 2x dx
0
2 9 x 2 3 2 3
1 x
x0 0
2 18 9 x 02 3 2 3
9 x 02 3 2 18 x0 9 18 2 3 1.460. Diameter: 2 9 18 2 3 2.920
2
−3 −2 −1 −1 −2 −3
1
2
3
y
Section 6.3
Volume: The Shell Method
r
38. V 4
r
R x r 2 x 2 dx
4 R
4 R
2r 2 23 r
r
r
r2 x2 dx 4
r
2
2
x r 2 x 2 dx
r
x 2 3 2
r r
2 2 r 2R
abn axn dx
xn1 xa n1
b
40. (a) Area region
(c) Disk Method:
0
abn1
a
abn1
V 2
b
R1n
0
b
2a
bn1 n1
xbn xn1 dx
0
1 n 1 abn1 n1 n1
b2 x
b 2
n
2a
2
2a
R2n
n 1 n→ n 1
(b) lim R1n lim n→
xabn axn dx
0
n n1 n abnb n1
abn1
b
abn
n2
abn2
n→
b 0
b2abn n→
n bn2 abn2 n2 n2
n n 2
(d) lim R2n lim
lim abnb
n→
xn2 n2
n n 2
n n 2 1
lim b2abn
n→
(e) As n → , the graph approaches the line x 1.
4
42. (a) V 2
x f x dx
0
2 40 0 41045 22040 43020 0 34
20 5800 121,475 cubic feet 3
(b) Top line: y 50
1 1 40 50 x 0 x ⇒ y x 50 20 0 2 2
Bottom line: y 40
20
V 2
0 20
2
0
2 2
0 40 x 20 2x 20 ⇒ y 2x 80 40 20
1 x x 50 dx 2 2
21 x
2
40
40
50x dx 2
x3 25x 2 6
20 0
x 2x 80 dx
20
2x 2 80x dx
20
2
2x 3 40x 2 3
40 20
32,000 2 26,000 3 3
121,475 cubic feet (Note that Simpson’s Rule is exact for this problem.)
281
282
Chapter 6
Applications of Integration
Section 6.4
Arc Length and Surfaces of Revolution
2. 1, 2 , 7, 10
4. y 2x3 2 3
(a) d 7 1 2 10 2 2 10
y
9
2 4 x 3 3
(b) y
s
4 3
272 1 9x
9
3 2
0
1
1
43
2
dx
53 x
7 1
10
3 6. y x2 3 4 2
2 823 2 1 54.929 27
8.
y x1 3, 1, 27
27
s
1
1
27
1
x1
32 23 x
1 4 1 x 4 2 2x
s
27
2
1
0
14. (a) y
1 x e ex 2
101 x
2 1
779 3.246 240
−5
dx
(b)
y 2x 1 1 y 2 1 4x 2 4x 1
1
2 0
L
1 1 e 2 2 3.627 2 e
(b)
y
1 1 x 2
1 y 2 1
y= 1 x+1
2 4x 4x 2 dx
L
1
0
(c) L 1.132
1 1 x 4
1
2
2
(c) L 5.653
5
−2
1 6x 3
2
0
1 ,0 ≤ x ≤ 1 1x
−1
5
2
e x ex dx
−3
1 x e ex 2
2
5
2
1 2
dx
12. (a) y x 2 x 2, 2 ≤ x ≤ 1
2 1 x e ex , 0, 2 1 y 2
1 4 1 x 4 2 2x
1 4 1 x 4 dx 2 2x
1
1 y e x ex , 0, 2 2
2
1 2 , 1, 2 2x 4
1 y 2 dx
1
1 y e x ex 2
s
2
2
4
a
103 2 23 2 28.794
10.
12 x b
1 3 2
2 3
y 1 y 2
x2 3
x5 1 10 6x 3
dx
1 3
2 1 1 3 dx 3x
1
y
2
x2 3 1 dx x2 3
27
3 2
1 9x dx
0
7
s
y 3x1 2, 0, 9
1 dx 1 x 4
Section 6.4
16. (a) y cos x,
≤ x ≤ 2 2
Arc Length and Surfaces of Revolution
y sin x
(b)
(c) 3.820
1 y 2 1 sin 2 x
2
L
y
1 x
y = cos x −2
2
2
2
1 sin 2 x dx
−2
18. (a) y ln x, 1 ≤ x ≤ 5
(b)
(c) L 4.367
2 −1
1 y 2 1
6
1 x2
5
L
1
1
−6
20. (a) x 36 y 2 , 0 ≤ y ≤ 3
(b)
dx 1 36 y 2 1 2 2y dy 2
y 36 x 2, 33 ≤ x ≤ 6
10
y
1
0
10
3
−5
(c) L 3.142 !
36 y 2 3
L −10
1 dx x2
0
y2 dy 36 y 2
6
36 y 2
dy
Alternatively, you can convert to a function of x. y 36 x 2 y
dy x dx 36 x 2
6
L
33
1 36 x x dx
6
2
2
33
6 36 x 2
dx
Although this integral is undefined at x 0, a graphing utility still gives L 3.142.
4
22.
0
1 dxd tan x
y
2
dx 2
s 1
( π4 , 1(
y = tan x 1
Matches (e)
(0, 0)
π 4
x 3π 8
24. f x x 2 4 2, 0, 4 (a) d 4 0 2 144 16 2 128.062 (b) d 1 0 2 9 16 2 2 1 2 0 9 2 3 2 2 25 0 2 4 3 2 144 25 2 160.151
4
(c) s
1 4x x 2 4 2 dx 159.087
0
(d) 160.287
283
284
Chapter 6
Applications of Integration
26. Let y ln x, 1 ≤ x ≤ e, y
1 and L1 x
Equivalently, x e y, 0 ≤ y ≤ 1,
e
1
1
dx e y, and L 2 dy
1 dx. x2
1
1
1 e 2y dy
0
1 e 2x dx.
0
Numerically, both integrals yield L 2.0035 y 31 10e x 20 ex 20
28.
1 y e x 20 ex 20 2
s
1 2
20
20
1 x 10 1 x 20 e 2 ex 10 e ex 20 4 2
1 y 2 1
12 e
x 20
ex 20
20
20
2
2
dx
e x 20 ex 20 dx 10e x 20 ex 20
20 20
20 e
1 47 ft e
Thus, there are 10047 4700 square feet of roofing on the barn. 30. y 693.8597 68.7672 cosh 0.0100333x y 0.6899619478 sinh 0.0100333x
299.2239
s
1 0.6899619478 sinh 0.0100333x 2 dx 1480
299.2239
(Use Simpson’s Rule with n 100 or a graphing utility.) 32.
34. y 2x
y 25 x 2 y 1 y 2
x
3 4
3
S 2
1 1x dx
2x
9
25 dx 25 x 2
5 arcsin 5 arcsin
, 4, 9
4
4
x 5
8 x 1 3 2 3
4
3
7.8540
4 3 arcsin 5 5
1 2 5 7.8540 s 4
x 1 dx
4
5 dx 25 x 2
1 x
9
25 25 x 2 4
s
y
25 x 2
9 4
8 3 2 10 53 2 171.258 3
Section 6.4
36.
y
x 2
y
1 2
y 2x
3
S 2
x 2
0
40.
1 y 2
x1 4x2 dx
0
54 dx
25 2 x 8
6 0
4
3
1 4x2 1 2 8x dx
0
6 1 4x
3
2 3 2
0
95
y ln x y
S 2
6
3 2 37 1 117.319 6
42. The precalculus formula is the distance formula between two points. The representative element is
1 x
xi 2 yi 2
y 1 x xi. i
2
i
x2 1 , 1, e x2
x x 1 dx
e
S 2
2
x
1
2
e
2
x 2 1 dx 22.943
1
44. The surface of revolution given by f1 will be larger. rx is larger for f1. 46. y r 2 x 2 y 1 y 2
285
38. y 9 x2, 0, 3
5 1 y 2 , 0, 6 4
Arc Length and Surfaces of Revolution
48. From Exercise 47 we have: S 2
r
2
r
a
r
0
r
r
r
rx
r 2 x 2
r2 2
r dx 2 rx
r r
x2
dx
4 r2
dx
r 2 x 2
0
r2 r2 x2
S 2
a
x r 2 x 2
2x dx r 2 x 2
2rr 2 x 2
2r 2
2rr 2
a 0
a2
2r r r 2 a 2 2 rh where h is the height of the zone y
r
h r−h
x
x2 + y2= r2
286
Chapter 6
Applications of Integration
50. (a) We approximate the volume by summing 6 disks of thickness 3 and circumference Ci equal to the average of the given circumferences: 6
r
V
i
2 3 4 C 6
3
2
i1
Ci
2
6
3
i
i1
2
i1
50 2 65.5 65.5 2 70 70 2 66 66 2 58 58 2 51 51 2 48 2
2
2
2
3 4
3 57.75 2 67.75 2 68 2 62 2 54.5 2 49.5 2 4
2
2
3 21813.625 5207.62 cubic inches 4 (b) The lateral surface area of a frustum of a right circular cone is sR r . For the first frustum,
S1 32
65.52 50 250 65.5 2 2 1 2
50 2 65.59 65.52 50
r s
2 1 2
h
.
Adding the six frustums together, S
R
50 2 65.59 15.5 2
2 1 2
65.5 2 709 4.5 2
2 1 2
70 2 669 24
66 2 589 28
58 2 519 27
51 2 489 23
2 1 2
2 1 2
2 1 2
2 1 2
224.30 208.96 208.54 202.06 174.41 150.37 1168.64
18
(c) r 0.00401 y 3 0.1416 y 2 1.232 y 7.943
(d) V
r 2 dy 5275.9 cubic inches
0
20
18
S
2 r y 1 r y 2 dy
0
1179.5 square inches −1
19 −1
52. Individual project, see Exercise 50, 51.
54. (a)
1 x9 , 0 ≤ x ≤ 3
x2 y2 1 9 4
1 x9
2
Ellipse: y1 2 y2 2
2
(b) y 2
1 x9
2
y 2
2x
1 9
x2
9
4
−5
2 1 2
121 x9 2x 9
5
L
3
0
1
2x 39 x2
4x2 dx 81 9x2
2
x2 + y = 1 9 4 −4
(c) You cannot evaluate this definite integral, since the integrand is not defined at x 3. Simpson’s Rule will not work for the same reason. Also, the integrand does not have an elementary antiderivative. 56. Essay
Section 6.5
Section 6.5
Work 4. W Fd 92000 2 5280 47,520,000 ft lb
2. W Fd 28004 11,200 ft lb
Work
1
b
6. W
Fx dx is the work done by a force F moving an object along a straight line from x a to x b.
a
9
8. (a) W
0
7
(b) W
10. W
0
9
20 dx
0
10
6 dx 54 ft lbs
5 5 2 x dx x 4 8
6
lb 3.33 ft lb
40 in
10x 90 dx 140 20
10
7
160 ft lbs
9
(c) W
0
1 2 x3 x dx 27 81
9
(d) W
9
2 32 x 3
x dx
0
9 ft lbs
0
9
0
2 27 18 ft lbs 3
12. Fx kx
14. Fx kx
800 k70 ⇒ k
70
W
80 7
70
Fx dx
0
0
16
kx dx
0
kx 2 2
524
W
540x dx 270x 2
16
4
80 40x 2 x dx 7 7
W2
70
16 0
524 16
15x dx 15x 2
0
0
4 0
240 ft lb
cm 280 Nm
28000 n
16. W 7.5
15 k1 k
k ⇒ k 540 72
h
18. W
80,000,000 80,000,000 dx 2 x x 4000
4.21875 ft lbs
h 4000
80,000,000 20,000 h
lim W 20,000 miton 2.1 1011 ft lb
h→
20. Weight on surface of moon: 16 12 2 tons Weight varies inversely as the square of distance from the center of the moon. Therefore, Fx 2
k x2 k 1100 2
k 2.42 10 6
1150
W
1100
2.42 10 6 2.42 10 6 dx x2 x
1150 1100
2.42 10 6
1 1 1100 1150
95.652 mi ton 1.01 10 9 ft lb
287
288
Chapter 6
Applications of Integration
22. The bottom half had to be pumped a greater distance then the top half. 24. Volume of disk: 4 y Weight of disk: 98004 y Distance the disk of water is moved: y W
12
y98004 dy 39,200
10
2
12
y2
10
39,200 22 862,400 newton–meters
26. Volume of disk:
23 y y 2
Weight of disk: 62.4
y 7
23 y y 2
5 4 3 2
Distance: y 4 (a) W 62.4 9 (b) W
4 62.4 9
x
2
y3
0
28. Volume of each layer:
4 1 dy 62.4 y 4 9 4
6
4
y 3 dy
2 0
49 62.4 14 y
6
4
4
110.9 ft lb
y3 3 y y 3 y 3
Distance: 6 y
−4 −3 −2 −1
2
4
3
y 4
(2, 3) 3
3
1
7210.7 ft lb
Weight of each layer: 55.6y 3 y
W
y
2
3
55.66 yy 3 dy 55.6
0
18 3y y2 dy
1
y = 3x − 3
0
x
55.6 18y
3y2 2
1
2
3
4
y3 3 3
0
3252.6 ft lb 30. Volume of layer: V 122254 y 2 y
y
Weight of layer: W 4224254 y 2 y
Ground
12
level
Distance:
6
19 y 2
19 25 1008 y 2 4 2.5
−y x
2.5
W
19 2
3 −9 −6 −3 −3
25 19 y2 y dy 4 2
4224
2.5
2.5
2.5
2
dy
2.5
3
6
9
−6
254 y y dy 2
The second integral is zero since the integrand is odd and the limits of integration are symmetric to the origin. The first integral represents the area of a semicircle of radius 52 . Thus, the work is W 1008
192 52 12 29,925 ft lb 94,012.16 ft lb. 2
Section 6.5
Work
32. The lower 10 feet of chain are raised 5 feet with a constant force. W1 3105 150 ft lb The top 5 feet will be raised with variable force. Weight of section: 3 y Distance: 5 y
5
W2 3
0
3 5 y dy 5 y 2 2
W W1 W2 150
5 0
75 ft lb 2
75 375 ft lb 2 2
34. The work required to lift the chain is 337.5 ft lb (from Exercise 31). The work required to lift the 500-pound load is W 50015 7500. The work required to lift the chain with a 100-pound load attached is W 337.5 7500 7837.5 ft lbs
6
36. W 3
0
3 12 2y dy 12 2y 2 4
6 0
3 12 2 108 ft lb 4
38. Work to pull up the ball: W1 50040 20,000 ft lb
40.
p
Work to pull up the cable: force is variable 2500
Weight per section: 1 y
40
0
k ⇒ k 2500 1
3
Distance: 40 x W2
k V
W
1 40 x dx 40 x 2 2
1
40
2500 dV 2500 ln V V
3 1
2500 ln 3 2746.53 ft lb
0
800 ft lb W W1 W2 20,000 800 20,800 ft lb 42. (a) W FD 80002 16,000 ft lbs (b) W
20 0 420,000 222,000 415,000 210,000 45000 0 36
24,88.889 ft lb (c) Fx 16,261.36x 4 85,295.45x 3 157,738.64x 2 104,386.36x 32.4675 25,000
0
2 0
(d) Fx 0 when x 0.524 feet. F x is a maximum when x 0.524 feet.
2
(e) W
0
4
44. W
0
Fx dx 25,180.5 ft lbs
ex 1 dx 11,494 ft lb 100 2
2
46. W
0
1000 sinh x dx 2762.2 ft lb
289
290
Chapter 6
Section 6.6 2. x
Applications of Integration
Moments, Centers of Mass, and Centroids
73 42 35 86 17 7438 11
4. x
126 14 62 30 118 0 12 1 6 3 11
8. 200x 5505 x (Person on left)
6. The center of mass is translated k units as well.
200x 2750 550x 750x 2750 x 323 feet
10.
101 25 54 0 x 10 2 5 y
12.
x
101 25 50 0 10 2 5
x, y 0, 0
y
y
15 6 152 2 93 62 15 40.5 27 15 12 6 2
122 61
15 8 152 2 96 64 15 40.5 27 15 12 6 2
123 65
8 6
m2 (5, 5)
4
m3 (− 4, 0)
x, y
2
64 , 62 27 27
y
−4
m3 (6, 8)
8 x
−2
m1 4 (1, −1)
−2
6
6
m2 (− 1, 5)
m1 (2, 3)
2
x −2
2 −2
2
14.
m
0
1 1 2 x 2 2
0
2
x
0
x
x, y
0
4 3
y
2
2
4 5 Mx 3 y m 4 5 3 My
2
12x dx 8 x
2
Mx
1 2 x3 x dx 2 6
1 1 2 x dx 2 2
2 My 3 m 4 2 3
32, 35
4
dx
40 x
2
5
0
0
32 4 40 5
2
1
( 32 , 35 ) x 1
2
0
x3 dx
8x
2
4
0
2
2
m4 (2, − 2)
6
8
Section 6.6
1
m
16.
x x dx
0
1
Mx
23 x
x x dx
2
2
1
0
1 0
6
y 1
x2 x3 dx 2 2 3
1
x
x2
0
12 0 1
3 4 1 2
Mx 1 6 m 12 2
My
x2 2
x x
0
y
3 2
Moments, Centers of Mass, and Centroids
( x, y )
1 4
1
x x x dx
x 3 2 x 2 dx
0
25 x
5 2
x3 3
1 0
15
x 1 4
My 2 6 x m 15 5
25 , 12
x, y
9
18. m
0
23x
3 2
9 x
Mx
0
2
9
0
x2 9 6
13x 1 dx x 31x dx 9
x 1
0
18
0
1 1 x1 1 3 x 1 x 1 dx 2 3 2
9 0
9
0
1 x x 1 x 1 dx 3
9
0
81 81 486 5 5
185, 25
y 5
(9, 4)
4 3
(185, 52 )
2 1 −2
(0, 1) x
−1
2
4
6
8
10
9
1 3
x x 2
0
9
0
x 31x dx
1 1 x x2 2 x dx 3 9
45 27 27 36 2 2 3
81 45 5 4 My 18 Mx 5 x ;y m 9 5 m 9 2 2 2
x, y
1 1 1 2 x x3 2 x3 2 x2 2 x x dx 3 3 9 3 2
x2 x3 4 x3 2 2 6 27 3
My
27 9 2 2
1 2 1 x3 2 x2 dx x5 2 x3 3 5 9
9 0
1 2
3 4
1
291
292
Chapter 6
20.
m 2
Applications of Integration
8
3 4 x 2 3 dx 2 4x x 5 3 5
0
8
0
128 5
y 12
By symmetry, My and x 0.
8
Mx 2
0
8
4 x 2 3 3 4 x 2 3 dx 16x x 7 3 2 7
8 0
512 5 20 y 7 128 7
20 7
x, y 0 ,
−8
m
2y y 2 dy y 2
2
0
y3 3
2
0
4 3
2y y 2 4y 3 y5 2y y 2 dy y4 2 2 3 5
2 0
8 15
2
( x, y ) 1
2
y 2y y 2 dy
0
2y 3 y 4 3 4
0
x
4 3
1
25 , 1
m
1
y 2 y 2 dy
y2 2y y3
3 2
2
1
9 2
2
y 2 y 2 y 2 y 2 dy 2 1 2
1
y 2 2 y 4 dy
2
y 2 3 y 5 2 3 5
2 1
36 5
My 36 2 8 m 5 9 5
x
1
y y 2 y 2 dy
2
85 , 12
4
A
1
Mx
1
4
My
x
1
1 dx ln x x 4
1 2
y3 y4 3 4
Mx 9 2 1 m 4 9 2
y
26.
1
2y y 2 y 3 dy y 2
4 1
ln 4
8 2 8
1 1 1 dx x2 2 x
1 dx x x
4 1
4 1
3
1
1
3
2 1
9 4
( x, y )
1
x 1 −1
2
Mx
y 3
2
My
x, y
2
2
24.
2
Mx 4 3 1 m 3 4
y
x, y
8
y
My 8 3 2 x m 15 4 5 Mx
4
0
My
x
−4 −4
2
22.
( x, y )
512 7
2
3
4
Section 6.6
2
28.
A Mx
2
2
x 2 4 dx 2
0
2
1 2
2
x 2 44 x 2 dx
2
1 x 5 8x 3 16x 2 5 3
4 x 2 dx 8x
2
1 2
2x 3 3
2 0
Moments, Centers of Mass, and Centroids
16
16 32 3 3
2
x 4 8x 2 16 dx
2
5
32
64 256 32 3 15
My 0 by symmetry.
4
30. m
3
xex 2 dx 2.3760
0
4
Mx
0
xex 2 xex 2 dx 2 2
( x, y )
4
dx 0.7619
x 2ex
−1 −1
4
My
5
0
x 2ex 2 dx 5.1732
0
x
My 2.2 m
y
Mx 0.3 m
Therefore, the centroid is 2.2, 0.3.
2
32. m
2
8 dx 6.2832 x2 4
2
Mx
1 8 2 2 2 x 4
x
3
8 dx 32 4
2
2
2
1 dx 5.14149 x 2 4 2
−3
3
Mx y 0.8 m
−1
x 0 by symmetry. Therefore, the centroid is 0, 0.8. 34.
A bh ac 1 1 A ac x
y
x, y
c
1 1 ac 2
1 2ac
y
0 c
0
y= c x b
b ya c
0
y
c 0
dy
( x, y ) y = c (x − a ) b x
(0, 0)
1 1 abc a 2c b a 2ac 2
b b ya y c c
(a + b, c)
(b, c)
2
c
b y c
2ab y a 2 dy c
1 ab 2 y a 2y 2ac c 1 ac
2
dy
1 y2 c 2
b 2 a , 2c
This is the point of intersection of the diagonals.
c 0
c 2
(a, 0)
293
294
Chapter 6
Applications of Integration
36. x 0 by symmetry
y
1 A r2 2
r
2 1 A r2 y
r
2 1 r2 2
r
1 x3 2 2 r x r 3
4r 3
x, y 0,
r r
1 4r 3 4r r2 3 3
1
38.
A
1 3
1 2x x 2 dx
0
1 3 A
1
x3
1 2x x 2 3 1 2x x 2 dx 2 2
0
x, y
3 2
x 2x 2 x 3 dx 3
0
1
1
x 1 2x x 2 dx 3
0
y3
x
r
r 2 x 2 2 dx
1
1 4x 2 4x 3 x 4 dx
0
x2 32 x 2
3
x4 4
1 0
1 4
1
1 2x x 2 2 dx
0
3 4 x5 x x3 x4 2 3 5
1 0
7 10
14, 107
40. (a) M y 0 by symmetry My
b b because there is more area above y 2 2 than below.
(b) y >
2n
b
2n b
x b x 2n dx 0
because bx x 2n1 is an odd function.
b x2nb x2n dx (c) M x 2 2n b
2n
b
1 2 x 4n1 b x 2 4n 1
b 2 b 1 2n
(d)
b
1 2 b x 4n dx 2 2n b
y
b 1 2n
2
3
4
y
5 9b
7 13 b
9 17 b
2n
b
(e) lim y lim
2n b
n→
x 2n1 A b x 2n dx 2 bx 2n 1 2n b 2 b
1 3 5b
b
b 4n1 2n 4n b 4n1 2n 4n 1 4n 1
2n
n
2n
n→
2n 1 1 b b 4n 1 2
(f) As n → , the figure gets narrower. y
2n
y = x 2n
b
0
b 2n1 2n 4n b 2n1 2n 2n 1 2n 1
Mx 4n b 4n1 2n 4n 1 2n 1 b A 4n b 24n1 2n 2n 1 4n 1
y=b x
−
2n
b
2n
b
Section 6.6
Moments, Centers of Mass, and Centroids
295
42. Let f x be the top curve, given by l d. The bottom curve is dx. x
0
0.5
1.0
1.5
2.0
f
2.0
1.93
1.73
1.32
0
d
0.50
0.48
0.43
0.33
0
2
(a) Area 2
(b) f x 0.1061x 4 0.06126x 2 1.9527
f x dx dx
0
2
dx 0.02648x 4 0.01497x 2 .4862
2 1.50 4 1.45 21.30 4 .99 0 34
1 13.86 4.62 3
y
(c)
Mx 4.9133 1.068 A 4.59998
x, y 0, 1.068
f x dx f x dx dx 2 2 2
Mx
2
f x 2 dx 2 dx
0
2 3.75 4 3.4945 22.808 4 1.6335 0 34
3
f
1 29.878 4.9797 6 y
d −2
Mx 4.9797 1.078 A 4.62
2 0
x, y 0, 1.078 44. Centroids of the given regions:
12, 32, 2, 12, and 72, 1
Area: A 3 2 2 7
x, y
33 2 21 2 21 15 2 15 7 7 14
2
m1 m3
m2 1
x 1
2
3
y
7 7 287 55 6 , P 0, 8 8 64 2 16
5 4 3
By symmetry, x 0. y
4
15 , 25 14 14
7 7 7 46. m1 2 , P1 0, 8 4 16 m2
4 3
31 2 22 27 2 25 2 25 x 7 7 14 y
y
2
7 47 16 287 6455 16 16,569 5523 7 4 287 64 6384 2128
x, y 0,
m2
5523 0, 2.595 2128
m1 − 23
−1
− 21
x 1 2
1
3 2
296
Chapter 6
Applications of Integration
48. Centroids of the given regions: 3, 0 and 1, 0
50. V 2 rA 2 34 24 2
Mass: 8 y0 x
x, y
81 3 8 3 8 8
883 , 0 1.56, 0
6
52.
A
2
6
My
4 2 x 2 dx x 23 2 3
6
2
32 3
y
6
6
x2 x 2 dx 2
2
x x 2 dx
Let u x 2, x u 2, du dx:
4
My 2
4
u 2 u du 2
0
645 323
2 x
2
5u
u3 2 2u1 2 du 2
0
(6, 4)
4
2
2
5 2
4 u3 2 3
( x, y )
4 0
x 2
4
6
704 15
My 704 15 22 A 32 3 5
rx
22 5
V 2rA 2
294.89 225323 1408 15
54. A planar lamina is a thin flat plate of constant density. The center of mass x, y is the balancing point on the lamina.
56. Let R be a region in a plane and let L be a line such that L does not intersect the interior of R. If r is the distance between the centroid of R and L, then the volume V of the solid of revolution formed by revolving R about L is V 2rA where A is the area of R. 58. The centroid of the circle is 1, 0. The distance traveled by the centroid is 2. The arc length of the circle is also 2. Therefore, S 22 4 2. y 2
C
1
d x
−1
1 −1 −2
3
Section 6.7
Section 6.7
Fluid Pressure and Fluid Force
Fluid Pressure and Fluid Force
2. F PA 62.4516 4992 lb
4. F 62.4h 448 62.4h48 62.4448 11,980.8 lb
6. h y 3 y
8. h y y L y 2 4 y 2
4 L y y 3 F 62.4
0
3
0
4 62.4 3
F 62.4
4 3 y y dy 3
62.4
3
3y y 2 dy
0
y2 4 y 2 dy
2
23 4 y
0
2 3 2
2
332.8 lb
y
y3 3
3y 2
4 62.4 3 2 3
374.4 lb
0
1 x −1
Force is one-third that of Exercise 5.
1
y −3
4
2 1 x −2
−1
1
2
10. h y y
y x
4 L y 9 y 2 3
−1
1 −1
0
F 62.4
−2
4 y 9 y 2 dy 3 3
62.4
0
2 3
3
−4
9 y 2 1 22y dy
4 62.4 9 y 2 3 2 9
0 3
748.8 lb
12. hy 1 3 2 y
y
1+ 3 2
L1y 2y lower part L2y 23 2 y upper part
3 2 2
F 29800
0
19,600
19,600
y2 2
1 3 2 yy dy
3 2y
y3 3 2 2 3
3 2 2
3
1 3 2 y3 2 y dy
3 2y 18y
0
9 2 2 1 9 2 1 4 4
44,1003 2 2 Newtons
3 2
3 2 3 2 2
y3 3
6 2 1 y 2
3 2
3 2 2
3 −3
−2
−1
x 1
2
3
297
298
Chapter 6
Applications of Integration 16. h y y
14. h y 6 y L y 1
L y 2
5
F 9800
16 y dy
y 9800 6y 2
0
2
0
0 2 5
43 9 y
F 140.7 171,500 Newtons
y
3
y2
43 9 y dy 2
140.74 3
140.73 4 23 9 y
6
0
3
9 y 2 2y dy 0
2 3 2
3
3376.8 lb
y 2 x −3 −2 −1
1
2
x
3
−2
2
−4 −6
18. h y y
y 1
5 L y 5 y 3
0
F 140.7
140.7
x −1
5 y 5 y dy 3 3
−2
0
−5
3
20. h y
4
6
(5, −3)
0 3
452 15 1055.25 lb
3 y 2
L y 2
12 9 4y
3 2
F 42
3
−4
5 5y y2 dy 3
5 5 140.7 y2 y3 2 9 140.7
−3
2
3 2
2
32 y 9 4y
3 2
2
dy 63
3 2
9 4y 2 dy
21 4
3 2
3 2
9 4y 2 8y dy
The second integral is zero since it is an odd function and the limits of integration are symmetric to the origin. 3 The first integral is twice the area of a semicircle of radius 2 .
9 4y 2 2 9 4 y 2 9 Thus, the force is 63 4 141.75 445.32 lb. 22. (a) F wk r2 62.47 22 1747.2 lbs (b) F wk r2 62.4532 2808 lbs
24. (a) F wkhb 62.4
112 35 5148 lbs
(b) F wkhb 62.4
175 510 10,608 lbs
Review Exercises for Chapter 6 26. From Exercise 21: F 64 15
12 753.98 lb 2
28. h y 3 y
30. h y 12 y
Solving y 5x 2 x 2 4 for x, you obtain x 4y 5 y.
L y 2
L y 2
716 y 2
2
716 y 2
4
F 62.4
4y 5y
3
F 62.4 2
y 3 y dy 546.265 lb 5y
3 y
0
3
2124.8
0
12 y716 y 2 dy
0
4y dy 5y
4
62.47
12 y16 y 2 dy 21373.7 lb
0
y
10 8
y
6 5 4 x −6 −4 −2
2
2
4
6
1 x −3 −2 −1 −1
1
2
3
32. Fluid pressure is the force per unit of area exerted by a fluid over the surface of a body.
34. The left window experiences the greater fluid force because its centroid is lower.
Review Exercises for Chapter 6
5
2. A
1 2
4x
1 x
5 1 2
1
1 dx x2
4
4. A
y 2 2y 1 dy
0 1
81 5
y 2 2y 1 dy
0
y
1
6
y 1 2 dy
0
5
( 21 , 4 )
(5, 4)
3
y 13 3
1 0
1 3
2
y
1
( 5, 251 ) 1
2
3
4
x
6 3 2
(−1, 1)
1 1 2
−2
− 23
− 21
x
299
300
Chapter 6
Applications of Integration
2
6. A
y 3 y 2 1 dy
1
8. A 2
2
1 1 2y y 2 y 3 2 3
2 1
2
6
2 csc x dx
2 y y 2 dy
1
2
6
2 2x ln csc x cot x
9 2
2 0
y
2
3
3 ln 2 3
23 ln 2 3 1.555
(5, 2) 2
y 3
1
( π6 , 2 )
x 2
3
4
5
( 5π6 , 2)
−1
(2, −1)
−2
1
10. A
5 3
3
12 cos y dy cos y 21 dy
x 2π 3
5π 6
7 3
5 3
y sin y 2
5 3
3
y sin y 2
7 3 5 3
23 3
12. Point of intersection is given by: x3 x2 4x 3 0 ⇒ x 0.783.
3 4x x 2 x 3 dx
0
1 1 3x 2x 2 x 3 x 4 3 4
3
0.783 0
1.189
( 21 , 7π3 ) ( 21 , 5π3 )
2
0.783
A
y
4
( 21 , π3 ) −2
π 2
π 3
π 6
−1
3
x
−1
−2 (.7832, .4804)
2
2
14. A 2
2x 2 x 4 2x 2 dx
16. y x 1 ⇒ x y 2 1
0 2
2
4x 2 x 4 dx
y
0
x1 ⇒ x 2y 1 2
2
4 1 2 x3 x5 3 5
2 0
128 8.5333 15
A
5
10
(−2, 8)
x 1
1
(2, 8)
−4
2y 1 y 2 1 dy
0
23 x 1
x1 dx 2
1 x 1 2 4
3 2
4 y −2
(0, 0) 3
(5, 2) 2 1 x
(1, 0) 2 −1 −2
3
4
5
5 1
4 3
Review Exercises for Chapter 6
1
18. A
5
2 dx
0
1
2 x 1 dx
301
y 5
x y2 1
4
2
A
3
y 2 1 dy
(5, 2)
0
3 y 1
3
y
2 0
14 3
1
(1, 0) x 2
20. (a) R1t 5.28341.2701t 5.2834 e0.2391t
3
4
5
(b) R2t 10 5.28 e0.2t
15
40
Difference
R1t R2t dt 171.25 billion dollars
10
0
10 0
22. (a) Shell
(b) Shell
2
V 2
y 3 dy
0
4 y 2
2 0
2
8
V 2
2 yy 2 dy
0 2
2
y
2y 2 y 3 dy
0
4
2
3
23 y
3
1 y4 4
2 0
8 3
2 y 1 4 x 1
2
3
4
3
1 x 1
(c) Disk
(d) Disk
2
V
y 4 dy
0
5 y 5
2 0
32 5
3
4
2
V
y 2 1 2 12 dy
0 2
y
2
y 4 2y 2 dy
0
4
3
15 y
5
2 y3 3
2 0
176 15
2 y 1
5 x 1
2
3
4
4 3 2 1 x 1
2
3
4
302
Chapter 6
24. (a) Shell
Applications of Integration (b) Disk
a
V 4
b x a 2 x 2 dx a
0
2 b a
a 2 x 2 1 22x dx
2 b 2 2 1 a x x3 a2 3
0
43ab a
2
x 2 3 2
0
b2 2 a x 2 dx a2
0
a
a
a
V 2
4 a 2b 3
V 2
1 1 x 2
0
2 arctan x 2
2 x2 + y =1 a2 b2
(0, b)
x
x
(a, 0)
(a, 0)
28. Disk
0
y 2 x2 + y =1 a2 b2
(0, b)
1
a
4 ab 2 3
y
26. Disk
1
2
dx
V
ex 2 dx
0 1
1
0
0
4 0
e2x dx e 2x 2
1 0
2e 2 2 1 e1 2
2 2
2
y
1
y 3 2
x 1
−2
x
−1
1
2
−1
30. (a) Disk
(b) Shell
0
V
1 0 1
x 2x 1 dx
u x 1
y
x u2 1
x3
x4 x3 4 3
dx
x2
12 1 0
x
−1
dx 2u du 0
V 2
1
x 2x 1 dx
−1
1
y
4
u 2 1 2 u 2du
0
x
−1
1
4
u 6 2u 4 u 2 du
0
−1
4
17 u
7
2 1 u5 u3 5 3
1 0
32 105
Review Exercises for Chapter 6 1 1 32. Ax bh 2a 2 x 2 3a 2 x 2 2 2
34.
3 a 2 x 2
3
a
1 x3 6 2x
1 1 y x 2 2 2 2x
a
V 3
y
303
a 2 x 2 dx 3 a 2x
x3 3
a a
4a3
1 y 2
3
12 x
3
s
1
2
1 2x 2
2
1 1 1 2 1 x 2 dx x 3 2 2x 6 2x
3 1
14 3
Since 43 a 3 3 10, we have a 3 53 2. Thus, a
5 2 3 1.630 meters. 3
2 a 2− x 2
2 a 2− x 2
2 a 2− x 2
36. Since f x tan x has f x sec 2 x, this integral represents the length of the graph of tan x from x 0 to x 4. This length is a little over 1 unit. Answers (b).
38. y 2x y
1 x
1 y 2 1
1 x1 x x
x x 1 d x 4
3
S 2
3
2x
0
4
40. F kx
0
3
3 2
0
56 3
8 dV 4 gal min 12 gal min ft3 min dt 7.481 gal ft3 7.481
50 9
50 x 9 9
W
23x 1
42. We know that
50 k9 ⇒ k F
x 1 dx
0
V r 2h
50 25 2 x dx x 9 9
225 in
9 0
lb 18.75 ft lb
19 h
dV dh dt 9 dt
dh 9 dV 9 8 3.064 ft min. dt dt 7.481 Depth of water: 3.064t 150 150 49 minutes 3.064 4912 588 gallons pumped
Time to drain well: t
Volume of water pumped in Exercise 41: 391.7 gallons 391.7 588 52 x x
58852 78 391.7
Work 78 ft ton
304
Chapter 6
Applications of Integration
44. (a) Weight of section of cable: 4 x Distance: 200 x
200
W4
200 x dx 2200 x 2
0
200 0
80,000 ft lb 40 ft ton
(b) Work to move 300 pounds 200 feet vertically: 200300 60,000 ft lb 30 ft ton Total work work for drawing up the cable work of lifting the load 40 ft ton 30 ft ton 70 ft ton
b
46.
W
Fx dx
a
Fx
x 6, 2 9 4 3x 16,
9
W
0 ≤ x ≤ 9 9 ≤ x ≤ 12
12
2 x 6 dx 9
0
9
32 x
1 x 2 6x 9
9
2
0
34 x 16 dx
16x
12 9
9 54 96 192 54 144 51 ft lbs
3
48.
A
3 1 A 32 x
1 2x 3 x 2 dx x 2 3x x 3 3 1
3 32
3
1
x2x 3 x 2 dx
323 12
3 32
1
17 x, y 1, 5
3 1
32 3
3x 2x 2 x 3 dx
1
2x 3 2 x 4 dx
3 4 1 9x 6x 2 x 3 x 5 64 3 5
1
3
3
y
3
3 64
3 3 2 2 3 1 4 x x x 32 2 3 4
3
1
1
3
1
y
9 12x 4x 2 x 4 dx 9
17 5
6
( x, y )
3
x
−3
8
50.
A
0
5 1 A 16 x
5 16
8
16 5
y 6
0
( x, y )
1 x x 2 3 x dx 2
165 12 x 8
4 3
0
x, y
0
2
8
6
4
5 3 8 3 1 3 x x 16 8 6 y
3 1 1 x 2 3 x dx x 5 3 x 2 2 5 4
3
37 x
1 5 2 16
40 103 , 21
7 3
8
x 2 −2
10 3
0
1 x 2 dx 4
1 3 x 12
8 0
40 21
4
6
8
Review Exercises for Chapter 6 26. From Exercise 21: F 64 15
12 753.98 lb 2
28. h y 3 y
30. h y 12 y
Solving y 5x 2 x 2 4 for x, you obtain x 4y 5 y.
L y 2
L y 2
716 y 2
2
716 y 2
4
F 62.4
4y 5y
3
F 62.4 2
y 3 y dy 546.265 lb 5y
3 y
0
3
2124.8
0
12 y716 y 2 dy
0
4y dy 5y
4
62.47
12 y16 y 2 dy 21373.7 lb
0
y
10 8
y
6 5 4 x −6 −4 −2
2
2
4
6
1 x −3 −2 −1 −1
1
2
3
32. Fluid pressure is the force per unit of area exerted by a fluid over the surface of a body.
34. The left window experiences the greater fluid force because its centroid is lower.
Review Exercises for Chapter 6
5
2. A
1 2
4x
1 x
5 1 2
1
1 dx x2
4
4. A
y 2 2y 1 dy
0 1
81 5
y 2 2y 1 dy
0
y
1
6
y 1 2 dy
0
5
( 21 , 4 )
(5, 4)
3
y 13 3
1 0
1 3
2
y
1
( 5, 251 ) 1
2
3
4
x
6 3 2
(−1, 1)
1 1 2
−2
− 23
− 21
x
299
300
Chapter 6
Applications of Integration
2
6. A
y 3 y 2 1 dy
1
8. A 2
2
1 1 2y y 2 y 3 2 3
2 1
2
6
2 csc x dx
2 y y 2 dy
1
2
6
2 2x ln csc x cot x
9 2
2 0
y
2
3
3 ln 2 3
23 ln 2 3 1.555
(5, 2) 2
y 3
1
( π6 , 2 )
x 2
3
4
5
( 5π6 , 2)
−1
(2, −1)
−2
1
10. A
5 3
3
12 cos y dy cos y 21 dy
x 2π 3
5π 6
7 3
5 3
y sin y 2
5 3
3
y sin y 2
7 3 5 3
23 3
12. Point of intersection is given by: x3 x2 4x 3 0 ⇒ x 0.783.
3 4x x 2 x 3 dx
0
1 1 3x 2x 2 x 3 x 4 3 4
3
0.783 0
1.189
( 21 , 7π3 ) ( 21 , 5π3 )
2
0.783
A
y
4
( 21 , π3 ) −2
π 2
π 3
π 6
−1
3
x
−1
−2 (.7832, .4804)
2
2
14. A 2
2x 2 x 4 2x 2 dx
16. y x 1 ⇒ x y 2 1
0 2
2
4x 2 x 4 dx
y
0
x1 ⇒ x 2y 1 2
2
4 1 2 x3 x5 3 5
2 0
128 8.5333 15
A
5
10
(−2, 8)
x 1
1
(2, 8)
−4
2y 1 y 2 1 dy
0
23 x 1
x1 dx 2
1 x 1 2 4
3 2
4 y −2
(0, 0) 3
(5, 2) 2 1 x
(1, 0) 2 −1 −2
3
4
5
5 1
4 3
Review Exercises for Chapter 6
1
18. A
5
2 dx
0
1
2 x 1 dx
301
y 5
x y2 1
4
2
A
3
y 2 1 dy
(5, 2)
0
3 y 1
3
y
2 0
14 3
1
(1, 0) x 2
20. (a) R1t 5.28341.2701t 5.2834 e0.2391t
3
4
5
(b) R2t 10 5.28 e0.2t
15
40
Difference
R1t R2t dt 171.25 billion dollars
10
0
10 0
22. (a) Shell
(b) Shell
2
V 2
y 3 dy
0
4 y 2
2 0
2
8
V 2
2 yy 2 dy
0 2
2
y
2y 2 y 3 dy
0
4
2
3
23 y
3
1 y4 4
2 0
8 3
2 y 1 4 x 1
2
3
4
3
1 x 1
(c) Disk
(d) Disk
2
V
y 4 dy
0
5 y 5
2 0
32 5
3
4
2
V
y 2 1 2 12 dy
0 2
y
2
y 4 2y 2 dy
0
4
3
15 y
5
2 y3 3
2 0
176 15
2 y 1
5 x 1
2
3
4
4 3 2 1 x 1
2
3
4
302
Chapter 6
24. (a) Shell
Applications of Integration (b) Disk
a
V 4
b x a 2 x 2 dx a
0
2 b a
a 2 x 2 1 22x dx
2 b 2 2 1 a x x3 a2 3
0
43ab a
2
x 2 3 2
0
b2 2 a x 2 dx a2
0
a
a
a
V 2
4 a 2b 3
V 2
1 1 x 2
0
2 arctan x 2
2 x2 + y =1 a2 b2
(0, b)
x
x
(a, 0)
(a, 0)
28. Disk
0
y 2 x2 + y =1 a2 b2
(0, b)
1
a
4 ab 2 3
y
26. Disk
1
2
dx
V
ex 2 dx
0 1
1
0
0
4 0
e2x dx e 2x 2
1 0
2e 2 2 1 e1 2
2 2
2
y
1
y 3 2
x 1
−2
x
−1
1
2
−1
30. (a) Disk
(b) Shell
0
V
1 0 1
x 2x 1 dx
u x 1
y
x u2 1
x3
x4 x3 4 3
dx
x2
12 1 0
x
−1
dx 2u du 0
V 2
1
x 2x 1 dx
−1
1
y
4
u 2 1 2 u 2du
0
x
−1
1
4
u 6 2u 4 u 2 du
0
−1
4
17 u
7
2 1 u5 u3 5 3
1 0
32 105
Review Exercises for Chapter 6 1 1 32. Ax bh 2a 2 x 2 3a 2 x 2 2 2
34.
3 a 2 x 2
3
a
1 x3 6 2x
1 1 y x 2 2 2 2x
a
V 3
y
303
a 2 x 2 dx 3 a 2x
x3 3
a a
4a3
1 y 2
3
12 x
3
s
1
2
1 2x 2
2
1 1 1 2 1 x 2 dx x 3 2 2x 6 2x
3 1
14 3
Since 43 a 3 3 10, we have a 3 53 2. Thus, a
5 2 3 1.630 meters. 3
2 a 2− x 2
2 a 2− x 2
2 a 2− x 2
36. Since f x tan x has f x sec 2 x, this integral represents the length of the graph of tan x from x 0 to x 4. This length is a little over 1 unit. Answers (b).
38. y 2x y
1 x
1 y 2 1
1 x1 x x
x x 1 d x 4
3
S 2
3
2x
0
4
40. F kx
0
3
3 2
0
56 3
8 dV 4 gal min 12 gal min ft3 min dt 7.481 gal ft3 7.481
50 9
50 x 9 9
W
23x 1
42. We know that
50 k9 ⇒ k F
x 1 dx
0
V r 2h
50 25 2 x dx x 9 9
225 in
9 0
lb 18.75 ft lb
19 h
dV dh dt 9 dt
dh 9 dV 9 8 3.064 ft min. dt dt 7.481 Depth of water: 3.064t 150 150 49 minutes 3.064 4912 588 gallons pumped
Time to drain well: t
Volume of water pumped in Exercise 41: 391.7 gallons 391.7 588 52 x x
58852 78 391.7
Work 78 ft ton
304
Chapter 6
Applications of Integration
44. (a) Weight of section of cable: 4 x Distance: 200 x
200
W4
200 x dx 2200 x 2
0
200 0
80,000 ft lb 40 ft ton
(b) Work to move 300 pounds 200 feet vertically: 200300 60,000 ft lb 30 ft ton Total work work for drawing up the cable work of lifting the load 40 ft ton 30 ft ton 70 ft ton
b
46.
W
Fx dx
a
Fx
x 6, 2 9 4 3x 16,
9
W
0 ≤ x ≤ 9 9 ≤ x ≤ 12
12
2 x 6 dx 9
0
9
32 x
1 x 2 6x 9
9
2
0
34 x 16 dx
16x
12 9
9 54 96 192 54 144 51 ft lbs
3
48.
A
3 1 A 32 x
1 2x 3 x 2 dx x 2 3x x 3 3 1
3 32
3
1
x2x 3 x 2 dx
323 12
3 32
1
17 x, y 1, 5
3 1
32 3
3x 2x 2 x 3 dx
1
2x 3 2 x 4 dx
3 4 1 9x 6x 2 x 3 x 5 64 3 5
1
3
3
y
3
3 64
3 3 2 2 3 1 4 x x x 32 2 3 4
3
1
1
3
1
y
9 12x 4x 2 x 4 dx 9
17 5
6
( x, y )
3
x
−3
8
50.
A
0
5 1 A 16 x
5 16
8
16 5
y 6
0
( x, y )
1 x x 2 3 x dx 2
165 12 x 8
4 3
0
x, y
0
2
8
6
4
5 3 8 3 1 3 x x 16 8 6 y
3 1 1 x 2 3 x dx x 5 3 x 2 2 5 4
3
37 x
1 5 2 16
40 103 , 21
7 3
8
x 2 −2
10 3
0
1 x 2 dx 4
1 3 x 12
8 0
40 21
4
6
8
Problem Solving for Chapter 6 54. F 62.4165 4992 lb
52. Wall at shallow end:
5
F 62.4
y20 dy 1248
0
y2 2
5
15,600 lb
0
Wall at deep end:
10
F 62.4
y20 dy 624 y 2
0
10
62,400 lb
0
Side wall:
5
F1 62.4
y 40 dy 1248 y 2
0 5
F2 62.4
5
31,200 lb
0
5
10 y 8y dy 62.4
0
80y 8y 2 dy
0
F F1 F2 72,800 lb y 20 15 10 5 x
−5
5 10 15 20 25 30 35 40 45
Problem Solving for Chapter 6
1
2. R
x1 x dx
0
x2 x3 2
3 1
0
1 1 1 2 3 6
Let c, mc be the intersection of the line and the parabola.
4. 8y2 x21 x2 y±
x 1 x2 2 2
Then, mc c1 c ⇒ m 1 c or c 1 m.
1 1 2 6
y
1m
x x2 mx dx
0.5
0
0.25
1 x2 x3 x2 m 12 2 3 2
1m 0
−1.5
x
− 0.5
0.5
1.5
− 0.25
1 m2 1 m3 1 m2 m 2 3 2
− 0.5
1 61 m2 41 m3 6m1 m2 1 m26 41 m 6m 1 m22 2m
13
12
S 22
13
x
0
1m
m1
1 2x2 2 2 1 x2
1
1 1 m3 2
12
For x > 0, y
0.2063
5 2 3
1
1 2x2 2 2 1 x2
dx 2
305
306
Chapter 6
Applications of Integration 8. fx2 ex
6. By the Theorem of Pappus,
fx e x2
V 2 r A
f x 2e x2 C
1 2 d w2 l2 lw 2
f 0 0 ⇒ C 2 f x 2e x2 2
10. Let f be the density of the fluid and 0 the density of the iceberg. The buoyant force is
0
F f g
h
Ay dy
where Ay is a typical cross section and g is the acceleration due to gravity. The weight of the object is
Lh
W 0 g
h
Ay dy.
FW
0
f g
h
Lh
Ay dy 0 g
h
Ay dy
0 submerged volume 0.92 103 0.893 or 89.3% f total volume 1.03 103 12. (a) y 0 by symmetry
6
My 2
x
1 6
m2
1
y
6
1 dx 2 x4
1
b
1 b
m2
1
2
−1
lim x
x −1 −2
63 x, y ,0 43
−3
2
3
4
5
y = − 14 x
b2 1 1 dx 3 x b2 1 2b3 1 dx 4 x 3b3
b2 1b2 3bb 1 x 2b3 13b3 2b2 b 1 b→
y = 14 x
1
215 1 dx x4 324
3536 63 x 215324 43 (b) My 2
3
1 35 dx x3 36
3 2
x, y
14. (a) Trapezoidal: Area (b) Simpson’s: Area
x, y
2b3bb b 11, 0 2
32, 0
160
0 250 254 282 282 273 275 280 0 9920 sq ft 28
160
0 450 254 482 282 473 275 480 0 10,41313 sq ft 38
16. Point of equilibrium: 1000 0.4x2 42x x 20, p 840
P0, x0 840, 20
20
1000 0.4x2 840 dx 2133.33
Consumer surplus
0
20
Producer surplus
0
840 42x dx 8400
PA R T
I C H A P T E R 6 Applications of Integration Section 6.1
Area of a Region Between Two Curves
. . . . . . . . . . .2
Section 6.2
Volume: The Disk Method . . . . . . . . . . . . . . . . . . 9
Section 6.3
Volume: The Shell Method . . . . . . . . . . . . . . . . . 17
Section 6.4
Arc Length and Surfaces of Revolution . . . . . . . . . . . 22
Section 6.5
Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
Section 6.6
Moments, Centers of Mass, and Centroids . . . . . . . . . 30
Section 6.7
Fluid Pressure and Fluid Force
. . . . . . . . . . . . . . . 37
Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
C H A P T E R 6 Applications of Integration Section 6.1
Area of a Region Between Two Curves
Solutions to Odd-Numbered Exercises
0 x 2 6x dx
x 2 2x 3 x 2 4x 3 dx
6
1. A
6
0
x 2 6x dx
0
3
3. A
3
0
0
1
0
3x 3 x dx 6
4
7.
2x 2 6x dx
0
5. A 2
0
6
or
x 3 x dx
0
6
x dx 2
x 1
1
1
x 3 x dx
9.
4 2x 3
0
y
3
x dx 6
11.
3
2 sec x dx y
y 6
5
3
5
4 3
3 2
2
1
2π 3
1
x 1
2
3
4
x
5
1
2
3
4
5
6
2
13. f x x 1 gx x 1
15. A
0
2
2
A 4
0
Matches (d)
1 3 x 2 x 1 dx 2
1 3 x x 1 dx 2
x8 x2 x 4
2
2
0
y
(3, 4)
3
168 24 2 0 2
2
y
(2, 6)
6 5
(0, 1) 4 3
x 1
2
1 −2
2
(2, 3)
(0, 2)
3
(0, 1) x 1
3
4
π
π
3
3
−1
2π 3
x
Section 6.1 17. The points of intersection are given by:
Area of a Region Between Two Curves
19. The points of intersection are given by:
x 2 4x 0
x 2 2x 1 3x 3
xx 4 0 when x 0, 4
x 2x 1 0 when x 1, 2
4
A
0
2
gx f x dx
A
1
4
2
x 2 4x dx
1
0
gx f x dx
x3 2x 3
3x 3 x 2 2x 1 dx
2
4
2
0
1
2 x x 2 dx
32 3
x2 x3 2 3
2x
2 1
9 2
y
(0, 0)
y
(4, 0)
1
2
3
x
5
10
−1
(2, 9)
8
−2
6
−3
4
( 1, 0)
−4
x
−5
4
21. The points of intersection are given by:
3
x0
1
A
2 y y dy 2y y 2
0
2
3x 1 x 1
x2
1
23. The points of intersection are given by:
x 2 x and x 0 and 2 x 0 x1
2
1 0
3x x when x 0, 3
3
1
A
f x gx dx
0
3
Note that if we integrate with respect to x, we need two integrals. Also, note that the region is a triangle.
3x 1 x 1 dx
0
3
y
3x1 2 x dx
0
3
2
29 3x
(1, 1)
1
3 2
x2 2
3 0
y
(2, 0) x
(0, 0)
2
5
3
4
(3, 4)
3 2
(0, 1) x
−2
25. The points of intersection are given by: y2 y 2
1
1
(4, 2)
1
x 1
g y f y dy
1
(1,
y 2 y 2 dy
2y
y2 y3 2 3
2 1
2
1
2
4
3
2
A
3
y
2
y 2 y 1 0 when y 1, 2
2
3
9 2
1)
3
4
5
3 2
3
4
Chapter 6
Applications of Integration
2
27. A
f y g y dy
1
29. y
10 10 ⇒ x x y
10 dy y
10
2
A
y 2 1 0 dy
1
2
y3 y 3
2 1
6
2
10ln 10 ln 2
y
10 ln 5 16.0944
3
(0, 2)
10
10 ln y
(5, 2)
y 1 12
x
2
3
4
5
(0, −1)
(0, 10)
6
(2, −1)
(1, 10)
8
2
6 4
(0, 2) −4
(5, 2) x
−2
2
4
6
8
31. The points of intersection are given by: x 3 3x 2 3x x 2
11
xx 1x 3 0 when x 0, 1, 3
1
A
3
f x gx dx
0
gx f x dx
1
1
1
(0, 0)
(1, 1) −1
x 2 x 3 3x 2 3x dx
1
3
x 3 4x 2 3x dx
0
−6
3
x 3 3x 2 3x x 2 dx
0
(3, 9)
x 3 4x 2 3x dx
1
x4 34 x 4
3
x4
3 2 x 2
1
4
0
3 4 x3 x2 3 2
3 1
37 12
Numerical Approximation: 0.417 2.667 3.083 33. The points of intersection are given by: x 4x 3 3 4x x 2
2xx 4 0 when x 0, 4
4
A
3 4x x 2 x 2 4x 3 dx
0 4
2x 2 8x dx
0
2x 3 4x 2 3
4 0
9
2
64 3
Numerical Approximation: 21.333
(0, 3) −6
(4, 3) 12
−3
12
Section 6.1
Area of a Region Between Two Curves
35. f x x 4 4x 2, gx x 2 4
2
The points of intersection are given by:
−4
x 4 4x 2 x 2 4
(− 2, 0)
x 4 5x 2 4 0
4
(2, 0)
(−1, − 3)
(1, − 3) −5
x 2 4x 2 1 0 when x ± 2, ± 1 By symmetry,
1
A2
0 1
2
2
x 4 4x 2 x 2 4 dx 2
x 2 4 x 4 4x 2 dx
1
2
x 4 5x 2 4 dx 2
0
x 4 5x 2 4 dx
1
2
5
2
5 3 4 2 5
x5 1
1
5x 3 4x 3
0
2
5
32
x 5 5x 3 4x 5 3
2 1
40 1 5 8 4 3 5 3
8.
Numerical Approximation: 5.067 2.933 8.0
39. 1 x 3 ≤
37. The points of intersection are given by: 1 x2 1 x2 2
x2
2
x2
Numerical approximation: 1.759
3
1 0
A
0
−3
5
x ±1
−1
0
−1
1 x2 dx 1 x2 2
2 arctan x 2
(0, 1)
−4
0
2
x3 6
1 0
4 16 2 13 1.237
Numerical Approximation: 1.237
41. A 2
3
f x gx dx
y
0
2
3
g 4
2 sin x tan x dx
π , 3
3
0
(2, 3)
(0, 2)
f x gx dx
1
1 x 2 1 x 3 dx 1.759 2
3
1
A2
2
(−1, ( ( 1, 12 ( 1 2
x4 x2 2 0
1 x 2 on 0, 2 2
2
3
0
2 2 cos x ln cos x
3 f
1
π
(0, 0)
2
x
π 2
21 ln 2 0.614 3 4
π , 3
3
5
5
6
Chapter 6
43. A
2
Applications of Integration
45. A
xex 0 dx 2
0
0
2
2
1
2 cos x cos x dx
0
2
2 x sin x
1 2 ex 2
1 cos x dx 4 12.566
1 0
1 1 1
0.316 2 e
y
0
y
1
3
(1, e1 )
(0, 1) 2
g
(2π, 1)
(0, 0)
x 1
f π 2
−1
47. A
π
x
2π
2 sin x sin 2x 0 dx
1 cos 2x 2
1
0
2 cos x
0
1 1 x e 0 dx x2
e 1 x
4.0
3 1
e e 1 3 1.323
4
3
( π, 0)
(0, 0)
0
3
49. A
0
4 x x , y 0, x (b) A dx, 4x 3
51. (a) y
3
6 0
0
x3
3
0
No, it cannot be evaluated by hand. (c) 4.7721
6
−1
4 −1
x
53. Fx
0
1 t2 t 1 dt t 2 4
(a) F0 0
x 0
x2 x 4 (b) F2
22 23 4
(c) F6
62 6 15 4
y y
y 6 5 4 3 2
−1 −1
6
6
5
5
4
4
3
3
2
2
t 1
2
3
4
5
6
−1 −1
t 1
2
3
4
5
6
−1 −1
t 1
2
3
4
5
6
Section 6.1
55. F
cos
1
2 d sin 2 2
1
Area of a Region Between Two Curves
2 2 sin 2 (b) F0
(a) F1 0
2
0.6366
(c) F
12 2
2
1.0868
y y 3 2
1 2
− 21
c
57. A
0
c
1 2
1 2
− 21
− 21
1 2
1
θ − 21
ba b y a y dy c c
a 2 y ay 2c
(b, c)
y= cx b
y=
c b − a (x − a )
c 0
ac ac ac 2 2
x
1 baseheight 2
(0, 0)
(a, 0)
59. f x x 3
y 8
f x 3x 2
6
2
Tangent line:
2
x 3 3x 2 dx
1
2
4
3
f (x) = x3 −6
(− 2, − 8)
The tangent line intersects f x x 3 at x 2. 1
(1, 1) x
−4 −3 −2
y 1 3x 1 or y 3x 2
y = 3x − 2
4
At 1, 1, f 1 3.
A
− 21
y
a y a dy c
0
3 2
θ
1
1 2
− 21
y
3 2
x4 3x2 4
2
2x
1 2
−8
27 4
61. The variable is y. 63. x 4 2x 2 1 ≤ 1 x 2 on 1, 1
y
1
A
1
1 x 2 x 4 2x 2 1 dx
2
(0, 1)
1
1
x2 x 4 dx
x3 x5 3 5
x
1 1
4 15
You can use a single integral because x 4 2x 2 1 ≤ 1 x 2 on 1, 1. 65. Offer 2 is better because the accumulated salary (area under the curve) is larger.
( 1, 0)
(1, 0)
1 2
1
θ
7
8
Chapter 6
Applications of Integration
3
A
67.
3
9 x 2 dx 36
y 10
9b
9b
9b
9 x 2 b dx 18
6 4
9 b x 2 dx 9
2
x
0
6
9 bx 3 x3
9b
0
(
9
2
9
2
b, b)
(
x
x2
6
b, b)
9
2 9 b 3 2 9 3
9 b 3 2 9b b9
n
x
69. lim
→0 i1
i
9 3
4
27 2 9 3
4
3.330
x i2 x
y
0.6
i 1 where xi and x is the same as n n
1
x
0
x2 x3 dx 2 3
x2
(1, 0) x
1 . 6 0
0.2 0.4
0.13t dt
0
73. (a) y1 275.06751.0537t 275.0675e0.0523t] 460
10
15
10
0.13t 2
2 5 0
$1.625 billion
(b) y2 239.94071.0417t 239.9407e0.0408t 460
0 240
0.6 0.8 1.0
(0, 0)
5
7.21 0.58t 7.21 0.45t dt
0
(c)
f (x)
0.2
1
5
71.
0.4
y1 y2 dt 649.5 billion dollars
0 240
10
(d) No, model y1 > y2 forever because 1.0537 > 1.0417. No, these models are not accurate. According to news reports, E > R eventually.
Section 6.2
Volume: The Disk Method
75. The total area is 8 times the area of the shaded region to the right. A point x, y is on the upper boundary of the region if
9
y
2
y=x
x 2 y 2 2 y
x 2 y 2 4 4y y 2
1
( x, y )
x 2 4 4y
x 1
4y 4 x 2 y1
2
x2 . 4
We now determine where this curve intersects the line y x. x1
x2 4
x 2 4x 4 0 x Total area 8
4 ± 16 16 2 ± 22 ⇒ x 2 22 2
222
1 x4 x dx 2
0
8 x
x3 x2 12 2
222 0
16 42 5 80.4379 3.503 3
1 31 5 x dx 5
77. (a) A 2
5.5
0
1 0 dx
5
x 92 5 x x 2 5 109 5 5.5 5 6.031 m
2
5
5.5
0
5
3 2
(b) V 2A 26.031 12.062 m 3
2
(c) 5000 V 500012.062 60,310 pounds
79. True 81. False. Let f x x and gx 2x x2. f and g intersect at 1, 1, the midpoint of 0, 2 . But
b
2
f x gx dx
a
x 2x x2 dx
0
Section 6.2
Volume: The Disk Method
1
1. V
1
x 1 2 dx
0
x
1
2 dx
4
x dx
1
1
5. V
0
7. y x 2 ⇒ x y
4
y
0
2 y2
2 x2
4 1
4 0
2 dy
8
x3 x
3
x
2
1 0
3
15 2
1
x 2 2 x 3 2 dx
0
V
x 2 2x 1 dx
0
4
3. V
2 0. 3
x 4 x 6 dx
x5 x7 5
7 1 0
2 35
9. y x 2 3 ⇒ x y 3 2
4
0
1
y dy
V
0
1
y 3 2 2 dy
0
y 3 dy
4 y4
1 0
4
10
Chapter 6
Applications of Integration
11. y x, y 0, x 4 (b) R y 4, r y y 2
(a) Rx x, rx 0
4
V
x
0
4
x dx
0
2
2 dx
V
16 y 4 dy
0
2 x
4
0
1 16y y 5 5
8
2
y
2 0
128 5
y
3
3
2
2
1
1 x 1
2
3
x
4
1
−1
2
3
4
−1
(c) R y 4 y 2, r y 0
(d) R y 6 y 2, r y 2
2
V
4 y 2 2 dy
6 y 2 2 4 dy
0
2
2
V
0
16 8y 2 y 4 dy
2
0
8 1 16y y 3 y 5 3 5
2 0
32 12y 2 y 4 dy
0
256 15
1 32y 4y 3 y 5 5
2 0
192 5
y y 3 4 3
2
2 1 1 x 1
2
x
3
1
2
3
4
5
−1
−1
−2
13. y x2, y 4x x 2 intersect at 0, 0 and 2, 4. (a) Rx 4x x 2 rx x 2
(b) Rx 6 x2, rx 6 4x x 2
2
V
V
0
2
16x 2 8x 3 dx
8
0
6 x 2 2 6 4x x 2 2 dx
0
2
2
4x x 2 2 x 4 dx
163 x
x 3 5x 2 6x dx
0
3
2x 4
2 0
32 3
8
x4 3 x
y
4
5
3
3x 2
y
4 5 3 4 3
2
2 1 1 −1
x 1
2
3
−2
−1
x 1
2
3
4
2 0
64 3
Section 6.2
17. Rx 4, rx 4
15. Rx 4 x, rx 1
3
V
4 x 2 1 2 dx
0
3
42 4
0
x 2 8x 15 dx
3
0
V
3
3 4x x3
2
Volume: The Disk Method
3
15x
0
0
18
1 1x 1 1x
dx 2
8 1 dx 1 x 1 x2
8 ln1 x
1 1x
y
5
8 ln 4
3
8 ln 4
2
1 1 4
3 32.485 4
y 1 x
−1
1
2
3
4
3 2 1 x
−1
1
2
3
4
2
3
4
5
−1
19. R y 6 y, r y 0
y 5
4
V
6 y 2 dy
4 3
0 4
y 2 12y 36 dy
2 1
0
x
y3 6y 2 36y 3
1
4
−1
0
208 3
21. R y 6 y 2, r y 2
23. Rx
2
V
6 y 2 2 2 2 dy
2
3
y4 12y 2 32 dy
0
0
y5 4y 3 32y 2 5
3
V
0
2
2
1 x 1
384 5
2
, rx 0
1 x 1
2
1 dx x1
0 ln 4
ln x 1
0
dx 3
y
2
1
x 1 −1
2
3
3 0
11
12
Chapter 6
Applications of Integration
1 25. Rx , rx 0 x
4
V
1 x
1
1 x
27. Rx ex, rx 0
1
V
2
dx
ex 2 dx
0 1
4
e2x dx
0
1
e2x 2
3 4
y
2
1 0
1 e2 1.358 2
y 1
2 x 1
2
3 1
−1
x 1
2
0 2
6
0
9
1 6 y 3
1 6 y 3
2
3 2
2
−1
33. V
sin x 2 dx 4.9348
0
y
dy 3
6
36 12y y 2 dy
2
0
6
8 y 6 5 4 3 2 1 x 3
4
5
6
1
0
216 216 216 9 3
(2, 5)
6
8 x3 x3 10x2 24x 3 0
2
y3 36y 6y 2 9 3
1
8
4x3 8x2 20x 24 dx
152 125 277 3 3 3
31. y 6 3x ⇒ x
y 10
2
8 x4 x3 10x2 24x 3
V
x2 12 5 2x x22 dx
2
3
4x3 8x2 20x 24 dx
0
3
5 2x x22 x2 12 dx
29. V
2
x 1
2
3
x −2
1
2
3
4
Section 6.2
2
35. V
Volume: The Disk Method
2
ex 2 dx 1.9686 2
37. V
0
39. A 3
1
e x 2 ex 2 2 dx 49.0218
41. Disk Method:
Matches (a)
b
V
y
d
Rx 2 dx or V
a
Ry 2 dy
c
Washer Method:
2
b
V
1
Rx 2 rx 2 dx or
a
d
V
x 1
2
Ry 2 ry 2 dy
c
43.
y
y 4
−2
3
3
2
2
1
1 x
−1
x
2
1
1
2
3
4
The volumes are the same because the solid has been translated horizontally. 1 45. Rx x, rx 0 2
r
6
V
47. Rx r 2 x 2, rx 0
0
V
1 2 x dx 4
r
r 2 x 2 dx
r
3 x 12
6 0
2
18
r 2 x 2 dx
0
1 2 r 2x x 3 3
1 Note: V r 2h 3
r 0
1 4 2 r 3 r 3 r 3 3 3
1 326 3
y
18
y = r2 − x2
y 4 3 2
(−r, 0)
1 x 1 −1 −2
2
3
4
5
6
x
(r, 0)
13
14
Chapter 6
Applications of Integration
h
V
r y y yr 1 , R y r 1 , r y 0 H H H
49. x r
r 1
0
y H
2
h
dy r 2
1
0
h2 h3 H 3H 2
r 2h 1
2
51. V
0
1 2 x 2 x 8
53. (a) Rx V
2
dx
1 2 1 3 y y H 3H 2
r2 h
64
H
2 1 y 2 y 2 dy H H
r2 y
y
h
h 0
−r
h h2 H 3H 2
2
x 4 2 x dx
0
2x 5 x 6 64 5 6
3 25 x 2 , rx 0 5
9 25
2 0
18 25
25 x 2 dx
V
5
25 x 2 dx
0
18 x3 25x 25 3
5 0
30
(b) R y
5
5
x
r
5 9 y 2, r y 0, x ≥ 0 3
25 9
3
9 y 2 dy
0
25 y3 9y 9 3
3 0
50
y
60 6
y 4 8 6
2
4 x 2 −6
−4
2
6
4
x
−2
2
4
6
−2 −4
55. Total volume: V
4 503 500,000 3 ft 3 3
y 60 40
Volume of water in the tank:
y0
2500
2 y2
50
dy
2500 y 2 dy
y3 3
20
y0
−60
50
2500y
2500y0
y03 3
−40
y0
−60
50
250,000 3
When the tank is one-fourth of its capacity: y 3 250,00 1 500,000 2500y0 0 4 3 3 3 125,000 7500y0 y03 250,000
−20
y03 7500y0 125,000 0 y0 17.36 Depth: 17.36 50 32.64 feet When the tank is three-fourths of its capacity the depth is 100 32.64 67.36 feet.
x 20
40
60
Section 6.2
b
h
57. (a)
(b)
r 2 dx (ii)
0
a
b
1
x2 b2
Volume: The Disk Method
2
is the volume of an ellipsoid with axes 2a and 2b.
is the volume of a right circular cylinder with radius r and height h.
r
(c)
dx (iv)
r
r 2 x 2
y=r
(0, a)
2 dx (iii)
is the volume of a sphere with radius r.
y
y
15
y 2 y=a 1− x b2
y=
r 2 − x2
(h, r)
x
(−r, 0)
x
(−b, 0)
(r, 0)
(b, 0)
x
h
(d)
0
rx h
r
2
(e)
dx (i)
is the volume of a right circular cone with the radius of the base as r and height h. y
R r 2 x 2 R r 2 x 2 dx (v) 2
r
2
is the volume of a torus with the radius of its circular cross section as r and the distance from the axis of the torus to the center of its cross section as R. y
(h, r) r 2 − x2
R+
y= r x h
R x
R−
r2− x2 −r
59.
x
r
y 4 3 2
x 2
3
4
Base of Cross Section x 1 x 2 1 2 x x 2 (b) Ax bh 2 x x 21
(a) Ax b 2 2 x x 2 2
V
2
V
1
4 4x 3x 2 2x 3 x 4 dx
1 1 4x 2x 3 x 3 x 4 x 5 2 5
2 + x − x2
2 + x − x2
2
4 4x 3x 2 2x 3 x 4
2 1
81 10
1
2 x x 2 dx 2x
1 2 + x − x2
x2 x3 2 3
2 1
9 2
16
Chapter 6
61.
y
Applications of Integration 3 y (a) A y b 2 1
1
1
V
3 4
0
1 2
2
3 y 1 dy 2
1−
1
1 4
1 2y 1 3 y 2 3 dy
1−
0
3 3 y y 4 3 y 5 3 2 5
x 1 4
1 2
3 4
1
1 0
1 10
3 Base of Cross Section 1 y
3 y 1 1 1 (b) A y r 2 2 2 2
V
1 8
1
0
3 y 1 dy 2
81 1 2
1
4
1
1
3
V
4
0
2
1 8 10 80
1 1 3 y (c) A y bh 1 23 2 2 3
3 y
3 y
1−
3
y
1−
3
1−
y
3
y
3 y 2
3 y 1 dy 2
3
4
101 403
1−
3
y
3 1 3 y 1 (d) A y ab 2 1 2 y 2 2
3 y 2 1 2
2
V
1
0
a
3 y 1 dy 2
1 2 10 20
b 1−
63. Let A1x and A 2x equal the areas of the cross sections of the two solids for a ≤ x ≤ b. Since A1x A 2x, we have
b
V1
65.
A1 x dx
y
1 4 4 25 r 2 3 2 125 3 2 3
25 r 2 3 2
b
a
3
A2 x dx V2
a
25 r 2
Thus, the volumes are the same.
25
125 2
125 2
2 3
25 r2 2 2 3
251 22 3 r 2 r 51 22 3 3.0415 67. (a) Since the cross sections are isosceles right triangles: 1 1 1 Ax bh r 2 y 2 r 2 y 2 r 2 y 2 2 2 2 V (b) Ax V
1 2
r
r
r
r 2 y 2 dy
r 2 y 2 dy r 2y
0
y3 3
r 0
2 r3 3
x
y
1 1 tan 2 bh r 2 y 2r 2 y 2 tan r y 2 2 2 2
tan 2
r
r
r
r 2 y 2 dy tan
As → 90 , V → .
0
r 2 y 2 dy tan r 2 y
y3 3
r 0
2 3 r tan 3
3
y
3
y
Section 6.3
Section 6.3
3. px x
hx x
hx x
2
xx dx
0
23x
3 2 0
2 x 3 3
2
0
16 3
4
V 2
16 3
4
2
45 x
4
52
0
128 5
7. px x
hx x
hx 4x x 2 x 2 4x 2x 2
2
2
2
x 3 dx
V 2
y
0
x4x 2x 2 dx
4
2 0
4
2
8
4
3
2x 2
dx
x3
4
1
23 x
3
2
1 x4 4
0
16 3
2 1
x
−1
1
9. px x
2
x
−1
3
1
2
11. px x
hx 4 4x x 2 x 2 4x 4
hx
1
ex 2 2
2
2
1
x 3 4x 2 4x dx
V 2
0
x
0
x4 34 x 4
3
2x 2
2 0
8 3
1 2
ex
dx
22
1
2
ex 2 x dx 2
0
y
2 ex 2
4
2
1 0
2 1
y
3 1
2
3 4
1 x
−1
1
2
1 2
3 1 4
x 1 4
13. p y y h y 2 y
2
y 2 y dy
0
2
2
3
0
2
V 2
y
0
4 x 2
2
x 32 dx
0
5. px x
V 2
xx dx
0
V 2
17
Volume: The Shell Method
1. px x
V 2
Volume: The Shell Method
2y y 2 dy
0
2 y 2
y3 3
2 0
8 3
1 2
3 4
1
1 e
0.986
3
18
Chapter 6
Applications of Integration
15. p y y and h y 1 if 0 ≤ y <
1 . 2
y 1
1 1 p y y and h y 1 if ≤ y ≤ 1. y 2
12
V 2
1 2
1
y dy 2
0
1 y dy
1 4
12
2 y2
2
3 4
12
y2 2
2 y
0
1 12
4 4 2
x 1 2
17. px 4 x
hx 4x x 2
2
V 2
V 2
4
2
x 3 6x 2 8x dx
x4 2x 4
3
4x 2
2
2
16
0
x 3 9x 2 20x dx
0
0
4
5 x4x x 2 dx
0
2
2 2
4
4 x4x 2x 2 dx
0
2
19. px 5 x
hx 4x x 2 x 2 4x 2x 2
3 2
1
y
x4 3x 4
3
10x 2
4 0
64
y
4
4 3
3
2
2
1 1 x 1
x 1
2
2
3
4
−1
3
21. (a) Disk
(b) Shell
Rx
px x
x3
rx 0
h x x 3
2
V
x 6 dx
0
x7
7 2 0
128 7
y
x5
5 2
y 8
6
6
4
4
2
2 x
−1
1
2
−1
3
(c) Shell px 4 x
y
h x x 3
8
2
4 xx 3 dx
6
0 4
2
2
x 4 dx 2
0
8
V 2
2
V 2
4x 3
dx
x4
0
1 2 x 4 x 5 5
2 0
96 5
2 x 1
2
3
4
x 1
2
3
0
64 5
Section 6.3 23. (a) Shell
Volume: The Shell Method
(c) Shell
p y y
px a x
h y a 12 y 12 2
hx a 12 x 12 2
a
V 2
V 2
a xa 12 x 12 2 dx
0
a
2
a
y a 2a 12 y 12 y dy
0
a
ay 2a 12 y32 y 2 dy
2
0
a2 2a 32 x 12 2a 12 x 32 x 2 dx
0
2
a2 y
2
a2 4a5
2
4a 12 52 y 3 y 5 3
3
3
3
a 3
a
2 a 2x
0
15a
3
4 32 32 4 12 52 1 3 a x a x x 3 5 3
a 0
4 a 3 15
y
(0, a)
y
(0, a)
(a, 0) x
(a, 0) x
(b) Same as part (a) by symmetry
d
25. V 2
b
p yh y dy
or V 2
pxhx dx
a
x
5
27.
5
x 1 dx
1
x 1
1
2 dx
This integral represents the volume of the solid generated by revolving the region bounded by y x 1, y 0, and x 5 about the x-axis by using the Disk Method.
29. (a)
1.5
y = (1 − x 4/3) 3/4
−0.25
1.5 −0.25
2
2
y 5 y 2 1 dy
0
represents this same volume by using the Shell Method.
(b) x 43 y 43 1, x 0, y 0 y 1 x 43 34
y
1
V 2
4
0
3 2 1 x 1
2
19
3
−1
Disk Method
4
5
x1 x 43 34 dx 1.5056
20
Chapter 6
31. (a)
Applications of Integration 33. y 2ex, y 0, x 0, x 2
7
y=
(x −
3
−
2) 2 (x
6) 2
Volume 7.5 Matches (d)
−1
7 y
−1
6
(b) V 2
3 x 22x 62 dx 187.249 x
2
2 1
x 1
2
35. px x 1 2 x 2
hx 2
2
V 2
x 2
0
2
1 2 x dx 2 2
Now find x0 such that
2
x0
0
2x
0
12
x2
1 3 x dx 2
2 0
4 total volume
y
2
1 x4 8
1 2x 02
1 3 1 x dx 2 x 2 x 4 2 8
2x
x0 0
1
1 4 x 4 0
x 2
1
x 04 8x 02 4 0 x 02 4 ± 23
(Quadratic Formula)
Take x0 4 23 since the other root is too large. Diameter: 24 23 1.464
1
37. V 4
1
2 x1 x 2 dx
1
8
1
39. Disk Method
R y r 2 y 2
1
1
x2
2 2
dx 4
1
x1
x2
dx
r y 0
1
8
1
x 1 x 2 122 dx
2
r 2 y 2 dy
rh
3 1 x
4 2 2
r
V
1
2 32
1
4 2
r 2y
y3 3
r rh
y
r
−r
x r
1 h 23r h 3
Section 6.3
r
41. (a) 2
hx 1
0
x dx (ii) r
is the volume of a right circular cone with the radius of the base as r and height h. y
r
(b) 2
r
Volume: The Shell Method
R x 2r 2 x 2 dx (v)
is the volume of a torus with the radius of its circular cross section as r and the distance from the axis of the torus to the center of its cross section as R. y
x=R y=h 1− x r
(
(0, h)
(
y=
r2 − x2
x
(r, 0) (r, 0)
(−r, 0)
x
y=−
r2 − x2
r
r
(c) 2
2xr 2 x 2 dx (iii) is the
(d) 2
hx dx (i) is the volume of a
0
0
volume of a sphere with radius r. y
r2 − x2
y=
right circular cylinder with a radius of r and a height of h. y
(r, h) (r, 0)
y=−
x
r2 − x2 x
b
(e) 2
2ax1 x 2b 2 dx (iv)
0
is the volume of an ellipsoid with axes 2a and 2b. y
y =a (0, a)
2 1 − x2 b
(b, 0)
(0, −a)
y = −a
x
2 1− x b2
200
43. (a) V 2
x f x dx
0
2 200 0 42519 25019 47517 210015 412514 215010 41756 0 38
1,366,593 cubic feet (b) d 0.000561x 2 0.0189x 19.39 24
−20
225 −6
200
(c) V 2
xd x dx 2 213,800 1,343,345 cubic feet
0
(d) Number gallons V 7.48 10,048,221 gallons
21
22
Chapter 6
Applications of Integration
Section 6.4
Arc Length and Surfaces of Revolution
1. 0, 0 , 5, 12
3. y
(a) d 5 0 2 12 0 2 13
y x 1 2, 0, 1
12 x 5
(b) y
1
0
y
1
13 x 5
5 0
13
2 8 1 1.219 3
2
y
x1 dx 2
1 y 2
1 3
8
3 2
x 2 3 1
1
0
1 x4 2 8 4x 1 3 1 x 3, 1, 2 2 2x
12 x
3
1 2 , 1, 2 2x 3
b
s
1 y 2 dx
a
2
2 dx 3x 1 3
3 2 2 3 x 1 3 2 2 3
1
3 2
y
x 2 3 1 dx x 2 3
1
dx
3 1 x
1 , 1, 8 x 1 3
8
2
7.
1
12 5
3 2 3 x 2
8
s
1 x dx
0
5
5. y
1
s
12 y 5 s
2 3 2 x 1 3
1
8 1
18 x
1 3 1 x 3 dx 2 2x
4
1 4x 2
2 1
33
2.063 16
55 22 8.352
y lnsin x ,
9.
y
3
4, 4
1 cos x cot x sin x
1 y 2 1 cot2 x csc2 x s
3 4
csc x dx
4
3 4
4
ln csc x cot x
ln2 1 ln2 1 1.763 11. (a) y 4 x 2, 0 ≤ x ≤ 2
y 2x
(b) 1
5
y 2
1 4x
(c) L 4.647 2
2
L
0
−1
3 −1
1 4x 2 dx
Section 6.4 1 13. (a) y , 1 ≤ x ≤ 3 x
1 x2
y
(b)
2
1 y 2 1 −1
1 x4
L
1
1
−1
15. (a) y sin x, 0 ≤ x ≤
1 dx x4
y cos x
(b)
(c) L 3.820
1 y 2 1 cos 2 x
2
L −
(c) L 2.147
3
4
Arc Length and Surfaces of Revolution
2
1 cos 2x dx
0
3 2 −0.5
17. (a) x ey, 0 ≤ y ≤ 2
y
(b)
y ln x
1 x
1 y 2 1
1 ≥ x ≥ e2 0.135
1
L
3
e
−1
2
(c) L 2.221 1 x2
1 x1 dx 2
3 −1
Alternatively, you can do all the computations with respect to y. (a) x e y 0 ≤ y ≤ 2
dx ey dy
(b) 1
(c) L 2.221
1e dx dy 2
2y
2
L
1 e2y dy
0
19. (a) y 2 arctan x, 0 ≤ x ≤ 1
(b) y
2 1 x2
1
3
L
1
0
−0.5
1.5
−3
(c) L 1.871 4 dx 1 x 2 2
23
24
Chapter 6
2
21.
Applications of Integration
dxd x
1
0
5 1
2
2
y
dx 5
(0, 5) y = 25 x +1
4
s 5
3
Matches (b)
2
(2, 1)
1
x −1
1
2
3
4
23. y x 3, 0, 4 (a) d 4 0 2 64 0 2 64.125 (b) d 1 0 2 1 0 2 2 1 2 8 1 2 3 2 2 27 8 2 4 3 2 64 27 2
64.525
4
(c) s
4
1 3x 2 2 dx
0
1 9x 4 dx 64.666
0
(d) 64.672
4
(c) y1 1, L1
y
25. (a) 4
y1
y4
1 −1
−1
y2
y2
3 2
2 dx 5.657
0
5
y3 1
2
y3
x
3
4
4
3 1 2 x , L2 4 1 x, L 3 2
5
1
0
4
1
0
5 3 2 x , L4 y4 16
1
0
1 3 2 x 3x 1 2 2 3 2
2
When x 0, y 3 . Thus, the fleeting object has traveled 3 units when it is caught. y
1 3 1 2 3 1 2 1 x1 x x 3 2 2 2 x 1 2
1 y 2 1
1
s
0
x 1 2 x 1 2 4x 4x
x1 1 dx 2x 1 2 2
1
x 1 2 x1 2 dx
0
1 2 3 2 x 2x 1 2 2 3
1 0
4 2 2 3 3
The pursuer has traveled twice the distance that the fleeing object has traveled when it is caught. 29.
y 20 cosh y sinh
x , 20 ≤ x ≤ 20 20
x 20
1 y 2 1 sinh 2
20
L
20
cosh
x x cosh 2 20 20 x dx 2 20
20
0
40 sinh1 47.008 m.
cosh
x x dx 220 sinh 20 20
20 0
x2 dx 5.916 4
4
(b) y1, y2, y3, y4
27. y
9x dx 5.759 16
25 3 x dx 6.063 256
Section 6.4
33. y
y 9 x 2
31.
y
x
Arc Length and Surfaces of Revolution x3 3
y x 2, 0, 3
9 x 2
3
1 y 2
2
s
0
2
S 2
9 9 x2
0
0
9 dx 9 x2
3 dx 9 x 2
3 arcsin
3 arcsin
y
x3 1 6 2x
y
x2 1 2 2 2x
1 y 2
x 3
2
2
1
2
2
1
2
9 1 x
8282 1 258.85 9
0
3 x 2 37. y
y
1 9x1
x
1
x2 1 2 dx 2 2x
x5 x 1 dx 12 3 4x 3
x6 x2 1 2 72 6 8x
8
x3 1 6 2x
1 , 1, 8 3x 2 3
S 2
2 1
47 16
39. y sin x
2 3
18
4 3
dx
8
x 1 39x 4 3 1 dx
1 8
9x 4 3 1 1 212x 1 3 dx
1
27 9x
145145 1010 199.48 27
4 3
1 3 2
8 1
41. A rectifiable curve is one that has a finite arc length.
y cos x, 0, S 2
3
4 3 2
x2 1 2 2 , 1, 2 2 2x
S 2
1 x 4 1 24x 3 dx
0
0
2 arcsin 0 3
6
3
2
2.1892 3
3 arcsin
35.
x3 1 x 4 dx 3
sin x1 cos 2 x dx
0
14.4236 43. The precalculus formula is the surface area formula for the lateral surface of the frustum of a right circular cone. The representative element is y 1 x x .
2 f di xi2 yi2 2 f di
2
i
i
i
25
26
Chapter 6
Applications of Integration
y
hx r
y
h r
45.
y 9 x 2
47.
y
r2 h2 r2
1 y 2
r
S 2
x
0
r 0
3 9 x 2
2
r2 h2 dx r2
2r 2 h 2 x 2 r 2
1 y 2
x 9 x 2
S 2
0
3x 9 x 2
2
3
r r 2 h 2
0
dx
2x 9 x 2
69 x 2
dx
2 0
6 3 5 14.40 See figure in Exercise 48. 1 1 2 x x 3 2 3
y
49.
1 1 2 3 1 2 1 1 2 x x x 9x 1 2 6 2 6
y
1 1 1 x 18 81x x1 2 9x 1 2 2 36 36
1 y 2 1
1 3
S 2
0
3
361 x
1 1 2 x x 3 2 3
1 2
9 1 2 2 dx
1 1 2x 9x 2 dx x x 2 3x 3 3 3 3
1 3
0
Amount of glass needed: V
2 6
1 3 0
1 3
0
13 x
1 2
x 3 2 x1 2 9x 1 2 dx
2 ft 0.1164 ft2 16.8 in 2 27
0.015
0.00015 ft 3 0.25 in 3 27 12
51. (a) y f x 0.0000001953x4 0.0001804x3 0.0496x2 4.8323x 536.9270
400
(b) Area
f x dx 131,734.5 square feet
0
3.0 acres (Answers will vary.)
400
(c) L
1 fx 2 dx 794.9 feet
0
(Answers will vary.)
b
53. (a) V
1
1 dx x2 x
b 1
1
1 b
(b) S 2 2
y = 1x
b
1
y
2
b
1
2
1
b
1
x 1
—CONTINUED—
b
1 x 1 x
1 x1 dx 2
2
1 x1 dx 4
x 4 1
x3
dx
Section 6.5
Work
27
53. —CONTINUED—
(c) lim V lim 1 b→
b→
1 b
(d) Since x 4 1
x 4
>
x3
x3
we have
b
x 4 1
x3
1
b
dx >
1
1 > 0 on 1, b x
1 dx ln x x
b 1
ln b
and lim ln b → . Thus, b→
b
lim 2
b→
55. (a) Area of circle with radius L: A L 2 Area of sector with central angle (in radians) 1 S A
L 2 L 2 2 2 2
x 4 1
x3
1
dx .
(b) Let s be the arc length of the sector, which is the circumference of the base of the cone. Here, s L 2 r, and you have S
1 2 1 1 1 s L L2 Ls L 2 r rL 2 2 L 2 2
(c) The lateral surface area of the frustum is the difference of the large cone and the small one. S r2 L L 1 r1L 1 L
r2 L L 1 r2 r1
r2
L1
L L1 L1 ⇒ Lr1 L 1 r2 r1 By similar triangles, r2 r1
r1
Hence, S r2 L L 1 r2 r1 r2 L Lr1 L r1 r2 .
Section 6.5
Work
1. W Fd 100 10 1000 ft lb
3. W Fd 112 4 448 joules (newton-meters)
5. Work equals force times distance, W FD.
7. Since the work equals the area under the force function, you have c < d < a < b .
9. F x kx
11. F x kx
5 k 4 k
5 4
50
7
W
250 k 30 ⇒ k
0
W
5 2 5 x dx x 4 8
7 0
245 in lb 8 30.625 in lb 2.55 ft lb
20
25 3
50
F x dx
20
25 25x 2 x dx 3 6
8750 n cm 87.5 joules or Nm
50 20
28
Chapter 6
Applications of Integration
1 3
13. F x kx
15. W 18
20 k 9
0
W
0
1 3
7 12
324x dx 162x 2
1 3
12
kx 2 2
7 12
20 k 9 W
kx dx
20 10 2 x dx x 9 9
12 0
Note:
40 ft lb 3
0
1 3
k ⇒ k 324 18
37.125 ft lbs
1 4 inches 3 foot
17. Assume that Earth has a radius of 4000 miles. k x2
F x
4000
k
4000 2
s
k 80,000,000
4100
(a) W
4000
4100
487.8 mi
4000
tons
80,000,000 dx 1395.3 mi ton x2 1.47 1010 ft lb
80,000,000 x2
F x
5.15 109 ft lb
4300
(b) W
80,000,000 80,000,000 dx x2 x
19. Assume that the earth has a radius of 4000 miles. k x2
F x
15,000
(a) W
4000
160,000,000 160,000,000 dx x2 x
15,000 4000
29,333.333 mi ton
k
4000 2
10
2.93 10 4 mi ton
k 160,000,000 160,000,000 F x x2
10,666.667 40,000
3.10 1011 ft lb
26,000
(b) W
4000
160,000,000 160,000,000 dx x2 x
26,000 4000
6,153.846 40,000 33,846.154 mi ton 3.38 10 4 mi ton 3.57 1011 ft lb
21. Weight of each layer: 62.4 20 y
y 6
Distance: 4 y
4
(a) W
2
4
(b) W
0
5
62.4 20 4 y dy 4992y 624y 2
62.4 20 4 y dy 4992y 624y 2
4 2 4 0
4
2496 ft
lb lb
Weight of disk of water: 9800 4 y Distance the disk of water is moved: 5 y
4
0
5 y dy
y2 2
4
5 y 9800 4 dy 39,200
0
39,200 5y
2 1
9984 ft
23. Volume of disk: 2 2 y 4 y
W
4−y
3
4 0
39,200 12 470,400 newton–meters
x 1
2
3
4
5
6
Section 6.5
25. Volume of disk:
23 y y 2
Weight of disk: 62.4
y
23 y y 2
4 3 2
x
6
6 y
y 2 dy
0
4 1 62.4 2y 3 y 4 9 4
6 0
−4 −3 −2 −1
2995.2 ft lb
1
3
4
10
Weight of disk: 62.4 36 y 2 y
8
4
6
y
y 36 y 2 dy
x −6 −4 −2 −2
0 6
62.4
2
y
2
W 62.4
6−y
5
27. Volume of disk: 36 y 2 y
Distance: y
29
7
Distance: 6 y 4 62.4 W 9
Work
0
1
36y y 3 dy 62.4 18y 2 y 4 4
2
4
6
6 0
20,217.6 ft lb 29. Volume of layer: V lwh 4 2 9 4 y 2 y
y
Weight of layer: W 42 8 9 4 y 2 y
Tractor
6 4
13 y Distance: 2
1.5
336
−y x
42 8 9 4 y 2 13 2
13 2
2
1.5
W
8
1.5
1.5
−6 −4 −2 −2
13 y dy 2
1.5
9 4 y 2 dy
1.5
2
4
6
−4
9 4 y 2 y dy
The second integral is zero since the integrand is odd and the limits of integration are symmetric to the origin. The first integral represents the area of a semicircle of radius 32 . Thus, the work is W 336
132 32 12 2457 ft lb 2
31. Weight of section of chain: 3 y Distance: 15 y
15
W3
15 y dy
0
3 15 y 2 2 337.5 ft
lb
33. The lower 5 feet of chain are raised 10 feet with a constant force. W1 3 5 10 150 ft lb The top 10 feet of chain are raised with a variable force.
15
Weight per section: 3 y
0
Distance: 10 y
10
W2 3
0
3
10 y dy 10 y 2 2 150 ft
W W1 W2 300 ft lb
lb
10 0
30
Chapter 6
Applications of Integration 37. Work to pull up the ball: W1 50015 7500 ft lb
35. Weight of section of chain: 3 y
Work to wind up the top 15 feet of cable: force is variable
Distance: 15 2y W3
7.5
0
3 15 2y dy 15 2y 2 4
7.5
Weight per section: 1 y
0
Distance: 15 x
3 152 168.75 ft 4
15
lb
W2
0
1 15 x dx 15 x 2 2
15 0
112.5 ft lb Work to lift the lower 25 feet of cable with a constant force: W3 12515 375 ft lb W W1 W2 W3 7500 112.5 375 7987.5 ft lb
39.
p
k V
41. Fx
1000
k 2
W
k 2 x 2
k 2000
3
W
2
5
0
1. x
32 810.93 ft lb
(b) x
0
3. x
126 3 14 3 62 3 30 3 118 3 99 3 12 1 6 3 11 33
125x 750 x 6 feet
y
x, y
17 18 112 115 118 12 11111
7 5 8 5 12 5 15 5 18 5 17 12 5 5
50x 750 75x
x
100x 125 x 3 dx 10,330.3 ft lb
Moments, Centers of Mass, and Centroids
7. 50x 75L x 7510 x
9.
5
45. W
65 31 53 6 635 7
5. (a) x
3
1000 1.8 lnx 1 dx 3249.44 ft lb
Section 6.6
2
2000 dV 2000 ln V V
2000 ln
43. W
1 k k 1 dx k 1 2 2 x 2 4 2 2 x 3k units of work 4 1
52 13 31 10 513 9 52 11 34 1 513 9
109 , 91
y
m1 (2, 2)
2
m2 1 (− 3, 1)
x −3 −2 −1 −1
1
2
3
−2 −3 −4
m3 (1, − 4)
Section 6.6
11.
x
7 32 41 27 10 63 34216 8
y
7 33 40 21 10 60 34216 16
7 7 x, y , 8 16
m
x dx
0 4
Mx y
m5 (− 3, 0) −4 −2 −2
23 x
4
32
0
x2
x dx 2 4
4 0
16 3
My
0
x 4
6
8
4
4
3 2
2
y
Mx 3 3 4 m 16 4 4
m3 (7, 1)
m4 (0, 0)
m1 (−2, −3)
x
0
31
y
m2 6 (− 1, 0)
4
13.
Moments, Centers of Mass, and Centroids
1
2 x x dx x 52 5
4
0
( x, y )
64 5
x 1
2
3
4
My 64 3 12 x m 5 16 5
x, y
125 , 34
1
15.
m
x 2 x 3 dx
0 1
Mx
Mx 12 12 m 35 35
1
My
x, y
0
12
y
1
x4
x6
0
xx 2 x 3 dx
x 3 x 4 dx
4
5 1 0
20
x 1 4
1 2
3 4
1
35 , 12 35
x 2 4x 2 x 2 dx
0
2
3
x 2 5x 4x 2 3x dx
0
x5
11x 3
3 0
2
x3
3x 2
3 0
9 2
y 6
( x, y )
1 −1
3
x x 2 4x 2 x 2 dx
0
32 , 225
2
99 5
3
0
3
x 4 8x 3 11x 2 12x dx
0
My 27 2 3 x m 4 9 2
5 4
3
Mx 99 2 22 m 5 9 5
3 2
x 2 4x 2 x 2 x 2 4x 2 x 2 dx 2
2x 4 6x 2 2 5 3
My
x, y
x4 x5
Mx
y
3 4
( x, y )
0
0 3
My 12 3 m 20 5
m
1
35 0 1
1 4
3
17.
x5 x7 dx 2 5 7
1 2
1
0
x
4 1
3
x 2 x 3 2 x x 3 dx 2 2
0
y
x3 x4
x 3 3x 2 dx
x4 x3 4
3 0
27 4
x 1
2
3
4
5
32
Chapter 6
Applications of Integration
8
19.
m
x 23 dx
0 8
Mx
35 x
96 5
0
x 23 23 3 73 x dx x 2 2 7
0
8
y
0
6
192 7
4
Mx 192 5 10 m 7 96 7
y
8
My
2
( x, y ) x
3 xx 23 dx x 83 8
0
2
8 0
96
4
6
8
−2
My 5 5 96 m 96
x
10 7
x, y 5 ,
2
21.
8
53
m 2
4 y 2 dy 2 4y
0
2
My 2
0
y3 3
2 0
32 3
y
4 y2 8 y5 4 y 2 dy 16y y 3 2 3 5
2 0
2
256 15
1
( x, y ) x
My 256 3 8 x m 15 32 5
1
−2
85 , 0
3
23.
m
2y y 2 y dy
0 3
My
0
2
3y2
3
y 4 4y 3 3y 2 dy
0
3 0
0
1
x 2 x 4 dx
0
1
x 2 x 3 dx
2
x3 3
1 0
1 x3 x5 2 3 5
1 6
1 0
1 1 1 1 2 3 5 15
x3 4 13 4 12 3
x4
y y 23y y 2 dy
( x, y )
0
27 10
1
3
3y 2 y 3 dy y 3
0
2 x
3
x x 2 dx
y 3
1
−3
−2
x
−1
1 −1
y 2y y 2 y dy
1
0
9 2
My
0
3 3 x, y , 5 2
1 2
3
Mx 27 2 3 m 4 9 2
y
Mx
y3 3
0
A
y5 y4 y3 2 5
3
Mx
2
2y y 2 y 2y y 2 y dy 2 2
My 27 2 3 m 10 9 5
x
25.
3
−1
By symmetry, M x and y 0.
x, y
2
1 0
1
1
y4 4
3 0
27 4
Section 6.6
3
27.
A
0
Mx
3
1 2
3 0
2
0
2x 2 8x 8 dx
0
2x 2 4x dx
0
3
2x 3
33
9 12 21
3
2x 4 dx
3
My
2x 4 dx x 2 4x
Moments, Centers of Mass, and Centroids
2x 2
3
3
2x 3
4x 2 8x
3 0
18 36 24 78
18 18 36
0
5
29. m
10x 125 x 3 dx 1033.0
0
5
Mx
10x 125 x 3 2
0
10x 125 x 3 dx
5
My
10x 2 125 x 3 dx
0
x
My 3.0 m
y
Mx 126.0 m
10 3
3,124,375 130,208 24
5
50
x 2 125 x 3 dx
0
5
125 x 3 3x 2 dx
0
12,500 5 3105.6 9
400
−1
Therefore, the centroid is 3.0, 126.0.
20
31. m
3 5 400 x 2 dx 1239.76
20 20
Mx
3 400 x 2 5 3 400 x 2 dx 5 2 20
25 2 y
20
20
400 x
2 23
dx 20064.27
6 −50
33.
1 A 2ac ac 2 1 1 A ac x
x 0 by symmetry. Therefore, the centroid is 0, 16.2.
y
50
−25
1 2ac
x, y
c
2
0
4ab 4ab y 2 y 2 dy c c
1 2ab 2 4ab 3 y 2y 2ac c 3c 1 ac 1 ac
25 −5
2
c
0
Mx 16.18 m
ac1 12 b c a y a b c a y a dy
c
y
0 c
0
1 2 b abc 2ac 3 3
b c a y a b c a y a dy
y
0
c
2a 2 y 2a dy c c
2 y2 y3 c 2 3c
c 0
c
0
y
y2 dy c
c 3
b3 , 3c
In Exercise 566 of Section P.2, you found that b3, c3 is the point of intersection of the medians. y
(b, c) y = c (x + a) b+a
y=
c b − a (x − a )
( x, y ) x
(−a, 0)
(a, 0)
34
Chapter 6
Applications of Integration
c 35. A a b 2 2 1 A ca b x y
2 c a b
c
b c a x a dx ca 2 b b c a x c
x
0
c2
2 1 ca b 2
c
0
ac 2 2
ca b 2
2 ba 1 x a dx c ca b
ba c
2bc 2
c
0
x3 2abc a x2 a x 2
3
2ac 2
2
3ac 2
6 ba c
c
x 2
2
c2b a
c 0
a 2bc
1 b a2c acb a a 2c ca b 3
1 b 2 2ab a 2c 3acb a 3a 2c 3ca b
a 2 ab b 2 1 b 2 2ab a 2 3ab 3a 2 3a 2 3a b 3a b
0
2ab a x a 2 dx c
1 ca b
3a b 3a b
2
2 b a x 3 ax 2 ca b c 3 2
ax dx
0
b a 2 ca b 3
2
y
a 2bc a 2 ab b 2 , . Thus, x, y 3a b 3a b
y = b −c a x + a
The one line passes through 0, a2 and c, b2. It’s equation is y
ba a x . 2c 2
The other line passes through 0, b and c, a b. It’s equation is y
a
(0, a) ( x, y )
( c, b)
a 2b x b. c
(0, 0) x
(c, 0)
b
x, y is the point of intersection of these two lines. 37. x 0 by symmetry
y
1 A ab 2 b
2 1 A ab 2 1 y ab 2
a
a
−a
b
a 2 x 2 a
a x x3
1 b2 ab a 2
4b 3
x, y 0,
39. (a)
3 a
2
a
a
dx
b 4a 3 4b a 3 3 3
(e) Mx
y
y=b
b x 2b x 2 dx 2 b
b
b2 x4 1 x5 d x b 2x 2 2 5 b
b
b 2 b −5 −4 −3 −2 −1
x 1 2 3 4 5
A (b) x 0 by symmetry (c) M y (d) y >
b
b
x
2
xb x 2 dx 0 because bx x 3 is odd
b b since there is more area above y than below 2 2
5
b b b b 24 3 3
b
b
5
b x 2 dx bx
b
b
4b 2 b
b b y
b 2 b
Mx 4b 2 b5 3 b. A 5 4b b3
x3 3
b
b
Section 6.6
Moments, Centers of Mass, and Centroids
41. (a) x 0 by symmetry
40
A2
f x dx
0
240 20 5560 30 429 226 420 0 278 34 3 3
f x 2 40 10 72160 dx 30 2 4 29 2 226 2 4 20 2 0 7216 2 34 3 3 40 40
Mx y
Mx 721603 72160 12.98 A 55603 5560
x, y 0, 12.98 (b) y 1.02 105 x 4 0.0019x 2 29.28 (c) y
Mx 23697.68 12.85 A 1843.54
x, y 0, 12.85 43. Centroids of the given regions: 1, 0 and 3, 0 Area: A 4
2
4 1 3 4 3 4 4
x
y
4 0 0 y 0 4
1 x 1
3
−1 −2
443 , 0 1.88, 0
x, y
32, 0, 5, and 0, 152
y
45. Centroids of the given regions: 0,
7
Area: A 15 12 7 34
6 5
150 120 70 0 x 34 y
1532 125 7152 135 34 34
x, y 0,
135 34
Mass: 4 2 4 1 2 3 2 3 4 2 2
y0
x, y
3 2 1 −4 −3 −2 −1
x 1
2
3
4
47. Centroids of the given regions: 1, 0 and 3, 0
x
4
223 , 0 2.22, 0
49. V 2 rA 2 516 160 2 1579.14
35
36
Chapter 6
Applications of Integration
1 51. A 44 8 2
1 1 y 8 2 ry
4
0
y 4
x3 1 4 x4 x dx 16x 16 3
4 0
8 3
2
8 3
1
V 2 rA 2
838 1283 134.04
53. m m1 . . . mn
x 1
2
55. (a) Yes. x, y
My m1x1 . . . mn xn Mx m1y1 . . . mn yn x
( x, y )
3
My Mx ,y m m
3
4
56, 185 2 56, 41 18
(b) Yes. x, y
56 2, 185 176, 185
(c) Yes. x, y
56, 185
(d) No. 57. The surface area of the sphere is S 4 r 2. The arc length of C is s r. The distance traveled by the centroid is
y
S 4 r 2 4r. s r
d
r
This distance is also the circumference of the circle of radius y.
(0, y) −r
x
r
d 2 y Thus, 2 y 4r and we have y 2r. Therefore, the centroid of the semicircle y r 2 x 2 is 0, 2r.
1
59.
A
x n dx
0
m A Mx
2
nx 1
1 0
1 n1
n1
1
x n 2 dx
0 1
My
n1
y
2 2nx 1 2n1
xx n dx
0
0
22n 1
1
(1, 1)
y=xn
n 2 0 n 2 x n2
x
My n 1 m n2
y
Mx n1 n1 m 22n 1 4n 2
Centroid:
1
1
nn 12, 4nn 12
As n → , x, y → 1, 14 . The graph approaches the x-axis and the line x 1 as n → .
x 1
Section 6.7
Section 6.7
Fluid Pressure and Fluid Force
Fluid Pressure and Fluid Force
1. F PA 62.453 936 lb
3. F 62.4h 26 62.4h6 62.426 748.8 lb
5. h y 3 y L y 4
7. h y 3 y L y 2
3
F 62.4
3 y4 dy
3y 1
3
3
3
124.8
0
y2 249.6 3y 2
3 0
3 y
0
3 y dy
249.6
F 262.4
0
y2 dy 3
3
0
1123.2 lb
124.8 3y
y
y3 9
3y 1 dy
3 0
748.8 lb
y 4 4
3 2
2
1
1
x 1
2
3
4 x −2
9. h y 4 y
−1
1
11. h y 4 y
L y 2 y
L y 2
4
F 262.4
4 y y dy
0
4y 1 2 y 3 2 dy
8y3
3 2
24 y dy
0
0
124.8
2
F 9800
4
124.8
2
2y 5 2 5
4 0
9800 8y y 2
1064.96 lb
2 0
117,600 Newtons
y
y 3
3
x −2
1 x −2
−1
1
2
−1
1
2
37
38
Chapter 6
Applications of Integration
13. h y 12 y
L y 10
2y 3
L y 6
9
F 9800
15. h y 2 y
0
2
2y 3 9800 72y 7y 2 9
9 0
2 y10 dy
F 140.7
2y 12 y 6 dy 3
0 2
1407
2,381,400 Newtons
2 y dy
0
1407 2y
y
y2 2
2
2814 lb
0
y 9 4 6
3
3 x −3
3
6
x
9
−6 −4 −2 −1
2
4
6
−2
17. h y 4 y L y 6
19. h y y L y 2
4
F 140.7
4 y6 dy
4
4 y dy
0
844.2 4y
y2 4 2
0
2
0
F 42
0
844.2
12 9 4y
6753.6 lb
3 2
y 9 4y 2 dy
0
42 8
214 23 9 4y
y
3 2
9 4y 2 1 2 8y dy 0
2 3 2
3 2
94.5 lb
y
5 2 3 1
x
1
−2
x −3 −2 −1 −1
1
2
−1
1
2
−1
3
−2
21. h y k y
y
water level
L y 2 r 2 y 2
r
r
Fw
r
k y r 2 y 2 2 dy
r
w 2k
r
−r
r
r 2 y 2 dy
r
r 2 y 2 2y dy
r
x
−r
The second integral is zero since its integrand is odd and the limits of integration are symmetric to the origin. The first integral is the area of a semicircle with radius r.
F w 2k
r2 0 wk r 2 2
Section 6.7 23. h y k y
Fluid Pressure and Fluid Force
39
25. From Exercise 22:
L y b
F 641511 960 lb
h 2
Fw
k yb dy
h 2
y2 2
wb ky
h 2 h 2
wbhk wkhb
y
water level h 2
−b 2
x
b 2 −h 2
27. h y 4 y
4
F 62.4
4 y L y dy
0
Using Simpson’s Rule with n 8 we have:
4380 0 43.53 235 42.58 229 41.510 2110.25 40.510.5 0
F 62.4
3010.8 lb 29. h y 12 y L y 2 4 2 3 y 2 33 2
4
F 62.4
212 y4 2 3 y 2 33 2 dy
0
31. (a) If the fluid force is one half of 1123.2 lb, and the height of the water is b, then h y b y L y 4
b
6448.73 lb
F 62.4
0
y
1 b y4 dy 1123.2 2
b
10 8
by 2 y2
6 4
x −6 −4 −2
b y dy 2.25
0
2
4
6
b2
b 0
2.25
b2 2.25 2 b2 4.5 ⇒ b 2.12 ft.
(b) The pressure increases with increasing depth.
d
33. F Fw w
c
hyLy dy, see page 471.
40
Chapter 6
Applications of Integration
Review Exercises for Chapter 6
5
1. A
1
1 1 dx x2 x
5 1
1 dx x2 1
1
4 5
3. A
y
1
arctan x (1, 1)
1
1 1
4 4 2
1 5,, 25
y
x 2
3
(5, 0)
4
(1, 0)
2
1,,
1
1 2
1,
1 (−1, 0)
1
5. A 2
1 (1, 0)
x
2
x x 3 dx
7. A
0
e 2 e x dx
0
12 x
2
1 2
2
1 x4 4
1
xe 2 e x
0
2 0
e2 1
1 2
y y
(0, e 2) (2, e 2)
6 1
(1, 1)
4
x
(0, 0) 1
1
(0, 1) x
1
1
2
3
1
( 1,
9. A
1)
5 4
4
11. A
3 8x x 2 x 2 8x 3 dx
0
cos x sin x
5 4
8
4
12 12 12 12
4 2 2 2
16x 2x 2 dx
0
8
sin x cos x dx
2 8x 2 x 3 3
8 0
512 170.667 3
20
(8, 3)
−4
(0, 3)
− 16
10
Review Exercises for Chapter 6 13. y 1 x
2
1
A
0
2
(0, 1)
1 x dx 2
−1
2
(1, 0)
1
1 2x 1 2 x dx
−1
0
4 1 x x3 2 x 2 3 2
1 0
1 0.1667 6
15. x y 2 2y ⇒ x 1 y 12 ⇒ y 1 ± x 1
0
A
1
1 x 1 1 x 1 dx
2
A
2
0 y 2 2y dy
0
0
y 3
0
1
2 x 1 dx
1 2y y 2 dy y 2 y 3 3
(0, 2)
2 0
1
4 3
(0, 0) −2
x
−1
2 −1
2
17. A
x 2
1 1
0
2
0
x dx 2
y1
dx
3
y
1 x 2 dx 3
2
3
3 x dx
2
2
(2, 0) x
y x 2 ⇒ x y 2, y 1
(3, 1)
(0, 1)
x ⇒ x 2 2y 2 1
3
1
A
y 2 2 2y dy
0 1
3y dy
0
32 y
1
2
0
3 2
19. Job 1 is better. The salary for Job 1 is greater than the salary for Job 2 for all the years except the first and 10th years. 21. (a) Disk
(b) Shell
4
V
x 2 dx
0
3x
3 4 0
64 3
x 2 dx
23 x
2
4
4
V 2
0
y
y
4
4
3
3
2
2
1
1 x 1
2
3
—CONTINUED—
4
x 1
3
4
3
0
128 3
41
42
Chapter 6
Applications of Integration
21. —CONTINUED— (c) Shell
(d) Shell
4
V 2
V 2
0 4
2
4
4 xx dx
6 xx dx
0 4
4x x 2 dx
2
0
6x x 2 dx
0
2 2x 2
x3 3
4 0
64 3
1 2 3x 2 x 3 3
y
4 0
160 3
y 5
4
4 3 3 2
2 1
1
x 1
x 1
23. (a) Shell
2
(b) Disk
4
V 4
x
0
3
212316 x
4
3 16 x 2 dx 4
3
2
V 2
0
4
2 3 2
0
64
3 16 x 2 4
4 0
4
2
2 1 x
−3 −2 −1
2
1
3
−3 −2 −1
−2
−2
−4
−4
x 1
2
3
y
1
V 2
0
1
0
x dx x4 1
1
2x dx x 2 2 1 x
arctan x
2
1
1
0
4 0 4
2
2
dx
48
y
4
1
5
9 x3 16x 8 3
y
25. Shell
4
−1
3
Review Exercises for Chapter 6 27. Shell
y 3
u x 2 x
43
2
2
u2
1 x
dx 2u du
1
6
V 2
2 2
4
0
1 x 2
1
dx 4
u 3 2u du 4 1u
3 u
4
3
2
x
0
u 2 2u du 1u
2
u2 u 3
0
1 u 2 3u 3 ln 1 u 2
2
3
4
5
6
−1
2 0
−2 −3
3 du 1u
4 20 9 ln 3 42.359 3
0
29. Since y ≤ 0, A
31. From Exercise 23(a) we have: V 64 ft 3
x x 1 dx.
1
1 V 16 4
ux1 xu1 dx du
1
A
1
u 1 u du
0
Disk:
1 0
16 9 y 2 dy 16 9
y0
3
9 y 2 dy 1
9y 31 y
4 15
y0
3
9y
y
0
3
9
1 y03 27 9 9 3 y03 27y0 27 0
x
−1
3
1 9
u 3 2 u 1 2 du
0
2 5 2 2 3 2 u u 5 3
y0
By Newton’s Method, y0 1.042 and the depth of the gasoline is 3 1.042 1.958 ft.
−1
33.
4 f x x 5 4 5
35.
f x x 1 4
y 300 cosh y
1 f x 2 1 x u 1 x x u 1 2 dx 2u 1 du
4
s
0
u u 1 du
1
3
u 3 2 u 1 2 du
2
1
2
25 u
5 2
2 u 3 2 3
3 1
4 3 2 u 3u 5 15
8 1 6 3 6.076 15
3 x sinh 20 2000
s
1 20
2000
2000
x 1 203 sinh 2000
2000
2000
2
dx
x 400 9 sinh 2000 dx 2
4018.2 ft (by Simpson’s Rule or graphing utility)
3
1 x dx 2
x 2000 280, 2000 ≤ x ≤ 2000
3 1
44
Chapter 6
Applications of Integration
3 37. y x 4
39. F kx y
1 y 2
4 k1
3 4
F 4x
25 16
W
5
4
S 2
0
4x dx 2x 2
0
3 x 4
15 dx 25 16 8 2
x2 4 0
50 in
15
5 0
lb 4.167 ft lb
y 4
(4, 3) 3
2 1 x 1
2
3
4
13 y 2
41. Volume of disk:
Weight of disk: 62.4
43. Weight of section of chain: 5 x Distance moved: 10 x
13 y 2
Distance: 175 y 62.4 9
W
10
W5
0
150
175 y dy
0
62.4 y2 175y 9 2
5 10 x dx 10 x 2 2 250 ft lb
150 0
104,000 ft lb 163.4 ft ton
b
45. W
Fx dx
a 4
80
ax 2dx
0
ax 3 3
4 0
64 a 3
380 15 a 3.75 64 4
a
47. A
a x
0
6 1 A a2 x
6 a2
a
0
a6 12
x, y
2
3 a2
a
0
0
4 1 a 2 a x 1 2 x dx ax a x 3 2 x 2 3 2
x a x 2 dx a
y
2 dx
6 a2
a 0
a2 6
a
ax 2 a x 3 2 x 2 dx
y
0
a x 4 dx
a
( x, y )
a
a 2 4a 3 2x 1 2 6ax 4a 1 2x 3 2 x 2 dx
0
3 2 8 8 1 a x a 3 2x 3 2 3ax 2 a 1 2x 5 2 x 3 a2 3 5 3
a5 , 5a
a
a 0
a 5
x
10 0
Review Exercises for Chapter 6 49. By symmetry, x 0.
y
1
A2
a 2 x 2 dx 2 a 2x
0
x3 3
a
4a 3 3
0
a2
1 3 A 4a 3
( x, y )
4a3 12
a
y
3
a
a 2 x 2 2 dx
−a
a 4 2a 2x 2 x 4 dx
0
a
2a 2 3 1 5 6 4 x x 3 a x 8a 3 5
6 2 5 1 5 2a 2 5 3 a a a 8a 3 5 5
x
a
6 8a 3
x, y 0,
a
0
2a 2 5
51. y 0 by symmetry
y 4
For the trapezoid:
6
x
0
6
1
1 1 x1 x1 6 6
x
x3
−1
dx
1 2 x 2x dx x2 3 9
0
6 0
−2 −3 −4
60
For the semicircle: m
122 2 2
8
My
x 4 x 6 2 4 x 6 2 dx 2
6
8
x 4 x 6 2 dx
6
Let u x 6, then x u 6 and dx du. When x 6, u 0. When x 8, u 2.
2
My 2
2
u 6 4 u 2 du 2
0
234 u 1
2
2
2 3 2
0
12
2
u 4 u 2 du 12
0
2
x
2 2 16 4 4 9 12 4 3 3
4 4 9 3
180 4 4 9 3
The centroid of the blade is
1
2 9
, 0. 239 49 9
4 u 2 du
0
Thus, we have: x 18 2 60
(6, 2)
2
m 46 16 18 My
y = 61 x + 1
3
29 49 3 9
1
2
3
4
5
7
(6, −2)
45
46
Chapter 6
Applications of Integration
53. Let D surface of liquid; weight per cubic volume.
d
F
D y f y g y dy
c
d
D f y g y dy
c
d
c
y f y g y dy
d
f y g y dy D
D
d
c
y
c
y f y g y dy
d
f y g y dy
c
d f
c
x
AreaD y Areadepth of centroid
Problem Solving for Chapter 6 1 1 1. T c c2 c3 2 2
c
R
cx x2 dx
0
cx2
2
x3 3
c
0
c3 c3 c3 2 3 6
1 3 c T 2 lim lim 3 c→0 R c→0 1 c3 6
3. (a)
1 V 2
1
2 1 y2 2 1 y2 dy
0
1
2
2
4 41 y2 1 y2 4 41 y2 1 y2 dy
0
1
8
1 y2 dy
Integral represents 1 4 area of circle
0
8
4 2
⇒ V 42
2
(b) x R2 y2 r2 ⇒ x R ± r2 y2 1 V 2
r
R r2 y2 R r 2 y2 dy
0
2
2
r
4Rr2 y2 dy
0
1 4R r2 2 r2 R 4 V 2 2 r 2 R
r
5. V 22 2
r2 h2 4
23r
2
xr2 x2 dx
x23 2
x2 + y2 = r2
r
r
r2 h2 4
4 h3 h3 which does not depend on r! 3 8 6
( x, y )
g
2 r2 − h 4 r
h 2
Problem Solving for Chapter 6 (b) Tangent at Aa, a3: y a3 3a2x a
7. (a) Tangent at A: y x3, y 3x2 y 1 3x 1
y 3a2x 2a 3
y 3x 2
x3 3a2x 2a3 0
To find point B:
x a2x 2a 0 ⇒
x3 3x 2
To find point B:
B 2a, 8a3
x3 3x 2 0
x 12x 2 0 ⇒ B 2, 8
y 8a3 12a2x 2a
Tangent at B:
y x3, y 3x2
Tangent at B:
y 12a2x 16a3
y 8 12x 2
x3 12a2x 16a3 0
To find point C:
x 2a2x 4a 0 ⇒
y 12x 16 x3 12x 16
To find point C:
x3 12x 16 0
x 2 x 4 0 ⇒ C 4, 64
Area of R
2
Area of S
27 4
x3 3x 2 dx
4
2
x3 3a2x 2a3 dx
27 4 a 4
2a
12a2x 16a3 x3 dx 108a4
Area of S 16area of R
12x 16 x3 dx 108 area S 16 area R
Area of S 16area of R
2a
4a
1
Area of S
C 4a, 64a3
a
2
Area of R
47
x
9. sx
1 ft2 dt
ds 1 fx2 dx
(a) sx
ds 1 fx2 dx
(b)
dydx dx 2
ds2 1 fx2dx2 1
x
(c) sx
1
1
2
(d) s2
32t
2
1 2
2
x
dt
9 1 t dt 4
1
9 8 9 1 t dt 1 t 4 27 4
1
dx2 dy2
This is the length of the curve y x
3 2
3 2 2 1
22 13 22 13 2.0858 27 27
from x 1 to x 2.
dx
6
My
1
6
1
x
5 3 12 35 36 7
2 1
x
−2 −3
3
4
5
6 1
6
(b) m 2
2 1 dx 2 x2 x
6 1
1
5 3
6
My 2
1
35 36
x, y
3
2
1
1 1 dx 2 x3 x
y
−1
6
1 1 x 3 3 x x
m2
b
11. (a) y 0 by symmetry
127, 0
x
1 b2 1 dx 3 x b2
1 2b 1 dx x2 b
2b 1 b 2b b2 1 b2 b 1
(c) lim x lim b→
b→
2b 2 b1
x, y
b 2b 1, 0
x, y 2, 0
48
Chapter 6
Applications of Integration
13. (a) W area 2 4 6 12 (b) W area 3 1 1 2
15. Point of equilibrium: 50 0.5x 0.125x x 80, p 10
1 1 7 2 2
P0, x0 10, 80
80
50 0.5x 10 dx 1600
Consumer surplus
0
80
Producer surplus
10 0.125x dx 400
0
17. (a) Wall at shallow end From Exercise 22: F 62.42420 9984 lb (b) Wall at deep end From Exercise 22: F 62.44820 39,936 lb 20
(c) Side wall From Exercise 22: F1 62.42440 19,968 lb
F2 62.4
15
y=8 10
4
8 y10y dy
4
8y y 2 dy 624 4y 2
0
26,624 lb Total force: F1 F2 46,592 lb
x = 40
5
0
624
y
y3 3
4
x 5 10 15 20 25
0
40 45 1x y = 10
C H A P T E R 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals Section 7.1
Basic Integration Rules . . . . . . . . . . . . . . . . . . . 308
Section 7.2
Integration by Parts . . . . . . . . . . . . . . . . . . . . . 312
Section 7.3
Trigonometric Integrals . . . . . . . . . . . . . . . . . . . 321
Section 7.4
Trigonometric Substitution . . . . . . . . . . . . . . . . . 328
Section 7.5
Partial Fractions
Section 7.6
Integration by Tables and Other Integration Techniques . . 343
Section 7.7
Indeterminate Forms and L’Hôpital’s Rule
Section 7.8
Improper Integrals
Review Exercises
. . . . . . . . . . . . . . . . . . . . . . 336
. . . . . . . . 348
. . . . . . . . . . . . . . . . . . . . . 353
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363
C H A P T E R 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals Section 7.1
Basic Integration Rules
Solutions to Even-Numbered Exercises
2. (a)
d x 2 ln x2 1 C 21 x2 2x dx 1 x 1
(b)
d 2x x2 122 2x2x2 12x_ 21 3x2 C 2 dx x2 12 x2 14 x 13
(c)
1 d arctan x C dx 1 x2
(d)
2x d lnx2 1 C 2 dx x 1
x dx matches (a). x2 1
4. (a)
d 2x sinx2 1 C 2xcosx2 12x 2 sinx2 1 22x2 cosx2 1 sinx2 1 dx
(b)
d 1 1 sinx2 1 C cosx2 12x x cosx2 1 dx 2 2
(c)
d 1 1 sinx2 1 C cosx2 12x x cosx2 1 dx 2 2
(d)
d 2x sinx2 1 C 2xcosx2 12x 2 sinx2 1 22x2 cosx2 1 sinx2 1 dx
x cosx2 1 dx matches (c).
6.
2t 1 dt t2 t 2
u t 2 t 2, du 2t 1 dt Use
12.
du . u
sec 3x tan 3x dx
u 3x, du 3 dx Use
308
sec u tan u du.
8.
2 dt 2t 12 4
10.
u 2t 1, du 2dt, a 2 Use
2x x2 4
u x2 4, du 2x dx, n
du . u2 a2
Use
14.
1 dx x x2 4
u x, du dx, a 2 Use
dx
du . u u2 a2
un du.
1 2
Section 7.1
309
18. Let u t 9, du dt.
16. Let u x 4, du dx.
Basic Integration Rules
6x 45 dx 6 x 45 dx 6
x 46 C 6
2 2 dt 2 t 92 dt C t 92 t9
x 46 C
20. Let u 4 2x2, du 4x dx.
x 4 2x2 dx
22.
x
3 dx 2x 32
1 4 2x21 24xdx 4
1 4 2x23 2 C 6
x dx
3 2
x2 3 2x 31 C 2 2 1
3 x2 C 2 22x 3
24. Let u x2 2x 4, du 2x 1 dx.
x1 x2 2x 4
dx
1 x2 2x 41 22x 1 dx 2
x2 2x 4 C
26.
28.
2x dx x4
2 dx
8 dx 2x 8 ln x 4 C x4
1 1 1 1 1 1 dx 3 dx 3 dx 3x 1 3x 1 3 3x 1 3 3x 1
30.
32.
x 1
1 x
3
sec 4x dx
1 4
x 1
1 1 1 3x 1 ln 3x 1 ln 3x 1 C ln C 3 3 3 3x 1
3 1 3 2 3 dx x x x
x3
3 1 1 1 2 dx x2 3x 3 ln x C x x 2 x
sec4x4 dx
34. Let u cos x, du sin x dx.
1 ln sec 4x tan 4x C 4
sin x cos x
dx cos x1 2sin x dx 2 cos x C
2
36. Let u cot x, du csc x dx.
csc2 xecot x dx ecot xcsc2 x dx ecot x C
38.
3ex
5 dx 5 2 5
3ex
1 2
x
ee dx x
ex dx 3 2ex
5 1 2ex dx 2 3 2ex
5 ln 3 2ex C 2
2
2x 3 2 dx
310
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals sin x dx cos x
40. Let u lncos x, du
tan xln cos x dx
1 cos d sin
csc d
cot d
ln cos xtan x dx
2 1 2 dx 3sec x 1 3 sec x 1
sec x 1
sec x 1 dx
46.
2 sec x 1 dx 3 tan2 x
2 sec x 2 dx cot2 x dx 3 tan2 x 3
2 cos x 2 dx csc2 x 1 dx 3 sin2 x 3
2 1 2 2 cot x x C 3 sin x 3 3
3 dt 3 arctan t C t2 1
2 csc x cot x x C 3 1 1 50. Let u , du 2 dt. t t
48. Let u 3 x, du 3 dx.
3 1 1 dx dx 4 3x2 3 4 3x 2
3 x 1 arctan C 2 2 3
52.
54.
56.
dy tan22x, 0, 0 dx
1
x 1 4x2 8x 3 1 1 4x x2
dx
dx
2
2x 1 2x 12 1 1
5 x2 4x 4
y
(a)
(b)
1
dx
tan22xdx
e1 t 1 dt e1 t 2 dt e1 t C t2 t
dx arcsec 2x 1 C
1 5 x 22
dx arcsin
sec22x 1 dx
x 52 C
a 5
1 tan2x x C 2 1.2
0, 0: 0 C x
−0.6
0.6
y
1 tan2x x 2
− 1.2
1.2
−1
58.
− 1.2
60. r
(0, 1) 5
−2
2 −2
ln csc cot ln sin C
lncos x2 C 2
tan x dx
44.
42.
1 et2 dt et
1 2et e2t dt et
et 2 et dt et 2t et C
Section 7.1 62. Let u 2x, du 2 dx. y
1 dx x 4x2 1
Basic Integration Rules
64. Let u sin t, du cos t dt.
2 dx 2x 2x2 1
sin2 t cos t dt
0
13 sin t 3
0
0
arcsec 2x C
66. Let u 1 ln x, du
e
1
1 ln x dx x
68.
1
e
1 ln x
1
4
0
72.
1 25 x2
dx arcsin
x 5
4 0
x2 dx x
e
1
arcsin
2
1
1
2 dx x
1x dx
1 1 ln x2 2
70.
2
1 dx. x
2
x 2 ln x
1
1 ln 4 0.386
1 2
4 0.927 5 6
4 x2 1 x2 dx lnx2 4x 13 arctan C x2 4x 13 2 3 3 `
−10
10
The antiderivatives are vertical translations of each other. −6
74.
ex ex 2
3
dx
1 3x e 9ex 9ex e3x C 24
sec u tan u du sec u C
76.
The antiderivatives are vertical translations of each other. 5
−5
5
−5
78. Arctan Rule:
du 4 1 arctan C a2 u2 a a
1 82. f x x3 7x2 10x 5
5
f x dx < 0 because
80. They differ by a constant: sec2 x C1 tan2 x 1 C1 tan2 x C.
2
84.
0
4 dx 4 x2 1
86. A
sin 2x dx
0
Matches (d).
0
2
y
more area is below the x-axis than above.
2
1 cos 2x 2
0
1
y
3
1
5 2 1 2
1 0
5 x 1
−5
2
3
4
x π 4
π 2
311
312
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
2
88.
2 x2 dx
0
2
A
(c) Let f x x over the interval 0, 2. Revolve this region about the y-axis.
(b) Let f x 2 x over the interval 0, 2. Revolve this region about the x-axis.
(a) Let f x 2 x2 over the interval 0, 2.
2 x2 dx
V
V 2
2
2 x dx
0
y
f ( x) =
2π x 2
20
xx dx
0
2
2
25
2
2
0
2 x2
dx
2 x2 dx
0
0
y 15
y
10 3
f ( x) =
2x
2
f ( x) = x
5 x 1
2
2
1
1 x 1
x 1
90. (a) (b)
1 n 0
n
sinnx dx
0 3
1
n
sinnxn dx
0
1 1 dx arctan x 3 3 31 x2 6 1
3 3
y
n
1 cosnx
94.
1
0
2
y x 2 3 y
x
1 y 2 1
1 x1 x x 2 x
0
2 3x1 3
1 y 2 1
x x 1 dx
9
S 2
3
1 arctan 3 3
y 2 x
92.
2
2
4 9x 2 3
8
s
1
1
4 dx 7.6337 9x2 3
9
2
2 x 1 dx
0
3 x 1
4
2
3 2
9 0
8 10 10 1 256.545 3
y 12 9 6 3 x 3
6
Section 7.2 2.
9
12
Integration by Parts
d 2 x sin x 2x cos x 2 sin x x2 cos x 2x sin x 2x sin x 2 cos x 2 cos x x2 cos x. Matches (d) dx
Section 7.2
Integration by Parts
4.
1 d x x ln x 1 x ln x ln x. Matches (a) dx x
6.
x2 e2x dx
8.
u x2, dv e2x dx
12. dv ex dx ⇒
ln 3x dx
10.
u ln 3x, dv dx
v
ex dx ex
ux
⇒ du dx
2
x dx 2 xex dx ex
14.
x2 cos x dx
u x2, dv cos x dx
e1 t 1 dt e1 t 2 dt e1 t C t2 t
ex dx 2xex ex C
2 xex
2xex 2ex C
16. dv x 4 dx ⇒
⇒ du
u ln x
x 4 ln x dx
20. dv
v
⇒ du
2
1 dx x ln x 3
x5 x5 1 1 4 dx ln x x dx 5 x 5 5
x5
5 ln x 1 C 25
1 1 dx x2 x
1 dx x
1 ln x 1 C dx x2 x x
v
x2 1 2 x dx
1 2 x2 1
⇒ du 2x3ex 2xex dx 2xex x2 1 dx 2
2
2
x3ex x2ex dx 2 2
1 2 x 1 x2
24. dv
1 dx ⇒ v x2
u ln 2x ⇒ du
x dx ⇒
x2 1 2
u x2ex
x5 1 ln x x5 C 5 25
ln x ln x dx x2 x
22. dv
18. Let u ln x, du
1 dx x
x5 ln x 5
1 dx ⇒ x2
u ln x
x5 5
v
2
2
2
xex dx 2
2
1 1 dx x2 x
1 dx x
ln 2x ln 2x dx x2 x
2
x2ex ex ex C C 2 2 2 x 1 2 2 x 1
ln 2x 1 1 ln 2x 1 dx C C x2 x x x
1 dx. x
ln x 3
1x dx 2 1 ln x
2
C
313
314
Chapter 7
26. dv
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
1 2 3x
dx ⇒
2
2 3x 1 2 dx 2 3x 3
⇒ du dx
ux
v
x 2 3x
2x 2 3x 2 2 3x dx 3 3
dx
2x 2 3x 4 2 2 3x 2 2 3x 2 3x 3 2 C 9x 2 2 3x C
3x 4 C 3 27 27 27
28. dv sin x dx ⇒
v cos x
⇒ du dx
ux
x sin dx x cos x
cos x dx x cos x sin x C
30. Use integration by parts twice. (2) u x, du dx, dv sin x dx, v cos x
(1) u x2, du 2x dx, dv cos x dx, v sin x
x2 cos x dx x2 sin x 2 x cos x
x2 cos x dx x2 sin x 2 x sin x dx
x2 sin x 2x cos x 2 sin x C
32. dv sec tan d ⇒
v
sec tan d sec
34. dv dx
⇒ du d
u
sec tan d sec
v
sec d
v
(2) dv e x dx ⇒
e x dx e x
v
ex cos 2x dx ex cos 2x 2 ex sin 2x dx ex cos 2x 2 ex sin 2x 2 ex cos 2x dx
5 e x cos 2x dx ex cos 2x 2ex sin 2x e x cos 2x dx
38. dv dx
⇒
ex
cos 2x 2 sin 2x C 5
vx
u ln x ⇒ du
1 dx x
y ln x y
ln x dx x ln x
x
x 1 x2
e x dx e x
u sin 2x ⇒ du 2 cos 2x dx
u cos 2x ⇒ du 2 sin 2x dx
dx
dx
4 x arccos x 1 x2 C
36. Use integration by parts twice. ⇒
1 1 x2
4 arccos x dx 4 x arccos x
sec ln sec tan C
(1) dv e x dx
dx x
u arccos x ⇒ du
⇒
1 dx x ln x x C x 1 ln x C x
cos x dx
Section 7.2 40. Use integration by parts twice. (1) dv x 1 dx ⇒
v
315
2
x 1 1 2 dx x 1 3 2 3
⇒ du 2x dx
u x2
(2) dv x 1 3 2dx ⇒
v
2
x 1 3 2 dx x 1 5 2 5
⇒ du dx
ux y
Integration by Parts
x 2 x 1 dx
2 4 4 2 2 2 x2 x 1 3 2 x x 1 3 2 dx x 2 x 1 3 2 x x 1 5 2 3 3 3 3 5 5
x 1 5 2 dx
2 x 1 3 2 2 8 16 x 2 x 1 3 2 x x 1 5 2
x 1 7 2 C
15x2 12x 8 C 3 15 105 105
⇒
42. dv dx u arctan y
v
dx x
1 2 1 x ⇒ du dx dx 2 1 x 2 2 2 4 x2
arctan
x x dx x arctan 2 2
y
44. (a)
(b)
4
x 2x dx x arctan ln 4 x 2 C 4 x2 2
y x
−6
dy 18 ex 3 sin 2x, 0, dx 37
ex 3 sin 2x dx
Use integration by parts twice.
4
(1) u sin 2x, du 2 cos 2x −4
dv ex 3 dx, v 3ex 3
ex 3 sin 2x dx 3ex 3 sin 2x
6ex 3 cos 2x dx
(2) u cos 2x, du 2 sin 2x dv ex 3 dx, v 3ex 3
ex 3 sin 2x dx 3ex 3 sin 2x 6 3ex 3 cos 2x
37 ex 3 sin 2x dx 3ex 3 sin 2x 18ex 3 cos 2x C y
ex 3 sin 2x dx
1 3ex 3 sin 2x 18ex 3 cos 2x C 37
18 1 : 0 18 C ⇒ C 0 0, 18 37 37 37 y
1 x 3 3e sin 2x 18ex 3 cos 2x 37
6ex 3 sin 2x dx C
316
46.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
dy x sin x, y 0 4 dx y
48. See Exercise 3.
(0, 4)
8
1
0
−5
x2e x dx x2e x 2xe x 2e x
1 0
e 2 0.718
10 −2
50. dv sin 2x dx ⇒
1 sin 2x dx cos 2x 2
v
⇒ du dx
ux
1 1 x sin 2x dx x cos 2x 2 2
52. dv x dx
Thus,
cos 2x dx
x arcsin x2 dx
1 1 x cos 2x sin 2x C 2 4
x sin 2x dx
0
v
u arcsin x2 ⇒ du
1 sin 2x 2x cos 2x C 4
⇒
14 sin 2x 2x cos 2x
0
. 2
2x
dx
1 x4
x3 dx 1 x 4
1 x2 arcsin x2 2 1 x 4 1 2 C 2 4
1 2 x arcsin x2 1 x 4 C 2
x arcsin x2dx
0
x2 2
x dx
x2 arcsin x2 2
1
Thus,
1 2 . 4 54. Use integration by parts twice. (1) dv ex, v ex, u cos x, du sin x dx
ex cos x dx ex cos x
ex sin x dx
(2) dv ex dx, v ex, u sin x, du cos x dx
ex cos x dx ex cos x ex sin x
ex cos x dx ⇒ 2 ex cos x dx ex sin x ex cos x
Thus,
2
ex cos x dx
0
⇒
56. dv dx
x
e
e 2
sin x ex cos x 2
2
v
u ln 1 x2 ⇒ du
1
0
0
1 2
dx x
2x dx 1 x2
x ln 1 x2 2
2
ln 1 x2 dx x ln 1 x2
Thus,
sin 2 cos 2
2x2 dx 1 x2 1
1 dx x ln 1 x2 2x 2 arctan x C 1 x2
ln 1 x2 dx x ln 1 x2 2x 2 arctan x
1 0
ln 2 2
. 2
1 2 x arcsin x2 1 x 4 2
1 0
Section 7.2 58. u x, du dx, dv sec2 x dx, v tan x
x sec2 x dx x tan x
60.
4
tan x dx
2
4 ln 22 0
1 ln 2 4 2
x3 cos 2x dx x3
u and its derivatives
x3
e2x
3x2
2e2x
6x
1 2x 4e
6
8e2x
0
1 2x 16 e
2 8 16 x2 x 2 3 2dx x2 x 2 5 2 x x 2 7 2
x 2 9 2 C 5 35 315
76.
5
0
68. Yes. u ln x, dv x dx
v and its antiderivatives
u and its derivatives
x3
3x2
1 2
6x
14
cos 2x
6
18
sin 2x
0
1 16
Alternate signs
u and its derivatives
x2
2x
2 5
x 2 5 2
2
4 35
x 2 7 2
0
8 315
x 2 9 2
2
x 2 5 2 35x2 40x 32 C 315
66. Answers will vary. See pages 488, 493.
4 sin d
1
Alternate signs
1
2
1 4x3 sin 2x 6x2 cos 2x 6x sin 2x 3 cos 2x C 8
74.
v and its antiderivatives
Alternate signs
12 sin 2x 3x 41 cos 2x 6x 81 sin 2x 6161 cos 2x C
3 3 3 1 x3 sin 2x x2 cos 2x x sin 2x cos 2x C 2 4 4 8
64.
4
0
1 1 2x 1 1 x 3e2x dx x 3 e2x 3x2 e2x 6x e2x 6 e C 2 4 8 16 1 e2x 4x3 6x2 6x 3 C 8
62.
x sec x dx x tan x ln cos x
0
317
Hence,
Integration by Parts
70. No. Substitution.
cos 2x sin 2x
cos 2x
v and its antiderivatives
x 2 3 2
72. No. Substitution.
1 4 cos 4 3 sin 12 2 cos 24 sin 24 cos C 5
x 4 25 x2 3 2 dx
1,171,875 arcsin x 5 x 2x2 25 25 x2 5 2 625x 25 x2 3 2 46,875x 25 x2 128 16 64 128
14,381.0699
5 0
318
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
78. (a) dv 4 x dx ⇒
v
2
4 x 1 2 dx 4 x 3 2 3
⇒ du dx
ux
x
2 2 4 x dx x 4 x 3 2 3 3
4 x 3 2 dx
4 2 2 x 4 x 3 2 4 x 5 2 C 4 x 3 2 3x 8 C 3 15 15 (b) u 4 x ⇒ x u 4 and dx du
x 4 x dx
u 4 u1 2du
u3 2 4u1 2 du
8 2 2 u5 2 u3 2 C u3 2 3u 20 C 5 3 15
2 2
4 x 3 23 4 x 20 C 4 x 3 2 3x 8 C 15 15
80. (a) dv 4 x dx ⇒
v
4 x 1 2 dx
82. n 0:
2 4 x 3 2 3 ⇒ du dx
ux
2 2 x 4 x dx x 4 x 3 2 3 3
n 1: n 2:
4 x
3 2 dx
4 2 x 4 x 3 2 4 x 5 2 C 3 15
(b) u 4 x ⇒ x 4 u and dx du
4u1 2 u3 2 du 2 8 u3 2 u5 2 C 3 5
u xn
xex dx xex ex C xex
ex dx
x2ex dx x2ex 2xex 2ex C
n 3:
x3ex dx x3ex 3x2ex 6xex 6ex C
x3ex 3 x2e x dx
n 4: x 4ex dx x 4ex 4x3ex 12x2ex 24xex 24ex C
x 4 x dx 4 u u du
84. dv cos x dx ⇒
ex dx ex C
x2ex 2 xex dx
2 4 x 3 25x 2 4 x C 15 2 4 x 3 2 3x 8 C 15
2 3 2 u 20 3u C 15
2
4 x 3 220 3 4 x C 15
2
4 x 3 2 3x 8 C 15
v sin x
⇒ du n x n1 dx
x n cos x dx xn sin x n xn1 sin x dx
x 4ex 4 x3ex dx In general, x nex dx xnex nx n1ex dx. (See Exercise 86)
1 86. dv eax dx ⇒ v eax a u x n ⇒ du nx n1 dx
x neax dx
x neax n a a
x n1eax dx
Section 7.2
Integration by Parts
88. Use integration by parts twice. 1 v eax a
(1) dv eax dx ⇒
u cos bx ⇒ du b sin bx
eax cos bx dx
Therefore, 1
eax cos bx b a a
u sin bx ⇒ du b cos bx
eax sin bx dx
eax cos bx beax sin bx b2 2 a a2 a b2 a2
e
ax
1 v eax a
(2) dv eax dx ⇒
eax cos bx b eax sin bx b a a a a
eax cos bx dx
eax cos bx dx
eax a cos bx b sin bx a2
cos bx dx
eax a cos bx b sin bx C. a2 b2
eax cos bx dx
90. n 2 (Use formula in Exercise 84.)
x2 cos x dx x2 sin x 2 x sin x dx (Use formula in Exercise 83.) n 1
x2 sin x 2 x cos x
cos x dx x2 sin x 2x cos x 2 sin x C
92. n 3, a 2 (Use formula in Exercise 86 three times.)
x3e2x dx
x3e2x 3 2 2
1 A 9
1 9
x2e2x dx n 3, a 2
x3e2x 3 x2e2x 2 2 2
x3e2x 3x2e2x 3 xe2x 1 2 4 2 2 2
e2x 3
4x 6x2 6x 3 C 8
xe2x dx n 2, a 2
94. dv ex 3 dx ⇒ ux
v 3ex 3
e2x dx
x3e2x 3x2e2x 3xe2x 3e2x C n 1, a 2 2 4 4 8
96. A
x 3
xe
0
⇒ du dx
3
See Exercise 83.
dx
3
0
3xe 3 e x 3
3
3
x 3
0
0
1 9 9ex 3 9 e
dx
0
1 1 2 1 1 0.264 e e e 0.4
−1
4
− 0.1
−1
4
3
−1
x sin x dx x cos x sin x
0
319
320
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
98. In Example 6, we showed that the centroid of an equivalent region was
1, 8 . By symmetry, the centroid of this region is 8, 1 . π 2
You can also solve this problem directly.
1
A
0
arcsin x dx x x arcsin x 1 x2 2 2
My A
x
Mx A
y
1
x
0
1
0
y
1 0
Example 3
x 1
2 2 0 1 1
arcsin x dx 2 8
2 arcsin x arcsin x dx 1 2 2
2
100. (a) Average
1 4
(b) Average
2
1.6t ln t 1 dt 0.8t2 ln t 0.4t2 t
1
1.6t ln t 1 dt 0.8t2 ln t 0.4t2 t
3
4 3
3.2 ln 2 0.2 2.018 12.8 ln 4 7.2 ln 3 1.8 8.035
102. c t 30,000 500t, r 7%, t1 5
5
30,000 500t e0.07t dt 500
P
0
5
60 t e0.07t dt
0
Let u 60 t, dv e0.07t dt, du dt, v
60 t 1007 e
P 500
0.07t
5 0
100 0.07t e . 7
5
100 7
e0.07t dt
0
e $131,528.68 60 t 1007 e 10,000 49
500
104.
0.07t
5
0.07t
0
0
x 2 cos nx dx
5
x2 2 2x sin nx 2 cos nx 3 sin nx n n n
2 2 cos n 2 cos n n2 n
4 cos n n2
44 n n ,, ifif nn isis even odd
1 n 4 n2
2 2
106. For any integrable function, f x dx C f x dx, but this cannot be used to imply that C 0.
2 , sin x ≤ 1 ⇒ x sin x ≤ x ⇒
108. On 0,
2
0
x sin x dx ≤
2
0
x dx.
Section 7.3
Trigonometric Integrals
321
110. fx cosx, f 0 2 (a) It cannot be solved by integration.
Section 7.3
(b) You obtain the points 4
n
xn
yn
0
0
2
1
0.05
2.05
2
0.10
2.098755
3
0.15
2.146276
80
2.8403565
4.0
0
4 0
Trigonometric Integrals
2. (a) y sec x ⇒ y sec x tan x sin x sec2 x.
(b) y cos x sec x ⇒ y sin x sec x tan x sin x sec2 x sin x
Matches (iii)
sin x1 sec2 x sin x tan2 x Matches i (c) y x tan x
1 3 tan x ⇒ y 1 sec2 x tan2 x sec2 x 3 tan2 x tan2 x1 tan2 x tan4 x Matches iv
(d) y 3x 2 sin x cos3 x 3 sin x cos x ⇒ y 3 2 cos xcos3 x 6 sin x cos2 xsin x 3 cos2 x 3 sin2 x 3 2 cos4 x 6 cos2 x1 cos2 x 3 cos2 x 31 cos2 x 8 cos4 x Matches ii
4.
cos 3 x sin4 x dx
cos x1 sin2 xsin4 x dx
sin4 x sin6 xcos x dx
sin5 x sin7 x C 5 7
6. Let u cos x, du sin x dx.
sin3 x dx
1 x x 8. Let u sin , du cos dx. 3 3 3
cos3
x dx 3
cos
3
x 3
1 sin 3x dx
1 sin2
2
x 3
13 cos 3x dx
3 sin
x 1 x sin3 C 3 3 3
3 sin
x x sin3 C 3 3
sin x1 cos2 x dx
cos2 xsin x dx
1 cos3 x cos x C 3
sin x dx
322
10.
Chapter 7
sin5 t dt cos t
12.
sin2 2x dx
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
sin t1 cos2 t2cos t12 dt
sin t1 2 cos2 t cos4 tcos t12 dt 4 2 cos t12 2cos t32 cos t72 sin t dt 2cos t12 cos t52 cos t92 C 5 9
1 cos 4x 1 1 dx x sin 4x C 2 2 4
14.
sin4 2 d
1 4x sin 4x C 8
1 cos 4 2
1 cos 4 d 2
1 1 2 cos 4 cos2 4 d 4 1 cos 8 d 2
1 4
1 2 cos 4
1 4
3 1 2 cos 4 cos 8 d 2 2
1 3 1 1 sin 4 sin 8 C 4 2 2 16
1 1 3 sin 8 C sin 4 8 8 64 16. Use integration by parts twice. dv sin2 x dx
1 cos 2x x sin 2x 1 2x sin 2x ⇒ v 2 2 4 4
u x2 ⇒ du 2x dx dv sin 2x dx ⇒
1 v cos 2x 2
⇒ du dx
ux
1 1 x2 sin2 x dx x22x sin 2x 4 2
2x2 x sin 2x dx
1 1 1 1 x3 x2 sin 2x x3 2 4 3 2
x sin 2x dx
1 1 1 1 1 x3 x2 sin 2x x cos 2x 6 4 2 2 2
cos 2x dx
1 1 1 1 x3 x2 sin 2x x cos 2x sin 2x C 6 4 4 8
1 4x3 6x2 sin 2x 6x cos 2x 3 sin 2x C 24
2
18. Let u sin x, du cos x dx.
2
cos5 x dx
0
20.
0
2
1 sin2 x2 cos x dx
0
2
1 2 sin2 x sin4 x cos x dx
0
sin x
8 15
sin2 x dx
2
2 3 1 sin x sin5 x 3 5
0
1 2
2
1 cos 2x dx
0
2
1 1 x sin 2x 2 2
0
4
Section 7.3
22.
sec22x 1 dx
1 tan2x 1 C 2
24.
sec6 3x dx
28.
30. Let u sec 2t, du 2 sec 2t tan 2t
32.
36.
26.
tan2 x dx
sec2 x 1dx tan x x C
tan3 2t sec3 2t dt
34.
sec2
42. y
sin2 x cos2 x
2 1 1 tan 3x tan3 3x tan5 3x C 3 9 15
tan5 2x sec2 2x dx
tan3 3x dx
x x x 1 x tan dx 2 tan sec2 dx 2 2 2 2 2
tan2 x sec5 x
1 2 tan2 3x tan4 3xsec2 3x dx
1 x x x sec2 dx C tan4 2 2 2 2
x C 2
tan2
38.
1 tan2 3x2 sec2 3x dx
1 tan6 2x C 12
sec5 2t sec3 2t C 10 6
sec4 2t sec2 2tsec 2t tan 2t dt
sec2 or
323
sec2 2t 1sec3 2t tan 2t dt
x x x 1 x x tan dx 2 sec sec tan dx 2 2 2 2 2 2
sec2
tan3
Trigonometric Integrals
sec2 3x 1tan 3x dx
1 3 sin 3x 1 tan 3x3 sec2 3x dx dx 3 3 cos 3x
1 2 1 tan 3x ln cos 3x C 6 3
x C 2
cos2 d 2 2
cos5 x dx
40. s
sin2 x cos3 x dx
sin2 x1 sin2 xcos x dx
sin2 x sin4 xcos x dx
sin 2 1 C 8 2
1 2 sin 2 C 16
sin3 x sin5 x C 3 5
tan x sec4 x dx
tan12 xtan2 x 1sec2 x dx
tan52 x tan12 x sec2 x dx
2 2 72 tan x tan32 x C 7 3
sin2
1 cos 2
1 4
1 2cos d 1 cos 4
sin2 d
1 8
1 cos 2 d
2
d
324
Chapter 7
44. (a)
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
y
(b)
1
y
y
tan3 x C 3
x
−1
1
( ) 1
0, 4
−1
1 dy sec2 x tan2 x, 0, dx 4
sec2 x tan2 x dx u tan x, du sec2 x dx
0, 41: 41 C ⇒ y 31 tan
3
46.
dy 3y tan2 x, y0 3 dx 8
48.
cos 4 cos3 d
(0, 3)
−1
1
x
1 4
cos 4 cos 3 d
1 cos 7 cos d 2
sin 7 sin C 14 2
−2
50.
x 1 x 52. Let u tan , du sec2 dx. 2 2 2
sin4x cos 3x dx sin 4x cos 3x dx
1 2
1 1 cos x cos 7x C 2 7
sin x sin 7xdx
tan4
x x sec4 dx 2 2
csc2 3x cot 3x dx
1 3
56.
tan4
2
1 7 cos x cos 7x C 14
54. u cot 3x, du 3 csc2 3x dx
cot3 t dt csc t
cot 3x3 csc2 3x dx
1 cot2 3x C 6
x x x tan2 1 sec2 dx 2 2 2
tan6
x x tan4 2 2
12 sec 2x dx 2
x 2 2 7x tan tan5 C 7 2 5 2
cos3 t dt sin2 t
cos t dt sin2 t
1 sin2 tcos t dt sin2 t
cos t dt
1 sin t C sin t
csc t sin t C
58.
60.
sin2 x cos2 x dx cos x
1 sec t dt cos t 1
1 2 cos 2 x dx cos x
sec x 2 cos x dx ln sec x tan x 2 sin x C
cos t 1 dt cos t 1 cos t
62.
0
3
0
sec2 ttan t dt
2
3 tan t 32
0
2 3
3
sec2 x 1 dx
0
3
tan x x
66. 4
tan2 x dx
sec t dt ln sec t tan t C
64. Let u tan t, du sec2 t dt. 4
0
sin 3 cos d
1 2
3
3
sin 4 sin 2 d
1 1 1 cos 4 cos 2 2 4 2
0
Section 7.3
2
68.
sin2 x 1 dx
2
70.
sin2 x cos2 x dx
2
2
2
2
Trigonometric Integrals
325
2x 1 dx 1 cos 2 2
32 21 cos 2x dx 32 x 41 sin 2x
1 4x sin 4x C 32
72.
2
3 2
tan31 x dx
tan21 x ln cos1 x C 2
2 2
−6
6
−3
3
−2
74.
−2
sec41 x tan1 x dx
2
sec41 x C 4
76.
1 cos 2 d
0
2
− 3.5
3
1
3 2 4
3.5
−2
2
78.
sin6 x dx
0
2
1 5x 3 1 2 sin 2x sin 4x sin3 2x 8 2 8 6
0
5 32
80. See guidelines on page 500. 82. (a) Let u tan x, du sec2 x dx.
sec2 x tan x dx
(b)
8
1 2 tan x C1 2 −4
Or let u sec x, du sec x tan x dx.
(c)
1 1 1 1 1 sec2 x C tan2 x 1 C tan2 x C tan2 x C2 2 2 2 2 2
84. Disks
y
Rx tan x rx 0 V 2
1 1 2
4 2
tan x dx
− 12
0
2
4 −2
1 sec xsec x tan x dx sec2 x C 2
−1
4
sec x 1 dx 2
0
2 1
4
2 tan x x
0
1.348 4
π 8
π 4
x
2
2 2 sin 4 sin 2
0
326
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
2
86. (a) V
cos2 x dx
0
(b) A
2
2
2
0
2
cos x dx sin x
0
1 cos 2x dx
0
1 x sin 2x 2 2
2
0
2 4
1
Let u x, dv cos x dx, du dx, v sin x. x
2
0
y
1 2 1 4
0
2
2
sin x dx x sin x cos x
0
0
2 1 2 2
2
cos2 x dx
y
0
2
1
1 cos 2x dx
0
2
1 1 x sin 2x 4 2
x, y
2
x cos x dx x sin x
0
( π 2− 2 , π8 (
1 2
8
π 4
2 2, 8
88. dv cos x dx ⇒
π 2
x
v sin x
u cosn1 x ⇒ du n 1cosn2 x sin x dx
cosn x dx cosn1 x sin x n 1 cosn2 x sin2 x dx cosn1 x sin x n 1 cosn2 x1 cos2 x dx
cosn1 x sin x n 1 cosn2 x dx n 1 cosn x dx
Therefore, n cosn x dx cosn1 x sin x n 1 cosn2 x dx cosn x dx
cosn1 x sin x n 1 n n
cosn2 x dx.
90. Let u secn2 x, du n 2secn2 x tan x dx, dv sec2 x dx, v tan x.
secn x dx secn2 x tan x
n 2 secn2 x tan2 x dx
secn2 x tan x n 2 secn2 xsec2 x 1 dx secn2 x tan x n 2
secn x dx
secn2 x dx
n 1 secn x dx secn2 x tan x n 2 secn2 x dx secn x dx
92.
cos4 x dx
1 n2 secn2 x tan x n1 n1
cos3 x sin x 3 4 4
cos2 x dx
secn2 x dx
cos3 x sin x 3 cos x sin x 1 4 4 2 2
dx
1 3 3 1 cos3 x sin x cos x sin x x C 2 cos3 x sin x 3 cos x sin x 3x C 4 8 8 8
Section 7.3
94.
sin4 x cos2 x dx
cos3 x sin3 x 1 6 2
cos2 x sin2 x dx
cos3 x sin x 1 cos3 x sin3 x 1 6 2 4 4
cos2 x dx
x 1 1 cos x sin x 1 C cos3 x sin3 x cos3 x sin x 6 8 8 2 2
1 8 cos3 x sin3 x 6 cos3 x sin x 3 cos x sin x 3x C 48
96. (a) n is odd and n ≥ 3.
2
cosn x dx
0
cosn1 x sin x n
2
0
2
n1 n
n1 n
n 2
n1 n
n2n4
n1 n
n2n4...
cosn3 x sin x n2
n3 n3 n3
2
0
n3 n2
cosn5 x sin x n4
n5
cosn2 x dx
0
2
0
0
n5
2
cos x dx
0
n3
n5
n2n4...1
0
(Reverse the order)
234567. . .n n 1
234567. . .n n 1 n1 n
n3
n5
n2n4...
n1 n
n3
2
cos2 x dx
(From part (a).)
0
n5
2
n 2 n 4 . . . 2 4 sin 2x 0
n1 n
2 123456. . .n n 1
123456. . .n n 12
n3
n5
x
n2n4... 4
1
(Reverse the order)
cosn6 x dx
0
2
(b) n is even and n ≥ 2.
0
n5 n4
2
n1 n
cosn x dx
cosn4 x dx
0
cosn6 x dx
2
2
2
n n 1 nn 32 nn 54 . . . sin x
1
n1 n
Trigonometric Integrals
327
328
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
Section 7.4 2.
Trigonometric Substitution
1 x 1 d x x2 16 1 1 x 8 ln x2 16 x xx2 16 C 8 x2 16 2 2 dx 2 2 2 x 16 x x 16
4.
x2 x2 16
x2 16 8 x x2 16 x2 2 2 2 x 16 x 16 x 2x 16
16 x2 x2 16 2x2 16
Indefinite integral:
2
x2 x2 16
Matches (d)
x 3 x 37 6x x2 1 d 8 arcsin C 8 dx 4 2 1 x 34 2
8 16 x 3
2
4 2 x 3 1
3x
1
7 6x x2
16 x 32 x 32 2 2 216 x 3
16 x2 6x 9 16 x2 6x 9 216 x 32
2 16 x 32 216 x 32
16 x 32
Indefinite integral:
7 6x x2
7 6x x2 dx
Matches (c)
6. Same substitution as in Exercise 5.
10 x225
x2
dx 10
5 cos d
25
sin2
5 cos
2 2 225 x2 csc2 d cot C C 5 5 5x
8. Same substitution as in Exercise 5
x2 dx 25 x2
25 sin2 25 5 cos d 5 cos 2
1 25 25 sin 2 C sin cos C 2 2 2
25 x x arcsin 2 5 5
25 x2
5
10. Same substitution as in Exercise 9
x2 4
x
1 cos 2 d
dx
2 tan 2 sec tan d 2 2 sec
2tan C 2
x2 4
2
C 21 25 arcsin 5x x
25 x2 C
tan2 d 2 sec2 1 d arcsec
1 7 6x x2 2
2x C x
2
4 2 arcsec
2x C
Section 7.4 12. Same substitution as in Exercise 9
x3 dx 4
x2
8 sec3 2 sec tan d 8 sec4 d 2 tan
8 1 tan2 sec2 d 8 tan
8 x2 4 3 2
3 x 4 4 C 31 2
14. Same substitution as in Exercise 13.
Trigonometric Substitution
tan3 8 C tan 3 tan2 C 3 3
1 3
x2 4 12 x2 4 C x2 4 x2 8 C
9x3 tan3 sec3 dx 9 sec2 d 9 sec2 1sec tan d 9 sec C 2 1 x sec 3
3 sec sec2 3 C 31 x2 1 x2 3 C 31 x2x2 2 C 16. Same substitution as in Exercise 13
x2 dx 1 x22
x2 dx 1 x2 4
1 2
tan2 sec2 d sec4
1 cos 2 d
sin2 d
1 sin 2 1 sin cos C 2 2 2
x 1 arctan x 2 1 x2
1 1 x2
C 21 arctan x 1 x x C 2
18. Let u x, a 1, and du dx.
1 x2 dx
20.
1 x1 x2 ln x 1 x2 C 2
x 1 dx 9 x21 22x dx 2 9 x 2 9 x21 2 C
22.
1 25 x2
dx arcsin
x C 5
(Power Rule)
24. Let u 16 4x2, du 8x dx.
x16 4x2 dx
1 8
16 4x21 28x dx
1 2 16 4x23 2 C 4 x23 2 C 12 3
26. Let u 1 t 2, du 2t dt.
t 1 dt 1 t 23 2 2
28. Let 2x 3 tan , dx
4x2 9
x4
1 t 23 22t dt
C
3 sec2 d, 4x2 9 3 sec . 2
dx
8 9
8 C 27 sin3
3 sec 3 2 sec2 d 3 24 tan4
1 1 t 2
cos d sin4
8 csc3 C 27
4x2 93 2 C 27x3
4x 2 + 9 2x
θ 3
329
330
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
30. Let 2x 4 tan , dx 2 sec2 d, 4x2 16 4 sec .
1 dx x4x2 16
1 4
2 sec2 d 2 tan 4 sec
sec 1 d tan 4
x2 + 4
x
csc d
θ
2
1 1 x2 4 2 ln csc cot C ln C 4 4 x
32. Let x 3 tan , dx 3 sec2 d, x2 3 3 sec2 .
x2
1 dx 33 2
1 3
3 sec2 d
33 sec3
x2 + 3
x
cos d
θ 3
1 sin C 3
x C 3x2 3
34. Let u x2 2x 2, du 2x 2 dx.
x 1x2 2x 2 dx
1 2
1 x2 2x 21 22x 2 dx x2 2x 23 2 C 3
36. Let x sin , x sin2 , dx 2 sin cos d, 1 x cos .
1 x x
dx
cos 2 sin cos d sin 1
2 cos2 d
x
θ 1− x
1 cos 2 d
sin cos C arcsinx x1 x C 38. Let x tan , dx sec2 d, x2 1 sec2 .
1 x3 x 1 dx x 4 2x2 1 4
x4
4x3 4x dx 2x2 1
1 lnx 4 2x2 1 4
1 1 lnx2 1 2 2
1 dx x2 12
d sec4
θ 1
1 cos 2 d
1 1 lnx2 1 sin cos C 2 2
1 x lnx2 1 arctan x 2 C 2 x 1
x2 + 1 x
sec2
Section 7.4
40. u arcsin x, ⇒ du
x arcsin x dx
1 x2 dx, dv x dx ⇒ v 2 2 1 x
x2 1 x2 arcsin x dx 2 2 1 x2
x sin , dx cos d, 1 x2 cos
x arcsin x dx
Trigonometric Substitution
x2 1 arcsin x 2 2
x2 1 sin2 cos d arcsin x cos 2 4
1 cos 2 d
1 1 1 x2 x2 arcsin x sin 2 C arcsin x sin cos C 2 4 2 2 4
1 1 x2 arcsin x arcsin x x1 x2 C 2x2 1 arcsin x x1 x2 C 2 4 4
42. Let x 1 sin , dx cos d, 1 x 12 2x x2 cos .
x2 2x x2
dx
x2
1 x 12
dx 1
1 sin 2cos d cos
1 2 sin
sin2
x−1
θ 1 − (x − 1)2
d
3 1 2 sin cos 2 d 2 2
1 3 2 cos sin 2 C 2 4 1 3 2 cos sin cos C 2 2
1 3 arcsinx 1 22x x2 x 12x x2 C 2 2
3 1 arcsinx 1 2x x2x 3 C 2 2
44. Let x 3 2 sec , dx 2 sec tan d, x 32 4 2 tan .
x dx x 6x 5 2
x dx x 32 4
2 sec 3 2 sec tan d 2 tan
2 sec2 3 sec d
2 tan 3 ln sec tan C1
2
x 33 4
2
3 lnx 2 3
x 32 4
2
C1
x2 6x 5 3 ln x 3 x2 6x 5 C
331
332
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
46. Same substitution as in Exercise 45 (a)
1 dt 1 t25 2
3 2
Thus,
0
cos d cos5
sec4 d
tan2 1 sec2 d
1 3 1 t tan tan C 3 3 1 t 2
t 1 t3 2 5 2 dt 1 t 31 t 23 2 1 t 2
3
t 1 t 2
C
3 2
0
3 2 33 8 3 3 23 3.464. 31 43 2 1 4
(b) When t 0, 0. When t 3 2, 3. Thus,
3 2
0
1 1 3 dt tan tan 1 t25 2 3
48. (a) Let 5x 3 sin , dx
1 3 3 3 23 3.464. 3
3 cos d, 9 25x2 3 cos . 5
9 1 cos 2 d 5 2
0
3 3 cos cos d 5
9 25x2 dx
9 1 sin 2 C 10 2
9 sin cos C 10
5x 5x 9 arcsin 10 3 3
3 5
Thus,
3
9 25x2 dx
0
9 25x2
3
C
9 5x 5x9 25x2 arcsin 10 3 9
3 5
0
9 9 . 10 2 20
3 (b) When x 0, 0. When x , . 5 2
3 5
Thus,
9 25x2 dx
2
109 sin cos
0
0
9 9 . 10 2 20
50. (a) Let x 3 sec , dx 3 sec tan d, x2 9 3 tan .
x2 9
x2
dx
3 tan 3 sec tan d 9sec2 tan2 d sec sin2 d cos 1 cos2 d cos
sec cos d
ln sec tan sin C ln —CONTINUED—
x 3
x2 9 9 C 3 x
x2
Section 7.4
Trigonometric Substitution
333
50. —CONTINUED—
6
Hence,
x2 9
x2
3
dx ln
. 3
(b) When x 3, 0; when x 6,
6
Hence,
x2 9
x2
3
52.
54.
x2 9 x2 9 x 3 3 x
dx ln sec tan sin
3
0
6
3
ln 2 3
3
ln 2 3
2
3
2
.
.
1 75 x2 2x 113 2 dx x 1x2 2x 26x2 2x 11 ln x2 2x 11 x 1 C 4 2
x2x2 4 dx
1 3 2 1 x x 4 xx2 4 2 ln x x2 4 C 4 2
56. (a) Substitution: u x2 1, du 2x dx (b) Trigonometric substitution: x sec 58. (a) x2 y k2 25
1 (b) Area square circle 4
y
Radius of circle 5 k 2 52 52 50
1 25 52 25 1 4 4
(0, k)
k 52
5
1 (c) Area r 2 r 2 r 2 1 4 4
5 5
x
60. (a) Place the center of the circle at 0, 1; x2 y 12 1. The depth d satisfies 0 ≤ d ≤ 2. The volume is
V32 6
d
1 y 12 dy
0
1 arcsiny 1 y 11 y 12 2
d
(Theorem 7.2 (1)) 0
3 arcsind 1 d 11 d 12 arcsin1 (b)
3 3 arcsind 1 3d 12d d 2. 2
d
10
V6
(d)
1 y 12 dy
0
dV dV dt dd 0
dd 61 d 12 dt
1
d t 4
2 0
(c) The full tank holds 3 9.4248 cubic meters. The horizontal lines y
⇒ d t (e)
1 241 d 12
0.3
9 3 3 , y , y 4 2 4
intersect the curve at d 0.596, 1. 0, 1.404. The dipstick would have these markings on it.
0
2 0
The minimum occurs at d 1, which is the widest part of the tank.
334
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
62. Let x h r sin , dx r cos d, r 2 x h2 r cos .
y
Shell Method:
hr
V 4
xr 2 x h2 dx
x
hr
4
2
2
h2
2
4 r 2
2
1 cos 2 d r
2
2 r 2h
h r sin r cos r cos d 4 r 2
1 sin 2 2
2
2
h−r
2
4 r 3
h r sin cos2 d
sin cos2 d
3
h+r
2
2
cos3
h
2
r
2 2 r 2h
2
x−h
θ r 2 − (x − h)2
1 64. y x2, y x, 1 y 2 1 x2 2
4
s
1 x2 dx
0
12xx
2
4
1 ln x x2 1
(Theorem 7.2)
0
1 417 ln4 17 9.2936 2
66. (a) Along line: d1 a2 a4 a1 a2
y
Along parabola: y x2, y 2x
(a, a 2)
a
d2
1 4x2 dx
0
y = x2
a
1 2x4x2 1 ln 2x 4x2 1 4
(Theorem 7.2)
0
x
(0, 0)
1 2a4a2 1 ln2a 4a2 1 4
(b) For a 1, d1 2 and d2
5
2
1 ln2 5 1.4789. 4
For a 10, d1 10101 100.4988 d2 101.0473. (c) As a increases, d2 d1 → 0. 68. (a)
(b) y 0 for x 72
25
− 10
80 −5
(c) y x
x2 x x , y 1 , 1 y 2 1 1 72 36 36
72
s
1 1
0
36 2
x 36
2
72
dx 36
18
x 36
0
1 1
x 36
2
x 36
1 1
2 ln 1 2
1
2
ln 1
361 dx 2
x 36
1 1
36
2 ln 1 2
x 36
2 18 ln
2
72 0
2 1 2 1
82.641
Section 7.4
Trigonometric Substitution y
70. First find where the curves intersect. y 2 16 x 42
6
1 4 x 16
(4, 4) 4 2
16 2 16x 4 2 x4
x −2 −2
16 2 16x 2 128x 162 x4
x4
16x2
2
4
0
−6
128x 0
1 1 1 2 x dx 42 x3 4 4 12
4
My
x
0
x4 16
1 2 x dx 4
4 0
4
y=
16 4 3
8
x16 x 42 dx
4
8
4
0
8
x 416 x 42 dx
4
16
416 x 42 dx
4
13 16 x 4
8
2 3 2
4
2 16 arcsin
x4 x 416 x 42 4
1 16 163 2 2 16 3 2
16 16 643 16 112 3
4
Mx
0
dx
8
2
1 1 2 x 2 4
4
1 16 x 42 dx 2
321 x5 8x x 6 4
64 416 32 64 32 5 6 15
5 4
3 8
0
4
x
My 112 3 16 112 48 28 12 4.89 A 16 3 4 16 12 4 3
y
Mx 416 15 104 1.55 A 16 3 4 54 3
x, y 4.89, 1.55 72. Let r L tan , dr L sec2 d, r 2 L2 L2 sec2 . 1 R
R
0
b
2mL 2mL dr r 2 L23 2 R 2m RL
a
L sec2 d L3 sec3
b
cos d
RL sin
2m RL
b
a
r r 2 L2
2m LR 2 L2
r 2 + L2 r
θ L
a
2m
6
10
−4
xx 4x2 4x 32 ⇒ x 0, 4 A
4
R 0
8 4
16 − (x − 4)
2
335
336
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
0.8
74. (a) Finside 48
1
0.8 y21 y 2 dy
0.8
96 0.8
1
0.8
1 y2 dy
1
y1 y 2 dy
2 arcsin y y1 y 31 y 0.8
96
0.8
1
2
961.263 121.3 lbs
2 32
1
0.4
(b) Foutside 64
1
0.4 y21 y 2 dy
0.4
128 0.4 128
1
0.4
1 y 2 dy
y1 y 2 dy
1
0.4 arcsin y y1 y 2 31 1 y 232 2
76. S 1520.4 111.2t 15.8t2 (a)
0.4
1
92.98
78. False
60
x2 1
x
dx
0
7
tan sec tan d sec
tan2 d
0
1 (b) St 1520.4 111.2t 15.8t212111.2 31.6t 2 S5 2.71 (c) Average value
80. True
1
1
St dt 68.24
x21 x2 dx 2
1 2
12
10
1
x21 x2 dx 2
0
Section 7.5
sin2 cos cos d 2
0
2
sin2 cos2 d
0
Partial Fractions
2.
4x 2 3 A B C x 53 x 5 x 52 x 53
6.
2x 1 A Bx C Dx E 2 2 xx 2 12 x x 1 x 12
8.
1 1 A B 4x 2 9 2x 32x 3 2x 3 2x 3
4.
1 A2x 3 B2x 3 3 2,
1 1 dx 4x2 9 6
2x 1 3 dx 2x 1 3 dx
1 ln 2x 3 ln 2x 3 C 12
1 2x 3 ln C 12 2x 3
10.
x2 x2 A B x 2 4x 3 x 1x 3 x 1 x 3
x1 dx x2 4x 3
1 6.
When x 1 6A, A When x 32 , 1 6B, B 16 .
2
x 1 dx x 1x 3
1 dx ln x 3 C x3
Section 7.5
12.
Partial Fractions
337
5x2 12x 12 A B C xx 2x 2 x x2 x2 5x2 12x 12 Ax2 4 Bxx 2 Cxx 2 When x 0, 12 4A ⇒ A 3. When x 2, 16 8B ⇒ B 2. When x 2, 32 8C ⇒ C 4.
14.
5x2 12x 12 dx x3 4x
3 dx x
2 dx x2
4 dx 3 ln x 2 ln x 2 4 ln x 2 C x2
x3 x 3 2x 1 A B x1 x1 x2 x 2 x 2x 1 x2 x1 2x 1 Ax 1 Bx 2 When x 2, 3 3A, A 1. When x 1, 3 3B, B 1.
x3 x 3 dx x2 x 2
16.
x1
1 1 dx x2 x1
x2 x2 x ln x 2 ln x 1 C x ln x 2 x 2 C 2 2
x2 A B xx 4 x 4 x
18.
2x 3 A B x 12 x 1 x 12
x 2 Ax Bx 4
2x 3 Ax 1 B
When x 4, 6 4A, A 32 . When x 0, 2 4B, B 12 .
x2 dx x 2 4x
When x 1, B 1. When x 0, A 2.
32 12 dx x4 x
2x 3 dx x 12
2 1 dx x 1 x 12
2 ln x 1
3 1 ln x 4 ln x C 2 2
20.
1 C x1
4x2 4x2 4x2 A B C x3 x2 x 1 x2x 1 x 1 x2 1x 1 x 1 x 1 x 12 4x2 Ax 12 Bx 1x 1 Cx 1 When x 1, 4 2C ⇒ C 2. When x 1, 4 4A ⇒ A 1. When x 0, 0 1 B 2 ⇒ B 3.
x3
4x2 dx x1 x2
1 dx x1
3 dx x1
ln x 1 3 ln x 1
22.
2 dx x 12
2 C x 1
6x A 6x Bx C x3 8 x 2x2 2x 4 x 2 x2 2x 4 6x Ax2 2x 4 Bx Cx 2 When x 2, 12 12A ⇒ A 1. When x 0, 0 4 2C ⇒ C 2. When x 1, 6 7 B 21 ⇒ B 1.
6x dx x3 8
1 dx x2
x 2 dx x2 2x 4
1 dx x2
3 arctan
x 1 dx x2 2x 4
1 3 x1 arctan C ln x2 2x 4 3 2 3
1 ln x2 2x 4 2
ln x 2 ln x 2
3x 1
3
C
3 dx x2 2x 1 3
338
24.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
x2 x 9 Ax B Cx D 2 2 x2 92 x 9 x 92 x2 x 9 Ax Bx2 9 Cx D Ax3 Bx2 9A Cx 9B D By equating coefficients of like terms, we have A 0, B 1, D 0, and C 1.
x2 x 9 x2 92
26.
1 dx x2 9
x dx x2 92
x 1 1 arctan C 3 3 2x2 9
x 2 4x 7 A Bx C x 1x 2 2x 3 x 1 x 2 2x 3 x 2 4x 7 Ax 2 2x 3 Bx Cx 1 When x 1, 12 6A. When x 0, 7 3A C. When x 1, 4 2A 2B 2C. Solving these equations we have A 2, B 1, C 1.
x 2 4x 7 dx 2 x3 x 2 x 3
1 dx x1
2 ln x 1
28.
x 1 dx x 2 2x 3
1 ln x 2 2x 3 C 2
x 2 x 3 Ax B Cx D 2 2 x 2 32 x 3 x 32 x 2 x 3 Ax Bx 2 3 Cx D Ax 3 Bx 2 3A Cx 3B D By equating coefficients of like terms, we have A 0, B 1, 3A C 1, 3B D 3. Solving these equations we have A 0, B 1, C 1, D 0.
x2 x 3 dx x4 6x 2 9
30.
1 x dx x 2 3 x 2 32
1 3
arctan
x 3
1 C 2x 2 3
x1 A B C 2 x 2x 1 x x x1 x 1 Axx 1 Bx 1 Cx 2 When x 0, B 1. When x 1, C 2. When x 1, 0 2A 2B C. Solving these equations we have A 2, B 1, C 2.
5
1
x1 dx 2 x 1
x2
5
1
1 dx x
5
1
5
1 dx 2 x2
1
1
1 2 ln x 2 ln x 1 x
2 ln 2 ln
x 1 x1 x
5 4 3 5
5 1
1 dx x1 5
Section 7.5
x2
x1 2 7 6x 2 1 1 dx 3 ln C x 2x 13 x x x 1 2x 12
1
32.
0
34.
x2 x dx x1
1
0
0
x2
2x 1 dx x ln x 2 x 1 x1
0 1 ln 3
1
2, 1: 3 ln
1
dx
Partial Fractions
4
1 1 2 7 1 C 1⇒ C 2 3 ln 2 2 1 2 2
(2, 1)
− 4.7
4.7
−4
36.
x2
3, 4:
x3 1 2 C dx ln x 2 4 2 4 2 2 x 4
8
2 22 1 1 ln 5 C 4 ⇒ C ln 5 2 5 5 2
(3, 4) −8
8 −2
38.
x2x 9 1 5 dx 2 ln x 2 C x 3 6x 2 12x 8 x 2 x 22
10
3, 2: 0 1 5 C 2 ⇒ C 4
(3, 2) − 10
10 −3
40.
x2 x 2 dx arctan x ln x 1 C x3 x2 x 1
10
2, 6: arctan 2 0 C 6 ⇒ C 6 arctan 2
(2, 6) −2
5 −2
42. Let u cos x, du sin x dx.
44.
1 A B uu 1 u u1
1 Au 1 Bu
1 Au 1 Bu When u 0, A 1. When u 1, B 1, u cos x. du sin dx.
sin x 1 dx du cos x cos2 x uu 1
1 A B , u tan x, du sec2 x dx uu 1 u u1
1 du u
ln u 1 ln u C
ln
u1 C u
ln
cos x 1 C cos x
sec2 x dx tan xtan x 1
1 du u1
When u 0, A 1. When u 1, 1 B ⇒ B 1.
ln 1 sec x C
1 du uu 1
1 1 du u u1
ln u ln u 1 C
ln
u C u1
ln
tan x C tan x 1
339
340
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
46. Let u ex, du ex dx. 1 A Bu C 2 u2 1u 1 u 1 u 1 1 Au2 1 Bu Cu 1 When u 1, A 12 . When u 0, 1 A C. 1 1 1 When u 1, 1 2A 2B 2C. Solving these equations we have A 2 , B 2 , C 2 , u e x, du e x dx.
48.
e
2x
ex dx 1e x 1
1 2
1 1 ln u 1 ln u2 1 arctan u C 2 2
1 2 ln e x 1 ln e 2x 1 2 arctan e x C 4
1 du u2 1u 1
u 1 1 du uu 11 du 2
1 A B a2 x2 a x a x
1 A C B 2 x2a bx x x a bx
50.
1 Axa bx Ba bx Cx 2
1 Aa x Ba x
When x 0, 1 Ba ⇒ B 1a. When x ab, 1 Ca2b2 ⇒ C b2a2. When x 1, 1 a bA a bB C ⇒
When x a, A 12a. When x a, B 12a.
1 1 dx a2 x2 2a
1 1 dx ax ax
1 ln a x ln a x C 2a
A ba 2. 1 dx x a bx 2
1 ax ln C 2a a x
52.
dy 4 , y0 5 dx x2 2x 3
54. (a)
8
(b)
−1
56. A 2
0
y
3
2
3
dx 14
0
2x 6
0
5 2
2 3 2
1 dx 16 x2
14 4 x ln 8 4x
7 ln 7 2.595 4
3 0
x
(From Exercise 46)
−3 −2 −1
1 b b ln x ln a bx C a2 ax a2
a bx b 1 ln C ax a2 x
1 x b ln C ax a2 a bx
Nx A1 B1x An Bn x . . . Dx ax2 bx c ax2 bx cn
−2
7 1 dx 16 x2
ba2 1a b2a2 2 dx x x a bx
Nx A1 A2 Am . . . Dx px q px q2 px qm
3
3
1
2
3
Section 7.5 58. (a)
Partial Fractions
341
(e) k 1, L 3
y 6
i y0 5: y
5 4
15 5 2e3t
1 32 3 ii y0 : y 2 12 52e3t 1 5e3t
3 2 1
5
t 1
2
3
4
5
6
(b) The slope is negative because the function is decreasing. (c) For y > 0, lim yt 3.
0
B A dy yL y y Ly
(d)
dy kyL y dt
(f )
dy dy d 2y k y L y 0 dt 2 dt dt
1 1 1 AL y By ⇒ A , B L L
dy yL y
1 L
1 dy y
5 0
t →
⇒ y
k dt
1 dy Ly
⇒
L 2
From the first derivative test, this is a maximum.
1 ln y ln L y kt C1 L
ln
y k L t LC1 Ly C2ekLt
When t 0,
y Ly
y y0 y0 C2 ⇒ ekLt. L y0 L y L y0 y0L . y0 L y0ekLt
Solving for y, you obtain y
3
60. (a) V
0
2x 2 dx 4 2 x 1
3
0
3
4
0
x2
x2 dx 12
1 1 dx x2 1 x2 12
(partial fractions)
1 x arctan x 2 2 x 1
x x2 1
4 arctan x 2 arctan x —CONTINUED—
dy dy L y dt dt y
k dt
3 0
3
(trigonometric substitution)
0
2 arctan 3
3 5.963 10
342
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
60. —CONTINUED—
3
(b) A
0
x
1 A
x2 3
0
2x2 1 dx 2 x 1 ln 10
2 ln 10
3
0
3
0
0
ln 10
3
2
0
3
1 2x 2 arctan x ln 10 1 1 y A 2
3
2x dx lnx2 1 1
0
y
2
2 dx x2 1
(1.521, 0.412) 1
2 3 arctan 3 1.521 ln 10
2 2x 2 dx x2 1 ln 10
(partial fractions)
1 x 2 arctan x arctan x 2 ln 10 2 x 1
2 1 x arctan x ln 10 2 2x2 1
3 0
3
x2 dx x2 12
1 1 dx x2 1 x2 12
2
−1
3
0
x 1
3
(trigonometric substitution)
0
1 x arctan x 2 ln 10 x 1
3 0
1 3 arctan 3 0.412 ln 10 10
x, y 1.521, 0.412 B 1 1 A 1 ,A ,B y0 xz0 x y0 x z0 x z0 y0 z0 y0
62. (a) 1 z0 y0
(Assume y0 z0)
1 1 dx kt C y0 x z0 x
1 z x ln 0 kt C, when t 0, x 0 z0 y0 y0 x C
z 1 ln 0 z0 y0 y0
kt
1 z x z ln 0 ln 0 z0 y0 y0 x y0
y0z0 x z0 y0k t 0 0 x
z y
ln
y0z0 x ez0 y0kt z0 y0 x x (b) (1) If y0 < z0, lim x y0. t →
(2) If y0 > z0, lim x z0. t →
y0z0 ez0 y0kt 1 z0ez0 y0kt y0 (c) If y0 z0, then the original equation is
1 dx y0 x 2
k dt
y0 x1 kt C1 x 0 when t 0 ⇒
1 C1 y0
kt y0 1 1 1 kt y0 x y0 y0 y0 x
y0 kt y0 1
x y0
y0 kt y0 1
As t → , x → y0 x0.
Section 7.6
Section 7.6
Integration by Tables and Other Integration Techniques
Integration by Tables and Other Integration Techniques
2. By Formula 13: b 2, a 5
2 2 1 4 1 5 4x x dx ln 3 x22x 52 3 25 x5 2x 5 2x 5
2 4x 5 x 8 ln C 375 2x 5 75 x2x 5
4. By Formula 29: a 3
C
6. By Formula 41:
1 1 x 9 x dx x2 9 arcsec C 3 x 3 3 2
8. By Formulas 51 and 47:
cos3 x dx 2
x
u x, du 10. By Formula 71:
1 1 dx 1 tan 5x 5
2
x
9 x 4
2 1 x dx
x sin x
3
1 dx 2 x
1 1 u ln cos u sin u C 5 2
1 5x ln cos 5x sin 5x C 10
u 5x, du 5 dx
12. By Formula 85:
a 21, b 2
ex 2 sin 2x dx
ex 2 1 sin 2x 2 cos 2x C 1 4 4 2
4 x 2 1 e sin 2x 2 cos 2x C 17 2
14. By Formulas 90 and 91:
ln x3 dx xln x3 3 ln x2 dx xln x3 3x 2 2 ln x ln x2 C x ln x3 3ln x2 6 ln x 6 C
16. (a) By Formula 89:
x 4 ln x dx
x5 x5 1 5 x ln x C 2 1 4 1 ln x C 5 25 5
(b) Integration by parts: u ln x, du
x 4 ln x dx
x5 ln x 5
x5 1 dx, dv x 4 dx, v x 5
x5 1 x5 x5 dx ln x C 5 x 5 25
1 2x dx 2 32 x 22 1 x2 arcsin C 2 3
2 1 2 cos x dx sin x cos 2 x 2 C 3 3 2 x
1 5 dx 1 tan 5x
dx
cos3 x
cos
2
343
344
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
18. (a) By Formula 24: a 75, x u, and
(b) Partial fractions:
1 1 x 75 dx ln C x 2 75 2 75 x 75
3
30
ln
A 1 B x 2 75 x 75 x 75 1 A x 75 B x 75
x 75 C x 75
x 75: 1 2A 75 ⇒ A
3 1 1 30 2 75 10 3
x 75: 1 2B 75 ⇒ B
1 dx x 2 75
20. By Formula 21:
22. By Formula 79:
x
1 x
30
3 30
3 30 dx x 75 x 75
3
30
3
ln
x 75 C x 75
2 dx 2 x 1 x C 3
arcsec 2x dx
1 2x arcsec 2x ln 2x 4x 2 1 C 2
u 2x, du 2 dx
24. By Formula 52:
26. By Formula 7:
28. By Formula 14:
x sin x dx sin x x cos x C
1 sin
3
1 2 2x 2 dx arctan C arctanx 1 C x 2 2x 2 2
4
30. By Formula 56:
2
C
x2 1 25 10 ln 3x 5 dx 3x 3x 52 27 3x 5
d
32. By Formula 71:
1 1 3 2 d 3 1 sin 3
1 dt t 1 ln t2
u ln t, du
u e x, du e x dx
1 tan 3 sec 3 C 3
34. By Formula 23:
ex 1 dx e x ln cos e x sin e x C 1 tan e x 2
1 1 dt arctanln t C 1 ln t2 t
1 dt t
36. By Formula 26:
1 x x 2 3 3 ln x x 2 3 C 2
38. By Formula 27:
1 3x2 2 2 3x2 3 dx 27
3 x 2 dx
x 2 2 3x2 dx
1 3x18x 2 2 2 9x 2 4 ln 3x 2 9x 2 C 827
Section 7.6
40. By Formula 77:
2 3
x arctanx3 2 dx
42. By Formula 45:
arctanx3 2
Integration by Tables and Other Integration Techniques
32 x dx
2 3 2 x arctanx3 2 ln 1 x3 C 3
ex ex dx C 2x 3 2 1 e
1 e 2x
u e x, du e x dx 44. By Formula 27:
2x 32 2x 32 4 dx
1 2x 32 2x 3)2 42 dx 2
1 2x 3 2x 32 2 2x 32 4 ln 2x 3 2x 32 4 C 8
u 2x 3, du 2 dx
46. By Formula 31:
cos x
sin2 x 1
dx ln sin x sin2 x 1 C
u sin x, du cos x dx
48.
3x dx 3x
3x
9 x 2
3
1 dx
9 x 2
3 arcsin
50. By Formula 67:
dx
un
a bu
du
x 9 x 2 C 3
du
a bu
2 , v a bu b
a bu 2n n1 2u n
a bu u a bu du b b
a bu
2n au n1 bu n 2u n
a bu du b b
a bu
2u n 2na u n1 un
a bu du 2n du b b a bu
a bu
54.
2u n 2n n1
a bu u a bu du b b
Therefore, 2n 1
tan3 d
x dx
9 x 2
52. Integration by parts: w u n, dw nu n1 du, dv
un
a bu
un
a bu
du
2 n un1 u a bu na du and b
a bu
2 un1 un a bu na du . 2n 1b
a bu
uncos u du un sin u n un1sin u du
w un, dv cos u du, dw nun1 du, v sin u
tan2 2
tan d
tan2 ln cos x C 2
345
346
56.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
ln un du uln un
nln un1
1uu du uln u
w ln un, dv du, dw nln un1
58.
x x 2 2x dx
0, 0:
n
n ln un1 du
1u du, v u
1 2x 2 2x3 2 3x 1 x 2 2x 3 ln x 1 x 2 2x C 6
15
1 3 ln 1 C 0 ⇒ C 0 6
−6
6
(0, 0)
−15
60.
2 2x x 2
x1
0, 2 :
dx 2 2x
x2
3 ln
3 2 2x x 2
x1
3
(0, 2 )
C
2 3 ln 3 2 C 2 ⇒ C 3 ln 3 2
−2.5
1
−6
62.
0, 1: C 1 ⇒ y
64.
sin 1 sin 1 sin d ln cos 1 sin 2 1 sin cos
1 sin 1 sin ln 2 1 sin cos
C
10
1
sin sin d d 1 cos2 1 cos 2
−8
8
(0, 1) −2
2
66.
0
1 d 3 2 cos
arctancos C
1
0
1
2
0
u tan
68.
cos d 1 cos
cos 1 cos d 1 cos 1 cos
70.
2
1 d sec tan
2u 1 u2 21 u2 3 1 u2
1 du 5u2 1
25 arctan 5 u
1 0
2
5
arctan 5
1 d 1 cos sin cos
cos cos2 d sin2
csc cot cot2 d
ln 1 sin C
csc cot csc2 1 d
csc cot C
cos d 1 sin
u 1 sin , du cos d
Section 7.6
2
72. A
0
1 2
0
1 2
2x dx 2 1 ex
1 4
2
1 2 2 x ln1 e x 2
1 1 4 ln1 e4 ln 2 2 2
347
y
x 2 dx 1 ex 2
Integration by Tables and Other Integration Techniques
x 1
2
0
0.337 square units
74. Log Rule:
1 du, u ex 1 u
78. Formula 16 with u e2x
76. Integration by parts
5
82. W
80. A reduction formula reduces an integral to the sum of a function and a simpler integral. For example, see Formula 50, 54.
0
500x
26 x 2
dx
5
250
26 x 21 22x dx
0
500 26 x 2 500 26 1 2049.51 ft lbs
84.
2
1 20
0
2500 5000 dt 1 e4.81.9t 1.9
2
0
1.9 dt 1 e4.81.9t
2500 4.8 1.9t ln1 e4.81.9t 1.9
2500 1 ln1 e 4.8 ln1 e4.8 1.9
2500 1e 3.8 ln 1.9 1 e4.8
86. (a)
401.4
20
k
6x 2ex 2
dx 50
0
By trial and error, k 5.51897.
0
5.51897
(b)
0
88. True
6x 2ex 2 dx
2
−1
5.52
0
5 0
348
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
Section 7.7
Indeterminate Forms and L’Hôpital’s Rule
1 ex 1 x →0 x
2. lim
1
−1
0.1
0.01
0.001
0.001
0.01
0.1
0.9516
0.9950
0.9995
1.00005
1.005
1.0517
x f x
6x
4. lim
x →
3x2 2x
3.4641
exact: 63
5
1
10
102
103
104
105
f x
6
3.5857
3.4757
3.4653
3.4642
3.4641
x →1
(b) lim
x →1
8. (a) lim
x →0
x →
(b) lim
x →
0
100 0
2x2 x 3 2x 3x 1 lim lim 2x 3 5 x →1 x →1 x1 x1 2x2 x 3 ddx 2x2 x 3
4x 1 lim 5 lim x →1 x →1 x1 ddx x 1
1
sin 4x sin 4x lim 2 21 2 x →0 2x 4x
10. (a) lim
−2
x
6. (a) lim
(b) lim
x →0
sin 4x 4 cos 4x ddx sin 4x
lim lim 2 x →0 x →0 2x ddx 2x
2
2x 1 2x 1x2 0 lim 0 2 x → 4x x 4 1x 4 2x 1 ddx 2x 1
2 lim lim 0 4x2 x x → ddx 4x2 x x → 8x 1
x2 x 2 2x 1 lim 3 x →1 x →1 x1 1
12. lim
14. lim x →2
4 x2
x2
lim x →2
lim x →2
16. lim x →0
ex 1 x ex 1 lim 3 x →0 x 3x2 lim x →0
ex 6x
sin ax a cos ax a lim x→0 sin bx x→0 b cos bx b
20. lim
x →
x1 1 lim 0 x2 2x 3 x → 2x 2
x2 2x 2 lim x lim x 0 x→ ex x→ e x→ e
28. lim
x4 x2 1 x 4 x2
2 18. lim ln x lim 2 ln x x→1 x2 1 x→1 x2 1
lim
2x 2x
lim
1 1 x2
x→1
x→1
24. lim
1
arctan x 4 11 x2 1 lim x →1 x →1 x1 1 2
22. lim
26. lim
x→
x3 3x2 lim x→ x1 1
Section 7.7
30. lim
x→
x2 x lim x→ 1 1x2 1
Indeterminate Forms and L’Hôpital’s Rule
32. lim
x2
x→
sin x 0 x
Note: Use the Squeeze Theorem for x > .
e x2 12e x2 lim x→ x x→ 1
ln x 4 4 ln x 4x lim lim x→ x3 x→ x→ 3x2 x3
36. lim
34. lim
lim
x→
4 0 3x3
38. (a) lim x3 cot x 0 (b) lim x3 cot x lim (c)
x→0
40. (a) lim x tan x→
x→0
x→0
1 sin x 1 ≤ ≤ x x x
x3 tan x
lim x→0
3x2 sec2 x
0
(b) lim x tan x→
1 0 x
1 tan1x lim x x→ 1x
1
lim
x→
0
1x2 sec21x 1x2
3
lim sec2 x→
−1
(c)
1x 1
2
1
10
−1
42. (a) lim e x x2x 1 x→0
(b) Let y lim e x x2x. x→0
ln y lim x→0
2 ln
ex
x
1 1 (b) Let y lim 1 . x x→
1 x
x
x
x→
x
ln y lim
2e x 1e x x lim 4 x→0 1
x→
x ln1 1x lim ln 1 1x1x
x→
1 1x 1x2
Thus, ln y 4 ⇒ y e 4 54.598. (c)
44. (a) lim 1
lim
x→
60
1x2
lim
x→
1 1 1 1x
Thus, ln y 1 ⇒ y e1 e. Therefore,
lim 1
x→ 0
1 x
x
e.
2 0
(c)
5
0 −1
10
349
350
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals 48. (a) lim 3x 4 x4 00
46. (a) lim 1 x1x 0 x →
x→4
(b) Let y lim 3x 4 x4.
(b) Let y lim 1 x
1x.
x→
ln y lim
x→
lim
x→
x→4
ln y lim x 4ln 3x 4
ln1 x x
x→4
111 x 0
ln 3x 4
1x 4
lim
1x 4 1x 42
x→4
x→4
Thus, ln y 0 ⇒ y e0 1. Therefore, lim 1 x1x 1.
lim x 4 0
x →
(c)
lim
x→4
Hence, lim 3x 4 x4 1.
5
x→4
(c)
2
10
0 −1
4
7 0
2 x
x
50. (a) lim cos x→0
00
2 x . x
(b) Let y lim cos x→0
x 1 4 x x41 1 x1 1 x1 lim lim x 4 x 4 x 4
52. (a) lim (b)
x →2
2
x →2
2
2
2
x →2
2 x
ln y lim x ln cos x→0
lim x →2
000
2 x
x
Hence, lim cos x→0
(c)
lim x →2
1. (c)
2
1 2x 1 2x 1 4xx 1
0.25
2 4
2
− 0.25
0
3 0
54. (a) lim x→0
(b) lim x→0
10x x3
(c)
2
10x x3 lim 10xx 3 2
x→0
10
0
5
2
−20
56. (a)
(b) Let y sin xx, then ln y x lnsin x.
2
−
lim
x →0
2
−1
lnsin x cos xsin x x2 2x lim lim lim 0 2 x →0 x →0 tan x x →0 sec2 x 1x 1x
Therefore, since ln y 0, y 1 and lim sin xx 1. x →0
1 8
Section 7.7
58. (a)
Indeterminate Forms and L’Hôpital’s Rule x3 3x2 6x 6 lim 2x lim 2x lim 2x 0 2x x→ e x→ 2e x→ 4e x→ 8e
(b) lim
1 −1
5
−3
62. Let f x x 25 and gx x.
60. See Theorem 7.4.
64. lim
x3 3x2 6x 6 lim lim 0 lim e2x x→ 2e2x x→ 4e2x x→ 8e2x
66. lim
ln x2 2 ln xx lim x→ x3 3x2
x→
x→
x→
xm mxm1 lim enx x→ nenx mm 1xm2 n2enx
lim
2 ln x 3x3
lim
lim
2x 2 lim 3 0 x→ 9x 9x2
. . . lim
x→
x→
70.
68. lim
x→
x→
m! 0 nmenx
x
1
5
10
20
30
40
50
100
ex x5
2.718
0.047
0.220
151.614
4.40 105
2.30 109
1.66 1013
2.69 1033
ln x x Horizontal asymptote: y 0 (See Exercise 29)
74. y
72. y x x, x > 0 lim x x and lim x x 1
x→
x→0
dy x1x ln x1 1 ln x 0 dx x2 x2
No horizontal asymptotes ln y x ln x
Critical number:
1 1 dy x ln x y dx x
0, e Sign of dydx: y f x: Increasing 1 Relative maximum: e, e
dy x x1 ln x 0 dx
e1, 0 Increasing 1 1 1 Relative minimum: e1, e1e , e e
e, Decreasing
x e1
Critical number:
xe
Intervals:
0, e1 Sign of dydx: y f x: Decreasing Intervals:
1 −1
4
(e, 1e (
1e
−4
4
(
−1
((
1
1, 1 e e e
( 4
−1
sin x 1 0 x→ x
76. lim
(Numerator is bounded)
Limit is not of the form 00 or . L’Hôpital’s Rule does not apply.
ex 0 0 x → 1 ex 10
78. lim
Limit is not of the form 00 or . L’Hôpital’s Rule does not apply.
351
352
Chapter 7
80. A P 1
r n
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals nt
r ln A ln P nt ln 1 ln P n
lim
n →
ln 1
r n
lim
r n
1 nt
1 r n2 1 rn 1 2 nt
n →
1 nt
ln 1
lim
n →
rt
1
1
r n
rt
Since lim ln A ln P rt, we have lim A eln Prt eln Pert Pert. Alternatively, n→
r lim A lim P 1 n → n → n
n→
nt
1 n
lim P n →
r
nr rt
Pert.
82. Let N be a fixed value for n. Then
N 1x N2 N 1N 2x N3 N 1! x N1 lim lim . . . lim 0. x x x→ e x→ x→ x→ e ex ex
lim
84. f x
xk 1 k
k 1,
k=1
f x x 1 1 10x 0.1 1 0.1
x 0.1
f x
k 0.01,
x 0.01 1 100x 0.01 1 f x 0.01
k →0
(See Exercise 68)
6
k 0.1,
lim
k = 0.1 k = 0.01 −2
10 −2
xk 1 xk ln x lim ln x k →0 k 1
1 86. f x , gx x2 4, 1, 2
x f c f 2 f 1 g2 g1 g c
88. f x ln x, gx x3, 1, 4
f 4 f 1 f c g4 g1 g c
12 1c2 3 2c
ln 4 1c 1 2 3 63 3c 3c 3c3 ln 4 63
1 1 3 6 2c
c3
2c3 6 3 3 c
90. False. If y exx2, then y
x2ex 2xex xex x 2 ex x 2 . x4 x4 x3
c
21 ln 4
ln214 2.474 3
92. False. Let f x x and gx x 1. Then lim
x →
x 1, but lim x x 1 1. x → x1
Section 7.8
e1x , 0,
94. gx
2
Improper Integrals
x0
y
x0
1.5
gx g0 e1x lim g0 lim x →0 x →0 x0 x
2
e1x e1x , then ln y ln Let y x x
2
2
0.5
1 1 x2 ln x 2 ln x . Since x x2
x −2
−1
1
2
− 0.5
ln x 1x x2 lim lim 0 lim x2 ln x lim 2 3 x →0 x →0 1x x →0 2x x →0 2
1 x x
2
we have lim
x →0
ln x
2
. Thus, lim y e
x →0
0 ⇒ g0 0.
Note: The graph appears to support this conclusion—the tangent line is horizontal at 0, 0. 98. lim x ln 21ln x
96. lim f xgx
x →0
x →a
Let y x ln 21ln x, then:
y f xgx ln y gx ln f x
ln y
lim gx ln f x
ln 2 1 ln x
x →a
As x → a, ln y ⇒
, and hence y . Thus,
lim ln y lim
x →0
lim f xgx .
x →0
ln 2ln x ln 2x lim x →0 1 ln x 1x
x→0
Thus, lim y eln 2 2. x →
Improper Integrals
2. Infinite discontinuity at x 3.
4
3
1 dx lim b →3 x 332
x 332 dx
2 b 3
4
b
lim 2x 312 b →3
lim 2 b →3
4 b
Diverges 4. Infinite discontinuity at x 1.
2
0
1 dx x 123
1
0
1 dx x 123
b
lim b→1
0
b→1
2
1
1 dx x 123
1 dx lim c→1 x 123
3 x 1 lim 3
Converges
ln 2ln x 1 ln x
lim ln 2 ln 2
x →a
Section 7.8
ln x
b 0
2
c
1 dx x 123
3 x 1 lim 3 c→1
2 c
0 3 3 0 6
353
354
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
6. Infinite limit of integration.
0
8.
0
e2x dx lim
b→
ex dx 0. You need to evaluate the limit.
0
lim
b→
0
1 2x e 2
lim
b→
b
b
e2x dx
b
e
ex dx lim b→
0
1 1 0 2 2
lim
b→
b
x
0
e
1 1
b
Converges
10.
5
1
b
dx lim b→ x3
1
b→
xex2 dx lim
x 1ex dx lim
eax sin bx dx lim
b→
0
b→
18.
e
x2
c→
0
x 1ex dx lim xex b→
0
ax
e
a sin bx b cos bx a2 b2
x2
0
x3 dx lim b→ 12
b
0
x2
b→
x dx lim b→ 1
1 lnx 2
lim
2
b
0
1
1
lim
b 0
b→
0 0 by L’Hôpital’s Rule. b e b
20.
1
ln x dx lim b→ x b→
b
0
1 2 1 x2
x
x2 1 2
dx
Diverges
b→
lim
3 4
0
b b 0 2 a b2 a 2 b2
22.
163x
lim eb22b 4 4 4
b
c
4x14 dx
1
2x 4
b
0
b→
b→
xex2 dx lim
b
lim
b
b→
dx lim
5 2
1
0
16.
b
4 4 x
1
2
14.
12.
25x
lim
5 dx x3
24.
0
b
1
ln x dx x
ln x2 2
b 1
Diverges
b
ex dx lim ln1 ex b→ 1 ex
ln 2
0
b
Diverges
0
1 2
Diverges
26.
6
30.
x x b dx lim 2 cos b→ 2 2 0 0 x Diverges since cos does not approach a limit as x → . 2 sin
0
4 dx lim b→6 6 x
0
4
b
8 dx lim 8 ln x b→0 x
4 b
32.
0
b
e
ln x2 dx lim b→0
2 ln x dx
0
lim 2x ln x 2x b→0
0
e b
80 8 6
lim 2e 2e 2b ln b 2b
8 6
0
sec d lim
Diverges
e
46 x12 dx
8 dx lim b→0 x
Diverges
0
2
34.
0
b
lim 86 x12 b→6
4
28.
b→ 2
b→0
ln sec tan
b 0
2
,
36.
0
1 x dx lim arcsin b→2 2 4 x2
b 0
2
Section 7.8
2
38.
0
1 dx lim b→2 4 x2
b
0
1 1 1 1 2x dx lim ln b→2 4 2x 2x 4 2x
Diverges
3
40.
1
2
2 dx x 283
b 0
Improper Integrals
355
0
3
2x 283 dx
1
2x 283 dx
2
b
lim b→2
1
2x 283 dx
3
2x 283 dx lim c→2
6 lim x 253 b→2 5
b
c
6 lim x 253 c→2 5
1
3 c
Diverges
42.
Thus,
1
1 dx x ln x
0
1 dx x ln x 1
lim lnln x b→1
44. If p 1,
e
e
lim
1
1
1 dx ln ln x C x ln x
c→
1 dx lim ln x a→0 x
a
lim ln a . a→0
Diverges. If p 1,
e
1
1 dx x ln x
0
lnln x
.
1 x1p dx lim a→0 xp 1p
1 a
lim a→0
1 1 p 1a p . 1p
1 if 1 p > 0 or p < 1. 1p
This converges to
e
Diverges
1
46. (a) Assume
gx dx L (converges).
a
Since 0 ≤ f x ≤ gx on a, , 0 ≤
f x dx ≤
a
gx dx diverges, because otherwise, by part (a), if
a
1
48.
0
3 x
56.
f x dx converges.
a
gx dx converges, then so does
dx
f x dx.
1 3 converges. 1 13 2
1 1 ≥ on 2, and x x 1
a
50.
x 4ex dx converges.
0
See Exercise 44, p 13 .
(See Exercise 45.)
1
2
x
1 1 ≤ 32 on 1, and x x1 x
54. Since
gx dx L and
a
1
52. Since
a
(b)
dx diverges by Exercise 43,
2
1
1 x32
1 x 1
dx diverges.
dx converges by Exercise 43,
1 1 ≥ since x ln x < x on 2, . Since x x ln x
1
1
2
x
1 x1 x
dx diverges by Exercise 43,
2
dx converges.
1 x ln x
dx diverges.
58. See the definitions, pages 540, 543. 60. Answers will vary.
(a)
1e
Converges
62. f t t
ex 2x
dx
(b)
x dx
(Example 4)
Diverges
Fs
st 1e
s
test dt lim b→
0
1
2
1 ,s > 0 s2
st
b 0
356
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
64. f t eat Fs
66. f t sin at
eat est dt
0
etas dt
0
1 e as
b
t as
0
1 1 ,s > a as sa
68. f t sinh at
est sinh at dt
0
est sin at dt
b→
2
est s sin at a cos at a2
a a ,s > 0 s2 a2 s2 a2
eat 1 dt 2 2
at
s
0
et sa et sa dt
0
b
1 1 1 et sa et sa b→ 2 s a s a
lim
0
e
est
0
lim
0
lim b→
Fs
Fs
0
0
1 1 1 2 s a s a
1 1 a 1 2 ,s > a 2 s a s a s a2
70. (a) A
1
1 1 dx x2 x
1
1
(b) Disk: V
1
(c) Shell: V 2
1
x
x1 dx lim 2 ln x
b
2
b→
1
1
dx lim 3 b→ x4 3x
b 1
Diverges 72. x 22 y2 1 2x 2 2yy 0 y
x 2 y
1 y 2 1
3
S 4
1
x dx 4
y
x 2 2y 2
3
1
1 (Assume y > 0. y
x dx 4
1 x 22
3
1
74. (a) Fx W (b)
2 1 x 22
dx
4
0 2 arcsin1 2 arcsin1 8
lim 4 1 x 22 2 arcsinx 2 a→1 b→3
x2 1 x 22 b
a
K K ,5 , K 80,000,000 x2 40002
80,000,000 80,000,000 dx lim b→ x2 x 4000
80,000,000 W 10,000 2 x
b 4000
80,000,000 10,000 b b 8000 Therefore, 4000 miles above the earth’s surface.
b 4000
20,000 mi-ton
80,000,000 20,000 b
2
3
b 0
Section 7.8
76. (a)
2 2t5 e dt 5 4
(b)
0
0
b 2 2t5 e dt lim e2t5 1 b→ 5 0
4
2 2t5 e dt e2t5 5
0
e85 1
0.7981 79.81%
(c)
Improper Integrals
25 e dt lim te 2t5
t
0
2t5
b→
5 e2t5 2
b 0
5 2
5
78. (a) C 650,000
25,0001 0.08te0.06t dt
0
650,000 25,000
$778,512.58
$905,718.14
1 0.06t t 0.06t 1 e 0.08 e e0.06t 0.06 0.06 0.062
5 0
10
(b) C 650,000
25,0001 0.08te0.06t dt
0
650,000 25,000 (c) C 650,000
1 0.06t t 0.06t 1 e e 0.08 e0.06t 0.06 0.06 0.062
10 0
25,0001 0.08te0.06t dt
0
t e 0.06
650,000 25,000 lim
0.06t
b→
80. (a)
1
1
xn
1
1 e0.06t 0.062
b 0
$1,622,222.22
(b) It would appear to converge. y
1
1.00
dx will converge if n > 1 and will diverge if n ≤ 1.
0.50 0.25
u
1
0.06t
0.75
(c) Let dv sin x dx ⇒
b
t e 0.06
1
1 1 dx lim b→ x2 x
1
1
b
1 dx lim ln x b→ x
0.08
1 x
v cos x
⇒ du
sin x cos x dx lim b→0 x x cos 1
1
b 1
x −5 − 0.25
1 dx x2
1
15
20
cos x dx x2
cos x dx x2
Converges 82. (a) Yes, the integral is not defined at x 2.
(b)
5
(c) As n → , the integral approaches 4 4 . (d) In
2
0
4 dx 1 tan xn
I2 3.14159 I4 3.14159 I8 3.14159 I12 3.14159
0 −2
2
357
358
Chapter 7
84. (a) f x
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
1 2 ex70 18 32
(b) P72 ≤ x <
0.2525
(c) 0.5 P70 ≤ x ≤ 72 0.5 0.2475 0.2525
90
f x dx 1.0
These are the same answers because by symmetry,
50
P70 ≤ x <
0.4
0.5
and 50
0.5 P70 ≤ x <
90
P70 ≤ x ≤ 72 P72 ≤ x <
− 0.2
86. False. This is equivalent to Exercise 85.
.
88. True
Review Exercises for Chapter 7 2.
2
xe x1 dx
2 1 x1 e 2x dx 2
4.
x 1 dx 1 x2122x dx 2 1 x 2
1 2 e x1 C 2
1 1 x212 C 2 12
1 x2 C
6.
u4 3u2 du
2x2x 3 dx
10.
8.
x x 4 2x2 x 1 1 2 x 4 2x2 1 x 12
22x 332 x 1 C 5
u 2x 3, x
u5 u3 C 5
x4 2x2 x 1 dx x2 12
u2 3 , dx u du 2
dx
x
1 2
2x dx x2 12
1 C 2x2 1
x2 1ex dx x2 1ex 2 xex dx x2 1ex 2xex 2 ex dx exx2 2x 1 1
(1) dv ex dx
⇒
v ex
u x2 1 ⇒ du 2x dx (2) dv ex dx ⇒ ux
⇒ du dx
12. u arctan 2x, du
v ex
2 dx, dv dx,v x 1 4x2
arctan 2x dx x arctan 2x x arctan 2x
14.
lnx2 1 dx
2x dx 1 4x2
1 ln1 4x2 C 4
1 2
lnx2 1 dx
1 x ln x2 1 2
1 x ln x2 1 2
x2
x2 dx 1
dx
1 dx x2 1
1 x1 1 C x ln x2 1 x ln 2 2 x1
dv dx
⇒
vx
u lnx2 1 ⇒ du
2x dx x2 1
Review Exercises for Chapter 7
16.
ex arctanex dx ex arctanex ex arctanex ⇒
dv ex dx
20.
x dx 2
sin2
ex dx 1 e2x
or
tan sec4 d
22.
tan3 tan sec2 d
sec3 sec tan d
cos 2sin cos 2 d
x2 9
x
dx
24.
1 ln1 e2x C 2
1 1 1 1 1 cos x dx x sin x C x sin x C 2 2 2
tan sec4 d
e2x dx 1 e2x
v ex
u arctan ex ⇒ du
18.
359
1 4 1 tan tan2 C1 4 2
1 sec4 C2 4
cos 2 sin2 sin cos 2 d
1 sin cos 3cos sin d sin cos4 C 4
3 tan 3 sec tan d 3 sec
x
3 tan2 d
x2 − 9
θ 3
3 sec2 1 d 3tan C x2 9 3 arcsec
3x C
x 3 sec , dx 3 sec tan d, x2 9 3 tan
26.
9 4x2 dx
1 9 2x2 2 dx 2
2 9 arcsin
1 2
9 2x x arcsin 9 4x2 C 4 3 2
1
28.
2x 2x9 4x2 C 3
sin 1 d 1 2 cos2 2
1 2
1 2 sin d 1 2 cos2
arctan 2 cos C
u 2 cos , du 2 sin d
360
Chapter 7
30. (a)
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
x4 x dx 64 tan3 sec3 d
(b)
x4 x dx 2 u4 4u2 du
64 sec4 sec2 sec tan d
64 sec3 3 sec3 5 C 15
24 x32 3x 8 C 15
2u3 2 3u 20 C 15
24 x32 3x 8 C 15
u2 4 x, dx 2u du
x 4 tan2 , dx 8 tan sec2 d, 4 x 2 sec
(c)
x4 x dx
u32 4u12 du
(d)
x4 x dx
4 x32 dx
2u32 3u 20 C 15
4 2x 4 x32 4 x52 C 3 15
24 x32 3x 8 C 15
24 x32 3x 8 C 15
dv 4 x dx ⇒
2 v 4 x32 3
⇒ du dx
ux 2x3 5x2 4x 4 4 3 2x 3 x2 x x x1
32.
34.
u 4 x, du dx
2x 2 4 x32 3 3
2x3 5x2 4x 4 dx x2 x
2x 3
3 4 dx x2 3x 4 ln x 3 ln x 1 C x x1
4x 2 A B 3x 12 x 1 x 12 4x 2 3Ax 1 3B Let x 1: 2 3B ⇒ B
2 3
Let x 2: 6 3A 3B ⇒ A
36.
4x 2 4 dx 3x 12 3
sec2 d tan tan 1
4 3
2 1 dx x1 3
1 du uu 1
1 du u1
ln u 1 ln u C ln u tan , du sec2 d 1 A B uu 1 u u1 1 Au 1 Bu Let u 0: 1 A ⇒ A 1 Let u 1: 1 B
4 1 2 2 1 dx lnx 1 C 2 lnx 1 C x 12 3 3x 1 3 x1
1 du u
tan 1 C ln 1 cot C tan
Review Exercises for Chapter 7
38.
x 2 3x
dx
24 3x 2 3x C (Formula 21) 27
40.
x 1 1 dx 2 dx 1 ex 2 1 eu
6x 8 2 3x C 27
u x2
1 u ln1 eu C 2
(Formula 84)
1 2 x2 ln1 e x C 2
42.
3 3 1 dx 3 dx 2 3x3x2 1 2x9x2 1
44.
(Formula 33)
tann x dx
50.
u 3x
1 1 1 dx dx 1 tan x 1 tan x
46.
3 arcsec 3x C 2
u x
11 x lncos x sin x C (Formula 71) 2
tann2xsec2 x 1 dx
tann2 x sec2 x dx
1 tann1 x n1
1 x dx
48.
4u4 4u2 du
4u5 4u3 4 C 1 x 323x 2 C 5 3 15
u 1 x, x u4 2u2 1, dx 4u3 4u du
52.
3x3 4x Cx D Ax B 2 2 x2 12 x 1 x 12 3x3 4x Ax Bx2 1 Cx D Ax3 Bx2 A Cx B D A 3, B 0, A C 4 ⇒ C 1, BD0 ⇒ D0
3x3 4x x dx dx 3 2 x2 12 x 1
54.
x dx x2 12
3 1 lnx2 1 C 2 2x2 1
sin cos 2 d
sin2 2 sin cos cos2 d
1 sin 2 d
1 dx u 2x, du 2x
tann2 x dx
u4u3 4u du
csc2x 1 dx 2 csc2x dx x 2x
2 ln csc2x cot2x C
tann2 x dx
1 1 cos 2 C 2 cos 2 C 2 2
361
362
Chapter 7
56. y
4 x2
2x
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
dx
2 cos 2 cos d 4 sin
csc sin d
ln csc cos cos C ln
4 x2 2 4 x2 C x 2
x 2 sin , dx 2 cos d, 4 x 2 2 cos
1 cos d
58. y
sin 1 cos
d 1 cos 12sin d 21 cos C
u 1 cos, du sin d
1
60.
x
x 2x 4
0
2
0
dx 2 ln x 4 ln x 2
1
62.
xe3x dx
0
e9 3x 1
2
3x
0
1 5e6 1 224.238 9
2 ln 3 2 ln 4 ln 2 ln
3
64.
0
9 0.118 8
x 22 x 1 x dx 3 1 x
3 0
4 4 8 3 3 3
1 dx 25 x2
1 x5 ln 10 x 5
4
66. A
0
68. By symmetry, y 0. A 4 5 1 44 x 4
17 3.4 5
70. s
y
4 0
1 1 1 ln ln 9 0.220 10 9 10
1 sin2 2x dx 3.82
0
3 2 1
(3.4, 0) x
−1
1
3
4
5
−2 −3
x, y 3.4, 0
72. lim
x →0
76.
sin x cos x 1 lim sin 2 x x →0 2 cos 2 x 2 2
74. lim xex lim 2
x →
x →
y lim x 1ln x x →1
ln y lim ln x lnx 1 x →1
lim x →1
lnx 1 lim x →1 1 ln x
lim 2xln x 0 x →1
Since ln y 0, y 1.
1 1 2 ln x x1 ln2 x x lim lim x →1 x →1 1 1 x1 1 x ln2 x x x2
x 1 2 lim 2 0 x → 2xe x ex
358
Chapter 7
84. (a) f x
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
1 2 ex70 18 32
(b) P72 ≤ x <
0.2525
(c) 0.5 P70 ≤ x ≤ 72 0.5 0.2475 0.2525
90
f x dx 1.0
These are the same answers because by symmetry,
50
P70 ≤ x <
0.4
0.5
and 50
0.5 P70 ≤ x <
90
P70 ≤ x ≤ 72 P72 ≤ x <
− 0.2
86. False. This is equivalent to Exercise 85.
.
88. True
Review Exercises for Chapter 7 2.
2
xe x1 dx
2 1 x1 e 2x dx 2
4.
x 1 dx 1 x2122x dx 2 1 x 2
1 2 e x1 C 2
1 1 x212 C 2 12
1 x2 C
6.
u4 3u2 du
2x2x 3 dx
10.
8.
x x 4 2x2 x 1 1 2 x 4 2x2 1 x 12
22x 332 x 1 C 5
u 2x 3, x
u5 u3 C 5
x4 2x2 x 1 dx x2 12
u2 3 , dx u du 2
dx
x
1 2
2x dx x2 12
1 C 2x2 1
x2 1ex dx x2 1ex 2 xex dx x2 1ex 2xex 2 ex dx exx2 2x 1 1
(1) dv ex dx
⇒
v ex
u x2 1 ⇒ du 2x dx (2) dv ex dx ⇒ ux
⇒ du dx
12. u arctan 2x, du
v ex
2 dx, dv dx,v x 1 4x2
arctan 2x dx x arctan 2x x arctan 2x
14.
lnx2 1 dx
2x dx 1 4x2
1 ln1 4x2 C 4
1 2
lnx2 1 dx
1 x ln x2 1 2
1 x ln x2 1 2
x2
x2 dx 1
dx
1 dx x2 1
1 x1 1 C x ln x2 1 x ln 2 2 x1
dv dx
⇒
vx
u lnx2 1 ⇒ du
2x dx x2 1
Review Exercises for Chapter 7
16.
ex arctanex dx ex arctanex ex arctanex ⇒
dv ex dx
20.
x dx 2
sin2
ex dx 1 e2x
or
tan sec4 d
22.
tan3 tan sec2 d
sec3 sec tan d
cos 2sin cos 2 d
x2 9
x
dx
24.
1 ln1 e2x C 2
1 1 1 1 1 cos x dx x sin x C x sin x C 2 2 2
tan sec4 d
e2x dx 1 e2x
v ex
u arctan ex ⇒ du
18.
359
1 4 1 tan tan2 C1 4 2
1 sec4 C2 4
cos 2 sin2 sin cos 2 d
1 sin cos 3cos sin d sin cos4 C 4
3 tan 3 sec tan d 3 sec
x
3 tan2 d
x2 − 9
θ 3
3 sec2 1 d 3tan C x2 9 3 arcsec
3x C
x 3 sec , dx 3 sec tan d, x2 9 3 tan
26.
9 4x2 dx
1 9 2x2 2 dx 2
2 9 arcsin
1 2
9 2x x arcsin 9 4x2 C 4 3 2
1
28.
2x 2x9 4x2 C 3
sin 1 d 1 2 cos2 2
1 2
1 2 sin d 1 2 cos2
arctan 2 cos C
u 2 cos , du 2 sin d
360
Chapter 7
30. (a)
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
x4 x dx 64 tan3 sec3 d
(b)
x4 x dx 2 u4 4u2 du
64 sec4 sec2 sec tan d
64 sec3 3 sec3 5 C 15
24 x32 3x 8 C 15
2u3 2 3u 20 C 15
24 x32 3x 8 C 15
u2 4 x, dx 2u du
x 4 tan2 , dx 8 tan sec2 d, 4 x 2 sec
(c)
x4 x dx
u32 4u12 du
(d)
x4 x dx
4 x32 dx
2u32 3u 20 C 15
4 2x 4 x32 4 x52 C 3 15
24 x32 3x 8 C 15
24 x32 3x 8 C 15
dv 4 x dx ⇒
2 v 4 x32 3
⇒ du dx
ux 2x3 5x2 4x 4 4 3 2x 3 x2 x x x1
32.
34.
u 4 x, du dx
2x 2 4 x32 3 3
2x3 5x2 4x 4 dx x2 x
2x 3
3 4 dx x2 3x 4 ln x 3 ln x 1 C x x1
4x 2 A B 3x 12 x 1 x 12 4x 2 3Ax 1 3B Let x 1: 2 3B ⇒ B
2 3
Let x 2: 6 3A 3B ⇒ A
36.
4x 2 4 dx 3x 12 3
sec2 d tan tan 1
4 3
2 1 dx x1 3
1 du uu 1
1 du u1
ln u 1 ln u C ln u tan , du sec2 d 1 A B uu 1 u u1 1 Au 1 Bu Let u 0: 1 A ⇒ A 1 Let u 1: 1 B
4 1 2 2 1 dx lnx 1 C 2 lnx 1 C x 12 3 3x 1 3 x1
1 du u
tan 1 C ln 1 cot C tan
Review Exercises for Chapter 7
38.
x 2 3x
dx
24 3x 2 3x C (Formula 21) 27
40.
x 1 1 dx 2 dx 1 ex 2 1 eu
6x 8 2 3x C 27
u x2
1 u ln1 eu C 2
(Formula 84)
1 2 x2 ln1 e x C 2
42.
3 3 1 dx 3 dx 2 3x3x2 1 2x9x2 1
44.
(Formula 33)
tann x dx
50.
u 3x
1 1 1 dx dx 1 tan x 1 tan x
46.
3 arcsec 3x C 2
u x
11 x lncos x sin x C (Formula 71) 2
tann2xsec2 x 1 dx
tann2 x sec2 x dx
1 tann1 x n1
1 x dx
48.
4u4 4u2 du
4u5 4u3 4 C 1 x 323x 2 C 5 3 15
u 1 x, x u4 2u2 1, dx 4u3 4u du
52.
3x3 4x Cx D Ax B 2 2 x2 12 x 1 x 12 3x3 4x Ax Bx2 1 Cx D Ax3 Bx2 A Cx B D A 3, B 0, A C 4 ⇒ C 1, BD0 ⇒ D0
3x3 4x x dx dx 3 2 x2 12 x 1
54.
x dx x2 12
3 1 lnx2 1 C 2 2x2 1
sin cos 2 d
sin2 2 sin cos cos2 d
1 sin 2 d
1 dx u 2x, du 2x
tann2 x dx
u4u3 4u du
csc2x 1 dx 2 csc2x dx x 2x
2 ln csc2x cot2x C
tann2 x dx
1 1 cos 2 C 2 cos 2 C 2 2
361
362
Chapter 7
56. y
4 x2
2x
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
dx
2 cos 2 cos d 4 sin
csc sin d
ln csc cos cos C ln
4 x2 2 4 x2 C x 2
x 2 sin , dx 2 cos d, 4 x 2 2 cos
1 cos d
58. y
sin 1 cos
d 1 cos 12sin d 21 cos C
u 1 cos, du sin d
1
60.
x
x 2x 4
0
2
0
dx 2 ln x 4 ln x 2
1
62.
xe3x dx
0
e9 3x 1
2
3x
0
1 5e6 1 224.238 9
2 ln 3 2 ln 4 ln 2 ln
3
64.
0
9 0.118 8
x 22 x 1 x dx 3 1 x
3 0
4 4 8 3 3 3
1 dx 25 x2
1 x5 ln 10 x 5
4
66. A
0
68. By symmetry, y 0. A 4 5 1 44 x 4
17 3.4 5
70. s
y
4 0
1 1 1 ln ln 9 0.220 10 9 10
1 sin2 2x dx 3.82
0
3 2 1
(3.4, 0) x
−1
1
3
4
5
−2 −3
x, y 3.4, 0
72. lim
x →0
76.
sin x cos x 1 lim sin 2 x x →0 2 cos 2 x 2 2
74. lim xex lim 2
x →
x →
y lim x 1ln x x →1
ln y lim ln x lnx 1 x →1
lim x →1
lnx 1 lim x →1 1 ln x
lim 2xln x 0 x →1
Since ln y 0, y 1.
1 1 2 ln x x1 ln2 x x lim lim x →1 x →1 1 1 x1 1 x ln2 x x x2
x 1 2 lim 2 0 x → 2xe x ex
Problem Solving for Chapter 7 78. lim x →1
ln2x x 2 1
lim
x →1
1
0
2 2x
x 11x ln x
lim
2x 2 2 lim 1 x 1 x ln x x→1 1 1 ln x
x→1
lim x →1
80.
2x 2 2 ln x ln xx 1
0 b
6 dx lim 6 ln x 1 b→1 x1
82.
e1x
x
0
2
a→0 b →
Diverges
84. V
xex 2dx
0
x2e2x dx
0
lim
b→
e4
2x
86.
2
b→
1 1 1 dx < x5 x10 x15
4x1
lim
1 1 9x9 14x14
4
b
2x2 2x 1
0
2
b
<
2
2
0.015846 <
2
4
1 dx < x5 1
x5
2
1 1 2 dx x5 x10 x15
1 1 1 1 dx < lim 4 9 14 b→ 1 4x 9x 7x
1 dx < 0.015851 x5 1
Problem Solving for Chapter 7
1
2. (a)
0
ln x dx lim x ln x b→0
1 b
1 lim b ln b b 1 b→0
Note: lim b ln b lim
b→0
b→0
1
0
ln b 1b lim 0 b1 b→0 1b2
ln x2 dx lim xln x2 2x ln x 2x b→0
1 b
2 lim bln b2 2b ln b 2b 2 b→0
(b) Note first that lim bln bn 0 (Mathematical induction). Also,
b→0
ln x
n1
dx xln xn1 n 1 ln xn dx.
1
Assume
ln xn dx 1n n!.
0
1
Then,
0
ln xn1 dx lim xln xn1 b→0
1 b
1
n 1
ln xn dx
0
0 n 11n n! 1n1n 1!.
dx lim e1x
b 2
b a
101
363
364
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
lim
4.
x→
xx cc
lim x ln
x→
lim
x→
x
1 4
xx cc ln 41
lnx c lnx c ln 4 1x 1 1 xc xc lim ln 4 x→ 1 2 x
lim
x→
2c x2 ln 4 x cx c lim
x→
2cx2 ln 4 c2
x2
2c ln 4 2x 2 ln 2 c ln 2 6. sin BD, cos OD 1 1 Area DAB DABD 1 cos sin 2 2 Shaded area R
1 1 1BD sin 2 2 2 2 DAB 121 cos sin Shaded area 12 sin
lim R lim
→0
→0
1 cos sin 1 cos cos sin2 lim →0 sin 1 cos
lim
1 cos sin cos sin 2 sin cos sin
lim
sin 4 cos sin 4 cos 1 lim 3 →0 sin 1
→0
→0
1 u2 x 1 u2 3 u2 8. u tan , cos x , 2 cos x 2 2 2 1u 1 u2 1 u2 dx
2
0
2 du 1 u2
1 dx 2 cos x
1
0
1
0
1 2 u du 2
2 du 3 u2
2 1 arctan 3 3
2
1 u2 3 u2
1 u arctan 3 3
2 3 6
1
0
3 0.6046 9
Problem Solving for Chapter 7 10. Let u cx, du c dx.
b
cb
ec
2 x2
dx
0
du 1 c c
eu
2
0
As b → , cb → . Hence,
cb
eu du 2
0
dx
ec
2 x2
0
1 c
ex dx. 2
0
x 0 by symmetry.
y
Mx m
ec2x2
2
0
dx
2
ec x dx
2
2 2
0
1 2
0
e2c x dx 2 2
0
ec x dx 2 2
1 2 ex dx 1 2c 0 2 1 x2 e dx c 0
1 2 2
Thus, x, y 0,
2
4
2
4
.
12. (a) Let y f 1x, f y x, dx fy dy.
f 1x dx
y fy dy
y f y
u y, du dy
dv fy dy, v f y
f y dy
x f 1x
f y dy
(b) f 1x arcsin x y, f x sin x
arcsin x dx x arcsin x
sin y dy 1
x arcsin x cos y C x arcsin x 1 x2 C
y 1 − x2
(c) f x ex, f 1x ln x y
e
1
ln x dx x ln x
1
e
1
e y dy
0
e ey
1 0
e e 1 1
x 1 ⇔ y 0; x e ⇔ y 1
x
365
366
Chapter 7
14. (a) Let x
I
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
u, dx du. 2
2
0
sin x dx cos x sin x
2
2
0
Hence, 2I
2
sin x dx cos x sin x
0
2
1 dx
0
0
(b) I
2
2
0
cos x dx sin x cos x
⇒ I . 2 4
2 u du cos u sin u 2 2 n
n
cosn u du sinn u cosn u
Thus, 2I
2
0
16.
cos u du sin u cos u
sinn
2
0
2 u du cos u sin u 2 2 sin
0
1 dx
⇒ I . 2 4
P1 P2 Pn Nx ... Dx x c1 x c2 x cn Nx P1x c2x c3. . .x cn P2x c1x c3. . .x cn . . . Pnx c1x c2. . .x cn1 Let x c1: Nc1 P1c1 c2c1 c3. . .c1 cn P1
Nc1 c1 c2c1 c3. . .c1 cn
Let x c2: Nc2 P2c2 c1c2 c3. . .c2 cn P2
Nc2 c2 c1c2 c3. . .c2 cn
Let x cn: Ncn Pncn c1cn c2. . .cn cn1 Pn
Ncn cn c1cn c2. . .cn cn1
If Dx x c1x c2x c3. . .x cn, then by the Product Rule Dx x c2x c3. . .x cn x c1x c3. . .x cn . . . x c1x c2x c3. . .x cn1 and Dc1 c1 c2c1 c3. . .c1 cn Dc2 c2 c1c2 c3. . .c2 cn
Dcn cn c1cn c2. . .cn cn1. Thus, Pk Nck Dck for k 1, 2, . . ., n.
Problem Solving for Chapter 7
18.
st
50,000 dt 50,000 400t
32t 12,000 ln
16t 2 12,000 ln 50,000 ln50,000 400t dt
16t 2 12,000t ln 50,000 12,000 t ln50,000 400t
400t dt 50,000 400t
16t 2 12,000t ln
50,000 12,000t 50,000 400t
16t 2 12,000t ln
50,000 12,000t 1,500,000 ln50,000 400t C 50,000 400t
1
50,000 dt 50,000 400t
s0 1,500,000 ln 50,000 C 0 C 1,500,000 ln 50,000
st 16t 2 12,000t 1 ln
50,000 50,000 400t 1,500,000 ln 50,000 400t 50,000
When t 100, s100 557,168.626 feet 20. Let u x ax b, du x a x b dx, dv f x dx, v fx.
b
a
b
x ax b dx x ax b fx
b
a
b
a
x a x b fx dx u 2x a b
dv fx dx
2x a b fx dx
2x a b f x 2
b
a
a
f x dx
b a
b
a
2 f x dx
367
C H A P T E R 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals Section 7.1
Basic Integration Rules
Section 7.2
Integration by Parts
Section 7.3
Trigonometric Integrals . . . . . . . . . . . . . . . . . . . 65
Section 7.4
Trigonometric Substitution
Section 7.5
Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . 84
Section 7.6
Integration by Tables and Other Integration Techniques
Section 7.7
Indeterminate Forms and L’Hôpital’s Rule . . . . . . . . . 96
Section 7.8
Improper Integrals
Review Exercises
. . . . . . . . . . . . . . . . . . . 50
. . . . . . . . . . . . . . . . . . . . . 55
. . . . . . . . . . . . . . . . . 74
. . 90
. . . . . . . . . . . . . . . . . . . . . 102
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
C H A P T E R 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals Section 7.1
Basic Integration Rules
Solutions to Odd-Numbered Exercises 1. (a) (b)
d x2 1 C 21x2 11 22x 2x dx x 1
(c)
1 1 2 x d 1 2 x 1 C x 11 22x dx 2 2 2 2x2 1
(d)
2x d lnx2 1 C 2 dx x 1
x x2 1
3. (a)
dx matches (b).
d 2x x2 122 2x2x2 12x 21 3x2 2 2 2 C dx x 1 x2 14 x 13
(c)
1 d arctan x C dx 1 x2
(d)
2x d lnx2 1 C 2 dx x 1
1 dx matches (c). x2 1
3x 24 dx
u 3x 2, du 3 dx, n 4 Use
11.
t sin t 2 dt
Use
7.
sin u du.
1
x 1 2x
dx
9.
u 1 2x, du
un du.
u t 2, du 2t dt
50
x d 2 lnx2 1 C 21 x2 2x dx 1 x 1
(b)
5.
d 2x2 1 C 2 12 x2 11 22x 2x dx x2 1
Use
1 x
dx
3 1 t2
u t, du dt, a 1 Use
du . u
13.
cos xesin x dx
u sin x, du cos x dx Use
eu du.
dt
du a 2 u 2
Section 7.1
17. Let u z 4, du dz
15. Let u 2x 5, du 2 dx.
2x 53 2 dx
1 2x 53 22 dx 2
5 z 44 C dz 5 z 45 dx 5 5 z 4 4
1 2x 55 2 C 5
19. Let u t3 1, du 3t2 dt.
3 t3 1 dt t2
1 3
Basic Integration Rules
21.
v
5 C 4z 44
1 dv 3v 13
t3 11 33t2 dt
1 t3 14 3 C 3 4 3
t3 14 3 C 4
v dv
1 3v 133dv 3
1 1 v2 C 2 63v 12
23. Let u t3 9t 1, du 3t2 9 dt 3t2 3 dt.
25.
t2 3 1 3t 2 3 1 dt dt ln t 3 9t 1 C t 3 9t 1 3 t 3 9t 1 3
x2 dx x1
29.
x 1 dx
1 dx x1
1 2 x x ln x 1 C 2
1 2x22 dx
27. Let u 1 ex, du ex dx.
ex dx ln1 ex C 1 ex
4 4 x 4x 4 4x2 1dx x5 x3 x C 12x 4 20x2 15 C 5 3 15
31. Let u 2 x2, du 4 x dx.
xcos 2 x2 dx
1 cos 2 x24 x dx 4 1 sin 2 x 2 C 4
33. Let u x, du dx.
cscx cot x dx
1 csc x cot x dx 1 cscx C
35. Let u 5x, du 5 dx.
e5x dx
1 5x 1 e 5 dx e5x C 5 5
37. Let u 1 e x, du e x dx.
2 dx 2 ex 1
2
1 ex 1
ee dx x x
ex dx 2 ln1 e x C 1 ex
51
52
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
39.
ln x2 1 ln x 2 dx 2 ln x dx 2 C ln x C x x 2
41.
1 sin x dx cos x
2
sec x tan x dx ln sec x tan x ln sec x C ln sec xsec x tan x C
1 1 cos 1 cos 1
43.
csc 1 d cos 1
cos 1
cos 1
cos 1 cos2 1
cos 1 sin2
cot csc2
csc cot csc2 d
csc cot C
45.
1 cos C sin sin
1 cos C sin
3z 2 3 2z dz dz dz 2 2 z2 9 2 z2 9 z 9
47. Let u 2t 1, du 2 dt.
3 2 z lnz2 9 arctan C 2 3 3
1 1 2t 12
dt
1 2 dt 2 1 2t 12
1 arcsin2t 1 C 2
49. Let u cos
2t , du 2 sint 2 t dt. 2
1 tan2 t 1 2 sin2 t dt dt t2 2 cos2 t t2
1 2 ln cos 2 t
51.
6x x2
53.
4 dx 4x2 4x 65
55.
ds t 1 , 0, dt 2 1 t 4
3
dx 3
C
1 9 x 32
dx 3 arcsin
(b) u t 2, du 2t dt
1
t
−1
x 3 3 C
1 1 1 x 1 2 2x 1 dx arctan C arctan C x 1 22 16 4 4 4 8
s
(a)
1
0,
t 1 t 4
dt
1 1 2t dt arcsin t 2 C 2 2 1 t22 0.8
1 1 1 1 : arcsin 0 C ⇒ C 2 2 2 2 −1.2
−1
1.2
1 1 s arcsin t 2 2 2 − 0.8
Section 7.1
57.
59. y
10
Basic Integration Rules
1 ex2 dx
53
e2x 2ex 1 dx
1 e2x 2ex x C 2
−10
10 −2
y 3e0.2x
61.
dy sec2 x dx 4 tan2 x
63. Let u 2x, du 2 dx.
4
Let u tan x, du sec2 x dx. y
0
sec2 x 1 tan x dx arctan C 4 tan2 x 2 2
1
2
x
xe
dx
0
1 2
1
0
1 2 2x dx ex 2
2
x
e
4
1
2 3
0
1 1 dx 4 9x2 3
2 3
0
4
cos 2x2 dx
0
4
2 sin 2x
0
0
2x dx x2 9
1 1 e1 0.316 2
69. Let u 3x, du 3 dx.
1 2
1
0
1 2
67. Let u x2 9, du 2x dx.
65. Let u x2, du 2x dx.
cos 2x dx
3 4 3x2
x2 91 22x dx
4
0
2x2 9
71.
1 3x arctan 6 2
dx
4 0
4
1 1 x2 dx arctan C x2 4x 13 3 3
The antiderivatives are vertical translations of each other. 1
2 3 0 −7
0.175 18
5
−1
73.
1 2 d tan sec C or 1 sin 1 tan 2
The antiderivatives are vertical translations of each other.
75. Power Rule:
un du
un1 C, n 1. n1
u x2 1, n 3
6
− 2
7 2
−6
77. Log Rule:
du ln u C, u x2 1. u
79. The are equivalent because exC1 e x eC1 Ce x, C eC1
54
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
81. sin x cos x a sinx b sin x cos x a sin x cos b a cos x sin b sin x cos x a cos b sin x a sin b cos x Equate coefficients of like terms to obtain the following. 1 a cos b
1 a sin b
and
Thus, a 1 cos b. Now, substitute for a in 1 a sin b.
cos1 b sin b
1
1 tan b ⇒ b Since b
2
83.
0
4
1 ,a 2. Thus, sin x cos x 2 sin x . 4 cos 4 4
dx sin x cos x
dx 2 sinx 4
1 2
csc x
1 dx ln csc x cot x 4 4 4 2
C
1 a
4x dx 3 x2 1
85. Let u 1 x2, du 2x dx.
0
12x
1 6a2
x1 x2 dx
A4
y
x ax2 dx
0
1
Matches (a).
87.
2
a x3 3
1
2
3
1 x21 22x dx
0
2
4 1 x23 2 3
1
1 0
Let
4 3
1 2 1 , 12a 2 3, a . 6a 2 3 2 y
( a1 , a1)
2
y
y=x
x 1
2
3 1
1
y = ax 2
1 2
x
x − 12
1
1 2
− 12
2
−1
89. (a) Shell Method: Let u x2, du 2x dx.
1
V 2
(b) Shell Method:
y
V 2
dx
1 2
ex
0 1
2
x
2x dx
e
0
xex dx 2
0
2
x
xe
b
1
x 1 2
2
b 0
1
1 eb 2
e
x2
1 0
1 e 1.986 1
eb 2
b
4 3
3 4 3
ln 33 4
0.743
1 a 0
Section 7.2
4
91. A
0
x
1 A
x 5 dx 5 arcsin 5
25 x2
4
x
0
4
y 4
3
25 x 4
2
2 12
2x dx
(2.157, y )
1
0
x
1 5 25 x212 5 arcsin45
4 5
5 dx
25 x2
1 5 5 arcsin45 2
5 arcsin
0
Integration by Parts
1
4
2
3
4
0
1 3 5 arcsin45
2 2.157 arcsin45 y tan x
93.
y sec2 x 1 y 2 1 2 sec4 x
14
s
1 2 sec4x dx
0
1.0320
Section 7.2
Integration by Parts
1.
d sin x x cos x cos x x sin x cos x x sin x. Matches (b) dx
3.
d 2 x x e 2xex 2ex x2ex 2xex 2xex 2ex 2ex x2ex. Matches (c) dx
5.
xe2x dx
7.
ux
ln x2 dx
u ln x 2, dv dx
u x, dv e2x dx
11. dv e2x dx ⇒
v
1 e2x dx e2x 2
⇒ du dx
1 xe2x dx xe2x 2
1 e2x dx 2
1 1 1 xe2x e2x C 2x 2x 1 C 2 4 4e
9.
x sec2 x dx
u x, dv sec2 x dx
55
56
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
13. Use integration by parts three times. (1) dv ex dx ⇒
v
(2) dv ex dx ⇒
ex dx ex
⇒ du 3x2 dx
u x3
u x2
v
(3) dv ex dx ⇒
ex dx ex
⇒ du 2x dx
v
⇒ du dx
ux
x3ex dx x3ex 3 x2ex dx x3ex 3x2ex 6 xex dx x3ex 3x2ex 6xex 6ex C exx3 3x2 6x 6 C
15.
x2 ex dx 3
17. dv t dt
1 3
1 3 3 ex 3x2dx ex C 3
⇒
v
t dt
t2 2
1 dt t1
u lnt 1 ⇒ du
t lnt 1 dt
t2 1 lnt 1 2 2
t2 dt t1
t2 1 lnt 1 2 2
t1
t2 2
lnt 1
1 dt t1
t2
1 t lnt 1 C 2 2
1 2t 2 1 ln t 1 t 2 2t C 4
19. Let u ln x, du
ln x2 dx x
1 dx. x
21. dv
ln x2
1x dx ln3x
3
1 dx ⇒ 2x 12
v
u x2
xe2x xe2x dx 2x 12 22x 1
⇒ du 2x dx
x2 1ex dx
x2ex dx
ux
e2x dx 2
e2x xe2x C 22x 1 4
e2x C 42x 1
v
ex dx ex
⇒ du dx
exdx x2ex 2 xex dx ex
x2ex 2 xex
(2) dv ex dx ⇒
ex dx ex
1 22x 1
e2x2x 1 dx
23. Use integration by parts twice. v
2x 12 dx
⇒ du 2xe2x e2x dx
u xe2x
(1) dv ex dx ⇒
C
ex dx ex x2ex 2xex ex C x 12ex C
ex dx ex
Section 7.2
25. dv x 1 dx ⇒
2 x 112dx x 132 3
v
27. dv cos x dx ⇒
⇒ du dx
ux
2 2 x x 1 dx xx 132 3 3
x 132 dx
cos x dx sin x
x cos x dx x sin x
sin x dx x sin x cos x C
4 2 xx 132 x 152 C 3 15
2x 132 3x 2 C 15
29. Use integration by parts three times. (1) u x3, du 3x2, dv sin x dx, v cos x
x3 sin dx x3 cos x 3 x2 cos x dx
(2) u x2, du 2x dx, dv cos x dx, v sin x
x3 sin x dx x3 cos x 3 x2 sin x 2 x sin x dx x3 cos x 3x2 sin x 6 x sin x dx
(3) u x, du dx, dv sin x dx, v cos x
x3 sin x dx x3 cos x 3x2 sin x 6 x cos x
cos x dx
x3 cos x 3x2 sin x 6x cos x 6 sin x C
t csc t cot t dt t csc t
csc t dt
⇒
33. dv dx
31. u t, du dt, dv csc t cot dt, v csc t
v
u arctan x ⇒ du
t csc t ln csc t cot t C
dx x
1 dx 1 x2
arctan x dx x arctan x x arctan x
35. Use integration by parts twice. (1) dv e2xdx ⇒
v
1 e2x dx e2x 2
1 1 e2x sin x dx e2x sin x 2 2
5 4
(2) dv e2x dx ⇒
1 1 1 2x 1 e2x cos x dx e2x sin x e cos x 2 2 2 2
1 1 e2x sin x dx e2x sin x e2x cos x 2 4 1 e2x sin x dx e2x2 sin x cos x C 5
v
x dx 1 x2
1 ln1 x2 C 2
1 e2x dx e2x 2
u cos x ⇒ du sin x dx
u sin x ⇒ du cos x dx
57
⇒ du dx
ux
v
Integration by Parts
e2x sin x dx
58
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
37. y xex
2
y
1 2 2 xe x dx ex C 2
39. Use integration by parts twice. (1) dv
1
2 3t
dt ⇒
v
(2) dv 2 3t dt ⇒
41.
v
2 2 3t12 dt 2 3t32 9
⇒ du dt
ut
2 2 3t12 dt 2 3t 3
⇒ du 2t dt
u t2
y
2t 2 2 3t 4 t2 dt 3 3
2 3t
t 2 3t dt
2 2t 2 2 3t 4 2t 2 3t32 3 3 9 9
8t 16 2t 2 2 3t 2 3t32 2 3t52 C 3 27 405
2 2 3t 27t 2 24t 32 C 405
2 3t32 dt
cos yy 2x
cos y dy
2x dx
sin y x 2 C 43. (a)
dy x y cos x, 0, 4 dx
(b)
y
8
6
dy
y
y12 dy
2
2
2
x cos x dx
4
u x, du dx, dv cos x dx, v sin x
x cos x dx
2y12 x sin x
x 4
sin x dx
6
x sin x cos x C
0, 4: 2412 0 1 C ⇒ C 3 2 y x sin x cos x 3
45.
dy x x8 e , y0 2 dx y
10
−10
10 −2
−6
6 −2
Section 7.2
Integration by Parts
47. u x, du dx, dv ex2 dx, v 2ex2
xex2 dx 2xex2
4
Thus,
2ex2 dx 2xex2 4ex2 C
xex2 dx 2xex2 4ex2
0
8e2 4e2 4
4 0
12e2 4 2.376. 49. See Exercise 27.
2
0
1
51. u arccos x, du
1 x2
0
1 2
dx, dv dx, v x
arccos x dx x arccos x
x
1 x2
dx x arccos x 1 x2 C
12
Thus,
2
x cos x dx x sin x cos x
arccos x x arccos x 1 x2
0
12 0
34 1
1 1 arccos 2 2
3 1 0.658. 6 2
53. Use integration by parts twice. (1) dv e x dx ⇒
v
(2) dv ex dx ⇒
e x dx e x
u sin x ⇒ du cos x dx
e x sin x dx e x sin x
e x cos x dx e x sin x e x cos x
v
ex dx ex
u cos x ⇒ du sin x dx
e x sin x dx
2 e x sin x dx e xsin x cos x e x sin x dx
ex sin x cos x C 2
1
Thus,
e x sin x dx
0
55. dv x2 dx, v
x2 ln x dx
x
1 0
e 1 esin 1 cos 1 1 sin 1 cos 1 0.909. 2 2 2
1 x3 , u ln x, du dx 3 x
x3 ln x 3
x3 1 dx 3 x
1 2 x3 ln x x dx 3 3
2
Hence,
e2 sin x cos x
x2 ln x dx
x3 ln x 9x 3
1
2
3
1
1
8 1 8 7 8 ln 2 ln 2 1.071. 3 9 9 3 2
59
60
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
57. dv x dx, v
x2 1 dx , u arcsec x, du
2 x x2 1
x2 x arcsec x dx arcsec x 2
Hence,
4
x22 dx x x2 1
2 arcsec x 2 x x2
x arcsec x dx
2
1
x2 2x 1 dx arcsec x 2 4 x2 1
8 arcsec 4
x2 1 arcsec x x2 1 C 2 2
8 arcsec 4
2
15
2
15
2
1
4 2
23 23
3
2
2 3
7.380.
59.
x2e2x dx x2
12 e 2x 14 e 2 18 e C 2x
2x
2x
u and its derivatives
x2
e2x
2x
1 2x 2e
2
1 2x 4e
0
1 2x 8e
1 1 1 x2e2x xe2x e2x C 2 2 4 1 e2x2x2 2x 1 C 4
61.
63.
u and its derivatives
x3 cos x 3x2 sin x 6x cos x 6 sin x C
x3
sin x
3x2 6 sin x x3 6x cos x C
3x2
cos x
6x
sin x
6
cos x
0
sin x
x sec2 x dx x tan x ln cos x C
71. Yes. Let u x and du
v and its antiderivatives
Alternate signs
u and its derivatives
x
sec2 x
1
tan x
0
ln cos x
67. No. Substitution.
1 , dx.
x 1
73.
t 3e4t dt
Substitution also works. Let u x 1
2
0
v and its antiderivatives
Alternate signs
x3 sin x dx x3cos x 3x2sin x 6x cos x 6 sin x C
65. Integration by parts is based on the product rule.
75.
v and its antiderivatives
Alternate signs
e2x sin 3x dx
e2x2 sin 3x 3 cos 3x 13
2
0
1 2e 3 0.2374 13
69. Yes. u x2, dv e2x dx
e4t 32t3 24t2 12t 3 C 128
Section 7.2 77. (a) dv 2x 3 dx ⇒
v
Integration by Parts
1 2x 312 dx 2x 332 3
⇒ du 2 dx
u 2x
2 2 2x 2x 3 dx x2x 332 3 3
2x 332 dx
2 2 x2x 332 2x 352 C 3 15
2 2 2x 3323x 3 C 2x 332x 1 C 15 5 1 u3 and dx du 2 2
(b) u 2x 3 ⇒ x
2x 2x 3 dx
2
1 u 3 12 1 u du 2 2 2
u32 3u12 du
1 2 52 u 2u32 C 2 5
1 1 2 u32u 5 C 2x 3322x 3 5 C 2x 332x 1 C 5 5 5 x
79. (a) dv
4 x2
dx ⇒ v
4 x212x dx 4 x2
u x2 ⇒ du 2x dx
x3 dx x2 4 x2 2 x 4 x2 dx
4 x2 2 1 x2 4 x2 4 x232 C 4 x2 x2 8 C 3 3
(b) u 4 x2 ⇒ x2 u 4 and 2x dx du ⇒ x dx
x3 dx
4 x2
1 2
x2 x dx
4 x2
1 du 2
u41 du
u 2
u12 4u12 du
1 2 32 u 8u12 C 2 3
1 1 1 u12u 12 C 4 x2 4 x2 12 C 4 x2 x2 8 C 3 3 3
81. n 0: n 1: n 2: n 3: n 4:
ln x dx xln x 1 C
x ln x dx
x2 2 ln x 1 C 4
x2 ln x dx
x3 3 ln x 1 C 9
x3 ln x dx
x4 4 ln x 1 C 16
x 4 ln x dx
x5 5 ln x 1 C 25
In general,
xn ln x dx
xn1 n 1ln x 1 C. (See Exercise 85.) n 12
61
62
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
83. dv sin x dx ⇒
⇒ du nx n1 dx
u xn
x n sin
85. dv x n dx ⇒
v cos x
x dx
x n
cos x n
u ln x
x n1
cos x dx
v
1 dx x
⇒ du
x n ln x dx
x n1 n1
x n1 ln x n1
xn dx n1
x n1 x n1 C ln x n1 n 12
x n1 n 1ln x 1 C n 12
87. Use integration by parts twice. 1 v eax a
(1) dv eax dx ⇒
u cos bx ⇒ du b sin bx dx
u sin bx ⇒ du b cos bx dx
eax cos bx dx
eax sin bx b eax cos bx b a a a a
Therefore, 1
eax sin bx b a a
eax sin bx dx
b2 a2
e
ax
sin bx dx
eax sin bx dx
eax sin bx dx
x3 ln x dx
eax sin bx b2 2 a a
eax sin bx dx
eaxa sin bx b cos bx a2 eaxa sin bx b cos bx C. a2 b2 91. a 2, b 3 (Use formula in Exercise 88.)
89. n 3 (Use formula in Exercise 85.)
1 v eax a
(2) dv eax dx ⇒
x4 4 ln x 1 C 16
e2x cos 3x dx
e2x2 cos 3x 3 sin 3x C 13
1
93. dv ex dx ⇒
4
A
xex dx xex
0
1
4
0
4
0
ex dx
4 ex e4
4
0
5 0.908 e4
exsin x cos x 1 2
1 1 1 1 2 e 1 2 e
0.395 See Exercise 87.
3
−1
ex sin x dx
0
⇒ du dx
ux
95. A
v ex
1
7 −1
0
1.5 0
1 0
Section 7.2
e
97. (a) A
e
ln x dx x x ln x
1
1
1 See Exercise 4.
y
(b) Rx ln x, rx 0
Integration by Parts
2
(e, 1)
e
V
ln x2 dx
1
1 e
xln x2 2x ln x 2x
1
x
Use integration by parts twice, see Exercise 7.
1
2
3
e 2 2.257 e
(c) px x, hx ln x
(d)
e
y
e 2 1 13.177 See Exercise 85. 2
x, y
x ln x dx 2
1
e
V 2
99. Average value
1
x
x2 1 2 ln x 4
1
4 sin 2t 2 cos 2t 1 4t 4 cos 2t 2 sin 2t e 5e4t 20 20
7 1 e4 0.223 10
10
100,000 4000te0.05t dt 4000
0
25 te0.05t dt
0
Let u 25 t, dv e0.05t dt, du dt, v
100 0.05t e 5
e dt
100 e e 4000 25 t 10,000 $931,265 5 25
25 t
100 0.05t e 5 0.05t
10 0
10
cos n cos n n n
2 cos n n
22nn, , ifif nn isis even odd
10
0.05t
0
10 0
x 1 x sin nx dx cos nx 2 sin nx n n
100 5
0.05t
0
103.
ln x2dx e 2 0.359 1 2
e
1 e2 , 2.097, 0.359 4 2
2
e4t cos 2t 5 sin 2t dt
10
P 4000
1 e 2 1
0
101. ct 100,000 4000t, r 5%, t1 10 P
1 x ln x dx e2 1 2.097 1 4
0
From Exercises 87 and 88
63
64
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
105. Let u x, dv sin
1
I1
x sin
0
n2 x dx, du dx, v n2 cos n2 x .
n 2x n x dx cos x 2 n 2
2
x 2 sin
1
2 n
0
2 n cos n 2
n 2 2 cos n 2 n
Let u x 2, dv sin I2
1
1
cos
0
n2 x dx
n2 sin n2 x
1
2
0
sin n2 2
n2 x dx, du dx, v n2 cos n2 x .
n 2x 2 n x dx cos x 2 n 2
2 n cos n 2
2 n 2 cos n 2 n
2 1
2 n
2
1
n2 sin n2 x
cos
n2 x dx
2
2
1
sin n2 2
n2 sin n2 n2 sin n2 n sin n2 2
hI1 I2 bn h
8h
2
2
107. Shell Method:
b
V 2
x f x dx
y
a
y = f ( x)
f (a )
dv x dx ⇒
x2 v 2
f (b )
u f x ⇒ du fx dx V 2
x2 f x x2 fxdx 2
b
2
x a
a
b
b2f b a2f a
b
x2 fx dx
a
Disk Method: V
f a
b2 a2 dy
0
f b
f a
b2 f 1 y 2 dy
b2 a2 f a b2 f b f a
b2f b a2f a
f b
f a
f b
f a
f 1y2 dy
f 1 y2 dy
Since x f 1 y, we have f x y and fxdx dy. When y f a, x a. When y f b, x b. Thus,
f b
f a
f 1 y 2 dy
b
x 2fx dx
a
and the volumes are the same.
Section 7.3
Trigonometric Integrals
109. fx xex (a) f x
xex dx xex ex C
(b)
1
Parts: u x, dv ex dx f 0 0 1 C ⇒ C 1 0
f x xex ex 1 (c) You obtain the points n
xn
4 0
(d) You obtain the points
yn
n
xn
yn
0
0
0
0
0
0
1
0.05
0
1
0.1
0
2
0.10
2.378 103
2
0.2
0.0090484
3
0.15
0.0069
3
0.3
0.025423
4
0.20
0.0134
4
0.4
0.047648
80
0.9064
40
0.9039
4.0
1
4.0
1
0
4
0
0
4 0
(e) f 4 0.9084 The approximations are tangent line approximations. The results in (c) are better because x is smaller.
Section 7.3
Trigonometric Integrals
1. f x sin4 x cos4 x (a) sin4 x cos4 x
2x 2x 1 cos
1 cos
2 2 2
2
1 1 2 cos 2x cos2 2x 1 2 cos 2x cos2 2x 4
1 1 cos 4x 22 4 2
1 3 cos 4x 4 (b) sin4 x cos4 x sin2 x2 cos4 x 1 cos2 x2 cos4 x 1 2 cos 2 x 2 cos4 x (c) sin4 x cos4 x sin4 x 2 sin2 x cos2 x cos4 x 2 sin2 x cos2 x sin2 x cos2 x2 2 sin2 x cos2 x 1 2 sin2 x cos2 x —CONTINUED—
65
66
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
1. —CONTINUED— (d) 1 2 sin2 x cos2 x 1 2 sin x cos xsin x cos x 1 sin 2x 1
12 sin 2x
1 2 sin 2x 2
(e) Four ways. There is often more than one way to rewrite a trigonometric expression. 3. Let u cos x, du sin x dx.
5. Let u sin 2x, du 2 cos 2x dx.
cos3 x sin x dx cos3 xsin x dx
sin5 2x cos 2x dx
1 cos4 x C 4
1 2
sin5 2x2 cos 2xdx
1 sin6 2x C 12
7. Let u cos x, du sin x dx.
sin x1 cos2 x2 cos2 x dx
sin5 x cos2 x dx
cos2 x 2 cos4 x cos6 xsin x dx
9.
cos3 sin d
cos 1 sin2 sin 1 2 d
sin 1 2 sin 5 2cos d
2 2 sin 3 2 sin 7 2 C 3 7
13.
sin2
cos2 d
1 cos 2 2
1 cos 2 d 2
1 1 cos2 2 d 4 1 4
1
1 cos 4 d 2
1 1 cos 4 d 8
1 1 sin 4 C 8 4
1 4 sin 4 C 32
11.
1 2 1 cos3 x cos5 x cos7 x C 3 5 7
cos2 3x dx
1 cos 6x dx 2
1 1 x sin 6x C 2 6
1 6x sin 6x C 12
Section 7.3
Trigonometric Integrals
15. Integration by parts. 1 cos 2x x sin 2x 1 ⇒ v 2x sin 2x 2 2 4 4
dv sin2 x dx u x ⇒ du dx
x sin2 x dx
1 1 x2x sin 2x 4 4
2x sin 2x dx
1 1 1 1 x2x sin 2x x2 cos 2x C 2x2 2x sin 2x cos 2x C 4 4 2 8 17. Let u sin x, du cos x dx.
2
2
cos3 x dx
0
1 sin2 x cos x dx
0
sin x
2
1 3 sin x 3
0
2 3
19. Let u sin x, du cos x dx.
2
2
cos7 x dx
0
1 sin2 x3 cos x dx
0
2
1 3 sin2 x 3 sin4 x sin6 x cos x dx
0
sin x sin3 x
21.
sec3x dx
1 ln sec 3x tan 3x C 3
25. dv sec2 x dx ⇒
sec3 x dx
sec3 x dx
27.
tan5
x dx 4
sec4 5x dx
0
16 35
1 tan2 5x sec2 5x dx
tan3 5x 1 tan 5x C 5 3
tan 5x 3 tan2 5x C 15
1 tan x
1 sec x tan x
2 sec3 x dx
⇒ du sec x tan x dx
u sec x
v
23.
2
3 5 1 sin x sin7 x 5 7
sec x tan2 x dx
1 sec x tan x
sec xsec2 x 1 dx
1 sec x tan x ln sec x tan x C1
1 sec x tan x ln sec x tan x C 2
sec2
tan3
x x 1 tan3 dx 4 4
x x sec2 dx 4 4
29. u tan x, du sec2 x dx
tan3
x dx 4
tan4
x 4
tan4
x x x 2 tan2 4 ln cos C 4 4 4
sec2
x x 1 tan dx 4 4
sec2 x tan x dx
1 2 tan x C 2
67
68
31.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
tan2 x sec2 x dx
tan3 x C 3
33.
sec6 4x tan 4x dx
35. Let u sec x, du sec x tan x dx.
sec3 x tan x dx
39. r
37.
sec2 xsec x tan x dx
1 4
sec5 4x4 sec 4x tan 4x dx
sec6 4x C 24
sec2 x 1 dx sec x
sec x cos x dx
ln sec x tan x sin x C
1 cos2 2 d
41. y
1 4
1 4
1 1 1 sin2 sin4 C 4 2 8
1 12 8 sin2 sin4 C 32
1 2 cos2 cos22 d 1 2 cos2
1 sec3 x C 3
sin4 d
1 cos4 d 2
43. (a)
tan2 x dx sec x
1 4
(b)
y 4
dy sin2 x, 0, 0 dx y
sin2 x dx
tan3 3x sec 3x dx
sec2 3x 1sec 3x tan 3x dx
1 1 sec2 3x3 sec 3x tan 3x dx 3 sec 3x tan 3x dx 3 3 1 1 sec3 3x sec 3x C 9 3
4
1 cos 2x dx 2
−6
6
x
sin 2x 1 C x 2 4
4
−4
45.
−4
1 sin 2x 0, 0: 0 C, y x 2 4
dy 3 sin x , y0 2 dx y
47.
sin 3x cos 2x dx
8
−9
9
1 2
sin 5x sin x dx
1 1 cos 5x cos x C 2 5
1 cos 5x 5 cos x C 10
−4
49.
sin sin 3 d
1 2
cos 2 cos 4 d
1 1 1 sin 2 sin 4 C 2 2 4
1 2 sin 2 sin 4 C 8
51.
cot3 2x dx
csc2 2x 1 cot 2x dx
1 2 cos 2x 1 cot 2x2csc2 2x dx dx 2 2 sin 2x
1 1 cot2 2x ln sin 2x C 4 2
1 ln csc2 2x cot2 2x C 4
Section 7.3
53. Let u cot , du csc2 d.
csc4 d
1 dx sec x tan x
cot2 t dt csc t
csc2 1 cot2 d
csc2 d
cot
57.
55.
Trigonometric Integrals
csc2 t 1 dt csc t
csc t sin tdt
ln csc t cot t cos t C
csc2 cot2 d
1 3 cot C 3
cos2 x dx sin x
1 sin2 x dx sin x
csc x sin x dx
ln csc x cot x cos x C
59.
tan4 t sec4 t dt
tan2 t sec2 ttan2 t sec2 t dt
tan2 t sec2 t 1
tan2 t sec2 t dt 2 sec2 t 1 dt 2 tan t t C
61.
sin2 x dx 2
0
x
1 sin 2x 2
0
4
1 cos 2x dx 2
63.
tan3 x dx
0
4
sec2 x 1 tan x dx
0
4
sec2 x tan x dx
0
4
0
sin x dx cos x
4
0
1 2 tan x ln cos x 2
1 1 ln 2 2 65. Let u 1 sin t, du cos t dt.
2
0
67. Let u sin x, du cos x dx. 2
0
cos t dt ln 1 sin t 1 sin t
2
ln 2
2
2
cos3 x dx 2
1 sin2 x cos x dx
0
2 sin x
69.
2
1 3 sin x 3
0
1 x cos4 dx 6x 8 sin x sin 2x C 2 16 6
−9
9
−6
71.
sec5 x dx
3
1 3 sec3 x tan x sec x tan x ln sec x tan x C 4 2
−3
3
−3
4 3
69
70
73.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
sec5 x tan x dx
4
1 sec5 x C 5
75.
sin 2 sin 3 d
0
1 1 sin sin 5 2 5
4
0
3 2 10
5
−2
2
−5
2
77.
sin4 x dx
0
2
1 3x 1 sin 2x sin 4x 4 2 8
79. (a) Save one sine factor and convert the remaining sine factors to cosine. Then expand and integrate.
0
3 16
(b) Save one cosine factor and convert the remaining cosine factors to sine. Then expand and integrate. (c) Make repeated use of the power reducing formula to convert the integrand to odd powers of the cosine.
81. (a) Let u tan 3x, du 3 sec2 3x dx.
(b)
sec4 3x tan3 3x dx
1 3
1 3
tan6 3x tan4 3x C1 18 12
sec2 3x tan3 3x sec2 3x dx
0.05
− 0.5
tan2 3x 1 tan3 3x3 sec2 3x dx
0.5
− 0.05
tan5 3x tan3 3x3 sec2 3x dx
Or let u sec 3x, du 3 sec 3x tan 3x dx.
(c)
sec4 3x tan3 3x dx
1 3
sec6 3x sec4 3x C 18 12
sec3 3x tan2 3x sec 3x tan 3x dx
sec3 3xsec2 3x 13 sec 3x tan 3x dx
sec6 3x sec4 3x 1 tan2 3x3 1 tan2 3x2 C C 18 12 18 12
1 1 1 1 1 1 1 tan6 3x tan4 3x tan2 3x tan4 3x tan2 3x C 18 6 6 18 12 6 12
tan6 3x tan4 3x 1 1 C 18 12 18 12
tan6 3x tan4 3x C2 18 12
1
83. A
sin2 x dx
y
0 1
0
1 cos2 x dx 2
x 1 sin2 x 2 4 1 2
1
1
1 2
0
x 1 2
1
Section 7.3
85. (a) V
2
sin2 x dx
0
(b) A
0
sin x dx cos x
0
1 x sin 2x 2 2
1 cos 2x dx
0
0
Trigonometric Integrals
2 2
112
Let u x, dv sin x dx, du dx, v cos x. x
1 A
x sin x dx
0
y
0
0
sin2 x dx
0
2
y
1
1 cos 2x dx
0
1 2
( π2 , π8 (
8 0
π 2
2 , 8
87. dv sin x dx ⇒
1 x cos x sin x 2
cos x dx
0
1 1 x sin 2x 8 2
x, y
x cos x
1 2A 1 8
1 2
x
π
v cos x
u sinn1 x ⇒ du n 1sinn2 x cos x dx
sinn x dx sinn1 x cos x n 1 sinn2 x cos2 x dx sinn1 x cos x n 1 sinn2 x1 sin2 x dx
sinn1 x cos x n 1 sinn2 x dx n 1 sinn x dx
Therefore, n sinn x dx sinn1 x cos x n 1 sinn2 x dx sinn x dx
sinn1 x cos x n 1 n n
sinn2 x dx.
89. Let u sinn1 x, du n 1sinn2 x cos x dx, dv cosm x sin x dx, v
mn m1
cosm1 x . m1
cosm x sinn x dx
sinn1 x cosm1 x n1 m1 m1
sinn2 x cosm2 x dx
n1 sinn1 x cosm1 x m1 m1
sinn2 x cosm x1 sin2 x dx
n1 sinn1 x cosm1 x m1 m1
sinn2 x cosm x dx
cosm x sinn x dx
sinn1 x cosm1 x n1 m1 m1
sinn2 x cosm x dx
cosm x sinn x dx
cosm1 x sinn1 n1 mn mn
cosm x sinn2 x dx
n1 m1
sinn x cosm x dx
71
72
91.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
sin5 x dx
sin4 x cos x 4 5 5
sin3 x dx
sin2 x cos x 2 sin4 x cos x 4 5 5 3 3
sin x dx
4 8 1 sin2 x cos x cos x C sin4 x cos x 5 15 15
93.
sec4
cos x 3 sin4 x 4 sin2 x 8 C 15
2 5 x dx 25 sec 2 5 x 25 dx 4
2 2 x 2 x 5 1 sec2 tan 2 3 5 5 3
2 x 2 x 2 x 5 sec2 tan 2 tan 6 5 5 5
2 x 5 tan 6 5
a1
1 b1 6 a0
sec2
2 5 x 25 dx
C
sec 2 5 x 2 C
2
t t b1 sin where: 6 6
12
1 12 1 6
95. (a) f t a0 a1 cos a0
f t dt
0
12
f t cos
0
12
f t sin
0
t dt 6 t dt 6
12 0 30.9 432.2 241.1 453.7 264.6 474.0 278.2 477.0 271.0 3122 460.1 247.1 435.7 30.9 55.46
a1
12 0 2 30.9 cos 0 4 32.2 cos 2 41.1 cos 4 53.7 cos 2 64.6 cos 6312 6 3 2 3
5 7 4 278.2 cos 4 77.0 cos 2 71.0 cos 6 6 3
3 5 11 2 47.1 cos 4 35.7 cos 30.9 cos 2 23.88 2 3 6
4 74.0 cos 4 60.1 cos b1
12 0 2 30.9 sin 0 4 32.2 sin 2 41.1 sin 4 53.7 sin 2 64.6 sin 6312 6 3 2 3
5 7 4 278.2 sin 4 77.0 sin 2 71.0 sin 6 6 3
3 5 11 2 47.1 sin 4 35.7 sin 30.9 sin 2 3.34 2 3 6
4 74.0 sin 4 60.1 sin Ht 55.46 23.88 cos —CONTINUED—
t t 3.34 sin 6 6
Section 7.3
Trigonometric Integrals
95. —CONTINUED— (b) a0
12 0 18.0 417.7 225.8 436.1 245.4 455.2 259.9 459.4 253.1 3122 443.2 234.3 424.2 18.0 39.34
a1
12 0 2 18.0 cos 0 4 17.7 cos 2 25.8 cos 4 36.1 cos 2 45.4 cos 6312 6 3 2 3
5 7 4 259.9 cos 4 59.4 cos 2 53.1 cos 6 6 3
3 5 11 2 34.3 cos 4 24.2 cos 18 cos 2 20.78 2 3 6
4 55.2 cos 4 43.2 cos b1
12 0 2 18.0 sin 0 4 17.7 sin 2 25.8 sin 4 36.1 sin 2 45.4 sin 6312 6 3 2 3
5 7 4 259.9 sin 4 59.4 sin 2 53.1 sin 6 6 3
3 5 11 2 34.3 sin 4 24.2 sin 18 sin 2 4.33 2 3 6
4 55.2 sin 4 43.2 sin Lt 39.34 20.78 cos
t t 4.33 sin 6 6
(c) The difference between the maximum and minimum temperatures is greatest in the summer. 85
H L
0
12 15
97.
cosmx cosnx dx
sinmx sinnx dx
sinmx cosnx dx
1 sinm nx sinm nx 2 mn mn
1 2
0, m n
sinm nx sinm nx dx
1 cosm nx cosm nx 2 mn mn
1 2
, m n
cosm n cosm n cosm n cosm n mn mn mn mn
0, since cos cos. sinmx cosmx dx
0, m n
cosm nx cosm nx dx
1 sinm nx sinm nx 2 mn mn 1 2
1 sin2mx m 2
0
73
74
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
Section 7.4 1.
Trigonometric Substitution
Indefinite integral:
x2 16
x
dx
2
x x2 16
2
4x
x2 16 x2 16 4
4x2
xx x 16 164 4x
4
3.
x2 16 4 d d 4 ln x2 16 C 4 ln x2 16 4 4 ln x x2 16 C dx x dx
4 x x x2 16
xx2 16 x2 16 4
4x2 4x2 16 16x2 16 x2x2 16 4x2 xx2 16 x2 16 4 x2 16x2 16 4x2 16
xx2 16 x2 16 4
x2 16 x2 16 x2 16 4 2 2 x x x 16 x 16 4
Matches (b)
x x16 x2 14 x1216 x2122x 16 x2 d C 8 8 arcsin dx 4 2 2 1 x42
16 x2 8 x2 2 2 2 16 x 216 x
16 x2 16 x2 x2 2 2 2 16 x2 216 x 216 x 216 x
Matches (a) 5. Let x 5 sin , dx 5 cos d, 25 x2 5 cos .
1 dx 25 x232
5 cos d 5 cos 3
1 25
5 x
sec d 2
θ 25 − x 2
1 tan C 25
x 2525 x2
7. Same substitution as in Exercise 5
25 x2
x
dx
C
25 cos2 d 1 sin2 5 d 5 csc sin d 5 sin sin
4x2 16 x2 16 4 x2 x2 16 4
5 ln csc cot cos C 5 ln
5 25 x2 25 x2 C x
Section 7.4
Trigonometric Substitution
9. Let x 2 sec , dx 2 sec tan d, x2 4 2 tan .
1 dx x2 4
2 sec tan d 2 tan
ln
x 2
x2 4
2
x
x2 − 4
sec d ln sec tan C1 θ 2
C1
ln x x 4 ln 2 C1 ln x x2 4 C 2
11. Same substitution as in Exercise 9
x3x2 4 dx
8 sec3 2 tan 2 sec tan d 32 tan2 sec4 d
32 tan2 1 tan2 sec2 d 32
tan3 tan5 C 3
5
32 x2 432 x2 4 32 3 tan 5 3 tan2 C 53 C 15 15 8 4
1 1 2 x 432 20 3x2 4 C x2 4323x2 8 C 15 15
13. Let x tan , dx sec2 d, 1 x2 sec .
x1 x2 dx
1 + x2
sec3 1 C 1 x232 C 3 3
tan sec sec2 d
x
θ
Note: This integral could have been evaluated with the Power Rule.
1
15. Same substitution as in Exercise 13
1 dx 1 x22
1 1 x2
4
dx 1 + x2 x
d sec4
sec2
θ
cos2 d
sin 2 1 2 2
1
sin cos C 2
x 1 arctan x 2 1 x2
1 x arctan x C 2 1 x2
1
1 1 cos 2 d 2
1 1 x2
C
17. Let u 3x, a 2, and du 3 dx.
4 9x2 dx
1 22 3x2 3 dx 3
1 1 3x4 9x2 4 ln 3x 4 9x2 C 3 2
2 1 x4 9x2 ln 3x 4 9x2 C 2 3
75
76
19.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
x 1 dx 9 2
x2
x2 9122x dx
x2
21.
1 16
x2
dx arcsin
4 C x
9C
(Power Rule) 23. Let x 2 sin , dx 2 cos d, 4 x2 2 cos .
16 4x2 dx 2 4 x2 dx
2 x
θ
2 2 cos 2 cos d
4 − x2
8 cos2 d 4 1 cos 2 d
4
1 sin 2 C 2
4 4 sin cos C 4 arcsin
2x x4 x
2
C
27. Let x sin , dx cos d, 1 x2 cos .
25. Let x 3 sec , dx 3 sec tan d, x2 9 3 tan .
1 x2 9
dx
3 sec tan d 3 tan
1 x2
x4
dx
sec d
cos cos d sin4
cot2 csc2 d
1 cot3 C 3
ln sec tan C1
x2 9 x ln C1 3 3
ln x x2 9 C
1
1 x232 C 3x3
x
θ
x
x2 − 9
1 − x2
θ 3
29. Same substitutions as in Exercise 28
1 dx x4x2 9
1 3
32 sec2 d 32 tan 3sec
1 1 4x2 9 3 csc d ln csc cot C ln C 3 3 2x
Section 7.4 31. Let x 5 tan , dx 5 sec2 d, x2 5 5 sec2 .
5x dx x2 532
55 tan 5 sec2 d 5 sec2 32
5
x2 + 5 x
tan d sec
θ 5
5 sin d 5 cos C 5
5
x2 5
C
5 C 5
x2
33. Let u 1 e2x, du 2e2x dx.
e2x1 e2x dx
1 2
1 1 e2x122e2x dx 1 e2x32 C 3
35. Let ex sin , ex dx cos d, 1 e2x cos .
ex1 e2x dx
cos2 d
1 2
1
1 cos 2 d
sin 2 1 2 2
ex
θ 1 − e 2x
1 1 sin cos C arcsin ex ex1 e2x C 2 2 37. Let x 2 tan , dx 2 sec2 d, x2 2 2 sec2 .
1 dx 4 4x2 x4
1 dx x2 22 2
2
4
sec2
4 sec4
cos2
x2 + 2 x
d
θ 2
d
2 1 cos 2 d
2 1
4 2
8 2
8
21 sin 2 C sin cos C
1 x 1 x arctan C 4 x2 2 2 2
Trigonometric Substitution
77
78
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
1 39. Since x > , 2 1 dx, dv dx ⇒ v x x4x2 1
u arcsec 2x, ⇒ du
arcsec 2x dx x arcsec 2x
2x sec , dx
arcsec 2x dx x arcsec 2x
1 4x x2
dx
1 sec tan d, 4x2 1 tan 2
x arcsec 2x
41.
1 4x2 1
dx
1 12 sec tan d x arcsec 2x tan 2
sec d
1 1 ln sec tan C x arcsec 2x ln 2x 4x2 1 C. 2 2
1 4 x 22
dx arcsin
x 2 2 C
43. Let x 2 2 tan , dx 2 sec2 d, x 22 4 2 sec .
x dx x 4x 8 2
x dx x 22 4
2 tan 22 sec2 d 2 sec
2 tan 1sec d
2 sec ln sec tan C1
2
x 22 4
2
ln
x 4x 8 2 ln
x 22 4
2
x2 2
C1
x 4x 8 x 2 C
x2 4x 8 2 ln x2 4x 8 x 2 ln 2 C1
2
2
45. Let t sin , dt cos d, 1 t 2 cos2 . (a)
t2 dt 1 t232
sin2 cos d cos3 1
tan2
d
t
θ 1 − t2
sec2 1 d
tan C
32
Thus,
0
t 1 t2
arcsin t C
32
t2 t dt arcsin t 1 t232 1 t 2
0
32 14
(b) When t 0, 0. When t 32, 3. Thus,
32
0
t2 dt tan 1 t232
3
0
3
0.685. 3
arcsin
3
2
3
0.685. 3
Section 7.4
Trigonometric Substitution
47. (a) Let x 3 tan , dx 3 sec2 d, x2 9 3 sec .
x3 x2 9
dx
27 tan3 3 sec2 d 3 sec
27 sec2 1 sec tan d
13 sec
27
3
9
3
Thus,
0
sec C 9 sec3 3 sec C
3
x2 9
3
3
C 31 x 9
x2 9
2
3
13 54
9x2 9 C
3
x3 1 dx x2 932 9x2 9 3 x2 9
32
0
9 27
2 272
18 92 9 2 2 5.272. (b) When x 0, 0. When x 3, 4. Thus,
3
0
x3 dx 9 sec3 3 sec 9
x2
4
0
9 22 32 91 3 9 2 2 5.272.
49. (a) Let x 3 sec , dx 3 sec tan d, x2 9 3 tan .
x2 x2 9
dx
9 sec2 3 sec tan d 3 tan
9
sec3
x
d
θ 3
12 sec tan 21sec d
9
(7.3 Exercise 90)
9 sec tan ln sec tan 2
9 x 2 3
x2 9
3
Hence,
6
4
ln
x2 9 x 3 3
x2 9 x2 9 xx2 9 x dx ln x2 9 2 9 3 3
9 2
27 627 ln 2 9 3
6
4
4 9 7 ln 43 37
93 27
6 27 4 7 9 ln ln 2 3 3
93 27
9 6 33 ln 2 4 7
93 27
9 4 7 2 3 12.644. ln 2 3
—CONTINUED—
x2 − 9
79
80
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
49. —CONTINUED— (b) When x 4, arcsec
43 .
When x 6, arcsec2
6
x2 9 dx sec tan ln sec tan 9 2
x2
4
. 3
9
2 2
x2
53.
x2
9 4 7 4 7 ln 3 3 3 3
3 ln 2 3 2
93 27
51.
3
arcsec43
9 6 33 12.644 ln 2 4 7
1 x2 dx x2 10x 9 x 15 33 ln x 5 x2 10x 9 C 2 10x 9
x2 1 dx xx2 1 ln x x2 1 C 2 1
a
57. A 4
0
55. (a) u a sin
b a2 x2 dx a
y
a
4b a
b y= a
(b) u a tan
a2 − x2
b
a2 x2 dx
0
4b 1 a 2
a2 arcsin
x xa2 x2 a
−a
a
x
a
0
−b
2b 2 a a 2
ab Note: See Theorem 7.2 for a2 x2 dx. 59. x2 y2 a2 x ± a2 y2
a
A2
a2 y2 dy a2 arcsin
h
ay ya
2
2 a
a2
2
arcsin
y2
a
(Theorem 7.2)
h
ha ha
2
h2
h a2 a2 arcsin ha2 h2 2 a
61. Let x 3 sin , dx cos d, 1 x 32 cos .
y
Shell Method:
2
4
1
V 4
x1 x 32 dx
x
2
4
1
2
−1
3 sin cos 2 d
2
−2
4
32
4
2 2 sin 2 3 cos
2
1 cos 2 d
2
3
1
1
2
cos2 sin d
2
2
3
2
6 2
3
(c) u a sec
Section 7.4
Trigonometric Substitution
1 1 x2 1 63. y ln x, y , 1 y 2 1 2 x x x2 Let x tan , dx sec2 d, x2 1 sec .
5
s
1
sec sec2 d tan
b
x2 1 dx x2
a
5
x2 1
x
1
b
a
x2 + 1 x
dx θ
sec 1 tan2 d tan
1
b
csc sec tan d
a
ln csc cot sec
b
a
x 26 1 ln
26 ln 2 1 2 5 5 2 1 26 1 26 2 4.367 or ln ln 26 1 5 2 1 x2 1
ln
1 x2 1 x
1
5
26 2
65. Length of one arch of sine curve: y sin x, y cos x L1
1 cos 2 x dx
0
Length of one arch of cosine curve: y cos x, y sin x L2
2
1 sin2 x dx
2
2
1 cos x 2 dx
ux
2
2
, du dx 2
0
1 cos 2 u du
1 cos 2 u du L1
0
67. (a)
(b) y 0 for x 200 (range)
60
−25
250 −10
(c) y x 0.005x 2, y 1 0.01x, 1 y2 1 1 0.01x 2 Let u 1 0.01x, du 0.01 dx, a 1. (See Theorem 7.2.)
200
s
200
1 1 0.01x2 dx 100
0
1 0.01x2 1 0.01 dx
0
50 1 0.01x1 0.01x2 1 ln 1 0.01x 1 0.01x2 1
50 2 ln 1 2 2 ln 1 2 1002 50 ln
2 1 2 1
229.559
200
0
81
82
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
69. Let x 3 tan , dx 3 sec2 d, x2 9 3 sec .
4
A2
0
4
3 dx 6 2 x 9
b
0
sec d 6 ln sec tan
6
a
b
dx 6 x2 9
a b
a
3 sec2 d 3 sec
6 ln
x 0 (by symmetry) y
4
1 1 2 A
4
3 9
x2
3
4
6 ln 3 0 y
dx 2
3 4 1 2
4
9 12 ln 3
3 1 x arctan 4 ln 3 3 3
1 dx 2 4 x 9
2 4 arctan 0.422 4 ln 3 3
x2 9 x
(0, 0.422) 1 4
x
4
−4
−2
2
4
4
1 4 arctan 0, 0.422 2 ln 3 3
x, y 0,
1 y 1 4x 2
y 2x,
71. y x2,
2x tan , dx
1 sec2 d, 1 4x2 sec 2
(For sec5 d and sec3 d, see Exercise 80 in Section 7.3)
2
S 2
a
0
4
b
x21 4x2 d x 2
b
sec3 tan2 d
a
4
tan 2 1 sec sec2 d 2 2
sec d sec d b
b
5
3
a
a
3 1 1 sec3 tan sec tan ln sec tan sec tan ln sec tan 4 4 2 2
1 1
1 4x 2322x 1 4x 2122x ln 1 4x2 2x 4 4 8
51 2 ln 3 2 2
32 102 4 4 8
b a
2
0
542 62 1 ln 3 22 4 4 6 8
2 ln 3 22
13.989
73. (a) Area of representative rectangle: 21 y2 y
y
Pressure: 262.43 y1 y 2 y
2
1
F 124.8
1
x=
3 y1 y d y
1
124.8 3
1
1
1 y 2 dy
x −2
y1 y 2 dy
1
32 arcsin y y1 y 21 23 1 y
1
124.8
2
2 32
1
62.43 arcsin 1 arcsin1 187.2 lb
1
(b) F 124.8
1
1 − y2
2
1
d y1 y 2 dy 124.8 d 124.8
1
1
1 y 2 dy 124.8
1
d2 arcsin y y1 y
y1 y 2 dy
1
2
1
124.80 62.4 d lb
2
Section 7.4
75. (a) m
dy y y 144 x2 dx x0
y
( 0, y +
Trigonometric Substitution
144 − x 2 (
12
144 x2
x
12
144 − x 2
( x, y ) x
y
x 2
(b) y
144 x2
x
4
6
8
10 12
dx 12
Let x 12 sin , dx 12 cos d, 144 x2 12 cos . y
x
θ
12 cos 1 sin2 12 cos d 12 d 12 sin sin
144 − x 2
12 csc sin d 12 ln csc cot 12 cos C
C
12 ln
144 x2 12 144 x2 12 x x 12
12 ln
12 144 x2 144 x2 C x
0
12
144 x2
x
144 x2.
12 144 x2 > 0 for 0 < x ≤ 12 x
(c) Vertical asymptote: x 0 (d) y 144 x2 12 ⇒ y 12 144 x2 Thus, 12 144 x2 12 ln
12
12
1 ln
144 x2
x
144 x2
x
144 x2
xe1 12 144 x2
xe1 122 144 x2
2
x2e2 24xe1 144 144 x2 x2e2 1 24xe1 0 x xe2 1 24e1 0 x 0 or x
24e1 7.77665. e2 1
Therefore,
12
s
7.77665 12
7.77665
1
144 x2
x
2
12
dx
7.77665
x
2
144 x2 dx x2
7.77665 12ln 12 ln 7.77665 5.2 meters.
12 dx 12 ln x x
12
12 0
When x 12, y 0 ⇒ C 0. Thus, y 12 ln Note:
30
83
84
Chapter 7
77. True
dx 1 x2
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
cos d cos
79. False
3
d
0
dx 1 x2 3
3
0
sec2 d sec3
3
cos d
0
81. Let u a sin , du a cos d, a 2 u 2 a cos .
a2 u2 du
a2 cos2 d a2
1 cos 2 d 2
1 a2 a2 sin 2 C sin cos C 2 2 2
a2 u u arcsin 2 a a
a2 u2
a
C 21 a arcsin ua u
a2 u2 C
2
Let u a sec , du a sec tan d, u2 a2 a tan .
u 2 a 2 du
a tan a sec tan d a 2 tan2 sec d
a 2 sec2 1 sec d a2 sec3 sec d
12 sec tan 21 sec d a sec d a 12 sec tan 21 lnsec tan u a a u ln 2 a a ua u a a C a2
2
2
2
2
2
2
2
1
1 uu2 a2 a2 ln u u2 a2 C 2
Let u a tan , du a sec2 d, u2 a2 a sec d.
u2 a2 du
a sec a sec2 d
12 sec tan 21 lnsec tan C
a2 sec3 d a2
Section 7.5
a2 u2 a2 2 a
u
1
a ln
u2 a2
a
u a
C1
1 uu2 a2 a2 ln u u2 a2 C 2
Partial Fractions
1.
5 A B 5 x 2 10x xx 10 x x 10
3.
2x 3 2x 3 A Bx C 2 x3 10x xx 2 10 x x 10
5.
16x 16x A C B 2 x3 10x 2 x 2x 10 x x x 10
7.
1 1 A B x 2 1 x 1x 1 x 1 x 1 1 Ax 1 Bx 1 When x 1, 1 2A, A 12 . When x 1, 1 2B, B 12 .
1 1 dx x2 1 2
1 1 dx x1 2
1 dx x1
1 1 ln x 1 ln x 1 C 2 2
1 x1 ln C 2 x1
Section 7.5
9.
3 3 A B x 2 x 2 x 1x 2 x 1 x 2
5 x Ax 1 B2x 1
When x 1, 3 3A, A 1. When x 2, 3 3B, B 1. 3 dx x2 x 2
1 9 3 When x 2 , 2 2 A, A 3.
1 dx x1
1 dx x2
When x 1, 6 3B, B 2.
ln x 1 ln x 2 C
ln
13.
5x dx 3 2x 2 x 1
x1 C x2
1 1 dx 2 dx 2x 1 x1
3 ln 2x 1 2 ln x 1 C 2
B C x 2 12x 12 A xx 2x 2 x x2 x2 x 2 12x 12 Ax 2x 2 Bxx 2 Cxx 2 When x 0, 12 4A, A 3. When x 2, 8 8B, B 1. When x 2, 40 8C, C 5.
x 2 12x 12 dx 5 x3 4x
1 dx x2
1 dx 3 x2
1 dx x
5 ln x 2 ln x 2 3 ln x C
15.
2x3 4x 2 15x 5 x5 A B 2x 2x x 2 2x 8 x 4x 2 x4 x2 x 5 Ax 2 Bx 4 When x 4, 9 6A, A 32 . When x 2, 3 6B, B 12 .
2x3 4x 2 15x 5 dx x2 2x 8
2x
x2
17.
12 32 dx x4 x2
3 1 ln x 4 ln x 2 C 2 2
B 4x 2 2x 1 A C 2 x 2x 1 x x x1 4x 2 2x 1 Axx 1 Bx 1 Cx 2 When x 0, B 1. When x 1, C 1. When x 1, A 3.
4x 2 2x 1 dx x3 x 2
1 3 1 1 2 dx 3 ln x ln x 1 C x x x1 x
1 ln x 4 x 3 C x
2 2 C B 19. x 3x 4 x 3x 4 A 3 2 x 4x 4x xx 22 x x 2 x 22
x2 3x 4 Ax 22 Bxx 2 Cx When x 0, 4 4A ⇒ A 1. When x 2, 6 2C ⇒ C 3. When x 1, 0 1 B 3 ⇒ B 2.
x2 3x 4 dx x3 4x2 4x
1 dx x
85
5x 5x A B 2x 2 x 1 2x 1x 1 2x 1 x 1
11.
3 x 2 Bx 1
Partial Fractions
2 dx x 2
ln x 2 ln x 2
3 dx x 22
3 C x 2
86
21.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
x2 1 A Bx C 2 xx 2 1 x x 1 x 2 1 Ax 2 1 Bx Cx When x 0, A 1. When x 1, 0 2 B C. When x 1, 0 2 B C. Solving these equations we have A 1, B 2, C 0.
x2 1 dx x3 x
1 dx x
2x dx x2 1
ln x 2 1 ln x C
ln
23.
x2
1 C x
A B Cx D x2 2 x 4 2x 2 8 x 2 x 2 x 2 x 2 Ax 2x 2 2 Bx 2x 2 2 Cx Dx 2x 2 When x 2, 4 24A. When x 2, 4 24B. When x 0, 0 4A 4B 4D, and when x 1, 1 9A 3B 3C 3D. Solving these equations we have A 16 , B 16 , C 0, D 13 .
x4
x 1 2 dx x 1 2 dx 2 x 1 x2 x ln 2 arctan C 6 x 2 2
x2 1 dx 2x 2 8 6
25.
2
1 dx 2
x A B Cx D 2x 12x 14x 2 1 2x 1 2x 1 4x 2 1 x A2x 14x 2 1 B2x 14x 2 1 Cx D2x 12x 1 When x 12 , 12 4A. When x 12 , 12 4B. When x 0, 0 A B D, and when x 1, 1 15A 5B 3C 3D. Solving these equations we have A 18 , B 18 , C 12 , D 0.
x 1 dx 16x 4 1 8
27.
2x 1 1 dx 2x 1 1 dx 4 4x x 1 dx 2
1 4x 2 1 ln C 16 4x 2 1
x2 5 A Bx C x 1x 2 2x 3 x 1 x 2 2x 3 x 2 5 Ax 2 2x 3 Bx Cx 1 A Bx 2 2A B Cx 3A C When x 1, A 1. By equating coefficients of like terms, we have A B 1, 2A B C 0, 3A C 5. Solving these equations we have A 1, B 0, C 2.
x3
x2 5 dx x2 x 3
1 dx 2 x1
1 dx x 12 2
ln x 1 2 arctan
x21 C
Section 7.5
29.
Partial Fractions
3 A B 2x 1x 2 2x 1 x 2 3 Ax 2 B2x 1 When x
1
0
1 2 ,
A 2. When x 2, B 1.
3 dx 2x 5x 2 2
1
0
2 dx 2x 1
1 dx x2
0
1
0
1
ln 2x 1 ln x 2 ln 2
31.
x1 A Bx C 2 xx 2 1 x x 1 x 1 Ax 2 1 x Cx When x 0, A 1. When x 1, 2 2A B C. When x 1, 0 2A B C. Solving these equations we have A 1, B 1, C 1.
2
1
x1 dx xx 2 1
2
1
1 dx x
ln x
2
1
x2
x dx 1
2
1
1 dx x2 1
1 lnx 2 1 arctan x 2
2 1
1 8 ln arctan 2 2 5 4
0.557
33.
3x dx 9 3 ln x 3 C x 2 6x 9 x3
4, 0: 3 ln4 3
35.
9 C0⇒C9 43
2 x2 x 2 x 1 arctan dx C x 2 22 2 2x 2 2 2
0, 1: 0
30
1 5 C1⇒C 4 4 3
(0, 1) −6
(4, 0)
−3
10
3 −1
−10
37.
2x 2 2x 3 1 2x 1 dx ln x 2 ln x 2 x 1 3 arctan C x2 x 2 2 3
x3
3, 10: 0
39.
20
(3, 10)
7 1 7 1 C 10 ⇒ C 10 ln 13 3 arctan ln 13 3 arctan 2 2 3 3
1 1 x2 dx ln C x2 4 4 x2
6, 4:
1 4 1 1 1 ln C 4 ⇒ C 4 ln 4 ln 2 4 8 4 2 4
10
(6, 4) − 10
10 −3
−2
6 −5
87
88
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
41. Let u cos x du sin x dx.
43.
3 cos x 1 dx 3 2 du sin2 x sin x 2 u u2
A B 1 uu 1 u u1
1 Au 1 Bu When u 0, A 1. When u 1, B 1, u cos x, du sin x dx.
u1 C u2
ln
1 sin x C 2 sin x
(From Exercise 9 with u sin x, du cos x dx)
sin x 1 dx du cos xcos x 1 uu 1
ln
1 du u
1 du u1
ln u ln u 1 C
u C ln u1 ln
cos x C cos x 1
45. Let u ex, du ex dx.
47.
1 A B u 1u 4 u 1 u 4
1 A B xa bx x a bx 1 Aa bx Bx When x 0, 1 a A ⇒ A 1a. When x ab, 1 abB ⇒ B ba.
1 Au 4 Bu 1
1 1 When u 1, A 5 . When u 4, B 5 , u ex, x dx. du e
ex dx x e 1ex 4
1 5
49.
1
u 1u 4
1 du u1
du
1 u1 ln C 5 u4
1 ex 1 ln C 5 ex 4
1 du u4
x A B a bx2 a bx a bx2
51.
−2
1b ab dx a bx a bx2
1 b
1 a dx a bx b
1 dx a bx2
a 1 1 ln a bx 2 C b2 b a bx
1 a ln a bx b2 a bx
1 x ln C a a bx
C
2 −4
1 b dx x a bx
1 ln x ln a bx C a
10
When x ab, B ab. When x 0, 0 aA B ⇒ A 1b.
6 dy , y0 3 dx 4 x2
x Aa bx B
x dx a bx2
1 1 dx xa bx a
Section 7.5 53. Dividing x3 by x 5.
Partial Fractions
89
55. (a) Substitution: u x2 2x 8 (b) Partial fractions (c) Trigonometric substitution (tan) or inverse tangent rule
57. Average Cost
1 80 75
1 5
80
75
80
75
124p dp 10 p100 p
124 1240 dp 10 p11 100 p11
1 124 1240 ln10 p ln100 p 5 11 11
80 75
1 24.51 4.9 5 Approximately $490,000.
3
59. A
1
10 dx 3 xx2 1
Matches (c) y 5 4 3 2 1 x 1
2
3
5
A B 1 1 ,AB x 1n x x 1 n x n1
61. 1 n1
4
1 1 dx kt C x1 nx
x1 1 ln kt C n1 nx When t 0, x 0, C
1 1 ln . n1 n
x1 1 1 1 ln kt ln n1 nx n1 n
x1 1 1 ln ln kt n1 nx n
ln
nx n n 1k t nx nx n en1kt nx x
nen1kt 1 n en1kt
Note: lim x n t →
90
63.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
Cx D x Ax B 1 x4 x2 2 x 1 x2 2 x 1 x Ax B x2 2 x 1 Cx D x2 2 x 1 A Cx3 B D 2 A 2 Cx2 A C 2 B 2 Dx B D 0 A C ⇒ C A 0 B D 2 A 2 C 1 A C 2 B 2 D
22 A 0 ⇒ A 0 and C 0 22 B 1 ⇒ B
2
and D
4
2
4
0 B D ⇒ D B Thus,
1
0
x dx 1 x4
1
0
1
2
4
2
4
2 4 2 4 2 dx x 2 x 1 x 2 x 1 2
0
1 1 dx 2 2 x 2 2 1 2 x 2 2 1 2
1 x 2 2 x 2 2 arctan arctan 1 2 1 2 1 2
1 0
1
1 arctan 2 x 1 arctan 2 x 1 2
1
arctan 2 1 arctan 2 1 arctan 1 arctan1 2
1 arctan 2 1 arctan 2 1 . 2 4 4
0
Since arctan x arctan y arctan x y 1 xy , we have:
1
0
1 x 2 1 2 1 1 1 2 dx arctan arctan 1 x4 2 1 2 1 2 1 2 2 2 2 2 4 2 8
Section 7.6 1. By Formula 6:
3. By Formula 26:
Integration by Tables and Other Integration Techniques
x x2 dx 2 x ln 1 x C 1x 2
e x1 e 2x dx
1 x
e e 2x 1 ln e x e 2x 1 C 2
u e x, du e x dx
5. By Formula 44:
1 x 2 1 dx C x x 21 x 2
Section 7.6
7. By Formulas 50 and 48:
sin42x dx
Integration by Tables and Other Integration Techniques
1 sin42x2 dx 2
1 sin 2x cos2x 3 2x sin 2x cos 2x C 2 4 8
1 sin32x cos2x 3 sin22x2 dx 2 4 4 3
9. By Formula 57:
1
x 1 cos x
u x, du
1 6x 3 sin 2x cos 2x 2 sin3 2x cos 2x C 16
dx 2
2 cot x csc x C
1 dx 2x
11. By Formula 84:
1 1 dx 1 cos x 2x
13. By Formula 89:
1 1 dx x ln1 e 2x C 1 e 2x 2
x 3 ln x dx
x4 4 ln x 1 C 16
x 2e x dx x 2e x 2 xe x dx
15. (a) By Formulas 83 and 82:
x 2e x 2 x 1e x C1 x 2e x 2xe x 2e x C (b) Integration by parts: u x 2, du 2x dx, dv e x dx, v e x
x 2e x dx x 2ex
2xe x dx
Parts again: u 2x, du 2 dx, dv e x dx, v e x
x 2e x dx x 2e x 2 xe x
2e x dx x 2e x 2xe x 2e x C
17. (a) By Formula: 12, a b 1, u x, and
1 1 1 1 x ln dx x 2x 1 1 x 1 1x
(b) Partial fractions:
C
1 x C ln x 1x
1 x1 ln C x x
1 A B C 2 x 2x 1 x x x1 1 Axx 1 Bx 1 Cx 2 x 0: 1 B x 1: 1 C x 1: 1 2A 2 1 ⇒ A 1
1 dx x 2x 1
1 1 1 2 dx x x x1
ln x
1 ln x 1 C x
1 x ln C x x1
91
92
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
19. By Formula 81:
21. By Formula 79:
1 2 2 xex e x C 2
x arcsecx 2 1 dx
1 arcsecx 2 1(2x dx 2 1 2
x 1 arcsecx 2 1 ln x 2 1 x 4 2x 2 C 2
u x 2 1, du 2x dx
23. By Formula 89:
27. By Formula 4:
x 2 ln x dx
x3 1 3 ln x C 9
25. By Formula 35:
31. By Formula 73:
e x arccos e x dx e x arccos e x 1 e 2x C
u e x, du e x dx
cos x dx arctansin x C 1 sin2 x
2 cos 1 sin d arctan C 3 2 sin sin2 2 2
u sin , du cos d
1 dx 3 x 22 9x 2
3
3x 2 3x2 2
2
32 9x 2 C 6x 2 9x 2
2x
C
39. By Formulas 54 and 55:
x 2 4
4x
C
x 1 2x dx dx 1 sec x 2 2 1 sec x 2 1 x 2 cot x 2 csc x 2 C 2
u sin x, du cos x dx
37. By Formula 35:
4
dx
35. By Formula 14:
1 x 2x 2
2x x 2 1 dx 2 dx ln1 3x C 1 3x2 1 3x2 9 1 3x
29. By Formula 76:
33. By Formula 23:
t3 cos t dt t3 sin t 3 t2 sin t dt
t3 sin t 3 t2 cos t 2 t cos t dt
t3 sin t 3t2 cos t 6 t sin t
sin t dt
t3 sin t 3t2 cos t 6t sin t 6 cos t C
dx
Section 7.6
41. By Formula 3:
Integration by Tables and Other Integration Techniques
ln x 1 dx 2 ln x 3 ln 3 2 ln x C x3 2 ln x 4
1 dx x
u ln x, du
43. By Formulas 1, 25, and 33:
x 1 2x 6 6 dx dx x 2 6x 102 2 x 2 6x 102
3 x3 1 arctanx 3 C 2x 2 6x 10 2 x 2 6x 10
45. By Formula 31:
x x 4 6x 2 5
3x 10 3 arctanx 3 C 2x 2 6x 10 2
1 2x dx 2 x 2 32 4
dx
1 1 x 2 6x 1022x 6 dx 3 dx 2
x 32 1 2
1 ln x 2 3 x 4 6x 2 5 C 2
u x 2 3, du 2x dx
47.
x3 dx 4 x 2
8 sin3 2 cos d 2 cos
x
8 1
cos2
sin d
θ 4 − x2
8 sin
cos 2
8 cos
8 cos3 C 3
2
sin d
4 x 2 2 x 8 C 3
x 2 sin , dx 2 cos d, 4 x 2 2 cos
49. By Formula 8:
e 3x dx 1 e x3
e x2 e x dx 1 e x3
2 1 ln 1 e x C 1 e x 21 e x 2
u e x, du e x dx
51.
1 2a bu a2 b2 1 A B u2 2 2 2 a bu b a bu2 b a bu a bu2
2a a2 u 2 Aa bu B aA B bAu b b
Equating the coefficients of like terms we have aA B a2 b2 and bA 2a b. Solving these equations we have A 2a b2 and B a2 b2.
a 1 bub du ba 1ba 1bu b du b1 u 2ab lna bu ba a 1 bu C
u2 1 2a 1 du 2 du 2 a bu2 b b b
2 2
C
a2 1 bu 2a ln a bu b3 a bu
2
2
2
3
3
93
94
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
53. When we have u2 a2:
When we have u2 a2:
u a tan du a
u2
a2
1 du u2 a23 2
sec2
a2
u a sec
d
sec2
du a sec tan d
a sec2 d a3 sec3
a2 a2 tan2
1 du u2 a23 2
a sec tan d a3 tan3
1 cos d a2
1 sin C a2
u C a2u2 a2
u
u
θ
1 cos d a2 sin2
u2 + a2
1 csc C a2
u C a2u2 a2
u2 + a2
θ a
a
55.
u2
arctan u du u arctan u u arctan u
1 2u du 2 1 u2 1 ln1 u2 C 2
u arctan u ln1 u2 C w arctan u, dv du, dw
57.
du ,vu 1 u2
21 x 1 dx C x x3 21 x
8
12, 5: 21 21 2 C 5 ⇒ C 7
( 12, 5) −0.5
1.5
21 x 7 y x 59.
1 1 x3 dx tan1x 3 2 C x2 6x 102 2 x 6x 10
3, 0: y
61.
−2
2
1 0 0 C0⇒C0 2 10
1 x3 tan1x 3 2 2 x 6x 10
−8
−2
1 d csc C sin tan
4 , 2: 22 C 2 ⇒ C 2 2 y csc 2 2
(3, 0)
10
(π4, 2) −
2
2 −2
8
Section 7.6
63.
1 d 2 3 sin
2
2 du 1 u2 2u 23 1 u2
Integration by Tables and Other Integration Techniques
2
65.
0
1 d 1 sin cos
2 du 1 u2 2u 1 u2 1 1 u2 1 u2
1 du 1u
0
1 du u 3u 1
0
ln 1 u
1
ln 2
1 du 3 2 5 u 2 4
u 23 25 u 23 25
u tan
1 ln 5
0
1
2 du 1 u2 6u 2
1
95
2
C
2u 3 5 C 2u 3 5 2 tan 3 5 2 1 ln C 5 2 tan 3 5 2
1
5
ln
u tan
67.
2
1 sin d 3 2 cos 2
2 sin d 3 2 cos
69.
cos 1 d 2 cos d 2
2 sin C
1 ln u C 2
u , du
1 ln3 2 cos C 2
1 d 2
u 3 2 cos , du 2 sin d
8
71. A
0
x dx x 1
y
73. Arctangent Formula, Formula 23,
4
22 x x 1 3
8
3
0
2
4 12 3
1 x 2
1 du, u ex u2 1
4
6
8
40 13.333 square units 3
75. Substitution: u x2, du 2x dx Then Formula 81.
77. Cannot be integrated.
79. Answers will vary. For example,
2xe2x dx
can be integrated by first letting u 2x and then using Formula 82.
96
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
5
81. W
2000xex dx
0
5
2000
xex dx
0
5
2000
xex1 dx
0
2000 xex ex
2000
6 1 e5
5 0
1919.145 ft lbs
3
83. (a) V 202
2
1 y 2
0
80 ln y 1 y 2
80 ln 3 10
W 148 80 ln 3 10
dy 3
11,840 ln 3 10
0
21,530.4 lb
145.5 cubic feet (b) By symmetry, x 0.
3
M 2
0
3
Mx 2
0
y
2
1 y 2
dy 4 ln y 1 y 2
2y dy 4 1 y 2
1 y 2
3 0
0 4 ln 3 10 3
4 10 1
Mx 4 10 1 1.19 M 4 ln 3 10
Centroid: x, y 0, 1.19
4
85. (a)
0
k
4
0
4
k dx 10 2 3x
(b)
0
15.417 dx 2 3x
8
10 10 1 0.6486 dx 2 3x
15.417
4
0 −1
ln307
87. False. You might need to convert your integral using substitution or algebra.
Section 7.7 1. lim
x →0
Indeterminate Forms and L’Hôpital’s Rule
sin 5x 5 2.5 exact: sin 2x 2
x f x
3
0.1
0.01
0.001
0.001
0.01
0.1
2.4132
2.4991
2.500
2.500
2.4991
2.4132
−1
1 −1
Section 7.7
Indeterminate Forms and L’Hôpital’s Rule
3. lim x5ex 100 0
10,000,000
x →
x f x
1
10
102
103
104
105
0.9901
90,484
3.7 109
4.5 1010
0
0 0
100 0
5. (a) lim
x →3
(b) lim
x →3
7. (a) lim
1 2x 3 2x 3 2 lim lim x →3 x 3x 3 x →3 x 3 x2 9 3 2 1 2x 3 2 d dx 2x 3 lim lim x →3 d dx x2 9 x →3 2x x2 9 6 3
x 1 2
x3
x →3
(b) lim
x 1 2
x →3
x3
lim
x 1 2
x3
x →3
lim
x →3
x 1 2
x 1 2
lim
x →3
1 x 1 4 1 lim x 3 x 1 2 x →3 x 1 2 4
1 2 x 1 1 d dx x 1 2 lim x →3 d dx x 3 1 4
5x2 3x 1 5 3 x 1 x2 5 lim 2 x → x → 3x 5 3 5 x2 3
9. (a) lim
5x2 3x 1 d dx 5x2 3x 1 10x 3 d dx 10x 3 10 5 lim lim lim lim 2 x → x → x → x → x → 6 3x 5 d dx 3x2 5 6x d dx 6x 3
(b) lim
x2 x 2 2x 1 lim 3 x →2 x →2 x2 1
11. lim
15. lim
x →0
13. lim
4 x2 2
x →0
x
x 4 x2 0 x →0 1
lim
ex 1 x ex 1 lim 2 x →0 x 1
17. Case 1: n 1 lim
x →0
ex 1 x ex 1 lim 0 x →0 x 1
Case 2: n 2 lim
x →0
ex 1 x ex 1 ex 1 lim lim 2 x →0 x →0 2 x 2x 2
Case 3: n ≥ 3 lim
x →0
ex 1 x ex 1 ex lim n1 lim n x →0 nx x →0 nn 1xn2 x
sin 2x 2 cos 2x 2 lim x→0 sin 3x x→0 3 cos 3x 3
19. lim
arcsin x 1 1 x2 lim 1 x →0 x →0 x 1
3x2 2x 1 6x 2 lim x→ x→ 2x2 3 4x
21. lim
23. lim
lim
x→
x2 2x 3 2x 2 lim x→ x → x1 1
25. lim
6 3 4 2
x3 3x2 lim x 2 x→ e x→ 1 2e x 2
27. lim
lim
x→
6x 6 lim 0 1 4e x 2 x→ 1 8e x 2
97
98
Chapter 7 x
29. lim
x→
x2 1
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
lim
1
x→
1 1 x2
1
31. lim
x→
cosx x ≤ 1x
Note: L’Hôpital’s Rule does not work on this limit. See Exercise 79. ln x 1 x 1 lim lim 2 0 x→ x2 x→ 2x x→ 2x
cos x 0 by Squeeze Theorem x
ex ex ex lim lim 2 x→ x x→ 2x x→ 2
33. lim
35. lim
37. (a) lim x ln x 0 0
39. (a) lim x sin
x →0
ln x (b) lim x ln x lim x →0 x →0 1 x lim x →0
x→
(b) lim x sin x→
1 x 1 x2
1 sin1 x lim x x→ 1 x
1 x2 cos1 x x→ 1 x2
lim
lim x 0 x →0
(c)
1 0 x
lim cos x→
2 0
(c)
10
1.5
−1
−12
1x 1
1 − 0.5
41. (a) lim x1 x 0 0, not indeterminant x →0
43. (a) lim x1 x 0 x →
(b) Let y lim x1 x.
(See Exercise 95)
x→
y x1 x
(b) Let
ln y ln
x1 x
ln y lim
1 ln x. x
x→
Thus, ln y 0 ⇒ y e0 1. Therefore,
1 ln x → . Hence, x
Since x → 0 ,
ln y → ⇒ y → 0.
ln x 1 x lim 0 x→ x 1
lim x1 x 1.
x →
(c)
2
Therefore, lim x1 x 0. x→0
(c)
−5
2
20 −0.5
− 0.5
2 − 0.5
45. (a) lim 1 x1 x 1
(c)
x →0
6
(b) Let y lim 1 x1 x. x →0
ln1 x x 1 1 x 1 lim x →0 1
ln y lim x →0
Thus, ln y 1 ⇒ y e1 e. Therefore, lim 1 x1 x e. x →0
−1
4 −1
Section 7.7 47. (a) lim 3xx 2 00
(c)
x→0
Indeterminate Forms and L’Hôpital’s Rule
99
7
(b) Let y lim 3xx 2. x→0
x ln x 2
ln x 2 x
ln y lim ln 3 x→0
lim ln 3 x→0
−6
lim ln 3 lim
1 x 2 x2
lim ln 3 lim
x 2
x→0
x→0
x→0
x→0
6 −1
ln 3 Hence, lim 3xx 2 3. x→0
49. (a) lim ln xx1 00
ln x (b) Let y lim
51. (a) lim
x
(b) lim
x
2
x →2
x→1
x1
x1
lim x 1ln x 0
2
x →2
x→1
x 8 4 x2 8 8 xx 2 x lim x →2 4 x2 x2 4
Hence, lim ln xx1 1 x→1
(c)
lim
2 x4 x x 2x 2
lim
x 4 3 x2 2
x →2
6
x →2
−4
8
(c)
4
−2 −7
5
−4
53. (a) lim x →1
(b) lim x →1
ln3x x 2 1 ln3x x 2 1
lim
x →1
lim x →1
55. (a)
3x 3 2 ln x x 1ln x
3
−1
7
3 2 x
x 1 x ln x
−1
(b) lim
(c)
x →3
8
x3 1 lim ln2x 5 x →3 2 2x 5 lim
x →3
−1
2x 5 1 2 2
4
−4
57. (a)
(b) lim x2 5x 2 x lim x2 5x 2 x
10
x→
x→
x2 5x 2 x x2 5x 2 x
x2 5x 2 x2 x→ x2 5x 2 x
lim −8
10 −2
lim
x→
lim
x→
5x 2
x2 5x 2 x
5 2 x
1 5 x 2 x2 1
5 2
100
59.
Chapter 7 0 , ,0 0
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
, 1, 00,
61. (a) Let f x x2 25 and gx x 5. (b) Let f x x 52 and gx x2 25. (c) Let f x x2 25 and gx x 53.
63. lim
x2 2x 2 lim lim 0 e5x x → 5e5x x → 25e5x
65. lim
ln x3 3ln x21 x lim x→ x 1
x →
x→
67. lim
x→
ln xn nln xn1 x lim x→ xm mxm1
3ln x2 x
lim
6ln x1 x x→ 1
lim
lim
x→
x→
nn 1ln xn2 x→ m2xm
lim
lim
x→
69.
x
ln x4 x
6ln x 6 lim 0 x→ x x
. . . lim
x→
10
102
104
106
108
1010
2.811
4.498
0.720
0.036
0.001
0.000
71. y x1 x, x > 0
lim
x →
1 ln x x
2x 2 lim x 0 x → e ex
Horizontal asymptote: y 0
1 1 dy 1 1 ln x 2 y dx x x x
1 dy x1 x 2 1 ln x x1 x 21 ln x 0 dx x Critical number:
n! 0 mnxm
73. y 2xex
Horizontal asymptote: y 1 (See Exercise 37) ln y
nln xn1 mxm
xe
0, e Sign of dy dx: y f x: Increasing Relative maximum: e, e1 e Intervals:
e, Decreasing
dy 2xex 2ex dx 2ex 1 x 0 Critical number:
x1
Intervals: , 1 Sign of dy dx:
y f x: Increasing Relative maximum:
4
1, Decreasing
1, 2e
3
(1, 2e ) −2
(e, e1/e) 0
6
−5
0
e2x 1 0 0 x →0 ex 1
75. lim
Limit is not of the form 0 0 or . L’Hôpital’s Rule does not apply.
10
77. lim x cos x →
1 1 x
Limit is not of the form 0 0 or . L’Hôpital’s Rule does not apply.
Section 7.7
79. (a) lim
x→
x
x2 1
x x
lim
x→
x→
lim
x→
(c)
(b) lim
x2 1 x
lim
Indeterminate Forms and L’Hôpital’s Rule
x→
x
x2 1
1
x 1 x2
1 x x2 1
x2 1
x
x→
1
lim
1 1 x2
1
x→
lim
2
1 0
lim
x→
lim
Applying L’Hôpital’s rule twice results in the original limit, so L’Hôpital’s rule fails.
1
−1.5
81. lim
v0kekt 32
k
k→0
321 ekt lim v0ekt k→0 k→0 k
lim lim
k→0
v0 320 tekt lim kt 32t v0 k→0 e 1
1 2x1 cos x x x cos x 2 Shaded area: Area of rectangle Area under curve
83. Area of triangle:
x
2x1 cos x 2
1 cos t dt 2x1 cos x 2 t sin t
0
x 0
2x1 cos x 2x sin x 2 sin x 2x cos x Ratio: lim
x →0
x x cos x 1 x sin x cos x lim 2 sin x 2x cos x x →0 2 cos x 2x sin x 2 cos x lim
1 x sin x cos x 2x sin x
lim
x cos x sin x sin x 2x cos x 2 sin x
lim
x cos x 2 sin x 2x cos x 2 sin x
lim
x 2 tan x 2x 2 tan x
x →0
x →0
x →0
x →0
1 cos x
1 cos x
1 2 sec2 x 3 x →0 2 2 sec2 x 4
lim
85. f x x3, gx x2 1, 0, 1 f c f b f a gb ga g c f 1 f 0 3c2 g1 g0 2c 1 3c 1 2 c
2 3
x x2 1 1
x
6
32 1 ekt
x→
x2 1
1.5
−6
101
2
87. f x sin x, gx cos x, 0, f c f 2 f 0 g 2 g0 g c 1 cos c 1 sin c 1 cot c c
4
102
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
89. False. L’Hôpital’s Rule does not apply since
91. True
lim x2 x 1 0.
x →0
x2 x 1 1 1 lim x 1 x →0 x →0 x x
lim
93. (a) sin BD cos DO ⇒ AD 1 cos 1 1 1 1 Area ABD bh 1 cos sin sin sin cos 2 2 2 2 (b) Area of sector:
1 1 1 1 1 Area OBD cos sin sin cos 2 2 2 2 2
Shaded area: (c) R
1 2
12 sin 12 sin cos sin sin cos 12 12 sin cos sin cos sin 12 sin 2 12 sin 2
(d) lim R lim →0
→0
cos cos 2 sin 2 sin 2 cos 4 cos 2 3 lim lim →0 →0 1 cos 2 2 sin 2 4 cos 2 4
lim
→0
95. lim f xgx x →a
y f xgx ln y gx ln f x lim gx ln f x
x →a
As x → a, ln y ⇒ , and hence y 0. Thus, lim f xgx 0.
x →a
b
97. f ab a
f tt b dt f ab a
a
f tt b
b a
b
f ab a f aa b f t dv f tdt ⇒
v f t
u t b ⇒ du dt
Section 7.8
Improper Integrals
1. Infinite discontinuity at x 0.
4
0
1 x
4
dx lim b →0
b
1 x
lim 2 x b →0
dx 4 b
lim 4 2 b 4 b →0
Converges
f t dt
a b a
f b f a
Section 7.8 3. Infinite discontinuity at x 1.
2
0
1 dx x 12
1
1 dx x 12
0
lim
lim
b
b→1
b→1
0
2
1
b
lim
0
103
1 dx x 12
1 dx lim c→1 x 12
1 x1
Improper Integrals
c→1
2
c
1 dx x 12
1 x1
2
1 1
c
Diverges
1
5. Infinite limit of integration.
7.
b
ex dx lim
b→
0
ex dx
because the integrand is not defined at x 0. Diverges
0
e
lim
x
b→
1 dx 2 x2
1
b 0
011
Converges
9.
1
1 dx lim b→ x2
b
1
1x
b
lim b→
1 dx x2
11.
3 x
1
dx lim
b→
b→
3x13 dx
1
9x 2
b
lim
1
1
b
3
23
1
Diverges
0
13.
0
xe2x dx lim
b→
0
1 2x 1e2x b→ 4
xe2x dx lim
b
b
lim
b→
1
1 2b 1e2b (Integration by parts) 4
Diverges
15.
b→
0
Since lim b→
b
x2ex dx lim
b→
0
b 0
lim b→
b2 2b 2 2 2 eb
ex cos x dx lim
0
x2 2x 2
x
b2 2b 2 0 by L’Hôpital’s Rule. eb
17.
e
x2ex dx lim
b→
1 x e cos x sin x 2
b 0
1 1 0 1 2 2
19.
4
1 dx lim b→ xln x3
1 ln x3 dx x 4
b
21.
2 2 dx 4 x
0
2 2 dx 4 x
0
2 ln x
lim
1 1 ln b2 ln 42 2 2
lim
1 1 1 2 2 ln 22 8ln 22
0 2
lim b→
1
2
b 4
b→
b→
b
0
2 dx 4 x2
2 dx lim c→ 4 x2
c
2 dx 4 x2 0
arctan2 lim arctan2 x
0 b
x
c→
0
2
c 0
104
Chapter 7
23.
0
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
1 1 dx lim b →0 x2 x
cos x dx lim
b→
0
sin x
b
1
0
Diverges since sin x does not approach a limit as x → .
arctane
0
0
25.
2 4 4
1
0
ex dx 1 e2x x
b→
b
b
lim
27.
1 dx lim b→ ex ex
1
1
b
Diverges
8
29.
1 dx lim 3 b→8 0 8 x
1
31.
0
x ln x dx lim b→0
b
0
tan d
0
x2 lnx 4 2
1
x2
b
2
33.
1 3 dx lim 8 x 23 b→8 2 8x
3
lnsec
b
lim
b→ 2
lim b→0
14
b 0
6
b2 ln b b2 1 since lim b2 ln b 0 by L’Hôpital’s Rule. b→0 2 4 4
4
35.
2
0
2 dx lim b→2 x x2 4
2 dx x x2 4
x 2
4
b
lim arcsec
Diverges
b→2
4 b
lim arcsec 2 arcsec b→2
4
37.
2
4
1 lim ln x x2 4 b→2 x 4 2
0 3 3
b
ln 4 2 3 ln 2 ln 2 3 1.317
2
39.
0
1 dx 3 x 1
1
0
1 dx 3 x 1
lim b→1
41.
0
4 dx xx 6
4 xx 6
Thus,
0
dx 4
xx 6
1
23
0
1
0
1 dx x1
3
b
3
2
2 x 1
lim c→1
4 dx xx 6
Let u x, u2 x, 2u du dx.
2 x 1
2
3
23
c
4 xx 6
1
3 3 0 2 2
dx
C
x 42u du du 4 8 8 8 2 arctan C arctan 2 uu 6 u 6 6 6 6 6
dx lim b→0
8 6
6
arctan
x
86 arctan 16
2 6 8
. 2 6 3
1
lim
b
8 6
c→
86 2
0
c
x 8 arctan 6 6
8 6
1
16
arctan
b2
Section 7.8
43. If p 1,
1
1 dx lim b→ x
b
1
Improper Integrals
105
b 1 dx lim ln x . b→ x 1
Diverges. For p 1,
1
1 x1p lim p dx b→ x 1p
This converges to
b 1
b 1 p. 1p 1p
lim b→
1
1 if 1 p < 0 or p > 1. p1
45. For n 1 we have
b
xex dx lim
b→
0
lim
b→
xex dx
0
e
xx
b
ex
Parts: u x, dv ex dx
0
lim ebb eb 1 b→
lim
b→
b
e
b
1 1 1 (L’Hôpital’s Rule) eb
Assume that
x nex dx converges. Then for n 1 we have
0
x n1ex dx x n1ex n 1 x nex dx
by parts u xn1, du n 1xn dx, dv ex dx, v ex. Thus,
b→
0
1
47.
0
x
xn1ex dx lim
n1ex
b 0
n 1
0
1 dx diverges. x3
53. Since
x2
49.
1
x3
1
1
x2
1
dx converges by Exercise 43,
1 1 ≥ 3 2 on 2, and 3 x xx 1
2
1 1 converges. 31 2
dx
(See Exercise 43, p 3.
1 1 ≤ 2 on 1, and 5 x
55. Since ex ≤ ex on 1, and
xnex dx, which converges.
0
(See Exercise 44, p 3 1.
51. Since
xnex dx 0 n 1
1
2
1 dx converges. x2 5
1 dx diverges by Exercise 43, 3 2 x
ex dx converges (see Exercise 5),
0
2
1 3 xx 1
dx diverges.
ex dx converges. 2
0
1
57. Answers will vary. See pages 540, 543.
59.
1 3 dx x 1
0
1 3 dx x 1
1
0
1 dx x3
These two integrals diverge by Exercise 44. 61. f t 1 Fs
63. f t t2
0
s e
est dx lim b→
1
st
b
1 ,s > 0 s 0
Fs
s s
t 2est dx lim b→
0
1
3
2 ,s > 0 s3
2 t2
2st 2est
b 0
106
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
65. f t cos at Fs
est cos at dt
0 st
e s cos at a sin at s a
lim b→
b
2
2
0
s s ,s > 0 s2 a2 s2 a2
0
67. f t cosh at Fs
est cosh at dt
0
est
0
e
at
etsa etsa dt
0
1 1 1 et sa et sa b→ 2 s a s a
lim
eat 1 dt 2 2
b 0
0
1 1 1 2 s a s a
1 1 s 1 2 ,s > a 2 s a s a s a2
69. (a) A
ex dx
(b) Disk:
0
lim
b→
ex
b
V
0 1 1
ex 2 dx
0
0
1 lim e2x b→ 2 (c) Shell: V 2
xex dx
0
2
lim 2 exx 1 b→
b 0
x23 y23 4
71.
2 13 2 13 x y y 0 3 3 y 1 y 2
y13 x13
1 xy
8
s4
0
y
8
2
(− 8, 0) −8
(0, 8)
−2
−8
(8, 0) x
2
(0, −8)
8
2 13
x
23 23
x
y23 x 23
2 x 23b 48
23
dx lim 8 b→0
3
8
x 4
23
2 x1/3
b 0
2
Section 7.8
Improper Integrals
73. n
xn1ex dx
0
(a) 1
b→
0
2
1
e
x 1
b→
0
x
(b) n 1
b→
x e
x 2ex dx lim
0
b
x
xex dx lim
0
3
e
ex dx lim
b→
1
x e
xnex dx lim
0
0
b
2xex 2ex
2 x
b
n x
b 0
0
2
b
lim n b→
xn1ex dx 0 n n
u xn, dv ex dx
0
(c) n n 1!
1 t7 e dt 7
75. (a)
0
1 t7 e dt lim et7 b→ 7
b 0
4
1
(b)
0
1 t7 e dt et7 7
4 0
et7 1
0.4353 43.53%
(c)
17 e dt lim te t7
t
0
t7
b→
7et7
b 0
077
5
77. (a) C 650,000
25,000 e0.06t dt 650,000
0
e 25,000 0.06
5
0.06t
0
$757,992.41
10
(b) C 650,000
25,000e0.06t dt $837,995.15
0
(c) C 650,000
25,000e0.06t dt 650,000 lim
b→
0
e 25,000 0.06
0.06t
b 0
$1,066,666.67
79. Let x a tan , dx a sec2 d, a2 x2 a sec .
1 dx a2 x232
a sec2 d 1 2 cos d a3 sec3 a
Pk
1
81.
x 1 k dx 2 lim a x232 a b→ a2 x2 2
x
θ a
1 1 x 2 sin 2 a a a2 x2 Hence,
a2 + x 2
b 1
k 1 k a2 1 1 1 . a2 a2 1 a2 a2 1
10 10 ⇒ x 0, 2. x2 2x xx 2 You must analyze three improper integrals, and each must converge in order for the original integral to converge.
3
0
1
f x dx
0
2
f x dx
1
3
f x dx
2
f x dx
107
108
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
83. For n 1, I1
x2
0
x 1 dx lim b→ 2 14
b
0
1 x2 142x dx lim 1 b→ 6 x2 13
For n > 1,
In
x2n1 x2n2 lim n3 dx b→ 2 n 2 x 1 2n 2x 1
2
0
0
(b)
0
(c)
0
1 x dx lim 2 b→ x2 14 6x 13
3
1 x dx x2 15 4
2 x5 x 16 5 2
0
0
0
n1 n2
0
b 0
1 . 6
n1 x2n3 dx 0 I x 1n2 n 2 n1 2
x 1 dx, v x2 1n3 2n 2x2 1n2
u x 2n2, du 2n 2x 2n3 dx, dv (a)
b
b 0
1 6
1 1 1 x dx x2 14 4 6 24
2 1 1 x3 dx x 15 5 24 60 2
85. False. f x 1 x 1 is continuous on 0, , lim 1 x 1 0, but x →
0
0 .
1 dx lim ln x 1 b→ x1
Diverges 87. True
Review Exercises for Chapter 7 1.
x x2 1 dx
1 x2 11 22x dx 2
3.
x 1 2x dx dx x2 1 2 x2 1
1 x2 13 2 C 2 3 2
1 ln x2 1 C 2
1 x2 13 2 C 3
5.
ln2x ln 2x2 dx C x 2
1 2 e2x sin 3x dx e2x cos 3x 3 3
9.
13 9
7.
16 16 x2
dx 16 arcsin
e2x cos 3x dx
2 1 2x 2 1 e sin 3x e2x cos 3x 3 3 3 3
e2x sin 3x dx
1 2 e2x sin 3x dx e2x cos 3x e2x sin 3x 3 9 e2x sin 3x dx
e2x 2 sin 3x 3 cos 3x C 13
(1) dv sin 3x dx ⇒ u e2x
1 v cos 3x 3
⇒ du 2e2x dx
(2) dv cos 3x dx ⇒ u e2x
v
1 sin 3x 3
⇒ du 2e2x dx
4x C
b
Review Exercises for Chapter 7 2 11. u x, du dx, dv x 51 2 dx, v x 53 2 3
2 x x 5 dx xx 53 2 3
1 x2 sin 2x dx x2 cos 2x 2
13.
2 x 53 2 dx 3
4
(1) dv sin 2x dx ⇒
156 x 34 C
u x2
x 53 2
x2 arcsin 2x 2
x arcsin 2x dx
19.
21.
dx
22x2 dx 1 2x2
1 1 x2 arcsin 2x
2x 1 4x2 arcsin 2x C (by Formula 43 of Integration Tables) 2 8 2
1
8x2 1 arcsin 2x 2x 1 4x2 C 16
v
cos3 x 1 dx
sec4
⇒ du dx
1 sin 2x 2
1 x2 arcsin 2x 2 8
u arcsin 2x ⇒ du
17.
x2
1 4x2
v
⇒
dv x dx
ux
1 v cos 2x 2
⇒ du 2x dx
(2) dv cos 2x dx ⇒
2 x 53 2 3x 10 C 15
x2 2 2 dx 1 4x2
1 sin2 x 1 cos x 1 dx
1 1 sin x 1 sin3 x 1 C 3
1 sin x 1 3 sin2 x 1 C 3
1 sin x 1 3 1 cos 2 x 1 C 3
1 sin x 1 2 cos 2 x 1 C 3
2x dx tan 2x 1 sec 2x dx 2
2
x x x 2 3 x 2 tan 2 tan C tan3 3 tan 3 2 2 3 2 2
tan2
1 d 1 sin
2x sec 2x dx sec 2x dx 2
2
1 sin d cos2
sin 2x dx
1 x 1 x2 cos 2x sin 2x cos 2x C 2 2 4
23x 15 x 5 C
15.
x cos 2x dx
1 1 1 x2 cos 2x x sin 2x 2 2 2
2 4 xx 53 2 x 55 2 C 3 15 x 53 2
109
C
sec2 sec tan d tan sec C
110
23.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
12 dx x2 4 x2
24 cos d 4 sin2 2 cos
x
csc2
3
2
d
θ 4 − x2
3 cot C 3 4 x2 C x
x 2 sin , dx 2 cos d, 4 x2 2 cos x 2 tan
25.
x2 + 4
dx 2 sec2 d
x
θ
4 x2 4 sec2
x3 dx 4 x2
2
8 tan3 2 sec2 d 2 sec
8 tan3 sec d 8 sec2 1tan sec d
sec3 sec C 3
8
x2 43 2 x2 4 C 24 2
8
x2 4
13 x
2
4 4 C
1 8 x2 x2 4 x2 4 C 3 3 1 x2 41 2x2 8 C 3
27.
4 x2 dx
2 cos 2 cos d 2 x
2 1 cos 2 d
θ 4 − x2
1 2 sin 2 C 2 2 sin cos C 2 arcsin
2x 2x
4 x2
2
C
1 x 4 arcsin x 4 x2 C 2 2
x 2 sin , dx 2 cos d, 4 x2 2 cos
Review Exercises for Chapter 7
29. (a)
x3 4 x2
dx 8
sin3 d cos4
(b)
x3 4 x2
8 sec sec2 3 C 3 4 x2
3
(c)
x3 dx x2 4 x2 4 x2
x dx ⇒ 4 x2
2x 4 x2 dx
31.
v 4 x2
⇒ du 2x dx
u x2
x 28 A B x2 x 6 x 3 x 2 x 28 Ax 2 Bx 3 x 2 ⇒ 30 B5 ⇒ B 6 ⇒ 25 A5
x3
33.
x 28 dx x2 6 6
⇒ A 5
5 6 dx 5 ln x 3 6 ln x 2 C x3 x2
x2 2x A Bx C 2 x 1x2 1 x 1 x 1 x2 2x Ax2 1 Bx Cx 1 Let x 1: 3 2A ⇒ A
3 2 3 2
Let x 0: 0 A C ⇒ C
Let x 2: 8 5A 2B C ⇒ B
x2 2x 3 dx x x2 x 1 2 3
1 2
1 1 dx x1 2 1 1 dx x1 4
x3 dx x2 1 2x 3 dx x2 1 2
1 dx x2 1
3 2
1 3 3 ln x 1 ln x2 1 arctan x C 2 4 2
1 6 ln x 1 lnx2 1 6 arctan x C 4
4 x2
3
x2 8 C
u2 4 x2, 2u du 2x dx
4 x2 2 2 x2 4 x2 4 x23 2 C x 8 C 3 3
dv
u2 4 du
u u2 12 C 3
x2 8 C
x 2 tan , dx 2 sec2 d
1 u3 4u C 3
8 cos4 cos2 sin d
dx
111
112
35.
Chapter 7
x2
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
15 2x x2 1 2 2x 15 x 2x 15
B 15 2x A x 3x 5 x 3 x 5 15 2x Ax 5 Bx 3 Let x 3:
9 8A ⇒ A
9 8
Let x 5: 25 8B ⇒ B
x2 dx x2 2x 15
dx
x
37.
9 8
25 8
25 1 dx x3 8
1 dx x5
9 25 ln x 3 ln x 5 C 8 8
x 1 2 dx ln 2 3x C 2 3x2 9 2 3x
39.
x 1 1 dx du 1 sin x2 2 1 sin u 1 tan u sec u C 2
(Formula 4)
1 tan x2 sec x2 C 2
41.
x 1 1 dx ln x2 4x 8 4 2 dx x2 4x 8 2 x 4x 8
(Formula 15)
1 2 2x 4 ln x2 4x 8 2 arctan 32 16 32 16 2
43.
C
(Formula 14)
1 x ln x2 4x 8 arctan 1 C 2 2
1 1 1 dx dx sin x cos x sin x cos x
1 ln tan x C
u x
(Formula 58)
45. dv dx
⇒
vx
1 u ln xn ⇒ du nln xn1 dx x
ln xn dx xln xn n ln xn1 d x
47.
sin cos d
1 2
sin 2 d
1 1 cos 2 4 4 dv sin 2 d ⇒ u
1 v cos 2 2
⇒ du d
1 1 1 cos 2 d cos 2 sin 2 C sin 2 2 cos 2 C 4 8 8
u x2 (Formula 56)
Review Exercises for Chapter 7
49.
x1 4 uu3 dx 4 du 1 x1 2 1 u2 4 4
u2 1
13 u
3
51.
1 cos x dx
1 du u2 1
sin x 1 cos x
113
dx
1 cos x1 2sin x dx
2 1 cos x C
u arctan u C
u 1 cos x, du sin x dx
4 x3 4 3x1 4 3 arctanx1 4 C 3 4 x , x u4, dx 4u3 du y
53.
cos x lnsin x dx sin x lnsin x
3 x3 9 dx ln C x2 9 2 x3 (by Formula 24 of Integration Tables)
sin x lnsin x sin x C dv cos x dx ⇒
v sin x
u lnsin x ⇒ du
57. y
55. y
cos x dx
cos x dx sin x
lnx2 xdx x ln x2 x x ln x 2 x x ln x2 x
5
2x2 x dx x2 x
59.
xx2 43 2 dx
2
15 x
2
45 2
5
2
1 5
2x 1 dx x1
2dx
1 dx x1
x ln x2 x 2x ln x 1 C ⇒
dv dx
vx
u lnx2 x ⇒ du
4
61.
1
ln x 1 dx ln x2 x 2
4
65. A
4 1
2x 1 dx x2 x
1 ln 42 2ln 22 0.961 2
0
x 4 x dx
0
4 u2u2u du
2
63.
0
2u4 4u2 du
y
2
u5 4u3
2
5
3
0 2
128 15
1 1 x x3 3
0
1 67. By symmetry, x 0, A . 2
0
x sin x dx x cos x sin x
1
2 1 2
1
x, y 0,
4 3
1 x22 dx
u 4 x, x 4 u2, dx 2u du
69. s
1 cos2 x dx 3.82
0
e2x 2e2x 4e2x lim lim 2 x → x x → 2x x → 2
73. lim
71. lim
x →1
ln x2
x 1 lim x →1
21 xln x 0 1
75. y lim ln x2 x x →
ln y lim
x →
2 x ln x 2 lnln x 0 lim x → x 1
Since ln y 0, y 1.
1 1
4 3
108
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
83. For n 1, I1
x2
0
x 1 dx lim b→ 2 14
b
0
1 x2 142x dx lim 1 b→ 6 x2 13
For n > 1,
In
x2n1 x2n2 lim n3 dx b→ 2 n 2 x 1 2n 2x 1
2
0
0
(b)
0
(c)
0
1 x dx lim 2 b→ x2 14 6x 13
3
1 x dx x2 15 4
2 x5 x 16 5 2
0
0
0
n1 n2
0
b 0
1 . 6
n1 x2n3 dx 0 I x 1n2 n 2 n1 2
x 1 dx, v x2 1n3 2n 2x2 1n2
u x 2n2, du 2n 2x 2n3 dx, dv (a)
b
b 0
1 6
1 1 1 x dx x2 14 4 6 24
2 1 1 x3 dx x 15 5 24 60 2
85. False. f x 1 x 1 is continuous on 0, , lim 1 x 1 0, but x →
0
0 .
1 dx lim ln x 1 b→ x1
Diverges 87. True
Review Exercises for Chapter 7 1.
x x2 1 dx
1 x2 11 22x dx 2
3.
x 1 2x dx dx x2 1 2 x2 1
1 x2 13 2 C 2 3 2
1 ln x2 1 C 2
1 x2 13 2 C 3
5.
ln2x ln 2x2 dx C x 2
1 2 e2x sin 3x dx e2x cos 3x 3 3
9.
13 9
7.
16 16 x2
dx 16 arcsin
e2x cos 3x dx
2 1 2x 2 1 e sin 3x e2x cos 3x 3 3 3 3
e2x sin 3x dx
1 2 e2x sin 3x dx e2x cos 3x e2x sin 3x 3 9 e2x sin 3x dx
e2x 2 sin 3x 3 cos 3x C 13
(1) dv sin 3x dx ⇒ u e2x
1 v cos 3x 3
⇒ du 2e2x dx
(2) dv cos 3x dx ⇒ u e2x
v
1 sin 3x 3
⇒ du 2e2x dx
4x C
b
Review Exercises for Chapter 7 2 11. u x, du dx, dv x 51 2 dx, v x 53 2 3
2 x x 5 dx xx 53 2 3
1 x2 sin 2x dx x2 cos 2x 2
13.
2 x 53 2 dx 3
4
(1) dv sin 2x dx ⇒
156 x 34 C
u x2
x 53 2
x2 arcsin 2x 2
x arcsin 2x dx
19.
21.
dx
22x2 dx 1 2x2
1 1 x2 arcsin 2x
2x 1 4x2 arcsin 2x C (by Formula 43 of Integration Tables) 2 8 2
1
8x2 1 arcsin 2x 2x 1 4x2 C 16
v
cos3 x 1 dx
sec4
⇒ du dx
1 sin 2x 2
1 x2 arcsin 2x 2 8
u arcsin 2x ⇒ du
17.
x2
1 4x2
v
⇒
dv x dx
ux
1 v cos 2x 2
⇒ du 2x dx
(2) dv cos 2x dx ⇒
2 x 53 2 3x 10 C 15
x2 2 2 dx 1 4x2
1 sin2 x 1 cos x 1 dx
1 1 sin x 1 sin3 x 1 C 3
1 sin x 1 3 sin2 x 1 C 3
1 sin x 1 3 1 cos 2 x 1 C 3
1 sin x 1 2 cos 2 x 1 C 3
2x dx tan 2x 1 sec 2x dx 2
2
x x x 2 3 x 2 tan 2 tan C tan3 3 tan 3 2 2 3 2 2
tan2
1 d 1 sin
2x sec 2x dx sec 2x dx 2
2
1 sin d cos2
sin 2x dx
1 x 1 x2 cos 2x sin 2x cos 2x C 2 2 4
23x 15 x 5 C
15.
x cos 2x dx
1 1 1 x2 cos 2x x sin 2x 2 2 2
2 4 xx 53 2 x 55 2 C 3 15 x 53 2
109
C
sec2 sec tan d tan sec C
110
23.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
12 dx x2 4 x2
24 cos d 4 sin2 2 cos
x
csc2
3
2
d
θ 4 − x2
3 cot C 3 4 x2 C x
x 2 sin , dx 2 cos d, 4 x2 2 cos x 2 tan
25.
x2 + 4
dx 2 sec2 d
x
θ
4 x2 4 sec2
x3 dx 4 x2
2
8 tan3 2 sec2 d 2 sec
8 tan3 sec d 8 sec2 1tan sec d
sec3 sec C 3
8
x2 43 2 x2 4 C 24 2
8
x2 4
13 x
2
4 4 C
1 8 x2 x2 4 x2 4 C 3 3 1 x2 41 2x2 8 C 3
27.
4 x2 dx
2 cos 2 cos d 2 x
2 1 cos 2 d
θ 4 − x2
1 2 sin 2 C 2 2 sin cos C 2 arcsin
2x 2x
4 x2
2
C
1 x 4 arcsin x 4 x2 C 2 2
x 2 sin , dx 2 cos d, 4 x2 2 cos
Review Exercises for Chapter 7
29. (a)
x3 4 x2
dx 8
sin3 d cos4
(b)
x3 4 x2
8 sec sec2 3 C 3 4 x2
3
(c)
x3 dx x2 4 x2 4 x2
x dx ⇒ 4 x2
2x 4 x2 dx
31.
v 4 x2
⇒ du 2x dx
u x2
x 28 A B x2 x 6 x 3 x 2 x 28 Ax 2 Bx 3 x 2 ⇒ 30 B5 ⇒ B 6 ⇒ 25 A5
x3
33.
x 28 dx x2 6 6
⇒ A 5
5 6 dx 5 ln x 3 6 ln x 2 C x3 x2
x2 2x A Bx C 2 x 1x2 1 x 1 x 1 x2 2x Ax2 1 Bx Cx 1 Let x 1: 3 2A ⇒ A
3 2 3 2
Let x 0: 0 A C ⇒ C
Let x 2: 8 5A 2B C ⇒ B
x2 2x 3 dx x x2 x 1 2 3
1 2
1 1 dx x1 2 1 1 dx x1 4
x3 dx x2 1 2x 3 dx x2 1 2
1 dx x2 1
3 2
1 3 3 ln x 1 ln x2 1 arctan x C 2 4 2
1 6 ln x 1 lnx2 1 6 arctan x C 4
4 x2
3
x2 8 C
u2 4 x2, 2u du 2x dx
4 x2 2 2 x2 4 x2 4 x23 2 C x 8 C 3 3
dv
u2 4 du
u u2 12 C 3
x2 8 C
x 2 tan , dx 2 sec2 d
1 u3 4u C 3
8 cos4 cos2 sin d
dx
111
112
35.
Chapter 7
x2
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
15 2x x2 1 2 2x 15 x 2x 15
B 15 2x A x 3x 5 x 3 x 5 15 2x Ax 5 Bx 3 Let x 3:
9 8A ⇒ A
9 8
Let x 5: 25 8B ⇒ B
x2 dx x2 2x 15
dx
x
37.
9 8
25 8
25 1 dx x3 8
1 dx x5
9 25 ln x 3 ln x 5 C 8 8
x 1 2 dx ln 2 3x C 2 3x2 9 2 3x
39.
x 1 1 dx du 1 sin x2 2 1 sin u 1 tan u sec u C 2
(Formula 4)
1 tan x2 sec x2 C 2
41.
x 1 1 dx ln x2 4x 8 4 2 dx x2 4x 8 2 x 4x 8
(Formula 15)
1 2 2x 4 ln x2 4x 8 2 arctan 32 16 32 16 2
43.
C
(Formula 14)
1 x ln x2 4x 8 arctan 1 C 2 2
1 1 1 dx dx sin x cos x sin x cos x
1 ln tan x C
u x
(Formula 58)
45. dv dx
⇒
vx
1 u ln xn ⇒ du nln xn1 dx x
ln xn dx xln xn n ln xn1 d x
47.
sin cos d
1 2
sin 2 d
1 1 cos 2 4 4 dv sin 2 d ⇒ u
1 v cos 2 2
⇒ du d
1 1 1 cos 2 d cos 2 sin 2 C sin 2 2 cos 2 C 4 8 8
u x2 (Formula 56)
Review Exercises for Chapter 7
49.
x1 4 uu3 dx 4 du 1 x1 2 1 u2 4 4
u2 1
13 u
3
51.
1 cos x dx
1 du u2 1
sin x 1 cos x
113
dx
1 cos x1 2sin x dx
2 1 cos x C
u arctan u C
u 1 cos x, du sin x dx
4 x3 4 3x1 4 3 arctanx1 4 C 3 4 x , x u4, dx 4u3 du y
53.
cos x lnsin x dx sin x lnsin x
3 x3 9 dx ln C x2 9 2 x3 (by Formula 24 of Integration Tables)
sin x lnsin x sin x C dv cos x dx ⇒
v sin x
u lnsin x ⇒ du
57. y
55. y
cos x dx
cos x dx sin x
lnx2 xdx x ln x2 x x ln x 2 x x ln x2 x
5
2x2 x dx x2 x
59.
xx2 43 2 dx
2
15 x
2
45 2
5
2
1 5
2x 1 dx x1
2dx
1 dx x1
x ln x2 x 2x ln x 1 C ⇒
dv dx
vx
u lnx2 x ⇒ du
4
61.
1
ln x 1 dx ln x2 x 2
4
65. A
4 1
2x 1 dx x2 x
1 ln 42 2ln 22 0.961 2
0
x 4 x dx
0
4 u2u2u du
2
63.
0
2u4 4u2 du
y
2
u5 4u3
2
5
3
0 2
128 15
1 1 x x3 3
0
1 67. By symmetry, x 0, A . 2
0
x sin x dx x cos x sin x
1
2 1 2
1
x, y 0,
4 3
1 x22 dx
u 4 x, x 4 u2, dx 2u du
69. s
1 cos2 x dx 3.82
0
e2x 2e2x 4e2x lim lim 2 x → x x → 2x x → 2
73. lim
71. lim
x →1
ln x2
x 1 lim x →1
21 xln x 0 1
75. y lim ln x2 x x →
ln y lim
x →
2 x ln x 2 lnln x 0 lim x → x 1
Since ln y 0, y 1.
1 1
4 3
114
Chapter 7
77. lim 1000 1 n →
Let y lim
n →
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals 0.09 n
n
1000 lim
n →
1 0.09 n
n
. 1 0.09 n n
0.09 ln y lim n ln 1 lim n → n → n
ln 1
0.09 n
n →
16
0
1
dx lim
4 x
b→0
3 x 4
16
34
b
lim
n →
1 n
Thus, ln y 0.09 ⇒ y e0.09 and lim 1000 1
79.
0.09 n
n
t0
n →
81.
x2 ln x dx lim
b→
1
0.09 0.09 0.09 1 n
9 1 3 ln x
Diverges
Converges
83.
lim
1000e0.09 1094.17.
32 3
0.09n2 1 0.09n 1 2 n
500,000e0.05t dt
0
e
500,000 0.05
0.05t
t0 0
500,000 0.05t0 e 1 0.05
10,000,0001 e0.05t0 (a) t0 20: $6,321,205.59 (b) t0 → : $10,000,000
1 0.95 2
(b) P15 ≤ x < 20
1 0.95 2
85. (a) P13 ≤ x <
ex12.9 20.95 dx 0.4581 2
2
13
ex12.9 20.95 dx 0.0135 2
2
15
Problem Solving for Chapter 7
1
1. (a)
1
1
1
x3 3
1 x2 dx x
1 1
2 1
1 4 3 3
1
1 x22 dx
1
1 2x2 x4 dx x
2x3 x5 3 5
1 1
2 1
(b) Let x sin u, dx cos u du, 1 x2 1 sin2 u cos2 u.
1
1
1 x2n dx
2
cos2 un cos u du
2
2
cos2n1 u du
2
2n
3 5 7 . . . 2n 1
2
2
4
6
2 3 4 5. . . 2n2n 1
2
22
42
62 . . . 2n2
22n1n!2 222nn!2 2n 1! 2n 1!
2 1 16 3 5 15
(Wallis’s Formula)
x3
b 1
Problem Solving for Chapter 7
lim
3.
x→
xx cc
lim x ln
x→
lim
x→
x
9
xx cc ln 9
lnx c lnx c ln 9 1x 1 1 xc xc lim ln 9 x→ 1 2 x
lim
x→
2c x2 ln 9 x cx c
x 2cx c ln 9 2
lim
x→
2
2
2c ln 9 2c 2 ln 3 c ln 3
5. sin
PB PB, cos OB OP
AQ AP BR OR OB OR cos The triangles AQR and BPR are similar: AR OR 1 OR cos BR ⇒ AQ BP sin sin OR sin OR cos OR lim OR lim
→0
→0
cos sin sin
sin cos cos cos 1
lim
sin cos 1
lim
sin cos sin
→0
→0
lim
→0
2
Q
cos sin sin
lim →0
y
cos cos sin cos
P
θ
R
O
B
A (1, 0)
x
115
116
Chapter 7
7. (a)
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals Area 0.2986
0.2
0
4 0
(b) Let x 3 tan , dx 3 sec2 d, x2 9 9 sec2 .
x2 dx 9 x2
Area
9 tan2 3 sec2 d 9 sec2 32
x2 + 9 x
tan2 d sec
θ 3
sin2 d cos 1 cos2 d cos
ln sec tan sin C
4
0
x2
x2
dx ln sec tan sin
9
32
ln
tan14 3
0
9 x x x2 9 3 3
x2
4
0
4 5 4 4 ln ln 3 3 3 5 5 (c) x 3 sinh u, dx 3 cosh u du, x2 9 9 sinh2 u 9 9 cosh2 u
4
A
0
x2 dx 2 x 932
sin14 3
0
9 sinh2 u 3 cosh u du 9 cosh2 u32
sinh14 3
tanh2 u du
0
sinh14 3
1 sech2 u du
0
sinh14 3
u tanh u
0
sinh1
4 4 tanh sinh1 3 3
43 169 1 tanh ln43 169 1
ln ln
43 35 tanhln43 35
ln 3 tanhln 3 ln 3
3 13 3 13
ln 3
4 5
Problem Solving for Chapter 7 9. y ln1 x2, y
1 x2 4x2 1 2x2 x4 4x2 1 x22 1 x22 1 x2
1 y 2 1
2x 1 x2
2
12
Arc length
1 y 2 dx
0 12
2 2
0 12
11 xx dx 1 1 2 x dx 2
0 12
1 x 1 1 1 1 x dx x ln1 x ln1 x
0
12 0
3 1 1 ln ln 2 2 2
1 ln 3 ln 2 ln 2 2 ln 3
11. Consider
1 dx. ln x
Let u ln x, du
If
1 0.5986 2
1 dx, x eu. Then dx
1 dx were elementary, then ln x
Hence,
1 dx ln x
1 u e du u
eu du. u
eu du would be too, which is false. u
1 dx is not elementary. ln x
13. x4 1 x2 ax bx2 cx d x4 a cx3 ac b dx2 ad bcx bd a c, b d 1, a 2
1
0
x4 1 x2 2 x 1x2 2x 1 Ax B dx x2 2x 1
1 2 x 1 2 4 dx 2 0 x 2 x 1
1 dx x4 1
1
0
2
4 2
4
1
0
Cx D dx x2 2 x 1
1
0
1 2 x 2 4 dx x2 2 x 1
arctan 2x 1 arctan 2x 1
1
0
arctan 2 1 arctan 2 1
0.5554 0.3116 0.8670
2
8
2
8
lnx
2
2x 1 lnx2 2x 1
ln2 2 ln2 2
2
4
1 0
4 4
2
8
0
117
118
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
15. Using a graphing utility,
1 x
1 0 x
1 x
(a) lim cot x x→0
(b) lim cot x x→0
(c) lim cot x x→0
cot x 1x 32.
Analytically,
1 x
1 x cot x 1 x cos x sin x lim lim x→0 x→0 x x x sin x
(a) lim cot x x→0
(b) lim cot x x→0
lim
cos x x sin x cos x x sin x lim x→0 sin x x cos x sin x x cos x
lim
sin x x cos x 0. cos x cos x x sin x
x→0
x→0
(c)
cot x 1x cot x 1x cot
2
lim
x→0
x
x2 cot2 x 1 x2
x2 cot2 x 1 2x cot2 x 2x2 cot x csc2 x lim 2 x→0 x 2x lim
cot2 x x cot x csc2 x 1
lim
cos2 x sin x x cos x sin3 x
lim
1 sin2 xsin x x cos x sin3 x
lim
sin x x cos x 1 sin3 x
x→0
x→0
x→0
x→0
Now, lim x→0
sin x x cos x cos x cos x x sin x lim x→0 sin3 x 3 sin2 x cos x lim
x 3 sin x cos x
lim
sinx x3 cos1 x 31.
x→0
x→0
Thus, lim cot x x→0
1 x2
1 x
cot x 1x 31 1 32.
The form 0 is indeterminant.
Problem Solving for Chapter 7
17.
P1 P2 P3 P4 x3 3x2 1 ⇒ c1 0, c2 1, c3 4, c4 3 2 x 13x 12x x x1 x4 x3 4
Nx x3 3x2 1 D x 4x3 26x 12 P1
N0 1 D 0 12
P2
N1 1 1 D 1 10 10
P3
N4 111 111 D 4 140 140
P4
N3 1 D 3 42
x3 3x2 1 112 110 111140 142 . x 13x2 12x x x1 x4 x3
Thus,
4
19. By parts,
b
f xg x dx f xg x
a
b
a
b
f xg x dx
a
f xgx
b
a
f xg x dx
b
a
f xgx dx.
b
a
gx f x dx
119
C H A P T E R Infinite Series
8
Section 8.1
Sequences . . . . . . . . . . . . . . . . . . . . . 369
Section 8.2
Series and Convergence . . . . . . . . . . . . . . 373
Section 8.3
The Integral Test and p-Series
Section 8.4
Comparisons of Series
Section 8.5
Alternating Series . . . . . . . . . . . . . . . . . 385
Section 8.6
The Ratio and Root Tests . . . . . . . . . . . . . 389
Section 8.7
Taylor Polynomials and Approximations . . . . . 393
Section 8.8
Power Series . . . . . . . . . . . . . . . . . . . . 398
Section 8.9
Representation of Functions by Power Series
. . . . . . . . . . 378
. . . . . . . . . . . . . . 381
Section 8.10 Taylor and Maclaurin Series
. . 403
. . . . . . . . . . . 408
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 414 Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . 421
C H A P T E R Infinite Series Section 8.1
8
Sequences
Solutions to Even-Numbered Exercises
32
2. an
2n n3
a1
2 1 4 2
a1
a2
4 5
a2
6 1 6
8 a3 27
a4
8 7
16 a4 81
a5
10 5 8 4
a5
a3
8. an 1n1 a1
4. an
2n
a3
2 1 a4 4 2 2 a5 5
k 2 1a
k
n 2
0 2
a2 cos 1
4 9
a3 cos
3 0 2
a4 cos 2 1 a5 cos
5 0 2
32 243 6 2 n n2
12. an
3n! 3n n 1!
a1 10 2 6 18
a1 31 3
3 25 2 2
a2 32 6
2 2 34 a3 10 3 3 3
2 3
14. a1 4, ak1
a1 cos
a2 10 1
2 a2 1 2
6. an cos
2 3
10. an 10
2 2 1
n
a4 10
1 3 87 2 8 8
a5 10
2 6 266 5 25 25
a3 33 9 a4 34 12 a5 35 15
1 16. a1 6, ak1 ak2 3
a2
1 2 1a
4
1 1 a2 a12 62 12 3 3
a3
2 2 1a
6
1 1 a3 a22 122 48 3 3
a4
3 2 1a
12
1 1 a4 a32 482 768 3 3
a5
4 2 1a
30
1 1 a5 a42 7682 196,608 3 3
1
2
3
4
369
370
Chapter 8
Infinite Series
18. Because the sequence tends to 8 as n tends to infinity, it matches (a). 22.
24.
4
−1
20. This sequence increases for a few terms, then decreases a2 16 2 8. Matches (b). 26.
10
12 −1
−1
12
4 an 2 , n 1, . . . , 10 n 28. an
n6 2
an 80.75n1, n 1, 2, . . . , 10
5 6 11 2 2
a6
66 6 2
32.
a5 240 80
36.
2425 600
2n 2! 2n!2n 12n 2 2n! 2n!
n 1n 2
n→
1 505 n2
2n 12n 2
40. lim
n→
5n 5 lim n→ 1 4n2 n2 4
44.
12
42. lim cos n→
2n 1
5 5 1 46.
2
−1
3n2 , n 1, . . . , 10 n2 1
25! 23!2425 23! 23!
a6 280 160
n 2! n!n 1n 2 n! n!
38. lim 5
an
30. an1 2an, a1 5
a5
12 −1
−1
−3
34.
4
4
−1
12 −1
−1
The graph seems to indicate that the sequence converges to 0. Analytically, lim an lim
n→
n→
The graph seems to indicate that the sequence converges to 3. Analytically,
1 1 lim 0. n32 x→ x32
48. lim 1 1n n→
lim an lim 3
n→
n→
3 n
50. lim
3 n n→
1
1 3 0 3. 2n
1, converges
does not exist, (alternates between 0 and 2), diverges.
52. lim
n→
1 1n 0, converges n2
54. lim
n→
lnn 12 ln n lim n→ n n
lim
n→
1 0, converges 2n
(L’Hôpital’s Rule) 56. lim 0.5n 0, converges n→
58. lim
n→
n 2! 1 lim 0, converges n→ nn 1 n!
Section 8.1
Sequences
2nn 1 2nn 1 lim 4n2n 1 21, converges 2
60. lim
n→
2
2
n→
2
1 n
62. an n sin
1 Let f x x sin . x lim x sin
x→
1 sin1x 1x 2 cos1x 1 lim lim lim cos cos 0 1 (L’Hôpital’s Rule) x→ x→ x x→ 1x 1x 2 x
or, lim
x→
sin1x sin y 1 lim 1. Therefore lim n sin 1. y→0 n→ 1x y n
64. lim 21n 20 1, converges n→
70. an
1n1 n2
76. an 1
2n 1 2n
66. lim
cos n 0, converges n2
68. an 4n 1
72. an
n2 3n 1
74. an 1n
78. an
1 n!
80. an
n→
3n2 2n1
xn1 n 1!
2n1 1 2n
82. Let f x
3x 6 . Then fx . x2 x 22
Thus, f is increasing which implies an is increasing.
an < 3, bounded
n2 84. an ne
a1 0.6065 a2 0.7358 a3 0.6694
Not monotonic; an ≤ 0.7358, bounded 86. an 23
88. an 32 <
n
n
Monotonic; lim an , not bounded
a1 23 a2
n→
4 9
8 a3 27
2 Not monotonic; an ≤ 3 , bounded
90. an
cos n n
a1 0.5403 a2 0.2081 a3 0.3230 a4 0.1634
32 n1 an1
Not monotonic; an ≤ 1, bounded
371
372
Chapter 8
Infinite Series
92. (a) an 4
4
3 n
(b)
3 < 4 ⇒ bounded n −1
3 4 < 3 an1 ⇒ monotonic n n1
an 4
Therefore, an converges.
94. (a) an 4
4
8
1 2n
12 0
lim 4
n→
(b)
3 4 n
6
1 ≤ 4.5 ⇒ an bounded 2n −1
1 1 an 4 n > 4 n1 2 2 an1 ⇒ an monotonic
(a) A1 $101.00 A2 $203.01
(b) A60 $8248.64
lim 4
n→
Therefore, an converges. 96. An 1001011.01n 1
12 −1
1 4 2n
98. The first sequence because every other point is below the x-axis.
(c) A240 $99,914.79
A3 $306.04 A4 $410.10 A5 $515.20 A6 $621.35 100. Impossible. The sequence converges by Theorem 8.5.
102. Impossible. An unbounded sequence diverges.
104. Pn 16,0001.045n
106. (a) an 410.9212n 6003.8545
P1 $16,720.00
12,000
P2 $17,472.40 P3 $18,258.66 P4 $19,080.30 P5 $19,938.91
108. an 1
1 n
lim sn L > 0,
n→
there exists for each > 0, an integer N such that sn L < for every n > N. Let L > 0 and we have,
a3 2.3704 a5 2.4883
< L, L < sn L < L, or 0 < sn < 2L
for each n > N.
a6 2.5216 n→
sn L
a4 2.4414
1 n
(b) For 2004, n 14 and an 11,757, or $11,757,000,000. 110. Since
a2 2.2500
10 0
n
a1 2.0000
lim 1
0
n
e
Section 8.2
Series and Convergence
373
112. If an is bounded, monotonic and nonincreasing, then a1 ≥ a2 ≥ a3 ≥ . . . ≥ an ≥ . . . . Then a1 ≤ a2 ≤ a3 ≤ . . . ≤ an ≤ . . . is a bounded, monotonic, nondecreasing sequence which converges by the first half of the theorem. Since an converges, then so does an. 114. True
116. True 1 1 , n 1, 2, . . . x 2 n1 xn1
118. x0 1, xn x1 1.5
x6 1.414214
x2 1.41667
x7 1.414214
x3 1.414216
x8 1.414114
x4 1.414214
x9 1.414214
x5 1.414214 x10 1.414214 The limit of the sequence appears to be 2. In fact, this sequence is Newton’s Method applied to f x x 2 2.
Section 8.2
Series and Convergence
2. S1 16 0.1667 S2
1 6
1 6
4. S1 1
0.3333
S2 1 13 1.3333
3 S3 16 16 20 0.4833
S3 1 13 15 1.5333
3 2 S4 16 16 20 15 0.6167
S4 1 13 15 19 1.6444
1 1 3 2 5 S5 6 6 20 15 42 0.7357
1 S5 1 13 15 19 11 1.7354
6. S1 1
8.
S2 1
0.5
S3 1
1 2
16 0.6667
S5 1
10.
21.03
n
1 24
1 120
Geometric series
Diverges by Theorem 8.6
0.6333
12.
Geometric series
n
2n 3
n1
n0
r 1.03
n
4 > 1 3
r
1 S4 1 12 16 24 0.6250 1 6
4
n0
1 2
1 2
3
> 1
lim
n→
Diverges by Theorem 8.6
n 1 0 2n 3 2
Diverges by Theorem 8.9
14.
n
n1
lim
n→
n 1
16.
2
n n2 1
n!
2
n1
lim
n→
1 1 1 n2
Diverges by Theorem 8.9
10
lim
n→
n
n! 2n
Diverges by Theorem 8.9
374
18.
Chapter 8
3
2
n
Infinite Series
1
n0
3 9
17
8
n
n0
17 8 64 . . . 1 3 9 81
S0
Matches graph (b).
Matches (d).
Analytically, the series is geometric:
Analytically, the series is geometric:
2
n
n0
1 1 3 1 2 3 1 3
17 , S 0.63, S3 5.1, . . . 3 1
17 8 9 n0 3
n
17 3 17 3 3 1 8 9 17 9
nn 2 2n 2n 2 2 6 4 8 6 10 8 12 10 14 . . . 1
n1
1
1
1
1
1
1
1
1
1
1
1
1
n1
1
nn 2
n1
24.
20.
5 S0 1, S1 , S2 2.11, . . . . 3
3
22.
2 4 . . . 3 9
2 2
1
lim Sn lim
n→
n→
12 41 2n 1 1 2n 2 2 41 43 1
1
n
26.
n0
n
n0
0.6
1 Geometric series with r 2 < 1.
Geometric series with r 0.6 < 1.
Converges by Theorem 8.6
Converges by Theorem 8.6
nn 4 n n 4
28. (a)
4
n1
1 1 (b)
(c)
1
1
n1
1 1 1 1 1 1 1 1 1 1 1 . . . 5 2 6 3 7 4 8 5 9 6 10
1 1 1 25 2.0833 2 3 4 12
n
5
10
20
50
100
Sn
1.5377
1.7607
1.9051
2.0071
2.0443
4
0
11 0
(d) The terms of the series decrease in magnitude slowly. Thus, the sequence of partial sums approaches the sum slowly.
30. (a)
30.85
n1
n1
(b)
(c)
3 20 1 0.85
(Geometric series)
n
5
10
20
50
100
Sn
11.1259
16.0625
19.2248
19.9941
19.999998
24
0
11 0
Section 8.2
5 3
32. (a)
1
n1
n1
(b)
5 15 3.75 1 1 3 4
(c)
n
5
10
20
50
100
Sn
3.7654
3.7499
3.7500
3.7500
3.7500
Series and Convergence
7
0
11 0
(d) The terms of the series decrease in magnitude rapidly. Thus, the sequence of partial sums approaches the sum rapidly.
34.
nn 2 2 n n 2 2 1 3 2 4 3 5 . . . 2 1 2 3 4
n1
36.
1
1
1
1
1
1
6 5
n
n
4
1
3 4
8
0.7
n
6 30 1 4 5
1
1
1
(Geometric)
1
1
1
1
0.9n
44.
81
1
n
n
4 2
7
n
10
9
n
2
n0
1
1 1
2 6 1 2 3 5
1
4 8 1 1 2 3
1 1 10 34 10 2 2 1 7 10 1 9 10 3 3
n
50. 0.21515
n0
81 1 Geometric series with a 100 and r 100
81 81 100 9 a 1 r 1 1 100 99 11
S
2
n0
10
100 100
2 3
n0
n0
48. 0.8181
40.
8 32 1 3 4
n1
52.
1
n1
n0
46.
1
2n 12n 3 2 2n 1 2n 3 2 3 5 5 7 7 9 . . . 2 3 6
n0
42.
1
n1
n1
38.
1
3 1 1 5 n0 200 100
3 1 Geometric series with a 200 and r 100
S
a 1 3 200 71 1 5 1 r 5 99 100 330
n1
2n 1
n1
n1 1 0 2n 1 2
lim
n→
Diverges by Theorem 8.9
54.
1
n1
56.
nn 3 3 n n 3
1
1
1
n1
1 41 12 51 13 61 14 71 15 81 16 91 . . .
1 3
1 1 1 11 1 , converges 3 2 3 18
3n
n
n1
3
ln 23n ln 22 3n ln n3 3n 3n lim lim lim 3 2 n→ n n→ n→ n→ 3n 6n 6 lim
(by L’Hôpital’s Rule) Diverges by Theorem 8.9
n
1
375
376
58.
Chapter 8
Infinite Series
1
4
60.
n
n0
2n
100
62.
n1
Geometric series with r
1 4
k
n
n1
Geometric series with r 2
Converges by Theorem 8.6
1 n
lim 1
n→
Diverges by Theorem 8.6
k n
n
ek 0
Diverges by Theorem 8.9
64. lim an 5 means that the limit of the sequence an is 5. n→
a
n
66. If lim an 0, then n→
a
n
diverges.
n1
a1 a2 . . . 5 means that the limit of the
n1
partial sums is 5. 68. (a) x 2 is the common ratio. (c) y1
(b) 1
2 2x
−5
Geometric series:
12
1 1 x 2
2 , x < 2 2x
1 < 0.0001 2n
72.
y = 10
Horizontal asymptote: y 10 4
x x a 1, r , < 1 ⇒ x < 2 2 2
1 0.8x
1 0.8
n
5
−5
2 5
5
x y2 1 2
70. f x 2
x2 x3 x x 2 4 8 2 n0
10,000 < 2n This inequality is true when n 14.
n −2
n0
16
0.01n < 0.0001
−2
2 S 10 1 4 5
10,000 < 10n This inequality is true when n 5. This series converges at a faster rate.
The horizontal asymptote is the sum of the series. f n is the nth partial sum. 74. Vt 225,0001 0.3n 0.7n225,000
n1
76.
1000.60 i
i0
V5 0.75225,000 $37,815.75
1001 0.6n 1 0.6
2501 0.6n Sum 250 million dollars
78. The ball in Exercise 77 takes the following times for each fall. s1 16t 2 16
s1 0 if t 1
s2 16t 2 160.81
s2 0 if t 0.9
s3 16t 160.81
s3 0 if t 0.92
2
2
sn 16t 160.81 2
n1
sn 0 if t 0.9n1
Beginning with s2, the ball takes the same amount of time to bounce up as it takes to fall. The total elapsed time before the ball comes to rest is t12
0.9
n1
1
n
1 2
0.9
n0
2 19 seconds. 1 0.9
n
80. Pn
1 2 3 3
P2
1 2 3 3
3 3
n0
1 2
n
n
2
4 27
1 3 1 1 2 3
million dollars
Section 8.2
Series and Convergence
377
82. (a) 64 32 16 8 4 2 126 in.2
64 2
(b)
1
n
n0
64 128 in.2 1 1 2 16 in.
Note: This is one-half of the area of the original square! 16 in.
13 9 2
84. Surface area 4 12 9 4
P 1 12
12t1
86.
r
n
12 P r P
Per 12n
n0
4 9
1
2
. . . 4 . . .
12r r 1 1
12 12t
P 1 1
n0
12t1
2
88. P 75, r 0.05, t 25
12t
r 1 1 12
12r
1 12r
12t
(a) A 75 (b) A
1
(b) A
1250
1 $75,743.82
75e0.0525 1 $44,732.85 e0.05 12 1
39
40,0001.04
n
n0
20e0.0650 1 $76,151.45 e0.06 12 1
40,000
111.04 1.04
40
$3,801,020 94. x 0.a1a2a3 . . . aka1a2a3 . . . ak
0.a1a2a3 . . . ak 1
101
1 1 10k 10k
10
1
n0
0.a1a2a3 . . . ak
2
k
3
. . .
n
k
1 11 10 a rational number k
96. Let Sn be the sequence of partial sums for the convergent series lim Sn L and since Rn
n→
a
n
L. Then
n1
ak L Sn,
kn1
we have lim Rn lim L Sn lim L lim Sn L L 0.
n→
1 $44,663.23
92. T 40,000 40,0001.04 . . . 40,0001.0439
12 0.06
1 0.06 12
0.a1a2a3 . . . ak
1225
P1 er 1212t Pert 1 r 12 1 er 12 e 1
90. P 20, r 0.06, t 50 (a) A 20
12 0.05
1 0.05 12
n→
n→
n→
98. If an bn converged, then an bn an bn would converge, which is a contradiction. Thus, an bn diverges.
378
Chapter 8
Infinite Series
100. True
104.
102. True;
1 1 1 1 1 2 3. . . r r r n0 r r
n
1r 1 1 1r r 1
This is a geometric series which converges if
Section 8.3 2.
since
1 < 1 r
4.
n2
Let f x xex2.
2 . 3x 5
f is positive, continuous, and decreasing for x ≥ 3.
f is positive, continuous, and decreasing for x ≥ 1.
1
ne
n1
n1
The Integral Test and p-Series
2
n 1 0 1000n 1 1000
1 < 1 ⇔ r > 1. r
3n 5
Let f x
lim
n→
2 2 dx ln3x 5 3x 5 3
1
Since fx
2x < 0 for x ≥ 3. 2e x2
Diverges by Theorem 8.10
xex2 dx 2x 2ex2
3
10e32
3
Converges by Theorem 8.10
6.
1
2n 1
8.
1 . 2x 1
1
n 3
Let f x
1 dx ln 2x 1 2x 1
1
x . x2 3
f x is positive, continuous, and decreasing for x ≥ 2 since
f is positive, continuous, and decreasing for x ≥ 1.
2
n1
n1
Let f x
n
3 x2 < 0 for x ≥ 2. x 2 3
fx
Diverges by Theorem 8.10
1
x dx ln x2 3 x2 3
1
Diverges by Theorem 8.10
10.
ne
k n
12.
1
n
13
n1
n1
xk Let f x x . e
Let f x
f is positive, continuous, and decreasing for x > k since
f is positive, continuous, and decreasing for x ≥ 1.
fx
1
for x > k. We use integration by parts.
1
xk ex dx xk ex
1
k
1 x13
dx
xk1 ex dx
1
2 x 3
23
Diverges by Theorem 8.10
1 k kk 1 . . . k! e e e e
Converges by Theorem 8.10
xk1k x < 0 ex
1 . x13
1
Section 8.3
14.
3
n
n1
2
n1
Divergent p-series with p 23 < 1
22.
1
n
20.
23
n1
2
Convergent p-series with p 2 > 1
5 > 1 3
1
n
1
n1
Convergent p-series with p
18.
n
16.
53
2
2
The Integral Test and p-Series
Convergent p-series with p > 1
2
n123. . .
2
n
24.
2
2
2 2 . . . 22 32
n1
n1
S1 2
S1 2
S2 3
S2 2.5
S3 3.67
S3 2.722
Matches (d)
Matches (c)
Diverges—harmonic series
Converges—p-series with p 2 > 1.
26. (a)
8
n
5
10
20
50
100
Sn
3.7488
3.75
3.75
3.75
3.75 0
11 0
The partial sums approach the sum 3.75 very rapidly. (b)
5
n
5
10
20
50
100
Sn
1.4636
1.5498
1.5962
1.6251
1.635 0
11 0
The partial sums approach the sum
28. x
n
x
n1
2
1.6449 slower than the series in part (a). 6
1
n
x
n1
Converges for x > 1 by Theorem 8.11.
30.
ln n p n2 n
If p 1, then the series diverges by the Integral Test. If p 1,
2
ln x dx xp
xp ln x dx
2
xp1
p 1
2
1 p 1 ln x
. (Use Integration by Parts.)
2
Converges for p 1 < 0 or p > 1.
32. A series of the form
n1
1 is a p-series, p > 0. np
The p-series converges if p > 1 and diverges if 0 < p ≤ 1.
34. The harmonic series
1
n.
n1
379
380
Chapter 8
Infinite Series
38. S4 1
36. From Exercise 35, we have: 0 ≤ S SN ≤
f x dx
R4 ≤
N
SN ≤ S ≤ SN N
≤ S ≤
n
n1
N
a
n
f x dx
n1
1 1 dx 4 x5 4x
1
n
1.0363 ≤
0.0010
4
≤ 1.0363 0.0010 1.0373
5
n1
f x dx
N
1 1 1 1 . . .
1.9821 2ln 23 3ln 33 4ln 43 11ln 113
40. S10
R10 ≤
10
1 1 dx x 1lnx 13 2lnx 12
1
n 1lnn 1
1.9821 ≤
10
1 1 1 1 2 3 4 0.5713 e e e e
R4 ≤
ex dx ex
4
e
0.5713 ≤
n
4
1
0.0870 2ln 113
≤ 1.9821 0.0870 2.0691
3
n1
42. S4
4
N
a
1 1 1 1.0363 25 35 45
44. 0 ≤ RN ≤
1 x32
N
dx
2 x12
N
2 N
< 0.001
N12 < 0.0005
0.0183
N > 2000
N ≥ 4,000,000
≤ 0.5713 0.0183 0.5896
n0
46.
RN ≤
ex2 dx 2ex2
N
2 e N2 e N2
N
2 e N2
< 0.001
48. Rn ≤
N
< 0.001
2 N arctan < 0.001 5 2 5
2 1 x dx 2 arctan x2 5 5 5
N
N arctan < 0.001118 2 5
> 2000
N > ln 2000 2
1.56968 < arctan
N > 2 ln 2000 15.2 N
N ≥ 16
5
N5
> tan 1.56968
N ≥ 2004
50. (a)
10
1 xp1 dx p x p 1
(b) f x
10
1 .p > 1 p 110 p1
1 xp
R10p
1 under the graph of f over the interval 10, p ≤ Area n n11
(c) The horizontal asymptote is y 0. As n increases, the error decreases.
52.
ln1 n ln
n2
1
2
n2
n2 1 n 1(n 1 ln lnn 1 lnn 1 2 ln n 2 2 n n n2 n2
ln 3 ln 1 2 ln 2 ln 4 ln 2 2 ln 3 ln 5 ln 3 2 ln 4 ln 6 ln 4 2 ln 5 ln 7 ln 5 2 ln 6 ln 8 ln 6 2 ln 7 ln 9 ln 7 2 ln 8 . . . ln 2
Section 8.4
54.
1
nn
2
n2
Let f x
56. 3
1
1
n
n1
Comparisons of Series
0.95
p-series with p 0.95
1 . xx 2 1
Diverges by Theorem 8.11
f is positive, continuous, and decreasing for x ≥ 2.
2
1 dx arcsec x xx 2 1
2
2 3
Converges by Theorem 8.10
58.
1.075
n
60.
n0
1
n1
Geometric series with r 1.075
2
1 1 1 3 2 3 n n1 n n1 n
62.
Since these are both convergent p-series, the difference is convergent.
Diverges by Theorem 8.6
64.
n
lnn
n2
lim lnn
n→
Diverges by Theorem 8.9
ln n 3 n2 n
Let f x
ln x . x3 1 3 ln x < 0 for x ≥ 2. x4
f is positive, continuous, and decreasing for x ≥ 2 since fx
2
ln x ln x dx 2 x3 2x
2
1 2
2
1 ln 2 1 dx 2 x3 8 4x
2
ln 2 1 (Use Integration by Parts.) 8 16
Converges by Theorem 8.10. See Exercise 14.
Section 8.4 2. (a)
2
Comparisons of Series 2
n 2 2 . . .
S1 2
an
an =
n1
4
2 2 2 . . . S1 4 0.5 2 0.5 n1 n 0.5
3
2
4 4 4 . . . S1 3.3 n 0.5 1.5 2.5 n1
(b) The first series is a p-series. It diverges p
2 n − 0.5
1 2
4 n + 0.5
an =
an = 2 n
1
< 1.
n 2
4
6
8
10
Σ
4 n + 0.5
(c) The magnitude of the terms of the other two series are greater than the corresponding terms of the divergent p-series. Hence, the other two series diverge. (d) The larger the magnitude of the terms, the larger the magnitude of the terms of the sequence of partial sums.
Sn
20
Σ
16
2 n − 0.5
12 8 4
Σ
2 n
8
10
n 2
4
6
381
382
Chapter 8
Infinite Series
4.
1 1 < 3n2 2 3n2
6.
Therefore,
1
2
n1
1 2
diverges by comparison with the divergent p-series
3n 3 < 4 5 4 n
n
10.
1
converges by comparison with the convergent p-series
3 n . 4
1 1 > 3 n 1 4 n 4 4 Therefore,
14.
3n
4n 4n > n 1 3
Therefore,
1
3
diverges by comparison with the divergent p-series
n1
2 3n 5 2 3n 2 lim n n n→ n→ 3 5 1 3
Therefore,
2 5
3 . n
n1
5n 3 n2 2n 5 5n2 3n 20. lim lim 2 5 n→ n→ n 2n 5 1 n Therefore,
n
2
n1
1
n.
n1
n1
3 n2 4 3n 3 lim n→ n→ n2 4 1 n
18. lim
3 4
n
2
n3
diverges by a limit comparison with the divergent harmonic series
1
n.
n3
1 nn2 1 n3 22. lim lim 3 1 n→ n→ n n 1 n3 Therefore,
5n 3 2n 5
diverges by a limit comparison with the divergent p-series
n
4
Therefore,
converges by a limit comparison with the convergent geometric series
4n 1
3 .
16. lim
3n
diverges by comparison with the divergent geometric series
1 1 . 4 n 4n1
1
3 2 .
n1
4 n 1
1
n
n1
1 1
3
n1
converges by comparison with the convergent geometric series
n
n
1 n3 2
Therefore,
3n n 4 5 n0
3
<
n3 1
n1
1
n.
n2
Therefore,
12.
1
n2
1 1 . 3 n1 n2
n0
for n ≥ 2.
n 1
converges by comparison with the convergent p-series
n
Therefore,
3n
8.
1
>
n 1
nn
2
n1
1 1
converges by a limit comparison with the convergent p-series
1
n.
n1
3
Section 8.4
24. lim
n→
n n 12n1 n lim 1 n→ n 1 1 2n1
26. lim
n→
Therefore,
2 1
5
n n
n1
2
n1
4
diverges by a limit comparison with the divergent harmonic series
converges by a limit comparison with the convergent geometric series
n1
1
n.
.
n1
n1
n→
5 n n2 4 5n 5 lim n→ n n2 4 2 1 n
n
n1
28. lim
383
Therefore,
n 12
Comparisons of Series
1 n2 sec21 n 1 tan1 n lim lim sec2 1 n→ n→ 1 n 1 n2 n
Therefore,
tan n 1
n1
diverges by a limit comparison with the divergent p-series
1
n.
n1
30.
5 5 1
n
32.
1 2n 15
Converges
Converges Geometric series with r
2
n4
n0
34.
3n
15
Limit comparison with
1
n
n1
2
n 1 n 2 2 3 3 4 4 5 . . . 2 1
1
1
1
1
1
1
1
1
n1
Converges; telescoping series
36.
3
nn 3
n1
Converges; telescoping series
n n 3 1
1
n1
38. If j < k 1, then k j > 1. The p-series with p k j converges and since lim
n→
Pn Qn L > 0, 1 nkj
the series
Pn
Qn converges by the limit comparison test. Similarly, if j ≥ k 1, then k j ≤ 1 which implies that
n1
Pn Q n n1
diverges by the limit comparison test.
40.
1 1 1 1 1 1 . . . , 2 3 8 15 24 35 n 1 n2
which converges since the degree of the numerator is two less than the degree of the denominator.
42.
n
n1
3
n2 1
diverges since the degree of the numerator is only one less than the degree of the denominator.
384
Chapter 8
Infinite Series
n 1 lim lim n 0 ln n n→ 1 n n→
44. lim
n→
Therefore,
1
ln n diverges.
n2
48. This is not correct. The beginning terms do not affect the convergence or divergence of a series.
46. See Theorem 8.13, page 585. One example is
1
n 1 diverges because lim n→
n2
1 n 1 1 1 n
In fact, 1 1 1 . . . diverges (harmonic) 1000 1001 n1000 n
and
1
50.
and
diverges (p-series).
n2 n
1
1 1 1 1 . . . , 200 210 220 n0 200 10n
52.
diverges
54. (a)
1 1 1 . . . 2 converges (p-series). 4 9 n1 n
1 1 1 1 1 . . . 3, 201 208 227 264 n1 200 n
converges
1
2n 1
2
n1
1 4n 1
4n
2
n1
converges since the degree of the numerator is two less than the degree of the denominator. (See Exercise 38.) (b)
(c)
n
5
10
20
50
100
Sn
1.1839
1.02087
1.2212
1.2287
1.2312
1
2n 1
2
n3
(d)
2 S2 0.1226 8
1 2 S9 0.0277 2 2n 1 8 n10
56. True 58. False. Let an 1 n, bn 1 n, cn 1 n2. Then, an ≤ bn cn, but
c
converges.
n
n1
60. Since
a
n
a
converges, then
n1
n an
n1
a
2
n
1
62.
n
(b)
a
n1
2
converge, and hence so does
n 1
2
2
1
n . 4
converges by Exercise 59.
64. (a)
a
n
1
n
3
and
b
n
1
n . Since 2
an 1 n3 1 lim lim 0 n→ bn n→ 1 n2 n→ n lim
converges, so does
1
n . 3
and
1
n
2
lim
n
n→
1
n and b
n
1
n. Since
an 1 n lim n lim n→ bn n→ 1 n
diverges, so does
1
n.
and
1
n
Section 8.5
Section 8.5 2.
6.
1n1 6 6 6 6 . . . n2 1 4 9 n1
4.
1n1 10 10 10 . . . n2n 2 8 n1
S1 6, S2 4.5
S1 5, S2 3.75
Matches (d)
Matches (a)
1n1
385
Alternating Series
Alternating Series
1
n 1! e 0.3679
n1
(a)
(b)
n
1
2
3
4
5
Sn
1
0
0.5
0.3333
0.375
6 0.3667
7
8
9
10
0.3681
0.3679
0.3679
0.3679
2
0
11 0
(c) The points alternate sides of the horizontal line that represents the sum of the series. The distance between successive points and the line decreases. (d) The distance in part (c) is always less than the magnitude of the next series.
8.
1n1
2n 1! sin1 0.8415
n1
(a)
(b)
n
1
2
3
4
5
6
7
8
9
10
Sn
1
0.8333
0.8417
0.8415
0.8415
0.8415
0.8415
0.8415
0.8415
0.8415
(c) The points alternate sides of the horizontal line that represents the sum of the series. The distance between successive points and the line decreases.
2
0
11 0
10.
1n1 n n1 2n 1
lim
n→
(d) The distance in part (c) is always less than the magnitude of the next series.
12.
1n
lnn 1
n1
n 1 2n 1 2
an1
Diverges by the nth Term Test. lim
n→
1 1 < an lnn 2 lnn 1
1 0 lnn 1
Converges by Theorem 8.14
14.
1n1n 2 n1 n 1
an1 lim
n→
n1 n < 2 an n 12 1 n 1
n 0 n2 1
Converges by Theorem 8.14
16.
1n1n2 2 n1 n 5
lim
n→
n2 1 n2 5
Diverges by nth Term Test
386
18.
Chapter 8
Infinite Series
1n1 lnn 1 n1 n1
lim
n sin 1
n1
lnn 1 lnn 1 1 < for n ≥ 2 n 1 1 n1
an1
n→
20.
1n1 2n 1 2 n n1
Converges; (see Exercise 9)
lnn 1 1n 1 lim 0 n→ n1 1
Converges by Theorem 8.14
22.
1n n n1
1
n cos n
n1
24.
1n
2n 1!
n0
Converges; (see Exercise 9)
1 1 an < 2n 3! 2n 1!
an1 lim
n→
1 0 2n 1!
Converges by Theorem 8.14
26.
1n1 n 3 n n1
28.
n12 lim n16 n→ n13 n→
1n12en 21n1 n en e e2n 1 n1 n1
Let f x
lim
Diverges by the nth Term Test
fx
2ex . Then 1
e2x
2e2x 1 e2x < 0 for x > 0. e2x 12
Thus, f x is decreasing for x > 0 which implies an1 < an. lim
n→
2en 2en 1 lim 2n lim n 0 n→ e 1 n→ 2e
e2n
The series converges by Theorem 8.14.
30. S6
41n1
6
lnn 1 2.7067
n1
4
R6 S S6 ≤ a7 ln 8 1.9236; 0.7831 ≤ S ≤ 4.6303 32. S6
1n1n 0.1875 2n n1 6
7
R6 S S6 ≤ a7 27 0.05469; 0.1328 ≤ S ≤ 0.2422 34.
1n n n0 2 n!
36.
(a) By Theorem 8.15,
Rn
≤ aN1
(a) By Theorem 8.15, 1 < 0.001. 2N1N 1!
This inequality is valid when N 4. (b) We may approximate the series by
1 1 1 1 1 n n! 1 2 8 48 348 0.607. 2 n0 4
1n n0 2n!
n
(5 terms. Note that the sum begins with n 0.)
RN
≤ aN1
1 < 0.001. 2N 2!
This inequality is valid when N 3. (b) We may approximate the series by
1n 1 1 1 1 0.540. 2n ! 2 24 720 n0 3
(4 terms. Note that the sum begins with n 0.)
Section 8.5
38.
387
1n1 4n n n1
(b) We may approximate the series by
(a) By Theorem 8.15,
RN
≤ aN1
1n1 1 1 1 0.224. 4n n 4 32 192 n1 3
1 < 0.001. N 1
4N1
This inequality is valid when N 3.
40.
Alternating Series
1n1 n4 n1
(3 terms)
42.
By Theorem 8.15, RN ≤ aN1
1 < 0.001. N 14
1n1 n1 n 1
The given series converges by the Alternating Series Test, but does not converge absolutely since the series
This inequality is valid when N 5.
1
n1
n1
diverges by the Integral Test. Therefore, the series converge conditionally.
44.
1n1 n n n1
46.
1
1
n n n
n1
n1
32
1n n2 n0 e
n→
50.
1n1 n1.5 n1
1
1
n
n2
n1
converges by a comparison to the convergent geometric series
2n 3 2 Therefore, the series diverges by the n 10
nth Term Test.
e
n0
lim
which is a convergent p-series.
Therefore, the given series converges absolutely.
48.
1n12n 3 n 10 n1
1.5
is a convergent p-series.
Therefore, the given series converge absolutely.
e . n
1
n0
Therefore, the given series converges absolutely.
52.
1n
n 4
n0
The given series converges by the Alternating Series Test, but
1 4
n0 n
diverges by a limit comparison to the divergent p-series
1
n.
n1
Therefore, the given series converges conditionally.
54.
1
n1
arctan n
n1
lim arctan n
n→
0 Therefore, the series diverges by 2
the nth Term Test.
388
56.
Chapter 8
Infinite Series
1n1 sin2n 12 n n n1 n1
58. S Sn Rn ≤ an1 (Theorem 8.15)
The given series converges by the Alternating Series Test, but
n1
1 sin2n 12 n n1 n
is a divergent p-series. Therefore, the series converges conditionally.
60.
1 1 1 1n1 1 . . . (Alternating Harmonic Series) n 2 3 4 n1
62.
1n np n1
64.
1 1n1 converges, but diverges n n1 n1 n
If p 0, then lim
n→
1 1 np
and the series diverges. If p > 0, then lim
n→
1 1 1 0 and < p. np n 1p n
Therefore, the series converge by the Alternating Series Test.
66. (a)
xn n1 n
(b) When x 1, we have the convergent alternating series
1n . n n1
converges absolutely (by comparison) for 1 < x < 1,
When x 1, we have the divergent harmonic series
since
xn < xn and n
1 . n
x
n
Therefore,
is a convergent geometric series for 1 < x < 1.
xn n1 n
converges conditionally for x 1.
68. True, equivalent to Theorem 8.16
70.
n
n1
2
3 5
converges by limit comparison to convergent p-series 1
n . 2
3 2
72. Converges by limit comparison to convergent geometric 1 . series 2n
74. Diverges by nth Term Test. lim an
76. Converges (conditionally) by Alternating Series Test.
78. Diverges by comparison to Divergent Harmonic Series:
n→
ln n 1 > for n ≥ 3. n n
Section 8.6
Section 8.6 2.
The Ratio and Root Tests
389
The Ratio and Root Tests
2k 2! 2k 2! 1 2k! 2k2k 12k 2! 2k2k 1
4. Use the Principle of Mathematical Induction. When k 3, the formula is valid since
1 233!35 1. Assume that 1 6!
1 2n n!2n 32n 1 . . . 135 2n 5 2n! and show that 1 2n1n 1!2n 12n 1 . . . . 135 2n 52n 3 2n 2! To do this, note that:
35.
1
1 . . 2n 52n 3 1
1
1 . . 2n 5
2n 3
2n n!2n 32n 1 2n!
2n 3
2n n!2n 1 2n!
2n 2n 1n!2n 12n 1 2n!2n 12n 2
2n1n 1!2n 12n 1 2n 2!
35.
1
2n 12n 2
2n 12n 2
The formula is valid for all n ≥ 3.
6.
4 n! 4 16 2 . . . 3
n
3
1
9
1
8.
n1
1n1 4 4 4 . . . (2n! 2 24 n1
S1 2
3 S1 , S2 1.03 4
Matches (b)
Matches (c)
10.
4e
n
4
n0
4 . . . e
S1 4 Matches (e)
12. (a) Ratio Test: lim
n→
(b)
(c)
an1 an
n 12 1 n 1! n2 2n 2 lim lim 2 n→ n→ n 1 n2 1 n!
n
5
10
15
20
25
Sn
7.0917
7.1548
7.1548
7.1548
7.1548
n 1 1 0 < 1. Converges
(d) The sum is approximately 7.15485
10
(e) The more rapidly the terms of the series approach 0, the more rapidly the sequence of the partial sums approaches the sum of the series. 0
11 0
390
14.
Chapter 8
Infinite Series
3n
n!
n0
16.
n1
an1 3 lim n→ n 1! an
lim
n→
lim
n→
n!
3n
3 0 n1
lim
n→
lim
n→
3n 1 3 2n 2
1n1n 2 nn 1 n1
an 1 n 13 2n1 lim n→ an n3 2n n→
2n
n3n
Therefore, by the Ratio Test, the series diverges.
20.
lim
an1 n 13n1 lim n→ an 2n1 n→
n
n1
n
lim
n3
2
3
n1
Therefore, by the Ratio Test, the series converges.
18.
n 2
an1
n 13 1 2n3 2
lim
n→
Therefore, by the Ratio Test, the series converges.
n3 n2 ≤ an n 1n 2 nn 1
n2 0 nn 1
Therefore, by Theorem 8.14, the series converges. Note: The Ratio Test is inconclusive since lim
n→
The series converges conditionally.
22.
1n13 2n n2 n1
lim
n→
24.
an1 3 2n1 lim 2 n→ n 2n 1 an lim
n→
n2
3 2n
n1
n→
3n2 3 > 1 2n2 2n 1 2
nn
n!
n1
lim
n→
an1 n 1 lim n→ an n 1!
n1
n!
nn
n→
lim
n 1n 1n n! n 1n!nn
lim
n n 1
n→
n
lim
an1
n 1! 2 lim n→ 3n 3! an lim
n→
lim
n→
an1 24n4 lim n→ 2n 3! an
3n!
n!2
n 12 0 3n 33n 23n 1
Therefore, by the Ratio Test, the series converges.
2n 22n 1n5 n 15
n!2
n→
e > 1
1n 24n n0 2n 1!
3n!
Therefore, by the Ratio Test, the series diverges.
30.
n5
2n!
Therefore, by the Ratio Test, the series diverges.
n0
an1 2n 2! lim n→ n 15 an lim
28.
n→
2n! n5
lim
Therefore, by the Ratio Test, the series diverges.
26.
an1 1. an
2n 1! 24 lim 0 n→ 2n 32n 2 24n
Therefore, by the Ratio Test, the series converges.
Section 8.6
32.
1n 2 4 6 . . . 2n
2 5 8 . . . 3n 1
n1
lim
n→
an1 2 4 . . . 2n2n 2 lim n→ 2 5 . . . 3n 13n 2 an
2
5 . . . 3n 1 2 4 . . . 2n
Therefore, by the Ratio Test, the series converges. 2 24 24 Note: The first few terms of this series are 2 25 25
lim
n→
an1 1 lim n→ n 14 an
4
p
n4 n lim n→ n 1 1
np n lim n→ n 1 1
1
n lim
n→
an1 1 lim n→ n 1p an
n 1 2n
6. 8
. .
1
n
38.
3n
2n 1
n 2n 1
n
n
n →
lim
n→
n a n lim
2n 2 n→ n 1
n→
lim
lim
n→
2n3n 1 n
3n
2n3n 1 32 3
e
n
n0
n an lim lim
n→
n→
e1 1e n
n
Therefore, by the Root Test, the series converges.
42.
n1 3n n0
44.
n→
n 3 1 lim n
n→
n
n→
n n 1
3
x y lim x1
Let
n→
x ln y lim ln x 1 n→
lim
1 lnx 1 x
lim
lnx 1 1 0. x x1
n→
n→
Since ln y 0, y e0 1, so lim
n→
n 1
3
1 . 3
Therefore, by the Root Test, the series converges.
5
1
n 5 n
n1
n lim an lim
3
27 8
Therefore, by the Root Test, the series diverges.
Therefore, by the Root Test, the series diverges.
3n
n1
n a lim n lim
40.
2n 2 2 3n 2 3
1
n1
n→
n→
p
n1
lim
4
n1
(b)
1
n
34. (a)
36.
The Ratio and Root Tests
n1
This is the divergent harmonic series.
391
392
46.
Chapter 8
4
Infinite Series
n
48.
2n
2
n1
n1
Since 4 < 1,this is convergent geometric series.
n 1
n 2n2 1 n2 1 lim > 0 n→ 2n2 1 1 n 2
lim
n→
This series diverges by limit comparison to the divergent harmonic series
1
n.
n1
50.
10
3 n
52.
3
n1
lim
3 2
1 n
10 3
lim
n→
Therefore, the series converges by a limit comparison test with the p-series
1
n
n1
54.
3 2
.
1n n2 n lnn
lim
n→
56.
lnn 2 n1 n
1 lnn ≤ 3 2 n2 n
1 1 ≤ an n 1 lnn 1 n lnn
Therefore, the series converges by comparison with the p-series
1 0 n lnn
1n 3n n2n n1
1
n
Therefore, by the Alternating Series Test, the series converges.
58.
2n ln 22n ln 222n lim lim n→ n→ 1 8n 8
4n2
Therefore, the series diverges by the nth Term Test for Divergence.
an1
2n 1
2
n1
10 3n3 2
n→
4n
n1
3 2
.
lim
n→
an1 3n1 lim n→ n 12n1 an
n2n 3n 3 lim n→ 2n 1 3n 2
Therefore, by the Ratio Test, the series diverges. 60.
3
n1
lim
n→
57. 18n
. . 2n 1 2n 1n!
an1 3 5 7 . . . 2n 12n 3 lim n→ an 18n12n 12n 1n!
3
18n 2n 1n! 2n 32n 1 1 2 lim 5 7 . . . 2n 1 n→ 182n 12n 1 18 9
Therefore, by the Ratio Test, the series converge.
64. (a) and (b) are the same.
62. (b) and (c)
n 14
n0
3
n
n 4 3
n1
n1
12
34 334
1n
n 12
n2 2
4
34
3
. . .
n1
1 1 1 . . . 2 2 22 3 23
1n1 1 1 1 . . . n2n 2 2 22 3 23 n1
Section 8.7
66. Replace n with n 2.
68.
3k
3k 2k k!
1 3 5 . . . 2k 1 2k!2k 1
k0
2n2 2n n2 n 2! n0 n!
Taylor Polynomials and Approximations
k0
6k k!
2k 1!
k0
0.40967 (See Exercise 3 and use 10 terms, k 9.)
70. See Theorem 8.18.
72. One example is
100 n. 1
n1
74. Assume that
lim an1an L > 1 or that lim an1an .
n→
n→
Then there exists N > 0 such that an1an > 1 for all n > N. Therefore,
an1 > an,
n > N ⇒ lim an 0 ⇒ n→
a
n
diverges
76. The differentiation test states that if
U
n
n1
is an infinite series with real terms and f x is a real function such that f 1n Un for all positive integers n and d 2 fdx 2 exists at x 0, then
U
n
n1
converges absolutely if f 0 f0 0 and diverges otherwise. Below are some examples. Convergent Series 1
Divergent Series 1
n , f x x
n, f x x
1 cos n, f x 1 cos x
sin n, f x sin x
3
3
1
Section 8.7
1
Taylor Polynomials and Approximations 4. y e12 13 x 13 x 1 1
2. y 18 x 4 12 x 2 1
Cubic Matches (b)
y-axis symmetry Three relative extrema Matches (c)
6. f x
4 3 x
4x13
4 fx x43 3
f 8 2 f8
P1x f 8 f8x 8
1 2 x 8 12 P1x
1 8 x 12 3
8
1 12
(8, 2) −4
14
−4
393
394
Chapter 8
Infinite Series
8. f x tan x
1 4
f
fx sec2 x
f
12 x P1x 2x 1
− 2
2 4
P1 f f 4 4
3
4
x 4
−3
2
f
4 2
fx sec x tan x
f
4 2
f x sec3 x sec x tan2 x
f
10. f x sec x
P2x f
2
4
4 3 2
−3
4 f 4 x 4 f 24 x 4
P2x 2 2 x
3 2 x 4 2 4
3 0
2
2
0.585
0.685
4
0.885
0.985
1.785
f x
1.8270 1.1995
1.2913
1.4142
1.5791
1.8088
4.7043
P2x
15.5414
1.2936
1.4142
1.5761
1.7810
4.9475
2.15
x
1.2160
12. f x x 2ex, f 0 0 (a)
fx x 2 2xex f x
f0 0 ex
f 0 2
f x x 2 6x 6ex
f 0 6
x2
4x 2
f 4x x 2 8x 12ex
P4x x 2 x3
14.
12x 4 x4 x 2 x3 4! 2
f x ex
f 0 1
fx
ex
f0 1
f x
ex
f 0 1
f x
ex
f 0 1
f 4x
ex
f 5x ex
P2
f −3
3
P3 −1
(c)
6x3 x 2 x3 3!
2
P4
f 40 12
2x 2 P2x x2 2! P3x x 2
(b)
f 0 2 P2 0 f 0 6 P3 0 f 40 12 P440
(d) f n0 Pnn0
f 40 1 f 50 1
P5x f 0 f0x
f0 2 f 0 3 f 40 4 f 50 5 x 2 x3 x4 x5 x x x x 1x 2! 3! 4! 5! 2 6 24 120
Section 8.7 16.
f x e3x
f 0 1
fx
3e3x
f0 3
f x
9e3x
f 0 9
Taylor Polynomials and Approximations
f x 27e3x f 0 27 f 4x 81e3x f 40 81 P4x 1 3x
18.
9 9 2 27 3 81 4 9 27 4 x x x 1 3x x 2 x3 x 2! 3! 4! 2 2 8
f x sin x
f 0 0
20.
f x x 2ex
f 0 0
f0
fx
2xex
f x 2 sin x
f 0 0
f x
2ex
f x 3 cos x
f 0 3
f x 6ex 6xex x 2ex
fx cos x
P3x 0 x
4xex
f x
x x11 1 x 11 x1 x1
f 0 0 f0 1
f x 2x 13
f 0 2
fx 6x 14
f0 6
24x 1
f 40 24
5
2 6 24 P4x 0 1x x2 x3 x4 x x2 x3 x4 2 6 24 24. f x tan x
f 0 0
fx sec2 x
f0 1
f x 2
sec2
x tan x
fx 4
sec2
tan2
x
f 0 0
x2
sec4
f0 2
x
2 1 P3x 0 1x 0 x3 x x3 6 3
26.
1 2
f x 2x2
f 2
fx 4x3
f2
f x 12x4
f 2
1 2
3 4
3 2 15 f 4x 240x6 f 4x 4 1 1 3 1 5 P4x x 2 x 22 x 23 x 24 2 2 8 4 32 fx 48x5
fx
f 0 2 f 0 6 f 40 12
2 2 6 3 12 4 x x x 2! 3! 4! 1 x 2 x3 x 4 2
fx x 12
f 4x
x2ex
f 4x 12ex 8xex x 2ex
0 2 3 3 3 3 x x x x 2! 3! 6
P4x 0 0x
22.
f0 0
x 2ex
395
396
Chapter 8
Infinite Series
28. f x x13
f 8 2
1 fx x23 3
f8
2 f x x53 9
f 8
fx
10 83 x 27
P3x 2
32.
f x
f8
30. f x x 2 cos x fx cos x x 2 sin x
1 12
x2
P2x 2 2 x
1
cos x f 2 2
2 2 x 2 2
5
28 3456
1 1 5 x 8 x 82 x 83 12 288 20,736
1 x2 1
2
P4
2x fx 2 x 12
f −3
3
Q4 P2
23x 2 1 f x 2 x 13
−2
24x1 x 2 x 2 14
fx f 4x
f 2
f x 2 cos x 4x sin x
1 144
10 27
f 2
245x 4 10x 2 1 x 2 15 (b) n 4, c 0
(a) n 2, c 0 P2x 1 0x
2 2 x 1 x2 2!
P4x 1 0x
2 2 0 24 4 x x3 x 1 x2 x 4 2! 3! 4!
(c) n 4, c 1 Q4x
34.
1 1 1 1 12 0 3 1 1 x 1 x 12 x 13 x 14 x 1 x 12 x 14 2 2 2! 3! 4! 2 2 4 8
f x ln x P1x x 1 P4x x 1 12 x 12 13 x 13 14 x 14 (a)
x
1.00
1.25
1.50
1.75
2.00
ln x
0.0000
0.2231
0.4055
0.5596
0.6931
P1x
0.0000
0.2500
0.5000
0.7500
1.0000
P4x
0.0000
0.2230
0.4010
0.5303
0.5833
(b)
2
P1 f −1
5
P4 −2
(c) As the distance increases, the accuracy decreases. 36. (a)
f x arctan x P3x x
(b)
y
(c) π 2
x3 3
π 4
x
0.75
0.50
0.25
0
0.25
0.50
0.75
f x
0.6435
0.4636
0.2450
0
0.2450
0.4636
0.6435
P3x
0.6094
0.4583
0.2448
0
0.2448
0.4583
0.6094
x
−1
f
P3
−π 4 −π 2
1 2
1
Section 8.7
40. f x 4xex 4
38. f x arctan x y
P7
Taylor Polynomials and Approximations 2
P5 P1
y
2 y = 4xe(−x /4)
4
2
2
1
f (x) = arctan x x
−3 −2
1
x
−4
3
4
−2
P9 P13
P11 P3
f x x 2ex x 2 x3
42. f
1 4 x 2
15 0.0328 f x x 2 cos x 2 2 x
44. f
2
2 x 2 2
78 6.7954
46. f x ex; f 6x ex ⇒ Max on 0, 1 is e1. R5x ≤
e1 6 1 0.00378 3.78 103 6!
48. f x arctan x; f 4x
24xx 2 1 1 x 24
50.
f x ex f n1x ex
⇒Max on 0, 0.4 is f 40.4 22.3672.
Max on 0, 0.6 is e0.6 1.8221.
22.3672 0.44 0.0239 R3x ≤ 4!
Rn ≤
1.8221 0.6n1 < 0.001 n 1!
By trial and error, n 5.
52.
f x cos x2
54.
gx cos x 1
x2 x 4 x6 . . . 2! 4! 6!
f x g x 2
f 0.6 1
2x 4 4x8 6x12 . . . 2! 4! 6! 2 4 6 0.64 0.68 0.612 . . . 2! 4! 6!
Since this is an alternating series, Rn ≤ an1
R3x
2n 0.64n < 0.0001. 2n!
By trial and error, n 4. Using 4 terms f 0.6 0.4257.
x
x3 3!
x4 sin z 4 x ≤ < 0.001 4! 4!
x 4 < 0.024
x 22 x 24 x 26 . . . 1 2! 4! 6! 1
f x sin x x
< 0.3936
0.3936 < x < 0.3936
397
398
Chapter 8
Infinite Series
56. f c P2c, fc P2c, and f c P2 c
58. See Theorem 8.19, page 611.
60.
62. (a) P5x x
y 10
P2
f P1
8
P5x 1
6 4 2 −20
10
−2
x2 x 4 2! 4!
This is the Maclaurin polynomial of degree 4 for gx cos x. x6 x2 x 4 (b) Q6x 1 for cos x 2 4! 6!
x
P3
x5 x3 for f x sin x 3! 5!
20
−4
x5 x3 P5x 3! 5! x3 x4 x2 (c) Rx 1 x 2! 3! 4! x3 x2 Rx 1 x 2! 3! Q6x x
The first four terms are the same! 64. Let f be an odd function and Pn be the nth Maclaurin polynomial for f. Since f is odd, f is even: fx lim
h→0
f x h f x f x h f x f x h f x lim lim fx. h→0 h→0 h h h
Similarly, f is odd, f is even, etc. Therefore, f, f , f 4, etc. are all odd functions, which implies that f 0 f 0 . . . 0. Hence, in the formula Pnx f 0 f0x
f 0x2 . . . all the coefficients of the even power of x are zero. 2!
f ic . 66. Let Pn x a0 a1x c a2x c2 . . . an x cn where ai i! Pnc a0 f c For 1 ≤ k ≤ n, Pnkc an k!
Section 8.8
f k!ck! f k
Power Series 4. Centered at
2. Centered at 0
6.
2x
n
n0
L lim
n→
kc.
8.
un 1 2xn1 lim 2x n→ un 2xn
2x < 1 ⇒ R
1 2
1n xn 2n n0
L lim
n→
1 x 2
un1 1n1xn1 lim n→ un 2n1
1 x < 1⇒R2 2
2n
1n xn
Section 8.8
10.
2n!x2n n! n0
L lim
n→
un 1 2n 2 n 1! lim n→ un 2n!x2nn! lim
n→
!x2n2
2n 22n 1 n 1
x2
The series only converges at x 0. R 0.
14.
1
n0
lim
n→
n1
k
12.
un 1 1 n 2x lim n→ un 1nn 1xn lim
n→
n 2x x n1
Interval: 1 < x < 1
When x 1, the series
1
n
Since the series is geometric, it converges only if xk < 1 or k < x < k.
3xn
2n!
16. n1
n0
lim
n→
un1 3xn1 lim n→ 2n 1! un lim
n→
n 1 diverges.
n 1 diverges.
n0
Therefore, the interval of convergence is 1 < x < 1.
18.
1n xn
n 1n 2
n0
lim
n→
un1 1n1xn1 lim n→ n 2n 3 un
Interval: 1 < x < 1 When x 1, the alternating series
n 1n 2 n 1x lim x n→ 1n xn n3 1n
n 1n 2 converges.
n0
When x 1, the series
1
1
n 1n 2 converges by limit comparison to n .
n0
n1
2
Therefore, the interval of convergence is 1 ≤ x ≤ 1.
20.
1n n!x 4n 3n n0
lim
n→
3x 0 2n 22n 1
n0
When x 1, the series
2n!
3xn
Therefore, the interval of convergence is < x <
n1
un1 1n1n 1!x 4n1 lim n→ un 3n1
R0
Center: x 4 Therefore, the series converges only for x 4.
399
n0
n 1xn n2
x
Power Series
3n
1n n!x 4n
lim
n→
n 1x 4 3
.
400
22.
Chapter 8
Infinite Series
x 2n1
n 14
n1
n0
lim
n→
un1 x 2n2 lim n2 n→ un n 24
n 14n1
x 2n1
R4
lim
n→
1 x 2n 1 x 2 4n 2 4
Center: x 2 Interval: 4 < x 2 < 4 or 2 < x < 6 When x 2, the alternating series
1n1
n 1 converges.
n0
When x 6, the series
1
n 1 diverges.
n0
Therefore, the interval of convergence is 2 ≤ x < 6.
24.
1n1x cn ncn n1
lim
n→
un1 1n2x cn1 lim n→ un n 1cn1
ncn
1n1x cn
Rc
lim
n→
nx c 1 xc cn 1 c
Center: x c Interval: c < x c < c or 0 < x < 2c When x 0, the p-series
1 diverges. n1 n
When x 2c, the alternating series
26.
1n1 converges. Therefore, the interval of convergence is 0 < x ≤ 2c. n n1
1nx2n1 n0 2n 1
lim
n→
un1 1n1x2n3 lim n→ un 2n 3
2n 1
1nx2n1
R1
lim
n→
2n 1 2 x x2 2n 3
Interval: 1 < x < 1
1n
When x 1,
2n 1 converges.
n0
When x 1,
1n1 converges. n0 2n 1
Therefore, the interval of convergence is 1 ≤ x ≤ 1.
28.
1n x 2n n! n0
lim
n→
30.
un1 1 lim n→ un n 1!
n1x 2n2
lim
n →
x2 0 n1
n!
1n x 2n
n1
Therefore, the interval of convergence is < x <
n!xn
2n! lim
n→
un1 n 1!xn1 lim n→ un 2n 2! lim
n→
.
2n! n!xn
n 1x 0 2n 22n 1
Therefore, the interval of convergence is < x <
.
Section 8.8
32.
2
4 6.
. . 2n
3 5 7 . . . 2n 1 x
n1
lim
n→
2n1
Power Series
401
un1 2 4 . . . 2n2n 2x 2n3 lim n→ un 3 5 7 . . . 2n 12n 3
3
2
5 . . . 2n 1 4 . . . 2nx 2n1
R1
lim
n→
2n 2x 2 x 2 2n 3
When x ± 1, the series diverges by comparing it to
1
2n 1
n1
which diverges. Therefore, the interval of convergence is 1 < x < 1.
34.
n!x cn
1 3 5 . . . 2n 1
n1
lim
n→
un1 lim n→ 1 un
n 1!x cn1 1 3 5 . . . 2n 1 n 1x c 1 lim x c . . 2n 12n 1 n→ n!x c 2n 1 2
3 5.
R2
Interval: 2 < x c < 2 or c 2 < x < c 2 The series diverges at the endpoints. Therefore, the interval of convergence is c 2 < x < c 2.
1 3n! c5. .2 .2nc 1 1 3 5 .n!.2. 2n 1 1 23 45 .6. . 2n2n 1 > 1
n
36. (a) f x
1n1x 5n , 0 < x ≤ 10 n5n n1
(b) fx
1n1n 1x 5n2 , 0 < x < 10 5n n2
(c) f x
f x dx
40. g2
1n1x 2n ,1 < x ≤ 3 n n1
38. (a) f x
1n1x 5n1 , 0 < x < 10 5n n1
(c) f x
. . .
(b) fx
(d)
2
3 2
n
1n1x 5n1 , 0 ≤ x ≤ 10 nn 15n n1
1
n0
2 4 . . . 3 9
1
x 2n1, 1 < x < 3
n1
n1
(d)
1
n1
f x dx
42. g2
n 1x 2n2, 1 < x < 3
n2
1n1x 2n1 ,1 ≤ x ≤ 3 nn 1 n1
3 alternating. Matches (d) 2
n
n0
S1 1, S2 1.67. Matches (a) 44. The set of all values of x for which the power series converges is the interval of convergence. If the power series converges for all x, then the radius of convergence is R . If the power series converges at only c, then R 0. Otherwise, according to Theorem 8.20, there exists a real number R > 0 (radius of convergence) such that the series converges absolutely for x c < R and diverges for x c > R.
46. You differentiate and integrate the power series term by term. The radius of convergence remains the same. However, the interval of convergence might change.
48. (a) f x
xn
n!, < x <
(See Exercise 11)
n0
(b) fx
xn
x2
x3
x4
n! 1 x 2! 3! 4! . . .
n1
xn nx n1 x n1 f x n! n 1 ! n1 n1 n0 n!
(c) f x
f 0 1 (d) f x ex
402
Chapter 8
Infinite Series
y1
50.
n1
22n n!
1n 4nx 4n1 7 11 . . . 4n 1
2
y
2n n!
n1
2
y
2n
n1
y x 2y x 2
3
1n 4n4n 1 x 4n2 1n 4nx 4n2 x 2 2n . . . . . . 4n 5 n! 3 7 11 4n 1 n2 2 n! 3 7 11
2
2n n!
n2
n1
52. J1x x (a) lim
k→
22n2
1n 4nx 4n2 1n x 4n2 x2 2n . . . 4n 5 n1 2 n! 3 7 11 . . . 4n 1 3 7 11
1n1 4n 1x4n2 1n1 x 4n2 22n 1 0 2n . . . . . . n 1! 3 7 11 4n 1 n1 2 n! 3 7 11 4n 1 22n 1
1k x 2k 1k x 2k1 k 1! k0 22k1k!k 1!
2
k0
1n x 4n 3 7 11 . . . 4n 1
2k1k!
uk1 1k1 x 2k3 lim 2k3 k→ 2 uk k 1!k 2!
22k1k!k 1!
1k x 2k1
Therefore, the interval of convergence is < x < (b) J1x
1
2
k
k0
k→
1x 2 0 22 k 2k 1
x2k1
k 1!
2k1k!
J1x
1k 2k 1x2k 2k1k!k 1! k0 2
J1x
1k 2k 12kx 2k1 22k1k!k 1! k1
x 2J1 xJ1 x 2 1J1
1k2k 1x 2k1 1k 2k 12kx 2k1 2k1 2 k!k 1! 22k1k!k 1! k1 k0
2
k0
1k2k 1x 2k1 1k2k 12kx 2k1 x 2k1k!k 1! 2 2 22k1k!k 1! k1 k1
1kx2k1 1kx2k3 x 2k1k!k 1! 2 k1 22k1k!k 1! 2 k0
1kx 2k1 2k 12k 2k 1 1 1kx 2k3 2k1 2k1 2 k!k 1! k!k 1! k1 k0 2
1kx2k14kk 1 1kx2k3 2k1 2k1 2 k!k 1! k!k 1! k1 k0 2
2
1kx 2k1 1k x2k3 k 1!k! k0 22k1k!k 1!
2k1
1k1x 2k3 1kx 2k3 2k1 k!k 1! k0 22k1k!k 1!
2
k0
1kx 2k3 1 1 0 22k1k!k 1! k0
x 1 3 1 5 1 x x x7 2 16 384 18,432
(d)
J0x
1k1x 2k1 1k12k 1x 2k1 2k2 k 1!k 1! k0 22k1k!k 1! k0 2
J1x
4
6
−4
k1
−6
1kx 2k3 1k x 2k1 2k1 k 1! k0 2 k!k 1!
2k1k!
(c) P7x
.
lim
1k1x 2k1 1kx 2k1 k!k 1! k0 22k1k!k 1! k0 2
2k1
Note: J0x J1x
Section 8.9
54. f x
1
n
n0
x 2n1 sin x 2n 1!
Representation of Functions by Power Series
56. f x
1
n
n0
(See Exercise 47.)
2
−
−2.5
2.5
−2
x 2n1 arctan x, 1 ≤ x ≤ 1 2n 1
(See Exercise 38 in Section 8.7.)
2
58.
403
−
x 2n1
x 2n1
2n 1! 2n 1!
n0
2
60. True; if
n1
a
Replace n with n 1.
n
xn
n0
converges for x 2, then we know that it must converge on 2, 2. 62. True
1
1
f x dx
0
0
Section 8.9 2. (a) f x
an xn dx
n0
an xn1 n0 n 1
0
n0
n
Representation of Functions by Power Series
4 4 5 a 5 x 1 x 5 1 r
n a 1 1
4 x n0 5 5
n
4xn n1 n0 5
4. (a)
1 a 1 1 x 1 x 1 r
x
n0
This series converges on 5, 5. 4x3 4 4 4 2 x x . . . 5 25 125 625 (b) 5 x ) 4 4 4 x 5 4 x 5 4 4 x x2 5 25 4 2 x 25 4 2 4x3 x 25 125 4x3 125 4x3 4x 4 125 625
n
1
n
xn
n0
This series converges on 1, 1. 1 x x 2 x3 . . . (b) 1 x ) 1 1x x x x 2 x2 x 2 x3 x3 x3 x 4
404
Chapter 8
Infinite Series
6. Writing f x in the form
a , we have 1r
8. Writing f x in the form a 1 r, we have 3 3 1 a 2x 1 3 2x 2 1 2 3x 2 1 r
4 4 7 a 4 . 5 x 7 x 2 1 1 7x 2 1 r
which implies that a 1 and r 2 3x 2. Therefore, the power series for f x is given by
Therefore, the power series for f x is given by 4 1 4 ar n x 2 5 x n0 n0 7 7
n
n 2 3 ar n x 2 , 2x 1 n0 3 n0
4x 2 . 7n1 n0
x 2 < 7
or 5 < x < 9
x 2
10. Writing f x in the form a 1 r, we have
5 x 2
n
2nx 2n , 3n n0
<
3 1 7 or < x < . 2 2 2
4 4 1 2 a 3x 2 8 3x 2 1 3 8x 2 1 r
which implies that a 1 5 and r 2 5x. Therefore, the power series for f x is given by
which implies that a 1 2 and r 3 8x 2. Therefore, the power series for f x is given by 1 3 4 arn x 2 3x 2 n0 2 8 n0
2n xn
5
n0
n1 ,
5 5 5 x < or < x < . 2 2 2
x 2 14.
<
2 8 14 or < x < . 3 3 3
Writing f x as a sum of two geometric series, we have
2 x 1
n
22x
n
n0
n0
1 31n 1 1 2n1 xn, x < or < x < . 2n1 2 2 2
16. First finding the power series for 4 4 x, we have 1 1 x 1 1 4x n0 4
n
1n xn 4n n0
Now replace x with x2. 1n x 2n 4 . 4 x2 n0 4n
The interval of convergence is x 2 < 4 or 2 < x < 2 since lim
n→
18. hx
un1 1 x lim n→ un 4n1
n1 2n2
4n
1n x 2n
x2 x2 . 4 4
x 1 1 1 1 n 1nxn x x 1 21 x 21 x 2n0 2n0
2
1 1
1n 1xn 0 2x 0x2 2x3 0x4 2x5 . . . 2n0 2
1 2x2n1 x2n1, 1 < x < 1 2n0 n0
n
1 3n x 2n , 2 n0 8n
4x 7 3 2 3 2 3 2 2 2x2 3x 2 x 2 2x 1 2 x 1 2x 1 1 2x 1 2x 3 4x 7 2x 2 3x 2 n0 2
12. Writing f x in the form a 1 r, we have
1 1 1 5 a 2x 5 5 2x 1 2 5x 1 r
1 1 ar n 2x 5 n0 5 n0
n
Section 8.9
1 x
n n
n0
1 nx 1 nn 1x
d dx
n
n1
n1
n
n2
n2
22. By integrating, we have
d2 1 2 . Therefore, dx 2 x 1 x 13
20. By taking the second derivative, we have d2 2 2 3 x 1 dx
Representation of Functions by Power Series
1 dx ln1 x C1 and 1x
1 n 2n 1 x , 1 < x < 1. n
n
n0
1 dx ln1 x C2. 1x
f x ln1 x 2 ln1 x ln1 x. Therefore, ln1 x 2
1 dx 1x
1 dx 1x
1n x n dx
n0
x n dx C1
n0
x n1 1n x n1 C2 n1 n0 n0 n 1
2x 2n2 1x 2n2
1n 1 xn1 C C n1 n1 n0 n0 2n 2 n0
C
To solve for C, let x 0 and conclude that C 0. Therefore, ln1 x2
x2n2
n 1, 1 < x < 1
n0
24.
x2
2x 2x 1n x 2n (See Exercise 23.) 1 n0
1 2x n
2n1
n0
Since
2x d lnx 2 1 2 , we have dx x 1
lnx 2 1
1n 2x 2n1 dx C
n0
1n x 2n2 , 1 ≤ x ≤ 1. n1 n0
To solve for C, let x 0 and conclude that C 0. Therefore, lnx 2 1
26. Since
1n x 2n2 , 1 ≤ x ≤ 1. n1 n0
1 1 dx arctan2x, we can use the result of Exercise 25 to obtain 4x 2 1 2
arctan2x 2
1 dx 2 4x 2 1
n0
To solve for C, let x 0 and conclude that C 0. Therefore,
1n 4nx 2n1 1 1 , < x ≤ . 2n 1 2 2 n0
arctan2x 2
1n 4n x 2n1 1 1 , < x ≤ . 2n 1 2 2 n0
1n 4n x 2n dx C 2
405
406
Chapter 8
28. x
Infinite Series
x 2 x3 x 4 ≤ lnx 1 2 3 4
5
S5 f
x 2 x3 x 4 x5 ≤ x 2 3 4 5
−4
8
S4 −3
x x
2
x3
x4
x 2 3 4
lnx 1 x
x 2 x3 x 4 x5 2 3 4 5
0.0
0.2
0.4
0.6
0.8
1.0
0.0
0.18227
0.33493
0.45960
0.54827
0.58333
0.0
0.18232
0.33647
0.47000
0.58779
0.69315
0.0
0.18233
0.33698
0.47515
0.61380
0.78333
1
In Exercise 35–38, arctan x
n
n0
30. gx x
x 2n1 . 2n 1
x3 , cubic with 3 zeros. 3
32. gx x
Matches (d)
1
n0
arctan x 2
34. The approximations of degree 3, 7, 11, . . . 4n 1, n 1, 2, . . . have relative extrema.
Matches (b)
In Exercises 36 and 38, arctan x
36.
x3 x5 x7 , 3 5 7
n
x 4n2 2n 1
n
x 4n3 C, C 0 4n 32n 1
1
n0
3 4
arctan x 2 dx
0
1
3 44n3 4n 32n 1
34n3 4n 32n 144n3
n
n0
1n
n0
x2n1 . 2n 1
1
n0
arctan x 2 dx
n
27 2187 177,147 192 344,064 230,686,720
Since 177,147 230,686,720 < 0.001, we can approximate the series by its first two terms: 0.13427
x 2 arctan x
38.
1
n
x 2n3 2n 1
1
n
x 2n4 2n 42n 1
1 1 1 . . . 2n 42n 122n4 64 1152
n0
x 2 arctan x dx
n0
1 2
x 2 arctan x dx
0
Since
1
n0
n
1 < 0.001, we can approximate the series by its first term: 1152
1 2
0
x 2 arctan x dx 0.015625.
Section 8.9
In Exercises 40 and 42,
n1
n1
42. (a)
407
1 x n, x 1. 1 x n0
40. Replace n with n 1.
nx
Representation of Functions by Power Series
n 1x
n
(b)
n0
1 2 n 3 n1 3
n
9 1 n 10 n1 10
n
2 2 n 9 n1 3
n1
9 9 n 100 n1 10
9 100
2 1 2 9 1 2 32
n1
1
1 9 102 9
46. Integrate the series and multiply by 1.
44. Replace x with x2. 48. (a) From Exercise 47, we have
1 120 1 120 arctan arctan arctan 119 239 119 239
arctan
arctan
1 239 28,561 arctan arctan 1 1120 119 120 1191 239
28,561 4
10 1 1 21 5 5 1 arctan arctan arctan arctan arctan 5 5 5 24 12 1 1 52
(b) 2 arctan
4 arctan
120 1 1 5 5 25 12 1 2 arctan 2 arctan arctan arctan arctan arctan 5 5 5 12 12 119 1 5 122
4 arctan
1 1 120 1 arctan arctan arctan (see part (a).) 5 239 119 239 4
1 1 1 2 1 3 5 6 arctan arctan arctan 2 3 1 1 21 3 5 6 4
50. (a) arctan
(b) 4 arctan
1 1 arctan 2 3
12 1 23
3
4
1 25 1 27 1 1 33 1 35 1 37 4 5 7 3 3 5 7
40.4635 40.3217 3.14
54. From Example 5, we have arctan x
52. From Exercise 51, we have
1
n1
n1
13 1 ln 34 0.2877.
56. From Exercise 54, we have
1
n1
n1
1 1 1n 2n1 32n12n 1 n0 3 2n 1
1
n0
arctan
n
n0
1n11 3n 1 3n n n1 n
ln
1
n
1 32n1 2n 1
1 0.3218. 3
1
n0
n
1 1 1n 2n 1 n0 2n 1
arctan 1
2n1
0.7854 4
x 2n1 . 2n 1
408
Chapter 8
Infinite Series
58. From Example 5, we have arctan x
1n
n0
1n
x 2n1 . 2n 1
3 1n 2n 2n 1 3
3 2n 1 3
n0
n
n0
1n 13 2n1 2n 1 n0
3
3 arctan
6 2 3
3
Section 8.10
1 3
Taylor and Maclaurin Series
2. For c 0, we have f x e3x f nx 3ne 3x ⇒ f n0 3n e 3x 1 3x
3xn 9x2 27x3 . . . 2! 3! n0 n!
4. For c 4, we have: f x sin x
f
4
2
fx cos x
f
4
2
f x sin x
f
4 22
f x cos x
f
4 22
f 4
4
f 4x sin x
2 2
2
2
and so on. Therefore we have: sin x
f n4 x 4 n n! n0
2
2 2
2
1 x 4
1
n0
x 4 2 x 4 3 x 4 4 . . . 2! 3! 4!
x 4 n1 1 n 1!
nn12
6. For c 1, we have: f x ex f nx ex ⇒ f n1 e ex
x 1n f n1x 1n x 12 x 13 x 14 . . . e 1 x 1 e n! 2! 3! 4! n! n0 n0
Section 8.10 8. For c 0, we have: f x lnx 2 1
f 0 0
fx
2x x2 1
f0 0
f x
2 2x 2 x2 12
f 0 2
f x
4xx 2 3 x2 13
f 0 0
f 4x
12x 4 6x 2 1 x 2 14
f 40 12
f 5x
48xx 4 10x 2 5 x 2 15
f 50 0
f 6x
2405x6 15x 4 15x 2 1 x 2 16
f 60 240
and so on. Therefore, we have: lnx 2 1
f n0xn 2x 2 0x3 12x 4 0x5 240x6 . . . 0 0x n! 2! 3! 4! 5! 6! n0
x2
1n x 2n2 x 4 x6 . . . 2 3 n1 n0
10. For c 0, we have; f x tanx
f 0 0
fx sec2x
f0 1
f x 2 sec2xtanx
f 0 0
f x 2 sec4x 2 sec2xtan2x
f 0 2
f 4x 8 sec4xtanx sec2xtan3x
f 40 0
f 5x 8 2 sec6x 11 sec4xtan2x 2 sec2xtan4x
f 50 16
tanx
f n0xn 2x3 16x5 . . . x3 2 x x x5 . . . n! 3! 5! 3 15 n0
12. The Maclaurin Series for f x e2x is
2xn . n! n0
f n1x 2n1e2x. Hence, by Taylor’s Theorem,
0 ≤ Rnx Since lim
n→
f n1z n1 2n1e2z n1 x x . n 1! n 1!
2n1xn1 2xn1 lim 0, it follows that Rnx → 0 as n → . n→ n 1! n 1!
Hence, the Maclaurin Series for e2x converges to e2x for all x.
Taylor and Maclaurin Series
409
410
Chapter 8
Infinite Series kk 1x2 kk 1k 2x3 . . . , we have 2! 3! 1 1232x 2 123252x3 . . . 1 x 2 2! 3!
14. Since 1 xk 1 kx 12
1 x
1
13x 2 135x3 . . . x 2 222! 233!
1
35.
1
n1
kk 1x2 kk 1k 2x3 . . . , we have 2! 3!
16. Since 1 xk 1 kx
1 x13 1
x 13x 1323 2!
2
x 2x2 2 5x3 2 5 8x4 . . . 2 3 3 3 2! 3 3! 344!
1
1n12 x 3 n2
58.
1n1 1 x 2 n2
we have 1 x312 1
ex
132353x3 . . . 3!
1
18. Since 1 x12 1
20.
. . 2n 1xn 2n n!
xn
35.
x3
x4
.
. . 2n 3xn
2n n!
1n1 1 x3 2 n2
x2
. . 3n 4
3nn!
35.
(Exercise 14)
. . 2n 3x3n
2n n!
.
x5
n! 1 x 2! 3! 4! 5! . . .
n0
e3x
22.
1n 3n xn 3xn 9x 2 27x3 81x 4 243x5 . . . 1 3x n! n! 2! 3! 4! 5! n0 n0
cos x cos 4x
1nx2n x2 x4 x6 1 . . . 2n ! 2! 4! 6! n0
1n42nx2n 1n4x2n 2n! 2n! n0 n0
1
26.
16x2 256x4 . . . 2! 4!
ex 1 x
x2 x3 . . . 2! 3!
ex 1 x
x2 x3 . . . 2! 3!
ex ex 2
2x2 2x4 . . . 2! 4!
2 cos hx ex ex
x2n
2 2n!
n0
24.
sin x
1nx2n1 n0 2n 1!
1nx32n1 n0 2n 1!
2 sin x3 2
x9 x15 . . . 3! 5!
2x3
2x9 2x15 . . . 3! 5!
2 x3
Section 8.10
28. The formula for the binomial series gives 1 x12 1
1 x 212 1 ln x x 2 1
x
x
411
1n 1 3 5 . . . 2n 1xn , which implies that 2n n! n1
1n 1 3 5 . . . 2n 1x 2n 2n n! n1
1 dx 1
x 2
x
1n1 3 5 . . . 2n 12n1 2n2n 1n! n1
x3 2
3
x2 x4 . . . 2! 4!
30. x cos x x 1
Taylor and Maclaurin Series
1 3x5 1 3 5x7 . . .. 245 2467
32.
arcsin x 2n!x2n1 nn!22n 1 x 2 n0
x3 x5 . . . 2! 4!
1
x
2n!x2n
2 n! 2n 1, x 0
n0
n
2
1nx2n1 2n! n0
2x 2 2x 4 2x6 . . . (See Exercise 33.) 2! 4! 6!
34. eix eix 2
1n x 2n eix eix x2 x 4 x6 1 . . . cosx 2 2! 4! 6! 2n! n0
36. gx ex cos x
8
x2
1x
2
4
4
x x . . . 6 24
1 2 24x . . . x2
−6
4
6
g P4
x2 x2 x3 x3 x4 x4 x4 x3 x 4 . . . 1x . . .1x 2 2 6 2 24 4 24 3 6
−40
38. f x ex ln1 x
1x
x 2 x3 x4 . . . 2 6 24
x x2 x
3
−3
P5 3
−3
2
3
4
5
x2 x3 x3 x3 x4 x4 x4 x4 x5 x5 x5 x5 x5 . . . 2 3 2 2 4 3 4 6 5 4 6 12 24
x 2 x3 3x5 . . . 2 3 40 f
x x2 x3 x4 x5 . . .
412
Chapter 8
40. f x
Infinite Series
ex . Divide the series for ex by 1 x. 1x
x2 x3 2 3 x2 1x 1x 2 1x x2 0 2 x2 2 1
f x 1
3x 4 . . . 8 x3 x4 x5 . . . 6 24 120
x 2 x 3 3x 4 . . . 2 3 8 6
P4
−6
f
6 −6
x3 6 x3 2 x3 x4 3 24 x3 x4 3 3 3x 4 x5 8 120 3x4 3x5 8 8
42. y x
x5 x2 x 4 x3 x 1 x cos x. 2! 4! 2! 4!
44. y x 2 x3 x 4 x 21 x x 2 x 2 Matches (d)
Matches (b)
x
46.
x
1 t3 dt
0
1
0
1n11 t3 2 n2
35.
. . 2n 3t3n
2n n!
1n11 3 5 . . . 2n 3t3n1 t4 8 3n 12n n! n2
x
1n11 3 5 . . . 2n 3x3n1 x4 8 3n 12n n! n2
dt
x
0
1n x 2n1 x3 x5 x7 x . . . , we have 3! 5! 7! n0 2n 1!
48. Since sinx
1n
t
sin1
1 1 x.
1
1
1
x2
x3
x4
2n 1! 1 3! 5! 7! . . . 0.8415.
(4 terms)
n0
50. Since ex
xn
x5
n! 1 x 2! 3! 4! 5! . . . , we have e
1
n0
and
11
1 1 1 1 . . . 2! 3! 4! 5!
1n1 1 1 1 1 1 e1 1 e1 1 . . . 0.6321. (6 terms) e 2! 3! 4! 5! 7! n! n1
52. Since sin x
1n x 2n1 x3 x5 x7 x . . . 2n 1 ! 3! 5! 7! n0
1n x 2n x2 x 4 x6 sin x 1 . . . x 3! 5! 7! n0 2n 1!
we have lim
x→0
1n x 2n sin x lim 1. x→0 n0 2n 1! x
Section 8.10
12
54.
0
arctan x dx x
Since 1
12
0
3
x5 x7 . . . 2 2 5 7
12 0
< 0.0001, we have
x→0
1
1
cos x dx
0.5
1
arctan x 1. x
1 2!x 4!x 6!x 8!x . . . dx x 2x2! 3x4! 4x6! 5x8! . . . 2
0.5
1 210, 600
Since
3
4
2
3
4
5
1
< 0.0001, we have
cos x dx 1 0.5
0.5
1 4
14
x lnx 1 dx
0
x
2
0
1 1 1 1 1 0.52 1 0.53 1 0.54 1 0.55 0.3243. 4 72 2880 201,600
x3 x4 x5 . . . dx 2 3 4
x3 4 x 2 5 x 3 6 x 4 . . . 3
4
5
14
6
0
14 < 0.0001, 15 5
Since
143 144 0.00472. 3 8
14
x lnx 1 dx
0
60. From Exercise 19, we have 1 2
2
ex 2dx 2
1
1 2
2
1
1n x 2n 1 dx nn! 2 2 n0
1n x 2n1
2 n!2n 1 n0
2
n
1
1 1n 2n1 1 2 n0 2nn!2n 1
1 2
1 2 17 3 2
31
2
62. f x sin
8191
26
6! 13
2! 5
127
23
3! 7
511
24
4! 9
25
5! 11
x ln1 x 2
4
f −4
8
P5
The polynomial is a reasonable approximation on the interval 0.60, 0.73.
−4
3 x f x arctan x, c 1
3
P5x 0.7854 0.7618x 1 0.3412
x 2! 1 0.0424 x 3! 1
x 14 x 15 1.3025 5.5913 4! 5!
2047
32,767 131,071 524,287 0.1359. 27 7! 15 28 8! 17 29 9! 19
x 2 x3 7x 4 11x5 P5x 2 4 48 96
64.
0.5
0.55
1
6
2
Note: We are using lim
58.
4
1 arctan x 1 1 1 1 dx 0.4872. x 2 3223 5225 7227 9229
0
2
9229
12
56.
1 x3 x5 x7 . . . dx x 3x
Taylor and Maclaurin Series
2
The polynomial is a reasonable approximation on the interval 0.48, 1.75.
3
f P5 −2
4 −1
413
414
Chapter 8
Infinite Series
66. a2n1 0 (odd coefficients are zero)
70. 60, v0 64, k y 3x
1 , g 32 16
32x 2 11632x3 116232x 4 . . . 2642122 3643123 4644124
2264x 2 2
3x 32
2
3x 32
24x 4 23x3 . . . 3 364 16 4644162
2n xn
n64 16 n
n2
72. (a) f x
68. Answers will vary.
3x 32
n2
xn
n32 16
n2
n
n2
lnx 2 1 . x2 1.5
From Exercise 8, you obtain P
(b)
1n x 2n 1 1n x 2n2 2 x n0 n1 n0 n 1
0
x2 x4 x6 x8 P8 1 2 3 4 5
−0.5
lnt2 1 dt t2
x
(c) Fx
2
0
x
Gx
P8t d t
0
x
0.25
0.50
0.75
1.00
1.50
2.00
Fx
0.2475
0.4810
0.6920
0.8776
1.1798
1.4096
Gx
0.2475
0.4810
0.6920
0.8805
5.3064
652.21
(d) The curves are nearly identical for 0 < x < 1. Hence, the integrals nearly agree on that interval. 74. Assume e pq is rational. Let N > q and form the following.
e 11
1 1 1 1 . . . . . . 2! N! N 1! N 2!
Set a N! e 1 1 . . .
1 N!
, a positive integer. But,
N 1 1! N 1 2! . . . N 1 1 N 11 N 2 . . . < N 1 1 N 1 1
a N!
2
1 1 1 1 1 . . . N1 N 1 N 12 N1
1
1 1 N1
. . .
1 , a contradiction. N
Review Exercises for Chapter 8 2. an
n2
n 1
n 4. an 4 : 3.5, 3, . . . 2 Matches (c)
32
6. an 6
Matches (b)
n1
: 6, 4, . . .
Review Exercises for Chapter 8 n 2
8. an sin
10. lim
n→
1 n
0
Converges
2
0
12
−2
The sequence seems to diverge (oscillates). n : 1, 0, 1, 0, 1, 0, . . . 2
sin
n 1 lim n→ lnn n→ 1n
n→
1 2n
n
1 1
k
lim k→
k 12
e12
Converges; k 2n
Diverges 16. Let
14. lim 1
12. lim
y bn cn1n ln y
18. (a) Vn 120,0000.70n, n 1, 2, 3, 4, 5 (b) V5 120,0000.705 $20,168.40
lnbn cn n
lim ln y lim
n→
n→
1 bn ln b cn ln c bn cn
Assume b ≥ c and note that the terms bn ln b cn ln c bn ln b cn ln c n n n n n b c b c b cn converge as n → . Hence an converges. 20. (a)
(b)
k
5
10
15
20
25
Sk
0.3917
0.3228
0.3627
0.3344
0.3564
(c) The series converges by the Alternating Series Test.
1
0
12 0
22. (a)
(b)
k
5
10
15
20
25
Sk
0.8333
0.9091
0.9375
0.9524
0.9615
(c) The series converges, by the limit comparison test with
1
n .
2 2n2 n 4 3 n0 n0 3
n
0
2
12 0
2 26. Diverges. nth Term Test, lim an . n→ 3
24. Diverges. Geometric series, r 1.82 > 1.
28.
1
43 12
See Exercise 27.
30.
3
n0
2
n
2 1 n 1n 2 n0 3
n
n 1 n 2 1
1
n0
1 1 23
1 2 2 3 3 4 . . . 3 1 2 1
1
1
1
1
415
416
Chapter 8
Infinite Series
32. 0.923076 0.9230761 0.000001 0.0000012 . . .
0.9230760.000001
n
n0
39
32,0001.055
34. S
n
0.923076 923,076 1276,923 12 1 0.000001 999,999 1376,923 13
32,0001 1.05540 1 1.055
n0
36. See Exercise 86 in Section 8.2. AP
$4,371,379.65
12r 1 12r
100
12t
1
12
0.065 1 0.065 12
120
1
$16,840.32
38.
1
n 4
n1
3
n1
1 n34
40.
1 2
n1
Divergent p-series, p
42.
n
3 4
< 1
1 1 1 2 n 2n n n1 n1 2
The first series is a convergent p-series and the second series is a convergent geometric series. Therefore, their difference converges.
n1
nn 2
44. Since
n1
n1
1
3
n
converges,
3
n1
n
1 converges by the 5
Limit Comparison Test.
n 1nn 2 n1 lim 1 n→ n→ n 2 1n lim
1
n, the series diverges.
By a limit comparison test with
n1
46.
1nn n1 n 1
an1 lim
n→
48. Converges by the Alternating Series Test.
n 1
n2
≤
n
n1
an1
an
3 lnn 1 3 ln n 3 ln n < an, lim 0 n→ n1 n n
n 0 n1
By the Alternating Series Test, the series converges.
50.
n!
e
n
n1
lim
n→
an1 n 1! lim n→ an en1 lim
n→
en
n!
n1 e
By the Ratio Test, the series diverges. 52.
1 3
5.
. . 2n 1
2 5 8 . . . 3n 1
n1
lim
n→
an1 1 lim n→ 2 an
3. 5.
. . 2n 12n 1 . . 3n 13n 2
By the Ratio Test, the series converges.
2
1
5. 3.
. . 3n 1 2n 1 2 lim . . 2n 1 n→ 3n 2 3
Review Exercises for Chapter 8
417
54. (a) The series converges by the Alternating Series Test. (b)
(c)
x
5
10
15
20
25
Sn
0.0871
0.0669
0.0734
0.0702
0.0721
(d) The sum is approximately 0.0714.
0.3
0
12 0
56. No. Let an
3937.5 3937.5 is a convergent p-series. , then a75 0.7. The series 2 2 n n1 n
4 1
58. f x tan x
f
f x 2 sec2 x tan x
f
f
x 4 sec2 x tan2 x 2 sec4 x
f
4 4
62. e0.25 1 0.25
P6x 1 P10x 1
68.
4 16
2 x 4 4
2
8 x 3 4
66.
x2 x4 2! 4!
−10
x2 x4 x6 2! 4! 6!
lim
n
Geometric series which converges only if 2x < 1 or 1 1 2 < x < 2 .
10
f P6
2x
n0
P4
P10
−10
x2 x4 x6 x8 x10 2! 4! 6! 8! 10!
3nx 2n n n1 n→
3
10
0.252 0.253 0.254 0.779 2! 3! 4!
f x cos x P4x 1
0.752 0.754 0.756 0.7317 2! 4! 6!
4 2
f x sec2 x
P3x 1 2 x
64.
60. cos0.75 1
f
70.
un1 3n1x 2n1 lim n→ un n1
3x2
n
3nx 2n
1 3 Center: 2 R
5
Since the series converges at 3 and diverges at 73 , the interval of convergence is 53 ≤ x < 73 .
x2 x 2n n 2 2 n0 n0
n
Geometric series which converges only if
x2 < 1 or 2
0 < x < 4.
418
Chapter 8
Infinite Series
y
72.
y
3n12n 2x2n1 3n2nx2n1 n 2 n! 2n1n 1! n1 n0
y
3n12n 22n 1x2n 2n1n 1! n0
y 3xy 3y
74.
3nx 2n 2nn! n0
1n13n22n 2x2n2 1n3n1x2n 3n12n 22n 1x2n n1 n1 2 n 1! 2 n 1! 2nn! n0 n0 n0
1n13n2x2n2 1n3n1x2n 1n13n12n 2x 2n n n 2 n! 2 n! 2nn! n0 n0 n0
1n13n2x 2n2 1n3n1x2n 2n 1 1 n 2 n! 2nn! n0 n0
1n13n2x 2n2 1n3n1x2n 2n n 2 n! 2nn! n0 n0
1n3n1x 2n 2n 1n13n1x2n 2n n n1n 1! 2n 2 n! n1 n1 2
1n3n1x2n 2n 2n 0 2nn! n1
3 32 a 32 2 x 1 x2 1 x2 1 r
2 2 x
3
n
n0
76. Integral:
n1
n0
1n3xn n1 n0 2
1 1 x 3 78. 8 2x 3 x 32 x 33 . . . 8 2 8 4 n0
80.
1n3xn1
n 12
n
8 1 x 34
32 32 , 1 < x < 7 4 x 3 1 x
f x cos x f x sin x f x cos x f x sin x cos x
82.
2 f n 4x 4 n 2 2 x x n! 2 2 4 2 2! 4 n0
1nn12x 4 n1 n 1! n1
1 x 2 4
2
f x cscx f x cscx cotx f x csc3x cscx cot2x f x 5 csc3x cotx cscx cot3x f 4x 5 csc5x 15 csc3x cot2x cscx cot4x cscx
1 f n2x 2 n 1 x n! 2! 2 n0
2
5 x 4! 2
4
. . .
2
2
2
3!
x 4
3
2
2
4!
x 4
4
. . .
Review Exercises for Chapter 8 f x x12
84.
1 f x x12 2 f x
32
12 12 32 x
f x
52
f 4x x
12 12 x
12 12 32 52 x
72,
. . .
f n4x 4n n! n0
2
x 4 x 42 1 3x 43 1 3 5x 44 . . . 22 252! 283! 2114!
2
1n11 3 5 . . . 2n 3x 4n x 4 2 2 23n1n! n2
hx 1 x3
86.
h x 31 x4 h x 121 x5 h x 601 x6 h4x 3601 x7 h5x 25201 x8 1nn 2!x n 1nn 2n 1x 12x2 60x3 360x4 2520x5 . . . 1 1 3x 3 1 x 2! 3! 4! 5! 2n! 2 n0 n0
1
ln x
88.
n1
n1
ln
x 1n , n
0 < x ≤ 2 n
e23
n1
n1
1
n1
n1
92.
sin x
sin
n
x2n1 , 2n 1!
n
1 0.3272 32n12n 1!
13 1 n0
1 0.1823 5nn
1
n0
xn
< x <
x2
n! 1 x 2! . . .
94. ex
n0
x n1 x3 x x2 . . . 2! n0 n!
xex
1
xex dx xex ex
0
0
0
e e 0 1 1
x n1 x n2 dx n0 n! n0 n 2n!
1
1
ex
1 0
1
n 2n! 1
n0
xn
n!,
n0
65 1 65n 1
90.
< x <
2n 23n 1.9477 n n! n0 n0 3 n!
419
420
Chapter 8
Infinite Series
f x sin 2x
f 0 0
f x 2 cos 2x
f 0 2
f x 4 sin 2x
f 0 0
f x 8 cos 2x
f 0 8
96. (a)
f 4x 16 sin 2x
f 40 0
f 5x 32 cos 2x
f 50 32
f 6x 64 sin 2x
f 60 0
f 7x 128 cos 2x
f 70 128
4 0x2 8x3 0x4 32x5 0x6 128x7 . . . 4 8 7 . . . 2x x3 x5 x 2! 3! 4! 5! 6! 7! 3 15 315
sin 2x 0 2x (b) sin x sin 2x
1nx2n1 n0 2n 1!
1n2x2n1 2x3 2x5 2x7 . . . 2x 2n 1 ! 3! 5! 7! n0
2x
8x3 32x5 128x7 . . . 4 4 8 7 . . . 2x x3 x5 x 6 120 5040 3 15 315
(c) sin 2x 2 sin x cos x
x5 x7 x3 . . . 6 120 5040
4 x7 2x3 2x5 4 4 8 7 . . . . . . 2x x3 x5 x 3 15 315 3 15 315
2 x
2 x 2 x
cos t
98.
cos
x
cos
0
t
t
2
2
dt
4
1n t 2n 2n! n0
et
100.
2
et 1
1 2n!n 1
2n
2
et
x
0
x
1nx n1 2n!n 1
2n
x3 2
3
tn
tn
n!
n1 n t n1
2
n!
n0
n n
0
1 3x5 1 3 5x7 . . . 2 4 5 2 4 6 7
x2 1 3x4 1 3 5x6 arcsin x 1 . . . 2 3 2 4 5 2 4 6 7 x lim
1 t 2n 2n! n0
arcsin x x
x→0
6
x5 x5 x5 x7 x7 x7 x3 x3 x7 . . . 2 6 24 12 120 720 144 240 5040
n0
102.
2
n0
x . . . 1 x2 24x 720
arcsin x 1 x
1 arcsin x 1 x2 lim 1. By L’Hôpital’s Rule, lim x→0 x→0 x 1
tn1 1 t n1 n!
et 1 dt t
tn
x
n n! n1
0
xn
n n!
n1
414
Chapter 8
Infinite Series
66. a2n1 0 (odd coefficients are zero)
70. 60, v0 64, k y 3x
1 , g 32 16
32x 2 11632x3 116232x 4 . . . 2642122 3643123 4644124
2264x 2 2
3x 32
2
3x 32
24x 4 23x3 . . . 3 364 16 4644162
2n xn
n64 16 n
n2
72. (a) f x
68. Answers will vary.
3x 32
n2
xn
n32 16
n2
n
n2
lnx 2 1 . x2 1.5
From Exercise 8, you obtain P
(b)
1n x 2n 1 1n x 2n2 2 x n0 n1 n0 n 1
0
x2 x4 x6 x8 P8 1 2 3 4 5
−0.5
lnt2 1 dt t2
x
(c) Fx
2
0
x
Gx
P8t d t
0
x
0.25
0.50
0.75
1.00
1.50
2.00
Fx
0.2475
0.4810
0.6920
0.8776
1.1798
1.4096
Gx
0.2475
0.4810
0.6920
0.8805
5.3064
652.21
(d) The curves are nearly identical for 0 < x < 1. Hence, the integrals nearly agree on that interval. 74. Assume e pq is rational. Let N > q and form the following.
e 11
1 1 1 1 . . . . . . 2! N! N 1! N 2!
Set a N! e 1 1 . . .
1 N!
, a positive integer. But,
N 1 1! N 1 2! . . . N 1 1 N 11 N 2 . . . < N 1 1 N 1 1
a N!
2
1 1 1 1 1 . . . N1 N 1 N 12 N1
1
1 1 N1
. . .
1 , a contradiction. N
Review Exercises for Chapter 8 2. an
n2
n 1
n 4. an 4 : 3.5, 3, . . . 2 Matches (c)
32
6. an 6
Matches (b)
n1
: 6, 4, . . .
Review Exercises for Chapter 8 n 2
8. an sin
10. lim
n→
1 n
0
Converges
2
0
12
−2
The sequence seems to diverge (oscillates). n : 1, 0, 1, 0, 1, 0, . . . 2
sin
n 1 lim n→ lnn n→ 1n
n→
1 2n
n
1 1
k
lim k→
k 12
e12
Converges; k 2n
Diverges 16. Let
14. lim 1
12. lim
y bn cn1n ln y
18. (a) Vn 120,0000.70n, n 1, 2, 3, 4, 5 (b) V5 120,0000.705 $20,168.40
lnbn cn n
lim ln y lim
n→
n→
1 bn ln b cn ln c bn cn
Assume b ≥ c and note that the terms bn ln b cn ln c bn ln b cn ln c n n n n n b c b c b cn converge as n → . Hence an converges. 20. (a)
(b)
k
5
10
15
20
25
Sk
0.3917
0.3228
0.3627
0.3344
0.3564
(c) The series converges by the Alternating Series Test.
1
0
12 0
22. (a)
(b)
k
5
10
15
20
25
Sk
0.8333
0.9091
0.9375
0.9524
0.9615
(c) The series converges, by the limit comparison test with
1
n .
2 2n2 n 4 3 n0 n0 3
n
0
2
12 0
2 26. Diverges. nth Term Test, lim an . n→ 3
24. Diverges. Geometric series, r 1.82 > 1.
28.
1
43 12
See Exercise 27.
30.
3
n0
2
n
2 1 n 1n 2 n0 3
n
n 1 n 2 1
1
n0
1 1 23
1 2 2 3 3 4 . . . 3 1 2 1
1
1
1
1
415
416
Chapter 8
Infinite Series
32. 0.923076 0.9230761 0.000001 0.0000012 . . .
0.9230760.000001
n
n0
39
32,0001.055
34. S
n
0.923076 923,076 1276,923 12 1 0.000001 999,999 1376,923 13
32,0001 1.05540 1 1.055
n0
36. See Exercise 86 in Section 8.2. AP
$4,371,379.65
12r 1 12r
100
12t
1
12
0.065 1 0.065 12
120
1
$16,840.32
38.
1
n 4
n1
3
n1
1 n34
40.
1 2
n1
Divergent p-series, p
42.
n
3 4
< 1
1 1 1 2 n 2n n n1 n1 2
The first series is a convergent p-series and the second series is a convergent geometric series. Therefore, their difference converges.
n1
nn 2
44. Since
n1
n1
1
3
n
converges,
3
n1
n
1 converges by the 5
Limit Comparison Test.
n 1nn 2 n1 lim 1 n→ n→ n 2 1n lim
1
n, the series diverges.
By a limit comparison test with
n1
46.
1nn n1 n 1
an1 lim
n→
48. Converges by the Alternating Series Test.
n 1
n2
≤
n
n1
an1
an
3 lnn 1 3 ln n 3 ln n < an, lim 0 n→ n1 n n
n 0 n1
By the Alternating Series Test, the series converges.
50.
n!
e
n
n1
lim
n→
an1 n 1! lim n→ an en1 lim
n→
en
n!
n1 e
By the Ratio Test, the series diverges. 52.
1 3
5.
. . 2n 1
2 5 8 . . . 3n 1
n1
lim
n→
an1 1 lim n→ 2 an
3. 5.
. . 2n 12n 1 . . 3n 13n 2
By the Ratio Test, the series converges.
2
1
5. 3.
. . 3n 1 2n 1 2 lim . . 2n 1 n→ 3n 2 3
Review Exercises for Chapter 8
417
54. (a) The series converges by the Alternating Series Test. (b)
(c)
x
5
10
15
20
25
Sn
0.0871
0.0669
0.0734
0.0702
0.0721
(d) The sum is approximately 0.0714.
0.3
0
12 0
56. No. Let an
3937.5 3937.5 is a convergent p-series. , then a75 0.7. The series 2 2 n n1 n
4 1
58. f x tan x
f
f x 2 sec2 x tan x
f
f
x 4 sec2 x tan2 x 2 sec4 x
f
4 4
62. e0.25 1 0.25
P6x 1 P10x 1
68.
4 16
2 x 4 4
2
8 x 3 4
66.
x2 x4 2! 4!
−10
x2 x4 x6 2! 4! 6!
lim
n
Geometric series which converges only if 2x < 1 or 1 1 2 < x < 2 .
10
f P6
2x
n0
P4
P10
−10
x2 x4 x6 x8 x10 2! 4! 6! 8! 10!
3nx 2n n n1 n→
3
10
0.252 0.253 0.254 0.779 2! 3! 4!
f x cos x P4x 1
0.752 0.754 0.756 0.7317 2! 4! 6!
4 2
f x sec2 x
P3x 1 2 x
64.
60. cos0.75 1
f
70.
un1 3n1x 2n1 lim n→ un n1
3x2
n
3nx 2n
1 3 Center: 2 R
5
Since the series converges at 3 and diverges at 73 , the interval of convergence is 53 ≤ x < 73 .
x2 x 2n n 2 2 n0 n0
n
Geometric series which converges only if
x2 < 1 or 2
0 < x < 4.
418
Chapter 8
Infinite Series
y
72.
y
3n12n 2x2n1 3n2nx2n1 n 2 n! 2n1n 1! n1 n0
y
3n12n 22n 1x2n 2n1n 1! n0
y 3xy 3y
74.
3nx 2n 2nn! n0
1n13n22n 2x2n2 1n3n1x2n 3n12n 22n 1x2n n1 n1 2 n 1! 2 n 1! 2nn! n0 n0 n0
1n13n2x2n2 1n3n1x2n 1n13n12n 2x 2n n n 2 n! 2 n! 2nn! n0 n0 n0
1n13n2x 2n2 1n3n1x2n 2n 1 1 n 2 n! 2nn! n0 n0
1n13n2x 2n2 1n3n1x2n 2n n 2 n! 2nn! n0 n0
1n3n1x 2n 2n 1n13n1x2n 2n n n1n 1! 2n 2 n! n1 n1 2
1n3n1x2n 2n 2n 0 2nn! n1
3 32 a 32 2 x 1 x2 1 x2 1 r
2 2 x
3
n
n0
76. Integral:
n1
n0
1n3xn n1 n0 2
1 1 x 3 78. 8 2x 3 x 32 x 33 . . . 8 2 8 4 n0
80.
1n3xn1
n 12
n
8 1 x 34
32 32 , 1 < x < 7 4 x 3 1 x
f x cos x f x sin x f x cos x f x sin x cos x
82.
2 f n 4x 4 n 2 2 x x n! 2 2 4 2 2! 4 n0
1nn12x 4 n1 n 1! n1
1 x 2 4
2
f x cscx f x cscx cotx f x csc3x cscx cot2x f x 5 csc3x cotx cscx cot3x f 4x 5 csc5x 15 csc3x cot2x cscx cot4x cscx
1 f n2x 2 n 1 x n! 2! 2 n0
2
5 x 4! 2
4
. . .
2
2
2
3!
x 4
3
2
2
4!
x 4
4
. . .
Review Exercises for Chapter 8 f x x12
84.
1 f x x12 2 f x
32
12 12 32 x
f x
52
f 4x x
12 12 x
12 12 32 52 x
72,
. . .
f n4x 4n n! n0
2
x 4 x 42 1 3x 43 1 3 5x 44 . . . 22 252! 283! 2114!
2
1n11 3 5 . . . 2n 3x 4n x 4 2 2 23n1n! n2
hx 1 x3
86.
h x 31 x4 h x 121 x5 h x 601 x6 h4x 3601 x7 h5x 25201 x8 1nn 2!x n 1nn 2n 1x 12x2 60x3 360x4 2520x5 . . . 1 1 3x 3 1 x 2! 3! 4! 5! 2n! 2 n0 n0
1
ln x
88.
n1
n1
ln
x 1n , n
0 < x ≤ 2 n
e23
n1
n1
1
n1
n1
92.
sin x
sin
n
x2n1 , 2n 1!
n
1 0.3272 32n12n 1!
13 1 n0
1 0.1823 5nn
1
n0
xn
< x <
x2
n! 1 x 2! . . .
94. ex
n0
x n1 x3 x x2 . . . 2! n0 n!
xex
1
xex dx xex ex
0
0
0
e e 0 1 1
x n1 x n2 dx n0 n! n0 n 2n!
1
1
ex
1 0
1
n 2n! 1
n0
xn
n!,
n0
65 1 65n 1
90.
< x <
2n 23n 1.9477 n n! n0 n0 3 n!
419
420
Chapter 8
Infinite Series
f x sin 2x
f 0 0
f x 2 cos 2x
f 0 2
f x 4 sin 2x
f 0 0
f x 8 cos 2x
f 0 8
96. (a)
f 4x 16 sin 2x
f 40 0
f 5x 32 cos 2x
f 50 32
f 6x 64 sin 2x
f 60 0
f 7x 128 cos 2x
f 70 128
4 0x2 8x3 0x4 32x5 0x6 128x7 . . . 4 8 7 . . . 2x x3 x5 x 2! 3! 4! 5! 6! 7! 3 15 315
sin 2x 0 2x (b) sin x sin 2x
1nx2n1 n0 2n 1!
1n2x2n1 2x3 2x5 2x7 . . . 2x 2n 1 ! 3! 5! 7! n0
2x
8x3 32x5 128x7 . . . 4 4 8 7 . . . 2x x3 x5 x 6 120 5040 3 15 315
(c) sin 2x 2 sin x cos x
x5 x7 x3 . . . 6 120 5040
4 x7 2x3 2x5 4 4 8 7 . . . . . . 2x x3 x5 x 3 15 315 3 15 315
2 x
2 x 2 x
cos t
98.
cos
x
cos
0
t
t
2
2
dt
4
1n t 2n 2n! n0
et
100.
2
et 1
1 2n!n 1
2n
2
et
x
0
x
1nx n1 2n!n 1
2n
x3 2
3
tn
tn
n!
n1 n t n1
2
n!
n0
n n
0
1 3x5 1 3 5x7 . . . 2 4 5 2 4 6 7
x2 1 3x4 1 3 5x6 arcsin x 1 . . . 2 3 2 4 5 2 4 6 7 x lim
1 t 2n 2n! n0
arcsin x x
x→0
6
x5 x5 x5 x7 x7 x7 x3 x3 x7 . . . 2 6 24 12 120 720 144 240 5040
n0
102.
2
n0
x . . . 1 x2 24x 720
arcsin x 1 x
1 arcsin x 1 x2 lim 1. By L’Hôpital’s Rule, lim x→0 x→0 x 1
tn1 1 t n1 n!
et 1 dt t
tn
x
n n! n1
0
xn
n n!
n1
Problem Solving for Chapter 8
Problem Solving for Chapter 8 2. Let S
1
2n 1
2
n1
Then
1 1 1 2 2. . . 2 1 3 5
1 1 1 1 2 2 2 2 2. . . 6 1 2 3 4 S
1 1 . . . 22 42
S
1 1 1 1 2 2. . . 22 2 3
S
1 2 . 22 6
Thus, S
2 1 2 2 3 2 . 6 4 6 6 4 8
4. (a) Position the three blocks as indicated in the figure. The bottom block extends 1 6 over the edge of the table, the middle block extends 1 4 over the edge of the bottom block, and the top block extends 1 2 over the edge of the middle block. 1 6
The centers of gravity are located at bottom block:
1 1 1 6 2 3
middle block:
1 1 1 1 6 4 2 12
top block:
0 1 5 6 12
5 1 1 1 1 . 6 4 2 2 12
The center of gravity of the top 2 blocks is
121 125 2 61, which lies over the bottom block. The center of gravity of the 3 blocks is
31 121 125 3 0 which lies over the table. Hence, the far edge of the top block lies 1 1 1 11 6 4 2 12 beyond the edge of the table. n
(b) Yes. If there are n blocks, then the edge of the top block lies
1
2i from the
c1
edge of the table. Using 4 blocks, 4
1
1
1
1
1
25
2i 2 4 6 8 24
c1
which shows that the top block extends beyond the table. (c) The blocks can extend any distance beyond the table because the series diverges:
1
1
2i 2
c1
c1
1 . i
1 2 1 4
11 12
421
422
Chapter 8
6. a
Infinite Series
1n1a b a b b a b . . . 2 3 4 2n n1
If a b,
1n1 1n12a converges conditionally. a 2n n n1 n1
If a b,
ab 1n1a b diverges. 2n n1 n1 2n
No values of a and b give absolute convergence. a b implies conditional convergence. ex 1 x
8.
x2 . . . 2!
ex 1 x2 2
x4 . . . x12 . . . 2! 6!
f 120 12! 1 ⇒ f 120 665,280 12! 6! 6!
10. (a) If p 1,
2
If p > 1,
2
1 dx ln ln x x ln x
diverges.
2
1 ln b1p ln 21p converges. dx lim p b→ xln x 1p 1p
If p < 1, diverges. (b)
n4
1 1 1 diverges by part (a). n lnn2 2n4 n ln n
12. Let bn an r n.
bn
1 n
14. (a)
an
r n 1 n
r → Lr as n → .
an
1 n
1 1 0.01n 0.99 1 0.01 n0
1 0.01 0.012 . . .
1 Lr < r 1. r
1.010101 . . .
By the Root Test,
b
n
converges ⇒
a
n
rn
converges.
(b)
1 1 0.02n 0.98 1 0.02 n0
1 0.02 0.022 . . . 1 0.02 0.0004 . . . 1.0204081632 . . .
1
16. (a) Height 2 1 2
1
n
n1
2
(b) S 4 1
1 2
1 3
. . .
p-series, p 21 < 1
1 1 1 . . . 4 2 3 n1 n
4 1 1 (c) W 1 3 2 3 2 . . . 3 2 3 4 1 converges. 3 n1 n3 2
C H A P T E R Infinite Series
8
Section 8.1
Sequences . . . . . . . . . . . . . . . . . . . . . 121
Section 8.2
Series and Convergence . . . . . . . . . . . . . . 126
Section 8.3
The Integral Test and p-Series
Section 8.4
Comparisons of Series
Section 8.5
Alternating Series . . . . . . . . . . . . . . . . . 138
Section 8.6
The Ratio and Root Tests . . . . . . . . . . . . . 142
Section 8.7
Taylor Polynomials and Approximations . . . . . 147
Section 8.8
Power Series . . . . . . . . . . . . . . . . . . . . 152
Section 8.9
Representation of Functions by Power Series
. . . . . . . . . . 131
. . . . . . . . . . . . . . 135
Section 8.10 Taylor and Maclaurin Series
. . 157
. . . . . . . . . . . 160
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 167 Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . 172
C H A P T E R Infinite Series Section 8.1
8
Sequences
Solutions to Odd-Numbered Exercises 1. an 2n a1 21 2 a2 22 4 a3 23 8 a4 2 16 4
a5 2 32 5
21
3. an
1
21
2
3
1 8
4
1 16
21
5
a2
1 a3 2
a5
1nn12 n2
a1
11 1 12
1 13 a2 22 4
16 1 a3 32 9 1 110 a4 42 16 1 115 a5 52 25
13. a1 3, ak1 2ak 1 a2 2a1 1 23 1 4 a3 2a2 1 24 1 6 a4 2a3 1 26 1 10 a5 2a4 1
5. an sin
21
a1
1 a4 2
7. an
n
9. an 5
1 2
a1 sin
a3 sin
a3 5 a4 5 a5 5
3 1 2
a4 sin 2 0 a5 sin
5 1 2
1 32
1 1 n n2
11. an
a1 5 1 1 5 a2 5
1 2
a2 sin 0
1 4
n 2
1 1 19 2 4 4 1 1 43 3 9 9 1 1 77 4 16 16 1 1 121 5 25 25
3n n!
a1
3 3 1!
a2
32 9 2! 2
a3
33 27 3! 6
a4
34 81 4! 24
a5
35 243 5! 120
1 15. a1 32, ak1 ak 2 1 1 a2 a1 32 16 2 2 1 1 a3 a2 16 8 2 2 1 1 a4 a3 8 4 2 2 1 1 a5 a4 4 2 2 2
210 1 18
121
122
Chapter 8
Infinite Series 19. This sequence decreases and a1 4, a2 40.5 2. Matches (c).
17. Because a1 81 1 4 and a2 82 1 83 , the sequence matches graph (d). 21.
23.
8
25.
18
−1 −1
12
3
−1
12
12 −1
− 10
2 an n, n 1, . . . , 10 3
an 160.5
29. an
27. an 3n 1 a5 35 1 14 a6 36 1 17 Add 3 to preceeding term.
−1
n1,
n 1, . . . , 10
an
3 2n1
31.
2n , n 1, 2, . . . , 10 n1
10! 8!910 8! 8! 910 90
an
3 3 24 16
a6
3 3 25 32
Multiply the preceeding term by 12.
33.
n 1! n!n 1 n! n! n1
35.
2n 1! 2n 1! 2n 1! 2n 1!2n2n 1
5n2 5 2
37. lim
39. lim
n→
n2
n→
2n n2 1
lim
n→
43.
2
41. lim sin
1 1n2
n→
2
−1 −1
1n 0
2 2 1 45.
3
1 2n2n 1
12
12 −1
−2
The graph seems to indicate that the sequence converges to 1. Analytically, lim an lim
n→
n→
n1 x1 lim lim 1 1. x→ x→ n x
The graph seems to indicate that the sequence diverges. Analytically, the sequence is
an 0, 1, 0, 1, 0, 1, . . . . Hence, lim an does not exist. n→
47. lim 1n n→
n n 1
does not exist (oscillates between 1 and 1), diverges.
51. lim
n→
1 1n 0, converges n
49. lim
n→
3n2 n 4 3 , converges 2n2 1 2
lnn3 3 lnn lim n→ 2n n→ 2 n
53. lim
lim
n→
(L’Hôpital’s Rule)
3 1 0, converges 2 n
Section 8.1
34
55. lim
n→
59. lim
n→
n
0, converges
57. lim
n→
n n1 n 12 n2 lim n→ n n1 nn 1 n→
k n
63. an 1
61. lim
1 2n 0, converges n2 n
n
lim 1
n→
65. lim
n→
k n
n
n 1! lim n 1 , diverges n→ n!
np 0, converges n→ en p > 0, n ≥ 2
lim
Sequences
sin n 1 lim sin n 0, converges n→ n n
lim 1 u1u k ek u→0
k where u , converges n 69. an n2 2
67. an 3n 2
73. an
1n1 2n2
79. an
1
83.
75. an 1
1n1 1n1 2nn! . . 2n 1 2n!
35.
n ? n1 ≥ n1 2 2n2 2 ? n2 n3 2 n ≥ 2 n 1 ? 2n ≥ n 1 n ≥ 1 n ≥ 1
Hence,
2n ≥ n 1
1 n1 n n
71. an
n1 n2
77. an
n n 1n 2
1 1 < 4 an1, n n1 monotonic; an < 4 bounded.
81. an 4
85. an 1n
1n
a1 1 a2
1 2
a3
1 3
Not monotonic; an ≤ 1, bounded
2n3n ≥ 2n2n 1 n n1 ≥ n1 2 2n2 2 an ≥ an1
True; monotonic; an ≤ 18 , bounded 87. an 23 > n
23 n1 an1
Monotonic; an ≤ 23 , bounded
89. an sin
n6
a1 0.500 a2 0.8660 a3 1.000 a4 0.8660
Not monotonic; an ≤ 1, bounded
123
124
Chapter 8
Infinite Series
91. (a) an 5
5
1 n
1 ≤ 6 ⇒ an bounded n
an 5
1 1 1 n 3 3
1 1 > 5 n n1
n1
0.4
−1
12 −1
12 −0.1
lim 5
n→
< 131 3 1
Therefore, an converges. (b)
−1
1 ⇒ an bounded 3
an1 ⇒ an monotonic
Therefore, an converges. 7
<
1 1 1 n 3 3
an
an1 ⇒ an monotonic
(b)
1 1 1 n 3 3
93. (a) an
95. An P 1
1 5 n
r 12
lim
n→
1
n
n
97. (a) A sequence is a function whose domain is the set of positive integers.
(a) lim An , divergent. The amount will grow n→ arbitrarily large over time.
(b) An 9000 1
0.115 12
n
A1 $9086.25
A6 $9530.06
A2 $9173.33
A7 $9621.39
A3 $9261.24
A8 $9713.59
A4 $9349.99
A9 $9806.68
A5 $9439.60
A10 $9900.66
99. an 10
31 31 31
1 n
103. (a) An 0.8n 2.5 billion
(b) A sequence converges if it has a limit. (c) A bounded monotonic sequence is a sequence that has nondecreasing or nonincreasing terms, and an upper and lower bound.
101. an
3n 4n 1
105. (a) an 3.7262n2 75.9167n 684.25
(b) A1 $2 billion
1500
A2 $1.6 billion A3 $1.28 billion A4 $1.024 billion (c) lim 0.8n 2.5 0 n→
−1
8 0
(b) For 2004, n 14 and a14 1017, or $1017.
Section 8.1
107. an
Sequences
n n n1n 109. an
10n n!
a1 111 1
109 (a) a9 a10 9!
a2 2 1.4142
1,000,000,000 362,880
1,562,500 567
3 a3 3 1.4422 4 a4 4 1.4142 5 a5 5 1.3797 6 a6 6 1.3480
(b) Decreasing (c) Factorials increase more rapidly than exponentials.
Let y lim n1n. n→
ln y lim
n→
lim
n→
1n ln n ln n 1n lim 0 n→ 1 n
Since ln y 0, we have y e0 1. Therefore, n lim n 1. n→
111. an2 an an1 (a) a1 1
a7 8 5 13
a2 1
a8 13 8 21
a3 1 1 2
a9 21 13 34
a4 2 1 3
a10 34 21 55
a5 3 2 5
a11 55 34 89
a6 5 3 8
a12 89 55 144
(b) bn b1 b2 b3 b4 b5
an1 ,n ≥ 1 an 1 b6 1 1 2 b7 2 1 3 b8 2 5 b9 3 8 b10 5
(c) 1
1 1 1 bn1 anan1 1
an1 an
an an1 an1 bn an an
(d) If lim bn , then lim 1 n→
n→
1 . bn1
Since lim bn lim bn1 we have, n→
13 8 21 13 34 21 55 34 89 55
113. True
1 1 .
n→
1 2 0 2 1
1 ± 1 4 1 ± 5 2 2
Since an, and thus bn, is positive,
1 52 1.6180. 115. True
117. a1 2 1.4142 a2 2 2 1.8478 a3 2 2 2 1.9616 a4
2 2 2 2 1.9904
a5
2 2 2 2 2 1.9976
an is increasing and bounded by 2, and hence converges to L. Letting lim an L implies that 2 L L ⇒ L 2. n→ Hence, lim an 2. n→
125
126
Chapter 8
Infinite Series
Section 8.2
Series and Convergence
1. S1 1
3. S1 3
S2 1 14 1.2500
S2 3 92 1.5
S3 1 14 19 1.3611
S3 3 92 27 4 5.25
1 S4 1 14 19 16 1.4236
81 S4 3 92 27 4 8 4.875
1 1 1 1 S5 1 4 9 16 25 1.4636
9 27 81 243 S5 3 2 4 8 16 10.3125
5. S1 3 S2 3 32 4.5 S3 3 32 34 5.250 S4 3 32 34 38 5.625 3 S5 3 32 34 38 16 5.8125
7.
3 2 3
n
9.
Geometric series
n
11.
Geometric series
r 1.055 > 1
3 > 1 2
lim
n→
Diverges by Theorem 8.6
n
2
n1
lim
n→
n2 1
n2
15.
4 4 9 1
n
n0
Diverges by Theorem 8.9
5
9
15
1
n
n0
21
4 16, S2 4 16 2.95, .
S0
. .
15 1 1 . . . 1 4 4 16
15 45 , S , S 3.05, . . . 4 1 16 2
Matches graph (a).
Analytically, the series is geometric:
Analytically, the series is geometric:
44 9
1
n
n0
94 94 3 1 14 34
15 1 4 n0 4
n
154 154 3 1 14
54
n n 1 n n 1 1 2 2 3 3 4 4 5 . . . 1
n1
1
n1
1
1
1
1
1
1
1
1
1
n1
n n 1
23.
45
4 4
19.
Matches graph (c).
21.
2n 1 n1 n0 2
lim
1 9 1 . . . 1 4 4 16
9 9 S 0 , S1 4 4
n 10 n1
2n 1 1 2n 1 0 lim n1 n→ 2 n→ 2 2
n2 10 1
Diverges by Theorem 8.9
17.
n
Diverges by Theorem 8.9
Diverges by Theorem 8.6
13.
n1
n1
n0
n0
r
1000 1.055
2 4 3
lim Sn lim 1
n→
n→
1 1 n1
n
25.
0.9
n
n0
n0
Geometric series with r 34 < 1.
Geometric series with r 0.9 < 1.
Converges by Theorem 8.6
Converges by Theorem 8.6
Section 8.2
27. (a)
Series and Convergence
n n 3 2 n n 3 6
n1
1
1
n1
1 4 2 5 3 6 4 7 . . . 1
2
2 1 (b)
(c)
1
1
1
1
1
1
1 1 11 3.667 2 3 3
n
5
10
20
50
100
Sn
2.7976
3.1643
3.3936
3.5513
3.6078
5
0
11 0
(d) The terms of the series decrease in magnitude slowly. Thus, the sequence of partial sums approaches the sum slowly.
29. (a)
2 0.9
n1
n1
(b)
2 0.9
n
n0
2 20 1 0.9
(c)
n
5
10
20
50
100
Sn
8.1902
13.0264
17.5685
19.8969
19.9995
22
0
11 0
(d) The terms of the series decrease in magnitude slowly. Thus, the sequence of partial sums approaches the sum slowly.
31. (a)
10 0.25
n1
n1
(b)
10 40 13.3333 1 0.25 3
(c)
n
5
10
20
50
100
Sn
13.3203
13.3333
13.3333
13.3333
13.3333
15
11
0 0
(d) The terms of the series decrease in magnitude rapidly. Thus, the sequence of partial sums approaches the sum rapidly.
33.
n
n2
35.
2
12 12 1 1 1 1 1 n2 n 1 n 1 2 n2 n 1 n 1
1 2
1 3 1 1 2 2 4
n 1 n 2 8 n 1 n 2 8 2 3 3 4 4 5 . . . 82 4 8
2
1
n
10
n0
1
1
1
1
1
1
1
39.
2
1
1
n1
n0
41.
1 31 12 41 13 51 14 61 . . .
n1
37.
1
n
1 2 1 12
1 10 1 110
9
1
n
n0
43.
3 3
n0
1
n
1 2 1 12 3
3 9 1 13 4
127
128
45.
Chapter 8
Infinite Series
1 1 1 2 3 2 n
n0
n
n
n0
1 3
n
47. 0.4
n0
S
3 1 2 2 2
40100 3
1
n
51.
53.
3 40
and r
1 100
n 10
lim
1
1
n 10 1 0 10n 1 10
Diverges by Theorem 8.9
n n 2 1 3 2 4 3 5 4 6 . . . 1
1
1
1
1
1
1
1
n1
55.
3n 1
a 410 4 1 r 1 110 9
10n 1
n→
340 5 a 1 r 99100 66
S
n
n1
n0
Geometric series with a
4 1 10 10 n0
4 1 Geometric series with a 10 and r 10
1 1 1 12 1 13
49. 0.07575
2n 1
57.
4
2
n
4
2 1
1 3 , converges 2 2
n
59.
1.075
n
n1
n0
3n 1 3 0 n→ 2n 1 2
Geometric series with r 12
Geometric series with r 1.075
Converges by Theorem 8.6
Diverges by Theorem 8.6
lim
n0
n0
Diverges by Theorem 8.9
61.
n
ln n
63. See definition, page 567.
n2
lim
n→
n 1 lim ln n n→ 1n
(by L’Hôpital’s Rule) Diverges by Theorem 8.9 65. The series given by
67. (a) x is the common ratio. (b) 1 x x2 . . .
ar n a ar ar 2 . . . ar n . . . , a 0
n0
n0
x
n
1 , x < 1 1x
Geometric series: a 1, r x, x < 1
is a geometric series with ratio r. When 0 < r < 1, the a series converges to . The series diverges if r ≥ 1. 1r
(c) y1
3
1 1x
y2 1 x
−1.5
1.5 −1
11 0.5 0.5 x
69. f x 3
7
y=6
Horizontal asymptote: y 6
3 2 1
n
n0
S
3 6 1 12
−2
10 −1
The horizontal asymptote is the sum of the series. f n is the nth partial sum.
Section 8.2
71.
Series and Convergence
1 < 0.001 n n 1
10,000 < n2 n 0 < n2 n 10,000 n
1 ± 12 4 1 10,000
2
Choosing the positive value for n we have n 99.5012. The first term that is less than 0.001 is n 100.
18
n
< 0.001
10,000 < 8n This inequality is true when n 5. This series converges at a faster rate. n1
73.
8000 0.9 i
i0
8000 1 0.9 n1 1 1 0.9
80,000 1
n1
75.
100 0.75 i
i0
, n > 0
0.9n
100 1 0.75 n1 1 1 0.75
400 1 0.75n million dollars. Sum 400 million dollars
77. D1 16 D2 0.81 16 0.81 16 32 0.81
down
up
D3 16 0.81 2 16 0.81 2 32 0.81 2
D 16 32 0.81 32 0.81 2 . . . 16
32 0.81
n
16
n0
79. P n P 2
1 1 2 2
1 1 n0 2 2
1 1 2 2 n
2
n
2
32 152.42 ft 1 0.81
81. (a)
1
n
n1
n0
1 1
n
1 1 1 2 1 12
(b) No, the series is not geometric.
1 8
(c)
12 1 1 12
83. Present Value
2 2
19
50,000
n1
1 1.06
50,000 1 1.06 n0 1.06
1
n
2
n1
n
18
n 2 n1
0.01 2 i
i0 n
50,000 1 1.0619 1.06 1 1.061
85. w 1 , r 1.06
$557,905.82 The present value is less than $1,000,000. After accruing interest over 20 years, it attains its full value.
0.01 1 2n
0.01 2n 1
12
(a) When n 29: w $5,368,709.11 (b) When n 30: w $10,737,418.23 (c) When n 31: w $21,474,836.47
129
130
Chapter 8
Infinite Series
87. P 50, r 0.03, t 20 (a) A 50 (b) A
89. P 100, r 0.04, t 40
12 0.03 1 0.03 12
12 20
1 $16,415.10
(a) A 100
50 e0.03 20 1
$16,421.83 e0.0312 1
(b) A
91. (a) an 6110.1832 1.0544 x 6110.1832e0.05297n
12 0.04 1 0.04 12
12 40
1 $118,196.13
100 e0.04 40 1
$118,393.43 e0.0412 1
93. x 0.749999 . . . 0.74
0.009 0.1
n
n0 10,000
0.74
0.009 1 0.1
0.74 0.01 0.75 0 6,000
10
(b) 78,530 or $78,530,000,000 9
a
(c) Total
n
78,449 or $78,449,000,000
n0
95. By letting S0 0, we have an
n
a
k
k1
a
n
n1
97. Let
a
n
S
n
Sn1
n1
n
a
k
Sn Sn1. Thus,
k1
Sn1 c c
n1
1 and b
n
n0
S
n1
1 .
c S
c Sn .
n1
n1
99. False. lim
n0
n→
1 1 0, but diverges. n n1 n
Both are divergent series.
a
n
bn
n0
n0
1 1 1 1 0
101. False
ar
n
n1
1 a r a
The formula requires that the geometric series begins with n 0. 103. Let H represent the half-life of the drug. If a patient receives n equal doses of P units each of this drug, administered at equal time interval of length t, the total amount of the drug in the patient’s system at the time the last dose is administered is given by Tn P Pekt Pe2kt . . . Pe n1 kt where k ln 2 H. One time interval after the last dose is administered is given by Tn1 Pekt Pe2kt Pe3kt . . . Penkt. Two time intervals after the last dose is administered is given by Tn2 Pe2kt Pe3kt Pe4kt . . . Pe n1 kt and so on. Since k < 0, Tns→0 as s → , where s is an integer.
Section 8.3
Section 8.3 1.
1
3.
n1
e
1
n
n1
Let f x ex.
1 . x1
f is positive, continuous, and decreasing for x ≥ 1.
f is positive, continuous and decreasing for x ≥ 1.
131
The Integral Test and p-Series
n1
Let f x
The Integral Test and p-Series
1 dx lnx 1 x1
1
ex dx ex
1
1
1 e
Converges by Theorem 8.10
Diverges by Theorem 8.10
5.
n
2
n1
1 1
7.
n1
1 . x2 1 f is positive, continuous, and decreasing for x ≥ 1. Let f x
1
1 dx arctan x x2 1
1
lnn 1 n1
Let f x
lnx 1 x1
f is positive, continuous, and decreasing for x ≥ 2 since
4
fx
Converges by Theorem 8.10
1
1 lnx 1 < 0 for x ≥ 2. x 12
lnx 1 ln2x 1 dx x1 2
1
Diverges by Theorem 8.10
9.
nk1 k c
n
n1
Let f x
11.
1 3
n1
x k1 . xk c
Let f x
ck 1 x k c2
x k2
xk
1 . x3
f is positive, continuous, and decreasing for x ≥ 1.
f is positive, continuous, and decreasing for k ck 1 since x > fx
n
1
< 0
1 1 dx 2 x3 2x
1
1 2
Converges by Theorem 8.10
k for x > ck 1.
1
x k1 1 dx lnx k c xk c k
1
Diverges by Theorem 8.10
13.
1
1
n n
n1
5
n1
Divergent p-series with p
17.
1
n
1 2
n1
1 5
1 Divergent p-series with p 2 < 1
< 1
1
n
n1
15.
1 5
3 2
3 Convergent p-series with p 2 > 1
19.
1
n
n1
1.04
Convergent p-series with p 1.04 > 1
132 21.
Chapter 8
2
n
n1
4
2 2 2 . . . 1 23 4 33 4
3
Infinite Series 23.
2
nn 2 2 2
3 2
2 33 2 . . .
n1
S1 2
S1 2
S2 3.189
S2 2.707
S3 4.067
S3 3.092 Matches (b)
Matches (a) 3 4
Diverges—p-series with p
Converges—p-series with p 3 2 > 1
< 1
25. No. Theorem 8.9 says that if the series converges, then the terms an tend to zero. Some of the series in Exercises 21-24 converge because the terms tend to 0 very rapidly. N
27.
1
1
1
1
1
n1234. . .N
> M
n1
(a)
29.
M
2
4
6
8
N
4
31
227
1674
(b) No. Since the terms are decreasing (approaching zero), more and more terms are required to increase the partial sum by 2.
1
nln n
p
n2
If p 1, then the series diverges by the Integral Test. If p 1,
1 dx xln xp
2
ln xp
2
1 ln xp1 dx x p 1
.
2
Converges for p 1 < 0 or p > 1. 31. Let f be positive, continuous, and decreasing for x ≥ 1 and an f n. Then,
33. Your friend is not correct. The series
a
n
and
n1
f x dx
n10,000
1
is the harmonic series, starting with the 10,000th term, and hence diverges.
either both converge or both diverge (Theorem 8.10). See Example 1, page 578.
35. Since f is positive, continuous, and decreasing for x ≥ 1 and an f n, we have, RN S SN
a
n
0 ≤ RN ≤
n
n1
n1
Also, RN S SN
N
a
an > 0.
nN1
an ≤ aN1
nN1
N1
f x dx.
N
37. S6 1 R6 ≤
6
1.0811 ≤
1 1 1 1 1 1.0811 24 34 44 54 64
1 1 dx 3 x4 3x
1
n
n1
4
6
0.0015
≤ 1.0811 0.0015 1.0826
f x dx ≤
1 1 1 . . . n 10,000 10,001
N
f x dx. Thus,
Section 8.3 1 1 1 1 1 1 1 1 1 1 0.9818 2 5 10 17 26 37 50 65 82 101
39. S10
R10 ≤
1
n
0.9818 ≤
10
arctan 10 0.0997 2
≤ 0.9818 0.0997 1.0815
5
n1
1 dx arctan x x 1 2
10
1 2 3 4 4 9 16 0.4049 e e e e
41. S4
R4 ≤
The Integral Test and p-Series
x2
xe
4
ne
0.4049 ≤
1 2 dx ex 2 n2
4
43. 0 ≤ RN ≤
N
1 1 dx 3 x4 3x
N
1 < 0.001 3N 3
1 < 0.003 N3
5.6 108
N3 > 333.33
≤ 0.4049 5.6 108
n1
N > 6.93 N ≥ 7
45. RN ≤
e5x
N
1 dx e5x 5
N
e5N < 0.001 5
47. RN ≤
N
x2
1 < 0.005 e5N
N
arctan N < 0.001 2
arctan N < 1.5698
e5N > 200
arctan N > 1.5698
5N > ln 200
N > tan 1.5698
ln 200 5
N >
1 dx arctan x 1
N ≥ 1004
N > 1.0597 N ≥ 2
49. (a)
1
n
1.1 .
n2
This is a convergent p-series with p 1.1 > 1.
1
n ln n is a divergent series. Use the Integral Test.
n2
1 dx ln ln x 2 2 x ln x 6 1 1 1 1 1 1 (b) 1.1 21.1 31.1 41.1 51.1 61.1 0.4665 0.2987 0.2176 0.1703 0.1393 n n2
6
1
1
1
1
1
1
n ln n 2 ln 2 3 ln 3 4 ln 4 5 ln 5 6 ln 6 0.7213 0.3034 0.1803 0.1243 0.0930
n2
The terms of the convergent series seem to be larger than those of the divergent series! (c)
1 1 < n1.1 n ln n n ln n < n1.1 ln n < n0.1 This inequality holds when n ≥ 3.5 1015. Or, n > e40. Then ln e40 40 < e400.1 e4 55.
133
134
Chapter 8
Infinite Series
51. (a) Let f x 1 x. f is positive, continuous, and decreasing on 1, .
n
Sn 1 ≤
1
y
1 dx x
1
Sn 1 ≤ ln n
1 2
Hence, Sn ≤ 1 ln n. Similarly,
n1
Sn ≥
1
1 dx lnn 1. x
1
2
3 ... n −1
x n
n+1
Thus, lnn 1 ≤ Sn ≤ 1 ln n. (b) Since lnn 1 ≤ Sn ≤ 1 ln n, we have lnn 1 ln n ≤ Sn ln n ≤ 1. Also, since ln x is an increasing function, lnn 1 ln n > 0 for n ≥ 1. Thus, 0 ≤ Sn ln n ≤ 1 and the sequence an is bounded.
n1
(c) an an1 Sn ln n Sn1 lnn 1
n
1 1 dx ≥ 0 x n1
Thus, an ≥ an1 and the sequence is decreasing. (d) Since the sequence is bounded and monotonic, it converges to a limit, . (e) a100 S100 ln 100 0.5822 (Actually 0.577216.)
53.
1
2n 1
n1
1 . 2x 1
Let f x
f is positive, continuous, and decreasing for x ≥ 1.
1
1 dx ln 2x 1 2x 1
1
Diverges by Theorem 8.10
55.
nn n
1
p-series with p
5 4
1 4
n1
n1
2
n
n0
Geometric series with r 23 Converges by Theorem 8.6
Converges by Theorem 8.11
59.
3
57.
5 4
n 2 1 n n1
1 1 n
61.
n
n1
n 1 lim 10 lim n→ n2 1 n→ 1 1 n2
n→
Diverges by Theorem 8.9
Fails nth Term Test
lim 1
1 n
n
e0
Diverges by Theorem 8.9
63.
1
nln n
3
n2
Let f x
1 . xln x3
f is positive, continuous and decreasing for x ≥ 2.
2
1 dx xln x3
2
ln x3
1 ln x2 dx x 2
Converges by Theorem 8.10. See Exercise 13.
2
1 2ln x2
2
1 2ln 22
Section 8.4
Section 8.4 1. (a)
6
n
n1
32
Comparisons of Series
Comparisons of Series
6 6 . . . S1 6 1 232
an
6 an = 3/2 n
6
6 6 6 3 . . . S1 3 4 232 3 2
n
5
32
n1
an =
4 3
6 6 6 6 . . . S1 4.9 2 0.5 1.5 11.5 24.5
nn
n1
135
6 n n 2 + 0.5
an = 3/26 n +3
2 1
n
(b) The first series is a p-series. It converges p 32 > 1. (c) The magnitude of the terms of the other two series are less than the corresponding terms at the convergent p-series. Hence, the other two series converge. (d) The smaller the magnitude of the terms, the smaller the magnitude of the terms of the sequence of partial sums.
Sn
n
Σ
k=1
12
6
4
2
8
10
6 k 3/2 n
Σ
10
k=1
6 k k 2+ 0.5
8 6 4
n
Σ
2
k=1
6 k 3/2 + 3 n
2
3.
1 1 < 2 n2 1 n
5.
Therefore,
n
2
n1
diverges by comparison with the divergent p-series
Therefore,
1 1
n1
diverges by comparison with the divergent series
1
n 1.
n0
ln n
n1
converges by comparison with the convergent geometric series
1 ln n > . n1 n1
9. For n ≥ 3,
Therefore,
n0
1
n.
n2
1 1 < n 3n 1 3
n
1
n2
2
10
n1
1
3
8
1 1 > for n ≥ 2 n1 n
1 1
n.
7.
6
Therefore,
converges by comparison with the convergent p-series
n1
4
1 n . 3
n1
Note:
1
n 1 diverges by the integral test.
n1
11. For n > 3,
1 1 > . n2 n!
Therefore,
1
n!
n0
converges by comparison with the convergent p-series
1
n.
n1
2
13.
1 1 2 ≤ en en Therefore,
1
e
n0
n2
converges by comparison with the convergent geometric series
e .
n0
1
n
136
Chapter 8
15. lim
n→
Infinite Series
nn2 1 n2 lim 2 1 n→ n 1 1n
17. lim
n→
Therefore,
Therefore,
n 1
n
2
n1
n
diverges by a limit comparison with the divergent p-series
1
n.
n1
n3 nn 2 n2 3n lim 2 1 21. lim n→ n→ n 2n 1n
2n2 1 2 2n 1 2n5 n3 lim 3 5 n→ 3n 2n 1 1n 3
3n5
n→
Therefore,
Therefore,
3n
nn 2
n1
diverges by a limit comparison with the divergent p-series
converges by a limit comparison with the convergent p-series
1 3. n1 n
n1
1nn2 1 n2 lim 1 n→ n→ nn2 1 1n2
nk1nk 1 nk lim k 1 n→ n→ n 1 1n
25. lim
23. lim
Therefore,
Therefore,
2
nk1 k 1
n
1
nn
n1
1
n.
n3
2n2 1 2n 1
5
n1
1
n.
n1
19. lim
1 1
2
n0
diverges by a limit comparison with the divergent p-series
1n2 1 n 1 lim n→ n2 1 1n
1
n1
diverges by a limit comparison with the divergent p-series
converges by a limit comparison with the convergent p-series
1
n.
1 . 2 n n1
n1
sin1n 1n2 cos1n lim n→ n→ 1n 1n2
27. lim
29.
n
n1
n
1
n
n1
Diverges
1 lim cos 1 n→ n
1
p-series with p 2
Therefore,
sinn 1
n1
diverges by a limit comparison with the divergent p-series
1
n.
n1
31.
3
n1
n
1 2
33.
Converges Direct comparison with
n1
1 3
n
2n 3
35.
n
2
n 12
n1
n1
Diverges; nth Term Test
Converges; integral test
n
lim
n→
n 1 0 2n 3 2
Section 8.4 an lim nan by given conditions lim nan is finite n→ 1n n→ n→ and nonzero.
39.
37. lim
Comparisons of Series
1 2 3 4 5 n . . . , 2 2 5 10 17 26 n1 n 1
which diverges since the degree of the numerator is only one less than the degree of the denominator.
Therefore,
a
n
n1
diverges by a limit comparison with the p-series
1
n.
n1
41.
n
3
n1
1 1
5n n 3 lim 5n n 3 51 0 3
43. lim n n→
converges since the degree of the numerator is three less than the degree of the denominator.
1 45. See Theorem 8.12, page 583. One example is 2 1 n n1 converges because
47.
1 1 < and 1 n2
1.0
4
n3 diverges. 3
4
Terms of ∞ Σ an
0.8
n=1
0.6
1
n
n1
n→
Therefore,
n1
n2
4
4
5n
Terms of ∞ 2 Σ an
0.4
2
n=1
0.2
converges ( p-series).
n 4
8
12
16
20
For 0 < an < 1, 0 < an2 < an < 1. Hence, the lower terms are those of an2.
49.
1 1 1 1 . . . , diverges 200 400 600 n1 200n
51.
1 1 1 1 1 , converges 201 204 209 216 n1 200 n2
55. False. Let an 1n3 and bn 1n2. 0 < an ≤ bn and both
53. Some series diverge or converge very slowly. You cannot decide convergence or divergence of a series by comparing the first few terms.
1
n
n1
3
and
n1
57. True
b
n
converges, lim bn 0. There exists N such that bn < 1 for n > N. Thus, n→
n1
anbn < an for n > N and
a
n bn
n1
converges by comparison to the convergent series
a
n.
i1
61.
1
n
2
and
1
n
3
both converge, and hence so does
n n n . 1
1
1
2
3
5
1
n
converge.
59. Since
137
2
138
Chapter 8
Infinite Series
63. (a) Suppose bn converges and an diverges. Then there exists N such that 0 < bn < an for n ≥ N. This means that 1 < an bn for n ≥ N. Therefore, lim an bn 0. Thus, an must also converge. n→
(b) Suppose bn diverges and an converges. Then there exists N such that 0 < an < bn for n ≥ N. This means that 0 < an bn < 1 for n ≥ N. Therefore, lim an bn . Thus, an must also diverge. n→
65. Start with one triangle whose sides have length 9. At the nth step, each side is replaced by four smaller line segments each having 13 the length of the original side.
3 3
#Sides
Length of sides
3
3
9
3
42
9
9
3 4n
9 13
34 3
1 3
1 2 3
n
At the nth step there are 3 4n sides, each of length 9 13 . At the next step, there are 3 4n new triangles of side 9 13 area of an equilateral triangle of side x is 14 3 x 2. Thus, the new triangles each have area n
1 4 3
3
9
n1
The area of the 3
2
3 1
4 32n
. The
.
4n new triangles is
4
3 4n
n1
3 1
32n
33 4 4 9
n .
The total area is the infinite sum 33 4 93 4 4 9 n0
n
93 33 1 93 33 9 183 . 4 4 1 49 4 4 5 5
The perimeter is infinite, since at step n there are 3 4n sides of length 9 13 . Thus, the perimeter at step n is 27 43 → . n
n
Section 8.5 1.
6
n
2
n1
5.
Alternating Series
6 6 6 . . . 1 4 9
3.
10
n2
n1
n
10 10 . . . 2 8
S1 6, S2 7.5
S1 5, S2 6.25
Matches (b)
Matches (c)
1n1 0.7854 4 n1 2n 1
(a)
(b)
n
1
2
3
4
5
6
7
8
9
10
Sn
1
0.6667
0.8667
0.7238
0.8349
0.7440
0.8209
0.7543
0.8131
0.7605
1.1
0
11 0.6
(c) The points alternate sides of the horizontal line that represents the sum of the series. The distance between successive points and the line decreases. (d) The distance in part (c) is always less than the magnitude of the next term of the series.
Section 8.5
7.
Alternating Series
1n1 2 0.8225 n2 12 n1
(a)
(b)
n
1
2
3
4
5
6
7
8
9
10
Sn
1
0.75
0.8611
0.7986
0.8386
0.8108
0.8312
0.8156
0.8280
0.8180
1.1
0
11 0.6
(c) The points alternate sides of the horizontal line that represents the sum of the series. The distance between successive points and the line decreases. (d) The distance in part (c) is always less than the magnitude of the next term in the series.
9.
1n1 n n1
an1 lim
n→
11.
1 1 < an n1 n
an1
1 0 n
lim
n→
Converges by Theorem 8.14.
13.
1n n2 2 n1 n 1
lim
n→
n2
1n1 n1 2n 1
1 1 < an 2n 1 1 2n 1
1 0 2n 1
Converges by Theorem 8.14
15.
n2 1 1
1n n1 n
an1
Diverges by the nth Term Test
lim
n→
1 n 1
1 n
<
1 n
an
0
Converges by Theorem 8.14
17.
1n1n 1 lnn 1 n1
lim
n→
19.
sin
n1
n1 1 lim lim n 1 lnn 1 n→ 1n 1 n→
2n 1 1n1 2 n1
Diverges by the nth Term Test
Diverges by the nth Term Test
21.
n1
n1
cos n 1
n
Diverges by the nth Term Test
23.
1n n! n0
an1 lim
n→
1 1 < an n 1! n!
1 0 n!
Converges by Theorem 8.14
139
140
25.
Chapter 8
Infinite Series
1n1n n2 n1
n 1 n < for n ≥ 2 n 1 2 n2
an1
n
lim
n→
27.
n2
1n12en 1n12 n n e e e2n 1 n1 n1
Let f x
0
2ex . Then e 1
f x
2x
2ex e2x 1 < 0. e2x 12
Thus, f x is decreasing. Therefore, an1 < an , and
Converges by Theorem 8.14
lim
n→
2en 2en 1 lim 0. lim e2n 1 n→ 2e2n n→ en
The series converges by Theorem 8.14.
29. S6
31n1 2.4325 n2 n1 6
3
R6 S S6 ≤ a7 49 0.0612; 2.3713 ≤ S ≤ 2.4937 31. S6
21n 0.7333 n! n0 5
2
R6 S S6 ≤ a7 6! 0.002778; 0.7305 ≤ S ≤ 0.7361 33.
1n n! n0
35.
1n
n0
(a) By Theorem 8.15,
RN
2n 1!
≤ aN1
(a) By Theorem 8.15, 1 < 0.001. N 1!
RN
This inequality is valid when N 6.
≤ aN1
1 < 0.001.
2N 1 1!
This inequality is valid when N 2.
(b) We may approximate the series by
(b) We may approximate the series by
1n 1 1 1 1 1 11 n! 2 6 24 120 720 n0 6
2
1n
1
1
2n 1! 1 6 120 0.842.
n0
0.368.
(3 terms. Note that the sum begins with n 0.)
(7 terms. Note that the sum begins with n 0.)
37.
1n1 n n1
39.
(a) By Theorem 8.15,
RN
1 ≤ aN1 < 0.001. N1
This inequality is valid when N 1000. (b) We may approximate the series by
1n1 1 1 1 1 1 . . . n 2 3 4 1000 n1 1000
0.693. (1000 terms)
1n1 3 n1 2n 1
By Theorem 8.15,
RN
≤ aN1
1 < 0.001. 2N 13 1
This inequality is valid when N 7.
Section 8.5
41.
1n1 2 n1 n 1
1
n 1
2
n1
43.
1n1 n n1
The given series converges by the Alternating Series Test, but does not converge absolutely since
converges by comparison to the p-series
1
n
1
2
n1
is a divergent p-series. Therefore, the series converges conditionally.
Therefore, the given series converge absolutely.
45.
141
n.
n1
Alternating Series
1n1 n2 2 n1 n 1
47.
n2 1 n→ n 12
1n n2 lnn
The given series converges by the Alternating Series Test, but does not converge absolutely since the series
lim
Therefore, the series diverges by the nth Term Test.
n 2
1 ln n
diverges by comparison to the harmonic series
1
n.
n1
Therefore, the series converges conditionally.
49.
1n n 3 n2 n 1
n
3
n2
51.
n0
n 1
n0
is convergent by comparison to the convergent geometric series
2
1
n. 2
n2
1
2n 1!
converges by a limit comparison to the convergent p-series
1n
2n 1!
1
n
n0
Therefore, the given series converges absolutely.
since 1 1 < n for n > 0. 2n 1! 2 Therefore, the given series converges absolutely.
53.
1n cos n n0 n 1 n0 n 1
55.
The given series converges by the Alternating Series Test, but
n 1 n 1
cos n
n0
1
n0
diverges by a limit comparison to the divergent harmonic series,
1
n.
n1
lim
n→
cos n n 1 1, therefore the series 1n
converges conditionally.
1n cos n 2 n n2 n1 n1
1
n
n1
2
is a convergent p-series. Therefore, the given
series converges absolutely.
142
Chapter 8
Infinite Series
57. An alternating series is a series whose terms alternate in sign. See Theorem 8.14.
59.
a converges. is conditionally convergent if a diverges, but a converges.
a a
n
is absolutely convergent if
n
n
n
n
61. (b). The partial sums alternate above and below the horizontal line representing the sum.
63. Since
a converges we have
65.
n
n1
n1
1
n
2
converges, hence so does
1
n.
n1
4
lim an 0.
n→
Thus, there must exist an N > 0 such that aN < 1 for all n > N and it follows that an2 ≤ an for all n > N. Hence, by the Comparison Test,
a
2
n
n1
converges. Let an 1n to see that the converse is false.
67. False Let an
69.
n1
1n . n
71. Diverges by nth Term Test. lim an n→
75. Convergent Geometric Series r
1 e
or Integral Test
10
n
32
10
1
n
n1
32
convergent p-series
73. Convergent Geometric Series r 78 < 1
77. Converges (absolutely) by Alternating Series Test
79. The first term of the series is zero, not one. You cannot regroup series terms arbitrarily.
Section 8.6 1.
The Ratio and Root Tests
n 1! n 1nn 1n 2! n 2! n 2! n 1nn 1
3. Use the Principle of Mathematical Induction. When k 1, the formula is valid since 1 13
5.
. . 2n 1 2n! 2n n!
and show that 13
5.
. . 2n 12n 1
—CONTINUED—
2n 2! . 2n1n 1!
21! . Assume that 21 1!
Section 8.6
The Ratio and Root Tests
143
3. —CONTINUED— To do this, note that:
35.
1
. . 2n 12n 1 1
35.
. . 2n 1 2n 1
2n! 2n 1 2n n!
2n!2n 1 2n 2 2n 1 2n n!
2n!2n 12n 2 2n1n!n 1
2n 2! 2n1n 1
The formula is valid for all n ≥ 1.
5.
n 4 3
n
1
n1
34 2169 . . .
7.
3 S1 , S2 1.875 4
33 3n1 9 . . . n! 2 n1
5n 3
9.
4n
n
n1
S1 9
S1 2
Matches (f)
Matches (a)
4 8 2 7
2
. . .
Matches (d)
11. (a) Ratio Test: lim
n→
(b)
(c)
an1 n 1258n1 n1 lim lim n→ n→ an n258n n
n
5
10
15
20
25
Sn
9.2104
16.7598
18.8016
19.1878
19.2491
2
5 5 < 1. Converges 8 8
20
0
12 0
(d) The sum is approximately 19.26. (e) The more rapidly the terms of the series approach 0, the more rapidly the sequence of the partial sums approaches the sum of the series.
13.
n!
3
n0
lim
n→
15.
n
an1 n 1! lim n→ an 3n1 lim
n→
3n
n!
n1 3
n
2
lim
n→
n1
lim
n→
an1 n1 lim n→ 2n1 an lim
n→
2n n
n1 1 2n 2
Therefore, by the Ratio Test, the series converges.
n
an 1 n 134n1 lim n→ an n34n lim
n→
3n 1 3 4n 4
Therefore, by the Ratio Test, the series converges.
19.
n
3
n1
Therefore, by the Ratio Test, the series diverges.
17.
n 4
2n
n
n1
lim
n→
2
an1 2n1 lim n→ n 12 an lim
n→
2n2
n 12
n2
2n
2
Therefore, by the Ratio Test, the series diverges.
144
21.
Chapter 8
Infinite Series
1n 2n n! n0
an1 2n1 lim n→ n 1! an
lim
n→
n!
2n
2 0 n→ n 1
n!
n3
23.
n
n1
lim
n→
an1 n 1! lim n→ n 13n1 an n→
4n
n!
n0
lim
n→
an1 4n1 lim n→ n 1! an
n!
4n
4 0 n1
lim
n→
Therefore, by the Ratio Test, the series converges.
27.
3n
n 1
n
n0
lim
n→
an1 3n1 lim n→ n 2n1 an
n 1n 3n 1n 3 n1 lim lim n n→ n 2n1 n→ n 2 n 2 3
n→
n
0
1e 0
nn 12 , let y lim nn 12 . Then, n
To find lim
n→
ln y lim
n→
n
n→
ln y lim n ln
1n 2 0 nn 12 lim ln n 1n 0 n→
1n 1 1n 2 1 by L’Hôpital’s Rule 1n2
1 y e1 . e Therefore, by the Ratio Test, the series converges.
29.
3
4n 1
lim
an1 4n1 lim n1 n→ an 1 3
n0
n→
n
3n 1 43n 1 41 13n 4 lim n1 lim n n→ n→ 4 1 3 3 3 13n
Therefore, by the Ratio Test, the series diverges.
31.
1n1n!
1 3 5 . . . 2n 1
n0
lim
n→
an1 lim n→ 1 an
3
n 1! 5 . . . 2n 12n 3
1
35.
. . 2n 1 n!
Therefore, by the Ratio Test, the series converges. Note: The first few terms of this series are 1
n3n n!
Therefore, by the Ratio Test, the series diverges.
Therefore, by the Ratio Test, the series converges.
25.
n 3
lim
lim
1 1
3
2!
1
35
3!
1
lim
n→
357
n1 1 2n 3 2
. . .
Section 8.6
32
n1
lim
n→
32
12
n32 n lim n→ n 1 1
n12 n lim n→ n 1 1
1
1
12
n1
lim
n→
35.
an1 1 lim n→ n 132 an
n
(b)
an1 1 lim n→ n 112 an
2n 1 n
1
n
37.
2n n 1
n lim an lim
n
n
n→
n 1 2n 1 2
n→
n→
lim
n→
Therefore, by the Root Test, the series converges.
1n
n an lim lim
n
n→
lim
39.
ln n
n2
n1
n→
145
1
n
33. (a)
The Ratio and Root Tests
ln1n
n
n
1
ln n
n
0
Therefore, by the Root Test, the series converges.
2n 1
n
n
n1
n 2 n n 1 lim 2 n n 1n n→
n a lim n lim
n→
n→
n n, let y lim x x. Then To find lim n→
n→
ln y lim
n→
x x ln
1 ln x 1x lim ln x lim lim 0. n→ n→ x n→ 1 x
n n 1 21 1 3. Therefore, by the Root Test, the series diverges. Thus, ln y 0, so y e0 1 and lim 2 n→
41.
1
ln n
43.
n
n3
ln1n
n a lim n lim
n→
n
n→
n
lim
n→
1 0 ln n
1n1 5 n n1
an1
Therefore, by the Root Test, the series converges.
5 5 < an n1 n
lim
n→
5 0 n
Therefore, by the Alternating Series Test, the series converges (conditional convergence).
45.
3
1
nn 3 n
n1
n1
47.
32
2n
n1
n1
This is convergent p-series.
lim
n→
2n 20 n1
This diverges by the nth Term Test for Divergence.
49.
1n 3n 32 1 1n 3n2 3 n n 2 2 9 2 n1 n1 n1
n
3 Since r 2 > 1, this is a divergent geometric series.
51.
10n 3 n2n n1
lim
n→
10n 3n2n 10n 3 10 lim n→ 12n n
Therefore, the series converges by a limit comparison test with the geometric series
2 .
n0
1
n
146
53.
Chapter 8
Infinite Series
cosn 2n n1
55.
cosn 1 ≤ n 2n 2
n1
n7n n1 n! lim
n→
Therefore, the series
an1 n 17n1 lim n→ an n 1!
n!
n7n
lim
n→
7 0 n
Therefore, by the Ratio Test, the series converges.
cosn 2n
converges by comparison with the geometric series
2 . 1
n
n0
57.
1n 3n1 n! n1
lim
n→
an1 3n lim n→ an n 1!
n!
3n1
lim
n→
3 0 n1
Therefore, by the Ratio Test, the series converges.
59.
3n
3 5 7 . . . 2n 1
n1
lim
n→
an1 lim n→ 3 an
3n1 . . 5 7 . 2n 12n 3
3
57.
. . 2n 1 3 0 lim n→ 2n 3 3n
Therefore, by the Ratio Test, the series converges.
63. (a) and (b) are the same.
61. (a) and (c) n 15n1 n5n n 1! n1 n! n0
5
252 353 454 . . . 2! 3! 4!
65. Replace n with n 1.
n
4
n1
n
n1 n1 n0 4
67. Since
69. See Theorem 8.17, page 597.
310 1.59 105, 210 10! use 9 terms.
3k 0.7769 k k1 2 k! 9
71. No. Let an The series
1 . n 10,000
1
n 10,000 diverges.
n1
73. The series converges absolutely. See Theorem 8.17.
Section 8.7 75. First, let
147
Second, let
n a lim n r < 1
n→
n a lim n r > R > 1.
n→
and choose R such that 0 ≤ r < R < 1. There must exist n a some N > 0 such that n < R for all n > N. Thus, for n > N, we an < Rn and since the geometric series
Taylor Polynomials and Approximations
n a Then there must exist some M > 0 such that n > R for all n > M. Thus, for n > M, we have an > Rn > 1 which implies that lim an 0 which in turn implies that
n→
R
a
n
n
n0
diverges.
n1
converges, we can apply the Comparison Test to conclude that
a n
n1
converges which in turn implies that
a
n
converges.
n1
Section 8.7
Taylor Polynomials and Approximations 3. y e12x 1 1
1
1. y 2 x 2 1 Parabola
Linear
Matches (d)
Matches (a)
5. f x
4 x
fx 2x32
f 1 2
fx sec x tan x
P1x f 1 f1x 1 P1x f
4 2x 1 P1x 2x 6
4 2
f
4 2
4 f 4 x 4
P1x 2 2 x
10
P1
f
7. f x sec x
f 1 4
4x12
4
5
(1, 4)
f
f −2
( π4 , 2)
6 −2
−
P1
4
2 −1
9. f x
4 x
f 1 4
4x12
x
0
0.8
0.9
1.0
1.1
1.2
2
fx 2x
f1 2
f x
Error
4.4721
4.2164
4.0
3.8139
3.6515
2.8284
f x 3x
f 1 3
P2x
7.5
4.46
4.215
4.0
3.815
3.66
3.5
32
52
f 1 P2 f 1 f1x 1 x 12 2 3 4 2x 1 x 12 2 10
P2 (1, 4) f −2
6 −2
148 11.
Chapter 8
Infinite Series
f x cos x
(b)
P2x 1
1 2 2x
P4x 1
1 2 2x
P6x 1
1 2 2x
(a)
1 4 24 x
1 4 24 x
P2x x
f x cos x
P2 x 1
f 0 P2 0 1 1 6 720 x
f x sin x
P4 x x
f 4x cos x
2
P6
−3
fx sin x
f 40 1 P440 f 5x sin x P65x x
P4 3
f
P44x 1
P2
f 6x cos x
−2
P6x 1
f 60 1 P660 (c) In general, f n0 Pnn0 for all n. 13.
f x ex
f 0 1
fx ex
f x e2x
f 0 1
f0 1
fx 2e2x
f0 2
f x ex
f 0 1
f x 4e2x
f 0 4
fx ex
f0 1
fx 8e2x
f0 8
f 4x 162x
f 40 16
P3x f 0 f0x 1x
17.
2
f 0 2 f0 3 x x 2! 3!
6
1 2x 2x 2
f 0 0
fx cos x
f0 1
f x sin x fx cos x
f 0 0
f x
xex
f0 1
fx
xex xex
0
f 50 1
P5x 0 1x
0 2 1 3 0 1 x x x 4 x5 2! 3! 4! 5!
f x
f 4x
f 0 1
1 x 12
f 0 0
f0 1
2ex
f 0 2
3ex
f0 3
4ex
f 40 4
P4x 0 x
2 2 3 4 x x3 x 4 2! 3! 4!
x x2
fx sec x tan x
P2x 1 0x
f 0 2
fx
6 x 14
f0 6
f 4x
24 x 15
P4x 1 x
f 0 1
f x
2 x 12
f 40 24 2 2 6 3 24 4 x x x 2! 3! 4!
1 x x 2 x3 x 4
1 3 1 4 x x 2 6
23. f x sec x
f0 1
f x
4 3 2 4 x x 3 3
ex
1 5 1 3 x x 6 120
1 x1
fx
f x xex fx
f 5x cos x
x
19.
xex
f 40
sin x
4 2 8 16 4 x x3 x 2! 3! 4!
P4x 1 2x
x3
f x sin x
f 4x
21.
x2
15.
sec3 x
f0 0
sec x
tan2
x
f 0 1
1 2 1 x 1 x2 2! 2
Section 8.7 f x
25.
1 x
fx f x
f 4x
149
f 1 1 1 x2
f1 1
2 x3
fx
Taylor Polynomials and Approximations
f 1 2 6 x4
f1 6
24 x5
f 41 24
P4x 1 x 1
2 6 24 x 12 x 13 x 14 2! 3! 4!
1 x 1 x 12 x 13 x 14 f x x
27.
fx
f 1 1
1
f1
2x
1 f x 4xx fx
1 2
3
f1
15 16x3x
3 8
f 41
15 16
fx
fx 4 sec2 x tan2 x 2 sec4 x f 4x 8 sec2 x tan3 x 16 sec4 x tan x f 5x 16 sec2 x tan4 x 88 sec4 x tan2 x 16 sec6 x 6
2
f
Q3 −6
f 1 1 f1 2
6 x4
f 41 6
1 P4x 0 x 1 x 12 2 1 1 x 13 x 14 3 4
P3x 0 x
f x 2 sec2 x tan x
2 P3 P5
1 x2
(a) n 3, c 0
sec2 x
−
f1 1
2 x3
f 4x
5 1 x 13 x 14 16 128
f x tan x
31.
f 1 0
1 x
f x fx
1 1 P4x 1 x 1 x 12 2 8
f x ln x fx
1 f 1 4
8x 2x
f 4x
29.
0 2 2 1 x x3 x x3 2! 3! 3
(b) n 5, c 0 0 2 2 0 16 5 x x3 x 4 x 2! 3! 4! 5! 2 5 1 x x x3 3 15
P5x 0 x
(c) n 3, c
4
4 x 4 2! 4
2 x 4 4
2
Q3x 1 2 x 12 x
2
16 x 3! 4
8 x 3 4
3
3
150
Chapter 8
Infinite Series
f x sin x
33.
P1x x P3x x 16 x3 1 5 P5x x 16 x3 120 x 1 1 1 P7x x 6 x3 120 x5 5040 x7
(a)
(b)
x
0.00
0.25
0.50
0.75
1.00
sin x
0.0000
0.2474
0.4794
0.6816
0.8415
P1x
0.0000
0.2500
0.5000
0.7500
1.0000
P3x
0.0000
0.2474
0.4792
0.6797
0.8333
P5x
0.0000
0.2474
0.4794
0.6817
0.8417
P7x
0.0000
0.2474
0.4794
0.6816
0.8415
3
P1
P3
P7 f − 2
2
P5 −3
(c) As the distance increases, the accuracy decreases 35. f x arcsin x (a) P3x x (b)
x3 6
(c)
y
π 2
x
0.75
0.50
0.25
0
0.25
0.50
0.75
f x
0.848
0.524
0.253
0
0.253
0.524
0.848
P3x
0.820
0.521
0.253
0
0.253
0.521
0.820
x
−1
1
P3 −
P8
y
y
P4 4
6
x
−4 −3 −2
8
2
−4 −3
−6
41. f x ex 1 x
x 2 x3 2 6
12 0.6042 f x ln x x 1 12 x 12 13 x 13 14 x 14
f 1.2 0.1823 45. f x cos x; f 5x sin x ⇒ Max on 0, 0.3 is 1. R4x ≤
f(x) = ln (x 2 + 1)
1 x
P6 P2
43.
P2
2
f(x) = cos x
2
f
P6
3
6
−6
π 2
39. f x) lnx2 1
37. f x cos x
1 0.35 2.025 105 5!
P8
P4
3
4
f
Section 8.7
47. f x arcsin x; f 4x
151
x6x 2 9 ⇒ Max on 0, 0.4 is f 40.4 7.3340. 1 x 272
7.3340 0.44 0.00782 7.82 103 4!
R3x ≤
49. gx sin x gn1x Rnx ≤
Taylor Polynomials and Approximations
51. f x lnx 1
≤ 1 for all x
f n1x
1 0.3n1 < 0.001 n 1!
By trial and error, n 3.
Rn ≤
1n1n! ⇒ Max on 0, 0.5 is n!. x 1n1
0.5n1 n! < 0.0001 0.5n1 n 1! n1
By trial and error, n 9. (See Example 9.) Using 9 terms, ln1.5 0.4055.
53.
f x ex 1 x
x 2 x3 , x < 0 2 6
ez 4 x < 0.001 4!
R3x
ez x 4 < 0.024
55. The graph of the approximating polynomial P and the elementary function f both pass through the point c, f c and the slopes of P and f agree at c, f c. Depending on the degree of P, the nth derivatives of P and f agree at c, f c.
xe z4 < 0.3936 x <
0.3936 < 0.3936, z < 0 e z4
0.3936 < x < 0 57. See definition on page 607.
59. The accuracy increases as the degree increases (for values within the interval of convergence).
61. (a) f x e x
63. (a) Q2x 1
2x 22 32
(b) R2x 1
2x 62 32
P4x 1 x
1 2 1 3 1 4 x x x 2 6 24
gx xe x Q5x x
x2
1 1 1 5 x x3 x 4 2 6 24
(c) No. The polynomial will be linear. Translations are possible at x 2 8n.
Q5x x P4x (b) f x sin x P5x x
x5 x3 3! 5!
gx x sin x Q6x x P5x x 2 (c) gx
x6 x4 3! 5!
sin x 1 x2 x 4 P5x 1 x x 3! 5!
65. Let f be an even function and Pn be the nth Maclaurin polynomial for f. Since f is even, f is odd, f is even, f is odd, etc. (see Exercise 45). All of the odd derivatives of f are odd and thus, all of the odd powers of x will have coefficients of zero. Pn will only have terms with even powers of x. 67. As you move away from x c, the Taylor Polynomial becomes less and less accurate.
152
Chapter 8
Infinite Series
Section 8.8
Power Series
1. Centered at 0
5.
xn n1
1
n
n0
n→
n1
1n xn
n1 x x n2
lim
n→
9.
7.
un1 1n1xn1 lim n→ un n2
L lim
x
3. Centered at 2
L lim
n→
lim
n→
< 1⇒R1
2x2n n0 2n!
n→
11.
un 1 2x 2n 2! lim n→ un 2x2n2n! lim
n→
2n2
1n x n n n1 n→
n 12
un1 1 lim n→ un n1
n1xn1
n→
n
1n xn
nx x n1
Interval: 1 < x < 1
n
xn
lim
n→
un1 x n1 lim n→ un n 1! lim
n→
n!
n0
1
n diverges.
Therefore, the interval of convergence is 1 < x ≤ 1.
2n!2
n0
lim
n→
x
n
un1 2n 2!x n1 lim n→ un 2n1
2n
2n!x n
lim
n→
Therefore, the series converges only for x 0.
19.
2n 22n 1 x 2
1n1x n 4n n1
n!
xn
x 0 n1
The series converges for all x. Therefore, the interval of convergence is < x < .
n1
17.
1 2
1n When x 1, the alternating series converges. n n1 When x 1, the p-series
x
2x
Since the series is geometric, it converges only if x2 < 1 or 2 < x < 2.
15.
lim
2
n0
lim
2n2x
n2
2xn
2x 0 2n 22n 1 2
Thus, the series converges for all x. R .
un1 2xn1 lim n→ n 12 un
2 x < 1⇒R
L lim
13.
2xn 2 n1 n
Since the series is geometric, it converges only if x4 < 1 or 4 < x < 4.
Section 8.8
21.
1n1x 5n n5n n1
Power Series
153
lim
n→
un1 1n2x 5n1 lim n→ un n 15n1
n5n
1n1x 5n
R5
lim
n→
nx 5 1 x5 5n 1 5
Center: x 5 Interval: 5 < x 5 < 5 or 0 < x < 10 When x 0, the p-series
n1
1 diverges. n
When x 10, the alternating series
1n1 converges. n n1
Therefore, the interval of convergence is 0 < x ≤ 10.
23.
1n1x 1n1 n1 n0
lim
n→
un1 1n2x 1n2 lim n→ un n2
n1
1n1x 1n1
R1
lim
n→
n 1x 1 x 1 n2
Center: x 1 Interval: 1 < x 1 < 1 or 0 < x < 2
1
n 1 diverges by the integral test.
When x 0, the series
n0
When x 2, the alternating series
1n1 converges. n0 n 1
Therefore, the interval of convergence is 0 < x ≤ 2.
25.
x cn1 cn1 n1
lim
n→
27.
un1 x cn lim n→ un cn
cn1
x cn1
Rc
n
n 1 2x
n1
n1
1 xc c
lim
n→
un1 n 12xn lim n→ un n2 lim
Center: x c
n→
Interval: c < x c < c or 0 < x < 2c
When x 0, the series
1
n1
R diverges.
n1
When x 2c, the series
1 diverges.
n1
Therefore, the interval of convergence is 0 < x < 2c.
2xn 12 2x nn 2
1 2
Interval:
n1
n2xn1
1 1 < x < 2 2
1 n When x , the series diverges 2 n 1 n1 by the nth Term Test.
1n1n 1 When x , the alternating series diverges. 2 n1 n1
Therefore, the interval of convergence is
1 1 < x < . 2 2
154
29.
Chapter 8
Infinite Series
x 2n1
2n 1!
n0
lim
n→
un1 x 2n3 lim n→ 2n 3! un lim
n→
2n 1! x 2n1
x2
2n 22n 3
0
Therefore, the interval of convergence is < x < 31.
.
kk 1 . . . k n 1x n n! n1
lim
n→
un1 kk 1 . . . k n 1k nx n1 lim n→ un n 1!
kk 1 .
n! k nx lim x . . k n 1xn n→ n1
R1
When x ± 1, the series diverges and the interval of convergence is 1 < x < 1.
kk 11 2 . .k .n n 1 ≥ 1
. . .
33.
1n1 3 7 11 . . . 4n 1x 3n 4n n1
lim
n→
un1 1n2 3 lim n→ un lim
n→
7 11 .
. . 4n 14n 3x 3n1 4n1
1n1
4n 3 7 11 . . . 4n 1x 3n
4n 3x 3 4
R0
Center: x 3 Therefore, the series converges only for x 3.
35. (a) f x
2 , 2 < x < 2 n
x
(Geometric)
37. (a) f x
n0
(b) fx
22 n
x
n1
, 2 < x < 2
1n1x 1n1 ,0 < x ≤ 2 n1 n0
(d)
2 n
n2
39. g1
n1 2
2x
2
x
n2
, 2 < x < 2
1
n
1
n0
(c) f x
n1
, 2 ≤ x < 2
n0
3
1
n1
x 1n, 0 < x < 2
n0
n 1 2
f x dx
(b) fx
n1
(c) f x
1 1 . . . 3 9
(d)
1
n1
nx 1n1, 0 < x < 2
n1
f x dx
41. g3.1
1n1x 1n2 ,0 ≤ x ≤ 2 n 1n 2 n1
3 3.1
n
diverges. Matches (b)
n0
S1 1, S2 1.33. Matches (c) 43. A series of the form
a x c n
n
n0
is called a power series centered at c.
45. A single point, an interval, or the entire real line.
Section 8.8
1n x 2n1 , < x < n0 2n 1!
47. (a) f x
Power Series
(See Exercise 29.)
1n x 2n , < x < 2n! n0
gx
1n x 2n gx 2n! n0
(b) fx
(c) gx
1n1x 2n1 1n x2n1 1n x 2n1 f x 2n 1! n1 2n 1! n0 n0 2n 1!
(d) f x sin x and gx cos x
y
49.
x 2n n n!
2
n0
y
2nx 2n1 n n1 2 n!
y
2n2n 1x 2n2 2n n! n1
y xy y
2nx 2n x 2n 2n2n 1x 2n2 n n n 2 n! n1 n1 2 n! n0 2 n!
2n 1x 2n 2n2n 1x 2n2 n n! 2 2n n! n1 n0
(a) lim
n0
51. J0x
2n 22n 1x2n 2n 1x2n 2n 1 2n 1 2n1n 1! 2n n!
2n 1x 2n 2n 1 2n 1 0 2n1n 1! n0
1k x 2k 2k 2 k0 2 k!
k→
uk1 1k1 x 2k2 lim 2k2 k→ 2 uk k 1! 2
22k k!2
1k x 2k
lim
Therefore, the interval of convergence is < x < J0
(b)
1
k
k0
J0
4k k!2
2k2k 1x 2k2 2k 22k 1x 2k 1k1 4k k!2 4k1 k 1! 2 k0
k
1
k
1k1
k0
—CONTINUED—
x 2k 2kx 2k1 2k 2 x 2k1 1k1 k1 k 2 4 k! 4 k 1! 2 k0
k1
x 2J0 xJ0 x 2J0
.
1x 2 0 22k 12
1
k1
J0
k→
22k 1 x 2k2 2x 2k2 x 2k2 1k1 k1 1k k 2 k1 4 k 1!k! 4 k 1!k! k0 4 k! k0
1k x2k2 22k 1 2 1 1 1 4k k!2 4k 1 4k 1 k0
1k x 2k2 4k 2 2 4k 4 0 4k k!2 4k 4 4k 4 4k 4 k0
155
156
Chapter 8
Infinite Series
51. —CONTINUED— x2 x4 x6 4 64 2304
(c) P6x 1
1
(d)
0
0
3
−6
1
J0dx
1k x 2k k 2 dx k0 4 k!
x 4 1 k! 2k 1
k
k
k0
6
1
2k1
2
0
1k k 4 k! 22k 1 k0
−5
1
1 1 0.92 12 320
(exact integral is 0.9197304101)
1
53. f x
n
n0
x 2n cos x 2n!
2
x
n
n0
1 1 for 1 < x < 1 1 x 1 x 3
−2
2
−1
−2
n n
n0
(See Exercise 47.)
57.
1 x
55. f x
2 x
1 0
n
n0
(a)
2 34
n
n0
8 3
n
(b)
n0
n0
1 8 1.6 1 38 5
n
8 3
n
n0
1.80
−1
34 2
1 8 0.7272 1 38 11
1.10
−1
6 0
2 N
(c) The alternating series converges more rapidly. The partial sums of the series of positive terms approach the sum from below. The partial sums of the alternating series alternate sides of the horizontal line representing the sum.
(d)
1nx n n2n n0
converges for x 2 but diverges for x 2.
3
n
> M
n0
M 10 N
59. False;
6 0.60
4
100
1000
10,000
9
15
21
61. True; the radius of convergence is R 1 for both series.
Section 8.9
Section 8.9 1. (a)
Representation of Functions by Power Series
Representation of Functions by Power Series
1 12 a 2 x 1 x2 1 r
2 2
1 x
n
3. (a)
1 12 a 2 x 1 x2 1 r
xn
2
n0
n1
n0
2 2 1
x
n
n0
This series converges on 2, 2.
1n xn 2n1 n0
This series converges on 2, 2.
1 x x2 x3 . . . 2 4 8 16 (b) 2 x ) 1 x 1 2 x 2 x x2 2 4 x2 4 x 2 x3 4 8 x3 8 x3 x4 8 16
1 x x2 x3 . . . 2 4 8 16 (b) 2 x ) 1 x 1 2 x 2 x x2 2 4 x2 4 x 2 x3 4 8 x3 8 x3 x4 8 16
5. Writing f x in the form a1 r, we have
7. Writing f x in the form a1 r, we have
1 13 1 2 x 3 x 5 1 13x 5
3 3 a 2x 1 1 2x 1 r
which implies that a 13 and r 13x 5.
which implies that a 3 and r 2x.
Therefore, the power series for f x is given by
Therefore, the power series for f x is given by
1 1 1 ar n x 5 2 x n0 3 3 n0
x 5n
3
n0
n1
3 ar n 32xn 2x 1 n0 n0
n
<
1
n
< x <
1 . 2
3 3 32 a x 2 2 x 1 12x 1 r which implies that a 32 and r 12x. Therefore, the power series for f x is given by
which implies that a 111 and r 211x 3. Therefore, the power series for f x is given by
2x , 2x < 1 or 2
11. Writing f x in the form a1 r, we have
111 a 1 211x 3 1 r
1 1 ar n 2x 5 n0 11 n0
n0
1 1 2x 5 11 2x 3
3
, x 5 < 3 or 2 < x < 8.
9. Writing f x in the form a1 r, we have
x 3
157
3 1 3 ar n x x 2 n0 2 2 n0
2nx 3n , 11n1 n0
17 11 5 or < x < . 2 2 2
n
1n x n 3 x n , n1 2 2 2 n0 n0
112 x 3
n
3
x
< 2 or 2 < x < 2.
158
13.
Chapter 8
Infinite Series
3x 2 1 2 1 1 1 x 2 x 2 x 2 x 1 2 x 1 x 1 12x 1 x Writing f x as a sum of two geometric series, we have 3x 1 x x x 2 n0 2
2
n
1x
n
2
n0
1
n
n0
The interval of convergence is 1 < x < 1 since lim
n→
15.
un1 1 2n1x n1 lim n→ un 2n1
2n
1 2n xn
1 x n.
lim
n→
1 2n1x x. 2 2n1
1 1 2 1 x2 1 x 1 x Writing f x as a sum of two geometric series, we have 2 xn xn 1 1n x n 2x 2n. 1 x 2 n0 n0 n0 n0
The interval of convergence is x 2 < 1 or 1 < x < 1 since lim
17.
n→
un1 2x 2n2 lim x2 . n→ un 2x 2
1 1n x n 1 x n0
1 1n xn 12n x n xn 1 x n0 n0 n0
hx
2 1 1 1n x n xn 1n 1 x n x 2 1 1 x 1 x n0 n0 n0
2 0x 2x2 0x3 2x4 0x5 2x6 . . .
2x
2n,
1 < x < 1 (See Exercise 15.)
n0
19. By taking the first derivative, we have d 1 x 12 dx
d 1 1 . Therefore, dx x 1 x 12
1 x 1 nx n
n
n0
n
n1
n1
1
n1
n 1 x n, 1 < x < 1.
n0
21. By integrating, we have lnx 1
1 dx lnx 1. Therefore, x1
1n x n dx C
n0
1n x n1 , 1 < x ≤ 1. n1 n0
To solve for C, let x 0 and conclude that C 0. Therefore, lnx 1
23.
1n x n1 , 1 < x ≤ 1. n1 n0
1 1n x 2n 1n x 2n, 1 < x < 1 x 2 1 n0 n0
25. Since,
1 1 1 1 1n x n, we have 2 1n 4x 2n 1n 4n x 2n 1n 2x2n, < x < . x 1 n0 4x 1 n0 2 2 n0 n0
Section 8.9 x2 x 2 x3 ≤ lnx 1 ≤ x 2 2 3
27. x
x x
5
S3 f −4
x2 2
lnx 1 8
x
S2 −3
29. gx x, line, Matches (c)
31. gx x
1
In Exercises 35 and 37, arctan x
n
n0
35. arctan
Representation of Functions by Power Series
x2 2
x3 3
0.0
0.2
0.4
0.6
0.8
1.0
0.000
0.180
0.320
0.420
0.480
0.500
0.000
0.180
0.336
0.470
0.588
0.693
0.000
0.183
0.341
0.492
0.651
0.833
x3 x5 , Matches (a) 3 5
33. f x arctan x is an odd function (symmetric to the origin)
x2n1 . 2n 1
1 1 1 142n1 1n 1 . . . 1n 2n1 4 4 n0 2n 1 2n 1 4 192 5120 n0
1 1 Since 5120 < 0.001, we can approximate the series by its first two terms: arctan 14 14 192 0.245.
arctan x 2 x 4n1 1n x 2n 1 n0
37.
12
0
arctan x 2 x 4n2 dx 1n x 4n 22n 1 n0
arctan x 2 1 1 1 . . . dx 1n 4n2 8 1152 x 4n 2 2n 1 2 n0
Since
1 < 0.001, we can approximate the series by its first term: 1152
In Exercises 39 and 41, use
39. (a)
12
0
arctan x 2 dx 0.125 x
1 x n, x < 1. 1 x n0
d d 1 1 1 x2 dx 1 x dx
x nx , x < 1 n
n0
n1
n1
(b)
x x nx n1 nx n, x < 1 2 1 x n1 n1
(c)
1x 1 x nx n1 x n, x < 1 1 x2 1 x2 1 x2 n1
2n 1x , x < 1 n
n0
(d)
x1 x x 2n 1x n 2n 1x n1, x < 1 1 x2 n0 n0
41. Pn En
12
n1
n
nPn
1 2 n
n1
n
1 1 n 2 n1 2
159
n1
1 1 2 2 1 12 2 Since the probability of obtaining a head on a single toss is 12 , it is expected that, on average, a head will be obtained in two tosses.
160
Chapter 8
Infinite Series
43. Replace x with x.
45. Replace x with x and multiply the series by 5.
47. Let arctan x arctan y . Then, tanarctan x arctan y tan tanarctan x tanarctan y tan 1 tanarctan x tanarctan y xy tan 1 xy arctan
49. (a) 2 arctan
1x xyy . Therefore, arctan x arctan y arctan1x xyy for xy 1.
21 2 4 1 1 1 arctan arctan arctan arctan 3 2 2 2 1 1 22
(b) 8 arctan
1 1 1 0.5 0.5 0.5 4 arctan 8 2 7 2 3 5 7
3
5
lnx 1
1
417 1 73
3
1n1xn 1n xn1 n 1 n n0 n1
1
n1
1 75 1 77 3.14 5 7
n1
1n12 5n 2n 5n n n1 n
1n1xn . n n1
n1
n1
7
53. From Exercise 51, we have
51. From Exercise 21, we have
Thus,
1 1 4 1 4 3 1 7 25 arctan arctan arctan arctan arctan 1 arctan 2 7 3 7 1 4 31 7 25 4
2 arctan
ln
25 1 ln 57 0.3365.
1n11 2n 1 n 2 n n1 n
ln
12 1 ln 23 0.4055
55. From Exercise 54, we have
1
n
n0
1 1 22n1 1 1n arctan 0.4636. 22n12n 1 n0 2n 1 2
57. The series in Exercise 54 converges to its sum at a slower rate because its terms approach 0 at a much slower rate.
59. f x
1
n1
n1
f 0.5
n1
n1
Section 8.10
Taylor and Maclaurin Series
1. For c 0, we have: f x e2x f nx 2n e2x ⇒ f n0 2n e2x 1 2x
2xn 4x 2 8x3 16x 4 . . . 2! 3! 4! n0 n!
x 1n , 0 < x ≤ 2 n
1n1
0.5n 1 2n n n n1
1 2n 0.6931 n
Section 8.10
Taylor and Maclaurin Series
3. For c 4, we have: f x cosx
f
4
f x sinx
f
4 22
f x cosx
f
4 22
f x sinx
f
f 4x cosx
f 4
2
2
4
2
4
2
2 2
and so on. Therefore, we have: cos x
f n 4 x 4 n n! n0
2
2
1 x 4 x 2! 4
2
x 4 3 x 4 4 . . . 3! 4!
1nn1 2 x 4 n . n! n0
2
2
[Note: 1nn1 2 1, 1, 1, 1, 1, 1, 1, 1, . . .] 5. For c 1, we have, f x ln x f x
1 x
f 1 1 1 x2
f x
f 1 1
2 x3
f x
f 4x f 5x
f 1 0
f 1 2
6 x4
f 41 6
24 x5
f 51 24
and so on. Therefore, we have: ln x
f n1x 1n n! n0
0 x 1 x 1
1
n0
n
x 12 2x 13 6x 14 24x 15 . . . 2! 3! 4! 5!
x 12 x 13 x 14 x 15 . . . 2 3 4 5 x 1n1 n1
161
162
Chapter 8
Infinite Series
7. For c 0, we have: f x sin 2x
f 0 0
f x 2 cos 2x
f 0 2
f x 4 sin 2x
f 0 0
f x 8 cos 2x
f 0 8
f 4x
f 40 0
16 sin 2x
f 5x 32 cos 2x
f 50 32
f 6x 64 sin 2x
f 60 0
f 7x 128 cos 2x
f 70 128
and so on. Therefore, we have: sin 2x
f n0xn 0x 2 8x3 0x 4 32x5 0x6 128x7 . . . 0 2x n! 2! 3! 4! 5! 6! 7! n0
1n2x2n1 8x3 32x5 128x7 . . . 3! 5! 7! 2n 1! n0
2x
9. For c 0, we have: f x secx
f 0 1
f x secxtanx
f 0 0
f x sec3x secxtan2x f x 5 f 4x
5
secx
f 0 1
xtanx secx
sec3
x
f 0 0
tan3
x 18
sec5
sec3
x
x secx
tan2
x
tan4
f 40 5
f n0xn x 2 5x 4 . . . 1 n! 2! 4! n0
11. The Maclaurin series for f x cos x is
1x2n . n0 2n!
Because f n1x ± sin x or ± cos x, we have f n1z ≤ 1 for all z. Hence by Taylor’s Theorem,
0 ≤ Rnx Since lim
n→
f n1z
n 1!
xn1 ≤
x . n 1! n1
xn1 0, it follows that Rnx → 0 as n → . Hence, the Maclaurin series for cos x converges to cos x for all x. n 1!
kk 1x 2 kk 1k 2x3 . . . , we have 2! 3! 23x 2 234x3 2345x 4 . . . 1 x2 1 2x 1 2x 3x 2 4x3 5x 4 . . . 2! 3! 5!
13. Since 1 xk 1 kx
1 n 1x .
n0
n
n
Section 8.10
15.
1
2 1 2
121 2x
4 x 2
and since 1 x1 2 1
1n 1 1 1 1 2 2 4 x n1
17. Since 1 x1 2 1
ex
xn
3 5.
x3
x4
. . 2n 3xn
2n n!
1n1 1 x2 2 n2
x2
1n 1 3 5 . . . 2n 1xn , we have 2n n! n1
. . 2n 1x 22n 1n 1 1 2 n! 2 n1 n
1n1 1 x 2 n2
we have 1 x 21 2 1
19.
3 5.
Taylor and Maclaurin Series
3 5.
3 5.
. . 2n 1x2n . n!
3n1
2
(Exercise 14)
. . 2n 3x 2n
2n n!
.
x5
n! 1 x 2! 3! 4! 5! . . .
n0
ex 2 2
x 2n x 2 2n x2 x4 x6 x8 1 2 3 4 . . . n n! 2 2 2! 2 3! 2 4! n0 n0 2 n!
21. sin x sin 2x
23.
1n x 2n1 x3 x5 x7 x . . . 3! 5! 7! n0 2n 1!
1n 22n1x2n1 1n2x2n1 8x3 32x5 128x7 . . . 2x 2n 1! 2n 1! 3! 5! 7! n0 n0
1n x 2n x2 x 4 . . . 1 2n! 2! 4! n0
cos x
1n x3n 1n x3 22n x3 x6 1 . . . 2n! 2n! 2! 4! n0 n0
cos x3 2
25.
ex 1 x
x2 x3 x4 x5 . . . 2! 3! 4! 5!
ex 1 x
x2 x3 x4 x5 . . . 2! 3! 4! 5!
e x ex 2x
2x3 2x5 2x7 . . . 3! 5! 7!
1 x x3 x2n1 x5 x7 e ex x . . . 2 3! 5! 7! n0 2n 1!
sinhx
1 27. cos2x 1 cos2x 2
31.
29. x sin x x x
1 2x2 2x4 2x6 . . . 11 2 2! 4! 6!
x2
1n2x2n 1 1 2 2n! n0
sin x x x3 3! x5 5! . . . x x 1
x2 x4 . . . 2! 4!
1nx2n
2n 1!, x 0
n0
x3 x5 . . . 3! 5!
x6 x4 . . . 3! 5!
1nx2n2 n0 2n 1!
163
164
Chapter 8
Infinite Series
eix 1 ix
33.
eix 1 ix
ix2 ix3 ix4 . . . x 2 ix3 x 4 ix5 x6 1 ix . . . 2! 3! 4! 2! 3! 4! 5! 6! ix2 ix3 ix4 . . . x 2 ix3 x 4 ix5 x6 1 ix . . . 2! 3! 4! 2! 3! 4! 5! 6!
2ix3 2ix5 2ix7 . . . 3! 5! 7!
eix eix 2ix
1n x2n1 x3 x5 x7 eix eix x . . . sinx 2i 3! 5! 7! n0 2n 1!
35. f x ex sin x
14
1x
x2 2
x3 6
x4
. . .
24
x x2
x x2
x3 x5 . . . 3 30
x 6 120 . . . x3
x5
P5 f
x5 x5 x3 x3 x4 x4 x5 . . . 2 6 6 6 120 12 24
−6
37. hx cos x ln1 x
1
x2 2
x4 24
4
. . .
x
x2 2
x3 3
x4 4
x5 5
. . .
P5
−3
x2 x3 x3 x4 x4 x5 x5 x5 . . . 2 3 2 4 4 5 6 24
x
x2
39. gx
2
6
3x5 40
9
h
x
x3
6 −2
−4
. . .
sin x . Divide the series for sin x by 1 x. 1x
5x2 5x4 6 6 x3 x5 0x4 . . . 1 x x 0x2 6 120 x x2 x3 x2 6 x2 x3 5x3 0x4 6 5x4 5x3 6 6 x5 5x4 6 120 5x4 5x5 6 6
gx x x2 4
x x2
5x 3 5x 4 . . . 6 6
g −6
6
P4 −4
41. y x 2
x4 x3 x x x sin x. 3! 3!
Matches (a)
43. y x x 2 Matches (c)
x3 x2 x 1x xex. 2! 2!
Section 8.10
x
45.
0
0
1n1t2n2 dt n 1! n0
x
0
1n1 t2n3
2n 3n 1!
x
0
n0
1n1x 2n3
2n 3n 1!
n0
1n x 1n1 x 12 x 13 x 14 . . . x 1 n1 2 3 4 n0
47. Since ln x
1 1 1 . . . 1 1n1 0.6931. 2 3 4 n n1
we have ln 2 1
49. Since ex
1n t2n 1 dt n! n0
x
et 1 dt 2
Taylor and Maclaurin Series
xn
x2
(10,001 terms)
x3
n! 1 x 2! 3! . . . ,
n0
we have e2 1 2
2n 22 23 . . . 7.3891. 2! 3! n0 n!
(12 terms)
51. Since cos x
1n x 2n x2 x 4 x6 x8 1 . . . 2n ! 2! 4! 6! 8! n0
1 cos x
1n x 2n2 x6 x8 x2 x 4 . . . 2! 4! 6! 8! n0 2n 2!
1n x 2n1 x x3 x5 x7 1 cos . . . x 2! 4! 6! 8! n0 2n 2!
we have lim
x→0
1
53.
0
1x 2n1 1 cos x lim 0. x→0 n0 2n 2! x
sin x dx x
1
0
1nx 2n dx n0 2n 1!
1n x 2n1
2n 12n 1! n0
1
0
1n
2n 12n 1!
n0
Since 1 7 7! < 0.0001, we have
1
0
sin x 1 1 dx 1 . . . 0.9461. x 3 3! 5 5!
Note: We are using lim x→0
2
55.
sin x 1. x
2
x cos x dx
0
0
1nx4n1 2 dx 2n! n0
Since 219 2 766,080 < 0.0001, we have
1
x cos x dx 2
0
0.3
57.
0.3
1 x3 dx
0.1
0.1
Since
1 7 56 0.3
n0
1n x4n3 2 4n 3 2n! 2
1n 2x4n3 2 n0 4n 32n!
0
23 2 27 2 211 2 215 2 219 2 0.7040. 3 14 264 10,800 766,080
0.3
x 5x . . . 1 x2 x8 16x 5x128 . . . dx x x8 56x 160 1664 3
6
9
12
4
7
< 0.0001, we have
1 x3 dx
2
0
0.17
0.3
0.1
2
0.3 0.1 80.3 1
4
0.14
1 0.37 0.17 0.2010. 56
10
13
0.1
165
166
Chapter 8
Infinite Series
59. From Exercise 19, we have 1 2
1
ex 2 dx 2
0
1 2
1
0
1 1n x 2n dx n n! 2 2 n0
61.
f x x cos 2x P5x x
2x3
1 2
x 21 n!2n 1
n
2n1
1
n
n0
0
1 2 1 3 2 1
1n 4n x2n1 2n! n0
63.
1n 1 n 2 n0 2 n!2n 1
1
2! 5
2
1
0.3414.
f x x ln x, c 1
x 13 x 14 71x 15 24 24 1920
P5x x 1
2x5 3
3! 7
23
3
2
g
P5
f −3
−2
3
4
P5 −2
−2
The polynomial is a reasonable approximation on the interval 14 , 2 .
The polynomial is a reasonable approximation on the interval 34 , 34 .
67. (a) Replace x with x.
65. See Guidelines, page 636.
(b) Replace x with 3x. (d) Replace x with 2x, then replace x with 2x, and add the two together.
(c) Multiply series by x.
69. y tan
g g kx x 2 ln 1 kv0 cos k v0 cos
2
3
gx g kx 1 kx kv0 cos k2 v0 cos 2 v0 cos
tan x
gx gx gx 2 gkx3 gk2x 4 . . . kv0 cos kv0 cos 2v02 cos2 3v03 cos3 4v04 cos4
tan x
gx2 kgx3 k2gx 4 . . . 2 3 3 2v0 cos 3v0 cos 4v04 cos4
71. f x
e0,
1 kx 3 v0 cos
tan x
1 kx 4 vo cos
4
. . .
2
x0 x0
1 x2
,
f x f 0 e1 x 0 lim x→0 x0 x 2
(a)
(b) f 0 lim
y
x→0
2
e1 x . Then x→0 x 2
Let y lim
1 −3 −2 −1
x 1
2
3
ln y lim ln x→0
e x lim x1 ln x lim 1 x x 1 x2
x→0
2
x→0
Thus, y e 0 and we have f 0 0. (c)
f n0 n f 0x f 0x 2 . . . x f 0 0 f x n! 1! 2! n0
This series converges to f at x 0 only.
73. By the Ratio Test: lim
n →
xn1 n! x 0 which shows that xn converges for all x. nlim → n 1 n 1! xn n0 n!
2
2
ln x
.
Review Exercises for Chapter 8
Review Exercises for Chapter 8 1. an
7. an
1 n!
2 3. an 4 : 6, 5, 4.67, . . . n Matches (a)
5n 2 n
9. lim
n→
8
5. an 100.3n1: 10, 3, . . . Matches (d)
n1 0 n2
11. lim
n→
n3 n 1 2
Converges
0
12 0
The sequence seems to converge to 5. lim an n→ lim
n→
5n 2 n
2 5 n
n→ lim 5
13. lim n 1 n lim n 1 n n→
n→
n 1 n n 1 n
sinn 0 n Converges
lim
n→
1 n 1 n
17. An 5000 1
15. lim
n→
n 1, 2, 3
0
0.05 4
n
Converges
50001.0125n
(a) A1 5062.50
A5 5320.41
A2 5125.78
A6 5386.92
A3 5189.85
A7 5454.25
A4 5254.73
A85522.43
(b) A40 8218.10 19. (a)
(b)
k
5
10
15
20
25
Sk
13.2
113.3
873.8
6448.5
50,500.3
(c) The series diverges geometric r 32 > 1
21. (a)
0
12
− 10
k
5
10
15
20
25
Sk
0.4597
0.4597
0.4597
0.4597
0.4597
(c) The series converges by the Alternating Series Test.
120
23. Converges. Geometric series, r 0.82, r < 1.
(b)
1
0
12
− 0.1
25. Diverges. nth Term Test. lim an 0. n→
167
168
27.
Chapter 8
3 2
Infinite Series
n
29.
n0
2
1 n
n0
2 Geometric series with a 1 and r 3 .
S
1 n 1 n 1 n 3 n0 2 n0 3 1 1 3 1 2 1 1 2 1 1 3 2 2
1 1 a 3 1 r 1 2 3 1 3
0.090.01
31. 0.09 0.09 0.0009 0.000009 . . . 0.091 0.01 0.0001 . . .
n
n0
33. D1 8
35. See Exercise 86 in Section 8.2.
D2 0.78 0.78 160.7
A
D 8 160.7 160.72 . . . 160.7n . . . 8
160.7
n
8
n0
37.
b→
16 4513 meters 1 0.7
ln x 1 3x 9x 3
3
39.
1
1 3 2n n n1
43.
1 n3 2n n3 2 lim 1 n→ n→ n3 2n 1 n3 2
n1
1
3 2 ,
2
1 1 1 2 n n n1 n1 n
3 5 . . . 2n 1 2 4 6 . . . 2n
1
1 3 5 . . . 2n 1 2 4 6 . . . 2n
an
1 1 1 > 32 54 . . . 2n 2n 2 2n 2n
By a limit comparison test with the convergent p-series
1
n1
lim
n
200e0.062 1 e0.06 12 1
Since the second series is a divergent p-series while the first series is a convergent p-series, the difference diverges.
By the Integral Test, the series converges.
n
n1
1 1 9 9
0
Pe rt 1 er 12 1
$5087.14
b
x4 lnx dx lim
1
41.
0.09 1 1 0.01 11
1
1
1
2n 2 n diverges (harmonic series),
Since
the series converges.
n1
n1
so does the original series.
45. Converges by the Alternating Series Test (Conditional convergence)
49.
n
e
n1
lim
n→
47. Diverges by the nth Term Test
51.
n2
an1 n1 lim n12 n→ e an lim
n→
lim
n→
n 1 2n1n
2 en
2
en
2
en n
e 1 n n 1 2n1
01 0 < 1 By the Ratio Test, the series converges.
2n
n
n1
lim
n→
3
an1 2n1 lim n→ n 13 an lim
n→
3
n3
2n
2n 2 n 13
Therefore, by the Ratio Test, the series diverges.
Review Exercises for Chapter 8
53. (a) Ratio Test: lim
n→
an1 n 13 5n1 lim n→ an n3 5n
n n 135 53 < 1
lim
n→
Converges (b)
(c)
x
5
10
15
20
25
Sn
2.8752
3.6366
3.7377
3.7488
3.7499
(d) The sum is approximately 3.75.
4
0
12
−1
55. (a)
1
N
x2
1x
dx
N
1 N
N
5
10
20
30
40
n
1.4636
1.5498
1.5962
1.6122
1.6202
0.2000
0.1000
0.0500
0.0333
0.0250
5
10
20
30
40
n
1.0367
1.0369
1.0369
1.0369
1.0369
0.0004
0.0000
0.0000
0.0000
0.0000
N
1 2
n1
N
(b)
N
1 1 dx 4 x5 4x
N
1 4N4
1 dx x2
N N
1 5
n1
N
1 dx x5
The series in part (b) converges more rapidly. The integral values represent the remainders of the partial sums. 57. f x ex 2
f 0 1
1 fx ex 2 2 1 f x ex 2 4 1 fx ex 2 8
f0 f 0
1 4
f0
P3x f 0 f0x f 0
1 2
1 8
x2 x3 f 0 2! 3!
1 1 x2 1 x3 1 x 2 4 2! 8 3! 1 1 1 1 x x2 x3 2 8 48
59. sin95 sin
95 95 95 95 0.996 95180 95180 180 3! 180 5! 180 7! 180 9!
61. ln1.75 0.75
3
3
5
5
7
7
9
9
0.752 0.753 0.754 0.755 0.756 . . . 0.7515 0.560 2 3 4 5 6 15
169
170
Chapter 8
Infinite Series
63. f x cos x, c 0 Rnx
f n1z n1 x n 1!
f n1z
x n1 n 1!
≤ 1 ⇒ Rnx ≤
(a) Rnx ≤
0.5n1 < 0.001 n 1!
(b) Rnx ≤
This inequality is true for n 6.
This inequality is true for n 4. (c) Rnx ≤
0.5n1 < 0.0001 n 1!
(d) Rnx ≤
10 x
2n1 < 0.0001 n 1!
This inequality is true for n 10.
This inequality is true for n 5.
65.
1n1 < 0.001 n 1!
n
n0
Geometric series which converges only if x 10 < 1 or 10 < x < 10.
67.
1nx 2n n 12 n0
lim
n→
69.
un1 1n1x 2n1 lim n→ un n 22
n 12
1nx 2n
x2
n!x 2
n
n0
lim
n→
which implies that the series converges only at the center x 2.
R1 Center: 2 Since the series converges when x 1 and when x 3, the interval of convergence is 1 ≤ x ≤ 3.
y
71.
x2n n!2
1
n
4n
n0
y
1n12n 2x2n1 1n2nx2n1 n 2 4 n! 4n1n 1! 2 n1 n0
y
1n12n 22n 1x2n 4n1n 1! 2 n0
x2y xy x2y
1n12n 2x2n2 1n12n 22n 1x2n2 x2n1 1n n 2 n1n 1! 2 n1n 1! 2 4 4 4 n! n0 n0 n0
1
n1
n0
1n14n 12 1 1n n x 2n2 4n1n 1! 2 4 n! 2
1n11 1 1n n x 2n2 0 4nn! 2 4 n! 2
n0
73.
2 2 3 a 3 x 1 x 3 1 r
3 3
n0
2 x
n
n0
n0
2n 22n 1 1n12n 2 1n n1 n 2 x 2n2 n1 2 2 4 n 1! 4 n 1! 4 n!
1n12n 22n 1 1 1 1n n x 2n2 4n1n 1! 2 4 n! 2
n0
2x n 3n1
un1 n 1!x 2n1 lim n→ un n!x 2n
75. Derivative:
2nx n1 n1 n1 3
Review Exercises for Chapter 8 2x 4 8 2 77. 1 x x2 x3 . . . 3 9 27 n0 3
n
1 3 , 1 2x 3 3 2x
3 3 < x < 2 2
f x sinx
79.
fx cosx f x sinx f x cosx, . . . sinx
f nxx 3 4 n n! n0
2
2
2
2
x 34 2 22!x 34
81. 3x eln3x ex ln3 and since ex
2
2 1nn1 2x 3 4 n . . . 2 n0 n!
xn
n!, we have
f x
83.
n0
3x
x ln 3n n! n0
1 x ln 3
1 x
fx x2 ln2 3 x3 ln3 3 x 4 ln4 3 . . . . 2! 3! 4!
f x
1 x2
2 x3
6 fx 4, . . . x f n1x 1 n 1 x n0 n!
85. 1 xk 1 kx
n!x 1n x 1n n! n0 n0
kk 1x2 kk 1k 2x3 . . . 2! 3!
x 1 54 5x2 1 54 59 5x3 . . . 5 2! 3!
1 x1 5 1
1 1 4x2 1 4 9x3 . . . 1 x 2 5 5 2! 533!
ln x
87.
1n14 x 5 n2
1
x 2 6 3 . . . x2 x 5 25 125
1
n1
n1
ln
x 1n , n
54 1 5 4n 1
n1
n1
89.
ex
xn
n!,
n0 n
n1
1
. . 5n 6xn
5nn!
0 < x ≤ 2
n1
9 14 .
1
1 0.2231 4nn
e1 2
1
< x <
2 n! 1.6487
n0
n
171
172
Chapter 8
cos x
91.
Infinite Series
1
n
x2n , < x < 2n!
n
22n 0.7859 32n2n!
n0
cos
23 1 n0
95. (a) f x e2x
f 0 1
(b) ex
f0 2
f x 4e2x
f 0 4
f x 8e2x
f 0 8
e2x 1 2x
4x2 8x3 . . . 2! 3!
1 2x 2x2
x
0
e2x
1 x x2 x6 . . . 2
3
x2 x2 4 x3 x3 x3 x3 . . . 1 2x 2x 2 x 3 . . . 2 2 6 6 2 2 3
1nt 2n1 n0 2n 1!
1 1n t n 1 t n0
99.
1 sin t t n0 2n 1!
sin t dt t
n t 2n
ln1 t
1n t 2n1
x
2n 12n 1!
0
n0
1nx 2n1
2n 12n 1!
x
n0
arctan x x
101.
0
x3 x5 x7 x9 . . . 3 5 7 9
x 52 x 92 x132 x172 . . . arctan x x 3 5 7 9 x lim
x→0
arctan x 0 x
1 arctan x 2x 1 x2 lim lim 0. By L’Hôpital’s Rule, lim x→0 x→0 x→0 1 x2 1 x 2x
Problem Solving for Chapter 8 1. (a) 1
3 29 427 . . . 3 3 1
1
1
1 2
n0
1 2 (b) 0, , , 1, etc. 3 3 (c) lim Cn 1 n→
3 3
n0
2xn 4x 2 8x 3 . . . 1 2x n! 2! 3! n0
4 1 2x 2x 2 x 3 . . . 3
x2 x3 . . . 2 6
1 x x x2
xn
4 3 . . . x 3
(c) e2x ex ex 1 x
sin t
n!
n0
fx 2e2x
97.
93. The series for Exercise 41 converges very slowly because the terms approach 0 at a slow rate.
1 2
n
110
n
13 1 1 23
1n t n1 1 dt 1t n1 n0
1n t n lnt 1 t n0 n 1
lnt 1 dt t
1n t n1 2 n0 n 1
x
0
1n x n1 2 n0 n 1
Problem Solving for Chapter 8 nn 1 . 2
3. If there are n rows, then an For one circle,
3 2
1
a1 1 and r1
1 3 3 2
3
6
1
r1
23 1 2
For three circles, a2 3 and 1 23r2 2r2 r2
1 2 23
2r2 1
r2
3 r2
For six circles, a3 6 and 1 23r3 4r3 r3
1 23 4
2r3 1
Continuing this pattern, rn Total Area rn2an An lim An
n→
5. (a)
a x n
n
2
1
4
r3
3 r3
1 . 23 2n 1
23 12n 1
2
nn 1 2
nn 1 2 2 3 2n 12
8
1 2x 3x2 x3 2x4 3x5 . . . 1 x3 x6 . . . 2x x4 x7 . . . 3x2 x5 x8 . . . 1 x3 x6 . . . 1 2x 3x2 1 2x 3x2
1 1 x3
R 1 because each series in the second line has R 1. (b)
a x n
n
a0 a1x . . . ap1x p1 a0 x p a1x p1 . . . . . . a01 x p . . . a1 x1 x p . . . . . . a p1 x p11 x p . . . a0 a1 x . . . a p1x p11 x p . . . a0 a1 x . . . ap1 x p1
R1
1 . 1 xp
173
174
Chapter 8
Infinite Series
ex 1 x
7.
xex x x2
x2 . . . 2! x n1 x3 . . . 2! n0 n!
xex dx xex ex C
x n2
n 2n!
n0
Letting x 0, C 1. Letting x 1,
1
1
1
n 2n! 2 n 2n!.
1
n0
n1
1
1
n 2n! 2.
Thus,
n1
9. Let a1
0
sin x dx, a2 x
2
sin x dx, a3 x
3
2
sin x dx, etc. x
Then,
0
sin x dx a1 a2 a3 a4 . . . . x
Since lim an 0 and an1 < an, this series converges. n→
11. (a) a1 3.0 a2 1.73205 a3 2.17533 a4 2.27493 a5 2.29672 a6 2.30146 lim an
n→
1 13 [See part (b) for proof.] 2
(b) Use mathematical induction to show the sequence is increasing. Clearly, a2 a a1 aa > a a1. Now assume an > an1. Then an a > an1 a an a > an1 a
an1 > an. Use mathematical induction to show that the sequence is bounded above by a. Clearly, a1 a < a. Now assume an < a. Then a > an and a 1 > 1 implies aa 1 > an1 a2 a > an a2 > an a a > an a an1. Hence, the sequence converges to some number L. To find L, assume an1 an L: L a L ⇒ L2 a L ⇒ L2 L a 0 L
1 ± 1 4a . 2
Hence, L
1 1 4a . 2
Problem Solving for Chapter 8
13. (a)
1
2
n1
n 1n
S1
1 1 1 1 1 . . . 211 221 231 241 251 1 1 20
S1 1
1 9 8 8
S3
9 1 11 8 4 8
S4
11 1 45 8 32 32
S5
1 47 45 32 16 32
2n 1 21 an1 n1 1n 1 1 1n 1 an 2 2 n
(b)
n
This sequence is 18, 2, 18, 2, . . . which diverges. (c)
2 n
1 n 1n
2 2
21 1
n
1n
n
1 1 n 1 1 n 12 → 1 and n 2 → 1. → < 1 converges because 21 2, 2, 2, 2, . . . and n 1 2 2 n
15. S6 130 70 40 240 S7 240 130 70 440 S8 440 240 130 810 S9 810 440 240 1490 S10 1490 810 440 2740
175
C H A P T E R 9 Conics, Parametric Equations, and Polar Coordinates Section 9.1
Conics and Calculus . . . . . . . . . . . . . . . . . . . . 424
Section 9.2
Plane Curves and Parametric Equations . . . . . . . . . . 434
Section 9.3
Parametric Equations and Calculus
Section 9.4
Polar Coordinates and Polar Graphs . . . . . . . . . . . . 444
Section 9.5
Area and Arc Length in Polar Coordinates . . . . . . . . .452
Section 9.6
Polar Equations of Conics and Kepler’s Laws . . . . . . . 458
Review Exercises
. . . . . . . . . . . . 439
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 469
C H A P T E R 9 Conics, Parametric Equations, and Polar Coordinates Section 9.1
Conics and Calculus
Solutions to Even-Numbered Exercises 2. x2 8y
4.
Vertex: 0, 0
Center: 2, 1 Ellipse Matches (b)
p2 > 0 Opens upward Matches graph (a).
6.
x 22 y 12 1 16 4
x2 y2 1 9 9
8.
Circle radius 3. Matches (g)
x 22 y2 1 9 4 Hyperbola Center: 2, 0 Horizontal transverse axis. Matches (d)
10. x2 8y 0
12. x 12 8 y 2 0
x2 42y Vertex: 0, 0 Focus: 0, 2
x 12 42 y 2
y
(0, 0) −8
x
−4
8
4
y
Focus: 1, 4
−4
Directrix: y 2
Vertex: 1, 2 Directrix: y 0
−8
−8
x
−4
8
4 −4
(1, − 2) −12
−8 −12
14. y 2 6y 8x 25 0
16. y 2 4y 8x 12 0
y 2 6y 9 8x 25 9
y 2 4y 4 8x 12 4
y 32 42x 2
y 22 42x 2
Vertex: 2, 3 Focus: 4, 3
8
Focus: 0, 2
Directrix: x 0
4
Directrix: x 4
−20 −16 −12
−8
x
−4
(− 2, − 3)
y 4 2 −6
−4
−2
−12
x 4
(2, − 2) −4
−8
424
Vertex: 2, 2
y
−6 −8
6
Section 9.1 y 16x2 8x 6 16x2 8x 16 10
18.
x2 2x 1 8y 9 1
x 12 42 y 1
6y 10 x 42
x 42 6y 53 x 42 4
425
20. x2 2x 8y 9 0
6y x 42 10
32
Conics and Calculus
Vertex: 1, 1
2
Focus: 1, 3
y 5 3
Vertex: 4, 53
−8
10
Directrix: y 1
4
Focus: 4, 16
−10
Directrix: y
−2
19 6
10
−4
x 12 42 y 2
22.
24. Vertex: 0, 2
y 22 42x 0
x2 2x 8y 15 0
y 2 8x 4y 4 0 y 4 x 22 4x x2
26.
28. From Example 2: 4p 8 or p 2
x 4x y 0
Vertex: 4, 0
2
x 42 8 y 0 x2 8x 8y 16 0 30. 5x2 7y 2 70
y 6
y2 x2 1 14 10 a2
14,
b2
32.
1 3 a2 1, b2 , c2 4 4
4 2
10,
c2
4
−6
x
−2
2
Center: 0, 0
−4
Foci: ± 2, 0
−6
x 22 y 42 1 1 14
4
Center: 2, 4
6
Foci:
Vertices: ± 14, 0
2
, 4
3
e
2
16x2 25y2 64x 150y 279 0 16x2 4x 4 25y2 6y 0 279 64 225 10
x 22 y 32 1 58 25
y 1 x
−1
1
2
3
4
−1
5 2 9 a2, , b2 , c2 a2 b2 8 5 40
−2
Center: 2, 3
−4
Foci:
2
Vertices: e
±
2
c 3 a 5
310 , 3 20 ±
10
4
, 3
−3
(2, −3)
−5
−4
−3
−2
x
−1 −1
(− 2, − 4)
−2 −3 −4
Vertices: 1, 4, 3, 4
14 2 e 7 14
34.
2 ±
3
y
−5
426
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
36x2 9y 2 48x 36y 43 0
36.
4 4 9 y 2 4y 4 43 16 36 36 x2 x 3 9 9
x 23 y 2 1 14 1 2
2
1 3 a2 1, b2 , c2 4 4 Center: Foci:
32, 2
32, 2 ± 23
32, 3, 32, 1
Vertices:
3
Solve for y: 9 y 2 4y 4 36x2 48x 43 36
y 22
−4
2
36x2 48x 7 9
−1
1 y 2 ± 36x2 48x 7 3
(Graph each of these separately.)
2x2 y 2 4.8x 6.4y 3.12 0
38.
50x2 25y 2 120x 160y 78 0
12 32 36 256 x 25 y 2 y 78 72 256 250 5 25 5 25
50 x2
x 65 2 y 165 2 1 5 10 a2 10, b2 5, c2 5 Center: Foci:
56, 165
56, 165 ± 5
Vertices:
7
56, 165 ± 10
Solve for y: y2 6.4y 10.24 2x2 4.8x 3.12 10.24
−7
5 −1
y 3.22 7.12 4x 2x2 y 3.2 ± 7.12 4x 2x2 40. Vertices: 0, 2, 4, 2
(Graph each of these separately.) 42 Foci: 0, ± 5
1 2 Horizontal major axis
Major axis length: 14
Center: 2, 2
Center: 0, 0
a 2, c 1 ⇒ b 3
c 5, a 7 ⇒ b 24
x 22 y 22 1 4 3
y2 x2 1 24 49
Eccentricity:
Vertical major axis
Section 9.1
44. Center: 1, 2
46.
Vertical major axis
Conics and Calculus
y2 x2 1 25 9 a 5, b 3, c a2 b2 34
Points on ellipse: 1, 6, 3, 2
Center: 0, 0
From the sketch, we can see that h 1, k 2, a 4, b 2
y
y
Foci: ± 34, 0 (1, 2) x
−2
−8
3 Asymptotes: y ± x 5
(3, 2) −4
10 8 6 4 2
Vertices: ± 5, 0
(1, 6) 6
x 12 y 22 1. 4 16
−4 −2 −4 −6 −8 −10
4 −2
48.
427
y 1 2 x 4 2 1 122 52
x 4
8 10
50. y 2 9x2 36x 72 0 y 2 9x2 4x 4 72 36 36
a 12, b 5, c a 2 b2 13
x 22 y2 1 36 4
Center: 4, 1 Vertices: 4, 11, 4, 13 Foci: 4, 14, 4, 12 12 Asymptotes: y 1 ± x 4 5
a 6, b 2, c a2 b2 210 Center: 2, 0 Vertices: 2, 6, 2, 6
Foci: 2, 210, 2, 210
y 20
Asymptotes: y ± 3x 2 y
5 x −5
1
2
6
7
8 5 x
−2 −1
− 20
4
5
−5
52. 9x2 6x 9 4 y2 2y 1 78 81 4 1
54.
9x 2 y 2 54x 10y 55 0 9x 2 6x 9 y 2 10y 25 55 81 25
9x 32 4 y 12 1
y 12 x 32 1 14 19
1
13 1 1 a ,b ,c 2 3 6
Foci:
3, 21, 3, 23
1 3, 1 ± 13 6
2
10 1 a , b 1, c 3 3
1
Center: 3, 5
3
Center: 3, 1 Vertices:
x 32 y 52 1 19 1
y
3 Asymptotes: y 1 ± x 3 2
−3
x
−1 −1
3 ± 13, 5 10 , 5 Foci: 3 ± 3
10
Vertices:
−8
Solve for y: y2 10y 25 9x2 54x 55 25
y 52 9x2 54x 80 y 5 ± 9x2 54x 80 (Graph each curve separately.)
2 0
428
Chapter 9
Conics, Parametric Equations, and Polar Coordinates 58. Vertices: 0, ± 3
3y 2 x 2 6x 12y 0
56.
Asymptotes: y ± 3x
3 y 2 4y 4 x 2 6x 9 0 12 9 3
y 2 x 3 1 1 3 2
2
a 1, b 3, c 2
Vertical transverse axis a3 a Slopes of asymptotes: ± ± 3 b Thus, b 1. Therefore,
6
Center: 3, 2 Vertices: 3, 1, 3, 3
−4
10
Foci: 3, 0, 3, 4
y 2 x2 1. 9 1
−4
Solve for y: 3 y 2 4y 4 x2 6x 12
y 22
x2 6x 12 3
y2 ±
x
2
6x 12 3
(Graph each curve separately.) 62. Center: 0, 0
60. Vertices: 2, ± 3 Foci: 2, ± 5
Vertex: 3, 0
Vertical transverse axis
Focus: 5, 0
Center: 2, 0
Horizontal transverse axis
a 3, c 5, b2 c2 a2 16
a 3, c 5, b2 c2 a2 16
y2 x 22 Therefore, 1. 9 16
Therefore,
64. Focus: 10, 0
66. (a)
3 Asymptotes: y ± x 4
y2 x2 1. 9 16
y 2 x2 1, y 2 2x2 4, 2yy 4x 0, 4 2 y
4x 2x 2y y
Horizontal transverse axis Center: 0, 0 since asymptotes intersect at the origin. c 10
At x 4: y ± 6, y
± 24
6
±
4 3
4 At 4, 6: y 6 x 4 or 4x 3y 2 0 3
b 3 3 Slopes of asymptotes: ± ± and b a a 4 4 c2 a2 b2 100 Solving these equations, we have a2 64 and b2 36. Therefore, the equation is x2 y2 1. 64 36
4 At 4, 6: y 6 x 4 or 4x 3y 2 0 3 (b) From part (a) we know that the slopes of the normal lines must be 34. 3 At 4, 6: y 6 x 4 or 3x 4y 36 0 4 3 At 4, 6: y 6 x 4 or 3x 4y 36 0 4
68. 4x2 y 2 4x 3 0
70. 25x2 10x 200y 119 0
72. y2 x 4y 5 0
A 4, C 1
A 25, C 0
A 0, C 1
AC < 0
Parabola
Parabola
Hyperbola
Section 9.1 2x2 2xy 3y y 2 2xy
74.
Conics and Calculus
9x2 54x 81 36 4 y2 4y 4
76.
2x2 y 2 3y 0
9x2 4y2 54x 16y 61 0
A 2, C 1, AC > 0
A 9, C 4, AC > 0
Ellipse
Ellipse
78. (a) An ellipse is the set of all points x, y, the sum of whose distance from two distinct fixed points (foci) is constant.
x h y k x h y k 1 or 1 a2 b2 b2 a2 2
(b)
2
2
2
82. Assume that the vertex is at the origin. x2 4py
(a)
82 4p
c 80. e , c a2 b2 a
For e 1, the ellipse is elongated.
84. (a) Without loss of generality, place the coordinate system so that the equation of the parabola is x2 4py and, hence,
3 100
y
6400 y y 1600 3 3
2 ⇒ x± 100
(b)
1283 ± 6.53 meters.
dy 1 x1 dx 2
x 2x 9
4 100
)
( 0, 1003 )
( 8, 1003 )
3x 9
2 100
x3
1 100
−8
dy 0 dx
At 0, 0, the slope is 1: y x. At 6, 3, the slope is 2: y 2x 9. Solving for x,
y
(
x2 4x 4y 0 2x 4 4
5 100
3 −8, 100
2p1 x.
Therefore, for distinct tangent lines, the slopes are unequal and the lines intersect.
(b) The deflection is 1 cm when y
0 < e < 1
For e 0, the ellipse is nearly circular.
1600 p 3 x2 4
y 3.
x
−4
4
8
Point of intersection: 3, 3
86. The focus of x2 8y 42y is 0, 2. The distance from a point on the parabola, x, x28, and the focus, 0, 2, is d
x 0 x8 2 . 2
2
2
Since d is minimized when d 2 is minimized, it is sufficient to minimize the function f x x2
4
x2 2 8
3
16
xx
x 2 = 8y
(0, 2)
x x3 x. 4 16
fx 0 implies that x3
y
2 x2 2 . 8
fx 2x 2
429
2
x, x 8
( )
1 −3
−2
x
−1
1 −2
16x 1 0 ⇒ x 0. 2
This is a minimum by the First Derivative Test. Hence, the closest point to the focus is the vertex, 0, 0.
2
3
430
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
88. (a) C 0.0853t2 0.2917t 263.3559 (b)
1 x y2 4
90.
320
1 x y 2
0
1 x2 1
18 0
s
dC 0.1706t 0.2971 (c) dt
4
1
0
y
3
y4 dy 21 4 y 2
2
x −1
2
3
4
1 y4 y2 4 ln y 4 y 2 4
1 420 4 ln 4 20 4 ln 2 4
4 0
25 ln2 5 5.916
1 1
dy
0
2
−1
4
5 4
y2 4
5
The consumption of fruits is increasing at a rate of 0.1706 pounds/year.
h
92. x2 20y y y
94. A 2
x 20 x 10
y12 dy
0
x
0
1
x 10
2
r
dx 2
0
10 23100 x
r
h
4p
r
S 2
4py dy
0
2
2 32
0
4p x100 x2 dx 10
(b) The thumbtacks are located at the foci and the length of string is the constant sum of the distances from the foci.
98.
e 0.0167
Focus Vertex
h
32
0
8 ph32 3
100 r 232 1000 15
96. (a) At the vertices we notice that the string is horizontal and has a length of 2a.
23y
c a c 149,570, 000
c 2,497,819
Focus Vertex
Least distance: a c 147,072,181 km Greatest distance: a c 152,067,819 km
100. e
AP AP
122,000 4000 119 4000 122,000 4000 119 4000
121,881
0.9367 130,119
102.
y2 x2 1 a2 b2 y2 x2 2 2 2 1 2 a a b a y2 x2 2 2 1 2 a a a c2a2 x2 y2 1 a2 a21 e2 As e → 0, 1 e2 → 1 and we have x2 y2 2 1 or the circle x2 y 2 a2. 2 a a
Section 9.1
104.
x2 y2 1 2 4.5 2.52
Conics and Calculus
y
y2 2.5 2
x2 4.52 1
5 ft x
9 x ± 2.52 y 2 5
3 ft
9 ft
V Area of bottomLength Area of topLength V
4.52.5 16 16 2
90
144 5
0.5
9 2.52 y 2 dy (Recall: Area of ellipse is ab.) 5
0
2 y2.52 y 2 2.52 arcsin 2.5 0
0.5
y
1
90
72 1 0.56 2.52 arcsin
318.5 ft3 5 5
106. 9x2 4y 2 36x 24y 36 0 18x 8yy 36 24y 0
8y 24y 18x 36 y
18x 36 8y 24
y 0 when x 2. y undefined when y 3. At x 2, y 0 or 6. Endpoints of major axis: 2, 0, 2, 6 At y 3, x 0 or 4. Endpoints of minor axis: 0, 3, 4, 3 Note: Equation of ellipse is
4
108. (a) A 4
0
x 22 y 32 1 4 9
3 x 3 16 x2 dx x16 x2 16 arcsin 4 2 4
4
V 2
(b) Disk:
0
9 9 16 x2 dx 16 8
4 0
12
16x 31x
4
3
0
48
3 y 16 x2 4 y 1 y2
4
S 22
0
3x 416 x2
1 16169x x 2
2
16161616 x x 9x dx 4 43
3 16 x2 4
2
7x 3 7x256 7x2 256 arcsin 16 87
—CONTINUED—
4
2
2
0
4 0
16 x2
256 7x2
416 x2
dx
3 4
7 3 487 256 arcsin
138.93 4 87
4
0
256 7x2 dx
431
432
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
108. —CONTINUED—
4
V 4
(c) Shell:
0
416 x dx 3 2316 x
x
3
1
2
4
2
2 32
0
64
4 x 9 y 2 3 4y x 3 9 y2 1 x 2
S 22 4
3
0
110. (a)
3
0
1 9916y y 2
2
4 9 y 2 3
99 99y y 16y 2
2
2
dy
4 81 7y 2 dy 9
16 9 27
8 3712 81 ln37 12 81 ln 9 168.53 97
7y81 7y
2
81 ln 7y 81 7y 2
x2 y2 21 2 a b
At P, y
b2 a2
Slope of line through c, 0 and x0, y0: m2
y0 x0 c
x
y0 m. 0
y0 b2x0 2 x0 c a y0 a2y 2 b2x0x0 c 2 0 2 y0 bx a y0x0 c b2x0 y0 1 2 0 x0 c a y0
a2y02 b2x02 b2x0c a2b2 b2x0c b2a2 x0c b2 2 2 2 2 2 2 x0y0a b a y0c x0y0c a y0c y0cx0c a y0c
arctan
b2 b2 arctan y0c y0c
m m tan 1 1 m1m
y0 x0 c
xb2 ya2
m m (c) tan 2 1 m2m
0
(b) Slope of line through c, 0 and x0, y0: m1
2x 2yy 2 0 a2 b y
3
y0 b2x 2 0 x0 c a y0 a2y 2 b2x0x0 c 2 0 2 y0 bx a y0x0 c b2x0y0 1 2 0 x0 c a y0
a2y02 b2x02 b2x0c a2b2 b2x0c b2a2 x0c b2 2 2 2 2 2 2 2 a x0y0 a cy0 b x0 y0 x0y0a b a cy0 y0cx0c a y0c
arctan
yb c 2
0
Since , the tangent line to an ellipse at a point P makes equal angles with the lines through P and the foci.
Section 9.1
112. (a) e
(b)
Conics and Calculus
114. The transverse axis is vertical since 3, 0 and 3, 3 are the foci.
a2 b2 c ⇒ ea2 a2 b2. Hence, a a
3, 23
x h2 y k2 1 a2 b2
Center:
x h2 y k2 2 1. 2 a a 1 e2
3 5 c , 2a 2, b2 c2 a2 2 4 Therefore, the equation is
x 22 y 32 1 4 41 e2
y 32 2 x 32 1. 1 54
7
−3
433
9 −1
(c) As e approaches 0, the ellipse approaches a circle. 116. Center: 0, 0 Horizontal transverse axis Foci: ± c, 0 Vertices: ± a, 0 The difference of the distances from any point on the hyperbola is constant. At a vertex, this constant difference is
a c c a 2a. Now, for any point x, y on the hyperbola, the difference of the distances between x, y and the two foci must also be 2a. x c2 y 02 x c2 y 02 2a x c2 y 2 2a x c2 y 2
x c2 y 2 4a2 4ax c2 y 2 x c2 y 2 4xc 4a2 4ax c2 y 2 xc a2 ax c2 y 2
y
x2c2 2a2cx a4 a2x2 2cx c2 y 2
x2
c2
x2 a2
a2
a2y2
a2
c2
( x, y )
a2
(c, 0) (− c, 0) (− a, 0) (a, 0)
2
y 1 c2 a2
Since a2 b2 c2, we have x2a2 y 2b2 1.
118. c 150, 2a 0.001186,000, a 93, b 1502 932 13,851 x2 y2 1 2 93 13,851
752 13,851
x 110.3 miles.
y2 x2 21 2 a b 2x 2yy b2x 2 0 or y 2 2 a b ay y y0
When y 75, we have x2 932 1
120.
b2x0 x x0 a2y0
a2y0y a2y02 b2x0x b2x02 b2x02 a2y02 b2x0x a2y0 y a2b2 b2x 0 x a2y 0 y x0x y0 y 2 1 a2 b
x
434
Chapter 9
Conics, Parametric Equations, and Polar Coordinates Ax2 Cy2 Dx Ey F 0
122.
A x2
(Assume A 0 and C 0; see (b) below)
D E x C y 2 y F A C
D D2 E E2 D2 E2 x 2 C y2 y 2 F R A 4A C 4C 4A 4C
A x2
x 2AD y 2CE 2
2
C
A
(a) If A C, we have
R AC (b) If C 0, we have
x 2AD y 2CE 2
2
R A
A x
(c) If AC > 0, we have
C y
2
D
E
R A
2
F Ey
D2 . 4A
E 2C
2
F Dx
E2 . 4C
2
These are the equations of parabolas.
If A 0, we have
which is the standard equation of a circle.
x 2A y 2C
D 2A
1
R C
(d) If AC < 0, we have
x 2AD y 2CE 2
which is the equation of an ellipse.
R A
2
R C
±1
which is the equation of a hyperbola. 124. True
126. False. The y4 term should be y2.
Section 9.2
Plane Curves and Parametric Equations
2. x 4 cos2
y 2 sin
0 ≤ x ≤ 4
2 ≤ y ≤ 2
(a)
2
x
0
y
2
0
4
2
2
4
2
0
2
0
2
2
4
(c)
3
−1
(b)
(d) 3 2 1 x 1
2
3
5
−3
y
−2
128. True
x cos2 4 y2 sin2 4
5
y2 x 1 4 4 x 4 y2, 2 ≤ y ≤ 2 (e) The graph would be oriented in the opposite direction.
Section 9.2 4. x 3 2t
435
6. x 2t2
y 2 3t y23
Plane Curves and Parametric Equations
y t4 1
3 2 x
2x
y
2y 3x 13 0
2
x2 1, x ≥ 0 4
1
For t < 0, the orientation is right to left.
y
For t > 0, the orientation is left to right. y
6
6 5
4
4 3
2
2 x 2
6
4
1
8
x
−1 −1
1
2
3
4
5
6
8. x t 2 t, y t 2 t Subtracting the second equation from the first, we have x y 2t or t
xy 2
2
t
1
0
1
y
2 4
x y 2 x y y 4 2
x
2
0
0
2
6
y
6
2
0
0
2
3 2
Since the discriminant is
x
−1
B2 4AC 2 2 41 1 0,
3
2
4
−1
the graph is a rotated parabola. 4 t, t ≥ 0 10. x
12. x 1
y3t y3
x 4,
x ≥ 0
x1
3
y
2 1 −2
−1
1
2
yt2
1 1 implies t t x1
1 1 x1
x −1
14. x t 1
yt1
y
−3
1 t
x y 2 1 y 3 y 5 4 3
y
3 2 1
−2
1
−3
x
−2
x
2 1
2
3
4
5
−3
16. x et, x > 0 y e2t 1 y x2 1
18.
y 3 2
1 1, x > 0 x2
1 −3
−2
−1
x −1 −2 −3
1
3
x tan2
y
y sec2
4
sec2 tan2 1
3
yx1
2
x ≥ 0
1 x 1
2
3
4
436
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
x 2 cos
20.
x cos
22.
y 6 sin
2x 6y 2
2
x 4 2 cos
24.
y 2 sin 2
y 1 2 sin
y 4 sin cos
cos2 sin2 1
x 4 2 4 cos2
1 x2 sin2
y2 x2 1 ellipse 4 36
y 1 2 4 sin2
y ± 4x1 x2
x 4 2 y 1 2 4 3
3
y
−1
−2
4
7
2
2 −6 −4
4
−2
6
x
26. x sec
28.
2
y tan
x cos3
1.5
y sin3 −3
x2 sec2
3
y2 tan2
−2
x2 3 cos2
2
y2 3 sin2
−2
30. x ln 2t
−1.5
32. x e2t
3
4
y et
y t2 t
−5
−3
ex 2
−2
3 −1
e 2x 1 2x e y r 4
y2 x
−1
y > 0
3 −1
y x, x > 0
34. By eliminating the parameters in (a) – (d), we get x2 y 2 4. They differ from each other in orientation and in restricted domains. These curves are all smooth. 4t 2 1 1 1 4 2 y (b) x (a) x 2 cos , y 2 sin t t t
y
x ≥ 0, x 2
3
y0
y 1 −3
2 x
−1
1
3
1 x
−1
−3
1
3
−1 −2
(c) x t
y 4 t
x ≥ 0
y ≥ 0
(d) x 4 e2t
y et
2 < x ≤ 0
y > 0
y
y
3
3
2
1
1
x 1
2
3
−3
−2
−1
x
Section 9.2
Plane Curves and Parametric Equations
36. The orientations are reversed. The graphs are the same. They are both smooth. 38. The set of points x, y corresponding to the rectangular equation of a set of parametric equations does not show the orientation of the curve nor any restriction on the domain of the original parametric equations.
x h r cos
40.
x h a sec
42.
y k r sin
y k b tan
cos
xh r
xh sec a
sin
yk r
yk tan b
cos2 sin2
x h 2 y k 2 1 r2 r2
x h 2 y k 2 1 a2 b2
x h 2 y k 2 r 2 44. From Exercise 39 we have
46. From Exercise 40 we have
x 1 4t
x 3 3 cos
y 4 6t.
y 1 3 sin .
Solution not unique
48. From Exercise 41 we have a 5, c 3 ⇒ b 4 x 4 5 cos y 2 4 sin .
Solution not unique
Center: 4, 2 Solution not unique
50. From Exercise 42 we have a 1, c 2 ⇒ b 3 x 3 tan
52. y
Center: 0, 0 Solution not unique The transverse axis is vertical, therefore, x and y are interchanged. 56. x sin
54. y x2 Example
Example x t, y
y sec .
2 t1
x t, y
y t2
x t 3,
y t6
60. x 2 sin
y 2 4 cos
6
x t,
2 t 1
58. x 2 4 sin
y 1 cos
−6
2 x1
y 2 cos
9
6 −2
Not smooth at x 2n 1
−9
4
9 −3
−
0
Smooth everywhere
5
437
438
Chapter 9
62. x
3t 1 t3
y
3t 2 1 t3
Conics, Parametric Equations, and Polar Coordinates
64. Each point x, y in the plane is determined by the plane curve x f t , y gt . For each t, plot x, y . As t increases, the curve is traced out in a specific direction called the orientation of the curve.
2
−3
3
−2
Smooth everywhere 66. (a) Matches (ii) because 1 ≤ x ≤ 0 and 1 ≤ y ≤ 2.
(b) Matches (i) because x y 2 2 1 for all y.
68. x cos3
70. x cot
y 2 sin
y 4 sin cos
Matches (a)
Matches (c)
2
72. Let the circle of radius 1 be centered at C. A is the point of tangency on the line OC. OA 2, AC 1, OC 3. P x, y is the point on the curve being traced out as the angle changes AP . AB 2 and AP ⇒ 2. Form the right triangle CDP. The angle AB and OCE 2 DCP
3 . 2 2 2
x OE Ex 3 sin
y EC CD 3 sin cos 3
3
C 2
A
1
1
3 sin sin 3 2
Hence, x 3 cos cos 3, y 3 sin sin 3. 74. False. Let x t 2 and y t. Then x y 2 and y is not a function of x. 76. (a) x v0 cos t y h v0 sin t 16t 2 t
x x x ⇒ y h v0 sin 16 v0 cos v0 cos v0 cos y h tan x
(b) y 5 x 0.005x 2 h tan x h 5, tan 1 ⇒ 0.005 v02
2
16 sec2 2 x v02
16 sec2 2 x v02
(c)
80
, and 4
16 16 sec2 4 22 v02 v0 32 6400 ⇒ v0 80. 0.005
Hence, x 80 cos45 t y 5 80 sin45 t 16t 2.
0
250
−5
(d) Maximum height: y 55 at x 100 Range: 204.88
α D P = ( x, y )
θ
sin 3 3 cos cos 3 2 2
y
x
B E
x
Section 9.3
Section 9.3 2.
Parametric Equations and Calculus
Parametric Equations and Calculus
dy dydt 1 3t23 dx dxdt 13t23
4.
dy dyd 12e2 1 1 e32 32 dx dxd 2e 4 4e
8. x t2 3t 2, y 2t
6. x t, y 3t 1
dy 2 2 when t 0. dx 2t 3 3
dy 3 6t 6 when t 1. dx 12t
4 4 d 2y 222t 3 when t 0. dx2 2t 3 2t 32 9
3t d 2y 6 concave upwards dx2 12t
concave downward 10. x cos , y 3 sin dy 3 cos 3 cot dx sin
dx is undefined when 0.
d 2y 3 csc2 3 3 2 dx sin sin
d 2y is undefined when 0. dx2
dy
14. x sin , y 1 cos
12. x t, y t 1 dy 1 2t 1 dx 1 2t
t t 1
dy sin 0 when . dx 1 cos
1 cos cos sin2 1 cos 2 d 2 dx 1 cos
2 when t 2.
2y
d 2y t 1 2t t 12t 1 t 1 dx2 1 2t
1 1 when t 2. t 132
1 1 when . 1 cos 2 4
concave downward
concave downward 16. x 2 3 cos , y 3 2 sin
18. x t 1, y
2 cos 2 dy cot dx 3 sin 3 At 1, 3, 0, and
(a)
dy is undefined. dx
−3
(b) At t 1, x, y 0, 2, and dy dx dy 1, 1, 1 dt dt dx
Tangent line: y 5 3
5
−4
dy 0. At 2, 5, , and 2 dx
4 23
1 1, t 1 t
4
Tangent line: x 1
At
439
,2 ,
7 dy 23 , and . 6 dx 3
(c)
Tangent line: 23 4 33 y2 x 3 2
dy 1. At 0, 2, y 2 1x 0 dx y x 2
(d)
4
−3
5
23x 3y 43 3 0 −4
440
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
20. x 4 cos , y 3 sin , (a)
3 4 (b) At
4
−6
(d)
dy 3 4 3 . At , , dx 4 2 2
dx dy 32 dy 3 22, , dt dt 2 dx 4
6
−4
(c)
3 4 3 , x, y , , and 4 2 2
y
3 2
3 4 x 4 2
4
−6
6
3 y x 32 4
−4
22. x t2 t, y t3 3t 1 crosses itself at the point x, y 2, 1. At this point, t 1 or t 2. dy 3t2 3 dx 2t 1 At t 1, At t 2,
dy 0 and y 1. Tangent Line dx
dy 9 3 and y 1 3x 2 or y 3x 5. Tangent Line dt 3
24. x 2, y 21 cos Horizontal tangents:
dy 2 sin 0 when 0, ± , ± 2, . . . . d
Points: 4n, 0, 22n 1, 4 where n is an integer. Points shown: 0, 0, 2, 4, 4, 0 Vertical tangents:
dx 2 0; none d
26. x t 1, y t 2 3t Horizontal tangents: Point:
28. x t2 t 2, y t 3 3t
dy 3 2t 3 0 when t . dt 2
Vertical tangents:
dx 1 0; none dt
Point:
30. x cos , y 2 sin 2 Horizontal tangents: Points:
dy 3 5 7 4 cos 2 0 when , , , . d 4 4 4 4
22, 2 , 22, 2 , 22, 2 , 22, 2
Vertical tangents:
dy 3t 2 3 0 when t ± 1. dt
Points: 2, 2, 4, 2
21, 49
Vertical tangents:
Horizontal tangents:
dx sin 0 when 0, . d
Points: 1, 0, 1, 0
74, 118
dx 1 2t 1 0 when t . dt 2
Section 9.3 32. x 4 cos2 , y 2 sin
Since dxd 0 at 2 and 32, exclude them. dx 8 cos sin 0 when d
Vertical tangents:
441
34. x cos2 , y cos
dy 3 2 cos 0 when , . d 2 2
Horizontal tangents:
Parametric Equations and Calculus
Horizontal tangents:
dy sin 0 when x 0, . d
Since dxd 0 at these values, exclude them. Vertical tangents:
0, .
dx 2 cos sin 0 when d
3 , . 2 2
Point: 4, 0
Exclude 0, . Point: 0, 0
36. x t2 1, y 4t3 3, 1 ≤ t ≤ 0
dydt 2
dx dy dx 2t, 12t2, dt dt dt
0
s
1
2
38. x arcsin t, y ln1 t 2, 0 ≤ t ≤
4t2 144t4
0
4t2 144t4 dt
1 36t232 54
1
2t1 36t2 dt
1 1 3732 4.149 54
0 1
1 2
dx 1 t dy 1 2t , dt 2 1 t2 1 t2 1 t 2 dt
12
s
0 12
dxdt dydt dt 1 1 t dt 2
2
12
2 2
0
1 t1 ln 2 t1
0
1 dt 1 t2
12 0
1 1 1 ln3 0.549 ln 2 3 2 t5 1 dx 1 dy t 4 3, 1, 4 10 6t dt dt 2 2t
40. x t, y
2
S
1
1
2
1
2
1
t2 2t1 dt 4
t4 1 4 2 2t
4a
2
2 1
779 240
44. x cos sin , y sin cos , dy sin d S
2
2 cos2 2 sin2 d
0
2
0
d
2 2
2
0
2 2
a2 sin2 a2 cos2 d
2
0
dt
3
2
dx dy a sin , a cos d d
0
t4 1 4 dt 2 2t 1
S4
4
10t 6t 5
2
42. x a cos , y a sin ,
dx cos d
2
d 4a
0
2a
442
Chapter 9
46. x
Conics, Parametric Equations, and Polar Coordinates
4t 4t 2 , y 1 t3 1 t3
(a) x 3 y 3 4xy
(b)
dy 1 t 38t 4t 23t2 dt 1 t 32
4
−6
4t2 t3 3 2. 0 when t 0 or t 1 t32
6
Points: 0, 0,
4 3 2, 4 3 4 1.6799, 2.1165 3
3
−4
1
(c) s 2
0
1
8
41 2t 3 1 t 32
2
4t2 t 3 1 t 32
2
t 8 4t 6 4t 5 4t 3 4t 2 1
1 t 32
0
1
dt 2
0
16 t 8 4t 6 4t 5 4t 3 4t 2 1 dt 1 t 34
dt 6.557
48. x 3 cos , y 4 sin
4
dy dx 3 sin , 4 cos d d s
2
−6
6
9 sin2 16 cos2 d 22.1
−4
0
50. x t, y 4 2t,
1 52. x t3, y t 1, 1 ≤ t ≤ 2, y-axis 3
dx dy 1, 2 dt dt
2
(a) S 2
dx dy t2, 1 dt dt
4 2t1 4 dt
0
254t t 2
2 0
2
(b) S 2
(a) S 4
1
t1 4 dt 5 t 2
0
54. x a cos , y b sin ,
2
2 0
45
2
ab sin
1
a2 b2 4ab cos2 d a2 e
2ab e cos 1 e2 cos2 arcsine cos e
2b2
e
32 17 232 23.48 9
b sin a2 sin2 b2 cos2 d
0
13 4 4 t t 1 dt x 132 3 9
dx dy a sin , b cos d d
0
4
2
S 2
85
2aabb arcsin
a2 b2
a
—CONTINUED—
2
2
2
a2 b2
a
2
c : eccentricity a
2b
2
0
2
2
e sin 1 e2 cos2 d
0
ab e1 e2 arcsine e
abe arcsine
2 1
Section 9.3
Parametric Equations and Calculus
443
54. —CONTINUED— (b) S 4
2
a cos a2 sin2 b2 cos2 d
0
4
2
4a c
a cos b2 c2 sin2 d
0
2
c cos b2 c 2 sin2 d
0
2a c sin b2 c2 sin2 b2 ln c sin b2 c 2 sin2 c
2a c b2 c2 b2 ln c b2 c 2 b2 ln b c
2
0
2ab2 a a2 b2 b2 1e ln 2a 2 ln 2 2 b e 1e a b
2a 2
56. (a) 0
58. One possible answer is the graph given by x t, y t.
(b) 4
y
4 3 2 1 x
−4 −3 −2
1
2
3
4
−2 −3
−4
b
60. (a) S 2
dxdt dydt dt dx dy f t dt dt dt
gt
a
b
(b) S 2
a
2
2
2
2
62. Let y be a continuous function of x on a ≤ x ≤ b. Suppose that x f t, y gt, and f t1 a, f t 2 b. Then using integration by substitution, dx f t dt and
b
y dx
a
64. x 4 t, y t,
0
A
t
4
t2
gtf t dt.
t1
dx 1 , 0 ≤ t ≤ 4 dt 24 t
1 dt 24 t
2
4 u2 du
0
1 u u4 u2 4 arcsin 2 2
2 0
Let u 4 t, then du 1 24 t dt and t 4 u2. x y
x, y
1
0
4 tt
4
1 2
0
t
4
2
1 1 dt 2 24 t
1 1 dt 4 24 t
0
4
0
t dt
4
1 2 32 t 2 3
dx sin d
0
V 2
2
3 sin 2sin d
0
18
sin3 d 18 cos
2
0 4
8 3
t 1 28 t 4 t dt 4 3 4 t
38, 38
66. x cos , y 3 sin ,
cos3 3
0
2
12
0 4
8 3
444
Chapter 9
Conics, Parametric Equations, and Polar Coordinates dx 2 csc2 d
68. x 2 cot , y 2 sin2 ,
0
A2
0
2 sin2 2 csc2 d 8
2
12
144 x2
x
0
2
4
72. 2a2 is area of deltoid (c).
3 70. 8 a2 is area of asteroid (b).
76. (a) y 12 ln
2
d 8
144 x
74. 2ab is area of teardrop (e).
(b) x 12 sech
2
0 < x ≤ 12
t t , y t 12 tanh , 0 ≤ t 12 12
60
60
0
12 0
0
12
Same as the graph in (a), but has the advantage of showing the position of the object and any given time t.
0
(c)
dy 1 sech2t12 t sinh dx secht12 tant12 12
y 24
t t t Tangent line: y t0 12 tanh 0 sinh 0 x 12 sech 0 12 12 12
y t0 sinh
(0, y0)
16 12
t0 x 12
8
( x, y )
4
y-intercept: 0, t0
x 2
Distance between 0, t0 and x, y: d
12 sech 12t 12 tanh 12t
0
2
0
2
4
6
10
12
12
d 12 for any t ≥ 0. 78. False. Both dxdt and dydt are zero when t 0. By eliminating the parameter, we have y x 23 which does not have a horizontal tangent at the origin.
Section 9.4 2.
Polar Coordinates and Polar Graphs
7
2, 4
4.
7 0 6
7 0 6
74 2
y 0 sin
x 3 cos1.57 0.0024
y 3 sin1.57 3
x, y 0.0024, 3
π 2
x, y 0, 0
(− 0.0024, 3)
π 2
π
2
2,
x, y 2, 2 (−
6. 3, 1.57
x 0 cos
x 2 cos y 2 sin
7 2 4
0, 76
2) (0, 0) 0 1
2
0 1
0 1
2
Section 9.4
8. r, 2,
11 6
Polar Coordinates and Polar Graphs
10. r, 8.25, 1.3
12. x, y 0, 5 r ±5
x, y 2.2069, 7.9494
x, y 1.7321, 1
tan undefined
y
3 3 , , 5, , 5, 2 2 2 2
y 8
2
6
(−1.7321, 1)
−2
(2.2069, 7.9494)
4
1
y
2 −2
x
−1
1
x
−1
1
−2
2
2
−2
x
−1
1
−2
2
3
−1
−4
−1
−6
−2
−8
−3 −4
(0, −5)
−5
14. x, y 4, 2
y
r ± 16 4 ± 2 5
x 1
2
3
4
5
−1
2 1 tan 4 2
−2
0.464
−4
(4, −2)
−3
2 5, 0.464, 2 5, 2.678
−5
16. x, y 3 2, 3 2
18. x, y 0, 5
r, 5, 1.571
r, 6, 0.785
20. (a) Moving horizontally, the x-coordinate changes. Moving vertically, the y-coordinate changes. (b) Both r and values change. (c) In polar mode, horizontal (or vertical) changes result in changes in both r and . 22. x2 y 2 2ax 0
π 2
24.
r 2 2ar cos 0 rr 2a cos 0
π 2
r 10 sec
xy 4
26.
x 10 r cos 10
r 2a cos
0
a
0
2a
2
28.
8
r 2 r 2 9cos 2 0 r 2 9 cos 2
8 csc 2 π 2
π 2
0
0 4
6
x2 y 22 9x2 y 2 0
r 2 4 sec csc
2
4
r 22 9r 2 cos2 r 2 sin2 0
r cos r sin 4
445
1
2
12
446
Chapter 9
30.
r 2
Conics, Parametric Equations, and Polar Coordinates
r 5 cos
32.
r2 4
r 2 5r cos
x2 y 2 4
x2 y2 5x
x 25
1
2
y2
52
5 6
3 y x 3 2
y
3
3
x
−1 −1
x
y
y
1
5 6
tan tan
25 25 y2 4 4
x2 5x
y
34.
4
2
3 2
1
1 −2 −1
x 1
2
3
4
6
−2
x
−1
1
−2
−1
−3 −4
−2
38. r 51 2 sin
r 2 csc
36.
40. r 4 3 cos
0 ≤ < 2
r sin 2 y2
0 ≤ < 2
3 −10
y20
2
6
10 −4
y
3
−18
10
−6
1
x
−1
42. r
1
2
2 4 3 sin
44. r 3 sin
Traced out once on 0 ≤ ≤ 2
1 46. r2 .
5 2
0 ≤ < 4
Graph as
4
r1
3
−6
6
1
, r2
1
It is traced out once on 0, . 1.5
−3
3 −4 −1
.
−2
2
−1.5
Section 9.4
Polar Coordinates and Polar Graphs
48. (a) The rectangular coordinates of r1, 1 are r1 cos 1, r1 sin 1. The rectangular coordinates of r2, 2 are r2 cos 2, r2 sin 2. d 2 x2 x12 y2 y12 r2 cos 2 r1 cos 12 r2 sin 2 r1 sin 12 r22 cos2 2 2r1r2 cos 1 cos 2 r12 cos2 1 r22 sin2 22 2r1r2 sin 1 sin 2 r12 sin2 1 r22 cos2 2 sin2 2 r12 cos2 1 sin2 1 2 r1r2cos 1 cos 2 sin 1 sin 2 r12 r22 2r1r2 cos1 2 d r12 r22 2r1r2 cos1 2 (b) If 1 2, the points lie on the same line passing through the origin. In this case, d r12 r22 2r1r2 cos0
r1 r22 r1 r2
(c) If 1 2 90, then cos1 2 0 and d r12 r22, the Pythagorean Theorem! (d) Many answers are possible. For example, consider the two points r1, 1 1, 0 and r2, 2 2, 2. d
1 2
2
212 cos 0
5 2
Using r1, 1 1, and r2, 2 2, 52 , d
1
2
22 212 cos
5 5. 2
You always obtain the same distance.
50.
10, 76 , 3, d
102
52. 4, 2.5, 12, 1
32
7 2103 cos 6
d 42 122 2412 cos2.5 1 160 96 cos 1.5 12.3
109 60 cos 6 109 30
3 7.6
54. r 21 sin
56. (a), (b) r 3 2 cos
dy 2 cos sin 2 cos 1 sin dx 2 cos cos 2 sin 1 sin At 2, 0,
At 3,
dy 1. dx
7 dy , is undefined. 6 dx
3 dy At 4, , 0. 2 dx
4
−8
4
−4
r, 1, 0 ⇒ x, y 1, 0 Tangent line: x 1 (c) At 0,
dy does not exist (vertical tangent). dx
447
448
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
58. (a), (b) r 4
r a sin
60.
dy a sin cos a cos sin d
6
−8
2a sin cos 0
8
0,
−6
4 ⇒ x, y 2 2, 2 2
dx a sin2 a cos2 a1 2 sin2 0 d
at r, 4,
Tangent line: y 2 2 1 x 2 2
sin ±
y x 4 2
,
3 5 7 , , , 4 4 4 4
Horizontal: 0, 0, a,
a 2 2, 4 , a 2 2, 34
Vertical:
62.
1 2
2
dy , 1. 4 dx
(c) At
3 , , 2 2
r a sin cos2
64. r 3 cos 2 sec
dy a sin cos3 2a sin2 cos a cos3 sin d
2
2a sin cos3 sin3 cos
−2
4
2a sin cos cos2 sin2 0 −2
3 0, tan2 1, , 4 4 Horizontal:
2a
,
4
4
,
Horizontal tangents: 2.133, ± 0.4352
2a 3
4
,
4
, 0, 0
66. r 2 cos3 2
r 3 cos
68.
π 2
r 2 3r cos
2
x2 y2 3x −3
3
3 x 2
2
0
−2
Circle: r
Horizontal tangents:
1.894, 0.776, 1.755, 2.594, 1.998, 1.442 Center:
y2 3 2
32, 0
Tangent at pole:
70. r 31 cos
1
9 4
2
π 2
Cardioid Symmetric to polar axis since r is a function of cos . 0 1
0
3
2
2 3
r
0
3 2
3
9 2
6
2
4
Section 9.4 72. r sin5
Polar Coordinates and Polar Graphs
74. r 3 cos 2
Rose curve with five petals
Rose curve with four petals
Symmetric to 2
Symmetric to the polar axis,
Relative extrema occur when
Relative extrema: 3, 0, 3,
dr 3 5 7 9 5 cos5 0 at , , , , . d 10 10 10 10 10
Tangents at the pole:
2 3 4 Tangents at the pole: 0, , , , 5 5 5 5
0
3 , 4 4
0 2
76. r 2
80. r 5 4 sin
78. r 1 sin
Circle radius: 2
3 , 3, , 3, 2 2
π 2
1
y2
, and pole 2
7 5 and given the same tangents. 4 4
π 2
x2
449
Limaçon
Cardioid
4
Symmetric to
π 2
2
π 2
0
0
1
1
2
r
9
6
0
6
2
7
5
3
1
2 π 2
0 2
r
82.
6 2 sin 3 cos
2r sin 3r cos 6 2y 3x 6 Line π 2
84. r
1
Hyperbolic spiral
4
2
3 4
5 4
3 2
r
4
2
4 3
1
4 5
2 3
π 2
0 1
0 1
4
450
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
86. r2 4 sin
π 2
Lemniscate Symmetric to the polar axis,
Relative extrema: ± 2,
2
, and pole 2 0
2
0
6
2
5 6
r
0
± 2
±2
± 2
0
Tangent at the pole: 0 90. r 2 cos 2 sec
88. Since r 2 csc 2
1 , sin
the graphs has symmetry with respect to 2. Furthermore, r ⇒ as ⇒ 0
Strophoid r ⇒ as ⇒ r ⇒ as ⇒
r cos 4 cos2 2
r r 1 2 2 sin sin y
x 4 cos2 2 lim 4 cos2 2 2
ry 2y r r
2
r 2 cos 2 sec 22 cos2 1 sec
r ⇒ as ⇒ . Also, r 2
2
→ ±2
2y . y1
x = −2
Thus, r ⇒ ± as y ⇒ 1.
2
−3
3
4
−2 −4
4
y=1 −2
92. x r cos , y r sin x2 y2 r 2, tan
y x
94. Slope of tangent line to graph of r f at r, is dy f cos f sin . dx f sin f cos If f 0 and f 0, then is tangent at the pole.
96. r 4 cos 2
98. r 2 sec
Rose curve
Line
Matches (b)
Matches (d)
Section 9.4 100. r 6 1 cos (a) 0, r 6 1 cos
(c)
2
2 6 1 cos cos sin sin 2 2
9
r 6 1 cos
−9
15
6 1 sin
−9
(b)
Polar Coordinates and Polar Graphs
, r 6 1 cos 4 4
15
12 −12
12 −3
−9
The graph of r 6 1 cos is rotated through the angle 2.
15
−6
The graph of r 6 1 cos is rotated through the angle 4.
102. (a) sin
sin cos cos sin 2 2 2
(b) sin sin cos cos sin sin
cos
r f sin
2
r f sin
f sin
f cos
(c) sin
3 3 3 sin cos cos sin 2 2 2
cos
r f sin
3 2
f cos
104. r 2 sin 2 4 sin cos
(a) r 4 sin
cos 6 6
(b) r 4 sin
2
−3
−3
−2
3
−2
2 2 cos 3 3
(d) r 4 sin cos 4 sin cos 2
2 −3 −3
3
3 −2 −2
2
3
(c) r 4 sin
cos 4 sin cos 2 2
451
452
Chapter 9
Conics, Parametric Equations, and Polar Coordinates π 2
106. By Theorem 9.11, the slope of the tangent line through A and P is f cos f sin f sin f cos
Radial line
Polar curve r = f (θ)
P = (r, θ)
This is equal to tan
ψ
sin cos tan tan tan . 1 tan tan cos sin tan
Tangent line
θ
0
A
Equating the expressions and cross-multiplying, you obtain
f cos f sin cos sin tan sin cos tan f sin f cos f cos2 f cos sin tan f sin cos f sin2 tan f sin2 f sin cos tan f sin cos f cos2 tan f cos2 sin2 f tan cos2 sin2 tan
108. tan At
r 31 cos drd 3 sin
110. tan
3 1 22 2 2 , tan . 4 2 2
arctan
f r . f drd
At
2 2 2 1.041 59.64
r 4 sin 2 drd 8 cos 2
3 sin 3 , tan . 6 2 cos 3 2
arctan
−6
2
6
−4
−5
112. tan
4
5
−8
23 0.7137 40.89
r 5 undefined ⇒ . drd 0 2 6
−9
9
−6
114. True
Section 9.5 2. (a) r 3 cos
116. True
Area and Arc Length in Polar Coordinates (b) A 2
π 2
12 3 cos
9
2
2
d
0
2
cos2 d
0
0 2
A
32
4
2
9 4
9 2
2
1 cos 2 d
0
9 sin 2 2 2
2
0
9 4
Section 9.5
4
6 sin 22 d 36
0
sin2 2 d
6. A 2
0
4
36
4
1 2
4. A 2
Area and Arc Length in Polar Coordinates
0
1 cos 4 d 2
18
sin 4 4
1 2
10
cos 5 d 2
0
10
1 1 sin10 2 10
453
0
20
4
0
4 92
18
2
1 2
8. A 2
1 sin d 2
0
3 1 2 cos sin 2 2 4
10. A 2 2
0
3 8 4
1 2
2
4 6 sin 2 d
arcsin23
2
16 48 sin 36 sin2 d
arcsin23
2
2 16 48 sin 361 cos d 2
arcsin23
34 48 cos 9 sin 2
2
arcsin23
1.7635
2 −8
8
−12
12. Four times the area in Exercise 11, A 4 33 . More specifically, we see that the area inside the outer loop is
12
2
2
6
21 2 sin 2 d
2
6
4 16 sin 16 sin2 d 8 63.
6
The area inside the inner loop is 2
1 2
3 2
7 6
21 2 sin 2 d 4 63.
−4
4 −1
Thus, the area between the loops is 8 63 4 63 4 123. 14. r 31 sin
16. r 2 3 cos
r 31 sin
r cos
Solving simultaneously,
Solving simultaneously,
31 sin 31 sin 2 sin 0
0, . Replacing r by r and by in the first equation and solving, 31 sin 31 sin , sin 1, 2. Both curves pass through the pole, 0, 3 2, and 0, 2, respectively. Points of intersection: 3, 0, 3, , 0, 0
2 3 cos cos 1 2 5 , . 3 3 Both curves pass through the pole, (0, arccos 23), and 0, 2, respectively. cos
Points of intersection:
12, 3 , 12, 53 , 0, 0
454
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
20.
18. r 1 cos r 3 cos
4
π 2
r2
Solving simultaneously,
Line of slope 1 passing through the pole and a circle of radius 2 centered at the pole.
1 cos 3 cos cos
1 2
0 1
3
Points of intersection:
2, 4 , 2, 4
5 , . 3 3 Both curves pass through the pole, 0, , and 0, 2, respectively. Points of intersection:
32, 3 , 32, 53 , 0, 0
22. r 3 sin
π 2
Points of intersection:
r 2 csc
0 1
2
17 3
2
17 3
2
, arcsin
17 3
, arcsin
2
,
17 3
2
,
3.56, 0.596, 3.56, 2.545 The graph of r 3 sin is a limaçon symmetric to 2, and the graph of r 2 csc is the horizontal line y 2. Therefore, there are two points of intersection. Solving simultaneously, 3 sin 2 csc sin2 3 sin 2 0 sin
3 ± 17 2
arcsin
17 3
2
0.596.
24. r 31 cos r
r=
6 1 cos
The graph of r 31 cos is a cardioid with polar axis symmetry. The graph of
6 1 − cos θ
−10
5
r 61 cos is a parabola with focus at the pole, vertex3, , and polar axis symmetry. Therefore, there are two points of intersection. Solving simultaneously, 31 cos
5
r = 3(1 − cos θ )
6 1 cos
1 cos 2 2 cos 1 ± 2
arccos 1 2 . Points of intersection: 32, arccos 1 2 4.243, 1.998, 32, 2 arccos 1 2 4.243, 4.285
−5
Section 9.5
Area and Arc Length in Polar Coordinates
26. r 4 sin r 21 sin
2
1 2
28. A 4
2
18
Points of intersection: 0, 0, 4, 2
91 sin d 2
0
455
1 sin 2 d
0
9 3 8 2
(from Exercise 14)
7
The graphs reach the pole at different times ( values). 6
−7
7
r = 4 sin θ −7 −6
6 −2
r = 2 (1 + sin θ)
32. A 2
30. r 5 3 sin and r 5 3 cos intersect at 4 and 5 4.
1 A2 2
5 4
5 3 sin
4
2 d
59 9 30 cos sin 2 2 4
6
2
6
3 sin 2 d
4
4
−4
59 302 50.251 2
4
−4
8
−12
12
−8
34. Area Area of r 2a cos Area of sector twice area between r 2a cos and the lines , . 3 2
2 1 A a 2 a 2 3 2
2
2
3
2 a 2 2a 2 3
sin 2 2 a 2 2a 2 3 2
3
2 a 2 3 π 2
2a 2
tan 1, 4
2a cos 2 d
a2
3
6
33a 2
2a
4
1 cos 2 d 2
1 1 a2 2 4 2
0
π 2
r = a sin θ
0
0
a π θ =−3
4
1 1 a2 a2 4 8 a
a
a cos 2 d
0
2
2
1 sin 2 a2 2 2
2 a 2
1 2
0
1 cos 2 d
3 2 3 4
π θ = 3
36. r a cos , r a sin
A2
1 2
2
6
2 sin 2 d
4 cos 2 4 sin d
2 sin2 4 cos
5 4
2 2 59 5 9 59 9 30 30 2 4 2 4 2 4 2 4
2
1 2
r = a cos θ
2
6
33
456
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
38. By symmetry, A1 A2 and A3 A4.
6
1 2
A1 A2
2a cos 2 a2 d
3
6
a2 2
a2 sin 2 2
3
4 cos2 1 d 2a2
6
3
1 5 2 1 a 2 2 6 2
5 a 2 2a 2 12
A6 2
5 6
12
5 6
cos 2 d
3 a2 3 a 2 1 a2 1 2 2 2 4
5 6
0
6
6
4
6
4
6
r=2
6
0
2a
4 6
2a sin 2 a2 d
4 sin2 1 d a 2 sin 2
4
6
12 1 23
a2
[Note: A1 A6 A7 A4 a2 area of circle of radius a]
r sec 2 cos ,
40.
< < 2 2
y
1
r cos 1 2 cos2 x12
r 2 cos2 x2 12 2 2 r x y2
x 2 y 2x x 2 y 2 2x 2
x
−1
y 2x 1 x 2 x3 y2
1 2
A2
4
0
x 21 x 1x
4
sec 2 cos 2 d
0
sec2 4 4 cos2 d
4
0
sec2 4 21 cos 2 d tan 2 sin 2
0
r = 2a cos θ
π θ = −3
a2 d
a 2 3 a 2 5 3 a2 a2 12 3 2 12 12 2
π θ= 6
A1 a
4
A6 A4
π θ= 4 A7
a
A5
12
1 cos 2 d a 2
12
A2
5π θ= 6
3 5 a 2 3 a2 a2 12 3 2 12 2
π θ= 3
2a
r = 2a sin θ
1 cos 2 d
2a sin 2 d 2
a 2 2 sin 2
a2
6
2a cos 2 2a sin 2 d
2a sin 2 d
0
A7 2
4
6
A3
6
2a 2
6
4
π 2
5 a 2 a 2 2 sin 2 12
4
1 2 a 2 a 2 2 4
A3 A4 A5
a 2 sin 2
1 2
4
0
2
2
Section 9.5 42. r 2a cos
s
2
2
r 8 sin
2a cos 2 2a sin 2 d
2
457
44. r 81 cos , 0 ≤ ≤ 2
r 2a sin 2
Area and Arc Length in Polar Coordinates
s2
81 cos 2 8 sin 2 d
0
2a d 2
2
2
2 a
16
1 2 cos cos2 sin2 d
0
162
1 cos d
0
162
1 cos
0
162
0
sin 1 cos
3221 cos
1 cos 1 cos
d
d
0
64
46. r sec , 0 ≤ ≤
3
48. r e, 0 ≤ ≤
50. r 2 sin2 cos , 0 ≤ ≤
10
2
3
−2
−3
4
−25
−2
−5
−3
Length 7.78
Length 31.31
Length 1.73 exact 3
4
5
52. r a cos r a sin S 2
2
0
2 a2
a cos cos a2 cos a2 sin2 d
2
cos2 d a2
0
a2
2
1 cos 2 d
0
sin 2 2
2
0
2a2 2
54. r a1 cos r a sin S 2
a1 cos sin a21 cos 2 a2 sin2 d 2 a2
0
22 a2
1 cos 32sin d
sin 1 cos 2 2 cos d
42 a2 1 cos 52 5
0
32 a2 5
58. The curves might intersect for different values of :
56. r
S 2
0
0
r 1
0
See page 696.
sin 2 1 d 42.32
458
Chapter 9
60. (a) S 2
(b) S 2
Conics, Parametric Equations, and Polar Coordinates
f sin f 2 f2 d
f cos f 2 f2 d
62. r 8 cos , 0 ≤ ≤ (a) A
1 2
r 2 d
0
1 2
64 cos2 d 32
0
0
1 cos 2 sin 2 d 16 2 2
16
0
(Area circle r2 42 16) (b)
0.2
0.4
0.6
0.8
1.0
1.2
1.4
A
6.32
12.14
17.06
20.80
23.27
24.60
25.08
1 (c), (d) For 4 of area 4 12.57: 0.42 1 For 2 of area 8 25.13: 1.57 2 3 For 4 of area 12 37.70: 2.73
(e) No, it does not depend on the radius. 64. False. f 0 and g sin 2 have only one point of intersection.
Section 9.6 2. r
Polar Equations of Conics and Kepler’s Laws
2e 1 e cos
(a) e 1, r
4. r
2 , parabola 1 cos
2e 1 e sin
(a) e 1, r
2 , parabola 1 sin
(b) e 0.5, r
1 2 , ellipse 1 0.5 cos 2 cos
(b) e 0.5, r
1 2 , ellipse 1 0.5 sin 2 sin
(c) e 1.5, r
3 6 , hyperbola 1 1.5 cos 2 3 cos
(c) e 1.5, r
3 6 , hyperbola 1 1.5 sin 2 3 sin
4
e = 1.5
9
e=1
−9
e = 1.5
3
e=1 −9
9
e = 0.5 −4
6. r
e = 0.5
−3
4 1 0.4 cos (b) r
(a) Because e 0.4 < 1, the conic is an ellipse with vertical directrix to the left of the pole. (c)
9
7
−10
10 −8
8
4 1 0.4 cos
The ellipse is shifted to the left. The vertical directrix is to the right of the pole 4 r . 1 0.4 sin The ellipse has a horizontal directrix below the pole.
−7
8. Ellipse; Matches (f)
−5
10. Parabola; Matches (e)
12. Hyperbola; Matches (d)
Section 9.6
14. r
6 1 cos
16. r
Parabola since e 1
Polar Equations of Conics and Kepler’s Laws
1 5 5 3 sin 1 3 5sin
18. r3 2 cos 6 r
3 Ellipse since e < 1 5
Vertex: 3, 0 π 2
Vertices:
5 5 3 , , , 8 2 2 2
2 1 2 3 cos
Ellipse since e
π 2 0 4
6 3 2 cos
Vertices: 6, 0,
8
0 1
2
2 < 1 3
65,
π 2
0 1
20. r
6 2 3 7 sin 1 7 3sin
Hyperbola since e Vertices:
22. r
43, 0 , 4,
π 2
0
0
2
2
Hyperbola
4
−6
6
26.
Hyperbola
4
−6
−4
28. r
6
−4
6
1 cos 3
30. r
6 Rotate the graph of r 1 cos
counterclockwise through the angle . 3
6 3 7 sin 2 3
Rotate graph of r
6 . 3 7 sin
Clockwise through angle of 2 3. 4
6
−6 −14
6
10
−4 −10
4
4 1 2 cos
Vertices:
π 2
24.
3
Hyperbola since e 2 > 1
7 > 1. 3
3 3 3 , , , 5 2 2 2
2
5
459
460
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
32. Change to
:r 6
2
1 sin
6
34. Parabola
e 1, y 1, d 1 r
36. Ellipse
38. Hyperbola
3 e , y 2, d 2 4 r
40. Parabola Vertex: 5,
3 e , x 1, d 1 2
ed 1 e sin
r
e 1, d 10
ed 1 e cos
r
23 4 1 3 4 sin
3 2 1 3 2 cos
6 4 3 sin
3 2 3 cos
42. Ellipse
10 ed 1 e cos 1 cos
44. Hyperbola
3 Vertices: 2, , 4, 2 2
Vertices: 2, 0, 10, 0
3 10 e ,d 2 3
1 e ,d8 3
r
ed r 1 e sin 8 3 1 1 3 sin 8 3 sin
46. r
1 ed 1 e sin 1 sin
ed 1 e cos
5 1 3 2 cos
10 2 3 cos
4 is a parabola with horizontal directrix above the pole. 1 sin
(a) Parabola with vertical directrix to left pole.
(b) Parabola with horizontal directrix below pole.
(c) Parabola with vertical directrix to right of pole.
(d) Parabola (b) rotated counterclockwise 4.
x2 y2 1 a2 b2
48. (a)
x 2b 2 y 2a 2 a 2b 2 b 2 r 2 cos 2 a2r 2 sin2 a 2b 2 r 2 b 2 cos 2 a 21 cos 2 a 2b 2 r 2 a 2 cos 2 b 2 a 2 a 2b 2 r2
a2
a 2b 2 a 2b 2 2 2 2 a cos a c 2 cos 2 b2
b2 b2 1 c a 2 cos 2 1 e 2 cos 2
y2 x2 21 2 a b
(b)
x 2b 2 y 2a 2 a 2b 2 b 2r 2 cos 2 a 2r 2 sin 2 a 2b 2 r 2 b 2 cos 2 a 2 1 cos 2 a 2b2 r 2 a 2 cos 2 a 2 b 2 a 2b 2 r2
a 2b 2 b2 a 2 c 2 cos 2 1 c 2 a 2 cos 2 b 2 1 e 2 cos 2
Review Exercises for Chapter 9
50. a 4, c 5, b 3, e r2
5 4
52. a 2, b 1, c 3, e
9 1 2516 cos 2
54. A 2
1 2
2
r2
2
1 1 34 cos 2
2
1 d 3.37 3 2 sin 2 2
ed 1 e cos
56. (a) r
3
3 22 sin d 4 2
2
(b) The perihelion distance is a c a ea a1 e .
1 e2 a a1 e . 1e
When 0, r c a ea a a1 e .
When , r
Therefore,
The aphelion distance is a c a ea a1 e . ed a1 e 1e
When 0, r
a1 e 1 e ed
1 e2 a a1 e . 1e
a1 e 2 ed. Thus, r
1 e2 a . 1 e cos
58. a 1.427 109 km
60. a 36.0 10 6 mi, e 0.206
e 0.0543 r
r
1.422792505 109 1 e 2 a 1 e cos 1 0.0543 cos
34.472 10 6 1 e 2 a
1 e cos 1 0.206 cos
Perihelion distance: a1 e 28.582 10 6 mi Aphelion distance: a1 e 43.416 10 6 mi
Perihelion distance: a1 e 1.3495139 109 km Aphelion distance: a1 e 1.5044861 109 km r a sin b cos
62.
461
r 2 ar sin br cos x 2 y 2 ay bx x 2 y 2 bx ay 0 represents a circle.
Review Exercises for Chapter 9 2. Matches (b) - hyperbola
4. Matches (c) - hyperbola
6. y 2 12y 8x 20 0
y
y 2 12y 36 8x 20 36
16
y 6 2 42 x 2
12
Parabola Vertex: 2, 6 −4
x 8
12
462
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
8. 4x 2 y 2 16x 15 0
y
4x 2 4x 4 y 2 15 16
1
(2, 0)
x 2 2 y 2 1 14 1
x 1
2
3
−1
Ellipse Center: 2, 0
−2
Vertices: 2, ± 1 4x 2 4y 2 4x 8y 11 0
10.
y 4
1 4 y 2 2y 1 11 1 4 4
4 x2 x
x 12 2 y 1 2 1 2 2
3
1 x
−2
1
Hyperbola
−1
12, 1 1 Vertices: ± 2, 1 2 1 Asymptotes: y 1 ± x 2
−2
3
4
Center:
12. Vertex: 4, 2 Focus: 4, 0 Parabola opens downward p 2
x 4 2 42 y 2 x 2 8x 8y 0 14. Center: 0, 0
16. Foci: 0, ± 8
Solution points: 1, 2 , 2, 0
Asymptotes: y ± 4x
Substituting the values of the coordinates of the given points into
Center: 0, 0
x2 y2 2 1, b2 a
c8 a y x 4x asymptote → a 4b b
we obtain the system
b1 a4 1, 4b 2
2
2
1.
18.
16 x2 3y 2 1. and b 2 4, 3 4 16
21 y2 x2 1, a 5, b 2, c 21, e 4 25 5
By Example 5 of Section 9.1, C 20
2
0
b2 c2 a2 64 4b 2 ⇒ 17b2 64 ⇒ b2
Solving the system, we have a2
Vertical transverse axis
1 2521 sin d 23.01. 2
64 1024 ⇒ a2 17 17
x2 y2 1 102417 6417
Review Exercises for Chapter 9
463
1 2 x 200
20. y
(a) x 2 200y
y
1 2 x 200
y
1 x 100
(b)
x2 450 y Focus: 0, 50
x 1 10,000 x dx 38,294.49 S 2 x 1 10,000 2
1 y 2
100
2
0
a
22. (a) A 4
0
b 4b 1 a 2 x 2 dx a a 2
b
(b) Disk: V 2
0
b
S 4
0
4 a b2
b
b 2 y 2 dy
0
a 0
ab
2 a 2 2 1 b y y3 b2 3
b
4 a 2b 3
0
b
b 4 c 2y 2 dy
0
2 a cy b 4 c 2y 2 b 4 ln cy b 4 c 2y 2 b 2c
0
b
2 a 2 b c b 2 c 2 b 4 ln cb b b 2 c 2 b 4 lnb 2 b 2c
ab 2 ca ln c e
a
(c) Disk: V 2
0 a
S 22
0
a2 2 2 a 2 b y 2 dy 2 b2 b
x a
b 4 a 2 b 2 y 2 a b 2 y 2 dy b b b 2 y 2
2 a 2
x a 2 x 2 a 2 arcsin
4 b a2
2
2 a 2
b2 2 2 b 2 a x 2 dx 2 a2 a
eb ln11 ee 2
a
a 2 x 2 dx
0
a 4 a 2 b 2 x 2 b a 2 x 2 a a a 2 x 2
a
a 4 c 2x 2 dx
0
2 b
t x 4 ⇒ y x 4 2 Parabola
0
4 ab 2 3
2
2
a 0
abe arcsine
26. x 3 3 cos , y 2 5 sin
24. x t 4, y t2
a
dx
2 b cx cx a 4 c 2x 2 a 4 arcsin 2 a 2c a
c a b 2 a c a 2 c 2 a 4 arcsin a 2c a
2 b 2 2 1 a x x3 a2 3
x 3 3 y 5 2 2
2
1
28. x 5 sin3 , y 5 cos3
5x
23
x 3 2 y 2 2 1 9 25
y 7 6
23
1
y 6 4
4
y
3
2
2 1 −1
5y
x23 y23 523
Ellipse
5
x 1
2
3
4
5
6
7
7 6 5 4 3 2 1 −2 −1 −2 −3
−6
x
−4
2 −4 −6
x 1 2 3 4 5 6 7 8
4
6
464
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
30. x h 2 y k 2 r 2
y2 x2 1 16 9
32. a 4, c 5, b2 c2 a2 9,
x 5 2 y 3 2 2 2 4 Let
x2 y2 sec 2 and tan 2 . 16 9
Then x 3 tan and y 4 sec . 34. x a b cos t b cos y a b sin t b sin
a b b t
a b b t (b) a 3, b 1
(a) a 2, b 1
(c) a 4, b 1
x cos t cos t 2 cos t
x 2 cos t cos 2t
x 3 cos t cos 3t
y sin t sin t 0
y 2 sin t sin 2t
y 3 sin t sin 3t
4
2
4
y=0 −2 ≤ x ≤ 2 −3
−6
3
−6
6
6
−4
−2
−4
(e) a 3, b 2
(d) a 10, b 1 x 9 cos t cos 9t
(f) a 4, b 3
x cos t 2 cos
y 9 sin t sin 9t
y sin t 2 sin
10
−15
x cos t 3 cos
t 2
y sin t 3 sin
4
15
−6
−10
t 2
rcos sin
−6
6
6
−4
38. x t 4 y t2
y v w r sin r cos
(a)
r sin cos
dy 2t 2t 0 when t 0. dx 1 Point of horizontal tangency: 4, 0
y
(b) t x 4 y x 4 2
θ t
v
(c)
rθ
θ w
r
u
y 6 5
( x, y ) x
t 3
4
−4
36. x t u r cos r sin
t 3
4 3 2 1 x 1
2
3
4
5
6
Review Exercises for Chapter 9 40. x
1 t
42. x 2t 1
y t2 (a)
dy 2t 2t 3 dx 1t 2 No horizontal tangents
y
1 t 2 2t
(a)
dy t 2 2t 2 2t 2 dx 2
t 0
1 (b) t x
1t 0 when t 1. t 2t 2 2
Point of horizontal tangency: 1, 1
1 y 2 x
x1 2
(b) t
(c)
y
1 4 x 1 2 2 2 x 1 2 x 3 x 1
y
4 3
(c)
y
2
2
1 x
−2 −2
2
4
x
−1
1
2
44. x 6 cos
46. x e t
y 6 sin (a)
465
y et
dy 6 cos 3 cot 0 when , . dx 6 sin 2 2
(a)
Points of horizontal tangency: 0, 6 , 0, 6 (b)
6x 6y
(c)
y
2
2
1 1 dy et 2t 2 dx et e x No horizontal tangents
(b) t ln x
1
1 y eln x e ln1x , x > 0 x (c)
4
y
3
2 −4
x
−2
2
2
4
−2 1
−4
x 1
48. x 2 sin
50.
3
x 6 cos y 6 sin
y 2 cos (a), (c)
2
dx 6 sin d
8
−8
dy 6 cos d
8
−4
3 dx (b) At ,
1.134, 2 , 6 d 2
dy dy 0.5, and
0.441 dt dx
s
36 sin 2 36 cos 2 d 6
0
(one-half circumference of circle)
0
6
466
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
52. x, y 1, 3
y
(−1, 3)
r 1 2 3 2 10
2
arctan 3 1.89 108.43
1
r, 10, 1.89 , 10, 5.03
−3
−2
x
−1
1
2
3
−1 −2 −3
r 10
54.
r
56.
r 2 100
1 2 cos
2r r cos 1
x 2 y 2 100
2 ± x 2 y 2 x 1 4x 2 y 2 x 1 2 3x 2 4y 2 2x 1 0
58.
r 4 sec
4 3 cos 3
60.
3 4
tan 1
4 12 cos 32 sin
r cos 3 sin 8
y 1 x
x 3 y 8
y x
64. x 2 y 2 arctan
62. x 2 y 2 4x 0 r 2 4r cos 0
12
2
a2
r 2 2 a 2
r 4 cos
66.
y x
π 2
68. r 3 csc , r sin 3, y 3
π 2
Horizontal line
Line 0 1
2 0 1
70. r 3 4 cos Limaçon Symmetric to polar axis π 2
2
0
3
2
2 3
r
1
1
3
5
7
0
2
3
4
Review Exercises for Chapter 9
467
72. r 2 Spiral Symmetric to 2 π 2
0 2
4
8
0
r
0
4 5
2
3 4 3 2
5 4 5 2
2
3 2 3
74. r cos 5
π 2
Rose curve with five petals Symmetric to polar axis
2 3 4 , 1, , 1, , 1, 5 5 5 5 3 7 9 Tangents at the pole: , , , , 10 10 2 10 10
Relative extrema: 1, 0 , 1,
76. r 2 cos2 Lemniscate Symmetric to the polar axis Relative extrema: ± 1, 0
0
r
±1
6 ±
0 1
π 2
4
2
2
0 0 1 2
3 Tangents at the pole: , 4 4
78. r 2 sin cos 2
80. r 4sec cos
0.75
Bifolium
3
Semicubical parabola
Symmetric to 2 −1
1 −0.25
Symmetric to the polar axis r ⇒ as ⇒ 2 r ⇒ as ⇒ 2
−1
5
−3
82. r 2 4 sin2 (a) 2r
ddr 8 cos2
(b)
4 cos2 dr d r
Tangents at the pole: 0, (c)
2
−3
dy r cos 4 cos 2 sin r dx r sin 4 cos 2 cos r
3
2
cos2 sin sin2 cos cos2 cos sin2 sin
Horizontal tangents: dy 0 when cos2 sin sin 2 cos 0, dx
tan tan 2 , 0, , 0, 0 , ± 2 3, 3 3
Vertical tangents when cos 2 cos sin 2 sin 0: −2
tan 2 tan 1, 0, , 0, 0 , ± 2 3, 6 6
468
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
84. False. There are an infinite number of polar coordinate representations of a point. For example, the point x, y 1, 0 has polar representations r, 1, 0 , 1, 2 , 1, , etc. 86. r a sin , r a cos The points of intersection are a 2, 4 and 0, 0 . For r a sin , dy a cos sin a sin cos 2 sin cos . dx a cos 2 a sin 2 cos 2
m1
At a 2, 4 , m1 is undefined and at 0, 0 , m1 0. For r a cos , dy a sin 2 a cos 2 cos 2 . dx a sin cos a cos sin 2 sin cos
m2
At a 2, 4 , m2 0 and at 0, 0 , m2 is undefined. Therefore, the graphs are orthogonal at a 2, 4 and 0, 0 . 90. r 4 sin 3
88. r 51 sin
1 A2 2
32
2
51 sin
d 117.81 75 2
2
A3
1 2
3
4 sin 3 2 d
0
12.57 4
4 −8
4
8
−6
6
−12 −4
92. r 3, r 2 18 sin 2
4
9 r 2 18 sin 2 −6
sin 2
−4
12 A2
1 2
6
1 2
12
18 sin 2 d
0
1 2
512
12
9 d
1 2
2
512
18 sin 2 d
1.2058 9.4248 1.2058 11.84
94. r e, 0 ≤ ≤ A
1 2
96. r a cos 2,
e 2 d 133.62
0
s8
4
dr 2a sin 2 d
a 2 cos 2 2 4a 2 sin 2 2 d
0
10
8a
4
1 3 sin 2 2 d (Simpson’s Rule: n 4)
0
−25
5
−5
a 1 41.1997 21.5811 41.8870 2 6
9.69a
Review Exercises for Chapter 9
50. a 4, c 5, b 3, e r2
5 4
52. a 2, b 1, c 3, e
9 1 2516 cos 2
54. A 2
1 2
2
r2
2
1 1 34 cos 2
2
1 d 3.37 3 2 sin 2 2
ed 1 e cos
56. (a) r
3
3 22 sin d 4 2
2
(b) The perihelion distance is a c a ea a1 e .
1 e2 a a1 e . 1e
When 0, r c a ea a a1 e .
When , r
Therefore,
The aphelion distance is a c a ea a1 e . ed a1 e 1e
When 0, r
a1 e 1 e ed
1 e2 a a1 e . 1e
a1 e 2 ed. Thus, r
1 e2 a . 1 e cos
58. a 1.427 109 km
60. a 36.0 10 6 mi, e 0.206
e 0.0543 r
r
1.422792505 109 1 e 2 a 1 e cos 1 0.0543 cos
34.472 10 6 1 e 2 a
1 e cos 1 0.206 cos
Perihelion distance: a1 e 28.582 10 6 mi Aphelion distance: a1 e 43.416 10 6 mi
Perihelion distance: a1 e 1.3495139 109 km Aphelion distance: a1 e 1.5044861 109 km r a sin b cos
62.
461
r 2 ar sin br cos x 2 y 2 ay bx x 2 y 2 bx ay 0 represents a circle.
Review Exercises for Chapter 9 2. Matches (b) - hyperbola
4. Matches (c) - hyperbola
6. y 2 12y 8x 20 0
y
y 2 12y 36 8x 20 36
16
y 6 2 42 x 2
12
Parabola Vertex: 2, 6 −4
x 8
12
462
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
8. 4x 2 y 2 16x 15 0
y
4x 2 4x 4 y 2 15 16
1
(2, 0)
x 2 2 y 2 1 14 1
x 1
2
3
−1
Ellipse Center: 2, 0
−2
Vertices: 2, ± 1 4x 2 4y 2 4x 8y 11 0
10.
y 4
1 4 y 2 2y 1 11 1 4 4
4 x2 x
x 12 2 y 1 2 1 2 2
3
1 x
−2
1
Hyperbola
−1
12, 1 1 Vertices: ± 2, 1 2 1 Asymptotes: y 1 ± x 2
−2
3
4
Center:
12. Vertex: 4, 2 Focus: 4, 0 Parabola opens downward p 2
x 4 2 42 y 2 x 2 8x 8y 0 14. Center: 0, 0
16. Foci: 0, ± 8
Solution points: 1, 2 , 2, 0
Asymptotes: y ± 4x
Substituting the values of the coordinates of the given points into
Center: 0, 0
x2 y2 2 1, b2 a
c8 a y x 4x asymptote → a 4b b
we obtain the system
b1 a4 1, 4b 2
2
2
1.
18.
16 x2 3y 2 1. and b 2 4, 3 4 16
21 y2 x2 1, a 5, b 2, c 21, e 4 25 5
By Example 5 of Section 9.1, C 20
2
0
b2 c2 a2 64 4b 2 ⇒ 17b2 64 ⇒ b2
Solving the system, we have a2
Vertical transverse axis
1 2521 sin d 23.01. 2
64 1024 ⇒ a2 17 17
x2 y2 1 102417 6417
Review Exercises for Chapter 9
463
1 2 x 200
20. y
(a) x 2 200y
y
1 2 x 200
y
1 x 100
(b)
x2 450 y Focus: 0, 50
x 1 10,000 x dx 38,294.49 S 2 x 1 10,000 2
1 y 2
100
2
0
a
22. (a) A 4
0
b 4b 1 a 2 x 2 dx a a 2
b
(b) Disk: V 2
0
b
S 4
0
4 a b2
b
b 2 y 2 dy
0
a 0
ab
2 a 2 2 1 b y y3 b2 3
b
4 a 2b 3
0
b
b 4 c 2y 2 dy
0
2 a cy b 4 c 2y 2 b 4 ln cy b 4 c 2y 2 b 2c
0
b
2 a 2 b c b 2 c 2 b 4 ln cb b b 2 c 2 b 4 lnb 2 b 2c
ab 2 ca ln c e
a
(c) Disk: V 2
0 a
S 22
0
a2 2 2 a 2 b y 2 dy 2 b2 b
x a
b 4 a 2 b 2 y 2 a b 2 y 2 dy b b b 2 y 2
2 a 2
x a 2 x 2 a 2 arcsin
4 b a2
2
2 a 2
b2 2 2 b 2 a x 2 dx 2 a2 a
eb ln11 ee 2
a
a 2 x 2 dx
0
a 4 a 2 b 2 x 2 b a 2 x 2 a a a 2 x 2
a
a 4 c 2x 2 dx
0
2 b
t x 4 ⇒ y x 4 2 Parabola
0
4 ab 2 3
2
2
a 0
abe arcsine
26. x 3 3 cos , y 2 5 sin
24. x t 4, y t2
a
dx
2 b cx cx a 4 c 2x 2 a 4 arcsin 2 a 2c a
c a b 2 a c a 2 c 2 a 4 arcsin a 2c a
2 b 2 2 1 a x x3 a2 3
x 3 3 y 5 2 2
2
1
28. x 5 sin3 , y 5 cos3
5x
23
x 3 2 y 2 2 1 9 25
y 7 6
23
1
y 6 4
4
y
3
2
2 1 −1
5y
x23 y23 523
Ellipse
5
x 1
2
3
4
5
6
7
7 6 5 4 3 2 1 −2 −1 −2 −3
−6
x
−4
2 −4 −6
x 1 2 3 4 5 6 7 8
4
6
464
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
30. x h 2 y k 2 r 2
y2 x2 1 16 9
32. a 4, c 5, b2 c2 a2 9,
x 5 2 y 3 2 2 2 4 Let
x2 y2 sec 2 and tan 2 . 16 9
Then x 3 tan and y 4 sec . 34. x a b cos t b cos y a b sin t b sin
a b b t
a b b t (b) a 3, b 1
(a) a 2, b 1
(c) a 4, b 1
x cos t cos t 2 cos t
x 2 cos t cos 2t
x 3 cos t cos 3t
y sin t sin t 0
y 2 sin t sin 2t
y 3 sin t sin 3t
4
2
4
y=0 −2 ≤ x ≤ 2 −3
−6
3
−6
6
6
−4
−2
−4
(e) a 3, b 2
(d) a 10, b 1 x 9 cos t cos 9t
(f) a 4, b 3
x cos t 2 cos
y 9 sin t sin 9t
y sin t 2 sin
10
−15
x cos t 3 cos
t 2
y sin t 3 sin
4
15
−6
−10
t 2
rcos sin
−6
6
6
−4
38. x t 4 y t2
y v w r sin r cos
(a)
r sin cos
dy 2t 2t 0 when t 0. dx 1 Point of horizontal tangency: 4, 0
y
(b) t x 4 y x 4 2
θ t
v
(c)
rθ
θ w
r
u
y 6 5
( x, y ) x
t 3
4
−4
36. x t u r cos r sin
t 3
4 3 2 1 x 1
2
3
4
5
6
Review Exercises for Chapter 9 40. x
1 t
42. x 2t 1
y t2 (a)
dy 2t 2t 3 dx 1t 2 No horizontal tangents
y
1 t 2 2t
(a)
dy t 2 2t 2 2t 2 dx 2
t 0
1 (b) t x
1t 0 when t 1. t 2t 2 2
Point of horizontal tangency: 1, 1
1 y 2 x
x1 2
(b) t
(c)
y
1 4 x 1 2 2 2 x 1 2 x 3 x 1
y
4 3
(c)
y
2
2
1 x
−2 −2
2
4
x
−1
1
2
44. x 6 cos
46. x e t
y 6 sin (a)
465
y et
dy 6 cos 3 cot 0 when , . dx 6 sin 2 2
(a)
Points of horizontal tangency: 0, 6 , 0, 6 (b)
6x 6y
(c)
y
2
2
1 1 dy et 2t 2 dx et e x No horizontal tangents
(b) t ln x
1
1 y eln x e ln1x , x > 0 x (c)
4
y
3
2 −4
x
−2
2
2
4
−2 1
−4
x 1
48. x 2 sin
50.
3
x 6 cos y 6 sin
y 2 cos (a), (c)
2
dx 6 sin d
8
−8
dy 6 cos d
8
−4
3 dx (b) At ,
1.134, 2 , 6 d 2
dy dy 0.5, and
0.441 dt dx
s
36 sin 2 36 cos 2 d 6
0
(one-half circumference of circle)
0
6
466
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
52. x, y 1, 3
y
(−1, 3)
r 1 2 3 2 10
2
arctan 3 1.89 108.43
1
r, 10, 1.89 , 10, 5.03
−3
−2
x
−1
1
2
3
−1 −2 −3
r 10
54.
r
56.
r 2 100
1 2 cos
2r r cos 1
x 2 y 2 100
2 ± x 2 y 2 x 1 4x 2 y 2 x 1 2 3x 2 4y 2 2x 1 0
58.
r 4 sec
4 3 cos 3
60.
3 4
tan 1
4 12 cos 32 sin
r cos 3 sin 8
y 1 x
x 3 y 8
y x
64. x 2 y 2 arctan
62. x 2 y 2 4x 0 r 2 4r cos 0
12
2
a2
r 2 2 a 2
r 4 cos
66.
y x
π 2
68. r 3 csc , r sin 3, y 3
π 2
Horizontal line
Line 0 1
2 0 1
70. r 3 4 cos Limaçon Symmetric to polar axis π 2
2
0
3
2
2 3
r
1
1
3
5
7
0
2
3
4
Review Exercises for Chapter 9
467
72. r 2 Spiral Symmetric to 2 π 2
0 2
4
8
0
r
0
4 5
2
3 4 3 2
5 4 5 2
2
3 2 3
74. r cos 5
π 2
Rose curve with five petals Symmetric to polar axis
2 3 4 , 1, , 1, , 1, 5 5 5 5 3 7 9 Tangents at the pole: , , , , 10 10 2 10 10
Relative extrema: 1, 0 , 1,
76. r 2 cos2 Lemniscate Symmetric to the polar axis Relative extrema: ± 1, 0
0
r
±1
6 ±
0 1
π 2
4
2
2
0 0 1 2
3 Tangents at the pole: , 4 4
78. r 2 sin cos 2
80. r 4sec cos
0.75
Bifolium
3
Semicubical parabola
Symmetric to 2 −1
1 −0.25
Symmetric to the polar axis r ⇒ as ⇒ 2 r ⇒ as ⇒ 2
−1
5
−3
82. r 2 4 sin2 (a) 2r
ddr 8 cos2
(b)
4 cos2 dr d r
Tangents at the pole: 0, (c)
2
−3
dy r cos 4 cos 2 sin r dx r sin 4 cos 2 cos r
3
2
cos2 sin sin2 cos cos2 cos sin2 sin
Horizontal tangents: dy 0 when cos2 sin sin 2 cos 0, dx
tan tan 2 , 0, , 0, 0 , ± 2 3, 3 3
Vertical tangents when cos 2 cos sin 2 sin 0: −2
tan 2 tan 1, 0, , 0, 0 , ± 2 3, 6 6
468
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
84. False. There are an infinite number of polar coordinate representations of a point. For example, the point x, y 1, 0 has polar representations r, 1, 0 , 1, 2 , 1, , etc. 86. r a sin , r a cos The points of intersection are a 2, 4 and 0, 0 . For r a sin , dy a cos sin a sin cos 2 sin cos . dx a cos 2 a sin 2 cos 2
m1
At a 2, 4 , m1 is undefined and at 0, 0 , m1 0. For r a cos , dy a sin 2 a cos 2 cos 2 . dx a sin cos a cos sin 2 sin cos
m2
At a 2, 4 , m2 0 and at 0, 0 , m2 is undefined. Therefore, the graphs are orthogonal at a 2, 4 and 0, 0 . 90. r 4 sin 3
88. r 51 sin
1 A2 2
32
2
51 sin
d 117.81 75 2
2
A3
1 2
3
4 sin 3 2 d
0
12.57 4
4 −8
4
8
−6
6
−12 −4
92. r 3, r 2 18 sin 2
4
9 r 2 18 sin 2 −6
sin 2
−4
12 A2
1 2
6
1 2
12
18 sin 2 d
0
1 2
512
12
9 d
1 2
2
512
18 sin 2 d
1.2058 9.4248 1.2058 11.84
94. r e, 0 ≤ ≤ A
1 2
96. r a cos 2,
e 2 d 133.62
0
s8
4
dr 2a sin 2 d
a 2 cos 2 2 4a 2 sin 2 2 d
0
10
8a
4
1 3 sin 2 2 d (Simpson’s Rule: n 4)
0
−25
5
−5
a 1 41.1997 21.5811 41.8870 2 6
9.69a
Problem Solving for Chapter 9
98. r
2 ,e1 1 cos
100. r
4 45 3 ,e 5 3 sin 1 35sin 5
Ellipse
Parabola
π 2
π 2
0 2 0 1
102. r
8 4 5 ,e 2 5 cos 1 52cos 2
Hyperbola
2
104. Line Slope: 3 Solution point: 0, 0
π 2
y 3 x, r sin 3 r cos , tan 3,
3
0 1
2
106. Parabola
Vertex: 2, 2 Focus: 0, 0 e 1, d 4 4 r 1 sin
108. Hyperbola Vertices: 1, 0, 7, 0 Focus: 0, 0 4 7 a 3, c 4, e , d 3 4
4374 7 r 4 3 4 cos 1 cos 3
Problem Solving for Chapter 9 2. Assume p > 0. Let y mx p be the equation of the focal chord.
y
x2
= 4py (0, p)
First find x-coordinates of focal chord endpoints: x
x2 4py 4pmx p x2 4pmx 4p2 0
y = −p
4pm ± 16p2m2 16p2 2pm ± 2pm2 1 2 x x2 4py, 2x 4py ⇒ y . 2p
x
(a) The slopes of the tangent lines at the endpoints are perpendicular because 1 1 2pm 2pm2 1 2p 2pm 2pm2 1 4p1 2 4p2m2 4p2m2 1 4p1 2 4p2 1 2p —CONTINUED—
469
470
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
2. —CONTINUED— (b) Finally, we show that the tangent lines intersect at a point on the directrix y p. Let b 2pm 2pm2 1 and c 2pm 2pm2 1. b2 8p2m2 4p2 8p2mm2 1 c2 8p2m2 4p2 8p2mm2 1 b2 2pm2 p 2pmm2 1 4p c2 2pm2 p 2pmm2 1 4p Tangent line at x b: y
b2 b bx b2 x b ⇒ y 4p 2p 2p 4p
Tangent line at x c: y
c2 c c2 cx x c ⇒ y 4p 2p 2p 4p bx b2 cx c2 2p 4p 2p 4p
Intersection of tangent lines:
2bx b2 2cx c2 2xb c b2 c2 2x4pm2 1 16p2mm2 1 x 2pm Finally, the corresponding y-value is y p, which shows that the intersection point lies on the directrix.
4.
y2 x2 2 1, a2 b2 c2, MF2 Mf1 2a 2 a b y
y
β
b2x a2y
α
F2(−c, 0)
M (x 0 , y 0 ) x
b2x y y0 2 0 x x0 a y0
Tangent line at Mx0, y0:
β
F1(c, 0) Q
yy0 y02 x0 x x02 b2 a2 x0 x y0 y x02 y02 2 2 2 a2 b a b x0 x y0 y 2 1 a2 b
QF2 QF1 MQ
b2 b2 ⇒ Q 0, . y0 y0
At x 0, y
x
2
0
c2
b4 2d y0
y0
b2 y0
2
f
By the Law of Cosines,
F2Q2 MF2 2 MQ2 2MF2 MQcos d 2 MF2 2 f 2 2 f MF2 cos
F1Q2 MF12 f 2 2 f MF1cos d 2 MF12 f 2 2 f MF1cos . cos
F2 2 f 2 d 2 MF12 f 2 d2 , cos 2 f MF2 2 f MF1
MF2 MF1 2a. Let z MF1. Slopes: MF1: —CONTINUED—
y0 b2 b2 ; QF1: ; QF2: x0 c y0c y0c
Problem Solving for Chapter 9
471
4. —CONTINUED— To show , consider
MF2 2 f 2 d2 2 f MF1 MF12 f 2 d 2 2 f MF2 ⇔
z 2a2 f 2 d 2 z z2 f 2 d 2 z 2a
⇔
z2 2az f 2 d 2
⇔
x0 c2 y02 2az x02 y0
⇔
b2 y0
c 2
2
b4 y02
az x0c a2 0
⇔
ax0 c2 y02 x0c a2
⇔
x02b2 a2y02 a2b2 x02 y02 2 1. a2 b
⇔
Thus, and the reflective property is verified.
6. (a) y2
t21 t22 2 1 t22 ,x 1 t22 1 t22
11 tt 2t 1x 1x 1t 2 1 1t 2 2
2
r cos sin2 sin2 cos2 r cos3
t2
r cos sin2 cos2 cos2 sin2
2
r cos cos 2
11 xx.
r cos 2
(d) r 0 for
π 2
(c)
11 rr cos cos
sin2 1 r cos cos2 1 r cos
2
1
Thus, y2 x2
r 2 sin2 r 2 cos2
(b)
sec
3 , . 4 4
Thus, y x and y x are tangent lines to curve at the origin. 0 1
(e) yt
2
1 t21 3t2 t t32t 1 4t2 t4 0 1 t22 1 t22
t4 4t2 1 0 ⇒ t2 2 ± 5 ⇒ x
5 1
2
,±
5 1
2
2 5
1 2 ± 5 3 5 1 2 ± 5 1 ± 5 5 1 3 5 1 5 2
472
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
10. r
8. (a)
ab ,0 ≤ ≤ a sin b cos 2
r a sin b cos ab ay bx ab y x 1 b a Line segment
Generated by Mathematica
u2 du, (b) x, y cos 2 0 t
u2 sin du is on 2 0 t
1 Area ab 2
the curve whenever x, y is on the curve. (c) xt cos
t2 t2 , yt sin , xt2 yt2 1 2 2
a
Thus, s
dt a.
0
On , , s 2. 12. Let r, be on the graph. r 2 1 2r cos r 2 1 2r cos 1
r 2 12 4r 2 cos2 1 r 4 2r 2 1 4r 2 cos2 1 r 2r 2 4 cos2 2 0 r 2 4 cos2 2 r 2 22 cos2 1 r 2 2 cos 2 14. (a) r 2
(c) r 2 2 cos
4
4
Cardioid
Circle radius 2 −6
−6
6
−4
−4
(b) r 2 cos
(d) r 2 3 cos
4
Convex limaçon −6
6
−4
16. The curve is produced over the interval 0 ≤ ≤ 9.
6
Limaçon with inner loop
4
−6
6
−4
C H A P T E R 9 Conics, Parametric Equations, and Polar Coordinates Section 9.1
Conics and Calculus . . . . . . . . . . . . . . . . . . . . 177
Section 9.2
Plane Curves and Parametric Equations . . . . . . . . . . 188
Section 9.3
Parametric Equations and Calculus
Section 9.4
Polar Coordinates and Polar Graphs . . . . . . . . . . . . 198
Section 9.5
Area and Arc Length in Polar Coordinates
Section 9.6
Polar Equations of Conics and Kepler’s Laws . . . . . . . 210
Review Exercises
. . . . . . . . . . . . 192
. . . . . . . . 205
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222
C H A P T E R 9 Conics, Parametric Equations, and Polar Coordinates Section 9.1
Conics and Calculus
Solutions to Odd-Numbered Exercises 3. x 32 2 y 2
1. y 2 4x
5.
Vertex: 0, 0
Vertex: 3, 2
p1 > 0
p 12 < 0
Opens to the right Matches graph (h).
Opens downward Matches graph (e).
x2 y 2 1 9 4
7.
Center: 0, 0 Ellipse Matches (f)
Hyperbola Center: 0, 0 Vertical transverse axis. Matches (c)
9. y 2 6x 4 32 x
11. x 3 y 22 0
Vertex: 0, 0
y 22 4 14 x 3
y
3 Focus: 2 , 0
Directrix: x
y2 x2 1 16 1
8
3 2
Vertex: 3, 2 (0, 0)
12
8
y
Focus: 3.25, 2
4 x
4
4
Directrix: x 2.75
4
(− 3, 2) −8
−6
2
−4
x
−2 −2
8
−4
13. y2 4y 4x 0
15. x2 4x 4y 4 0
y 2 4y 4 4x 4
x 2 4x 4 4y 4 4
y 22 41x 1
x 22 41 y 2
Vertex: 1, 2
Vertex: 2, 2
y
Focus: 0, 2
6
Directrix: x 2
4
y
Focus: 2, 1
4
(− 2, 2)
Directrix: y 3 (−1, 2) −6
x
2
2 2
4
−4
x
−2
2 −2
6 −4
177
178
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
17. y 2 x y 0
19. y 2 4x 4 0
y2 y 14 x 14
y 12 2 4 14 x 14 1 1 Vertex: 4 , 2 1 Focus: 0, 2
y 2 4x 4 41x 1 Vertex: 1, 0
2
4
Focus: 0, 0
−5
2
Directrix: x 2
1
Directrix: x 2
−6
6
−3
−4
y 22 42x 3
21.
23.
x h2 4p y k x2 46 y 4
y 2 4y 8x 20 0
x2 24y 96 0 y 4 x2
25.
x2 y 4 0 27. Since the axis of the parabola is vertical, the form of the equation is y ax2 bx c. Now, substituting the values of the given coordinates into this equation, we obtain
29. x2 4y 2 4 x2 y 2 1 4 1
3 c, 4 9a 3b c, 11 16a 4b c. Solving this system, we have a Therefore,
5 3, b
14 3,
c 3.
Foci: ± 3, 0
−2
3
2
y
33.
12
(1, 5)
Center: 1, 5
9x2 4y 2 36x 24y 36 0 9x2 4x 4 4 y2 6y 9 36 36 36 36
4
x 22 y 32 1 4 9
x 8
4
4
8
Vertices: 1, 10, 1, 0 4 e 5
x
1
Vertices: ± 2, 0
a2 25, b2 9, c2 16 Foci: 1, 9, 1, 1
(0, 0) −1
Center: 0, 0
e
x 12 y 52 1 9 25
2
a2 4, b2 1, c2 3
2 y 53 x2 14 3 x 3 or 5x 14x 3y 9 0.
31.
y
a2 9, b2 4, c2 5 y
Center: 2, 3 Foci: 2, 3 ± 5
6
Vertices: 2, 6, 2, 0 e
(− 2, 3) 2
5
3
x 6
4
2
2
Section 9.1 12x2 20y 2 12x 40y 37 0
35.
12 x2 x
x
2
3x
a2 4, b2 2, c 2 2
a2 5, b2 3, c2 2 Center: Foci:
Center:
12, 1
Foci:
12 ± 2, 1
Vertices:
9 1 9 2 y 2 2y 1 2 4 4 4 4
x 32 2 y 12 1 4 2
60
x 12 2 y 12 1 5 3
32, 1
32 ± 2, 1 21, 1, 72, 1
Vertices:
12 ± 5, 1
Solve for y: 2 y 2 2y 1 x2 3x
Solve for y:
y 12
20 y 2 2y 1 12x2 12x 37 20
y 12
57 12x 12x2 20
y 1 ±
57 12x20 12x
y 1 ±
2
1 −2
−3
4
3
−3 −3
41. Vertices: 3, 1, 3, 9 Minor axis length: 6 Vertical major axis Center: 3, 5
39. Center: 0, 0 Focus: 2, 0 Vertex: 3, 0 Horizontal major axis a 3, c 2 ⇒ b 5
a 4, b 3
x2
x 32 y 52 1 9 16
9
y2 5
1
43. Center: 0, 0 Horizontal major axis Points on ellipse: 3, 1, 4, 0 Since the major axis is horizontal,
ax by 1. 2
2
2
2
Substituting the values of the coordinates of the given points into this equation, we have
a9 b1 1, and 16a 1. 2
2
2
The solution to this system is a2 16, b2 167. Therefore, x2 y2 x2 7y 2 1, 1. 16 167 16 16
1 2 4
1 7 3x x2 2 4
7 12x8 4x
(Graph each of these separately.)
(Graph each of these separately.) 1
179
x2 2y 2 3x 4y 0.25 0
37.
1 20y 2 2y 1 37 3 20 4
Conics and Calculus
2
180
45.
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
y 2 x2 1 1 4
47.
x 12 y 22 1 4 1
a 1, b 2, c 5
a 2, b 1, c 5
Center: 0, 0
Center: 1, 2
Vertices: 0, ± 1
Vertices: 1, 2, 3, 2
1 Asymptotes: y ± x 2
1 Asymptotes: y 2 ± x 1 2
Foci: 1 ± 5, 2
Foci: 0, ± 5
y
y
4
1 x
2
1 x
4
2
2
1
2
3
2
4
2 4 5
4
9x2 y 2 36x 6y 18 0
49.
x2 9y 2 2x 54y 80 0
51.
9x2 4x 4 y 2 6y 9 18 36 9
x2 2x 1 9 y 2 6y 9 80 1 81 0 x 12 9 y 32 0
x 22 y 32 1 1 9
1 y 3 ± x 1 3
a 1, b 3, c 10 Center: 2, 3
Degenerate hyperbola is two lines intersecting at 1, 3.
Vertices: 1, 3, 3, 3 Foci: 2 ± 10, 3
y
Asymptotes: y 3 ± 3x 2
x
4
y
2
2 2
x
2
2
4
4
6
2
6
4 6
53.
9y 2 x2 2x 54y 62 0
55.
9 y 2 6y 9 x2 2x 1 62 1 81 18
3x2 2y 2 6x 12y 27 0 3x2 2x 1 2 y 2 6y 9 27 3 18 12
y 32 x 12 1 2 18 a 2, b 32, c 25 Center: 1, 3
x 12 y 32 1 4 6
−5
7
Vertices: 1, 3 ± 2
9 y2 6y 9 x2 2x 62 81 x2 2x 19 9
1 y 3 ± x2 2x 19 3 (Graph each curve separately.)
1 −5
7
Foci: 1 ± 10, 3 −7
Solve for y:
Center: 1, 3 Vertices: 1, 3, 3, 3
Foci: 1, 3 ± 25
y 32
a 2, b 6, c 10
1
−7
Solve for y:
2 y 2 6y 9 3x2 6x 27 18
y 32
3x2 6x 9 2
y 3 ±
3x
(Graph each curve separately.)
2
2x 3 2
Section 9.1 57. Vertices: ± 1, 0 Asymptotes: y ± 3x Horizontal transverse axis Center: 0, 0
181
59. Vertices: 2, ± 3 Point on graph: 0, 5 Vertical transverse axis Center: 2, 0 a3
b b a 1, ± ± ± 3 ⇒ b 3 a 1 Therefore,
Conics and Calculus
Therefore, the equation is of the form y2 x 22 1. 9 b2
x2 y 2 1. 1 9
Substituting the coordinates of the point 0, 5, we have 25 4 21 9 b
9 or b2 . 4
Therefore, the equation is
61. Center: 0, 0 Vertex: 0, 2 Focus: 0, 4 Vertical transverse axis
63. Vertices: 0, 2, 6, 2 2 2 Asymptotes: y x, y 4 x 3 3 Horizontal transverse axis
a 2, c 4, b2 c2 a2 12 Therefore,
y2 x 22 1. 9 94
Center: 3, 2 a3
y2 x2 1. 4 12
b 2 Slopes of asymptotes: ± ± a 3 Thus, b 2. Therefore,
x 32 y 22 1. 9 4
65. (a)
x2 2x x y 2 1, 2yy 0, y 9 9 9y At x 6: y ± 3, y At 6, 3 : y 3
±6
93
(b) From part (a) we know that the slopes of the normal lines must be 9 23 .
± 23
At 6, 3 : y 3
9
23 x 6 9
or 9x 23y 60 0 At 6, 3 : y 3
or 2x 33y 3 0 At 6, 3 : y 3
9 x 6 23
23 x 6 9
9 23
x 6
or 9x 23y 60 0
or 2x 33y 3 0 67. x2 4y 2 6x 16y 21 0
69. y2 4y 4x 0
71. 4x2 4y 2 16y 15 0
A 1, C 4
A 0, C 1
AC4
AC 4 > 0
Parabola
Circle
Ellipse 73. 9x2 9y 2 36x 6y 34 0
3x2 6x 3 6 2y2 4y 2
75.
AC9
3x2 2y2 6x 4y 5 0
Circle
A 3, C 2, AC < 0 Hyperbola
182
Chapter 9
Conics, Parametric Equations, and Polar Coordinates 79. (a) A hyperbola is the set of all points x, y for which the absolute value of the difference between the distances from two distance fixed points (foci) is constant.
77. (a) A parabola is the set of all points x, y that are equidistant from a fixed line (directrix) and a fixed point (focus) not on the line. (b) x h2 4py k or y k2 4px h
(b)
(c) See Theorem 9.2.
y k2 x h2 x h2 y k2 1 or 1 2 2 a b a2 b2
b a (c) y k ± x h or y k ± x h a b 83. y ax2
81. Assume that the vertex is at the origin.
y 2ax
x2 4py
3 4p1 2
The equation of the tangent line is y ax02 2ax0x x0 or y 2ax 0 x ax 02.
9 p 4
Let y 0. Then:
The pipe is located 94 meters from the vertex.
ax02 2ax0x 2ax02
y
ax02 2ax0x
3
Focus
Therefore,
2
(− 3, 1)
x0 x is the x-intercept. 2
y
(3, 1) 1
−3
−2
−1
x 2
1
3
(x0, ax02 )
y = ax 2
x
( x2 , 0) 0
85. (a) Consider the parabola x2 4py. Let m0 be the slope of the one tangent line at x1, y1 and therefore, 1m0 is the slope of the second at x2, y2. From the derivative given in Exercise 32 we have: m0
1 x or x1 2pm0 2p 1
1 1 2p x2 or x2 m0 2p m0 Substituting these values of x into the equation x2 4py, we have the coordinates of the points of tangency 2pm0, pm02 and 2pm0, pm02 and the equations of the tangent lines are
y pm02 m0x 2pm0 and
2p x . y mp 1 m m 2
0
0
0
The point of intersection of these lines is
pm m 1, p and is on the directrix, y p. 2
0
y
x 2 = 4py
0
2p p − , 2 m0 m0
(
) (2pm0, pm02) x
y = −p
( p(mm− 1) , − p) 0
0
—CONTINUED—
Section 9.1
Conics and Calculus
85. —CONTINUED— (b) x2 4x 4y 8 0
x 22 4 y 1. Vertex 2, 1 2x 4 4
dy 0 dx dy 1 x1 dx 2
At 2, 5, dydx 2. At 3, 54 , dydx 12 . Tangent line at 2, 5: y 5 2x 2 ⇒ 2x y 1 0. Tangent line at 3, 54 : y 54 12 x 3 ⇒ 2x 4y 1 0. Since m1m2 2 12 1, the lines are perpendicular. 1 1 Point of intersection: 2x 1 x 2 4 5 5 x 2 4 x
1 2
y0 1 Directrix: y 0 and the point of intersection 2 , 0 lies on this line.
87.
y x x2 dy 1 2x dx At x1, y1 on the mountain, m 1 2x1. Also, m
y1 1 . x1 1
y1 1 1 2x1 x1 1
x1 x12 1 1 2x1x1 1 x12 x1 1 2x12 x1 1 x12 2x1 2 0 x1
2 ± 22 412 2 ± 23 1 ± 3 21 2
Choosing the positive value for x1, we have x1 1 3. m 1 2 1 3 3 23 m
01 1 x0 1 x0 1
1 3 23 Thus, x0 1 1 x0 1 3 23 3 23 1 x0 3 23 x0. 3 The closest the receiver can be to the hill is 233 1 0.155.
y 2
(− 1, 1) −2
1
−1
( x1 , y1 ) (x0, 0) 1
−1 −2
x
183
184
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
89. Parabola Vertex: 0, 4
Circle Center: 0, k Radius: 8
x2 4p y 4
x2 y k2 64
y
42 4p0 4
42 0 k2 64
p 1 −6
x2 4 y 4
−4
−2
4
6
k 43
8
−4
x2 y4 4
k2 48
x 2
−2
(Center is on the negative y-axis.)
x2 y 43 64
−6
2
−8
y 43 ± 64 x2 Since the y-value is positive when x 0, we have y 43 64 x2.
4
A2
4
0
x3 1 x 43x x64 x2 64 arcsin 12 2 8
64 1 163 248 32 arcsin 12 2
2 4x 2 16
x2 43 64 x2 dx 4
4 0
16 4 33 2
15.536 square feet 3
91. (a) Assume that y ax2.
y
(− 60, 20)
(60, 20) 20
2 1 1 2 x ⇒ y 20 a60 ⇒ a 360 180 180 2
(b) f x
1 2 1 x , fx x 180 90
60
1 x 1 90
S2
0
15 10
2
2 dx 90
5
60
902
dx
−60 −45 −30 −15
0
x2
60
2 1 x902 x2 902 ln x 902 x2 90 2
1 6011,700 902 ln 60 11,700 902 ln 90 90
1 180013 902 ln 60 3013 902 ln 90 90
2013 90 ln
60 9030
10 213 9 ln
13
(formula 26)
0
2 3 13 128.4 m
1 1 3 93. x2 4py, p , , 1, , 2 4 2 2
95.
y
As p increases, the graph becomes wider. 1
y
p=
1 4
p=2 p= p=
−8
3 2
1 2
x
8
3
p=1
24
−16
2
16
17
16
15
14
13
12
11
4
10
5
9
6
8
7
7
8
6
9
5
10
4
11
3
12
2
13
1
14
15
16
17
x
x 15 30 45 60
Section 9.1 5 97. a , b 2, c 2
52
2
22
Conics and Calculus
3 2
The tacks should be placed 1.5 feet from the center. The string should be 2a 5 feet long. c a
e
99.
y
A P 2a c
AP a 2
x
caP
103.
AP AP P 2 2
c A P2 A P a A P2 A P
e
101. e
(a, 0) P
A
35.34au 0.59au AP
0.9672 AP 35.34au 0.59au
y2 x2 1 102 52
105. 16x2 9y 2 96x 36y 36 0 32x 18yy 96 36y 0
2yy 2x 2 0 102 5 y
y18y 36 32x 96
52x x 102y 4y
At 8, 3: y
y
32x 96 18y 36
y 0 when x 3. y is undefined when y 2.
8 2 12 3
At x 3, y 2 or 6. Endpoints of major axis: 3, 2, 3, 6
The equation of the tangent line is y 3 3 x 8. It 2 25 will cross the y-axis when x 0 and y 3 8 3 3 . 2
At y 2, x 0 or 6. Endpoints of minor axis: 0, 2, 6, 2 Note: Equation of ellipse is
2
1 x 4 x2 dx x4 x2 4 arcsin 2 2
107. (a) A 4
0
2
V 2
(b) Disk:
0
2 0
or, A ab 21 2
2
1 1 1 4 x2 dx 4x x3 4 2 3
x 32 y 22 1 9 16
2 0
8 3
1 y 4 x2 2 y 1 y2
2
S 22
y
x 24 x2
1 16 x 4x 16 4y 3x
16 3x2
0
—CONTINUED—
4y
2
2
2
dx 2 3 3x16 3x
2
16 arcsin
3x
4
2 0
2 9 43 21.48 9
185
186
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
107. —CONTINUED—
2
V 2
(c) Shell:
2
x4 x2 dx
0
2x4 x212 dx
0
2 4 x232 3
2 0
16 3
x 21 y 2 x 1 x2
2y 1 y 2
1 1 4y y
1
S 22
21 y 2
2
1
0
3y2
1 y 2
2
1 3y 2 1 y 2
1
dy 8
1 3y 2 dy
0
436
8 3y1 3y 2 ln 3y 1 3y 2 23
0
1 e2 sin2 d
2100 10a ⇒ a 20
0
For
x2 25
y2 49
Hence, the length of the major axis is 2a 40.
1, we have
a 7, b 5, c 49 25 26, e
C 47
34.69
Area ellipse ab a10
2
C 4a
3 ln 2 3
111. Area circle r 2 100
109. From Example 5,
1
c 26 . a 7
2
0
1 4924 sin d 2
281.3558 37.9614 113. The transverse axis is horizontal since 2, 2 and 10, 2 are the foci (see definition of hyperbola).
115. 2a 10 ⇒ a 5 c 6 ⇒ b 11
Center: 6, 2 c 4, 2a 6, b2 c2 a2 7 Therefore, the equation is
x 62 y 22 1. 9 7
117. Time for sound of bullet hitting target to reach x, y: Time for sound of rifle to reach x, y:
2c x c2 y 2 vm vs
y
( x, y )
x c2 y 2
vs
2c x c2 y 2 x c2 y 2 Since the times are the same, we have: vm vs vs 4c2 4c x c2 y 2 x c2 y 2 x c2 y 2 2 vm vmvs vs2 vs2 x c2 y2
1 vv x 2 m 2 s
x2 c2vs2vm2
c2
vm2
2
y2
vm2x vs2c vsvm
vv
y2 1 vs2vm2
2 s 2 m
1 c2
(− c, 0) rifle
x
(c, 0) target
Section 9.1
Conics and Calculus
187
119. The point x, y lies on the line between 0, 10 and 10, 0. Thus, y 10 x. The point also lies on the hyperbola x236 y264 1. Using substitution, we have: x2 10 x2 1 36 64 16x2 910 x2 576 7x2 180x 1476 0 x
180 ± 1802 471476 27
180 ± 1922 90 ± 962 14 7
Choosing the positive value for x we have: x
90 962 160 962
6.538 and y
3.462 7 7 2y 2 x2 2y 2 x2 2 1 ⇒ 2 1 2, c 2 a2 b2 a2 b b a
121.
x2 2y 2 2y 2 x2 2 1 ⇒ 2 2 1 a2 b2 b b a b2 1
x2 x2 1 1 2 1 ⇒ 2 x2 2 2 2 a a b2 a a b2 x2
2aa2 b2 2ac 2a2a2 b2 ⇒ x± ± 2 2 2a b 2a2 b2 2a2 b2
2y2 1 2a2c 2 1 2 2 b a 2a2 b2 y2
⇒ 2yb
2
2
2a2
b2 b2
b4 b2 ⇒ y± 22a2 b2 22a2 b2
There are four points of intersection:
2ac 2a2
b2
,±
b2 , 22a2 b2
2ac 2a2
b2
,±
b2 22a2 b2
2y2 2x 4yy b2x x2 2 1 ⇒ 2 2 0 ⇒ ye 2 2 a b a b 2a y a2 At
2y 2 2x 4yy b 2x x2 2 1 ⇒ 2 2 0 ⇒ yh 2 2 b b c b 2c y 2ac
2a2
ye
, b2
b2 2a2
b2 , the slopes of the tangent lines are: 22a2 b2 2a b 2ac
2
22a
2
b2
2
b2
c a
b2 and
yh 2c 2
2ac 2a2 b2
22a
b2 2
b2
a c
Since the slopes are negative reciprocals, the tangent lines are perpendicular. Similarly, the curves are perpendicular at the other three points of intersection. 123. False. See the definition of a parabola.
125. True
127. False. y2 x2 2x 2y 0 yields two intersecting lines.
129. True
188
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
Section 9.2
Plane Curves and Parametric Equations
1. x t, y 1 t (b)
(a) t
0
1
2
3
4
x
0
1
2
3
2
y
1
y
1
1
0
2
x
−1
3
1
2
3
−1 −2
(c)
2
−3 −1
3
(d) x2 t y 1 x2, x ≥ 0
−4
5. x t 1
3. x 3t 1 y 2t 1 y2
x 3 1 1
7. x t 3
y t2
y 12 t 2
y x 12
x t 3 implies t x13 y 12 x 23
y
2x 3y 5 0
y
4 y
1 4 x
3 2
2
1
1
2
3
x
2
2
4
x
4
2 2
9. x t, t ≥ 0
11. x t 1
yt2
t t1
y t2
y
x1 x
y
4
x4 x 2 2 2 y
y
3
8
2
4 4
1
x −1
y
y x2 2, x ≥ 0 y
13. x 2t
2
3
4
5
2
6
x
4
−2
x
2
2
15. x et, x > 0
y 5
y e3t 1
4
y x3 1, x > 0
3 2 1 −2 −1
x −1
1
2
3
4
4
8
12
Section 9.2 17. x sec
19. x 3 cos , y 3 sin
y cos
x
y 2 cos 2 2
x sin2 2 16
x2 y 2 9.
xy 1 y
x 4 sin 2
21.
Squaring both equations and adding, we have
, < ≤ 2 2
0 ≤ <
Plane Curves and Parametric Equations
y
y2 cos2 2 4
4
1 x
2
≥ 1, y ≤ 1
y2 x2 1 16 4
x
4
2
2
4 4
2
y
4
3
−6
2
6
1 x
1
2
−4
3
2 3
x 4 2 cos
23.
x 4 2 cos
25.
y 1 sin
y 1 4 sin
x 4 cos2 4
x 42 cos2 4
y 12 sin2 1
y 12 sin2 16
2
x 42 y 12 1 4 1
x 42 y 12 1 4 16
2
3
−1
8
−2
−4
−5
x 4 sec
27.
10
29. x t 3
y 3 tan x2 sec2 16
x et
31.
y 3 ln t
y e3t
3 x ln x y 3 ln
et
2
y2 tan2 9
−1
x2 y2 1 16 9
1 x
3 y et 3 y
5
1 x
y
−2
6
1 x3
x > 0 y > 0
−9
9 3
−6 −1
5 −1
189
190
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
33. By eliminating the parameters in (a) – (d), we get y 2x 1. They differ from each other in orientation and in restricted domains. These curves are all smooth except for (b). (a) x t, y 2t 1
(b) x cos
3
1 ≤ y ≤ 3
dx dy 0 when 0, ± , ± 2, . . . . d d
2
y
1
−2
y 2 cos 1
1 ≤ x ≤ 1
y
x
−1
1
3
2
−1
2 1
−2
x
−1
1
2
−1
(c) x et
(d) x e t
y 2et 1
x > 0
y > 1
x > 0
y
y > 1
y
4
4
3
3
2
2
1
1 x
−1
y 2et 1
1
2
−1
3
x 1
2
3
35. The curves are identical on 0 < < . They are both smooth. Represent y 21 x2 37. (a)
4
(b) The orientation of the second curve is reversed.
4
(c) The orientation will be reversed. −6
−6
6
(d) Many answers possible. For example, x 1 t, y 1 2t, and x 1 t, x 1 2t.
−4
−4
39.
6
x x1 tx2 x1
x h a cos
41.
y y1 t y2 y1
y k b sin
x x1 t x2 x1 y y1
xh cos a
xx xx y 1
2
2
yk sin b
y1
1
x h2 y k2 1 a2 b2
y y1 y y1 2 x x1 x2 x1 y y1 mx x1 43. From Exercise 39 we have
45. From Exercise 40 we have
x 5t
x 2 4 cos
y 2t.
y 1 4 sin .
Solution not unique
Solution not unique
47. From Exercise 41 we have a 5, c 4 ⇒ b 3 x 5 cos y 3 sin . Center: 0, 0 Solution not unique
Section 9.2
Plane Curves and Parametric Equations
51. y 3x 2
49. From Exercise 42 we have a 4, c 5 ⇒ b 3
53. y x3
Example
Example
x 4 sec
x t,
y 3t 2
x t,
y t3
y 3 tan .
x t 3,
y 3t 11
3 t, x
yt
x tan t,
y tan3 t
Center: 0, 0 Solution not unique 55. x 2 sin
57. x 32 sin
59. x 3 cos3
y 21 cos
y 1 32 cos
y 3 sin3
5
5
4
−6 −2
−2
16
6
7 −1
−1
−4
Not smooth at 2n
61. x 2 cot
Not smooth at x, y ± 3, 0 and 0, ± 3, or 12 n. 63. See definition on page 665.
4
y 2 sin 2
−6
6
−4
Smooth everywhere 65. A plane curve C, represented by x f t, y gt, is smooth if f and g are continuous and not simultaneously 0. See page 670. 67. x 4 cos
69. x cos sin
y 2 sin 2
y sin cos
Matches (d)
Matches (b)
71. When the circle has rolled radians, we know that the center is at a, a. sin sin180
191
C BD b
AP cos cos180 b
b
or
or
BD b sin
AP b cos
Therefore, x a b sin and y a b cos . 73. False x t2 ⇒ x ≥ 0 x t2 ⇒ y ≥ 0 The graph of the parametric equations is only a portion of the line y x.
y
P b A
B
C θ a D
x
192
Chapter 9
75. (a) 100 mihr
Conics, Parametric Equations, and Polar Coordinates
1005280 440 ftsec 3600 3
x v0 cos t
(d) We need to find the angle (and time t) such that x
440 cos t 3
y3
y h v0 sin t 16t 2 3 (b)
sin t 16t 440 3
cos t 400 440 3 sin t 16t 440 3
10 3
1200 1200 sin 16 440 3 440 cos 440 cos
7 400 tan 16 400 0
It is not a home run—when x 400, y ≤ 20. (c)
10.
From the first equation t 1200440 cos . Substituting into the second equation,
2
30
0
2
400 tan 16
60
120 44
2
sec2
tan 120 44 2
2
1.
We now solve the quadratic for tan : 16
0
120 44
2
tan2 400 tan 7 16
120 44
2
tan 0.35185 ⇒ 19.4
400 0
Yes, it’s a home run when x 400, y > 10.
Section 9.3 1.
Parametric Equations and Calculus
dy dydt 4 2 dx dxdt 2t t
3.
dy dydt 2 cos t sin t 1 dx dxdt 2 sin t cos t
Note: x y 1 ⇒ y 1 x and dydt 1 5. x 2t, y 3t 1
7. x t 1, y t 2 3t
dy dydt 3 dx dxdt 2
dy 2t 3 1 when t 1. dx 1
d 2y 0 Line dx2
d 2y 2 concave upwards dx2
9. x 2 cos , y 2 sin dy 2 cos cot 1 when . dx 2 sin 4 d 2y csc2 csc3 2 when . 2 dx 2 sin 2 4 concave downward
11. x 2 sec , y 1 2 tan 2 sec2 dy dx sec tan
2 sec 2 csc 4 when . tan 6
d 2y 2 csc cot dx2 sec tan 2 cot3 6 3 when concave downward
. 6
0
2
Section 9.3 13. x cos3 , y sin3
15. x 2 cot , y 2 sin2
3 sin2 cos dy dx 3 cos2 sin
4 sin cos dy 2 sin3 cos dx 2 csc2
tan 1 when
. 4
At
sec2 1 d 2y dx2 3 cos 2 sin 3 cos4 sin
Parametric Equations and Calculus
2
,
3
3 2
, 23, and dydx 3 8 3.
y
Tangent line:
sec4 csc 4 2 when . 3 3 4
3 3 3 2 x 2 8
3
3 3x 8y 18 0 At 0, 2,
concave upward
dy , and 0. 2 dx
Tangent line: y 2 0
At 2 3,
3 1 dy , , and . 2 6 dx 8
y
Tangent line:
3 1 x 2 3 2 8
3x 8y 10 0
17. x 2t, y t2 1, t 2 (a)
19. x t 2 t 2, y t3 3t, t 1 (a)
10
−6
5
−1
6
−4
8
−3
(b) At t 2, x, y 4, 3, and
(b) At t 1, x, y 4, 2, and
dx dy dy 2, 4, 2 dt dt dx
dx dy dy 3, 0, 0 dt dt dx
(c)
dy 2. At 4, 3, y 3 2x 4 dx y 2x 5
(d)
(c)
dy 0. At 4, 2, y 2 0x 4 dx y2
(d)
10
5
(4, 2)
(4, 3) −5
5 −4
21. x 2 sin 2t, y 3 sin t crosses itself at the origin, x, y 0, 0. At this point, t 0 or t . dy 3 cos t dx 4 cos 2t At t 0:
3 dy 3 and y x. Tangent Line dx 4 4
At t ,
3 dy 3 and y x Tangent Line dx 4 4
−1
8
−3
193
194
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
23. x cos sin , y sin cos Horizontal tangents:
dy sin 0 when 0, , 2, 3, . . . . d
Points: 1, 2n 1 , 1, 2n where n is an integer. Points shown: 1, 0, 1, , 1, 2 dx 3 5 cos 0 when , , ,. . .. d 2 2 2
Vertical tangents:
1
2n 1 , 1n1 2
3 5 Points shown: , 1, , 1, , 1 2 2 2 n1
Points:
25. x 1 t, y t 2 Horizontal tangents:
27. x 1 t, y t 3 3t dy 2t 0 when t 0. dt
Horizontal tangents:
dy 3t 2 3 0 when t ± 1. dt Points: 0, 2, 2, 2
dx 1 0; none dt
Vertical tangents:
Point: 1, 0 Vertical tangents: 3
dx 1 0; none dt
3
(2, 2) −4 −2
5
4
(1, 0)
(0, − 2)
−1
−3
29. x 3 cos , y 3 sin Horizontal tangents:
dy 3 3 cos 0 when , . d 2 2
31. x 4 2 cos , y 1 sin
Points: 0, 3, 0, 3 dx 3 sin 0 when 0, . d
Vertical tangents:
dy 3 cos 0 when , . d 2 2
Horizontal tangents:
Points: 4, 0, 4, 2 dx 2 sin 0 when x 0, . d
Vertical tangents:
Points: 3, 0, 3, 0
Points: 6, 1, 2, 1 2
4
(0, 3) (4, 0) 0 −6
(− 3, 0)
(3, 0)
9
(2, − 1)
6
(6, − 1)
(4, − 2) (0, − 3) −4
−4
35. x t 2, y 2t, 0 ≤ t ≤ 2
33. x sec , y tan Horizontal tangents:
dy sec2 0; none d
Vertical tangents:
dx sec tan 0 when x 0, . d
Points: 1, 0, 1, 0 4
−6
(− 1, 0)
(1, 0)
−4
6
dydt
dx dy dx 2t, 2, dt dt dt
2
2
4t 2 4 4t 2 1
2
s2
t 2 1 dt
0
0
t t 2 1 ln t t2 1 2 5 ln 2 5 5.916
2
Section 9.3
37. x et cos t, y et sin t, 0 ≤ t ≤
2
2
0
dx dt
2
dy dt
2
0
0
2
dt
2e2t dt 2
2
et1 dt
2et
0
21 e2 1.12
1 1 9 dt 4t 2
dx 3a cos2 sin , d
dy 3a sin2 cos d
1 u2 du
0
1 ln 37 6 6 37 3.249 12
3
t
dt
dx dy a1 cos , a sin d d
S2
a2 1 cos 2 a2 sin2 d
sin cos cos2 sin2 d
2 2a
0
6a
0
0
2
6
0
dt
t
0
43. x a sin , y a1 cos ,
9a2 cos4 sin2 9a2 sin4 cos2 d
12a
1 36t
1 ln 1 u2 u u 1 u2 12
2
S4
1
6
1 6
u 6 t, du
41. x a cos3 , y a sin3 ,
0
2
1
S
195
dx 1 dy , 3 dt 2 t dt
39. x t, y 3t 1,
dx dy etsin t cos t, etcos t sin t dt dt s
Parametric Equations and Calculus
1 cos d
0
2
sin 2 d 3a cos 2
0
2
0
6a
2 2a
0
sin
1 cos
4 2a 1 cos
d
0
8a
45. x 90 cos 30t, y 90 sin 30t 16t 2 (a)
(d) y 0 ⇒ 90 sin t 16t2 ⇒ t
35
x 90 cos t 0
240 0
(b) Range: 219.2 ft (c)
dx dy 90 cos 30, 90 sin 30 32t. dt dt y 0 for t
45 . 16
90 cos 302 90 sin 30 32t2 dt
902 2 cos 2 0 ⇒ 45 32 By the First Derivative Test, 45 4 maximizes the range.
dx 90 cos , dt
90 dy 90 sin 32 90 sin 32 sin 90 sin dt 16 s
0
230.8 ft
902 902 cos sin sin 2 16 32
x
4516
s
90 sin 16
9016sin
90 cos 2 90 sin 2 dt
0
9016sin
0
0
9016sin
90 dt 90t
2
90 sin 16
ds 902 cos 0 ⇒ d 16 2 By the First Derivative Test, 90 maximizes the arc length.
196
Chapter 9
47. (a)
Conics, Parametric Equations, and Polar Coordinates
x t sin t
x 2t sin2t
y 1 cos t
y 1 cos2t
0 ≤ t ≤ 2
0 ≤ t ≤
3
The time required for the particle to traverse the same path is t 4.
−1
49. x t, y 2t,
dx dy 1, 2 dt dt
51. x 4 cos , y 4 sin ,
4
4
2t 1 4 dt 4 5
0
S 2
t dt
0
4
2 5 t 2
0
0
32
4
2
dx dy 3a cos2 sin , 3a sin2 cos d d
−4
x
−2
2
4
−2 −4
61. x r cos , y r sin
y
r sin
r 2 sin2 r 2 cos2 d
0
sin d
θ
0
2r 2 cos
2
0
1 cos
r 2
32
59. s
a
2
dx dt
dydt 2
0
12 2 a 5
2
dt
See Theorem 9.8, page 678.
x t, y t.
2
0
12a2 sin5 5
sin4 cos d
b
4
2
57. One possible answer is the graph given by
y
0
See Theorem 9.7, page 675.
2r 2
2
t dt
a sin3 9a2 cos4 sin2 9a2 sin4 cos2 d 12a2
dy dydt dx dxdt
cos d 32 sin
0
0
2
16 5
0
53. x a cos3 , y a sin3 ,
S 2
4 cos 4 sin 2 4 cos 2 d
0
5 t 2
55.
2
4
t 1 4 dt 2 5
dx dy 4 sin , 4 cos d d
0
32 5
4
S 4
1
3
−1
(b) S 2
y 1 cos 2 t
−
3
(c) x 12 t sin 12 t
3
−
(a) S 2
(b) The average speed of the particle on the second path is twice the average speed of a particle on the first path.
x
Section 9.3
Parametric Equations and Calculus
197
63. x t, y 4 t, 0 ≤ t ≤ 4
4
A
4 t
0
x y
3 16 3 32
1 2 t
dt
4
1 2
4
4t12 t12 dt
0
28 t 3t t 1
0 4
4 t 2
0
0
2 1 t dt 323 4 t dt 323 4t t2 4
4 t t
4
2
1 2 t
dt
3 64
2
0
0
4
4
16t12 8t12 t32 dt
0
16 3
3 4
4
3 16 2 32 t t t t 2 t 64 3 5
0
8 5
3 8 x, y , 4 5
65. x 3 cos , y 3 sin ,
dx 3 sin d
0
V 2
2
3 sin 23 sin d
0
54
sin3 d
2 0
54
1 cos2 sin d
2
54 cos
67.
cos3 3
0
2
36
x 2 sin2
y
y 2 sin2 tan
2
dx 4 sin cos d A
2
π 0≤θ< 2
2 sin2 tan 4 sin cos d 8
0
sin4 d
0
3 sin3 cos 3 8 sin cos 4 8 8
69. ab is area of ellipse (d).
75. (a) x
−2
2
2
0
1 x
−1
1 −1 −2
3 2
71. 6a2 is area of cardioid (f).
73. 83 ab is area of hourglass (a).
1 t2 2t , y , 20 ≤ t ≤ 20 1 t2 1 t2
2
The graph is the circle x 2 y 2 1, except the point 1, 0. Verify: x2 y 2
11 tt 1 2t t 2 2 2
2
2
−3
1 2t 2 t 4 4t 2 1 t 22 1 1 t 22 1 t 22
(b) As t increases from 20 to 0, the speed increases, and as t increases from 0 to 20, the speed decreases. 77. False d g t d 2y dt f t f tg t g t f t dx2 f t f t 3
3
−2
198
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
Section 9.4 1.
Polar Coordinates and Polar Graphs
4, 2
3.
5. 2, 2.36
0 2
x 4 cos
2 4
y 4 sin
x 4 cos y 4 sin
4, 3
x 2 cos2.36 1.004
2 3
y 2 sin2.36 0.996
x, y 1.004, 0.996
3 23
π 2
x, y 2, 23
x, y 0, 4 π 2
π 2
(4, 36π )
(
2, 2.36 )
0
(− 4, − π3 )
1
0 1
2
3
0 1
7. r, 5,
3 4
9. r, 3.5, 2.5
11. x, y 1, 1 r ± 2
x, y 2.804, 2.095
x, y 3.5355, 3.5355
tan 1
y
5 5 , , 2, , 2, 4 4 4 4
y
1 4
(−3.54, 3.54)
3
−1
1 −3
−2
−2 x
−1
1
2
3
y
−1
2
−4
x
2
(2.804, − 2.095)
−3
1
(1, 1)
−1
1
x 1
13. x, y 3, 4
y
r ± 9 16 ± 5 tan
2
5
(− 3, 4)
43
4 3
2.214, 5.356, 5, 2.214, 5, 5.356
2 1 −4
−3
−2
−1
x 1
5 4 17. x, y 2 , 3
15. x, y 3, 2
r, 3.606, 0.588
r, 2.833, 0.490 y
19. (a) x, y 4, 3.5
π 2
(b) r, 4, 3.5
4
(4, 3.5) 0
3
1
2
(4, 3.5) 1 x 1
2
3
4
Section 9.4 21. x2 y 2 a2
π 2
Polar Coordinates and Polar Graphs y4
23.
π 2
r sin 4
ra
r 4 csc 0 a 0 2
3x y 2 0
25.
y 2 9x
27.
3r cos r sin 2 0
r 2 sin2 9r cos
r3 cos sin 2 r
4
r
2 3 cos sin
9 cos sin2
r 9 csc2 cos
π 2
π 2
0 1
0 1
r3
29.
y2
3
4
5
r sin
31.
6
7
r
33.
r 2 r sin
r2 9 x2
2
2
y2
9
x2 y
x2
y
2
1 2
tan r tan
y 2
tanx2 y 2 1 4
x2 y 2 arctan
x2 y 2 y 0
1
y x y x
y x
2
1
1
y 2π
2
1
π
2
π
1 2
x
2π
−π x 1 2
35.
r 3 sec
37. r 3 4 cos
y
r cos 3
− 2π
1 2
6
0 ≤ < 2
3
− 12
x3
6
2
x30
−6
1
x 1
39. r 2 sin
2
41. r
4
0 ≤ < 2 −4
5
2 1 cos
Traced out once on < <
5
−10
5
−5 −2
199
200
Chapter 9
43. r 2 cos
Conics, Parametric Equations, and Polar Coordinates
32
45. r2 4 sin 2
2
0 ≤ < 4
−3
0 ≤ <
3
2
2
−3
3
−2
−2
r 2h cos k sin
47.
Radius: h2 k 2 Center: h, k
r 2 2rh cos k sin r 2 2hr cos kr sin x2 y 2 2hx ky x2 y 2 2hx 2ky 0
x2 2hx h2 y 2 2ky k2 0 h2 k 2 x h2 y k2 h2 k 2
49.
4, 23, 2, 6
51. 2, 0.5, 7, 1.2 d 22 72 227 cos0.5 1.2
20 16 cos 2 5 4.5 2
d
42
22
2 242 cos 3 6
53 28 cos0.7 5.6
53.
55. (a), (b) r 31 cos
r 2 3 sin dy 3 cos sin cos 2 3 sin dx 3 cos cos sin 2 3 sin
4
2 cos 3 sin 1 2 cos 3 sin 1 3 cos 2 2 sin 6 cos2 2 sin 3
2 ⇒ x, y 0, 3
r, 3,
Tangent line: y 3 1x 0
y x 3
dy (c) At , 1.0. 2 dx
57. (a), (b) r 3 sin
5
−4
5 −1
r,
3 2 3, 3 ⇒ x, y 3 4 3, 94
Tangent line: y
33 9 3 x 4 4 y 3x
(c) At
4
−4
dy , 0. 2 dx dy 2 At 2, , . dx 3 3 dy , 0. At 1, 2 dx At 5,
−8
9 2
dy , 3 1.732. 3 dx
Section 9.4 59.
r 1 sin
61.
dy 1 sin cos cos sin d
201
r 2 csc 3 dy 2 csc 3 cos 2 csc cot sin d
cos 1 2 sin 0
3 cos 0
1 3 5 cos 0, sin ⇒ , , , 2 2 2 6 6 Horizontal tangents:
Polar Coordinates and Polar Graphs
3 , 2 2
2, 32, 12, 6 , 12, 56
5, 2 , 1, 32
Horizontal:
dx 1 sin sin cos cos d sin sin2 sin2 1 2 sin2 sin 1 2 sin 1sin 1 0 sin 1, sin Vertical tangents:
1 7 11 ⇒ , , 2 2 6 6
32, 76, 32, 116 65. r 2 csc 5
63. r 4 sin cos2
10
2
−3
3
−12
12
−6
−2
Horizontal tangents:
2 , 3, 32
Horizontal tangents: 7,
0, 0, 1.4142, 0.7854, 1.4142, 2.3562 r 3 sin
67.
69. r 21 sin
π 2
r2 3r sin
Cardioid 0
x 2 y 2 3y
x2 y
3 2
Circle r
3 2
2
π 2
0
9 4
1
2
Symmetric to y-axis, 2
3
32
Center: 0,
Tangent at the pole: 0 71. r 2 cos3 π 2
Rose curve with three petals Symmetric to the polar axis
Relative extrema: 2, 0, 2,
2 , 2, 3 3
0 2
0
6
4
3
2
2 3
5 6
r
2
0
2
2
0
2
0
Tangents at the pole:
5 , , 6 2 6
2
1
2
3
202
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
73. r 3 sin 2 π 2
Rose curve with four petals
Symmetric to the polar axis, , and pole 2
Relative extrema: ± 3,
5 , ± 3, 4 4
Tangents at the pole: 0,
0 3
2
, 3 2 give the same tangents. 75. r 5
77. r 41 cos
π 2
π 2
Cardioid
Circle radius: 5 x2 y2 25
0
0 2
4
79. r 3 2 cos
2
6
r 3 csc
81.
π 2
r
1
10
π 2
y3
Symmetric to polar axis 0
6
r sin 3
Limaçon
4
0
3
2
2 3
2
3
4
5
Horizontal line
2
0 1
83. r 2
π 2
Spiral of Archimedes Symmetric to
2
0
4
2
3 4
5 4
3 2
r
0
2
3 2
2
5 2
3
0 1
Tangent at the pole: 0 85. r2 4 cos2 Lemniscate
Symmetric to the polar axis, , and pole 2
π 2
Relative extrema: ± 2, 0 0
0
6
r
±2
± 2
4 0
Tangents at the pole:
1
3 , 4 4
2
3
2
Section 9.4
Polar Coordinates and Polar Graphs 2
89. r
87. Since r 2 sec 2
1 , cos
Hyperbolic spiral r ⇒ as ⇒ 0
the graph has polar axis symmetry and the lengths at the pole are
, . 3 3
r
2 2 2 sin 2 sin ⇒ r r sin y
y
2 sin
x = −1 4
Furthermore,
−6
r ⇒ as ⇒ 2
6
lim
→0
2 sin 2 cos lim 2 →0 1
−4 3
r ⇒ as ⇒ . 2 Also, r 2
y=2
1 r r 2 2 cos r cos x
−3
3 −1
rx 2x r r
2x . 1x
Thus, r ⇒ ± as x ⇒ 1. 91. The rectangular coordinate system consists of all points of the form x, y where x is the directed distance from the y-axis to the point, and y is the directed distance from the x-axis to the point. Every point has a unique representation. The polar coordinate system uses r, to designate the location of a point. r is the directed distance to the origin and is the angle the point makes with the positive x-axis, measured clockwise. Point do not have a unique polar representation. 97. r 31 cos
95. r 2 sin circle
93. r a circle
b line
Cardioid
Matches (c)
Matches (a) 99. r 4 sin (a) 0 ≤ ≤
2
(b)
≤ ≤ 2
(c)
π 2
π 2
0 1
2
≤ ≤ 2 2 π 2
0 1
2
0 1
2
203
204
Chapter 9
Conics, Parametric Equations, and Polar Coordinates π 2
101. Let the curve r f be rotated by to form the curve r g. If r1, 1 is a point on r f , then r1, 1 is on r g. That is,
(r, θ +φ )
g1 r1 f 1. Letting 1 , or 1 , we see that
(r, θ )
g g1 f 1 f .
φ θ 0
103. r 2 sin
(a) r 2 sin
2 2 sin cos 4 2
(b) r 2 cos 2 cos 4
4
−6
−6
6
6
−4
−4
(d) r 2 cos
(c) r 2 sin 2 sin
4
4
−6
−6
6
6
−4
−4
105. (a) r 1 sin
(b) r 1 sin
π 2
4
π 2
Rotate the graph of r 1 sin
0 1
107. tan
2
r 21 cos dr d 2 sin
At , tan is undefined ⇒
109. tan
. 2
At
3
−6
r 2 cos 3 dr d 6 sin 3
, tan 0 ⇒ 0. 6 2
3
−3
0 1
through the angle 4.
−3
3
−2
2
Section 9.5
111.
r
Area and Arc Length in Polar Coordinates
6 dr 6 sin 61 cos 1 ⇒ 1 cos d 1 cos 2
7
ψ
6 r 1 cos 1 cos tan dr 6 sin sin d 1 cos 2 2 At , tan 3
21
1 3
θ −8
7
−3
3.
2
, 60 3 113. True
115. True
Section 9.5
Area and Arc Length in Polar Coordinates
1. (a) r 8 sin
(b) A 2
π 2
12 8 sin 2
64
2
d
0
2
sin2 d
0
32
2
1 cos 2 d
0
32
0 2
4
sin 2 2
2
16
0
A 42 16
3. A 2
1 2
6
1 sin 6 6
2 cos 32 d 2
0
6
0
3
5. A 2
7. A 2
2
1 2
2
1 sin 2 d
1 2
4
1 1 sin 4 2 4
9. A 2 2
32 2 cos 41 sin 2
2
3 2
cos 22 d
0
1 2
2 3
4
0
8
1 2 cos 2 d
3 4 sin sin 2
2 3
2 33 2
2
−1
4
−2
11. The area inside the outer loop is
12
2 3
2
2
1 2 cos 2 d 3 4 sin sin 2
0
2 3
0
4 33 . 2
−1
4
From the result of Exercise 9, the area between the loops is
4 33 2 33 33. A 2 2
−2
205
206
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
13. r 1 cos
15. r 1 cos
r 1 cos
r 1 sin
Solving simultaneously,
Solving simultaneously,
1 cos 1 cos
1 cos 1 sin
2 cos 0
cos sin tan 1
3 , . 2 2 Replacing r by r and by in the first equation and solving, 1 cos 1 cos , cos 1, 0. Both curves pass through the pole, 0, , and 0, 0, respectively.
2 , 1, 32, 0, 0
Points of intersection: 1,
3 7 , . 4 4
Replacing r by r and by in the first equation and solving, 1 cos 1 sin , sin cos 2, which has no solution. Both curves pass through the pole, 0, , and 0, 2, respectively. Points of intersection:
17. r 4 5 sin
19. r
r 3 sin
2 3
2 2
,
4
2 7
, 2 2
2
,
4
, 0, 0
π 2
r2
Solving simultaneously, 4 5 sin 3 sin sin
1 2
Solving simultaneously, we have 0
2 2, 4.
1
Points of intersection:
5 , . 6 6
2, 4, 2, 4
Both curves pass through the pole, 0, arcsin 4 5, and 0, 0, respectively. Points of intersection:
32, 6 , 32, 56, 0, 0
21. r 4 sin 2
π 2
r2 r 4 sin 2 is the equation of a rose curve with four petals and is symmetric to the polar axis, 2, and the pole. Also, r 2 is the equation of a circle of radius 2 centered at the pole. Solving simultaneously, 4 sin 2 2 2
5 , 6 6 5 , . 12 12
Therefore, the points of intersection for one petal are 2, 12 and 2, 5 12. By symmetry, the other points of intersection are 2, 7 12, 2, 11 12, 2, 13 12, 2, 17 12, 2, 19 12, and 2, 23 12.
0 1
3
Section 9.5 23. r 2 3 cos sec 2
r
25. r cos
r = sec θ 2
4
Area and Arc Length in Polar Coordinates
r 2 3 sin
−4
Points of intersection:
8
0, 0, 0.935, 0.363, 0.535, 1.006 −4
The graphs reach the pole at different times ( values).
r = 2 + 3 cos θ
The graph of r 2 3 cos is a limaçon with an inner loop b > a and is symmetric to the polar axis. The graph of r sec 2 is the vertical line x 1 2. Therefore, there are four points of intersection. Solving simultaneously, 2 3 cos
1
r = cos θ
−4
5
sec 2
−5
r = 2 − 3 sin θ
6 cos2 4 cos 1 0 cos
2 ± 10 6
arccos
2 6 10 1.376
arccos
2 6
10
2.6068.
Points of intersection: 0.581, ± 2.607, 2.581, ± 1.376 27. From Exercise 21, the points of intersection for one petal are 2, 12 and 2, 5 12. The area within one petal is
12
1 2
0
12
16
Total area 4
29. A 4
1 2
2
1 sin 4 4
12
0
12
5 12
sin22 d 2
12
0
8
5 12
1 2
4 sin 22 d
12
1 2
2
5 12
4 sin 22 d
4
d (by symmetry of the petal)
5 12
2
22 d
−6
4 3. 3
6
−4
43 3 163 43 34 4 33 3 2 sin 2 d
0
6
2
2 11 12 cos sin2
0
−9
9
11 24 −6
12
6
31. A 2
4 sin 2 d
0
6
12 41 sin2
16
0
1 2
2
6
8 2 23 4 33 3 3 5
−6
6
−3
2
4
22 d
6
33. A 2
a2
1 2
a 1 cos 2 d
0
32 2 sin sin42
3a 2 a 2 5a 2 2 4 4
0
a2 4
a 2 4
207
208
Chapter 9
35. A
Conics, Parametric Equations, and Polar Coordinates
a 2 1 8 2
a 2 a 2 8 2 a 2 8
a 2 8
a1 cos 2 d
2
2
π 2
32 2 cos cos22 d
sin 2 3 2 sin 2 2 4
a2
32 34 2 2 2
2
0
a2
a
2a
2
a2
37. (a) r a cos2
(b)
4
a=6
a=4
r3 ar2 cos2 −6
x 2 y 23 2 ax 2
6
−4
(c) A 4
12
2
6 cos2 2 4 cos2 2 d 40
0
10
2
cos4 d 10
0
2
2
1 cos 22 d
0
4 1 2 cos 2 1 cos d 10 32 sin 2 81 sin 4 2
2
0
0
15 2
39. r a cosn For n 1:
For n 2:
r a cos A
a 2
2
r a cos 2
a 2 4
A8
π 2
4
a cos 22 d
0
a 2 2
π 2
a
0
a
For n 3:
0
For n 4:
r a cos 3 A6
1 2
r a cos 4
12
6
a cos 32 d
0
π 2
a 2 4
A 16
12
8
a cos 42 d
0
a 2 2
π 2
a
0 a
0
In general, the area of the region enclosed by r a cosn for n 1, 2, 3, . . . is a 2 4 if n is odd and is a 2 2 if n is even.
Section 9.5
43. r 1 sin
41. r a r 0 s
Area and Arc Length in Polar Coordinates
r cos
2
a2 02 d a
0
2 0
2a
s2
(circumference of circle of radius a)
3 2
2
1 sin 2 cos 2 d
3 2
22
2
1 sin d
3 2
22
cos 1 sin
2
42 1 sin
d
3 2
2
42 2 0 8
2
45. r 2, 0 ≤ ≤
1 47. r , ≤ ≤ 2
49. r sin3 cos , 0 ≤ ≤ 1
4
0.5
−1
− 0.5 −1
2 −1
53. r ea r aea
r 6 sin
Length 4.39
Length 0.71
51. r 6 cos
S 2
−1
− 0.5
Length 4.16
2
S 2
6 cos sin 36 cos2 36 sin2 d
72
2
ea cos ea2 aea2 d
0
0
2
sin cos d
21 a2
0
2
e2a cos d
0
36 sin2
2
21 a2
0
36
21 4a2 1
a2
2a
2
ea 2a
r 8 sin 2 S 2
4
4 cos 2 sin 16 cos2 2 64 sin2 2 d
0
32
4
cos 2 sin cos2 2 4 sin2 2 d 21.87
0
Area Arc length
1 2
f 2d
1 2
59. (a) is correct: s 33.124.
r2 d
f 2 f 2 d
r2
ddr
2
d
2
4ae 1 2a cos sin
55. r 4 cos 2
57.
2
0.5
0
209
210
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
61. Revolve r a about the line r b sec where b > a > 0.`
π 2
f a f 0 S 2
2a a
2
b a cos a2 02 d
a
b
0
0
2a b a sin 2a2b
2 0
4 2ab
63. False. f 1 and g 1 have the same graphs. 65. In parametric form,
b
s
a
dx dt
dydt 2
2
dt.
Using instead of t, we have x r cos f cos and y r sin f sin . Thus, dx dy f cos f sin and f sin f cos . d d It follows that
ddx ddy 2
Therefore, s
Section 9.6 1. r
2
f 2 f2.
f 2 f2 d
Polar Equations of Conics and Kepler’s Laws
2e 1 e cos
(a) e 1, r
3. r
2 , parabola 1 cos
2e 1 e sin
(a) e 1, r
2 , parabola 1 sin
(b) e 0.5, r
1 2 , ellipse 1 0.5 cos 2 cos
(b) e 0.5, r
1 2 , ellipse 1 0.5 sin 2 sin
(c) e 1.5, r
3 6 , hyperbola 1 1.5 cos 2 3 cos
(c) e 1.5, r
3 6 , hyperbola 1 1.5 sin 2 3 sin
e = 1.0
4
e = 1.5
e = 0.5 −4
4
−9
9
8
e = 1.5 e = 1.0 −4
e = 0.5
−8
Section 9.6 5. r
Polar Equations of Conics and Kepler’s Laws
4 1 e sin
(a)
5
(b)
e = 0.1
− 30
5 − 30
30
30
e = 0.25 e = 0.5 e = 0.75 e = 0.9
− 40
− 40
The conic is a parabola.
The conic is an ellipse. As e → 1, the ellipse becomes more elliptical, and as e → 0 , it becomes more circular. (c)
e = 1.1
80
e=1 e=2 − 90
90
− 40
The conic is a hyperbola. As e → 1, the hyperbolas opens more slowly, and as e → , they open more rapidly. 7. Parabola; Matches (c)
13. r
9. Hyperbola; Matches (a)
1 1 sin
15. r
Parabola since e 1 1 3 Vertex: , 2 2
6 2 cos
17. r 2 sin 4 r
3 1 1 2 cos
Ellipse since e
π 2
11. Ellipse; Matches (b)
1 < 1 2
Vertices: 2, 0, 6, 0 1
2
4 2 sin 2 1 1 2 sin
Ellipse since e
π 2
Vertices:
1 < 1 2
43, 2 , 4, 32 π 2
0 1
3
0 1
19. r
5 5 1 2 cos 1 2 cos
Hyperbola since e 2 > 1
5 Vertices: 5, 0, , 3 π 2
21. r
3 3 2 2 6 sin 1 3 sin
Hyperbola since e 3 > 1 Vertices:
38, 2 , 43, 32 π 2
0 4
6
8
0 1
3
4
211
212
Chapter 9
23.
Conics, Parametric Equations, and Polar Coordinates Ellipse
1
−2
25.
Parabola
2
2 −2
1
−2
−2
27. r
1
1 sin 4
29. r
Rotate the graph of r
6
2 cos
6
Rotate the graph of
1 1 sin
r
counterclockwise through the angle . 4
6 2 cos
clockwise through the angle . 6 5
2 −2
10
−8
4
−6
31. Change to
−3
:r 4
5
5 3 cos
4
.
33. Parabola e 1, x 1, d 1 r
35. Ellipse 1 e , y 1, d 1 2 r
ed 1 e sin
1 2 1 1 2 sin
1 2 sin
41. Ellipse Vertices: 2, 0, 8, 3 16 e ,d 5 3 r
ed 1 e cos 16 5 1 3 5 cos 16 5 3 cos
ed 1 1 e cos 1 cos 39. Parabola
37. Hyperbola
2
e 2, x 1, d 1
Vertex: 1,
2 ed r 1 e cos 1 2 cos
e 1, d 2, r
43. Hyperbola
Vertices: 1,
3 3 , 9, 2 2
5 9 e ,d 4 5 r
ed 1 e sin
9 4 1 5 4 sin
9 4 5 sin
2 1 sin
45. Ellipse if 0 < e < 1, parabola if e 1, hyperbola if e > 1.
Section 9.6
1 < 1 2
213
4 49. a 5, c 4, e , b 3 5
47. (a) Hyperbola e 2 > 1 (b) Ellipse e
Polar Equations of Conics and Kepler’s Laws
r2
(c) Parabola e 1
9 1 16 25 cos 2
(d) Rotated hyperbola e 3 51. a 3, b 4, c 5, e
5 3
53. A 2
16 r2 1 25 9 cos 2
9
1 2
0
0
3 2 cos
d 2
1 d 10.88 2 cos 2
55. Vertices: 126,000, 0, 4119, a
126,000 4119 c 40,627 84,000 65,059.5, c 65,059.5 4119 60,940.5, e , d 4119 2 a 43,373 40,627
r
411984,000 43,373 345,996,000 ed 1 e cos 1 40,627 43,373 cos 43,373 40,627 cos
When 60 , r
345,996,000
15,004.49. 23,059.5
Distance between the surface of the earth and the satellite is r 4000 11,004.49 miles. 57. a 92.957 106 mi, e 0.0167 r
59. a 5.900 109 km, e 0.2481
92,931,075.2223 1 e2a 1 e cos 1 0.0167 cos
r
5.537 10 9 1 e 2 a
1 e cos 1 0.2481 cos
Perihelion distance: a1 e 91,404,618 mi
Perihelion distance: a1 e 4.436 10 9 km
Aphelion distance: a1 e 94,509,382 mi
Aphelion distance: a1 e 7.364 10 9 km
61. r
5.537 109 1 0.2481 cos
9
1 2
(a) A
0
1 2
9
9
1 2 248 1 2 (b)
10 1 5.537 0.2481 cos
0
2
0
2
d 9.341 1018 km2
10 d 1 5.537 0.2481 cos 5.537 10 1 0.2481 cos d 9
2
9
2
1 0.2481 cos 5.537 109
2
21.867 yr
d 9.341 1018
0.8995 rad In part (a) the ray swept through a smaller angle to generate the same area since the length of the ray is longer than in part (b). (c) r s
5.537 10 90.2481 sin 1 0.2481 cos 2
9
0
10 1.3737297 10 sin 1 5.537 0.2481 cos 1 0.2481 cos
9
2
9
2
2
d 2.559 109 km
2.559 109 km
1.17 108 km yr 21.867 yr s
0.899
10 1.3737297 10 sin d 4.119 1 5.537 0.2481 cos 1 0.2481 cos
9
4.119 109 km
1.88 108 km yr 21.867 yr
2
2
9
2
109 km
214
Chapter 9
63. r1
Conics, Parametric Equations, and Polar Coordinates
ed ed and r2 1 sin 1 sin
Points of intersection: ed, 0, ed, ed ed cos cos sin dy 1 sin 1 sin 2 r1: dx ed ed cos sin cos 1 sin 1 sin 2
At ed, 0,
dy dy 1. At ed, , 1. dx dx
ed ed cos cos sin dy 1 sin 1 sin 2 r2: dx ed ed cos sin cos 1 sin 1 sin 2
At ed, 0,
dy dy 1. At ed, , 1. dx dx
Therefore, at ed, 0 we have m1m2 11 1, and at ed, we have m1m2 11 1. The curves intersect at right angles.
Review Exercises for Chapter 9 1. Matches (d) - ellipse
3. Matches (a) - parabola
16x 2 16y 2 16x 24y 3 0
5.
x
2
x
y
1 3 9 3 1 9 y2 y 4 2 16 16 4 16
1 x
1 x 2
2
3 y 4
1 2
1
2
1 , 2
2
Circle Center:
1
3 4
12, 43
Radius: 1 3x 2 2y 2 24x 12y 24 0
7.
3x 2 2y 2 12x 12y 29 0
9.
3x 2 8x 16 2 y 2 6y 9 24 48 18
3x 2 4x 4 2 y 2 6y 9 29 12 18
x 4 2 y 3 2 1 2 3
x 2 2 y 3 2 1 13 12
Hyperbola
Ellipse
Center: 4, 3
Center: 2, 3
Vertices: 4 ± 2, 3 Asymptotes: y 3 ±
Vertices:
32 x 4
2, 3
±
2
2
y
x
1
y
1
2
1 6
2
4
3 4
2 x
6
4
2
(2, − 3)
3
Review Exercises for Chapter 9 13. Vertices: 3, 0, 7, 0
11. Vertex: 0, 2
15. Vertices: ± 4, 0
Directrix: x 3
Foci: 0, 0, 4, 0
Foci: ± 6, 0
Parabola opens to the right
Horizontal major axis
Center: 0, 0
p3
Center: 2, 0
Horizontal transverse axis
y 2 4 3x 0
a 5, c 2, b 21
a 4, c 6, b 36 16 25
y 2 4y 12x 4 0
x 22 y 2 1 25 21
x2 y2 1 16 20
2
17.
215
5 x2 y2 1, a 3, b 2, c 5, e 9 4 3
19. y x 2 has a slope of 1. The perpendicular slope is 1. y x 2 2x 2
By Example 5 of Section 9.1, C 12
2
0
dy 1 5 2x 2 1 when x and y . dx 2 4
1
5 sin2 d 15.87. 9
y
Perpendicular line:
5 1 1 x 4 2
4x 4y 7 0 21. (a) V abLength 12 16 192 ft 3
3
(b) F 262.4
8 4 3 y 9 y 2 dy 62.4 3 3 3 3
3
3
8 3 9 3 9 62.4 3 2 2 2 2
3
y9 y 2 dy
y 1 3 8 y 9 y 2 9 arcsin 9 y 2 32 62.4 3 2 3 3
3
9 y 2 dy 3 3
38 62.4272 7057.274
3
(c) You want 4 of the total area of 12 covered. Find h so that
h
2
x=
4 9 y 2 dy 3 3
0
h
9 y 2 dy
0
y 1 y 9 y 2 9 arcsin 2 3
h 0
y 4 3
9−
y2
4
h
9 8
Area of filled tank above x-axis is 3π.
2 x
−2
2 −2
9 8
−4
Area of filled tank below x-axis is 6π.
h3 94.
h9 h 2 9 arcsin
By Newton’s Method, h 1.212. Therefore, the total height of the water is 1.212 3 4.212 ft. (d) Area of ends 212 24 Area of sides PerimeterLength 16
2
1
0
256
167 sin d 16 from Example 5 of Section 9.1 2
122 1 167 sin 0 41 167 sin 8 21 167 sin 4 2
2
2
1 167 sin 38 1 167 sin 2 353.65
4
Total area 24 353.65 429.05
2
2
216
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
23. x 1 4t, y 2 3t t
y
x1 x1 ⇒ y23 4 4
4
2 1
11 3 y x 4 4
−1
4y 3x 11 0
x −1
1
2
3
5
−2
Line 25. x 6 cos , y 6 sin
6x 6y 2
x2
y2
2
27. x 2 sec , y 3 tan
y
1
x 22 sec2 1 tan2 1 y 32
4
x 22 y 32 1
2
36
−4 −2
Circle
y
x −2
2
4
8
Hyperbola
−4
4 2
x
−4
−2
2
4
8
−4
29. x 3 3 2t 3 5t
31.
x 3 2 y 4 2 1 16 9
y 2 2 6t 2 4t Let
(other answers possible)
x 3 2 y 4 2 cos 2 and sin 2 . 16 9
Then x 3 4 cos and y 4 3 sin . 35. (a) x 2 cot , y 4 sin cos , 0 < <
33. x cos 3 5 cos y sin 3 5 sin
4
5
−12 −7
12
8
−4
(b) 4 x2y 4 4 cot2 4 sin cos
−5
16 csc2 16
sin cos
cos sin
82 cot 8x 37. x 1 4t y 2 3t (a)
dy 3 dx 4
(b) t
x1 4
y
(c) 5
No horizontal tangents
3 3x 11 y 2 x 1 4 4
4
2 1 x
1
2
3
5
Review Exercises for Chapter 9
217
39. x 1 t y 2t 3 (a)
dy 2 2t 2 dx 1t 2
(b) t
No horizontal tangents t 0
1 x
y
(c)
y
6
2 3 x
4 2 x
4
41. x
1 2t 1
y
1 t 2t
2
y 2 5 sin (a)
2t 2 dy t 2 2t 2 t 12t 1 2 0 when t 1. (a) dx 2 t 2t 2 2 2t 1 2 Point of horizontal tangency: (b) 2t 1
13 , 1
1 1 1 1 ⇒ t x 2 x
dy 5 cos 3 2.5 cot 0 when , . dx 2 sin 2 2 Points of horizontal tangency: 3, 7, 3, 3
(b)
x 3 2 y 2 2 1 4 25 y
(c)
8
4
1 1 1x 2 x
12 1 x x 2
x
4
4x 2 4x 2 2 1 x 4x1 x 5x 1x 1
(c)
y 3 2
−2
x
−1
2
3
−1 −2
45. x cos 3 y 4 sin 3 (a)
dy 12 sin 2 cos 4 sin 4 tan 0 when 0, . dx 3 cos 2 sin cos But,
dy dx 0 at 0, . Hence no points of horizontal tangency. dt dt
(b) x 23
4y
23
1
y
(c) 4
x
4
4
43. x 3 2 cos
2
y
2
2
2
4
4
8 4
218
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
47. x cot
x r cos sin
49.
y rsin cos
y sin 2 2 sin cos (a), (c)
dx r cos d
2
−3
3
dy r sin d
−2
sr
dx 1 dy dy (b) At , 4, 1, and 6 d d dx 4
2 cos 2 2 sin 2 d
0
r
d
0
51. x, y 4, 4
1 2 r 2
x −1
1
2
3
4
5
−2
r, 42,
7 , 4
42,
3 4
−3
−4 −5
r 3 cos r2
(4, − 4)
r 2 2r 1 cos
3r cos
x 2 y 2 2 ± x 2 y 2 2x
x 2 y 2 2x 2 4x 2 y 2
x 2 y 2 3x 0 r 2 cos 2 cos 2 sin 2
r 21 cos
55.
x 2 y 2 3x
r4
0
1
7 4
57.
y
r 4 2 4 2 42
53.
r 2 2
r2
cos 2
59.
r 2 sin 2
r 4 cos 2 sec 4 2 cos 2 1
x 2 y 2 2 x 2 y 2
cos1
r cos 8 cos 2 4 x8
x
2
x2 4 y2
x 3 xy 2 4x 2 4y 2 y2 x2
61. x 2 y 2 2 ax 2y
63. x 2 y 2 a 2 arctan
r 4 a r 2 cos 2 r sin ra
cos 2
y x
2
r2 a22
sin
65. r 4
67. r sec
π 2
Circle of radius 4
1 cos
π 2
r cos 1, x 1
Centered at the pole Symmetric to polar axis,
2, and pole
44 xx
0 2
6
Vertical line
0 1
Review Exercises for Chapter 9 69. r 21 cos
71. r 4 3 cos
Cardioid
Limaçon
Symmetric to polar axis
Symmetric to polar axis
π 2
π 2
0
0
1
2
0
3
2
2 3
0
r
4
3
2
1
0
r
1
73. r 3 cos 2
4
3 5 2
2
2 3 11 2
4
π 2
Rose curve with four petals Symmetric to polar axis,
, and pole 2
0 4
3 Relative extrema: 3, 0, 3, , 3, , 3, 2 2
Tangents at the pole:
3 , 4 4
75. r 2 4 sin 2 2
π 2
r ± 2 sin 2 Rose curve with four petals
, and pole 2 3 Relative extrema: ± 2, , ± 2, 4 4 Tangents at the pole: 0, 2
0
Symmetric to the polar axis,
77. r
3 cos 4
79. r 4 cos 2 sec
Graph of r 3 sec rotated through an angle of 4 5
−1
2
Strophoid Symmetric to the polar axis r ⇒ as ⇒ 2 r ⇒ as ⇒ 2
8
4
−1
−6
6
−4
7
219
220
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
81. r 1 2 cos (a) The graph has polar symmetry and the tangents at the pole are
, . 3 3 (b)
2 sin 2 1 2 cos cos dy dx 2 sin cos 1 2 cos sin Horizontal tangents: 4 cos 2 cos 2 0, cos When cos
3 4
3 4
3 4
3 4
1 ± 33 1 33 ,r12 8 8
1 ± 1 32 1 ± 33 8 8
3 4 33,
1 8 33 0.686, 0.568 33 1 33 , arccos 0.686, 0.568 8 33 1 33 , arccos 2.186, 2.206 8 33 1 33 , arccos 2.186, 2.206. 8
33
, arccos
Vertical tangents: 1 sin 4 cos 1 0, sin 0, cos , 4
0, , ± arccos
14, 1, 0, 3,
12, ± arccos 41 0.5, ± 1.318 (c)
2.5
−5
1
−2.5
83. Circle: r 3 sin sin 2 dy 3 cos sin 3 sin cos dy tan 2 at , 3 dx 3 cos cos 3 sin sin cos 2 sin 2 6 dx Limaçon: r 4 5 sin dy 5 cos sin 4 5 sin cos dy 3 at , dx 5 cos cos 4 5 sin sin 6 dx 9 Let be the angle between the curves: tan
23.
3 39
1 13
Therefore, arctan
3
2 3 3 49.1 .
Review Exercises for Chapter 9 85. r 1 cos , r 1 cos The points 1, 2 and 1, 32 are the two points of intersection (other than the pole). The slope of the graph of r 1 cos is m1
r sin r cos dy sin 2 cos 1 cos . dx r cos r sin sin cos sin 1 cos
At 1, 2, m1 11 1 and at 1, 32, m1 11 1. The slope of the graph of r 1 cos is m2
sin 2 cos 1 cos dy . dx sin cos sin 1 cos
At 1, 2, m2 11 1 and at 1, 32, m 2 11 1. In both cases, m 1 1m 2 and we conclude that the graphs are orthogonal at 1, 2 and 1, 32. 87. r 2 cos A2
1 2
89. r sin
2 cos 2 d 14.14
0
9 2
A2
3
1 2
0.10
−3
cos 2 2
sin cos 2 2 d
0
32
6
0.5
−3 − 0.5
0.5 − 0.1
91. r 2 4 sin 2 A2
93. r 4 cos , r 2
12
2
4 sin 2 d 4
0
A2
12
−3
3
2
3
22 a
−3
1 cos d 22 a
0
0
2 ,e1 1 sin
Parabola
sin d 42 a1 cos 12 1 cos
99. r
0
8a
6 2 2 ,e 3 2 cos 1 23cos 3
Ellipse π 2
π 2
0
2
0
2
4 6
8
4 cos 2 d 4.91
a 21 cos 2 a 2 sin 2 d
0
97. r
6
−2
1 2
3
−3
4 d
0
2
95. s 2
3
221
214
Chapter 9
63. r1
Conics, Parametric Equations, and Polar Coordinates
ed ed and r2 1 sin 1 sin
Points of intersection: ed, 0, ed, ed ed cos cos sin dy 1 sin 1 sin 2 r1: dx ed ed cos sin cos 1 sin 1 sin 2
At ed, 0,
dy dy 1. At ed, , 1. dx dx
ed ed cos cos sin dy 1 sin 1 sin 2 r2: dx ed ed cos sin cos 1 sin 1 sin 2
At ed, 0,
dy dy 1. At ed, , 1. dx dx
Therefore, at ed, 0 we have m1m2 11 1, and at ed, we have m1m2 11 1. The curves intersect at right angles.
Review Exercises for Chapter 9 1. Matches (d) - ellipse
3. Matches (a) - parabola
16x 2 16y 2 16x 24y 3 0
5.
x
2
x
y
1 3 9 3 1 9 y2 y 4 2 16 16 4 16
1 x
1 x 2
2
3 y 4
1 2
1
2
1 , 2
2
Circle Center:
1
3 4
12, 43
Radius: 1 3x 2 2y 2 24x 12y 24 0
7.
3x 2 2y 2 12x 12y 29 0
9.
3x 2 8x 16 2 y 2 6y 9 24 48 18
3x 2 4x 4 2 y 2 6y 9 29 12 18
x 4 2 y 3 2 1 2 3
x 2 2 y 3 2 1 13 12
Hyperbola
Ellipse
Center: 4, 3
Center: 2, 3
Vertices: 4 ± 2, 3 Asymptotes: y 3 ±
Vertices:
32 x 4
2, 3
±
2
2
y
x
1
y
1
2
1 6
2
4
3 4
2 x
6
4
2
(2, − 3)
3
Review Exercises for Chapter 9 13. Vertices: 3, 0, 7, 0
11. Vertex: 0, 2
15. Vertices: ± 4, 0
Directrix: x 3
Foci: 0, 0, 4, 0
Foci: ± 6, 0
Parabola opens to the right
Horizontal major axis
Center: 0, 0
p3
Center: 2, 0
Horizontal transverse axis
y 2 4 3x 0
a 5, c 2, b 21
a 4, c 6, b 36 16 25
y 2 4y 12x 4 0
x 22 y 2 1 25 21
x2 y2 1 16 20
2
17.
215
5 x2 y2 1, a 3, b 2, c 5, e 9 4 3
19. y x 2 has a slope of 1. The perpendicular slope is 1. y x 2 2x 2
By Example 5 of Section 9.1, C 12
2
0
dy 1 5 2x 2 1 when x and y . dx 2 4
1
5 sin2 d 15.87. 9
y
Perpendicular line:
5 1 1 x 4 2
4x 4y 7 0 21. (a) V abLength 12 16 192 ft 3
3
(b) F 262.4
8 4 3 y 9 y 2 dy 62.4 3 3 3 3
3
3
8 3 9 3 9 62.4 3 2 2 2 2
3
y9 y 2 dy
y 1 3 8 y 9 y 2 9 arcsin 9 y 2 32 62.4 3 2 3 3
3
9 y 2 dy 3 3
38 62.4272 7057.274
3
(c) You want 4 of the total area of 12 covered. Find h so that
h
2
x=
4 9 y 2 dy 3 3
0
h
9 y 2 dy
0
y 1 y 9 y 2 9 arcsin 2 3
h 0
y 4 3
9−
y2
4
h
9 8
Area of filled tank above x-axis is 3π.
2 x
−2
2 −2
9 8
−4
Area of filled tank below x-axis is 6π.
h3 94.
h9 h 2 9 arcsin
By Newton’s Method, h 1.212. Therefore, the total height of the water is 1.212 3 4.212 ft. (d) Area of ends 212 24 Area of sides PerimeterLength 16
2
1
0
256
167 sin d 16 from Example 5 of Section 9.1 2
122 1 167 sin 0 41 167 sin 8 21 167 sin 4 2
2
2
1 167 sin 38 1 167 sin 2 353.65
4
Total area 24 353.65 429.05
2
2
216
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
23. x 1 4t, y 2 3t t
y
x1 x1 ⇒ y23 4 4
4
2 1
11 3 y x 4 4
−1
4y 3x 11 0
x −1
1
2
3
5
−2
Line 25. x 6 cos , y 6 sin
6x 6y 2
x2
y2
2
27. x 2 sec , y 3 tan
y
1
x 22 sec2 1 tan2 1 y 32
4
x 22 y 32 1
2
36
−4 −2
Circle
y
x −2
2
4
8
Hyperbola
−4
4 2
x
−4
−2
2
4
8
−4
29. x 3 3 2t 3 5t
31.
x 3 2 y 4 2 1 16 9
y 2 2 6t 2 4t Let
(other answers possible)
x 3 2 y 4 2 cos 2 and sin 2 . 16 9
Then x 3 4 cos and y 4 3 sin . 35. (a) x 2 cot , y 4 sin cos , 0 < <
33. x cos 3 5 cos y sin 3 5 sin
4
5
−12 −7
12
8
−4
(b) 4 x2y 4 4 cot2 4 sin cos
−5
16 csc2 16
sin cos
cos sin
82 cot 8x 37. x 1 4t y 2 3t (a)
dy 3 dx 4
(b) t
x1 4
y
(c) 5
No horizontal tangents
3 3x 11 y 2 x 1 4 4
4
2 1 x
1
2
3
5
Review Exercises for Chapter 9
217
39. x 1 t y 2t 3 (a)
dy 2 2t 2 dx 1t 2
(b) t
No horizontal tangents t 0
1 x
y
(c)
y
6
2 3 x
4 2 x
4
41. x
1 2t 1
y
1 t 2t
2
y 2 5 sin (a)
2t 2 dy t 2 2t 2 t 12t 1 2 0 when t 1. (a) dx 2 t 2t 2 2 2t 1 2 Point of horizontal tangency: (b) 2t 1
13 , 1
1 1 1 1 ⇒ t x 2 x
dy 5 cos 3 2.5 cot 0 when , . dx 2 sin 2 2 Points of horizontal tangency: 3, 7, 3, 3
(b)
x 3 2 y 2 2 1 4 25 y
(c)
8
4
1 1 1x 2 x
12 1 x x 2
x
4
4x 2 4x 2 2 1 x 4x1 x 5x 1x 1
(c)
y 3 2
−2
x
−1
2
3
−1 −2
45. x cos 3 y 4 sin 3 (a)
dy 12 sin 2 cos 4 sin 4 tan 0 when 0, . dx 3 cos 2 sin cos But,
dy dx 0 at 0, . Hence no points of horizontal tangency. dt dt
(b) x 23
4y
23
1
y
(c) 4
x
4
4
43. x 3 2 cos
2
y
2
2
2
4
4
8 4
218
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
47. x cot
x r cos sin
49.
y rsin cos
y sin 2 2 sin cos (a), (c)
dx r cos d
2
−3
3
dy r sin d
−2
sr
dx 1 dy dy (b) At , 4, 1, and 6 d d dx 4
2 cos 2 2 sin 2 d
0
r
d
0
51. x, y 4, 4
1 2 r 2
x −1
1
2
3
4
5
−2
r, 42,
7 , 4
42,
3 4
−3
−4 −5
r 3 cos r2
(4, − 4)
r 2 2r 1 cos
3r cos
x 2 y 2 2 ± x 2 y 2 2x
x 2 y 2 2x 2 4x 2 y 2
x 2 y 2 3x 0 r 2 cos 2 cos 2 sin 2
r 21 cos
55.
x 2 y 2 3x
r4
0
1
7 4
57.
y
r 4 2 4 2 42
53.
r 2 2
r2
cos 2
59.
r 2 sin 2
r 4 cos 2 sec 4 2 cos 2 1
x 2 y 2 2 x 2 y 2
cos1
r cos 8 cos 2 4 x8
x
2
x2 4 y2
x 3 xy 2 4x 2 4y 2 y2 x2
61. x 2 y 2 2 ax 2y
63. x 2 y 2 a 2 arctan
r 4 a r 2 cos 2 r sin ra
cos 2
y x
2
r2 a22
sin
65. r 4
67. r sec
π 2
Circle of radius 4
1 cos
π 2
r cos 1, x 1
Centered at the pole Symmetric to polar axis,
2, and pole
44 xx
0 2
6
Vertical line
0 1
Review Exercises for Chapter 9 69. r 21 cos
71. r 4 3 cos
Cardioid
Limaçon
Symmetric to polar axis
Symmetric to polar axis
π 2
π 2
0
0
1
2
0
3
2
2 3
0
r
4
3
2
1
0
r
1
73. r 3 cos 2
4
3 5 2
2
2 3 11 2
4
π 2
Rose curve with four petals Symmetric to polar axis,
, and pole 2
0 4
3 Relative extrema: 3, 0, 3, , 3, , 3, 2 2
Tangents at the pole:
3 , 4 4
75. r 2 4 sin 2 2
π 2
r ± 2 sin 2 Rose curve with four petals
, and pole 2 3 Relative extrema: ± 2, , ± 2, 4 4 Tangents at the pole: 0, 2
0
Symmetric to the polar axis,
77. r
3 cos 4
79. r 4 cos 2 sec
Graph of r 3 sec rotated through an angle of 4 5
−1
2
Strophoid Symmetric to the polar axis r ⇒ as ⇒ 2 r ⇒ as ⇒ 2
8
4
−1
−6
6
−4
7
219
220
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
81. r 1 2 cos (a) The graph has polar symmetry and the tangents at the pole are
, . 3 3 (b)
2 sin 2 1 2 cos cos dy dx 2 sin cos 1 2 cos sin Horizontal tangents: 4 cos 2 cos 2 0, cos When cos
3 4
3 4
3 4
3 4
1 ± 33 1 33 ,r12 8 8
1 ± 1 32 1 ± 33 8 8
3 4 33,
1 8 33 0.686, 0.568 33 1 33 , arccos 0.686, 0.568 8 33 1 33 , arccos 2.186, 2.206 8 33 1 33 , arccos 2.186, 2.206. 8
33
, arccos
Vertical tangents: 1 sin 4 cos 1 0, sin 0, cos , 4
0, , ± arccos
14, 1, 0, 3,
12, ± arccos 41 0.5, ± 1.318 (c)
2.5
−5
1
−2.5
83. Circle: r 3 sin sin 2 dy 3 cos sin 3 sin cos dy tan 2 at , 3 dx 3 cos cos 3 sin sin cos 2 sin 2 6 dx Limaçon: r 4 5 sin dy 5 cos sin 4 5 sin cos dy 3 at , dx 5 cos cos 4 5 sin sin 6 dx 9 Let be the angle between the curves: tan
23.
3 39
1 13
Therefore, arctan
3
2 3 3 49.1 .
Review Exercises for Chapter 9 85. r 1 cos , r 1 cos The points 1, 2 and 1, 32 are the two points of intersection (other than the pole). The slope of the graph of r 1 cos is m1
r sin r cos dy sin 2 cos 1 cos . dx r cos r sin sin cos sin 1 cos
At 1, 2, m1 11 1 and at 1, 32, m1 11 1. The slope of the graph of r 1 cos is m2
sin 2 cos 1 cos dy . dx sin cos sin 1 cos
At 1, 2, m2 11 1 and at 1, 32, m 2 11 1. In both cases, m 1 1m 2 and we conclude that the graphs are orthogonal at 1, 2 and 1, 32. 87. r 2 cos A2
1 2
89. r sin
2 cos 2 d 14.14
0
9 2
A2
3
1 2
0.10
−3
cos 2 2
sin cos 2 2 d
0
32
6
0.5
−3 − 0.5
0.5 − 0.1
91. r 2 4 sin 2 A2
93. r 4 cos , r 2
12
2
4 sin 2 d 4
0
A2
12
−3
3
2
3
22 a
−3
1 cos d 22 a
0
0
2 ,e1 1 sin
Parabola
sin d 42 a1 cos 12 1 cos
99. r
0
8a
6 2 2 ,e 3 2 cos 1 23cos 3
Ellipse π 2
π 2
0
2
0
2
4 6
8
4 cos 2 d 4.91
a 21 cos 2 a 2 sin 2 d
0
97. r
6
−2
1 2
3
−3
4 d
0
2
95. s 2
3
221
222
Chapter 9
101. r
Conics, Parametric Equations, and Polar Coordinates
4 2 3 ,e 2 3 sin 1 32sin 2
103. Circle Center:
Hyperbola
5, 2 0, 5 in rectangular coordinates
Solution point: 0, 0
π 2
x 2 y 5 5 25 x 2 y 2 10y 0
0
2
3
4
r 2 10r sin 0 r 10 sin
105. Parabola
107. Ellipse
Vertex: 2,
Vertices: 5, 0, 1,
Focus: 0, 0
Focus: 0, 0
e 1, d 4
2 5 a 3, c 2, e , d 3 2
4 r 1 cos
2352 5 r 2 3 2 cos 1 cos 3
Problem Solving for Chapter 9 y
1. (a) 10 8 6 4
(− 1, 14 )
(4, 4)
2
−6 −4 −2
x 2
−2
4
6
(b) x2 4y 2x 4y 1 y x 2 ⇒ y 2x 4
y 4 2x 4
1 1 1 1 y x 1 ⇒ y x 4 2 2 4
Tangent line at 4, 4
Tangent line at 1,
Tangent lines have slopes of 2 and 12 ⇒ perpendicular. (c) Intersection: 1 1 2x 4 x 2 4 8x 16 2x 1 10x 15 x
3 ⇒ 2
32, 1
Point of intersection, 32, 1, is on directrix y 1.
1 4
Problem Solving for Chapter 9 3. Consider x2 4py with focus 0, p.
y
A
B
Let pa, b be point on parabola. zx 4py ⇒ y yb
F
x 2p
P(a, b) x
Q
a x a Tangent line 2p
For x 0, y b
a 4pb a2 b b. a b 2p 2p 2p
Thus, Q 0, b. FQP is isosceles because
FQ p b FP a 02 b p2 a2 b2 2bp p2 4pb b2 2bp p2 b p2 b p. Thus, FQP BPA FPQ.
5. (a) In OCB, cos
2a ⇒ OB 2a sec . OB
r 2a tan sin
(c)
r cos 2a sin2
OA In OAC, cos ⇒ OA 2a cos . 2a
r 3 cos 2a r 2 sin2
x2 y2x 2ay2
r OP AB OB OA 2asec cos 2a
cos1 cos
2a
y2
x3 2a x
sin2 cos
2a tan sin (b) x r cos 2a tan sin cos 2a sin2 y r sin 2a tan sin sin 2a tan
sin2 , 2
2
< <
1 + t2
θ
Let t tan , < t < . t2 t2 t3 Then sin2 and x 2a , y 2a . 2 2 1t 1t 1 t2
7. y a1 cos ⇒ cos
arccos
t
1
ay a
a a y a
x a sin
a a y sinarccosa a y
a a y
a arccos
a arccos
x a arccos
2ay y2
a
a a y 2ay y , 0 ≤ y ≤ 2a 2
θ
a−y
2ay − y 2
223
224
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
3 5 7 , , , ,. . . 2 2 2 2
9. For t y
11. (a) Area
0
2 2 2 2 , , , ,. . . 3 5 7
Hence, the curve has length greater that
(b) tan
2 2 2 2 S . . . 3 5 7
2 1 1 1 1 . . . 3 5 7
2 1 1 1 1 . . . > 2 4 6 8
1 2
1 2 r d 2
x=1 r = sec θ
α
sec2 d
1
0
h 1 ⇒ Area 1tan
1 2 ⇒ tan
sec2 d
0
(c) Differentiating,
d tan sec2 . d
.
13. If a dog is located at r, , then its neighbor is at r,
: 2
x, y r cos , r sin and x, y r sin , r cos . The slope joining these points is r cos r sin sin cos slope of tangent line at r, . r sin r cos sin cos dr sin r cos sin cos d dr sin cos cos r sin d dr r d
⇒
dr d r ln r C1 r e C1 r Ce r
4 d2 ⇒ r Ce
4
Finally, r
d 2
d 2
⇒ C
d 2
e4
e4 .
15. (a) The first plane makes an angle of 70 with the positive x-axis, and is 150 miles from P:
(c)
280
x1 cos 70 150 375t y1 sin 70 150 375t
0
1 0
Similarly for the second plane, x2 cos 135 190 450t
The minimum distance is 7.59 miles when t 0.4145.
cos 45 190 450t y2 sin 135 190 450t sin 45 190 450t (b) d x2 x12 y2 y12
cos 45190 450t cos 70150 375t 2 sin 45190 450t sin 70150 375t 2 12
Problem Solving for Chapter 9 17.
4
−6
4
6
−6
4
4
6
−6
6
−6
6
−4
−4
−4
−4
4
4
4
4
−6
6
−6
6
−6
6
−4
−4
−4
4
4
4
−6
6
−4
−6
6
−4
−6
n 1, 2, 3, 4, 5 produce “bells”; n 1, 2, 3, 4, 5 produce “hearts”.
6
−4
6
−4
−6
225
PA R T
I C H A P T E R P Preparation for Calculus Section P.1
Graphs and Models . . . . . . . . . . . . . . . . . . . . . . 2
Section P.2
Linear Models and Rates of Change . . . . . . . . . . . . . 7
Section P.3
Functions and Their Graphs . . . . . . . . . . . . . . . . . 14
Section P.4
Fitting Models to Data . . . . . . . . . . . . . . . . . . . . 18
Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
C H A P T E R P Preparation for Calculus Section P.1
Graphs and Models
Solutions to Odd-Numbered Exercises
1 1. y 2 x 2
3. y 4 x2
x-intercept: 4, 0
x-intercepts: 2, 0, 2, 0
y-intercept: 0, 2
y-intercept: 0, 4
Matches graph (b)
Matches graph (a) 7. y 4 x2
5. y 32x 1 x
4
2
0
2
4
x
3
2
0
2
3
y
5
2
1
4
7
y
5
0
4
0
5
y
y 8
(4, 7)
6
2
(−2, 0)
(0, 1)
−8 −6 −4
2 −4
(0, 4)
(2, 4)
4
(−4, − 5)
6
x
4
6
−6
8
4
(− 3, − 5)
(3, − 5)
−4 −6
11. y x 4
9. y x 2 x
5
4
3
2
1
0
1
x
0
1
4
9
16
y
3
2
1
0
1
2
3
y
4
3
2
1
0
y
y 10 8 6 4 2
6 4
(−5, 3)
(−4, 2) 2 −6
−4
(1, 3) (0, 2)
−2
(−1, 1)
(−3, 1)
x
(− 2, 0) −2
2
6
−2
−8
(2, 0) x
−4
(− 2, −2)
−6
2
2
−4 −6 −8 − 10
(4, − 2)
(16, 0) x
2
(1, − 3) (0, − 4)
12 14 16 18
(9, − 1)
Section P.1 13.
15.
Xmin = -3 Xmax = 5 Xscl = 1 Ymin = -3 Ymax = 5 Yscl = 1
3
5
(− 4.00, 3) (2, 1.73) −6
6
−3
(a) 2, y 2, 1.73
Note that y 4 when x 0.
(b) x, 3 4, 3
y 02 0 2
y-intercept:
y 0225 02
y-intercept:
y 0; 0, 0
y 2; 0, 2 0 x2 x 2
x-intercepts:
y 5 2 3 1.73 3 5 4
19. y x225 x2
17. y x2 x 2
21. y
Graphs and Models
x-intercepts:
0 x225 x2
0 x 2x 1
0 x25 x5 x
x 2, 1; 2, 0, 1, 0
x 0, ± 5; 0, 0; ± 5, 0
32 x x
23. x2y x2 4y 0
y-intercept:
None. x cannot equal 0.
x-intercepts:
32 x 0 x
y-intercept: 02y 02 4y 0 y 0; 0, 0
0 2 x
x-intercept:
x 4; 4, 0
x20 x2 40 0 x 0; 0, 0
25. Symmetric with respect to the y-axis since
27. Symmetric with respect to the x-axis since
y x 2 x 2. 2
y2 y2 x3 4x.
2
31. y 4 x 3
29. Symmetric with respect to the origin since
xy xy 4.
No symmetry with respect to either axis or the origin.
x y x2 1 y
35. y x3 x is symmetric with respect to the y-axis
33. Symmetric with respect to the origin since
since y x3 x x3 x x3 x .
x . x2 1
37. y 3x 2
y
Intercepts:
23 , 0, 0, 2
2
(0, 2)
1
2 3,
Symmetry: none
0 x
1
2 1
3
4
Chapter P
39. y
Preparation for Calculus
x 4 2
43. y x 32
41. y 1 x2
Intercepts:
Intercepts:
Intercepts:
1, 0, 1, 0, 0, 1
8, 0, 0, 4
3, 0, 0, 9
Symmetry: y-axis
Symmetry: none
Symmetry: none
y y
y 12
2
2
(8, 0)
2
2
8
4
10
( 1, 0)
2
(0,
10
(0, 1)
x
x
−2
4)
2
6
1
8
2
2 − 10 − 8 − 6
10
47. y xx 2
45. y x3 2 3 2, 0 , 0, 2
−2
Symmetry: origin
Symmetry: none
y 4
Domain: x ≥ 2
y
3 2
5 y
4
2, 0)
(0, 2)
2
1
2
1
(− 2, 0) −4 −3
−1
3
4
−4
2
3
2
−3
3
1
1
−2
4
x
x
−4 −3 −2 −1
5
1 3
(0, 0)
6
3 3
4
Intercepts: 0, 0
0, 0, 2, 0
Symmetry: none
x 2
(− 3, 0)
49. x y3
Intercepts:
Intercepts:
(
(0, 9)
8
(1, 0)
(0, 0) x 1
2
3
4
−2
1 x Intercepts: none
51. y
53. y 6 x
y 3
y 8
Intercepts:
6
2
0, 6, 6, 0, 6, 0
1
Symmetry: origin
x 1
2
Symmetry: y-axis
3
(0, 6)
4 2
(− 6, 0) −8
−4 −2 −2
(6, 0) x 2
4
6
8
−4 −6 −8
57. x 3y2 6
55. y2 x 9 y2 x 9
3y2 6 x
y ± x 9
4
Intercepts:
0, 3, 0, 3, 9, 0 Symmetry: x-axis
y±
(0, 3) (−9, 0) − 11
1
(0, − 3) −4
2 3x
Intercepts:
3
(0, 2 ) (6, 0)
−1
8
6, 0, 0, 2, 0, 2 Symmetry: x-axis
(0, − 2 ) −3
Section P.1 59. y x 2x 4x 6 (other answers possible)
Graphs and Models
61. Some possible equations: yx y x3 y 3x3 x 3 x y
63.
xy2⇒y2x
xy7⇒y7x
65.
2x y 1 ⇒ y 2x 1
3x 2y 11 ⇒ y
2 x 2x 1 7x
3 3x 1x
3x 11 2
3x 11 2
14 2x 3x 11
The corresponding y-value is y 1.
5x 25
Point of intersection: 1, 1
x5 The corresponding y-value is y 2. Point of intersection: 5, 2
67. x2 y 6 ⇒ y 6 x2
69. x2 y 2 5 ⇒ y 2 5 x 2 xy1⇒yx1
xy4⇒y4x
5 x2 x 12
6 x2 4 x 0 x2 x 2
5 x2 x2 2x 1
0 x 2x 1
0 2x2 2x 4 2x 1x 2
x 2, 1
x 1 or x 2 The corresponding y-values are y 2 and y 1.
The corresponding y-values are y 2 (for x 2) and y 5 (for x 1).
Points of intersection: 1, 2, 2, 1
Points of intersection: 2, 2, 1, 5 71.
y x3
y x3 2x2 x 1
73.
yx
y x2 3x 1
x3 x
x3 2x2 x 1 x2 3x 1
x3 x 0 xx 1x 1 0
x3 x2 2x 0 xx 2x 1 0 x 1, 0, 2
x 0, x 1, or x 1 The corresponding y-values are y 0, y 1, and y 1.
1, 5, 0, 1, 2, 1 4
Points of intersection: 0, 0, 1, 1, 1, 1 −4
y = x 3 − 2x 2 + x − 1 (2, 1)
(0, −1)
6
(−1, −5) −8
y = −x 2 + 3 x − 1
5
6
Chapter P
Preparation for Calculus
75. 5.5x 10,000 3.29x
5.5x 2 3.29x 10,0002 30.25x 10.8241x2 65,800x 100,000,000 0 10.8241x2 65,830.25x 100,000,000
Use the Quadratic Formula.
x 3133 units The other root, x 2949, does not satisfy the equation R C. This problem can also be solved by using a graphing utility and finding the intersection of the graphs of C and R. 77. (a) Using a graphing utility, you obtain
(b)
250
y 0.0153t2 4.9971t 34.9405 (c) For the year 2004, t 34 and y 187.2 CPI.
−5
35 − 50
79.
400
0
100 0
If the diameter is doubled, the resistance is changed by approximately a factor of 14. For instance, y20 26.555 and y40 6.36125. 81. False; x-axis symmetry means that if 1, 2 is on the graph, then 1, 2 is also on the graph. 83. True; the x-intercepts are
b
± b2 4ac
2a
,0 .
85. Distance to the origin K Distance to 2, 0 x2 y2 Kx 22 y2, K 1
x2 y 2 K 2x2 4x 4 y2
1 K 2 x 2 1 K 2y 2 4K 2x 4K 2 0 Note: This is the equation of a circle!
Section P.2
Section P.2
Linear Models and Rates of Change
1. m 1
3. m 0
7.
9. m
y
m=1
5 4
2
1
5. m 12
2 4 53
11. m
6 3 2
y
m = − 32
m is undefined
1
−1
(2, 3)
3
Linear Models and Rates of Change
51 22 4 0
undefined
3
x 3
4
2
5
y
(5, 2)
1
m = −2 −1
6 x 1
2
3
5
6
4
−2
3
−3 −4
(2, 5)
5
7
2
(3, − 4)
(2, 1)
1
−5
−2 −1 −1
x 1
3
4
5
6
−2
13. m
23 16 12 34
y 3 2
12 2 14
(− 12 , 23 ) −3
−2
(− 34 , 16 ) x 1
2
3
−1 −2 −3
15. Since the slope is 0, the line is horizontal and its equation is y 1. Therefore, three additional points are 0, 1, 1, 1, and 3, 1. 17. The equation of this line is y 7 3x 1 y 3x 10 . Therefore, three additional points are 0, 10, 2, 4, and 3, 1. 19. Given a line L, you can use any two distinct points to calculate its slope. Since a line is straight, the ratio of the change in y-values to the change in x-values will always be the same. See Section P.2 Exercise 93 for a proof.
7
8
Chapter P
Population (in millions)
21. (a)
Preparation for Calculus (b) The slopes of the line segments are
270
255.0 252.1 2.9 21
260
257.7 255.0 2.7 32
250
260.3 257.7 2.6 43
1 2 3 4 5 6 7 8 9
Year (0 ↔ 1990)
262.8 260.3 2.5 54 265.2 262.8 2.4 65 267.7 265.2 2.5 76 270.3 267.7 2.6 87 The population increased most rapidly from 1991 to 1992.
m 2.9 23. x 5y 20 y
25. x 4
15 x
4
Therefore, the slope is m 0, 4. 27.
y 34 x 3
1 5
The line is vertical. Therefore, the slope is undefined and there is no y-intercept.
and the y-intercept is
y 23 x
29.
4y 3x 12
y 2 3x 3
31.
y 2 3x 9
3y 2x
0 3x 4y 12
2x 3y 0
y 3x 11 y 3x 11 0
y
y 5
4
4
3
y 3
(0, 3) 2
2
2
1
1
(0, 0) x
x
−4 −3 −2 −1
1
1
2
3
4
−2 −1 −1
x 1
2
4
5
6
(3, − 2)
−2
−1
3
−3 −4 −5
33. m
60 3 20
35. m
y 0 3x 0
1 3 2 20
y 1 2x 4
8
3y 8x 40 0
y
6
(2, 6)
4
2
y
(2, 1)
1
(0, 0) x 2
4
6
8
−2 −1 −1
x 2
−2
−8
40 8 y x 3 3
0 2x y 3
y
−8 −6 −4 −2
80 8 25 3 8 y 0 x 5 3
y 1 2x 2
y 3x
2
37. m
−3
−5
(0, −3)
3
4
5
9 8 7 6 5 4 3 2 1 −1 −2
(2, 8)
(5, 0) x 1 2 3 4
6 7 8 9
Section P.2 81 55
39. m
Undefined.
Vertical line x 5
41. m
Linear Models and Rates of Change
72 34 114 11 12 0 12 2 y
x3
43.
x30
3 11 x 0 4 2
y
y 9 8 7 6 5 4 3 2 1 −1
y
(5, 8)
11 3 x 2 4
2 1
22x 4y 3 0 −1
y
(5, 1) x 1 2 3 4
(3, 0) 1
4
6 7 8 9
3
−2
−2
( 12 , 72 )
2 1 −4 −3 −2 −1
y x 1 2 3
45.
( 0, 34 ) x 1
2
3
4
47.
3x 2y 6 0
y x 1 a a 1 2 1 a a 3 1 a a3⇒xy3 xy30
49.
51. y 2x 1
y 3 y30
y 3
y 2 1 x
−3 −2 −1
1
2
3
4
5
−2
−2
x
−1
1
2
−1
−4 −5 −6
y 2 32x 1
53.
y
3 2x
55. 2x y 3 0 y 2x 3
1 2
2y 3x 1 0
y 1
y
x
4
2
3
1
2 1
2
2
1 x
−4 −3 −2
1 −2 −3 −4
2
3
4
3
3
2
4
x
9
10
Chapter P
57.
Preparation for Calculus
10
10
− 10
− 15
10
− 10
15
− 10
The lines do not appear perpendicular.
The lines appear perpendicular.
The lines are perpendicular because their slopes 1 and 1 are negative reciprocals of each other. You must use a square setting in order for perpendicular lines to appear perpendicular. 61. 5x 3y 0
59. 4x 2y 3 y 2x 2
y 53x
m2
m 53
3
y 1 2x 2
(a)
(a)
24y 21 40x 30
y 1 2x 4
24y 40x 9 0
2x y 3 0 y1
(b)
y 78 53x 34
1 2 x
2
(b)
y 78 35x 34 40y 35 24x 18
2y 2 x 2
40y 24x 53 0
x 2y 4 0 63. (a) x 2 ⇒ x 2 0 (b) y 5 ⇒ y 5 0 65. The slope is 125. Hence, V 125t 1 2540 125t 2415 67. The slope is 2000. Hence, V 2000t 1 20,400 2000t 22,400 69.
5
(2, 4)
−3
(0, 0)
6
−1
You can use the graphing utility to determine that the points of intersection are 0, 0 and 2, 4. Analytically, x2 4x x2 2x2 4x 0 2xx 2 0 x 0 ⇒ y 0 ⇒ 0, 0 x 2 ⇒ y 4 ⇒ 2, 4. The slope of the line joining 0, 0 and 2, 4 is m 4 02 0 2. Hence, an equation of the line is y 0 2x 0 y 2x.
Section P.2
71. m1 m2
Linear Models and Rates of Change
10 1 2 1 2 0 2 2 1 3
m1 m2 The points are not collinear. y
73. Equations of perpendicular bisectors: y y
c ab ab x 2 c 2
c ba ab x 2 c 2
Letting x 0 in either equation gives the point of intersection:
0, a
2
(b, c)
( b −2 a , 2c )
( a +2 b , 2c )
(− a, 0)
x
(a, 0)
b2 c2 . 2c
This point lies on the third perpendicular bisector, x 0. 75. Equations of altitudes: y
y
ab x a c (b, c)
xb y
ab x a c
(a, 0) x
(− a, 0)
Solving simultaneously, the point of intersection is
b, a
2
b2 . c
77. Find the equation of the line through the points 0, 32 and 100, 212. 9 m 180 100 5 9 F 32 5 C 0
F 95 C 32 5F 9C 160 0 For F 72, C 22.2. 79. (a) W1 0.75x 12.50
(b)
50
W2 1.30x 9.20 (c) Both jobs pay $17 per hour if 6 units are produced. For someone who can produce more than 6 units per hour, the second offer would pay more. For a worker who produces less than 6 units per hour, the first offer pays more.
(6, 17) 0
30 0
Using a graphing utility, the point of intersection is approximately 6, 17. Analytically, 0.75x 12.50 1.30x 9.20 3.3 0.55x ⇒ x 6 y 0.756 12.50 17.
11
12
Chapter P
Preparation for Calculus
81. (a) Two points are 50, 580 and 47, 625. The slope is m
(b)
50
625 580 15. 47 50
p 580 15x 50
0
p 15x 750 580 15x 1330
1 If p 655, x 15 1330 655 45 units.
1 or x 15 1330 p
83. 4x 3y 10 0 ⇒ d
85. x y 2 0 ⇒ d
1500 0
1 (c) If p 595, x 15 1330 595 49 units.
40 30 10 10 2 42 32
5
12 11 2 1 1 2
2
5 5 2 2 2
87. A point on the line x y 1 is 0, 1. The distance from the point 0, 1 to x y 5 0 is d
10 11 5 1 5 12 12
2
4 2
2 2.
89. If A 0, then By C 0 is the horizontal line y CB. The distance to x1, y1 is
By C Ax By C. C B A B B
Ax C Ax By C. C A A B A
d y1
1
1
1
2
2
If B 0, then Ax C 0 is the vertical line x CA. The distance to x1, y1 is d x1
1
1
1
2
2
(Note that A and B cannot both be zero.) The slope of the line Ax By C 0 is AB. The equation of the line through x1, y1 perpendicular to Ax By C 0 is: y y1
B x x1 A
Ay Ay1 Bx Bx1 Bx1 Ay1 Bx Ay The point of intersection of these two lines is: Ax By C
⇒
Bx Ay Bx1 Ay1 ⇒
A2x ABy AC B
2x
ABy
B2x
1
(1) ABy1 (2)
A2 B2x AC B2x1 ABy1 (By adding equations (1) and (2)) x Ax By C
⇒
AC B2x1 ABy1 A2 B2
ABx B2y BC
Bx Ay Bx1 Ay1⇒ ABx
A2 y
ABx1
(3) A2 y1
(4)
A2 B2y BC ABx1 A2y1 (By adding equations (3) and (4)) y
—CONTINUED—
BC ABx1 A2y1 A2 B2
Section P.2
Linear Models and Rates of Change
89. —CONTINUED— Ay AC A Bx B ABy , BC A ABx point of intersection B 2
1
2
1
2
1
2
2
1
2
The distance between x1, y1 and this point gives us the distance between x1, y1 and the line Ax By C 0. Ay AC A Bx B ABy x BC A ABx B AC ABy A x By
BC A ABx A B B 2
d
1
2
1
2
2
1
2
1
2
1
1
2
2
2
2
1
2
2
2
1
1
2
2
2
2
1
2
1
2
2
1 2 2
1
y1
2
2
2
AC A ByB Ax BC A AxB By A B C Ax By A B
1
1
2
2
2
Ax1 By1 C A2 B2
91. For simplicity, let the vertices of the rhombus be 0, 0, a, 0, b, c, and a b, c, as shown in the figure. The slopes of the diagonals are then m1
y
(b, c)
(a + b , c )
c c . and m2 ab ba
x
(0, 0)
Since the sides of the Rhombus are equal, a2 b2 c2, and we have m1m2
c ab
c
c2
(a, 0)
c2
b a b2 a2 c2 1.
Therefore, the diagonals are perpendicular.
93. Consider the figure below in which the four points are collinear. Since the triangles are similar, the result immediately follows. y2 y1 y2 y1 x2 x1 x2 x1
95. True. a c a ax by c1 ⇒ y x 1 ⇒ m1 b b b b c bx ay c2 ⇒ y x 2 a a
y
m2 (x 2 , y2 )
(x *2 , y*2 )
(x1, y1 ) (x *1, y*1 )
x
1 m1
⇒ m2
b a
13
14
Chapter P
Preparation for Calculus
Section P.3
Functions and Their Graphs 3. (a) g0 3 02 3
1. (a) f 0 20 3 3 (b) f 3 23 3 9
(b) g3 3 3 3 3 0
(c) f b 2b 3
(c) g2 3 22 3 4 1
(d) f x 1 2x 1 3 2x 5
(d) gt 1 3 t 12 t2 2t 2
5. (a) f 0 cos20 cos 0 1 (c) f
2
4 cos2 4 cos 2 0
(b) f
3 cos23 cos23 12
7.
f x x f x x x3 x3 x3 3x2x 3xx2 x3 x3 3x2 3xx x2, x 0 x x x
9.
f x f 2 1x 1 1 x2 x2
1 x 1 1 x 1 2x 1 ,x2 x 2x 1 1 x 1 x 2x 11 x 1 x 11 x 1
11. hx x 3
13. f t sec
Domain: x 3 ≥ 0 ⇒ 3, Range: , 0
t 4
t 2k 1 ⇒ t 4k 2 4 2 Domain: all t 4k 2, k an integer Range: , 1, 1,
15. f x
1 x
Domain: , 0, 0, Range: , 0, 0,
17. f x
2x2x 1,2, xx ≤ 11 2
20. f x
2
x x54,, xx >≤ 55
2
(a) f 2 22 2 6
(a) f 3 3 4 1 1
(b) f 0 02 2 2
(b) f 0 0 4 2 (c) f 5 5 4 3
(c) f 1 12 2 3 (d) f
s2
2 2
s2
2 2
2s4
(d) f 10 10 52 25
10
8s2
(Note: s2 2 > 1 for all s)
Domain: 4,
Domain: ,
Range: 0,
Range: 2, 22. gx
4 x
1 24. f x 2 x3 2
y
Domain: ,
6
Domain: , 0, 0,
4
8 6
Range: ,
2
Range: , 0, 0,
y
4
x 2
4
6
x
−6 −4
26. f x x 4 x2 4
Domain: 2, 2 Range:
2, 22 2, 2.83
28. h 5 cos
y
(0, 2) 2, 0(
−4 −3 −2
y-intercept: 0, 2 x-intercept: 2, 0
x −1
1
2
3
6
5 4 3 2 1
Range: 5, 5 −2 π
4
4
y
Domain: ,
3
(−
2
2
2π
θ
−2 −3 −4
30. x2 4 y 0 ⇒ y x2 4 y is a function of x. Vertical lines intersect the graph at most once.
34. x2 y 4 ⇒ y 4 x2 y is a function of x since there is one value of y for each x.
−5
32. x2 y2 4 y ± 4 x2 y is not a function of x. Some vertical lines intersect the graph twice. x2 4 y is a function of x since there is one value of y for each x.
36. x2y x2 4y 0 ⇒ y
x2
294
Chapter P
Preparation for Calculus
38. p1x x3 x 1 has one zero. p2x x3 x has three zeros. Every cubic polynomial has at least one zero. Given px Ax3 Bx2 Cx D, we have p → as x → and p → as x → if A > 0. Furthermore, p → as x → and p → as x → if A < 0. Since the graph has no breaks, the graph must cross the x-axis at least one time.
2
P1 −3
−2
40. The function is f x cx. Since 1, 1 4 satisfies the equation, c 14. Thus, f x 14x.
3
P2
42. The function is hx c x . Since 1, 3 satisfies the equation, c 3. Thus, hx 3 x .
20 1 mimin during the first 40 2 4 minutes. The student is stationary for the following 62 2 minutes. Finally, the student travels 1 mimin 10 6 during the final 4 minutes.
44. The student travels
46. (a)
A 500 400 300 200 100 t 10
20
30
40
50
(b) A15 345 acresfarm 48. (a) gx f x 4
y 4
4
g6 f 2 1
Shift f left 2 units
3
3 2
2
g0 f 4 3 Shift f right 4 units
(b) gx f x 2
y
(6, 1)
1
x
−1
1
2
3
5
6
−7 −6 −5 −4 −3
7
(0, 1) x
−1
1
−2
−2
−3
(− 6, −3)
(0, −3)
−4
−4
(c) gx f x 4
(d) gx f x 1
y 6
Vertical shift upwards 4 units
y 2
(2, 5)
Vertical shift down 1 unit
5 4
1
(2, 0) x
−5 −4 −3 −2 −1
2
3
2
(− 4, 1)
−3
1
x
−5 −4 −3 −2 −1
1
2
(− 4, − 4)
3
(e) gx 2f x
1 (f) gx 2 f x
y
(2, 2) 2
g2 2 f 2 2 −5 −4 −3 −2 −1
−3
y 2
g2 12 f 2 12
1
(2, 12 )
1
x 1
2
3
g4 12 f 4 32
−5 −4 −3
(− 4, − 32 )
−4
(− 4, − 6 )
−5 −6
−2
g4 2f 4 6
−4
−5 −6
x −1 −2 −3 −4 −5 −6
50. (a) hx sinx 2 1 is a horizontal shift 2 units to the left, followed by a vertical shift 1 unit upwards. (b) hx sinx 1 is a horizontal shift 1 unit to the right followed by a reflection about the x-axis.
1
2
3
Section P.3 52. (a) f g1 f 0 0
Functions and Their Graphs
54. f x x2 1, gx cos x
(b) g f 1 g1 0
f gx f gx f cos x cos2 x 1
(c) g f 0 g0 1
Domain: ,
(d) f g4 f 15 15
g f x gx 2 1 cosx 2 1
(e) f gx f x2 1 x2 1
Domain: ,
(f) g f x g x x 1 x 1 x ≥ 0
No, f g g f.
2
56. f gx f x 2
1 x 2
Domain: 2,
g f x g
1x 1x 2 1 x 2x
You can find the domain of g f by determining the intervals where 1 2x and x are both positive, or both negative. + −2
+
+ − − +
−1 − 1 2
0
+
+ 1
+
x 2
Domain: , 12 , 0,
58. (a)
3 x 3 x f x 60. f x
25
Odd
100
0 0
(b) H1.6x 0.0021.6x2 0.0051.6x 0.029 0.00512x 2 0.008x 0.029 62. f x sin2x sinx sinx sin xsin x sin2x Even 64. (a) If f is even, then 4, 9 is on the graph. 66. f x a2nx2n a2n2x2n2 . . . a2x2 a0 a2nx2n a2n2x2n2 . . . a2x2 a0 f x Even 68. Let F x f xgx where f is even and g is odd. Then F x f xgx f x gx f xgx F x. Thus, F x is odd.
(b) If f is odd, then 4, 9 is on the graph.
295
296
Chapter P
Preparation for Calculus
70. (a) Let F x f x ± gx where f and g are even. Then, F x f x ± gx f x ± gx F x. Thus, F x is even. (b) Let F x f x ± gx where f and g are odd. Then, F x f x ± gx f x gx F x. Thus, F x is odd. (c) Let F x f x ± gx where f is odd and g is even. Then, F x f x ± gx f x ± gx. Thus, F x is neither odd nor even.
72. By equating slopes,
02 y2 03 x3 y2 y
L x2 y2
74. True
6 x3 6 2x 2 x3 x 3,
x x 2x 3 . 2
2
76. False; let f x x2. Then f 3x 3x2 9x2 and 3f x 3x2. Thus, 3f x f 3x.
Section P.4
Fitting Models to Data
2. Trigonometric function
4. No relationship
6. (a)
8. (a) s 9.7t 0.4
20
(b)
0
5
20 0 −1
No, the relationship does not appear to be linear. (b) Quiz scores are dependent on several variables such as study time, class attendance, etc. These variables may change from one quiz to the next. 10. (a) Linear model: H 0.3323t 612.9333 (b)
The model fits well. (c) If t 2.5, s 24.65 meterssecond. 12. (a) S 180.89x 2 205.79x 272 (b)
600
0
1300 0
45 −1
25000
14
0 0
The fit is very good. (c) When t 500, H 0.3323500 612.9333 446.78.
(c) When x 2, S 583.98 pounds.
Review Exercises for Chapter P 14. (a) t 0.00271s2 0.0529s 2.671 (b)
16. (a) T 2.9856 104 p3 0.0641 p2 5.2826p 143.1 (b)
21
100
20
350
0 150
0
110
(c) The curve levels off for s < 20.
(c) For T 300F, p 68.29 pounds per square inch.
(d) t 0.002s2 0.0346s 0.183
(d) The model is based on data up to 100 pounds per square inch.
21
100
0 0
The model is better for low speeds.
18. (a) Ht 84.4 4.28 sin
6t 3.86
(c)
100
One model is Ct 58 27 sin (b)
6t 4.1.
13
0 0
100
(d) The average in Honolulu is 84.4. The average in Chicago is 58. 0
(e) The period is 12 months (1 year).
13 0
(f) Chicago has greater variability 27 > 4.28.
20. Answers will vary.
Review Exercises for Chapter P 2. y x 1x 3 x 0 ⇒ y 0 10 3 3 ⇒ 0, 3
297
y-intercept
y 0 ⇒ 0 x 1x 3 ⇒ x 1, 3 ⇒ 1, 0, 3, 0 4. xy 4 x 0 and y 0 are both impossible. No intercepts.
x-intercepts 6. Symmetric with respect to y-axis since y x4 x2 3 y x4 x2 3.
298
Chapter P
Preparation for Calculus
8. 4x 2y 6 y 2x 3
2x 15y 25
y
2 y 15 x 53
Slope: 2
10 8
2 Slope: 15
y-intercept: 3
y-intercept:
y
6
5 3
4 2
1 −2
12. y x6 x
10. 0.02x 0.15y 0.25
y x
−1
2
−2
3
x 2
4
8
3
−1 −2
1
−3
x
−4
4
8
12
−1 −2
14. y x 4 4
3 x 6 16. y 8
18.
x 1 x2 7
y
Xmin = -40 Xmax = 40 Xscl = 10 Ymin = -40 Ymax = 40 Yscl = 10
x
−1 −1
2
3
4
5
6
−2 −3 −4
yx1 0 x2 x 6 No real solution No points of intersection The graphs of y x 1 and y x2 7 do not intersect.
−5 −6
20. y kx3 (a) 4 k13 ⇒ k 4 and y 4x3
(b) 1 k23 ⇒ k 18 and y 18 x3
(c) 0 k03 ⇒ any k will do!
(d) 1 k13 ⇒ k 1 ⇒ y x3
22.
24.
y 14
3 4 3 t 11
(7, 12)
12 10 8 6
44 9 3t
4 2 −2
36 3 1 3 t 3 8
1
2
3
4
5
6
(7, −1)
x
53 3t t
The line is vertical and has no slope.
53 3
Review Exercises for Chapter P 26. y 6 0x 2
28. m is undefined. Line is vertical. x5
y 6 Horizontal line
y
y 6
8
(−2, 6)
4
2 −4
−2
(5, 4)
2
4
−6
299
−4 x 2
−2
4
6
−2
x 2
4
6
8
−2 −4 −6
−4
2 y 3 x 1 3
30. (a)
32. (a) C 9.25t 13.50t 36,500 22.75t 36,500
3y 9 2x 2
(b) R 30t
2x 3y 11 0 (b) Slope of perpendicular line is 1.
(c)
30t 22.75t 36,500 7.25t 36,000
y 3 1x 1
t 5034.48 hours to break even.
yx2 0xy2 m
(c)
43 1 21
y 3 1x 1 yx2 0xy2 y3
(d)
y30 34. x2 y 0
36. x 9 y2
Function of x since there is one value for y for each x. y
Not a function of x since there are two values of y for some x. y
6 4
5 4
2
3
1
2
−12 −9 −6 −3 −1
1 −3
−2
−1
x 3
6
12
−2
x 1
2
3 −4
38. (a) f 4 42 2 18
f 1 1 2 1
(because 4 < 0)
40. f x 1 x2 and gx 2x 1
(b) f 0 0 2 2
(a) f x gx 1 x2 2x 1 x2 2x
(c)
(b) f xgx 1 x22x 1 2x3 x2 2x 1 (c) g f x g1 x2 21 x2 1 3 2x2
Review Exercises for Chapter P 15. (a) y 1.81x3 14.58x2 16.39x 10 (b)
17. (a) Yes, y is a function of t. At each time t, there is one and only one displacement y.
300
(b) The amplitude is approximately
2.35 1.652 0.35. 0
The period is approximately
7 0
20.375 0.125 0.5.
(c) If x 4.5, y 214 horsepower.
(c) One model is y 0.35 sin4 t 2. (d)
4
0.9
0 0
19. Answers will vary.
Review Exercises for Chapter P 1. y 2x 3 x 0 ⇒ y 20 3 3 ⇒ 0, 3 3 3 y 0 ⇒ 0 2x 3 ⇒ x 2 ⇒ 2 , 0
3. y
y-intercept x-intercept
x1 x2
5. Symmetric with respect to y-axis since
01 1 1 ⇒ 0, x0⇒y 02 2 2
y0⇒0
x2y x2 4y 0 y-intercept
x1 ⇒ x 1 ⇒ 1, 0 x2
7. y 12 x 32
x2y x2 4y 0.
x-intercept
11. y 7 6x x2
1 5 9. 3 x 6 y 1
25 x y 65
y
y
y 25 x 65
3 2
2 5
Slope:
1
y-intercept:
6 5
5
x 1
2
3
y x
1
10 3
2
2
x 3
2
1
1 1
5
5
19
20
Chapter P
Preparation for Calculus 17. 3x 4y 8
15. y 4x2 25
13. y 5 x
4x 4y 20
Domain: , 5
Xmin = -5 Xmax = 5 Xscl = 1 Ymin = -30 Ymax = 10 Yscl = 5
y 5 4 3
28
7x
x 4 y 1 Point: 4, 1
2 1
x 1
2
3
4
5
19. You need factors x 2 and x 2. Multiply by x to obtain origin symmetry y xx 2x 2. x3 4x.
21.
23.
y
15 1t 1 0 1 2
5
1t
4 5 2
( 5, )
3 2
t
1
( 32 , 1)
4 3
7 3
x 1
2
Slope
3
4
5
52 1 32 3 5 32 72 7
y 5 32x 0
25.
y 0 23x 3
27.
y 32x 5 2y 3x 10 0
y 23x 2 3y 2x 6 0 y
y 4
4
2
2 −4
−2
(−3, 0) x
−2 −4
−8
2
(0, −5)
4
6
8
−8
−6
−4
x −2 −4 −6 −8
2
4
Review Exercises for Chapter P
y4
29. (a)
7 x 2 16
5 (b) Slope of line is . 3
16y 64 7x 14
5 y 4 x 2 3
0 7x 16y 78
3y 12 5x 10
40 2 2 0
m
(c)
y 2x
0 5x 3y 22 x 2
(d)
2x y 0
21
x20
31. The slope is 850. V 850t 12,500. V3 8503 12,500 $9950 33. x y2 0
35. y x2 2x
y ± x
Function of x since there is one value of y for each x.
Not a function of x since there are two values of y for some x.
y 4 3
y 3 2
x −1
x
−2 −1
1 1
2
3
4
5
3
4
−2
6
−2 −3
37. f x
1 x
39. (a) Domain: 36 x2 ≥ 0 ⇒ 6 ≤ x ≤ 6 Range: 0, 6
(a) f 0 does not exist. 1 1 1 x 1 1 1 x f 1 x f 1 (b) x x 1 xx 1 , x 1, 0 1 x 41. (a) f x x3 c, c 2, 0, 2 y
c
Range: all y 0
or , 0, 0,
,
(c) Domain: all x or Range: all y or
,
(b) f x x c3, c 2, 0, 2 y
c
c
or , 5, 5,
(b) Domain: all x 5
0
3
1
c
0
2
1
2 x
3
2
2
3
x 2
2 2
c
2
—CONTINUED—
or 6, 6
3
c
2
3
22
Chapter P
Preparation for Calculus
41. —CONTINUED— (c) f x x 23 c, c 2, 0, 2
(d) f x cx3, c 2, 0, 2 y
y
2
c
3
2
c
2
2
c
1
0
1
c x 2
3
4
2
1
1
0 2
x 3
1 2
c
c
2
43. (a) Odd powers: f x x, gx x3, hx x5
Even powers: f x x2, gx x4, hx x6
g
2
2
g
4
h
h
−3
f
3
f −3 −2
3 0
The graphs of f, g, and h all rise to the left and to the right. As the degree increases, the graph rises more steeply. All three graphs pass through the points 0, 0, 1, 1, and 1, 1.
The graphs of f, g, and h all rise to the right and fall to the left. As the degree increases, the graph rises and falls more steeply. All three graphs pass through the points 0, 0, 1, 1, and 1, 1. (b) y x7 will look like hx x5, but rise and fall even more steeply. y x8 will look like hx x6, but rise even more steeply. 45. (a)
(b) Domain: 0 < x < 12
y
x
40
x
y
2x 2y 24
0
12 0
y 12 x A xy x12 x 12x x2
47. (a) 3 (cubic), negative leading coefficient (b) 4 (quartic), positive leading coefficient (c) 2 (quadratic), negative leading coefficient
(c) Maximum area is A 36. In general, the maximum area is attained when the rectangle is a square. In this case, x 6. 49. (a) Yes, y is a function of t. At each time t, there is one and only one displacement y. (b) The amplitude is approximately
0.25 0.252 0.25.
(d) 5, positive leading coefficient
The period is approximately 1.1. (c) One model is y (d)
1 2 1 cos t cos5.7t 4 1.1 4
0.5
0
−0.5
2.2
Problem Solving for Chapter P
23
Problem Solving for Chapter P x2 6x y2 8y 0
1. (a)
4 (b) Slope of line from 0, 0 to 3, 4 is Slope of tangent line 3 3 is Hence, 4
x2 6x 9 y2 8y 16 9 16 x 32 y 42 25 Center: 3, 4
3 3 y 0 x 0 ⇒ y x 4 4
Radius: 5
(c) Slope of line from 6, 0 to 3, 4 is
40 4 . 36 3
3 Slope of tangent line is . Hence, 4
3 3 9 (d) x x 4 4 2 3 9 x 2 2
3 9 3 y 0 x 6 ⇒ y x 4 4 2
Tangent line
x3 Intersection:
3. Hx
10
x ≥ 0 x < 0
Tangent line
3, 49
y 4 3 2 1 x
−4 −3 −2 −1 −1
1
2
3
4
−2 −3 −4
(a) Hx 2
(b) Hx 2
y
y
4
4
3
3
2
2
1
1 x
−4 −3 −2 −1 −1
1
2
3
x
−4 −3 −2 −1 −1
4
1
2
3
4
1
2
3
4
1
2
3
4
−2 −3
−3
−4
−4
(c) Hx
(d) Hx
y
y
4
4
3
3
2
2
1 x
−4 −3 −2 −1 −1
1
2
3
−2
−2
−3
−3
−4
−4
1 (e) 2Hx
(f ) Hx 2 2
y
x
−4 −3 −2 −1 −1
4
y
4
4
3
3
2 1 −4 −3 −2 −1 −1
1 x 1
2
3
4
−4 −3 −2 −1 −1
−2
−2
−3
−3
−4
−4
x
24
Chapter P
Preparation for Calculus
5. (a) x 2y 100 ⇒ y Ax xy x
100 x 2
7. The length of the trip in the water is 22 x2, and the length of the trip over land is 1 3 x2. Hence, the total time is
1002 x x2 50x 2
T
Domain: 0 < x < 100 (b)
4 x2
2
1 3 x2
1600
0
110 0
Maximum of 1250 m 2 at x 50 m, y 25 m. 1 (c) Ax x2 100x 2 1 x2 100x 2500 1250 2 1 x 502 1250 2 A50 1250 m 2 is the maximum. x 50 m, y 25 m. 9. (a) Slope
94 5. Slope of tangent line is less than 5. 32
(b) Slope
41 3. Slope of tangent line is greater than 3. 21
(c) Slope
4.41 4 4.1. Slope of tangent line is less than 4.1. 2.1 2
(d) Slope
f 2 h f 2 2 h 2
2 h2 4 h
4h h2 h
4 h, h 0 (e) Letting h get closer and closer to 0, the slope approaches 4. Hence, the slope at 2, 4 is 4. 11. (a) At x 1 and x 3 the sounds are equal. (b)
I x2 y2
3
2I x 32 y2
−6
3
x 32 y2 4x2 y2 3x2 3y2 6x 9 x2 2x y2 3
x 12 y2 4 Circle of radius 2 centered at 1, 0
−3
4
hours.
Problem Solving for Chapter P d1d2 1
13.
y
x 12 y2 x 12 y2 1 x 12x 12 y2x 12 x 12 y4 1 x2 12 y22x2 2 y4 1 x4 2x2 1 2x2y2 2y2 y4 1
x4 2x2y2 y4 2x2 2y2 0 x2 y22 2x2 y2 Let y 0. Then x4 2x2 ⇒ x 0
or
x2 2.
Thus, 0, 0, 2, 0 and 2, 0 are on the curve.
2
(− 2 , 0)
1
( 2 , 0) x
−2
2 −1 −2
(0, 0)
25
Review Exercises for Chapter P 14. (a) t 0.00271s2 0.0529s 2.671 (b)
16. (a) T 2.9856 104 p3 0.0641 p2 5.2826p 143.1 (b)
21
100
20
350
0 150
0
110
(c) The curve levels off for s < 20.
(c) For T 300F, p 68.29 pounds per square inch.
(d) t 0.002s2 0.0346s 0.183
(d) The model is based on data up to 100 pounds per square inch.
21
100
0 0
The model is better for low speeds.
18. (a) Ht 84.4 4.28 sin
6t 3.86
(c)
100
One model is Ct 58 27 sin (b)
6t 4.1.
13
0 0
100
(d) The average in Honolulu is 84.4. The average in Chicago is 58. 0
(e) The period is 12 months (1 year).
13 0
(f) Chicago has greater variability 27 > 4.28.
20. Answers will vary.
Review Exercises for Chapter P 2. y x 1x 3 x 0 ⇒ y 0 10 3 3 ⇒ 0, 3
297
y-intercept
y 0 ⇒ 0 x 1x 3 ⇒ x 1, 3 ⇒ 1, 0, 3, 0 4. xy 4 x 0 and y 0 are both impossible. No intercepts.
x-intercepts 6. Symmetric with respect to y-axis since y x4 x2 3 y x4 x2 3.
298
Chapter P
Preparation for Calculus
8. 4x 2y 6 y 2x 3
2x 15y 25
y
2 y 15 x 53
Slope: 2
10 8
2 Slope: 15
y-intercept: 3
y-intercept:
y
6
5 3
4 2
1 −2
12. y x6 x
10. 0.02x 0.15y 0.25
y x
−1
2
−2
3
x 2
4
8
3
−1 −2
1
−3
x
−4
4
8
12
−1 −2
14. y x 4 4
3 x 6 16. y 8
18.
x 1 x2 7
y
Xmin = -40 Xmax = 40 Xscl = 10 Ymin = -40 Ymax = 40 Yscl = 10
x
−1 −1
2
3
4
5
6
−2 −3 −4
yx1 0 x2 x 6 No real solution No points of intersection The graphs of y x 1 and y x2 7 do not intersect.
−5 −6
20. y kx3 (a) 4 k13 ⇒ k 4 and y 4x3
(b) 1 k23 ⇒ k 18 and y 18 x3
(c) 0 k03 ⇒ any k will do!
(d) 1 k13 ⇒ k 1 ⇒ y x3
22.
24.
y 14
3 4 3 t 11
(7, 12)
12 10 8 6
44 9 3t
4 2 −2
36 3 1 3 t 3 8
1
2
3
4
5
6
(7, −1)
x
53 3t t
The line is vertical and has no slope.
53 3
Review Exercises for Chapter P 26. y 6 0x 2
28. m is undefined. Line is vertical. x5
y 6 Horizontal line
y
y 6
8
(−2, 6)
4
2 −4
−2
(5, 4)
2
4
−6
299
−4 x 2
−2
4
6
−2
x 2
4
6
8
−2 −4 −6
−4
2 y 3 x 1 3
30. (a)
32. (a) C 9.25t 13.50t 36,500 22.75t 36,500
3y 9 2x 2
(b) R 30t
2x 3y 11 0 (b) Slope of perpendicular line is 1.
(c)
30t 22.75t 36,500 7.25t 36,000
y 3 1x 1
t 5034.48 hours to break even.
yx2 0xy2 m
(c)
43 1 21
y 3 1x 1 yx2 0xy2 y3
(d)
y30 34. x2 y 0
36. x 9 y2
Function of x since there is one value for y for each x. y
Not a function of x since there are two values of y for some x. y
6 4
5 4
2
3
1
2
−12 −9 −6 −3 −1
1 −3
−2
−1
x 3
6
12
−2
x 1
2
3 −4
38. (a) f 4 42 2 18
f 1 1 2 1
(because 4 < 0)
40. f x 1 x2 and gx 2x 1
(b) f 0 0 2 2
(a) f x gx 1 x2 2x 1 x2 2x
(c)
(b) f xgx 1 x22x 1 2x3 x2 2x 1 (c) g f x g1 x2 21 x2 1 3 2x2
300
Chapter P
Preparation for Calculus
42. f x x3 3x2
44. (a) f x x2x 62
6
−6
100
(0, 0) 6 −4
(2, − 4)
10
−6
−25
(a) The graph of g is obtained from f by a vertical shift down 1 unit, followed by a reflection in the x-axis:
(b) gx x3x 62 300
gx f x 1 x3 3x2 1 −2
(b) The graph of g is obtained from f by a vertical shift upwards of 1 and a horizontal shift of 2 to the right. gx f x 2 1
10 −100
(c) hx x3x 63
x 2 3x 2 1 3
2
200 −4
10
− 800
46. For company (a) the profit rose rapidly for the first year, and then leveled off. For the second company (b), the profit dropped, and then rose again later.
48. (a) y 1.204x 64.2667 (b)
70
0
33 0
(c) The data point 27, 44 is probably an error. Without this point, the new model is y 1.4344x 66.4387.
Problem Solving for Chapter P 2. Let y mx 1 be a tangent line to the circle from the point 0, 1. Then x2 y 12 1 x2 mx 1 12 1
m2 1x2 4mx 3 0 Setting the discriminant b2 4ac equal to zero, 16m2 4m2 13 0 16m2 12m2 12 4m2 12 m ± 3 Tangent lines: y 3x 1 and y 3x 1.
Problem Solving for Chapter P 4. (a) f x 1
(b) f x 1
y
y
4
−3
4
x
−1
1
x
−4
3
4
−2
−2
−4
−4
(c) 2f x
(d) f x
y
y
4
4 2
−4
x
−2
2
−4
4
−2
−2
−4
−4
(e) f x
(f) f x
y 4
4
2
4
y
2 x
−2
2
−4
4
x
−2
−2
−2
−4
−4
(g) f x
2
4
2
−4
x
−2
y 4 2
−4
x
−2
2
4
−2 −4
6. (a) 4y 3x 300 ⇒ y
300 3x 4
3x2 300x 300 3x Ax x2y x 2 2
y
(b) 4000 3500 3000 2500 2000
Domain: 0 < x < 100
1500 1000 500 x 25
50
75
100
Maximum of 3750 ft 2 at x 50 ft, y 37.5 ft. 3 (c) Ax x2 100x 2 3 x2 100x 2500 3750 2 3 x 502 3750 2 A50 3750 square feet is the maximum area, where x 50 ft and y 37.5 ft.
301
302
Chapter P
Preparation for Calculus
8. Let d be the distance from the starting point to the beach. Average velocity
distance time
2d d d 120 60
2 1 1 120 60
80 km hr y
10. 4 3 2
(4, 2)
1 x 1
2
3
4
5
−1
(a) Slope
1 32 1 . Slope of tangent line is greater than . 94 5 5
(b) Slope
21 1 1 . Slope of tangent line is less than . 41 3 3
(c) Slope
10 2.1 2 10 . Slope of tangent line is greater than . 41 4.1 4 41
(d) Slope
f 4 h f 4 4 h 4
(e)
4 h 2
4 h 2
h
h
4 h 2
h
4 h 2 4 h 2
4 h 4 h4 h 2 1 4 h 2
,h0
1 1 As h gets closer to 0, the slope gets closer to . The slope is at the point 4, 2. 4 4
Problem Solving for Chapter P I
12. (a)
kI
x2 y2
x 42 y2
x 42 y2 k2x2 y2 k2 1x2 k2 1y2 8x 16 If k 1, then x 2 is a vertical line. So, assume k2 1 0. Then x2 y2
8x 16 k2 1 k2 1
x k
4 1
2
x k
4 1
2
2
2
y2
16 16 k2 1 k2 12
y2
k 4k 1 , Circle
1 2
(b) If k 3, x
2
2
2
y2
32
2
(c) For large k, the center of the circle is near 0, 0, and the radius becomes smaller. y 2 1 x −2
(− 12 , 0)
1
2
−2
14. f x y
1 1x
(a) Domain: all x 1 Range: all y 0 (b) f f x f
1 1 x
1
1 1 1x
1 1x x1 1x1 x x 1x
Domain: all x 0, 1 (c) f f f x f
x x 1
1 1 x x1 1 1 x x
Domain: all x 0, 1 (d) The graph is not a line. It has holes at 0, 0 and 1, 1. y 2 1 x −2
1
−2
2
303
A Preview of Calculus . . . . . . . . . . . . . . . . . . . 305
Section 1.2
Finding Limits Graphically and Numerically . . . . . . . 305
Section 1.3
Evaluating Limits Analytically . . . . . . . . . . . . . . . 309
Section 1.4
Continuity and One-Sided Limits
Section 1.5
Infinite Limits
Review Exercises
. . . . . . . . . . . . . 315
. . . . . . . . . . . . . . . . . . . . . . . 320
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327
C H A P T E R 1 Limits and Their Properties Section 1.1
A Preview of Calculus
Solutions to Even-Numbered Exercises 4. Precalculus: rate of change slope 0.08
2. Calculus: velocity is not constant Distance 20 ftsec15 seconds 300 feet 6. Precalculus: Area 2
8. Precalculus: Volume 326 54
2
2 5 5 5 10.417 2 3 4 5 5 5 5 5 5 5 1 9.145 Area 5 2 1.5 2 2.5 3 3.5 4 4.5
10. (a) Area 5
(b) You could improve the approximation by using more rectangles.
Section 1.2 2.
x
1.9
1.99
1.999
2.001
2.01
2.1
f x
0.2564
0.2506
0.2501
0.2499
0.2494
0.2439
lim
x→2
4.
x2 0.25 x2 4 3.1
3.01
3.001
2.999
2.99
2.9
f x
0.2485
0.2498
0.2500
0.2500
0.2502
0.2516
lim
1 x 2
x3
Actual limit is 14 .
0.25
x
3.9
3.99
3.999
4.001
4.01
4.1
f x
0.0408
0.0401
0.0400
0.0400
0.0399
0.0392
lim
x→4
8.
Actual limit is 14 .
x
x→3
6.
Finding Limits Graphically and Numerically
xx 1 45 0.04 x4 0.1
x f x lim
x→0
0.0500
0.01 0.0050
cos x 1 0.0000 x
Actual limit is 251 .
0.001 0.0005
0.001
0.01
0.0005
0.0050
0.1 0.0500
(Actual limit is 0.) (Make sure you use radian mode.)
305
306
Chapter 1
Limits and Their Properties
10. lim x2 2 3
12. lim f x lim x2 2 3
x→1
x→1
1 does not exist since the x3 function increases and decreases without bound as x approaches 3.
x→1
16. lim sec x 1
14. lim
x→3
18. lim sinx 0
x→0
x→1
20. Ct 0.35 0.12 t 1 (a)
1
0
5 0
(b)
t Ct
3
3.3
3.4
3.5
3.6
3.7
4
0.59
0.71
0.71
0.71
0.71
0.71
0.71
lim Ct 0.71
t→3.5
(c)
3
2.5
2.9
3
3.1
3.5
4
0.47
0.59
0.59
0.59
0.71
0.71
0.71
t Ct
lim Ct does not exist. The values of C jump from 0.59 to 0.71 at t 3.
t→3.5
22. You need to find such that 0 < x 2 < implies f x 3 x2 1 3 x2 4 < 0.2. That is,
0.2 4 0.2 3.8 3.8 3.8 2
< x2 4 < x2 < x2 < x < x2
< < < < <
0.2 4 0.2 4.2 4.2 4.2 2
So take 4.2 2 0.0494.
Then 0 < x 2 < implies 4.2 2 < x 2 < 4.2 2 3.8 2 < x 2 < 4.2 2.
Using the first series of equivalent inequalities, you obtain
f x 3 x2 4 < 0.2.
24. lim 4 x→4
x 2 2
4
x 2 < 0.01 2 2
x < 0.01 2
1 x 4 < 0.01 2
Hence, if 0 < x 4 < 0.02, you have 0 < x 4 < 0.02
1 x 4 < 0.01 2 2
4
x < 0.01 2
x 2 < 0.01 2
f x L < 0.01
Section 1.2 26. lim x2 4 29
28. lim 2x 5 1
x→5 2
x→3
x 4 29 < 0.01
Finding Limits Graphically and Numerically
Given > 0:
x2 25 < 0.01
2x 5 1 <
2x 6 < 2 x 3 <
x 5x 5 < 0.01 0.01
x 5 < x 5
x 3
If we assume 4 < x < 6, then 0.0111 0.0009.
0.01 , you have 11
x5 <
0.01 1 < 0.01 11 x5
Hence, if 0 < x 5 <
x 5
x 5 < 0.01
x2 25 < 0.01
x2 4 29 < 0.01
f x L < 0.01
Hence, if 0 < x 3 < , you have 2
9
2 3
x 1
2 3x
29 3
2 3
2x 6 <
2x 5 1 <
f x L < 32. lim 1 1 x→2
Given > 0:
0, you have
<
1 1 <
f x L <
< 32
Hence, let 32.
x 3 < 2
x→1
2 3x
2
Hence, let 2.
2 2 29 30. lim 3 x 9 3 1 9 3
Given > 0:
<
3 Hence, if 0 < x 1 < 2, you have
2 3x
x 1 < 32
2 3x
9
2 3
29 3
< <
f x L <
34. lim x 4 2 x→4
Given > 0:
x 2
x 2 <
x 4 <
x→3
x 2 <
x 2
x 3 0
x 3
0 < x 4 < 3 ⇒ x 4 < x 2
Given > 0:
x 2
Assuming 1 < x < 9, you can choose 3. Then,
36. lim x 3 0
⇒ x 2 < .
< <
Hence, let .
Hence for 0 < x 3 < , you have
x 3 <
x 3 0 <
f x L <
307
308
Chapter 1
Limits and Their Properties
40. f x
38. lim x2 3x 0 x→3
Given > 0:
x2 3x 0
xx 3
x2
x3 4x 3 −3
1 2
lim f x
<
4
x→3
5
−4
<
The domain is all x 1, 3. The graphing utility does not show the hole at 3, 12 .
x 3 < x If we assume 4 < x < 2, then 4.
Hence for 0 < x 3 < , you have 4
1
1
x 3 < 4 < x
xx 3 <
x 3x 0 <
f x L < 2
42. f x
x3 x2 9
lim f x
x→3
1 6
44. (a) No. The fact that f 2 4 has no bearing on the existence of the limit of f x as x approaches 2.
3
−9
(b) No. The fact that lim f x 4 has no bearing on the x→2 value of f at 2.
3
−3
The domain is all x ± 3. The graphing utility does not show the hole at 3, 16 . 46. Let px be the atmospheric pressure in a plane at altitude x (in feet).
48.
0.002
(1.999, 0.001) (2.001, 0.001)
lim px 14.7 lbin2
x→0
1.998 0
2.002
Using the zoom and trace feature, 0.001. That is, for
0 < x 2 < 0.001,
50. True
x2 4 4 < 0.001. x2
52. False; let f x
x10, 4x, 2
x4 x4
.
lim f x lim x2 4x 0 and f 4 10 0
x→4
x2 x 12 7 x→4 x4
54. lim
n
4 0.1 n
f 4 0.1 n
n
4 0.1 n
x→4
f 4 0.1 n
1
4.1
7.1
1
3.9
6.9
2
4.01
7.01
2
3.99
6.99
3
4.001
7.001
3
3.999
6.999
4
4.0001
7.0001
4
3.9999
6.9999
Section 1.3
56. f x mx b, m 0. Let > 0 be given. Take
If 0 < x c <
. m
xc
, then m
That is, 12L < gx L < 12L 1 2L
gx
<
3
< 2L
Hence for x in the interval c , c , x c, 1 gx > 2L > 0.
x→c
Evaluating Limits Analytically (a) lim gx 2.4
10
x→4
(b) lim gx 4 x→0
0
4.
(a) lim f t 0
10
t→4
(b) lim f t 5
−5
t→1
10
10
−5
− 10
gx
12 x 3 x9
8. lim 3x 2 33 2 7 x→3
x→2
10. lim x2 1 12 1 0 x→1
14. lim
x→3
18. lim
x→3
2 2 2 x 2 3 2
x 1
x4
3 1
34
f t t t 4
6. lim x3 23 8
2
22. lim 2x 13 20 13 1 x→0
12. lim 3x3 2x2 4 313 212 4 5 x→1
16. lim
x→3
2x 3 23 3 3 x5 35 8
3 3 x 4 442 20. lim x→4
24. (a) lim f x 3 7 4 x→3
(b) lim gx 42 16 x→4
(c) lim g f x g4 16 x→3
26. (a) lim f x 242 34 1 21 x→4
3 21 6 3 (b) lim gx
28. lim tan x tan 0 x→
x→21
(c) lim g f x g21 3 x→4
30. lim sin x→1
34.
1 such that 0 < x 0 < implies gx L < 2L.
which shows that lim mx b mc b.
2.
x sin 1 2 2
lim cos x cos
x→53
5 1 3 2
309
1 58. lim gx L, L > 0. Let 2L. There exists > 0
mx c < mx mc < mx b mc b <
Section 1.3
Evaluating Limits Analytically
32. lim cos 3x cos 3 1 x→
36. lim sec x→7
6x sec 76 23
3
310
Chapter 1
Limits and Their Properties
38. (a) lim 4f x 4 lim f x 4 x→c
x→c
32 6
3 3 3 lim f x 27 3 f x 40. (a) lim x→c
(b) lim f x gx lim f x lim gx x→c
x→c
x→c
(c) lim f xgx lim f x lim gx x→c
x→c
x→c
x→c
lim f x x→c f x 27 3 (b) lim x→c 18 lim 18 18 2
3 1 2 2 2
x→c
(c) lim f x lim f x2 272 729
32 12 43
2
x→c
lim f x 32 f x x→c 3 x→c gx lim gx 12
x→c
(d) lim f x23 lim f x23 2723 9
(d) lim
x→c
x→c
x→c
42. f x x 3 and hx
x2 3x agree except at x 0. x
44. gx
1 x and f x 2 agree except at x 0. x1 x x
(a) lim hx lim f x 5
(a) lim f x does not exist.
(b) lim hx lim f x 3
(b) lim f x 1
x→2
x→1
x→2
x→0
x→0
x→0
2x2 x 3 and gx 2x 3 agree except at x1 x 1.
x3 1 and gx x2 x 1 agree except at x1 x 1.
46. f x
48. f x
lim f x lim gx 5
x→1
lim f x lim gx 3
x→1
x→1
x→1
7
4
−8
4
−4
50. lim
x→2
2x x 2 lim x2 4 x→2 x 2x 2 lim
x→2
54. lim
x
lim
x→0
x 1 2
x→3
x3
x→4
x2 5x 4 x 4x 1 lim x2 2x 8 x→4 x 4x 2
lim
x→3
lim
x→4
2 x 2
x
x→0
lim
56. lim
52. lim
1 1 x2 4
2 x 2
x→0
4 −1
−8
2x2
2 x 2 2 x 2
2 x 2 x
x 1 2
x3
x 1 3 1 x 2 6 2
lim
x→0
1 2 x 2
x 1 2 x 1 2
lim
x→3
2 1 4 22
x3 1 1 lim x 3x 1 2 x→3 x 1 2 4
1 1 4 x 4 x4 4 4x 4 1 1 lim lim 58. lim x→0 x→0 x→0 4x 4 x x 16
60. lim
x→0
x x2 x2 x2 2x x x2 x2
x2x x lim lim lim 2x x 2x
x→0
x→0
x→0
x
x
x
Section 1.3
62. lim
x→0
Evaluating Limits Analytically
311
x x3 x3 x3 3x2 x 3x x2 x3 x3 lim
x→0
x
x lim
x→0
64. f x
x3x2 3x x x2 lim 3x2 3x x x2 3x2
x→0
x
4 x x 16
1
0
x
15.9
15.99
15.999
16
16.001
16.01
16.1
f x
.1252
.125
.125
?
.125
.125
.1248
20
−1
It appears that the limit is 0.125.
4 x 4 x lim x→16 x 16 x→16 x 4 x 4
Analytically, lim
lim
x→16
1 x 4
1 . 8
x5 32 80 x→2 x 2
100
66. lim
x f x
1.9
1.99
72.39
79.20
1.999 79.92
1.9999
2.0
79.99
?
2.0001 80.01
2.001 80.08
2.01
2.1
80.80
88.41
x5 32 x 2x4 2x3 4x2 8x 16 lim x→2 x 2 x→2 x2
Analytically, lim
lim x4 2x3 4x2 8x 16 80. x→2
(Hint: Use long division to factor x5 32.)
68. lim
x→0
31 cos x 1 cos x lim 3 x→0 x x
30 0
sin x tan2 x sin2 x lim lim 2 x→0 x→0 x cos x x→0 x x
72. lim
cos2 x sin x
10 0
76. lim
x→ 4
1 tan x cos x sin x lim sin x cos x x→4 sin x cos x cos2 x sin x cos x lim x→ 4 cos xsin x cos x 1 lim x→ 4 cos x lim sec x x→ 4
2
78. lim
x→0
sin 2x sin 2x lim 2 sin 3x x→0 2x
13 sin3x3x 21 13 1 32
70. lim
→0
cos tan sin lim 1 →0
74. lim sec 1 →
−4
3 −25
312
Chapter 1
Limits and Their Properties
80. f h 1 cos 2h 0.1
h f h
4
0.01
1.98
0.001
1.9998
2
0
0.001
0.01
0.1
?
2
1.9998
1.98
−5
5
−4
Analytically, lim 1 cos 2h 1 cos0 1 1 2.
The limit appear to equal 2.
h→0
82. f x
sin x 3 x
2
−3
0.1
x f x
0.01
0.215
0.0464
0.001
0
0.001
0.01
0.1
0.01
?
0.01
0.0464
0.215
−2
The limit appear to equal 0.
sin x 3 2 sin x lim x 01 0. 3 x→0 x→0 x x
Analytically, lim
84. lim
h→0
x h x x h x f x h f x lim lim h→0 h→0 h h h
lim
h→0
3
x h x x h x
xhx 1 1 lim h→0 h x h x 2x x h x
f x h f x x2 2xh h2 4x 4h x2 4x x h2 4x h x2 4x lim lim h→0 h→0 h→0 h h h
86. lim
lim
h→0
h2x h 4 lim 2x h 4 2x 4 h→0 h
88. lim b x a ≤ lim f x ≤ lim b x a x→a
x→a
x→a
90. f x x sin x 6
b ≤ lim f x ≤ b x→a
Therefore, lim f x b. x→a
− 2
2 −2
lim x sin x 0
x→0
92. f x x cos x
94. hx x cos
6
− 2
0.5
2
− 0.5
0.5
−6
− 0.5
lim x cos x 0
x→0
1 x
lim x cos
x→0
1 0 x
Section 1.3 x2 1 and gx x 1 agree at all points x1 except x 1.
96. f x
Evaluating Limits Analytically
98. If a function f is squeezed between two functions h and g, hx ≤ f x ≤ gx, and h and g have the same limit L as x → c, then lim f x exists and equals L. x→c
100. f x x, gx sin2 x, hx
sin2 x x When you are “close to” 0 the magnitude of g is “smaller” than the magnitude of f and the magnitude of g is approaching zero “faster” than the magnitude of f. Thus, g f 0 when x is “close to” 0
2
g −3
3
h f −2
102. st 16t2 1000 0 when t s lim
t→5102
5 210 st
5 10 seconds 1000 16 2
510 2
t
104. 4.9t2 150 0 when t
lim
t→5102
0 16t2 1000 510 t 2
16 t2 lim
t→5102
lim
t→5102
125 2
510 t 2
16 t
t 5 210
t 5 10
2
16 t lim
t→5102
510 2
510 8010 ftsec 253 ftsec 2
1500
5.53 seconds. 150 4.9 49
The velocity at time t a is sa st 4.9a2 150 4.9t2 150 4.9a ta t lim lim t→a t→a t→a at at at
lim
lim 4.9a t 2a4.9 9.8a msec. t→a
Hence, if a 150049, the velocity is 9.8150049 54.2 msec. 106. Suppose, on the contrary, that lim gx exists. Then, since lim f x exists, so would lim f x gx, which is a x→c
x→c
contradiction. Hence, lim gx does not exist.
x→c
x→c
108. Given f x x n, n is a positive integer, then lim x n lim x x n1 lim xlim x n1
x→c
x→c
x→c
x→c
c lim x x n2 clim xlim x n2 x→c
x→c
cclim x→c
x→c
. . . c n.
x x n3
110. Given lim f x 0: x→c
For every > 0, there exists > 0 such that f x 0 < whenever 0 < x c < .
Now f x 0 f x
313
f x 0 < for x c < . Therefore, lim f x 0. x→c
314
Chapter 1
Limits and Their Properties
f x ≤ f x ≤ f x lim f x ≤ lim f x ≤ lim f x x→c x→c x→c
112. (a) If lim f x 0, then lim f x 0. x→c
x→c
0 ≤ lim f x ≤ 0 x→c
Therefore, lim f x 0. x→c
(b) Given lim f x L: x→c
For every > 0, there exists > 0 such that f x L < whenever 0 < x c < . Since f x L ≤ f x L < for x c < , then lim f x L .
x→c
116. False. Let f x
114. True. lim x3 03 0 x→0
3x xx 11
,c1
Then lim f x 1 but f 1 1. x→1
1 118. False. Let f x 2 x2 and gx x2. Then f x < gx for all x 0. But lim f x lim gx 0. x→0
120. lim
x→0
1 cos x 1 cos x lim x→0 x x
x→0
1 cos x
1 cos x
1 cos2 x sin2 x lim x→0 x1 cos x x →0 x1 cos x
lim lim
x→0
sin x x
lim
x→0
sin x
1 cos x
sin x x
lim 1 sincosx x x→0
10 0
122. f x
sec x 1 x2
(a) The domain of f is all x 0, 2 n. (b)
x→0
2
(d) − 3 2
3 2
1 2
(c) lim f x
sec x 1 sec x 1 x2 x2
−2
The domain is not obvious. The hole at x 0 is not apparent.
Hence, lim
x→0
sec x 1
1 sin2 x 1 tan2 x x2sec x 1 cos2 x x2 sec x 1
sec x 1 1 sin2 x 1 lim 2 x→0 x cos2 x x2 sec x 1
11
124. The calculator was set in degree mode, instead of radian mode.
sec2 x 1
sec x 1 x2sec x 1
12 21.
Section 1.4
Section 1.4 2. (a) (b)
Continuity and One-Sided Limits
315
Continuity and One-Sided Limits
lim f x 2
4. (a)
lim f x 2
(b)
x→2
x→2
lim f x 2
6. (a)
lim f x 2
(b)
x→2
x→2
lim f x 0
x→1
lim f x 2
x→1
(c) lim f x 2
(c) lim f x 2
(c) lim f x does not exist.
The function is continuous at x 2.
The function is NOT continuous at x 2.
The function is NOT continuous at x 1.
x→2
8. lim x→2
x→2
2x 1 1 lim x2 4 x→2 x 2 4
x→1
10. lim x→4
x 2
x4
lim x→4
lim x→4
lim x→4
12. lim x→2
x 2
14. lim x→0
x2
lim
x→2
x 2
x4
x 2 x 2
x4 x 4 x 2 1 x 2
1 4
x2 1 x2
x x2 x x x2 x x2 2xx x2 x x x2 x lim x→0 x x lim x→0
2xx x2 x x
lim 2x x 1 x→0
2x 0 1 2x 1 16. lim f x lim x2 4x 2 2 x→2
x→2
lim f x lim
x→2
x→2
x2
18. lim f x lim 1 x 0 x→1
x→1
4x 6 2
lim f x 2
x→2
20. lim sec x does not exist since x→ 2
lim
x→ 2
sec x and
lim
x→ 2
24. lim 1 x→1
x→2
sec x do not exist.
2x 1 1 2
22. lim 2x x 22 2 2
26. f x
x2 1 x1
has a discontinuity at x 1 since f 1 is not defined.
x, 28. f x 2, 2x 1,
x < 1 x 1 has discontinuity at x 1 since f 1 2 lim f x 1. x→1 x > 1
30. f t 3 9 t2 is continuous on 3, 3.
32. g2 is not defined. g is continuous on 1, 2.
316
Chapter 1
Limits and Their Properties
x is continuous for all real x. 2
34. f x
1 is continuous for all real x. x2 1
38. f x
x has nonremovable discontinuities at x 1 and x 1 since lim f x and lim f x do not exist. x→1 x→1 x2 1
36. f x cos
x3 has a nonremovable discontinuity at x 3 since lim f x does not exist, and has a removable discontinuity x→3 x2 9 at x 3 since
40. f x
lim f x lim
x→3
42. f x
x→3
1 1 . x3 6
x1 x 2x 1
44. f x
has a nonremovable discontinuity at x 2 since lim f x does not exist, and has a removable discontinux→2 ity at x 1 since lim f x lim
x→1
46. f x
x→1
x 3
x3 has a nonremovable discontinuity at x 3 since lim f x x→3 does not exist.
1 1 . x2 3
3,
2x x,
x < 1 x ≥ 1
2
has a possible discontinuity at x 1. 1. f 1 12 1 2.
lim f x lim 2x 3 1
x→1
x→1
x→1
lim f x 1 x→1
lim f x lim x2 1 x→1
3. f 1 lim f x x→1
f is continuous at x 1, therefore, f is continuous for all real x.
48. f x
2x, x 4x 1, 2
x ≤ 2 has a possible discontinuity at x 2. x > 2
1. f 2 22 4 2.
lim f x lim 2x 4
x→2
x→2
x→2
x→2
lim f x does not exist. x→2
lim f x lim x2 4x 1 3
Therefore, f has a nonremovable discontinuity at x 2.
50. f x
csc x , 6 2,
1. f 1 csc
2 6
2. lim f x 2 x→1
3. f 1 lim f x x→1
x 3 x 3
≤ 2 > 2
csc x , 6 2,
f 5 csc
1 ≤ x ≤ 5 x < 1 or x > 5
has possible discontinuities at x 1, x 5.
5 2 6
lim f x 2
x→5
f 5 lim f x x→5
f is continuous at x 1 and x 5, therefore, f is continuous for all real x.
Section 1.4
x has nonremovable discontinuities at each 2 2k 1, k is an integer.
58. lim g(x lim
20
x→0
x→0
lim fx 0
x→0
4 sin x 4 x
lim gx lim a 2x a
x→0
f is not continuous at x 4
x→0
−8
x→0
8
Let a 4. −10
x2 a2 x→a x a
60. lim gx lim x→a
lim x a 2a x→a
Find a such that 2a 8 ⇒ a 4.
62. f gx
1 x 1
Nonremovable discontinuity at x 1. Continuous for all x > 1. Because f g is not defined for x < 1, it is better to say that f g is discontinuous from the right at x 1. 66. hx
64. f gx sin x2 Continuous for all real x
1 x 1x 2
Nonremovable discontinuity at x 1 and x 2. 2
−3
4
−2
68. f x
cos x 1 , x < 0
5x, x
lim f x lim
x→0
x→0
3
x ≥ 0 −7
f 0 50 0
cos x 1 0 x
2
−3
lim f x lim 5x 0
x→0
x→0
Therefore, lim f x 0 f 0 and f is continuous on the entire real line. (x 0 was the only possible discontinuity.) x→0
70. f x xx 3 Continuous on 3,
317
54. f x 3 x has nonremovable discontinuities at each integer k.
52. f x tan
56. lim f x 0
Continuity and One-Sided Limits
72. f x
x1 x
Continuous on 0,
318
Chapter 1
74. f x
Limits and Their Properties
x3 8 x2
76. f x x3 3x 2 is continuous on 0, 1. f 0 2 and f 1 2
14
By the Intermediate Value Theorem, f x 0 for at least one value of c between 0 and 1. −4
4 0
The graph appears to be continuous on the interval 4, 4. Since f 2 is not defined, we know that f has a discontinuity at x 2. This discontinuity is removable so it does not show up on the graph.
78. f x
4 x tan is continuous on 1, 3. x 8
f 1 4 tan
3 4 < 0 and f 3 tan > 0. 8 3 8
By the Intermediate Value Theorem, f 1 0 for at least one value of c between 1 and 3.
82. h 1 3 tan
80. f x x3 3x 2 f x is continuous on 0, 1. f 0 2 and f 1 2 By the Intermediate Value Theorem, f x 0 for at least one value of c between 0 and 1. Using a graphing utility, we find that x 0.5961. 84. f x x2 6x 8
h is continuous on 0, 1.
f is continuous on 0, 3.
h0 1 > 0 and h1 2.67 < 0.
f 0 8 and f 3 1
By the Intermediate Value Theorem, h 0 for at least one value between 0 and 1. Using a graphing utility, we find that 0.4503.
1 < 0 < 8 The Intermediate Value Theorem applies. x2 6x 8 0
x 2x 4 0 x 2 or x 4 c 2 (x 4 is not in the interval.) Thus, f 2 0.
86. f x
x2 x x1
The Intermediate Value Theorem applies.
f is continuous on 2 , 4. The nonremovable discontinuity, x 1, lies outside the interval. 5
f
5 35 20 and f 4 2 6 3
20 35 < 6 < 6 3
x2 x 6 x1 x2 x 6x 6 x2 5x 6 0
x 2x 3 0 x 2 or x 3 c 3 (x 2 is not in the interval.) Thus, f 3 6.
Section 1.4 88. A discontinuity at x c is removable if you can define (or redefine) the function at x c in such a way that the new function is continuous at x c. Answers will vary. (a) f x
x 2
Continuity and One-Sided Limits
1, 0, (c) f x 1, 0,
x2 sinx 2 (b) f x x2
319
if x ≥ 2 if 2 < x < 2 if x 2 if x < 2
y 3 2 1 −3
−2
−1
x −1
1
2
3
−2 −3
90. If f and g are continuous for all real x, then so is f g (Theorem 1.11, part 2). However, fg might not be continuous if gx 0. For example, let f x x and gx x2 1. Then f and g are continuous for all real x, but fg is not continuous at x ± 1.
1.04, 92. C 1.04 0.36t 1, 1.04 0.36t 2,
0 < t ≤ 2 t > 2, t is not an integer t > 2, t is an integer
You can also write C as C
Nonremovable discontinuity at each integer greater than 2.
1.04, 1.04 0.362 t,
0 < t ≤ 2 . t > 2
C 4 3 2 1 t 1
2
3
4
94. Let st be the position function for the run up to the campsite. s0 0 (t 0 corresponds to 8:00 A.M., s20 k (distance to campsite)). Let rt be the position function for the run back down the mountain: r0 k, r10 0. Let f t st rt. When t 0 (8:00 A.M.),
f 0 s0 r0 0 k < 0.
When t 10 (8:10 A.M.), f 10 s10 r10 > 0. Since f 0 < 0 and f 10 > 0, then there must be a value t in the interval 0, 10 such that f t 0. If f t 0, then st rt 0, which gives us st rt. Therefore, at some time t, where 0 ≤ t ≤ 10, the position functions for the run up and the run down are equal. 96. Suppose there exists x1 in a, b such that f x1 > 0 and there exists x2 in a, b such that f x2 < 0. Then by the Intermediate Value Theorem, f x must equal zero for some value of x in x1, x2 (or x2, x1 if x2 < x1). Thus, f would have a zero in a, b, which is a contradiction. Therefore, f x > 0 for all x in a, b or f x < 0 for all x in a, b. 98. If x 0, then f 0 0 and lim f x 0. Hence, f is x→0 continuous at x 0. If x 0, then lim f t 0 for x rational, whereas t→x
lim f t lim kt kx 0 for x irrational. Hence, f is not
t →x
t →x
continuous for all x 0.
100. True 1. f c L is defined. 2. lim f x L exists. x→c
3. f c lim f x x→c
All of the conditions for continuity are met.
320
Chapter 1
Limits and Their Properties
102. False; a rational function can be written as PxQx where P and Q are polynomials of degree m and n, respectively. It can have, at most, n discontinuities.
104. (a)
S 60 50 40 30 20 10 t 5
10
15 20 25 30
(b) There appears to be a limiting speed and a possible cause is air resistance. 106. Let y be a real number. If y 0, then x 0. If y > 0, then let 0 < x0 < 2 such that M tan x0 > y (this is possible since the tangent function increases without bound on 0, 2). By the Intermediate Value Theorem, f x tan x is continuous on 0, x0 and 0 < y < M, which implies that there exists x between 0 and x0 such that tan x y. The argument is similar if y < 0. 108. 1. f c is defined. 2. lim f x lim f c x f c exists. x→0
x→c
Let x c x. As x → c, x → 0 3. lim f x f c. x→c
Therefore, f is continuous at x c. 110. Define f x f2x f1x. Since f1 and f2 are continuous on a, b, so is f. f a f2a f1a > 0 and
f b f2b f1b < 0.
By the Intermediate Value Theorem, there exists c in a, b such that f c 0. f c f2c f1c 0 ⇒ f1c f2c
Section 1.5 2.
Infinite Limits
1 x2 1 lim x→2 x 2 lim
4.
x→2
6. f x x
x 4
lim sec
x 4
x→2
x x2 9 3.5
f x 1.077
3.1
3.01
3.001
2.999
2.99
2.9
2.5
5.082 50.08
500.1
499.9
49.92
4.915
0.9091
lim f x
x→3
lim f x
x→3
lim sec
x→2
Section 1.5
8. f x sec
321
x 6
3.5
x
Infinite Limits
f x 3.864
3.1
3.01
19.11
191.0
3.001 2.999 2.99 2.9
2.5
1910
3.864
1910
191.0
19.11
lim f x
x→3
lim f x
x→3
10. lim x→2
lim
x→2
4 x 23
12. lim x→0
2x 2x lim 2 x→0 x 1 x x 1 x 2
Therefore, x 0 is a vertical asymptote.
4 x 23
lim
2x x21 x
lim
2x x21 x
x→1
Therefore, x 2 is a vertical asymptote.
x→1
Therefore, x 1 is a vertical asymptote.
14. No vertical asymptote since the denominator is never zero.
16.
lim hs and lim hs .
s→5
s→5
Therefore, s 5 is a vertical asymptote. lim hs and lim hs .
s→5
s→5
Therefore, s 5 is a vertical asymptote.
18. f x sec x x
1 has vertical asymptotes at cos x
2n 1 , n any integer. 2
20. gx
12x3 x2 4x 1 xx2 2x 8 3x2 6x 24 6 x2 2x 8
1 x, 6 x 2, 4 No vertical asymptotes. The graph has holes at x 2 and x 4.
22. f x
x
x3
4x2 x 6 4x 3x 2 4 , x 3, 2 2x2 9x 18 xx 2x2 9 xx 3
Vertical asymptotes at x 0 and x 3. The graph has holes at x 3 and x 2.
24. hx
x 2x 2 x2 4 x3 2x2 x 2 x 2x2 1
has no vertical asymptote since lim hx lim
x→2
x→2
4 x2 . x2 1 5
26. ht
tt 2 t ,t2 t 2t 2t 2 4 t 2t 2 4
Vertical asymptote at t 2. The graph has a hole at t 2.
322
Chapter 1
28. g
Limits and Their Properties
tan sin has vertical asymptotes at cos
x2 6x 7 lim x 7 8 x→1 x→1 x1
30. lim
2n 1 n, n any integer. 2 2
2 −3
3
There is no vertical asymptote at 0 since lim
→0
tan 1.
−12
Removable discontinuity at x 1
sinx 1 1 x1
32. lim
x→1
2
34. lim x→1
Removable discontinuity at x 1
−3
2x 1x
3
−2
36. lim x→4
40. lim
x→3
44.
48.
x2
x2 1 16 2
38.
x2 1 x2 9
lim
x→ 2
lim
x→ 12
lim
x→12
6x2 x 1 3x 1 5 lim 4x2 4x 3 x→12 2x 3 8
42. lim x2 x→0
2 cos x
46. lim
x→0
x2 tan x and
lim
x→ 12
x2 tan x .
Therefore, lim x2 tan x does not exist. x→ 12
1 x
x 2 lim x 2tan x 0 x→0 cot x
50. f x
x2
x3 1 x1
lim f x lim x 1 0
x→1
x→1
4
−8
8
−4
52. f x sec
x 6
54. The line x c is a vertical asymptote if the graph of f approaches ± as x approaches c.
lim f x
x→3
6
−9
9
−6
56. No. For example, f x
1 has no x2 1
58. P
vertical asymptote. lim
V→0
k V k k (In this case we know that k > 0.) V
Section 1.5
200 ftsec 6 3 (b) r 50 sec2 200 ftsec 3
60. (a) r 50 sec2
(c)
lim
→ 2
62. m
Total distance Total time
50
2d dx dy
50
2xy yx
m0
lim m lim
v→c
(b)
v→c
m0 1 v2c2
x
30
40
50
60
y
150
66.667
50
42.857
(c) lim x→25
25x x 25
As x gets close to 25 mph, y becomes larger and larger.
50y 50x 2xy 50x 2xy 50y 50x 2yx 25 25x y x 25 Domain: x > 25 1 1 1 1 66. (a) A bh r2 1010 tan 102 2 2 2 2
(b)
(c)
f
50 tan 50 Domain:
0.3
0.6
0.9
1.2
1.5
0.47
4.21
18.0
68.6
630.1
0, 2 (d)
100
0
lim A
→ 2
1.5 0
68. False; for instance, let f x
70. True
x2 1 . x1
The graph of f has a hole at 1, 2, not a vertical asymptote.
72. Let f x
1 1 and gx 4, and c 0. x2 x
1 1 2 and lim 4 , but x→0 x x→0 x lim
x1 x1 lim x x 1 0. 2
lim
x→0
2
4
x→0
4
323
1 v2c2
50 sec2
64. (a) Average speed
Infinite Limits
74. Given lim f x , let g x 1. then lim x →c
by Theorem 1.15.
x →c
gx 0 fx
324
Chapter 1
Limits and Their Properties
Review Exercises for Chapter 1 2. Precalculus. L 9 12 3 12 8.25 4.
0.1
x f x
0.01
0.358
0.5
0.001
0.001
0.01
0.1
0.354
0.354
0.353
0.349
0.354
−1
1
lim f x 0.2
x→0
6. gx
−0.5
3x x2
(b) lim gx 0
(a) lim gx does not exist.
x→0
x→2
Assuming 4 < x < 16, you can choose 5.
8. lim x 9 3. x→9
Let > 0 be given. We need
x→5
x 9 < x 3
x 9 < 5 < x 3
x 3 < ⇒ x 3 x 3 < x 3
10. lim 9 9. Let > 0 be given. can be any positive
Hence, for 0 < x 9 < 5, you have
x 3 <
f x L <
12. lim 3 y 1 3 4 1 9 y→4
number. Hence, for 0 < x 5 < , you have
9 9 < f x L < 14. lim t→3
t2 9 lim t 3 6 t→3 t3
16. lim
4 x 2
x→0
x
lim lim
x→0
18. lim
s→0
x 1 4 x 2
11 s 1 lim 11 s 1 11 s 1 s→0
s s 11 s 1 lim
s→0
20. lim
4 x 2
x→0
x→2
11 s 1 1 1 lim s 11 s 1 s→0 1 s 11 s 1 2
x2 4 x 2x 2 lim x3 8 x→2 x 2x2 2x 4 lim
x→2
x2 x2 2x 4
4 1 12 3
22.
lim
x→ 4
4x 44 tan x 1
4 x 2 4 x 2
1 4
Review Exercises for Chapter 1
24. lim
x→0
cos x 1 cos cos x sin sin x 1 lim x→0 x x lim
x→0
cos x x 1
lim sin
x→0
sin x x
0 01 0 3 2 7 26. lim f x 2gx 4 23 12 x→c
28. f x (a)
3 x 1 x1
x f x
lim
x→1
(c) lim x→1
1.1
1.01
1.001
0.3228
0.3322
0.3332
3 x 1 0.333 x1
0.3333
3
−3
3 x 1 3 x 3 3 x 2 1 x
1x 3 3 x 11 x x2
lim
1 3 x 3 x2 1
−3
2
lim
x→1
2
Actual limit is 13 .
3 x 3 x 1 1 lim x→1 x1 x1
x→1
(b)
1.0001
1 3
30. st 0 ⇒ 4.9t2 200 0 ⇒ t2 40.816 ⇒ t 6.39 sec When t 6.39, the velocity is approximately lim t→a
sa st lim 4.9a t t→a at lim 4.96.39 6.39 62.6 msec. t→6.39
32. lim x 1 does not exist. The graph jumps from 2 to 3 x→4
34. lim gx 1 1 2. x→1
at x 4.
36. lim f s 2 s→2
38. f x
3x2 x 2 , x1 0,
lim f x lim
x→1
x→1
x 1 x1
3x x 2 x1 2
lim 3x 2 5 0 x→1
Removable discontinuity at x 1 Continuous on , 1 1,
325
326
Chapter 1
52xx,3,
40. f x
Limits and Their Properties
x x 1 1 1x 1 lim 1
x
x ≤ 2 x > 2
42. f x
lim 5 x 3
x→2
x→0
lim 2x 3 1
Domain: , 1 , 0,
x→2
Nonremovable discontinuity at x 2 Continuous on , 2 2,
Nonremovable discontinuity at x 0 Continuous on , 1 0,
x1 2x 2
44. f x
46. f x tan 2x Nonremovable discontinuities when
x1 1 x→1 2x 1 2 lim
x
Removable discontinuity at x 1 Continuous on , 1 1,
2n 1 4
Continuous on
2n 4 1, 2n 4 1 for all integers n. 48. lim x 1 2 x→1
lim x 1 4
x→3
Find b and c so that lim x2 bx c 2 and lim x2 bx c 4. x→1
x→3
1bc2
Consequently we get Solving simultaneously,
b
and 9 3b c 4.
3 and
50. C 9.80 2.50 x 1 , x > 0
c 4. 52. f x x 1x (a) Domain: , 0 1,
9.80 2.50 x 1
(b) lim f x 0
C has a nonremovable discontinuity at each integer.
x→0
30
(c) lim f x 0 x→1
0
5 0
54. hx
4x 4 x2
56. f x csc x Vertical asymptote at every integer k
Vertical asymptotes at x 2 and x 2
58.
62.
lim
x→ 12
lim
x→1
66. lim x→0
x 2x 1
x2 2x 1
x1
sec x
x
60.
lim
x→1
64. lim x→2
68. lim x→0
x1 1 1 lim x4 1 x→1 x2 1x 1 4 1
3 2 x 4
cos2 x
x
Problem Solving for Chapter 1 tan 2x x
70. f x (a)
327
0.1
x f x lim
x→0
0.01
2.0271
0.001
2.0003
2.0000
0.001
0.01
0.1
2.0000
2.0003
2.0271
tan 2x 2 x
(b) Yes, define f x
2, x
tan 2x ,
x0 x0
.
Now f x is continuous at x 0.
Problem Solving for Chapter 1 1 1 x 2. (a) Area PAO bh 1x 2 2 2
4. (a) Slope
1 1 y x2 Area PBO bh 1y 2 2 2 2 (b) ax
40 4 30 3
(b) Slope
3 Tangent line: y 4 x 3 4
3 4
x22 Area PBO x Area PAO x2
3 25 y x 4 4
x
4
2
1
0.1
0.01
Area PAO
2
1
12
120
1200
Area PBO
8
2
12
1200
120,000
ax
4
2
1
110
(c) Let Q x, y x, 25 x2 mx
x3
(d) lim mx lim x→3
1100
x→3
25 x2 4
x3
25 x2 4 25 x2 4
25 x2 16 x→3 x 3 25 x2 4
lim
(c) lim ax lim x 0 x→0
25 x2 4
x→0
lim
x→3
lim
x→3
3 x3 x x 325 x2 4 3 x 25 x2 4
3 6 44 4
This is the slope of the tangent line at P.
6.
a bx 3
x
a bx 3
x
a bx 3 a bx 3
x
x→0
Setting
b 3 3
b 3 bx 3
3, you obtain b 6.
Thus, a 3 and b 6.
ax tan x a because lim 1 x→0 tan x x
a2 2 a
bx lim x→0 x 3 bx 3 lim
x→0
Thus,
Thus, lim
x→0
x→0
Letting a 3 simplifies the numerator.
x→0
x→0
lim f x lim
a bx 3 xa bx 3
3 bx 3
8. lim f x lim a2 2 a2 2
.
a2 a 2 0
a 2a 1 0 a 1, 2
328
Chapter 1
Limits and Their Properties 1 (a) f 4 4 4
y
10.
f 3
1 3
3 2
0
f 1 1 1
1 x
−1
1
−1
(b) lim f x 1
(c) f is continuous for all real numbers except
x→1
lim f x 0
x→1
x 0, ± 1, ± 12, ± 13, . . .
lim f x
x→0
lim f x
x→0
−2
v2
12. (a)
192,000 v02 48 r
192,000 v2 v02 48 r r lim r
v→0
192,000 v v02 48 192,000 48 v02
Let v0 48 43 feetsec. (b)
v2
1920 v02 2.17 r
1920 v2 v02 2.17 r r lim r
v→0
1920 v2 v02 2.17 1920 2.17 v02
Let v0 2.17 misec 1.47 misec. r
(c)
lim r
v→0
10,600 v2 v02 6.99 10,600 6.99 v02
Let v0 6.99 2.64 misec. Since this is smaller than the escape velocity for earth, the mass is less.
14. Let a 0 and let > 0 be given. There exists 1 > 0 such that if 0 < x 0 < , then f x L < . Let 1 a . Then for 0 < x 0 < 1 a , you have
x < a1
ax < 1 f ax L < .
As a counterexample, let f x Then lim f x 1 L, x→0
but lim f ax lim f 0 2. x→0
x→0
12
x0 . x0
C H A P T E R 1 Limits and Their Properties Section 1.1
A Preview of Calculus . . . . . . . . . . . . . . . . . . . . 27
Section 1.2
Finding Limits Graphically and Numerically . . . . . . . . 27
Section 1.3
Evaluating Limits Analytically
Section 1.4
Continuity and One-Sided Limits . . . . . . . . . . . . . . 37
Section 1.5
Infinite Limits . . . . . . . . . . . . . . . . . . . . . . . . 42
. . . . . . . . . . . . . . . 31
Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
C H A P T E R 1 Limits and Their Properties Section 1.1
A Preview of Calculus
Solutions to Odd-Numbered Exercises 1. Precalculus: 20 ftsec15 seconds 300 feet
3. Calculus required: slope of tangent line at x 2 is rate of change, and equals about 0.16.
5. Precalculus: Area 12 bh 12 53 15 2 sq. units
7. Precalculus: Volume 243 24 cubic units
9. (a)
6
(1, 3) −4
8 −2
(b) The graphs of y2 are approximations to the tangent line to y1 at x 1. (c) The slope is approximately 2. For a better approximation make the list numbers smaller:
0.2, 0.1, 0.01, 0.001 11. (a) D1 5 12 1 52 16 16 5.66 5 5 5 5 5 5 (b) D2 1 2 1 2 3 1 3 4 1 4 1 2
2
2
2
2.693 1.302 1.083 1.031 6.11 (c) Increase the number of line segments.
Section 1.2 1.
x
1.9
1.99
1.999
2.001
2.01
2.1
f x
0.3448
0.3344
0.3334
0.3332
0.3322
0.3226
lim
x→2
3.
Finding Limits Graphically and Numerically
x2 0.3333 x2 x 2
0.1
x f x lim
x→0
0.01
0.2911
0.001
0.2889
x 3 3
x
Actual limit is 13 .
0.2887
0.2887
0.001
0.01
0.1
0.2887
0.2884
0.2863
Actual limit is 1 23.
27
28
Chapter 1
5.
x
2.9
f x
0.0641
lim
x→3
7.
Limits and Their Properties
2.999
3.001
3.01
3.1
0.0627
0.0625
0.0625
0.0623
0.0610
1x 1 14 0.0625 x3
0.1
x
2.99
f x
0.01
0.9983
lim
x→0
0.001
0.99998
sin x 1.0000 x
Actual limit is 161 .
1.0000
0.001
0.01
0.1
1.0000
0.99998
0.9983
(Actual limit is 1.) (Make sure you use radian mode.)
11. lim f x lim 4 x 2
9. lim 4 x 1
x→2
x→3
13. lim
x→2
x 5 does not exist. For values of x to the left of 5, x 5 x 5 equals 1,
x5 whereas for values of x to the right of 5, x 5 x 5 equals 1. x→5
15. lim tan x does not exist since the function increases and x→ 2
17. lim cos1x does not exist since the function oscillates x→0
between 1 and 1 as x approaches 0.
decreases without bound as x approaches 2. 19. Ct 0.75 0.50 t 1 (a)
(b)
3
t
3
3.3
3.4
3.5
3.6
3.7
4
C 1.75
2.25
2.25
2.25
2.25
2.25
2.25
2.5
2.9
3
3.1
3.5
4
1.75
1.75
1.75
2.25
2.25
2.25
lim Ct 2.25
t→3.5
(c)
5
0 0
t
2
C 1.25
lim Ct does not exist. The values of C jump from 1.75 to 2.25 at t 3. t→3
21. You need to find such that 0 < x 1 < implies 1 f x 1 1 < 0.1. That is, x
0.1 <
1 1 < 0.1 x
1 0.1 <
1 x
< 1 0.1
9 < 10
1 x
11 < 10
10 > 9
x
>
10 11
10 10 1 > x 1 > 1 9 11 1 1 > x 1 > . 9 11
So take
1 . Then 0 < x 1 < implies 11
1 1 < x1 < 11 11
1 1 < x1 < . 11 9
Using the first series of equivalent inequalities, you obtain
f x 1
1 1 < < 0.1. x
Section 1.2 23. lim 3x 2 8 L x→2
3x 2 8
3x 6 3 x 2
< 0.01 < 0.01 < 0.01
0.01 0.0033 3 0.01 Hence, if 0 < x 2 < , you have 3
0 < x2 <
3x 6 < 0.01
3x 2 8 < 0.01
f x L < 0.01 3 x 2 < 0.01
27. lim x 3 5
Finding Limits Graphically and Numerically
25. lim x2 3 1 L x→2
x2 3 1
x2 4
x 2x 2
x 2 x 2
< 0.01 < 0.01
If we assume 1 < x < 3, then 0.015 0.002.
Hence, if 0 < x 2 < 0.002, you have 1
x 2
x 2 < 0.01
x2 4 < 0.01
x2 3 1 < 0.01
f x L < 0.01
12 x 1 12 4 1 3
x 1 3 <
x 2 <
<
1 2
<
Hence, let .
1 2
1 2
Hence, if 0 < x 2 < , you have
x2 <
x 3 5 <
f x L <
x 4
x 4
< < 2
Hence, let 2.
Hence, if 0 < x 4 < 2, you have
x 4 < 2
x 2
x 1 3 1 2
1 2
< <
f x L <
31. lim 3 3
3 33. lim x0
x→6
x→0
Given > 0:
x <
3 x 0 < Given > 0:
33 <
3
x
0 <
< 3
Hence, any > 0 will work.
Hence, let 3.
Hence, for any > 0, you have
Hence for 0 < x 0 < 3, you have
1
x 2 < 0.002 50.01 < x 2 0.01
Given > 0:
x 3 5
x 2
< 0.01
0.01
x→4
Given > 0:
< 0.01
x 2 < x 2
29. lim
x→2
f x L < 33 <
29
x <
0:
Given > 0:
x 2 4
x 2 4
x 2 x 2 x 2
x2 1 2
x2 1
x 1x 1
<
x 2 < 0
< <
Hence, .
x 2 <
x 2 4 <
x 2 4 <
f x L <
If we assume 0 < x < 2, then 3.
Hence for 0 < x 1 < , you have 3
1
1
x2 1 <
x2 1 2 <
f x 2 < x9
0.5
41. f x
x4 1 6
x 1 < 3 < x 1
(because x 20)
x 5 3
lim f x
<
x2 <
x→4
<
x 1 < x 1
Hence for 0 < x 2 < , you have
39. f x
<
−6
10
x 3
lim f x 6
6
x→9
−0.1667
0
10 0
The domain is 5, 4 4, . 1 The graphing utility does not show the hole at 4, 6 .
The domain is all x ≥ 0 except x 9. The graphing utility does not show the hole at 9, 6.
43. lim f x 25 means that the values of f approach 25 as x gets closer and closer to 8. x→8
45. (i) The values of f approach different numbers as x approaches c from different sides of c:
(ii) The values of f increase without bound as x approaches c:
(iii) The values of f oscillate between two fixed numbers as x approaches c:
y
y
y
6
4
5
3
4
2
3
4 3
2
1
1
x
−4 −3 −2 −1 −1
1
2
3
4
−3 −2 −1 −1
3
4
−3
−4
−4
47. f x 1 x1x lim 1 x1x e 2.71828
x→0
y 7
3
(0, 2.7183)
2 1 −3 −2 −1 −1
x 1
2
2
5
−2
−3
x
−4 −3 −2
x 2
3
4
5
x
f x
x
f x
0.1
2.867972
0.1
2.593742
0.01
2.731999
0.01
2.704814
0.001
2.719642
0.001
2.716942
0.0001
2.718418
0.0001
2.718146
0.00001
2.718295
0.00001
2.718268
0.000001
2.718283
0.000001
2.718280
3
4
Section 1.3 49. False; f x sin xx is undefined when x 0. From Exercise 7, we have lim
x→0
51. False; let f x
sin x 1. x
Evaluating Limits Analytically
31
53. Answers will vary.
x10, 4x, 2
x4 . x4
f 4 10 lim f x lim x2 4x 0 10
x→4
x→4
55. If lim f x L1 and lim f x L2, then for every > 0, there exists 1 > 0 and 2 > 0 such that x c < 1 ⇒ f x L1 < and x→c
x→c
x c < 2 ⇒ f x L2 < . Let equal the smaller of 1 and 2. Then for x c L1 L2 L1 f x f x L2 ≤ L1 f x f x L2 < . Therefore, L1 L2 < 2. Since > 0 is arbitrary, it follows that L1 L2.
< , we have
57. lim f x L 0 means that for every > 0 there exists > 0 such that if x→c
0 < x c < , then
f x L 0
< .
This means the same as f x L < when 0 < x c < .
Thus, lim f x L. x→c
Section 1.3 1.
Evaluating Limits Analytically (a) lim hx 0
7
x→5
−8
3.
(b) lim hx 6
x→0
π
−π
x→1
13
(a) lim f x 0
4
−7
(b) lim f x 0.524
−4
hx x2 5x
f x x cos x 7. lim 2x 1 20 1 1
5. lim x4 24 16 x→2
x→0
9. lim x2 3x 32 33 9 9 0 x→3
11. lim 2x2 4x 1 232 43 1 18 12 1 7 x→3
13. lim
x→2
17. lim
x→7
1 1 x 2 5x x 2
15. lim
x→1
57 7 2
35 9
35 3
x3 13 2 2 x2 4 12 4 5 5
19. lim x 1 3 1 2 x→3
x→ 3
6
32
Chapter 1
Limits and Their Properties
21. lim x 32 4 32 1
23. (a) lim f x 5 1 4
x→4
x→1
(b) lim gx 43 64 x→4
(c) lim g f x g f 1 g4 64 x→1
25. (a) lim f x 4 1 3
27. lim sin x sin x→ 2
x→1
(b) lim gx 3 1 2
1 2
x→3
(c) lim g f x g3 2 x→1
29. lim cos x→2
33.
x 2 1 cos 3 3 2
lim sin x sin
x→56
31. lim sec 2x sec 0 1 x→0
5 1 6 2
35. lim tan x→3
37. (a) lim 5gx 5 lim gx 53 15 x→c
x→c
(b) lim f x gx lim f x lim gx 2 3 5 x→c
x→c
x→c
(c) lim f xgx lim f x lim gx 23 6 x→c
x→c
x→c
lim f x
f x 2 x→c (d) lim x→c gx lim gx 3
39. (a) lim f x3 lim f x3 43 64 x→c
x→c
(b) lim f x lim f x 4 2 x→c
x→c
(c) lim 3 f x 3 lim f x 34 12 x→c
x→c
(d) lim f x32 lim f x32 432 8 x→c
x→c
41. f x 2x 1 and gx x 0.
4x tan 34 1
2x2 x agree except at x
x→c
43. f x xx 1 and gx
x3 x agree except at x 1. x1
(a) lim gx lim f x 1
(a) lim gx lim f x 2
(b) lim gx lim f x 3
(b) lim gx lim f x 0
x→0
x→1
45. f x
x→1
x→0
x→1
x→1
x2 1 and gx x 1 agree except at x 1. x1
lim f x lim gx 2
x→1
x→1
47. f x x 2.
x→1
x→1
x3 8 and gx x2 2x 4 agree except at x2
lim f x lim gx 12
x→2
3
x→2
12 −3
4
−4
−9
9 0
49. lim
x→5
x5 x5 lim x2 25 x→5 x 5x 5 lim
x→5
1 1 x 5 10
51. lim
x→3
x2 x 6 x 3x 2 lim x→3 x 3x 3 x2 9 lim
x→3
x 2 5 5 x 3 6 6
Section 1.3
53. lim
x 5 5
lim
x
x→0
lim
x→0
55. lim
x 5 3
x4
x→4
x 5 5
lim
x→4
x 5 5 x 5 5
5 x 5 5 1 1 lim x x 5 5 x→0 x 5 5 2 5 10
x 5 3
x4
x→4
lim
x
x→0
Evaluating Limits Analytically
x 5 3 x 5 3
1 x 5 9 1 1 lim x 4 x 5 3 x→4 x 5 3 9 3 6
1 1 2 2 x 1 1 2x 2 22 x 57. lim lim lim x→0 x→0 x→0 22 x x x 4
59. lim
x→0
2x x 2x 2x 2 x 2x lim lim 2 2
x→0
x→0
x
x
x x2 2x x 1 x2 2x 1 x2 2x x x2 2x 2 x 1 x2 2x 1 lim
x→0
x→0
x
x
61. lim
lim 2x x 2 2x 2
x→0
63. lim
x 2 2
x→0
x
x
0.1
f x
0.354 0.01
0.358
2
0.001
0
0.001
0.01
0.1
0.345
?
0.354
0.353
0.349
0.354
−3
3
−2
Analytically, lim
x 2 2
x
x→0
lim
x 2 2
x
x→0
lim
x→0
x 2 2
x 2 2
x22
x x 2 2
lim
x→0
1 x 2 2
1 2 2
2
4
1 1 2x 2 1 65. lim x→0 x 4
0.354
3
−5
x
0.1
0.01
0.001
0
0.001
f x
0.263
0.251
0.250
?
0.250
1 1 2x 2 2 2 x Analytically, lim lim x→0 x→0 x 22 x
1
0.01
0.1
0.249
0.238
x
1
1
−2
1
1
lim . x x→0 lim 22 x x x→0 22 x 4
33
34
Chapter 1
67. lim
x→0
Limits and Their Properties
sin x lim x→0 5x
sin x x
15 1 15 51
69. lim
x→0
1 sin x1 cos x lim x→0 2 2x2
sin x x
1 cos x x
1 10 0 2
1 cos h2 1 cos h 1 cos h lim h→0 h→0 h h
sin2 x sin x lim sin x 1 sin 0 0 x→0 x→0 x x
71. lim
73. lim
00 0
75. lim
x→ 2
cos x lim sin x 1 x→ 2 cot x
79. f t
f t
t→0
sin 3t sin 3t lim t→0 2t 3t
4
0.01
2.96
0.001
2.9996
3
0
0.001
0.01
0.1
?
3
2.9996
2.96
− 2
2 −1
The limit appear to equal 3.
sin 3t sin 3t lim 3 31 3. t→0 t→0 t 3t
Analytically, lim
81. f x
32 1 32 23
sin 3t t 0.1
t
77. lim
sin x2 x
1
− 2
x
0.1
0.01 0.001 0 0.001 0.01 0.1
f x
0.099998 0.01 0.001 ? 0.001 0.01 0.099998
2
−1
sin x2 sin x2 lim x 01 0. x→0 x→0 x x2
Analytically, lim
83. lim
h→0
f x h f x 2x h 3 2x 3 2x 2h 3 2x 3 2h lim lim lim 2 h→0 h→0 h→0 h h h h
4 4 4 4 f x h f x 4x 4x h xh x 85. lim lim lim lim 2 h→0 x hx h→0 h→0 h→0 h h x hxh x 87. lim 4 x2 ≤ lim f x ≤ lim 4 x2 x→0
x→0
x→0
89. f x x cos x 4
4 ≤ lim f x ≤ 4 x→0
Therefore, lim f x 4. x→0
− 3 2
3 2
−4
lim x cos x 0
x→0
Section 1.3
91. f x x sin x
93. f x x sin
6
2
1 x
−0.5
0.5
−6
−0.5
lim x sin x 0
lim x sin
x→0
95. We say that two functions f and g agree at all but one point (on an open interval) if f x gx for all x in the interval except for x c, where c is in the interval.
1 0 x
97. An indeterminant form is obtained when evaluating a limit using direct substitution produces a meaningless fractional expression such as 00. That is, lim
x→c
f x gx
for which lim f x lim gx 0 x→c
99. f x x, gx sin x, hx
When you are “close to” 0 the magnitude of f is approximately equal to the magnitude of g. Thus, g f 1 when x is “close to” 0.
f h
−5
x→c
sin x x
3
g
5
−3
101. st 16t2 1000 lim t→5
s5 st 600 16t2 1000 16t 5t 5 lim lim lim 16t 5 160 ftsec. t→5 t→5 t→5 5t 5t t 5
Speed 160 ftsec 103. st 4.9t2 150 s3 st 4.932 150 4.9t2 150 4.99 t2 lim lim t→3 t→3 t→3 3t 3t 3t
lim
lim
x→3
4.93 t3 t lim 4.93 t 29.4 msec x→3 3t
105. Let f x 1x and gx 1x. lim f x and lim gx do not exist. x→0
x→0
lim 0 0
1 1 lim f x gx lim x→0 x→0 x x
x→0
107. Given f x b, show that for every > 0 there exists a > 0 such that f x b < whenever x c < . Since f x b b b 0 < for any > 0, then any value of > 0 will work.
35
0.5
− 2
x→0
Evaluating Limits Analytically
109. If b 0, then the property is true because both sides are equal to 0. If b 0, let > 0 be given. Since lim f x L,
x→c
there exists > 0 such that f x L < b whenever 0 < x c < . Hence, wherever 0 < x c < , we have
b f x L
<
or
bf x bL
which implies that lim bf x bL. x→c
<
36
Chapter 1
Limits and Their Properties
M f x ≤ f xgx ≤ M f x
111.
113. False. As x approaches 0 from the left,
lim M f x ≤ lim f xgx ≤ lim M f x
x→c
x→c
x→c
x 1. x
2
M0 ≤ lim f xgx ≤ M0 x→c
−3
0 ≤ lim f xgx ≤ 0
3
x→c
Therefore, lim f xgx 0.
−2
x →c
115. True.
117. False. The limit does not exist. 4
−3
6
−2
119. Let f x
4,4,
if x ≥ 0 if x < 0
lim f x lim 4 4.
x→0
x→0
lim f x does not exist since for x < 0, f x 4 and for x ≥ 0, f x 4.
x→0
rational 0,1, ifif xx isis irrational 0, if x is rational g x x, if x is irrational
121. f x
lim f x does not exist.
x→0
No matter how “close to” 0 x is, there are still an infinite number of rational and irrational numbers so that lim f x does not x→0 exist. lim gx 0.
x→0
When x is “close to” 0, both parts of the function are “close to” 0.
123. (a) lim
x→0
1 cos x 1 cos x lim x→0 x2 x2 lim
x→0
1 cos2 x x 1 cos x 2
sin2 x x→0 x2
lim 1
1 cos x
1 cos x
1
1 cos x
12 21
(b) Thus,
1 cos x 1 1 ⇒ 1 cos x x2 x2 2 2 1 ⇒ cos x 1 x2 for x 0. 2
1 (c) cos0.1 1 0.12 0.995 2 (d) cos0.1 0.9950, which agrees with part (c).
Section 1.4
Section 1.4
3. (a) lim f x 0
(b) lim f x 1
(b) lim f x 0
x→3
x→4
x→3
The function is continuous at x 3.
x→0
x→4
(c) lim f x does not exist
(c) lim f x 0
x→3
x
lim
x→0
9.
does not exist because
x x2 9
grows
x
1
1
lim x x→0 xx x x
x→0
x→3
x x2 9
x 1. x
lim
x→3
lim
x→3
without bound as x → 3 .
1 1 x x x x x x lim 13. lim x→0 x→0 x xx x
15. lim f x lim
The function is NOT continuous at x 4.
The function is NOT continuous at x 3.
x5 1 1 lim x2 25 x→5 x 5 10
x
lim f x 2
x→4
(b) lim f x 2
x→3
(c) lim f x 1
11. lim
5. (a)
x→3
x→3
x→5
37
Continuity and One-Sided Limits
1. (a) lim f x 1
7. lim
Continuity and One-Sided Limits
1 xx x
1 1 2 xx 0 x
x2 5 2 2
17. lim f x lim x 1 2 x→1
x→1
lim f x lim x3 1 2
x→1
x→1
lim f x 2
x→1
21. lim 3x 5 33 5 4
19. lim cot x does not exist since x→
x→4
x 3 for 3 < x < 4
lim cot x and lim cot x do not exist.
x→
x→
23. lim 2 x does not exist x→3
because lim2 x 2 3 5
x→3
and
25. f x
1 x2 4
27. f x
has discontinuities at x 2 and x 2 since f 2 and f 2 are not defined.
x x 2
has discontinuities at each integer k since lim f x lim f x. x→k
x→k
lim 2 x 2 4 6.
x→3
29. gx 25 x2 is continuous on 5, 5 .
31. lim f x 3 lim f x. x→0
x→0
f is continuous on 1, 4 .
33. f x x2 2x 1 is continuous for all real x.
38
Chapter 1
Limits and Their Properties
35. f x 3x cos x is continuous for all real x.
37. f x
x is not continuous at x 0, 1. Since x2 x
x 1 for x 0, x 0 is a removable x2 x x 1 discontinuity, whereas x 1 is a nonremovable discontinuity.
39. f x
x is continuous for all real x. x2 1
41. f x
x2 x 2x 5
has a nonremovable discontinuity at x 5 since lim f x x→5 does not exist, and has a removable discontinuity at x 2 since lim f x lim
x→2
43. f x
x 2 has a nonremovable discontinuity at x 2 since
45. f x
x,x ,
x2
x→2
1 1 . x5 7
lim f x does not exist.
x→2
x ≤ 1 x > 1
2
has a possible discontinuity at x 1. 1. f 1 1 2.
lim f x lim x 1
x→1
x→1
x→1
x→1
lim f x 1
lim f x lim x2 1
x→1
3. f 1 lim f x x→1
f is continuous at x 1, therefore, f is continuous for all real x. x 1, 47. f x 2 3 x,
x ≤ 2
1. f 2
x > 2
2 12 2
lim f x lim
x→2
2.
has a possible discontinuity at x 2.
x→2
2x 1 2
lim f x lim 3 x 1
x→2
x→2
lim f x does not exist.
x→2
Therefore, f has a nonremovable discontinuity at x 2.
49. f x
x tan 4 , x,
x x
1. f 1 1 2. lim f x 1 x→1
3. f 1 lim f x x→1
x < 1 tan 4 , ≥ 1 x,
1 < x < 1 has possible discontinuities at x 1, x 1. x ≤ 1 or x ≥ 1
f 1 1 lim f x 1
x→1
f 1 lim f x x→1
f is continuous at x ± 1, therefore, f is continuous for all real x.
Section 1.4
Continuity and One-Sided Limits
39
51. f x csc 2x has nonremovable discontinuities at integer multiples of 2.
53. f x x 1 has nonremovable discontinuities at each integer k.
55. lim f x 0
57. f 2 8
50
x→0
lim f x 0
Find a so that lim ax2 8 ⇒ a
x→0
x→2
f is not continuous at x 2.
−8
8 2. 22
8 −10
59. Find a and b such that lim ax b a b 2 and lim ax b 3a b 2. x→1
x→3
a b 2
3a b 2 4
4a
a 1 b
2, f x x 1, 2,
x ≤ 1 1 < x < 3 x ≥ 3
2 1 1
61. f gx x 12
63. f gx
Continuous for all real x.
Nonremovable discontinuities at x ± 1
67. f x
65. y x x Nonremovable discontinuity at each integer 0.5
1 1 x2 5 6 x2 1
2xx 2x,4, 2
x ≤ 3 x > 3
Nonremovable discontinuity at x 3 5
−3
3
−5
7
−1.5 −5
69. f x
x x2 1
71. f x sec
Continuous on ,
73. f x
sin x x
Continuous on: . . . , 6, 2, 2, 2, 2, 6, 6, 10, . . . 1 75. f x 16x4 x3 3 is continuous on 1, 2 .
f 1 33 16 and f 2 4. By the Intermediate Value Theorem, f c 0 for at least one value of c between 1 and 2.
3
−4
x 4
4
−2
The graph appears to be continuous on the interval 4, 4 . Since f 0 is not defined, we know that f has a discontinuity at x 0. This discontinuity is removable so it does not show up on the graph.
40
Chapter 1
Limits and Their Properties
77. f x x2 2 cos x is continuous on 0, . f 0 3 and f 2 1 > 0. By the Intermediate Value Theorem, f c 0 for the least one value of c between 0 and .
81. gt 2 cos t 3t
79. f x x3 x 1 f x is continuous on 0, 1 . f 0 1 and f 1 1 By the Intermediate Value Theorem, f x 0 for at least one value of c between 0 and 1. Using a graphing utility, we find that x 0.6823. 83. f x x2 x 1
g is continuous on 0, 1 .
f is continuous on 0, 5 .
g0 2 > 0 and g1 1.9 < 0.
f 0 1 and f 5 29
By the Intermediate Value Theorem, gt 0 for at least one value c between 0 and 1. Using a graphing utility, we find that t 0.5636.
1 < 11 < 29 The Intermediate Value Theorem applies. x2 x 1 11 x2 x 12 0
x 4x 3 0 x 4 or x 3 c 3 (x 4 is not in the interval.) Thus, f 3 11. 85. f x x3 x2 x 2
87. (a) The limit does not exist at x c.
f is continuous on 0, 3 .
(b) The function is not defined at x c.
f 0 2 and f 3 19
(c) The limit exists at x c, but it is not equal to the value of the function at x c.
2 < 4 < 19 The Intermediate Value Theorem applies.
(d) The limit does not exist at x c.
x3 x2 x 2 4 x3 x2 x 6 0
x 2x2 x 3 0 x2 (x2
x 3 has no real solution.) c2
Thus, f 2 4. 89.
91. The functions agree for integer values of x:
y 5 4 3 2 1 −2 −1
gx 3 x 3 x 3 x f x 3 x 3 x x 1
3 4 5 6 7
−2 −3
However, for non-integer values of x, the functions differ by 1. f x 3 x gx 1 2 x.
The function is not continuous at x 3 because lim f x 1 0 lim f x.
x→3
for x an integer
x→3
1 1 For example, f 2 3 0 3, g2 3 1 4.
Section 1.4
t 2 2 t
Continuity and One-Sided Limits
N
93. Nt 25 2 t
0
1
1.8
2
3
3.8
Nt
50
25
5
50
25
5
Number of units
50 40 30 20 10
t 2
Discontinuous at every positive even integer.The company replenishes its inventory every two months.
4
6
8
10 12
Time (in months)
95. Let V 43 r 3 be the volume of a sphere of radius r. V1 43 4.19 V5 3 53 523.6 4
Since 4.19 < 275 < 523.6, the Intermediate Value Theorem implies that there is at least one value r between 1 and 5 such that Vr 275. (In fact, r 4.0341.) 97. Let c be any real number. Then lim f x does not exist since there are both rational and x→c
irrational numbers arbitrarily close to c. Therefore, f is not continuous at c. y
1, if x < 0 0, if x 0 99. sgnx 1, if x > 0
4 3 2 1
(a) lim sgnx 1
−4 −3 −2 −1
x→0
x 1
2
3
4
−2
(b) lim sgnx 1
−3
x→0
−4
(c) lim sgnx does not exist. x→0
101. True; if f x gx, x c, then lim f x lim gx and x→c
x→c
103. False; f 1 is not defined and lim f x does not exist. x→1
at least one of these limits (if they exist) does not equal the corresponding function at x c.
105. (a) f x
b
0 ≤ x < b b < x ≤ 2b
0
(b) gx
y
2b
x 2
0 ≤ x ≤ b
b
x 2
b < x ≤ 2b
y
2b
b
x b
b
2b
NOT continuous at x b.
x b
2b
Continuous on 0, 2b .
41
42
Chapter 1
107. f x
Limits and Their Properties
x c2 c
x
, c > 0
Domain: x c2 ≥ 0 ⇒ x ≥ c2 and x 0, c2, 0 0, x c2 c
lim
x
x→0
lim
x c2 c
x→0
x
x c2 c x c2 c
x c2 c2 1 1 lim x→0 xx c2 c x→0 x c2 c 2c
lim
Define f 0 12c to make f continuous at x 0. 109. hx xx
15
h has nonremovable discontinuities at x ± 1, ± 2, ± 3, . . . . −3
3 −3
Section 1.5 1.
Infinite Limits
lim 2
x x2 4
lim 2
x x2 4
x→2
x→2
3.
lim tan
x 4
lim tan
x 4
x→2
x→2
5. f x
1 x2 9
x
3.5
3.1
3.01
3.001
2.999
2.99
2.9
2.5
f x
0.308
1.639
16.64
166.6
166.7
16.69
1.695
0.364
lim f x
x→3
lim f x
x→3
7. f x
x2 x2 9
x
3.5
3.1
3.01
3.001
2.999
2.99
2.9
2.5
f x
3.769
15.75
150.8
1501
1499
149.3
14.25
2.273
lim f x
x→3
lim f x
x→3
Section 1.5
9. lim x→0
1 1 lim x2 x→0 x2
Therefore, x 0 is a vertical asymptote.
Infinite Limits
43
x2 2 x 2x 1
11. lim x→2
x2 2 x 2x 1
lim
x→2
Therefore, x 2 is a vertical asymptote. lim
x2 2 x 2x 1
lim
x2 2 x 2x 1
x→1
x→1
Therefore, x 1 is a vertical asymptote.
13.
lim
x→2
x2
x2 x2 and lim 2 x→2 x 4 4
15. No vertical asymptote since the denominator is never zero.
Therefore, x 2 is a vertical asymptote. lim
x→2
x2 x2 and lim 2 x→2 x 4 x2 4
Therefore, x 2 is a vertical asymptote.
17. f x tan 2x x
21.
lim
x→2
lim
x→2
sin 2x has vertical asymptotes at cos 2x
2n 1 n , n any integer. 4 4 2 x
x 2x 1
x x 2x 1
Therefore, x 2 is a vertical asymptote. lim
x x 2x 1
lim
x x 2x 1
x→1
x→1
19. lim 1 t→0
4 4 lim 1 2 t→0 t2 t
Therefore, t 0 is a vertical asymptote.
x3 1 x 1x2 x 1 x1 x1
23. f x
has no vertical asymptote since lim f x lim x2 x 1 3
x→1
x→1
Therefore, x 1 is a vertical asymptote.
25. f x
x 5x 3 x3 ,x5 x 5x2 1 x2 1
No vertical asymptotes. The graph has a hole at x 5.
27. st
t has vertical asymptotes at t n, n sin t
a nonzero integer. There is no vertical asymptote at t 0 since lim t→0
t 1. sin t
44
Chapter 1
Limits and Their Properties
x2 1 lim x 1 2 x→1 x 1 x→1
29. lim
31.
lim
x2 1 x1
lim
x2 1 x1
x→1
2
−3
x→1
3
8
−3
3
Vertical asymptote at x 1
−8
−5
Removable discontinuity at x 1
33. lim x→2
37.
x3 x2
lim
x→3
45. lim
x→
x→3
x2 2x 3 x1 4 lim x→3 x 2 x2 x 6 5
41. lim 1 x→0
35. lim
1 x
x2 x x 1 lim 2 x→1 x 1x 1 x→1 x 1 2
39. lim
43. lim x→0
x lim x sin x 0 csc x x→
x2 x 3x 3
47.
2
2 sin x
lim
x→ 12
x sec x and
lim
x→ 12
x sec x .
Therefore, lim x sec x does not exist. x→ 12
49. f x
x2 x 1 x3 1
lim f x lim
x→1
x→1
51. f x
1 x1
1 x2 25
lim f x
x→5
0.3
3
−8
−4
8
5
−0.3
−3
53. A limit in which f x increases or decreases without bound as x approaches c is called an infinite limit. is not a number. Rather, the symbol
55. One answer is f x
x3 x3 . x 6x 2 x2 4x 12
lim f x
x→c
says how the limit fails to exist. 57.
k , 0 < r < 1. Assume k 0. 1r k lim S lim (or if k < 0) r→1 r→1 1 r
59. S
y 3 2 1 −2
x
−1
1 −1 −2
3
Section 1.5
61. C
528x , 0 ≤ x < 100 100 x
63. (a) r
(a) C25 $176 million
(b) r
(b) C50 $528 million (c) C75 $1584 million
x
x→25
f x
1
0.5
0.2
0.1
0.01
0.001
0.0001
0.1585
0.0411
0.0067
0.0017
0
0
0
0.5
lim
x→0 − 1.5
x sin x 0 x
1.5
− 0.25
(b)
x f x
1
0.5
0.2
0.1
0.01
0.001
0.0001
0.1585
0.0823
0.0333
0.0167
0.0017
0
0
0.001
0.0001
0.1667 0.1667
0.1667
0.25
− 1.5
lim
1.5
x→0
x sin x 0 x2
− 0.25
(c)
x f x
1
0.5
0.2
0.1
0.1585
0.1646
0.1663
0.1666
0.01
0.25
− 1.5
lim
1.5
x→0
x sin x 0.1167 16 x3
− 0.25
(d)
x f x
1
0.5
0.2
0.1
0.1585
0.3292
0.8317
1.6658
0.01
0.001
0.0001
16.67
166.7
1667.0
1.5
− 1.5
1.5
− 1.5
For n ≥ 3, lim x→0
x sin x . xn
lim
x→0
x sin x x4
7 ftsec 12
215 3 ftsec 2 625 225
(c) lim
528 Thus, it is not possible. (d) lim x→100 100 x 65. (a)
27 625 49
2x 625 x2
Infinite Limits
45
46
Chapter 1
Limits and Their Properties (b) The direction of rotation is reversed.
67. (a) Because the circumference of the motor is half that of the saw arbor, the saw makes 17002 850 revolutions per minute.
(d)
(c) 220 cot 210 cot : straight sections. The angle subtended in each circle is
2 2 2
(e)
2 .
0.3
0.6
0.9
1.2
1.5
L
306.2
217.9
195.9
189.6
188.5
450
Thus, the length of the belt around the pulleys is 20 2 10 2 30 2 .
0
Total length 60 cot 30 2
(f)
0, 2
Domain:
2
0
lim
→ 2
L 60 188.5
(All the belts are around pulleys.) (g) lim L →0
71. False; let
69. False; for instance, let f x
1, f x x 3,
x2 1 or x1
x0 x 0.
The graph of f has a vertical asymptote at x 0, but f 0 3.
x gx 2 . x 1 73. Given lim f x and lim gx L: x→c
x→c
(2) Product:
If L > 0, then for L2 > 0 there exists 1 > 0 such that gx L < L2 whenever 0 < x c < 1. Thus, L2 < gx < 3L2. Since lim f x then for M > 0, there exists 2 > 0 such that f x > M2L whenever
x c
x→c
< 2. Let be the smaller of 1 and 2. Then for 0 < x c < , we have f xgx > M2LL2 M.
Therefore lim f xgx . The proof is similar for L < 0. x→c
(3) Quotient: Let > 0 be given.
There exists 1 > 0 such that f x > 3L2 whenever 0 < x c < 1 and there exists 2 > 0 such that gx L <
L2 whenever 0 < x c < 2. This inequality gives us L2 < gx < 3L2. Let be the smaller of 1 and 2. Then for 0 <
x c
gx f x
<
3L2 . 3L2
Therefore, lim
x→c
75. Given lim
x→c
< , we have
gx 0. f x
1 0. f x
Suppose lim f x exists and equals L. Then, x→c
lim 1 1 1 x→c 0. x→c f x lim f x L lim
x→c
This is not possible. Thus, lim f x does not exist. x→c
Review Exercises for Chapter 1
Review Exercises for Chapter 1 1. Calculus required. Using a graphing utility, you can estimate the length to be 8.3. Or, the length is slightly longer than the distance between the two points, 8.25. 3.
x
0.1
0.01 0.001
f x
0.26
0.25
0.250
1
0.001
0.01
0.1
0.2499
0.249
0.24
−1
lim f x 0.25
x→0
5. hx
1
−1
x2 2x x
7. lim 3 x 3 1 2
(a) lim hx 2
x→1
x→0
Let > 0 be given. Choose . Then for
(b) lim hx 3
0 < x 1 < , you have
x→1
x 1 < 1 x < 3 x 2 < f x L < 9. lim x2 3 1 x→2
1
x 2.
Let > 0 be given. We need x2 3 1 < ⇒ x2 4 x 2x 2 < ⇒ x 2 < Assuming, 1 < x < 3, you can choose 5. Hence, for 0 < x 2 < 5 you have
1
x 2 < 5 < x 2
x 2x 2 < x2 4 < x2 3 1 < f x L <
11. lim t 2 4 2 6 2.45
13. lim
t→2
t→4
15. lim
x→4
x 2
x4
lim
x→4
lim
x→4
x 2
17. lim
x 2x 2 1 x 2
1 4 2
x→0
1 4
x3 125 x 5x2 5x 25 lim x→5 x 5 x→5 x5
19. lim
lim x2 5x 25 x→5
75
21. lim
x→0
1 t2 1 lim t2 4 t→2 t 2 4
1x 1 1 1 x 1 lim x→0 x xx 1 1 1 lim x→0 x 1
1 cos x x lim x→0 sin x sin x
1 xcos x 10 0
47
48
Chapter 1
23. lim
x→0
Limits and Their Properties
sin 6 x 12 sin 6 cos x cos 6 sin x 12 lim x→0 x x 1 2
lim
x→0
0
3 sin x cos x 1 lim x x→0 2 x
3
2
1
3
2
25. lim f x gx 4 3 2 3
2
1
x→c
27. f x (a)
2x 1 3
x1
x
1.1
1.01
1.001
1.0001
f x
0.5680
0.5764
0.5773
0.5773
lim
2x 1 3
x1
x→1
(c) lim
2x 1 3
x→1
x1
Actual limit is 33.
0.577 lim
2x 1 3
x1
x→1
lim
2 2x 1 3
2 23
sa st 4.942 200 4.9t2 200 lim t→a t→4 at 4t t→4
2 0
1 3 3 3
29. lim
lim
−1
2x 1 3
2x 1 3 x 1 2x 1 3
x→1
2
2x 1 3
lim x→1
(b)
31. lim x→3
x 3 x3
lim
x→3
x 3 1 x3
4.9t 4t 4 4t
lim 4.9t 4 39.2 msec t→4
33. lim f x 0 x→2
35. lim ht does not exist because lim ht 1 1 2 and t→1
t→1
lim ht 121 1 1.
t→1
37. f x x 3 lim x 3 k 3 where k is an integer.
x→k
lim x 3 k 2 where k is an integer.
x→k
Nonremovable discontinuity at each integer k Continuous on k, k 1 for all integers k 41. f x lim
x→2
1 x 22
1 x 22
Nonremovable discontinuity at x 2 Continuous on , 2 2,
39. f x
3x2 x 2 3x 2x 1 x1 x1
lim f x lim 3x 2 5
x→1
x→1
Removable discontinuity at x 1 Continuous on , 1 1, 43. f x
3 x1
lim f x
x→1
lim f x
x→1
Nonremovable discontinuity at x 1 Continuous on , 1 1,
Problem Solving for Chapter 1 45. f x csc
x 2
49
47. f 2 5 Find c so that lim cx 6 5.
Nonremovable discontinuities at each even integer. Continuous on
x→2
c2 6 5
2k, 2k 2
2c 1
for all integers k. c
x2 4 x2 x 2 x2 x2
51. f x
49. f is continuous on 1, 2. f 1 1 < 0 and f 2 13 > 0. Therefore by the Intermediate Value Theorem, there is at least one value c in 1, 2 such that 2c3 3 0.
1 2
(a) lim f x 4 x→2
(b) lim f x 4 x→2
(c) lim f x does not exist. x→2
53. gx 1
2 x
55. f x
Vertical asymptote at x 10
Vertical asymptote at x 0
57.
lim
x→2
2x2 x 1 x2
61. lim
x2 2x 1 x1
65. lim
sin 4x 4 sin 4x lim x→0 5x 5 4x
69. C
80,000p , 0 ≤ 0 < 100 100 p
x→1
x→0
8 x 102
59.
lim
x→1
x1 1 1 lim x3 1 x→1 x2 x 1 3
63. lim x x→0
54
67. lim x→0
(a) C15 $14,117.65
(b) C50 $80.000
(c) C90 $720,000
(d)
lim
p→100
1 x3
csc 2x 1 lim x →0 x sin 2x x
80,000p 100 p
Problem Solving for Chapter 1 1. (a) Perimeter PAO x2 y 12 x2 y2 1
x2
x2
1
Perimeter PBO x 1 2
2
y2
x2
x2
x4
y2
1
1
x 12 x4 x2 x4 1
(c) lim r x x→0
101 2 1 101 2
(b) rx
x2 x2 12 x2 x4 1 x 12 x4 x2 x4 1
x
4
2
1
0.1
0.01
Perimeter PAO
33.02
9.08
3.41
2.10
2.01
Perimeter PBO
33.77
9.60
3.41
2.00
2.00
rx
0.98
0.95
1
1.05
1.005
50
Chapter 1
Limits and Their Properties
3. (a) There are 6 triangles, each with a central angle of 60 3. Hence,
5. (a) Slope
12bh 6 121 sin 3
Area hexagon 6
(b) Slope of tangent line is
3 3 2.598. 2
y 12 h = sin θ
h = sin 60°
y
1
1
mx Error:
3 3 0.5435. 2
12bh n 121 sin 2n n sin 22 n.
An
12
2.598
3
24
48
3.106
An
12 169 x2 12 169 x2
lim
x2 25 x 512 169 x2
lim
x 5 12 169 x2
3.139
Letting x 2 n,
144 169 x2 x 512 169 x2
x→5
(d) As n gets larger and larger, 2 n approaches 0.
12 169 x2 x5
lim
x→5
96
3.133
x→5
x→5
An n
6
5 169 x Tangent line 12 12
(d) lim mx lim
(b) There are n triangles, each with central angle of 2 n. Hence,
n
5 x 5 12
169 x2 12 x5
x→5
(c)
5 . 12
(c) Q x, y x, 169 x2
θ
60°
12 5
5 10 12 12 12
This is the same slope as part (b).
sin2 n sin2 n sin x 2 n 2 n x
which approaches 1 .
7. (a) 3 x1 3 ≥ 0
(b)
(c)
0.5
x1 3 ≥ 3
x ≥ 27
− 30
Domain: x ≥ 27, x 1 (d) lim f x lim x→1
12 − 0.1
3 x1 3 2
x1
x→1
3 x1 3 2 3 x1 3 2
3 x1 3 4 x→1 x 1 3 x1 3 2
lim lim
x→1 x1 3
lim
x→1
1
x2 3
x1 3 1 x1 3 1 3 x1 3 2
1 x2 3 x1 3 1 3 x1 3 2
1 1 1 1 12 2 12
9. (a) lim f x 3: g1, g4 x→2
(b) f continuous at 2: g1 (c) lim f x 3: g1, g3, g4 x→2
lim f x
x→27
3 271 3 2
27 1
2 1 0.0714 28 14
Problem Solving for Chapter 1 11.
y
13. (a)
y
4 3
2
2 1 −4 −3 −2 −1
x 1
2
3
4
1
−2 −3
x a
−4
(a)
f 1 1 1 1 1 0
(b) (i) lim Pa, bx 1 x→a
f 0 0
(ii) lim Pa, bx 0
f 12 0 1 1
(iii) lim Pa, bx 0
f 2.7 3 2 1 (b)
b
lim f x 1
x→1
lim f x 1
x→1
lim f x 1
x→1 2
(c) f is continuous for all real numbers except x 0, ± 1, ± 2, ± 3, . . .
x→a
x→b
(iv) lim Pa, bx 1 x→b
(c) Pa, b is continuous for all positive real numbers except x a, b. (d) The area under the graph of u, and above the x-axis, is 1.
51
C H A P T E R 2 Differentiation Section 2.1
The Derivative and the Tangent Line Problem . . . 53
Section 2.2
Basic Differentiation Rules and Rates of Change . 60
Section 2.3
The Product and Quotient Rules and Higher-Order Derivatives . . . . . . . . . . . . . . 67
Section 2.4
The Chain Rule . . . . . . . . . . . . . . . . . . . 73
Section 2.5
Implicit Differentiation . . . . . . . . . . . . . . . 79
Section 2.6
Related Rates . . . . . . . . . . . . . . . . . . . . 85
Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . 92
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . 98
C H A P T E R Differentiation Section 2.1
2
The Derivative and the Tangent Line Problem
Solutions to Odd-Numbered Exercises
1. (a) m 0
(c) y
3. (a), (b)
(b) m 3 y
f )4) 4
f )1) )x 1
f )1)
1)
x
1
y
f 4 f 1 x 1 f 1) 41 3 x 1 2 3
1x 1 2
6
f )4)
5
x1
5 )4, 5)
4
f )4)
f )1)
3
3 2
f )1) )1, 2)
2
1
x 1
5. f x 3 2x is a line. Slope 2
2
3
4
5
6
7. Slope at 1, 3 lim
x→0
g1 x g1 x
1 x2 4 3 x→0 x
lim
lim
x→0
1 2x x2 1 x
lim 2 2x 2 x→0
9. Slope at 0, 0 lim
t→0
f 0 t f 0 t
3t t2 0 t→0 t
lim
lim 3 t 3 t→0
11. f x 3 f x lim
x→0
lim
x→0
f x x f x x 33 x
lim 0 0 x→0
13. f x 5x fx lim
x→0
lim
x→0
15. hs 3 f x x f x x 5x x 5x x
lim 5 5 x→0
hs lim
2 s 3
s→0
hs s hs s
2 2 3 s s 3 s 3 3 lim s→0 s
2 s 3 2 lim s→0 s 3
53
54
Chapter 2
Differentiation
17. f x 2x2 x 1 f x x f x x
fx lim
x→0
lim
x→0
2x x2 x x 1 2x2 x 1 x
2x2 4xx 2x2 x x 1 2x2 x 1 x→0 x
lim
4xx 2x2 x lim 4x 2x 1 4x 1 x→0 x→0 x
lim
19. f x x3 12x f x x f x x
fx lim
x→0
lim
x→0
x x3 12x x x3 12x x
x3 3x2x 3xx2 x3 12x 12x x3 12x x→0 x
lim
3x2x 3xx2 x3 12x x→0 x
lim
lim 3x2 3xx x2 12 3x2 12 x→0
21. f x
1 x1
fx lim
x→0
f x x f x x
1 1 x x 1 x 1 lim x→0 x lim
x 1 x x 1 xx x 1x 1
lim
x xx x 1x 1
lim
1 x x 1x 1
x→0
x→0
x→0
1 x 12
23. f x x 1 fx lim
x→0
lim
x→0
lim
x→0
lim
x→0
f x x f x x x x 1
x
x 1
x x 1 x 1
xx x 1 x 1 1 x x 1 x 1
1 1 x 1 x 1 2x 1
x x 1 x 1 x x 1 x 1
Section 2.1
25. (a) f x x2 1 fx lim
x→0
The Derivative and the Tangent Line Problem
17. (b)
8
f x x f x x
(2, 5) −5
x x2 1 x2 1 lim x→0 x
5 −2
2xx x2 x→0 x
lim
lim 2x x 2x x→0
At 2, 5, the slope of the tangent line is m 22 4. The equation of the tangent line is y 5 4x 2 y 5 4x 8 y 4x 3. 27. (a) f x x3 fx lim
x→0
18. (b)
10
(2, 8)
f x x f x x
−5
5
x x3 x3 lim x→0 x
−4
3x2x 3xx2 x3 x→0 x
lim
lim 3x2 3xx x2 3x2 x→0
At 2, 8, the slope of the tangent is m 322 12. The equation of the tangent line is y 8 12x 2 y 12x 16. 29. (a) f x x fx lim
x→0
lim
18. (b) f x x f x x x x x
x
x→0
lim
x→0
lim
x→0
(1, 1)
x x x x x x
x x x x x x x 1 x x x
1 2x
At 1, 1, the slope of the tangent line is m
1 1 . 21 2
The equation of the tangent line is y1 y
1 x 1 2 1 1 x . 2 2
3
−1
5 −1
55
56
Chapter 2
31. (a) f x
Differentiation
(b)
4 x
(4, 5)
f x x f x fx lim x→0 x
lim
x x
x→0
− 12
4 4 x x x x x
xx xx x 4x x 2x x 4x x xxx x
lim
x3 2x 2x xx2 x3 x 2x 4x xxx x
x→0
12
−6
lim
x→0
10
x2x xx2 4x x→0 xxx x
lim
x2 xx 4 x→0 xx x
lim
x2 4 4 1 2 x2 x
At 4, 5, the slope of the tangent line is m1
4 3 16 4
The equation of the tangent line is 3 y 5 x 4 4 3 y x2 4 33. From Exercise 27 we know that fx 3x2. Since the slope of the given line is 3, we have
fx
3x2 3 x ± 1.
1 . 2xx
Since the slope of the given line is 12 , we have
Therefore, at the points 1, 1 and 1, 1 the tangent lines are parallel to 3x y 1 0. These lines have equations y 1 3x 1
35. Using the limit definition of derivative,
and
y 3x 2
y 1 3x 1 y 3x 2.
1 1 2xx 2 x 1.
Therefore, at the point 1, 1 the tangent line is parallel to x 2y 6 0. The equation of this line is 1 y 1 x 1 2 1 1 y1 x 2 2 1 3 y x . 2 2
37. g5 2 because the tangent line passes through 5, 2 g5
20 2 1 5 9 4 2
39. f x x ⇒ fx 1
b
Section 2.1
The Derivative and the Tangent Line Problem
43.
41. f x x ⇒ fx matches (a)
57
y 4
decreasing slope as x →
3 2 1 −4 −3 −2 −1 −1
x 1
2
3
4
−2 −3 −4
Answers will vary. Sample answer: y x 45. (a) If fc 3 and f is odd, then fc fc 3 (b) If fc 3 and f is even, then fc f c 3 47. Let x0, y0 be a point of tangency on the graph of f. By the limit definition for the derivative, fx 4 2x. The slope of the y line through 2, 5 and x0, y0 equals the derivative of f at x0: 7
5 y0 4 2x0 2 x0
6
(2, 5)
5 4
5 y0 2 x04 2x0
3 2
5 4x0 x02 8 8x0 2x02
(3, 3) (1, 3)
1 x
−2
0 x0 4x0 3 2
1
2
3
6
0 x0 1x0 3 ⇒ x0 1, 3 Therefore, the points of tangency are 1, 3 and 3, 3, and the corresponding slopes are 2 and 2. The equations of the tangent lines are y 5 2x 2
y 5 2x 2
y 2x 1
y 2x 9
49. (a) g0 3 (b) g3 0 (c) Because g1 3 , g is decreasing (falling) at x 1. 8
(d) Because g4 3 , g is increasing (rising) at x 4. 7
(e) Because g4 and g6 are both positive, g6 is greater than g4, and g6 g4 > 0. (f) No, it is not possible. All you can say is that g is decreasing (falling) at x 2. 1 51. f x 4 x3
2
By the limit definition of the derivative we have fx
3 2 4x . −2
2
1.5
1
0.5
0
0.5
1
1.5
2
f x
2
27 32
14
1 32
0
1 32
1 4
27 32
2
fx
3
27 16
3 4
3 16
0
3 16
3 4
27 16
3
x
2
−2
58
Chapter 2
Differentiation
55. f 2 24 2 4, f 2.1 2.14 2.1 3.99
f x 0.01 f x 0.01
53. gx
2x 0.01 x 0.012 2x x2 100
f2
3.99 4 0.1 Exact: f2 0 2.1 2
3
g f −2
4 −1
The graph of gx is approximately the graph of fx. 57. f x
1 x
and fx
1 . 2x3 2
5
As x → , f is nearly horizontal and thus f 0.
f −2
5
f′ −5
59. f x 4 x 32 Sx x
f 2 x f 2 x 2 f 2 x 4 2 x 32 3 1 x 12 x 2 3 x 2 3 x 2x 2 3 x x 1: Sx x 2 3 x 1
(a) x
5
S 0.1
3 3 x 2 3 x x 0.5: Sx 2 2 x 0.1: Sx
19 19 4 x 2 3 x 10 10 5
S1 −2
7
S 0.5 −1
(b) As x → 0, the line approaches the tangent line to f at 2, 3. 61. f x x2 1, c 2 f x f 2 x2 1 3 x 2x 2 lim lim lim x 2 4 x→2 x →2 x→2 x2 x2 x2
f2 lim
x→2
63. f x x3 2x2 1, c 2 f x f 2 x2x 2 x3 2x2 1 1 lim lim lim x2 4 x →2 x→2 x→2 x2 x2 x2
f2 lim
x→2
67. f x x 62 3, c 6
65. gx x , c 0 g0 lim
x→0
As x → 0 , As x → 0 ,
x gx g0 lim . Does not exist. x→0 x0 x
f6 lim
x→6
f x f 6 x6
1 →
lim
x 62 3 0 x6
lim
1 x 61 3
x
x
x
x
1
x
x
→
x→6
x→6
f
Does not exist.
Section 2.1
69. h x x 5, c 5 h5 lim
x→5
x 5 0
lim
x 5
x→5
x→5
71. f x is differentiable everywhere except at x 3. (Sharp turn in the graph.)
hx h5 x 5
lim
The Derivative and the Tangent Line Problem
x5
x5
Does not exist. 73. f x is differentiable everywhere except at x 1. (Discontinuity)
75. f x is differentiable everywhere except at x 3. (Sharp turn in the graph)
77. f x is differentiable on the interval 1, . (At x 1 the tangent line is vertical)
79. f x is differentiable everywhere except at x 0. (Discontinuity)
83. f x
81. f x x 1
The derivative from the left is lim
x→1
xx 11 ,, xx >≤ 11 3 2
The derivative from the left is
f x f 1 x1 0 lim 1. x→1 x1 x1
lim
x→1
f x f 1 x 13 0 lim x→1 x1 x1 lim x 12 0.
The derivative from the right is lim
x→1
x→1
f x f 1 x1 0 lim 1. x→1 x1 x1
The one-sided limits are not equal. Therefore, f is not differentiable at x 1.
The derivative from the right is lim
x→1
f x f 1 x 12 0 lim x→1 x1 x1 lim x 1 0. x→1
These one-sided limits are equal. Therefore, f is differentiable at x 1. f1 0 85. Note that f is continuous at x 2. f x
4xx 1,3, xx >≤ 22 2
f x f 2 x2 1 5 lim lim x 2 4. x→2 x→2 x2 x2
The derivative from the left is lim x→2
The derivative from the right is lim x→2
f x f 2 4x 3 5 lim lim 4 4. x→2 x→2 x2 x2
The one-sided limits are equal. Therefore, f is differentiable at x 2. f2 4 87. (a) The distance from 3, 1 to the line mx y 4 0 is d
y
Ax1 By1 C A2 B2
3
m3 11 4 3m 3 . m2 1
m2 1
2
1 x
(b)
1
5
2
3
4
The function d is not differentiable at m 1. This corresponds to the line y x 4, which passes through the point 3, 1.
−4
4 −1
59
60
Chapter 2
Differentiation
89. False. the slope is lim
x→0
f 2 x f 2 . x
91. False. If the derivative from the left of a point does not equal the derivative from the right of a point, then the derivative does not exist at that point. For example, if f x x , then the derivative from the left at x 0 is 1 and the derivative from the right at x 0 is 1. At x 0, the derivative does not exist.
93. f x
0,x sin1x,
x0 x0
Using the Squeeze Theorem, we have x ≤ x sin1x ≤ x , x 0. Thus, lim x sin1x 0 f 0 and f is continuous at x→0 x 0. Using the alternative form of the derivative we have lim
x→0
f x f 0 x sin1x 0 1 lim lim sin . x→0 x→0 x0 x0 x
Since this limit does not exist (it oscillates between 1 and 1), the function is not differentiable at x 0. gx
x0, sin1x, xx 00 2
Using the Squeeze Theorem again we have x2 ≤ x2 sin1x ≤ x2, x 0. Thus, lim x2 sin1x 0 f 0 and f is continux→0 ous at x 0. Using the alternative form of the derivative again we have lim
x→0
f x f 0 x2 sin1x 0 1 lim lim x sin 0. x→0 x→0 x0 x0 x
Therefore, g is differentiable at x 0, g0 0.
Section 2.2
Basic Differentiation Rules and Rates of Change
y x12
1. (a)
y 12 x12 y1 12 3. y 8 y 0
11. f x x 1 fx 1
19. y y
sin cos 2 cos sin 2
(b)
y x32
y x2
(c)
y 32 x12
(d)
y 2x
y1 32
y 3x2
y1 2
5. y x6
7. y
y 6x5
y1 3
1 x7 x7
y 7x8
5 x x15 9. y
1 1 y x45 45 5 5x
7 x8
15. gx x 2 4x3
13. f t 2t2 3t 6
17. st t3 2t 4
gx 2x 12x 2
fx 4t 3
21. y x2
1 cos x 2
y 2x
1 sin x 2
y x3
st 3t2 2
23. y
1 3 sin x x
y
1 3 cos x x2
Section 2.2
Function
Rewrite
Derivative
Simplify
5 y x2 2
y 5x3
y
5 x3
3 2x3
3 y x3 8
y
y
9 8x4
x
y x12
1 y x32 2
5 25. y 2 2x 27. y 29. y
Basic Differentiation Rules and Rates of Change
x
31. f x
3 3x2, 1, 3 x2
fx
6x3
9 4 x 8
y
1 7 33. f x x3, 2 5
6 3 x
1 2x32
0, 21
4x2 4x 1
21 fx x2 5
y 8x 4 y0 4
f0 0
f1 6
2 2 39. f x x 5 3x
37. f 4 sin , 0, 0
41. gt t2
fx 2x 6x3 2x
f 4 cos 1
6 x3
4 t2 4t3 t3
gt 2t 12t4 2t
f0 41 1 3 43. f x
y 2x 12, 0, 1
35.
x3 3x2 4 x 3 4x2 x2
45. y xx2 1 x3 x y 3x2 1
8 x3 8 fx 1 3 x x3 3 x x12 6x13 47. f x x 6
1 1 2 fx x12 2x23 2 2 x x23
49. hs s45 s23 4 2 4 2 h(s s45 s13 15 13 5 3 5s 3s
51. f x 6 x 5 cos x 6x12 5 cos x fx 3x12 5 sin x
3 x
5 sin x
55. (a) f x
53. (a) y x4 3x2 2 y 4x3 6x
fx
At 1, 0: y 413 61 2. y 0 2x 1
Tangent line:
3
−2
2x34
3 74 3 x 74 2 2x
At 1, 2, f1
2x y 2 0 (b)
2 4 3 x
Tangent line:
3 2 3 y 2 x 1 2 3 7 y x 2 2
2
(1, 0)
3x 2y 7 0
−1
(b)
5
(1, 2) −2
7 −1
12 t4
61
62
Chapter 2
Differentiation
59. y
57. y x4 8x2 2 y 4x3 16x
1 x2 x2
y 2x3
4xx2 4 4xx 2x 2
2 cannot equal zero. x3
Therefore, there are no horizontal tangents.
y 0 ⇒ x 0, ± 2 Horizontal tangents: 0, 2, 2, 14, 2, 14 61. y x sin x, 0 ≤ x < 2
63. x 2 kx 4x 9
y 1 cos x 0
2x k 4
cos x 1 ⇒ x
x 2 2x 4x 4x 9 ⇒ x 2 9 ⇒ x ± 3.
Horizontal tangent: ,
3 k x3 x 4
k 3 x2 4
Equate derivatives
Hence, k 2x 4 and
At x , y .
65.
Equate functions
For x 3, k 2 and for x 3, k 10.
Equate functions
Equate derivatives
3 2 x 3 2 3 3 4 3 3 x 3 ⇒ x x 3 ⇒ x 3 ⇒ x 2 ⇒ k 3. Hence, k x and 4 x 4 4 4 2 67. (a) The slope appears to be steepest between A and B.
(c)
y
(b) The average rate of change between A and B is greater than the instantaneous rate of change at B.
f B C A
D
E x
69. gx f x 6 ⇒ gx fx
y
71. 3
f f
1
x 3
2
1
1
2
3
2
If f is linear then its derivative is a constant function. f x ax b fx a
Section 2.2
Basic Differentiation Rules and Rates of Change
73. Let x1, y1 and x2, y2 be the points of tangency on y x2 and y x2 6x 5, respectively. The derivatives of these functions are y 2x ⇒ m 2x1
and
y 2x 6 ⇒ m 2x2 6.
m 2x1 2x2 6 x1 x2 3 Since y1 x12 and y2 x22 6x2 5,
y
5
m
y2 y1 x22 6x2 5 x12 2x2 6. x2 x1 x2 x1
4 3
)2, 3)
2
)1, 1)
1
x22 6x2 5 x2 32 2x2 6 x2 x2 3
x 2
3
)1, 0) 2
3
−1
x22 6x2 5 x22 6x2 9 2x2 62x2 3 2x22 12x2 14 4x22 18x2 18 2x22 6x2 4 0
y
5
2x2 2x2 1 0
4
)2, 4)
3
x2 1 or 2
2
x2 1 ⇒ y2 0, x1 2 and y1 4
1
x −1
Thus, the tangent line through 1, 0 and 2, 4 is
−2
40 y0 x 1 ⇒ y 4x 4. 21
x2 2 ⇒ y2 3, x1 1 and y1 1 Thus, the tangent line through 2, 3 and 1, 1 is y1
32 11x 1 ⇒ y 2x 1.
75. f x x, 4, 0
77. f1 1
1 1 fx x12 2 2 x 1 2 x
3.64
0y 4 x 0.77 3.33
4 x 2 x y 4 x 2 x x 4 x 2x x 4, y 2 The point 4, 2 is on the graph of f. Tangent line:
y2
02 x 4 4 4
4y 8 x 4 0 x 4y 4
1.24
63
64
Chapter 2
Differentiation
79. (a) One possible secant is between 3.9, 7.7019 and 4, 8: y8
20
8 7.7019 x 4 4 3.9
(4, 8) −2
y 8 2.981x 4
12 −2
y Sx 2.981x3.924 3 3 (b) fx x12 ⇒ f4 2 3 2 2 Tx 3x 4 8 3x 4 Sx is an approximation of the tangent line Tx. (c) As you move further away from 4, 8, the accuracy of the approximation T gets worse. 20
f T
−2
12
−2
(d)
x
3
f 4 x
1
T4 x
1
2
1
0.5
2.828
5.196
6.548
2
5
6.5
81. False. Let f x x2 and gx x2 4. Then fx gx 2x, but f x gx.
0.1
0
0.1
0.5
7.702
8
8.302
9.546
11.180
14.697
18.520
7.7
8
8.3
9.5
11
14
17
1
2
83. False. If y 2, then dydx 0. 2 is a constant.
85. True. If gx 3f x, then gx 3fx.
87. f t 2t 7, 1, 2 ft 2 Instantaneous rate of change is the constant 2. Average rate of change: f 2 f 1 22 7 21 7 2 21 1 (These are the same because f is a line of slope 2.)
3
1 89. f x , 1, 2 x fx
1 x2
Instantaneous rate of change:
1, 1 ⇒ f1 1
2, 21 ⇒ f2 41 Average rate of change: f 2 f 1 12 1 1 21 21 2
Section 2.2
Basic Differentiation Rules and Rates of Change
91. (a) st 16t2 1362
st 4.9t2 v0t s0
93.
4.9t2 120t
vt 32t (b)
65
vt 9.8t 120
s2 s1 1298 1346 48 ftsec 21
v5 9.85 120 71 msec
(c) vt st 32t
v10 9.810 120 22 msec
When t 1: v1 32 ftsec. When t 2: v2 64 ftsec. (d) 16t2 1362 0 t2 (e) v
1362 1362 9.226 sec ⇒ t 16 4
1362
4
32
1362
4
8 1362 295.242 ftsec 2 97. v 40 mph 3 mimin
v
23 mimin6 min 4 mi
Velocity (in mph)
60 50 40
v 0 mph 0 mimin
30
0 mimin2 min 0 mi
20 10
v 60 mph 1mimin
t 2
4
6
8
s
Distance (in miles)
95.
10
Time (in minutes)
10 8
(10, 6) 6
(6, 4) 4
(8, 4) 2
(0, 0)
t 2
1 mimin2 min 2 mi
4
6
8
(The velocity has been converted to miles per hour) (b) Using a graphing utility, you obtain
99. (a) Using a graphing utility, you obtain R 0.167v 0.02.
B 0.00586v2 0.0239v 0.46.
(c) T R B 0.00586v2 0.1431v 0.44
(d)
60
T
dT 0.01172v 0.1431 (e) dv
B R
For v 40, T40 0.612.
0
For v 80, T80 1.081.
(f) For increasing speeds, the total stopping distance increases.
For v 100, T100 1.315.
101. A s2,
dA 2s ds
100
0
103.
When s 4 m, dA 8 square meters per meter change in s. ds
C
1,008,000 6.3Q Q
dC 1,008,000 6.3 dQ Q2 C351 C350 5083.095 5085 $1.91 When Q 350,
dC $1.93. dQ
105. (a) f1.47 is the rate of change of the amount of gasoline sold when the price is $1.47 per gallon. (b) f1.47 is usually negative. As prices go up, sales go down.
10
Time (in minutes)
66
Chapter 2
Differentiation
107. y ax2 bx c Since the parabola passes through 0, 1 and 1, 0, we have
0, 1: 1 a02 b0 c ⇒ c 1 1, 0: 0 a12 b1 1 ⇒ b a 1. Thus, y ax2 a 1x 1. From the tangent line y x 1, we know that the derivative is 1 at the point 1, 0. y 2ax a 1 1 2a1 a 1 1a1 a2 b a 1 3 Therefore, y 2x2 3x 1. 109. y x3 9x y 3x2 9 Tangent lines through 1, 9: y 9 3x2 9x 1
x3 9x 9 3x3 3x2 9x 9 0 2x3 3x2 x22x 3 x 0 or x 32 3 81 3 9 The points of tangency are 0, 0 and 32 , 81 8 . At 0, 0 the slope is y0 9. At 2 , 8 the slope is y 2 4 .
Tangent lines: y 0 9x 0
9 3 y 81 8 4 x 2
and
y 94 x 27 4
y 9x 9x y 0
111. f x
9x 4y 27 0
xax , b, 3
2
x ≤ 2 x > 2
f must be continuous at x 2 to be differentiable at x 2. lim f x lim ax3 8a
x→2
x→2
lim f x lim x2 b 4 b
x→2
x→2
fx
3ax , 2x, 2
8a 4 b 8a 4 b
x < 2 x > 2
For f to be differentiable at x 2, the left derivative must equal the right derivative. 3a22 22 12a 4 a 13 4 b 8a 4 3
Section 2.3
The Product and Quotient Rules and Higher-Order Derivatives
67
113. Let f x cos x. f x x f x x
fx lim
x→0
lim
cos x cos x sin x sin x cos x x
lim
cos xcos x 1 sin x lim sin x x→0 x x
x→0
x→0
0 sin x1 sin x
Section 2.3
The Product and Quotient Rules and Higher-Order Derivatives
1. gx x 2 1x 2 2x gx x 2 12x 2 x 2 2x2x 2x3 2x 2 2x 2 2x3 4x 2 4x3 6x 2 2x 2
3 tt2 4 t13t2 4 3. ht
1 ht t132t t 2 4 t 23 3 2t 43 7t2 4 3t23
5. f x x 3 cos x
x x2 1
7. f x
fx x 3sin x cos x3x 2 3x cos x 2
9. hx
x3
sin x
3 x x13 3 x3 1 x 1
1 x 3 1 x23 x133x 2 3 hx x 3 12
1 x2 x 2 11 x2x 2 2 2 x 1 x 12
fx
11. gx
t2 4 3t23
sin x x2
gx
x 2cos x sin x2x x cos x 2 sin x x 22 x3
x3 1 x9x 2 3x23x 3 12 1 8x 3 x 3 12
3x23
13. f x x3 3x2x2 3x 5 fx x3 3x4x 3 2x2 3x 53x2 3 10x4 12x3 3x2 18x 15
15. f x fx
x2 4 x3
x 32x x 2 41 2x 2 6x x 2 4 x 32 x 32
f0 15
f1
f x x cos x
17.
fx xsin x cos x1 cos x x sin x f
4
2
2
2 2 4 4 2 8
164 1 1 32 4
x 2 6x 4 x 32
68
Chapter 2
Differentiation
Function
Rewrite
Derivative
Simplify 2x 2 3
19. y
x 2 2x 3
1 2 y x2 x 3 3
2 2 y x 3 3
y
21. y
7 3x3
y
7 3 x 3
y 7x4
y
23. y
4x32 x
y 4x, x > 0
y 2x12
y
25. f x fx
7 x4
2 x
3 2x x2 x2 1
x2 12 2x 3 2x x22x x2 12
2x2 4x 2 2x 12 2 x2 12 x 12
2 ,x1 x 12
27. f x x 1 fx 1
4 4x x x3 x3
x 34 4x1 x 2 6x 9 12 x 32 x 32
29. f x
2x 5 2x12 5x1 2 x
5 5 fx x12 x32 x32 x 2 2
x 2 6x 3 x 32
31. hs s3 22 s6 4s3 4 hs 6s5 12s2 6s2s3 2 1 2x 1 x 2x 1 33. f x x3 xx 3 x 2 3x 2
fx
x 2 3x2 2x 12x 3 2x 2 6x 4x 2 8x 3 x 2 3x2 x 2 3x2 2x 2 2x 3 2x 2 2x 3 2 x 2 3x2 x x 32
35. f x 3x3 4xx 5x 1 fx 9x2 4x 5x 1 3x3 4x1x 1 3x3 4xx 51 9x2 4x2 4x 5 3x 4 3x 3 4x 2 4x 3x 4 15x3 4x2 20x 9x 4 36x3 41x 2 16x 20 6x 4 12x 3 8x 2 16x 15x 4 48x 3 33x 2 32x 20
37. f x fx
x2 c2 x2 c2
x2 c22x x2 c22x x2 c22 4xc x2 c22 2
39. f x t2 sin t ft t2 cos t 2t sin t tt cos t 2 sin t
2x 5 2x 5 2x32 2xx
Section 2.3
41. f t ft
The Product and Quotient Rules and Higher-Order Derivatives
cos t t
43. f x x tan x
t sin t cos t t sin t cos t t2 t2
4 t 8 sec t t14 8 sec t 45. gt
1 1 gt t34 8 sec t tan t 34 8 sec t tan t 4 4t
fx 1 sec2 x tan2 x
47. y
31 sin x 3 sec x tan x 2 cos x 2
3 3 y sec x tan x sec2 x sec xtan x sec x 2 2 3 sec x tan x tan2 x 1 2 51. f x x2 tan x
49. y csc x sin x
fx x2 sec2 x 2x tan x
y csc x cot x cos x
xx sec2 x 2 tan x
cos x cos x sin2 x
cos xcsc2 x 1 cos x cot2 x 53. y 2x sin x x 2 cos x
55. gx
y 2x cos x 2 sin x x 2sin x 2x cos x 4x cos x 2 sin x x 2 sin x 57. g g
1 sin cos
sin 12
y y
61.
(form of answer may vary)
1 csc x 1 csc x
1 csc xcsc x cot x 1 csc xcsc x cot x 2 csc x cot x 1 csc x2 1 csc x2
6 212 2 3 43
2
ht ht h
2x2 8x 1 x 22
1 sin
y
59.
gx
xx 122x 5
sec t t tsec t tan t sec t1 t2 sec tt tan t 1 t2 sec tan 1 1 2 2
(form of answer may vary)
69
70
Chapter 2
Differentiation
63. (a) f x x3 3x 1x 2, 1, 3
53. (b)
10
fx x3 3x 11 x 23x2 3 − 10
4x3 6x2 6x 5
10
(1, − 3)
f1 1 slope at 1, 3.
− 10
Tangent line: y 3 1x 1 ⇒ y x 2 65. (a) f x
x , 2, 2 x1
51. (b)
fx
1 x 11 x1 x 12 x 12
f2
1 1 slope at 2, 2. 2 12
6
(2, 2) −3
6
−3
Tangent line: y 2 1x 2 ⇒ y x 4
4 , 1
f x tan x,
67. (a)
fx sec2 x f
55. (b)
4
( ( π ,1 4
−
4 2 slope at 4 , 1.
−4
Tangent line:
y12 x y 1 2x
4
2
4x 2y 2 0
69. f x fx
x2 x1
6 x 23 3x1 x 22 x 22
71. fx
x 12x x21 x 12
gx
x2 2x xx 2 x 12 x 12
gx
fx 0 when x 0 or x 2.
6 x 25 5x 41 x 22 x 22
5x 4 3x 2x 4 f x 2 x 2 x 2 x 2
f and g differ by a constant.
Horizontal tangents are at 0, 0 and 2, 4. 73. f x x n sin x fx x n cos x nx n1 sin x x
n1
x cos x n sin x
When n 1: fx x cos x sin x.
75. Area At 2t 1t 2t32 t12 At 2
When n 4: fx x3x cos x 4 sin x. For general n, fx x n1 x cos x n sin x.
12
3t12
When n 2: fx xx cos x 2 sin x. When n 3: fx x2x cos x 3 sin x.
32t 21t
12
1 12 t 2
6t 1 2 cm sec 2t
Section 2.3
77.
C 100
x , 200 x x 30 2
The Product and Quotient Rules and Higher-Order Derivatives
1 ≤ x
dC 400 30 100 3 dx x x 302
79.
Pt 500 1
4t 50 t2
50 t24 4t2t 50 t22
Pt 500
200 4t 2
50 t
(a) When x 10:
dC $38.13. dx
500
(b) When x 15:
dC $10.37. dx
2000
(c) When x 20:
dC $3.80. dx
2 2
50 t2
50 t 2 2
P2 31.55 bacteria per hour
As the order size increases, the cost per item decreases. 1 cos x
sec x
81. (a)
d d 1 cos x0 1sin x sin x 1 sec x dx dx cos x cos x2 cos x cos x cos x
d d 1 sin x0 1cos x cos x 1 csc x dx dx sin x sin x2 sin x sin x sin x
cot x
(c)
sin x
cos x sec x tan x
1 sin x
csc x
(b)
cos x csc x cot x sin x
cos x sin x
d d cos x sin xsin x cos xcos x sin2 x cos2 x 1 2 csc2 x cot x 2 dx dx sin x sin x sin2 x sin x
85. f x
83. f x 4x32 fx 6x12 f x 3x12
3
fx
1 x 11 x1 x 12 x 12
f x
2 x 13
x
87. f x 3 sin x
89. fx x2
fx 3 cos x
f x 2x
f x 3 sin x
93.
95. f x 2gx hx
y 4
fx 2gx hx
3
f2 2g2 h2
2
22 4
1 x 1
2
3
4
f 2 0 One such function is f x x 22.
0
x x1
91. f x 2x f 4x
97. f x
1 1 2x12 2 x
gx hx
fx
hxgx gxhx hx 2
f2
h2g2 g2h2 h2 2
12 34 12
10
71
72
Chapter 2
Differentiation
101. vt 36 t2, 0 ≤ t ≤ 6
y
99. f′
at 2t
2
f
1
v3 27 msec x
2
1
1
a3 6 msec
2
The speed of the object is decreasing. f″
It appears that f is cubic; so f would be quadratic and f would be linear. 103. vt at
100t 2t 15
(a) a5
2t 15100 100t2 2t 152
(b) a10
1500
1.2 ftsec2 210 15 2
1500 2t 152
(c) a20
1500
0.5 ftsec2 220 15 2
1500 2.4 ftsec2 25 15 2
105. f x gxhx fx gxhx hxgx
(a)
f x gxh x gxhx hxg x hxgx gxh x 2gxhx hxg x f x gxh x gxh x 2gxh x 2g xhx hxg x hxg x gxh x 3gx h x 3g xhx g xhx f
4x
gxh4x gxh x 3gxh x 3g xh x 3g xh x 3g xhx g xhx g4xhx gxh4x 4gxh x 6g xh x 4g xhx g4xhx
(b) f nx gxhnx
nn 1n 2 . . . 21 nn 1n 2 . . . 21 gxhn1x g xhn2x 1 n 1n 2 . . . 21 21 n 2n 3 . . . 21
nn 1n 2 . . . 21 g xhn3x . . . 321 n 3n 4 . . . 21
nn 1n 2 . . . 21 n1 g xhx gnxhx n 1n 2 . . . 21 1
gxhnx
n! n! gxhn1x g xhn2x . . . 1!n 1! 2!n 2! n! gn1xhx gnxhx n 1!1!
Note: n! nn 1 . . . 3
2 1 (read “n factorial.”)
Section 2.4
f
3 cos 3 12
fx sin x
f
3 sin 3 23
f x cos x
f
3 cos 3 21
107. f x cos x
(a) P1x fax a f a
3
2
x 3 21
(b)
1 x 4 3
2
3
2
2
P2
1 P2x f ax a2 f ax a f a 2
The Chain Rule
− 2
x 3 21
−2
(c) P2 is a better approximation.
109. False. If y f xgx, then
P1
f
(d) The accuracy worsens as you move farther away from x a 3. 111. True
dy f xgx gxfx. dx
113. True
hc f cgc gcfc f c0 gc0 0
115. f x x x
x , if x ≥ 0 x , if x < 0 2
2
2x, if x ≥ 0 2 x
2x, if x < 0 2, if x > 0 f x 2, if x < 0 fx
f 0 does not exist since the left and right derivatives are not equal.
Section 2.4
The Chain Rule
y f gx
u gx
y f u
1. y 6x 54
u 6x 5
y u4
3. y x2 1
u x2 1
y u
5. y csc3 x
u csc x
y u3
7. y 2x 73 y 32x 722 62x 72 11. f x 9 x223 2 4x fx 9 x2132x 3 39 x213
9. gx 34 9x4 gx 124 9x39 1084 9x3 13. f t 1 t12 1 1 ft 1 t121 2 21 t
73
74
Chapter 2
Differentiation
17. y 24 x 214
15. y 9x2 413 1 6x y 9x2 42318x 3 9x2 423
y 2
2 34
2x
x
4 4 x 23
21. f t t 32
19. y x 21 y 12 x21 23.
144 x
1 x 22
ft 2t 33
2 t 33
25. f x x2x 24
y x 212
fx x2 4x 231 x 242x
dy 1 1 x 232 dx 2 2x 232
2xx 23 2x x 2 2xx 233x 2
27. y x1 x2 x1 x212
121 x
y x
2 12
29. y
2x 1 x2121
x21 x212 1 x212 1 x
2
xx 52
x2 132 x2 x2 1
35. y y
1 x2 132
2
33. f v
2
gx 2
x5 x2 2
x 2 2 x 52x x 2 22
x2x2 132 x2 112
2
1 2x2 1 x2 31. gx
xx2 112
1 y x x2 1322x x2 1121 2
x 1 x
2 12
x x2 1
112vv
fv 3
2x 52 10x x 2 x 2 23
3
1 2v 1v
1 2v 1 v2 1 v 2
2
91 2v2 1 v4
3t 2 37. gt 2 t 2t 1
x 1
x 1 2
1 3x2 4x32 2xx2 12
gt
The zero of y corresponds to the point on the graph of y where the tangent line is horizontal.
3tt2 3t 2 t2 2t 132
The zeros of g correspond to the points on the graph of g where the tangent lines are horizontal. 24
2
y −1
g′
5
y′ −2
g
−5
3 −2
Section 2.4
x x 1
39. y
y
41. st
x 1x
st
2xx 1
The Chain Rule
22 t1 t 3 t 1 t
The zero of st corresponds to the point on the graph of st where the tangent line is horizontal.
y has no zeros. 4
3
y s′
−5
4
−3
6
y′ s −2 −3
43.
y
cos x 1 x
3
y
dy x sin x cos x 1 dx x2
−5
5
y′
x sin x cos x 1 x2
−3
The zeros of y correspond to the points on the graph of y where the tangent lines are horizontal. 45. (a)
(b) y sin 2x
y sin x y cos x
y 2 cos 2x
y0 1
y0 2
1 cycle in 0, 2
2 cycles in 0, 2 The slope of sin ax at the origin is a.
47.
49. gx 3 tan 4x
y cos 3x
gx 12 sec2 4x
dy 3 sin 3x dx 51. y sin x2 sin 2 x 2 y cos x2 2 2x 2 2x cos 2x 2
55. f x
53. hx sin 2x cos 2x hx sin 2x2 sin 2x cos 2x2 cos 2x 2
cos2
75
2x 2
sin2
2x
2 cos 4x. Alternate solution:
fx
hx
1 sin 4x 2
hx
1 cos 4x4 2 cos 4x 2
cos x cot x 2 sin x sin x sin2 xsin x cos x2 sin x cos x sin4 x sin2 x 2 cos2 x 1 cos2 x sin3 x sin3 x
76
Chapter 2
Differentiation
57. y 4 sec2 x y 8 sec x sec x tan x 8 sec2 x tan x
59. f 14 sin2 2 14 sin 2 2 f 2 14 sin 2 cos 2 2 sin 2 cos 2 12 sin 4
61. f x 3 sec2 t 1
63.
ft 6 sec t 1 sec t 1 tan t 1 6 sin t 1 6 sec t 1 tan t 1 cos3 t 1
y x
1 sin2x 2 4
x
1 sin4x2 4
2
dy 1 12 1 x cos4x28x dx 2 4 1 2x cos2x2 2x
65.
y sincos x
67.
dy coscos x sin x dx
st t 2 2t 812, 2, 4 st
sin x coscos x
s2
69.
f x
3 3x3 41, x3 4
fx 3x3 423x2 f1
1, 53 9x2 x3 42
71. f t ft
1 2 t 2t 8122t 2 2 t1 t 2 2t 8
3 4 3t 2 , 0, 2 t1 5 t 13 3t 21 t 12 t 12
f0 5
9 25
y 37 sec32x, 0, 36
73.
y 3 sec22x 2 sec(2x tan2x 6 sec32x tan2x y0 0 75. (a) f x 3x2 2 , 3, 5
77. (a) f x sin 2x, , 0 fx 2 cos 2x
1 fx 3x2 2126x 2
f 2
3x 3x2 2 f3
Tangent line: y 2x ⇒ 2x y 2 0
9 5
63. (b)
2
Tangent line: 9 y 5 x 3 ⇒ 9x 5y 2 0 5 (b)
−2
7
(3, 5) −5
5
−3
0
(π , 0)
2
Section 2.4
79. f x 2x2 13
The Chain Rule
77
81. f x sin x 2
fx 6x2 122x
fx 2x cos x 2
12xx 4 2x2 1
f x 2x 2xsin x2 2 cos x2
12x5 24x3 12x
2 cos x2 2x2 sin x2
f x 60x 4 72x2 12 125x2 1x2 1
83.
y
85.
y
f′
3
3
2
2
f
1
x
x 2
2
3
3
1
2
f 2
f′
3
The zeros of f correspond to the points where the graph of f has horizontal tangents.
The zeros of f correspond to the points where the graph of f has horizontal tangents.
87. gx f 3x gx f3x3 ⇒ gx 3 f3x 89. (a) f x gxhx
73. (b) f x ghx
fx gxhx gxhx
fx ghxhx
f5 32 63 24
f5 g32 2g3 Need g3 to find f5.
73. (c) f x
gx hx
fx
hxgx gxhx hx 2
f5
36 32 12 4 32 9 3
73. (d) f x gx 3 f x 3 gx 2gx f5 3326 162
(b) f 132,400331 v1
91. (a) f 132,400331 v1
f 1132,400331 v21
f 1132,400331 v21
132,400 331 v 2
When v 30, f 1.016.
When v 30, f 1.461. 93. 0.2 cos 8t The maximum angular displacement is 0.2 (since 1 ≤ cos 8t ≤ 1. d
0.2 8 sin 8t 1.6 sin 8t dt When t 3, d dt 1.6 sin 24 1.4489 radians per second.
132,400 331 v 2
95.
S CR 2 r 2
dR dr dS C 2R 2r dt dt dt
Since r is constant, we have drdt 0 and dS 1.76 10521.2 102105 dt 4.224 102 0.04224.
78
Chapter 2
Differentiation
97. (a) x 1.6372t3 19.3120t2 0.5082t 0.6161 (b) C 60x 1350 601.6372t3 19.3120t2 0.5082t 0.6161 1350 dC 604.9116t2 38.624t 0.5082 dt 294.696t2 2317.44t 30.492 The function
dC is quadratic, not linear. The cost function levels off at the end of the day, perhaps due to fatigue. dt
99. f x sin x
101. (a) rx fgxgx r1 fg1g1
(a) fx cos x f x 2 sin x
Note that g1 4 and f4
f x 3 cos x
Also, g1 0. Thus, r1 0
f 4 4 sin x (b) f x 2 f x 2 sin x 2sin x 0 f 2kx 1k 2k sin x
(c)
f 2k1x 1k1 2k1 cos x
(b) sx gf xf x s4 gf 4f 4 5 64 1 5 and Note that f 4 , g 2 2 62 2 5 f4 . 4
Thus, s4
103.
g xx n
2x n 2x2 nx
2x n2 xx n
x x n 2 xx n
a g
107. hx x cos x
2x 3 , gx 2 2x 3
dg 1 2 x nx122x n dx 2
hx x sin x
1 5 5 . 2 4 8
105. gx 2x 3
nx
x2
x
x
50 5 62 4
cos x,
x 0
x
3 2
Section 2.5
109. (a) f x tan
x 4
f 1 1
f x
x sec2 4 4
f x
x sec2 2 4
tan
x 4 4
P1x f1x 1 f 1 P2x (b)
Implicit Differentiation
f1
2 4 2
f 1
21 8 4
x 1 1. 2
1 x 12 f1x 1 f 1 x 12 x 1 1 2 4 8 2
2
P1 P2 f 0
3 0
(c) P2 is a better approximation than P1 (d) The accuracy worsens as you move away from x c 1. 111. False. If y 1 x12, then y 12 1 x121.
Section 2.5 1.
113. True
Implicit Differentiation
x2 y2 36
3.
2x 2yy 0 y
1 12 1 12 x y y 0 2 2
x y
y
x3 xy y2 4
5.
7.
3x2 xy y 2yy 0
y
3x 2 3x 2y 6xy 4xyy 2y2 0
4xy 3x 2y 6xy 3x 2 2y 2 y1
6xy 3x 4xy 3x 2
2y 2
yx
x3y3 y x 0
3x3y2 1y 1 3x2y3
y 3x2 2y x
9. x3 3x 2 2xy 2 12
x12 y12
3x3y2y 3x2y3 y 1 0
2y xy y 3x2
2
x12 y12 9
y
11.
1 3x2y3 3x3y2 1
sin x 2cos 2y 1 cos x 4sin 2yy 0 y
cos x 4 sin 2y
79
80
Chapter 2
Differentiation
13. sin x x1 tan y cos x x
sec2
y
y sinxy
15.
yy 1 tan y1
y xy y cosxy y x cosxyy y cosxy
cos x tan y 1 x sec2 y
y 17. (a) x2 y2 16 y2
16
(b)
y
x2
6
y ± 16 x2
(c) Explicitly:
6
x
y=−
16 − x 2
2x 2yy 0 y
x x x 2 y 16 x2 ± 16 x 19. (a) 16y2 144 9x2
(b)
x y
y 6
1 9 144 9x2 16 x2 16 16
4
y = 34
16 − x2
2
3 y ± 16 x2 4
−6
−2
x 2
6
−4 −6
y = − 43
16 − x2
(d) Implicitly:
(c) Explicitly:
18x 32yy 0
dy 3 ± 16 x2122x dx 8
y
3x 3x 9x 416 x2 443y 16y
xy 4
23.
xy y1 0 x y y y x
At 4, 1: y
2
(d) Implicitly:
dy 1 ± 16 x2122x dx 2
y
16 − x 2
−2
−6
21.
y=
2 −6
y2
y cosxy 1 x cosxy
1 4
y2
9x 16y
x2 4 x2 4
2yy
x 2 42x x 2 42x x 2 42
2yy
16x x 2 42
y
8x yx 2 42
At 2, 0, y is undefined.
Section 2.5
25.
x23 y23 5
1 y sec2x y 1 1 sec2x y y sec2x y tan2x y sin2x y tan2x y 1 x2 2 x 1
yx
x13 y13
81
tanx y x
27.
2 13 2 13 x y y 0 3 3 y
Implicit Differentiation
3
1 At 8, 1: y . 2
At 0, 0: y 0.
29.
x2 4y 8
x2 y22 4x2y
31.
2x2 y22x 2yy 4x2y y8x
x2 4y y2x 0 y
4x3 4x2yy 4xy2 4y3y 4x2y 8xy
2xy x2 4
4x2yy 4y3y 4x2y 8xy 4x3 4xy2
2x8x2 4 x2 4
4yx2y y3 x2 42xy x3 xy2 y
16x 2 x 42 At 2, 1: y
At 1, 1: y 0.
32 1 64 2
Or, you could just solve for y: y x
2
33.
8 4
tan y x
35.
ysec2 y 1
x 2 y 2 36 2x 2yy 0
1 y cos2 y, < y < sec2 y 2 2
y
sec2 y 1 tan2 y 1 x2
x2 y2 16 2x 2yy 0 y
x y
x yy 0 1 yy y2 0
xy
39.
y
3x2 3x2 2y 2y
y
2x3y 3y2 4x2
y2 y3y x2
y
y2 x2 16 3 y3 y
y2
y 2 x 2 36 3 y3 y
2yy 3x2
2
yx
y x
y2 x3
0
1 yy
x y
y1 xy y y2
1 y 1 x2 37.
2xy x3 xy2 x2y y3 x2
2x3
xy
3y
3y2x 6y 4x2
3y 3x 4x2 4y
x3
3y
xy 2x y2 2x
82
Chapter 2
Differentiation
41. x y 4
9
1 12 1 12 x y y 0 2 2 y
(9, 1)
y x
−1
14 −1
At 9, 1, y
1 3
1 Tangent line: y 1 x 9 3 1 y x4 3 x 3y 12 0 43. x2 y2 25 y
x y
At 4, 3:
6
Tangent line: y 3
4 x 4 ⇒ 4x 3y 25 0 3
(4, 3) −9
9
3 Normal line: y 3 x 4 ⇒ 3x 4y 0. 4
−6
At 3, 4:
6
3 Tangent line: y 4 x 3 ⇒ 3x 4y 25 0 4 Normal line: y 4
45.
4 x 3 ⇒ 4x 3y 0. 3
(−3, 4) −9
9
−6
x2 y2 r 2 2x 2yy 0 y
x slope of tangent line y
y slope of normal line x Let x0, y0 be a point on the circle. If x0 0, then the tangent line is horizontal, the normal line is vertical and, hence, passes through the origin. If x0 0, then the equation of the normal line is y y0 y
y0 x x0 x0 y0 x x0
which passes through the origin.
Section 2.5
47. 25x2 16y2 200x 160y 400 0
Implicit Differentiation
y
(− 4, 10)
50x 32yy 200 160y 0 y
200 50x 160 32y
2516
6
(− 8, 5)
(0, 5) 4
(− 4, 0)
Horizontal tangents occur when x 4: 16y2
10
x
−10 − 8 − 6 − 4
2004 160y 400 0
−2
2
y y 10 0 ⇒ y 0, 10 Horizontal tangents: 4, 0, 4, 10. Vertical tangents occur when y 5: 25x2 400 200x 800 400 0 25xx 8 0 ⇒ x 0, 8 Vertical tangents: 0, 5, 8, 5. 49. Find the points of intersection by letting y2 4x in the equation 2x2 y2 6. 2x2 4x 6
x 3x 1 0
and
The curves intersect at 1, ± 2.
y 2 = 4x
Parabola:
Ellipse: 4x 2yy 0 y
4
(1, 2) −6
2yy 4
2x y
y
6
(1, − 2) 2x 2 + y 2 = 6
2 y
−4
At 1, 2, the slopes are: y 1
y 1.
At 1, 2, the slopes are: y 1
y 1.
Tangents are perpendicular. 51. y x and x sin y
4
x = sin y
Point of intersection: 0, 0
−6
y x:
x sin y:
y 1
1 y cos y
6
(0, 0)
−4
x+y=0
y sec y At 0, 0, the slopes are: y 1
y 1.
Tangents are perpendicular. 53.
xy C
x2 y2 K
xy y 0
2x 2yy 0
y
y x
y
x y
At any point of intersection x, y the product of the slopes is yxxy 1. The curves are orthogonal.
2
2
C=4 −3
3
C=1 K = −1 −2
−3
3
K=2 −2
83
84
Chapter 2
Differentiation
55. 2y2 3x4 0 (a) 4yy 12x3 0
(b) 4y
4yy 12x3
dx dy 12x3 0 dt dt y
12x3 3x3 y 4y y
dx dy 3x3 dt dt
57. cos y 3 sin x 1 (b) sin y
(a) sin yy 3 cos x 0 y
3 cos x sin y
dy dx 3 cos x 0 dt dt dy dx sin y 3 cos x dt dt
59. A function is in explicit form if y is written as a function of x: y f x. For example, y x3. An implicit equation is not in the form y f x. For example, x 2 y 2 5. 61. (a) x4 44x2 y2
10
4y2 16x2 x4
− 10
10
1 y2 4x2 x4 4 y±
(b)
− 10
1 4x2 x4 4
1 y 3 ⇒ 9 4x 2 x4 4 36 16x2 x4 x4 16x2 36 0 x2
16 ± 256 144 8 ± 28 2
Note that x2 8 ± 28 8 ± 27 1 ± 7 2. Hence, there are four values of x: 1 7, 1 7, 17, 1 7 To find the slope, 2yy 8x x3 ⇒ y
x8 x2 . 23
1 For x 1 7, y 3 7 7, and the line is 1 1 y1 37 7x 1 7 3 37 7x 87 23 . 1 For x 1 7, y 3 7 7, and the line is
y2 137 7x 1 7 3 137 7x 23 87 . 1 For x 1 7, y 3 7 7, and the line is
y3 137 7x 1 7 3 137 7x 23 87 . 1 For x 1 7, y 3 7 7, and the line is 1 1 y4 37 7x 1 7 3 37 7x 87 23 .
—CONTINUED—
10
− 10
10
y1
y3
y2 − 10
y4
Section 2.6
Related Rates
61. —CONTINUED— (c) Equating y3 and y4,
1 7 7 x 1 7 3 13 7 7 x 1 7 3 3
7 7 x 1 7 7 7 x 1 7 7x 7 7 7x 7 77 7x 7 7 7x 7 77
167 14x x If x
87 7
87 87 , then y 5 and the lines intersect at ,5 . 7 7
63. Let f x xn xpq, where p and q are nonzero integers and q > 0. First consider the case where p 1. The derivative of f x x1q is given by d 1q f x x f x f t f x x lim lim x→0 t→x dx x tx where t x x. Observe that f t f x t1q x1q t1q x1q 1q q tx tx t x1qq
t1q
x1q
t11q
t1q x1q . . . t1qx12q x11q
t12qx1q
1 . t11q t12qx1q . . . t1qx12q x11q
As t → x, the denominator approaches qx11q. That is, d 1q 1 1 x 11q x1q1. dx qx q Now consider f x xpq xp1q. From the Chain Rule,
1 1 d p p p fx xp1q1 xp xp1q 1pxp1 xpqp p1 xpq 1 nxn1 n . q dx q q q q
Section 2.6 1.
Related Rates
y x
xy 4
3.
1 dx dy dt 2x dt
x
dx dy y 0 dt dt
dx dy 2x dt dt
y dx dy dt x dt
(a) When x 4 and dxdt 3,
dx x dy dt y dt
dy 1 3 3 . dt 24 4 (b) When x 25 and dydt 2, dx 225 2 20. dt
(a) When x 8, y 12, and dxdt 10, dy 12 5 10 . dt 8 8 (b) When x 1, y 4, and dydt 6, 1 3 dx 6 . dt 4 2
85
86
5.
Chapter 2
Differentiation
y x2 1
7.
y tan x
dx 2 dt
dx 2 dt
dy dx 2x dt dt
dy dx sec2 x dt dt
(a) When x 1,
(a) When x 3,
dy 212 4 cmsec. dt (b) When x 0,
dy 222 8 cmsec. dt (b) When x 4,
dy 202 0 cmsec. dt (c) When x 1, dy 212 4 cmsec. dt 9. (a)
dy dx negative ⇒ positive dt dt
(b)
dy dx positive ⇒ negative dt dt
13.
dy 2 22 4 cmsec. dt (c) When x 0, dy 122 2 cmsec. dt dy dx a . 11. Yes, y changes at a constant rate: dt dt No, the rate
dy dx is a multiple of . dt dt
D x2 y2 x2 x2 12 x4 3x2 1 dx 2 dt dx 2x3 3x dx 4x3 6x dD 1 4 x 3x2 1 124x3 6x 4 2 dt 2 dt x 3x 1 dt x4 3x2 1
15.
A r2
17. (a) sin
dr 3 dt
cos
12b ⇒ b 2s sin 2 s 2 h ⇒ h s cos 2 s 2 A
dA dr 2 r dt dt (a) When r 6,
1 1 bh 2s sin 2 2 2
s cos 2
s2 s2 2 sin cos sin 2 2 2 2
dA 2 63 36 cm2min. dt θ
(b) When r 24,
s
s h
dA 2 243 144 cm2min. dt
b
(b)
dA s2 d d 1 cos where radmin. dt 2 dt dt 2 When
12
dA s2 3 , 6 dt 2 2
3s 2
8
dA s2 1 1 s2 , 3 dt 2 2 2 8 (c) If d dt is constant, dAdt is proportional to cos . When
Section 2.6
dV 4 800 V r 3, 3 dt
19.
21.
ds dx 12x dt dt
dr 1 dV 1 800 dt 4 r 2 dt 4 r 2 (a) When r 30,
dr 1 2 800 cmmin. dt 4 302 9
(b) When r 60,
dr 1 1 800 cmmin. dt 4 60 2 18
1 1 9 V r 2h h2 h 3 3 4
s 6x2 dx 3 dt
dV dr 4 r 2 dt dt
23.
(a) When x 1, ds 1213 36 cm2sec. dt (b) When x 10, ds 12103 360 cm2sec. dt
since 2r 3h
3 3 h 4
h
dV 10 dt
r
dh 4dVdt dV 9 2 dh h ⇒ dt 4 dt dt 9h2 When h 15, 25.
8 410 dh ftmin. dt 9 152 405 12
6
Related Rates
1
3 1
(a) Total volume of pool
1 2126 1612 144 m3 2
Volume of 1m. of water
1 166 18 m3 2
2 h=1
(see similar triangle diagram) 18 % pool filled 144 100% 12.5%
(b) Since for 0 ≤ h ≤ 2, b 6h, you have 1 V bh6 3bh 36hh 18h2 2 dV dh 1 dh 1 1 1 36h ⇒ mmin. dt dt 4 dt 144h 1441 144
12 b=6
87
88
Chapter 2
Differentiation
x2 y2 252
27. 2x
dx dy 2y 0 dt dt dy x dt y
dx dx 2x since 2. dt y dt
25
y
x
(a) When x 7, y 576 24,
dy 27 7 ftsec. dt 24 12
When x 15, y 400 20, When x 24, y 7,
dy 215 3 ftsec. dt 20 2
dy 224 48 ftsec. dt 7 7
(b)
1 A xy 2
dx dA 1 dy x y dt 2 dt dt
From part (a) we have x 7, y 24, and
dx 2, dt
dy 7 . dt 12
7 dA 1 Thus, dt 2 7 12 242 527 21.96 ft2sec. 24 tan
(c)
sec2
x y
d 1 dt y
dx x 2 dt y
d 1 cos2 dt y
Using x 7, y 24,
dy dt
dx x 2 dt y
θ
dy dt
x
d dx dy 7 24 24 2, and cos , we have dt dt 12 25 dt 25
241 2 247 127 121 rad sec.
29. When y 6, x 122 62 63, and 12 − y
s x2 12 y2
2x
2
2
s x
108 36 12. x2
25
y
( x, y )
y 12
12 y 2
s2
dy ds dx 212 y1 2s dt dt dt x
dx dy ds y 12 s dt dt dt
Also, x2 y2 122 2x
dy dy x dx dx 2y 0⇒ . dt dt dt y dt
Thus, x
x dx ds dx y 12 s dt y dt dt
12x ds dx sy dx xx s ⇒ dt y dt dt 12x dy x dx 63 dt y dt 6
ds 126 1 3 0.2 msec (horizontal) dt 15 12 63 53
3 1 msec (vertical). 15
5
Section 2.6
s2 x2 y2
31. (a)
dx 450 dt
)
les
in e(
dy 600 dt
mi
y
c 200 tan
s
Di
s
100
x 200
100
dx dy ds 2x 2y 2s dt dt dt
Distance (in miles)
ds xdxdt ydydt dt s When x 150 and y 200, s 250 and ds 150450 200600 750 mph. dt 250 (b) t
250 1 hr 20 min 750 3
s2 902 x2
33.
2nd
x 30 30 ft
dx 28 dt 2s
3rd
x
1st s
ds ds x dx 2x ⇒ dt dt dt s
dx dt
90 ft Home
When x 30, s 902 302 3010 ds 28 30 28 8.85 ftsec. dt 10 3010 35. (a)
15 y ⇒ 15y 15x 6y 6 yx y
5 x 3
15
dx 5 dt dy 5 dt 3 (b)
6
dx 5 25 5 ftsec dt 3 3
10 d y x dy dx 25 5 ftsec dt dt dt 3 3
x y
Related Rates
89
90
Chapter 2
37. xt
Differentiation
39. Since the evaporation rate is proportional to the surface area, dVdt k4 r 2. However, since V 43 r 3, we have
1 t sin , x 2 y 2 1 2 6
(a) Period:
2 12 seconds 6
1 (b) When x , y 2 Lowest point:
1 12 2
dr dV 4 r 2 . dt dt 2
3
2
m.
0, 23
1 (c) When x , y 4
Therefore, k4 r 2 4 r 2
1 14
2
15
4
dr dr ⇒k . dt dt
and t 1
dx 1 t t cos cos dt 2 6 6 12 6
x2 y2 1 2x
dx dy dy x dx . 2y 0⇒ dt dt dt y dt
Thus, dy 14 dt 154 Speed
12 cos 6
1 23 24 15 1205.
15 12
5 5 msec 120 120
pV1.3 k
41. 1.3 pV 0.3
dV dp V1.3 0 dt dt
dV dp V 0 dt dt
V 0.3 1.3p
1.3p tan
43.
dV dp V dt dt
y 30
y
dy 3 msec. dt sec2
d 1 dy dt 30 dt 1 d cos2 dt 30
y
θ
dy dt
When y 30, 4 and cos 22. Thus, d 1 1 1 3 radsec. dt 30 2 20
x
30
Section 2.6
y tan , y 5 x
45.
L
dx 600 mihr dt d 5 sec2 2 dt x
y=5
θ x
dx dt
d 5 dx x2 5 dx cos2 2 2 2 dt x dt L x dt
L5 15dxdt sin 15600 120 sin
47.
2
2
2
2
(a) When 30 ,
d 120 1 30 radhr radmin. dt 4 2
(b) When 60 ,
d 3 3 120 90 radhr radmin. dt 4 2
(c) When 75 ,
d 120 sin2 75 111.96 radhr 1.87 radmin. dt
d 10 revsec2 radrev 20 radsec dt x (a) cos 30
P 30
θ x
d 1 dx sin dt 30 dt
x
dx d 30 sin 30 sin 20 600 sin dt dt (b)
2000
4π
0
− 2000
is least when dxdt n or n 180 .
(c) dxdt 600 sin is greatest when sin 1 ⇒ 2 n or 90 n
(d) For 30 , For 60 ,
49. tan
dx 1 600 sin30 600 300 cmsec. dt 2 3 dx 600 sin60 600 3003 cmsec dt 2
x ⇒ x 50 tan 50 d dx 50 sec2 dt dt 2 50 sec2
d dt
d 1 cos2 , ≤ ≤ dt 25 4 4
180
Related Rates
91
92
Chapter 2
Differentiation
51. x2 y2 25; acceleration of the top of the ladder
First derivative: 2x
d 2y dt 2
dy dx 2y 0 dt dt dy dx y 0 dt dt
x Second derivative: x
d 2x dx dt 2 dt
d 2y dy dx y 2 dt dt dt
dy 0 dt
xddt x dxdt dydt
1 d 2y dt 2 y When x 7, y 24,
2
2
2
2
dy 7 dx dx d 2x , and 2 (see Exercise 27). Since is constant, 2 0. dt 12 dt dt dt
d 2y 1 7 70 22 dt 2 24 12
49 1 625 241 4 144 24 144 0.1808 ft sec 2
2
53. (a) Using a graphing utility, you obtain ms 0.881s2 29.10s 206.2 (b)
dm dm ds ds 1.762s 29.10 dt ds dt dt
(c) If t s 1995, then s 15.5 and Thus,
ds 1.2. dt
dm 1.76215.5 29.101.2 2.15 million. dt
Review Exercises for Chapter 2 1. f x x2 2x 3 fx lim
x→0
f x x f x x
x x2 2x x 3 x2 2x 3 x→0 x
lim
x2 2xx x2 2x 2x 3 x2 2x 3 x→0 x
lim
2xx x2 2x lim 2x x 2 2x 2 x→0 x→0 x
lim
5. f is differentiable for all x 1.
3. f x x 1 fx lim
x→0
lim
x→0
lim
x→0
lim
x→0
lim
x→0
f x x f x x
x x 1 x 1 x x x x
x
x x x x x x
x x x x x x x 1 x x x
1 2 x
Review Exercises for Chapter 2
7. f x 4 x 2
1 4 9. Using the limit definition, you obtain gx x . 3 6
(a) Continuous at x 2. (b) Not differentiable at x 2 because of the sharp turn in the graph.
4 1 3 At x 1, g1 3 6 2
y 7 6 5 4 3 2 x
−1
1 2 3 4 5 6
−2 −3
11. (a) Using the limit defintion, fx 3x 2.
13. g2 lim
x→2
At x 1, f1 3. The tangent line is y 3x 1
x3 x 2 4 x→2 x2
lim
0
−4
x 2x 1 4 x→2 x2
lim
y 2 3x 1
(b)
gx g2 x2
2
lim
(−1, −2)
x→2
x 2x 2 x 2 x2
lim x 2 x 2 8 x→2
−4
15.
19. f x x8
17. y 25
y
f′
f
fx 8x7
y 0
2
1
x −1
1
21. ht 3t 4
23. f x x3 3x2
ht 12t 3
fx 3x2 6x 3xx 2
3 x 6x1 2 3x1 3 25. hx 6 x 3
hx 3x1 2 x2 3 29. f 2 3 sin f 2 3 cos
3 x
2 27. gt t2 3
1 3 x2
gx
4 3 4 t 3 3 3t
31. f 3 cos
sin 4
f 3 sin
cos 4
93
94
Chapter 2
Differentiation
F 200 T
33.
Ft
35.
st 16t2 s0 s9.2 169.22 s0 0
100 T
s0 1354.24
(a) When T 4, F4 50 vibrations/sec/lb.
The building is approximately 1354 feet high (or 415 m).
(b) When T 9, F9 3313 vibrations/sec/lb. 37. (a)
(c) Ball reaches maximum height when x 25.
y
y x 0.02x2
(d)
15
y 1 0.04x
10
y0 1
5
y10 0.6
x 20
40
60
Total horizontal distance: 50 (b) 0 x 0.02x2
y25 0 y30 0.2 y50 1
x 0x 1 implies x 50. 50 39. xt t2 3t 2 t 2t 1 (a) vt xt 2t 3
(e) y25 0
3 (b) vt < 0 for t < 2 .
(d) xt 0 for t 1, 2.
at vt 2 3 (c) vt 0 for t 2 . 3 3 1 1 1 x 2 2 2 1 2 2 4
v1 21 3 1
v2 22 3 1 The speed is 1 when the position is 0. 43. hx x sin x x1 2 sin x
41. f x 3x2 7x2 2x 3 fx 3x2 72x 2 x2 2x 36x 2
6x3
9x2
16x 7
1 sin x x cos x 2 x
x2 x 1 x2 1 x2 12x 1 x2 x 12x fx x2 12 2 x 1 2 x 12
45. f x 2x x2
47. f x
fx 2 2x3 2 1
hx
1 x3
2x3 1 x3
51. y
49. f x 4 3x21 fx 4 3x226x 53. y 3x 2 sec x y 3x 2 sec x tan x 6x sec x
6x 4 3x22
y
x2 cos x cos x 2x x 2sin x 2x cos x x 2 sin x cos2 x cos2 x
55. y x tan x y x sec2 x tan x
Review Exercises for Chapter 2
59. gt t3 3t 2
57. y x cos x sin x
gt 3t2 3
y x sin x cos x cos x x sin x
g t 6t 61. f 3 tan
y 2 sin x 3 cos x
63.
f 3 sec2
y 2 cos x 3 sin x
f 6 sec sec tan 6 sec tan
y 2 sin x 3 cos x
2
y y 2 sin x 3 cos x 2 sin x 3 cos x 0 65. f x 1 x31 2 1 fx 1 x31 23x2 2
3x2
2 1 x3
s2
1
s
s2
1
3s2
2
xx 31x
2
2
11 x 32x x 2 12
2x 3x 2 6x 1 x 2 13
71. y 3 cos3x 1
5 1 2s
5 2
x3 x2 1
hx 2
69. f s s2 15 2s3 5 fs
67. hx
s3
5 2
s2
3 2
y 9 sin3x 1
3s 1 5 5
3 2
s2
s3
ss2 13 28s3 3s 25
73. y
1 csc 2x 2
1 y csc 2x cot 2x2 2
y
csc 2x cot 2x
77. y
x sin 2x 2 4
75. y
1 1 cos 2x2 2 4
1 1 cos 2x sin2 x 2
2 3 2 2 sin x sin7 2x 3 7
y sin1 2 x cos x sin5 2 x cos x
79. y y
cos x sin x1 sin2 x
sin x x2
x 2 cos x sin x x 22
cos3 x sin x 81. f t t2t 15
83. gx 2xx 11 2
ft tt 147t 2 The zeros of f correspond to the points on the graph of f where the tangent line is horizontal. 0.1
gx
g does not equal zero for any value of x in the domain. The graph of g has no horizontal tangent lines.
f′
4
−0.1
1.3
g′
f −0.1
x2 x 13 2
−2
7
g −2
95
96
Chapter 2
Differentiation
85. f t t 11 2t 11 3 t 15 6 ft
87. y tan 1 x
5 6t 11 6
y
f does not equal zero for any x in the domain. The graph of f has no horizontal tangent lines.
sec2 1 x 2 1 x
y does not equal zero for any x in the domain. The graph has no horizontal tangent lines.
5
5
y
f − 20
f′ −2
7
2
y′
−1
−4
91. f x cot x
89. y 2x2 sin 2x
93. f t
t 1 t2
ft
t1 1 t3
f t
2t 2 1 t4
fx csc2 x
y 4x 2 cos 2x
f 2 csc xcsc x cot x
y 4 4 sin 2x
2 csc2 x cot x
95. g tan 3 sin 1 g 3 sec2 3 cos 1 g 18 sec2 3 tan 3 sin 1 97. T 700t2 4t 101 T
1400t 2 t2 4t 102
(a) When t 1, T
(b) When t 3,
14001 2 18.667 deg hr. 1 4 102
T
(d) When t 10,
(c) When t 5, T
14003 2 7.284 deg hr. 9 12 102
14005 2 3.240 deg hr. 25 30 102
T
140010 2 0.747 deg hr. 100 40 102
x2 3xy y3 10
99.
2x 3xy 3y 3y2y 0 3x y2y 2x 3y 2x 3y 3x y2
y
y x x y 16
101. y
12x x 1 2
y x
1 2
12y
1 2
y y1 2 0
x 2 x yy y 2 y x 2 xy x 2 xy y y 2 y 2 x y
2 xy y 2 x
2 y 2 xy x
2y x y y 2x y x x
Review Exercises for Chapter 2
x sin y y cos x
103.
105.
x cos yy sin y y sin x y cos x
6
(2, 4)
2x 2yy 0 −9
yx cos y cos x y sin x sin y y
x2 y2 20
97
9
x y y
y sin x sin y cos x x cos y
−6
1 At 2, 4: y 2 1 Tangent line: y 4 x 2 2 x 2y 10 0 Normal line: y 4 2x 2 2x y 0
107.
y x dy 2 units sec dt dx dy 1 dx dy ⇒ 2 x 4 x dt dt dt 2 x dt 1 dx 2 2 units/sec. (a) When x , 2 dt
109.
(b) When x 1,
dx 4 units/sec. dt
(c) When x 4,
dx 8 units/sec. dt
s 1 2 h 2
111. st 60 4.9t2 st 9.8t
1 s h 4
s 35 60 4.9t2 4.9t2
dV 1 dt
w 2 2s 2 2
14h 4 2 h
tan 30
dV 5 dh 4 h dt 2 dt 2dV dt dh dt 54 h When h 1,
dh 2 m min. dt 25
1 2
1 3
5 4.9 st xt
xt 3 st
4h 5 5 2 h 8 hh 2 2 4
25
t
Width of water at depth h:
V
s (t)
ds 5 dx 3 39.8 dt dt 4.9 1 2
2
38.34 m sec
s 2 h 2
30˚ x(t )
C H A P T E R 2 Differentiation Section 2.1
The Derivative and the Tangent Line Problem . . 330
Section 2.2
Basic Differentiation Rules and Rates of Change 338
Section 2.3
The Product and Quotient Rules and Higher-Order Derivatives . . . . . . . . . . . . . 344
Section 2.4
The Chain Rule . . . . . . . . . . . . . . . . . . 350
Section 2.5
Implicit Differentiation . . . . . . . . . . . . . . 356
Section 2.6
Related Rates . . . . . . . . . . . . . . . . . . . 361
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 367 Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . 373
C H A P T E R Differentiation Section 2.1
2
The Derivative and the Tangent Line Problem
Solutions to Even-Numbered Exercises
2. (a) m
1 4
4. (a)
f 4 f 1 5 2 1 41 3 f 4 f 3 5 4.75 0.25 43 1
(b) m 1
Thus,
f 4 f 1 f 4 f 3 > 41 43
(b) The slope of the tangent line at 1, 2 equals f1. This slope is steeper than the slope of the line through 1, 2 and 4, 5. Thus, f 4 f 1 < f1. 41 6. g x
3 3 x 1 is a line. Slope 2 2
8. Slope at 2, 1 lim
x→0
g2 x g2 x
5 2 x2 1 x→0 x
lim
5 4 4x x2 1 x→0 x
lim
lim 4 x 4 x→0
10. Slope at 2, 7 lim
t→0
h2 t h2 t
2 t2 3 7 t→0 t
lim
4 4t t2 4 t→0 t
lim
lim 4 t 4 t→0
14. f x 3x 2 fx lim
x→0
lim
x→0
lim
x→0
f x x f x x
330
gx lim
x→0
gx x gx x
lim
5 5 x
lim
0 0 x
x→0
x→0
1 16. f x 9 2x fx lim
x→0
f x x f x x
3x x 2 3x 2 x
lim
9 1 2x x 9 1 2x x
3x x
lim
21 21
lim 3 3 x→0
12. gx 5
x→0
x→0
Section 2.1
The Derivative and the Tangent Line Problem
331
18. f x 1 x2 fx lim
x→0
f x x f x x
1 x x2 1 x2 x→0 x
lim lim
x→0
1 x2 2xx x2 1 x2 x
2xx x2 lim 2x x 2x x→0 x→0 x
lim
20. f x x3 x2 fx lim
x→0
f x x f x x
x x3 x x2 x3 x2 x→0 x
lim
x3 3x2x 3xx2 x3 x2 2xx x2 x3 x2 x→0 x
lim lim
x→0
3x2x 3xx2 x3 2xx x2 x
lim 3x2 3xx x2 2x x 3x2 2x x→0
22. f x
1 x2
24. f x
fx lim
x→0
f x x f x x
1 1 x x2 x2 lim x→0 x x x x xx x2x2 2
fx lim
x→0
f x x f x x 4
lim
x→0
2
x x
4
x
x
lim
4 x 4 x x x x x x
2xx x2 2 2 x→0 xx x x
lim
4x 4x x x x x x x x x
2x x x x2x2
lim
4
x x x x x x
lim
x→0
lim lim
x→0
4
x
2x x4
2 x3
x→0
x→0
x→0
4 2
x x x x x x
x x x
x x x
332
Chapter 2
Differentiation
26. (a) f x x2 2x 1 fx lim
x→0
18. (b)
5
f x x f x x
(−3, 4)
−6
x x2 2x x 1 x2 2x 1 x→0 x
3
lim
lim
x→0
−1
2xx x2 2x x
lim 2x x 2 2x 2 x→0
At 3, 4, the slope of the tangent line is m 23 2 4. The equation of the tangent line is y 4 4x 3 y 4x 8. 28. (a) f x x3 1 f x x f x fx lim x→0 x
18. (b)
4
(1, 2) −6
6
x x3 1 x3 1 lim x→0 x lim
x3
−4
x 3xx x 1 x
3x 2
x→0
2
3
x3
1
lim 3x 2 3xx x2 3x 2 x→0
At 1, 2, the slope of the tangent line is m 312 3. The equation of the tangent line is y 2 3x 1 y 3x 1. (b)
30. (a) f x x 1
(5, 2)
f x x f x fx lim x→0 x lim
x x 1 x 1
x
x→0
lim
x→0
lim
x→0
−2
x x 1 x 1
x x 1 x 1
x x 1 x 1 x x x 1 x 1 1
x x 1 x 1
1 2 x 1
At 5, 2, the slope of the tangent line is m
1 1 2 5 1 4
The equation of the tangent line is 1 y 2 x 5 4 1 3 y x 4 4
4
10
−4
Section 2.1
The Derivative and the Tangent Line Problem
1 x1
32. (a) f x
(b)
f x x f x x→0 x 1 1 x x 1 x 1 lim x→0 x x→0
−6
3
−3
x 1 x x 1 xx x 1x 1
lim x→0
3
(0, 1)
fx lim
lim
333
1 x x 1x 1
1 x 12
At 0, 1, the slope of the tangent line is m
1 1. 0 12
The equation of the tangent line is y x 1. 34. Using the limit definition of derivative, fx 3x 2. Since the slope of the given line is 3, we have
36. Using the limit definition of derivative, fx
1 Since the slope of the given line is , we have 2
3x 2 3 x 2 1 ⇒ x ± 1.
1 1 2x 132 2
Therefore, at the points 1, 3 and 1, 1 the tangent lines are parallel to 3x y 4 0. These lines have equations
1 x 132 1x1⇒x2
y 3 3x 1 and y 1 3x 1 y 3x
1 . 2x 13 2
At the point 2, 1, the tangent line is parallel to x 2y 7 0. The equation of the tangent line is
y 3x 4
1 y 1 x 2 2 1 y x2 2 38. h1 4 because the tangent line passes through 1, 4 h1
40. f x x 2 ⇒ fx 2x
(d)
64 2 1 3 1 4 2
42. f does not exist at x 0. Matches (c)
44.
Answers will vary.
y 4
Sample answer: y x
3 2 1 x
−4 −3 −2
1 −2 −3 −4
2
3
4
334
Chapter 2
46. (a) Yes. lim
x→0
Differentiation
f x 2x f x f x x f x lim fx x→0 2Dx x
(b) No. The numerator does not approach zero. (c) Yes. lim
x→0
f x x f x x f x x f x f x x f x lim x→0 2x 2x lim
x→0
(d) Yes. lim
x→0
f x f x x f x f x x 2x 2x
1 1 fx fx fx 2 2
f x x f x fx x
48. Let x0, y0 be a point of tangency on the graph of f. By the limit definition for the derivative, fx 2x. The slope of the line y through 1, 3 and x0, y0 equals the derivative of f at x0: 3 y0 2x0 1 x0
10
(3, 9)
8 6
3 y0 1 x02x0
4
(−1, 1)
3 x02 2x0 2x02
−8 −6 −4 −2 −2
x02 2x0 3 0
−4
x 2
4
6
(1, −3)
x0 3x0 1 0 ⇒ x0 3, 1 Therefore, the points of tangency are 3, 9 and (1, 1, and the corresponding slopes are 6 and 2. The equations of the tangent lines are y 3 6x 1
y 3 2x 1
y 6x 9
y 2x 1
50. (a) f x x2 fx lim
x→0
(b) gx lim
x→0
f x x f x x
g x x g x x
x x3 x3 x→0 x
lim
x x2 x2 x→0 x
lim
x2 2xx x2 x2 x→0 x
x3 3x2x 3xx2 x3 x3 x
lim
x3x2 3xx x2 x
lim
x→0
lim
lim
x→0
x2x x x
x→0
lim 3x2 3xx x2 3x2 x→0
lim 2x x 2x
At x 1, g1 3 and the tangent line is
x→0
At x 1, f1 2 and the tangent line is y 1 2x 1
or
y 2x 1.
At x 0, f0 0 and the tangent line is y 0. At x 1, f1 2 and the tangent line is y 2x 1.
y 1 3x 1
y 3x 2.
At x 1, g1 3 and the tangent line is y 1 3x 1 or y 3x 2. 2
2
−3
or
At x 0, g0 0 and the tangent line is y 0.
3
−3
For this function, the slopes of the tangent lines are always distinct for different values of x.
−3
3
−2
For this function, the slopes of the tangent lines are sometimes the same.
Section 2.1
The Derivative and the Tangent Line Problem
1 52. f x 2 x2
3
By the limit definition of the derivative we have fx x. 2
x f x
2
fx
2
1.5
1
1.125 1.5
0.5
0.5 1
0.125 0.5
0
0.5
1
1.5
2
0
0.125
0.5
1.125
2
0
0.5
1
1.5
2
f x 0.01 f x 0.01
54. gx
−2
2 −1
1 56. f 2 23 2, f 2.1 2.31525 4
3 x 0.01 3 3 100
f2
8
2.31525 2 3.1525 Exact: f2 3 2.1 2
f
g −1
8 −1
The graph of gx is approximately the graph of fx. 58. f x
x3 3 3x and fx x2 3 4 4 6
f′ −9
9
f −6
60. f x x
1 x
f 2 x f 2 x 2 f 2 Sx x x
1 5 2 x 2 5 x 2 x 2
2 x
5 2x 3 5 22 x2 2 52 x x 2 x 2 22 x x 2 22 x 2
(a) x
1: Sx
5 5 5 5 x 2 x 6 2 6 6
4 5 4 9 x 0.5: Sx x 2 x 5 2 5 10 x 0.1: Sx
16 5 16 41 x 2 x 21 2 21 42
(b) As x → 0, the line approaches the tangent line to f at 2, 52 . 62. g x xx 1 x 2 x, c 1 gx g1 x2 x 0 xx 1 lim lim x→1 x→1 x→1 x 1 x1 x1
g1 lim
lim x 1 x→1
4
−6
6
S0.1 S0.5 S1
f −4
335
336
Chapter 2
Differentiation
64. f x x3 2x, c 1
x 1x2 x 3 f x f 1 x3 2x 3 lim lim lim x2 x 3 5 x→1 x →1 x→1 x1 x1 x1
f1 lim
x→1
1 66. f x , c 3 x f x f 3 1 x 1 3 3x lim lim x→3 x→3 x3 x3 3x
f3 lim
x→3
lim x 3 x→3 3x 9 1
1
1
68. g x x 31 3, c 3 gx g3 1 x 31 3 0 lim lim x→3 x→3 x→3 x 32 3 x 3 x3
g3 lim
Does not exist.
70. f x x 4 , c 4 f4 lim
x→4
x4 0 x4 f x f 4 lim lim x→4 x→4 x 4 x4 x4
Does not exist. 72. f x is differentiable everywhere except at x ± 3. (Sharp turns in the graph.) 74. f x is differentiable everywhere except at x 1. (Discontinuity) 76. f x is differentiable everywhere except at x 0. (Sharp turn in the graph) 78. f x is differentiable everywhere except at x ± 2. (Discontinuities) 80. f x is differentiable everywhere except at x 1. (Discontinuity) 82. f x 1 x2 The derivative from the left does not exist because lim
x→1
1 x2 0
1 x2 f x f 1 lim lim x→1 x→1 x1 x1 x1
1 x2
1 x2
lim x→1
1x
1 x2
. (Vertical tangent)
The limit from the right does not exist since f is undefined for x > 1. Therefore, f is not differentiable at x 1. 84. f x
xx,, xx >≤ 11 2
The derivative from the left is lim
x→1
f x f 1 x1 lim lim 1 1. x→1 x 1 x→1 x1
The derivative from the right is lim
x→1
f x f 1 x2 1 lim lim x 1 2. x→1 x 1 x→1 x1
These one-sided limits are not equal. Therefore, f is not differentiable at x 1.
Section 2.1
86. Note that f is continuous at x 2. f x
1 2x
The Derivative and the Tangent Line Problem
337
1, x < 2 x ≥ 2
2x,
The derivative from the left is lim
x→2
f x f 2 1 x 1 2 lim 12 x 2 1. lim 2 x→2 x→2 x 2 x2 x2 2
The derivative from the right is lim
x→2
2x 2 f x f 2 lim x→2 x2 x2
lim x→2
2x 2
2x 2
2x 4 2x 2 2 1 lim lim x 2 2x 2 x→2 x 2 2x 2 x→2 2x 2 2.
The one-sided limits are equal. Therefore, f is differentiable at x 2. f2 12 88. (a) f x x2 and fx 2x
72. (b) gx x3 and gx 3x2
y
y
5
3
4
f
g′
3
2
2
−4 −3 −2 −1
f'
g
1
1 x 1
2
3
4
−2
x
−1
1
2
−1 −3
72. (c) The derivative is a polynomial of degree 1 less than the original function. If hx x n, then hx nx n1. 72. (d) If f x x4, then fx lim
x→0
f x x f x x
lim
x x4 x4 x
lim
x4 4x3x 6x2x2 4xx3 x4 x4 x
lim
x4x3 6x2x 4xx2 x3 x
x→0
x→0
x→0
lim 4x3 6x2x 4xx2 x3 4x3 x→0
Hence, if f x x , then fx 4x3 which is consistent with the conjecture. However, this is not a proof, since you must verify the conjecture for all integer values of n, n ≥ 2. 4
90. False. y x 2 is continuous at x 2, but is not differentiable at x 2. (Sharp turn in the graph) 92. True—see Theorem 2.1
94.
3
−3
3 −1
As you zoom in, the graph of y1 x2 1 appears to be locally the graph of a horizontal line, whereas the graph of y2 x 1 always has a sharp corner at 0, 1. y2 is not differentiable at 0, 1.
338
Chapter 2
Differentiation
Section 2.2
Basic Differentiation Rules and Rates of Change
y x12
2. (a)
y x1
(b)
y x2
y 12 x32
5 15 y 3x4 2 sin x 2 sin x 8 8x4
4 y x3 3
y
4 3x3
y
2 9x3
28. y
3x2
y
30. y
4 x3
y 4x3
2 x 9
35, 2
y
2 3 x 9
y 12x2
y 12x2 34.
y 3x3 6, 2, 18
36. f x 35 x2, 5, 0
y 9x2
3 ft 2 5t
38.
5 5 2 cos x x3 2 cos x 2x3 8
Simplify
2 y x2 3
f
fx 6x2 2x 3
Derivative
2 3x2
3 , 5t
18. f x 2x3 x2 3x
Rewrite
26. y
f t 3
24. y
y cos x
Function
1 1 y x34 34 4 4x
8 x9
y 3x 2
22. y 5 sin x
gt sin t
4 x x14 10. y
16. y 8 x3
y 2t 2
20. gt cos t
y1 2
1 x8 x8
y 8x9
14. y t2 2t 3
gx 3
32.
8. y
y 8x7
f x 0
3x2 30x 75
y2 36
fx 6x 30 f5 0
35 35 gt 2 3 cos t, , 1
40. f x x2 3x 3x2
gt 3 sin t g 0
44. hx
y 2x3
y1 32
6. y x8
4. f x 2
y x2
(d)
y 32 x52
y1 1
y1 12
12. gx 3x 1
y x32
(c)
2x2 3x 1 2x 3 x1 x
1 2x2 1 hx 2 2 x x2 3 x 5 x x13 x15 48. f x
1 1 1 1 fx x23 x45 23 45 3 5 3x 5x
42. f x x x2
fx 2x 3 6x3 2x 3
fx 1 2x3
6 x3
1
2 x3
46. y 3x6x 5x 2 18x 2 15x3 y 36x 45x 2
50. f t t23 t13 4 2 1 2 1 ft t13 t23 13 23 3 3 3t 3t
Section 2.2
52. f x fx
2 3 x
Basic Differentiation Rules and Rates of Change
339
3 cos x 2x13 3 cos x
2 43 2 x 3 sin x 43 3 sin x 3 3x
54. (a) y x3 x
(b)
y 3x2 1
5
−5
5
At 1, 2: y 312 1 4. y 2 4x 1
Tangent line:
−7
4x y 2 0 56. (a) y x2 2xx 1
(b)
12
x3 3x2 2x y 3x2 6x 2 −3
At 1, 6: y 312 61 2 11.
3 −2
Tangent line: y 6 11x 1 0 11x y 5 60. y x2 1
58. y x3 x y 3x2 1 > 0 for all x.
y 2x 0 ⇒ x 0
Therefore, there are no horizontal tangents.
At x 0, y 1. Horizontal tangent: 0, 1 64. k x 2 4x 7
62. y 3x 2 cos x, 0 ≤ x < 2 y 3 2 sin x 0 sin x
3
2
⇒ x
2x 4
2 or 3 3
At x
3 3 ,y . 3 3
At x
2 23 3 ,y . 3 3
Horizontal tangents:
3 ,
3 3
3
66. kx x 4
Equate functions
k 1 2x
Equate derivatives
Hence, k 2x and
Equate functions Equate derivatives
Hence, x 2 and k 4 8 7 ⇒ k 3
, 23, 2
3 3
3
2x x x 4 ⇒ 2x x 4 ⇒ x 4 ⇒ k 4
68. The graph of a function f such that f > 0 for all x and the rate of change the function is decreasing i.e. f < 0 would, in general, look like the graph at the right.
y
x
340
Chapter 2
Differentiation
72.
70. gx 5f x ⇒ gx 5fx
y 2
f
1 −2
x
−1
1
3
4
f′ −3 −4
If f is quadratic, then its derivative is a linear function. f x ax2 bx c fx 2ax b 74. m1 is the slope of the line tangent to y x. m2 is the slope of the line tangent to y 1x. Since y x ⇒ y 1 ⇒ m1 1 and y
1 1 1 ⇒ y 2 ⇒ m2 2 . x x x
The points of intersection of y x and y 1x are 1 ⇒ x2 1 ⇒ x ± 1. x
x
At x ± 1, m2 1. Since m2 1m1, these tangent lines are perpendicular at the points intersection. 2 76. f x , 5, 0 x fx
78. f4 1 16
2 x2
2 0y x2 5 x
−10
10 2x x2y 10 2x x2
2x
10 2x 2x 4x 10 5 4 x ,y 2 5 The point 52 , 45 is on the graph of f. The slope of the
8 tangent line is f 52 25 .
Tangent line:
19 −1
y
8 5 4 x 5 25 2
25y 20 8x 20 8x 25y 40 0
Section 2.2
Basic Differentiation Rules and Rates of Change
(b) fx 3x2
80. (a) Nearby point: 1.0073138, 1.0221024 Secant line: y 1
Tx 3x 1 1 3x 2
1.0221024 1 x 1 1.0073138 1
(c) The accuracy worsens at you move away from 1, 1.
y 3.022x 1 1 (Answers will vary.)
341
2
2
(1, 1) −3
(1, 1) −3
3
f
T
3 −2 −2
(d)
x
3
2
1
f x
8
1
0
Tx
8
5
2
0.5 0.125 0.5
0.1
0
0.1
0.5
1
2
3
0.729
1
1.331
3.375
8
27
64
0.7
1
1.3
2.5
4
7
10
The accuracy decreases more rapidly than in Exercise 59 because y x3 is less “linear” than y x32. 82. True. If f x gx c, then fx gx 0 gx. 86. False. If f x
1 n xn, then fx nxn1 n1 xn x
88. f t t2 3, 2, 2.1 ft 2t
2, 1 ⇒ f2 22 4 2.1, 1.41 ⇒ f2.1 4.2 Average rate of change: f 2.1 f 2 1.41 1 4.1 2.1 2 0.1
st 16t2 22t 220 vt 32t 22 v3 118 ftsec st 16t2 22t 220 112 height after falling 108 ft 16t2
6
90. f x sin x, 0, fx cos x
Instantaneous rate of change:
92.
84. True. If y x 1 x, then dydx 11 1.
22t 108 0
2t 28t 27 0 t2 v2 322 22 86 ftsec
Instantaneous rate of change:
0, 0 ⇒ f0 1
6 , 12 ⇒ f6
3
2
0.866
Average rate of change: f 6 f 0 12 0 3 0.955 6 0 6 0 94. st 4.9t2 v0t s0 4.9t2 s0 0 when t 6.8. s0 4.9t2 4.96.82 226.6 m
342
Chapter 2
96.
Differentiation
98. This graph corresponds with Exercise 75.
v
s 10
40
Distance (in miles)
Velocity (in mph)
50
30 20 10 t 2
4
6
8
8
4 2
10
Time (in minutes)
(10, 6)
6
(0, 0)
(4, 2)
(6, 2) t
2
4
6
8
10
Time (in minutes)
(The velocity has been converted to miles per hour) 1 100. st at2 c and st at. 2 Average velocity:
st0 t st0 t 12at0 t2 c 12at0 t2 c t0 t t0 t 2t
12at02 2t0t t2 12at02 2t0t t2 2t
2at0t 2t
at0 st0 Instantaneous velocity at t t0 102. V s3,
dV 3s2 ds
When s 4 cm, 104.
dV 48 cm2. ds
C gallons of fuel usedcost per gallon
18,750 1.25 15,000 x x
dC 18,750 dx x2 x
10
C
1875
dC dx
187.5
15 1250 83.333
20 537.5 46.875
25 750 30
30 625 20.833
35
40
535.71
468.75
15.306
11.719
The driver who gets 15 miles per gallon would benefit more from a 1 mile per gallon increase in fuel efficiency. The rate of change is larger when x 15. 106.
dT KT Ta dt
Section 2.2
Basic Differentiation Rules and Rates of Change
1 108. y , x > 0 x y
y
1 x2
2
( )
(a, b) = a, a1
At a, b, the equation of the tangent line is
1
x 1 1 2 2x a or y 2 . a a a a
y
343
x 1
2
3
The x-intercept is 2a, 0.
2a.
The y-intercept is 0,
1 2 1 2. The area of the triangle is A bh 2a 2 2 a 110. y x2 y 2x (a) Tangent lines through 0, a: y a 2xx 0 x2 a 2x2 a x2 ± a x
The points of tangency are ± a, a. At a, a the slope is y a 2a. At a, a the slope is y a 2a. Tangent lines: y a 2a x a and y a 2a x a y 2a x a
y 2a x a
Restriction: a must be negative. (b) Tangent lines through a, 0: y 0 2xx a x2 2x2 2ax 0 x2 2ax xx 2a The points of tangency are 0, 0 and 2a, 4a2. At 0, 0 the slope is y0 0. At 2a, 4a2 the slope is y2a 4a. Tangent lines: y 0 0x 0 and y 4a2 4ax 2a y0
y 4ax 4a2
Restriction: None, a can be any real number.
f2x sin x is differentiable for all x 0.
112. f1x sin x is differentiable for all x n, n an integer.
You can verify this by graphing f1 and f2 and observing the locations of the sharp turns.
344
Chapter 2
Differentiation
Section 2.3
The Product and Quotient Rules and Higher-Order Derivatives
3 2. f x 6x 5x 2
4. gs s4 s2 s124 s2
fx 6x 53x 2 x3 26
4 s2 1 gs s122s 4 s 2 s12 2s 32 2 2 s12
18x3 15x 2 6x3 12 24x3 15x2 12
6. gx x sin x
8. gt
gx x cos x sin x
10. hs
hs
21 x x cos x 21 x sin x
gt
s s 1
12. f t
s 11 s
12s
ft
12
s 12 1 2 s 12
s 1 s
16. f x
fx x2 2x 13x2 x3 12x 2
x 1 2x 1 2
x 1 2
5x2
t2 2 2t 7
2t 72t t2 22 2t2 14t 4 2t 72 2t 72 cos t t3 t 3sin t cos t3t 2 t sin t 3 cos t t 32 t4
s 2 2 2s 1
14. f x x2 2x 1x3 1
3x2
4 5s 2 2s12
2
x2
fx
x 1
2x 2
f1 0
x1 x1
x 11 x 11 x 12 x1x1 x 12
f2 18.
f x
sin x x
fx
xcos x sin x1 x2
f
2 x 12 2 2 2 12
x cos x sin x x2
6 6 3236 12
2
33 18 2
3 3 6 2
Function 20. y
5x 2 3 4
Rewrite
Derivative
3 5 y x2 4 4
y
10 x 4
Simplify y
5x 2
Section 2.3
Function
The Product and Quotient Rules and Higher-Order Derivatives
Rewrite
Derivative
Simplify
22. y
4 5x2
4 y x2 5
8 y x3 5
y
24. y
3x2 5 7
5 3 y x2 7 7
y
6x 7
6 y x 7
26. f x fx
2 4 4 x 1 28. f x x 1 x 1 x x 1
x3 3x 2 x2 1
x2 13x2 3 x3 3x 22x x2 12 x4 6x2 4x 3 x2 12
12x x 12
5 16 x x23 6
5 1 6x16 x23
12
fx x4
3
13x 23
x 1x 1x 1 xx 11 4x
2x3
3 x x 3 x13x12 3 30. f x
fx x13
8 5x3
3
2
2xxx1 2 2
2
32. hx x2 12 x4 2x2 1 hx 4x3 4x 4xx2 1
Alternate solution: 3 x x 3 f x
x56 3x13 5 fx x16 x23 6
5 1 6x16 x23
34. gx x 2
2x x 1 1 2x x x 1
gx 2
2
x 12x x 21 2x 2 2x 1 x 2 2x x 2 2x 2 x 12 x 12 x 12
36. f x x2 xx2 1x2 x 1 fx 2x 1x2 1x2 x 1 x2 x2xx2 x 1 x2 xx2 12x 1 2x 1x 4 x3 2x2 x 1 x2 x2x3 2x2 2x x2 x2x3 x2 2x 1 2x5 x 4 3x3 x 1 2x5 2x2 2x5 x 4 x3 x2 x 6x5 4x3 3x2 1
38. f x
c2 x2 c2 x2
c2 x22x c2 x22x fx c2 x22
4xc2 c2 x22
40. f 1 cos f 1sin cos 1 cos 1 sin
345
346
Chapter 2
42. f x fx 46. hs
Differentiation
44. y x cot x
sin x x
y 1 csc2 x cot2 x
x cos x sin x x2 1 10 csc s s
hs
48. y
1 10 csc s cot s s2
y
sec x x x sec x tan x sec x x2 sec xx tan x 1 x2
52. f x sin x cos x
50. y x sin x cos x
fx sin xsin x cos xcos x
y x cos x sin x sin x x cos x
cos 2x 54. h 5 sec tan h 5 sec tan 5 sec
56. f x sec2
tan
x x x 1 3x
fx 2 58. f f
sin 1 cos
2
2
2
x 1
x5 2x3 2x2 2 (form of answer may vary) x2 12
60. f x tan x cot x 1
1 cos 1 cos 1 1 cos 2
fx 0 f1 0
(form of answer may vary) f x sin xsin x cos x
62.
fx sin xcos x sin x sin x cos xcos x sin x cos x sin2 x sin x cos x cos2 x sin 2x cos 2x f
4 sin 2 cos 2 1
64. (a) f x x 1x2 2, 0, 2
51. (b)
4
fx x 12x x2 21 3x2 2x 2 −4
f0 2 slope at 0, 2.
4
Tangent line: y 2 2x ⇒ y 2x 2 66. (a) f x
x1 , x1
2, 13
fx
x 11 x 11 2 x 12 x 12
f2
2 1 slope at 2, . 9 3
Tangent line: y
1 2 1 2 x 2 ⇒ y x 3 9 9 9
−4
54. (b)
4
−3
6
−4
Section 2.3
f x sec x,
68. (a)
3 , 2
The Product and Quotient Rules and Higher-Order Derivatives
. (b)
347
6
fx sec x tan x f
3 23 slope at 3 , 2.
−
−2
Tangent line:
y 2 23 x
3
63x 3y 6 23 0 70. f x fx
x2 x2 1
72. f x
xcos x 3 sin x 3x1 x cos x sin x x2 x2
gx
xcos x 2 sin x 2x1 x cos x sin x x2 x2
x2 12x x22x 2x 2 x2 12 x 12
fx 0 when x 0.
sin x 2x sin x 3x 5x f x 5 x x
gx
Horizontal tangent is at 0, 0.
f and g differ by a constant.
74. f x
cos x xn cos x xn
76. V r 2h t 2
fx xn sin x nxn1 cos x
xn1
1 t32 2t12 2
x sin x n cos x
Vt
x sin x n cos x xn1
When n 1: fx
x sin x cos x . x2
When n 2: fx
x sin x 2 cos x . x3
When n 3: fx
x sin x 3 cos x . x4
When n 4: fx
x sin x 4 cos x . x5
For general n, fx
x sin x n cos x . x n1
78.
P
fx gx 1 sin x sec x tan x cos x cos x 1 ⇒ sec x tan x csc x cot x ⇒ 1 ⇒ csc x cot x 1 cos x sin x sin x sin3 x 3 1 ⇒ tan x 1 ⇒ tan x 1 cos3 x 3 7 , 4 4
dP k 2 dV V
gx csc x, 0, 2
x
1 3 12 3t 2 cubic inchessec t t12 2 2 4t12
k V
f x sec x
80.
12t
348
Chapter 2
Differentiation
82. (a) nt 9.6643t2 90.7414t 77.5029 vt 276.4643t2 2987.6929t 1809.9714 (b) A
vt 276.46t2 2987.69t 1809.97 nt 9.66t2 90.74t 77.50
86. f x
fx 1
64 x3
192 x4
40.46x 2 2.09x 17.83 x 2 9.39x 8.022
1 x2 2x 1 x2 x x
fx 1 f x
32 x2
f x
A represents the average retail value (in millions of dollars) per 1000 motor homes. (c) At
84. f x x
88. f x sec x fx sec x tan x
1 x2
f x sec xsec2 x tan xsec x tan x sec xsec2 x tan2 x
2 x3 92. f 4x 2x 1
90. f x 2 2x1 f x 2x2
94. The graph of a differentiable function f such that f > 0 and f < 0 for all real numbers x would in general look like the graph below.
f 5x 2
2 x2
f 6x 0
y
f x
98. f x gxhx
96. f x 4 hx
100.
fx hx
fx gxhx hxgx
f2 h2 4
f2 g2h2 h2g2
y 3
f′
f f ′′
34 12
−2
14
−1
x −1
2
3
4
−2
It appears that f is quadratic; so f would be linear and f would be constant. 102. st 8.25t2 66t
Average velocity on:
vt 16.50t 66 at 16.50 tsec st ft
0
1
2
3
0
57.75
99
123.75
vt st ftsec
66
49.5
33
16.5
at vt ftsec2
16.5
16.5
16.5
16.5
4
0, 1 is
57.75 0 57.75. 10
1, 2 is
99 57.75 41.25. 21
2, 3 is
123.75 99 24.75. 32
3, 4 is
132 123.75 8.25. 43
132 0 16.5
Section 2.3
104. (a)
The Product and Quotient Rules and Higher-Order Derivatives
f x x n
f x
86. (b)
f n x nn 1n 2 . . . 21 n! f nx Note: n! nn 1 . . . 3
1 x
1nnn 1n 2 . . . 21 x n1 1nn! x n1
2 1 (read “n factorial.”)
106. xf x xfx f x
xf x xf x f x f x xf x 2f x
xf x xf x f x 2f x xf x 3f x In general, xf x n xf nx nf n1x. f
2 1
fx cos x
f
2 0
f x sin x
f
2 1
(a) P1x fax a f a 0 x P2x
11 2
(b)
1 1 f ax a2 fax a f a 1 x 2 2 2
1
1 x 2 2
2
2
1
2
(π2 , 1
(
108. f x sin x
− 2
3 2
−2
(c) P2 is a better approximation than P1. (d) The accuracy worsens as you move farther away from x a 110. True. y is a fourth-degree polynomial.
112. True
114. True. If vt c then at vt 0.
dny 0 when n > 4. dx n 116. (a) fg fg fg fg fg f g fg f g
True
(b) fg fg fg fg fg fg f g fg 2f g f g
fg f g
. 2
False
349
350
Chapter 2
Differentiation
Section 2.4
The Chain Rule
y f gx 2. y
1 x 1
4. y 3 tan x 2 6. y cos
3x 2
u gx
y f u
ux1
y u12
u x2
y 3 tan u
u
3x 2
y cos u
8. y 2x3 12
10. y 34 x 25
y 22x3 16x 2 12x22x3 1
y 154 x 22x 30x4 x 2 14. gx 5 3x 5 3x12
12. f t 9t 223
1 3 gx 5 3x123 2 25 3x
2 6 ft 9t 2139 3 3 9t 2
16. gx x2 2x 1 x 12 x 1 gx
3 27 fx 2 9x349 4 42 9x34
1,1, xx >< 11
20. st t 2 3t 11
22. y 5t 33
st 1t 2 3t 122t 3
18. f x 32 9x14
y 15t 34
2t 3 t2 3t 12
24. gt t2 212
15 t 34
26. f x x3x 93
1 t gt t2 2322t 2 2 t 232
fx x33x 923 3x 931 3x 929x 3x 9 27x 324x 3
1 28. y x 216 x 2 2
30. y
1 1 y x 2 16 x 2122x x16 x 212 2 2
x3 x16 x 2 216 x 2
x3x 2 32 216 x 2
32. ht
ht 2
t2 t3 2
t
3
t2 2
t
3
22t t23t2 t3 22
2t24t t4 2t34 t3 3 t3 23 t 23
1 x 4 4121 x x 4 4124x 3 2 y x4 4
2
x x4 4
x 4 4 2x 4 4 x4 4 32 x 4 x 4 432
Section 2.4
34. gx
gx 3 36. y y
3x 2 2 2x 3
2
2
2
22
33x 2 226x 2 18x 4 63x 2 223x 2 9x 2 2x 34 2x 34
x 2x 1
38. f x x2 x2 fx
1 2xx 132
x 25x 2 2x
The zeros of f correspond to the points on the graph of f where the tangent lines are horizontal.
y has no zeros. 7
4
y f′
y′ −6
6
f
−3
6
−1 −2
40. y t 2 9t 2 y
42. gx x 1 x 1
5t2 8t 9 2t 2
gx
The zero of y corresponds to the point on the graph of y where the tangent line is horizontal.
1 1 2x 1 2x 1
g has no zeros. 6
15
g
y′
g′
y −3
6
−2
10 −2
− 15
44.
y x2 tan
1 x
6
y
1 1 dy 2x tan sec2 dx x x
−4
y sin 3x
5
y′
The zeros of y correspond to the points on the graph of y where the tangent lines are horizontal.
46. (a)
351
3
3x2x 32 2x 36x2x 33x 2
The Chain Rule
−6
(b)
y sin
y 3 cos 3x y0 3 3 cycles in 0, 2
y y0
2x
12 cos 2x 1 2
Half cycle in 0, 2 The slope of sin ax at the origin is a.
352
48.
Chapter 2
Differentiation
y sin x
50. hx secx2 hx 2x secx2 tanx2
dy cos x dx 52. y cos 1 2x2 cos 1 2x2 y sin 1 2x221 2x2 41 2x sin1 2x2 54. g sec g sec
12 tan 12
12 sec 12 12 tan 12 sec 12 tan 12 12 2
sec 12 tan 12
1 1 sec 2 2
2
2
cos v cos v sin v csc v
56. gv
gv cos vcos v sin vsin v cos2 v sin2 v cos 2v 60. gt 5 cos 2 t 5cos t2
58. y 2 tan3 x y 6 tan2 x sec2 x
gt 10 cos tsin t 10 sin tcos t 5 sin 2 t
62. ht 2 cot2 t 2
64.
ht 4 cot t 2csc2 t 2
y 3x 5 cos x 2 3x 5 cos 2 x 2
4 cot t 2 csc2 t 2
dy 3 5 sin 2 x 22 2x dx 3 10 2x sinx2
66. y sin x13 sin x13 y cos x13
68.
13x 31sin x 23
cos x 1 cos x13 3 x23 sin x23
23
cos x
y 3x3 4x15, 2, 2 1 y 3x3 4x459x2 4 5 y2
70. f x
x2
1 x2 3x2, 3x2
fx 2x2 3x32x 3 f4 74. y
y
22x 3 x2 3x3
72. f x fx
1 2 x1 , 2, 3 2x 3
2x 31 x 12 5 2x 32 2x 32
f2 5
5 32
1 cos x, x
4, 161
9x2 4 53x3 4x45
2 , 2
1 sin x x2 2cos x
y2 is undefined.
Section 2.4
1 76. (a) f x xx2 5 , 2, 2 3
f x tan2 x,
78. (a)
f
x2 1 x2 5 3x2 5 3
62. (b)
4 , 1
4 212 4
Tangent line:
4 1 13 f2 3 33 3 9
y2
353
fx 2 tan x sec2 x
1 1 1 fx x x2 5122x x2 512 3 2 3
y14 x 64. (b)
Tangent line:
The Chain Rule
13 x 2 ⇒ 13x 9y 8 0 9
⇒ 4x y 1 0 4
4
−
6 −4 −9
9
−6
80. f x x 21 fx x 22 f x 2x 23
1 x 22
2 x 23
82. f x sec2 x fx 2 sec x sec x tan x 2 sec2 x tan x f x 2 sec2 xsec2 x 2 tan x2 sec2 x tan x 2 2 sec4 x 4 2 sec2 x tan2 x 2 2 sec2 xsec2 x 2 tan2 x 2 2 sec2 x3 sec2 x 2 84.
y
f
86.
3
f′
y
4
f
3
2
2
1 −3
f′
−2
−1
x −1
1
2
3
−2 −3
f is decreasing on , 1 so f must be negative there. f is increasing on 1, so f must be postive there. 88. gx f x 2 gx fx 22x ⇒ gx 2x fx 2
f x
−1 −2
4
f′
−3 −4
The zeros of f correspond to the points where the graph of f has horizontal tangents.
354
Chapter 2
Differentiation
90. (a) gx sin2 x cos 2 x 1 ⇒ gx 0
92. y 13 cos 12t 14 sin 12t v y 13 12 sin 12t 14 12 cos 12t
gx 2 sin x cos x 2 cos xsin x 0
4 sin 12t 3 cos 12t
(b) tan2 x 1 sec2 x
When t 8, y 0.25 feet and v 4 feet per second.
gx 1 f x Taking derivatives of both sides, gx fx. Equivalently, fx 2 sec x sec x tan x and gx 2 tan x sec2 x, which are the same. 94. y A cos t (a) Amplitude: A
3.5 1.75 2
(b) v y 1.75
y 1.75 cos t Period: 10 ⇒
0.35 sin
2 10 5
y 1.75 cos
t 5
t 5
96. (a) Using a graphing utility, or by trial and error, you obtain a model of the form
t 1 T t 64.18 22.15 sin 6
(b)
t sin 5 5
(c) Tt 22.15 cos
6t 1 6
11.60 cos
6t 1
20
100 0
13
−20 0
13 0
(d) The temperature changes most rapidly when t 4.1 (April) and t 10.1 (October). The temperature changes most slowly Tt 0 when t 1.1 (January) and t 7.1 (July). 98. (a) gx f x2 ⇒ gx fx (b) hx 2 f x ⇒ hx 2 fx (c) rx f 3x ⇒ rx f3x3 3 f3x Hence, you need to know f3x. 1 r0 3 f0 3 3 1
r1 3 f3 34 12 (d) sx f x 2 ⇒ sx fx 2 Hence, you need to know fx 2. s2 f0 13 , etc.
x fx gx hx
2
1
0
1
2
3
4
2 3
13
1
2
4
4
2 3
13
1
2
4
8
4 3
23
2
4
8
12
1
1
2
rx sx
1
3
4
Section 2.4
100. f x p f x for all x.
The Chain Rule
102. If f x f x, then
(a) Yes, fx p fx, which shows that f is periodic as well.
d d f x f x dx x fx1 fx
(b) Yes, let gx f 2x, so gx 2 f2x. Since f is periodic, so is g.
fx fx. Thus, fx is even.
104.
u u2 d 1 uu d u u u 2 u2122uu u , u 0 dx dx 2 u2 u
fx 2x
xx
2 2
108. f x sin x
106. f x x2 4
4 , x ±2 4
fx cos x
110. (a) f x sec2x
(b)
x , x k
sin sin x
6
P2
fx 2sec 2xtan 2x
P1
f x 22sec 2xtan 2x tan 2x 2sec 2xsec2 2x2 4sec 2xtan2 2x sec3 2x f
6 sec 3 2
f
6 2 sec 3 tan 3 43
f
6 423 2 56
f 0.78
0 0
3
P1x 43 x P2x
2 6
1 56 x 2 6
28 x
6
2
2
43 x
43 x
2 6
2 6
(c) P2 is a better approximation than P1. 112. False. If f x sin2 2x, then fx 2sin 2x2 cos 2x. 114. False. First apply the Product Rule.
(d) The accuracy worsens as you move away from x 6.
355
356
Chapter 2
Differentiation
Section 2.5 2.
Implicit Differentiations
x2 y2 16 2x 2yy 0 y
x3 y3 8
4.
3x2 3y2y 0
x y
y x2y y2x 2
6.
x2 y2
xy12 x 2y 0
8.
1 xy12xy y 1 2y 0 2
x2y 2xy y2 2yxy 0
x2 2xyy y2 2xy
x y y 1 2y 0 2xy 2xy
yy 2x y xx 2y
xy y 2xy 4xy y 0 y
2 sin x cos y 1
10.
2sin xsin yy cos ycos x 0 y
2xy y 4xy x
sin x cos y2 2
12.
2sin x cos y cos x sin yy 0
cos x sin yy 0
cos x cos y sin x sin y
y
cot x cot y 14.
cot y x y
16. x sec
csc2 yy 1 y y
1
1 1 csc2 y
y
1 tan2 y cot 2 y 18. (a) x2 4x 4 y2 6y 9 9 4 9
x 22 y 32 4 (Circle) y 32 4 x 22 y 3± 4 x 22 (c) Explicitly:
1 y
y 1 1 sec tan y2 y y
y2 1 1 y2 cos cot sec1y tan1y y y
(b)
y x 1
2
3
−1
4
5 2
y = −3 + 4 − (x − 2)
−2 −3 −4 −5
dy 1 ± 4 x 22122x 2 dx 2
x 2
4 x 22
x 2 ± 4 x 22
x 2 3 ± 4 x 22 3
x 2 y3
y = −3 −
4 − (x − 2)2
(d) Implicitly: 2x 2yy 4 6y 0
2y 6y 2x 2 y
x 2 y3
cos x sin y
Section 2.5
20. (a) 9y 2 x 2 9 y2 y
x2 9
(b) x2
1
9 9
4
−6
6
± x 2 9
−4
3
dy (c) Explicitly: dx (d) Implicitly:
9y 2
Implicit Differentiation
1 2
± x 2 9122x
3
±x
3x 9 2
x ±x 3± 3y 9y
x 9 2
18yy 2x 0 18yy 2x y
2x x 18y 9y
x2 y3 0
22.
x y3 x3 y3
24.
2x 3y2y 0
x3 3x2y 3xy2 y3 x3 y3
2x 3y2
3x2y 3xy2 0
2 At 1, 1: y . 3
x 2y 2xy 2xyy y 2 0
y
x 2 y xy 2 0
x2 2xyy y2 2xy y
y y 2x xx 2y
At 1, 1: y 1. x3 y3 4xy 1
26.
3x 2 3y 2y 4xy 4y
3y 2
4xy 4y y
At 2, 1, y
30.
x cos y 1
28.
xy sin y cos y 0
3x 2
y
4y 3x 2 3y 2 4x
4 12 8 38 5
1 cot y cot y x x
3 : y 21 3.
At 2,
4 xy2 x3 4 x2yy y21 3x2 y
cos y x sin y
x3 y3 6xy 0
32.
3x2 3y2y 6xy 6y 0
3x2 y2 2y4 x
y3y2 6x 6y 3x2 y
At 2, 2: y 2. At
6y 3x2 2y x2 3y2 6x y2 2x
169 32 4 . 43, 83 : y 163 649 83 40 5
357
358
Chapter 2
Differentiation
cos y x
34.
sin y y 1 y
1 ,0 < y < sin y
sin2 y cos2 y 1 sin2 y 1 cos2 y sin y 1 cos2 y 1 x 2 y
1 1 x 2
, 1 < x < 1 x2y2 2x 3
36.
2x2yy 2xy2 2 0 x2yy xy2 1 0 y
1 xy2 x2y
2xyy x2 y2 x2yy 2xyy y2 0 4xyy x2y2 x2yy y2 0 4 4xy2 1 xy22 x2yy y2 0 x x2y2 4xy2 4x2y4 1 2xy2 x2y4 x 4y3y x2y4 0 x 4y3y 2x2y4 2xy2 1 y 38. 1 xy x y
2x2y4 2xy2 1 x4y3
y xy x 1 y
y2 4x
40.
2yy 4
x1 1 1x
y
y 0
y 2y2y
y 0 42.
y2 2yy y
At 2,
2 y
x1 x2 1
x2 11 x 12x x2 12 x2 1 2x2 2x x2 12 1 2x x2 2yx2 12 5
5
: y 2 15544 4 1 10
Tangent line: y
1
5
5
2
1 x 2 105
105y 10 x 2 x 105y 8 0
5
.
1
(2, ) 5 5
−1
5
−1
2 4 y
2 y y 2
3
Section 2.5
Implicit Differentiation
44. x2 y2 9 y
x y
4
(0, 3)
At 0, 3:
−6
6
Tangent line: y 3 Normal line: x 0.
−4
At 2, 5 :
4
Tangent line: y 5 Normal line: y 5 46.
2 x 2 ⇒ 2x 5y 9 0 5 5
2
(2, 5 ) −6
6
x 2 ⇒ 5x 2y 0.
−4
y2 4x 2yy 4 y
2 1 at 1, 2 y
Equation of normal at 1, 2 is y 2 1x 1, y 3 x. The centers of the circles must be on the normal and at a distance of 4 units from 1, 2. Therefore,
x 12 3 x 22 16 2x 12 16 x 1 ± 22 . Centers of the circles: 1 22, 2 22 and 1 22, 2 22 Equations: x 1 22 2 y 2 22 2 16
x 1 22 2 y 2 22 2 16 48. 4x2 y2 8x 4y 4 0 8x 2yy 8 4y 0 y
8 8x 4 4x 2y 4 y2
Horizontal tangents occur when x 1: 41 2
y2
81 4y 4 0
y2 4y y y 4 0 ⇒ y 0, 4 Horizontal tangents: 1, 0, 1, 4. Vertical tangents occur when y 2: 4x2 22 8x 42 4 0 4x2 8x 4xx 2 0 ⇒ x 0, 2 Vertical tangents: 0, 2, 2, 2.
y
(1, 0) −1
1
x 2
3
−1
(2, − 2)
(0, − 2) −3 −4 −5
(1, − 4)
4
359
360
Chapter 2
Differentiation
50. Find the points of intersection by letting y2 x3 in the equation 2x2 3y2 5. 2x2 3x3 5
and
3x3 2x2 5 0
2
2x2 + 3y2 = 5
Intersect when x 1.
(1, 1) −2
Points of intersection: 1, ± 1 y2 x3:
2x 2 3y 2 5:
2yy 3x2
4x 6yy 0
y
4
(1, − 1)
3x2 2y
y
−2
y 2= x 3
2x 3y
At 1, 1, the slopes are: y
3 2
2 y . 3
At 1, 1, the slopes are: y
3 2
2 y . 3
Tangents are perpendicular. 52. Rewriting each equation and differentiating, x3 3y 1 y
x3y 29 3
3
x 1 3
y
y x2
x (3y − 29) = 3 15
1 3 29 3 x
y
x 3 = 3y − 3
1 . x2
−15
12 −3
For each value of x, the derivatives are negative reciprocals of each other. Thus, the tangent lines are orthogonal at both points of intersection. 54.
x2 y2 C2 2x 2yy 0
y Kx
2
2
K=1
y K −3
x y y
K = −1
C=1
−3
3
3
C=2
At the point of intersection x, y the product of the slopes is xyK xKxK 1. The curves are orthogonal.
−2
−2
56. x2 3xy2 y3 10 (a) 2x 3y2 6xyy 3y2y 0
(b) 2x
dx dy dy dx 3y 2 6xy 3y2 0 dt dt dt dt
6xy 3y2y 3y2 2x
2x 3y2
3y2 2x y 2 3y 6xy 58. (a) 4 sin x cos y 1
(b) 4 sin xsin y
4 sin xsin yy 4 cos x cos y 0 y
cos x cos y sin x sin y
cos x cos y
dx dy 6xy 3y2 dt dt
dy dx 4 cos x cos y 0 dt dt
dx dy sin x sin y dt dt
Section 2.6
60. Given an implicit equation, first differentiate both sides with respect to x. Collect all terms involving y on the left, and all other terms to the right. Factor out y on the left side. Finally, divide both sides by the left-hand factor that does not contain y.
Related Rates
62.
18
00
1671
B 1994
A 18
00
Use starting point B.
x y c
64. 1 2x
1
dy
2y dx
0
y dy dx x
Tangent line at x0, y0: y y0
y0
x x0
x0
x-intercept: x0 x0y0, 0 y-intercept: 0, y0 x0y0 Sum of intercepts:
x0 x0y0 y0 x0y0 x0 2x0y0 y0 x0 y0 2 c 2 c.
Section 2.6 2.
Related Rates
y 2x 2 3x dy dx 4x 6 dt dt
2x
dy dx 2y 0 dt dt
x dx dy dt y dt
dx 1 dy dt 4x 6 dt (a) When x 3 and
x2 y2 25
4.
dx dy 2, 43 62 12 dt dt
(b) When x 1 and
dy dx 1 5 5, 5 dt dt 41 6 2
dx y dy dt x dt (a) When x 3, y 4, and dxdt 8, dy 3 8 6 dt 4 (b) When x 4, y 3, and dydt 2, 3 3 dx 2 . dt 4 2
361
362
6.
Chapter 2
y
Differentiation
8.
1 1 x2
dx 2 dt
dx 2 dt
dx dy cos x dt dt
dy 2x dx dt 1 x22 dt
y sin x
(a) When x 6,
(a) When x 2,
dy cos 2 3 cmsec. dt 6
dy 222 8 cmsec. dt 25 25
(b) When x 4,
(b) When x 0,
dy cos 2 2 cmsec. dt 4
dy 0 cmsec. dt
dy cos 2 1 cmsec. dt 3
dy 222 8 cmsec. dt 25 25
(b) 14.
dx dy negative ⇒ negative dt dt
(c) When x 3,
(c) When x 2,
10. (a)
12. Answers will vary. See page 145.
dx dy positive ⇒ positive dt dt
D x2 y2 x2 sin2 x dx 2 dt dx x sin x cos x dx 2 2 sin x cos x dD 1 2 x sin2 x122x 2 sin x cos x dt 2 dt x2 sin2 x dt x2 sin2x
16.
A r2 dA dr 2 r dt dt If drdt is constant, dAdt is not constant. dA dr depends on r and . dt dt
18.
4 V r3 3 dr 2 dt dr dV 4 r 2 dt dt (a) When r 6,
dV 4 622 288 in3min. dt
When r 24,
dV 4 2422 4608 in3min. dt
(b) If drdt is constant, dVdt is proportional to r2.
20.
V x3 dx 3 dt dV dx 3x 2 dt dt (a) When x 1, dV 3123 9 cm3sec. dt
(b) When x 10, dV 310 2 3 900 cm3sec. dt
Section 2.6
22.
1 1 V r 2h r 23r r 3 3 3
Related Rates
1 1 25 3 25 3 V r 2h h h 3 3 144 3144
24.
dr 2 dt
By similar triangles, 5r 12h ⇒ r 125 h.
dV dr 3 r 2 dt dt
dV 10 dt
(a) When r 6,
dV 25 2 dh dh 144 dV h ⇒ dt 144 dt dt 25 h 2 dt
dV 3 622 216 in3min. dt
363
When h 8,
(b) When r 24,
dh 144 9 ftmin. 10 dt 2564 10 5
dV 3 2422 3456 in3min. dt r 12
h
1 26. V bh12 6bh 6h2 since b h 2 (a)
dV dh dh 1 dV 12h ⇒ dt dt dt 12h dt When h 1 and
(b) If
x2 y2 25
28. 2x
dy dx 2y 0 dt dt dx y dt x
dV dh 1 1 2, 2 ftmin dt dt 121 6
dh 3 dV 3 and h 2, then 122 9 ft3min. dt 8 dt 8
0.15y dy dy since 0.15. dt x dt
When x 2.5, y 18.75,
18.75 dx 0.15 0.26 msec dt 2.5
12 ft 3 ft
5
y
h ft
3 ft
x
30. Let L be the length of the rope. (a)
L2 144 x2 dL dx 2L 2x dt dt dx L dL 4L dL since 4 ftsec. dt x dt x dt When L 13, x L2 144 169 144 5 413 52 dx 10.4 ftsec. dt 5 5 Speed of the boat increases as it approaches the dock.
(b) If
dx 4, and L 13, dt
dL x dx dt L dt
5 4 13
20 ftsec 13
4 ft/sec 13 ft 12 ft
As L → 0,
dL increases. dt
364
Chapter 2
32.
x 2 y 2 s2 2x
Differentiation
34. s2 902 x2
since dydt 0
dx ds 0 2s dt dt dx s ds dt x dt
When s 10, x 100 25 75 53 dx 10 480 240 1603 277.13 mph. dt 3 53
2nd
30 ft
x 60
x
dx 28 dt ds x dt s
3rd
1st
s 90 ft
dx dt
Home
When x 60, s 902 602 3013 ds 60 56 28 15.53 ftsec. dt 3013 13
y
x
s
5 mi
x
20 y 6 yx
36. (a)
20y 20x 6y 20
14y 20x 10 x y 7
6 x y
dx 5 dt 50 dy 10 dx 10 5 ftsec dt 7 dt 7 7 (b)
50 35 15 d y x dy dx 50 5 ftsec dt dt dt 7 7 7 7
38. xt
3 sin t, x2 y2 1 5
(a) Period:
(c) When x
2 2 seconds
3 (b) When x , y 5 Lowest point:
3 1 5
2
3 ,y 10
1 14
2
15
4
and
3 1 1 3 sin t ⇒ sin t ⇒ t 10 5 2 6
4 m. 5
dx 3 cos t dt 5
4 0, 5
x2 y2 1 2x Thus,
dx dy dy x dx 2y 0⇒ . dt dt dt y dt 310 dy dt 154
5 cos 6 3
95 9 . 125 255
Speed
95 0.5058 msec 125
Section 2.6
1 1 1 R R1 R2
40.
42.
rg tan v2
32r sec2
dR2 1.5 dt 1 dR 2 dt R1
365
32r tan v2, r is a constant.
dR1 1 dt
1 R2
Related Rates
d dv 2v dt dt d dv 16r sec2 dt v dt
dR1 1 2 dt R2
dR2 dt
Likewise,
d v dv cos2 . dt 16r dt
When R1 50 and R2 75, R 30
1 1 dR 302 1 1.5 dt 502 752 0.6 ohmssec. sin
44.
10 x
dx 1ftsec dt cos
x 10
dx dt ddt 10 x
θ
2
d 10 dx 2 sec dt x dt
10 10 1 2 221 25 1 0.017 radsec 252 102 252 25 521 2521 525 Police
x 50
tan
46.
d 302 60 radmin radsec dt sec2
θ 50 ft
ddt 501 dxdt dx d 50 sec2 dt dt
(a) When 30 ,
dx 200 ftsec. dt 3
(c) When 70 ,
dx 427.43 ftsec. dt
48. sin 22
x y
(b) When 60 ,
dx 200 ftsec. dt
50. (a) dydt 3dxdt means that y changes three times as fast as x changes.
0 dx x dt y
x
x y2
dy 1 dt y
dx dt
dy sin 22 240 89.9056 mihr dt
y x 22˚
(b) y changes slowly when x 0 or x L. y changes more rapidly when x is near the middle of the interval.
366
Chapter 2
Differentiation
52. L2 144 x2; acceleration of the boat
First derivative: 2L L
d 2x . dt 2
dx dL 2x dt dt dx dL x dt dt
Second derivative: L
d 2L dL dt 2 dt
dL d 2x dx x 2 dt dt dt
dx dt
L ddtL dLdt dxdt
1 d 2x dt 2 x When L 13, x 5,
2
2
2
2
dx dL d 2L dL is constant, 2 0. 10.4, and 4 (see Exercise 30). Since dt dt dt dt
d 2x 1 130 42 10.42 dt 2 5 1 1 16 108.16 92.16 18.432 ftsec2 5 5 54.
yt 4.9t2 20
y
dy 9.8t dt y1 4.9 20 15.1 y1 9.8
20
y 20 By similar triangles, x x 12 20x 240 xy. When y 15.1, 20x 240 x15.1
20 15.1x 240 x
240 . 4.9
20x 240 xy 20
dx dy dx x y dt dt dt dx x dy dt 20 y dt
At t 1,
dx 2404.9 9.8 97.96 msec. dt 20 15.1
y 12
(0, 0)
x
x
92
Chapter 2
Differentiation
51. x2 y2 25; acceleration of the top of the ladder
First derivative: 2x
d 2y dt 2
dy dx 2y 0 dt dt dy dx y 0 dt dt
x Second derivative: x
d 2x dx dt 2 dt
d 2y dy dx y 2 dt dt dt
dy 0 dt
xddt x dxdt dydt
1 d 2y dt 2 y When x 7, y 24,
2
2
2
2
dy 7 dx dx d 2x , and 2 (see Exercise 27). Since is constant, 2 0. dt 12 dt dt dt
d 2y 1 7 70 22 dt 2 24 12
49 1 625 241 4 144 24 144 0.1808 ft sec 2
2
53. (a) Using a graphing utility, you obtain ms 0.881s2 29.10s 206.2 (b)
dm dm ds ds 1.762s 29.10 dt ds dt dt
(c) If t s 1995, then s 15.5 and Thus,
ds 1.2. dt
dm 1.76215.5 29.101.2 2.15 million. dt
Review Exercises for Chapter 2 1. f x x2 2x 3 fx lim
x→0
f x x f x x
x x2 2x x 3 x2 2x 3 x→0 x
lim
x2 2xx x2 2x 2x 3 x2 2x 3 x→0 x
lim
2xx x2 2x lim 2x x 2 2x 2 x→0 x→0 x
lim
5. f is differentiable for all x 1.
3. f x x 1 fx lim
x→0
lim
x→0
lim
x→0
lim
x→0
lim
x→0
f x x f x x
x x 1 x 1 x x x x
x
x x x x x x
x x x x x x x 1 x x x
1 2 x
Review Exercises for Chapter 2
7. f x 4 x 2
1 4 9. Using the limit definition, you obtain gx x . 3 6
(a) Continuous at x 2. (b) Not differentiable at x 2 because of the sharp turn in the graph.
4 1 3 At x 1, g1 3 6 2
y 7 6 5 4 3 2 x
−1
1 2 3 4 5 6
−2 −3
11. (a) Using the limit defintion, fx 3x 2.
13. g2 lim
x→2
At x 1, f1 3. The tangent line is y 3x 1
x3 x 2 4 x→2 x2
lim
0
−4
x 2x 1 4 x→2 x2
lim
y 2 3x 1
(b)
gx g2 x2
2
lim
(−1, −2)
x→2
x 2x 2 x 2 x2
lim x 2 x 2 8 x→2
−4
15.
19. f x x8
17. y 25
y
f′
f
fx 8x7
y 0
2
1
x −1
1
21. ht 3t 4
23. f x x3 3x2
ht 12t 3
fx 3x2 6x 3xx 2
3 x 6x1 2 3x1 3 25. hx 6 x 3
hx 3x1 2 x2 3 29. f 2 3 sin f 2 3 cos
3 x
2 27. gt t2 3
1 3 x2
gx
4 3 4 t 3 3 3t
31. f 3 cos
sin 4
f 3 sin
cos 4
93
94
Chapter 2
Differentiation
F 200 T
33.
Ft
35.
st 16t2 s0 s9.2 169.22 s0 0
100 T
s0 1354.24
(a) When T 4, F4 50 vibrations/sec/lb.
The building is approximately 1354 feet high (or 415 m).
(b) When T 9, F9 3313 vibrations/sec/lb. 37. (a)
(c) Ball reaches maximum height when x 25.
y
y x 0.02x2
(d)
15
y 1 0.04x
10
y0 1
5
y10 0.6
x 20
40
60
Total horizontal distance: 50 (b) 0 x 0.02x2
y25 0 y30 0.2 y50 1
x 0x 1 implies x 50. 50 39. xt t2 3t 2 t 2t 1 (a) vt xt 2t 3
(e) y25 0
3 (b) vt < 0 for t < 2 .
(d) xt 0 for t 1, 2.
at vt 2 3 (c) vt 0 for t 2 . 3 3 1 1 1 x 2 2 2 1 2 2 4
v1 21 3 1
v2 22 3 1 The speed is 1 when the position is 0. 43. hx x sin x x1 2 sin x
41. f x 3x2 7x2 2x 3 fx 3x2 72x 2 x2 2x 36x 2
6x3
9x2
16x 7
1 sin x x cos x 2 x
x2 x 1 x2 1 x2 12x 1 x2 x 12x fx x2 12 2 x 1 2 x 12
45. f x 2x x2
47. f x
fx 2 2x3 2 1
hx
1 x3
2x3 1 x3
51. y
49. f x 4 3x21 fx 4 3x226x 53. y 3x 2 sec x y 3x 2 sec x tan x 6x sec x
6x 4 3x22
y
x2 cos x cos x 2x x 2sin x 2x cos x x 2 sin x cos2 x cos2 x
55. y x tan x y x sec2 x tan x
Review Exercises for Chapter 2
59. gt t3 3t 2
57. y x cos x sin x
gt 3t2 3
y x sin x cos x cos x x sin x
g t 6t 61. f 3 tan
y 2 sin x 3 cos x
63.
f 3 sec2
y 2 cos x 3 sin x
f 6 sec sec tan 6 sec tan
y 2 sin x 3 cos x
2
y y 2 sin x 3 cos x 2 sin x 3 cos x 0 65. f x 1 x31 2 1 fx 1 x31 23x2 2
3x2
2 1 x3
s2
1
s
s2
1
3s2
2
xx 31x
2
2
11 x 32x x 2 12
2x 3x 2 6x 1 x 2 13
71. y 3 cos3x 1
5 1 2s
5 2
x3 x2 1
hx 2
69. f s s2 15 2s3 5 fs
67. hx
s3
5 2
s2
3 2
y 9 sin3x 1
3s 1 5 5
3 2
s2
s3
ss2 13 28s3 3s 25
73. y
1 csc 2x 2
1 y csc 2x cot 2x2 2
y
csc 2x cot 2x
77. y
x sin 2x 2 4
75. y
1 1 cos 2x2 2 4
1 1 cos 2x sin2 x 2
2 3 2 2 sin x sin7 2x 3 7
y sin1 2 x cos x sin5 2 x cos x
79. y y
cos x sin x1 sin2 x
sin x x2
x 2 cos x sin x x 22
cos3 x sin x 81. f t t2t 15
83. gx 2xx 11 2
ft tt 147t 2 The zeros of f correspond to the points on the graph of f where the tangent line is horizontal. 0.1
gx
g does not equal zero for any value of x in the domain. The graph of g has no horizontal tangent lines.
f′
4
−0.1
1.3
g′
f −0.1
x2 x 13 2
−2
7
g −2
95
96
Chapter 2
Differentiation
85. f t t 11 2t 11 3 t 15 6 ft
87. y tan 1 x
5 6t 11 6
y
f does not equal zero for any x in the domain. The graph of f has no horizontal tangent lines.
sec2 1 x 2 1 x
y does not equal zero for any x in the domain. The graph has no horizontal tangent lines.
5
5
y
f − 20
f′ −2
7
2
y′
−1
−4
91. f x cot x
89. y 2x2 sin 2x
93. f t
t 1 t2
ft
t1 1 t3
f t
2t 2 1 t4
fx csc2 x
y 4x 2 cos 2x
f 2 csc xcsc x cot x
y 4 4 sin 2x
2 csc2 x cot x
95. g tan 3 sin 1 g 3 sec2 3 cos 1 g 18 sec2 3 tan 3 sin 1 97. T 700t2 4t 101 T
1400t 2 t2 4t 102
(a) When t 1, T
(b) When t 3,
14001 2 18.667 deg hr. 1 4 102
T
(d) When t 10,
(c) When t 5, T
14003 2 7.284 deg hr. 9 12 102
14005 2 3.240 deg hr. 25 30 102
T
140010 2 0.747 deg hr. 100 40 102
x2 3xy y3 10
99.
2x 3xy 3y 3y2y 0 3x y2y 2x 3y 2x 3y 3x y2
y
y x x y 16
101. y
12x x 1 2
y x
1 2
12y
1 2
y y1 2 0
x 2 x yy y 2 y x 2 xy x 2 xy y y 2 y 2 x y
2 xy y 2 x
2 y 2 xy x
2y x y y 2x y x x
Review Exercises for Chapter 2
x sin y y cos x
103.
105.
x cos yy sin y y sin x y cos x
6
(2, 4)
2x 2yy 0 −9
yx cos y cos x y sin x sin y y
x2 y2 20
97
9
x y y
y sin x sin y cos x x cos y
−6
1 At 2, 4: y 2 1 Tangent line: y 4 x 2 2 x 2y 10 0 Normal line: y 4 2x 2 2x y 0
107.
y x dy 2 units sec dt dx dy 1 dx dy ⇒ 2 x 4 x dt dt dt 2 x dt 1 dx 2 2 units/sec. (a) When x , 2 dt
109.
(b) When x 1,
dx 4 units/sec. dt
(c) When x 4,
dx 8 units/sec. dt
s 1 2 h 2
111. st 60 4.9t2 st 9.8t
1 s h 4
s 35 60 4.9t2 4.9t2
dV 1 dt
w 2 2s 2 2
14h 4 2 h
tan 30
dV 5 dh 4 h dt 2 dt 2dV dt dh dt 54 h When h 1,
dh 2 m min. dt 25
1 2
1 3
5 4.9 st xt
xt 3 st
4h 5 5 2 h 8 hh 2 2 4
25
t
Width of water at depth h:
V
s (t)
ds 5 dx 3 39.8 dt dt 4.9 1 2
2
38.34 m sec
s 2 h 2
30˚ x(t )
98
Chapter 2
Differentiation
Problem Solving for Chapter 2 1. (a) x 2 y r2 r2 Circle
3
x 2 y Parabola Substituting,
−3
y r2 r2 y
3 −1
y 2 2r y r 2 r 2 y y2 2r y y 0 y y 2r 1 0 Since you want only one solution, let 1 2r 0 ⇒ r 12 Graph y x 2 and x 2 y 12 2 14
(b) Let x, y be a point of tangency: x 2 y b2 1 ⇒ 2x 2 y by 0 ⇒ y
x circle. by
y x 2 ⇒ y 2x (parabola). Equating, 2x
x by
3
2b y 1 by
−3
1 1 ⇒by 2 2
3 −1
Also, x 2 y b2 1 and y x 2 imply
y y b2 1 ⇒ y y y
1 2
1 ⇒ y 21 1 ⇒ y 43 and b 45.
54
Center: 0,
Graph y x 2 and x 2 y
3. (a)
5 4
2
1
f x cos x
P2x a0 a1x a2x 2
P10 a0 ⇒ a0 1
f 0 1
P20 a0 ⇒ a0 1
P10 a1 ⇒ a1 0
f0 0
P20 a1 ⇒ a1 0
f 0 1
1 P20 2a2 ⇒ a2 2
f x cos x
P1x a0 a1x
f 0 1 f0 0
(b)
P1x 1
1 P2x 1 2x 2
(c)
x
1.0
0.1
0.001
0
0.001
0.1
1.0
cos x
0.5403
0.9950
1
1
1
0.9950
0.5403
P2x
0.5
0.9950
1
1
1
0.9950
0.5
P2x is a good approximation of f x cos x when x is near 0. (d)
f x sin x
P3x a0 a1x a2 x 2 a3x3
f 0 0
P30 a0 ⇒ a0 0
f0 1
P30 a1 ⇒ a1 1
f 0 0
P30 2a2 ⇒ a2 0
f 0 1
P30 6a3 ⇒ a3 16
P3x x 6 x3 1
Problem Solving for Chapter 2 5. Let px Ax3 Bx 2 Cx D px 3Ax 2 2Bx C At 1, 1: A B C D 1
Equation 1
3A 2B C
Equation 2
14
At 1, 3: A B C D 3 3A 2B C
Equation 3
2
Equation 4
Adding Equations 1 and 3: 2B 2D 2 Subtracting Equations 1 and 3: 2A 2C 4 Adding Equations 2 and 4: 6A 2C 12 Subtracting Equations 2 and 4: 4B 16 1 Hence, B 4 and D 22 2B 5 1 Subtracting 2A 2C 4 and 6A 2C 12, you obtain 4A 8 ⇒ A 2. Finally, C 24 2A 0
Thus, px 2x3 4x 2 5. x4 a2 x 2 a 2 y 2
7. (a)
a2 y 2 a2x 2 x 4 y
± a2x 2 x 4
a
Graph: y1 (b)
a2x 2 x4
a
and y2
a2x 2 x4
a
2
a = 12 −3
3
a=2 a=1 −2
± a, 0 are the x-intercepts, along with 0, 0. (c) Differentiating implicitly, 4x 3 2a2 x 2a2 y y y
a2 2
±a 2a2x 4x3 xa2 2x 2 0 ⇒ 2x 2 a2 ⇒ x . 2 2a2y a2y
2
a2
a2 a y 2
2 2
a4 a4 a2y 2 4 2 a2y 2
a4 4
y2
a2 4
y±
a 2
Four points:
a2, a2, a2, 2a, a2, a2, a2, 2a
99
100
Chapter 2
9. (a)
Differentiation Line determined by 0, 30 and 90, 6:
y
(0, 30)
30
y 30
(90, 6) (100, 3) x 90
100
30 6 24 4 4 x 0 x x ⇒ y x 30 0 90 90 15 15
When x 100, y
4 10 100 30 > 3 ⇒ Shadow determined by man. 15 3
Not drawn to scale
(b)
Line determined by 0, 30 and 60, 6:
y
30
(0, 30)
y 30
(60, 6) (70, 3) x 60
70
Not drawn to scale
30 6 2 2 x 0 x ⇒ y x 30 0 60 5 5
When x 70, y
2 70 30 2 < 3 ⇒ Shadow determined by child. 5
(c) Need 0, 30, d, 6, d 10, 3 collinear. 30 6 63 24 3 ⇒ ⇒ d 80 feet 0d d d 10 d 10 (d) Let y be the length of the street light to the tip of the shadow. We know that For x > 80, the shadow is determined by the man. dy 5 dx 25 5 y yx ⇒ y x and . 30 6 4 dt 4 dt 4 For x < 80, the shadow is determined by the child. y y x 10 10 100 dy 10 dx 50 . ⇒y x and 30 3 9 9 dt 9 dt 9 Therefore,
25 dy 4 dt 50 9
x > 80 0 < x < 80
dy is not continuous at x 80. dt
11. Lx lim
x→0
Lx x Lx x
lim
Lx Lx Lx x
lim
Lx x
x→0
x→0
Also, L0 lim
x→0
Lx L0 x
But, L0 0 because L0 L0 0 L0 L0 ⇒ L0 0. Thus, Lx L0, for all x. The graph of L is a line through the origin of slope L0.
dx 5. dt
Problem Solving for Chapter 2 13. (a)
z (degrees)
0.1
0.01
0.0001
sin z z
0.0174524
0.0174533
0.0174533
(b) lim
z→0
sin z 0.0174533 z
In fact, lim
z→0
(c)
sin z z 180
d sin z z sin z sin z lim z →0 dz z lim
sin z cos z sin z cos z sin z z
lim
sin z
z →0
z →0
cos z 1 z
sin z0 cos z (d) S90 sin
lim
z →0
cos z
sin z z
cos z 180 180
90 sin 1; C180 cos 180 1 180 2 180
d d Sz sincz c coscz Cz dz dz 180 (e) The formulas for the derivatives are more complicated in degrees. 15. jt at (a) jt is the rate of change of the acceleration. (b) From Exercise 102 in Section 2.3, st 8.25t 2 66t vt 16.5t 66 at 16.5 at jt 0
101
Review Exercises for Chapter 2
Review Exercises for Chapter 2
2. f x
x1 x1
x x 1 x 1 f x x f x x x 1 x 1 fx lim lim x→0 x→0 x x
x x 1x 1 x x 1x 1 xx x 1x 1
lim
x→0
x2 xx x x x 1 x2 xx x x x 1 x→0 xx x 1x 1
lim
2x 2 2 lim xx x 1x 1 x→0 x x 1x 1 x 12
lim
x→0
2 4. f x x
6. f is differentiable for all x 3.
fx lim
x→0
f x x f x x
2 2 x x x lim x→0 x lim
2x 2x 2x xx xx
lim
2x xx xx
lim
2 2 2 x xx x
x→0
x→0
x→0
8. f x
x1 4x4xx2,,
if x < 2 if x ≥ 2
2
2
10. Using the limit defintion, you obtain hx
(a) Nonremovable discontinuity at x 2.
At x 2, h2
(b) Not differentiable at x 2 because the function is discontinuous there. y 5 4
1 −5 −4
−2 −1 −1
x 1
2
−2
12. (a) Using the limit definition, fx
2 . x 12
At x 0, f0 2. The tangent line is y 2 2x 0 y 2x 2
(b)
4
(0, 2) −6
6
−4
3 67 42 . 8 8
3 4x. 8
367
368
Chapter 2
14. f2 lim
x→2
Differentiation
f x f 2 x2
16.
y
1
1 1 x1 3 lim x→2 x2 lim
3x1 x 2x 13
lim
1 1 x 13 9
x→2
x→2
18. y 12
−
π 2
x
π 2
−1
f′
22. f t 8t 5
20. gx x12
y 0
f
ft 40t 4
gx 12x11
24. gs 4s4 5s2
2 28. hx x2 9
26. f x x12 x12
gs 16s3 10s
1 1 x1 fx x12 x32 32 2 2 2x
30. g 4 cos 6
32. g
g 4 sin g
hx
4 3 4 x 3 9 9x
5 sin 2 3 5 cos 2 3
34. s 16t2 s0 First ball: 16t2 100 0 t
10 2.5 seconds to hit ground 100 16 4
Second ball: 16t2 75 0 t2
7516 5 4 3 2.165 seconds to hit ground
Since the second ball was released one second after the first ball, the first ball will hit the ground first. The second ball will hit the ground 3.165 2.5 0.665 second later. 36. st 16t2 14,400 0 16t2 14,400 t 30 sec 1 1 Since 600 mph 6 mi/sec, in 30 seconds the bomb will move horizontally 6 30 5 miles.
Review Exercises for Chapter 2
38.
y
v02 64
(
v02 v02 , 64 128
) 2 0
( v32 , 0 ) x v02 128
(a) y x
32 32 2 x x 1 2x v02 v0
0 if x 0 or x
(b) y 1
v02 . 32
64 x v02
When x
v02 64 v02 , y 1 2 0. 64 v0 64
Projectile strikes the ground when x v0232. Projectile reaches its maximum height at x v0264. (one-half the distance) (c) y x
32 2 32 x x 1 2x 0 v02 v0
(d) v0 70 ftsec
when x 0 and x x0232. Therefore, the range is x v0232. When the initial velocity is doubled the range is x
Range: x
Maximum height: y
2v0 4v0 32 32 2
v02 702 153.125 ft 32 32
2
v02 702 38.28 ft 128 128
50
or four times the initial range. From part (a), the maximum height occurs when x v0264. The maximum height is 0
v02 v02 32 v02 2 y 64 64 v0 64
2
v02 v02 v02 . 64 128 128
0
160
If the initial velocity is doubled, the maximum height is
y
v02 2v02 2v02 4 64 128 128
or four times the original maximum height. 40. (a) y 0.14x2 4.43x 58.4
(b)
320
0
60 0
(c)
(d) If x 65, y 362 feet.
12
0
60 0
(e) As the speed increases, the stopping distance increases at an increasing rate.
369
370
Chapter 2
Differentiation
42. gx x3 3xx 2
44. f t t 3 cos t
gx x3 3x1 x 23x 2 3
ft t3sin t cos t3t2
x3 3x 3x3 6x 2 3x 6
t3 sin t 3t2 cos t
4x3 6x 2 6x 6 46. f x fx
x1 x1
48. f x
x 11 x 11 x 12
fx
2 x 12
50. f x 93x2 2x1 fx 93x2 2x26x 2 54. y 2x x 2 tan x
52. y 181 3x 3x2 2x2
y 56. y
y 2 x 2 sec2 x 2x tan x y 58. vt 36 t2, 0 ≤ t ≤ 6
sin x x2
x2 cos x sin x2x x cos x 2 sin x x4 x3 1 sin x 1 sin x
1 sin x cos x 1 sin xcos x 1 sin x2 2 cos x 1 sin x2
f x
a4 8 msec
h t 4 sin t 5 cos t
23 5x 3x2 x2 12
fx 3x34
v4 36 16 20 msec
ht 4 cos t 5 sin t
x2 16 6x 52x x2 12
60. f x 12x14
at vt 2t
62. ht 4 sin t 5 cos t
6x 5 x2 1
9 74 9 x 74 4 4x
y
64.
10 cos x x
xy cos x 10 xy y sin x 0 xy sin x y xy y sin x y y sin x
66. f x x2 113 1 fx x2 1232x 3 2x 3x2 123
68. f x x2
5
1 x
2x x1
1 x
fx 5 x2
4
2
Review Exercises for Chapter 2
70. h
371
72. y 1 cos 2x 2 cos2 x
1 3
y 2 sin 2x 4 cos x sin x
1 3 31 21 h 1 6
2 2 sin x cos x 4 sin x cos x 0
1 21 3 2 1 1 6 1 4 74. y csc 3x cot 3x
76. y
y 3 csc 3x cot 3x 3 csc2 3x 3 csc 3xcot 3x csc 3x
sec7 x sec5 x 7 5
y sec6 xsec x tan x sec4 xsec x tan x sec5 x tan xsec2 x 1 sec5 x tan3 x
3x
78. f x
80. y
x 2 1
1 3x 2 112 3x x 2 1122x 2 fx x2 1
y
x 1 sinx 1 cosx 11 x 12
3x 2 1 3x 2 x 2 132
cosx 1 x1
1 x 1 sinx 1 cosx 1 x 12
3 x 2 132
82. f x x 2x 4 2 x2 2x 82
100
f′
fx 4x3 3x2 6x 8 4x 2x 1x 4
−7
The zeros of f correspond to the points on the graph of f where the tangent line is horizontal.
gx
2x2 1 x2 1
y
g does not equal zero for any value of x. The graph of g has no horizontal tangent lines.
y does not equal zero for any x in the domain. The graph has no horizontal tangent lines. 75
y′
y
g′ 6
−3
3
g −3
− 60
3x 227x 2 23x
5
−6
5
86. y 3xx 23
84. gx xx2 112
− 25
f
372
Chapter 2
Differentiation
88. y 2 csc3 x y
3 x
10
csc3 x cot x
y
The zero of y corresponds to the point on the graph of y where the tangent line is horizontal.
−4
2 sec xsec x tan x
y 2 cos 2x
2 2 sec2 x tan x x3
96. hx xx2 1
x2x2 3 x2 132
x2 9y2 4x 3y 0
23x2 5x 3 x2 12
g x
26x3 15x2 18x 5 x2 13
(a) When h 9,
4 dv ftsec. dh 3
(b) When h 4,
dv 2 ftsec. dh
y2 x3 x2y xy y2
102.
2x 18yy 4 3y 0
0 x3 x2y xy 2y2
36y 1y 4 2x y
0 3x2 x2y 2xy xy y 4yy
x2 x 4yy 3x2 2xy y
4 2x 36y 1
cosx y x
104.
gx
4 dv dh h
x2 1
h x
6x 5 x2 1
v 2gh 232h 8h
98.
2x2 1
hx
100.
94. gx
y 2 sin x cos x sin 2x
y x2 sec2 x y 2x
8
92. y sin2 x
90. y x1 tan x 3
y′
−1
106.
1 y sinx y 1
3x2 2xy y x2 x 4y
x2 y2 16
10
2x 2yy 0
y sinx y 1 sinx y y
y
1 sinx y sinx y
cscx 1 1
y
−10
x y
At 5, 3: y
5 3
5 Tangent line: y 3 x 5 3 5x 3y 16 0 3 Normal line:y 3 x 5 5 3x 5y 30 0 108. Surface area A 6x 2, x length of edge. dx 5 dt da dx 12x 124.55 270 cm2sec dt dt
10
−10
Problem Solving for Chapter 2 tan x
110.
d 32 radmin dt sec2
1
θ
ddt dxdt
x
dx tan2 16 6x2 1 dt 1 dx 1 15 When x , 6 1 kmmin 450 kmhr. 2 dt 4 2
Problem Solving for Chapter 2 2.
Let a, a2 and b, b2 2b 5 be the points of tangency.
y 10 8 6 4 −8 −6 −4 −2
For y x 2, y 2x and for y x 2 2x 5, y 2x 2. Thus, 2a 2b 2 ⇒ a b 1, or a 1 b. Furthermore, the slope of the common tangent line is
x 2 4 6 8 10
−4 −6
a2 b2 2b 5 1 b2 b2 2b 5 2b 2 ab 1 b b ⇒
1 2b b2 b2 2b 5 2b 2 1 2b
⇒ 2b2 4b 6 4b2 6b 2 ⇒ 2b2 2b 4 0 ⇒ b2 b 2 0 ⇒ b 2b 1 0 b 2, 1 For b 2, a 1 b 1 and the points of tangency are 1, 1, 2, 5. The tangent line has slope 2: y 1 2x 1 ⇒ y 2x 1 For b 1, a 1 b 2 and the points of tangency are 2, 4 and 1, 8. The tangent line has slope 4: y 4 4x 2 ⇒ y 4x 4. 4. (a) y x 2, y 2x. Slope 4 at 2, 4. Tangent line: y 4 4x 2 y 4x 4 1
(b) Slope of normal line: 4. 1 Normal line: y 4 4x 2
y 14x 92 y 14x 92 x 2 ⇒ 4x 2 x 18 0 ⇒ 4x 9x 2 0 x 2, 94. Second intersection point: 94, 81 16 (c) Tangent line: y 0 Normal line: x 0 —CONTINUED—
373
374
Chapter 2
Differentiation
4. —CONTINUED— (d) Let a, a2, a 0, be a point on the parabola y x 2. Tangent line at a, a2 is y 2ax a a2. Normal line at a, a2 is 1 y x a a2. To find points of intersection, solve 2a x2 x2 x2
1 x a a2 2a
1 1 x a2 2a 2
1 1 1 1 x a2 2a 16a2 2 16a2
x 4a1 a 4a1 2
x
2
1 1 ± a 4a 4a
x
1 1 a ⇒ x a Point of tangency 4a 4a
x
1 1 1 2a2 1 a ⇒ x a 4a 4a 2a 2a
The normal line intersects a second time at x
2a2 1 . 2a
6. f x a b cos cx fx bc sin cx At 0, 1: a b 1 At
Equation 1
4 , 32: a b cosc4 23
Equation 2
c4 1
Equation 3
bc sin
From Equation 1, a 1 b. Equation 2 becomes 1 b b cos From Equation 3, b
1 cos
c4 23 ⇒ b b cos c4 12
1 1 1 1 c . Thus cos c c c 4 2 c sin c sin c sin 4 4 4
c4 21 c sinc4
Graphing the equation gc
1 c c c sin cos 1, you see that many values of c will work. 2 4 4
1 3 3 1 One answer: c 2, b , a ⇒ f x cos 2x 2 2 2 2
Problem Solving for Chapter 2 8. (a) b2y 2 x3a x; a, b > 0 y2
(b) a determines the x-intercept on the right: a, 0.
x a x b2 3
b affects the height.
x3a x
Graph y1
b
and y2
x3a x
b
(c) Differentiating implicitly. 2b 2 y y 3x 2a x x 3 3ax 2 4x 3 y
3ax 2 4x3 0 2b2y
⇒ 3ax 2 4x3 3a 4x x b2y 2 y2
3a . 4
1 a 3a4 a 3a4 27a 64 4 3
3
27a 4 33a2 ⇒y± 256b 2 16b
Two points:
10. (a) y x13 ⇒
3a4 , 3 16b3a , 3a4 , 316b3a
2
2
dy 1 23 dx x dt 3 dt
1 dx 1 823 3 dt dx 12 cmsec dt (b) D x 2 y 2 ⇒
dD 1 2 dx dy x y 2 2x 2y dt 2 dt dt
dx dy y dt dt x 2 y 2 x
y (c) tan ⇒ sec2 x
812 21 98 49 cmsec. 64 4 68 17
dy dx x y d dt dt dt x2
68 2
θ 8
From the triangle, sec
68
8
. Hence
d 81 212 16 4 radsec dt 68 68 17 64 64
375
376
Chapter 2
Differentiation
12. Ex lim
Ex x Ex
x
lim
ExE x Ex
x
x→0
x→0
lim Ex
E x x 1
Ex lim
E x 1
x
x→0
x→0
But, E0 lim
x→0
E x E0 E x 1 lim 1.
x→0
x
x
Thus, Ex ExE0 Ex exists for all x. For example: Ex e x.
14. (a) vt
27 t 27 ftsec 5
at (b) vt
27 ftsec2 5 27 27 t 27 0 ⇒ t 27 ⇒ t 5 seconds 5 5
S5
27 2 5 275 6 73.5 feet 10
(c) The acceleration due to gravity on Earth is greater in magnitude than that on the moon.
C H A P T E R 3 Applications of Differentiation Section 3.1
Extrema on an Interval
. . . . . . . . . . . . . . 103
Section 3.2
Rolle’s Theorem and the Mean Value Theorem
Section 3.3
Increasing and Decreasing Functions and the First Derivative Test . . . . . . . . . . . . . . 113
Section 3.4
Concavity and the Second Derivative Test . . . . 121
Section 3.5
Limits at Infinity
Section 3.6
A Summary of Curve Sketching
Section 3.7
Optimization Problems . . . . . . . . . . . . . . 145
Section 3.8
Newton’s Method . . . . . . . . . . . . . . . . . 155
Section 3.9
Differentials . . . . . . . . . . . . . . . . . . . . 160
. 107
. . . . . . . . . . . . . . . . . 129 . . . . . . . . . 136
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 163 Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . 172
C H A P T E R 3 Applications of Differentiation Section 3.1
Extrema on an Interval
Solutions to Odd-Numbered Exercises
1. f x fx
x2 x2 4
3. f x x
x2 42x x22x 8x 2 x2 42 x 42
f0 0
5.
fx 1 27x3 1 f3 1
f x x 223
27 27 x x2 2x2 x 27 x3
27 110 33
7. Critical numbers: x 2 x 2: absolute maximum
2 fx x 213 3 f2 is undefined. 9. Critical numbers: x 1, 2, 3
11. f x x2x 3 x3 3x2
x 1, 3: absolute maximum
fx 3x2 6x 3xx 2
x 2: absolute minimum
Critical numbers: x 0, x 2 15. hx sin2 x cos x, 0 < x < 2
13. gt t4 t, t < 3
124 t
gt t
1 4 t12
12
1 4 t12t 24 t 2
hx 2 sin x cos x sin x sin x2 cos x 1 On 0, 2, critical numbers: x
5 , x , x 3 3
8 3t 24 t
8 Critical number is t . 3 17. f x 23 x, 1, 2
19. f x x2 3x, 0, 3
fx 2 ⇒ No critical numbers
fx 2x 3
Left endpoint: 1, 8 Maximum
Left endpoint: 0, 0 Minimum
Right endpoint: 2, 2 Minimum
Critical number:
32 , 94 Maximum
Right endpoint: 3, 0 Minimum
103
104
Chapter 3
Applications of Differentiation
3 21. f x x3 x2, 1, 2 2 fx 3x2 3x 3xx 1 Left endpoint:
1, 25 Minimum
23. f x 3x23 2x, 1, 1 fx 2x13 2
3 x 2 1 3 x
Left endpoint: 1, 5 Maximum
Right endpoint: 2, 2 Maximum
Critical number: 0, 0 Minimum
Critical number: 0, 0
Right endpoint: 1, 1
1, 21
Critical number:
25. gt gt
t2
t2 , 1, 1 3
27. hs
6t t2 32
Left endpoint:
hs
1, 41 Maximum
Critical number: 0, 0 Minimum Right endpoint:
1 s 22
Left endpoint:
0, 21 Maximum
Right endpoint: 1, 1 Minimum
1, 14 Maximum 6
29. f x cos x, 0,
1
31. y
fx sin x Left endpoint: 0, 1 Maximum Right endpoint:
1 , 0, 1 s2
1 3 Minimum , 6 2
y
4 x tan , 1, 2 x 8 4 x 0 sec2 x2 8 8
x 4 sec2 2 8 8 x On the interval 1, 2, this equation has no solutions. Thus, there are no critical numbers. Left endpoint: 1, 2 3 1, 4.4142 Maximum Right endpoint: 2, 3 Minimum
33. (a) Minimum: 0, 3 Maximum: 2, 1 (b) Minimum: 0, 3
35. f x x2 2x (a) Minimum: 1, 1 Maximum: 1, 3
(c) Maximum: 2, 1
(b) Maximum: 3, 3
(d) No extrema
(c) Minimum: 1, 1 (d) Minimum: 1, 1
Section 3.1
37. f x
2x4x , 2, 2
0 ≤ x ≤ 1 1 < x ≤ 3
Extrema on an Interval
105
3 , 1, 4 x1 Right endpoint: 4, 1 Minimum
39. f x
Left endpoint: 0, 2 Minimum
4
Right endpoint: 3, 36 Maximum 36 −1
4 −1
−1
3 −4
41. (a)
f x 3.2x5 5x3 3.5x, 0, 1
(b)
5
(1, 4.7)
0
fx 16x4 15x2 3.5 16x4 15x2 3.5 0
1
(0.4398, − 1.0613)
x2
−2
Maximum: 1, 4.7 (endpoint)
Minimum: 0.4398, 1.0613
x
15 ± 152 4163.5 216 15 ± 449 32
15 32
449
0.4398
f 0 0 f 1 4.7 Maximum (endpoint) f
15 32
449
1.0613
Minimum: 0.4398, 1.0613 43. f x 1 x312, 0, 2 3 fx x21 x312 2 3 f x x4 4x1 x332 4 3 fx x6 20x3 81 x352 8 Setting f 0, we have x6 20x3 8 0. x3 x
20 ± 400 418 2 3 10
± 108 3 1
In the interval 0, 2, choose x
3 10
± 108 3 1 0.732.
3 10 108 f 1.47 is the maximum value.
45.
f x x 123, 0, 2 fx
2 x 113 3
2 f x x 143 9 8 fx x 173 27 f 4x f 5x
56 x 1103 81
560 x 1133 243 56
f 40 81 is the maximum value.
106
Chapter 3
Applications of Differentiation
47. f x tan x f is continuous on 0, 4 but not on 0, . y
49.
51. (a) Yes
53. (a) No
(b) No
(b) Yes
5 4 3
lim tan x .
x → 2
f
2 1
x
−2 −1
1
3
4
5
6
−2 −3
55. P VI RI 2 12I 0.5I 2, 0 ≤ I ≤ 15
S 6hs
57.
P 0 when I 0.
3s2 3 cos , ≤ ≤ 2 sin 6 2
dS 3s2 3csc cot csc2 d 2
P 67.5 when I 15. P 12 I 0
Critical number: I 12 amps
3s 2 csc 3cot csc 0 2
csc 3cot
When I 12 amps, P 72, the maximum output.
sec 3
No, a 20-amp fuse would not increase the power output. P is decreasing for I > 12.
arcsec3 0.9553 radians S
6 6hs 3s2 3
S
2 6hs 3s2 3
2
2
S arcsec3 6hs
3s 2 2 2
S is minimum when arcsec3 0.9553 radians. 59. (a) y ax2 bx c
y
A
y 2ax b
B
The coordinates of B are 500, 30, and those of A are 500, 45. From the slopes at A and B,
9% −500
6%
x 500
1000a b 0.09 1000a b 0.06. Solving these two equations, you obtain a 340000 and b 3200. From the points 500, 30 and 500, 45, you obtain 30
3 3 c 5002 500 40000 200
45
3 3 c. 5002 500 40000 200
In both cases, c 18.75 y
75 . Thus, 4
3 75 3 x2 x . 40000 200 4
—CONTINUED—
Section 3.2
Rolle’s Theorem and the Mean Value Theorem
107
59. —CONTINUED— (b)
x
500
400
300
200
100
000
100
200
300
400
500
d
0
.75
3
6.75
12
18.75
12
6.75
3
.75
0
For 500 ≤ x ≤ 0, d ax2 bx c 0.09x. For 0 ≤ x ≤ 500, d ax2 bx c 0.06x. (c) The lowest point on the highway is 100, 18, which is not directly over the point where the two hillsides come together. 61. True. See Exercise 25.
Section 3.2
63. True.
Rolle’s Theorem and the Mean Value Theorem
1. Rolle’s Theorem does not apply to f x 1 x 1 over 0, 2 since f is not differentiable at x 1.
3. f x x2 x 2 x 2x 1 x-intercepts: 1, 0, 2, 0 1 fx 2x 1 0 at x . 2
5. f x x x 4
7. f x x2 2x, 0, 2
x-intercepts: 4, 0, 0, 0
f 0 f 2 0
1 fx x x 41 2 x 41 2 2
f is continuous on 0, 2. f is differentiable on 0, 2. Rolle’s Theorem applies.
x 41 2
2x x 4
3 8 fx x 4 x 41 2 0 at x 2 3 9. f x x 1x 2x 3, 1, 3
fx 2x 2 2x 2 0 ⇒ x 1 c value: 1 11. f x x2 3 1, 8, 8
f 1 f 3 0
f 8 f 8 3
f is continuous on 1, 3. f is differentiable on 1, 3. Rolle’s Theorem applies.
f is continuous on 8, 8. f is not differentiable on 8, 8 since f0 does not exist. Rolle’s Theorem does not apply.
f x x3 6x2 11x 6 fx 3x2 12x 11 3x2 12x 11 0 ⇒ x c
6 3 6 3 ,c 3 3
6 ± 3 3
108
Chapter 3
13. f x
Applications of Differentiation
x2 2x 3 , 1, 3 x2
f 1 f 3 0 f is continuous on 1, 3. (Note: The discontinuity, x 2, is not in the interval.) f is differentiable on (1, 3. Rolle’s Theorem applies. fx
x 22x 2 x2 2x 31 0 x 22 x2 4x 1 0 x 22 x
4 ± 2 5 2 ± 5 2
c value: 2 5
15. f x sin x, 0, 2
17. f x
f 0 f 2 0
f 0 f
f is continuous on 0, 2. f is differentiable on 0, 2. Rolle’s Theorem applies.
0, 6
6 0
f is continuous on 0, 6. f is differentiable on 0, 6. Rolle’s Theorem applies.
fx cos x c values:
6x 4 sin2 x,
3 , 2 2
fx
6 8 sin x cos x 0
6 8 sin x cos x 1 3 sin 2x 4 2 3 sin 2x 2
3 1 arcsin x 2 2 x 0.2489 c value: 0.2489 19. f x tan x, 0,
f x x 1, 1, 1
21.
f 0 f 0
f 1 f 1 0
f is not continuous on 0, since f 2 does not exist. Rolle’s Theorem does not apply.
f is continuous on 1, 1. f is not differentiable on 1, 1 since f0 does not exist. Rolle’s Theorem does not apply. 1
−1
1
−1
Section 3.2
23.
4, 4 1 1
f x 4x tan x,
25. f t 16t2 48t 32
(b) v ft must be 0 at some time in 1, 2.
f is continuous on 1 4, 1 4. f is differentiable on 1 4, 1 4. Rolle’s Theorem applies.
ft 32t 48 0 t
fx 4 sec2 x 0 sec2 x
109
(a) f 1 f 2 64
1 1 f 0 4 4
f
Rolle’s Theorem and the Mean Value Theorem
3 seconds 2
4
sec x ± x±
2
1 2 1
arcsec ± arccos
2
± 0.1533 radian c values: ± 0.1533 radian 0.5
− 0.25
0.25
− 0.5
27.
29. f x
y
tangent line
1 , 0, 6 x3
f has a discontinuity at x 3.
(c2, f(c2)) f
(a, f(a)) (b, f(b)) (c1, f(c1)) a tangent line
b secant line
x
31. f x x2 is continuous on 2, 1 and differentiable on 2, 1. f 1 f 2 1 4 1 1 2 3 1 fx 2x 1 when x . Therefore, 2 1 c . 2
33. f x x2 3 is continuous on 0, 1 and differentiable on 0, 1. f 1 f 0 1 10 2 fx x1 3 1 3 x
23
c
8 27
3
8 27
110
Chapter 3
Applications of Differentiation
35. f x 2 x is continuous on 7, 2 and differentiable on 7, 2. f 2 f 7 0 3 1 2 7 9 3 fx
37. f x sin x is continuous on 0, and differentiable on 0, . f f 0 0 0 0 0
1 1 3 2 2 x
fx cos x 0 c
2 2 x 3
39. f x (a)
2 x
3 2
2x
9 4
x
1 4
c
1 4
2
x 1 on , 2 . x1 2 (c) fx
1
f
tangent − 0.5
2
secant
1 2 x 12 3
x 12
−1
3 2
x 1 ±
(b) Secant line: f 2 f 1 2 2 3 1 2 slope 2 1 2 5 2 3 y
2 2 x 2 3 3
6
2
In the interval 1 2, 2, c 1 6 2. f c
1 6 2
1 6 2 1
Tangent line: y 1
3y 2 2x 4 3y 2x 2 0
32 1 ±
y1
2 2 6 1
6
6
6 2 2 1 x 3 2
6
6
3
6 2 2 x 3 3 3
3y 2x 5 2 6 0
Section 3.2
Rolle’s Theorem and the Mean Value Theorem
111
41. f x x, 1, 9
1, 1, 9, 3 m (a)
31 1 91 4 (c)
3
tangent
1 2 x
f 9 f 1 1 91 4
secant
f
fx
1
9 1
1 (b) Secant line: y 1 x 1 4 y
3 1 x 4 4
1 1 2 c 4
c 2
c4
c, f c 4, 2
0 x 4y 3
m f4
1 4
1 Tangent line: y 2 x 4 4 y
1 x1 4
0 x 4y 4 43. st 4.9t 2 500 (a) Vavg
45. No. Let f x x2 on 1, 2.
s3 s0 455.9 500 14.7 m sec 30 3
(b) st is continuous on 0, 3 and differentiable on 0, 3. Therefore, the Mean Value Theorem applies.
fx 2x f0 0 and zero is in the interval (1, 2 but f 1 f 2.
vt st 9.8t 14.7 m sec t
14.7 1.5 seconds 9.8
47. Let St be the position function of the plane. If t 0 corresponds to 2 P.M., S0 0, S5.5 2500 and the Mean Value Theorem says that there exists a time t0, 0 < t0 < 5.5, such that St0 vt0
2500 0 454.54. 5.5 0
Applying the Intermediate Value Theorem to the velocity function on the intervals 0, t0 and t0, 5.5, you see that there are at least two times during the flight when the speed was 400 miles per hour. 0 < 400 < 454.54
112
Chapter 3
Applications of Differentiation
49. (a) f is continuous on 10, 4 and changes sign, f 8 > 0, f 3 < 0. By the Intermediate Value Theorem, there exists at least one value of x in 10, 4 satisfying f x 0.
(b) There exist real numbers a and b such that 10 < a < b < 4 and f a f b 2. Therefore, by Rolle’s Theorem there exists at least one number c in 10, 4 such that fc 0. This is called a critical number.
y
(c)
y
(d)
8
8
4
4 x
−8
−4
x −8
4
−4
4
−4
−4
−8
−8
(e) No, f did not have to be continuous on 10, 4. 51. f is continuous on 5, 5 and does not satisfy the conditions of the Mean Value Theorem. ⇒ f is not differentiable on 5, 5. Example: f x x
53. False. f x 1 x has a discontinuity at x 0.
y
8
f )x)
x
6
)5, 5)
) 5, 5) 4 2
x 4
2
2
4
2
55. True. A polynomial is continuous and differentiable everywhere. 57. Suppose that px x2n1 ax b has two real roots x1 and x2. Then by Rolle’s Theorem, since px1 px2 0, there exists c in x1, x2 such that pc 0. But px 2n 1x2n a 0, since n > 0, a > 0. Therefore, px cannot have two real roots. 59. If px Ax2 Bx C, then px 2Ax B
f b f a Ab2 Bb C Aa2 Ba C ba ba
Ab2 a2 Bb a ba
b aAb a B ba
Ab a B. Thus, 2Ax Ab a and x b a 2 which is the midpoint of a, b. 61. f x 12 cos x differentiable on , . fx 12 sin x 12 ≤ fx ≤
1 2
⇒ fx < 1 for all real numbers.
Thus, from Exercise 60, f has, at most, one fixed point. x 0.4502
Section 3.3
Section 3.3
Increasing and Decreasing Functions and the First Derivative Test
Increasing and Decreasing Functions and the First Derivative Test
1. f x x2 6x 8
3. y
Increasing on: 3,
Decreasing on: 2, 2 2 7. gx x 2x 8
5. f x 1 x2 x2
gx 2x 2
2 x3
Critical number: x 1
Discontinuity: x 0 Test intervals: Sign of fx:
< x < 0
Conclusion:
0 < x <
f > 0
f < 0
Increasing
Decreasing
Test intervals:
< x < 1
Sign of gx:
g < 0
g > 0
Decreasing
Increasing
Conclusion:
1 < x <
Increasing on: 1,
Increasing on , 0
Decreasing on: , 1
Decreasing on 0, Domain: 4, 4
9. y x16 x2 y
x3 3x 4
Increasing on: , 2, 2,
Decreasing on: , 3
fx
113
2 8 2 x 22x 22 16 x2 16 x2 x2
Critical numbers: x ± 22 Test intervals:
4 < x < 22
22 < x < 22
22 < x < 4
y < 0
y > 0
y < 0
Decreasing
Increasing
Decreasing
Sign of y: Conclusion:
Increasing on 22, 22 Decreasing on 4, 22, 22, 4 11. f x x2 6x
13. f x 2x2 4x 3
fx 2x 6 0
fx 4x 4 0
Critical number: x 3
Critical number: x 1
Test intervals: Sign of fx:
< x < 3
Conclusion:
3 < x <
Test intervals:
f < 0
f > 0
Sign of fx:
Decreasing
Increasing
Conclusion:
< x < 1
1 < x <
f > 0
f < 0
Increasing
Decreasing
Increasing on: 3,
Increasing on: , 1
Decreasing on: , 3
Decreasing on: 1,
Relative minimum: 3, 9
Relative maximum: 1, 5
114
Chapter 3
Applications of Differentiation
15. f x 2x3 3x2 12x fx 6x2 6x 12 6x 2x 1 0 Critical numbers: x 2, 1 Test intervals: Sign of fx:
< x < 2
2 < x < 1
f > 0
f < 0
f > 0
Increasing
Decreasing
Increasing
Conclusion:
1 < x <
Increasing on: , 2, 1, Decreasing on: 2, 1 Relative maximum: 2, 20 Relative minimum: 1, 7 17. f x x23 x 3x2 x3 fx 6x 3x2 3x2 x Critical numbers: x 0, 2 Test intervals:
< x < 0
0 < x < 2
Sign of fx:
f < 0
f > 0
f < 0
Conclusion:
Decreasing
Increasing
Decreasing
2 < x <
Increasing on: 0, 2 Decreasing on: , 0, 2, Relative maximum: 2, 4 Relative minimum: 0, 0
19. f x
x5 5x 5
fx x4 1 Critical numbers: x 1, 1 Test intervals: Sign of fx: Conclusion:
< x < 1
1 < x < 1
f > 0
f < 0
f > 0
Increasing
Decreasing
Increasing
Increasing on: , 1, 1, Decreasing on: 1, 1 Relative maximum: 1, 45 Relative minimum: 1, 45
1 < x <
Section 3.3
Increasing and Decreasing Functions and the First Derivative Test
21. f x x13 1
23. f x x 123
1 1 fx x23 23 3 3x
fx
Critical number: x 0
Critical number: x 1
< x < 0
Test intervals: Sign of fx: Conclusion:
0 < x <
2 3x 113
< x < 1
Test intervals:
f > 0
f > 0
Sign of fx:
Increasing
Increasing
Conclusion:
Decreasing
Increasing
No relative extrema
Decreasing on: , 1 Relative minimum: 1, 0
1, x < 5 x5 x5 1, x > 5
Critical number: x 5 Test intervals: Sign of fx:
< x < 5
Conclusion:
5 < x <
f > 0
f < 0
Increasing
Decreasing
Increasing on: , 5 Decreasing on: 5, Relative maximum: 5, 5 27. f x x fx 1
1 x 1 x2 1 2 x x2
Critical numbers: x 1, 1 Discontinuity: x 0 Test intervals: Sign of fx: Conclusion:
< x < 1
1 < x < 0
0 < x < 1
f > 0
f < 0
f < 0
f > 0
Increasing
Decreasing
Decreasing
Increasing
Increasing on: , 1, 1, Decreasing on: 1, 0, 0, 1 Relative maximum: 1, 2 Relative minimum: 1, 2
1 < x <
f > 0
Increasing on: 1,
25. f x 5 x 5
1 < x <
f < 0
Increasing on: ,
fx
115
116
Chapter 3
29. f x fx
Applications of Differentiation
x2 x2 9 18x x2 92x x22x 2 x2 92 x 92
Critical number: x 0 Discontinuities: x 3, 3 Test intervals: Sign of fx:
< x < 3
3 < x < 0
0 < x < 3
f > 0
f > 0
f < 0
f < 0
Increasing
Increasing
Decreasing
Decreasing
Conclusion:
3 < x <
Increasing on: , 3, 3, 0 Decreasing on: 0, 3, 3, Relative maximum: 0, 0 31. f x fx
x2 2x 1 x1
x 12x 2 x2 2x 11 x2 2x 3 x 3x 1 x 12 x 12 x 12
Critical numbers: x 3, 1 Discontinuity: x 1 Test intervals: Sign of fx:
< x < 3
3 < x < 1
1 < x < 1
f > 0
f < 0
f < 0
f > 0
Increasing
Decreasing
Decreasing
Increasing
Conclusion:
1 < x <
Increasing on: , 3, 1, Decreasing on: 3, 1, 1, 1 Relative maximum: 3, 8 Relative minimum: 1, 0 33. f x fx
x cos x, 0 < x < 2 2 1 sin x 0 2
Critical numbers: x
Test intervals:
5 , 6 6
0 < x <
6
5 < x < 6 6
5 < x < 2 6
Sign of fx:
f > 0
f < 0
f > 0
Conclusion:
Increasing
Decreasing
Increasing
Increasing on: Decreasing on:
0, 6 , 56, 2
6 , 56
6 , 126 3 5 5 6 3 Relative minimum: , 6 12 Relative maximum:
Section 3.3
Increasing and Decreasing Functions and the First Derivative Test
117
35. f x sin2 x sin x, 0 < x < 2 fx 2 sin x cos x cos x cos x2 sin x 1 0
7 3 11 , , , 2 6 2 6
Critical numbers: x
0 < x <
Test intervals:
2
7 < x < 2 6
7 3 < x < 6 2
11 3 < x < 2 6
11 < x < 2 6
Sign of fx:
f > 0
f < 0
f > 0
f < 0
f > 0
Conclusion:
Increasing
Decreasing
Increasing
Decreasing
Increasing
0, 2 , 76, 32 , 116, 2
Increasing on: Decreasing on:
2 , 76 , 32, 116
Relative minima:
76, 41 , 116, 14
Relative maxima:
2 , 2 , 32, 0
37. f x 2x9 x2, 3, 3 (a) fx
29 2x2 9 x2
(c)
y
(b) f′
Critical numbers: x ±
f
10 8
3 2
±
32 2
(d) Intervals:
4 2
3, 3 2 2 3 2 2, 3 2 2 3 2 2, 3
x 1
29 2x2 0 9 x2
1
2
8 10
fx < 0
fx > 0
fx < 0
Decreasing
Increasing
Decreasing
f is increasing when f is positive and decreasing when f is negative. 39. f t t 2 sin t, 0, 2 (a) ft t2 cos t 2t sin t tt cos t 2 sin t (b)
(c) tt cos t 2 sin t 0 t 0 or t 2 tan t t cot t 2
y 40
t 2.2889, 5.0870 (graphing utility)
f′
30
Critical numbers: t 2.2889, t 5.0870
20 10
−10 −20
t
π 2
2π
f
(d) Intervals:
0, 2.2889
2.2889, 5.0870
5.0870, 2
ft > 0
ft < 0
ft > 0
Increasing
Decreasing
Increasing
f is increasing when f is positive and decreasing when f is negative.
118
41.
Chapter 3
f x
Applications of Differentiation
x5 4x3 3x x2 1x3 3x x3 3x, x ± 1 x2 1 x2 1
y
f x gx x3 3x for all x ± 1.
(− 1, 2)
fx 3x2 3 3x2 1, x ± 1
fx 0 −4 −3
f symmetric about origin
x
−1
1 2 3 4 5
−2 −3 −4 −5
zeros of f: 0, 0, ± 3, 0
(1, − 2)
Holes at 1, 2 and 1, 2
No relative extrema 43. f x c is constant ⇒ fx 0
45. f is quadratic ⇒ f is a line.
y
y
4
4
2
f′
2
f′ −4
5 4 3
−2
2
x
x −4
4
−2
2
−2
−2
−4
−4
4
47. f has positive, but decreasing slope y
4 2
f′ x −4
−2
2
4
−2 −4
In Exercises 49–53, f x > 0 on , 4, f x < 0 on 4, 6 and f x > 0 on 6, . 49. gx f x 5
51.
gx fx
gx f x
53. gx f x 10
gx fx
gx fx 10 g0 f10 > 0
g6 f6 < 0
g0 f0 < 0
> 0, x < 4 ⇒ f is increasing on , 4. 55. fx undefined, x 4 < 0, x > 4 ⇒ f is decreasing on 4, . Two possibilities for f x are given below. (a)
y
(b)
y
6
2
4
1
x 1
2
1 x 2 −2
6
8
3
3
4
5
Section 3.3
Increasing and Decreasing Functions and the First Derivative Test
57. The critical numbers are in intervals 0.50, 0.25 and 0.25, 0.50 since the sign of f changes in these intervals. f is decreasing on approximately 1, 0.40, 0.48, 1, and increasing on 0.40, 0.48.
y
1
x
−1
Relative minimum when x 0.40.
1
Relative maximum when x 0.48.
−1
59. f x x, gx sin x, 0 < x < (a)
0.5
x
1
1.5
2
2.5
3
f x
0.5
1
1.5
2
2.5
3
gx
0.479
0.841
0.997
0.909
0.598
0.141
f x seems greater than gx on 0, . (b)
(c) Let hx f x gx x sin x
5
hx 1 cos x > 0 on 0, . Therefore, hx is increasing on 0, . Since h0 0, hx > 0 on 0, . Thus,
0 −2
x sin x > 0
x > sin x on 0,
x > sin x f x > gx on 0, .
61. v kR rr2 kRr2 r3
63.
v k2Rr 3r2
P
dP R R22vR1 vR1R22R1 R21 1 dR2 R1 R24
kr2R 3r 0 2
r 0 or 3 R Maximum when r
vR1R2 , v and R1 are constant R1 R22
2 3 R.
vR1R1 R2 0 ⇒ R2 R1 R1 R23
Maximum when R1 R2 . 65. (a) B 0.1198t4 4.4879t3 56.9909t2 223.0222t 579.9541 (b)
1500
0
20 0
(c) B 0 for t 2.78, or 1983, (311.1 thousand bankruptcies) Actual minimum: 1984 (344.3 thousand bankruptcies) 3 1 (c) The solution is a0 a1 0, a2 , a3 : 2 2
67. (a) Use a cubic polynomial f x a3 x 3 a 2 x 2 a1x a0.
1 3 f x x 3 x 2. 2 2
(b) fx 3a 3 x 2 2a 2 x a1.
0, 0:
2, 2:
0 a0
f 0 0
0 a1
f0 0
2 8a 3 4a 2
f 2 2
0 12a3 4a 2
f2 0
(d)
4
(2, 2) −2
(0, 0)
−4
4
119
120
Chapter 3
Applications of Differentiation
69. (a) Use a fourth degree polynomial f x a4 x 4 a 3 x 3 a 2 x 2 a1 x a0. (b) fx 4a4x3 3a3x2 2a2x a1 (0, 0:
4, 0:
2, 4:
0 a0
f 0 0
0 a1
f0 0
0 256a4 64a3 16a2
f 4 0
0 256a4 48a3 8a2
f4 0
4 16a4 8a3 4a2
f 2 4
0 32a4 12a3 4a2
f2 0
1 (c) The solution is a0 a1 0, a2 4, a3 2, a4 . 4 1 f x x4 2x3 4x2 4 (d)
5
(2, 4)
−2
(0, 0)
(4, 0)
5
−1
71. True
73. False
Let hx f x gx where f and g are increasing. Then hx fx gx > 0 since fx > 0 and gx > 0.
Let f x x3, then fx 3x2 and f only has one critical number. Or, let f x x3 3x 1, then fx 3x2 1 has no critical numbers.
75. False. For example, f x x3 does not have a relative extrema at the critical number x 0. 77. Assume that fx < 0 for all x in the interval a, b and let x1 < x2 be any two points in the interval. By the Mean Value Theorem, we know there exists a number c such that x1 < c < x2, and fc
f x2 f x1 . x2 x1
Since fc < 0 and x2 x1 > 0, then f x2 f x1 < 0, which implies that f x2 < f x1. Thus, f is decreasing on the interval. 79. Let f x 1 xn nx 1. Then fx n1 xn1 n n1 xn1 1 > 0 since x > 0 and n > 1. Thus, f x is increasing on 0, . Since f 0 0 ⇒ f x > 0 on 0,
1 xn nx 1 > 0 ⇒ 1 xn > 1 nx.
Section 3.4
Section 3.4
Concavity and the Second Derivative Test
Concavity and the Second Derivative Test 3. f x
1. y x2 x 2, y 2 Concave upward: ,
24 1444 x2 , y x 12 x2 123 2
Concave upward: , 2, 2, Concave downward: 2, 2
5. f x
x2 1 43x2 1 , y 2 2 x 1 x 13
7. f x 3x2 x3 fx 6x 3x2
Concave upward: , 1, 1,
f x 6 6x
Concave downward: 1, 1
Concave upward: , 1 Concave downward: 1,
9. y 2x tan x,
2 , 2
11. f x x3 6x2 12x fx 3x2 12x 12
y 2 sec2 x
f x 6x 2 0 when x 2.
y 2 sec2 x tan x Concave upward:
,0 2
Concave upward: 2,
2
0,
Concave downward:
13.
The concavity changes at x 2. 2, 8 is a point of inflection. Concave downward: , 2
1 f x x4 2x2 4 fx x3 4x f x 3x2 4 f x 3x2 4 0 when x ±
Test interval:
< x <
Sign of f x: Conclusion: Points of inflection:
2 3
2 3
.
2 3
< x <
2
2
3
3
< x <
f x > 0
f x < 0
f x > 0
Concave upward
Concave downward
Concave upward
± 23, 209
121
122
Chapter 3
Applications of Differentiation
15. f x xx 43 fx x3x 4 2 x 43 x 424x 4 f x 4x 12x 4 4x 42 4x 42x 1 x 4 4x 43x 6 12x 4x 2 f x 12x 4x 2 0 when x 2, 4. < x < 2
2 < x < 4
f x > 0
f x < 0
f x > 0
Concave upward
Concave downward
Concave upward
Test interval: Sign of f x: Conclusion:
4 < x <
Points of inflection: 2, 16, 4, 0 17. f x xx 3, Domain: 3, fx x f x
12x 3
1 2
x 3
3x 2 2x 3
6x 3 3x 2x 31 2 3x 4 4x 3 4x 33 2
f x > 0 on the entire domain of f (except for x 3, for which f x is undefined). There are no points of inflection. Concave upward on 3, 19. f x
x x2 1
fx
1 x2 x2 12
f x
2xx2 3 0 when x 0, ± 3 x2 13
Test intervals:
< x < 3
3 < x < 0
0 < x < 3
Sign of fx:
f < 0
f > 0
f < 0
f > 0
Conclusion:
Concave downward
Concave upward
Concave downward
Concave upward
Test interval:
0 < x < 2
Points of inflection:
21. f x sin
3,
2x , 0 ≤ x ≤ 4
1 x fx cos 2 2
1 x f x sin 4 2
f x 0 when x 0, 2, 4. Point of inflection: 2, 0
3
4
, 0, 0,
3,
3
4
3 < x <
Sign of f x: Conclusion:
2 < x < 4
f < 0
f > 0
Concave downward
Concave upward
Section 3.4
, 0 < x < 4 2
tan x 2 2
23. f x sec x fx sec x
Concavity and the Second Derivative Test
f x sec3 x
sec x tan2 x 0 for any x in the domain of f. 2 2 2
Concave upward: 0, , 2, 3 Concave downward: , 2, 3, 4 No points of inflection 25. f x 2 sin x sin 2x, 0 ≤ x ≤ 2 f x 2 cos x 2 cos 2x f x 2 sin x 4 sin 2x 2 sin x1 4 cos x f x 0 when x 0, 1.823, , 4.460. Test interval:
0 < x < 1.823
1.823 < x <
< x < 4.460
4.460 < x < 2
Sign of f x:
f < 0
f > 0
f < 0
f > 0
Concave downward
Concave upward
Concave downward
Concave upward
Conclusion:
Points of inflection: 1.823, 1.452, , 0, 4.46, 1.452 27. f x x4 4x3 2
29. f x x 52
fx 4x3 12x2 4x2x 3
fx 2x 5
f x 12x 2 24x 12xx 2
f x 2
Critical numbers: x 0, x 3
Critical number: x 5
However, f 0 0, so we must use the First Derivative Test. fx < 0 on the intervals , 0 and 0, 3; hence, 0, 2 is not an extremum. f 3 > 0 so 3, 25 is a relative minimum. 31. f x x3 3x2 3
f 5 > 0 Therefore, 5, 0 is a relative minimum.
33. gx x26 x3
fx 3x2 6x 3xx 2
gx xx 6212 5x
f x 6x 6 6x 1
g x 46 x5x2 24x 18
Critical numbers: x 0, x 2
12 Critical numbers: x 0, 5 , 6
f 0 6 < 0 Therefore, 0, 3 is a relative maximum. f 2 6 > 0 Therefore, 2, 1 is a relative minimum.
g 0 432 > 0 Therefore, 0, 0 is a relative minimum. g 12 5 155.52 < 0 Therefore, 5 , 268.7 is a relative minimum. 12
g 6 0 Test fails by the First Derivative Test, 6, 0 is not an extremum.
123
124
Chapter 3
Applications of Differentiation 4 x
37. f x x
35. f x x2 3 3 fx
2 3x1 3
fx 1
f x
2 9x4 3
f x
Critical number: x 0
4 x2 4 2 x x2
8 x3
Critical numbers: x ± 2
However, f 0 is undefined, so we must use the First Derivative Test. Since fx < 0 on , 0 and fx > 0 on 0, , 0, 3 is a relative minimum.
f 2 < 0 Therefore, 2, 4 is a relative maximum. f 2 > 0 Therefore, 2, 4 is a relative minimum.
39. f x cos x x, 0 ≤ x ≤ 4 fx sin x 1 ≤ 0 Therefore, f is non-increasing and there are no relative extrema. 41. f x 0.2x2x 33, 1, 4 (a) fx 0.2x5x 6x 32
(c)
y
f x x 34x 9.6x 3.6 2
0.4x 310x 24x 9
2 1
(b) f 0 < 0 ⇒ 0, 0 is a relative maximum. f
6 5
x 2
> 0 ⇒ 1.2, 1.6796 is a relative minimum.
4
f is increasing when f > 0 and decreasing when f < 0. f is concave upward when f > 0 and concave downward when f < 0.
3, 0, 0.4652, 0.7049, 1.9348, 0.9049
1 1 sin 3x sin 5x, 0, 3 5
(a) fx cos x cos 3x cos 5x fx 0 when x
f x 0 when x
5 ,x ,x . 6 2 6 5 ,x , x 1.1731, x 1.9685 6 6
6 , 0.2667, 1.1731, 0.9638, 1.9685, 0.9637,
4
56, 0.2667
Note: 0, 0 and , 0 are not points of inflection since they are endpoints.
f
2
−2
2 < 0 ⇒ 2 , 1.53333 is a relative maximum.
Points of inflection:
y
(c)
f x sin x 3 sin 3x 5 sin 5x
(b) f
1
f
Points of inflection:
43. f x sin x
f ′′
f′
2
π 4
π 2
f′
π
x
−4 −6 −8
f ′′
The graph of f is increasing when f > 0 and decreasing when f < 0. f is concave upward when f > 0 and concave downward when f < 0.
Section 3.4 f < 0 means f decreasing
y
45. (a)
Concavity and the Second Derivative Test
f increasing means concave upward
4 3
f > 0 means f increasing
y
(b)
f increasing means concave upward
4 3
2
2
1
1 x 1
3
2
x 1
4
47. Let f x x4.
49.
3
2
4
y
f
f x 12x2
2
f′
f 0 0, but 0, 0 is not a point of inflection.
f ′′
y
x
−2
6
1 −1
5 4 3 2 1 −3
−2
x
−1
51.
1
2
3
53.
y
f ′′
f′
y
f
4
4
2
x 2
2
(2, 0) (4, 0) x
2
2
4
55.
4
6
2
y
57.
y
3 2
f
1
(2, 0)
(4, 0) x
1 1 2
2
3
4
5
x 4
8 8
125
12
f ′′
f is linear. f is quadratic. f is cubic. f concave upwards on , 3, downward on 3, .
126
Chapter 3
Applications of Differentiation
59. (a) n 1:
n 2:
n 3:
n 4:
f x x 2
f x x 22
f x x 23
f x x 24
fx 1
fx 2x 2
fx 3x 22
fx 4x 23
f x 0
f x 2
f x 6x 2
fx 12x 22
No inflection points
No inflection points
Inflection point: 2, 0
No inflection points:
Relative minimum: 2, 0
6
−9
9
Relative minimum: 2, 0
6
−9
6
9
6
Point of inflection −6
−9
9
−6
−6
−9
9
−6
Conclusion: If n ≥ 3 and n is odd, then 2, 0 is an inflection point. If n ≥ 2 and n is even, then 2, 0 is a relative minimum. (b) Let f x x 2n, fx nx 2n1, f x nn 1x 2n2. For n ≥ 3 and odd, n 2 is also odd and the concavity changes at x 2. For n ≥ 4 and even, n 2 is also even and the concavity does not change at x 2. Thus, x 2 is an inflection point if and only if n ≥ 3 is odd. 61. f x ax3 bx 2 cx d Relative maximum: 3, 3 Relative minimum: 5, 1 Point of inflection: 4, 2 fx 3ax 2 2bx c, f x 6ax 2b
f 3 27a 9b 3c d 3 98a 16b 2c 2 ⇒ 49a 8b c 1 f 5 125a 25b 5c d 1 f3 27a 6b c 0, f 4 24a 2b 0 49a 8b c 1
24a 2b
27a 6b c
22a 2b 1
22a 2b a
1 2,
0
1
b 6, c
45 2 ,
2a
d 24
1 45 f x 2 x3 6x 2 2 x 24
0
1
Section 3.4
Concavity and the Second Derivative Test
127
63. f x ax3 bx2 cx d Maximum: 4, 1 Minimum: 0, 0 (a) fx 3ax2 2bx c,
f x 6ax 2b
f 0 0 ⇒ d 0
(b) The plane would be descending at the greatest rate at the point of inflection.
f 4 1 ⇒ 64a 16b 4c 1 f4 0 ⇒
48a 8b c 0
f0 0 ⇒
c0
f x 6ax 2b
3 3 x 0 ⇒ x 2. 16 8
Two miles from touchdown.
1 3 Solving this system yields a 32 and b 6a 16 . 1 3 3 2 f x 32 x 16 x
65. D 2 x4 5L x3 3L 2x 2
C 0.5x 2 15x 5000
67.
D 8x3 15L x 2 6L 2x x8x 2 15L x 6L 2 0 x 0 or x
15 ± 33 15L ± 33L L 16 16
x
15 16 33 L 0.578L.
S
5000t2 8 t2
dC 5000 0.5 2 0 when x 100 dx x
St
80,000t 8 t22
S t
80,0008 3t2 8 t23
C 5000 0.5x 15 x x
C average cost per unit
By the Second Derivative Test, the deflection is maximum when
69.
C
By the First Derivative Test, C is minimized when x 100 units.
S t 0 for t 8 3 1.633. Sales are increasing at the greatest rate at t 1.633 years. 71.
4 22
f x 2sin x cos x,
f
fx 2cos x sin x,
f
f x 2sin x cos x,
P1
0 4
−2
22 4
1 22 x 4 2 4
2
f
22 f 4
P1x 22 0 x
4
P2 −4
P1x 0 P2x 22 0 x
P2x 22 x
4
2
22 2 x
4
2
P2x 22 The values of f, P1, P2, and their first derivatives are equal at x 4. The values of the second derivatives of f and P2 are equal at x 4. The approximations worsen as you move away from x 4.
128 73.
Chapter 3
Applications of Differentiation
f x 1 x, fx f x
f 0 1
1 , 21 x
f0
1 , 41 x3 2
f 0
5
P1
1 2
f −8
4
P2
1 4
−3
21x 0 1 2x
P1x 1 P1x
1 2
21x 0 21 41x 0
P2x 1
2
1
x x2 2 8
1 x P2x 2 4 P2x
1 4
The values of f, P1, P2, and their first derivatives are equal at x 0. The values of the second derivatives of f and P2 are equal at x 0. The approximations worsen as you move away from x 0. 75. f x x sin
fx x f x x
1x
1
sin1x 1x cos1x sin1x
1 1 cos x2 x
−1
x1 cos1x x1 cos1x x1 sin1x 0
1 1 1 sin x x2 x
2
2
1
( π1 , 0) −1
3
1
Point of inflection:
1 , 0
When x > 1 , f < 0, so the graph is concave downward. 77. Assume the zeros of f are all real. Then express the function as f x ax r1x r2x r3 where r1, r2, and r3 are the distinct zeros of f. From the Product Rule for a function involving three factors, we have fx ax r1x r2 x r1x r3 x r2x r3 f x ax r1 x r2 x r1 x r3 x r2 x r3 a6x 2r1 r2 r3. Consequently, f x 0 if x
2r1 r2 r3 r1 r2 r3 Average of r1, r2, and r3. 6 3
79. True. Let y ax3 bx2 cx d, a 0. Then y 6ax 2b 0 when x b 3a, and the concavity changes at this point.
Section 3.5
Limits at Infinity
83. False. Concavity is determined by f .
81. False. f x 3 sin x 2 cos x fx 3 cos x 2 sin x 3 cos x 2 sin x 0 3 cos x 2 sin x 3 2
tan x
Critical number: x tan132 f tan1 32 3.60555 is the maximum value of y.
Section 3.5 1. f x
Limits at Infinity
3x2 x2 2
x x2 2
3. f x
5. f x
4sin x x2 1
No vertical asymptotes
No vertical asymptotes
No vertical asymptotes
Horizontal asymptote: y 3
Horizontal asymptote: y 0
Horizontal asymptotes: y 0
Matches (f)
Matches (d)
Matches (b)
7. f x
4x 3 2x 1
x
100
101
102
103
104
7
2.26
2.025
2.0025
2.0003
f x
10
105 106 2
− 10
10
2 − 10
lim f x 2
x→
9. f x
6x
10
4x2 5
x
100
101
102
103
104
105
106
f x
2
2.98
2.9998
3
3
3
3
− 10
10
− 10
lim f x 3
x→
11. f x 5
1 x2 1
6
x
100
101
102
103
104
105
106
f x
4.5
4.99
4.9999
4.999999
5
5
5
lim f x 5
x→
−1
8 0
129
130
Chapter 3
Applications of Differentiation
f x 5x3 3x2 10 10 5x 3 2 x2 x2 x
13. (a) hx
lim hx
x→
x2 2 0 x→ x3 1
15. (a) lim
(Limit does not exist)
x2 2 1 x2 1
(c) lim
x2 2 x1
x→
f x 5x3 3x2 10 3 10 5 3 x3 x3 x x
(b) hx
(b) lim
x→
lim hx 5
x→
(Limit does not exist)
f x 5x3 3x2 10 5 10 3 2 4 x4 x4 x x x
(c) hx
lim hx 0
x→
5 2x32 0 x→ 3x2 4
19. lim
17. (a) lim
x→
2x 1 2 1x 2 0 2 lim 3x 2 x→ 3 2x 3 0 3
2 5 2x32 x→ 3x32 4 3
(b) lim (c) lim
x→
21. lim
x →
5 2x32 3x 4
(Limit does not exist)
x 1x 0 lim 0 x2 1 x → 1 1x2 1
23.
lim
x →
5x 2 5x lim x 3 x → 1 3x
Limit does not exist.
25.
lim
x →
x x2
x
1 , x x2
lim
lim
1
x →
1 1x
1 2 2x 1 x 27. lim lim x→ x2 x x→ x2 x x2
lim
x→
for x
x2
x →
2 1x x (1x
< 0 we have x x2
1
for x < 0, x x2
2
29. Since 1x ≤ sin2xx ≤ 1x for all x 0, we have by the Squeeze Theorem, lim
x →
1 sin2x 1 ≤ lim ≤ lim x → x → x x x
0 ≤ lim
x →
sin2x ≤ 0. x
Therefore, lim
x →
sin2x 0. x
31. lim
x →
1 0 2x sin x
Section 3.5
33. (a) f x
x x1
4
lim
y=1
y = −1
x 1 lim x → x 1 x →
Limits at Infinity
−6
6
x 1 x1
−4
Therefore, y 1 and y 1 are both horizontal asymptotes.
35. lim x sin x →
1 sin t lim 1 x t→0 t
Let x 1t.
37.
x x 3 lim x x2 3 x x2 3 x → x → 2 lim
39. lim x x2 x lim x →
x →
lim
x →
41.
x f x
x
x
x2 x
x
2
x x
lim
x →
2
101
102
103
104
105
106
1
0.513
0.501
0.500
0.500
0.500
0.500
x x2 x x → 1 x →
lim
x →
3 0 x x2 3
x 1 1 lim 2 x x2 x x → 1 1 1x
lim
f x
100
x →
x
x 3
x x2 x
lim x xx 1 lim
43.
2
x x2 x x x2 x
8
−2
x x x2 x 1 1 1 1x
1 2
100
101
102
103
104
105
106
0.479
0.500
0.500
0.500
0.500
0.500
0.500
Let x 1t. sint2 1 sint2 1 1 lim x sin lim lim x → t →0 t →0 2 2x t t2 2
−1
1
−2
2
−1
131
132
Chapter 3
45. (a)
Applications of Differentiation
47. Yes. For example, let f x
y 4 3
y
f′
2
1
8 x
−4
1
2
3
4 4
−3
2
−4
x
(b) lim f x 3
−4
lim fx 0
x →
−2
2
4
6
−2
x →
(c) Since lim f x 3, the graph approaches that of a x →
horizontal line, lim f x 0. x →
49. y
2x 1x
y 3
Intercepts: 2, 0, 0, 2
2
−3 −2 −1
Symmetry: none Horizontal asymptote: y 1 since 2x 2x 1 lim lim . x → 1 x x → 1 x
x 1
2
3
4
5
−2 −3 −4 −5
Discontinuity: x 1 (Vertical asymptote) 51. y
x2
x 4
53. y
x2
x2 9
Intercept: 0, 0
Intercept: 0, 0
Symmetry: origin
Symmetry: y-axis
Horizontal asymptote: y 0
Horizontal asymptote: y 1 since
Vertical asymptote: x ± 2
x →
y
x2
x2 x2 1 lim 2 . x → x 9 9
Relative minimum: 0, 0
5 4 3 2 1 −1
lim
y 4
x 2 3 4 5
3
−2 −3 −4 −5
2 1 −3 −2 −1
x −1 −2
1
2
3
6x2 . x 22 1
Section 3.5
55. y
2x2 4
Limits at Infinity
57. xy2 4
x2
Domain: x > 0
Intercept: 0, 0
Intercepts: none
Symmetry: y-axis
Symmetry: x-axis
Horizontal asymptote: y 2
Horizontal asymptote: y 0 since
Vertical asymptote: x ± 2
2
lim
y
x →
8
x
0 lim x →
2 x
.
Discontinuity: x 0 (Vertical asymptote)
6 4
y
2 −4
4
x
−2
4
2
6
3 2 1 −1 −1
x 1
2
3
4
5
−2 −3 −4
59. y
2x 1x
61. y 2
3 x2
Intercept: 0, 0
Intercepts: ± 32, 0
Symmetry: none
Symmetry: y-axis
Horizontal asymptote: y 2 since
Horizontal asymptote: y 2 since
lim
x →
2x 2x 2 lim . x → 1 x 1x
lim
x →
Discontinuity: x 1 (Vertical asymptote)
2
lim
x →
y
y 4
1
3 x 1
2
3
4
2
5
1
−2
−4 −3 −2
−3
x 2
3
4
−4 −5 −6
63. y 3
2 x
y
2 2 2 2 Intercept: y 0 3 ⇒ 3 ⇒ x , 0 x x 3 3
8 7 6 5 4 3 2 1
Symmetry: none Horizontal asymptote: y 3 Vertical asymptote: x 0
2 x3 . 2
Discontinuity: x 0 (Vertical asymptote)
2
−3 −2 −1
2 x3 2
−4 −3 −2 −1
x 1 2 3 4 5
133
134
Chapter 3
65. y
Applications of Differentiation
x3 x2 4
67. f x 5
1 5x2 1 2 x x2
Domain: , 2, 2,
Domain: , 0, 0,
Intercepts: none
fx
2 ⇒ No relative extrema x3
Symmetry: origin 6 ⇒ No points of inflection x4
Horizontal asymptote: none
f x
Vertical asymptotes: x ± 2 (discontinuities)
Vertical asymptote: x 0
y
Horizontal asymptote: y 5
20 16 12 8 4 −5 −4 −3 −2 −1
7
y=5 1 2 3 4 5
−8 − 12 − 16 − 20
69. f x fx f x
x=0
x
−6
6 −1
x x2 4
x2 4 x2x x2 42
3
−4
x2 4 0 for any x in the domain of f . x2 42
5
x = − 2 −3
x=2
x2 422x x2 42x2 42x x2 42 2xx2 12 0 when x 0. x2 43
Since f x > 0 on 2, 0 and f x < 0 on 0, 2, then 0, 0 is a point of inflection. Vertical asymptotes: x ± 2 Horizontal asymptote: y 0 x2 x2 71. f x x2 4x 3 x 1x 3
x2 4x 3 x 22x 4 x2 4x 5 fx 2 0 x2 4x 32 x 4x 32 f x
x2 4x 322x 4 x2 4x 52x2 4x 32x 4 x2 4x 34 2x3 6x2 15x 14 0 when x 2. x2 4x 33
Since f x > 0 on 1, 2 and f x < 0 on 2, 3, then 2, 0 is a point of inflection. Vertical asymptote: x 1, x 3 Horizontal asymptote: y 0
2
x=3 −1
5
y=0 x=1 −2
Section 3.5 3x 73. f x 2 4x 1 fx f x
Limits at Infinity
2
y= 3
3 ⇒ No relative extrema 4x2 132
−3
2
3
y= −3 2
36x 0 when x 0. 4x2 152
−2
Point of inflection: 0, 0 Horizontal asymptotes: y ±
3 2
No vertical asymptotes
75. gx sin
gx
x x 2, 3 < x <
2 cos
1.2
x x 2
y = sin(1)
x 22
3
Horizontal asymptote: y 1 Relative maximum:
( π2−π 2 , 1)
12 0
2
x ⇒ x 5.5039 x2 2
2
No vertical asymptotes 77. f x
x3 3x2 2 2 , gx x xx 3 xx 3
(a)
(c)
8
− 80
f=g −4
(b) f x
− 70
x3 3x2 2 xx 3 x2x 3 2 xx 3 xx 3
x
2 gx xx 3
79. C 0.5x 500 C x
C 0.5 lim
x →
80
8 −2
C
70
500 x
0.5 500x 0.5
The graph appears as the slant asymptote y x.
135
136
Chapter 3
Applications of Differentiation
83. (a) T1t 0.003t 2 0.677t 26.564
81. line: mx y 4 0 y
(b)
90
5
T1 y = mx + 4
3 − 10
2
130 − 10
(3, 1)
1
x −2 −1 −1
1
2
3
4
(c)
90
T2
(a) d
Ax1 By1 C m3 11 4 A2 B2
m2 1
− 10
3m 3
m2 1
(b)
120 − 10
T2
7
1451 86t 58 t
(d) T10 26.6 −6
T20 25.0
6 −1
(c) lim dm 3 lim dm m →
t→
m →
The line approaches the vertical line x 0. Hence, the distance approaches 3.
85. Answers will vary. See page 195.
Section 3.6
(e) lim T2
86 86 1
(f) The limiting temperature is 86. T1 has no horizontal asymptote.
87. False. Let f x
2x x2 2
. (See Exercise 2.)
A Summary of Curve Sketching
1. f has constant negative slope. Matches (D) 5. (a) fx 0 for x 2 and x 2
3. The slope is periodic, and zero at x 0. Matches (A) (c) f is increasing on 0, .
f > 0
f is negative for 2 < x < 2 (decreasing function). f is positive for x > 2 and x < 2 (increasing function). (b) f x 0 at x 0 (Inflection point). f is positive for x > 0 (Concave upwards). f is negative for x < 0 (Concave downward).
(d) fx is minimum at x 0. The rate of change of f at x 0 is less than the rate of change of f for all other values of x.
Section 3.6
7. y y
x2
x2 3
A Summary of Curve Sketching
y
y=1
6x 0 when x 0. x2 32
1
181 x2 0 when x ± 1. y 2 x 33
1,
1 4
1 4
1,,
x 4
2
(0, 0))
4
Horizontal asymptote: y 1 y < x < 1 1 4
x 1 1 < x < 0 x0
0
0 < x < 1 1 4
x1
1 < x <
9. y
y
Conclusion
Decreasing, concave down
0
Point of inflection
Decreasing, concave up
0
Relative minimum
Increasing, concave up
0
Point of inflection
Increasing, concave down
1 3 x2
y y
y
11. y
1 < 0 when x 2. x 22
2 x 23
No relative extrema, no points of inflection
73, 0, 0, 27
Intercepts:
Vertical asymptote: x 2 Horizontal asymptote: y 3 y
x
2x x2 1
y
2x2 1 < 0 if x ± 1. x2 12
y
4xx2 3 0 if x 0. x2 13
Inflection point: 0, 0 Intercept: 0, 0 Vertical asymptote: x ± 1 Horizontal asymptote: y 0 Symmetry with respect to the origin
2
x
1
7 , 0 3
y
x
1
4
x 4
y
0
2
x 2
4
0,,
7 2
y
3
(0, 0)
4
137
138
Chapter 3
13. gx x gx 1 g x
Applications of Differentiation
4 x2 1 8x x4 2x2 8x 1 0 when x 0.1292, 1.6085 2 2 x 1 x2 12
3 8 1 0 when x ± x2 13 3
3x2
4
3 , 2.423 3 )1.6085, 2.724) ) 1.3788, 0) x 3
2
1
g 0.1292 < 0, therefore, 0.1292, 4.064 is relative maximum.
2
3
2
y
g 1.6085 > 0, therefore, 1.6085, 2.724 is a relative minimum. Points of inflection:
)0.1292, 4.064) 3 , 3.577 3
y
)0, 4)
x
33, 2.423, 33, 3.577
Intercepts: 0, 4, 1.3788, 0 Slant asymptote: y x
15. f x
x2 1 1 x x x
fx 1
y 4
1 0 when x ± 1. x2
2 f x 3 0 x
y=x
2
(1, 2) −4
x
−2
2
x=0
−4
Relative maximum: 1, 2
4
(−1, −2)
Relative minimum: 1, 2 Vertical asymptote: x 0 Slant asymptote: y x
17. y
x2 6x 12 4 x2 x4 x4
y 1
4 x 42
y
x
4
8 6
(6, 6)
4
y
x 2x 6 0 when x 2, 6. x 42 8 x 43
y < 0 when x 2. Therefore, 2, 2 is a relative maximum. y > 0 when x 6. Therefore, 6, 6 is a relative minimum. Vertical asymptote: x 4 Slant asymptote: y x 2
(0, −3)
x
y
2
2 x
6
8
(2, −2)
10
Section 3.6
A Summary of Curve Sketching
19. y xx 4,
y
Domain: , 4
y y
( 83 ,
4
16 3 3
2
8 3x 8 0 when x and undefined when x 4. 3 24 x
(0, 0)
(4, 0) x
−2
2
4
3x 16 16 0 when x and undefined when x 4. 44 x3 2 3
Note: x
16 3
is not in the domain. y
y
Conclusion
Increasing, concave down
0
Relative maximum
Decreasing, concave down
Undefined
Undefined
y < x < x
8 3
8 3
16 33
8 < x < 4 3 x4
0
21. hx x9 x2
Endpoint
Domain: 3 ≤ x ≤ 3
y
9 2x2 3 32 hx 0 when x ± ± 9 x2 2 2
−5 −4
(
5 4 3 2 1
(− 3, 0)
x2x2 27 h x 0 when x 0 9 x23 2
(0, 0)
)
(3, 0)
1 2 3 4 5
(
−
32 9
3 2, 9 2 2
x
−2 −1
2 , 2 3 2 9 Relative minimum: , 2 2 Relative maximum:
3 2, 9 − 2 2
)
−5
Intercepts: 0, 0, ± 3, 0 Symmetric with respect to the origin Point of inflection: 0, 0 23. y 3x2 3 2x y 2x1 3 2
y
21 x1 3 x1 3
5
0 when x 1 and undefined when x 0. y
(1, 1)
2 < 0 when x 0. 3x4 3 y
< x < 0 x0
0
0 < x < 1 x1 1 < x <
1
(
( 278 , 0 ) x
(0, 0) 2
y
y
Undefined
Undefined
Increasing, concave down
0
Relative maximum
Decreasing, concave down
Conclusion Decreasing, concave down Relative minimum
1
2
3
5
139
140
Chapter 3
Applications of Differentiation
25. y x3 3x2 3
y
y 3x2 6x 3xx 2 0 when x 0, x 2
4
(−0.879, 0)
(0, 3)
y 6x 6 6x 1 0 when x 1 (1, 1)
y < x < 0 x0
3
0 < x < 1 x1
1
1 < x < 2 x2 2 < x <
1
y
y
Conclusion
Increasing, concave down
0
Relative maximum
Decreasing, concave down
0
Point of inflection
Decreasing, concave up
0
Relative minimum
Increasing, concave up
(2.532, 0) x
2
4
(2, (1.347, 0)
2
27. y 2 x x3
1)
y
5
y 1 3x2
4
No critical numbers (0, 2)
y 6x 0 when x 0.
1
(1, 0) x
y < x < 0 x0 0< x <
2
y
y
3
2
1
2
3
Conclusion Decreasing, concave up
0
Point of inflection
Decreasing, concave down
29. f x 3x3 9x 1
y
(−1.785, 0) 8 ( 1, 7)
fx 9x2 9 9x2 1 0 when x ± 1 f x 18x 0 when x 0 f x < x < 1 x 1
7
1 < x < 0 x0
1
0 < x < 1 x1 1 < x <
5
fx
f x
(0, 1) (1.674, 0)
Conclusion
Increasing, concave down
0
Relative maximum
Decreasing, concave down
0
Point of inflection
Decreasing, concave up
0
Relative minimum
Increasing, concave up
x 3
1
1
2
3
2 4 6
(1, 5) (0.112, 0)
Section 3.6
A Summary of Curve Sketching
31. y 3x4 4x3
y
y 12x3 12x2 12x2x 1 0 when x 0, x 1. y 36x2 24x 12x3x 2 0 when x 0, x 23 . y < x < 1 x 1
1
1 < x <
23
x 23
16
27
23 < x < 0 x0
0
0 < x <
y
y
Decreasing, concave up
0
Relative minimum
Increasing, concave up
0
Point of inflection
Increasing, concave down
0
0
Point of inflection
Increasing, concave up
2
1
(− 43 , 0)
Conclusion
(0, 0) 1
fx f x
12x 2
12x 2
y
16 4x 1x 2 0 when x 1, x 2.
15
24x 12xx 2 0 when x 0, x 2. fx
f x
Decreasing, concave up
0
Relative minimum
Increasing, concave up
0
Point of inflection
Increasing, concave down
0
0
Point of inflection
Increasing, concave up
< x < 1 x 1
11
1 < x < 0 x0
0
0 < x < 2 x2 2 < x <
20
2
f x
(− 23 , − 1627 (
(−1, −1)
33. f x x4 4x3 16x 4x3
16
Conclusion
35. y x5 5x
(2, 16)
10 5
(0, 0)
−3
y 4
1 < x < 0 0
0 < x < 1 1 < x <
4
3
4
y
< x < 1
x1
2
(−1.679, 0)
)
4 5, 0
6
( 1, 4)
4
y 20x3 0 when x 0.
x0
x
1
(−1, −11)
(
y 5x4 5 5x4 1 0 when x ± 1.
x 1
x
−2
(0, 0) 2
1
1
y
y
Conclusion
2
Increasing, concave down
4
0
Relative maximum
Decreasing, concave down
6
0
Point of inflection
Decreasing, concave up
0
Relative minimum
Increasing, concave up
( 4 5, 0 )
(1,
2
4)
x
141
142
Chapter 3
Applications of Differentiation
37. y 2x 3 y
y
22x 3 3 undefined at x . 2x 3 2
4
(0, 3)
3
y 0
2 1
y
y < x <
3 2
x 32 3 2
0
< x <
Undefined
Conclusion
x
Decreasing
Increasing
1 sin 3x, 0 ≤ x ≤ 2 18 1 3 y cos x cos 3x 0 when x , . 6 2 2
y 2 1
1 5 7 11 . y sin x sin 3x 0 when x 0, , , , , 2 6 6 6 6
19 , 2 18
Relative minimum:
32 , 19 18
Inflection points:
−1
π 2
π
x 3π 2
−2
4
Relative minimum
39. y sin x
Relative maximum:
3
3 , 0 2
6 , 94, 56 , 49, , 0, 76 , 49, 116 , 94
41. y 2x tan x,
< x < 2 2
43. y 2csc x sec x, 0 < x <
2
y 2 sec2 x 0 when x ± . 4
y 2sec x tan x csc x cot x 0 ⇒ x 4
y 2sec2 x tan x 0 when x 0.
Relative minimum:
Relative maximum:
4 , 2 1
Relative minimum:
4 , 1 2
Inflection point: 0, 0
Vertical asymptotes: x ± 2
Vertical asymptotes: x 0, x y 16 12 8 4
y −4
2 1
− π 2
−1 −2
π 4
π 2
x
4 , 42
4
2
x
2
Section 3.6
45. gx x tan x,
3 3 < x < 2 2
gx
x sin x cos x 0 when x 0 cos2 x
g x
2cos x x sin x cos3 x
Vertical asymptotes: x
47. f x
A Summary of Curve Sketching
20x 1 19x2 1 x 1 x xx2 1 2
10
− 15
15
− 10
3 3 , , , 2 2 2 2
x 0 vertical asymptote
Intercepts: , 0, 0, 0, , 0
y 0 horizontal asymptote
Symmetric with respect to y-axis.
Minimum: 1.10, 9.05 Maximum: 1.10, 9.05
2 and 2 , 32
Increasing on 0,
Points of inflection: 1.84, 7.86, 1.84, 7.86
Points of inflection: ± 2.80, 0 y 10 8 6 4 2 −π
49. y
π 4
−2 −4 −6 −8 − 10
π
3π 2
x
x
51. f is cubic.
x2 7
f is quadratic.
2
f is linear. −4
4
y
f ′′
f −2
0, 0 point of inflection
x 2
2 1
y ± 1 horizontal asymptotes
53.
f′
y
2
y
4
f ′′
4
f 2 x −4
−2
2
4
x −4
−2
2
−2 −4
(any vertical translate of f will do)
−4
4
143
144
Chapter 3
Applications of Differentiation
y
55.
y
4 2
4
f
2
f ′′
x −4
x −8
8
−4
4
−2
−2
−4
−4
8
(any vertical translate of f will do)
57. Since the slope is negative, the function is decreasing on 2, 8, and hence f 3 > f 5.
59. f x
4x 12 4x 5
x2
Vertical asymptote: none Horizontal asymptote: y 4 9
−6
9 −1
The graph crosses the horizontal asymptote y 4. If a function has a vertical asymptote at x c, the graph would not cross it since f c is undefined. 61. hx
6 2x 3x
63. f x
23 x 2, if x 3 3x Undefined, if x 3
The rational function is not reduced to lowest terms.
x2 3x 1 3 x 1 x2 x2
3
−3
6
3 −3
−2
The graph appears to approach the slant asymptote y x 1.
4 −1
hole at 3, 2
65. f x (a)
cos2 x , 0, 4 x2 1 (b) fx
1.5
0
4
− 0.5
On 0, 4 there seem to be 7 critical numbers: 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5
cos xx cos x 2 x2 1sin x 0 x2 13 2
1 3 5 7 Critical numbers , 0.97, , 1.98, , 2.98, . 2 2 2 2 The critical numbers where maxima occur appear to be integers in part (a), but approximating them using f shows that they are not integers.
Section 3.7 67. Vertical asymptote: x 5
71. f x
Slant asymptote: y 3x 2
1 x5
y 3x 2
(b) As b varies, the position of the vertical asymptote changes: x b. Also, the coordinates of the minimum a > 0 or maximum a < 0 are changed.
3xn x4 1
(a) For n even, f is symmetric about the y-axis. For n odd, f is symmetric about the origin. (b) The x-axis will be the horizontal asymptote if the degree of the numerator is less than 4. That is, n 0, 1, 2, 3.
(d) There is a slant asymptote y 3x if n 5: 3x 3x5 3x 4 . x4 1 x 1 (e)
(c) n 4 gives y 3 as the horizontal asymptote.
75. (a)
2750
1
8 0
(b) When t 10, N10 2434 bacteria. (c) N is a maximum when t 7.2 (seventh day). (d) Nt 0 for t 3.2 (e) lim Nt t→
Section 3.7 1. (a)
1 3x2 13x 9 x5 x5
ax x b2
(a) The graph has a vertical asymptote at x b. If a > 0, the graph approaches as x → b. If a < 0, the graph approaches as x → b. The graph approaches its vertical asymptote faster as a → 0.
73. f x
145
69. Vertical asymptote: x 5
Horizontal asymptote: y 0 y
Optimization Problems
13,250 1892.86 7
Optimization Problems
First Number, x
Second Number
Product, P
10
110 10
10110 10 1000
20
110 20
20110 20 1800
30
110 30
30110 30 2400
40
110 40
40110 40 2800
50
110 50
50110 50 3000
60
110 60
60110 60 3000
—CONTINUED—
n
0
1
2
3
4
5
M
1
2
3
2
1
0
N
2
3
4
5
2
3
146
Chapter 3
Applications of Differentiation
1. —CONTINUED— (b)
First Number, x
Second Number
Product, P
10
110 10
10110 10 1000
20
110 20
20110 20 1800
30
110 30
30110 30 2400
40
110 40
40110 40 2800
50
110 50
50110 50 3000
60
110 60
60110 60 3000
70
110 70
70110 70 2800
80
110 80
80110 80 2400
90
110 90
90110 90 1800
100
110 100
100110 100 1000
The maximum is attained near x 50 and 60. (c) P x110 x 110x x2 (d)
(e)
3500
(55, 3025)
dP 110 2x 0 when x 55. dx d 2P 2 < 0 dx 2
0
120 0
The solution appears to be x 55. 3. Let x and y be two positive numbers such that xy 192. Sxyx
192 x
192 dS 1 2 0 when x 192. dx x d 2S 384 3 > 0 when x 192. dx2 x S is a minimum when x y 192. 7. Let x be the length and y the width of the rectangle. 2x 2y 100 y 50 x A xy x50 x dA 50 2x 0 when x 25. dx d 2A 2 < 0 when x 25. dx2 A is maximum when x y 25 meters.
P is a maximum when x 110 x 55. The two numbers are 55 and 55.
5. Let x be a positive number. Sx
1 x
dS 1 1 2 0 when x 1. dx x 2 d 2S 3 > 0 when x 1. dx2 x The sum is a minimum when x 1 and 1x 1.
9. Let x be the length and y the width of the rectangle. xy 64 y
64 x
P 2x 2y 2x 2
64x 2x 128x
dP 128 2 2 0 when x 8. dx x d 2P 256 3 > 0 when x 8. dx2 x P is minimum when x y 8 feet.
Section 3.7 11. d x 42 x 02
x4 4x 174
Since d is smallest when the expression inside the radical is smallest, you need only find the critical numbers of
Since d is smallest when the expression inside the radical is smallest, you need only find the critical numbers of
f x x2 7x 16.
f x x4 4x 17 4 .
fx 2x 7 0
fx 4x3 4 0
x 72
x1
By the First Derivative Test, the point nearest to 4, 0 is 72, 72 .
By the First Derivative Test, the point nearest to 2, 12 is 1, 1. y
y 4
4
3
3
( x, x )
2
2
3
(2, 12(
d
x
x 2
( x, x 2 )
1
d
1
1
147
13. d x 22 x2 12 2
x2 7x 16
15.
Optimization Problems
−2
(4, 0)
dQ kxQ0 x kQ0x kx2 dx
−1
1
2
17. xy 180,000 (see figure)
S x 2y x
d 2Q kQ0 2kx dx2
360,000 where S is the length x
of fence needed.
kQ0 2x 0 when x
Q0 . 2
360,000 dS 1 0 when x 600. dx x2
d 3Q Q 2k < 0 when x 0. dx3 2
d 2S 720,000 > 0 when x 600. dx2 x3
dQdx is maximum when x Q02.
S is a minimum when x 600 meters and y 300 meters.
y
x
19. (a) A 4area of side 2area of Top
(b) V lengthwidthheight
(a) A 4311 233 150 square inches
(a) V 3311 99 cubic inches
(b) A 455 255 150 square inches
(b) V 555 125 cubic inches
(c) A 43.256 266 150 square inches
(c) V 663.25 117 cubic inches
(c) S 4xy 2x 2 150 ⇒ y V x 2y x 2
150 2x 2 4x
x x
150 4x 2x 752 x 21 x 2
75 3 2 V x 0 ⇒ x ±5 2 2
y
3
x x
By the First Derivative Test, x 5 yields the maximum volume. Dimensions: 5 5 5. (A cube!)
148
Chapter 3
Applications of Differentiation s 2
V xs 2x2, 0 < x <
21. (a)
dV 2xs 2x2 s 2x2 dx s s s 2xs 6x 0 when x , s2 is not in the domain. 2 6 d 2V 24x 8s dx2 s d 2V < 0 when x . dx2 6 V
5 2s 3 is maximum when x . 27 6
2 2 (b) If the length is doubled, V 27 2s3 8 27 s3. Volume is increased by a factor of 8.
23.
16 2y x
2x
32 4y 2x x y
32 2x x 4
A xy
x 2 2
x 2
32 2x4 x x 2
y
x2 8
x
1 8x x 2 x 2 x 2 2 4 8 dA 8x x x8x 1 dx 2 4 4
8 32 . 1 4 4
0 when x d 2A 1 dx 2 4
y
< 0 when x 4 32
32 2 324 324 16 4 4
The area is maximum when y
25. (a)
32 16 feet and x feet. 4 4
02 y2 01 x1 y2
(b)
2 x1
L x 2 y 2
10
x
2
—CONTINUED—
4
(2.587, 4.162)
x 2 x 2 1 2
2
8 4 , x > 1 x 1 x 12
0
10 0
L is minimum when x 2.587 and L 4.162.
Section 3.7
Optimization Problems
25. —CONTINUED—
1 1 2 x (c) Area Ax xy x 2 x 2 2 x1 x1 Ax 1
x 1 x 1 1 0 x 12 x 12
x 12 1 x 1 ±1 x 0, 2 (select x 2) Then y 4 and A 4. Vertices: 0, 0, 2, 0, 0, 4 27.
see figure
A 2xy 2x25 x2
y
252x x 225 x
1 dA 2x dx 2 2
8
2
6
2
25252xx 0 when x y 5 2 2 3.54.
±
Width:
(
x
2
−6 −4 −2 −2
2
4
6
−4
52 52 52 ,0 , ± , . 2 2 2 52 ; Length: 52 2
29. xy 30 ⇒ y
30 x
x+2 x
30 2 A x 2 x
see figure
dA 30 30 2x2 30 x 2 2 0 when x 30. 2 dx x x x2
y
25 − x 2
2
By the First Derivative Test, the inscribed rectangle of maximum area has vertices
( x,
30 30
y
y+2
30
By the First Derivative Test, the dimensions x 2 by y 2 are 2 30 by 2 30 (approximately 7.477 by 7.477). These dimensions yield a minimum area. 31. V r 2h 22 cubic inches or h (a)
22 r2
Radius, r
Height
0.2
22 0.22
2 0.2 0.2
0.4
22 0.42
2 0.4 0.4
0.6
22 0.62
2 0.6 0.6
0.8
22 0.82
2 0.8 0.8
—CONTINUED—
Surface Area
22 220.3 0.22
22 111.0 0.42
22 75.6 0.62
22 59.0 0.82
149
150
Chapter 3
Applications of Differentiation
31. —CONTINUED— (c) S 2 r 2 2 rh
(b) Radius, r
Height
0.2
22 0.22
2 0.2 0.2
0.4
22 0.42
2 0.4 0.4
Surface Area
2 rr h 2 r r
22 220.3 0.22
22 111.0 0.42
(d)
22 44 2 r 2 r2 r
100
(1.52, 43.46) −1
4 −10
0.6
22 0.62
2 0.6 0.6
0.8
22 0.82
22 2 0.8 0.8 59.0 0.82
22 50.3 1.0 2
22 45.7 1.22
22 43.7 1.42
22 43.6 1.62
22 44.8 1.82
22 47.1 2.02
22 1.02
2 1.0 1.0
1.2
22 1.22
2 1.2 1.2
1.4
22 1.42
2 1.4 1.4
1.6
22 1.62
2 1.6 1.6
1.8
22 1.82
2 1.8 1.8
2.0
22 2.02
2 2.0 2.0
1.0
22 75.6 0.62
The minimum seems to be 43.46 for r 1.52. (e)
44 dS 3 4 r 2 0 when r 11 1.52 in. dr r 22 3.04 in. r2 Note: Notice that 22 1113 22 2 13 2r. h 2 23 r 11 h
The minimum seems to be about 43.6 for r 1.6. 33. Let x be the sides of the square ends and y the length of the package. P 4x y 108 ⇒ y 108 4x V x 2y x 2108 4x 108x 2 4x3 dV 216x 12x2 dx 12x18 x 0 when x 18. d 2V 216 24x 216 < 0 when x 18. dx2 The volume is maximum when x 18 inches and y 108 418 36 inches.
Section 3.7
35.
Optimization Problems
151
1 1 V x 2h x 2 r r 2 x2 see figure 3 3 x3 x dV 1 2x r r2 x2 2r2 2rr2 x2 3x2 0 dx 3 r2 x2 3r2 x2
2r 2 2rr 2 x2 3x2 0
(0, r)
2rr 2 x2 3x2 2r 2 h
4r 2r 2 x2 9x4 12x 2 r 2 4r4
x 0,
r
x
0 9x4 8x2r 2 x29x2 8r 2 22r 3
(x, −
r 2 − x2
(
By the First Derivative Test, the volume is a maximum when x
4r 22r and h r r 2 x 2 . 3 3
Thus, the maximum volume is
4r3 3281 r
1 8r 2 V 3 9
3
cubic units.
37. No, there is no minimum area. If the sides are x and y, then 2x 2y 20 ⇒ y 10 x. The area is Ax x10 x 10x x2. This can be made arbitrarily small by selecting x 0. 4 V 12 r 3 r 2h 3
39.
h
41. Let x be the length of a side of the square and y the length of a side of the triangle.
12 43 r 3 12 4 r r2 r2 3
4x 3y 10
12 4 r S 4 r 2 2 rh 4 r 2 2 r r 2 3
1 3 A x2 y y 2 2
24 8 2 4 2 24 r r 4 r 2 r 3 3 r 24 dS 8 3 9 1.42 cm. r 2 0 when r dr 3 r 48 d 2S 8 3 9 cm. 3 > 0 when r dr2 3 r
10 3y2 3 2 y 16 4
3 dA 1 10 3y3 y0 dy 8 2
30 9y 43y 0 y
3 9
The surface area is minimum when r cm and h 0. The resulting solid is a sphere of radius r 1.42 cm.
30 9 43
d 2A 9 43 > 0 dy 2 8 A is minimum when
r h
y
30 103 . and x 9 43 9 43
152
Chapter 3
Applications of Differentiation
43. Let S be the strength and k the constant of proportionality. Given h2 w2 242, h2 242 w 2,
45.
R
v02 sin 2
g
3 dR 2v02 cos 2 0 when , . d
g 4 4
S kwh2 S kw576 w2 k576w w3 dS k576 3w3 0 when w 83, h 86. dw d 2S 6k w < 0 when w 83. dw 2
d 2R 4v 2 0 sin 2 < 0 when . d 2 g 4 By the Second Derivative Test, R is maximum when 4.
These values yield a maximum.
47. sin tan I
h h ⇒s , 0 < < s sin 2 h 2 tan ⇒ h 2 tan ⇒ s 2 sec 2 sin
k sin k sin k sin cos2 s2 4 sec2 4
α
α
dI k sin 2 sin cos cos2 cos d 4
k cos cos 2 2 sin2 4
k cos 1 3 sin2 4
0 when
s
h
4 ft
3 1 , , or when sin ± . 2 2 3
Since is acute, we have sin
1 3
⇒ h 2 tan 2
12 2 feet.
Since d 2Id 2 k4 sin 9 sin2 7 < 0 when sin 13, this yields a maximum.
49.
S x2 4, L 1 3 x2 Time T
x2 4
2
x2 6x 10
4
dT x x3 0 dx 2x2 4 4x2 6x 10
S=
2 x
x2 + 4 3−x 1
L=
1 + (3 − x 2(
Q
x2 9 6x x2 x2 4 4x2 6x 10 x4 6x3 9x2 8x 12 0 You need to find the roots of this equation in the interval 0, 3 . By using a computer or graphics calculator, you can determine that this equation has only one root in this interval x 1. Testing at this value and at the endpoints, you see that x 1 yields the minimum time. Thus, the man should row to a point 1 mile from the nearest point on the coast.
Section 3.7
51.
T
x2 4
v1
Optimization Problems
x2 6x 10
v2
dT x x3 0 dx v1x2 4 v2x2 6x 10
θ1
2
3−x
x
1
Since
θ2
x x2 4
sin 1 and
x3 x2 6x 10
Q
sin 2
we have sin 1 sin 2 sin 1 sin 2 0⇒ . v1 v2 v1 v2 Since d 2T 4 1 > 0 dx2 v1x2 432 v2x2 6x 1032 this condition yields a minimum time. 53. f x 2 2 sin x
(a) Distance from origin to y-intercept is 2. Distance from origin to x-intercept is 2 1.57.
y
(b) d x2 y2 x2 2 2 sin x2
3
3 2 1
−
π 4
−1
π 4
π 2
x
(0.7967, 0.9795)
− 4
2
−1
Minimum distance 0.9795 at x 0.7967. (c) Let f x d 2x x 2 2 2 sin x2. fx 2x 22 2 sin x2 cos x Setting fx 0, you obtain x 0.7967, which corresponds to d 0.9795. 55. F cos kW F sin
2 k +1
kW F cos k sin
dF kWk cos sin 0 d
cos k sin 2 k cos sin ⇒ k tan ⇒ arctan k Since cos k sin
1 k2 k2 1, 1 k2 1
k2
the minimum force is F
kW kW . cos k sin k2 1
1
θ
k
153
154
Chapter 3
57. (a)
Applications of Differentiation (b)
Base 1
Base 2
Altitude
Area
22.1
8
8 16 cos 10
8 sin 10
22.1
8 sin 20
42.5
8
8 16 cos 20
8 sin 20
42.5
8 16 cos 30
8 sin 30
59.7
8
8 16 cos 30
8 sin 30
59.7
8
8 16 cos 40
8 sin 40
72.7
8
8 16 cos 40
8 sin 40
72.7
8
8 16 cos 50
8 sin 50
80.5
8
8 16 cos 50
8 sin 50
80.5
8
8 16 cos 60
8 sin 60
83.1
8
8 16 cos 60
8 sin 60
83.1
8
8 16 cos 70
8 sin 70
80.7
8
8 16 cos 80
8 sin 80
74.0
8
8 16 cos 90
8 sin 90
64.0
Base 1
Base 2
Altitude
Area
8
8 16 cos 10
8 sin 10
8
8 16 cos 20
8
The maximum cross-sectional area is approximately 83.1 square feet. (c) A a b
h 2
(d)
8 8 16 cos
8 sin
2
64cos cos2 sin2 642 cos2 cos 1
641 cos sin , 0 < < 90 (e)
dA 641 cos cos 64 sin sin
d
642 cos 1cos 1
100
(60°, 83.1)
0 when 60 , 180 , 300 . The maximum occurs when 60 .
0
90 0
59. C 100
x ,1 ≤ x 200 x x 30 2
C 100
400 30 x3 x 302
Approximation: x 40.45 units, or 4045 units 61.
S1 4m 12 5m 62 10m 32 64 dS1 24m 14 25m 65 210m 310 282m 128 0 when m . dm 141 Line: y
S 4
63. S3
64 x 141
64 64 64 1 5 6 10 3 141 141 141
256 320 640 858 1 6 3 6.1 mi 141 141 141 141
4m 1 5m 6 10m 3
m2
1
m2
1
m2
S3
1
30
Using a graphing utility, you can see that the minimum occurs when x 0.3. Line: y 0.3x
20
10
40.3 1 50.3 6 100.3 3 S3 4.5 mi. 0.32 1
(0.3, 4.5) m 1
2
3
Section 3.8
Section 3.8
Newton’s Method
1. f x x2 3 fx 2x x1 1.7
n
xn
f xn
fxn
f xn fxn
1
1.7000
0.1100
3.4000
0.0324
1.7324
2
1.7324
0.0012
3.4648
0.0003
1.7321
3. f x sin x fx cos x x1 3
n
xn
1
3.0000
2
3.1425
5. f x x3 x 1 fx 3x2 1 Approximation of the zero of f is 0.682.
xn
f xn fxn
fxn
f xn fxn
0.1411
0.9900
0.1425
3.1425
0.0009
1.0000
0.0009
3.1416
f xn
xn
f xn fxn
n
xn
f xn
fxn
f xn fxn
1
0.5000
0.3750
1.7500
0.2143
0.7143
2
0.7143
0.0788
2.5307
0.0311
0.6832
3
0.6832
0.0021
2.4003
0.0009
0.6823
n
xn
f xn
fxn
f xn fxn
1
1.2000
0.1416
2.3541
0.0602
1.1398
2
1.1398
0.0181
3.0118
0.0060
1.1458
3
1.1458
0.0003
2.9284
0.0001
1.1459
n
xn
f xn
fxn
f xn fxn
1
1.5000
0.3750
6.7500
0.0556
7. f x 3x 1 x fx
Newton’s Method
3 1 2x 1
Approximation of the zero of f is 1.146.
xn
xn
f xn fxn
f xn fxn
Similarly, the other zero is approximately 7.854. 9. f x x3 3 fx
3x2
Approximation of the zero of f is 1.442.
1.4444
1.4444
0.0134
6.2589
0.0021
1.4423
3
1.4423
0.0003
6.2407
0.0001
1.4422
fx 3x2 7.8x 4.79
n
xn
f xn
fxn
f xn fxn
1
0.5000
0.3360
1.6400
0.2049
0.7049
2
0.7049
0.0921
0.7824
0.1177
0.8226
3
0.8226
0.0231
0.4037
0.0573
0.8799
4
0.8799
0.0045
0.2495
0.0181
0.8980
5
0.8980
0.0004
0.2048
0.0020
0.9000
6
0.9000
0.0000
0.2000
0.0000
0.9000
—CONTINUED—
f xn fxn
2
11. f x x3 3.9x2 4.79x 1.881
Approximation of the zero of f is 0.900.
xn
xn
f xn fxn
155
156
Chapter 3
Applications of Differentiation
11. —CONTINUED—
n
xn
f xn
fxn
f xn fxn
1
1.1
0.0000
0.1600
0.0000
xn
f xn fxn
1.1000
Approximation of the zero of f is 1.100. n
xn
f xn
fxn
f xn fxn
1
1.9
0.0000
0.8000
0.0000
xn
f xn fxn
1.9000
Approximation of the zero of f is 1.900. 13. f x x sinx 1 fx 1 cosx 1 Approximation of the zero of f is 0.489.
15. hx f x gx 2x 1 x 4
n
xn
f xn
fxn
f xn fxn
1
0.5000
0.0206
1.8776
0.0110
0.4890
2
0.4890
0.0000
1.8723
0.0000
0.4890
hxn
h xn hxn
0.6000
0.0552
1.7669
0.0313
0.5687
0.5687
0.0001
1.7661
0.0000
0.5687
n
xn
h xn
hxn
h xn hxn
1
4.5000
0.1373
21.5048
0.0064
4.4936
2
4.4936
0.0039
20.2271
0.0002
4.4934
xn
1 2
hx 1 sec2 x Point of intersection of the graphs of f and g occurs when x 4.493.
Point of intersection of the graphs of f and g occurs when x 0.569. 17. hx f x gx x tan x
19. f x x2 a 0
21. xi1
fx 2x xi2 a xi1 xi 2xi
23. xi1
2xi 2 xi2 a xi2 a xi a 2xi 2xi 2 2xi 3xi4 6 4xi3
i
1
2
3
4
xi
1.5000
1.5694
1.5651
1.5651
4 6 1.565
f xn fxn
h xn
n
1 hx 2 2x 4
xn
xn
xn
h xn hxn
h xn hxn
xi2 7 2xi
i
1
2
3
4
5
xi
2.0000
2.7500
2.6477
2.6458
2.6458
7 2.646
Section 3.8 25. f x 1 cos x
Newton’s Method
157
fx sin x
n
xn
f xn
fxn
f xn fxn
Approximation of the zero: 3.141
1
3.0000
0.0100
0.1411
0.0709
3.0709
2
3.0709
0.0025
0.0706
0.0354
3.1063
3
3.1063
0.0006
0.0353
0.0176
3.1239
4
3.1239
0.0002
0.0177
0.0088
3.1327
5
3.1327
0.0000
0.0089
0.0044
3.1371
6
3.1371
0.0000
0.0045
0.0022
3.1393
7
3.1393
0.0000
0.0023
0.0011
3.1404
8
3.1404
0.0000
0.0012
0.0006
3.1410
27. y 2x3 6x2 6x 1 f x
29. y x3 6x2 10x 6 f x
y 6x2 12x 6 fx
y 3x2 12x 10 fx
x1 1
x1 2
fx 0; therefore, the method fails.
x2 1
n
xn
f xn
fxn
1
1
1
0
xn
f xn fxn
x3 2 x4 1 and so on. Fails to converge y
31. Answers will vary. See page 222. Newton’s Method uses tangent lines to approximate c such that f c 0.
1
First, estimate an initial x1 close to c (see graph). Then determine x2 by x2 x1
−1
f x1 . fx1
Calculate a third estimate by x3 x2
x a 3
f(x)
x1
x2 c
2
b
x
−1 −2
f x2 . fx2
Continue this process until xn xn1 is within the desired accuracy. Let xn1 be the final approximation of c.
33. Let gx f x x cos x x gx sin x 1. The fixed point is approximately 0.74.
n
xn
g xn
gxn
g xn gxn
1
1.0000
0.4597
1.8415
0.2496
2
0.7504
0.0190
1.6819
0.0113
0.7391
3
0.7391
0.0000
1.6736
0.0000
0.7391
xn
g xn gxn
0.7504
158
Chapter 3
Applications of Differentiation
35. f x x3 3x2 3, fx 3x2 6x (a)
(b) x1 1
4
f x1 1.333 fx1
x2 x1 −4
5
Continuing, the zero is 1.347. −2
(d)
y
3x
1 (c) x1 4
4
y
f
x2 x1
f x1 2.405 fx1
3
Continuing, the zero is 2.532.
x 2
(e) If the initial guess x1 is not “close to” the desired zero of the function, the x-intercept of the tangent line may approximate another zero of the function.
1
y
4
1.313x
5
3.156
The x-intercepts correspond to the values resulting from the first iteration of Newton’s Method.
37. f x
1 a0 x
fx
1 x2
xn1 xn
1xn a 1 xn xn2 a xn xn xn2a 2xn xn2a xn2 axn 1xn2 xn
39. f x x cos x, 0,
y
fx x sin x cos x 0
1
)0.860, 0.561)
Letting F x fx, we can use Newton’s Method as follows.
x 2
Fx 2 sin x x cos x
1
n
xn
F xn
Fxn
F xn Fxn
1
0.9000
0.0834
2.1261
0.0392
0.8608
2
0.8608
0.0010
2.0778
0.0005
0.8603
Approximation to the critical number: 0.860
xn
F xn Fxn
2 3
Section 3.8
Newton’s Method
41. y f x 4 x2, 1, 0
y
d x 12 y 02 x 12 4 x22 x4 7x2 2x 17
5
d is minimized when D x4 7x2 2x 17 is a minimum.
2 1
gx 12x2 14 g xn gxn
2.0000
34.0000
0.0588
1.9412
1.9412
0.0830
31.2191
0.0027
1.9385
1.9385
0.0012
31.0934
0.0000
1.9385
1
2.0000
2 3
(1, 0) x
−3
gxn
xn
(1.939, 0.240)
3
gx D 4x3 14x 2
n
159
g xn
xn
−1 −1
1
3
g xn gxn
x 1.939 Point closest to 1, 0 is 1.939, 0.240. Minimize: T
43.
T T
Distance rowed Distance walked Rate rowed Rate walked x2 4
3
x 3x2 4
x2 6x 10
4 x3 4x2 6x 10
0
4xx2 6x 10 3x 3x2 4 16x2x2 6x 10 9x 32x2 4 7x4 42x3 43x2 216x 324 0 Let f x 7x4 42x3 43x2 216x 324 and fx 28x3 126x2 86x 216. Since f 1 100 and f 2 56, the solution is in the interval 1, 2. f xn fxn
f xn fxn
f xn
fxn
1.7000
19.5887
135.6240
0.1444
1.5556
1.5556
1.0480
150.2780
0.0070
1.5626
1.5626
0.0014
49.5591
0.0000
1.5626
n
xn
1 2 3
xn
Approximation: x 1.563 miles 2,500,000 76x3 4830x2 320,000
45.
76x3 4830x2 2,820,000 0 Let f x 76x3 4830x2 2,820,000 fx 228x2 9660x. From the graph, choose x1 40. f xn fxn
f xn fxn
n
xn
f xn
fxn
1
40.0000
44000.0000
21600.0000
2.0370
37.9630
2
37.9630
17157.6209
38131.4039
0.4500
38.4130
3
38.4130
780.0914
34642.2263
0.0225
38.4355
4
38.4355
2.6308
34465.3435
0.0001
38.4356
xn
The zero occurs when x 38.4356 which corresponds to $384,356.
160
Chapter 3
Applications of Differentiation
47. False. Let f x x2 1x 1. x 1 is a discontinuity. It is not a zero of f x. This statement would be true if f x pxqx is given in reduced form. 49. True 51. f x 14 x3 3x2 34 x 2
y
3 3 fx 4 x2 6x 4
60 40
Let x1 12.
20 x
n
xn
f xn
fxn
f xn fxn
1
12.0000
7.0000
36.7500
0.1905
11.8095
2
11.8095
0.2151
34.4912
0.0062
11.8033
3
11.8033
0.0015
34.4186
0.0000
11.8033
xn
−10 −5
f xn fxn
5
15
20
Approximation: x 11.803
Section 3.9
Differentials
1. f x x2
x f x
fx 2x Tangent line at 2, 4: y f 2 f2x 2
x2
T x 4x 4
1.9
1.99
2
3.6100
3.9601
3.6000
3.9600
2.01
2.1
4
4.0401
4.4100
4
4.0400
4.4000
y 4 4x 2 y 4x 4 3. f x x5
1.99
2
2.01
fx 5x4
x f x x5
24.7610
1.9
31.2080
32
32.8080
40.8410
2.1
Tangent line at 2, 32: y f 2 f2x 2
T x 80x 128
24.0000
31.2000
32
32.8000
40.0000
y 32 80x 2 y 80x 128 5. f x sin x fx cos x Tangent line at 2, sin 2: y f 2 f2x 2
x
1.9
1.99
2
2.01
2.1
f x sin x
0.9463
0.9134
0.9093
0.9051
0.8632
T x cos 2x 2 sin 2
0.9509
0.9135
0.9093
0.9051
0.8677
y sin 2 cos 2x 2 y cos 2x 2 sin 2 1 3 7. y f x 2 x3, fx 2 x2, x 2, x dx 0.1
y f x x f x f 2.1 f 2 0.6305
dy fxdx f20.1 60.1 0.6
Section 3.9
Differentials
9. y f x x4 1, fx 4x3, x 1, x dx 0.01 y f x x f x
11.
dy fx dx
f 0.99 f 1
f10.01
0.994 1 14 1 0.0394
40.01 0.04
13. y
y 3x2 4 dy 6x dx
dy
15. y x 1 x2
dy x
19.
y
17.
dy sin
3 dx 2x 12
y 2x cot2 x dy 2 2 cot x csc2 xdx
x 1 2x2 1 x2 dx dx 2 1 x 1 x2
6x 1 1 cos 3 2
x1 2x 1
2 2 cot x 2 cot3 xdx
6x2 1 dx
21. (a) f 1.9 f 2 0.1 f 2 f20.1
23. (a) f 1.9 f 2 0.1 f 2 f20.1 1 12 0.1 1.05
1 10.1 0.9 (b) f 2.04 f 2 0.04 f 2 f20.04
(b) f 2.04 f 2 0.04 f 2 f20.04 1 2 0.04 0.98
1 10.04 1.04 25. (a) g2.93 g3 0.07 g3 g30.07 8
12
1
27. (a) g2.93 g3 0.07 g3 g30.07
0.07 8.035
8 00.07 8 (b) g3.1 g3 0.1 g3 g30.1
(b) g3.1 g3 0.1 g3 g30.1 8 29.
A x2
8 00.1 8 A r 2 r 14
x dx
1 ± 64
r dr ± 14
dA 2x dx A dA 212
0.1 7.95 31.
x 12
3 ±8
12
1 ± 64
square inches
A dA 2 r dr 28 ± 14 ± 7 square inches
161
162
Chapter 3
33. (a)
Applications of Differentiation
x 15 centimeter
35.
x dx ± 0.05 centimeters A
r dr ± 0.02 inches
x2
4 (a) V r 3 3
dA 2x dx 215± 0.05
dV 4 r 2 dr 462± 0.02 ± 2.88 cubic inches
± 1.5 square centimeters
(b) S 4 r 2
Percentage error:
dS 8r dr 86± 0.02 ± 0.96 square inches
± 1.5 2 dA 0.00666. . . % A 152 3
(b)
r 6 inches
4 r 2 dr 3dr dV (c) Relative error: V 43 r 3 r
dA 2x dx 2 dx 2 ≤ 0.025 A x x
0.025 dx ≤ 0.0125 1.25% x 2
3 0.02 0.01 1% 6
dS 8 r dr 2dr Relative error: S 4 r 2 r
20.02 2 0.000666 . . . % 6 3
37. V r 2h 40 r 2, r 5 cm, h 40 cm, dr 0.2 cm V dV 80 r dr 8050.2 80 cm3 (b) 0.0025360024 216 seconds
39. (a) T 2 Lg dT
3.6 minutes
dL g Lg
Relative error: dT dL g Lg T 2 Lg
dL 2L
1 relative error in L 2
1 0.005 0.0025 2
Percentage error:
1 dT 100 0.25% % T 4
41. 2645 26.75 h
d ± 15 ± 0.25 (a)
h 9.5 csc
b
dh 9.5 csc cot d dh cot d h
9.5
θ
dh cot 26.750.25 h
Converting to radians, cot 0.46690.0044 0.0087 0.87% in radians.
(b)
dh cot d ≤ 0.02 h 0.02 0.02 tan d ≤ cot d 0.02 tan 26.75 0.02 tan 0.4669 ≤ 26.75 0.4669 0.0216 2.16% in radians
Review Exercises for Chapter 3
43. r
v02 sin 2 32
45. Let f x x, x 100, dx 0.6. f x x f x f x dx
v0 2200 ftsec
changes from 10 to 11 dr
x
22002 cos 2 d 16
10
f x x 99.4
180
d 11 10
1 dx 2x
100
1 0.6 9.97 2100
Using a calculator: 99.4 9.96995
180
r dr
22002 20 cos 16 180
180 4961 feet
4961 feet 49. Let f x x, x 4, dx 0.02, f x 1 2x .
4 x, x 625, dx 1. 47. Let f x 4 x f x x f x f x d x
Then
1 dx 44x3
f 4.02 f 4 f 4 dx
1 4 624 4 625 f x x 1 4 625 3 4 5
4.02 4
1 1 0.02 2 0.02. 4 24
1 4.998 500
4 624 4.9980. Using a calculator,
51. In general, when x → 0, dy approaches y. 53. True
55. True
Review Exercises for Chapter 3 1. A number c in the domain of f is a critical number if f c 0 or f is undefined at c.
y 4
f ′(c) is 3 undefined.
f ′(c) = 0
x −4 −3
−1 −2 −3 −4
3. gx 2x 5 cos x, 0, 2
18
(6.28, 17.57)
g x 2 5 sin x 2
0 when sin x 5 . Critical numbers: x 0.41, x 2.73 Left endpoint: 0, 5 Critical number: 0.41, 5.41 Critical number: 2.73, 0.88 Minimum Right endpoint: 2, 17.57 Maximum
(2.73, 0.88) − 4
2 −1
1
2
4
163
164
Chapter 3
Applications of Differentiation
5. Yes. f 3 f 2 0. f is continuous on 3, 2 , differentiable on 3, 2.
7. f x 3 x 4 y
(a)
f x x 33x 1 0 for x 13.
6 4
c 13 satisfies f c 0.
2 x
−2
2
4
6
10
−4 −6
f 1 f 7 0 (b) f is not differentiable at x 4.
9.
f x x23, 1 ≤ x ≤ 8
f x x cos x,
11.
2 f x x13 3
f x 1 sin x f b f a 2 2 1 ba 2 2
f b f a 4 1 3 ba 81 7
f c 1 sin c 1
2 3 f c c13 3 7 c
13.
149
3
≤ x ≤ 2 2
c0
2744 3.764 729
f x Ax2 Bx C f x 2Ax B f x2 f x1 Ax22 x12 Bx2 x1 x2 x1 x2 x1 Ax1 x2 B f c 2Ac B Ax1 x2 B 2Ac Ax1 x2 c
x1 x2 Midpoint of x1, x2
2
15. f x x 12x 3 f x x 1 1 x 32x 1 2
x 13x 7 7 Critical numbers: x 1 and x 3
17. hx xx 3 x32 3x12 Domain: 0, 3 3 h x x12 x12 2 2 3 3x 1 x12x 1 2 2x Critical number: x 1
Interval:
< x < 1
1 < x <
7 3
7 3
< x <
Sign of f x:
f x > 0
f x < 0
f x > 0
Conclusion:
Increasing
Decreasing
Increasing
Interval:
0 < x < 1
Sign of h x:
h x < 0
h x > 0
Conclusion:
Decreasing
Increasing
1 < x <
Review Exercises for Chapter 3 19. ht 14t 4 8t
Test Interval: < t < 2
h t t 3 8 0 when t 2. Relative minimum: 2, 12
165
2 < t <
Sign of h t:
h t < 0
h t > 0
Conclusion:
Decreasing
Increasing
1 1 cos12t sin12t 3 4
21. y
v y 4 sin12t 3 cos12t (a) When t
1 , y inch and v y 4 inches/second. 8 4
(b) y 4 sin12t 3 cos12t 0 when
sin12t 3 3 ⇒ tan12t . cos12t 4 4
3 4 Therefore, sin12t and cos12t . The maximum displacement is 5 5 y (c) Period:
1345 41 53 125 inch. 2 12 6 1 6 6
Frequency:
23. f x x cos x, 0 ≤ x ≤ 2 f x 1 sin x f x cos x 0 when x Points of inflection:
3 , . 2 2
Test Interval: Sign of f x: Conclusion:
3 < x < 2 2
3 < x < 2 2
f x < 0
f x > 0
f x < 0
Concave downward
Concave upward
Concave downward
0 < x <
2
2 , 2 , 32, 32
25. gx 2x21 x2
y
g x 4x2x2 1 Critical numbers: x 0, ±
(−
1 2
1, 1 2 2
)
1
−2
g x 4 24x
(
1, 1 2 2
)
(0, 0)
2
x
−1
2
−2
g 0 4 > 0
Relative minimum at 0, 0
1 1 Relative maximums at ± , 2 2
y
27. 6
(5, f(5))
5 4
(3, f(3))
2 1 −1
29. The first derivative is positive and the second derivative is negative. The graph is increasing and is concave down.
7
3
−3
1 g ± 8 < 0 2
(6, 0) (0, 0) 2 3 4 5
x 7
166
Chapter 3
Applications of Differentiation
31. (a) D 0.0034t4 0.2352t3 4.9423t2 20.8641t 94.4025 (b)
369
0
29 0
(c) Maximum at 21.9, 319.5 1992 Minimum at 2.6, 69.6 1972 (d) Outlays increasing at greatest rate at the point of inflection 9.8, 173.7 1979
33. lim
x →
2x2 2 2 lim 5 x → 3 5x2 3
35. lim
2x 3 x4
39. f x
3x2
37. hx
Discontinuity: x 4 lim
x →
x →
5 cos x 0, since 5 cos x ≤ 5. x
3 2 x
Discontinuity: x 0
2x 3 2 3x lim 2 x → 1 4x x4
lim
x →
3x 2 2
Vertical asymptote: x 4
Vertical asymptote: x 0
Horizontal asymptote: y 2
Horizontal asymptote: y 2
41. f x x3
243 x
43. f x
x1 1 3x2
Relative minimum: 3, 108
Relative minimum: 0.155, 1.077
Relative maximum: 3, 108
Relative maximum: 2.155, 0.077 0.2
200
−2
−5
5
5
− 1.4
− 200
Vertical asymptote: x 0
Horizontal asymptote: y 0
45. f x 4x x2 x4 x Domain: , ; Range: , 4 f x 4 2x 0 when x 2. f x 2 Therefore, 2, 4 is a relative maximum. Intercepts: 0, 0, 4, 0
y
5
)2, 4)
4 3 2 1
x 1
2
3
5
Review Exercises for Chapter 3 47. f x x16 x2, Domain: 4, 4 , Range: 8, 8
y
2
2, 8
8
Domain: 4, 4 ; Range: 8, 8
6 4
16 2x2 f x 0 when x ± 22 and undefined when x ± 4. 16 x2 f x
2xx2 24 16 x232
2
(− 4, 0)
(4, 0) x
8
6
2
2
4
6
8
(0, 0)
8
2
2,
8
f 22 > 0
Therefore, 22, 8 is a relative minimum. f 22 < 0
Therefore, 22, 8 is a relative maximum. Point of inflection: 0, 0 Intercepts: 4, 0, 0, 0, 4, 0 Symmetry with respect to origin 49. f x x 13x 32
y
Domain: , ; Range: ,
4
f x x 12x 35x 11 0 when x 1,
11 , 3. 5
f x 4x 15x2 22x 23 0 when x 1,
( 115 , 1.11(
(1, 0)
(2.69, 0.46) (3, 0)
2
x
−2
4 −2
11± 6 . 5
6
(1.71, 0.60)
−4
f 3 > 0 Therefore, 3, 0 is a relative minimum. f
115 < 0
Therefore,
is a relative maximum. 115, 3456 3125
Points of inflection: 1, 0,
11 5
6
11 5
, 0.60 ,
6
, 0.46
Intercepts: 0, 9, 1, 0, 3, 0 51. f x x13x 323
y
Domain: , ; Range: ,
4 3
x1 f x 0 when x 1 and undefined when x 3, 0. x 313x23 2 f x 53 is undefined when x 0, 3. x x 343 3 4 is By the First Derivative Test 3, 0 is a relative maximum and 1, a relative minimum. 0, 0 is a point of inflection.
Intercepts: 3, 0, 0, 0
2 1
) 3, 0)
)0, 0) x
5
4
2
) 1,
1
1
1.59) 3
2
167
168
Chapter 3
Applications of Differentiation
x1 x1
53. f x
x
1
y
Domain: , 1, 1, ; Range: , 1, 1, f x
2 < 0 if x 1. x 12
f x
4 x 13
4
y
1 2
x 2
2
4
2
Horizontal asymptote: y 1 Vertical asymptote: x 1 Intercepts: 1, 0, 0, 1 55. f x
4 1 x2
y 5
Domain: , ; Range: 0, 4
8x 0 when x 0. f x 1 x22 3 81 3x2 . 0 when x ± f x 1 x23 3
(0, 4)
4
(−
3,3 3
(
(
3,3 3
1
2
(
2 1 −3
−2 −1
x −1
3
f 0 < 0 Therefore, 0, 4 is a relative maximum. Points of inflection: ± 33, 3 Intercept: 0, 4 Symmetric to the y-axis Horizontal asymptote: y 0
57. f x x3 x
y
4 x
10
Domain: , 0, 0, ; Range: , 6 , 6, f x 3x2 1 f x 6x
4 x2
3x4
4 0 when x ± 1. x2 x2
8 6x4 8 0 x3 x3
f 1 < 0 Therefore, 1, 6 is a relative maximum. f 1 > 0 Therefore, 1, 6 is a relative minimum. Vertical asymptote: x 0 Symmetric with respect to origin
5
(1, 6) x
2
1
(−1, −6) −5
1
x
2
0
Review Exercises for Chapter 3
59. f x x2 9
y
Domain: , ; Range: 0, f x
2xx2 9 0 when x 0 and is undefined when x ± 3. x2 9
10
5
2x2 9 is undefined at x ± 3. f x 2 x 9
)0, 9)
) 3, 0)
)3, 0) x
4
f 0 < 0
2
2
4
Therefore, 0, 9 is a relative maximum. Relative minima: ± 3, 0 Points of inflection: ± 3, 0 Intercepts: ± 3, 0, 0, 9 Symmetric to the y-axis 61. f x x cos x
y
)2 , 2
Domain: 0, 2 ; Range: 1, 1 2
3 3 , 2 2
f x 1 sin x ≥ 0, f is increasing. f x cos x 0 when x Points of inflection:
1)
2
3 , . 2 2
3 3 , , , 2 2 2 2
)0, 1)
, 2 2 x
2
Intercept: 0, 1 63. x2 4y2 2x 16y 13 0 (a) x 2 2x 1 4y 2 4y 4 13 1 16
y
x 1 4y 2 4 x 12 y 22 1 4 1 The graph is an ellipse: 2
2
4
(1, 3) 3 2 1
Maximum: 1, 3
(1, 1) x −1
Minimum: 1, 1
1
2
3
(b) x2 4y2 2x 16y 13 0 2x 8y
dy dy 2 16 0 dx dx dy 8y 16 2 2x dx dy 2 2x 1x dx 8y 16 4y 8
The critical numbers are x 1 and y 2. These correspond to the points 1, 1, 1, 3, 2, 1, and 2, 3. Hence, the maximum is 1, 3 and the minimum is 1, 1.
169
170
Chapter 3
Applications of Differentiation
65. Let t 0 at noon.
(100 − 12t, 0) (0, 0)
L d 2 100 12t2 10t2 10,000 2400t 244t 2
A
(100, 0)
d
300 dL 2400 488t 0 when t 4.92 hr. dt 61
B (0, −10t)
Ship A at 40.98, 0; Ship B at 0, 49.18 d 2 10,000 2400t 244t 2 4098.36 when t 4.92 4:55 P.M.. d 64 km 67. We have points 0, y, x, 0, and 1, 8. Thus,
y
08 8x y8 or y . m 01 x1 x1
(0, y)
10
(1, 8)
8 6
Let f x L 2 x 2
x 8x 1 . 2
4 2
x f x 2x 128 x1 x
(x, 0)
x 1 x 0 x 12
x 2
4
6
8
10
64x 0 x 13
x x 13 64 0 when x 0, 5 minimum. Vertices of triangle: 0, 0, 5, 0, 0, 10 69.
A Average of basesHeight
x 2 s
3s2 2sx x2
2
s
see figure
s
dA 1 s xs x 3s2 2sx x2 dx 4 3s2 2sx x2
x−s 2
22s xs x 0 when x 2s. 43s2 2sx x2 A is a maximum when x 2s. 71. You can form a right triangle with vertices 0, 0, x, 0 and 0, y. Assume that the hypotenuse of length L passes through 4, 6. 60 6x y6 or y 04 4x x4
Let f x L2 x2 y2 x 2 f x 2x 72
x 6x 4 . 2
0 x x 4 x 4 4 2
3 x x 43 144 0 when x 0 or x 4 144.
L 14.05 feet
s
x−s 2 x
m
3s 2 + 2sx − x 2 2
Review Exercises for Chapter 3 csc
73. csc
L1 or L1 6 csc 6
2 9 or L L2
2
9 csc
see figure
L1 θ
2
L2
θ 9
L L1 L2 6 csc 9 csc
171
6
(π2 − θ(
2 6 csc 9 sec
dL 6 csc cot 9 sec tan 0 d tan3
3 2 2 ⇒ tan 3 3 3
sec 1 tan2 csc L6
1 23
23
323 223
313
sec 323 223 tan 213
323 22312 323 22312 9 3323 22332 ft 21.07 ft Compare to Exercise 72 using a 9 and b 6. 13 2 313
75. Total cost Cost per hourNumber of hours T
v 110 11v 550 5 600 v 60 v 2
dT 11 550 11v 2 33,000 2 dv 60 v 60v 2 0 when v 3000 1030 54.8 mph. d 2T 1100 3 > 0 when v 1030 so this value yields a minimum. dv 2 v 77. f x x3 3x 1 From the graph you can see that f x has three real zeros. f x 3x2 3 f xn
f xn
f xn f xn
1.5000
0.1250
3.7500
0.0333
1.5333
2
1.5333
0.0049
4.0530
0.0012
1.5321
n
xn
f xn
f xn
f xn f xn
1
0.5000
0.3750
2.2500
0.1667
0.3333
2
0.3333
0.0371
2.6667
0.0139
0.3472
3
0.3472
0.0003
2.6384
0.0001
0.3473
n
xn
f xn
f xn
f xn f xn
1
1.9000
0.1590
7.8300
0.0203
1.8797
2
1.8797
0.0024
7.5998
0.0003
1.8794
n
xn
1
xn
f xn f xn
xn
xn
f xn f xn
f xn f xn
The three real zeros of f x are x 1.532, x 0.347, and x 1.879.
C H A P T E R 3 Applications of Differentiation Section 3.1
Extrema on an Interval
. . . . . . . . . . . . . . 378
Section 3.2
Rolle’s Theorem and the Mean Value Theorem
Section 3.3
Increasing and Decreasing Functions and the First Derivative Test . . . . . . . . . . . . . . 387
Section 3.4
Concavity and the Second Derivative Test . . . . 394
Section 3.5
Limits at Infinity
Section 3.6
A Summary of Curve Sketching
Section 3.7
Optimization Problems . . . . . . . . . . . . . . 419
Section 3.8
Newton’s Method . . . . . . . . . . . . . . . . . 429
Section 3.9
Differentials . . . . . . . . . . . . . . . . . . . . 434
. 381
. . . . . . . . . . . . . . . . . 402 . . . . . . . . . 410
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 437 Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . 445
C H A P T E R 3 Applications of Differentiation Section 3.1
Extrema on an Interval
Solutions to Even-Numbered Exercises
2. f x cos
x 2
4.
f x 3x x 1
12 x 1
fx 3x
x fx sin 2 2
12
x 1 3
3 x 112x 2x 1 2
f0 0 f2 0
3 x 1123x 2 2
32 0
f
6. Using the limit definition of the derivative,
lim
f x f 0 4 x 4 lim 1 x→0 x0 x
lim
f x f 0 4 x 4 lim 1 x→0 x0 x0
x→0
x→0
8. Critical number: x 0. x 0: neither
f0 does not exist, since the one-sided derivatives are not equal. 10. Critical numbers: x 2, 5
12. gx x2x2 4 x4 4x2
x 2: neither
gx 4x3 8x 4xx2 2
x 5: absolute maximum
Critical numbers: x 0, x ± 2
14. f x fx
4x x2 1
x2 14 4x2x 41 x2 2 x2 12 x 12
Critical numbers: x ± 1
16. f 2 sec tan , 0 < < 2 f 2 sec tan sec2 sec 2 tan sec sin
cos cos
sec 2
1
sec2 2 sin 1 On 0, 2, critical numbers:
378
7 11 , 6 6
Section 3.1
18. f x fx
2x 5 , 0, 5 3
20. f x x2 2x 4, 1, 1 fx 2x 2 2x 1
2 ⇒ No critical numbers 3
Left endpoint:
Extrema on an Interval
Left endpoint: (1, 5 Minimum Right endpoint: 1, 1 Maximum
0, 53 Minimum
Right endpoint: 5, 5 Maximum 22. f x x3 12x, 0, 4
3 24. gx x, 1, 1
fx 3x2 12 3x2 4
gx
Left endpoint: 0, 0
Left endpoint: 1, 1 Minimum
Critical number: 2, 16 Minimum
Critical number: 0, 0
Right endpoint: 4, 16 Maximum
Right endpoint: 1, 1 Maximum
Note: x 2 is not in the interval.
26. y 3 t 3 , 1, 5
28. ht
From the graph, you see that t 3 is a critical number. 4
−1
1 3x23
ht
t , 3, 5 t2 2 t 22
Left endpoint: 3, 3 Maximum
5
Right endpoint: −4
5, 53 Minimum
Left endpoint: 1, 1 Minimum Right endpoint: 5, 1 Critical number: 3, 3 Maximum
30. g x sec x,
6 , 3
32. y x2 2 cos x, 1, 3 y 2x sin x
gx sec x tan x Left endpoint:
Right endpoint:
2 ,
, 1.1547 6 3 6
3 , 2 Maximum
Left endpoint: 1, 1.5403 Right endpoint: 3, 7.99 Maximum Critical number: 0, 3 Minimum
Critical number: 0, 1 Minimum 34. (a) Minimum: 4, 1 Maximum: 1, 4
36. (a) Minima: 2, 0 and 2, 0 Maximum: 0, 2
(b) Maximum: 1, 4
(b) Minimum: 2, 0
(c) Minimum: 4, 1
(c) Maximum: 0, 2
(d) No extrema
(d) Maximum: 1, 3
379
380
Chapter 3
38. f x
Applications of Differentiation
22 3x,x ,
1 ≤ x < 3 3 ≤ x ≤ 5
2
2 , 0, 2 2x Left endpoint: 0, 1 Minimum
40. f x
Left endpoint: 1, 1 Maximum
3
Right endpoint: 5, 13 Minimum 3
(1, 1)
−3
12
(0, 1)
−1
(3, − 7)
5
−1
(5, − 13) −15
42. (a)
Maximum:
3
(2, ) 8 3
0
2, 83
Minimum: 0, 0, 3, 0 3
0
4 (b) f x x 3 x, 0, 3 3 fx
4 1 x 3 x121 3 x121 3 2
4 1 3 x12 x 23 x 3 2
44. f x
1 1 , ,3 x2 1 2
fx
2x x2 12
f x
21 3x2 x2 13
fx
24x 24x3 x2 14
Setting f 0, we have x 0, ± 1. 1
f 1 2 is the maximum value.
26 3x 62 x 22 x 3 x 3 3 x 3 3 x
Critical number: x 2 f 0 0 Minimum f 3 0 Minimum f 2
8 3
Maximum:
46.
f x f x f 4x f 5x
2, 38
1 , 1, 1 x2 1 24x 24x3 See Exercise 44. x2 14
48. Let f x 1x. f is continuous on 0, 1 but does not have a maximum. f is also continuous on 1, 0 but does not have a minimum. This can occur if one of the endpoints is an infinite discontinuity.
245x4 10x2 1 x2 15 240x
x2 16 3x4
10x2
y 2
3
f 40 24 is the maximum value.
1
−2
x
−1
1 −1
2
Section 3.2 y
50. 5 4
Rolle’s Theorem and the Mean Value Theorem
52. (a) No
54. (a) No
(b) Yes
(b) Yes
381
f
3 2
x
−2 −1
1
2
3
4
5
6
−2 −3
56. x
v 2 sin 2 3 , ≤ ≤ 32 4 4
C 2x
58.
C1 300,002
d is constant. dt
C300 1600
dx d dx by the Chain Rule dt d dt
300,000 , 1 ≤ x ≤ 300 x
C 2
v 2 cos 2 d 16 dt
300,000 0 x2
2x2 300,000 x2 150,000
In the interval 4, 34, 4, 34 indicate minimums for dxdt and 2 indicates a maximum for dxdt. This implies that the sprinkler waters longest when 4 and 34. Thus, the lawn farthest from the spinkler gets the most water.
x 10015 387 > 300 outside of interval C is minimized when x 300 units. Yes, if 1 ≤ x ≤ 400, then x 387 would minimize C.
60. f x x
y 3
The derivative of f is undefined at every integer and is zero at any noninteger real number. All real numbers are critical numbers.
2 1 −2
x
−1
1
2
3
−2
64. False. Let f x x2. x 0 is a critical number of f.
62. True. This is stated in the Extreme Value Theorem.
gx f x k x k2 x k is a critical number of g.
Section 3.2
Rolle’s Theorem and the Mean Value Theorem 4. f x xx 3
2. Rolle’s Theorem does not apply to f x cotx2 over , 3 since f is not continuous at x 2.
x-intercepts: 0, 0, 3, 0 3 fx 2x 3 0 at x . 2
6. f x 3xx 1 x-intercepts: 1, 0, 0, 0
1 x fx 3x x 112 3x 112 3x 112 x 1 2 2 fx 3x 112
32x 1 0 at x 32.
382
Chapter 3
Applications of Differentiation
8. f x x2 5x 4, 1, 4
f x x 3x 12, 1, 3
10.
f 1 f 4 0
f 1 f 3 0
f is continuous on 1, 4. f is differentiable on 1, 4.
f is continuous on 1, 3. f is differentiable on 1, 3. Rolle’s Theorem applies.
Rolle’s Theorem applies.
fx x 32x 1 x 12
fx 2x 5 2x 5 0 ⇒ x c value:
x 12x 6 x 1
5 2
x 13x 5
5 2
c value:
12. f x 3 x 3 , 0, 6
5 3
f x
14.
x2 1 , 1, 1 x
f 0 f 6 0
f 1 f 1 0
f is continuous on 0, 6. f is not differentiable on 0, 6 since f3 does not exist. Rolle’s Theorem does not apply.
f is not continuous on 1, 1 since f 0 does not exist. Rolle’s Theorem does not apply.
16. f x cos x, 0, 2 f 0 f 2 1 f is continuous on 0, 2. f is differentiable on 0, 2. Rolle’s Theorem applies.
f x cos 2x,
18.
f f
fx sin x
c value:
12 , 6
3 12 2
6 21
f
f 12 0
Rolle’s Theorem does not apply.
20.
f x sec x,
4 , 4
f f 2 4 4
f is continuous on 4, 4. f is differentiable on 4, 4. Rolle’s Theorem applies.
22. f x x x13, 0, 1 f 0 f 1 0 f is continuous on 0, 1. f is differentiable on 0, 1. (Note: f is not differentiable at x 0.) Rolle’s Theorem applies. fx 1
fx sec x tan x sec x tan x 0
1
x0
3 x2
c value: 0
x2 x c value:
1 3 x2 3
1 3 1 27
271
3
9
1 0 3 x2 3
3
9
0.1925
1
0
−1
1
Section 3.2
f x
24.
x x sin , 1, 0 2 6
f 1 f 0 0 f is continuous on 1, 0. f is differentiable on 1, 0. Rolle’s Theorem applies. 1 x fx cos 0 2 6 6
x 3 cos 6 x
Rolle’s Theorem and the Mean Value Theorem
26. Cx 10
1x x x 3
(a) C3 C6
25 3
Cx 10
(b)
1 3 0 x2 x 32
1 3 x2 6x 9 x2 2x2 6x 9 0
6 3 arccos Value needed in 1, 0.
x
0.5756 radian
c value: 0.5756
6 ± 63 3 ± 33 4 2
In the interval 3, 6: c
0.02
−1
6 ± 108 4
3 33 4.098. 2
0
−0.01
28.
30. f x x 3 , 0, 6
y
f is not differentiable at x 3.
f x
a
b
32. f x xx2 x 2 is continuous on 1, 1 and differentiable on 1, 1. f 1 f 1 1 1 1 fx 3x2 2x 2 1
3x 1x 1 0 1 c 3
34. f x x 1x is continuous on 12, 2 and differentiable on 12, 2. f 2 f 12 32 3 1 2 12 32 fx
1 1 x2
x2 1 c1
383
384
Chapter 3
Applications of Differentiation
36. f x x3 is continuous on 0, 1 and differentiable on 0, 1.
38. f x 2 sin x sin 2x ferentiable on 0, .
f 1 f 0 1 0 1 10 1 fx 3x2 1 x±
f f 0 0 0 0 0 fx 2 cos x 2 cos 2x 0 2cos x 2 cos2 x 1 0
3
3
In the interval 0, 1: c
is continuous on 0, and dif-
22 cos x 1cos x 1 0 3
3
.
1 2
cos x
cos x 1
5 , , 3 3
x In the interval 0, : c
. 3
40. f x x 2 sin x on , (c) fx 1 2 cos x 1
2
(a) tangent − 2
secant 2 tangent
f
− 2
c± , 2
f
2 2 2
2 2 2
f
(b) Secant line: slope
cos x 0
f f 1 2
y 1x
Tangent lines: y
2 2 1 x 2 yx2
y
yx
2 1 x 2 2
yx2 42. f x x4 4x3 8x2 5, 0, 5, 5, 80 m
80 5 15 50
(a)
150
(c) First tangent line: y f c mx c y 9.59 15x 0.67
tangent f
secant
0 15x y 0.46
tangent 5
0 0
(b)
Secant line: y 5 15x 0 0 15x y 5 fx 4x3 12x2 16x f 5 f 1 15 51 4c3 12c2 16c 15 0 4c3 12c2 16c 15 c 0.67 or c 3.79
Second tangent line: y f c mx c y 131.35 15x 3.79 0 15x y 74.5
Section 3.2
44. St 200 5 (a)
9 2t
Rolle’s Theorem and the Mean Value Theorem
S12 S0 2005 914 2005 92 450 12 0 12 7 St 200
(b)
2 9 t 450 7 2
1 1 2 t2 28 2 t 27 t 27 2 3.2915 months St is equal to the average value in April. 46. f a f b and fc 0 where c is in the interval a, b. (a) gx f x k
gx f x k
(b)
(c)
ga k gb k f a
ga gb f a k gx fx ⇒ gc 0
gx fx k
gx f k x g
ak g bk f a
gx k fk x
Interval: a, b
gc k fc 0
Critical number of g: c
Interval: a k, b k Critical number of g: c k
g
ck k fc 0
Interval:
ak, bk
Critical number of g:
c k
48. Let Tt be the temperature of the object. Then T0 1500 and T5 390 . The average temperature over the interval 0, 5 is 390 1500 222 Fhr. 50 By the Mean Value Theorem, there exists a time to, 0 < t0 < 5, such that Tt0 222. 2 x 50. f x 3 cos 2 ,
fx 6 cos
2x sin 2x 2
3 cos (a)
2x sin 2x (b) f and f are both continuous on the entire real line.
7
−2
2
−7
(c) Since f 1 f 1 0, Rolle’s Theorem applies on 1, 1. Since f 1 0 and f 2 3, Rolle’s Theorem does not apply on 1, 2.
(d) lim fx 0 x→3
lim fx 0
x→3
385
386
Chapter 3
Applications of Differentiation
52. f is not continuous on 5, 5. Example: f x
0,1x,
54. False. f must also be continuous and differentiable on each interval. Let
x0 x0
f x
y
x3 4x . x2 1
f (x) = 1x
4 2
(5, 15) x 2
(− 5, − 15)
4
−5
56. True 58. Suppose f x is not constant on a, b. Then there exists x1 and x2 in a, b such that f x1 f x2. Then by the Mean Value Theorem, there exists c in a, b such that fc
f x2 f x1 0. x2 x1
This contradicts the fact that fx 0 for all x in a, b. 60. Suppose f x has two fixed points c1 and c2. Then, by the Mean Value Theorem, there exists c such that fc
f c2 f c1 c2 c1 1. c2 c1 c2 c1
This contradicts the fact that fx < 1 for all x. 62. Let f x cos x. f is continuous and differentiable for all real numbers. By the Mean Value Theorem, for any interval a, b, there exists c in a, b such that f b f a fc ba cos b cos a sin c ba cos b cos a sin cb a
cos b cos a sin c
b a
cos b cos a ≤ b a since sin c
≤ 1.
Section 3.3
Section 3.3
Increasing and Decreasing Functions and the First Derivative Test
Increasing and Decreasing Functions and the First Derivative Test
2. y x 12
4. f x x4 2x2
Increasing on: , 1
Increasing on: 1, 0, 1,
Decreasing on: 1,
Decreasing on: , 1, 0, 1
6. y y
x2 x1 xx 2 x 12
Critical numbers: x 0, 2
Discontinuity: x 1
Test intervals: < x < 2
2 < x < 1
1 < x < 0
0 < x <
Sign of fx:
y > 0
y < 0
y < 0
y > 0
Conclusion:
Increasing
Decreasing
Decreasing
Increasing
Increasing on , 2, 0, Decreasing on 2, 1, 1, 0 8. hx 27x x3 hx 27 3x2 33 x3 x hx 0 Critical numbers: x ± 3 Test intervals:
< x < 3
3 < x < 3
Sign of hx:
h < 0
h > 0
h < 0
Decreasing
Increasing
Decreasing
Conclusion:
3 < x <
Increasing on 3, 3 Decreasing on , 3, 3, 10. y x y
387
4 x
x 2x 2 x2
Critical numbers: x ± 2
Discontinuity: 0
Test intervals: < x < 2
2 < x < 0
0 < x < 2
2 < x <
Sign of y:
y > 0
y < 0
y < 0
y > 0
Conclusion:
Increasing
Decreasing
Decreasing
Increasing
Increasing: , 2, 2, Decreasing: 2, 0, 0, 2
388
Chapter 3
Applications of Differentiation
12. f x x2 8x 10
14. f x x2 8x 12
fx 2x 8 0
fx 2x 8 0
Critical number: x 4
Critical number: x 4
Test intervals: Sign of fx:
< x < 4 4 < x <
Conclusion:
f < 0
f > 0
Decreasing
Increasing
Increasing on: 4,
fx 3x2 12x 3xx 4 Critical numbers: x 0, 4 Test intervals:
< x < 0
0 < x < 4
Sign of fx:
f > 0
f < 0
f > 0
Conclusion:
Increasing
Decreasing
Increasing
4 < x <
Increasing on , 0, 4, Decreasing on 0, 4 Relative maximum: 0, 15 Relative minimum: 4, 17 18. f x x 22x 1 fx 3xx 2 Critical numbers: x 2, 0 Test intervals:
< x < 2
2 < x < 0
Sign of fx:
f > 0
f < 0
f > 0
Conclusion:
Increasing
Decreasing
Increasing
Relative minimum: 0, 4
f > 0
f < 0
Conclusion:
Increasing
Decreasing
Relative maximum: 4, 4
16. f x x3 6x2 15
Relative maximum: 2, 0
Sign of fx:
Decreasing on: 4,
Relative minimum: 4, 6
Decreasing on: 2, 0
< x < 4
Increasing on: , 4
Decreasing on: , 4
Increasing on: , 2, 0,
Test intervals:
0 < x <
4 < x <
Section 3.3
Increasing and Decreasing Functions and the First Derivative Test 22. f x x23 4
20. f x x4 32x 4 fx 4x3 32 4x3 8
2 2 fx x13 13 3 3x
Critical number: x 2 Test intervals: Sign of fx: Conclusion:
389
Critical number: x 0
< x < 2
2 < x <
f < 0
f > 0
Test intervals:
< x < 0
Decreasing
Increasing
Sign of fx:
f < 0
f > 0
Conclusion:
Decreasing
Increasing
Increasing on: 2, Decreasing on: , 2
Increasing on: 0,
Relative minimum: 2, 44
Decreasing on: , 0
0 < x <
Relative minimum: 0, 4
24. f x x 113
26. f x x 3 1
1 fx 3x 123
fx
Critical number: x 1
Critical number: x 3
Test intervals: Sign of fx: Conclusion:
< x < 1
1 < x <
1, x > 3 x3 x3 1, x < 3
Test intervals:
f > 0
f > 0
Sign of fx:
Increasing
Increasing
Conclusion:
< x < 3 Decreasing
Increasing
Increasing on: 3,
No relative extrema
Decreasing on: , 3 Relative minimum: 3, 1
fx
x x1
x 11 x1 1 x 12 x 12
Discontinuity: x 1 Test intervals: Sign of fx: Conclusion:
< x < 1
f > 0
f > 0
Increasing
Increasing
Increasing on: , 1, 1, No relative extrema
1 < x <
f > 0
Increasing on: ,
28. f x
3 < x <
f < 0
390
Chapter 3
30. f x
Applications of Differentiation
x3 1 3 2 x2 x x
fx
1 6 x 6 x2 x3 x3
Critical number: x 6 Discontinuity: x 0 Test intervals: Sign of fx:
< x < 6
6 < x < 0
f < 0
f > 0
f < 0
Decreasing
Increasing
Decreasing
Conclusion:
0 < x <
Increasing on: 6, 0 Decreasing on: , 6, 0, Relative minimum:
32. f x fx
6, 121
x2 3x 4 x2
x 22x 3 x2 3x 41 x2 4x 10 x 22 x 22
Discontinuity: x 2 < x < 2
Test intervals: Sign of fx: Conclusion:
2 < x <
f > 0
f > 0
Increasing
Increasing
Increasing on: , 2, 2, No relative extrema 34. f x sin x cos x
1 sin 2x, 0 < x < 2 2
fx cos 2x 0 Critical numbers: x
3 5 7 , , , 4 4 4 4
0 < x <
Test intervals:
4
3 < x < 4 4
3 5 < x < 4 4
5 7 < x < 4 4
7 < x < 2 4
Sign of fx:
f > 0
f < 0
f > 0
f < 0
f > 0
Conclusion:
Increasing
Decreasing
Increasing
Decreasing
Increasing
Increasing on: Decreasing on:
0, 4 , 34, 54, 74, 2 4 , 34, 54, 74
Relative maxima:
4 , 21, 54, 21
Relative minima:
34, 21, 74, 12
Section 3.3
36. f x fx
Increasing and Decreasing Functions and the First Derivative Test
sin x , 0 < x < 2 1 cos2x
Test intervals:
0 < x <
cos x2 sin2x 0 1 cos2x2
Sign of fx: Conclusion:
Critical numbers: x
3 , 2 2
0, 2 , 32, 2
Increasing on: Decreasing on:
2 , 32
3 < x < 2 2
3 < x < 2 2
f > 0
f < 0
f > 0
Increasing
Decreasing
Increasing
2
Relative maximum:
2 , 1
Relative minimum:
32, 1
38. f x 10 5 x2 3x 16 , 0, 5 (a) fx
52x 3 x2 3x 16
(b)
y 15
f
12
6
f′
3 −1
(c)
52x 3 0 x2 3x 16
x 3
1
−3
4
(d) Intervals:
3 Critical number: x 2
0, 23
32, 5
fx > 0
fx < 0
Increasing
Decreasing
f is increasing when f is positive and decreasing when f is negative.
40. f x
x x cos , 0, 4 2 2
(a) fx
1 1 x sin 2 2 2
(b)
y 8 6
f 4 2
f′ π
(c)
x 1 1 sin 0 2 2 2 sin
x 1 2 x 2 2
Critical number: x
2π
3π
4π
x
(d) Intervals:
0,
, 4
fx > 0
fx > 0
Increasing
Increasing
f is increasing when f is positive.
391
392
Chapter 3
Applications of Differentiation 44. f x is a line of slope 2 ⇒ fx 2.
42. f t cos2t sin2t 2 sin2t gt, 2 < t < 2 ft 4 sin t cos t 2 sin 2t
6
f symmetric with respect to y-axis zeros of f: ±
4
−6
6 −2
Relative maximum: 0, 1
2 , 1, 2 , 1
Relative minimum: 2
−3
3
−2
46. f is a 4th degree polynomial ⇒ f is a cubic polynomial.
48. f has positive slope y
y 6
4
f′
3 2
−6 −4 −2
x 2
4
f′
6
x
−3 −2 −1
1
2
3
−2
In Exercises 50–54, f x > 0 on , 4, f x < 0 on 4, 6 and f x > 0 on 6, . gx 3f x 3
50.
52. gx f x
54. gx f x 10
gx fx
gx fx 10
g0 f0 > 0
g8 f2 < 0
gx 3fx g5 3f5 > 0
58. st 4.9sin t 2
56. Critical number: x 5 f4 2.5 ⇒ f is decreasing at x 4.
(a) vt 9.8sin t
(b) If 2, the speed is maximum,
f6 3 ⇒ f is increasing at x 6.
5, f 5 is a relative minimum. 60. C
vt 9.8 t.
3t ,t ≥ 0 27 t3
(a)
t
0
0.5
1
1.5
2
2.5
3
Ct
0
0.055
0.107
0.148
0.171
0.176
0.167
The concentration seems greater near t 2.5 hours. (b)
(c) C
0.25
0
3 0
The concentration is greatest when t 2.38 hours.
speed 9.8 sin t
27 t 33 3t3t 2 27 t 32 327 2t 3 27 t 32
3 C 0 when t 3 2 2.38 hours.
By the First Derivative Test, this is a maximum.
Section 3.3 62. P 2.44x P 2.44
Increasing and Decreasing Functions and the First Derivative Test
x2 5000, 0 ≤ x ≤ 35,000 20,000
Test intervals: Sign of P:
0 < x < 24,400
24,400 < x < 35,000
P > 0
P < 0
x 0 10,000
x 24,400 Increasing when 0 < x < 24,400 hamburgers. Decreasing when 24,400 < x < 35,000 hamburgers. 64. R 0.001T 4 4T 100 (a) R
0.004T 3 4 0 2 0.001T 4 4T 100
(b)
125
T 10 , R 8.3666
−100
100 −25
The minimum resistance is approximately R 8.37 at T 10 . 66. f x 2 sin3x 4 cos3x 6
−
−6
The maximum value is approximately 4.472. You could use calculus by finding fx and then observing that the maximum value of f occurs at a point where fx 0. For instance, f0.154 0, and f 0.154 4.472. 68. (a) Use a cubic polynomial f x a3x3 a2x2 a1x a0. (b) fx 3a3 x 2 2a 2 x a1
0, 0:
4, 1000:
0 a0
f 0 0
0 a1
f0 0
1000 64a3 16a2
f 4 1000
0 48a3 8a2
f4 0
(c) The solution is a0 a1 0, a2 f x (d)
125 3 375 2 x x. 4 2
1200
(4, 1000)
−3
(0, 0) −400
8
375 125 , a3 2 4
393
394
Chapter 3
Applications of Differentiation
70. (a) Use a fourth degree polynomial f x a4 x4 a3 x3 a 2 x 2 a1x a0. (b) fx 4a4x3 3a3x2 2a2x a1
1, 2:
1, 4:
3, 4:
2 a4 a3 a2 a1 a0
f 1 2
0 4a4 3a3 2a2 a1
f1 0
4 a4 a3 a2 a1 a0
f 1 4
0 4a4 3a3 2a2 a1
f1 0
4 81a4 27a3 9a2 3a1 a0
f 3 4
0 108a4 27a3 6a2 a1
f3 0
(c) The solution is a0
23 8 ,
a1
32 ,
a2
1 4,
a3
1 2,
a4 18
f x 18 x 4 12 x3 14 x2 32 x 23 8 . (d)
6
(−1 , 4)
(3, 4)
(1, 2)
−4
6 −2
74. True
72. False Let hx f xgx where f x gx x. Then hx x2 is decreasing on , 0.
If f x is an nth-degree polynomial, then the degree of fx is n 1.
76. False. The function might not be continuous. 78. Suppose fx changes from positive to negative at c. Then there exists a and b in I such that fx > 0 for all x in a, c and fx < 0 for all x in c, b. By Theorem 3.5, f is increasing on a, c and decreasing on c, b. Therefore, f c is a maximum of f on a, b and thus, a relative maximum of f.
Section 3.4
Concavity and the Second Derivative Test
2. y x3 3x2 2, y 6x 6 Concave upward: , 1 Concave downward: 1,
6. y
1 2 3x5 40x3 135x, y xx 2x 2 270 9
Concave upward: , 2, 0, 2 Concave downward: 2, 0, 2,
4. f x
x2 1 6 , y 2x 1 2x 13
Concave upward: , 12 Concave downward: 12 , 8. hx x5 5x 2 hx 5x4 5 hx 20x3 Concave upward: 0, Concave downward: , 0
Section 3.4 10. y x 2 csc x, ,
fx 6x2 6x 12 f x 12x 6
2 cot xcsc x cot x
csc2 x
2
csc3 x
csc x
cot2 x
f x 12x 6 0 when x 12.
Concave upward: 0,
< x <
Test interval
Concave downward: , 0
Sign of f x Conclusion Point of inflection:
14.
395
12. f x 2x3 3x2 12x 5
y 1 2 csc x cot x y 2 csc x
Concavity and the Second Derivative Test
1 2
1 2
< x <
f x < 0
f x > 0
Concave downward
Concave upward
12 , 132
f x 2x4 8x 3 fx 8x3 8 f x 24x 2 0 when x 0. However, 0, 3 is not a point of inflection since f x ≥ 0 for all x. Concave upward on ,
16. f x x3x 4 fx x3 3x2x 4 x2x 3x 4 4x2x 3 f x 4x 2 8xx 3 4xx 2x 3 12xx 2 0 f x 12xx 2 0 when x 0, 2. Test interval Sign of f x Conclusion
< x < 0
0 < x < 2
f x > 0
f x < 0
f x > 0
Concave upward
Concave downward
Concave upward
2 < x <
Points of inflection: 0, 0, 2, 16 18. f x xx 1, Domain: 1, 3x 2 1 fx x x 112 x 1 2 2x 1 6x 1 3x 2x 112 3x 4 f x 4x 1 4x 132 f x > 0 on the entire domain of f (except for x 1, for which f x is undefined). There are no points of inflection. Concave upward on 1, 20. f x
x1 x
Domain: x > 0
x1 fx 32 2x 3x f x 52 4x Point of inflection:
3, 43 3, 4 3 3
3 < x <
Test intervals
0 < x < 3
Sign of f x
f > 0
f < 0
Conclusion
Concave upward
Concave downward
396
Chapter 3
22. f x 2 csc
3x , 0 < x < 2 2
fx 3 csc f x
Applications of Differentiation
3x 3x cot 2 2
9 3x 3x 3x csc3 0 for any x in the domain of f. csc cot2 2 2 2 2
Concave upward:
0, 23 , 43, 2 23, 43
Concave downward:
No points of inflection 24. f x sin x cos x, 0 ≤ x ≤ 2 fx cos x sin x f x sin x cos x f x 0 when x
3 7 , . 4 4 0 < x <
Test interval: Sign of f x: Conclusion:
3 4
3 7 < x < 4 4
7 < x < 2 4
f x < 0
f x > 0
f x < 0
Concave downward
Concave upward
Concave downward
3 < x < 2 2
3 3 < x < 2 2
Points of inflection:
34, 0 , 74, 0
26. f x x 2 cos x, 0, 2 fx 1 2 sin x f x 2 cos x f x 0 when x
3 , 2 2 2
Test intervals:
0 < x <
Sign of f x:
f < 0
f > 0
f < 0
Conclusion:
Concave downward
Concave upward
Concave downward
Points of inflection:
2 , 2 , 32, 32
28. f x x2 3x 8
30. f x x 52
fx 2x 3
fx 2x 5
f x 2
f x 2
3 Critical number: x 2
Critical number: x 5
f 32 > 0
f 5 < 0
Therefore, 2 , 4 is a relative minimum. 3
41
Therefore, 5, 0 is a relative maximum.
Section 3.4 32. f x x3 9x2 27x fx 3x 2 18x 27 3x 32 f x 6x 3 Critical number: x 3 However, f 3 0, so we must use the First Derivative Test. fx ≥ 0 for all x and, therefore, there are no relative extrema.
Concavity and the Second Derivative Test
1 34. gx x 22x 42 8 gx
x 4x 1x 2 2
3 g x 3 3x x2 2 Critical numbers: x 2, 1, 4 g 2 9 < 0
2, 0 is a relative maximum. g 1 92 > 0
1, 10.125 is a relative minimum. g 4 9 < 0
4, 0 is a relative maximum. 36. f x x2 1 fx
38. f x
x
fx
x2 1
Critical number: x 0 1 f x 2 x 132
x x1 1 x 12
There are no critical numbers and x 1 is not in the domain. There are no relative extrema.
f 0 1 > 0 Therefore, 0, 1 is a relative minimum. 40. f x 2 sin x cos 2x, 0 ≤ x ≤ 2 fx 2 cos x 2 sin 2x 2 cos x 4 sin x cos x 2 cos x1 2 sin x 0 when x f x 2 sin x 4 cos 2x f
6 < 0
f
2 > 0
f
56 < 0
f
32 > 0
Relative maxima:
6 , 23 , 56, 23
Relative minima:
2 , 1 , 32, 3
5 3 , , , . 6 2 6 2
397
398
Chapter 3
Applications of Differentiation
42. f x x26 x2, 6, 6 (a) fx
3x4 x2 6 x2
(c)
y
f
6
fx 0 when x 0, x ± 2. f x
6x4 9x2 12 6 x232
f x 0 when x ±
x
−3
9 2
33
.
3
f'
f ''
−6
(b) f 0 > 0 ⇒ 0, 0 is a relative minimum. The graph of f is increasing when f > 0 and decreasing when f < 0. f is concave upward when f > 0 and concave downward when f < 0.
f ± 2 < 0 ⇒ ± 2, 42 are relative maxima. Points of inflection: ± 1.2758, 3.4035 44. f x 2x sin x, 0, 2 (a) fx 2x cos x
sin x 2x
(c)
y 4
Critical numbers: x 1.84, 4.82
f f'
2
cos x cos x sin x f x 2x sin x 2x 2x 2x2x
2
f ''
−2
2cos x 4x2 1sin x 2x 2x2x
x
π
−4
4x cos x 4x2 1sin x 2x2x
f is increasing when f > 0 and decreasing when f < 0. f is concave upward when f > 0 and concave downward when f < 0.
(b) Relative maximum: 1.84, 1.85 Relative minimum: 4.82, 3.09 Points of inflection: 0.75, 0.83, 3.42, 0.72 46. (a)
f < 0 means f decreasing
y 4
(b) 4
f decreasing means concave downward
3 2
f > 0 means f increasing
y
f decreasing means concave downward
3 2
1
1 x 1
2
3
x
4
48. (a) The rate of change of sales is increasing. S > 0
1
50.
2
4
3
y
f
3
f'
f ''
(b) The rate of change of sales is decreasing. S > 0, S < 0 (c) The rate of change of sales is constant. S C, S 0 (d) Sales are steady. S C, S 0, S 0 (e) Sales are declining, but at a lower rate. S < 0, S > 0 (f) Sales have bottomed out and have started to rise. S > 0
−2
x
−1
3 −1
Section 3.4 52.
Concavity and the Second Derivative Test
54.
y
399
y
3
2
f ''
2
1
f
1
(0, 0) x
−1
1
(2, 0) x
−1
3
1
3
−1 −2
f'
−3
56.
58. (a)
y
d 12
3 2
(0, 0)
(2, 0) x
−1
1
3
−1
t 10
(b) Since the depth d is always increasing, there are no relative extrema. fx > 0 (c) The rate of change of d is decreasing until you reach the widest point of the jug, then the rate increases until you reach the narrowest part of the jug’s neck, then the rate decreases until you reach the top of the jug. 3 x 60. (a) f x
y 3
fx 13 x23
2
f x 29 x53
1
(0, 0)
Inflection point: 0, 0
−6
(b) f x does not exist at x 0.
−4
x
−2
2
4
6
−2 −3
62. f x ax3 bx 2 cx d Relative maximum: 2, 4 Relative minimum: 4, 2 Point of inflection: 3, 3 fx 3ax 2 2bx c, f x 6ax 2b
f 2 8a 4b 2c d 4 56a 12b 2c 2 ⇒ 28a 6b c 1 f 4 64a 16b 4c d 2 f2 12a 4b c 0, f4 48a 8b c 0, f 3 18a 2b 0 28a 6b c 1
18a 2b
12a 4b c
16a 2b 1
16a 2b 1 2,
b
f x
1 3 2x
a
1
92 ,
0
2a
c 12, d 6
9 2 2x
12x 6
0
1
400
Chapter 3
Applications of Differentiation
64. (a) line OA: y 0.06x
slope: 0.06
line CB: y 0.04x 50
y
slope: 0.04
150
f x ax3 bx 2 cx d (−1000, 60) A
fx 3ax2 2bx c
1000, 60:
C (0, 50)
60 10003a 10002b 1000c d 0.06 10002 3a 2000b c
1000, 90:
(1000, 90) B
100
−1000
O
x 1000
90 10003a 10002b 1000c d 0.04 10002 3a 2000b c
The solution to this system of 4 equations is a 1.25 108, b 0.000025, c 0.0275, and d 50. (b) y 1.25 108x3 0.000025x2 0.0275x 50
(c)
0.1
100 −1100
1100
− 0.1 −1100
1100 −10
(d) The steepest part of the road is 6% at the point A. 66. S
5.755T 3 8.521T 2 0.654T 0.99987, 0 < T < 25 108 106 104
(a) The maximum occurs when T 4 and S 0.999999. (b)
(c) S20 0.9982
S
1.001 1.000 0.999 0.998 0.997 0.996 T
5
68. C 2x C 2
10
15
20
25
30
300,000 x
300,000 0 when x 10015 387 x2
By the First Derivative Test, C is minimized when x 387 units.
70. S (a)
100t2 , t > 0 65 t2 100
0
35 0
(b) St St
13,000t 65 t22 13,00065 3t2 0 ⇒ t 4.65 65 t23
S is concave upwards on 0, 4.65, concave downwards on 4.65, 30. (c) St > 0 for t > 0. As t increases, the speed increases, but at a slower rate.
Section 3.4 72.
f x 2sin x cos x,
Concavity and the Second Derivative Test
f 0 2
401
4
P2
f
fx 2cos x sin x,
f0 2
f x 2sin x cos x,
f 0 2
−6
6
P1
P1x 2 2x 0 21 x
−4
P1x 2 1 P2x 2 2x 0 2 2x 0 2 2 2x x 2
P2x 2 2x P2x 2 The values of f, P1, P2, and their first derivatives are equal at x 0. The values of the second derivatives of f and P2 are equal at x 0. The approximations worsen as you move away from x 0. 74.
f x
x , x1
f 2 2
3
x 1 fx , 2xx 12
32 f2 4 22
3x2 6x 1 f x 32 , 4x x 13
23 232 f 2 16 82
P2x
5 −1
32 32 52 x 2 x 4 4 2
32 1 232 x 2 x 22 2 32 x 2 232 x 22 4 2 16 4 32
32 4
P2x 2 P2x
P2 f
−1
P1x 2 P1x
P1
3
32 232 x 2 4 16
232 16
The values of f, P1, P2 and their first derivatives are equal at x 2. The values of the second derivatives of f and P2 are equal at x 2. The approximations worsen as you move away from x 2. 76. f x xx 62 x3 12x2 36x fx 3x2 24x 36 3x 2x 6 0 f x 6x 24 6x 4 0 Relative extrema: 2, 32 and 6, 0 Point of inflection 4, 16 is midway between the relative extrema of f.
402
Chapter 3
Applications of Differentiation
px ax3 bx2 cx d
78.
px 3ax2 2bx c px 6ax 2b 6ax 2b 0 x
b 3a
The sign of px changes at x b3a. Therefore, b3a, pb3a is a point of inflection.
p
b b3 b2 b 2b3 bc d a b c d 3 2 3a 27a 9a 3a 27a 2 3a
When px x3 3x2 2, a 1, b 3, c 0, and d 2. x0
3 1 31
y0
233 30 2 2 0 2 0 2712 31
The point of inflection of px x3 3x2 2 is x0, y0 1, 0. 80. False. f x 1x has a discontinuity at x 0. 82. True y sinbx Slope: y b cosbx b ≤ y ≤ b
Assume b > 0
84. False. For example, let f x x 24.
Section 3.5 2. f x
Limits at Infinity
2x x2 2
4. f x 2
x4
x2 1
6. f x
2x2 3x 5 x2 1
No vertical asymptotes
No vertical asymptotes
No vertical asymptotes
Horizontal asymptotes: y ± 2
Horizontal asymptote: y 2
Horizontal asymptote: y 2
Matches (c)
Matches (a)
Matches (e)
8. f x x f x
2x2 x1 100
101
102
1
18.18
198.02
lim f x
x→
20
103 1998.02
(Limit does not exist.)
104 19,998
105 199,998
106 1,999,998
0 −2
10
Section 3.5
10. f x
8x
Limits at Infinity
10
x2 3
x
101
102
103
104
105
106
107
f x
8.12
8.001
8
8
8
8
8
0
15 0
lim f x 8
x→
12. f x 4 x f x
3 x2 2
10
100
101
102
103
104
105
106
5
4.03
4.0003
4.0
4.0
4
4
0
15 0
lim f x 4
x→
14. (a) hx
f x 5x2 3x 7 7 5x 3 x x x
lim hx
x→
(b) hx
f x x2
(Limit does not exist) 5x2
16. (a) lim
x→
(b) lim
3x 7 3 7 5 2 x2 x x
x→
(c) hx
3 2x 2 3x 1 3
3 2x2 x→ 3x 1
(c) lim
lim hx 5
x→
3 2x 0 3x3 1
(Limit does not exist)
f x 5x2 3x 7 5 7 3 2 3 x3 x3 x x x
lim hx 0
x→
18. (a) lim
x→
(b) lim
x→
5x32 0 4x2 1
5x32 x→ 4x 1
22. lim 4
26. lim
x→
3x3 2 3 1 9x3 2x2 7 9 3
24. lim
12x x4
x→
5x32 5 4x 32 1 4
(c) lim
x→
20. lim
(Limit does not exist)
3 404 x
x x2 1
lim
x→
lim
x→
x→
1
x2 1 x2
1 x (1x
for x < 1
0, x x2
2
(Limit does not exist)
403
404
28.
Chapter 3
lim
x →
Applications of Differentiation
3x 1 3 1x , for x < 0 we have x2 x lim x2 x x → x2 x x2 lim
3 1x
x →
30. lim
x →
1 1x
x cos x cos x lim 1 x → x x
3
32. lim cos x→
1x cos 0 1
101 Note: lim
x →
cos x 0 by the Squeeze Theorem since x
cos x 1 1 ≤ ≤ . x x x
34. f x
3x
36. lim x tan
6
x2 2
lim f x 3
x →
−9
x→
1 tan t sin t lim lim t →0 x t →0 t t
cos t 1
11 1
9
lim f x 3
Let x 1t.
x→
−6
Therefore, y 3 and y 3 are both horizontal asymptotes. 38. lim 2x 4x2 1 lim x →
40.
x →
2x
lim 4x 1
2x 4x2 1
4x2 1
2x
x →
3x 9x x lim 3x 9x2 x 3x 9x2 x x → x → 2 lim
lim
x →
3x
2
9x x
2x 4x2 1
0
x 3x 9x2 x 1
lim
x →
1
2
3
9x2 x
for x
< 0 we have x x2
x2 1 1 lim x → 3 9 1x 6 42.
x
100
101
102
103
104
105
f x
1.0
5.1
50.1
500.1
5000.1
50,000.1
lim
x →
x2 xx2 x 1
Limit does not exist.
30
106 500,000.1
x2 xx2 x x3 lim x2 xx2 x x → x2 xx2 x
0
50 0
Section 3.5 44.
f x lim
101
102
103
104
105
106
2.000
0.348
0.101
0.032
0.010
0.003
0.001
x1
x →
3
100
x
xx
Limits at Infinity
0
0
25
−1
46. x 2 is a critical number.
48. (a) The function is even:
fx < 0 for x < 2.
(b) The function is odd:
fx > 0 for x > 2.
lim f x 5
x→
lim f x 5
x→
lim f x lim f x 6
x →
x →
For example, let f x
6 6. 0.1x 22 1
y
8
4
x 2
2
4
6
x3 x2
50. y
2x 9 x2
52. y
Intercept: 0, 0
23
Intercepts: 3, 0, 0,
Symmetry: origin
Symmetry: none
Horizontal asymptote: y 0
Horizontal asymptote: y 1 since
Vertical asymptote: x ± 3
x3 x3 . 1 lim lim x → x 2 x → x 2
y 6 5 4 3 2 1
Discontinuity: x 2 (Vertical asymptote) y 5 4 3 2 −4 −3 −2 −1
−5−4
x 1
3 4 5 6
−1 −2 −3 −4 −5 −6
−2 −3 −4 −5
54. y
x2
x2 9
y 5 4 3 2
Intercept: 0, 0 Symmetry: y-axis Horizontal asymptote: y 1 since lim
x →
x2
x2 x2 1 lim 2 . x → x 9 9
Discontinuities: x ± 3 (Vertical asymptotes) Relative maximum: 0, 0
−5 −4
−1 −2 −3 −4 −5
x 1
4 5
x 1 2
6
405
406
Chapter 3
56. y
Applications of Differentiation
2x2 4
58. x2y 4
x2
Intercept: 0, 0
Intercepts: none
Symmetry: y-axis
Symmetry: y-axis
Horizontal asymptote: y 2
Horizontal asymptote: y 0 since
Relative minimum: 0, 0 y
lim
x →
4 4 0 lim 2 . x → x x2
Discontinuity: x 0 (Vertical asymptote)
5 4 3
y
1 x
−5 −4 −3 −2 −1
1 2 3 4 5
−2 −3 −4 −5
4 3 2 1 x
−5 −4 −3 −2 −1
60. y
2x 1 x2
1 2 3 4 5
1 x
62. y 1
Intercept: 0, 0
Intercept: 1, 0
Symmetry: origin
Symmetry: none
Horizontal asymptote: y 0 since
Horizontal asymptote: y 1 since
lim
x →
2x 2x 0 lim . x → 1 x2 1 x2
Discontinuities: x ± 1 (Vertical asymptotes)
lim
x →
1 1x 1
lim
x →
1 1x .
Discontinuity: x 0 (Vertical asymptote) y
y 5 4 3 2 1
5 4 3 2 x
−5 −4 −3 −2
2
x
−5 −4 −3 −2 −1
1 2 3 4 5
−3 −4 −5
64. y 4 1
1 x2
66. y
x x2 4
Intercepts: ± 1, 0
Domain: , 2, 2,
Symmetry: y-axis
Intercepts: none
Horizontal asymptote: y 4
Symmetry: origin
Vertical asymptote: x 0
Horizontal asymptotes: y ± 1 since
y
lim
x2 4
1, lim
x →
x x2 4
1.
Vertical asymptotes: x ± 2 (discontinuities)
3 2 1 −5 −4 −3 −2
x
x →
5
y
x 2 3 4 5
5 4 3 2 −5 −4 −3
−1 −2 −3 −4 −5
x 1
3 4 5
Section 3.5 68. f x fx f x
x2
x2 1
x = −1
2x x2 12x x22x 2 0 when x 0. x2 12 x 12
Limits at Infinity
4
x=1
−3
(0, 0)
y=1
x2 122 2x2x2 12x 23x2 1 2 x2 14 x 13
407
3
−4
Since f 0 < 0, then 0, 0 is a relative maximum. Since f x 0, nor is it undefined in the domain of f, there are no points of inflection. Vertical asymptotes: x ± 1 Horizontal asymptote: y 1 70. f x fx f x
1 1 x2 x 2 x 1x 2
x = −1
2x 1 1 0 when x . x2 x 22 2
x2
( 12 , − 49( −3
x 2 2 2x 12 x2 x 24 2
x=2
2
x2
3
y=0
x 22x 1
−2
6 x 1 x2 x 23 x2
Since f 12 < 0, then 12 , 49 is a relative maximum. Since f x 0, nor is it undefined in the domain of f, there are no points of inflection. Vertical asymptotes: x 1, x 2 Horizontal asymptote: y 0 72. f x fx f x
x1 x2 x 1 xx 2 0 when x 0, 2. x2 x 12 2x3 3x2 1 0 when x 0.5321, 0.6527, 2.8794. x2 x 13
f 0 < 0 Therefore, 0, 1 is a relative maximum. f 2 > 0 Therefore,
2, 31 is a relative minimum. Points of inflection: 0.5321, 0.8440, 0.6527, 0.4491 and 2.8794, 0.2931 Horizontal asymptote: y 0
(− 0.6527, 0.4491) 2
(0.5321, 0.8440)
−3
3
(
−2, − 1 3
( −2
(0, 1)
(− 2.8794, − 0.2931)
408
Chapter 3
74. gx gx
Applications of Differentiation
2x
4
3x2 1
y= 2
3
2 3x2 132
−6
6
y=− 2
18x gx 3x2 152
3 −4
No relative extrema. Point of inflection: 0, 0. Horizontal asymptotes: y ±
2 3
No vertical asymptotes
76. f x fx
2 sin 2x Hole at 0, 4 x
6
4x cos 2x 2 sin 2x x2
−
2
There are an infinite number of relative extrema. In the interval 2 , 2 , you obtain the following.
5
2 −2
Relative minima: ± 2.25, 0.869, ± 5.45, 0.365 Relative maxima: ± 3.87, 0.513 Horizontal asymptote: y 0 No vertical asymptotes
78. f x (a)
1 x3 2x2 2 1 , gx x 1 2 2x2 2 x f=g
(c)
4
−6
70
−80
6
80
−70
−4
(b) f x
x3 2x2 2 2x2
2xx
3 2
2x2 2 2x2 2x2
The graph appears as the slant asymptote y 12 x 1.
1 1 x 1 2 gx 2 x
80.
lim
v1v2 →
100 1
1 100 1 0 100% v1v2c
82. y
3.351t2 42.461t 543.730 t2
(a)
5
20
100 0
(b) Yes. lim y 3.351 t→
Section 3.5
Limits at Infinity
409
100t2 , t > 0 65 t2
84. S (a)
120
5
30 0
100 100 1
(b) Yes. lim S t→
px a xn . . . a1x a0 lim n m . . . x → qx x → bm x b1x b0
86. lim
Divide px and qx by xm. an a1 a0 . . . m1 m px x mn x x 0. . .00 0 lim Case 1: If n < m: lim . . .00b 0 x qx x→ b b b 1 m m bm . . . m1 m0 x x a1 a0 an . . . m1 m an . . . 0 0 an px x x lim Case 2: If m n: lim . . .00b . b b x → qx x → b 1 0 m m bm . . . m1 m x x px lim Case 3: If n > m: lim x → qx x →
a1 a0 an x nm . . . m1 m x x ± . . .0 . . . ± . b b b 0 1 m bm . . . m1 m0 x x
88. False. Let y1 x 1, then y10 1. Thus, y1 1 2x 1 and y10 12. Finally, 1 1 and y10 . 4x 132 4
y1
1 1 1 1 Let p ax2 bx 1, then p0 1. Thus, p 2ax b and p0 2 ⇒ b 2 . Finally, p 2a and p0 4 ⇒ a 8 . Therefore,
18x2 12x 1,
x < 0
x 1,
x ≥ 0
f x
fx
12
f x
14x 1
12 14 x,
and f 0 1,
x < 0
1 and f0 , and x > 0 2
x 1 ,
14, ,
32
x < 0
1 and f 0 . x > 0 4
f x < 0 for all real x, but f x increases without bound.
410
Chapter 3
Section 3.6
Applications of Differentiation
A Summary of Curve Sketching
2. The slope of f approaches as x → 0, and approaches as x → 0. Matches (C)
4. The slope is positive up to approximately x 1.5. Matches (B)
6. (a) x0, x2, x4
(b) x2, x3 (d) x1
(c) x1 (e) x2, x3
8. y y
x x2 1
y 1
1 x2 1 xx 1 0 when x ± 1. x2 12 x2 12
x 3
3
4
3 < x < 1 x 1
1 2
1 < x < 0 x0
0
0 < x < 1 1 2
x1 1 < x < 3
3
x 3 3 < x <
4
3, 3 4
)
(−1, − 12 ) 1
(0, 0)
(−
Horizontal asymptote: y 0
< x < 3
(
x
2x3 x2 y 2 0 when x 0, ± 3. x 13
y
(1, 12 )
y
y
Conclusion
Decreasing, concave down
0
Point of inflection
Decreasing, concave up
0
Relative minimum
Increasing, concave up
0
Point of inflection
Increasing, concave down
0
Relative maximum
Decreasing, concave down
0
Point of inflection
Decreasing, concave up
3, − 3 4
)
2
Section 3.6
10. y
x2 1 x2 9
12. f x
A Summary of Curve Sketching
x2 2 1 x x
y
20x 0 when x 0 x2 92
fx
2 < 0 when x 0. x2
y
60x2 3 < 0 when x 0 x2 93
f x
4 0 x3
Therefore, 0, Intercept:
1 is a relative maximum. 9
1 0, 9
Intercept: 2, 0 Vertical asymptote: x 0 Horizontal asymptote: y 1 y
Vertical asymptotes: x ± 3
5
Horizontal asymptote: y 1
y
x 1
2
3
4
x=0
y=1 x
−1 −2 −3 −4 −5
14. f x x
4 5 1
0, − 9
(
)
x=3
32 x2
16. f x
64 x 4x2 4x 16 0 when x 4. x3 x3
fx 1 f x
2
(−2, 0)
5 4 3 2
x = −3
3
y=1
Symmetric about y-axis
−5 − 4
4
192 > 0 if x 0. x4
x3 4x x 2 x2 4 x 4
fx
x2x2 12 0 x2 42
f x
8xx2 12 0 when x 0 x2 43
when x 0, ± 23
Therefore, 4, 6 is a relative minimum.
Intercept: 0, 0
3 4, 0 Intercept; 2
Relative maximum: 23, 33
Vertical asymptote: x 0
Relative minimum: 23, 33
Slant asymptote: y x
Inflection point: 0, 0 Vertical asymptotes: x ± 2
y
(−2 4, 0) 3
Slant asymptote: y x
(4, 6)
8 6
y
4
y=x
2
8 6 4
x
−8 −6
2 −4 −6
4
6
8
x=0 −8 − 6 − 4
x 4 6 8 10
411
412
Chapter 3
18. y
3 2x2 5x 5 2x 1 x2 x2
y 2 y
Applications of Differentiation
y
(4 +2 6, 7.899 (
12
3 2x2 8x 5 4 ± 6 . 0 when x 2 x 2 x 22 2
8
(
6 0 x 23
4+ 6 , − 1.899 2
(
−8
y = 2x − 1 x
−4
4
8
12
x=2
4 2 4 Relative minimum: 2
6
Relative maximum:
6
−8
, 1.8990
, 7.8990
Intercept: 0, 52 Vertical asymptote: x 2 Slant asymptote: y 2x 1 20. gx x9 x Domain: x ≤ 9
Domain: 4 ≤ x ≤ 4
22. y x16 x2
gx
36 x 0 when x 6 29 x
y
g x
3x 12 < 0 when x 6 49 x32
y
28
x2
0 when x ± 22
16 x2
2xx2 24 0 when x 0 16 x232
Relative maximum: 6, 63
Relative maximum: 22, 8
Intercepts: 0, 0, 9, 0
Relative minimum: 22, 8
Concave downward on , 9
Intercepts: 0, 0, ± 4, 0
y
(0,
10 8 6 4 0) 2
−8 − 6 − 4 −2
(6, 6
Symmetric with respect to the origin
3)
Point of inflection: 0, 0 y
(9, 0)
x
2 4 6 8 10
(−4, 0) −5
(−2
10 8 6 4 2
−3 −2 −1
2 , −8)
(2
2 , 8)
(0, 0)
(4, 0)
x
1 2 3 4 5
−6 −8 −10
24. y 3x 123 x 12 y
2 2 2x 143 2x 1 0 when x 0, 2 13 x 1 x 113
y undefined for x 1 y
y 5 4 3
2 2 < 0 for all x 1 3x 143
Concave downward on , 1 and 1, Relative maximum: 0, 2, 2, 2 Relative minimum: 1, 0 Intercepts: 0, 2, 1, 0, 1.280, 0, 3.280, 0
(2, 0)
(2, 2) x
−5 − 4 −3
−1 −2 −3 −4 −5
1 2 3 4 5
(1, 0)
Section 3.6
A Summary of Curve Sketching
26. y 13 x3 3x 2
y
y x 1 0 when x ± 1 2
2
y 2x 0 when x 0
1
(−2, 0)
(1, 0) x
y < x < 1 4
x 1
3
1 < x < 0 2
x0
3
0 < x < 1 x1
0
1 < x <
y
−1
y
Conclusion
Decreasing, concave up
0
Relative minimum
Increasing, concave up
0
Point of inflection
Increasing, concave down
0
Relative maximum
Decreasing, concave down
(
0,
(−1, − 43 (
28. f x 13 x 13 2
1 −2 3
2
(
−2
y
fx x 12 0 when x 1.
4
f x 2x 1 0 when x 1.
3
(1, 2) 2
f x < x < 1 x1
2
1< x <
fx
f x
Conclusion
Increasing, concave down
0
0
Point of inflection
Increasing, concave up
( 5) 0, 3
(−0.817, 0) x 1
30. f x x 1x 2x 5
3
3
y
fx x 1x 2 x 1x 5 x 2x 5 x2
2
12
(2 −
3, 6 3 ) (2, 0)
4x 1 0 when x 2 ± 3 . −12 −9 −6 −3
f x 6x 2 0 when x 2. f x < x < 2 3 x 2 3
63
2 3 < x < 2 x2
0
2 < x < 2 3 x 2 3 2 3 < x <
63
fx
f x
Conclusion
Increasing, concave down
0
Relative maximum
Decreasing, concave down
0
Point of inflection
Decreasing, concave up
0
Relative minimum
Increasing, concave up
Intercepts: 0, 10, 1, 0, 2, 0, 5, 0
x 3
(2 +
6
9 12
3, −6 3)
413
414
Chapter 3
Applications of Differentiation
32. y 3x4 6x2
5 3
y
y 12x3 12x 12xx2 1 0 when x 0, x ± 1. y 36x2 12 123x2 1 0 when x ± y < x < 1 x 1
43
1 < x < x
3
3
3
3
0
3 < x < 0
x0
53
0 < x < x 3
3
3
3
3
3
0
3 < x < 1
x1 1 < x <
43
3
3
(− 33 , 0(
.
y
y
Conclusion
Decreasing, concave up
0
Relative minimum
Increasing, concave up
0
Point of inflection
Increasing, concave down
0
Relative maximum
Decreasing, concave down
0
Point of inflection
Decreasing, concave up
0
Relative minimum
Increasing, concave up
−5 − 4 −3 −2 4 −1, − 3
(
34. f x x4 8x3 18x2 16x 5
f x 12x 2 48x 36 12x 3x 1 0 when x 3, x 1.
< x < 1 x1 x3
4 < x <
Conclusion
Decreasing, concave up
0
0
Point of inflection
Decreasing, concave down
16
0
Point of inflection
Decreasing, concave up
0
Relative minimum
Increasing, concave up
27
f x
3 < x < 4 x4
fx
0
1 < x < 3
(0, 53 ) ( 33 , 0( x 2 3 4 5 4 1, − 3
)−3 (
)
y
fx 4x3 24x2 36x 16 4x 4x 12 0 when x 1, x 4.
f x
7 6 5 4 3 2
4 −4
(0, 5) (1, 0) 1
(5, 0) 3
4
6
−12 −20 −28
(3, −16) (4, −27)
x 7
Section 3.6
A Summary of Curve Sketching
36. y x 15
y
y 5x 14 0 when x 1.
2
y 20x 13 0 when x 1.
1
y < x < 1 x1
0
1 < x <
(1, 0)
y
y
Conclusion
Increasing, concave down
0
0
Point of inflection
Increasing, concave up
2
3
y
2x 3x2 6x 5 2x 3x 5x 1 x2 6x 5 x 5x 1
1 −1
38. y x2 6x 5 y
x
−1
6
5
(3, 4)
4
0 when x 3 and undefined when x 1, x 5.
3 2
2x2 6x 5 2x 5x 1 y 2 undefined when x 1, x 5. x 6x 5 x 5x 1
y
y
0
Undefined
Undefined
40. y cos x
Increasing, concave down Relative maximum
Decreasing, concave down
0
Undefined
Undefined
5
x
6
Increasing, concave up
5 . y sin x sin 2x sin x1 2 cos x 0 when x 0, , , 3 3 y cos x 2 cos 2x cos x 22 cos2 x 1 1 ± 33 0.8431, 0.5931. 4 cos x cos x 2 0 when cos x 8
y 2 1
π −1 −2
Therefore, x 0.5678 or 5.7154, x 2.2057 or 4.0775.
3 , 43, 53 , 43
Relative minimum:
4
Relative minimum, point of inflection
1 cos 2x, 0 ≤ x ≤ 2 2
Relative maxima:
(5, 0) 3
Relative minimum, point of inflection
2
2
Decreasing, concave up
0
5 < x <
1
Conclusion
3 < x < 5 x5
y
4
1 < x < 3 x3
(1, 0)
< x < 1 x1
1
, 23
Inflection points: 0.5678, 0.6323, 2.2057, 0.4449, 5.7154, 0.6323, 4.0775, 0.4449
2π
x
415
416
Chapter 3
Applications of Differentiation
42. y 2x 2 cot x, 0 < x < y 2 csc2 x 0 when x
y 5 4 3 2 1
3 , 4 4
y 2 csc2 x cot x 0 when x
2
Relative maximum:
34 , 32 5
Relative minimum:
4 , 2 3
Point of inflection:
2 , 4
x
π
−1 −2 −3 −4 −5
Vertical asymptotes: x 0, 44. y sec2
8x 2 tan 8x 1, 3 < x < 3
y 2 sec2
y 5 4 3 2
8x tan 8x 8 2 sec 8x 8 0 ⇒ x 2 2
x
Relative minimum: 2, 1
−5 − 4 −3 −2 −1
3 4 5
46. gx x cot x, 2 < x < 2 gx
y 4
sin x cos x x sin2x
3 2
x 1. x→0 tan x
−2π
g0 does not exist. But lim x cot x lim x→0
π
−1
2π
x
−3 −4
3 3 ,0 , ,0 , ,0 , ,0 2 2 2 2
−π
−2
Vertical asymptotes: x ± 2 , ± Intercepts:
(2, −1)
−2 −3 −4 −5
Symmetric with respect to y-axis. Decreasing on 0, and , 2
48. f x 5
x 1 4 x 1 2
50. f x
6
4x
52. f is constant.
x2 15
f is linear.
6
f is quadratic. −9
9
−8
8
y
f −6
x 2, 4 vertical asymptote y 0 horizontal asymptote
f ''
−6
y ± 4 horizontal asymptotes
x
0, 0 point of inflection f'
Section 3.6 54.
y
A Summary of Curve Sketching
y
120 100
10
f
80
8
60
f ′′
6 4 2 x
−6 −3
3
6
9
x −4 −2 −2
12 15
2
4
6
8
10
(any vertical translate of f will do) 56.
y
y
2 1
1 x
−2
1
x −2
2
−1
1
−1
−1
−2
−2
2
(any vertical translate of the 3 segments of f will do) 58. If fx 2 in 5, 5 , then f x 2x 3 and f 2 7 is the least possible value of f 2. If fx 4 in 5, 5 , then f x 4x 3 and f 2 11 is the greatest possible value of f 2.
60. gx
3x4 5x 3 x4 1
62. gx
Vertical asymptote: none
Horizontal asymptote: y 3
x2 x 2 x1
x 2x 1 x 2, if x 1 x1 Undefined, if x 1
The rational function is not reduced to lowest terms.
7
4
−8
−6
4
6 −1 −4
The graph crosses the horizontal asymptote y 3. If a function has a vertical asymptote at x c, the graph would not cross it since f c is undefined.
64. gx
2x2 8x 15 5 2x 2 x5 x5 18
−10
20 −2
The graph appears to approach the slant asymptote y 2x 2.
hole at 1, 3
417
418
Chapter 3
Applications of Differentiation
66. f x tansin x (a)
(b) f x tansin x tansin x
3
−2
tansin x f x
2
Symmetry with repect to the origin (d) On 1, 1, there is a relative maximum at 2, tan 1 and a relative minimum at 12, tan 1. 1
−3
(c) Periodic with period 2 (e) On 0, 1, the graph of f is concave downward. 68. Vertical asymptote: x 3
70. Vertical asymptote: x 0
Horizontal asymptote: none
Slant asymptote: y x
y
x2 x3
y x
1 1 x2 x x
1 1 72. f x ax2 ax axax 2, a 0 2 2 1 fx a2x a aax 1 0 when x . a f x a2 > 0 for all x. (a) Intercepts: 0, 0,
2a, 0
(b)
y
a=2
a = −2
5
Relative minimum:
1a, 21
4
a=1
Points of inflection: none
x
−3
a = −1
2
−1
3
74. Tangent line at P: y y0 fx0x x0 (a) Let y 0: y0 fx0x x0
(b) Let x 0: y y0 fx0x0
fx0x x0 fx0 y0 x x0 x-intercept:
y0 f x0 x0 fx0 fx0
f x0 x0 ,0 fx0
(c) Normal line: y y0 Let y 0: y0
y y0 x0 fx0 y f x0 x0 fx0 y-intercept: 0, f x0 x0 fx0
(d) Let x 0: y y0
1 x x0 fx0
1 x x0 fx0
y-intercept:
x x0 y0 fx0 x0 f x0 fx0
0
(f) PC 2 y02
f x0 f x0 x0 fx0 fx0
0, y
x0 fx0
ffxx f x ffxx f x 0
0
0
f x01 fx0 2 PC 2 fx0
x-intercept: x0 f x0 fx0, 0
x0 fx0
y y0
y0 fx0 x x0
(e) BC x0
1 x0 fx0
(g) AB x0 x0 f x0 fx0 f x0 fx0
x0 y0 AP f x01 fx0 2
(h) AP f x0 2
2 f
2
2
2
0
2
0
0
2
2
Section 3.7
Section 3.7
Optimization Problems
Optimization Problems
2. Let x and y be two positive numbers such that x y S. P xy xS x Sx x2
4. Let x and y be two positive numbers such that xy 192. S x 3y
S dP S 2x 0 when x . dx 2
192 3y y
192 dS 3 2 0 when y 8. dy y
S d 2P 2 < 0 when x . dx 2 2
d 2S 384 3 > 0 when y 8. dy 2 y
P is a maximum when x y S2.
S is minimum when y 8 and x 24.
6. Let x and y be two positive numbers such that x 2y 100. P xy y100 2y 100y 2y2 dP 100 4y 0 when y 25. dy d 2P 4 < 0 when y 25. dy 2 P is a maximum when x 50 and y 25. 8. Let x be the length and y the width of the rectangle. 2x 2y P y
P 2x P x 2 2
A xy x
P2 x P2 x x
y
2
x
P dA P 2x 0 when x . dx 2 4 d 2A P 2 < 0 when x . dx2 4 A is maximum when x y P4 units. (A square!) 10. Let x be the length and y the width of the rectangle. xy A y
A x
P 2x 2y 2x 2
y
Ax 2x 2Ax
2A dP 2 2 0 when x A. dx x d 2P 4A 3 > 0 when x A . dx2 x P is minimum when x y A centimeters. (A square!)
419
x
420
Chapter 3
Applications of Differentiation 14. f x x 12, 5, 3
12. f x x 8, 2, 0
d x 52 x 12 3 2
From the graph, it is clear that 8, 0 is the closest point on the graph of f to 2, 0.
x2 10x 25 x2 2x 22
y
x2 10x 25 x4 4x3 8x 4 x4 4x3 x2 18x 29
4
Since d is smallest when the expression inside the radical is smallest, you need to find the critical numbers of
2
gx x4 4x3 x2 18x 29
x 2
4
6
8
10
12
gx 4x3 12x2 2x 18 2x 12x2 8x 9 By the First Derivative Test, x 1 yields a minimum. Hence, 1, 4 is closest to 5, 3.
16.
F
v 22 0.02v 2
18. 4x 3y 200 is the perimeter. (see figure)
dF 22 0.02v 2 dv 22 0.02v 2 2
A 2xy 2x
200 3 4x 38 50x x 2
dA 8 50 2x 0 when x 25. dx 3
0 when v 1100 33.166. By the First Derivative Test, the flow rate on the road is maximized when v 33 mph.
d 2A 16 < 0 when x 25. dx2 3 A is a maximum when x 25 feet and y
100 3
y x
20. (a)
(c)
Height, x
Length & Width
Volume
1
24 21
124 21 2 484
2
24 22
224 22 2 800
3
24 23
324 23 2 972
4
24 24
424 24 2 1024
5
24 25
524 25 2 980
6
24 26
624 26 2 864
x
(b) V x24 2x2, 0 < x < 12 (d)
1200
0
12 0
The maximum volume seems to be 1024.
dV 2x24 2x2 24 2x2 24 2x24 6x dx 1212 x4 x 0 when x 12, 4 12 is not in the domain. 2V
d 122x 16 dx2 d 2V < 0 when x 4. dx2 When x 4, V 1024 is maximum.
feet.
Section 3.7 22. (a) P 2x 2 r 2x 2
y 2
2y
Optimization Problems
y
x
2x y 200 ⇒y
200 2x 2 100 x
(b) Length, x
Width, y
Area, xy
10
2 100 10
2 10 100 10 573
20
2 100 20
20
2 100 20 1019
30
2 100 30
30
2 100 30 1337
40
2 100 40
40
2 100 40 1528
50
2 100 50
50
2 100 50 1592
60
2 100 60
60
2 100 60 1528
The maximum area of the rectangle is approximately 1592 m2. (c) A xy x (e)
2 2 100 x 100x x2
2000
0
100 0
Maximum area is approximately 1591.55 m2 x 50 m.
24. You can see from the figure that A xy and y
6x . 2
y 6
6x 1 6x x2. Ax 2 2
5 4
y=
6−x 2
3
dA 1 6 2x 0 when x 3. dx 2 d 2A 1 < 0 when x 3. dx2 A is a maximum when x 3 and y 32.
2
( x, y )
1 x 1
2
3
4
5
6
421
422
Chapter 3
26. (a)
A
Applications of Differentiation
1 base height 2 1 216 h2 4 h 2
4 4
h
16 h24 h
16 − h 2
dA 1 16 h2122h4 h 16 h212 dh 2 16 h212h4 h 16 h2
2h2 2h 8 2h 4h 2 16 h2 16 h2
dA 0 when h 2, which is a maximum by the First Derivative Test. dh Hence, the sides are 216 h2 43, an equilateral triangle. Area 123 sq. units. (b) cos tan
4h 84 h
4 h 8
16 h2
Area 2
8
α
4+h
4
4h
4
h
1216 h 4 h 2
16 − h 2
4 h2 tan 64 cos4 tan A 64cos4 sec2 4 cos3 sin tan 0 ⇒ cos4 sec2 4 cos3 sin tan 1 4 cos sin tan 1 sin2 4 sin
1 ⇒ 30 and A 123. 2
(c) Equilateral triangle
28. A 2xy 2xr 2 x 2 see figure 2r dA 2r2 2x2 0 when x . dx 2 r2 x2
By the First Derivative Test, A is maximum when the rectangle has dimensions 2r by 2r2.
y
(−x,
r 2 − x2
(−r, 0)
( ( x,
r 2 − x2
( x
(r, 0)
Section 3.7 30. xy 36 ⇒ y
Optimization Problems
36 x
x+3 x
36 A x 3y 3 x 3 3 x
y
y+3
108 36 3x 9 x dA 108 3 0 ⇒ 3x2 108 ⇒ x 6, y 6 dx x2 Dimensions: 9 9
32. V r 2h V0 cubic units or h
S 2 r 2 2 rh 2 r 2
V0 r
V dS 2 2r 20 0 when r dr r h
V0 r2
2V units. 0
3
V0 V0223 2V013 3 V 2 2 V 23 213 2r 0 0
By the First Derivative Test, this will yield the minimum surface area. V r 2x
34.
x 2r 108 ⇒ x 108 2 r V
see figure
108 2 r 108r 2 2r3
r 2
r
x
dV 216r 6 r 2 6 r36 r dr 0 when r
36 and x 36.
36 d 2V 216 12 r < 0 when r . dr 2 Volume is maximum when x 36 inches and r 36 11.459 inches. 36.
V x 2h x 2 2r2 x2 2 x 2r 2 x2
see figure
dV 1 2 2 x2 r x 2122x 2xr 2 x2 dx 2
2 x r2 x2
6r 2r 2 ⇒x . 3 3
By the First Derivative Test, the volume is a maximum when 6r
3
and h
2r 3
.
Thus, the maximum volume is V
23 r 2r3 43r3 . 3
2
r2 − x2
(
(x, −
r2 − x2
(
x r
h
2r 2 3x2
0 when x 0 and x 2
x
(x,
423
424
Chapter 3
Applications of Differentiation
38. No. The volume will change because the shape of the container changes when squeezed. 4 40. V 3000 r 3 r 2h 3 3000 4 r r2 3
h
Let k cost per square foot of the surface area of the sides, then 2k cost per square foot of the hemispherical ends.
C 2k4 r 2 k2 rh k 8 r 2 2 r
6000 32 dC k r 2 0 when r dr 3 r
16 4 r k r 3000 r 3 3
2
2
6000 r
5.636 feet and h 22.545 feet. 1125 2 3
By the Second Derivative Test, we have
d 2C 12,000 32 k dr 2 3 r3
. > 0 when r 1125 2 3
Therefore, these dimensions will produce a minimum cost. 42. (a) Let x be the side of the triangle and y the side of the square. 3 4 A cot x 2 cot y 2 where 3x 4y 20 4 3 4 4
A
x
3
4
43 0
3 x2 5 x 2 4
A
5 4 cot x 2 cot y 2 where 4x 5y 20 4 4 4 5
4 2 x2 1.7204774 4 x , 0 ≤ x ≤ 5. 5
3 20 x2 5 x , 0 ≤ x ≤ . 4 3
3
(b) Let x be the side of the square and y the side of the pentagon.
2
A 2x 2.75276384 4
4 x 0 5
x 2.62 When x 0, A 27.528, when x 2.62, A 13.102, and when x 5, A 25. Area is maximum when all 20 feet are used on the pentagon.
60 43 9
When x 0, A 25, when x 60 43 9, A 10.847, and when x 203, A 19.245. Area is maximum when all 20 feet are used on the square. (c) Let x be the side of the pentagon and y the side of the hexagon. A
6 5 cot x 2 cot y 2 where 5x 6y 20 4 5 4 6
2 3 20 5x 2 5 cot x 3 , 0 ≤ x ≤ 4. 4 5 2 6
5 5 A cot x 33 2 5 6
20 5x 0 6
x 2.0475 When x 0, A 28.868, when x 2.0475, A 14.091, and when x 4, A 27.528. Area is maximum when all 20 feet are used on the hexagon
(d) Let x be the side of the hexagon and r the radius of the circle. A
2 6 cot x r 2 where 6x 2 r 20 4 6
10 3x 2 10 33 2 x ,0 ≤ x ≤ . 2 3
A 33 6
10 3x 0
x 1.748 When x 0, A 31.831, when x 1.748, A 15.138, and when x 103, A 28.868. Area is maximum when all 20 feet are used on the circle. In general, using all of the wire for the figure with more sides will enclose the most area.
Section 3.7 44. Let A be the amount of the power line.
Optimization Problems
425
y
A h y 2x2 y2
(0, h) h−y
x 2y dA 0 when y . 1 dy x2 y2 3
y
2x2 x d 2A 2 > 0 for y . 2 dy x y232 3
x
(−x, 0)
(x, 0)
The amount of power line is minimum when y x3 . 46. f x 12 x2 (a)
1 4 gx 16 x 12 x2 on 0, 4
(c) fx x, Tangent line at 22, 4 is
9
y 4 22 x 22
f g −1
y 22x 4.
4
gx 4 x3 x, Tangent line at 22, 0 is 1
−3
1 y 0 4 22 3 22 x 22
1 4 1 4 (b) dx f x gx 12 x2 16 x 12 x2 x2 16 x
y 22x 8.
dx 2x 14 x3 0 ⇒ 8x x3 ⇒ x 0, 22 in 0, 4 The maximum distance is d 4 when x 22.
The tangent lines are parallel and 4 vertical units apart. (d) The tangent lines will be parallel. If dx f x gx, then dx 0 fx gx implies that fx gx at the point x where the distance is maximum.
48. Let F be the illumination at point P which is x units from source 1. F
kI2 kI1 x2 d x2
2kI2 2kI 2kI2 dF 2kI1 0 when 3 1 . dx x3 d x3 x d x3 3 I 1 3 I 2
x dx
3 I x 3 I d x 1 2 3 I x 3 I d 3 I1 1 2
x
3 I d 1 3 I 2
3 I 1
3 6kI2 d I1 d 2F 6kI1 4 > 0 when x 3 . 2 4 3 I dx x d x I1 2
This is the minimum point.
50. (a) T θ
x2 4
2
3 x 4
x
x2 + 4
x2 4
2 x
dT x 1 0 dx 2x2 4 4
(b)
3−x Q
1 2
2x2 x2 4 x2 4 x2 T2 2
—CONTINUED—
1 hours 4
426
Chapter 3
Applications of Differentiation
50. —CONTINUED— T
(c)
x2 4
v1
3 x v2
(d) Cost x2 4 C1 3 xC2
dT 1 x 0 dx v1x2 4 v2
T
3 x 1C2 1C1 C2 1C2 C1
x2 d12
v1
v1 only. v2
d22 a x2
54. Cx 2kx2 4 k4 x
v2
Cx
xa dT x 0 dx v1x2 d12 v2d22 a x2
3x2 4 x
we have sin 1 sin 2 sin 1 sin 2 0⇒ . v1 v2 v1 v2
2 3
Or, use Exercise 50(d): sin
Since 2
Thus, x
2
d1 d2 d > 0 d x 2 v1x2 d1232 v2d22 a x2 32
θ
1 1 V r 2h r 2144 r 2 3 3
dV 1 1 r2 144 r 2122r 2r144 r 2 dr 3 2
1 288r 3r3 r96 r 2 0 when r 0, 46. 3 144 r 2 144 r 2
By the First Derivative Test, V is maximum when r 46 and h 43. Area of circle: A 122 144 Lateral surface area of cone: S 46 46 2 43 2 486 Area of sector: 144 486
.
x2 + 4 x
2 3
2
this condition yields a minimum time.
k0
4x2 x2 4
x xa sin 1 and sin 2 2 2 2 x d1 d2 a x2
2T
2xk x2 4
2x x2 4
Since
56.
v1 v2
depends on
52.
1C1
From above, sin
v x 1 x2 4 v2 sin
x2 4
1 2
r 72
2
144 486 2 3 6 1.153 radians or 66 72 3
4−x
C2 1 ⇒ 30. C1 2
Section 3.7
Optimization Problems
58. Let d be the amount deposited in the bank, i be the interest rate paid by the bank, and P be the profit. P 0.12 d id d ki 2 since d is proportional to i 2 P 0.12ki 2 iki 2 k0.12i 2 i3 0.24 dP k0.24i 3i 2 0 when i 0.08. di 3 d 2P k0.24 6i < 0 when i 0.08 Note: k > 0. di 2 The profit is a maximum when i 8%. P
60.
(a)
1 3 s 6s 2 400 10
3 3 dP s 2 12s ss 40 0 when x 0, s 40. ds 10 10 3 d 2P s 12 ds 2 5 d 2P 0 > 0 ⇒ s 0 yields a minimum. ds 2 d 2P 40 < 0 ⇒ s 40 yields a maximum. ds2 The maximum profit occurs when s 40, which corresponds to $40,000 P $3,600,000.
(b)
3 d 2P s 12 0 when s 20. ds 2 5 The point of diminishing returns occurs when s 20, which corresonds to $20,000 being spent on advertising.
62. S2 4m 1 5m 6 10m 3
Using a graphing utility, you can see that the minimum occurs when m 0.3. Line y 0.3x
S2 40.3 1 50.3 6 100.3 3 4.7 mi. S2
30
20
10
(0.3, 4.5) m 1
2
3
427
428
Chapter 3
Applications of Differentiation
h
x and cos . The area A of the 2 r 2 r cross equals the sum of two large rectangles minus
64. (a) Label the figure so that r2 x2 h2. Then, the area A is 8 times the area of the region given by OPQR:
(b) Note that sin
the common square in the middle.
h
A 22x2h 4h2 8xh 4h2
θ 2
R x
O
8r2 sin
h Q
P r
cos 4r2 sin2 2 2 2
4r2 sin sin2
12h
A8
2
cos sin
x2 x r2 x2r2 x2
2
A 4r2 cos sin
x hh
12r
8
2
cos 0 2 2
cos 2 sin
2 2
tan 2
8xr2 x2 4x2 4r2 Ax 8r2 x2
arctan2 1.10715 or 63.4
8x2 8x 0 r2 x2
8x2 8x 8r2 x2 r2 x2 x2 xr2 x2 r2 x2 2x2 r2 xr2 x2 4x4 4x2r2 r4 x2r2 x2 5x4 5x2r2 r4 0 Quadratic in x2. 5r2 ± 25r4 20r4 r2 5 ± 5 . 10 10 Take positive value. x2
xr
5 10
5
(c) Note that x2
0.85065r
Critical number
r2 r2 5 5 and r2 x2 5 5. 10 10
Ax 8xr2 x2 4x2 4r2
10r 5 5 10r 5 5
8
2
10r 20
8
4
12
2
12
4
r2 5 5 4r2 10
2 2r2 5r2 4r2 5
8 25 2 r25 2r2 r 5 5
5 4 2r2 5 1 2r25 1 5 5
Using the angle approach, note that tan 2, sin
Thus, A 4r2 sin sin2
2
2 1 1 4r2 1 5 2 5
4r25 1 2r25 1 2
2 5
and sin2
2 211 cos 211 15.
Section 3.8
Section 3.8
Newton’s Method
Newton’s Method
2. f x 2x2 3 fx 4x x1 1
f xn fxn
n
xn
f xn
fxn
1
1
1
4
0.125
5.0
0.025
xn
1 4
f xn fxn 5 4
2
5 1.25 4
fx sec2 x
n
xn
f xn
fxn
f xn fxn
x1 0.1
1
0.1000
0.1003
1.0101
0.0993
0.0007
2
0.0007
0.0007
1.0000
0.0007
0.0000
4. f x tan x
6. f x x5 x 1 fx 5x 1 4
Approximation of the zero of f is 0.755.
Approximation of the zero of f is 4.8284.
xn
xn
f xn
fxn
f xn fxn
1
0.5000
0.4688
1.3125
0.3571
0.8571
2
0.8571
0.3196
3.6983
0.0864
0.7707
3
0.7707
0.0426
2.7641
0.0154
0.7553
4
0.7553
0.0011
2.6272
0.0004
0.7549
Approximation of the zero of f is 0.7937.
xn
n
xn
f xn
fxn
f xn fxn
1
5
0.1010
0.5918
0.1707
4.8293
2
4.8293
0.0005
0.5858
.00085
4.8284
n
xn
f xn
fxn
f xn fxn
1
1
1
6
0.1667
0.8333
2
0.8333
0.1573
4.1663
0.0378
0.7955
3
0.7955
0.0068
3.7969
0.0018
0.7937
4
0.7937
0.0000
3.7798
0.0000
0.7937
10. f x 1 2x3 fx 6x2
f xn fxn
n
8. f x x 2x 1 1 fx 1 x 1
1.225
xn
f xn fxn
xn
f xn fxn
f xn fxn
429
430
Chapter 3
Applications of Differentiation
1 12. f x x 4 3x 3 2 fx 2x3 3 Approximation of the zero of f is 0.8937.
Approximation of the zero of f is 2.0720.
14. f x x3 cos x fx 3x2 sin x Approximation of the zero of f is 0.866.
16. hx f x gx 3 x hx 1
n
xn
f xn
fxn
f xn fxn
1
1
0.5
5
0.1
0.9
2
0.9
0.0281
4.458
0.0063
0.8937
3
0.8937
0.0001
4.4276
0.0000
0.8937
xn
f xn fxn
n
xn
f xn
fxn
f xn fxn
1
2
1
13
0.0769
2.0769
2
2.0769
0.0725
14.9175
0.0049
2.0720
3
2.0720
0.0003
14.7910
0.0000
2.0720
xn
f xn fxn
n
xn
f xn
fxn
f xn fxn
1
0.9000
0.1074
3.2133
0.0334
0.8666
2
0.8666
0.0034
3.0151
0.0011
0.8655
3
0.8655
0.0001
3.0087
0.0000
0.8655
1 x2 1
2x x2 12
xn
f xn fxn
n
xn
h xn
hxn
h xn hxn
1
2.9000
0.0063
0.9345
0.0067
2.8933
2
2.8933
0.0000
0.9341
0.0000
2.8933
n
xn
h xn
hxn
h xn hxn
1
0.8000
0.0567
2.3174
0.0245
0.8245
2
0.8245
0.0009
2.3832
0.0004
0.8241
xn
h xn hxn
Point of intersection of the graphs of f and g occurs when x 2.893. 18. hx f x gx x2 cos x hx 2x sin x One point of intersection of the graphs of f and g occurs when x 0.824. Since f x x2 and gx cos x are both symmetric with respect to the y-axis, the other point of intersection occurs when x 0.824. 20. f x xn a 0 fx nxn1 xi1 xi
xin a nxin1
nxin xin a n 1xin a nxin1 nxin1
xn
h xn hxn
Section 3.8
22. xi1
xi2 5 2xi
24. xi1
Newton’s Method
431
2xi3 15 3xi2
i
1
2
3
4
i
1
2
3
4
xi
2.0000
2.2500
2.2361
2.2361
xi
2.5000
2.4667
2.4662
2.4662
3 15 2.466
5 2.236
26. f x tan x fx sec2 x
n
xn
f xn
fxn
f xn fxn
Approximation of the zero: 3.142
1
3.0000
0.1425
1.0203
0.1397
3.1397
2
3.1397
0.0019
1.0000
0.0019
3.1416
3
3.1416
0.0000
1.0000
0.0000
3.1416
y x1
3 2
f xn fxn
30. f x 2 sin x cos 2x
28. y 4x3 12x2 12x 3 f x 12x2
xn
fx 2 cos x 2 sin 2x
24x 12 fx
x1
Fails because fx1 0.
fx2 0; therefore, the method fails. n
xn
f xn
fxn
f xn fxn
1
3 2
3 2
3
2
1
1
0
3 2
f xn fxn
n
xn
f xn
fxn
1 2
1
1
3 2
3
0
—
—
g xn gxn
xn
32. Newton’s Method could fail if fc 0, or if the initial value x1 is far from c. 34. Let gx f x x cot x x gx csc2 x 1. The fixed point is approximately 0.86.
g xn gxn
n
xn
g xn
gxn
1
1.0000
0.3579
2.4123
0.1484
0.8516
2
0.8516
0.0240
2.7668
0.0087
0.8603
3
0.8603
0.0001
2.7403
0.0000
0.8603
xn
36. f x sin x, fx cos x (a)
(d)
2
y 2
−
−2
(b) x1 1.8 x2 x1
(3, 0.141) (6.086, 0)
1
π 2 −1
π
x
(3.143, 0)
−2
f x1 6.086 fx1
(c) x1 3 x2 x1
(1.8, 0.974)
f x1 3.143 fx1
The x-intercepts correspond to the values resulting from the first iteration of Newton’s Method. (e) If the initial guess x1 is not “close to” the desired zero of the function, the x-intercept of the tangent line may approximate another zero of the function.
432
Chapter 3
Applications of Differentiation (b) xn1 xn2 11xn
38. (a) xn1 xn2 3xn
1 3
i
1
2
3
4
i
1
2
3
4
xi
0.3000
0.3300
0.3333
0.3333
xi
0.1000
0.0900
0.0909
0.0909
1 11
0.333
0.091
40. f x x sin x, 0,
y
fx x cos x sin x 0
4 3
Letting Fx fx, we can use Newton’s Method as follows.
Fx 2 cos x x sin x n
xn
1
2.0000
2
2.0290
1
Fxn
F xn Fxn
0.0770
2.6509
0.0290
2.0290
0.0007
2.7044
0.0002
2.0288
F xn
(2.029, 1.820)
2
xn
F xn Fxn
π 4
π 2
x
3π 4
Approximation to the critical number: 2.029 42. y f x x2, 4, 3
y
d x 42 y 32 x 42 x2 32 x4 7x2 8x 25
3 2
d is minimum when D x4 7x2 8x 25 is minimum.
(0.529, 0.280)
1
x
gx D 4x3 14x 8
−2 −1 −1
gx 12x2 14
1
2
3
4
−2 −3
n
xn
g xn
gxn
g xn gxn
1
0.5000
0.5000
17.0000
0.0294
0.5294
2
0.5294
0.0051
17.3632
0.0003
0.5291
3
0.5291
0.0001
17.3594
0.0000
0.5291
xn
(4, − 3)
g xn gxn
x 0.529 Point closest to 4, 3 is approximately 0.529, 0.280.
44. Maximize: C C
3t 2 t 50 t3 3t4 2t3 300t 50 0 50 t32
n
xn
f xn
fxn
f xn fxn
1
4.5000
12.4375
915.0000
0.0136
4.4864
2
4.4864
0.0658
904.3822
0.0001
4.4863
Let f x 3t4 2t3 300t 50 fx 12t3 6t2 300. Since f 4 354 and f 5 575, the solution is in the interval 4, 5. Approximation: t 4.486 hours
xn
f xn fxn
Section 3.8
Newton’s Method
46. 170 0.808x3 17.974x2 71.248x 110.843, 1 ≤ x ≤ 5 Let f x 0.808x3 17.974x2 71.248x 59.157 fx 2.424x2 35.948x 71.248. From the graph, choose x1 1 and x1 3.5. Apply Newton’s Method. n
xn
f xn
fxn
f xn fxn
1
1.0000
5.0750
37.7240
0.1345
1.1345
2
1.1345
0.2805
33.5849
0.0084
1.1429
3
1.1429
0.0006
33.3293
0.0000
1.1429
n
xn
f xn
fxn
f xn fxn
1
3.5000
4.6725
24.8760
0.1878
3.6878
2
3.6878
0.3286
28.3550
0.0116
3.6762
3
3.6762
0.0009
28.1450
0.0000
3.6762
xn
f xn fxn
xn
f xn fxn
The zeros occur when x 1.1429 and x 3.6762. These approximately correspond to engine speeds of 1143 revmin and 3676 revmin. 48. True
50. True
52. f x 4 x2 sinx 2 Domain: 2, 2 x 2 and x 2 are both zeros. fx 4 x2 cosx 2 Let x1 1.
x 4 x2
sinx 2
n
xn
f xn
fxn
f xn fxn
1
1.0000
0.2444
1.7962
0.1361
2
1.1361
0.0090
1.6498
0.0055
1.1416
3
1.1416
0.0000
1.6422
0.0000
1.1416
Zeros: x ± 2, x 1.142 y 1 x −2
1
−2 −3
2
xn
f xn fxn
1.1361
433
434
Chapter 3
Applications of Differentiation
Section 3.9
Differentials
2. f x 6 6x2 x2 fx 12x3
x
y
1.99
2
2.01
2.1
6 x2
1.6620
1.5151
1.5
1.4851
1.3605
9 3 Tx x 2 2
1.65
1.515
1.5
1.485
1.35
f x
12 x3
Tangent line at 2,
1.9
3 : 2
3 12 3 x 2 x 2 2 8 2 9 3 y x 2 2
4. f x x fx
1 2x
x
1.9
1.99
2
2.01
2.1
f x x
1.3784
1.4107
1.4142
1.4177
1.4491
1.3789
1.4107
1.4142
1.4177
1.4496
T x
Tangent line at 2, 2 :
x 2
1 2
y f 2 f2x 2 y 2
1 x 2 22
y
x 1 22 2
6. f x csc x fx csc x cot x Tangent line at 2, csc 2: y f 2 f2x 2 y csc 2 csc 2 cot 2x 2 y csc 2 cot 2x 2 csc 2 x
1.9
1.99
2
2.01
2.1
f x csc x
1.0567
1.0948
1.0998
1.1049
1.1585
T x csc 2 cot 2x 2 csc 2
1.0494
1.0947
1.0998
1.1048
1.1501
8. y f x 1 2x2, fx 4x, x 0, x dx 0.1 y f x x f x
dy fx dx
f 0.1 f 0
f00.1
1 20.1 1 20 0.02 2
2
00.1 0
10. y f x 2x 1, fx 2, x 2, x dx 0.01 y f x x f x
dy fx dx
f 2.01 f 2
f20.01
22.01 1 22 1 0.02
20.01 0.02
Section 3.9 12.
y 3x2 3
14. y 9 x2
dy 2x1 3dx
16.
2 dx x1 3
dy
1
y x
18.
x
dy
1
sec2 x x2 1
x
2
x 1 9 x21 22xdx dx 9 x2 2
y x sin x dy x cos x sin x dx
1 x1 dx dx dy 2x 2xx 2xx
20. y
Differentials
22. (a) f 1.9 f 2 0.1 f 2 f20.1 1 10.1 1.1
12 sec2 x tan x sec2 x2x dx x2 12
(b) f 2.04 f 2 0.04 f 2 f20.04 1 10.04 0.96
x tan x x
2 sec xx tan dx x 1 2
2
2
2
24. (a) f 1.9 f 2 0.1 f 2 f20.1
26. (a) g2.93 g3 0.07 g3 g30.07
1 00.1 1
8 30.07 7.79
(b) f 2.04 f 2 0.04 f 2 f20.04
(b) g3.1 g3 0.1 g3 g30.1
1 00.04 1 28. (a) g2.93 g3 0.07 g3 g30.07 8 50.07 7.65 (b) g3.1 g3 0.1 g3 g30.1 8 50.1 8.5
8 30.1 8.3 30.
1 A 2 bh, b 36, h 50
db dh ± 0.25 dA 12 b dh 12 h db A dA 2 36± 0.25 2 50± 0.25 1
1
± 10.75 square centimeters 32.
x 12 inches
34. (a)
x dx ± 0.03 inch
C 56 centimeters C dC ± 1.2 centimeters
(a) V x3
C 2 r ⇒ r
dV 3x2 dx 3122± 0.03
A r2
± 12.96 cubic inches (b) S 6x2
dA
dS 12x dx 1212± 0.03
C 2
2C
2
1 2 C 4
1 1 33.6 C dC 56± 1.2 2 2
33.6 dA 0.042857 4.2857% A 1 4562
± 4.32 square inches
(b)
dA 1 2C dC 2dC ≤ 0.03 A 1 4C2 C dC 0.03 ≤ 0.015 1.5% C 2
435
436
36.
Chapter 3
Applications of Differentiation
P 500x x2
12x
2
77x 3000 , x changes from 115 to 120
dP 500 2x x 77dx 577 3x dx 577 3115120 115 1160 1160 dP 100 100 2.7% P 43517.50
Approximate percentage change:
38. V 43 r3, r 100 cm, dr 0.2 cm
E IR
40.
V dV 4 r dr 4 100 0.2 8000 cm 2
2
3
R
E I
dR
E dI I2
dR E I 2 dI dI R E I I
dR dI dI R I I
1 1 A baseheight 9.5cot 9.5 45.125 cot 2 2 dA 45.125
h 50 tan
44.
42. See Exercise 41.
csc2
0.0044 sin 0.4669cos 0.4669
0.0109 1.09% in radians 3 x, x 27, dx 1 46. Let f x 3 x f x x f x fxdx
3 26 3 27
d
dh 50 1.2479 d ≤ 0.06 x 50 tan1.2479
csc2 d d dA A cot sin cos 0.25 sin 26.75cos 26.75
dh 50 sec2
d
71.5 1.2479 radians
1 dx 3 x2 3
1 1 1 3 2.9630 3 272 3 27
3 26 2.9625 Using a calculator,
48. Let f x x3, x 3, dx 0.01. f x x f x fx dx x3 3x2 dx f x x 2.993 33 3320.01 27 0.27 26.73 Using a calculator: 2.993 26.7309
sec2
9.9316 d ≤ 0.06 2.9886
d ≤ 0.018
h
θ
50 ft
Review Exercises for Chapter 3 50. Let f x tan x, x 0, dx 0.05, fx sec2 x.
52. Propagated error f x x f x,
Then
relative error
f 0.05 f 0 f0dx
437
dy dy , and the percent error 100. y y
tan 0.05 tan 0 sec2 00.05 0 10.05.
54. True,
y dy a x dx
56. False Let f x x, x 1, and x dx 3. Then y f x x f x f 4 f 1 1 and dy fx dx
1 3 3 . 21 2
Thus, dy > y in this example.
Review Exercises for Chapter 3 2. (a) f 4 f 4 3 (c)
(b) f 3 f 3 4 4
y
(d) Yes. Since f 2 f 2 1 1 and f 1 f 1 2, the Mean Value says that there exists at least one value c in 2, 1 such that
6 4
x −6 −4 −2
4
fc
6
−4
(e) No, lim f x exists because f is continuous at 0, 0.
−6
x →0
At least six critical numbers on 6, 6. 4. f x
f 1 f 2 2 1 1. 1 2 12
x x2 1
, 0, 2
(f) Yes, f is differentiable at x 2.
6. No. f is not differentiable at x 2.
1 fx x x2 1322x x2 112 2 1 x2 132 No critical numbers Left endpoint: 0, 0 Minimum Right endpoint: 2, 25 Maximum
8. No; the function is discontinuous at x 0 which is in the interval 2, 1 .
10.
1 f x , 1 ≤ x ≤ 4 x fx
1 x2
f b f a 14 1 34 1 ba 41 3 4 fc
1 1 c2 4
c2
12.
f x x 2x, 0 ≤ x ≤ 4 fx
1 2x
2
f b f a 6 0 3 ba 40 2 fc
1 3 2 2 2c
c1
438
Chapter 3
Applications of Differentiation
f x 2x2 3x 1
14.
fx 4x 3 f b f a 21 1 5 ba 40 fc 4c 3 5 c 2 Midpoint of 0, 4 16. gx x 13
Critical number: x 1
18. f x sin x cos x, 0 ≤ x ≤ 2
Critical numbers: x
gx
5 ,x 4 4
1 < x <
Sign of gx
gx > 0
gx > 0
Conclusion
Increasing
Increasing
0 < x <
Interval
fx cos x sin x
20. gx
< x < 1
Interval
gx 3x 12
4
5 < x < 4 4
5 < x < 2 4
Sign of fx
fx > 0
fx < 0
fx > 0
Conclusion
Increasing
Decreasing
Increasing
3 x sin 1 , 0, 4 2 2
Test Interval
3 x 1 cos 2 2 2
Sign of gx
gx > 0
gx < 0
gx > 0
Conclusion
Increasing
Decreasing
Increasing
0 when x 1
2 2 ,3
Relative maximum:
1 2 , 32
Relative minimum:
3 2 , 23
22. (a) y A sinkm t B coskm t
A sinkm t A ⇒ tankm t . B coskm t B
Therefore, sin km t cos km t
A A2 B2
B A2 B2
.
When v y 0, yA
A A B B A B B A
2
2
2
2
2
(b) Period:
y Akm coskm t Bkm sinkm t 0 when
0 < x < 1
2
B2.
1
2 2 < x < 3
3
2 km
Frequency:
1 1 km 2 km 2
2 < x < 4
Review Exercises for Chapter 3 24. f x x 22x 4 x3 12x 16 fx 3x2 12 f x 6x 0 when x 0.
439
Test Interval
< x < 0
Sign of f x
f x < 0
f x > 0
Concave downward
Concave upward
Conclusion
0 < x <
Point of inflection: 0, 16 26.
ht t 4t 1 Domain: 1, ht 1
2 t 1
y
28. 7
0 ⇒ t3
6 5 4
1 h t t 132 h 3
30.
C
1 > 0 8
3, 5 is a relative minimum.
Qx s 2x r
r Qs dC 2 0 dx x 2
3 2 1 x
−1
1
2
3
4
5
6
7
32. (a) S 0.1222t3 1.3655t2 0.9052t 4.8429 (b)
r Qs x2 2
25
1
14 0
2Qs x2 r x
34. lim
x →
2Qs r
2x 2x lim 0 3x2 5 x → 3 5x2
38. gx
5x2 2
x2
lim
x →
5x2 5 lim 5 2 x → 1 2x2
x2
(c) St 0 when t 3.7. This is a maximum by the First Derivative Test. (d) No, because the t3 coefficient term is negative.
36. lim
3x
x →
x2 4
40. f x
lim
x →
3 1 4x2
3
3x x2 2
lim
x →
3x x2 2
Horizontal asymptote: y 5
lim
x →
lim
x →
lim
x →
3x x2 2
3xx x2 2x2
3 1 2x2
lim
x →
lim
x →
3
3xx
x2 2 x2
3 3 1 2x2
Horizontal asymptotes: y ± 3
440
Chapter 3
Applications of Differentiation
2 4 cos x cos 2x 3
44. gx
42. f x x3 3x2 2x xx 1x 2 Relative minima: (0, 0, 1, 0, 2, 0
Relative minima: 2 k, 0.29 where k is any integer.
Relative maxima: 1.577, 0.38, 0.423, 0.38
Relative maxima: 2k 1 , 8.29 where k is any integer.
3
10
−2
4 − 5 2
−1
46. f x 4x3 x4 x34 x
y 30
Domain: , ; Range: , 27
25 20
fx 12x 4x 4x 3 x 0 when x 0, 3. 2
3
5 2
0
2
15 10
f x 24x 12x 2 12x2 x 0 when x 0, 2.
5 x
f 3 < 0
−2
1
2
3
5
Therefore, 3, 27 is a relative maximum. Points of inflection: 0, 0, 2, 16 Intercepts: 0, 0, 4, 0 48. f x x2 42
y
Domain: , ; Range: 0,
24 20
fx 4xx2 4 0 when x 0, ± 2. 23 f x 43x2 4 0 when x ± . 3 f 0 < 0
(0, 16) (−2, 0)
(2, 0) 8 4 x
−3 −2 −1
Therefore, 0, 16 is a relative maximum.
1
2
3
f ± 2 > 0 Therefore, ± 2, 0 are relative minima. Points of inflection: ± 233, 649 Intercepts: 2, 0, 0, 16, 2, 0 Symmetry with respect to y-axis 50. f x x 3x 23 Domain: , ; Range:
y
16,875 256 ,
−4
fx x 33x 22 x 23 4x 7x 22 0 when x 2,
f 74 > 0 Therefore, 74 , 16,875 256 is a relative minimum. Points of inflection: 2, 0, 12 , 625 16 Intercepts: 2, 0, 0, 24, 3, 0
−2
2
x
4
− 20
(0,−24)
7 4.
f x 4x 72x 2 x 224 62x 1x 2 0 when x 2,
(3, 0)
(−2, 0)
− 40 − 60
1 2.
( 74 , − 16.875 256 )
Review Exercises for Chapter 3 52. f x x 213x 123
y
Graph of Exercise 39 translated 2 units to the right x replaces by x 2.
3 2
1, 0 is a relative maximum.
1,
1
(−1, 0)
4 is a relative minimum.
3
1
2, 0 is a point of inflection.
−2
2x 1 x2
(1, 41/3)
y 3
Domain: , ; Range: 1, 1
2
(1, 1)
21 x1 x 0 when x ± 1. fx 1 x22 f x
x 3
−3
Intercepts: 1, 0, 2, 0 54. f x
(2, 0)
−3 −2
1 x −1
1
2
3
(−1, −1)
2x3 x2 0 when x 0, ± 3. 1 x23
−2 −3
f 1 < 0 Therefore, 1, 1 is a relative maximum. f 1 > 0 Therefore, 1, 1 is a relative minimum. Points of inflection: 3, 32, 0, 0, 3, 32 Intercept: 0, 0 Symmetric with respect to the origin Horizontal asymptote: y 0 56. f x
x2 1 x4
2
Domain: , ; Range: 0,
1
fx
1 x42x x24x3 2x1 x1 x1 x2 0 when x 0, ± 1. 1 x42 1 x42
f x
1 x422 10x4 2x 2x521 x44x3 21 12x4 3x8 0 when x ± 1 x44 1 x43
6 ± 3
33
4
f ± 1 < 0
Therefore, ± 1,
1 are relative maxima. 2
y 1
f 0 > 0 Therefore, 0, 0 is a relative minimum. Points of inflection:
±
4
6 33 , 0.29 , ± 3
Intercept: 0, 0 Symmetric to the y-axis Horizontal asymptote: y 0
3 4 1 2
(−1, ) 1 2
4
6 33 , 0.40 3
(1, 21) (0, 0) x
−3 −2 −1 − 21
1
2
3
.
441
442
Chapter 3
58. f x x2
Applications of Differentiation
1 x3 1 x x
Domain: , 0, 0, ; Range: ,
2x 4, 60. f x x 1 x 3 2, 2x 4,
1 2x3 1 1 fx 2x 2 0 when x 3 . x x2 2 2 2x3 1 0 when x 1. f x 2 3 x x3
Domain: , Range: 2,
Intercept: 0, 4 y 4
>0
1 f 3 2
(0, 4)
3 2
1 3 Therefore, 3 , 3 is a relative minimum. 2 4
1
Point of inflection: 1, 0
x 1
2
3
4
Intercept: 1, 0 Vertical asymptote: x 0 y 3 2 1
(−1, 0) −3 −2
62. f x
( 12 , 34 ) 3
3
x 1
2
3
1 2 sin x sin 2 x
Domain: 1, 1 ; Range:
y 1
32 3, 32 3
fx 2cos x cos 2 x 22 cos x 1cos x 1 0
1 2
x − 12
− 12
1 2
−1
2 Critical Numbers: x ± , 0 3
f x 2 sin x 2 sin 2 x 2 sin x1 4 cos x 0 when x 0, ± 1, ± 0.420. By the First Derivative Test:
32, 32 3 is a relative minimum. 23, 32 3 is a relative maximum.
Points of inflection: 0.420, 0.462, 0.420, 0.462, ± 1, 0, 0, 0 Intercepts: (1, 0, 0, 0, 1, 0 Symmetric with respect to the origin 64. f x xn, n is a positive integer. (a) fx nxn1 The function has a relative minimum at 0, 0 when n is even. (b) f x nn 1xn2 The function has a point of inflection at 0, 0 when n is odd and n ≥ 3.
x ≤ 1 1 < x ≤ 3 x > 3
Review Exercises for Chapter 3 x2 y2 1 1, y 144 x2 144 16 3
66. Ellipse:
A 2x
y 12
2 4 144 x2 x144 x2 3 3
144 − x 2
−12
(
12 −8 −12
4 144 2x2 0 when x 72 62. 3 144 x2
1 3
x
x2 dA 4 144 x2 dx 3 144 x2
( x,
8
2 The dimensions of the rectangle are 2x 122 by y 144 72 42. 3 68. We have points 0, y, x, 0, and 4, 5. Thus, m
50 5x y5 or y . 04 4x x4
Let f x L 2 x 2
x 5x 4
L (4, 5)
2
5 (x, 0)
x fx 2x 50 x4 x
(0, y)
x4x 0 x 42
4
100x 0 x 43
3 100 . x x 43 100 0 when x 0 or x 4
L
x
2
3 100 4 25x2 x x 42 25 10023 25 12.7 feet 2 3 100 x 4 x4
70. Label triangle with vertices 0, 0, a, 0, and b, c. The equations of the sides of the triangle are y cbx and y cb a x a. Let x, 0 be a vertex of the inscribed rectangle. The coordinates of the upper left vertex are x, cbx. The y-coordinate of the upper right vertex of the rectangle is cbx. Solving for the x-coordinate x of the rectangle’s upper right vertex, you get c c x x a b ba
y= cx b
b ax bx a
(
ba ab x xaa x. b b
(0, 0)
Finally, the lower right vertex is
a a b b x, 0 . Width of rectangle: a Height of rectangle:
ab xx b
c x see figure b
A WidthHeight a
A
ab xx b
bc x a ba x bc x
b x 0 when x . ba acb 2ac b 2
a c dA c a x x dx b b b
x, c x b
2
b2 a ba b2 bc b2 a2 2c 41 ac 21 12 ac 21 Area of triangle
(
(x, 0)
(b, c) c ( x − a) b−a
y=
(a − a b− b x, bc x( (a, 0)
(a − a b− b x, 0(
443
444
Chapter 3
Applications of Differentiation
72. You can form a right triangle with vertices 0, y, 0, 0, and x, 0. Choosing a point a, b on the hypotenuse (assuming the triangle is in the first quadrant), the slope is m
b0 bx yb . ⇒y 0a ax ax
Let f x L2 x2 y2 x2
. abx x 2
fx 2x 2
ab abx x a x 2
2x a x3 ab2 3 ab2. 0 when x 0, a a x3 3 Choosing the nonzero value, we have y b a 2b. 3 ab2 2 b L a 3 a2b 2
a2 3a43b23 3a23b43 b212 a23 b2332 meters 74. Using Exercise 73 as a guide we have L1 a csc and L2 b sec . Then dLd a csc cot b sec tan 0 when 3 ab, sec tan
a23 b23
b13
, csc
L L1 L2 a csc b sec a
a23 b23
a13
and
a23 b2312 a23 b2312 b a23 b2332. a13 b13
This matches the result of Exercise 72. 76. Total cost Cost per hourNumber of hours T
11v 825 v 110 7.50 500 v 50 v 2
dT 11 825 11v 2 41,250 2 dv 50 v 50v 2 0 when v 3750 256 61.2 mph. d 2T 1650 3 > 0 when v 256 so this value yields a minimum. dv 2 v 78. f x x3 2x 1 From the graph, you can see that f x has one real zero. fx 3x2 2 f changes sign in 1, 0 . n
xn
f xn
f xn
f xn fxn
1
0.5000
0.1250
2.7500
0.0455
0.4545
2
0.4545
0.0029
2.6197
0.0011
0.4534
On the interval 1, 0 : x 0.453.
xn
f xn fxn
Review Exercises for Chapter 3
43. r
v02 sin 2 32
45. Let f x x, x 100, dx 0.6. f x x f x f x dx
v0 2200 ftsec
changes from 10 to 11 dr
x
22002 cos 2 d 16
10
f x x 99.4
180
d 11 10
1 dx 2x
100
1 0.6 9.97 2100
Using a calculator: 99.4 9.96995
180
r dr
22002 20 cos 16 180
180 4961 feet
4961 feet 49. Let f x x, x 4, dx 0.02, f x 1 2x .
4 x, x 625, dx 1. 47. Let f x 4 x f x x f x f x d x
Then
1 dx 44x3
f 4.02 f 4 f 4 dx
1 4 624 4 625 f x x 1 4 625 3 4 5
4.02 4
1 1 0.02 2 0.02. 4 24
1 4.998 500
4 624 4.9980. Using a calculator,
51. In general, when x → 0, dy approaches y. 53. True
55. True
Review Exercises for Chapter 3 1. A number c in the domain of f is a critical number if f c 0 or f is undefined at c.
y 4
f ′(c) is 3 undefined.
f ′(c) = 0
x −4 −3
−1 −2 −3 −4
3. gx 2x 5 cos x, 0, 2
18
(6.28, 17.57)
g x 2 5 sin x 2
0 when sin x 5 . Critical numbers: x 0.41, x 2.73 Left endpoint: 0, 5 Critical number: 0.41, 5.41 Critical number: 2.73, 0.88 Minimum Right endpoint: 2, 17.57 Maximum
(2.73, 0.88) − 4
2 −1
1
2
4
163
164
Chapter 3
Applications of Differentiation
5. Yes. f 3 f 2 0. f is continuous on 3, 2 , differentiable on 3, 2.
7. f x 3 x 4 y
(a)
f x x 33x 1 0 for x 13.
6 4
c 13 satisfies f c 0.
2 x
−2
2
4
6
10
−4 −6
f 1 f 7 0 (b) f is not differentiable at x 4.
9.
f x x23, 1 ≤ x ≤ 8
f x x cos x,
11.
2 f x x13 3
f x 1 sin x f b f a 2 2 1 ba 2 2
f b f a 4 1 3 ba 81 7
f c 1 sin c 1
2 3 f c c13 3 7 c
13.
149
3
≤ x ≤ 2 2
c0
2744 3.764 729
f x Ax2 Bx C f x 2Ax B f x2 f x1 Ax22 x12 Bx2 x1 x2 x1 x2 x1 Ax1 x2 B f c 2Ac B Ax1 x2 B 2Ac Ax1 x2 c
x1 x2 Midpoint of x1, x2
2
15. f x x 12x 3 f x x 1 1 x 32x 1 2
x 13x 7 7 Critical numbers: x 1 and x 3
17. hx xx 3 x32 3x12 Domain: 0, 3 3 h x x12 x12 2 2 3 3x 1 x12x 1 2 2x Critical number: x 1
Interval:
< x < 1
1 < x <
7 3
7 3
< x <
Sign of f x:
f x > 0
f x < 0
f x > 0
Conclusion:
Increasing
Decreasing
Increasing
Interval:
0 < x < 1
Sign of h x:
h x < 0
h x > 0
Conclusion:
Decreasing
Increasing
1 < x <
Review Exercises for Chapter 3 19. ht 14t 4 8t
Test Interval: < t < 2
h t t 3 8 0 when t 2. Relative minimum: 2, 12
165
2 < t <
Sign of h t:
h t < 0
h t > 0
Conclusion:
Decreasing
Increasing
1 1 cos12t sin12t 3 4
21. y
v y 4 sin12t 3 cos12t (a) When t
1 , y inch and v y 4 inches/second. 8 4
(b) y 4 sin12t 3 cos12t 0 when
sin12t 3 3 ⇒ tan12t . cos12t 4 4
3 4 Therefore, sin12t and cos12t . The maximum displacement is 5 5 y (c) Period:
1345 41 53 125 inch. 2 12 6 1 6 6
Frequency:
23. f x x cos x, 0 ≤ x ≤ 2 f x 1 sin x f x cos x 0 when x Points of inflection:
3 , . 2 2
Test Interval: Sign of f x: Conclusion:
3 < x < 2 2
3 < x < 2 2
f x < 0
f x > 0
f x < 0
Concave downward
Concave upward
Concave downward
0 < x <
2
2 , 2 , 32, 32
25. gx 2x21 x2
y
g x 4x2x2 1 Critical numbers: x 0, ±
(−
1 2
1, 1 2 2
)
1
−2
g x 4 24x
(
1, 1 2 2
)
(0, 0)
2
x
−1
2
−2
g 0 4 > 0
Relative minimum at 0, 0
1 1 Relative maximums at ± , 2 2
y
27. 6
(5, f(5))
5 4
(3, f(3))
2 1 −1
29. The first derivative is positive and the second derivative is negative. The graph is increasing and is concave down.
7
3
−3
1 g ± 8 < 0 2
(6, 0) (0, 0) 2 3 4 5
x 7
166
Chapter 3
Applications of Differentiation
31. (a) D 0.0034t4 0.2352t3 4.9423t2 20.8641t 94.4025 (b)
369
0
29 0
(c) Maximum at 21.9, 319.5 1992 Minimum at 2.6, 69.6 1972 (d) Outlays increasing at greatest rate at the point of inflection 9.8, 173.7 1979
33. lim
x →
2x2 2 2 lim 5 x → 3 5x2 3
35. lim
2x 3 x4
39. f x
3x2
37. hx
Discontinuity: x 4 lim
x →
x →
5 cos x 0, since 5 cos x ≤ 5. x
3 2 x
Discontinuity: x 0
2x 3 2 3x lim 2 x → 1 4x x4
lim
x →
3x 2 2
Vertical asymptote: x 4
Vertical asymptote: x 0
Horizontal asymptote: y 2
Horizontal asymptote: y 2
41. f x x3
243 x
43. f x
x1 1 3x2
Relative minimum: 3, 108
Relative minimum: 0.155, 1.077
Relative maximum: 3, 108
Relative maximum: 2.155, 0.077 0.2
200
−2
−5
5
5
− 1.4
− 200
Vertical asymptote: x 0
Horizontal asymptote: y 0
45. f x 4x x2 x4 x Domain: , ; Range: , 4 f x 4 2x 0 when x 2. f x 2 Therefore, 2, 4 is a relative maximum. Intercepts: 0, 0, 4, 0
y
5
)2, 4)
4 3 2 1
x 1
2
3
5
Review Exercises for Chapter 3 47. f x x16 x2, Domain: 4, 4 , Range: 8, 8
y
2
2, 8
8
Domain: 4, 4 ; Range: 8, 8
6 4
16 2x2 f x 0 when x ± 22 and undefined when x ± 4. 16 x2 f x
2xx2 24 16 x232
2
(− 4, 0)
(4, 0) x
8
6
2
2
4
6
8
(0, 0)
8
2
2,
8
f 22 > 0
Therefore, 22, 8 is a relative minimum. f 22 < 0
Therefore, 22, 8 is a relative maximum. Point of inflection: 0, 0 Intercepts: 4, 0, 0, 0, 4, 0 Symmetry with respect to origin 49. f x x 13x 32
y
Domain: , ; Range: ,
4
f x x 12x 35x 11 0 when x 1,
11 , 3. 5
f x 4x 15x2 22x 23 0 when x 1,
( 115 , 1.11(
(1, 0)
(2.69, 0.46) (3, 0)
2
x
−2
4 −2
11± 6 . 5
6
(1.71, 0.60)
−4
f 3 > 0 Therefore, 3, 0 is a relative minimum. f
115 < 0
Therefore,
is a relative maximum. 115, 3456 3125
Points of inflection: 1, 0,
11 5
6
11 5
, 0.60 ,
6
, 0.46
Intercepts: 0, 9, 1, 0, 3, 0 51. f x x13x 323
y
Domain: , ; Range: ,
4 3
x1 f x 0 when x 1 and undefined when x 3, 0. x 313x23 2 f x 53 is undefined when x 0, 3. x x 343 3 4 is By the First Derivative Test 3, 0 is a relative maximum and 1, a relative minimum. 0, 0 is a point of inflection.
Intercepts: 3, 0, 0, 0
2 1
) 3, 0)
)0, 0) x
5
4
2
) 1,
1
1
1.59) 3
2
167
168
Chapter 3
Applications of Differentiation
x1 x1
53. f x
x
1
y
Domain: , 1, 1, ; Range: , 1, 1, f x
2 < 0 if x 1. x 12
f x
4 x 13
4
y
1 2
x 2
2
4
2
Horizontal asymptote: y 1 Vertical asymptote: x 1 Intercepts: 1, 0, 0, 1 55. f x
4 1 x2
y 5
Domain: , ; Range: 0, 4
8x 0 when x 0. f x 1 x22 3 81 3x2 . 0 when x ± f x 1 x23 3
(0, 4)
4
(−
3,3 3
(
(
3,3 3
1
2
(
2 1 −3
−2 −1
x −1
3
f 0 < 0 Therefore, 0, 4 is a relative maximum. Points of inflection: ± 33, 3 Intercept: 0, 4 Symmetric to the y-axis Horizontal asymptote: y 0
57. f x x3 x
y
4 x
10
Domain: , 0, 0, ; Range: , 6 , 6, f x 3x2 1 f x 6x
4 x2
3x4
4 0 when x ± 1. x2 x2
8 6x4 8 0 x3 x3
f 1 < 0 Therefore, 1, 6 is a relative maximum. f 1 > 0 Therefore, 1, 6 is a relative minimum. Vertical asymptote: x 0 Symmetric with respect to origin
5
(1, 6) x
2
1
(−1, −6) −5
1
x
2
0
Review Exercises for Chapter 3
59. f x x2 9
y
Domain: , ; Range: 0, f x
2xx2 9 0 when x 0 and is undefined when x ± 3. x2 9
10
5
2x2 9 is undefined at x ± 3. f x 2 x 9
)0, 9)
) 3, 0)
)3, 0) x
4
f 0 < 0
2
2
4
Therefore, 0, 9 is a relative maximum. Relative minima: ± 3, 0 Points of inflection: ± 3, 0 Intercepts: ± 3, 0, 0, 9 Symmetric to the y-axis 61. f x x cos x
y
)2 , 2
Domain: 0, 2 ; Range: 1, 1 2
3 3 , 2 2
f x 1 sin x ≥ 0, f is increasing. f x cos x 0 when x Points of inflection:
1)
2
3 , . 2 2
3 3 , , , 2 2 2 2
)0, 1)
, 2 2 x
2
Intercept: 0, 1 63. x2 4y2 2x 16y 13 0 (a) x 2 2x 1 4y 2 4y 4 13 1 16
y
x 1 4y 2 4 x 12 y 22 1 4 1 The graph is an ellipse: 2
2
4
(1, 3) 3 2 1
Maximum: 1, 3
(1, 1) x −1
Minimum: 1, 1
1
2
3
(b) x2 4y2 2x 16y 13 0 2x 8y
dy dy 2 16 0 dx dx dy 8y 16 2 2x dx dy 2 2x 1x dx 8y 16 4y 8
The critical numbers are x 1 and y 2. These correspond to the points 1, 1, 1, 3, 2, 1, and 2, 3. Hence, the maximum is 1, 3 and the minimum is 1, 1.
169
170
Chapter 3
Applications of Differentiation
65. Let t 0 at noon.
(100 − 12t, 0) (0, 0)
L d 2 100 12t2 10t2 10,000 2400t 244t 2
A
(100, 0)
d
300 dL 2400 488t 0 when t 4.92 hr. dt 61
B (0, −10t)
Ship A at 40.98, 0; Ship B at 0, 49.18 d 2 10,000 2400t 244t 2 4098.36 when t 4.92 4:55 P.M.. d 64 km 67. We have points 0, y, x, 0, and 1, 8. Thus,
y
08 8x y8 or y . m 01 x1 x1
(0, y)
10
(1, 8)
8 6
Let f x L 2 x 2
x 8x 1 . 2
4 2
x f x 2x 128 x1 x
(x, 0)
x 1 x 0 x 12
x 2
4
6
8
10
64x 0 x 13
x x 13 64 0 when x 0, 5 minimum. Vertices of triangle: 0, 0, 5, 0, 0, 10 69.
A Average of basesHeight
x 2 s
3s2 2sx x2
2
s
see figure
s
dA 1 s xs x 3s2 2sx x2 dx 4 3s2 2sx x2
x−s 2
22s xs x 0 when x 2s. 43s2 2sx x2 A is a maximum when x 2s. 71. You can form a right triangle with vertices 0, 0, x, 0 and 0, y. Assume that the hypotenuse of length L passes through 4, 6. 60 6x y6 or y 04 4x x4
Let f x L2 x2 y2 x 2 f x 2x 72
x 6x 4 . 2
0 x x 4 x 4 4 2
3 x x 43 144 0 when x 0 or x 4 144.
L 14.05 feet
s
x−s 2 x
m
3s 2 + 2sx − x 2 2
Review Exercises for Chapter 3 csc
73. csc
L1 or L1 6 csc 6
2 9 or L L2
2
9 csc
see figure
L1 θ
2
L2
θ 9
L L1 L2 6 csc 9 csc
171
6
(π2 − θ(
2 6 csc 9 sec
dL 6 csc cot 9 sec tan 0 d tan3
3 2 2 ⇒ tan 3 3 3
sec 1 tan2 csc L6
1 23
23
323 223
313
sec 323 223 tan 213
323 22312 323 22312 9 3323 22332 ft 21.07 ft Compare to Exercise 72 using a 9 and b 6. 13 2 313
75. Total cost Cost per hourNumber of hours T
v 110 11v 550 5 600 v 60 v 2
dT 11 550 11v 2 33,000 2 dv 60 v 60v 2 0 when v 3000 1030 54.8 mph. d 2T 1100 3 > 0 when v 1030 so this value yields a minimum. dv 2 v 77. f x x3 3x 1 From the graph you can see that f x has three real zeros. f x 3x2 3 f xn
f xn
f xn f xn
1.5000
0.1250
3.7500
0.0333
1.5333
2
1.5333
0.0049
4.0530
0.0012
1.5321
n
xn
f xn
f xn
f xn f xn
1
0.5000
0.3750
2.2500
0.1667
0.3333
2
0.3333
0.0371
2.6667
0.0139
0.3472
3
0.3472
0.0003
2.6384
0.0001
0.3473
n
xn
f xn
f xn
f xn f xn
1
1.9000
0.1590
7.8300
0.0203
1.8797
2
1.8797
0.0024
7.5998
0.0003
1.8794
n
xn
1
xn
f xn f xn
xn
xn
f xn f xn
f xn f xn
The three real zeros of f x are x 1.532, x 0.347, and x 1.879.
172
Chapter 3
Applications of Differentiation
79. Find the zeros of f x x4 x 3. fx 4x3 1 From the graph you can see that f x has two real zeros. f changes sign in 2, 1.
n
xn
f xn
f xn
f xn fxn
1
1.2000
0.2736
7.9120
0.0346
1.1654
2
1.1654
0.0100
7.3312
0.0014
1.1640
xn
f xn fxn
On the interval 2, 1: x 1.164. f changes sign in 1, 2. f xn
f xn fxn
0.5625
12.5000
0.0450
1.4550
1.4550
0.0268
11.3211
0.0024
1.4526
1.4526
0.0003
11.2602
0.0000
1.4526
n
xn
1
1.5000
2 3
f xn
xn
f xn fxn
On the interval 1, 2; x 1.453. 81.
y x1 cos x x x cos x dy 1 x sin x cos x dx dy 1 x sin x cos x dx
83.
S 4 r 2. dr r ± 0.025 dS 8r dr 89± 0.025 ± 1.8 square cm dS 8 r dr 2 dr 100 100 100 S 4 r 2 r
2± 0.025 100 ± 0.56% 9
4 V r3 3 dV 4 r 2 dr 492± 0.025 ± 8.1 cubic cm dV 4 r 2 dr 3 dr 100 100 100 V 43r 3 r
3± 0.025 100 ± 0.83% 9
Problem Solving for Chapter 3 1. Assume y1 < d < y2. Let gx f x dx a. g is continuous on a, b and therefore has a minimum c, gc on a, b. The point c cannot be an endpoint of a, b because ga fa d y1 d < 0 gb fb d y2 d > 0 Hence, a < c < b and gc 0 ⇒ fc d.
Problem Solving for Chapter 3 3. (a) For a 3, 2, 1, 0, p has a relative maximum at 0, 0. For a 1, 2, 3, p has a relative maximum at 0, 0 and 2 relative minima. (b) px 4ax3 12x 4xax2 3 0 ⇒ x 0, ±
3a
p x 12ax2 12 12ax2 1 For x 0, p 0 12 < 0 ⇒ p has a relative maximum at 0, 0. (c) If a > 0, x ±
3a are the remaining critical numbers.
3a 12 3a 12 24 > 0 ⇒ p has relative minima for a > 0.
p ±
(d) 0, 0 lies on y 3x2. Let x ±
3a. Then
px a
3a
2
6
3 a
2
a=1 a=3 y
2ax 2x
2x2
8 7 6 5 4 3 2
a
For a ≥ 0, there is one relative minimum at 0, 0.
(c) For a < 0, there are two relative minima at x ±
−2
2a.
(d) There are either 1 or 3 critical points. The above analysis shows that there cannot be exactly two relative extrema. c x2 x c c c 2x 0 ⇒ 2 2x ⇒ x3 ⇒ x x2 x 2
cx 3
2c 2 x3
If c 0, f x x2 has a relative minimum, but no relative maximum.
2c is a relative minimum, because f 2c > 0. c If c < 0, x is a relative minimum too. 2 If c > 0, x
3
3
Answer: all c.
a=2
a=0
a = −1 a = −2 a = −3
(b) For a < 0, there is a relative maximum at 0, 1.
f x
a = −1 a = −3
3x2 is satisfied by all the relative extrema of p.
p x 16x2 2a
fx
1
2
3
3a a9 18a a9.
5. px x 4 ax2 1
7. f x
a=1 x
−3
(a) px
a=3
2
9 Thus, y 3 ± a
4x4
y
a=2
3
x −1 −2
2
−8
a = −2 a=0
173
174
Chapter 3
9. Set
Applications of Differentiation
f b f a f ab a k. b a2
Define Fx f x f a fax a kx a2. Fa 0, Fb f b f a fab a kb a2 0 F is continuous on a, b and differentiable on a, b. There exists c1, a < c1 < b, satisfying Fc1 0. Fx fx fa 2kx a satisfies the hypothesis of Rolle’s Theorem on a, c1: Fa 0, Fc1 0. There exists c2, a < c2 < c1 satisfying F c2 0. Finally, F x f x 2k and F c2 0 implies that k
f c2 . 2
Thus, k
11. E E
f b f a f ab a f c2 1 ⇒f b f a fab a f c2b a2. b a2 2 2
tan 1 0.1 tan 10 tan tan2 0.1 tan 1 10 tan
1 10 tan 10 sec2 2 tan sec2 10 tan tan2 10 sec2 0 1 10 tan
⇒ 1 10 tan 10 sec2 2 tan sec2 10 tan tan2 10 sec2 ⇒ 10 sec2 2 tan sec2 100 tan sec2 20 tan2 sec2 100 tan sec2 10 tan2 sec2 ⇒ 10 2 tan 10 tan2 ⇒ 10 tan2 2 tan 10 0 tan
2 ± 4 400 0.90499, 1.10499 20
Using the positive value, 0.7356, or 42.14 . 13. v 2400 sin v 2400 cos 0
3 2n, 2n, n an integer 2 2
Problem Solving for Chapter 3 x y 4 1 or y x 4. 3 4 3
15. The line has equation Rectangle:
4 4 Area A xy x x 4 x2 4x. 3 3 8 8 3 Ax x 4 0 ⇒ x 4 ⇒ x 3 3 2 Dimensions:
3 2 2
Calculus was helpful.
Circle: The distance from the center r, r to the line
r
r r 1 3 4
x y 1 0 must be r: 3 4
12 7r 12 7r 12 5 12 5
1 1 9 16 5r 7r 12 ⇒ r 1 or r 6.
Clearly, r 1. x y 1 and satisfies x y r. 3 4
Semicircle: The center lies on the line
7 12 r r 1 ⇒ r 1 ⇒ r . No calculus necessary. 3 4 12 7
Thus
17. y 1 x21 y
2x 1 x22
y
3 23x2 1 1 0 ⇒ x± ± 3 x2 13 3
y′′:
+++ −−−− −−−− +++ −
3 3
0
3 3
The tangent line has greatest slope at
19. (a)
x sin x
3 3
3
,
4
and least slope at 33, 34 .
0.1
0.2
0.3
0.4
0.5
1.0
0.09983
0.19867
0.29552
0.38942
0.47943
0.84147
sin x ≤ x (b) Let f x sin x. Then fx cos x and on 0, x you have by the Mean Value Theorem, fc cosc Hence,
f x f 0 , 0 < c < x x0 sin x x
sin x cosc ≤ 1 x
⇒ sin x ≤ x ⇒ sin x ≤ x
175
Review Exercises for Chapter 3 50. Let f x tan x, x 0, dx 0.05, fx sec2 x.
52. Propagated error f x x f x,
Then
relative error
f 0.05 f 0 f0dx
437
dy dy , and the percent error 100. y y
tan 0.05 tan 0 sec2 00.05 0 10.05.
54. True,
y dy a x dx
56. False Let f x x, x 1, and x dx 3. Then y f x x f x f 4 f 1 1 and dy fx dx
1 3 3 . 21 2
Thus, dy > y in this example.
Review Exercises for Chapter 3 2. (a) f 4 f 4 3 (c)
(b) f 3 f 3 4 4
y
(d) Yes. Since f 2 f 2 1 1 and f 1 f 1 2, the Mean Value says that there exists at least one value c in 2, 1 such that
6 4
x −6 −4 −2
4
fc
6
−4
(e) No, lim f x exists because f is continuous at 0, 0.
−6
x →0
At least six critical numbers on 6, 6. 4. f x
f 1 f 2 2 1 1. 1 2 12
x x2 1
, 0, 2
(f) Yes, f is differentiable at x 2.
6. No. f is not differentiable at x 2.
1 fx x x2 1322x x2 112 2 1 x2 132 No critical numbers Left endpoint: 0, 0 Minimum Right endpoint: 2, 25 Maximum
8. No; the function is discontinuous at x 0 which is in the interval 2, 1 .
10.
1 f x , 1 ≤ x ≤ 4 x fx
1 x2
f b f a 14 1 34 1 ba 41 3 4 fc
1 1 c2 4
c2
12.
f x x 2x, 0 ≤ x ≤ 4 fx
1 2x
2
f b f a 6 0 3 ba 40 2 fc
1 3 2 2 2c
c1
438
Chapter 3
Applications of Differentiation
f x 2x2 3x 1
14.
fx 4x 3 f b f a 21 1 5 ba 40 fc 4c 3 5 c 2 Midpoint of 0, 4 16. gx x 13
Critical number: x 1
18. f x sin x cos x, 0 ≤ x ≤ 2
Critical numbers: x
gx
5 ,x 4 4
1 < x <
Sign of gx
gx > 0
gx > 0
Conclusion
Increasing
Increasing
0 < x <
Interval
fx cos x sin x
20. gx
< x < 1
Interval
gx 3x 12
4
5 < x < 4 4
5 < x < 2 4
Sign of fx
fx > 0
fx < 0
fx > 0
Conclusion
Increasing
Decreasing
Increasing
3 x sin 1 , 0, 4 2 2
Test Interval
3 x 1 cos 2 2 2
Sign of gx
gx > 0
gx < 0
gx > 0
Conclusion
Increasing
Decreasing
Increasing
0 when x 1
2 2 ,3
Relative maximum:
1 2 , 32
Relative minimum:
3 2 , 23
22. (a) y A sinkm t B coskm t
A sinkm t A ⇒ tankm t . B coskm t B
Therefore, sin km t cos km t
A A2 B2
B A2 B2
.
When v y 0, yA
A A B B A B B A
2
2
2
2
2
(b) Period:
y Akm coskm t Bkm sinkm t 0 when
0 < x < 1
2
B2.
1
2 2 < x < 3
3
2 km
Frequency:
1 1 km 2 km 2
2 < x < 4
Review Exercises for Chapter 3 24. f x x 22x 4 x3 12x 16 fx 3x2 12 f x 6x 0 when x 0.
439
Test Interval
< x < 0
Sign of f x
f x < 0
f x > 0
Concave downward
Concave upward
Conclusion
0 < x <
Point of inflection: 0, 16 26.
ht t 4t 1 Domain: 1, ht 1
2 t 1
y
28. 7
0 ⇒ t3
6 5 4
1 h t t 132 h 3
30.
C
1 > 0 8
3, 5 is a relative minimum.
Qx s 2x r
r Qs dC 2 0 dx x 2
3 2 1 x
−1
1
2
3
4
5
6
7
32. (a) S 0.1222t3 1.3655t2 0.9052t 4.8429 (b)
r Qs x2 2
25
1
14 0
2Qs x2 r x
34. lim
x →
2Qs r
2x 2x lim 0 3x2 5 x → 3 5x2
38. gx
5x2 2
x2
lim
x →
5x2 5 lim 5 2 x → 1 2x2
x2
(c) St 0 when t 3.7. This is a maximum by the First Derivative Test. (d) No, because the t3 coefficient term is negative.
36. lim
3x
x →
x2 4
40. f x
lim
x →
3 1 4x2
3
3x x2 2
lim
x →
3x x2 2
Horizontal asymptote: y 5
lim
x →
lim
x →
lim
x →
3x x2 2
3xx x2 2x2
3 1 2x2
lim
x →
lim
x →
3
3xx
x2 2 x2
3 3 1 2x2
Horizontal asymptotes: y ± 3
440
Chapter 3
Applications of Differentiation
2 4 cos x cos 2x 3
44. gx
42. f x x3 3x2 2x xx 1x 2 Relative minima: (0, 0, 1, 0, 2, 0
Relative minima: 2 k, 0.29 where k is any integer.
Relative maxima: 1.577, 0.38, 0.423, 0.38
Relative maxima: 2k 1 , 8.29 where k is any integer.
3
10
−2
4 − 5 2
−1
46. f x 4x3 x4 x34 x
y 30
Domain: , ; Range: , 27
25 20
fx 12x 4x 4x 3 x 0 when x 0, 3. 2
3
5 2
0
2
15 10
f x 24x 12x 2 12x2 x 0 when x 0, 2.
5 x
f 3 < 0
−2
1
2
3
5
Therefore, 3, 27 is a relative maximum. Points of inflection: 0, 0, 2, 16 Intercepts: 0, 0, 4, 0 48. f x x2 42
y
Domain: , ; Range: 0,
24 20
fx 4xx2 4 0 when x 0, ± 2. 23 f x 43x2 4 0 when x ± . 3 f 0 < 0
(0, 16) (−2, 0)
(2, 0) 8 4 x
−3 −2 −1
Therefore, 0, 16 is a relative maximum.
1
2
3
f ± 2 > 0 Therefore, ± 2, 0 are relative minima. Points of inflection: ± 233, 649 Intercepts: 2, 0, 0, 16, 2, 0 Symmetry with respect to y-axis 50. f x x 3x 23 Domain: , ; Range:
y
16,875 256 ,
−4
fx x 33x 22 x 23 4x 7x 22 0 when x 2,
f 74 > 0 Therefore, 74 , 16,875 256 is a relative minimum. Points of inflection: 2, 0, 12 , 625 16 Intercepts: 2, 0, 0, 24, 3, 0
−2
2
x
4
− 20
(0,−24)
7 4.
f x 4x 72x 2 x 224 62x 1x 2 0 when x 2,
(3, 0)
(−2, 0)
− 40 − 60
1 2.
( 74 , − 16.875 256 )
Review Exercises for Chapter 3 52. f x x 213x 123
y
Graph of Exercise 39 translated 2 units to the right x replaces by x 2.
3 2
1, 0 is a relative maximum.
1,
1
(−1, 0)
4 is a relative minimum.
3
1
2, 0 is a point of inflection.
−2
2x 1 x2
(1, 41/3)
y 3
Domain: , ; Range: 1, 1
2
(1, 1)
21 x1 x 0 when x ± 1. fx 1 x22 f x
x 3
−3
Intercepts: 1, 0, 2, 0 54. f x
(2, 0)
−3 −2
1 x −1
1
2
3
(−1, −1)
2x3 x2 0 when x 0, ± 3. 1 x23
−2 −3
f 1 < 0 Therefore, 1, 1 is a relative maximum. f 1 > 0 Therefore, 1, 1 is a relative minimum. Points of inflection: 3, 32, 0, 0, 3, 32 Intercept: 0, 0 Symmetric with respect to the origin Horizontal asymptote: y 0 56. f x
x2 1 x4
2
Domain: , ; Range: 0,
1
fx
1 x42x x24x3 2x1 x1 x1 x2 0 when x 0, ± 1. 1 x42 1 x42
f x
1 x422 10x4 2x 2x521 x44x3 21 12x4 3x8 0 when x ± 1 x44 1 x43
6 ± 3
33
4
f ± 1 < 0
Therefore, ± 1,
1 are relative maxima. 2
y 1
f 0 > 0 Therefore, 0, 0 is a relative minimum. Points of inflection:
±
4
6 33 , 0.29 , ± 3
Intercept: 0, 0 Symmetric to the y-axis Horizontal asymptote: y 0
3 4 1 2
(−1, ) 1 2
4
6 33 , 0.40 3
(1, 21) (0, 0) x
−3 −2 −1 − 21
1
2
3
.
441
442
Chapter 3
58. f x x2
Applications of Differentiation
1 x3 1 x x
Domain: , 0, 0, ; Range: ,
2x 4, 60. f x x 1 x 3 2, 2x 4,
1 2x3 1 1 fx 2x 2 0 when x 3 . x x2 2 2 2x3 1 0 when x 1. f x 2 3 x x3
Domain: , Range: 2,
Intercept: 0, 4 y 4
>0
1 f 3 2
(0, 4)
3 2
1 3 Therefore, 3 , 3 is a relative minimum. 2 4
1
Point of inflection: 1, 0
x 1
2
3
4
Intercept: 1, 0 Vertical asymptote: x 0 y 3 2 1
(−1, 0) −3 −2
62. f x
( 12 , 34 ) 3
3
x 1
2
3
1 2 sin x sin 2 x
Domain: 1, 1 ; Range:
y 1
32 3, 32 3
fx 2cos x cos 2 x 22 cos x 1cos x 1 0
1 2
x − 12
− 12
1 2
−1
2 Critical Numbers: x ± , 0 3
f x 2 sin x 2 sin 2 x 2 sin x1 4 cos x 0 when x 0, ± 1, ± 0.420. By the First Derivative Test:
32, 32 3 is a relative minimum. 23, 32 3 is a relative maximum.
Points of inflection: 0.420, 0.462, 0.420, 0.462, ± 1, 0, 0, 0 Intercepts: (1, 0, 0, 0, 1, 0 Symmetric with respect to the origin 64. f x xn, n is a positive integer. (a) fx nxn1 The function has a relative minimum at 0, 0 when n is even. (b) f x nn 1xn2 The function has a point of inflection at 0, 0 when n is odd and n ≥ 3.
x ≤ 1 1 < x ≤ 3 x > 3
Review Exercises for Chapter 3 x2 y2 1 1, y 144 x2 144 16 3
66. Ellipse:
A 2x
y 12
2 4 144 x2 x144 x2 3 3
144 − x 2
−12
(
12 −8 −12
4 144 2x2 0 when x 72 62. 3 144 x2
1 3
x
x2 dA 4 144 x2 dx 3 144 x2
( x,
8
2 The dimensions of the rectangle are 2x 122 by y 144 72 42. 3 68. We have points 0, y, x, 0, and 4, 5. Thus, m
50 5x y5 or y . 04 4x x4
Let f x L 2 x 2
x 5x 4
L (4, 5)
2
5 (x, 0)
x fx 2x 50 x4 x
(0, y)
x4x 0 x 42
4
100x 0 x 43
3 100 . x x 43 100 0 when x 0 or x 4
L
x
2
3 100 4 25x2 x x 42 25 10023 25 12.7 feet 2 3 100 x 4 x4
70. Label triangle with vertices 0, 0, a, 0, and b, c. The equations of the sides of the triangle are y cbx and y cb a x a. Let x, 0 be a vertex of the inscribed rectangle. The coordinates of the upper left vertex are x, cbx. The y-coordinate of the upper right vertex of the rectangle is cbx. Solving for the x-coordinate x of the rectangle’s upper right vertex, you get c c x x a b ba
y= cx b
b ax bx a
(
ba ab x xaa x. b b
(0, 0)
Finally, the lower right vertex is
a a b b x, 0 . Width of rectangle: a Height of rectangle:
ab xx b
c x see figure b
A WidthHeight a
A
ab xx b
bc x a ba x bc x
b x 0 when x . ba acb 2ac b 2
a c dA c a x x dx b b b
x, c x b
2
b2 a ba b2 bc b2 a2 2c 41 ac 21 12 ac 21 Area of triangle
(
(x, 0)
(b, c) c ( x − a) b−a
y=
(a − a b− b x, bc x( (a, 0)
(a − a b− b x, 0(
443
444
Chapter 3
Applications of Differentiation
72. You can form a right triangle with vertices 0, y, 0, 0, and x, 0. Choosing a point a, b on the hypotenuse (assuming the triangle is in the first quadrant), the slope is m
b0 bx yb . ⇒y 0a ax ax
Let f x L2 x2 y2 x2
. abx x 2
fx 2x 2
ab abx x a x 2
2x a x3 ab2 3 ab2. 0 when x 0, a a x3 3 Choosing the nonzero value, we have y b a 2b. 3 ab2 2 b L a 3 a2b 2
a2 3a43b23 3a23b43 b212 a23 b2332 meters 74. Using Exercise 73 as a guide we have L1 a csc and L2 b sec . Then dLd a csc cot b sec tan 0 when 3 ab, sec tan
a23 b23
b13
, csc
L L1 L2 a csc b sec a
a23 b23
a13
and
a23 b2312 a23 b2312 b a23 b2332. a13 b13
This matches the result of Exercise 72. 76. Total cost Cost per hourNumber of hours T
11v 825 v 110 7.50 500 v 50 v 2
dT 11 825 11v 2 41,250 2 dv 50 v 50v 2 0 when v 3750 256 61.2 mph. d 2T 1650 3 > 0 when v 256 so this value yields a minimum. dv 2 v 78. f x x3 2x 1 From the graph, you can see that f x has one real zero. fx 3x2 2 f changes sign in 1, 0 . n
xn
f xn
f xn
f xn fxn
1
0.5000
0.1250
2.7500
0.0455
0.4545
2
0.4545
0.0029
2.6197
0.0011
0.4534
On the interval 1, 0 : x 0.453.
xn
f xn fxn
Problem Solving for Chapter 3 80. Find the zeros of f x sin x x 1. fx cos x 1 From the graph you can see that f x has three real zeros.
n
xn
f xn
f xn
f xn fxn
1
0.2000
0.2122
3.5416
0.0599
0.2599
2
0.2599
0.0113
3.1513
0.0036
0.2635
3
0.2635
0.0000
3.1253
0.0000
0.2635
n
xn
f xn
f xn
f xn fxn
1
1.0000
0.0000
2.1416
0.0000
n
xn
f xn
f xn
f xn fxn
1
1.8000
0.2122
3.5416
0.0599
1.7401
2
1.7401
0.0113
3.1513
0.0036
1.7365
3
1.7365
0.0000
3.1253
0.0000
1.7365
xn
xn
f xn fxn
f xn fxn
1.0000
xn
f xn fxn
The three real zeros of f x are x 0.264, x 1, and x 1.737.
82.
y 36 x2
84.
x dy 1 36 x2122x dx 2 36 x2 dy
p 75
1 x 4
p p8 p7
75
x dx 36 x2
8 7 1 75 4 4 4
1 1 1 dp dx 1 4 4 4
p dp because p is linear
Problem Solving for Chapter 3 2. (a)
dV 3x2dx 3x2 x V x x3 x3 3x2x 3xx2 x3
V dV 3xx2 x3 3xx x2x
x, where → 0 as x → 0. (b) Let
y fx. Then → 0 as x → 0. x
Furthermore, y dy y fxdx x.
445
446
Chapter 3
Applications of Differentiation
4. Let hx gx f x, which is continuous on a, b and differentiable on a, b. ha 0 and hb gb f b.
y
By the Mean Value Theorem, there exists c in a, b such that
g
f
hb ha gb f b hc . ba ba
x a
b
Since hc gc fc > 0 and b a > 0, gb f b > 0 ⇒ gb > f b. 6. (a) f 2ax b, f 2a 0. No points of inflection. (b) f 3ax2 2bx c, f 6ax 2b 0 ⇒ x
b . One point of inflection. 3a
(c) y kyL y kLy ky2 y kLy 2kyy kyL 2y L If y , then y 0 and this is a point of inflection because of the analysis below. 2 y′′:
++++++
−−−−−−
L y= 2
8.
d 5 13
θ 12
x d 132 x2, sin . d Let A be the amount of illumination at one of the corners, as indicated in the figure. Then A
kI kIx sin 132 x2 132 x232
Ax kI
x2 169321 x
32 x
2
169122x
169 x23
⇒ x2 16932 3x2x2 16912 x2 169 3x2 2x2 169 x
13 2
9.19 feet
By the First Derivative Test, this is a maximum.
0
Problem Solving for Chapter 3 10. Let T be the intersection of PQ and RS. Let MN be the perpendicular to SQ and PR passing through T. Let TM x and TN b x. SN MR bx MR ⇒ SN bx x x NQ PM bx PM ⇒ NQ bx x x SQ
bx bx d MR PM x x
1 1 bx 1 b x2 1 2x2 2bx b2 Ax Area dx d b x d x d 2 2 x 2 x 2 x
1 x4x 2b 2x2 2bx b2 Ax d 2 x2
Ax 0 ⇒ 4x2 2xb 2x2 2bx b2 2x2 b2 x
b 2
bx b b2 d d 2 1d. x b2
Hence, we have SQ
Using the Second Derivative Test, this is a minimum. There is no maximum. S
Q
N b−x
b
T x P
d
M
R
12. (a) Let M > 0 be given. Take N M. Then whenever x > N M, you have f x x2 > M. (b) Let > 0 be given. Let M
1. Then whenever x > M 1,
you have x2 >
1 1 1 ⇒ 2 < ⇒ 2 0 < . x x
(c) Let > 0 be given. There exists N > 0 such that f x L < whenever x > N. 1 1 Let Let x . N. y If 0 < y <
f x L
1 1 1 , then < ⇒ x > N and N x N
f
1 L < . y
447
448
Chapter 3
Applications of Differentiation
14. Distance 42 x2 4 x2 42 f x fx
x 42 x2
4x 4 x2 42
0
x4 x2 42 x 442 x2 x2 16 8x x2 16 x2 8x 1616 x2 32x2 8x3 x 4 x 4 8x3 32x2 128x 256 128x 256 x2 The bug should head towards the midpoint of the opposite side. Without Calculus: Imagine opening up the cube: P x
Q
The shortest distance is the line PQ, passing through the midpoint.
v 16. (a) s
km m 1000 5 hr km v sec 18 3600 hr
v
20
40
60
80
100
s
5.56
11.11
16.67
22.22
27.78
d
5.1
13.7
27.2
44.2
66.4
dt 0.071s2 0.389s 0.727 (c)
(b) The distance between the back of the first vehicle and the front of the second vehicle is dt, the safe stopping distance. The first vehicle passes the given point in 5.5s seconds, and the second vehicle takes dss more seconds. Hence, T (d)
ds 5.5 . s s
T s 0.071s 0.389
6.227 s
10
Ts 0.071
6.227 6.227 ⇒ s2 s2 0.071 ⇒ s 9.365 msec
0
30
T 9.365 1.719 seconds
0
9.365 msec
1 5.5 T 0.071s2 0.389s 0.727 s s The minimum is attained when s 9.365 msec. 18. (a)
x
0
0.5
1
2
1 x
1
1.2247
1.4142
1.7321
1 x1 2
1
1.25
1.5
2
(e) d9.365 10.597 m
(b) Let f x 1 x. Using the Mean Value Theorem on the interval 0, x, there exists c, 0 < c < x, satisfying fc
1 f x f 0 1 x 1 . 21 c x0 x
Thus 1 x
3600 3.37 kmhr 1000
x x 1 < 1 because 1 c > 1. 21 c 2
Review Exercises for Chapter 4
Review Exercises for Chapter 4 2.
y
4.
u 3x du 3 dx
f′
2
3 3x
x
dx
2 3x133 dx 3x23 C 3
f
6.
x3 2x2 1 dx x2
x 2 x2 dx
8.
5 cos x 2 sec2 x dx 5 sin x 2 tan x C
1 1 x2 2x C 2 x 12. 45 mph 66 ftsec
10. f x 6x 1 fx
6x 1 dx 3x 12 C1
Since the slope of the tangent line at 2, 1 is 3, it follows that f2 3 C1 3 when C1 0. fx 3x 12 f x
3x 12 dx x 13 C2
f 2 1 C2 1 when C2 0.
30 mph 44 ftsec at a vt at 66 since v0 66 ftsec. a st t 2 66t since s0 0. 2 Solving the system vt at 66 44 a s(t t2 66t 264 2
f x x 13
we obtain t 245 and a 5512. We now solve 5512t 66 0 and get t 725. Thus, s
72 725 5512 2 5
2
66
725 475.2 ft.
Stopping distance from 30 mph to rest is 475.2 264 211.2 ft. 14. at 9.8 msec2 vt 9.8t v0 9.8t 40 st 4.9t2 40t s0 0 (a) vt 9.8t 40 0 when t
40 4.08 sec. 9.8
(b) s4.08 81.63 m (c) vt 9.8t 40 20 when t (d) s2.04 61.2 m
20 2.04 sec. 9.8
483
484
Chapter 4
Integration
16. x1 2, x2 1, x3 5, x4 3, x5 7 (a)
1 16 1 5 x 2 1 5 3 7 5 i1 i 5 5
(b)
x 2 1 5 3 7 210
5
1
i1
1
1
5
(c)
1
1
37
i
2x x
22 22 21 12 25 52 23 32 27 72 56
2 i
i
i1 5
(d)
x x i
1 2 5 1 3 5 7 3 5
i1
i2
1 18. y 9 x2, x 1, n 4 4
9 41 4 9 41 9 9 41 16 9 41 25
S4 1
22.5
9 419 9 41 16 9 41 25 9 9
s4 1
14.5 20. y x2 3, x Area lim
12
f ci x
8 6
lim
y
right endpoints 10
n
n→ i1
n→
2 n
n
2i n
i1
2
3
2 n
4 2 x 1
2 n 4i2 lim 3 n→ n i1 n2 lim
2 4 nn 12n 1 3n n n2 6
lim
43 n 1n2n 1 6 38 6 263
n→
n→
2
1 2 22. y x3, x 4 n Area lim
y 20
n
f ci x
15
n→ i1
n
1 2i lim 2 n→ i1 4 n
10
3
2 n
5 x
1 n 24i 24i2 8i3 lim 8 2 3 n→ 2n i1 n n n
4 n 3i 3i2 i3 1 2 3 n i1 n n n
lim
4 3 nn 1 3 nn 12n 1 1 n2n 12 n 2 3 n n 2 n 6 n 4
n→
4 6 4 1 15
1
lim
n→
2
2
3
4
Review Exercises for Chapter 4
24. (a) S m
b4b4 m2b4 b4 m3b4 b4 m4b4 b4 mb16 1 2 3 4 5mb8 2
s m0 (b) Sn
y
y = mx
b4 mb4b4 m2b4 b4 m3b4 b4 mb16 1 2 3 3mb8 2
f n n n n mn i n
bi
n
b
i1
sn
2
mbi
b
b
i1
n1 b bi m n n i0
bi f n
2 n
i1
n1 i0
mb2 nn 1 mb2n 1 2 n 2 2n
x=b
mb2 n 1n mb2n 1 i 2 n 2 2n i0
b b m n n
2
2 n1
mb2n 1 mb2n 1 1 2 1 1 lim mb bmb baseheight n→ n→ 2n 2n 2 2 2
(c) Area lim
b
(d)
mx dx
0
n
26.
lim
→ i1
12mx
b
2
0
1 mb2 2
3
3ci9 ci 2 xi
3x9 x2 dx
28.
y
1
6 5 3 2 1 x − 4 −3 −2 −1
1
2
3
4
−2
4
4
6
30. (a)
0
3
(b)
3
f x dx
0
16 x2 dx
1 42 8 2
6
f x dx
f x dx 4 1 3
3
6
f x dx
6
f x dx 1 1
3
4
(c)
f x dx 0
4
6
(d)
6
10 f x dx 10
3
12 12x2 dx x3 2
x4 2x2 5 dx
1
6 x 3
3
2
1
2
36.
2
2
38.
1
1 1 dx x2 x3
4
40.
1
5 x5
34.
1
1 1 x2 x3 dx 2 x 2x 1
4
4
t3
1 1 2
2 8 1 2 8 2 1
1
1
1
1 1
2
52 15
3 2t
2
2
t2 2 dt
2x3 5x 3
325 163 10 325 163 10
1
16 6 6 , (d) 9 3
sec2 t dt tan t
4
f x dx 101 10
3
3
32.
485
1
14 3
(semicircle)
x
486
Chapter 4
Integration
2
42.
x 4 dx
0
2
x2 4x 2
0
2
10
44.
1
x2 x 2 dx
6
5 4 y
2 1
3 x
−2 −1
2
1
3
4
2
5
x
−2
1
3
−1 −2
3
1
x1 x dx x12 x32 dx
48. Area
0
sec2 x dx
0
23x
2 2 4 3 5 15
32
3
2 x52 5
1
tan x
0
0
3
y 5 4
y
3 1
2
π 6
x 1
50.
1 20
2
x3 dx
0
8 x4
2 0
y
2
8
2
6
3 2 x
4
x3
2
( 3 2 , 2) x 1
52. Fx
1 x2
56.
x
2 1
8 1 1 2 24 3 3 2
y 7
46.
x3 x2 2x 3 2
1 x
2
2
54. Fx csc2 x
dx
x2 2 x2 dx
x3 1 2x C 3 x
π 3
x
10 7 9 3 6 2
Review Exercises for Chapter 4 58. u x3 3, du 3x2 dx
x2x3 3 dx
1 2 x3 312 3x2 dx x3 332 C 3 9
60. u x2 6x 5, du 2x 6 dx
x3 1 2x 6 1 2 1 dx x 6x 51 C C x2 6x 52 2 x2 6x 52 2 2x2 6x 5
62.
64.
66.
68.
70.
x sin 3x2 dx
cos x sin x
dx
1 1 sin 3x26x dx cos 3x2 C 6 6
sin x12 cos x dx 2sin x12 C 2sin x C
1 cot4 csc2 d cot 4csc2 d cot 5 C 5
1
x2x3 13 dx
0
6
72.
1 1 sec 2x tan 2x2 dx sec 2x C 2 2
sec 2x tan 2x dx
3
x3 133x2 dx
x2 8122x dx
1 3
x 1 dx 6 3x2 8
1
1 x3 14 12
0
6
3
13x
1 0
812
2
1 5 16 1 12 4
6 3
1 27 1 3
74. u x 1, x u 1, dx du When x 1, u 0. When x 0, u 1.
0
2
1
1
x2x 1 dx 2
u 12u du
0 1
2
u52 2u32 u12 du 2
0
4
76.
27u
72
4 2 u52 u32 5 3
1 0
32 105
sin 2x dx 0 since sin 2x is an odd function.
4
78. u 1 x, x 1 u, dx du When x a, u 1 a. When x b, u 1 b.
b
Pa, b
a
1155 3 1155 x 1 x32 dx 32 32
1155 32
1b
1 u3 u32 du
1a
1b
u92 3u72 3u52 u32 du
1a
1155 2u52 105u3 385u2 495u 231 32 1155
(a) P0, 0.25
16 105u
385u2 495u 231
(b) P0.5, 1
16 105u
385u2 495u 231
u52
u52
3
3
1155 2 11 2 2 92 6 72 2 52 u u u u 32 11 3 7 5
1b 1a
16 105u u52
0.025 2.5%
0.736 73.6%
0.75 1
0 0.5
3
1b 1a
385u2 495u 231
1b 1a
487
488
Chapter 4
2
80.
1.75 sin
0
Integration
t 2 t dt 1.75 cos 2 2
2
2 7 1.751 1 2.2282 liters
Increase is 7 5.1 1.9 0.6048 liters.
1
82. Trapezoidal Rule n 4:
0
1
Simpson’s Rule n 4:
0
x32 1 21432 21232 23432 1 0 0.172 2 dx 2 2 3x 8 3 14 3 12 3 342 2
x32 1 41432 21232 43432 1 0 0.166 2 dx 2 2 3x 12 3 14 3 12 3 342 2
Graphing utility: 0.166
84. Trapezoidal Rule n 4:
1 sin2 x dx 3.820
0
Simpson’s Rule n 4: 3.820 Graphing utility: 3.820
Problem Solving for Chapter 4
x
2. (a) Fx
sin t2 dt
2
x Fx (b) Gx x Gx
0
1.0
1.5
1.9
2.0
2.1
2.5
3.0
4.0
5.0
0.8048
0.4945
0.0265
0.0611
0
0.0867
0.3743
0.0312
0.0576
0.2769
1 x2
x
sin t2 dt
2
1.9
1.95
1.99
2.01
2.05
2.1
0.6106
0.6873
0.7436
0.7697
0.8174
0.8671
lim Gx 0.75
x→2
(c) F2 lim
x→2
Fx F2 x2
1 x→2 x 2
lim
x
sin t2 dt
2
lim Gx x→2
Since Fx sin x2, F2 sin 4 lim Gx. x→2
Note: sin 4 0.7568
Problem Solving for Chapter 4 4. Let d be the distance traversed and a be the uniform acceleration. We can assume that v0 0 and s0 0. Then at a vt at 1 s t at2. 2 st d when t
2da. 2da
2ad.
1 2ad 0 2
ad2 .
The highest speed is v a The lowest speed is v 0. The mean speed is
The time necessary to traverse the distance d at the mean speed is t
d ad2
2da
which is the same as the time calculated above. 6. (a)
y 100 80 60 40 20 x 0.2
0.4
0.6
0.8
1.0
(b) v is increasing (positive acceleration) on 0, 0.4 and 0.7, 1.0. (c) Average acceleration
v0.4 v0 60 0 150 mihr2 0.4 0 0.4
(d) This integral is the total distance traveled in miles.
1
vt dt
0
1 385 38.5 miles
0 220 260 240 240 65 10 10
(e) One approximation is a0.8
v0.9 v0.8 50 40 100 mihr2 0.9 0.8 0.1
(other answers possible)
489
490
Chapter 4
Integration
x
8.
x
f tx t dt
0
d dx
Thus,
0
x
x f t dt
x
t f t dt x
0
x
0
x
f tx t dt x f x
0
x
f t dt
t f t dt
0
x
f t dt x f x
0
f t dt
0
Differentiating the other integral, d dx
x
x
0
x
f v dv dt
0
f v dv.
0
Thus, the two original integrals have equal derivatives,
x
x
f tx t dt
0
t
0
f v dv dt C
0
Letting x 0, we see that C 0.
1
10. Consider
2 3
1
x dx x32
0
0
2 . The corresponding 3
Riemann Sum using right-hand endpoints is Sn Thus, lim
1 n
1 n
2 . . . n
1
1 2 . . . n n32
1 2 . . . n
n32
n→
3
12. (a) Area
nn
2 . 3
3
3
9 x2 dx 2
9 x2 dx
y
0
2 9x
x3 3
10
3
8 7 6 5 4 3 2 1
0
2 27 9 36 (b) Base 6, height 9. Area
2 2 bh 69 36. 3 3
−4
x
−2 −1
1 2
4 5
(c) Let the parabola be given by y b2 a2x2, a, b > 0.
ba
Area 2
b2 a2x2 dx
0
2
b2x
y
a2
x3 3
ba b2
0
a a3 a
2 b2
2
b
b
3
a 3 a 34 a
2 Base
b3
1 b3
b3 −
2b , height b2 a
Archimedes’ Formula: Area
2 2b 2 4 b3 b 3 a 3 a
b a
b a
x
Problem Solving for Chapter 4 14. (a) 1 i3 1 3i 3i2 i3 ⇒ 1 i3 i3 3i2 3i 1 3i2 3i 1 i 13 i3
(b) n
3i
3i 1
2
i1
n
i 1
i3
3
i1
23 13 33 23 . . . n 13 n3 n 13 1 n
Hence, n 13
3i
2
3i 1 1
i1
(c) n 13 1
n
3i
2
3i 1
i1 n
3i
⇒
2
n
i
2
2
i1
3nn 1 n 2
3nn 1 n 2
n3 3n2 3n
i1
⇒
n
3i
2n3 6n2 6n 3n2 3n 2n 2
2n3 3n2 n 2
nn 12n 1 2
nn 12n 1 6
i1
20
16. (a) C 0.1
12 sin
8
18
(b) C 0.1
12 sin
10
t 8 14.4 t 8 dt cos 12 12
0
0
b b
0 b
0
Then,
f x dx. f x f b x
f b u du f b u f u f b u du f b u f u f b x dx f b x f x
b
2A
0
f x dx f x f b x
b
0
f b x dx f b x f x
b
1 dx b.
0
b Thus, A . 2
1
(b) b 1 ⇒
0
14.4 1 1 $9.17 18 10
3 14.4 3 10.8 6 $3.14 14.4 2 2
Let u b x, du dx. A
8
Savings 9.17 3.14 $6.03. b
t 8 14.4 t 8 6 dt cos 0.6t 12 12
18. (a) Let A
20
sin x 1 dx sin1 x sin x 2
491
C H A P T E R Integration
4
Section 4.1
Antiderivatives and Indefinite Integration . . . . . . . . . 177
Section 4.2
Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
Section 4.3
Riemann Sums and Definite Integrals . . . . . . . . . . . 188
Section 4.4
The Fundamental Theorem of Calculus . . . . . . . . . . 192
Section 4.5
Integration by Substitution . . . . . . . . . . . . . . . . . 197
Section 4.6
Numerical Integration
Review Exercises
. . . . . . . . . . . . . . . . . . . 204
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
C H A P T E R Integration Section 4.1
4
Antiderivatives and Indefinite Integration
Solutions to Odd-Numbered Exercises
1.
d 3 d 9 C 3x3 C 9x4 4 dx x3 dx x
3.
d 1 3 x 4x C x2 4 x 2x 2 dx 3
5.
dy 3t2 dt
7.
dy x3 2 dx
y t3 C Check:
x x
13.
1 dx 2x3
15.
11.
Integrate
Simplify
x 4 3
3 4 3 x C 4
4 3
x1 2 C 1 2
1 x2 C 2 2
x2 3x C 2
x
17.
21.
x2 3 dx
3 x5 3 C x5 3 C 5 3 5
d 3 5 3 3 x2 x C x2 3 dx 5
C
2x 3x 2dx x 2 x3 C
Check:
d 1 4 x 2x C x3 2 dx 4
3 x2 dx
d 2 5 2 x C x3 2 dx 5
1 C 4x2
1 x3 2 dx x 4 2x C 4
Check:
2
d x2 3x C x 3 dx 2
C
x3 2 dx
1 3 x dx 2
x 3dx
Check:
23.
x1 3 dx
dx
Check:
19.
3 x dx
1
Check:
Rewrite
9.
2 y x5 2 C 5
d 3 t C 3t2 dt
Given
2 x3 2 2x 1 dx x5 2 x2 x C 5
Check:
25.
d 2 x x3 C 2x 3x 2 dx
1 dx x3
Check:
d 2 5 2 x x2 x C x3 2 2x 1 dx 5
x3 dx
x2 1 C 2C 2 2x
d 1 1 C 3 dx 2x2 x
177
178
27.
Chapter 4
x2 x 1 dx x
Check:
29.
Integration
2 2 2 x3 2 x1 2 x1 2 dx x5 2 x3 2 2x1 2 C x1 23x2 5x 15 C 5 3 15
d 2 5 2 2 3 2 x2 x 1 x x 2x1 2 C x3 2 x1 2 x1 2 dx 5 3 x
x 13x 2 dx
3x2 x 2 dx
31.
1 x3 x2 2x C 2 Check:
y2 y dy
2 y5 2 dy y7 2 C 7
d 2 7 2 y C y5 2 y2 y dy 7
Check:
d 3 1 2 x x 2x C 3x2 x 2 dx 2 x 13x 2
33.
37.
dx
1 dx x C
Check:
d x C 1 dx
1 csc t cot t dt t csc t C
39.
d t csc t C 1 csc t cot t dt
tan2 y 1 dy
2 sin x 3 cos x dx 2 cos x 3 sin x C d 2 cos x 3 sin x C 2 sin x 3 cos x dx
Check:
Check:
41.
35.
sec2 sin d tan cos C d tan cos C sec2 sin d
Check:
43. f x cos x
sec2 y dy tan y C
y
d Check: tan y C sec2 y tan2 y 1 dy
3 2
C
3
2C
0
x 2 2
3
45. f x 2
47. fx 1 x2
f x 2x C
f x x
f )x)
2x
2
y
x3 3
f )x)
5
f )x)
f )x)
2x
x
3 2
x 3
2
1
2
3
x 3
2
3 1
Answers will vary.
f
2
Answers will vary.
2x 1 dx x2 x C
1 12 1 C ⇒ C 1
x3 3
4
3
dy 2x 1, 1, 1 dx y
x y
2
49.
x3 C 3
4
f′
C
3
y x2 x 1
Section 4.1
51.
dy cos x, 0, 4 dx y
Antiderivatives and Indefinite Integration
179
53. (a) Answers will vary. y
5
cos x dx sin x C
4 sin 0 C ⇒ C 4 y sin x 4
x
−3
5
−3
(b)
dy 1 x 1, 4, 2 dx 2 x2 xC 4
y
6
−4
42 2 4C 4
8 −2
2C x2 x2 4
y
55. fx 4x, f 0 6 f x
4x dx 2x 2 C
57. ht 8t3 5, h1 4 ht
8t3 5dt 2t4 5t C
f0 6 202 C ⇒ C 6
h1 4 2 5 C ⇒ C 11
f x 2x 2 6
ht 2t4 5t 11 61. f x x3 2
59. f x 2 f2 5
f4 2
f 2 10
f 0 0
fx
fx
2 dx 2x C1
f2 4 C1 5 ⇒ C1 1 fx 2x 1 f x
2x 1 dx x2 x C2
f 2 6 C2 10 ⇒ C2 4 f x x x 4 2
x3 2 dx 2x1 2 C1
fx f x
2 x
1.5t 5 dt 0.75t 2 5t C
h0 0 0 C 12 ⇒ C 12 ht 0.75t2 5t 12 (b) h6 0.7562 56 12 69 cm
C1
3
2x1 2 3 dx 4x1 2 3x C2
f x 4x1 2 3x 4 x 3x
x
2 f4 C1 2 ⇒ C1 3 2
f 0 0 0 C2 0 ⇒ C2 0
63. (a) ht
2
180
Chapter 4
Integration
65. f 0 4. Graph of f is given. (a) f4 1.0
(f) f is a minimum at x 3.
(b) No. The slopes of the tangent lines are greater than 2 on 0, 2. Therefore, f must increase more than 4 units on 0, 4.
(g)
y 6 4
(c) No, f 5 < f 4 because f is decreasing on 4, 5. (d) f is an maximum at x 3.5 because f3.5 0 and the first derivative test.
2 x −2
(e) f is concave upward when f is increasing on , 1 and 5, . f is concave downward on 1, 5. Points of inflection at x 1, 5. 67. at 32 ft sec2 vt
v0 st 32t v0 0 when t time to reach 32 maximum height.
32t 60dt 16t 2 60t C2 s
st 16t2 60t 6 Position function
The ball reaches its maximim height when
15 16 15 18 8
v0
2
60
32 550 v0
v02 v02 550 64 32
73. From Exercise 71, f t 4.9t2 10t 2. v t 9.8t 10 0 (Maximum height when v 0.)
9.8 dt 9.8t C1
9.8t 10
v0 v0 C1 ⇒ vt 9.8t v0
t
9.8t v0 dt 4.9t v0t C2 2
f 0 s0 C2 ⇒ f t 4.9t 2 v0t s0
f
10 9.8
10 9.8 7.1 m
a 1.6
st
v0
158 6 62.26 feet
vt
2
15 seconds 8
71. at 9.8
75.
v0
v0 187.617 ft sec
32t 60 t
32 1632 v02 35,200
vt 32t 60 0
f t
8
st 16t2 v0t
32 dt 32t C1
vt
6
−6
s0 6 C2
s
4
69. From Exercise 68, we have:
v0 60 C1 st
2
1.6 dt 1.6t v0 1.6t, since the stone was dropped, v0 0.
1.6t dt 0.8t2 s0
s20 0 ⇒ 0.8202 s0 0 s0 320 Thus, the height of the cliff is 320 meters. vt 1.6t v20 32 m sec
Section 4.1
0 ≤ t ≤ 5
77. xt t3 6t2 9t 2
Antiderivatives and Indefinite Integration
79. vt
(a) vt xt 3t 12t 9 2
xt
3t2 4t 3 3t 1t 3
t1 2
t > 0
vt dt 2t1 2 C
x1 4 21 C ⇒ C 2
at vt 6t 12 6t 2 (b) vt > 0 when 0 < t < 1 or 3 < t < 5.
xt 2t1 2 2 position function
(c) at 6t 2 0 when t 2.
1 1 at vt t3 2 3 2 acceleration 2 2t
v2 311 3 81. (a) v0 25 km hr 25
1000 250 m sec 3600 36
v13 80 km hr 80
1000 800 m sec 3600 36
83. Truck: vt 30 st 30t Let s0 0. Automobile: at 6
at a constant acceleration
vt 6t Let v0 0.
vt at C
st 3t2 Let s0 0.
v0 v13
At the point where the automobile overtakes the truck:
250 250 ⇒ vt at 36 36
30t 3t2
800 250 13a 36 36
0 3t2 30t 0 3tt 10 when t 10 sec.
550 13a 36 a st a
(b)
s13 85.
1 t
(a) s10 3102 300 ft (b) v10 610 60 ft sec 41 mph
550 275
1.175 m sec2 468 234
t 2 250 t 2 36
s0 0
275 132 250 13 189.58 m 234 2 36
1 mi hr5280 ft mi 22 ft sec 3600 sec hr 15 (a)
t
0
5
10
15
20
25
30
V1ft sec
0
3.67
10.27
23.47
42.53
66
95.33
V2ft sec
0
30.8
55.73
74.8
88
93.87
95.33
(c) S1t S2t
V1t dt
0.1068 3 0.0416 2 t t 0.3679t 3 2
V2t dt
0.1208t3 6.7991t2 0.0707t 3 2
In both cases, the constant of integration is 0 because S10 S20 0 S130 953.5 feet S230 1970.3 feet The second car was going faster than the first until the end.
(b) V1t 0.1068t2 0.0416t 0.3679 V2t 0.1208t2 6.7991t 0.0707
181
182
Chapter 4
Integration
87. at k vt kt k st t2 since v0 s0 0. 2 At the time of lift-off, kt 160 and k2t2 0.7. Since k2t2 0.7,
1.4k 1.4 1.4 160 v k k k t
1.4k 1602 ⇒ k
1602 1.4
18,285.714 mihr2 7.45 ftsec2. 89. True
91. True
93. False. For example,
95. fx
1,3x,
x
x dx
x dx
x dx because
x2 x3 C C1 3 2
x2 C 2
2
0 ≤ x < 2 2 ≤ x ≤ 5
x C1, 0 ≤ x < 2 f x 3x2 C2, 2 ≤ x ≤ 5 2 f 1 3 ⇒ 1 C1 3 ⇒ C1 2 f is continuous: Values must agree at x 2: 4 6 C2 ⇒ C2 2
0 ≤ x < 2 x 2, f x 3x 2 2, 2 ≤ x ≤ 5 2 The left and right hand derivatives at x 2 do not agree. Hence f is not differentiable at x 2.
Section 4.2 1.
5
5
5
i1
i1
i1
2i 1 2 i 1 21 2 3 4 5 5 35 4
3.
k
2
k0 9
7.
Area
1 1 1 1 1 158 1 1 2 5 10 17 85 9.
i1 20
15.
i1
58 3 8
1
3i
j
4
5.
k1
i 2 20
i1
2021 420 2
n
11.
2 n n i1
17.
i 1
j1
2i 2
c c c c c 4c 2i
20
2
3
i1
2in
13.
19
i
2
i1
192039 2470 6
3 n 3i 2 1 n i1 n
2
15
19.
ii 1
2
i1
15
i
2
3
i1
15
i
2
15
i
i1
21. sum seqx
>
Section 4.2
2 3, x, 1, 20, 1 2930 (TI-82)
i1
20
i
152162 151631 1516 2 4 6 2
2
3
i1
14,400 2,480 120
2020 1220 1 320 6
202141 60 2930 6
12,040 9 33 23. S 3 4 2 51 2 16.5
s 1 3 4 1 9 2
25 2
25. S 3 3 51 11
12.5
s 2 2 31 7
14 14 12 14 34 14 114 1 2 8 3 2 0.768 1 2 3 1 1 1 1 1 3 1 s4 0 0.518 4 4 4 2 4 4 4 8
27. S4
29. S5 1 s5
n→
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0.646 65 5 75 5 85 5 95 5 2 5 6 7 8 9 10
81n n n 4 1 814 lim n 2
31. lim
1 1 1 1 1 1 1 1 15 65 5 75 5 85 5 95 5 51 61 71 81 91 0.746
2
4
4
n→
81 81 2n3 n2 1 n4 4 4
18n nn 2 1 182 lim n n n 1821 9 2
33. lim n→
2
2i 1 1 n 1 nn 1 n2 n Sn 2 2i 1 2 2 2 n n n 2 n i1 i1 n
35.
2
n→
S10
12 1.2 10
S100 1.02 S1000 1.002 S10,000 1.0002 6kk 1 6 n 2 6 nn 12n 1 nn 1 3 k k 3 3 n n k1 n 6 2 k1 n
37.
6 2n2 3n 1 3n 3 1 22n2 2 Sn n2 6 n
S10 1.98 S100 1.9998 S1000 1.999998 S10,000 1.99999998
n n
39. lim
n→ i1
16i 2
lim
n→
16 n 16 nn 1 n2 n i lim 2 lim 8 n→ n n→ n2 i1 2 n2
Area
8 lim 1 n1 8 n→
183
184
Chapter 4
Integration
1 1 n1 1 n 1n2n 1 i 12 lim 3 i 2 lim 3 3 n→ i1 n n→ n i1 n→ n 6 n
41. lim
lim
n→
1 2n3 3n2 n 1 2 3n 1n2 lim n→ 6 6 n3 1
1 nn 1 2
1 nn 2 lim n 1 n i 2 lim n n n n
43. lim
i
45. (a)
n
1
2
n→ i1
n→
1
i1
n
1
n→
i1
y
(e)
x
31
2 lim 1 n 2n n 21 21 3 2
2
n→
5
10
50
100
3
sn
1.6
1.8
1.96
1.98
2
Sn
2.4
2.2
2.04
2.02
1
i 12n 2n lim n4 i 1 n
1
(b) x
n
(f) lim
x
n→
3
n→
i1
Endpoints:
lim
n→
2n < 22n < . . . < n 12n < n2n 2 n
f x
i n n n
2
lim
n→
(c) Since y x is increasing, f mi f xi1 on xi1, xi. sn
2
i1
x
n4 nn 2 1
lim
2n 1 2 n
n→
i1
n 2 2 i 1 n n i1
2 n
2n n 1 n4 2
4 n i n→ n2 i1 lim
i1
2i 2 f n
lim
i1
n
i1
4 nn 1 n lim 2 n→ n 2
20 2 n n
0
49. Because the curve is concave upward, the midpoint approximation will be less than the actual area: <
dx
Section 4.3
51. f x
1 x4
53.
is not integrable on the interval 3, 5 and f has a discontinuity at x 4.
Riemann Sums and Definite Integrals
y
4 3 2 1 1
2
3
x
4
a. A 5 square units 55.
y
2 3 2
1 1 2
x 1 2
1
3 2
2
1
d.
1 2 sin x dx 12 1 2
0
3
57.
x 3 x dx
0
n
4
8
12
16
20
Ln
3.6830
3.9956
4.0707
4.1016
4.1177
Mn
4.3082
4.2076
4.1838
4.1740
4.1690
Rn
3.6830
3.9956
4.0707
4.1016
4.1177
4
8
12
16
20
Ln
0.5890
0.6872
0.7199
0.7363
0.7461
Mn
0.7854
0.7854
0.7854
0.7854
0.7854
Rn
0.9817
0.8836
0.8508
0.8345
0.8247
2
59.
sin2 x dx
0
n
61. True
63. True
65. False
2
0
67. f x x2 3x, 0, 8 x0 0, x1 1, x2 3, x3 7, x4 8 x1 1, x2 2, x3 4, x4 1 c1 1, c2 2, c3 5, c4 8 4
f c x f 1 x i
1
f 2 x2 f 5 x3 f 8 x4
i1
41 102 404 881 272
x dx 2
191
192
Chapter 4
69. f x
1,0,
Differentiation
x is rational x is irrational
is not integrable on the interval 0, 1. As → 0, f ci 1 or f ci 0 in each subinterval since there are an infinite number of both rational and irrational numbers in any interval, no matter how small. 71. Let f x x2, 0 ≤ x ≤ 1, and xi 1 n. The appropriate Riemann Sum is
f c x n n
n
i
i
2
i
i1
i1
lim
n→
1 n 1 3 i 2. n n i1
1 2 1 1 22 32 . . . n2 lim 3 n→ n n3
n→
1. f x
0
n2n 1n 1 6
1 1 1 2n2 3n 1 1 lim n→ 3 6n2 2n 6n2 3
lim
Section 4.4
The Fundamental Theorem of Calculus
4 x2 1
5
5
3. f x x x2 1
2
4 dx is positive. x2 1
−5
2
5
x x2 1 dx 0
−5
−5
−2
2x dx x2
t2 2 dt
t3 2t
2t 12 dt
0
101
0
1
9.
1
1
1
4
1
u2 du
u
34t
t1 3 t2 3 dt
1
3
23.
2t
1 1
43t
3
2t2 t
34t
4 3
3 2
1 0
4 1 21 3 3
23 4
4u1 2
3 t5 3 5
23u
1 x2 2 3 2 x 3 2 3
x x1 2 dx
0
0
0
1
3 2 0
3
1
23 4 32
4 4
1 0
1 1 2 1 3 2 3 18
2x 3 dx split up the integral at the zero x
3 2
3x x2
4
27 34 53 20
3
3 2x dx
0
0
34 2 34 2 4
3 2
2x 3 dx
0
4 3
1
0
21.
1
1
1
x x 1 dx 3 3
0
0
2
2
u1 2 2u1 2 du
1
x2 2x
13 2 31 2 103
4
x 2 dx
23 2 3 1 21
3 t 2 dt
1
19.
4t2 4t 1 dt
3 3 1 dx x x2 x
1
17.
1
0
2
15.
7.
1
0
13.
1
3
1
11.
0
1
1
5.
5
x2 3x
3 3 2
3 2
92 49 0 9 9 94 29 292 49 29
12 2 52
Section 4.4
3
25.
0
2
x 2 4 dx
3
4 x 2 dx
0
8
27.
x3 3
2 0
4 sec tan d 4 sec
6
6
3
10,000t 6 dt 10,000
0
3
37. A
3
3 3
3
2
3
233
3 0
2
0
x x2 dx
0
1
35. A
$135,000
x 5 x10 2x
2 3x1 2 x3 2 dx 2x3 2 x5 2 5 0
cos x dx sin x
0
3
42 42 0
t2 6t 2
3
3 x x dx
0
39. A
1 0 1 2
0
2
sec2 x dx tan x
3
33.
3
3
3
8 8 9 12 8 3 3
6
x3 4x
1 sin x dx x cos x
6
31.
23 3
0
29.
x 2 4 dx
2
4x
The Fundamental Theorem of Calculus
3
3
0
0
x2 x3
3 1
2
12 3 5
41. Since y ≥ 0 on 0, 2,
1
2
A
2
0
x3 x dx
2
43. Since y ≥ 0 on 0, 2,
45.
x4 x2 4 2
2 0
0
4 2 6.
3x2 1 dx x3 x
0
A
1 6
0
x 2 x dx
x2 4x3
3 2 2
2
0
f c2 0
6 8 2 3
c 2 c
3 4 2 3
c 2 c 1
2 0
8 2 10.
2
8 2 3
3 4 2 1 3
c 12 6 34
2
6 34 2 64 2 c 1 ± 3
c 1 ±
2
c 0.4380 or c 1.7908
193
194
Chapter 4
4
47.
Integration
4
2 sec2 x dx 2 tan x
4
21 21 4
4
4 4 4
f c
2 sec2 c
8
sec2 c
4
sec c ±
2
c ± arcsec
2
± arccos
49.
2
1 2 2
2
4 x2 dx
Average value 4 x2
51.
1 0
2
± 0.4817
2
1 1 4x x3 4 3
2
1 4
8 38 8 38 38
sin x dx
0
1 cos x
0
3 3
, 8 3
(
5
(
2
3 3
, 8 3
(0.690, π2 ( − 2
(2.451, π2 (
−1
6
57.
0
0
6
2 f x dx
0
0
f x dx Fb Fa.
a
3 2
f x dx area of region A 1.5
6
b
then
2
2
3
53. If f is continuous on a, b and F x f x on a, b,
2
2
0
6
f x dx f x dx
f x dx 1.5 5.0 6.5
2
6
2 dx
f x dx
0
12 3.5 15.5 61. (a) F x k sec2 x
(b)
F 0 k 500
1 3 0
3
500 sec2 x dx
0
F x 500 sec2 x
3
1500 tan x
0
1500 3 0
826.99 newtons 827 newtons
63.
1 50
5
0
(
−1
2
x 0.690, 2.451
59.
2
−3
8 8 2 3 ± 1.155. when x2 4 or x ± 3 3 3
sin x
55.
(−
8 3
2 Average value
0.1729t 0.1552t2 0.0374t3 dt
1 0.08645t2 0.05073t3 0.00935t4 5
5 0
0.5318 liter
Section 4.4
65. (a)
The Fundamental Theorem of Calculus
(b)
1
0
10
24
0
−1
24 0
The average value of S appears to be g.
The area above the x-axis equals the area below the x-axis. Thus, the average value is zero. 67. (a) v 8.61 104t3 0.0782t2 0.208t 0.0952 (b)
90
− 10
70 − 10
60
(c)
vt dt
0
8.61 4 10
4t 4
x
69. Fx
t 5 dt
0
t2 5t 2
x 0
0.0782t3 0.208t 2 0.0952t 3 2
x2 5x 2
60 0
2476 meters
x
71. Fx
x
10 2 dv 1 v
1
10 1 10 10 1 x x
F2
4 52 8 2
F5
25 25 55 2 2
F2 10
12 5
F8
64 58 8 2
F5 10
45 8
F8 10
78 354
x
73. Fx
cos d sin
1
x 1
x
sin x sin 1
75. (a)
t 2 dt
0
F2 sin 2 sin 1 0.0678
10 v
10v2 dv
x
t2 2t 2
0
x 1
1 x2 2x 2
d 1 2 (b) x 2x x 2 dx 2
F5 sin 5 sin 1 1.8004 F8 sin 8 sin 1 0.1479
77. (a)
x 3 t
dt
8
(b)
4t
3
x
4 3
8
79. (a)
x 4
d 3 4 3 3 x x 12 x1 3 dx 4
2
(b)
sec2 t dt tan t
x x 4
tan x 1
d tan x 1 sec2 x dx
x
x
81. F x
x
3 3 x 4 3 16 x 4 3 12 4 4
t 2 2t dt
F x x2 2x
83. F x
1
x
t 4 1 dt
F x x4 1
85. F x
t cos t dt
0
F x x cos x
195
196
Chapter 4
Integration
x2
87. Fx
4t 1 dt
Alternate solution:
x
x2
Fx
2t t 2
x2
4t 1 dt
x
x
2x 22 x 2 2x2 x
0
x
8x 10
x2
4t 1 dt
0
x
F x 8
4t 1 dt
x2
4t 1 dt
0
4t 1 dt
0
F x 4x 1 4x 2 1 8
sin x
89. Fx
t dt
0
23t 3 2
sin x 0
2 sin x3 2 3
F x sin x1 2 cos x cos x sin x
x3
91. Fx
sin t 2 dt
0
F x sinx32
3x2 3x2 sin x 6
Alternate solution
sin x
Fx
t dt
0
F x sin x
d sin x sin xcos x dx
x
93. gx
f t dt
0
1 1 g0 0, g1 , g2 1, g3 , g4 0 2 2
45t
x
5000 25
2
(b)
f
g
2
3
x
5 4
0
12 5 4 x 1000125 12x5 4 5
C1 1000125 121 $137,000 C5 1000125 1255 4 $214,721
x 1
t1 4 dt
0
5000 25 3
y
1
95. (a) Cx 5000 25 3
C10 1000125 12105 4 $338,394
4
−1 −2
g has a relative maximum at x 2.
1
99. False;
97. True
1
0
x2 dx
1
1
x2 dx
x2 dx
0
Each of these integrals is infinite. f x x2 has a nonremovable discontinuity at x 0.
1 x
101. f x
0
1 dt t 1 2
x
1 dt 2 t 1 0
By the Second Fundamental Theorem of Calculus, we have f x
1 1 1 2 1 x2 1 x2 x 1
1 1 0. 1 x2 x2 1
Since f x 0, f x must be constant.
Section 4.5 103. xt t 3 6t 2 9t 2 xt 3t 2 12t 9 3t 2 4t 3 3t 3t 1
5
Total distance
xt dt
0
5
3 t 3t 1 dt
0
1
3
t 2 4t 3 dt 3
3
0
5
t 2 4t 3dt 3
1
t 2 4t 3dt
3
4 4 20 28 units
4
105. Total distance
xt dt
1
4
vt dt
1 4 1
1
t
dt
2t12
4 1
22 1 2 units
Section 4.5
1. 3. 5.
7.
f g x
g x dx
u g x
du g x dx
5x2 1210x dx
5x2 1
10x dx
x2 1
2x dx
tan x
sec2 x dx
x
x2 1
dx
tan2 x sec2 x dx
1 2x4 2 dx
Check:
9.
Integration by Substitution
1 2x5 C 5
d 1 2x5 C 21 2x4 dx 5
9 x2122x dx
Check:
9 x232 2 C 9 x232 C 32 3
2 d 2 9 x232 C dx 3 3
3
29 x2122x 9 x22x
Integration by Substitution
197
198
11.
Chapter 4
x3x4 32 dx
Check:
13.
17.
t2 232 1 2 1 t2 232 C C t 2122t dt 2 2 32 3
32t2 2122t d t2 232 C t2 212t dt 3 3
5 5 1 x2132x dx 2 2
d 15 15 1 x243 C dx 8 8
1 x243 15 C 1 x243 C 43 8
4
3 1 x2 31 x2132x 5x1 x213 5x
1 x 1 d C 21 x232x dx 41 x22 4 1 x23
x2 1 1 1 x31 1 dx 1 x323x2 dx C C 1 x32 3 3 1 31 x3
x
1
d 1 1 x2 C 11 x323x2 3 dx 31 x 3 1 x32
1 x2
1 1 1 x22 x 1 dx 1 x232x dx C C 2 3 1 x 2 2 2 41 x22
Check:
27.
5x1 x213 dx
Check:
25.
Check:
23.
1 1 x3 15 x3 15 x3 143x2 dx C C 3 3 5 15
d x3 15 5x3 143x2 C x2x3 14 dx 15 15
tt2 2 dt
Check:
21.
Check:
19.
x2x3 14 dx
Check:
x4 33 1 1 x4 33 C C x4 324x3 dx 4 4 3 12
d x4 33 3x4 32 3 C 4x x4 32x3 dx 12 12
Check:
15.
Integration
dx
1 1 1 x212 1 x2122x dx C 1 x2 C 2 2 12
d 1 x 1 x212 C 1 x2122x 1 x2 dx 2 1 t
t1 dt 1 1t t1 dt 1 41t 3
3
2
4
2
1 1 1 1t 4 d C 4 1 dt 4 4 t
t1 t1 1 1t 3
2
1 1 1 2x12 dx 2x12 2 dx C 2x C 2 2 12 2x
Check:
d 2x C 212x122 1 dx 2x
C
2
3
Section 4.5
29.
31.
33.
x2 3x 7 dx x
2 2 x32 3x12 7x12 dx x52 2x32 14x12 C xx2 5x 35 C 5 5
Check:
d 2 52 x2 3x 7 x 2x32 14x12 C dx 5 x
2 dt t
Check:
d 14 2 t t2 C t 3 2t t 2 t dt 4 t
t2 t
1 t3 2t dt t 4 t2 C 4
9 yy dy
35. y
Check:
9y12 y32 dy 9
23y 52y 32
52
2 C y3215 y C 5
d 2 32 d 2 6y32 y52 C 9y12 y32 9 yy y 15 y C dy 5 dy 5
4x
4x 16 x2
dx
37. y
4 x dx 2 16 x2122x dx 4
Integration by Substitution
x2 2 16 12x C 2
2 12
2x2 416 x2 C
39. (a) 3
1 x2 2x 322x 2 dx 2
1 x2 2x 31 C 2 1
1 C 2x2 2x 3
x4 x2 dx
x 2
dy x4 x2, 2, 2 dx y
−2
x1 dx x2 2x 32
(b)
y
−1
1 2
1 4 x2122x dx 2
2
1
34 x232 C 34 x232 C
1 2, 2: 2 4 2232 C ⇒ C 2 3 1 y 4 x232 2 3 2
−2
2
−1
41.
45.
sin x dx cos x C
1 1 1 1 1 cos d cos 2 d sin C 2
43.
sin 2x dx
1 1 sin 2x2x dx cos 2x C 2 2
199
200
Chapter 4
Integration
1 2
49.
1 tan5 x C tan5 x C 5 5
51.
47.
1 1 sin 2x2 1 sin 2x2 cos 2x dx C sin2 2x C 2 2 2 4
sin 2x cos 2x dx
sin 2x cos 2x dx sin 2x cos 2x dx
tan4 x sec2 x dx
1 1 cos 2x2 1 cos 2x2 sin 2x dx C1 cos2 2x C1 2 2 2 4 2 sin 2x cos 2x dx
1 2
sin 4x dx
OR
1 cos 4x C2 8
csc2 x dx cot x3csc2 x dx cot3 x
53.
OR
cot2 x dx
cot x2 1 1 1 1 C C tan2 x C sec2 x 1 C sec2 x C1 2 2 cot2 x 2 2 2
csc2 x 1 dx cot x x C
55. f x
cos
x x dx 2 sin C 2 2
Since f 0 3 2 sin 0 C, C 3. Thus, f x 2 sin 57. u x 2, x u 2, dx du
xx 2 dx
u 2u du u32 2u12 du
4 2 u52 u32 C 5 3
2u32 3u 10 C 15
2 x 232 3x 2 10 C 15
2 x 2323x 4 C 15
59. u 1 x, x 1 u, dx du
x21 x dx 1 u2u du u12 2u32 u52 du
23u
2u32 35 42u 15u2 C 105
2 1 x32 35 421 x 151 x2 C 105
2 1 x3215x2 12x 8 C 105
32
4 2 u52 u72 C 5 7
x 3. 2
Section 4.5
1 1 61. u 2x 1, x u 1, dx du 2 2
x2 1 dx 2x 1
12u 1 2 1 1 du 2 u
1 12 2 u u 2u 1 4 du 8 1 u32 2u12 3u12 du 8
1 2 52 4 32 u u 6u12 C 8 5 3
u12 2 3u 10u 45 C 60
2x 1
60
32x 12 102x 1 45 C
1 2x 112x2 8x 52 C 60
1 2x 13x2 2x 13 C 15
63. u x 1, x u 1, dx du
x dx x 1 x 1
u 1 du u u
du
u 1 u 1
u u 1
1 u12 du u 2u12 C u 2u C x 1 2x 1 C x 2x 1 1 C x 2x 1 C1 where C1 1 C. 65. Let u x2 1, du 2x dx.
1
1
xx2 13 dx
1 2
1
1
x2 132x dx
8x
67. Let u x3 1, du 3x2 dx
2
1
2x2x3 1 dx 2
1 3
2
23 x
x3 1123x2 dx
1
3
132 32
2 1
4 3 x 132 9
2 1
4 27 22 12 982 9
1
2
14
1 1
0
Integration by Substitution
201
202
Chapter 4
Integration
69. Let u 2x 1, du 2 dx.
4
0
1 1 dx 2 2x 1
71. Let u 1 x, du
9
1
4
2x 1122 dx 2x 1
0
4
9 1 2
0
1 dx. 2x
1 dx 2 2 x 1 x
9
1
1 x 2
21 x dx 1 2x
9 1
1 1 1 2 2
73. u 2 x, x 2 u, dx du When x 1, u 1. When x 2, u 0.
2
1
23x dx 32 sin23x
0
0
u32 u12 du
1
2
cos
0
2 u 1 u du
1
2
75.
0
x 12 x dx
5u 2
52
2 u32 3
0 1
5 3 154 2
2
3 3 33 2 2 4
77. u x 1, x u 1, dx du When x 0, u 1. When x 7, u 8.
7
Area
8
3 x 1 dx x
0
3 u du u 1
1 8
u43 u13 du
1
79. A
2 sin x sin 2x dx 2 cos x
0
81. Area
4
83.
0
23
2
2x dx 2
23
sec2
2
sec2
73
3 3 12 384 7 7 4
3 u43 4
0
23
2
1
2 3 1
7
85.
xx 3 dx 28.8
3
144 5
2x 12 dx
2x 12 dx
0
−1 0
1 1 4 1 2x 12 2 dx 2x 13 C1 x3 2x2 x C1 2 6 3 6 4 4x2 4x 1 dx x3 2x2 x C2 3
1 They differ by a constant: C2 C1 . 6
d 7.377 6
4
8
0
−1
cos 5
5
3
87.
15
−1
1209 28
4
2x 12 dx 2 tan2x
x 10 dx 3.333 3 2x 1
8
1 cos 2x 2
3
89.
37u
−1
Section 4.5
91. f x x2x2 1 is even.
2
93. f x xx2 13 is odd.
2
2
x2x2 1 dx 2
x4 x2 dx 2
0
325 3 8
2
2
95.
0
x2 dx
3 x3
0
2 0
2
x2 dx
x2 dx
0
2
0
4
2
2
0
0
xx2 13 dx 0
272 15
2
8 3
(b)
2
(d)
2
4
x3 6x2 2x 3 dx
2
x2 dx 2
4
4
16 3
2
3x2 dx 3
x2 dx 8
0
4
x3 2x dx
x2 dx
0
0
8 x2 dx 3
4
97.
2
2
x2 dx
(c)
x5 x3 5 3
8 ; the function x2 is an even function. 3 2
(a)
Integration by Substitution
4
6x2 3 dx 0 2
6x2 3 dx 2 2x3 3x
0
4 0
232
99. Answers will vary. See “Guidelines for Making a Change of Variables” on page 292.
2
101. f x xx2 12 is odd. Hence,
2
xx2 12 dx 0.
k dV dt t 12
103.
Vt
k k C dt t 12 t1
V0 k C 500,000 1 V1 k C 400,000 2 Solving this system yields k 200,000 and C 300,000. Thus, Vt
200,000 300,000. t1
When t 4, V4 $340,000.
105.
1 ba (a) (b) (c)
b
74.50 43.75 sin
a
262.5 t 1 74.50t cos 3 6
t 1 262.5 t dt 74.50t cos 6 ba 6
262.5 t 1 74.50t cos 3 6
3 0
1 262.5 t 74.50t cos 12 6
6 3
b a
1 262.5 223.5 102.352 thousand units 3
1 262.5 447 223.5 102.352 thousand units 3
12 0
1 262.5 262.5 894 74.5 thousand units 12
203
204
Chapter 4
107.
1 ba (a) (b)
Integration
b
1 1 1 cos60 t sin120t b a 30 120
2 sin60 t cos120 t dt
a
1240 0
30 1 2 1201 301
240
2x 12 dx
1 1 2x 12 2 dx 2x 13 C 2 6
(c)
1 1 1 cos60 t sin120 t 130 0 30 120
109. False
10
10
10
ax3 bx2 cx d dx
30 301 0 amps
30
1
10
10
ax3 cx dx
10
Odd 113. True
2 5 2 2 1.382 amps
111. True
130 0
1
60
0
1 1 1 cos60 t sin120 t 1240 0 30 120
10
bx2 d dx 0 2
bx2 d dx
0
Even
4 sin x cos x dx 2 sin 2x dx cos 2x C 115. Let u x h, then du dx. When x a, u a h. When x b, u b h. Thus,
b
bh
f x h dx
a
x2 dx
0 2
Trapezoidal:
x2 dx
0 2
Simpson’s:
x2 dx
0
2
3. Exact:
x3 dx
0 2
Trapezoidal:
x3 dx
0 2
Simpson’s:
f x dx.
ah
Numerical Integration 2
1. Exact:
bh
f u du
ah
Section 4.6
0
a
30 0 301 4 1.273 amps
160
1 1 1 cos60 t sin120 t 160 0 30 120
b
x3 dx
3x 1
2
3
0
8 2.6667 3
2
2
1 1 02 4 2 1 1 04 6 2
212 2
32
2
212 4
32
2
213 2
32
3
213 4
32
3
4
4.000
3
3
x4
2 0
1 1 02 4 2 1 1 04 6 2
11 2.7500 4
8 2.6667 3
17 4.2500 4
24 4.0000 6
22 22
23 23
Section 4.6
2
5. Exact:
x3 dx
0 2
Trapezoidal:
x3 dx
0 2
Simpson’s:
x3 dx
0
14x
2 0
x dx
x dx
4
x dx
4
2
1 2
Trapezoidal:
23x
9
32
4
1
2
1
3
24
2 3
34
4
3
34
213 2 3
54
213 4
3
2
54
3
64
2
3
64
2 3
74
4
3
74
8 4.0625
3
8 4.0000
16 38 12.6667 3 3
18
378 2 214 2 478 2 264 2 578 2 314 2 678 3
378
1 1 dx x 12 x1
2 1
478
21 4
678 3 12.6667
31 4
1 1 1 0.1667 3 2 6
578
26 4
1 1 1 1 1 1 1 2 dx 2 2 x 12 8 4 54 12 32 12 74 12 9
1 1 32 8 32 1 0.1676 8 4 81 25 121 9
1 1 1 1 1 1 1 4 dx 2 4 x 12 12 4 54 12 32 12 74 12 9
1 x3 dx
0 2
1 x3 dx
0
1 1 64 8 64 1 0.1667 12 4 81 25 121 9
2
Simpson’s:
2
5 24 24
11. Trapezoidal:
3
24
5 22 16
Simpson’s:
2
12.6640
9
9. Exact:
3
1 1 04 12 4
9
Simpson’s:
1 1 02 8 4
4
Trapezoidal:
4.0000
4
9
7. Exact:
Numerical Integration
1 1 2 1 18 2 2 2 1 278 3 3.283 4 1 1 4 1 18 2 2 4 1 278 3 3.240 6
Graphing utility: 3.241
1
13.
1
x 1 x dx
0
x1 x dx
0
x1 x dx
x1 x dx
1
Trapezoidal:
0
1
Simpson’s:
0
Graphing utility: 0.393
1 02 8
1 04 12
121 21 2 341 43 0.342
1 1 1 2 4 4
121 21 4 341 43 0.372
1 1 1 2 4 4
205
206
Chapter 4
Integration
2
15. Trapezoidal:
cosx2 dx
2
8
0
2
2
cos 0 2 cos
4
2
2
2 cos
2
2 cos
2
2
2
2
2 cos
2
4 cos
2
4
2
2
2
cos
2
0.957
2
Simpson’s:
cosx2 dx
2
12
0
cos 0 4 cos
4
2
4
2
cos
2
0.978 Graphing utility: 0.977
1.1
17. Trapezoidal:
1 sin1 2 sin1.0252 2 sin1.052 2 sin1.0752 sin1.12 0.089 80
sin x2 dx
1 1.1
Simpson’s:
1 sin1 4 sin1.0252 2 sin1.052 4 sin1.0752 sin1.12 0.089 120
sin x2 dx
1
Graphing utility: 0.089
4
19. Trapezoidal:
x tan x dx
0
4
Simpson’s:
x tan x dx
0
2 2 3 3 02 tan 2 tan 2 tan 0.194 32 16 16 16 16 16 16 4
2 2 3 3 04 tan 2 tan 4 tan 0.186 48 16 16 16 16 16 16 4
Graphing utility: 0.186
21. (a)
23.
y
f x x3 fx 3x2
y = f ( x)
f x 6x f x 6 f 4x 0 x
a
b
(a) Trapezoidal: Error ≤
The Trapezoidal Rule overestimates the area if the graph of the integrand is concave up.
f x is maximum in 0, 2 when x 2. (b) Simpson’s: Error ≤ f 4x 0.
25. f x
2 in 1, 3 . x3
(a) f x is maximum when x 1 and f 1 2. Trapezoidal: Error ≤ f 4x
23 12n2
2 < 0.00001, n2 > 133,333.33, n > 365.15; let n 366.
24 in 1, 3 x5
(b) f 4x is maximum when x 1 and when f 41 24. Simpson’s: Error ≤
25 180n4
2 03 12 0.5 since 1242
24 < 0.00001, n4 > 426,666.67, n > 25.56; let n 26.
2 05 0 0 since 18044
Section 4.6
Numerical Integration
27. f x 1 x (a) f x
1 in 0, 2 . 41 x32 1
f x is maximum when x 0 and f 0 4.
< 0.00001, n
8 1 12n2 4
Trapezoidal: Error ≤
> 16,666.67, n > 129.10; let n 130.
2
15 in 0, 2 161 x72
(b) f 4x
15
f 4x is maximum when x 0 and f 40 16. Simpson’s: Error ≤
< 0.00001, n
32 15 180n4 16
4
> 16,666.67, n > 11.36; let n 12.
29. f x tanx2 (a) f x 2 sec2x2 1 4x2 tanx2 in 0, 1 .
f x is maximum when x 1 and f 1 49.5305. 1 03 49.5305 < 0.00001, n2 > 412,754.17, n > 642.46; let n 643. 12n2
Trapezoidal: Error ≤
(b) f 4x 8 sec2x2 12x2 3 32x4 tanx2 36x2 tan2x2 48x4 tan3x2 in 0, 1
f 4x is maximum when x 1 and f 41 9184.4734. Simpson’s: Error ≤
1 05 9184.4734 < 0.00001, n4 > 5,102,485.22, n > 47.53; let n 48. 180n4
31. Let f x Ax3 Bx2 Cx D. Then f 4x 0. Simpson’s: Error ≤
b a5 0 0 180n4
Therefore, Simpson’s Rule is exact when approximating the integral of a cubic polynomial.
1
Example:
x3 dx
0
1 1 04 6 2
3
1
1 4
This is the exact value of the integral. 33. f x 2 3x2 on 0, 4 . Ln
Mn
Rn
Tn
Sn
4
12.7771
15.3965
18.4340
15.6055
15.4845
8
14.0868
15.4480
16.9152
15.5010
15.4662
10
14.3569
15.4544
16.6197
15.4883
15.4658
12
14.5386
15.4578
16.4242
15.4814
15.4657
16
14.7674
15.4613
16.1816
15.4745
15.4657
20
14.9056
15.4628
16.0370
15.4713
15.4657
n
207
208
Chapter 4
Integration
35. f x sin x on 0, 4 . Ln
Mn
Rn
Tn
Sn
4
2.8163
3.5456
3.7256
3.2709
3.3996
8
3.1809
3.5053
3.6356
3.4083
3.4541
10
3.2478
3.4990
3.6115
3.4296
3.4624
12
3.2909
3.4952
3.5940
3.4425
3.4674
16
3.3431
3.4910
3.5704
3.4568
3.4730
20
3.3734
3.4888
3.5552
3.4643
3.4759
n
37. A
2
x cos x dx
0
Simpson’s Rule: n 14
2
x cos x dx
0
28 cos 28 2 14 cos 14 4 328 cos 328 . . . 2 cos 2
0 cos 0 4 84
0.701 y
1
1 2
π
π
4
2
x
5
39. W
100x 125 x 3 dx
0
Simpson’s Rule: n 12
5
100x 125 x3 dx
0
400
12
41.
0
6
1 x2
12 0 36
15 125 15 12
12
3
200
10 125 10 12
12
3
. . . 0 10,233.58 ft lb
Simpson’s Rule, n 6
6 46.0209 26.0851 46.1968 26.3640 46.6002 6.9282
1000 125 2125 2120 2112 290 290 295 288 275 235 89,250 sq m 210
t
45.
3
1 113.098 3.1416 36
43. Area
dx
125 125
5 5 0 400 312 12
sin x dx 2, n 10
0
By trial and error, we obtain t 2.477.
C H A P T E R Integration
4
Section 4.1
Antiderivatives and Indefinite Integration . . . . . . . . . 450
Section 4.2
Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456
Section 4.3
Riemann Sums and Definite Integrals . . . . . . . . . . . 462
Section 4.4
The Fundamental Theorem of Calculus . . . . . . . . . . 466
Section 4.5
Integration by Substitution . . . . . . . . . . . . . . . . . 472
Section 4.6
Numerical Integration
Review Exercises
. . . . . . . . . . . . . . . . . . . 479
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488
C H A P T E R Integration Section 4.1
4
Antiderivatives and Indefinite Integration
Solutions to Even-Numbered Exercises
2.
1 d 4 1 x C 4x3 2 dx x x
4.
d 2x2 3 d 2 32 x 2x12 C C dx dx 3 3x
x12 x32 6.
dr d
8.
r C Check:
2x2 1 C 2 c 2 x
Check:
d 1 C 2x3 dx x 2
Given
Rewrite
Integrate
Simplify
10.
1 dx x2
x1 C 1
1 C x
12.
xx2 3 dx
x3 3x dx
x4 x2 3 C 4 2
1 4 3 2 x x C 4 2
14.
1 x1 C 9 1
1 C 9x
16.
1 dx 3x 2
1 2 x dx 9
5 xdx 5x
x2 C 2
Check:
20.
18.
x4 2x 2 2x C 4
4x3 6x 2 1dx x 4 2x 3 x C d 4 x 2x 3 x C 4x 3 6x 2 1 dx
d x4 2x 2 2x C x3 4x 2 dx 4
x
dx x 2x 1
12
1 x32 1 x12 2 x12 dx C x32 x12 C 2 32 2 12 3
d 2 32 1 1 x x12 C x12 x12 x dx 3 2 2x
Check:
x 3 4x 2dx
Check:
450
x2 d 5x C 5 x dx 2
Check:
22.
x2 dx
x2 1 x32
dy 2x3 dx y
d C d
Section 4.1
24.
4 x3 1 dx
Check:
28.
4 x34 1 dx x74 x C 7
x2 2x3 3x4dx
x1 2x2 3x3 C 1 2 3
1 1 1 2 3C x x x
36.
40.
x4 dx
d 1 1 C 4 dx 3x3 x
2t2 12 dt
30.
4t4 4t2 1 dt
4 4 t5 t3 t C 5 3 Check:
d 45 43 t t t C 4t 4 4t2 1 dt 5 3 2t2 12
x 2 2x 3 x4
1 3 t2 3t3 dt t3 t4 C 3 4
3 dt 3t C
34.
d 13 34 t t C t2 3t3 1 3tt2 dt 3 4
Check:
1 t 2 sin t dt t3 cos t C 3
1 2 sec2 d 3 tan C 3
sec ytan y sec y dy
Check:
sec y tan y sec2 y dy
42.
d 3t C 3 dt
38.
d 13 t cos t C t2 sin t dt 3
d 1 3 tan C 2 sec2 d 3
cos x dx 1 cos2 x
sec y tan y C
d Check: sec y tan y C sec y tan y sec2 y dy
Check:
sec ytan y sec y
cos x dx sin2 x
46. fx x
y 6
f x C=3
x2 C 2 y
C=0 C = −2 4
6
d 1 csc x C csc x cot x dx sin x
y
2
x
f ( x) = 2
2
f′ f (x) = − 1x + 1
f′
4
−4
−4
x
−2
4 −2
−4
−2
cos x sin x
1 x2
6
−2
cos x 1 cos2 x
2 f ( x) = x + 2
8 8
x dx cos sin x
1 f x C x
x
−2
1 sin x
48. fx
4 2
csc x cot x dx csc x C
44. f x x
451
x3 1 C 3C 3 3x
1 3tt2 dt
Check:
d 1 1 1 2 3 C x2 2x3 3x4 dx x x x
Check:
1 dx x4
Check:
32.
26.
d 4 74 4 x3 1 x x C x34 1 dx 7
x 2 2x 3 dx x4
Check:
Antiderivatives and Indefinite Integration
x 2
4
−4
f (x) = − 1x
452
50.
Chapter 4
Integration
dy 2x 1 2x 2, 3, 2 dx y
52.
2x 1 dx x2 2x C
2 32 23 C ⇒ C 1 y x2 2x 1
54. (a)
y
(b)
5
4
3
1 C ⇒ C2 1
y
1 2, x > 0 x
x2 dx
1 C x
5
(−1, 3)
y
x3 xC 3
3
13 1 C 3
−4
4
−5
1 3 1C 3
−5
C y
56. gx 6x 2, g0 1 gx
y
dy x2 1, 1, 3 dx
x
−4
1 dy 2 x2, 1,3 dx x
6x 2 dx 2x3 C
7 3 7 x3 x 3 3 58. fs 6s 8s 3, f 2 3 f s
6s 8s 3ds 3s 2 2s 4 C
g0 1 203 C ⇒ C 1
f 2 3 322 224 C 12 32 C ⇒ C 23
gx 2x3 1
f s 3s 2 2s 4 23 62. f x sin x
60. f x x2 f0 6
f0 1
f 0 3 fx
f 0 6
1 x 2 dx x3 C1 3
fx
sin x dx cos x C1
f0 0 C1 6 ⇒ C1 6
f0 1 C1 1 ⇒ C1 2
1 fx x3 6 3
fx cos x 2
f x
1 3 1 x 6 dx x 4 6x C2 3 12
f 0 0 0 C2 3 ⇒ C2 3 f x
1 4 x 6x 3 12
f x
cos x 2 dx sin x 2x C2
f 0 0 0 C2 6 ⇒ C2 6 f x sin x 2x 6
Section 4.1
64.
dP kt, 0 ≤ t ≤ 10 dt Pt
2 kt12 dt kt32 C 3
Antiderivatives and Indefinite Integration
66. Since f is negative on , 0, f is decreasing on , 0. Since f is positive on 0, , f is increasing on 0, . f has a relative minimum at 0, 0. Since f is positive on , , f is increasing on , .
P0 0 C 500 ⇒ C 500
y 3
2 P1 k 500 600 ⇒ k 150 3 2 Pt 150t32 500 100t32 500 3
f′
2
−3
x
−2
1
70. v0 16 ftsec
f 0 s0
(a)
st 16t2 16t 64 0 16t2 t 4 0
32 dt 32t C1
t
f0 0 C1 v0 ⇒ C1 v0 ft 32t v0 f t st
1 ± 17 2
Choosing the positive value,
32t v0 dt 16t2 v0t C2
f 0 0 0 C2 s0 ⇒ C2 s0
3
−3
s0 64 ft
2
−2
f
f0 v0
ft vt
f ′′
1
P7 100732 500 2352 bacteria 68. f t at 32 ftsec2
453
t
1 17
2.562 seconds. 2 vt st 32t 16
(b)
f t 16t 2 v0t s0 v
1 2 17 321 2 17 16
1617 65.970 ftsec 72. From Exercise 71, f t 4.9t2 1600. (Using the canyon floor as position 0.) f t 0 4.9t2 1600 4.9t2 1600 t2
1600 ⇒ t 326.53 18.1 sec 4.9
74. From Exercise 71, f t 4.9t2 v0t 2. If f t 200 4.9t2 v0t 2, then vt 9.8t v0 0 for this t value. Hence, t v09.8 and we solve 4.9
9.8 v0
2
v0
9.8 2 200 v0
v2 4.9 v02 0 198 2 9.8 9.8 4.9 v02 9.8 v02 9.82 198 4.9 v02 9.82 198 v02 3880.8 ⇒ v0 62.3 msec.
454
76.
Chapter 4
v dv GM
Integration
1 dy y2
1 2 GM v C 2 y
15t 9
(b) vt > 0 when 0 < t <
5 and 3 < t < 5. 3
7 (c) at 6t 14 0 when t . 3
1 GM C v02 2 R 1 2 GM 1 2 GM v v0 2 y 2 R
v
73 373 573 3 2 32 34
2GM 2GM v02 y R
v2 v02 2GM
80. (a) at cos t
f t
7t2
at vt 6t 14
1 2 GM v C 2 0 R
vt
t3
(a) vt xt 3t2 14t 15 3t 5t 3
When y R, v v0.
v2
0 ≤ t ≤ 5
78. xt t 1t 32
1y R1
at dt
cos t dt sin t C1 sin t since v0 0
vt dt
sin t dt cos t C2
f 0 3 cos0 C2 1 C2 ⇒ C2 4 f t cos t 4 (b) vt 0 sin t for t k, k 0, 1, 2, . . . v0 45 mph 66 ftsec
(a) 16.5t 66 44
30 mph 44 ftsec
t
15 mph 22 ftsec at a
s
vt at 66
(b) 16.5t 66 22 t
ec
132 when a at 16.5 vt 16.5t 66
st 8.25t 2 66t
mp
h=
66
30
33 16.5. 2
44
ft/s
h=
(c)
66 66 a
mp
2
44 16.5 117.33 ft ft/s ec mp 0m h= ph 22 f t/se c
s
66 at 66 0 when t . a
44
2.667 16.5
0
15
vt 0 after car moves 132 ft.
66 a 66 s a 2 a
22
1.333 16.5
22 16.5 73.33 ft
a st t2 66t Let s0 0. 2
45
82.
132 73.33 feet
117.33 feet
It takes 1.333 seconds to reduce the speed from 45 mph to 30 mph, 1.333 seconds to reduce the speed from 30 mph to 15 mph, and 1.333 seconds to reduce the speed from 15 mph to 0 mph. Each time, less distance is needed to reach the next speed reduction.
Section 4.1
Antiderivatives and Indefinite Integration
30
84. No, car 2 will be ahead of car 1. If v1t and v2t are the respective velocities, then
0
86. (a) v 0.6139t3 5.525t2 0.0492t 65.9881
30
v2t dt >
455
v1t dt.
0
0.6139t4 5.525t3 0.0492t2 65.9881t 4 3 2 Note: Assume s0 0 is initial position
(b) st
vtdt
s6 196.1 feet
88. Let the aircrafts be located 10 and 17 miles away from the airport, as indicated in the figure. vAt kA t 150
vB kB t 250
1 sAt kA t2 150t 10 2
Airport
A
B
0
10
17
1 sB kB t2 250t 17 2
(a) When aircraft A lands at time tA you have vAtA kA tA 150 100 ⇒ kA
50 tA
1 sAtA kA t2A 150tA 10 0 2
1 50 2 t 150tA 10 2 tA A 125tA 10 tA
kA
10 . 125
50 125 625 2 t 150t 10 50 625 ⇒ SAt S1t tA 10 2
Similarly, when aircraft B lands at time tB you have vBtB kB tB 250 115 ⇒ kB
135 tB
1 sBtB kB t2B 250tB 17 0 2
1 135 2 t 250tB 17 2 tB B 365 t 17 2 B tB
kB (b)
34 . 365
135 365 49,275 49,275 2 135 ⇒ SBt S2t t 250t 17 tB 34 34 68 (c) d sBt sAt
20
20
Yes, d < 3 for t > 0.0505. sB
d 3
sA 0.1
0
0
0
90. True
0.1
0
92. True
456
Chapter 4
Integration
94. False. f has an infinite number of antiderivatives, each differing by a constant.
96.
d sx2 cx2 2sxsx 2cxcx dx 2sxcx 2cxsx 0 Thus, sx cx2 k for some constant k. Since, 2
s0 0 and c0 1, k 1. Therefore,
sx2 cx2 1. [Note that sx sin x and cx cos x satisfy these properties.]
Section 4.2
Area 5
6
2.
kk 2 31 42 53 64 50
4.
4
i 1
2
1
1
1
47
j3
k3
6.
1
j 3 4 5 60
i 13 0 8 1 27 4 64 9 125 238
i1
10.
2
2 n 2i 1 1 n i1 n
18.
i
10
15
2i 3 2 i 315
i1
12.
j1
15
16.
j 1 4
4
5 1 i i1
2
i1
i1
10
2
1
i1
10
i
3
i1
10
i
22. sum seqx 15
i
1011 102112 3080 4 2
3
2i
15 215 1 2 1515 1 2 4 2 152162 1516 14,160 4
2 1 55 3 2 6
2 1 1 9 4.5 3 2 3 2
26. S 5 2 1 s 21
14 2 14 12 2 14 34 2 14
i1
3 2x, x, 1, 15, 1 14,160 (TI-82)
i1
s 4 4 2 01 10
28. S8
1 2
14
54 2 14 32 2 14 74 2 14
2 2
5 6 7 1 1 2 3 16 1 2 6.038 4 2 2 2 2 2 2
1 s8 0 2 4
10
24. S 5 5 4 21 16
i 1 n
1 n1 n i0
1
i1
2
14.
101121 10 375 6
15216 45 195
ii
2
i1
2
20.
10
i
1
>
15
8.
14 2 14 12 2 14 . . . 74 2 14 5.685
14
2
Section 4.2
30. S5 1 s5
15 1 15 15 1 25 15 1 35 15 1 45 15 2
2
n→
2
24 21 16 9 1 1 0.859 5 5 5 5 5
1 15 15 1 25 15 1 35 15 1 45 15 0 0.659 2
2
2
64n nn 16 2n 1 646 lim 2n
3
32. lim
2
3
n→
2
3n2 n 64 64 2 n3 6 3
n1 nn 2 1 21 lim n n n 211 21 2
34. lim n→
2
4j 3 1 n 1 4nn 1 2n 5 3n Sn 2 4j 3 2 2 n j1 n 2 n j1 n n
36.
2
n→
S10
25 2.5 10
S100 2.05 S1000 2.005 S10,000 2.0005 4 n 3 4 n2n 12 nn 12n 1 4i2i 1 4 i i2 4 4 n n n 4 6 i1 i1 n
38.
4 n3 2n2 n 2n2 3n 1 n3 4 6
1 3n3 6n2 3n 4n2 6n 2 3n3
1 3n3 2n2 3n 2 Sn 3n3
S10 1.056 S100 1.006566 S1000 1.00066567 S10,000 1.000066657
40. lim
n n
42. lim
1 n n
n
2i
2
n→ i1
n
n→ i1
2i
2
lim
n→
2
4 n 4 nn 1 4 1 i lim 2 1 2 lim n→ n n→ 2 n2i1 2 n
lim
n→
2 lim
n n
2
i1
i 4 i
n
n
i1
i1
4n
2
2 3 nn 1 4nn 12n 1 n 4n n3 2 6
n→
2 n→ n3 n→
2 n n 2i2 n3i1
lim lim
1 2 n2 34 n2 3n2
2 12
2
4 26 3 3
Area
457
458
Chapter 4
1 n n 2 lim n n 2i n
3
2i
44. lim n→
Integration
2
n
1
n→
i1
3
4
i1
1 n 3 n 6n2i 12ni 2 8i 3 n→ n4 i1
2 lim 2 lim
1 4 nn 1 nn 12n 1 n2n 12 n 6n2 12n 8 n4 2 6 4
2 lim
1 3 n3 4 n6 n2 2 n4 n2
n→
2 lim 10
46. (a)
2
n→
n→
2
4 13 2 20 n n
(b) x
y
31 2 n n
4
Endpoints: 3
1 < 1
2 1
1 16,666.67, n > 129.10; let n 130. 12n2
Trapezoidal: Error ≤ f 4 x
24 in 0, 1 1 x 5
(b) f 4 x is maximum when x 0 and f 4 0 24. Simpson’s: Error ≤
1 24 < 0.00001, n4 > 13,333.33, n > 10.75; let n 12. (In Simpson’s Rule n must be even.) 180n4
28. f x x 1 23 (a) f x
2 in 0, 2 . 9x 1 43 2
f x is maximum when x 0 and f 0 9.
< 0.00001, n
8 2 12n4 9
Trapezoidal: Error ≤ (b) f 4 x
> 14,814.81, n > 121.72; let n 122.
2
56 in 0, 2 81x 1 103 56
f 4 x is maximum when x 0 and f 4 0 81.
< 0.00001, n
32 56 Simpson’s: Error ≤ 180n4 81 be even.)
4
> 12,290.81, n > 10.53; let n 12. (In Simpson’s Rule n must
30. f x sinx2 (a) f x 2 2x2 sinx2 cosx2 in 0, 1 .
f x is maximum when x 1 and f 1 2.2853. 1 0 3 2.2853 < 0.00001, n2 > 19,044.17, n > 138.00; let n 139. 12n2 (b) f 4 x 16x4 12 sinx2 48x2 cosx2 in 0, 1 Trapezoidal: Error ≤
f 4 x is maximum when x 0.852 and f 4 0.852 28.4285. Simpson’s: Error ≤
1 0 5 28.4285 < 0.00001, n4 > 15,793.61, n > 11.21; let n 12. 180n4
32. The program will vary depending upon the computer or programmable calculator that you use. 34. f x 1 x2 on 0, 1 . Ln
Mn
Rn
Tn
Sn
4
0.8739
0.7960
0.6239
0.7489
0.7709
8
0.8350
0.7892
0.7100
0.7725
0.7803
10
0.8261
0.7881
0.7261
0.7761
0.7818
12
0.8200
0.7875
0.7367
0.7783
0.7826
16
0.8121
0.7867
0.7496
0.7808
0.7836
20
0.8071
0.7864
0.7571
0.7821
0.7841
n
482
Chapter 4
Integration
sin x on 1, 2 . x
36. f x
Ln
Mn
Rn
Tn
Sn
4
0.7070
0.6597
0.6103
0.6586
0.6593
8
0.6833
0.6594
0.6350
0.6592
0.6593
10
0.6786
0.6594
0.6399
0.6592
0.6593
12
0.6754
0.6594
0.6431
0.6593
0.6593
16
0.6714
0.6594
0.6472
0.6593
0.6593
20
0.6690
0.6593
0.6496
0.6593
0.6593
n
38. Simpson’s Rule: n 8
2
8 3
1 32 sin d 2
0
3
1 32 sin 0 4 1 32 sin 16 2 1 32 sin 8 . . . 1 32 sin 2
2
2
2
2
6 17.476
40. (a) Trapezoidal:
2
f x dx
0
2 4.32 24.36 24.58 25.79 26.14 27.25 27.64 28.08 8.14 12.518 28
Simpson’s:
2
f x dx
0
2 4.32 44.36 24.58 45.79 26.14 47.25 27.64 48.08 8.14 12.592 38
(b) Using a graphing utility, y 1.3727x3 4.0092x2 0.6202x 4.2844
2
Integrating,
y dx 12.53
0
42. Simpson’s Rule: n 6
1
4
0
4 4 2 4 2 4 1 1 dx 1 1 x2 36 1 16 2 1 26 2 1 36 2 1 4 6 2 1 56 2 2
3.14159 44. Area
120 75 281 284 276 267 268 269 272 268 256 242 223 0 212
7435 sq m 46. The quadratic polynomial px
x x2 x x3 x x1 x x3 x x1 x x2 y y y x1 x2 x1 x3 1 x2 x1 x2 x3 2 x3 x1 x3 x2 3
passes through the three points.
Review Exercises for Chapter 4
Review Exercises for Chapter 4 1.
y
3.
7.
f
2 1 2x2 x 1 dx x3 x2 x C 3 2
f
x
5.
x3 1 dx x2
x
1 1 1 dx x2 C x2 2 x
9. fx 2x, 1, 1 f x
11.
2x dx x2 C
When x 1:
4x 3 sin x dx 2x2 3 cos x C
at a vt
a dt at C1
v0 0 C1 0 when C1 0.
y 1 C 1
vt at
C2
st
y 2 x2
a at dt t2 C2 2
s0 0 C2 0 when C2 0. a st t 2 2 a s30 302 3600 or 2 a
23600 8 ftsec2. 302
v30 830 240 ftsec 10
13. at 32
15. (a)
i1
vt 32t 96
n
(b)
st 16t2 96t (b) s3 144 288 144 ft
(d) s
i
3
i1
(a) vt 32t 96 0 when t 3 sec.
(c) vt 32t 96
2i 1
96 3 when t sec. 2 2
32 1694 9632 108 ft
10
(c)
4i 2
i1
209
210
Chapter 4
17. y
Integration
1 10 , x , n 4 x2 1 2
1 10 10 10 10 2 1 122 1 12 1 322 1
Sn S4
13.0385
10 10 10 10 1 2 122 1 1 1 322 1 22 1
s n s4
9.0385 9.0385 < Area of Region < 13.0385 4 19. y 6 x, x , right endpoints n
8
n
6
f ci x
Area lim n→
y
lim
4
i1
n
n→ i1
2
4i 4 n n
6
lim
4 4 nn 1 6n n n 2
lim
24 8
n→
n→
21. y 5 x2, x Area lim
x
−2
2
3 n
y 6
f ci x
n→ i1
4 3 2
3i 5 2 n
2
3 n 12i 9i2 2 1 n→ n i1 n n
lim
1
3 n
−4 −3
3 12 nn 1 9 nn 12n 1 n 2 n n 2 n 6
lim
3 18 n n 1 92 n 1n2n 1
n→
x
−1
1
2
3
4
−2
lim
n→
8
n1 24 8 16 n
n→ i1
lim
6
n
n
4
−2
2
3 18 9 12
23. x 5y y2, 2 ≤ y ≤ 5, y Area lim
3 n
y 6
52 n 2 n n n
3i
3i
2
3
4
n→ i1
lim
n→
n
9i2
3 15i i 10 4 12 2 n i1 n n n
3 n 3i 9i2 lim 6 2 n→ n i1 n n lim
n→
3
18
9 27 9 2 2
1 x 1
3 3 nn 1 9 nn 12n 1 6n 2 n n 2 n 6
2
2
3
4
5
6
Review Exercises for Chapter 4
n
25.
2ci 3 xi
lim
→
i1
6
2x 3 dx
211
y
27.
4
12 9 6 3 x
−3
3
6
9
−3
5
0
5
5 x 5 dx
0
0
(triangle)
6
29. (a)
2
2 6
(c)
gx dx 210 33 11
2
f x dx 510 50
43
2 x dx 2x
x2 2
xx dx
4
3
0
9
4 16 8 4 1
x
9
4
0
4
2x 1 dx x2 x
1
8
52
sin d cos
3
4
3
34
3
1
9
0
8
25x
x3 2 dx
4
34
41.
2
4x
3 x 1 dx
1
39.
6
2
8
37.
f x dx 3
6
5f x dx 5
2
33.
gx dx 10 3 7
2
6
6
31.
2f x 3gx dx 2
2
(d)
6
f x dx
2
gx dx 10 3 13
2
6
f x gx dx
6
f x dx
2
6
(b)
6
f x gx dx
35.
1
2
2
11
2
2
43.
x2 9 dx
3
y
6
4
2
x3
6
643 36 9 27
64 54 10 3 3 3
6 4 2 x
2
4
6
4
x 4
0
3
y
2
1
3 9x
8 2
1
2 4
6
4t 3 2t dt t 4 t 2
2 2
3 1
2 9 5 4 5 52243 32 422 5 5
0
73 (c) 4 1
16 16 2
8
5
5 5 x dx
x dx
25 2
212
Chapter 4
Integration
1
45.
x x3 dx
0
2 4 x2
x4
1 0
9
1 1 1 2 4 4
47. Area
1
4 x
dx
12 4x12
9 1
83 1 16
y
y 10 8
1
6 4
x 1
2 −2
1
49.
1 94
9
4
1 x
5 2x
9
1
dx
4
2 2 3 2 Average value 5 5
x 2
−2
4
6
y
1
x 2
51. Fx x21 x3
x2 13 dx
x2 x3 3
dx
4
6
8
10
53. Fx x2 3x 2
x6 3x4 3x2 1 dx
57. u x3 3, du 3x2 dx
25 2 , 4 5
5 2
25 x 4
55.
10
2
2 1 5 x x
8
x3 312 x2 dx
x7 3 5 x x3 x C 7 5
1 2 x3 312 3x2 dx x3 312 C 3 3
59. u 1 3x2, du 6x dx
x1 3x24 dx
61.
63.
65.
67.
69.
sin3 x cos x dx
sin 1 cos
1 1 1 1 3x246x dx 1 3x25 C 3x2 15 C 6 30 30
1 4 sin x C 4
1 cos 12 sin d 21 cos 12 C 21 cos C
dx
tann x sec2 x dx
tann1 x C, n 1 n1
1 sec x2 sec x tan x dx 2
1
xx2 4 dx
1 2
1 1 1 sec x2 sec x tan x dx 1 sec x3 C 3
2
1
x2 42x dx
1 x2 42 2 2
2
1 9 0 9 4 4 1
Review Exercises for Chapter 4
3
71.
0
3
1 dx 1 x
1 x12 dx 21 x12
0
3 0
213
422
73. u 1 y, y 1 u, dy du When y 0, u 1. When y 1, u 0.
1
2
0
y 11 y dy 2
0
1 u 1u du
1 0
2
u32 2u12 du 2
1
75.
cos
0
x dx 2 2
cos
0
2x 12 dx 2 sin2x
0
25u
52
0
4 u32 3
1
28 15
2
77. u 1 x, x 1 u, dx du When x a, u 1 a. When x b, u 1 b.
b
Pa, b
a
15 4
15 15 x1 x dx 4 4
1 uu du
1a
u32 u12 du
1a
1b
1b
(a) P0.50, 0.75 (b) P0, b
1 x32 3x 2 2
1 x32 3x 2 2
b 0
15 2 52 2 32 u u 4 5 3 0.75 0.50
1b 1a
15 2u32 3u 5 4 15
1b 1a
1 x32 3x 2 2
b a
0.353 35.3%
1 b32 3b 2 1 0.5 2
1 b 3b 2 1 32
b 0.586 58.6% 79. p 1.20 0.04t C
15,000 M
t1
p ds
t
(b) 2005 corresponds to t 15.
(a) 2000 corresponds to t 10. C
15,000 M
11
1.20 0.04t dt
C
10
11
15,000 1.20t 0.02t2 M
2
81. Trapezoidal Rule n 4:
1
2
Simpson’s Rule n 4:
1
10
15,000 1.20t 0.02t2 M
16 15
27,300 M
24,300 M
1 1 1 2 2 2 1 dx 0.257 1 x3 8 1 13 1 1.253 1 1.53 1 1.753 1 23
1 1 1 4 2 4 1 dx 0.254 1 x3 12 1 13 1 1.253 1 1.53 1 1.753 1 23
Graphing utility: 0.254 83. Trapezoidal Rule n 4:
2
x cos x dx 0.637
85. (a) R < I < T < L
0
Simpson’s Rule n 4: 0.685
(b) S4
Graphing Utility: 0.704
40 f 0 4f 1 2f 2 4f 3 f 4 34
1 1 1 4 42 21 4 5.417 3 2 4
214
Chapter 4
Integration
Problem Solving for Chapter 4
1
1. (a) L1
1
(b) Lx
1 dt 0 t
1 by the Second Fundamental Theorem of Calculus. x
L1 1
x
(c) Lx 1
1
2.718
1 dt for x 2.718 t
1 dt 0.999896 t
1
x1
(d) We first show that
1
To see this, let u
x1
Then
1
1 dt t
1 dt t
1
1 dt. 1x1 t
1 t and du dt. x1 x1
1
1 x1 du 1x1 ux1
Lx1x2
Now,
(Note: The exact value of x is e, the base of the natural logarithm function.)
x1x2
1
1
1 du 1x1 u x2
x2
1 du 1x1 u x1
1
1 du u
1
1
1x1 t
dt.
t 1 du using u u x 1x1 1
1 dt t
1
1
x2
1
1 du u
1 du u
Lx1 Lx2 .
x
3. Sx
0
sin
2t dt 2
y
(a)
y
(b)
2
3
1 2 x 1
3
1
−1
x
−2
1
2
The zeros of y sin
3
2
5
x2 x2 0 ⇒ n ⇒ x2 2n ⇒ x 2n, n integer. 2 2
Relative maximum at x 2 1.4142 and x 6 2.4495 Relative minimum at x 2 and x 8 2.8284 (d) S x cos
2x x 0 ⇒ 2x 2
2
n ⇒ x2 1 2n ⇒ x 1 2n, n integer 2
Points of inflection at x 1, 3, 5, and 7.
72 23
x2 correspond to the relative 2
extrema of Sx. (c) Sx sin
6
Problem Solving for Chapter 4 y
5. (a) 5 4 3 2 1
(b) (6, 2)
(8, 3) f
x
0
1
Fx
0
(0, 0)
2
1 2
3
2
4
7 2
5
4
7 2
215
6
7
8
2
1 4
3
x 2
−1 −2 −3 −4 −5
4 5 6 7 8 9
(2, −2)
1, 1, (d) F x fx 1, 2
0 ≤ x < 2 x, x 4, 2 ≤ x < 6 (c) f x 1 x 1, 6 ≤ x ≤ 8 2
0 ≤ x < 2 x22, Fx f t dt x22 4x 4, 2 ≤ x < 6 0 14x2 x 5, 6 ≤ x ≤ 8
x
0 < x < 2 2 < x < 6 6 < x < 8
x 2 is a point of inflection, whereas x 6 is not. ( f is not continuous at x 6.)
Fx f x. F is decreasing on 0, 4 and increasing on 4, 8. Therefore, the minimum is 4 at x 4, and the maximum is 3 at x 8.
1
7. (a)
1
cos x dx cos
1
1
1 3
1
cos x dx sin x
1
cos 13 2 cos 13 1.6758
2 sin1 1.6829
Error. 1.6829 1.6758 0.0071
1
(b)
1 1
1 1 3 1 dx x2 1 13 1 13 2
(Note: exact answer is 2 1.5708)
(c) Let px ax3 bx2 cx d.
1
1
ax4
4
px dx
p
1 3
b
a
1 Fx dx 2
b
f x fx dx
a
b
1
2b 2d 3
11. Consider
x5 dx
0
1 Fx 2
1
p 13 b3 d b3 d 2b3 2d 1
9. Consider Fx f x2 ⇒ Fx 2f x fx. Thus,
bx3 cx2 dx 3 2
x6 6
1 0
1 . 6
The corresponding Riemann Sum using right endpoints is Sn
1 n
1n 2n 5
5
n . . . n
5
a
1 Fb Fa 2 1 f b2 f a2 2
1 5 1 25 . . . n5 n6
Thus, lim Sn lim n→
n→
15 25 . . . n5 1 . n6 6
216
Chapter 4
Integration
b
13. By Theorem 4.8, 0 < f x ≤ M ⇒
b
f x dx ≤
a
b
Similarly, m ≤ f x ⇒ mb a
b
f x dx.
m dx ≤
a
M dx Mb a.
a
a
b
Thus, mb a ≤
f x dx ≤ Mb a. On the interval 0, 1, 1 ≤ 1 x4 ≤ 2 and b a 1.
a
1
Thus, 1 ≤
1 x4 dx ≤ 2.
0
15. Since f x ≤ f x ≤ f x ,
b
a
17.
1 365
365
0
b
f x dx ≤
a
100,000 1 sin
b
f x dx ≤
1
Note:
1 x4 dx 1.0894
0
f x dx ⇒
a
b
a
f x dx ≤
b
a
f x dx.
2t 60 100,000 365 2t 60 dt t cos 365 365 2 365
365
0
100,000 lbs.
C H A P T E R 5 Logarithmic, Exponential, and Other Transcendental Functions Section 5.1
The Natural Logarithmic Function: Differentiation . . . . 493
Section 5.2
The Natural Logarithmic Function: Integration . . . . . . 498
Section 5.3
Inverse Functions . . . . . . . . . . . . . . . . . . . . . . 503
Section 5.4
Exponential Functions: Differentiation and Integration . . 509
Section 5.5
Bases Other than e and Applications . . . . . . . . . . . . 516
Section 5.6
Differential Equations: Growth and Decay . . . . . . . . . 522
Section 5.7
Differential Equations: Separation of Variables
Section 5.8
Inverse Trigonometric Functions: Differentiation . . . . . 535
Section 5.9
Inverse Trigonometric Functions: Integration
. . . . . . 527
. . . . . . . 539
Section 5.10 Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . 543 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 548 Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554
C H A P T E R 5 Logarithmic, Exponential, and Other Transcendental Functions Section 5.1
The Natural Logarithmic Function: Differentiation
Solutions to Even-Numbered Exercises 2. (a)
(b)
y
3
3 0.2
4
2 1 −4 x 1
2
3
4
The graphs are identical.
−1
4. (a) ln 8.3 2.1163
8.3
(b)
1
6. (a) ln 0.6 0.5108
0.6
1 dt 2.1163 t
(b)
1
8. f x ln x
1 dt 0.5108 t
10. f x ln x
Reflection in the x-axis
Reflection in the y-axis and the x-axis
Matches (d)
Matches (c)
14. f x ln x
12. f x 2 ln x
16. gx 2 ln x
Domain: x 0
Domain: x > 0 y
Domain: x > 0 y
y 3
2
3
2 1 1
2
x 1
2
3
4
−3
−2
x
−1
1
2
3
−1
1
−2 −2
18. (a) ln 0.25 ln 14 ln 1 2 ln 2 1.3862
x
−3
1
2
3
20. ln23 ln 232 32 ln 2
(b) ln 24 3 ln 2 ln 3 3.1779 3 12 1 2 ln 2 ln 3 0.8283 (c) ln 3
1 (d) ln 72 ln 1 3 ln 2 2 ln 3 4.2765
22. ln xyz ln x ln y ln z
1 24. lna 1 lna 112 2 lna 1
26. ln 3e2 ln 3 2 ln e 2 ln 3
28. ln
1 ln 1 ln e 1 e
493
494
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
30. 3 ln x 2 ln y 4 ln z ln x3 ln y 2 ln z4 ln
x 3y 2 z4
32. 2 ln x lnx 1 lnx 1 2 ln
34.
x x ln 2 x 1x 1 x 1
3 3 x2 1 lnx2 1 lnx 1 lnx 1 ln ln 2 2 x 1x 1
36.
2
xx
1 1
2 2
3
38. lim ln6 x
3
40. lim ln x →5
x →6
x x 4
f=g −1
5 −1
42. y ln x32 y
3 ln x 2
44. y ln x12
3 2x
y
46. hx ln2x2 1
50.
48.
1 lnx2 4 2
1 dy x ln x 1 ln x dx x
52. f x ln
dy 1 2x x 2 dx 2 x2 4 x 4 ln t t
54. ht
58. y ln y
xx 11 31 lnx 1 lnx 1
fx
56.
t1t ln t 1 ln t ht t2 t2
3
1 1 1 1 2 2 3 x1 x1 3 x2 1 3x2 1
y x ln x
1 4x 4x 2 2x2 1 2x 1
y lnx2 4
1 2x
1 At 1, 0, y 2 .
3 At 1, 0, y 2.
hx
1 ln x 2
x 2x 3 ln 2x lnx 3
1 1 3 x x 3 xx 3
y lnln x 1 dy 1x dx ln x x ln x
60. f x ln x 4 x2 fx
1 x 1 4 x2 x 4 x2 1 4 x2
ln 5 1.6094
Section 5.1
62.
x2 4 1 2 x2 4 ln 2x2 4 x
y
x2 4
2x2
The Natural Logarithmic Function: Differentiation
dy 2x2 xx2 4 4xx2 4 1 1 dx 4x 4 4 2 x2 4
1 1 ln 2 x2 4 ln x 4 4
x x 4 4x1 2
Note that: 1 1 2 x2 4 2 x2 4 Hence,
2 x2 4 2 x2 4 x2 2 x2 4
x2 4 1 1 2 x2 4 x 1 dy 3 2 dx 2xx 4 x 4 x2 4x x2 4
1 12 2 x2 4 x2 4 1 3 2 x 4x 2xx 4
x2 4 x2 4 1 x2 4 3 2 x 4x x3 4xx 4
64. y ln csc x y
66.
csc x cot x cot x csc x
dy sec x tan x sec2 x dx sec x tan x
68.
y ln1 sin2 x
y ln sec x tan x
1 ln1 sin2 x 2
sec xsec x tan x sec x sec x tan x
ln x
70. gx
t 2 3dt
l
dy 1 2 sin x cos x sin x cos x dx 2 1 sin2 x 1 sin2 x
gx ln x2 3
d ln x2 3 ln x dx x
(Second Fundamental Theorem of Calculus)
72. (a)
y 4 x2 ln
12 x 1 ,
0, 4
1 1 dy 2x dx 12x 1 2 2x When x 0,
1 x2 dy 1 . dx 2
1 Tangent line: y 4 x 0 2 1 y x4 2 (b)
8
−4
4
−4
74.
lnxy 5x 30 ln x ln y 5x 30 1 1 dy 50 x y dx 1 dy 1 5 y dx x y 5xy y dy 5y dx x x
495
496
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
y xln x 4x
76.
y x
1x ln x 4 3 ln x
x y xy x x ln x 4x x3 ln x 0 78. y x ln x
6
Domain: x > 0 y 1 y
1 0 when x 1. x
(1, 1) −1
6 −1
1 > 0 x2
Relative minimum: 1, 1 80. y
ln x x
2
(e, e−1)
3 2
3 − 23
,2e
−1
Domain: x > 0 y
(e
) 6
x1x ln x 1 ln x 0 when x e. x2 x2
−4
x21x 1 ln x2x 2ln x 3 y 0 when x e32. x4 x3 Relative maximum: e, e1 3 Point of inflection: e32, 2 e32
x 82. y x2 ln . 4 y x2
Domain x > 0
x x 1 2x ln x 1 2 ln 0 when x 4 4
x x 1 1 2 ln ⇒ ln ⇒ x 4e12 4 4 2 y 1 2 ln
f x x ln x,
84.
x 1 x 2x 3 2 ln 4 x 4
fx 1 ln x, 1 f x , x
Point of inflection: 4e32, 24e3 4
(4e−3/2, −24e−3)
−4
−4
(4e−1/2, −8e−1)
f 1 1
P2x f 1 f1x 1 x 1 P1x 1,
1 x 1 2, 2
P11 0
1 f 1x 12 2 P21 0
P11 1
P2x 1 x 1 x, P2 x x,
8
f 1 1
P1x f 1 f1x 1 x 1,
y 0 when x 4e32 Relative minimum: 4e12, 8e1
f 1 0
P21 1
P2 1 1
The values of f, P1, P2, and their first derivatives agree at x 1. The values of the second derivatives of f and P2 agree at x 1. 3
f P1
P2 −2
4 −1
Section 5.1 86. Find x such that ln x 3 x.
The Natural Logarithmic Function: Differentiation
88.
f x x ln x 3 0 fx 1 xn1
f xn 4 ln xn xn xn fxn 1 xn
n
1
xn
2
2
f xn
0.3069
y x 1x 2x 3 ln y
1 x
1 lnx 1 lnx 2 lnx 3
2
1 dy 1 1 1 1 y dx 2 x1 x2 x3
3
2.2046
2.2079
0.0049
0.0000
497
1 3x2 12x 11 2 x 1x 2x 3
dy 3x2 12x 11 dx 2y
3x2 12x 11 2x 1x 2x 3
Approximate root: x 2.208
xx
2
y
90.
2
1 1
92.
dy dx
x
x2 2
1 2x 1 x4 1
x 1x 2 x 1x 2
ln y lnx 1 lnx 2 lnx 1 lnx 2
1 ln y lnx2 1 lnx2 1
2 1 dy 1 2x 2x y dx 2 x2 1 x2 1
y
1 1 dy 1 1 1 y dx x1 x2 x1 x2
dy 2 4 6x2 12 y 2 2 y 2 dx x 1 x 4 x 1x2 4
x2 1122x 2 x 112x2 1x2 1
x 1x 2 6x2 2 x 1x 2 x 1x 1x 2x 2
2x 2 x 132x2 112
6x2 2 x 12x 22
94. The base of the natural logarithmic function is e. 96. gx ln f x, f x > 0 gx
fx f x
(a) Yes. If the graph of g is increasing, then gx > 0. Since f x > 0, you know that fx gxf x and thus, fx > 0. Therefore, the graph of f is increasing. 98. t (a)
(b) No. Let f x x2 1 (positive and concave up). gx lnx2 1 is not concave up.
5.315 , 1000 < x 6.7968 ln x (d)
50
dt 1 5.3156.7968 ln x2 dx x
4000
1000 0
(b) t1167.41 20 years T 1167.412012 $280,178.40 (c) t1068.45 30 years T 1068.453012 $384,642.00
5.315 x6.7968 ln x2
When x 1167.41, dtdx 0.0645. When x 1068.45, dtdx 0.1585. (e) There are two obvious benefits to paying a higher monthly payment: 1. The term is lower 2. The total amount paid is lower.
498
Chapter 5
100. (a)
Logarithmic, Exponential, and Other Transcendental Functions (b) T p
350
34.96 3.955 p p
(c)
30
T10 4.75 deglbin2 0
T70 0.97 deglbin2
100
0
100 0
0
lim Tp 0
p→
102. y 10 ln (a)
10
100 x2
x
100 x
2
10 ln 10 100 x2 ln x 100 x2 (c) lim
20
x→10
0
dy 0 dx
10 0
(b)
dy x 1 x 10 dx x 100 x2 10 100 x2 100 x2
x
10
10
100 x2 10 100 x2
x
100 x2 10 100 x2
10x 100x x
2
1
10 x
100 x2 10 x 2 2 x 100 x 10 100 x
x 10 x 10 100 x2
2 x 10 100 x2 10 100 x x x x2
When x 5, dydx 3. When x 9, dydx 199. 104. y ln x y
106. False
is a constant.
1 > 0 for x > 0. x
Since ln x is increasing on its entire domain 0, , it is a strictly monotonic function and therefore, is one-to-one.
Section 5.2 2.
6.
d ln 0 dx
The Natural Logarithmic Function: Integration
10 1 dx 10 dx 10 ln x C x x
1 1 1 dx 3 dx 3x 2 3 3x 2 1 ln 3x 2 C 3
4. u x 5, du dx
1 dx ln x 5 C x5
8. u 3 x3, du 3x2 dx
x2 1 1 dx 3x2 dx 3 x3 3 3 x3 1 ln 3 x3 C 3
Section 5.2
The Natural Logarithmic Function: Integration
499
10. u 9 x2, du 2x dx
12.
x 9 x2
x3
1 9 x2122x dx 9 x2 C 2
dx
1 xx 2 3x2 6x dx dx u x3 3x2 4 2 3 3x 4 3 x 3x2 4
14.
1 ln x3 3x2 4 C 3
2x2 7x 3 dx x2
2x 11
19 dx x2
16.
x3 6x 20 dx x5
x2 11x 19 ln x 2 C
18.
x3 3x2 4x 9 dx x2 3
3 x
3x
3 ln 1 x13 C
x2 5x 19
1
x dx
1 ln ln x C 3
xx 2 dx x 13
x2 2x 1 1 dx x 13
x 12 dx x 13
1 dx x1
1 dx x 13
ln x 1
26. u 1 3x, du
1 dx 1 3x
3 2 dx ⇒ dx u 1 du 23x 3
12 u 1 du u3
2 3
1
1 du u
2 u ln u C 3
2 1 3x ln1 3x C 3
2 2 3x ln1 3x C1 3 3
115 dx x5
x3 5x2 19x 115 ln x 5 C 3 2
1 1 1 dx x lnx3 3 ln x
24.
1 1 1 dx 3 dx 1 x13 1 x13 3x23
x23
20.
x2 1 lnx2 3 C 2 2
1 dx 3x23
22. u 1 x13, du
x dx x2 3
1 dx x 13
1 C 2x 12
500
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
28. u x13 1, du
3 x
x 1 3
dx
1 dx ⇒ dx 3u 12 du 3x23
u1 3u 12 du u u1 2 u 2u 1 du u
3 3
u2 3u 3
u3 3
3
x
13
3
C
3u2 3u ln u 2
13 3x13 12 3x13 1 ln x13 1 3 2
tan 5 d
C
3 ln x13 1
30.
1 du u
3x23 3x13 x C1 2
1 5 sin 5 d 5 cos 5
32.
sec
1 ln cos 5 C 5
34. u cot t, du csc2 t dt
csc2
2 ln sec
36.
x x tan C 2 2
sec t tan t dt lnsec t tan t lncos t C
t dt ln cot t C cot t
x x 1 dx 2 sec dx 2 2 2
ln
sec t tan t C cos t
ln sec tsec t tan t C
38. y
2x dx x2 9
40. r
0, 4: 4 ln0 9 C ⇒ y lnx2 9 4 ln 9 ln x2 9 C
sec2 t dt tan t 1
ln tan t 1 C C 4 ln 9
, 4: 4 ln0 1 C ⇒ C 4 r ln tan t 1 4 10
8
(0, 4)
−9
9 −8
(π , 4) −2
−4
42.
dy ln x , 1, 2 dx x (a)
8
(b)
y
y
ln x ln x2 dx C x 2
2
y1 2 ⇒ 2
1 x
4 −1 −2
Hence, y
ln 12 C ⇒ C 2 2
ln x2 2. 2
Section 5.2
1
44.
The Natural Logarithmic Function: Integration
1 1
1 dx ln x 2 1 x 2
46. u ln x, du
2
e
ln 3 ln 1 ln 3
e
1
48.
0
x1 dx x1
1
1
1 dx
0
0
2
1 dx x ln x
e
e
e2
1 1 dx ln ln x ln x x
e
ln 2
2 dx x1
0 1 2 ln 2
1
x 2 ln x 1
0.2
50.
1 dx x
0.2
csc 2 cot 2 2 d
0.1
csc2 2 2 csc 2 cot 2 cot2 2 d
0.1 0.2
2 csc2 2 2 csc 2 cot 2 1 d
0.1
cot 2 csc 2
52. ln sin x C ln
ln
0.2 0.1
0.0024
1 C ln csc x C csc x
54. ln csc x cot x C ln
56.
csc x cot xcsc x cot x csc2 x cot2 x C ln C csc x cot x csc x cot x
1 C ln csc x cot x C csc x cot x
1 x 2 dx 1 x 6 1 x 4 ln 1 x C1 1 x 4x x 4 ln 1 x C where C C1 5.
58.
60.
tan2 2x 1 dx ln sec 2x tan 2x sin 2x C sec 2x 2
4
sin2 x cos2 x dx ln sec x tan x 2 sin x cos x 4
2
ln
2 1 2 1
4
4
2 1.066
Note: In Exercises 62 and 64, you can use the Second Fundamental Theorem of Calculus or integrate the function.
62. F x
tan t dt
0
Fx tan x
x2
x
64. F x
1
1 dt t
2x 2 Fx 2 x x
501
502
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
4
66.
68. A
y
1
x4 dx x
4
x 4 ln x
2 1
1
1
4 dx x 4 1
4 4 ln 4 1
x 1
3
2
4
3 4 ln 4 8.5452
−1 −2
6
A3 Matches (a) 9
0 0
4
70.
10 ln cos0.3x 3
10 10 ln cos1.2 1 ln cos0.3 11.7686 3 3
2x tan0.3x dx x2
1
16
1
4
8
0
5
−2
72. Substitution: u x2 4 and Power Rule
74. Substitution: u tan x and Log Rule
76. Answers will vary.
78. Average value
1 42
4
2
2
4
2
4x 1 dx x2
1 1 2 dx x x
1 x
1 1 ln 2 4 2
1 1 ln 4 1.8863 4 2
2 ln x 2 ln 4 2 ln 2
82. t
84.
10 ln 2
300
250
80. Average value
300 250
St
k dt k ln t C k ln t C since t > 1. t
S2 k ln 2 C 200 S4 k ln 4 C 300 Solving this system yields k 100ln 2 and C 100. Thus, St
100 ln t ln t 100 100 1 . ln 2 ln 2
x dx 6
3 ln2 3 ln1 0
3 ln2 3
4.1504 units of time
dS k dt t
sec
10 10 4 ln 200 ln 150 ln ln 2 ln 2 3
0
12 6 ln sec 6x tan 6x
1 dT T 100
10 lnT 100 ln 2
2
4 2
1 20
2 0
Section 5.3 86. k 1: f1x x 1 k 0.5: f0.5x k 0.1: f0.1x
y
x 1
0.5
2 x 1
f1 8 6
1 x 1 10 10 0.1
10 x
f
4
f
0.1 x
2
88. False
Section 5.3
(b)
10
8
f 4
3x 3x 34 x 4 4
2 x
−2
3 3 4x g f x g3 4x x 4
2
f x 1 x3
(b) 2
g −2
x
−1
16 12
16 16 x x
f
8
g
g f x g16 x2 16 16 x2
x 8
x2 x
f x
1 , x ≥ 0 1x
gx
1x , 0 < x ≤ 1 x
20
20
f gx f 16 x 16 16 x2
16
y
(b)
gx 16 x
1 g f x g 1x
3
−2
3 x3 x 1 x3
1 x x
2 −1
x3
f x 16 x2, x ≥ 0
f gx f
8
f 3
1 1 x x 3 1
g
y
3 1 x 1 f gx f 3 1 x 3
g f x g1
4
−2
3 1 x gx
8. (a)
8
y
3x 4
f gx f
6. (a)
6
Inverse Functions
f x 3 4x
4
90. False; the integrand has a nonremovable discontinuity at x 0.
d 1 ln x dx x
4. (a)
0.5
2
k→0
gx
503
10
lim fk x ln x
2. (a)
Inverse Functions
(b)
1 1x x 1 1x 1x
y
3
g
2
1 1 x 1x 1 1 x x 1
12
1
f x 1
1x x 1
2
3
504
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
10. Matches (b)
12. Matches (d) x2 4 Not one-to-one; does not have an inverse
16. F x
14. f x 5x 3 One-to-one; has an inverse y
18. gt
x2
1 t 2 1
Not one-to-one; does not have an inverse
y −2
3
x
−1
1
2 1
−1
1 2
−3
3
x
−1
−3
−
1 2
1 2
1
1
−1
− 2
−1
20. f x 5xx 1
22. hx x 4 x 4
One-to-one; has an inverse
Not one-to-one; does not have an inverse
12
9
−9
0
9
6 0
−9
24. f x cos
3x 2
3 3x 2 4 fx sin 0 when x 0, , , . . . 2 2 3 3 f is not strictly monotonic on , . Therefore, f does not have an inverse. 28. f x lnx 3, x > 3
26. f x x3 6x2 12x fx 3x2 12x 12 3x 22 ≥ 0 for all x.
fx
f is increasing on , . Therefore, f is strictly monotonic and has an inverse.
f x 3x y
30.
32.
f is increasing on 3, . Therefore, f is strictly monotonic and has an inverse.
f x x3 1 y
y 3
3 y 1 x
x y 3
3 x 1 f1x
x
1 > 0 for x > 3. x3
y
3 x
f x x2 y, 0 ≤ x
34.
x y y x
1 f
x x
1
y
x x 3
f 1
5 4 3 2
y 3 2
−3
1
−3
2
3
f
2
−1
1
f x
−2
3
f −1 2 3 4 5
f −1
1
f
4
x
−5 − 4 −3
f
y
−4 −5
x 1
2
3
4
Section 5.3 f x x2 4 y, x ≥ 2
36.
Inverse Functions
505
5 2x 1 y f x 3
38.
y5 243 486 x5 243 y 486
x y2 4
x
y x2 4 f 1x x2 4, x ≥ 0
f 1x
y
x5 243 486
5
f
4
−1
6
f
f
3
The graphs of f and f 1 are reflections of each other across the line y x.
f −1
2
−4
8
1 x 1
2
3
4
−6
5
f x x3 5 y
40.
f x
42.
x y5 3
x2 y x
x
2 y1
y
2 x1
f 1x
2 x1
y x5 3
44.
x
0
2
4
f 1x
6
2
0
y
f 1x x5 3 2
f −1 f −3
3
8 6 4
f
−1
(4, 0) f
The graphs of f and f 1 are reflections of each other across the line y x.
−6
(2, 2)
2
4 −2
(0, 6)
2
4
6
x 8
6
−4
The graphs of f and f 1 are reflections of each other across the line y x. 46. f x k2 x x3 is one-to-one for all k 0. Since f 13 2, f 2 3 k2 2 23 12k ⇒ k 14 .
x 2 1 1 f x
48. f x x 2 on 2, x2
> 0 on 2,
f is increasing on 2, . Therefore, f is strictly monotonic and has an inverse.
50. f x cot x on 0, fx csc2 x < 0 on 0, f is decreasing on 0, . Therefore, f is strictly monotonic and has an inverse.
2
52. f x sec x on 0,
2
fx sec x tan x > 0 on 0,
f is increasing on 0, 2. Therefore, f is strictly monotonic and has an inverse.
506
54.
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
3 y on 0, 10 x2
f x 2
f −1
2x2 3 x2y
f
x22 y 3
10
0 0
3 y ± 2x 3 2y
x±
f 1x
The graphs of f and f 1 are reflections of each other across the line y x.
5
2 3 x, x < 2
56. (a), (b)
58. (a), (b)
2
f −1 f
h −1
−3
10
h
−10
3
10
−10
−2
(c) Yes, f is one-to-one and has an inverse. The inverse relation is an inverse function.
(c) h is not one-to-one and does not have an inverse. The inverse relation is not an inverse function. 60. f x 3
62. f x ax b
Not one-to-one; does not have an inverse
f is one-to-one; has an inverse ax b y x
yb a
y
xb a
f 1x
64. f x 16 x4 is one-to-one for x ≥ 0. 16
x4
x3y
x4
xy3
yx
yx3
4 16 x y
f 1
x
f
fx
16 x, x ≤ 16 70. Yes, the area function is increasing and hence one-to-one. The inverse function gives the radius r corresponding to the area A.
1 5 1 x 2x3; f 3 243 54 11 a. 27 27 1 5x4 6x2 27
f 111
x x 3, x ≥ 0
1
4
68. No, there could be two times t1 t2 for which ht1 ht2.
72. f x
66. f x x 3 is one-to-one for x ≥ 3.
y
16 y 4 16
xb ,a 0 a
1 1 27 1 f f 111 f3 534 632 17
Section 5.3
Inverse Functions
f x cos 2x, f 0 1 a
74.
fx 2 sin 2x
f 11
1 1 1 1 which is undefined. f f 11 f0 2 sin 0 0
f x x 4, f 8 2 a
76.
fx
f 12
1 2x 4 1 1 1 1 4 f f 12 f8 1 28 4 1 4
78. (a) Domain f Domain f 1 ,
(d)
f x 3 4x, 1, 1
(b) Range f Range f 1 ,
fx 4
(c)
f1 4
y
f f
x
−5 −4 −3 −2 −1 −2 −3 −4 −5
2 3
80. (a) Domain f 0, , Domain f 1 0, 4 (b) Range f 0, 4, Range f 1 0, (c)
y 4
2
1 4
f 11
1 4
f x
4 1 x2
fx
8x , f1 2 x2 12
x 1
2
4 x x 2
f 1x
f −1 f
82.
(d)
f 1x
f 1x
3
1
3x , 1, 1 4
f 1x
3 2
−1
3
4
x2
4x x
, f 12
1 2
x 2 lny2 3 12
dy 1 2y y2 3 dx
dy 16 3 13 dy y2 3 . At 0, 4, . dx 4y dx 16 16 In Exercises 84 and 86, use the following. f x 18 x 3 and g x x3 3 f1 x 8 x 3 and g1 x x
84. g1 f 13 g1 f 13 g10 0
3 4 86. g1 g14 g1g14 g1
3 3 9 4 4
507
508
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
In Exercises 88 and 90, use the following. f x x 4 and g x 2x 5 f1 x x 4 and g1 x
x5 2
88. f 1 g1x f 1g1x f 1
90. g f x g f x
x 2 5
x5 4 2
x3 2
92. The graphs of f and f 1 are mirror images with respect to the line y x.
gx 4 2x 4 5 2x 3
Note: g f 1
94. Theorem 5.9: Let f be differentiable on an interval I. If f has an inverse g, then g is differentiable at any x for which fgx 0. Moreover, gx
96. f is not one-to-one because different x-values yield the same y-value.
34 53
Example: f 3 f
x3 2 f 1 g1
Hence, g f 1x
1 , fgx 0 fgx
98. If f has an inverse, then f and f 1 are both one-to-one. Let f11x y then x f 1 y and f x y. Thus, f 11 f.
Not continuous at ± 2. 100. If f has an inverse and f x1 f x2, then f 1 f x1 f 1 f x2 ⇒ x1 x2. Therefore, f is one-to-one. If f x is one-toone, then for every value b in the range, there corresponds exactly one value a in the domain. Define gx such that the domain of g equals the range of f and gb a. By the reflexive property of inverses, g f 1. 102. True; if f has a y-intercept.
104. False Let f x x or gx 1 x .
106. From Theorem 5.9, we have: gx
1 fgx
g x
fgx0 f gxgx fgx2
f gx 1 fgx fgx2
f gx fgx3
If f is increasing and concave down, then f > 0 and f < 0 which implies that g is increasing and concave up.
Section 5.4
Section 5.4
Exponential Functions: Differentiation and Integration
e2 0.1353. . .
2.
Exponential Functions: Differentiation and Integration
4.
ln 0.1353. . . 2
ln 0.5 0.6931. . .
6. eln 2x 12 2x 12
1 e0.6931. . . 2
x6 10. 6 3ex 8
8. 4ex 83 ex
3ex 14
83 4
ex
83 x ln 3.033 4
x ln
12. 200e4x 15 e4x
14 3
143 1.540
14. ln x2 10 x2 e10
15 3 200 40
x ± e5 ± 148.4132
3 4x ln 40 x
40 1 ln 0.648 4 3 18. lnx 22 12
16. ln 4x 1
x 22 e12
4x e e x
x 2 e6
e 0.680 4
x 2 e6 405.429
1 20. y 2 ex
22. y ex2
y
y
4
4
3
3
2
2
1 x
−1
1
24. (a)
2
3
−2
(b)
10
−8
−1
10 −2
Horizontal asymptotes: y 0 and y 8
x 1
2
10
−8
10 −2
Horizontal asymptote: y 4
509
510
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
28. y
26. y Ceax
C 1 eax
Horizontal asymptote: y 0 lim
Reflection in the y-axis
x →
Matches (d)
lim
C C 1 eax
x →
C 0 1 eax
Horizontal asymptotes: y C and y 0 Matches (b) 30. f x ex3 gx ln
32. f x ex1
y
f
4
6
3
4
2
2
1 x
−2
lim
x →
y
8
−2
34. In the same way,
gx 1 ln x
3 ln x
x3
g
2
36. 1 1
4
6
x
er for r > 0.
f
g
x
−1
8
1 xr
1
3
2
4
−1
1 1 1 1 1 1 2.71825396 2 6 24 120 720 5040 e 2.718281828 e > 11
1 1 1 1 1 1 2 6 24 120 720 5040
38. (a) y e2x
(b) y e2x
y 2e2x
y 2e2x
At 0, 1, y 2.
At 0, 1, y 2.
40. f x e1x
42.
fx e1x
y ex
2
44.
y x2ex dy x2ex 2xex dx
dy 2 2xex dx
xex 2 x
46. gt e3t
2
6 gt e3t 6t 3 3 3t2 t e 2
48.
y ln
11 ee x
50.
x
ln1 ex ln1 ex dy ex ex x dx 1 e 1 ex
52.
y
ex ex 2
dy ex ex dx 2
2ex 1 e2x
y ln
x
ex 2
lnex ex ln 2 dy ex ex dx ex ex
54.
e
y xex ex ex x 1 dy ex ex x 1 xex dx
e2x 1 e2x 1
Section 5.4
Exponential Functions: Differentiation and Integration
56. f x e3 ln x
58.
e3 x
fx
y ln ex x dy 1 dx
62. gx x ex ln x
exy x2 y2 10
60.
x dxdy ye
xy
2x 2y
dy 0 dx
gx
1 ex ex ln x x 2x
dy xy xe 2y yexy 2x dx
g x
dy yexy 2x xy dx xe 2y
1 xex ex ex ex ln x 32 4x x2 x ex2x 1 1 ex ln x x2 4xx
y ex3 cos 2x 4 sin 2x
64.
y ex6 sin 2x 8 cos 2x ex3 cos 2x 4 sin 2x ex10 sin 2x 5 cos 2x 5ex2 sin 2x cos 2x y 5ex4 cos 2x 2 sin 2x 5ex2 sin 2x cos 2x 5ex5 cos 2x 25ex cos 2x y 2y 25ex cos 2x 25ex2 sin 2x cos 2x 5ex3 cos 2x 4 sin 2x 5y Therefore, y 2y 5y ⇒ y 2y 5y 0.
66. f x
ex ex 2
2
fx
ex ex > 0 2
fx
ex ex 0 when x 0. 2
−3
(0, 0)
3
−2
Point of inflection: 0, 0
68. gx gx g x
1 2
1 2
1 2
ex3 2 2
x 3ex3 2 2
(
2,
x 2x 4ex3 2
e− 0.5 2π
( (
( 4,
0
e− 0.5 2π
( 6
0
3, 0.399 1 1 e , 4, e 2, 0.242, 4, 0.242 2 2
1
2
12
12
70. f x xex
2
fx xex ex ex1 x 0 when x 1. f x ex ex1 x exx 2 0 when x 2. Relative maximum: 1, e1 Point of inflection: 2,
1 2π
2
3, Points of inflection: 2, Relative maximum:
(
3,
0.8
2e2
(1, e −1) −2
(2, e−2) 4
−2
511
512
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
72. f x 2 e3x4 2x
( 53 , 96.942) ( 70.798 )
100
fx e3x2 3e3x4 2x e3x10 6x 0 when x
5 3.
4 , 3
fx e3x6 3e3x10 6x e3x24 18x 0 when x 43 . Relative maximum: Point of inflection:
53 , 96.942 43 , 70.798
2.5 0
e
(b) Ax xf c x 10
f c f c x
74. (a)
− 0.5
c cx cx ec e (d) c
cex c x
x ex 1
2
cex c x c
x ex 1
0
10x2 x1ex (c) Ax x e e 1
4 0
lim c 1
x →0
lim c 0
6
x →
(2.118, 4.591)
9
0 0
The maximum area is 4.591 for x 2.118 and f x 2.547. 76. Let x0, y0 be the desired point on y ex.
y
y ex
3 2
Slope of tangent line
y ex
1
1 ex y
Slope of normal line
y ex0 ex0x x0 We want 0, 0 to satisfy the equation: ex0 x0ex0 1 x0e2x0 x0e2x0 1 0 Solving by Newton’s Method or using a computer, the solution is x0 0.4263.
0.4263, e0.4263
( 0.4263, e − 0.4263 ) x
−1
1 −1
2
3
x x exe 1 1
10x2 x1ex e x e 1
10cec 10c xecx
cecx c xec
x
Section 5.4
Exponential Functions: Differentiation and Integration
513
78. V 15,000e0.6286t , 0 ≤ t ≤ 10 (a)
(b)
20,000
10
0 0
dV 9429e0.6286t dt
(c)
When t 1,
dV 5028.84. dt
When t 5,
dV 406.89. dt
20,000
10
0 0
80. 1.56e0.22t cos 4.9t ≤ 0.25 (3 inches equals one-fourth foot.) Using a graphing utility or Newton’s Method, we have t ≥ 7.79 seconds.
2
10
0
−2
82. (a) V1 1686.79t 23,181.79 V2 109.52t2 3220.12t 28,110.36
(b) The slope represents the rate of decrease in value of the car.
20,000
0
10 0
(c) V3 31,450.770.8592t 31,450.77e0.1518t
(d) Horizontal asymptote: lim V3t 0 t→
As t → , the value of the car approaches 0. (e)
84.
dV3 4774.2e0.1518t dt For t 5,
dV3 2235 dollarsyear. dt
For t 9,
dV3 1218 dollarsyear. dt
f x ex 2, f 0 1 2
2
P1
fx xex 2, f0 0 2
fx
2 x2ex 2
2 ex 2
2 ex 2
x2
1, f0 1
P1x 1 0x 0 1, P10 1
f
−3
3
P2 −2
P1x 0, P10 0 P2x 1 0x 0
1 x2 x 02 1 , P20 1 2 2
P2x x, P20 0 P2x 1, P20 1 The values of f, P1, P2 and their first derivatives agree at x 0. The values of the second derivatives of f and P2 agree at x 0.
514
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions 88. Let u x4, du 4x3 dx.
86. nth term is xnn! in polynomial: y4 1 x
Conjecture: ex 1 x
4
90.
94.
4
e1 1 1
1 e
92.
2
xex 2 dx 2
0
3
x2ex 2 dx
ex 2x dx ex 2 2
0
2
2
0
1 e1
3
1 1 e2x 2e2x dx dx ln1 e2x C 1 e2x 2 1 e2x 2
e1 e
2ex 2ex dx 2 ex ex2ex ex dx ex ex2
e2x 2ex 1 dx ex
ex 2 ex dx
ex 2x ex C
108.
lne2x1 dx
2x 1 dx
106.
1 e sec 2x sec 2x tan 2x dx e sec 2x C 2
u sec 2x, du 2 sec 2x tan 2x
1 sin x e2x dx cos x e2x C1 2
f0 1
1 1 C1 ⇒ C1 1 2 2
fx cos x f x
1 2x e 1 2
cos x
1 2x e 1 dx 2
1 sin x e2x x C2 4 f 0
1 1 C2 ⇒ C2 0 4 4
f x x sin x
1 2x e 4
2 C ex ex
110. y
x2 x C
112. fx
2 x32 3x2 2 3 e dx ex 2 C 3 2 3
102. Let u ex ex, du ex ex dx.
ex ex dx lnex ex C ex ex
C
96. Let u 1 e2x, du 2e2x dx.
100. Let u ex ex, du ex exdx.
104.
4
x
4x dx e
x2 , du x dx. 2
2
4
x2 x3 ... 2! 3!
3
2
x
e
e1x 1 1 2 2 2 dx e1x 3 dx e1x C x3 2 x 2
98. Let u
e3x dx e3x
3
x2 x3 x4 2! 3! 4!
ex ex 2 dx
e2x 2 e2x dx
1 1 2x e 2x e2x C 2 2
Section 5.4
114. (a)
Exponential Functions: Differentiation and Integration
(b)
y
4
y
xe0.2x dx
x
−4
4
−4
0, 23
dy 2 xe0.2x , dx 2
1 0.4
e0.2x 0.4x dx 2
1 0.2x2 2 e C 2.5e0.2x C 0.4
0, 23: 23 2.5e C 2.5 C ⇒ C 1 0
y 2.5e0.2x 1 2
b
116.
ex dx ex
a
b a
2
ea eb
118.
0
1 e2x 2 dx e2x 2x 2
2 0
1 1 e4 4 4.491 2 2 a
4
b
−2
4 0
4
120. (a)
x
x ex dx, n 12
122.
0
0.30.3t dt
0 x
e
Midpoint Rule: 92.1898
1 2
e0.3x 1
1 2
e0.3x
1 2
0
Simpson’s Rule: 92.7385 Graphing Utility: 92.7437
1 2
0.3t
Trapezoidal Rule: 93.8371
2
(b)
2xex dx, n 12
0.3x ln
0
Midpoint Rule: 1.1906 x
Trapezoidal Rule: 1.1827
1 ln 2 2
ln 2 2.31 minutes 0.3
Simpson’s Rule: 1.1880 Graphing Utility: 1.18799 124.
t
0
1
2
3
4
R
425
240
118
71
36
6.052
5.481
4.771
4.263
3.584
ln R
515
(a) ln R 0.6155t 6.0609 R e0.6155t6.0609 428.78e0.6155t (b)
450
−1
5 0
4
(c)
0
4
Rt dt
0
428.78e0.6155t dt 637.2 liters
516
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
126. The graphs of f x ln x and gx ex are mirror images across the line y x.
ln
130.
128. (a) Log Rule: u ex 1 (b) Substitution: u x2
ea ln ea ln eb a b eb
ln eab a b Therefore, ln
ea ea ln eab and since y ln x is one-to-one, we have b eab. eb e
Section 5.5 2. y
2 1
Bases Other than e and Applications
t8
4. y
At t0 16, y
8. loga
12
168
1 4
2 1
t5
6. log27 9 log27 2723 23
At t0 2, y
1 loga 1 loga a 1 a
10. (a)
25
0.7579
12. (a) log3 19 2
2723 9 log27 9
(b)
12
32 19
2 3
4912 7
(b)
1634 8
log49 7 12
3 log16 8 4 18. y 3x
16. y 2x
14. y 3x1
2
x
1
0
1
2
3
x
2
1
0
1
2
x
0
±1
±2
y
1 9
1 3
1
3
9
y
16
2
1
2
16
y
1
1 3
1 9
y
y
12
y
8 1
10 6 8 4
6 4
−1
20. (a) log3
1
2
2 −2
x
−1
x 1
2
3
1 x 81 3x
1 81
x 4 (b) log6 36 x 6x 36 x2
4
−2
−1
−1
x 1
2
22. (a) logb 27 3 b3 27 b3 (b) logb 125 3 b3 125 b5
Section 5.5 24. (a) log3 x log3x 2 1
Bases Other than e and Applications
(b) log10x 3 log10 x 1
log3 xx 2 1 xx 2 x2
log10
31
x3 1 x x3 101 x
2x 3 0
x 1x 3 0
x 3 10x
x 1 OR x 3
3 9x
x 3 is the only solution since the domain of the logarithmic function is the set of all positive real numbers. 56x 8320
26.
x
35x1 86
28.
6x ln 5 ln 8320 x
5x1
ln 8320 0.935 6 ln 5
86 3
x 1ln 5 ln
863
ln x1
863
ln 5 ln
x1
1 0.10 365
30.
365t
ln 5
3.085
t 3 102.6
0.10 365t ln 1 ln 2 365 1 365
863
32. log10t 3 2.6
2
t
t 3 102.6 401.107 ln 2 6.932 0.10 ln 1 365
34. log5 x 4 3.2 x 4 53.2
x 4 53.22 56.4 x 4 56.4 29,748.593 36. f t 3001.007512t 735.41 Zero: t 10
38. gx 1 2 log10 xx 3 Zeros: x 0.826, 3.826 5
10
(10, 0) 0
−10
20
1 3
−5
(− 0.826, 0) 5
(3.826, 0)
−5
517
518
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
40. f x 3x
x
gx log3 x
2
1
0
1
2
1 9
1 3
1
3
9
f x
y
9
f 6
1 9
1 3
1
3
9
2
1
0
1
2
x g x
42. gx 2x
g x
−3
3
6
9
−3
46. f t
y x62x
44.
gx ln 2 2x
3
dy x 2ln 662x 62x dx
ft
62x 2xln 6 1
62x1 2x ln 6 48. g 52 sin 2
50.
1 g 52 2 cos 2 2 ln 5 52 sin 2
52. hx log3
54.
1 3x 2 ln 3 2xx 1
56. f t t 32 log2 t 1 t 32 ft
1 lnt 1 2 ln 2
y log10 2x log10 2 log10 x
y log10
x2 1 x
dy 2x 1 dx x2 1 ln 10 x ln 10
1 0 x 1 ln 3
1 1 1 ln 3 x 2x 1
32t 2t ln 3 1 t2
log10x2 1 log10 x
1 log3 x 1 log3 2 2
1 1 hx x ln 3 2
t2 ln 3 32t 32t t2
dy 1 1 0 dx x ln 10 x ln 10
x x 1 2
log3 x
32t t
58.
1 1 2x ln 10 x2 1 x
x2 1 1 ln 10 xx2 1
y xx1 ln y x 1ln x
1 1 3 t32 t12 lnt 1 2 ln 2 t1 2
1 dy 1 x 1 ln x y dx x dy x1 y ln x dx x
x x2 x 1 x ln x
60.
y 1 x1x ln y
62.
1 ln1 x x
0
64.
2
0
33 52 dx
2
27 25 dx
2 dx
2
2x
lnx 1 1 x1x 1 x x1 x
5x C ln 5
0
5x dx
1 1 1 dy 1 ln1 x 2 y dx x 1x x dy y 1 lnx 1 dx x x 1 x
0 2
4
Section 5.5
66.
2
3 x 7 3x dx
70. (a)
1 3x2 23 x 7 dx 2
68.
sin x
2
Bases Other than e and Applications
519
cos x dx, u sin x, du cos x dx
1 sin x 2 C ln 2
1 2
73x C 2 ln 7
(b)
y
dy esin x cos x dx
6
y
4 2
sin x
e
, 2 sin x
cos x dx e
C
, 2: 2 esin C 1 C ⇒ C 1 x
y esin x 1
10 −2
72. logb x
ln x log10 x ln b log10 b
74. f x log10 x (a) Domain: x > 0
(d) If f x < 0, then 0 < x < 1.
y log10 x
(e) f x 1 log10 x log10 10
(b)
log1010x
10y x
x
f 1 (c)
10x
x must have been increased by a factor of 10.
log10 1000 log10 103 3 log1010,000 log10
104
(f) log10
4
x log x1
10 x1
log10 x2
2
3n n 2n
If 1000 ≤ x ≤ 10,000, then 3 ≤ f x ≤ 4.
Thus, x1x2 102n 100n . 76. f x ax (a) f u v auv au av f u f v
78. Vt 20,000 (a)
(b) f 2x a2x ax 2 f x 2
34
t
(b)
V
34
3 dV 20,000 ln dt 4
t
(c)
V ′(x) x
20,000
dV When t 1: 4315.23 dt
16,000 12,000 8,000
When t 4:
4,000 t 2
V2 20,000
4
6
8
10
34 $11,250 2
dV 1820.49 dt
−1000
2
4
6
12
− 2000 − 3000 − 4000 − 5000 − 6000
Horizontal asymptote: v 0 As the car ages, it is worth less each year and depreciates less each year, but the value of the car will never reach $0.
520
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
80. P $2500, r 6% 0.06, t 20
A 2500 1
0.06 n
20n
n
1
2
4
12
365
Continuous
A
8017.84
8155.09
8226.66
8275.51
8299.47
8300.29
A 2500e0.0620 8300.29 82. P $5000, r 7% 0.07, t 25
0.07 A 5000 1 n
n
1
2
4
12
365
Continuous
A
27,137.16
27,924.63
28,340.78
28,627.09
28,768.19
28,773.01
t
1
10
20
30
40
50
P
94,176.45
54,881.16
30,119.42
16,529.89
9071.80
4978.71
25n
A 5000e0.0725 84. 100,000 Pe0.06t ⇒ P 100,000e0.06t
86. 100,000 P 1
0.07 365
365t
⇒ P 100,000 1
0.07 365
365t
t
1
10
20
30
40
50
P
93,240.01
49,661.86
24,663.01
12,248.11
6082.64
3020.75
88. Let P $100, 0 ≤ t ≤ 20. A 100e0.03t
(a)
A = 100e0.06t A = 100e
0.05t
(b)
A = 100e0.03t
A 100e0.05t A20 271.83
n→
400
A20 182.21 (b)
90. (a) lim
P
or 86%
0.860.25e0.25n 0.215e0.25n 0.25n 1 e 1 e0.25n
P3 0.069
20
0
0.86 0.86 1 e0.25n
P10 0.016
0
A 100e0.06t
(c)
A20 332.01
92. (a)
(d)
12,000
p t
38,000 t5 1 19et5 e 0 5 1 19et53
19et5 1 0
40 0
(b) Limiting size: 10,000 fish (c)
pt
10,000 1 19et5
pt
et5 19 10,000 1 19et52 5
38,000et5 1 19et52
p1 113.5 fishmonth p10 403.2 fishmonth
t ln 19 5 t 5 ln 19 14.72
Section 5.5 94. (a) y1 6.0536x 97.5571
521
(c) The slope of 6.0536 is the annual rate of change in the amount given to philanthropy.
y2 100.0751 17.8148 ln x
(d) For 1996, x 6 and y1 6.0536, y2 2.9691,
y3 99.45571.0506x
y3 6.6015, y4 3.2321.
y4 101.2875x0.1471 (b)
Bases Other than e and Applications
y3 is increasing at the greatest rate in 1996.
150
0
8 100
y3 seems best.
3
96. A
x
x
3 dx
0
ln 3
3
3
0
98.
26 23.666 ln 3
y
x
1
101
102
104
106
1 x1x
2
2.594
2.705
2.718
2.718
30
20
10
x 1
100.
2
3
t
0
1
2
3
4
y
600
630
661.50
694.58
729.30
y Ckt When t 0, y 600 ⇒ C 600. y 600kt 630 661.50 694.58 729.30 1.05, 1.05, 1.05, 1.05 600 630 661.50 694.58 Let k 1.05. y 6001.05t
102. True.
104. True.
f en1 f en ln en1 ln en n1n 1
n
d y Cex dxn y for n 1, 2, 3, . . .
106. True. f x gxex 0 ⇒ gx 0 since ex > 0 for all x.
522
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
108. y x sin x ln y ln xsin x sin x ln x 1 y sin x cos x ln x y x
y xsin x At
sin x cos x ln x x
2 , 2 , y
2
sin2
sin2 cos ln 2 2 2
2 0 1 2
1 x 2 2
Tangent line: y
yx
Section 5.6 2.
Differential Equations: Growth and Decay
dy 4x dx y
dy 4y dx
4.
4 x dx 4x
x2 C 2
y
6.
dy dx 4y 1 dy 4y
4 y exC1 Cex
9y2 4x32 C
y 4 Cex y x1 y
xy y 100x
10.
y 100x x y x100 y
y x 1y y dx 1y
dy 1y
x dx x dx
x2 ln1 y C1 2 1 y ex 2 C1 2
y eC1 ex 2 1
y x 100 y y dx 100 y
1 dy 100 y
ln100 y
x dx x dx
x2 C1 2
2
Cex 2 1 2
x dx
3y2 2 32 x C1 2 3
ln 4 y dy x C1
8.
3y
3yy x
3yy dx
dx
x
ln100 y
x2 C1 2
100 y ex 2 C1 2
y eC1 ex 2 100 2
y 100 Cex 2 2
Section 5.6 dP k10 t dt
12.
dP dt dt
Differential Equations: Growth and Decay dy kxL y dx
14.
1 dy kx L y dx
k10 t dt
k dP 10 t2 C 2 k P 10 t2 C 2
1 dy dx L y dx
kx dx
kx 2 1 dy C1 Ly 2
lnL y
kx 2 C1 2
L y ekx 2 C1 2
y L eC1 ekx 2 2
y L Cekx 2 2
16. (a)
y
(b)
4
dy xy, dx dy x dx y
(0, 12 ) x
−4
0, 21
4
ln y
x2 C 2
y ex 2C C1ex 2
−4
2
2
0, 21: 21 C e 1
18.
dy 3 t, 0, 10 dt 4
dy
20.
15
3 t dt 4
0
1 y t32 C 2
dy y
10
ln y
−5
3 dt 4
(0, 10) −5
3 t C1 4
5 −5
10 Ce0 ⇒ C 10 y 10e3t4
dN kN dt
24.
dP kP dt P Cekt
(Theorem 5.16)
0, 250: C 250
(Theorem 5.16)
0, 5000: C 5000
1, 400: 400 250ek ⇒ k ln
400 8 ln 250 5
When t 4, N 250e4ln85 250eln85
4
250
40
eC1 e34t Ce3t4
1 y t 32 10 2
N Cekt
1 1 2 ⇒ y ex 2 2 2
y e34tC1
1 10 032 C ⇒ C 10 2
22.
⇒ C1
dy 3 y, 0, 10 dt 4
(0, 10)
0
85
4
8192 5
1, 4750: 4750 5000ek ⇒ k ln
19 20
When t 5, P 5000eln19205 5000
19 20
5
3868.905
523
524
Chapter 5
26. y Cekt, 0, 4,
Logarithmic, Exponential, and Other Transcendental Functions
5, 12
C4
3, 12, 4, 5
1 Ce3k 2
y 4ekt
5 Ce4k
1 4e5k 2 k
y Cekt,
28.
2Ce3k
ln18 0.4159 5
1 4k Ce 5
10e3k e4k
y 4e0.4159t
10 ek k ln 10 2.3026 y Ce2.3026t 5 Ce2.30264 C 0.0005 y 0.0005e2.3026t
30. y
dy ky dt
32.
dy 1 2 xy dx 2 dy > 0 when y > 0. Quadrants I and II. dx
34. Since y Ce ln121620t , we have 1.5 Ce ln1216201000 ⇒ C 2.30 which implies that the initial quantity is 2.30 grams. When t 10,000, we have y 2.30e ln12162010,000 0.03 gram. 36. Since y Ce ln125730t, we have 2.0 Ce ln12573010,000 ⇒ C 6.70 which implies that the initial quantity is 6.70 grams. When t 1000, we have y 6.70e ln1257301000 5.94 grams. 38. Since y Ce ln125730t, we have 3.2 Ce ln1257301000 ⇒ C 3.61. Initial quantity: 3.61 grams. When t 10,000, we have y 1.08 grams. 40. Since y Ce ln1224,360t, we have 0.4 Ce ln1224,36010,000 ⇒ C 0.53 which implies that the initial quantity is 0.53 gram. When t 1000, we have y 0.53e ln1224,3601000 0.52 gram.
42. Since
dy ky, y Cekt or y y0ekt. dx
1 y y0e5730k 2 0 k
ln 2 5730
0.15y0 y0eln 25730t ln 0.15 t
ln 2t 5730 5730 ln 0.15 15,682.8 years. ln 2
44. Since A 20,000e0.055t, the time to double is given by 40,000 20,000e0.055t and we have 2 e0.055t ln 2 0.055t t
ln 2 12.6 years. 0.055
Amount after 10 years: A 20,000e0.05510 $34,665.06
Section 5.6 46. Since A 10,000ert and A 20,000 when t 5, we have the following. 20,000 10,000e5r
48. Since A 2000ert and A 5436.56 when t 10, we have the following. 5436.56 2000e10r
ln 2 0.1386 13.86% 5
r
Differential Equations: Growth and Decay
Amount after 10 years: A 10,000e ln 2510 $40,000
r
ln5436.562000 0.10 10% 10
The time to double is given by 4000 2000e0.10t t
0.06 12
50. 500,000 P 1
ln 2 6.93 years. 0.10
1240
0.09 12
52. 500,000 P 1
P 500,0001.005480 $45,631.04
1225
P 500,000 1
0.09 12
300
$53,143.92
(c) 2000 1000 1
54. (a) 2000 10001 0.6t 2 1.06t
2 1
ln 2 t ln1.06 t
ln 2 11.90 years ln 1.06
(b) 2000 1000 1
2 1
0.06 12
0.06 12
12t
1 12
0.055 12
ln 2 12t ln 1 1 t 12
12t
0.055 365
0.055 365
ln 2 365t ln 1 t
12t
1 365
365t
365t
0.055 365
ln 2 12.60 years 0.055 ln 1 365
(d) 2000 1000e0.055t
0.055 12
ln 2 12.63 years 0.055 ln 1 12
ln 2 11.55 years 0.06
ln 2 12.95 years ln 1.055
t
2 1
0.055 2 1 12
0.06 365 1 ln 2 t 11.55 years 0.06 365 ln 1 365
ln 2 t ln1.055
ln 2 365t ln 1
(c) 2000 1000 1
2 1.055t
(b) 2000 1000 1
365t
ln 2 0.06t
56. (a) 2000 10001 0.055t
t
365t
2 e0.06t
ln 2 11.58 years 0.06 ln 1 12
(d) 2000 1000e0.06t
0.06 ln 2 12t ln 1 12 t
12t
0.06 365
0.06 365
2 e0.055t ln 2 0.055t t
ln 2 12.60 years 0.055
525
526
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
58. P Cekt Ce0.031t
60. P Cekt Ce0.004t
P1 11.6 Ce0.0311 ⇒ C 11.9652
P1 3.6 Ce0.0041 ⇒ C 3.5856
P 11.9652e0.031t
P 3.5856e0.004t
P10 16.31
or
P10 3.45
16,310,000 people in 2010
62. (a) N 100.15961.2455t
C 742,000 632,000 742,000e2k
Analytically, 400 100.15961.2455t 1.2455t
k
400 3.9936 100.1596
t
When t 4, y $538,372.
ln 3.9936 6.3 hours. ln 1.2455
20 301 e30k
e0.0366t
ln13 ln 3 0.0366 30 30 e0.0366t
25 301 e0.0366t
(b)
30e30k 10
N 301
ln632742 0.0802 2
y 742,000e0.0802t
t ln 1.2455 ln 3.9936
k
3,450,000 people in 2010
y Cekt, 0, 742,000, 2, 632,000
64.
(b) N 400 when t 6.3 hours (graphing utility)
66. (a)
or
t
1 6 ln 6 49 days 0.0366
68. S 251 ekt (a) 4 251 ek1 ⇒ 1 ek
4 21 21 ⇒ ek ⇒ k ln 0.1744 25 25 25
(b) 25,000 units lim S 25 t→
(c) When t 5, S 14.545 which is 14,545 units.
(d)
25
0
8 0
70. (a) R 979.39931.0694t 979.3993e0.0671t I 0.1385t4 2.1770t3 9.9755t2 23.8513t 266.4923 (b)
2000
Rate of growth Rt 65.7e0.0671t 0
10 0
(c)
(d) Pt
500
I R
1
0
10 0
0
10 0
Section 5.7
72.
93 10 log10
I 10log10 I 16 1016
Differential Equations: Separation of Variables
6.7 log10 I ⇒ I 106.7 I 80 10 log10 16 10log10 I 16 10
1 dy y 80
k dt
When t 0, y 1500. Thus, C ln 1420.
10 106.7
10
Percentage decrease:
dy ky 80 dt
lny 80 kt C.
8 log10 I ⇒ I 108 6.7
Since
74.
8
100 95%
When t 1, y 1120. Thus, k1 ln 1420 ln1120 80 k ln 1040 ln 1420 ln Thus, y 1420eln104142 t 80. When t 5, y 379.2.
76. True
78. True
Section 5.7
Differential Equations: Separation of Variables
2. Differential equation: y
2xy x2 y2
Solution: x 2 y 2 Cy Check: 2x 2yy Cy y
2x 2y C
y
2xy 2xy 2xy 2xy 2y 2 Cy 2y 2 x 2 y 2 y 2 x 2 x 2 y 2
4. Differential equation: y 2y 2y 0 Solution: y C1ex cos x C2ex sin x y C1 C2ex sin x C1 C2ex cos x
Check:
y 2 C1ex sin x 2C2ex cos x y 2y 2y 2C1ex sin x 2C2ex cos x 2 C1 C2ex sin x C1 C2ex cos x 2C1ex cos x C2ex sin x 2C1 2C1 2C2 2C2ex sin x 2C2 2C1 2C2 2C1ex cos x 0 2 6. y 3 e2x ex
y 23 2e2x ex y 23 4e2x ex 2 2 Substituting, y 2y 3 4e2x ex 2 3 2e2x ex 2ex.
104 . 142
527
528
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
In Exercises 8 –12, the differential equation is y 4 16y 0. 8.
y 3 cos 2x
y 5 ln x
10.
y4 48 cos 2x y4 16y 48 cos 2x 48 cos 2x 0,
y4 Yes. y4 16y
12.
30 x4 30 80 ln x 0, x4
No.
y 3e2x 4 sin 2x y4 48e2x 64 sin 2x y4 16y 24e2x 64 sin 2x 163e2x 4 sin 2x 0,
Yes
In 14–18, the differential equation is xy 2y x3ex. 14. y x 2ex, y x 2ex 2xex ex x2 2x xy 2y x ex
x2
2x 2
x2ex
x3ex,
16. y sin x, y cos x xy 2y xcos x 2sin x x3ex,
Yes.
18. y x2ex 5x2, y x2ex 2xex 10x xy 2y xx2ex 2xex 10x 2x2ex 5x2 x3ex,
Yes. 22. 2x 2 y 2 C passes through 3, 4
20. y A sin t
29 16 C ⇒ C 2
d 2y A 2 sin t dt 2
Particular solution: 2x 2 y 2 2
Since d 2ydt 2 16y 0, we have A 2 sin t 16A sin t 0. Thus, 2 16 and ± 4. 24. Differential equation: yy x 0 General solution: x 2 y 2 C Particular solutions: C 0, Point C 1, C 4, Circles
26. Differential equation: 3x 2yy 0 General solution: 3x2 2y2 C 6x 4yy 0 23x 2yy 0 3x 2yy 0 Initial condition: y1 3: 312 232 3 18 21 C Particular solution: 3x2 2y2 21
y 2 1
1
2
x
No.
Section 5.7
Differential Equations: Separation of Variables Initial conditions: y2 0, y2
28. Differential equation: xy y 0 General solution: y C1 C2 ln x
0 C1 C2 ln 2
1 1 y C2 , y C2 2 x x 1 1 xy y x C2 2 C2 0 x x
1 2
y
C2 x
1 C2 ⇒ C2 1, C1 ln 2 2 2 Particular solution: y ln 2 ln x ln
x 2
30. Differential equation: 9y 12y 4y 0 General solution: y e2x3C1 C2 x
2 2 2 y e2x3C1 C2x C2e2x3 e2x3 C1 C2 C2 x 3 3 3 y
2 2 2 2x3 2 2 2 2 e C C2 C2 x e2x3 C2 e2x3 C1 2C2 C2 x 3 3 1 3 3 3 3 3
9y 12y 4y 9
23 e 23 C 2x3
1
2C2
2 2 2 C x 12e2x3 C1 C2 C2 x 4e2x3C1 C2 x 0 3 2 3 3
Initial conditions: y0 4, y3 0 0 e2C1 3C2 4 1C1 0 ⇒ C1 4 0 e24 3C2 ⇒ C2
4 3
4 Particular solution: y e2x3 4 x 3
32.
dy x3 4x dx y
36.
34.
x4 2x2 C 4
x3 4x dx
dy x cos x 2 dx y
x cosx2 dx
1 sinx2 C 2
u x 2, du 2x dx
40.
dy x 5 x. Let u 5 x, u2 5 x, dx 2u du dx y
y
38.
x 5 x dx
5 u2u2u du
10u2 2u4 du
10u3 2u5 C 3 5
10 2 5 x32 5 x52 C 3 5
dy ex dx 1 ex
ex dx ln1 ex C 1 ex
dy tan2 x sec2 x 1 dx y
sec2 x 1 dx tan x x C
529
530
42.
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
dy 5ex2 dx y
dy x 2 2 dx 3y 2
44.
21 dx
5ex2 dx 52 ex2
3y 2 dy
10ex2 C
dr 0.05s ds
46.
y3
48.
dr
dy y
r 0.025s 2 C
x 2 2dx
x3 2x C 3
xy y
0.05s ds
y
50.
dx x
dy 6 cos x dx
y dy
y2
dy 5x dx
dy
4y
54. 5x
x2 9
y 5x2 912 C 56. x y y 0
dy 3ex dx
4y dy
dx
3ex dx
2y2 3ex C 58. 2xy ln x2 0
y12 dy x12 dx
2x
dy 2 ln x dx
2 32 2 y x32 C1 3 3
dy
ln x dx x
y
ln x2 C 2
y32 x32 C Initial condition: y 1 4, 432 132 8 1 9 C
y1 2: 2 C 1 y ln x2 2 2
Particular solution: y32 x32 9
y 1 x2
60.
dy x 1 y 2 dx
1 y212 y dy
1 x212 x dx
1 y212 1 x212 C y0 1:
0 1 C ⇒ C 1
1 y 2 1 x 2 1
62.
6 cos x dx
y2 6 sin x C1 2
ln y ln x ln C ln Cx y Cx
52. x2 9
dr er2s ds
erdr
e2s ds
1 er e2s C 2 1 1 r0 0: 1 C ⇒ C 2 2 1 1 er e2s 2 2 1 1 er e2s 2 2 r ln r ln
12e
2s
1 e2s 1 ln 2 2
1 2e 2s
12 sin x C
Section 5.7
64. dT kT 70 dt 0
Differential Equations: Separation of Variables dy 2y dx 3x
66.
dT k dt T 70
lnT 70 k t C1
3 dy y
2 dx x
ln y3 ln x2 ln C
T 70 Cekt
y3 Cx2
Initial condition: T0 140; 140 70 70 Ce0 C
Initial condition: y8 2, 23 C82, C
Particular solution: T 70 70ekt, T 701 ekt
Particular solution: 8y3 x2, y
m
68.
dy y 0 y dx x 0 x
dy y
70.
1 23 x 2
f x, y x3 3x2y2 2y2 f tx, ty t3x3 3t4x2y2 2t2y2
dx x
Not homogeneous
ln y ln x C1 ln x ln C ln Cx y Cx f x, y
72.
f t x, t y
xy
x2 y2
74.
f t x, t y tantx t y tant x y
tx ty x2 t 2 y 2
t 2
f x, y tanx y
Not homogeneous
2
xy t xy t
x 2 y 2 t x 2 y 2
Homogeneous of degree 1
f x, y tan
76.
f t x, t y tan
y x ty y tan tx x
Homogeneous of degree 0
78. y
x3 y3 xy2
y
80.
x y 2 dy x3 y3 dx y vx,
dy x dv v dx
xvx x dv v dx x3 vx3 dx 2
x4 v2 dv x3 v3 dx x3 dx v3 x3 dx xv 2 dv dx
v 2 dv
vx
1 dx x
v3 ln x C 3
yx 3 lnx C 3
y3 3x3 ln x Cx3
dv x 2 v 2 x 2 dx 2x 2 v
2v dx 2x dv
x2 y2 , y vx 2xy
1 v2 dx v
2v dx dv v2 1 x
lnv 2 1 ln x ln C ln v2 1
C x
y2 C 1 x2 x y 2 x2 Cx
C x
1 8
531
532
Chapter 5
82.
y
Logarithmic, Exponential, and Other Transcendental Functions
2x 3y , y vx x
dv 2x 3vx vx 2 3v dx x dv 2 2v ⇒ dx
x
ln 1 v ln
x2
x 2 v 2 dx x 2 x 2 vv dx x dv 0
dv dx 2 1v x
ln C ln
y 2 dx xx y dy 0, y vx
84.
x2C
1v dx dv v x
v ln v ln x ln C1 ln
C1 x
v ln
C1 xv
1vxC 2
1
y x2C x
C1 ev vx
y x2C 1 x
C1 eyx y
y Cx3 x
y Ceyx Initial condition: y1 1, 1 Ce1 ⇒ C e Particular solution: y e1yx
86. 2x 2 y 2 dx xy d y 0
88.
dy x dx y
Let y vx, dy x dv v dx.
v 2x 2
2x 2 v 2
2
2v 2
2x 2
y
dx xv xx d v v d x 0
2x 2
dx
x3v
dv 0
d x x v d v
x2 C1 v 212
12
C x2 y212 x
1 Cx2 y212 x y1 0: 1 C1 0 ⇒ C 1 1 x2 y2 x 1 x x2 y2
2
4
−4
ln x2 ln1 v 212 ln C
x
−2 −2
1 ln1 v 2 C1 2
1 y2 C 1 2 x2 x
2 −4
2 v dx dv x 1 v2 2 ln x
4
y dy x dx y 2 x 2 C1 2 2 y 2 x2 C
Section 5.7 dy 0.25x4 y dx
90.
y
8
dy 0.25x dx 4y
dy y4
Differential Equations: Separation of Variables
4
0.25x dx
−4
x
−2
2
4
−2
1 x dx 4
1 ln y 4 x 2 C1 8
y 4 eC1 18x Ce18x 2
2
y 4 Ce18x
2
dy 2 y, y0 4 dx
92.
94.
dy 0.2x2 y, y0 9 dx
9
10
−5
5
−5
−1
96.
dy ky, y Cekt dt Initial conditions: y0 20, y1 16 20 Ce C 0
98.
5 0
dy kx 4 dx The direction field satisfies dydx 0 along x 4: Matches (b).
16 20ek k ln
4 5
Particular solution: y 20et ln45 When 75% has been changed: 5 20et ln45 1 et ln45 4 t
100.
ln14 6.2 hr ln45
dy ky2 dx
The direction field satisfies dydx 0 along y 0, and grows more positive as y increases. Matches (d).
102. From Exercise 101, w 1200 Cekt, k 1 w 1200 Cet w0 w0 1200 C ⇒ C 1200 w0 w 1200 1200 w0et
533
534
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
104. Let the radio receiver be located at x0, 0. The tangent line to y x x2 joins 1, 1 and x0, 0.
y
2
Transmitter (−1, 1) 1
y=x−
x2
Radio
−1
(b) Now let the transmitter be located at 1, h.
(a) If x, y is the point of tangency on the y x x2, then
1 2x
y 1 x x2 1 1 2x x1 x1
x2 2x h 1 0
2x 2 0
2 ± 4 8 x 2
1
x
3
1 3 3 5
10 1 x0 1 1 3
3 1 x0 6 3 3
(c)
3 2 h h 4
3
Then,
h0 h 3 2 h h 4 1 x0 1 1 2 h
4 3 6 1.155 6 3 3
2h 4 3 2 h 2 h
x0 1
2 h h 2h 4 3 2 h
10
x0 0.25
2
y x x2
6 3 3
6 3 3 x0 6 3 3 x0
2 ± 4 4h 1
1 2 h
y x x2 3 3 5 Then
y h x x2 h x1 x1
x 2x 2 1 2x x x 2 h
x 2x2 1 2x x x2 1 x2
x
x 1 x0
h 2 h 1 2h 4 3 2 h
3
−2
There is a vertical asymptote at h 14 , which is the height of the mountain. x2 2y2 C
106. Given family (hyperbolas):
2x 4yy 0 y y
Orthogonal trajectory:
2yy 2C
x 2y
2y x
y
−2
2x y
y dy 2x dx y2 x 2 K1 2
k x2
2x 2 y 2 K 4
2
3
y
ln y 2 ln x ln k
−3
C y2 1 y y 2x y 2x
Orthogonal trajectory (ellipse):
dy 2 dx y x
y kx2
y 2 2Cx
108. Given family (parabolas):
−6
6
−4
Section 5.8
Inverse Trigonometric Functions: Differentiation
110. Given family (exponential functions): y Cex y
Cex
y
Orthogonal trajectory (parabolas):
535
4
y −6
1 y
6
−4
y dy dx y2 x K1 2 y2 2x K 114. Two families of curves are mutually orthogonal if each curve in the first family intersects each curve in the second family at right angles.
112. The number of initial conditions matches the number of constants in the general solution.
116. True dy x 2y 1 dx 118. True x 2 y 2 2Cy
x 2 y 2 2Kx dy K x dx y
x dy dx C y x Cy
Kx Kx x2 2Kx 2x2 x2 y 2 2x2 y 2 x2 2 2 1 2 2 2 2 y Cy y 2Cy 2y x y 2y x y2
Section 5.8
Inverse Trigonometric Functions: Differentiation
2. y arccos x (a)
1
x y
0.8
3.142
(b)
0.6
2.499
y
0.4
2.214
1.982
0.2 1.772
0 1.571
0.2
0.4
0.6
0.8
1
1.369
1.159
0.927
0.634
0
(d) Intercepts:
(c)
π
No symmetry −1
1 0
x
−1
4.
1
1, 4 4
,
,
3,
3 , 6 3 6
0, 2 and 1, 0
6. arcsin 0 0
3, 3
536
Chapter 5
8. arccos 0
12. arccos
Logarithmic, Exponential, and Other Transcendental Functions
2
3
2
10. arc cot 3
56
18. (a) tan arccos
2
2
14. arcsin0.39 0.40
tan4 1
16. arctan3 1.25
(b) cos arcsin 2
5 6
5 12 13 13
13 5
θ
2
12
θ 2
53
20. (a) sec arctan
34
5
65 5 1111
(b) tan arcsin
θ
5
−3
11
θ
34 −5
6
22. y secarc tan 4x
24. y cosarccot x
1 + 16x 2
arctan 4x
θ
y sec 1 16x2
1
y cos
26. y sec arcsinx 1
arcsinx 1
x−1
θ
1 y sec 2x x2
30.
2x −
arcsin x2
xh r
x
r
θ r 2 − (x − h)2
r 2 x h2
r
32. arctan2x 5 1 2x 5 tan1
f=g −2
2
1 x tan1 5 1.721 2
−5
Asymptote: x 0 x 2
2 4 − x2
x cos 2 tan
x x2 1
xh r
y cos
5
arccos
1
θ
28. y cos arcsin 1
x2 + 1
arccot x
4x
θ x
4 x2
x
x−h
Section 5.8
Inverse Trigonometric Functions: Differentiation
537
34. arccos x arcsec x x cosarcsec x x
1 x
x
x2 − 1
θ 1
x2 1 x ±1
36. (a) arcsinx arcsin x, x ≤ 1.
(b) arccosx arccos x, x ≤ 1.
Let y arcsinx. Then,
Let y arccosx. Then,
x sin y ⇒ x sin y ⇒ x siny.
x cos y ⇒ x cos y ⇒ x cos y.
Thus, y arcsin x ⇒ y arcsin x. Therefore, arcsinx arcsin x.
38. f x arctan x
x tan y 2
2
Thus, y arccos x ⇒ y arccos x. Therefore, arccosx arccos x.
y
40. f x arccos
π
π 2
−6 −4 −2
2
4
2t
fx
1 1 x21 x 2x11 x 2
52. y lnt2 4 y
1 x x4 x2 4 arcsin 2 2
56.
1 1 1 x 4 x212 2x 4 x2 2 1 x22 2 2 1 x2 4 4 x2 4 x2 2 4 x2
4
x2
2
2x 4x2 1
hx 2x arctan x
fx 0
2
4
x
6
1
x 4x2 1
48. hx x2 arctan x
50. f x arcsin x arccos x
y
−2
44. f x arcsec 2x
1 t 4
−6 −4
Range: 0,
46. f x arctan x
54. y
π
(4, 0)
Domain: 4, 4
6
42. f t arcsin t 2
fx
(−4, π )
x 4 cos y x
f x is the graph of arctan x shifted 2 units upward.
ft
y
x cos y 4
Domain: , Range: 0,
4x
2t 1 t2 4 2
x2 1 x2
1 t arctan 2 2 1
t 1 2
2
12
1 2t 1 2t 2 t2 4 t2 4 t 4
y x arctan 2x
1 ln1 4x2 4
dy 2x 1 8x arctan2x arctan2x dx 1 4x2 4 1 4x2
538
Chapter 5
58. y 25 arcsin y 5
Logarithmic, Exponential, and Other Transcendental Functions x x25 x2 5
1 1 x22
60. y arctan
1 25 x2 x 25 x212 2x 2
y
1 1 1 x2 422x 2 1 x22 2
25 x2 25 x2 2 2 25 x 25 x 25 x2
2 x x2 4 x2 42
2x2 25 x2
2x2 8 x x2 42
y
62. f x arctan x, a 1 fx
1 1 x2
f x
2x 1 x22
P1 (x) π 2
f
π 4 −4
−2
x 2
P2 (x)
P1x f 1 f 1x 1
1 x 1 4 2
P2x f 1 f 1x 1
1 1 1 f 1x 12 x 1 x 12 2 4 2 4
64. f x arcsin x 2x
66. f x arcsin x 2 arctan x
1 1 fx 2 0 when 1 x2 or 2 2 1 x x± f x f
1 x 2 2x2 4
3
2
1 1 x2
x4 6x2 3 0 x ± 0.681
Relative minimum:
3
2
2 , 0.68 3
23, 0.68
68. arctan 0 0. is not in the range of y arctan x. x 3
3x
3 dx d 2 dt x 9 dt If x 10,
70. The derivatives are algebraic. See Theorem 5.18.
74. cos
arccot
If x 3,
By the First Derivative Test, 0.681, 0.447 is a relative maximum and 0.681, 0.447 is a relative minimum.
0
(b)
fx
d 11.001 radhr. dt
d 66.667 radhr. dt
A lower altitude results in a greater rate of change of .
750 s
s
arccos d d dt ds
750s
θ 750
ds 1 750 ds 2 dt s2 dt 1 750s
750 ds ss2 7502 dt
Section 5.9 76. (a) Let y arcsin u. Then
Inverse Trigonometric Functions: Integration (b) Let y arctan u. Then
1
sin y u y
cos y y u
1 − u2
sec2 y
u u dy . dx cos y 1 u2
u y
dy u dx
1
dy u u . dx sec2 y 1 u2
(c) Let y arcsec u. Then
(d) Let y arccos u. Then
u
sec y u
u2− 1
1
cos y u
y
dy sec y tan y u dx
1 + u2
tan y u
u
539
1
1 − u2
y
sin y
dy u u . dx sec y tan y u u2 1
dy u dx
u
u u dy . dx sin y 1 u2
Note: The absolute value sign in the formula for the derivative of arcsec u is necessary because the inverse secant function has a positive slope at every value in its domain.
(f) Let y arccsc u. Then u
csc y u
1
y
(e) Let y arccot u. Then
csc y cot y
1 + u2
cot y u
1
dy u dx
u2 − 1
dy u u dx csc y cot y u u2 1
y
csc2 y
dy u dx
u
Note: The absolute value sign in the formula for the derivative of arccsc u is necessary because the inverse cosecant function has a negative slope at every value in its domain.
u u dy . dx csc2 y 1 u2
78. f x sin x
3
gx arcsinsin x
f
(a) The range of y arcsin x is 2 ≤ y ≤ 2.
− 2
2
g
(b) Maximum: 2
−3
Minimum: 2 82. False
80. False
The range of y arcsin x is , . 2 2
Section 5.9
1
6.
4 4 3 4 dx dx arctan3x C 1 9x2 3 1 9x2 3
10.
1 1 x1 dx arctan C 4 x 12 2 2
3 4x2
2
1
Inverse Trigonometric Functions: Integration
2.
2
arcsin2 0 arccos2 0 0
dx
3
2
2 1
4x2
dx
3 2
arcsin2x C
1
4.
0
8.
12.
1 x 2 dx arcsin 2 4 x
3
1 x 1 arctan 2 dx 3 3 3 9 x x4 1 dx x2 1
1
0
3 3
6
36
1 x2 1 dx x3 x C 3
540
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
14. Let u t 2, du 2t dt.
t4
16. Let u x2, du 2x dx.
t 1 1 t2 1 dt 2t dt arctan C 2 2 2 16 2 4 t 8 4
1 1 1 dx 2x dx 2 x2x22 22 xx 4 4
18. Let u arccos x, du
12
0
1 dx. 1 x2
12
arccos x dx 1 x2
0
2
22.
1
1 dx 3 x 22
2
24.
0
20. Let u 1 x2, du 2x dx. 0
12
2
1
arccos x dx 1 x2
1 arccos2 x 2
0
3
2
0
x 1 dx 1 x2 2
32 0.925 32
1 1 x2 dx arctan 32 x 22 3 3
cos x dx arctansin x 1 sin2 x
1 x2 arcsec C 4 2
4
26.
2 1
0
3
1 2x dx 1 x2
12 ln1 x
0
2
3
ln 2
3
18
3 1 dx. u x, du dx, dx 2u du 2x1 x 2x
3 2u du du 3 3 arctan u C 2 u1 u2 1 u2 3 arctanx C
28.
1 x2
30.
x2 2x 2 1 dx dx x 12 4 2 x 12 4
4x 3
dx 2
2
2
34.
36.
x2
2 x2 4x
dx
dx 41 x2 3 arcsin x C
3 dx x 12 4
2
dx 1 x2 arctan 2 3 3 2 x 2 9
2 2
4 1 arctan 3 3
2x 2 1 dx 7 dx ln x2 2x 2 7 arctanx 1 C x2 2x 2 1 x 12
2 4 x2 4x 4
2 dx 4 x 22
2 arcsin
40.
1 1 x2
3 x1 1 lnx2 2x 5 arctan C 2 2 2
dx 4x 13
2x 5 dx x2 2x 2
dx 3
32.
2x 1 x2
1 dx x 1x2 2x
dx
38. Let u x2 2x, du 2x 2 dx.
x1 x2 2x
x2 C 2
1 dx arcsecx 1 C x 1x 12 1
dx
1 x2 2x122x 2 dx 2
x2 2x C
Section 5.9
Inverse Trigonometric Functions: Integration
42. Let u x2 4, du 2x dx.
x 9 8x2 x4
1 2x 1 x2 4 dx arcsin C 2 25 x2 42 2 5
dx
44. Let u x 2, u2 2 x, 2u du dx
x 2
x1
dx
2u2 du 3
u2
2u
46. The term is
48. (a) (b)
50. (a) (b) (c)
32
2
2u2 6 6 1 du 2 du 6 2 du u2 3 u 3
u 6 arctan C 2x 2 23 arctan 3 3
9 9 9 3 : x2 3x x2 3x x 4 4 4 2
2
9 4
(c)
x 3 2 C
2
ex dx cannot be evaluated using the basic integration rules. xex dx
2
1 x2 e C, u x2 2
1 1x 1 e dx e1x C, u x2 x
1 dx cannot be evaluated using the basic integration rules. 1 x4
x 1 2x 1 dx dx arctanx2 C, u x2 1 x4 2 1 x22 2
x3 1 4x3 1 dx dx ln1 x 4 C, u 1 x 4 4 1x 4 1 x4 4
52. (a)
54.
y
4
dy 2y , y0 2 dx 16 x2 3
x
−1.25
1.25 −3 −4
(b)
−1
dy x16 y2, 0, 2 dx dy 16 y2
x dx
4y x2 C 2
arcsin
42 C ⇒ C 6
0, 2: arcsin
4y x2 6 2
arcsin
3
y x2 sin 4 2 6
y 4 sin
x2 6
2
541
542
Chapter 5
1
56. A
0
Logarithmic, Exponential, and Other Transcendental Functions
x 1 dx arcsin 2 4 x2
1 0
1
6
58.
arcsin x dx 0.571
0
y
y
π
1
2
π 4
x
60. Fx
1 2
x
1
1 2
x2
x
1 2
1
2 dt t2 1
(a) Fx represents the average value of f x over the interval x, x 2. Maximum at x 1, since the graph is greatest on 1, 1.
(b) Fx arctan t Fx
62.
1 6x x2
x2 x
arctanx 2 arctan x
1 1 4x 1 1 x2 x2 4x 5 2 0 when x 1. 2 2 1 x 2 1x x2 1x2 4x 5 x 1x2 4x 5
dx
(a) 6x x2 9 x2 6x 9 9 x 32
1 dx 6x x2
(c)
4
y2
dx x3 arcsin C 3 9 x 32
y1 −1
(b) u x, u2 x, 2u du dx
1 6u2 u4
2u du
7
−2
2 6 u2
du 2 arcsin
u6 C 2 arcsin 6 C x
The antiderivatives differ by a constant, 2. Domain: 0, 6
64. Let f x arctan x fx
x 1 x2
1 1 x2 2x2 > 0 for x > 0. 1 x2 1 x22 1 x2
y 5
x Since f 0 0 and f is increasing for x > 0, arctan x > 0 for x > 0. Thus, 1 x2 x arctan x > . 1 x2 Let gx x arctan x gx 1
1 x2 > 0 for x > 0. 2 1x 1 x2
Since g0 0 and g is increasing for x > 0, x arctan x > 0 for x > 0. Thus, x > arctan x. Therefore, x < arctan x < x. 1 x2
y3
4 3 2
y2
1
y1 2
4
6
x 8
10
Section 5.10
Section 5.10
Hyperbolic Functions
e0 e0 1 2
2. (a) cosh0
4. (a) sinh10 0 (b) tanh10 0
2 (b) sech1 0.648 e e1
6. (a) csch12 ln (b) coth13
8.
1 2 5 0.481
1 4 ln 0.347 2 2
1 cosh 2x 1 e2x e2x2 e2x 2 e2x ex ex 2 2 4 2
10. 2 sinh x cosh x 2
e
x
ex 2
e
x
2
cosh2 x
ex e2x e2x sinh 2x 2 2
xy2
12. 2 cosh
Hyperbolic Functions
x 2 y coshx 2 y 2 e e
x
2
exy2 2
xy2
e
exy2 2
ey ey ex ex ex ey ey 4 2 2
cosh x cosh y tanh x
14.
1 2
2
Putting these in order:
1 2
sech2 x 1 ⇒ sech2 x
3
3 ⇒ sech x 4 2
23 1 cosh x 3 32 1 2 coth x 12 sinh x tanh x cosh x csch x
1 33
122 3 3
3
3
csch x 3
cosh x
23 3
sech x
3
tanh x
1 2
coth x 2
3
3
3
16. y coth3x y 3
sinh x
18. gx lncosh x
3x
csch2
20. y x cosh x sinh x y x sinh x cosh x cosh x x sinh x
gx
1 sinh x tanh x cosh x
22. ht t coth t ht 1 csch2 t coth2 t
2
543
544
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
24. gx sech2 3x
28. y sechx 1
26. f x esinh x
gx 2 sech3x sech3x tanh3x3
y sechx 1 tanhx 1
fx cosh xesinh x
6 sech 3x tanh 3x 2
30. f x x sinhx 1 coshx 1
6
fx x coshx 1 sinhx 1 sinhx 1 x coshx 1 fx 0 for x 0. By the First Derivative Test,0, cosh1 0, 1.543 is a relative minimum.
−6
6 (0, −1.543) −2
32. hx 2 tanh x x
34.
y a cosh x y a sinh x
2
y a cosh x
(0.88, 0.53) −3
3
Therefore, y y 0.
(− 0.88, − 0.53) −2
Relative maximum: 0.88, 0.53 Relative minimum: 0.88, 0.53 36. f x cosh x
f 1 cosh0 1
f x sinh x
f 1 sinh0 0
f x cosh x
f 1 cosh0 1
3
f P2 P1 −2
P1x f 0 f0x 0 1
2 0
1 P2x 1 2 x2
38. (a) y 18 25 cosh
x , 25 ≤ x ≤ 25 25
(b) At x ± 25, y 18 25 cosh1 56.577. At x 0, y 18 25 43.
80
x (c) y sinh 25. At x 25, y sinh1 1.175 −25
25 −10
40. Let u x, du
1 dx. 2x
42. Let u cosh x, du sinh x dx.
cosh x 1 dx 2 cosh x dx 2 sinhx C x 2x
44. Let u 2x 1, du 2 dx.
sech22x 1 dx
1 sech22x 12 dx 2 1 tanh2x 1 C 2
sinh dx 1 sinh2 x
sinh x 1 dx C cosh2 x cosh x
sech x C 46. Let u sech x, du sech x tanh x dx.
sech3 x tanh x dx sech2 xsech x tanh x dx 1 sech3 x C 3
Section 5.10
48.
cosh2 x dx
4
1 cosh 2x dx 2
50.
0
Hyperbolic Functions
1 x dx arcsin 2 5 25 x
4 0
1 sinh 2x x C 2 2
1 1 x sinh 2x C 2 4
52.
2 1 1 1 4x2 dx 2 2 dx 2 ln C 2 2 2x x1 4x 2x1 2x
56. y tanh1
54. Let u sinh x, du cosh x dx.
cosh x 9 sinh2 x
dx arcsin arcsin
sinh3 x C e
x
58. y sech1cos 2x, 0 < x < y
e 6
x
y
C
2x
1 1 2 1 x22 2 4 x2
4
1 2 sin 2x 2 2 sin 2x 2 sec 2x, cos 2x sin 2x cos 2x cos 2x1 cos2 2x
since sin 2x ≥ 0 for 0 < x < 4. 60. y csch1 x2 csch x x 11 x 2 x 1x
y 2 csch1 x
1
2
2
62. y x tanh1 x ln1 x2 x tanh1 x y x
1 1 x tanh
1
2
x
1 ln1 x2 2
64. See page 401, Theorem 5.22.
x tanh1 x 1 x2
66. Equation of tangent line through P x0, y0: y a sech1
y
a2 x02 x0 a2 x02 x x0 a x0
When x 0,
Q
a P
y a sech1
x0 x a2 x02 a2 x02 a sech1 0. a a
Hence, Q is the point 0, a sech1x0a.
x02 2
a
x 1 2x 1 1 3 x2 dx dx ln C 4 2 2 9x 2 9 x 2 6 3 x2
1 3 ln C 12 3 x2 x2
x
L
Distance from P to Q: d x02 a2
68.
(a, 0)
arcsin
4 5
545
546
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
3 70. Let u x32, du x dx. 2
72.
x 1 x3
dx
2 2 1 3 2 x dx sinh1x32 C ln x32 1 x3 C 3 1 x322 2 3 3
dx x 2x2 4x 8
dx x 2x 22 4
2 x 22 4 1 C ln 2 x2
74.
1 dx x 12x2 4x 8
1 dx x 12x 12 6
3 x 12 3 1 1 dx 1 ln C x1 2 x 1x 12 32 6
76. Let u 2x 1, du 2 dx. y
78. y
1 dx x 14x2 8x 1 1 2x dx 4x x2
ln 4x x2
2 2x 1 3 2x 12 2
4 2x 1 dx 3 dx 4x x2 x 22 4
3
e2x
0
2
1 2
0
e2x dx e2x
2 lne
1 1 lne4 e4 ln 2 2 2
e
ln
4
dx
1 2e2x e2x dx e2x e2x 2x
6 x2 4
6 ln x x2 4
1
5
2 2x e
82. A
tanh 2x dx
0
3 4x2 8x 1 1 ln C 2x 1 3
3 x 2 2 3 x4 ln C ln 4x x2 ln C 4 x 2 2 4 x
2
80. A
dx
e2x
2 0
5 3
6 ln 5 21 6 ln 3 5 6 ln
53 215 3.626
e4 1.654 2
84. (a) vt 32t
(b) st
vt dt
32t dt 16t2 C
s0 1602 C 400 ⇒ C 400 st 16t2 400 —CONTINUED—
Section 5.10
Hyperbolic Functions
547
84. —CONTINUED— dv 32 kv2 dt
(c)
dv kv2 32
t→
dt
st
32 k v 1 ln t C 232 32 k v
Since v0 0, C 0. ln
32 k v 32 k v
32 k v 32 k v
v
k e2
32k t
e32k t e32k t
32 e 32 k
ex ex tan y sinh x. 2
1
k 32k
ln cosh 32k t C
s2t 0 when t 8.3 seconds
e32k t
32k t
When air resistance is not neglected, it takes approximately 3.3 more seconds to reach the ground.
tanh 32k t
x
e +e
−x x
e −e
−x
y 2
Thus, y arctansinh x. Therefore, arctansinh x arcsintanh x. y sech1 x sech y x sech ytanh yy 1 1 1 1 sech ytanh y sech y1 sech2 y x1 x2
y sinh1 x sinh y x
cosh yy 1 y
32
tanh 32k t dt
s1t 0 when t 5 seconds.
86. Let y arcsintanh x. Then, ex ex and ex ex
k
s1t 16t2 400.
e 32k t 1 e
32
s2t 400 100 ln cosh0.32 t 32k t
k
When k 0.01,
32 e232k t 1
90.
.
400 ⇒ 400 1k ln cosh 32k t.
y
k
When t 0, s0 C
e232k t
v k k e232k t 32e232k t 1
88.
32
1 ln cosh 32k t C. k
232k t
32 k v e232k t 32 k v
sin y tanh x
k
tanh32k t
(e) Since tanhct dt 1c ln coshct (which can be verified by differentiation), then
Let u k v, then du k dv. 1
32
The velocity is bounded by 32k.
dv dt 32 kv2
k
(d) lim
1 1 1 cosh y sinh2 y 1 x2 1
548
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
Review Exercises for Chapter 5 2. f x lnx 3
4. lnx2 1x 1 lnx2 1 lnx 1
y 3
Horizontal shift 3 units to the right Vertical asymptote: x 3
x=3
2 1
x 1
2
4
5
6
−1 −2 −3
x 25x 1 3
6. 3ln x 2 lnx2 1 2 ln 5 3 ln x 6 lnx2 1 ln 52 ln x3 lnx2 16 ln 25 ln
8. ln x lnx 3 0
10. hx ln
ln xx 3 0
hx
xx 3 e0
2
6
xx 1 ln x lnx 1 lnx 2 x2
1 1 1 x2 4x 2 x x 1 x 2 x3 3x2 2x
x2 3x 1 0 x
3 ± 13 2
x
3 13 3 13 < 0. only since 2 2
12. f x lnxx2 22 3 ln x fx
16.
2 lnx2 2 3
1 2 2x 7x2 6 3 x 3 x2 2 3x 6x
y
a bx bx 1 2 ax2a bx x a bx
lnx x
0
tan
dx
e
1 dx x
x 2x 1 1 dx dx ln x2 1 C x2 1 2 x2 1 2
1 b a 1 b 2 ax2 a2 xa bx ax axa bx
4
24.
20. u ln x, du
18. u x2 1, du 2x dx
1 a bx a lna bx b2
1 b lna bx ln x ax a2
y
dy 1 ab x b dx b2 a bx a bx
1 b a bx ln ax a2 x
1 b dy 1 b 1 2 2 dx a x a a bx x
14.
22.
1
1 1 1 2 ln x dx ln x C 2 x 4
4
4 x dx ln cos 4 x 0 ln
0
12 2 ln 2 1
ln x dx x
e
1
ln x1
e
1 1 dx ln x2 x 2
1
1 2
Review Exercises for Chapter 5 26. (a)
f x 5x 7
(b)
y 5x 7
6
f
−10
−1 6
y7 x 5 x7 y 5 x7 f 1x 5
28. (a)
f x x3 2
f
−10
(c) f 1 f x f 15x 7 f f 1x f
(b)
x 5 7 5 x 5 7 7 x f
f
−1
−4
3 y2x
5x 7 7 x 5
3
y x3 2
5
3 x2y 3 x 2 f 1x
549
−3
3 x3 2 2 x (c) f 1 f x f 1x3 2 3 x 2 f f 1x f 3 x 2 2 x 3
30. (a)
f x x2 5, x ≥ 0
(b)
4
f −1
y x2 5
−6
6
y 5 x
f
x 5 y
f 1
x x 5
−6
(c) f 1 f x f 1x2 5 x2 5 5 x for x ≥ 0. f f 1x f x 5 x 5 5 x 2
f x xx 3
32.
f x ln x
34.
f 4 4
x ex
f 1
f 1x ex
1 fx x 3 xx 31 2 2
f 10 e0 1
f4 1 2 3
f 14
36. (a)
1 1 f4 3
f x e1x
(b)
4
y e1x ln y 1 x x 1 ln y y 1 ln x f 1x 1 ln x
f
−4
f
−1
5
−2
(c) f 1 f x f 1 e1x 1 lne1x 1 1 x x f f 1x f 1 ln x e1 1ln x eln x x
550
Chapter 5
38. y 4ex
Logarithmic, Exponential, and Other Transcendental Functions 40. gx ln
2
y
1 e e
x
x
ln ex ln1 ex x ln1 ex
5 4
ex 1 1 ex 1 ex
gx 1 1 x
−5 − 4 −3 −2 −1
1 2 3 4 5
−2 −3 −4 −5
42. hz ez 2
1 46. f esin 2 2
44. y 3e3 t
2
hz zez 2 2
y 3e3 t3t2
2x sin x2 xey
dy ey dx
dy 2x sin x2 ey dx xey 52. Let u e2x e2x, du 2e2x e2x dx.
dx
3 1
x2ex
1 x3 1 2 1 3 e 3x dx ex 1 C 3 3
1 lne2x e2x C 2
e2x 1 1 dx 2e2x dx e2x 1 2 e2x 1
e1 x 1 dx e1 x 2 dx e1 x C x2 x
54. Let u x3 1, du 3x2 dx.
e2x e2x 1 2e2x 2e2x dx dx 2x 2x e e 2 e2x e2x
56.
f cos 2 esin 2
1 1 50. Let u , du 2 dx. x x
cos x2 xey
48.
9e3 t t2
58. (a), (c)
10,000
1 lne2x 1 C 2 0
5 0
V 8000e0.6t, 0 ≤ t ≤ 5
(b)
Vt 4800e0.6t V1 2634.3 dollars year V4 435.4 dollars year
2
60. Area
2ex dx 2ex
0
62. gx 62x
2 0
2e2 2 2
2 1.729 e2 64. y log4 x2
2
y
y
5 4 3 2 1
8 6
−5 − 4 −3 −2 −1 2
−4
−2
x 2
4
−2 −3 −4 −5
x 1 2 3 4 5
Review Exercises for Chapter 5 68. y x4x
66. f x 4xex
y 4x x 4x ln 4
fx 4xex ln 44xex 4xex1 ln 4 x log5 x log5x 1 x1
70. hx log5 hx
1 1 1 1 1 ln 5 x x1 ln 5 x x 1
74. t 50 log10
72.
21 t 1 1 t dt 2 C t2 ln 2
18,000 18,000 h
t 50 log10
(c)
(a) Domain: 0 ≤ h < 18,000 (b)
551
10t 50
t
18,000 18,000 h
18,000 18,000 h
18,000 h 18,00010t 50
100
h 18,0001 10t 50
80
As h → 18,000, t → .
60 40 20 h 4000
12,000
(d) t 50 log10 18,000 50 log1018,000 h dt 50 dh ln 1018,000 h 50 d 2t dh2 ln 1018,000 h2
Vertical asymptote: h 18,000
No critical numbers As t increases, the rate of change of the altitude is increasing.
76. 2P Pe10r
y5
78.
2 e10r y600 5
ln 2 10r r
dy 0.012y, s > 50 ds
1 0.012
dy y
ds
1 ln y s C1 0.012 y Ce0.012s When s 50, y 28 Ce0.01250 ⇒ C 28e0.6 y 28e0.60.012s, s > 50 e2x dy dx 1 e2x
dy
12
t 1620
600 1620
3.868 grams
ln 2 6.93% 10
80. (a)
82.
12
e2x 1 2e2x dx dx 2x 1e 2 1 e2x
1 y ln1 e2x C 2
(b) Speed(s)
50
55
60
65
70
Miles per Gallon (y)
28
26.4
24.8
23.4
22.0
552
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
84. y ey sin x 0 dy ey sin x dx
ey dy
sin x dx
ey cos x C1 ey
1 cos x C
y ln
86.
C C1
1 ln cos x C cos x C
dy 3x y (homogeneous differential equation) dx x 3x y dx x dy 0 Let y vx, dy x dv v dx. 3x vx dx xx dv v dx 0
3x 2vx dx x2 dv 0 3 2v dx x dv
1 dx x
ln x
1 dv 3 2v
1 ln 3 2v C1 ln3 2v1 2 ln C2 2
x C23 2v1 2
x2 C3 2v C 3 2
yx
x3 C3x 2y 3Cx 2Cy y
88.
dv kv 9.8 dt (a)
dv kv 9.8
x3 3Cx 2C
(b) lim vt t→
dt
9.8 k
1 9.8 kv0 9.8ekt dt k
(c) st
1 1 9.8t kv0 9.8ekt C k k
1 9.8t 2kv0 9.8ekt C k k
1 ln kv 9.8 t C1 k ln kv 9.8 kt C2 kv 9.8
ektC2
C3
ekt
1 v 9.8 C3ekt k At t 0,
1 v0 9.8 C3 ⇒ C3 kv0 9.8 k 1 v 9.8 kv0 9.8ekt k
Note that k < 0 since the object is moving downward.
s0
1 1 kv 9.8 C ⇒ C s0 2kv0 9.8 k2 0 k
st
9.8t 1 1 2kv0 9.8ekt s0 2kv0 9.8 k k k
9.8t 1 2kv0 9.8ekt 1 s0 k k
Review Exercises for Chapter 5 90. hx 3 arcsin2x
92. (a) Let arccot 2
5
cot 2
y
1
θ
4
1 tanarccot 2 tan . 2 −π 2
553
−π 4
π 4
x
π 2
2
(b) Let arcsec 5
−2
sec 5
5 2
−4
1 . cosarcsec 5 cos 5
θ 1
94. y arctanx2 1 y
96. y
2x 2x 1 x2 12 x4 2x2 2
y
1 arctan e2x 2
1 1 e2x 2e2x 4x 2 1e 1 e4x
x 98. y x2 4 2 arcsec , 2 < x < 4 2 y
x x 4 2
x2 4 1 4 x x2 4 2 2 2 2 x x 2x 2 1 x 4 x x 4 x x 4
100. Let u 5x, du 5 dx.
1 1 dx 3 25x2 5
102.
104.
3
1 5x 1 5 dx arctan C 3 5x2 53
1 1 x dx arctan C 16 x2 4 4
4x 1 1 x dx 4 dx 4 x21 22x dx 4 arcsin 4 x2 C 2 2 2 2 4 x 4 x
106. Let u arcsin x, du
2
1 dx. 1 x2
arcsin x 1 dx arcsin x2 C 2 1 x2
y
108. π 2
y = arcsin x π 4
x 0.25
0.5
0.75
1
2
Since the area of region A is 1
0
1
the shaded area is
arcsin x dx
0
110. y x tanh1 2x y x
1 0.571. 2
112. Let u x3, du 3x2 dx.
1 2 4x tanh 2
sin y dy ,
1
2x
2x tanh1 2x 1 4x2
x2sech x32 dx
1 1 sech x323x2 dx tanh x3 C 3 3
554
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
Problem Solving for Chapter 5 2. (a)
y
(b)
y
4
2
3
1
π 2
−1
π
3π 2
2
x
2π
1
−2
π 2
sin x dx
0
2
sin x dx ⇒
2
2
sin x dx 0
0
(d) 1
π 2
1 2
1
π 4
1
1
arccos x dx 2
2
y
4. y 0.5x and y 1.2x intersect y x. y
π 2
y = 0.5 x
4 , 12. 6
2
0
y = 2x y = x
does not intersect y x. y = 1.2 x −6
Suppose y x is tangent to y ax at x, y.
6 −2
ax x ⇒ a x1x. y ax ln a 1 ⇒ x ln x1x 1 ⇒ ln x 1 ⇒ x e, a e1e For 0 < a ≤ e1e 1.445, the curve y ax intersects y x. 6. (a) y f x arcsin x sin y x Area A
Area B
126 12 0.2618
4
6
22
(b)
arcsin x dx AreaC
12
6
2
2
3
2
3 2
2
4 22 A B
2 3 2 8 2 12
—CONTINUED—
4
sin y dy cos y
82 121
2 3
2
0.1346
x
1 is symmetric with respect to the 1 tan x2
point
2x
x
sin x 2 dx 22 4
x
2π
y
π
−1
3π 2
0
y
(c)
π
0.1589
1 1 dx 1 tan x2 2 2 4
Problem Solving for Chapter 5
555
6. —CONTINUED—
y
ln 3
(c) Area A
ey dy
y = ln x
0
ln 3
ey
0
ln 3
312
ey = x
A B
3
Area B
ln x dx 3ln 3 A 3 ln 3 2 ln 27 2 1.2958
x 1
2
3
1
(d)
tan y x Area A
3
tan y dy
4 3
4
ln cos y
y
y = arctan x
2 1 1 ln ln ln2 ln 2 2 2 2
3
Area C
arctan x dx
1
A
3 3 21 ln 2 4 1
C
B
x 1
3
43 3 21 ln 2 0.6818 12 10. Let u tan x, du sec2 x dx
y ex
8.
π 3 π 4
y
ex
yb
ea
Area
x a
4
0
1 dx sin x 4 cos2 x 2
y eax aea b Tangent line
bx ab b b ea
sec2 x dx tan2 x 4
du u2 4
12 arctanu2
1 0
xa1 ca1
Thus, a c a a 1 1. dy 1 y1 y, y0 dt 4
1 1 dy y 1y
0
0
eax aea b
4
1
If y 0,
12. (a)
1 1 arctan 2 2
y
(b)
1
dt
ln y ln 1 y t C
y ln tC 1y y etC C1et 1y y C1et yC1et y y0
C1et 1 1 C1et 1 C2et
1 1 ⇒ C2 3 4 1 C2
Hence, y
1 . 1 3et
—CONTINUED—
−6
−4
−2
x 2
4
6
dy y1 y y y2 dt d2y 1 y y 2yy ⇒ y 0 for y dt 2 2 d2y 1 1 d2y > 0 if 0 < y < and 2 < 0 if < y < 1. 2 dt 2 2 dt 1 Thus, the rate of growth is maximum at y , the 2 point of inflection.
556
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
12. —CONTINUED— (c) y y1 y, y0 2 1 1 C2et
As before, y y0 2 Thus, y
1 1 ⇒ C2 1 C2 2
1 2 . 1 t 2 et 1 e 2
The graph is different: y 6 4 2 −6
−4
−2
x 2
4
6
14. (a) u 985.93 985.93
v 985.93
0.095 120,0000.095 1 12 12
0.095 120,0000.095 1 12 12
12t
(d) t 12.7 years Again, the larger part goes for interest.
v
(c) The slopes are negatives of each other. Analytically,
u15 v15 14.06.
u
12t
(b) The larger part goes for interest. The curves intersect when t 27.7 years.
du dv u 985.93 v ⇒ dt dt
1000
0
35 0
C H A P T E R 5 Logarithmic, Exponential, and Other Transcendental Functions Section 5.1
The Natural Logarithmic Function: Differentiation . . . . 218
Section 5.2
The Natural Logarithmic Function: Integration . . . . . . 223
Section 5.3
Inverse Functions . . . . . . . . . . . . . . . . . . . . . . 227
Section 5.4
Exponential Functions: Differentiation and Integration . . 233
Section 5.5
Bases Other than e and Applications . . . . . . . . . . . . 240
Section 5.6
Differential Equations: Growth and Decay . . . . . . . . . 246
Section 5.7
Differential Equations: Separation of Variables
Section 5.8
Inverse Trigonometric Functions: Differentiation . . . . . 259
Section 5.9
Inverse Trigonometric Functions: Integration
. . . . . . 251
. . . . . . . 263
Section 5.10 Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . 267 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278
C H A P T E R 5 Logarithmic, Exponential, and Other Transcendental Functions Section 5.1
The Natural Logarithmic Function: Differentiation
Solutions to Odd-Numbered Exercises 1. Simpson’s Rule: n 10 x
0.5
x
1
1 dt t
0.6932
0.5
Note:
1
1 dt t
1.5
2
2.5
3
3.5
4
0.4055
0.6932
0.9163
1.0987
1.2529
1.3865
1
1 dt 0.5 t
3. (a) ln 45 3.8067
45
(b)
1
5. (a) ln 0.8 0.2231
0.8
1 dt 3.8067 t
(b)
1
7. f x ln x 2
1 dt 0.2231 t
9. f x ln x 1
Vertical shift 2 units upward
Horizontal shift 1 unit to the right
Matches (b)
Matches (a)
11. f x 3 ln x
13. f x ln 2x
Domain: x > 0
15. f x lnx 1
Domain: x > 0
Domain: x > 1 y
y
y 3
2
2 2
1 1
1
x 3
x 1
2
3
4
1
1
2
3
1
1
2
4
x
5
2
3
17. (a) ln 6 ln 2 ln 3 1.7917
2 19. ln 3 ln 2 ln 3
(b) ln 23 ln 2 ln 3 0.4055 (c) ln 81 ln 34 4 ln 3 4.3944 (d) ln 3 ln 312 12 ln 3 0.5493
21. ln
218
xy ln x ln y ln z z
3 a2 1 lna2 113 23. ln
1 lna2 1 3
Section 5.1
x x 1 2
25. ln
3
3
The Natural Logarithmic Function: Differentiation
27. ln zz 12 ln z lnz 12
3lnx2 1 ln x3
ln z 2 lnz 1
3lnx 1 lnx 1 3 ln x
29. lnx 2 lnx 2 ln
31.
x2 x2
1 xx 32 1 2 lnx 3 ln x lnx2 1 ln 2 ln 3 3 x 1
33 . 2 ln 3
35.
xxx 31
2
3
2
1 9 lnx2 1 ln 9 lnx2 1 ln 2 x2 1 37. lim lnx 3
3
x →3
f=g 9
0
−3
39. lim lnx23 x ln 4 1.3863
41. y ln x3 3 ln x
x →2
y
3 x
At 1, 0, y 3. 45. gx ln x 2 2 ln x
43. y ln x2 2 ln x y
2 x
gx
47.
y ln x4 dy 1 4ln x3 4ln x3 dx x x
2 x
At 1, 0, y 2.
49.
y ln xx2 1 ln x
1 lnx2 1 2
51. f x ln
dy 1 1 2x 2x2 1 dx x 2 x2 1 xx2 1
ln t t2
53. gt
55.
t 21t 2t ln t 1 2 ln t gt t4 t3
57.
xx 11 21 lnx 1 lnx 1
y ln
fx
1 dy 1 1 1 dx 2 x 1 x 1 1 x2
x ln x lnx2 1 x2 1
1 2x 1 x2 2 x x 1 xx2 1
y lnln x2 1 d 2xx2 2 1 dy ln x2 2 dx ln x dx ln x2 x ln x2 x ln x
59. f x ln fx
4 x2
x
1 ln4 x2 ln x 2
x 1 4 4 x2 x xx2 4
219
220 61.
Chapter 5 y
Logarithmic, Exponential, and Other Transcendental Functions
x2 1 lnx x2 1 x
dy xxx2 1 x2 1 1 dx x2 x x2 1
63.
1 1 x 1 x x2 1
2 2
x
x2 1 x x 1 2
y ln sin x
1
x
x x2 1
1 1 1 x2 x2 1 2 2 2 x 1 x 1 x x 1 x2
2 2
65.
y ln
dy cos x cot x dx sin x
cos x cos x 1
ln cos x ln cos x 1
dy sin x sin x sin x tan x dx cos x cos x 1 cos x 1
67.
y ln
1 sin x 2 sin x
69. f x sin 2x ln x2 2 sin 2x ln x
fx 2 sin 2x
ln 1 sin x ln 2 sin x dy cos x cos x dx 1 sin x 2 sin x
3 cos x sin x 1sin x 2 (b)
4
(1, 3)
1 dy 6x dx x When x 1,
2 sin 2x 2x cos 2x ln x x
2 sin 2x x cos 2x ln x2 x
y 3x2 ln x, 1, 3
71. (a)
1x 4 cos 2x ln x
−1
dy 5. dx
2
−3
Tangent line: y 3 5x 1 y 5x 2 0 5x y 2 73.
x2 3 ln y y2 10 2x
3 dy dy 2y 0 y dx dx 2x
y
dy 3 2y dx y
x2 2x 0
2
Domain: x > 0 1 x 1x 1 0 when x 1. x x
y 1
1 > 0 x2
1 Relative minimum: 1, 2
2 x2
xy y x
x2 ln x 2
y x
2 x
y
dy 2x 2xy dx 3y 2y 3 2y 2
77. y
y 2ln x 3
75.
(1, 12 ) 0
3 0
2
Section 5.1
The Natural Logarithmic Function: Differentiation
79. y x ln x
221
2
Domain: x > 0 y x
1x ln x 1 ln x 0 when x e
1.
0
3
(e−1 , −e−1 ) −1
1 > 0 x
y
Relative minimum: e1, e1
81. y
x ln x
4
( e, e )
Domain: 0 < x < 1, x > 1
(e2, e2/2)
0
y
ln x1 x1x ln x 1 0 when x e. ln x2 ln x2
y
ln x21x ln x 12x ln x 2 ln x 0 when x e 2. ln x4 xln x3
9
−4
Relative minimum: e, e Point of inflection: e2, e22 83.
f x ln x,
f 1 0
1 fx , x
The values of f, P1, P2, and their first derivatives agree at x 1. The values of the second derivatives of f and P2 agree at x 1.
f1 1
2
1 f x 2, x
f 1 1
P1 f
P1x f 1 f1x 1 x 1, P2x f 1 f1x 1 x 1 P1x 1,
P11 0
1 x 12, 2
P21 0
P21 1
P21 1
85. Find x such that ln x x.
87.
f x ln x x 0 1 1 x
n
1
xn
0.5 0.1931
y xx2 1 ln y ln x
f xn 1 ln xn xn xn fxn 1 xn
f xn
−2
P11 1
P2x 1,
xn1
5
P2
1 f 1x 12 2
P2x 1 x 1 2 x,
fx
−1
2
1 dy 1 x 2 y dx x x 1
2x2 1 2x2 1 dy y 2 dx xx 1 x2 1
3
0.5644
0.5671
0.0076
0.0001
Approximate root: x 0.567
1 lnx2 1 2
222
Chapter 5
y
89.
Logarithmic, Exponential, and Other Transcendental Functions
x23x 2 x 12
ln y 2 ln x
y
91.
1 ln3x 2 2 lnx 1 2
xx 132 x 1
ln y ln x
3 1 lnx 1 lnx 1 2 2
1 dy 3 2 2 y dx x 23x 2 x 1
1 3 1 1 dy 1 1 y dx x 2 x1 2 x1
dy 3x2 15x 8 y dx 2x3x 2x 1
dy y 2 3 1 dx 2 x x1 x1
3x3 15x2 8x 2x 133x 2
y 4x2 4x 2 2x2 2x 1x 1 2 xx2 1 x 132
93. Answers will vary. See Theorem 5.1 and 5.2.
95. ln ex x because f x ln x and gx ex are inverse functions.
97. (a) f 1 f 3
(b) fx 1
10 log10
99.
1010
10I ln1010 ln 10I ln1010ln I 16 ln 10 160 10 log 16
16
10
I
10 10 10 ln 1010 16 ln 10 10 ln 10 16 ln 10 6 ln 10 60 decibels ln 10 ln 10 ln 10
101. (a) You get an error message because ln h does not exist for h 0. (b) Reversing the data, you obtain h 0.8627 6.4474 ln p. (c)
2 0 for x 2. x
(d) If p 0.75, h 2.72 km. (e) If h 13 km, p 0.15 atmosphere. (f)
h 0.8627 6.4474 ln p 1 6.4474
25
1 dp (implicit differentiation) p dh
p dp dh 6.4474 0
1 0
For h 5, p 0.55 and dpdh 0.0853 atmos/km. For h 20, p 0.06 and dpdh 0.00931 atmos/km. As the altitude increases, the rate of change of pressure decreases.
103. (a) f x ln x, gx x
4 x (b) f x ln x, gx 15
25
g g
f f
0
500 0
20,000
0 0
1 1 fx , gx x 2x
1 1 fx , gx 4 3 x 4x
For x > 4, gx > fx. g is increasing at a faster rate than f for “large” values of x.
For x > 256, gx > fx. g is increasing at a faster rate than f for “large” values of x. f x ln x increases very slowly for “large” values of x.
Section 5.2
The Natural Logarithmic Function: Integration
105. False ln x ln 25 ln25x lnx 25
Section 5.2
1.
The Natural Logarithmic Function: Integration
5 1 dx 5 dx 5 ln x C x x
3. u x 1, du dx
5. u 3 2x, du 2 dx
1 dx ln x 1 C x1
7. u x2 1, du 2x dx
1 1 1 dx 2 dx 3 2x 2 3 2x
x 1 1 dx 2x dx x2 1 2 x2 1
1 ln 3 2x C 2
1 lnx2 1 C 2
lnx2 1 C
9.
x2 4 dx x
x
4 dx x
11. u x3 3x2 9x, du 3x2 2x 3 dx
x2 4 ln x C 2
x2 2x 3 1 3x2 2x 3 dx dx x3 3x2 9x 3 x3 3x2 9x
13.
x2 3x 2 dx x1
17.
x4 x 4 dx x2 2
6 dx x1
x4
x2 2
1
dx
x3 3x2 5 dx x3
x dx x2 2
21. u x 1, du dx x 1
x2 4x 6 ln x 1 C 2
19. u ln x, du
x3 1 2x lnx2 2 C 3 2
15.
23.
x 11 2 dx
2x 1
1 2
C
2x 1 C
1 ln x3 3x2 9x C 3
x2
5 dx x3
x3 5 ln x 3 C 3
1 dx x
1 ln x2 3 dx ln x C x 3
2x dx x 12
2
2x 2 2 dx x 12
2x 1 1 dx 2 dx x 12 x 12 1 1 dx dx 2 x1 x 12
2 ln x 1
2 C x 1
223
224
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions 1 dx ⇒ u 1 du dx 2x
25. u 1 2x, du
1 dx 1 2x
u 1 du u
u ln u C1
1 du u
u
1 2x ln 1 2x C1 2x ln 1 2x C where C C1 1.
27. u x 3, du
x x 3
dx 2
1 dx ⇒ 2u 3 du dx 2x
2 6u 9 lnu C
2
u 32 u2 6u 9 du 2 du 2 u u
u2
1
u6
9 du u
u2 12u 18 ln u C1
x 3 12 x 3 18 ln x 3 C1 2
x 6x 18 ln x 3 C where C C1 27.
29.
cos d ln sin C sin
31.
csc 2x dx
u sin , du cos d
33.
1 ln csc 2x cot 2x C 2
cos t dt ln 1 sin t C 1 sin t
37. y
35.
3 dx 2x
3
(1, 0) −10
3 ln x 2 C
tan2 d
(0, 2)
4
−3
3
−3
1 0, 2: 2 lncos 0 C ⇒ C 2 2
1 s ln cos 2 2 2
dy 1 , 0, 1 dx x 2 (a)
y 3 ln x 2
41.
1 tan22 d 2 1 ln cos 2 C 2
−10
sec x tan x dx ln sec x 1 C sec x 1
10
1, 0: 0 3 ln1 2 C ⇒ C 0
39. s
10
1 dx x2
1 csc 2x2 dx 2
(b)
y
(0, 1)
y
1 dx ln x 2 C x2
3
3
y0 1 ⇒ 1 ln 2 C ⇒ C 1 ln 2 x
−2
4
−3
Hence, y ln x 2 1 ln 2 ln
x2 1. 2
−3
6
−3
Section 5.2
4
43.
0
47.
0
2 x
2
x1
0
1
49.
cos d ln sin sin
21
1
0 ln 3 2
ln
e
1
7 3
1 2
2 sin 2 1.929 1 sin 1
1 C ln sec x C cos x
ln
1 dx x1
225
1 dx x
1 ln x2 1 dx 1 ln x3 x 3 1 e
53. ln sec x tan x C ln
55.
x ln x 1
2
51. ln cos x C ln
45. u 1 ln x, du
5 ln 13 4.275 3
2 dx x1
2 2 x
4
0
5 5 dx ln 3x 1 3x 1 3
The Natural Logarithmic Function: Integration
sec x tan xsec x tan x sec2 x tan2 x C ln C sec x tan x sec x tan x
1 C ln sec x tan x C sec x tan x
1 dx 2 1 x 2 ln 1 x C1 1 x 2 x ln 1 x C where C C1 2.
57.
cos1 x dx sin1 x C 2
59.
4
2
csc x sin x dx lncsc x cot x cos x
4
ln2 1
2
2
0.174
Note: In Exercises 61 and 63, you can use the Second Fundamental Theorem of Calculus or integrate the function.
63. F x
Fx
1 x
Fx
3x
x
61. F x
1 dt 1 t
x
1 dt t
3x
1
1 dt t
x
1
1 dt t
65.
y
2
3 1 0 3x x
x
−1
−1 2
A 1.25 Matches (d)
67. A
4 2 x
1
4 dx x
4
1
x
4 dx x
10
2 4 ln x
15 8 ln 2 13.045 square units 2
x2
4 1
8 4 ln 4
1 2 0
6 0
1 2
1
226
Chapter 5
2
69.
2 sec
0
Logarithmic, Exponential, and Other Transcendental Functions
x 12 dx 6
2
sec
0
6x6 dx
10
12 ln sec 6x tan 6x
12 12 ln sec tan ln 1 0 3 3
12 ln 2 3 5.03041
1 42
77. Average value
4
2
8 dx 4 x2
0
4 0
4
75. Divide the polynomials: x2 1 x1 x1 x1
79. Average value
x2dx
2
4 4
0
73. Substitution: u x2 4 and Log Rule
71. Power Rule
81. Pt
2
1 x
1 e1
e
1
ln x 1 ln x2 dx x e1 2
4 2
14 21 1
e
1
1 1 e1 2
1 0.291 2e 2
3000 0.25 dt 30004 dt 12,000 ln 1 0.25t C 1 0.25t 1 0.25t
P0 12,000 ln 1 0.250 C 1000 C 1000
Pt 12,000 ln 1 0.25t 1000 1000 12 ln 1 0.25t 1 P3 1000 12ln 1.75 1 7715
83.
1 50 40
50
40
50
40 $168.27
90,000 dx 3000 ln 400 3x 400 3x
85. (a) 2x2 y2 8
(b) y2 e1 xdx eln xC eln1 xeC
y2 2x2 8 Let k 4 and graph y2
y1 2x2 8 y2
2x2
8
4 x
1 k x
yy 2 2 x x 1
2
10
10 − 10 − 10
10
10 − 10 − 10
(c) In part (a),
2x2 y2 8 4x 2yy 0 y
2x . y
In part (b),
y2 2yy y
4 4x1 x 4 x2 2 2y 2y y 2 2 . yx 2 y x 4x 2x
Using a graphing utility the graphs intersect at 2.214, 1.344. The slopes are 3.295 and 0.304 1 3.295, respectively.
Section 5.3 87. False
Inverse Functions
89. True
1 ln x lnx12 ln x12 2
1 dx ln x C1 x
ln x ln C ln Cx , C 0
Section 5.3 1. (a)
Inverse Functions
f x 5x 1 gx
(b) 3
x1 5
f 2 1
x1 x1 5 1x f gx f 5 5
g( f x g5x 1
3. (a)
y
g x
−3
1
2
3
5x 1 1 x 5
f x x3
(b)
y
3 x gx
3
f
2
3 x f gx f 3 x 3 x
g
1
x
3 3 g f x gx3 x x
−3 −2
1
2
3
−2 −3
5. (a)
f x x 4
(b)
gx x2 4, x ≥ 0
g 10
f gx f x2 4
8 6
x2 4 4 x2 x
4
g f x g x 4
f x
1 x
f
2 x
x 4 2 4 x 4 4 x
7. (a)
y 12
2
4
6
(b)
8
10
12
y 3
1 gx x f gx
1 x 1x
g f x
1 x 1x
9. Matches (c)
2
f=g
1 x −1
11. Matches (a)
1
2
3
227
228
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
13. f x 34 x 6
1 3 s2 One-to-one; has an inverse
17. hs
15. f sin Not one-to-one; does not have an inverse
One-to-one; has an inverse y
1 −4
y
8
8 1 6 4
π 2
2
−8
−6
−4
−7
θ
3π 2
x
−2
−1
19. f x ln x
21. gx x 53
One-to-one; has an inverse
One-to-one; has an inverse
2
200
−1
5 −10
2 − 50
−2
23. f x x a3 b
25. f x
fx 3x a2 ≥ 0 for all x.
x4 2x2 4
f x x3 4x 0 when x 0, 2, 2.
f is increasing on , . Therefore, f is strictly monotonic and has an inverse.
f is not strictly monotonic on , . Therefore, f does not have an inverse.
27. f x 2 x x3 fx 1 3x2 < 0 for all x. f is decreasing on , . Therefore, f is strictly monotonic and has an inverse. 29.
f x 2x 3 y x y
31.
f x x5 y
y3 2 x3 2
f x x y
33.
5 y x
x y2
5 x y
y x2 f 1x x2, x ≥ 0
5 x x15 f 1x y
x3 x 2
f 1
y
f 2
f y
3
1
f
1
1
2 4
f
x 2
1
2
1
2
f
1
x 2
2 2
f
4
2
x 1
2
3
Section 5.3 f x 4 x2 y, 0 ≤ x ≤ 2
35.
Inverse Functions
229
3 x 1 y f x
37.
x 4 y2
x y3 1
y 4 x2
y x3 1 f 1x x3 1
f 1x 4 x2, 0 ≤ x ≤ 2
2
y
f −1 f
3
−3
f
2
3
1
f
The graphs of f and f 1 are reflections of each other across the line y x.
−2
1
x 1
2
3
f x x23 y, x ≥ 0
39.
f x
41.
x y32 y x32 y
f 1x x32, x ≥ 0 4
f −1
The graphs of f and f 1 are reflections of each other across the line y x.
f
0
f 1x
x x2 7
x
7y 1 y2
7x 1 x2 7x 1 x2
, 1 < x < 1
2
f −1 f
6 −3
0
y
3
The graphs of f and f 1 are reflections of each other across the line y x.
−2
43.
x
1
2
3
4
f 1x
0
1
2
4
y
45. (a) Let x be the number of pounds of the commodity costing 1.25 per pound. Since there are 50 pounds total, the amount of the second commodity is 50 x. The total cost is y 1.25x 1.6050 x
(4, 4)
4
0.35x 80
0 ≤ x ≤ 50.
3 2
(b) We find the inverse of the original function:
(3, 2)
y 0.35x 80
(2, 1)
1
(1, 0) 1
x 2
3
4
0.35x 80 y x 100 35 80 y 20 Inverse: y 100 35 80 x 7 80 x.
x represents cost and y represents pounds. (c) Domain of inverse is 62.5 ≤ x ≤ 80. (d) If x 73 in the inverse function, 100 y 100 35 80 73 5 20 pounds.
230
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
47. f x x 42 on 4,
4 on 0, x2
49. f x
fx 2x 4 > 0 on 4,
fx
f is increasing on 4, . Therefore, f is strictly monotonic and has an inverse.
8 < 0 on 0, x3
f is decreasing on 0, . Therefore, f is strictly monotonic and has an inverse.
51. f x cos x on 0, fx sin x < 0 on 0, f is decreasing on 0, . Therefore, f is strictly monotonic and has an inverse. x y on 2, 2 x2 4
f x
53.
Domain: all x Range: 2 < y < 2
x 2 y 4y x x2 y x 4y 0 a y, b 1, c 4y
f
1 ± 1 4 y4y 1 ± 1 16y2 x 2y 2y y f 1x
1 0,
55. (a), (b)
The graphs of f and f 1 are reflections of each other across the line y x.
2
−3
3
f
−1
−2
1 16x2 2x, if x 0
if x 0 57. (a), (b)
6
4
f
g
f −1 −5
−6
10
6
g−1 −4
(c) Yes, f is one-to-one and has an inverse. The inverse relation is an inverse function. 59. f x x 2, Domain: x ≥ 2 fx
1 > 0 for x > 2. 2x 2
f is one-to-one; has an inverse x 2 y
x 2 y2 x y2 2 yx 2
−4
(c) g is not one-to-one and does not have an inverse. The inverse relation is not an inverse function.
61. f x x 2 , x ≤ 2 x 2 2x f is one-to-one; has an inverse 2xy 2yx f 1x 2 x, x ≥ 0
2
f 1
x x2 2, x ≥ 0
63. f x x 32 is one-to-one for x ≥ 3.
x 3 y 2
x 3 y x y 3 y x 3 f 1x x 3, x ≥ 0 (Answer is not unique)
65. f x x 3 is one-to-one for x ≥ 3. x3y
xy3 yx3 f 1x x 3, x ≥ 0 (Answer is not unique)
Section 5.3 67. Yes, the volume is an increasing function, and hence one-to-one. The inverse function gives the time t corresponding to the volume V.
f x x3 2x 1, f 1 2 a
71.
1 1 1 2 f f 12 f1 312 2 5
6 21 a
f x sin x, f
73.
fx cos x
1
f 1
12 f f
1 1
1 32
12
1 1 f 6 cos 6
23 3
4 f x x3 , f 2 6 a x
75.
fx 3x2
f 16
4 x2
1 1 1 1 f f 16 f2 322 422 13
77. (a) Domain f Domain f 1 ,
79. (a) Domain f 4, , Domain f 1 0,
(b) Range f Range f 1 ,
(b) Range f 0, , Range f 1 4,
(c)
(c)
y
f f −1
8 6
x 1
2
3
4
x
−3
2
f x x3,
(d)
6
8
10
12
f x x 4, 5, 1 fx
1 2x 4
12 43
f5
1 2
3 x, f 1x
12, 18
4
fx 3x2 f
f 1
f
2
−2
(d)
f −1
10
2
−3 −2
y 12
3
1
x
f 1
18, 12
1 3 2 3 x
18 34
231
69. No, Ct is not one-to-one because long distance costs are step functions. A call lasting 2.1 minutes costs the same as one lasting 2.2 minutes.
fx 3x2 2 f 1
Inverse Functions
f 1x x2 4, 1, 5
f 1x 2x f 11 2
232 81.
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
x y3 7y2 2 1 3y2
dy dy 14y dx dx
1 1 1 dy dy . At 4, 1, . dx 3y2 14y dx 3 14 11 Alternate solution: let f x x3 7x2 2. Then fx 3x2 14x and f1 11. Hence,
1 1 dy . dx 11 11
In Exercises 83 and 85, use the following. f x 18 x 3 and g x x3 3 x f1 x 8 x 3 and g1 x
83. f 1 g11 f 1g11 f 11 32
85. f 1 f 16 f 1 f 16 f 172 600
In Exercises 87 and 89, use the following. f x x 4 and g x 2x 5 f1 x x 4 and g1 x
x5 2
87. g1 f 1x g1 f 1x g1x 4
x 4 5 2
x1 2
89. f gx f gx f 2x 5 2x 5 4 2x 1 x1 2 g1 f 1
Hence, f g1x
Note: f g1 91. Answers will vary. See page 335 and Example 3.
93. y x2 on , does not have an inverse.
95. f is not one-to-one because many different x-values yield the same y-value.
97. Let f gx y then x f g1y. Also,
Example: f 0 f 0
2n 1 Not continuous at , where n is an integer 2
f gx y f gx y gx f 1y x g1 f 1y g1 f 1y Since f and g are one-to-one functions, f g1 g1 f 1.
99. Suppose gx and hx are both inverses of f x. Then the graph of f x contains the point a, b if and only if the graphs of gx and hx contain the point b, a. Since the graphs of gx and hx are the same, gx hx. Therefore, the inverse of f x is unique.
Section 5.4
Exponential Functions: Differentiation and Integration 103. True
101. False Let f x x2.
x
105. Not true
f x
107.
Let f x
1 x, x,
0 ≤ x ≤ 1 1 < x ≤ 2
2
.
fx
f is one-to-one, but not strictly monotonic.
Section 5.4
dt 1 t 4
, f 2 0
1 1 x 4
f 10
1 1 17 f2 1 17
Exponential Functions: Differentiation and Integration
1. e0 1
3.
ln 2 0.6931
5. eln x 4 x4
e0.6931. . . 2
ln1 0
9. 9 2ex 7
7. ex 12
2ex 7
x ln 12 2.485
ex 1 x0 13. ln x 2
11. 50ex 30 ex
x e2 7.3891
3 5
x ln x ln
35 53 0.511
15. lnx 3 2 x3
17. lnx 2 1 x 2 e1 e
e2
x 2 e2
x 3 e2 10.389
x e2 2 5.389 19. y ex
21. y ex
y
2
Symmetric with respect to the y-axis Horizontal asymptote: y 0
4 3
y
2
x 1
1
2
3
x 1
1
233
234
Chapter 5
23. (a)
Logarithmic, Exponential, and Other Transcendental Functions (b)
7
(c)
3
g
f
7
f
f
q
−2 −5
4
h
7 −1
−4
Horizontal shift 2 units to the right
8 −1
−3
Vertical shift 3 units upward and a reflection in the y-axis
A reflection in the x-axis and a vertical shrink 27. y C1 eax
25. y Ceax Horizontal asymptote: y 0
Vertical shift C units
Matches (c)
Reflection in both the x- and y-axes Matches (a)
29. f x e2x
31. f x ex 1
gx lnx
gx lnx 1
1 ln x 2
y y 6
f
6
f
4
4
g
2 2
g
x 2
x
−2
2
4
4
6
6
−2
33.
35.
3
1 1 1,000,000
1,000,000
g
2.718280469
e 2.718281828
f −1
4
e >
−1
1 1 1,000,000
As x → , the graph of f approaches the graph of g. 0.5 x e0.5 lim 1 x → x
37. (a) y e3x
(b) y e3x
y 3e3x
y 3e3x
At 0, 1, y 3.
At 0, 1, y 3.
39. f x e2x
41. f x e2xx
2
fx 2e2x
45. gt et et 3 gt 3et et 2et et
43.
dy ex dx 2x
dy 2 2x 1e2xx dx
47.
y ln ex x2 2
dy 2x dx
y ex
49.
y ln1 e2x 2e2x dy dx 1 e2x
1,000,000
Section 5.4
51.
y
Exponential Functions: Differentiation and Integration
2 2ex ex 1 ex ex
dy ex2x 2 exx2 2x 2 x2ex dx
dy 2ex ex 2ex ex dx
y x2ex 2xex 2ex ex x2 2x 2
53.
2ex ex ex ex 2
55. f x ex ln x fx ex
y ex sin x cos x
57.
1x e
x
ln x ex
1x ln x
dy excos x sin x sin x cos xex dx ex2 cos x 2ex cos x 61. f x 3 2xe3x
xey 10x 3y 0
59. xey
fx 3 2x3e3x 2e3x
dy dy ey 10 3 0 dx dx
7 6xe3x
dy y xe 3 10 ey dx
fx 7 6x3e3x 6e3x 36x 5e3x
dy 10 ey y dx xe 3 y ex cos2x sin2x
63.
y ex 2 sin 2x 2 cos 2x ex cos2x sin 2x ex 1 2 cos2x 1 2 sin2x
y ex 2 2sin 2x 2 2cos 2x ex 1 2 cos2x 1 2sin 2x ex 1 22 sin 2x 1 22 cos 2x
2y 3y 2ex 1 2cos 2x 1 2sin 2x 3ex cos2x sin 2x ex 1 22 cos 2x 1 22 sin 2x y
Therefore, 2y 3y y ⇒ y 2y 3y 0. 65. f x fx
ex ex 2
6
ex ex 0 when x 0. 2 (0, 1)
−3
ex ex fx > 0 2
3
0
Relative minimum: 0, 1 67. gx
1 2
ex2 2 2
2,
1 2 gx x 2ex2 2 2
(
1,
1 2 g x x 1x 3ex2 2 2
2, Points of inflection: 1, Relative maximum:
(
0.8
0
2, 0.399 1 e , 3, 12 e 1, 0.242, 3, 0.242 2
1
2
1 2
1 2
e− 0.5 2π
1 2π
( (
(
3,
e− 0.5 2π
( 4
0
235
236
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
69. f x x2ex
3
fx x2ex 2xex xex2 x 0 when x 0, 2. f x ex2x x2 ex2 2x
(2, 4 e −2 )
(0, 0)
exx2 4x 2 0 when x 2 ± 2.
−1
5 0
Relative minimum: 0, 0
(2 ±
2, (6 ± 4
2)e− (2 ± 2)
Relative maximum: 2, 4e2 x 2 ± 2 y 2 ± 22e2 ± 2 Points of inflection: 3.414, 0.384, 0.586, 0.191 71. gt 1 2 tet
5
(− 1, 1 + e)
gt 1 tet g t tet
(0, 3)
−6
6
Relative maximum: 1, 1 e 1, 3.718
−3
Point of inflection: 0, 3 73.
A baseheight 2xex
2
y
dA 2 2 4x2ex 2ex dx
3 2
2ex 1 2x2 0 when x 2
2
2
(
. −2
A 2e1 2
75. y
a aL x b e L ex b b b y 1 aex b2 1 aex b2
y
1 aex b2
e aL aLb e 21 ae b x b
2
x b
x b
1 aex b4
1 aex b
e aL 2aLb e ab e b x b
2
x b
x b
1 aex b3
Laex b aex b 1 1 aex b3 b2
y 0 if aex b 1 ⇒ yb ln a
x 1 ln ⇒ x b ln a b a
L L L 1 aeb ln a b 1 a1 a 2
Therefore, the y-coordinate of the inflection point is L 2.
ab e x b
) x
−1
1 −1
L , a > 0, b > 0, L > 0 1 aex b
2 −1 2 ,e 2
2
)
Section 5.4
Exponential Functions: Differentiation and Integration
77. ex x ⇒ f x x ex
79. (a)
237
4
fx 1 ex xn1 xn
f xn x exn xn n fxn 1 exn
−4
−2
x1 1 x2 x1
f x1 0.5379 fx1
x3 x2
f x2 0.5670 fx2
x4 x3
f x3 0.5671 fx3
(b) When x increases without bound, 1 x approaches zero, and e1 x approaches 1. Therefore, f x approaches 2 1 1 1. Thus, f x has a horizontal asymptote at y 1. As x approaches zero from the right, 1 x approaches , e1 x approaches and f x approaches zero. As x approaches zero from the left, 1 x approaches , e1 x approaches zero, and f x approaches 2. The limit does not exist since the left limit does not equal the right limit. Therefore, x 0 is a nonremovable discontinuity.
We approximate the root of f to be x 0.567.
81.
h
0
5
10
15
20
P
10,332
5,583
2,376
1,240
517
ln P
9.243
8.627
7.773
7.123
6.248
(a)
5
(b) ln P ah b
12
P eahb ebeah P Ceah, C eb −2
For our data, a 0.1499 and C e9.3018 10,957.7
22 0
P 10,957.7e0.1499h
y 0.1499h 9.3018 is the regression line for data h, ln P. (c)
(d)
12,000
dP 10,957.710.1499e0.1499h dh 1642.56e0.1499h
0
22
For h 5,
0
83.
f x ex 2, f 0 1
dP dP 776.3. For h 18, 110.6. dh dh
7
f
1 1 fx ex 2, f0 2 2 1 1 f x ex 2, f 0 4 4 P1x 1
P1
P2 −6
6 −1
1 x x 0 1, P10 1 2 2
1 1 P1x , P10 2 2 1 1 x2 x P2x 1 x 0 x 02 1, P20 1 2 8 8 2 1 1 1 P2x x , P20 4 2 2 1 1 P2x , P20 4 4 The values of f, P1, P2 and their first derivatives agree at x 0. The values of the second derivatives of f and P2 agree at x 0.
238
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
85. (a) y ex
(b) y ex
y1 1 x
y2 1 x 4
x2 2
4
y
y
y1 −2
y2
2 −2
−1
2 −1
(c) y ex y3 1 x
x2 x3 2 6
4
y y 3
−2
2 −1
87. Let u 5x, du 5 dx.
5x
89. Let u 2x, du 2 dx.
1
5x
e 5dx e C
91.
0
xex dx 2
1 x2 1 2 e 2xdx ex C 2 2
93.
1 2
1
1 e2x 2 dx e2x 2 0 1 e2 1 1 e2 2 2e2
e2x dx
e x 1 dx 2 ex dx 2ex C x 2x
95. Let u 1 ex, du ex dx.
ex ex ex dx dx ln1 ex C ln x C x lnex 1 C x x 1e 1e e 1
3 3 97. Let u , du 2 dx. x x
3 e3 x
1
x2
dx
1 3
99. Let u 1 ex, du ex dx.
e3 x
1
1 e 3 x 3
3 1
105.
2 1 e x3 2 C 3
e e2 1 3
103.
ex ex dx ln ex ex C ex ex
e sin x cos x dx
ex1 ex dx 1 ex1 2ex dx
2
101. Let u ex ex, du ex ex dx.
x3 dx
3
1 sin x e cos x dx 1 sin x e C
5 ex dx e2x
5e2x dx
ex dx
5 e2x ex C 2
107.
ex tanex dx
tanex ex dx
ln cosex C
1 0
Section 5.4
109. Let u ax2, du 2ax dx. Assume a 0
y
Exponential Functions: Differentiation and Integration
111. fx
1 x 1 e ex dx ex ex C1 2 2
xeax dx
f0 C1 0
f x
2
1 2 1 ax2 e 2ax dx eax C 2a 2a
1 x 1 e ex dx ex ex C2 2 2
f 0 1 C2 1 ⇒ C2 0 1 f x ex ex 2
113. (a)
y
dy 2ex 2, dx
(b)
5
y (0, 1)
1 2ex 2 dx 4 ex 2 dx 2
4ex 2 C
x
−2
0, 1
5
0, 1: 1 4e0 C 4 C ⇒ C 5
−2
y 4ex 2 5 6
−4
8 −2
5
115.
ex dx ex
0
5 0
6
e5 1 147.413
117.
xex 4dx 2ex 4 2
0
150
2
6
0
2e3 2
2 1.554
3
0
−4.5
6
4.5
0
−3
119. (a) f u v euv eu ev (b) f k x e
kx
x
123.
eu f u ev f v
e f x . x k
k
60
121. 0.0665
e0.0139t48 dt 2
48
Graphing Utility: 0.4772 47.72%
x
et dt ≥
0
1 dt
0
x
e t
0
≥
t
x
0
ex 1 ≥ x ⇒ ex ≥ 1 x for x ≥ 0
125. f x ex. Domain is , and range is 0, . f is continuous, increasing, one-to-one, and concave upwards on its entire domain. lim ex 0 and lim ex .
x→
x→
2
127. Yes. f x Cex, C a constant.
129. ex > 0 ⇒
0
exdx > 0.
239
240
Chapter 5
131. f x
Logarithmic, Exponential, and Other Transcendental Functions
ln x x
(a) fx
y
1 ln x 0 when x e. x2
1 2
On 0, e, fx > 0 ⇒ f is increasing. On e, , fx < 0 ⇒ f is decreasing.
x
e
2
6
4
8
− 12
(b) For e ≤ A < B, we have: ln A ln B > A B B ln A > A ln B ln AB > ln BA AB > BA. (c) Since e < , from part (b) we have e > e.
Section 5.5 1. y
2 1
Bases Other than e and Applications
t3
3. y
At t0 6, y
12
63
1 4
2 1
t7
5. log2 18 log2 23 3
At t0 10, y
7. log7 1 0
12
107
0.3715
11. (a) log10 0.01 2
23 8
9. (a)
102 0.01
log2 8 3 (b)
31
(b) log0.5 8 3
1 3
0.53 8
1 log3 1 3
15. y
13. y 3x x
2
1
0
1
2
y
1 9
1 3
1
3
9
12 3 8
13 3 x
17. hx 5x2
x
x
2
1
0
1
2
y
9
3
1
1 3
1 9
y
x
1
0
1
2
3
y
1 125
1 25
1 5
1
5
y y
4
4 4
3
3
3 2
2 1
x 2
1
1
2
x 2
1
1
2
x 1
2
3
4
Section 5.5 19. (a) log10 1000 x
Bases Other than e and Applications
21. (a) log3 x 1
10x 1000
31 x x 13
x3
(b) log2 x 4
(b) log10 0.1 x 10x 0.1
24 x
x 1
x 16
1
x2 x log5 25
23. (a)
x2
x log5
52
(b) 3x 5 log2 64 2
3x 5 log2 26 6
x2 x 2 0
3x 1
x 1x 2 0
x 13
x 1 OR x 2 32x 75
25.
3 xln 2 ln 625
2x ln 3 ln 75 x
23x 625
27.
1 ln 75 1.965 2 ln 3
3x
ln 625 ln 2
x3
1 0.09 12
29.
12t
31. log2x 1 5
3
x 1 25 32
0.09 12t ln 1 ln 3 12 t
ln 625 6.288 ln 2
1 12
x 33
ln 3 12.253 0.09 ln 1 12
33. log3 x2 4.5 x2 3 4.5 x ± 34.5 ± 11.845 37. hs 32 log10s 2 15
35. gx 621x 25
Zero: s 2.340
Zero: x 1.059
40
30
(− 1.059, 0) −4
10 −1
−30
8
(2.340, 0) −20
241
242
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
39. f x 4x
x
gx log4 x
y
2
1
0
1 2
1
1 16
1 4
1
2
4
f x
f
3 2
g
1 16
1 4
1
2
4
2
1
0
1 2
1
x gx
41. f x 4x
x
43.
fx ln 4 4x
1
2
3
1
45. gt t2 2t
y 5x2
gt t 2 ln 2 2t 2t 2t
dy ln 5 5x2 dx
t 2t t ln 2 2 2t t2 t ln 2
47. h 2 cos
49.
h 2 sin ln 22 cos
2
51. f x log2
1 dy dx x ln 3
ln 2 cos sin
x2 x1
53.
2 log2 x log2 x 1 fx
55. gt gt
y log3 x
y log5 x2 1 dy 1 dx 2
2 1 x ln 2 x 1 ln 2
1 log5 x2 1 2
2x
x
x2 1ln 5 x2 1ln 5
x2 ln 2xx 1
10 log4 t 10 ln t t ln 4 t 10 t 1t ln t ln 4 t2
y x2x
57.
ln y
2 ln x x
2 1 2 2 1 dy ln x 2 2 1 ln x y dx x x x x
10 5 1 ln t 2 1 ln t t 2 ln 4 t ln 2
dy 2y 2 1 ln x 2x2x 21 ln x dx x
59.
y x 2x1
61.
ln y x 1 lnx 2
1 dy 1 x 1 lnx 2 y dx x2 dy x1 y lnx 2 dx x2
xx 21 lnx 2
x 2x1
x
x
3 dx
3 C ln 3
Section 5.5
2
63.
x
x
1
2 dx
ln 2
2
2
65.
1
1 1 4 ln 2 2
Bases Other than e and Applications
x
2
x5
dx
7 7 2 ln 2 ln 4
67.
1 x2 5 2x dx 2 1 5x C 2 ln 5
2
1 2 5x C 2 ln 5
32x dx, u 1 32x, du 2ln 332x dx 1 32x 1 2 ln 3
69.
243
2x 1 2 ln 33 2x ln1 3 C 2x dx 13 2 ln 3
1 dy 0.4x3, 0, dx 2 y
x3
0.4
(a)
x3
dx 3 0.4
(b)
4
4
13 dx
−6
6
x
−4
3 0.4x3 C 3ln 2.50.4x3 C ln 0.4 y 3 ln 2.50.4x3
y
(0, 12 (
4
−4
−4
1 3 ln 2.5 2
31 0.4x3 1 ln 2.5 2
71. Answers will vary. Example: Growth and decay problems. 73.
y
x
1
2
8
y
0
1
3
4
(8, 3)
3
(a) y is an exponential function of x: False (b) y is a logarithmic function of x: True; y log2 x (c) x is an exponential function of y: True, 2y x
2
(d) y is a linear function of x: False
(2, 1)
1
(1, 0) x 2
4
6
8
1 x ln 2 gx xx ⇒ gx xx 1 ln x
75. f x log2 x ⇒ fx
77. Ct P1.05t (a) C10 24.951.0510
[Note: Let y gx. Then: ln y ln xx x ln x 1 1 y x ln x y x y y1 ln x y x 1 ln x gx.
x
hx x ⇒ hx 2x 2
kx 2x ⇒ kx ln 22x From greatest to smallest rate of growth: gx, kx, hx, f x
$40.64 (b)
dC Pln 1.051.05t dt When t 1:
dC 0.051P dt
When t 8:
dC 0.072P dt
dC t (c) dt ln 1.05 P1.05
ln 1.05Ct The constant of proportionality is ln 1.05.
244
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
79. P $1000, r 312 % 0.035, t 10
0.035 A 1000 1 n
n
1
2
4
12
365
Continuous
A
1410.60
1414.78
1416.91
1418.34
1419.04
1419.07
n
1
2
4
12
365
Continuous
A
4321.94
4399.79
4440.21
4467.74
4481.23
4481.69
10n
A 1000e0.03510 1419.07 81. P $1000, r 5% 0.05, t 30
0.05 A 1000 1 n
30n
A 1000e0.0530 4481.69 83. 100,000 Pe0.05t ⇒ P 100,000e0.05t
85. 100,000 P 1
0.05 12
t
1
10
20
30
40
50
P
95,122.94
60,653.07
36,787.94
22,313.02
13,583.53
8208.50
12t
⇒ P 100,000 1
0.05 12
12t
t
1
10
20
30
40
50
P
95,132.82
60,716.10
36,864.45
22,382.66
13,589.88
8251.24
87. (a) A 20,000 1
0.06 365
3658
89. (a) lim 6.7e48.1t 6.7e0 6.7 million ft3
$32,320.21
t →
(b) A $30,000
(c) A 8000 1
(b) 0.06 365
3658
20,000 1
0.06 365
3654
1 0.06 365
3658
1
0.06 365
3654
322.27 48.1t e t2
V20 0.073 million ft3yr V60 0.040 million ft3yr
$12,928.09 25,424.48 $38,352.57 (d) A 9000
V
1
$34,985.11 Take option (c).
91. y (a)
300 3 17e0.0625x (c) If y 66.67%, then x 38.8 or 38,800 egg masses.
100
(d) y 3003 17e0.0625x 1
0
y
318.75e0.0625x 3 17e0.0625x2
y
19.921875e0.0625x 17e0.0625x 3 3 17e0.0625x 3
100 0
(b) If x 2 (2000 egg masses), y 16.67 16.7%.
17e0.0625x 3 0 ⇒ x 27.8 or 27,800 egg masses.
Section 5.5 93. (a) B 4.75396.7744d 4.7539e1.9132d (b)
Bases Other than e and Applications
245
Bd 9.0952e1.9132d
(c)
B0.8 42.03 tonsinch
120
B1.5 160.38 tonsinch
0
2 0
4
95. (a)
(c) The functions appear to be equal: f t gt ht
f t dt 5.67
0
Analytically,
4
gt dt 5.67
f t 4
0 4
ht dt 5.67
2t3
8 4 4 gt
4
3
913
23 t
t
ht 4e0.653886t 4 e0.653886 t 40.52002t
0
(b)
38
6
gt 4
−1
94 40.52002 13 t
t
No. The definite integrals over a given interval may be equal when the functions are not equal.
5 −1
10
97. P
2000e0.06t dt
99.
0
2000 e 0.06
0.06t
10
t
0
1
2
3
4
y
1200
720
432
259.20
155.52
0
y Ckt
$15,039.61
When t 0, y 1200 ⇒ C 1200. y 1200kt 720 432 259.20 155.52 0.6, 0.6, 0.6, 0.6 1200 720 432 259.20 Let k 0.6. y 12000.6t
101. False. e is an irrational number.
103. True.
105. True.
f gx 2 elnx2 2x2x g f x ln2 ex 2 ln ex x
d x d e ex and ex ex dx dx ex ex when x 0.
e0e0 1
246
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
dy 8 5 y y , y0 1 dt 25 4
107.
8 4 dy dt ⇒ y54 y 25 5 ln y ln ln
1 1 dy y 54 y
8 dt ⇒ 25
54 y 52 t C
54y y 52 t C y e25tC C1e25t 54 y
y 0 1 ⇒ C1 4 ⇒ 4e25t ⇒ 4e25t
54 y y ⇒ 5e
⇒ y
5e25t 5 1.25 1 4 e0.4t 1 0.25e0.4t
Section 5.6
1.
25t
4e25t y y 4e25t 1y
4e25t
Differential Equations: Growth and Decay
dy x2 dx y
y 54 y
dy y2 dx
3.
x 2dx
x2 2x C 2
dy dx y2 1 dy y2
dx
ln y 2 x C1 y 2 exC1 Cex y Cex 2
y
5.
5x y
yy 5x
yy dx y dy
5x dx 5x dx
1 2 5 2 y x C1 2 2
y x y
7.
y x y
y dx y
x dx
dy y
x dx
ln y
2 32 x C1 3
y e23x
32
y 2 5x 2 C
C1
eC1 e23x
32
Ce23x
32
Section 5.6
9. 1 x2y 2xy 0 y
2x y y 1 x2
k dQ 2 dt t
11.
2xy 1 x2
Differential Equations: Growth and Decay
dQ dt dt
dN k250 s ds
13.
k dt t2
dN ds ds
2x dx 1 x2
dy y
2x dx 1 x2
k250 s ds
k dN 250 s 2 C 2
k dQ C t
y dx y
247
k N 250 s2 C 2
k Q C t
ln y ln1 x2 C1 ln y ln1 x 2 ln C ln y ln C1 x2 y C1 x 2 15. (a)
dy x6 y, 0, 0 dx
(b)
y 9
7
dy x y6
ln y 6 x
−5
−1
−6
x 2 C 2
6 −1
y 6 ex 2C C1ex 2 2
5
(0, 0)
2
y 6 C1ex
22
0, 0: 0 6 C1 ⇒ C1 6 ⇒ y 6 6ex 2 2
17.
dy 1 t, 0, 10 dt 2
dy
19.
16
1 t dt 2 −4
dy y
4
21.
0, 4: 4
y 10et2
23.
(Theorem 5.16)
0, 20,000: C 20,000
C
3, 10: 10 4e3k ⇒ k
1 5 ln 3 2
4, 12,500: 12,500 20,000e4k ⇒ k
When x 6, y 4e13 ln526 4eln52
2
52
dV kV dt V Cekt
(Theorem 5.16)
4
10 −1
10 Ce0 ⇒ C 10
dy ky dx
Ce0
−1
y et2 C1 eC1 et2 Cet2
1 2 t 10 4
y Cekx
(0, 10)
1 dt 2
1 ln y t C1 2
−1
1 10 02 C ⇒ C 10 4 y
16
(0, 10)
1 y t2 C 4
dy 1 y, 0, 10 dt 2
2
25
1 5 ln 4 8
When t 6, V 20,000e14 ln586 20,000eln58
32
20,000
58
32
9882.118
248
Chapter 5
25. y Cekt,
Logarithmic, Exponential, and Other Transcendental Functions
0, 12, 5, 5
27.
1 Cek
1 C 2 y
y Cekt, 1, 1, 5, 5 5 Ce5k 5Cek Ce5k
1 kt e 2
5ek e5k
1 5 e5k 2 k
ln 10 0.4605 5
y
1 0.4605t e 2
5 e4k k
ln 5 0.4024 4
y Ce0.4024t 1 Ce0.4024 C 0.6687 y 0.6687e0.4024t
29. A differential equation in x and y is an equation that involves x, y and derivatives of y.
31.
dy 1 xy dx 2 dy > 0 when xy > 0. Quadrants I and III. dx
33. Since the initial quantity is 10 grams, y 10eln121620t . When t 1000, y 10eln1216201000 6.52 grams. When t 10,000, y 10eln12162010,000 0.14 gram. 35. Since y Celn121620t, we have 0.5 Celn12162010,000 ⇒ C 36.07. Initial quantity: 36.07 grams. When t 1000, we have y Celn1216201000 23.51 grams. 37. Since the initial quantity is 5 grams, we have y 5.0eln125730t. When t 1000, y 4.43 g. When t 10,000, y 1.49 g. 39. Since y Celn1224,360t, we have 2.1 Celn1224,3601000 ⇒ C 2.16. Thus, the initial quantity is 2.16 grams. When t 10,000, y 2.16eln1224,36010,000 1.63 grams.
41. Since
dy ky, y Cekt or y y0ekt. dx
1 y y0e1620k 2 0 k
ln 2 1620
y y0eln 2t 1620. When t 100, y y0eln 216.2 y00.9581. Therefore, 95.81% of the present amount still exists.
43. Since A 1000e0.06t, the time to double is given by 2000 1000e0.06t and we have 2 e0.06t ln 2 0.06t t
ln 2 11.55 years. 0.06
Amount after 10 years: A 1000e0.0610 $1822.12
Section 5.6 45. Since A 750ert and A 1500 when t 7.75, we have the following.
ln 2 0.0894 8.94% 7.75
Amount after 10 years: A 750e0.089410 $1833.67
r
ln1292.85500 0.0950 9.50% 10
The time to double is given by 1000 500e0.0950t t
0.075 12
49. 500,000 P 1
1220
0.075 12
P 500,000 1
240
2 1
ln 2 10.24 years ln 1.07
0.07 12
0.007 12
1 t 12
t
ln 2 9.90 years 0.07
0.085 12
12t
0.085 365
0.085 365
ln 2 365t ln 1 t
12t
1 365
365t
365t
0.085 365
ln 2 8.16 years 0.085 ln 1 365
(d) 2000 1000e0.085t
0.085 12
ln 2 8.18 years 0.085 ln 1 12
0.07 365
ln 2 9.90 years 365 ln1 0.07365
(c) 2000 1000 1
ln 2 8.50 years ln 1.085
365t
ln 2 0.07t
ln 2 t ln1.085
ln 2 12t ln 1
365t
2 1
0.085 2 1 12
0.07 365
2 e0.07t
2 1.085t
420
(d) 2000 1000e0.07t
55. (a) 2000 10001 0.085t
0.07 365
ln 2 365t ln 1 t
ln 2 9.93 years 12 ln1 0.0712
(b) 2000 1000 1
12t
t
0.08 12
12t
0.07 ln 2 12t ln 1 12 t
ln 2 t ln 1.07
2 1
1235
P 500,000 1
(c) 2000 1000 1
2 1.07t
$30,688.87
53. (a) 2000 10001 0.07t
(b) 2000 1000 1
0.08 12
51. 500,000 P 1
$112,087.09
t
ln 2 7.30 years. 0.095
249
47. Since A 500ert and A 1292.85 when t 10, we have the following. 1292.85 500e10r
1500 750e7.75r r
Differential Equations: Growth and Decay
2 e0.085t ln 2 0.085t t
ln 2 8.15 years 0.085
250
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
57. P Cekt Ce0.009t
59. P Cekt Ce0.036t
P1 8.2 Ce0.0091 ⇒ C 8.1265
P1 4.6 Ce0.0361 ⇒ C 4.7686
P 8.1265e0.009t
P 4.7686e0.036t
P10 7.43
or
P10 6.83
7,430,000 people in 2010
61. If k < 0, the population decreases.
or
6,830,000 people in 2010
P Cekx, 0, 760, 1000, 672.71
63.
C 760
If k > 0, the population increases.
672.71 760e1000x x
ln672.71760 0.000122 1000
P 760e0.000122x When x 3000, P 527.06 mm Hg. 65. (a)
19 301 e20k 30e20k 11 k
25 301 e0.0502t
(b)
e0.0502t
ln1130 0.0502 20
t
N 301 e0.0502t
1 6 ln 6 36 days 0.0502
67. S Cekt (a)
S 5 when t 1
(b) When t 5, S 20.9646 which is 20,965 units.
5
(c)
Cek
30
lim Cekt C 30
t →
5 30ek 0
k ln 16 1.7918
40 0
S 30e1.7918t 69. At Vte0.10t 100,000e0.8 t e0.10t 100,000e0.8 t0.10t
0.4 dA 100,000 0.10 e0.8 t0.10t 0 when 16. dt t The timber should be harvested in the year 2014, 1998 16. Note: You could also use a graphing utility to graph At and find the maximum of At. Use the viewing rectangle 0 ≤ x ≤ 30 and 0 ≤ y ≤ 600,000. I 71. I 10 log10 , I0 1016 I0 (a) 1014 10 log10 (b) 109 10 log10
1014 20 decibels 1016
109 70 decibels 1016
106.5 (c) 106.5 10 log10 16 95 decibels 10 (d) 104 10 log10
73. R
ln I 0 , I eR ln 10 10R ln 10
(a) 8.3
ln I 0 ln 10
I 108.3 199,526,231.5 (b) 2R
ln I 0 ln 10
I e2R ln 10 e2R ln 10 eR ln 102 10R 2
104 120 decibels 1016
Increases by a factor of e2R ln 10 or 10R. (c)
1 dR dI I ln 10
Section 5.7
Differential Equations: Separation of Variables
75. False. If y Cekt, y Ckekt constant.
Section 5.7
77. True
Differential Equations: Separation of Variables
1. Differential equation: y 4y Solution: y Ce4x Check: y 4Ce4x 4y 3. Differential equation: y y 0 Solution: y C1 cos x C2 sin x y C1 sin x C2 cos x
Check:
y C1 cos x C2 sin x y y C1 cos x C2 sin x C1 cos x C2 sin x 0
5. y cos x ln sec x tan x y cos x
1 sec x tan x sec2 x sin x ln sec x tan x sec x tan x
cos x sec xtan x sec x sin x ln sec x tan x sec x tan x
1 sin x ln sec x tan x
1 y sin x sec x tan x sec2 x cos x ln sec x tan x sec x tan x
sin xsec x cos x ln sec x tan x Substituting,
y y sin xsec x cos x ln sec x tan x cos x ln sec x tan x tan x. In Exercises 7–11, the differential equation is y4 16y 0. y 3 cos x
7.
y e2x
9.
y4 3 cos x
y4 16e2x
y4 16y 45 cos x 0,
y4 16y 16e2x 16e2x 0,
No.
Yes. y C1e2x C2e2x C3 sin 2x C4 cos 2x
11.
y4 16C1e2x 16C2e2x 16C3 sin 2x 16C4 cos 2x y4 16y 0, Yes.
251
252
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
In 13–17, the differential equation is xy 2y x3ex. 15. y x 22 ex , y x 2ex 2x2 ex
13. y x 2, y 2x xy 2y x2x 2x2 0 x3ex
xy 2y xx2ex 2xex 4x 2x2ex 2x2 x3ex,
No.
Yes.
1 x 1 xy 2y x 2 ln x x3ex, x
17. y ln x, y
19.
No.
21. y 2 Cx3 passes through 4, 4
y Cekx
16 C64 ⇒ C 14
dy Ckekx dx Since dy dx 0.07y, we have Thus, k 0.07.
Ckekx
Particular solution: y2 14 x3 or 4y 2 x 3
0.07Cekx.
23. Differential equation: 4yy x 0
2
C=0
General solution: 4y 2 x 2 C
−3
Particular solutions: C 0, Two intersecting lines C ± 1, C ± 4, Hyperbolas 2
−2
2
2
C = −1
C=1 −3
3
3
−3
−2
3
−2
2
C = −4
C=4 −3
−3
3
3
−2
−2
25. Differential equation: y 2y 0 General Solution: y Ce2x y 2y C2e2x 2Ce2x 0 Initial condition: y0 3, 3 Ce0 C Particular solution: y 3e2x
27. Differential equation: y 9y 0 General solution: y C1 sin 3x C2 cos 3x y 3C1 cos 3x 3C2 sin 3x, y 9C1 sin 3x 9C2 cos 3x y 9y 9C1 sin 3x 9C2 cos 3x 9C1 sin 3x C2 cos 3x 0
Initial conditions: y 2 C1 sin
6 2, y 6 1
2 C
2
cos
2 ⇒ C
1
2
y 3C1 cos 3x 3C2 sin 3x 1 3C1 cos
2 3C sin 2
2
3C2 ⇒ C2
1 3
Particular solution: y 2 sin 3x
1 cos 3x 3
Section 5.7
Differential Equations: Separation of Variables Initial conditions: y2 0, y2 4
29. Differential equation: x 2 y 3xy 3y 0 General solution: y C1 x C2 x 3
0 2C1 8C2
y C1 3C2 x2, y 6C2 x
y C1 3C2 x 2
x2y 3x y 3y x26C2 x 3xC1 3C2 x2
4 C1 12C2 C1 4C2 0
3C1x C2 x3 0
C1 12C2 4
1 C2 , C1 2 2
1 Particular solution: y 2x x3 2
31.
dy 3x2 dx y
33.
3x 2 dx x 3 C
x dy dx 1 x2 y
1 x dx ln1 x2 C 1 x2 2
u 1 x 2, du 2x dx
35.
2 dy x 2 1 dx x x y
1
37.
y
u 2x, du 2dx
dy xx 3 dx y
xx 3 dx
u2 3u2udu
dy 2 xex dx y
u5 u C 52 x 3 5
3
5 2
2x 33 2 C
43.
xex dx 2
1 sin 2x dx cos 2x C 2
Let u x 3, then x u2 3 and dx 2u du.
2 u4 3u 2 du 2
41.
2 dx x
x 2 ln x C x ln x 2 C
39.
dy sin 2x dx
1 x2 e C 2
u x 2, du 2x dx
dy x dx y
y dy
x dx
y2 x2 C1 2 2 y2 x2 C
45.
dr 0.05r ds
dr r
0.05 ds
ln r 0.05s C1 r e0.05sC1 Ce0.05s
47. 2 xy 3y
dy y
3 dx 2x
ln y 3 ln2 x ln C ln C2 x3 y Cx 23
253
254
Chapter 5
49.
yy sin x
y dy y2 2 y2
Logarithmic, Exponential, and Other Transcendental Functions
51. 1 4x2
sin x dx
dy x dx dy
x 1 4x2
cos x C1
dy
2 cos x C
dx
x 1 4x2
1 8
dx
1 4x21 28x dx
1 y 1 4x21 2 C 4 53. y ln x x y 0
dy y
55. yy ex 0 ln x dx x
u ln x, du dxx
y dy
2 C 1
ex dx
y2 ex C1 2
1 ln y ln x 2 C1 2 y e1 2lnx
y 2 2ex C
Celnx 2 2
Initial condition: y0 4, 16 2 C, C 14 Particular solution: y 2 2ex 14 57. yx 1 y 0
59. y1 x2
dy x 1 dx y
ln y
dy x1 y 2 dx
y x dy dx 1 y2 1 x2
x 12 C1 2
1 1 ln1 y 2 ln1 x 2 C1 2 2 ln1 y 2 ln1 x 2 ln C lnC1 x 2
y Cex1 2 2
Initial condition: y2 1, 1 Ce1 2, C e1 2 Particular solution: y e1 x1 2 ex 2
2
2 2x
1 y 2 C1 x2 y0 3: 1 3 C ⇒ C 4 1 y2 41 x2 y 2 3 4x 2
61.
du uv sin v 2 dv
63. dP kP dt 0
du u
v sin v 2 dv
dP k dt P
ln P kt C1
1 ln u cos v 2 C1 2
P Cekt
u Cecos v 2 2
Initial condition: P0 P0, P0 Ce0 C
Initial condition: u0 1, C
1 e1 2 e1 2
Particular solution: u e1cosv 2 2
Particular solution: P P0 ekt
Section 5.7 dy 9x dx 16y
65.
Differential Equations: Separation of Variables
67.
0y y dy dx x 2 x 2
dy y
16y dy 9x dx 8y 2
m
9 2 x C 2
1 dx 2
1 ln y x C1 2 y Cex 2
9 25 Initial condition: y 1 1, 8 C, C 2 2 Particular solution: 8y 2
9 2 25 x , 2 2
16y 2 9x 2 25
69.
71. f x, y
f x, y x 3 4xy 2 y 3 f t x, t y t 3 x 3 4t xt 2 y 2 t3 y3
f tx, ty
t 3x3 4xy 2 y3
f x, y 2 ln x y
75.
f t x, t y 2 ln t x t y
vx x
2
xy , y vx 2x
tx x 2 ln ty y
y
79.
dv x v x dx 2x
vx
dv 1 v v dx 2
dv 1v
x y
Homogeneous degree 0
Not homogeneous y
f x, y 2 ln f t x, t y 2 ln
2 ln t 2 x y 2ln t 2 ln x y
77.
t4x2y2 x2y2 t3 2 2 2 x y2 ty
t2x2
Homogeneous of degree 3
Homogeneous of degree 3
73.
x2y2 x2 y2
ln1 v 2 ln x ln C ln Cx 1 Cx 1 v 2
dv x xv dx x xv
v dx x dv
dx x
xy , y vx xy
1v dx 1v
v1 dx dv v2 2v 1 x
C
1 Cx 1 y x2
x2 Cx x y2
x Cx y
2
C1 1 ln v 2 2v 1 ln x ln C1 ln 2 x
v 2 2v 1 x 2
y2 y C 2 1 2 x2 x x
y 2 2xy x 2 C
255
256
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
y
81.
xy , y vx x2 y 2
xv dx x dv 2xev vx dx 0
x2 v dv 2 vx dx x x2 v 2 v dx x dv
1v dv v3 2
ev dv
v dx 1 v2
2 dx x
ev ln C1 x 2
dx x
ey x ln C1 ln x 2 ey x C ln x 2
1 2 ln v ln x ln C1 ln C1 x 2v
x dy 2xey x y dx 0, y v x
83.
Initial condition: y1 0, 1 C
1 ln C1x v 2v 2
Particular solution: ey x 1 ln x 2
x 2 ln C1 y 2y 2
y Cex 2y 2
2
x sec xy y dx x dy 0, y v x
85.
87.
dy x dx
x sec v xvdx xv dx x dv 0
cos v dv
y 4
sec v v dx v dx x dv dx x
x 2
2
sin v ln x ln C1 4
x Cesin v Cesiny x Initial condition: y1 0, 1
y Ce0
C
x dx
1 2 x C 2
Particular solution: x esiny x
89.
dy 4y dx
y 8
dy 4y
dx
ln 4 y x C1 4 y ex C1
4
y 4 Cex −4 −3
91.
x 1
2
3
4
dy 0.5y, y0 6 dx
93.
dy 0.02y10 y, y0 2 dx
12
−6
12
6 −4
− 12
48 −2
Section 5.7
95.
dy ky, y Cekt dt Initial conditions:
Differential Equations: Separation of Variables
97. y0 y0
dy k y 4 dx The direction field satisfies dy dx 0 along y 4; but not along y 0. Matches (a).
y0 y1620 2 C y0 y0 y0e1620k 2 k
ln1 2 1620
Particular solution: y y0etln 2 1620 When t 25, y 0.989y0, y 98.9% of y0.
99.
dy k yy 4 dx The direction field satisfies dy dx 0 along y 0 and y 4. Matches (c). dw k1200 w dt
101.
dw 1200 w
k dt
dv Wv
k dt
lnW v k t C1
ln1200 w kt C1 1200 w
dv kW v dt
103. (a)
ektC1
v W Cekt
Cekt
w 1200 Cekt
Initial conditions:
w0 60 1200 C ⇒ C 1200 60 1140
W 20, v 0 when t 0, and
w 1200 1140e
v 5 when t 1.
kt
(a)
1400
C 20, k ln3 4
1400
Particular solution: v 201 eln3 4t 201 e0.2877t 0
10 0
0
10 0
(b) s 1400
201 e0.2877t dt
20t 3.4761e0.2877t C Since S0 0, C 69.5 and we have s 20t 69.5e0.2877t 1. 0
10 0
(b) k 0.8: t 1.31 years k 0.9: t 1.16 years k 1.0: t 1.05 years (c) Maximum weight: 1200 pounds lim w 1200 t→0
257
258
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
105. Given family (circles): x 2 y 2 C
107. Given family (parabolas): x2 Cy 2x Cy
2x 2yy 0 y Orthogonal trajectory (lines):
y
x y
y
y x
y
Orthogonal trajectory (ellipses):
dy y
2x 2x 2y 2 C x y x
dx x
x 2y
2 y dy x dx
ln y ln x ln K
y2
y Kx
x2 K1 2
x 2 2y 2 K
4 4 −6
6 −6
6
−4 −4
109. Given family: y 2 Cx3
111. A general solution of order n has n arbitrary constants while in a particular solution initial conditions are given in order to solve for all these constants.
2yy 3Cx2 y
3Cx2 3x2 y2 3y 2y 2y x3 2x y
Orthogonal trajectory (ellipses):
2x 3y
3 y dy 2 x dx 3y 2 x 2 K1 2 3y 2 2x 2 K 4
−6
6
−4
113. Mx, ydx Nx, ydy 0, where M and N are homogeneous functions of the same degree. 117. False f t x, t y t 2x 2 t 2 xy 2 t 2 f x, y
115. False. Consider Example 2. y x3 is a solution to xy 3y 0, but y x3 1 is not a solution.
Section 5.8
Section 5.8
Inverse Trigonometric Functions: Differentiation
Inverse Trigonometric Functions: Differentiation
1. y arcsin x (a)
x
1
0.8
0.6
0.4
0.2
0
0.2
0.4
0.6
0.8
1
y
1.571
0.927
0.644
0.412
0.201
0
0.201
0.412
0.644
0.927
1.571
(b)
y
(c)
π 2
(d) Symmetric about origin: arcsinx arcsin x
2
−1
Intercept: 0, 0
1
x
−1
1 −2 −π 2
3. False.
5. arcsin
arccos
1 2 6
1 2 3
since the range is 0, .
7. arccos
1 2 3
11. arccsc 2
9. arctan
4
3
3
6
13. arccos0.8 2.50
15. arcsec1.269 arccos
1 1.269
0.66
3 3 17. (a) sin arctan 4 5
21 cot 6 3
19. (a) cot arcsin 5 3
3
θ
θ 4
1
2
4 5 (b) sec arcsin 5 3
(b) csc arctan
5
4
5 12
135
12
θ
θ
3
5 13
23. y sinarcsec x
21. y cosarcsin 2x 1
arcsin 2x
2x
y cos 1 4x
2
θ
arcsec x, 0 ≤ ≤ ,
1− 4 x 2
y sin
x2 1
x
2
x x2 − 1
θ 1
The absolute value bars on x are necessary because of the restriction 0 ≤ ≤ , 2, and sin for this domain must always be nonnegative.
259
260
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
x 3
25. y tan arcsec
x
θ
x arcsec 3 y tan
27. y csc arctan
x2 − 9
arctan
3
x2 9
2x
29. sinarctan 2x
x2 + 2 x
θ 2
2 x2 2
x
31. arcsin3x 12
1 4x2
3x sin 12
2
x 13 sin 12 1.207
f=g
−2
x
y csc
3
x 2
2
−2
Asymptotes: y ± 1 arctan 2x
1 + 4 x2
tan 2x sin
2x
θ
2x 1 4x2
1
33. arcsin2x arccos x
2x sin arccos x
2x 1 x, 0 ≤ x ≤ 1
2x 1 x 1
3x 1
1−x
θ
1 x 3
x
1 35. (a) arccsc x arcsin , x ≥ 1 x
(b) arctan x arctan
Let y arccsc x. Then for
1 ,x > 0 x 2
Let y arctan x arctan1 x. Then,
≤ y < 0 and 0 < y ≤ , 2 2
tan y
csc y x ⇒ sin y 1 x. Thus, y arcsin1 x. Therefore, arccsc x arcsin1 x.
tanarctan x tanarctan1 x 1 tanarctan x tanarctan1 x
x 1 x 1 x1 x
x 1 x (which is undefined). 0
Thus, y 2. Therefore, arctan x arctan1 x 2. 37.
f x arcsinx 1
y
π
x 1 sin y Domain: 0, 2 Range:
f x is the graph of arcsin x shifted 1 unit to the right.
( ) x
−1
1 −π 2 −π
(0, − π2 )
2
f x arcsec 2x 2x sec y
2, π 2
π 2
x 1 sin y
, 2 2
39.
x
3
Domain: Range:
y
(− 12 , π (
1 sec y 2
, 21 , 12,
0, 2 , 2 ,
π 2
( 12 , 0 ( −2
−1
x
1
2
Section 5.8
41. f x 2 arcsinx 1 2 2 f x 1 x 12 2x x2
45. f x arctan f x
Inverse Trigonometric Functions: Differentiation
43. gx 3 arccos g x
x a
47. gx
1 a a 1 x2 a2 a2 x2
g x
49. ht sinarccos t 1 t 2
x 2
31 2 3 2 1 x 4 4 x2 arcsin 3x x x 3 1 9x2 arcsin 3x x2 3x 1 9x2 arcsin 3x x21 9x2
51. y x arccos x 1 x2
1 t h t 1 t 21 22t 2 1 t2
y arccos x
x 1 x2
1 1 x21 22x 2
arccos x
53.
y
1 1 x1 ln arctan x 2 2 x1
55.
x 1 dy x arcsin x arcsin x 2 dx 1 x 1 x2
1 1 lnx 1 lnx 1 arctan x 4 2
y x arcsin x 1 x2
dy 1 1 1 2 1 1 dx 4 x 1 x 1 1 x2 1 x4
57. y 8 arcsin y 2
x x16 x2 4 2
59. y arctan x
16 x2 1 x 16 x21 22x 2 1 x 4 2 4
y
x 1 x2
1 1 x2 x2x 2 1x 1 x22
16 x2 8 x2 2 16 x 2 216 x2
1 x2 1 x2 1 x22
16 16 x2 x2 x2 2 16 x2 216 x
2 1 x22
61. f x arcsin x, a
1 2
y 1.5 1.0
1 f x 1 x2
0.5 x
1 1 P1x f f 2 2 P2x f
0.5 1.0 1.5
P2
x f x 1 x23 2
P1
− 1.0
f
1 23 1 x x 2 6 3 2
12 f 12 x 21 21 f 12 x 21
2
− 1.5
23 1 23 1 x x 6 3 2 9 2
2
261
262
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
f x arcsec x x 1 f x 1 x x2 1
63.
65.
f x
1 1 0 1 x2 1 x 42
1 x2 1 x 42
0 when x x2 1 1. x2x2 1 1
0 8x 16
x4 x2 1 0 when x2 x±
f x arctan x arctanx 4
x2
1 5 or 2
By the First Derivative Test, 2, 2.214 is a relative maximum.
1 5 ± 1.272 2
Relative maximum: 1.272, 0.606 Relative minimum: 1.272, 3.747 69. y arccot x, 0 < y <
67. The trigonometric functions are not one-to-one on , , sot their domains must be restricted to intervals on which they are one-to-one.
x cot y tan y
1 x
So, graph the function y arctan
d (b) dt
x 71. (a) cot 5
arccot
x 5
ht 16t 2 256
73. (a)
16t 2 256 0 when t 4 sec. (b) tan
h 16t2 256 500 500
500t
arctan
16
2
h
θ 500
16
d 8t 125 1000t dt 1 4 125t 2 162 15,625 1616 t 22 When t 1, d dt 0.0520 rad sec. When t 2, d dt 0.1116 rad sec.
1x for x > 0 and y arctan1x for x < 0.
1 5 x 1 5
2
dx 5 dx 2 dt x 25 dt
If
d dx 400 and x 10, 16 rad hr. dt dt
If
dx d 400 and x 3, 58.824 rad hr. dt dt
Section 5.9
75. tanarctan x arctan y
Inverse Trigonometric Functions: Integration
263
tanarctan x tanarctan y xy , xy 1 1 tanarctan x tanarctan y 1 xy
Therefore, arctan x arctan y arctan
1x xyy , xy 1.
Let x 12 and y 13 . arctan
13 56 56 arctan arctan arctan 1 12 arctan13 arctan 1 12 12 13 1 16 56 4
77. f x kx sin x f x k cos x ≥ 0 for k ≥ 1 fx k cos x ≤ 0 for k ≤ 1 Therefore, f x kx sin x is strictly monotonic and has an inverse for k ≤ 1 or k ≥ 1. 81. True
79. True
d sec2 x sec2 x arctantan x 1 dx 1 tan2 x sec2 x
1 d arctan x > 0 for all x. dx 1 x2
Section 5.9 1.
5
9
x2
Inverse Trigonometric Functions: Integration
dx 5 arcsin
3 C x
3. Let u 3x, du 3 dx.
16
0
5.
1 1 dx 3
1 9x2
16
0
16
1 1 3 dx arcsin3x 3
1 3x2
0
18
7 7 x dx arctan C 16 x2 4 4
7. Let u 2x, du 2 dx.
32
0
1 1 dx 1 4x2 2
9.
1 dx x 4x2 1
11.
x2
13.
1 x 12
x3 dx 1
1
32
0
32
2 1 2x
2
dx
12 arctan2x
0
6
2 dx arcsec 2x C 2x 2x2 1
x
x dx x2 1
x dx
dx arcsinx 1 C
1 2x 1 1 dx x2 lnx2 1 C (Use long division.) 2 x2 1 2 2
15. Let u t 2, du 2t dt.
t
1 t 4
dt
1 1 1 2t dt arcsint 2 C 2 1 t 22 2
264
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions 1
17. Let u arcsin x, du
1 2
0
19. Let u 1 x2, du 2x dx.
dx.
1 x2
0
1 2
arcsin 1 dx arcsin2 x 2
1 x2
0
x 1 dx 2 2
1x 12
2 0.308 32
0
12
1 x2122x dx
1 x2
0
12
3 2
2
0.134 21. Let u e2x, du 2e2x dx.
23. Let u cos x, du sin x dx.
1 1 e2x 2e2x e2x arctan C 4x dx 2x 2 dx 4e 2 4 e 4 2
sin x dx 2 2 1 cos x
2 1
sin x dx cos2 x
25.
1
x 1 x
dx. u x, x u2, dx 2u du
27.
arctancos x
2
x3 1 2x 1 dx dx 3 2 dx x2 1 2 x2 1 x 1
1 du 2u du 2 2 arcsin u C
1 u2 u 1 u2
1 lnx2 1 3 arctan x C 2
2 arcsin x C
29.
x5
9 x 32
dx
x 3
9 x 32
dx
8
9 x 32
9 x 32 8 arcsin 6x x2 8 arcsin
3x 1 C
1 dx arctanx 1 1 x 12
2x dx x2 6x 13
2x 6 1 dx 6 2 dx x2 6x 13 x 6x 13
2
0
1
x2 4x
dx
2 0
x 2 3 C
dx arcsin
x 2 2 C
ln x2 6x 13 3 arctan
35.
x 3 3 C
1 dx x2 2x 2
0
33.
dx
2
31.
1
4 x 22
2
2x 6 1 dx 6 dx x2 6x 13 4 x 32
37. Let u x2 4x, du 2x 4 dx.
x2
x2 4x
3
39.
2
dx
2x 3 dx
4x x2
3
2
1 x2 4x122x 4 dx x2 4x C 2
2x 4 dx
4x x2
3
2
1 dx
4x x2
2 4x x2 arcsin
x 2 2
3 2
3
4x x2124 2x dx
2
4 2 3
4
3
2
1.059 6
1
4 x 22
dx
Section 5.9
Inverse Trigonometric Functions: Integration
265
41. Let u x2 1, du 2x dx.
1 x 2x dx dx 1 arctanx2 1 C x4 2x2 2 2 x2 12 1 2
43. Let u et 3. Then u2 3 et, 2u du et dt, and
et 3 dt
2u2 du u2 3
2u 2 3 arctan
2 du
6
2u du dt. u2 3
1 du u2 3
u C 2 et 3 2 3 arctan
3
e 3 3 C t
45. A perfect square trinomial is an expression in x with three terms that factor as a perfect square. Example: x2 6x 9 x 32
47. (a) (c)
49. (a)
1
1 x2
dx arcsin x C, u x
(b)
x
1 x2
dx 1 x2 C, u 1 x2
1 dx cannot be evaluated using the basic integration rules. x 1 x2
x 1 dx
2 x 132 C, u x 1 3
(b) Let u x 1. Then x u2 1 and dx 2u du.
x x 1 dx
u2 1u2u du 2 u4 u2 du 2
u5 u3 C 5
3
2 2 2 3 2 u 3u 5 C x 1323x 1 5 C x 1323x 2 C 15 15 15
(c) Let u x 1. Then x u2 1 and dx 2u du.
x dx
x 1
u2 1 u3 2 2 2u du 2 u2 1 du 2 u C uu2 3 C x 1x 2 C u 3 3 3
Note: In (b) and (c), substitution was necessary before the basic integration rules could be used. 51. (a)
y
(b)
5
dy 3 , 0, 0 dx 1 x2
y3 x
−5
5
3 2
dx 3 arctan x C 1 x2
−8
8
0, 0: 0 3 arctan0 C ⇒ C 0 −5
53.
y 3 arctan x
(0, 0)
dy 10 , y 3 0 dx x x2 1 4
−6
55. A
3
1
12
−8
− 3 2
x2
1 dx 2x 1 4
2 arctan 1
x1 2
3 1
3
1
1 dx x 12 22
1 arctan1 0.3927 2 8
266
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
57. Area 11 1
y 2
Matches (c)
3 2
−1
x −1
1 2
2
1
59. (a)
0
4 dx 4 arctan x 1 x2
1 0
1
4 arctan 1 4 arctan 0 4
4 40
(b) Let n 6.
1
4
0
4 1 3.1415918 1 1 4136 1 219 1 414 1 249 1 2536 2
4 1 dx 4 1 x2 36
(c) 3.1415927
61. (a)
1 u u u d arcsin C 2 2 2 dx a a
1 u a
a u2
Thus, (b)
du
a2 u2
arcsin
ua C.
1 1 u u ua u d 1 arctan C 2 2 2 dx a a a 1 ua2 a a u2a2 a u2
Thus,
du a2 u2
1 u u dx arctan C. a2 u2 a a
(c) Assume u > 0. 1 1 u u ua u d 1 arcsec C . The case u < 0 is handled in a simidx a a a ua ua2 1 a u u2 a2a2
u u2 a2 lar manner.
Thus,
du u u2 a2
63. (a) vt 32t 500
u u u2 a2
dx
1 u arsec C. a a
(b) st
550
vt dt
32t 500 dt
16t 2 500t C s0 160 5000 C 0 ⇒ C 0 0
20 0
st 16t 2 500t When the object reaches its maximum height, vt 0. vt 32t 500 0 32t 500 t 15.625 s15.625 1615.6252 50015.625 3906.25 ft Maximum height
—CONTINUED—
Section 5.10
Hyperbolic Functions
63. —CONTINUED—
(c)
1 dv 32 kv2
dt
32k v t C k arctan v 32k t C 32
1
arctan
32k
1
32k v tanC 32k t 32 v tan C k
32k t
When t 0, v 500, C arctan 500k 32 , and we have vt
32k tan arctan 50032k
32k t .
(d) When k 0.001, v(t 32,000 tanarctan 5000.00003125 0.032 t. 500
0
7 0
vt 0 when t0 6.86 sec.
6.86
(e) h
32,000 tan arctan 5000.00003125 0.032 t dt
0
Simpson’s Rule: n 10; h 1088 feet (f) Air resistance lowers the maximum height.
Section 5.10 1. (a) sinh 3
Hyperbolic Functions
e3 e3 10.018 2
(b) tanh2
3. (a) cschln 2
sinh2 e2 e2 2 0.964 cosh2 e e2
(b) cothln 5
2 2 4 eln 2 eln 2 2 1 2 3 coshln 5 eln 5 eln 5 ln 5 sinhln 5 e eln 5
5. (a) cosh12 ln 2 3 1.317 (b) sech1
23 ln 1
7. tanh2 x sech2 x
ee
x x
1 4 9
2 3
ex ex
0.962
e 2
x
2 ex
2
e2x 2 e2x 4 e2x 2 e2x 2x 1 ex ex2 e 2 e2x
5 1 5 13 5 1 5 12
267
268
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
9. sinh x cosh y cosh x sinh y
e
x
ex 2
e
y
ey ex ex 2 2
e
y
ey 2
1 exy exy exy exy exy exy exy exy 4 exy exy 1 sinhx y 2exy exy 4 2
11. 3 sinh x 4 sinh3 x sinh x3 4 sinh2 x
e
x
ex 2
e
x
ex 1 3 e2x 2 e2x ex exe2x e2x 1 2 2
3 4 e
x
ex 2
2
e3x e3x 1 sinh3x e3x ex ex ex e3x ex 2 2
sinh x
13.
cosh2 x
32
2
3 2
1 ⇒ cosh2 x
tanh x
3 2 13 2
13 13 ⇒ cosh x 4 2
313 13
csch x
2 1 3 2 3
sech x
213 1 13 13 2
coth x
13 1 3 3 13
15. y sinh1 x2
17. f x lnsinh x
y 2x cosh1 x2
19. y ln tanh y
x 2
fx
21. hx
x 1 2 1 sech2 tanhx 2 2 2 sinhx 2 coshx 2
23. f t arctansinh t ft
1 cosh t 1 sinh2 t
cosh t sech t cosh2 t
hx
1 cosh x coth x sinh
1 x sinh2x 4 2 1 1 cosh2x 1 cosh2x sinh2 x 2 2 2
1 csch x sinh x 25. Let y gx. y xcosh x ln y cosh x ln x
1 dy cosh x sinh x ln x y dx x dy y cosh x xsinh x ln x dx x
xcosh x cosh x xsinh x ln x x
Section 5.10
Hyperbolic Functions
27. y cosh x sinh x2 y 2cosh x sinh xsinh x cosh x 2cosh x sinh x2 2e2x 29. f x sin x sinh x cos x cosh x, 4 ≤ x ≤ 4
(−π , cosh π )
12
(π , cosh π )
fx sin x cosh x cos x sinh x cos x sinh x sin x cosh x 2 sin x cosh x 0 when x 0, ± . − 2
Relative maxima: ± , cosh Relative minimum: 0, 1
31. gx x sech x
x cosh x
33.
y a sinh x y a cosh x
1
y a sinh x
(1.20, 0.66) −
2
(0, − 1) −2
y a cosh x
Therefore, y y 0. (−1.20, − 0.66) −1
Relative maximum: 1.20, 0.66 Relative minimum: 1.20, 0.66 35. f x tanh x
f 1 tanh1 0.7616
2
P1
1 0.4200 cosh21
fx sech2 x
f1
fx 2 sech2 x tanh x
f 1 0.6397
f −3
4
P2 −2
P1x f 1 f1x 1 0.7616 0.42x 1 P2x 0.7616 0.42x 1
37. (a) y 10 15 cosh
0.6397 x 12 2
x , 15 ≤ x ≤ 15 15
(b) At x ± 15, y 10 15 cosh1 33.146. At x 0, y 10 15 cosh1 25.
y
(c) y sinh
30
x . At x 15, y sinh1 1.175 15
20 10
− 10
x 10
20
39. Let u 1 2x, du 2 dx.
sinh1 2x dx
1 sinh1 2x2 dx 2
1 cosh1 2x C 2
41. Let u coshx 1, du sinhx 1 dx.
cosh2x 1 sinhx 1 dx
1 cosh3x 1 C 3
269
270
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
45. Let u
43. Let u sinh x, du cosh x dx.
cosh x dx ln sinh x C sinh x
x csch2
1 1 47. Let u , du 2 dx. x x
0
csch2
x2 x2 x dx coth C 2 2
1 5x 1 ln dx 25 x2 10 5 x
4 0
51. Let u 2x, du 2 dx.
1 1 ln 9 ln 3 10 5
2 4
0
x 2x 1 1 dx dx arctanx2 C x4 1 2 x22 1 2
y
57. y sinh1tan x 1 sec2 x sec x y tan2 x 1
3 9x2 1
59. y coth1sin 2x
y
1 2 cos 2x 2 sec 2x 1 sin2 2x
61. y 2x sinh12x 1 4x2 y 2x
1 2 4x 2 sinh
1
2
2x
4x 1 4x2
2 sinh12x
63. See page 395.
65. y a sech1
ax a
2
x2
x a2 x x2 a2 a2 x2 1 dy 2 2 2 2 2 2 2 2 2 2 dx x a x x a1 x a a x xa x xa x
67.
1 dx 1 e2x
69. Let u x, du
71.
1 x1 x
1 dx. 2x
dx 2
1 dx 4x x2
ex 1 1 e2x dx csch1ex C ln C x 2 ex e
ex1
1
dx 2 sinh 2 x 1 x 1
1
2
x C 2 ln x 1 x C
1 1 x4 1 x 2 2 ln C dx ln x 22 4 4 x 2 2 4 x
2 4
1 2 dx arcsin2x 1 2x2
55. y cosh13x
53. Let u x2, du 2x dx.
x2 dx 2
csch1 x coth1 x 1 1 1 1 dx csch coth 2 dx csch C x2 x x x x 4
49.
x2 , du x dx. 2
0
4
Section 5.10
73.
1 dx 1 4x 2x2
1 1 dx 3 2x 12 2
77. y
1 80 8x 16x2
x3 21x dx 5 4x x2
2
2
2
2x 1 3
dx
2x 1 3 2x 1 3 1 1 ln C ln C 26 2x 1 3 26 2x 1 3
75. Let u 4x 1, du 4 dx. y
4 4x 1 1 1 dx arcsin C 4 81 4x 12 4 9
dx
x 4
20 dx 5 4x x2
x 4 dx 20
4
sech
0 4
2
0
0
2
x dx 2
81. A
0
2
dx
2 dx ex 2
ex 2 dx 1
52 ln x
5 ln 4 17 5.237 2
ex 2 2
8 arctanex 2
4 0
5 2
5x
x4 1
ex 2
4
4
0
2x x22 1 2
8 arctane 2 5.207 2
1 dx 32 x 22
x2 x2 x2 20 x 2 3 10 x 1 10 x 5 C 4x C C 4x ln ln 4x ln 2 6 x 2 3 2 3 x5 2 3 x1
79. A 2
83.
Hyperbolic Functions
3k dt 16 3kt 16
1 dx x2 12x 32
1 x 6 2 1 x8 1 dx ln C ln C x 62 4 22 x 6 2 4 x4
When x 0:
t0 1 C ln2 4
When x 1:
t 10
30k 1 7 7 1 1 ln ln2 ln 16 4 3 4 4 6 k When t 20:
7 2 ln 15 6
163 152 ln 76 20 41 ln 2xx 88 ln
76
2
ln
x8 2x 8
x8 49 36 2x 8 62x 104 x
104 52 1.677 kg 62 31
dx
x4 1
2 0
271
272
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
85. As k increases, the time required for the object to reach the ground increases. ex ex 2
87. y cosh x y
y cosh1 x
89.
cosh y x
ex ex sinh x 2
sinh y y 1 y
91. y sech x
1 1 1 sinh y cosh2 y 1 x2 1
2 ex ex
y 2ex ex2ex ex
e
x
2 ex
ee
x x
ex sech x tanh x ex
Review Exercises for Chapter 5 1. f x ln x 3
y
Vertical shift 3 units upward Vertical asymptote: x 0
5 4
x=0
3 2 1 x 1
4x4x
2
3. ln
5
2
5. ln 3
2
3
4
5
1 1 2x 12x 1 1 ln ln2x 1 ln2x 1 ln4x2 1 1 5 4x2 1 5
3 4 x2 3 1 3 4 x2 ln x ln ln4 x2 ln x ln 3 ln 3 x
9. gx lnx
7. lnx 1 2 x 1 e2
gx
x 1 e4
1 ln x 2
1 2x
x e4 1 53.598 11. f x xln x
13.
x 1 fx ln x1 2 ln x 2 x 1 2 ln x 1 ln x 2 ln x 2ln x
15.
1 a bx 1 y ln lna bx ln x a x a
dy 1 b 1 1 dx a a bx x xa bx
y
1 a lna bx b2 a bx
b x dy 1 ab dx b2 a bx a bx2 a bx2
17. u 7x 2, du 7dx
1 1 1 1 dx 7 dx ln 7x 2 C 7x 2 7 7x 2 7
Review Exercises for Chapter 5
19.
4
sin x sin x dx dx 1 cos x 1 cos x
21.
1
x1 dx x
4
1
1
4
1 3 ln 4
1 dx x ln x x
ln 1 cos x C
3
23.
sec d ln sec tan
0
3
0
ln 2 3
f x 12 x 3
25. (a)
y
1 2x
(b)
7
3
f
−1
− 11
2 y 3 x
10
f
2x 3 y f 1x 2x 6
−7
1 1 (c) f 1 f x f 1 2 x 3 2 2 x 3 6 x 1 f f 1x f 2x 6 2 2x 6 3 x
f x x 1
27. (a)
(b)
4
f −1
y x 1
f
y2 1 x
−3
x2 1 y f 1
x
x2
6
−2
1, x ≥ 0
(c) f 1 f x f 1 x 1 x2 12 1 x f f 1x f x2 1 x2 1 1 x2 x for x ≥ 0.
3 x 1 f x
29. (a)
(b)
4
f −1
3 x 1 y
f
y3 1 x x3
−4
5
1y
f 1x x3 1
−2
3 x 1 (c) f 1 f x f 1 3 x 1 1 x 3
3 x3 1 1 x f f 1x f x3 1
31.
f x x3 2
f x tan x
33.
f 1x x 21 3
f
1 f 1x x 22 3 3
1 0.160 35 3
3
3
fx sec2 x
1 1 f 11 1 22 3 3 332 3
6
f
f 1
6 34
33 f1 6 43
273
274
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
f x ln x
35. (a)
(b)
2
f −1
y lnx
f −3
ey x e2y x
3
−2
e2x y
(c) f 1 f x f 1 ln x e2 ln x e ln x x
f 1x e2x
f f 1x f e2x ln e2x ln ex x
37. y ex 2
39. f x lnex x2 2
fx 2x
y 6 4 2 x
−2
2
4
−2
41. gt t2et
43. y e2x e2x
gx t2et 2tet tett 2
45. gx
x2 ex
e2x e2x 1 y e2x e2x1 22e2x 2e2x 2 e2x e2x yln x y2 0
47.
ex 2x x2ex x2 x gx e2x ex
y
1x ln xdydx 2ydydx 0 2y ln x
dy y dx x y dy dx x2y ln x
49. Let u 3x2, du 6x dx.
xe3x dx 2
51.
e4x e2x 1 dx ex
1 3x2 1 2 e 6x dx e3x C 6 6
xe1x dx 2
1 1x2 e 2x dx 2
1 2 e1x C 2 57.
e3x ex ex dx
1 e3x ex ex C 3
53.
e4x 3e2x 3 C 3ex
55. Let u ex 1, du ex dx.
ex dx ln ex 1 C ex 1
y exa cos 3x b sin 3x y ex3a sin 3x 3b cos 3x exa cos 3x b sin 3x ex3a b sin 3x a 3b cos 3x y ex33a b cos 3x 3a 3b sin 3x ex3a b sin 3x a 3b cos 3x ex6a 8b sin 3x 8a 6b cos 3x y 2y 10y ex6a 8b 23a b 10b sin 3x 8a 6b 2a 3b 10a cos 3x 0
Review Exercises for Chapter 5
4
59. Area
0
1 2 2 xex dx ex 2
4 0
1 e16 1 0.500 2 63. y log2x 1
61. y 33 2
y
y
6 5
4 3
4 3
2 1
2
x
−1 1
2
2
3 4
5
6 7
−2 −3
x
−4 −3 −2 −1
1
3 4
−4
−2
65. f x 3x1
y x2x1
67.
fx 3x1 ln 3
ln y 2x 1 ln x y 2x 1 2 ln x y x y y
69. gx log31 x gx
1 log31 x 2
71.
2x x 1 2 ln x x 2x x 1 2 ln x 2x1
x 15x1 dx 2
1 1 x12 5 C 2 ln 5
1 1 1 2 1 xln 3 2x 1ln 3
73. (a) y x a y ax a1
(b) y ax
(c)
y xx
(d) y aa
ln y x ln x
y ln aax
y 0
1 1 y x 1 ln x y x y y1 ln x y xx1 ln x 75. 10,000 Pe0.0715 P
77.
Ph 30ekh P18,000 30e18,000k 15
10,000 $3499.38 e1.05
k
ln 2 ln1 2 18,000 18,000
Ph 30eh ln 2 18,000 P35,000 30e35,000 ln 2 18,000 7.79 inches
79.
P Ce0.015t 2C Ce0.015t 2 e0.015t ln 2 0.015t t
ln 2 46.21 years 0.015
81.
dy x2 3 dx x
dy y
x
3 dx x
x2 3 ln x C 2
275
276
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
83. y 2xy 0
dy 2xy dx 1 dy y
2x dx
ln y x2 C1 2 C 1
ex
y
y Cex
2
85.
dy x2 y2 (homogeneous differential equation) dx 2xy
x2 y2 dx 2xy dy 0 Let y vx, dy x dv v dx.
x2 v2x2 dx 2xvxx dv v dx 0 x2 v2x2 2x2v2 dx 2x3v dv 0 x2 x2v2 dx 2x3v dv 1 v2 dx 2x dv
dx x
2v dv 1 v2
ln x ln 1 v 2 C1 ln 1 v 2 ln C Cx2
x
C C 1 v 2 1 y x2 x2 y2
1
Cx x2 y2
or C1
x x2 y2
87. y C1x C2 x3 y C1 3C2 x2 y 6C2 x x2 y 3xy 3y x26C2 x 3xC1 3C2 x2 C1x C2 x3 6C2 x3 3C1x 9C2 x3 3C1x 3C2 x3 0 x 2, y 0: 0 2C1 8C2 ⇒ C1 4C2 x 2, y 4: 4 C1 12C2 1 4 4C2 12C2 8C2 ⇒ C2 , C1 2 2 1 y 2x x3 2 89. f x 2 arctanx 3
y
4
−6
−4
x
−2
2 −2 −4
Review Exercises for Chapter 5
Let arcsin
91. (a)
277
1 2 2
1 2
sin
1
θ
1 1 sin arcsin sin . 2 2 Let arcsin
(b)
3 1 cos . 2 2
cos arcsin
x
93. y tanarcsin x y
1 2
1 2
sin
3
95. y x arcsec x
1 x2
y
1 x21 2 x21 x21 2 1 x23 2 1 x2
x arcsec x x x2 1
97. y xarcsin x2 2x 21 x2 arcsin x y
1 x2 2x arcsin x 2x 2 2 2 arcsin x arcsin x arcsin x2 1 x2 1 x2 1 x2
99. Let u e2x, du 2e2x dx.
1 dx e2x e2x
1 1 e2x 1 dx 2e2x dx arctane2x C 1 e4x 2 1 e2x2 2
101. Let u x2, du 2x dx.
x 1 x4
2x , du 4 2 x
2
arctanx 2 1 dx 4 x2 2
107.
dy A2 y2
1 1 1 2x dx arcsin x2 C 2 1 x22 2
dx
105. Let u arctan
103. Let u 16 x2, du 2x dx.
y arcsin A
arctan
x 2
dx.
4 2 x dx 41arctan 2x 2
k dt m
k tC m
2
C
109. y 2x coshx y 2
Since y 0 when t 0, you have C 0. Thus,
mk t Ay
sin
mk t
y A sin 111. Let u x2, du 2x dx.
x x4 1
dx
1 1 x 1 dx 2x dx ln16 x2 C 16 x2 2 16 x2 2
1 1 1 2x dx ln x2 x4 1 C 2 x22 1 2
1 sinh x 2 sinh x 2x 2x
272
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
85. As k increases, the time required for the object to reach the ground increases. ex ex 2
87. y cosh x y
y cosh1 x
89.
cosh y x
ex ex sinh x 2
sinh y y 1 y
91. y sech x
1 1 1 sinh y cosh2 y 1 x2 1
2 ex ex
y 2ex ex2ex ex
e
x
2 ex
ee
x x
ex sech x tanh x ex
Review Exercises for Chapter 5 1. f x ln x 3
y
Vertical shift 3 units upward Vertical asymptote: x 0
5 4
x=0
3 2 1 x 1
4x4x
2
3. ln
5
2
5. ln 3
2
3
4
5
1 1 2x 12x 1 1 ln ln2x 1 ln2x 1 ln4x2 1 1 5 4x2 1 5
3 4 x2 3 1 3 4 x2 ln x ln ln4 x2 ln x ln 3 ln 3 x
9. gx lnx
7. lnx 1 2 x 1 e2
gx
x 1 e4
1 ln x 2
1 2x
x e4 1 53.598 11. f x xln x
13.
x 1 fx ln x1 2 ln x 2 x 1 2 ln x 1 ln x 2 ln x 2ln x
15.
1 a bx 1 y ln lna bx ln x a x a
dy 1 b 1 1 dx a a bx x xa bx
y
1 a lna bx b2 a bx
b x dy 1 ab dx b2 a bx a bx2 a bx2
17. u 7x 2, du 7dx
1 1 1 1 dx 7 dx ln 7x 2 C 7x 2 7 7x 2 7
Review Exercises for Chapter 5
19.
4
sin x sin x dx dx 1 cos x 1 cos x
21.
1
x1 dx x
4
1
1
4
1 3 ln 4
1 dx x ln x x
ln 1 cos x C
3
23.
sec d ln sec tan
0
3
0
ln 2 3
f x 12 x 3
25. (a)
y
1 2x
(b)
7
3
f
−1
− 11
2 y 3 x
10
f
2x 3 y f 1x 2x 6
−7
1 1 (c) f 1 f x f 1 2 x 3 2 2 x 3 6 x 1 f f 1x f 2x 6 2 2x 6 3 x
f x x 1
27. (a)
(b)
4
f −1
y x 1
f
y2 1 x
−3
x2 1 y f 1
x
x2
6
−2
1, x ≥ 0
(c) f 1 f x f 1 x 1 x2 12 1 x f f 1x f x2 1 x2 1 1 x2 x for x ≥ 0.
3 x 1 f x
29. (a)
(b)
4
f −1
3 x 1 y
f
y3 1 x x3
−4
5
1y
f 1x x3 1
−2
3 x 1 (c) f 1 f x f 1 3 x 1 1 x 3
3 x3 1 1 x f f 1x f x3 1
31.
f x x3 2
f x tan x
33.
f 1x x 21 3
f
1 f 1x x 22 3 3
1 0.160 35 3
3
3
fx sec2 x
1 1 f 11 1 22 3 3 332 3
6
f
f 1
6 34
33 f1 6 43
273
274
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
f x ln x
35. (a)
(b)
2
f −1
y lnx
f −3
ey x e2y x
3
−2
e2x y
(c) f 1 f x f 1 ln x e2 ln x e ln x x
f 1x e2x
f f 1x f e2x ln e2x ln ex x
37. y ex 2
39. f x lnex x2 2
fx 2x
y 6 4 2 x
−2
2
4
−2
41. gt t2et
43. y e2x e2x
gx t2et 2tet tett 2
45. gx
x2 ex
e2x e2x 1 y e2x e2x1 22e2x 2e2x 2 e2x e2x yln x y2 0
47.
ex 2x x2ex x2 x gx e2x ex
y
1x ln xdydx 2ydydx 0 2y ln x
dy y dx x y dy dx x2y ln x
49. Let u 3x2, du 6x dx.
xe3x dx 2
51.
e4x e2x 1 dx ex
1 3x2 1 2 e 6x dx e3x C 6 6
xe1x dx 2
1 1x2 e 2x dx 2
1 2 e1x C 2 57.
e3x ex ex dx
1 e3x ex ex C 3
53.
e4x 3e2x 3 C 3ex
55. Let u ex 1, du ex dx.
ex dx ln ex 1 C ex 1
y exa cos 3x b sin 3x y ex3a sin 3x 3b cos 3x exa cos 3x b sin 3x ex3a b sin 3x a 3b cos 3x y ex33a b cos 3x 3a 3b sin 3x ex3a b sin 3x a 3b cos 3x ex6a 8b sin 3x 8a 6b cos 3x y 2y 10y ex6a 8b 23a b 10b sin 3x 8a 6b 2a 3b 10a cos 3x 0
Review Exercises for Chapter 5
4
59. Area
0
1 2 2 xex dx ex 2
4 0
1 e16 1 0.500 2 63. y log2x 1
61. y 33 2
y
y
6 5
4 3
4 3
2 1
2
x
−1 1
2
2
3 4
5
6 7
−2 −3
x
−4 −3 −2 −1
1
3 4
−4
−2
65. f x 3x1
y x2x1
67.
fx 3x1 ln 3
ln y 2x 1 ln x y 2x 1 2 ln x y x y y
69. gx log31 x gx
1 log31 x 2
71.
2x x 1 2 ln x x 2x x 1 2 ln x 2x1
x 15x1 dx 2
1 1 x12 5 C 2 ln 5
1 1 1 2 1 xln 3 2x 1ln 3
73. (a) y x a y ax a1
(b) y ax
(c)
y xx
(d) y aa
ln y x ln x
y ln aax
y 0
1 1 y x 1 ln x y x y y1 ln x y xx1 ln x 75. 10,000 Pe0.0715 P
77.
Ph 30ekh P18,000 30e18,000k 15
10,000 $3499.38 e1.05
k
ln 2 ln1 2 18,000 18,000
Ph 30eh ln 2 18,000 P35,000 30e35,000 ln 2 18,000 7.79 inches
79.
P Ce0.015t 2C Ce0.015t 2 e0.015t ln 2 0.015t t
ln 2 46.21 years 0.015
81.
dy x2 3 dx x
dy y
x
3 dx x
x2 3 ln x C 2
275
276
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
83. y 2xy 0
dy 2xy dx 1 dy y
2x dx
ln y x2 C1 2 C 1
ex
y
y Cex
2
85.
dy x2 y2 (homogeneous differential equation) dx 2xy
x2 y2 dx 2xy dy 0 Let y vx, dy x dv v dx.
x2 v2x2 dx 2xvxx dv v dx 0 x2 v2x2 2x2v2 dx 2x3v dv 0 x2 x2v2 dx 2x3v dv 1 v2 dx 2x dv
dx x
2v dv 1 v2
ln x ln 1 v 2 C1 ln 1 v 2 ln C Cx2
x
C C 1 v 2 1 y x2 x2 y2
1
Cx x2 y2
or C1
x x2 y2
87. y C1x C2 x3 y C1 3C2 x2 y 6C2 x x2 y 3xy 3y x26C2 x 3xC1 3C2 x2 C1x C2 x3 6C2 x3 3C1x 9C2 x3 3C1x 3C2 x3 0 x 2, y 0: 0 2C1 8C2 ⇒ C1 4C2 x 2, y 4: 4 C1 12C2 1 4 4C2 12C2 8C2 ⇒ C2 , C1 2 2 1 y 2x x3 2 89. f x 2 arctanx 3
y
4
−6
−4
x
−2
2 −2 −4
Review Exercises for Chapter 5
Let arcsin
91. (a)
277
1 2 2
1 2
sin
1
θ
1 1 sin arcsin sin . 2 2 Let arcsin
(b)
3 1 cos . 2 2
cos arcsin
x
93. y tanarcsin x y
1 2
1 2
sin
3
95. y x arcsec x
1 x2
y
1 x21 2 x21 x21 2 1 x23 2 1 x2
x arcsec x x x2 1
97. y xarcsin x2 2x 21 x2 arcsin x y
1 x2 2x arcsin x 2x 2 2 2 arcsin x arcsin x arcsin x2 1 x2 1 x2 1 x2
99. Let u e2x, du 2e2x dx.
1 dx e2x e2x
1 1 e2x 1 dx 2e2x dx arctane2x C 1 e4x 2 1 e2x2 2
101. Let u x2, du 2x dx.
x 1 x4
2x , du 4 2 x
2
arctanx 2 1 dx 4 x2 2
107.
dy A2 y2
1 1 1 2x dx arcsin x2 C 2 1 x22 2
dx
105. Let u arctan
103. Let u 16 x2, du 2x dx.
y arcsin A
arctan
x 2
dx.
4 2 x dx 41arctan 2x 2
k dt m
k tC m
2
C
109. y 2x coshx y 2
Since y 0 when t 0, you have C 0. Thus,
mk t Ay
sin
mk t
y A sin 111. Let u x2, du 2x dx.
x x4 1
dx
1 1 x 1 dx 2x dx ln16 x2 C 16 x2 2 16 x2 2
1 1 1 2x dx ln x2 x4 1 C 2 x22 1 2
1 sinh x 2 sinh x 2x 2x
278
Chapter 5
Logarithmic, Exponential, and Other Transcendental Functions
Problem Solving for Chapter 5 1. tan 1 3 x tan 2
6 10 x
6
Minimize 1 2:
3
f x 1 2 arctan fx
1 9 1 2 x
3 x 2
3 6 arctan x 10 x
1
θ1 0
θ
x
θ2 a
10
6 0 36 10 x2 1 10 x2 3 6 x2 9 10 x2 36
10 x2 36 2x2 9 100 20x x2 36 2x2 18 x2 20x 118 0 x a 10 218 4.7648
20 ± 202 4118 10 ± 218 2
f a 1.4153
1 2 1.7263 or 98.9 Endpoints: a 0: 1.0304 a 10: 1.2793 Maximum is 1.7263 at a 10 218 4.7648. 3. f x sinln x or 0,
(a) Domain: x > 0
(f)
2k. 2 2 2 2 . Two values are x e , e
(b) f x 1 sinln x ⇒ ln x
(c) f x 1 sinln x ⇒ ln x
3 2k. 2
1 (e) fx cosln x x
k ⇒ 2
x e2 on 1, 10
f 10 0.7440
1 lim f x seems to be . (This in incorrect.) 2
(g) For the points x e2, e32, e72, . . . we have f x 1. For the points x e2, e52, e92, . . .
fx 0 ⇒ cosln x 0 ⇒ ln x
−2
x→0
(d) Since the range of the sine function is 1, 1 , parts (b) and (c) show that the range of f is 1, 1 .
f 1 0
5
0
Two values are x e2, e32.
f e2 1
2
we have f x 1. That is, as x → 0, there is an infinite number of points where f x 1, and an infinite number where f x 1. Thus lim sinln x does not exist. x→0
Maximum is 1 at x
e2
4.8105
You can verifiy this by graphing f x on small intervals close to the origin.
Problem Solving for Chapter 5 5. (a)
279
Area sector t t t ⇒ Area sector Area circle 2 2 2
1 (b) Area AOP baseheight 2 At
1 cosh t sinh t 2
cosh t
x2 1 dx
1 cosh t
x2 1 dx
1
1 At cosh2 t sinh2 t cosh2 t 1 sinh t 2 1 cosh2 t sinh2 t sinh2 t 2 1 1 cosh2 t sinh2 t 2 2 1 At t C. But, A0 C 0 ⇒ C 0 2 1 Thus, At t or t 2 At. 2 9. Let u 1 x, x u 1, x u2 2u 1,
y ln x
7.
y
dx 2u 2du.
1 x
Area
1 y b x a a
4
1
1 dx x x
1 y x b 1 Tangent line a
3
2
If x 0, c b 1. Thus, b c b b 1 1.
u 1 du u2 u
32
2
2u 2 du u 1 u2 2u 1
32
2
u
du
2 ln u
3 2
2 ln 3 2 ln 2 2 ln
32
0.8109
11. (a)
dy y1.01 dt
y1.01 dy
(b)
y1 dy
y0.01 y0.01
y kt C y
0.01t C 1 C 0.01t
1 y C 0.01t100 y0 1: 1 Hence, y
1 ⇒ C1 C100
1 . 1 0.01t100
For T 100, lim y . t→T
k dt
y kt C1
dt
y0.01 t C1 0.01 1
1 C kt1
y0 y0
1 1 1 ⇒ C1 ⇒ C C1 y0 y0
Hence, y
1 1 kt y0
For t →
1 , y → . y0 k
1
.
280
Chapter 5
13. Since
Logarithmic, Exponential, and Other Transcendental Functions
dy k y 20, dt 1 dy y 20
k dt
ln y 20 k t C y Cekt 20. When t 0, y 72. Therefore, C 52. When t 1, y 48. Therefore, 48 52ek 20, ek 2852 713, and k ln713. Thus, y 52e ln713 t 20. When t 5, y 52e5 ln713 20 22.35.
15. (a)
dS k1SL S dt S
(b)
dS 4 ln S100 S dt 9
S dSdt 100 S dSdt
4 d 2S ln dt2 9
L is a solution because 1 Cekt
dS L1 Cekt 2C kekt dt
LC kekt 1 Cekt 2
Lk 1 LCe
1 Cekt
kt
k L L 1 Cekt
ln
49100 2S dSdt
0 when S 50 or
Choosing S 50, we have:
C Lekt
L L 1 Cekt
50
L 100. Also, S 10 when t 0 ⇒ C 9. And, S 20 when t 1 ⇒ k ln49.
100 1 9eln49t
2 1 9eln49t ln19 t ln49
k k1SL S, where k1 . L
100 Particular Solution. S 1 9eln49t
dS 0. dt
t 2.7 months (d)
S 140 120 100
100 1 9e0.8109t (c)
80 60 40 20
125
t 1
2
3
4
(e) Sales will decrease toward the line S L. 0
10 0
PA R T
I I C H A P T E R 6 Applications of Integration Section 6.1
Area of a Region Between Two Curves . . . . . . . . . . 264
Section 6.2
Volume: The Disk Method . . . . . . . . . . . . . . . . . 271
Section 6.3
Volume: The Shell Method
Section 6.4
Arc Length and Surfaces of Revolution . . . . . . . . . . 282
Section 6.5
Work
Section 6.6
Moments, Centers of Mass, and Centroids . . . . . . . . . 290
Section 6.7
Fluid Pressure and Fluid Force . . . . . . . . . . . . . . . 297
Review Exercises
. . . . . . . . . . . . . . . . 278
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 287
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305
C H A P T E R 6 Applications of Integration Section 6.1
Area of a Region Between Two Curves
Solutions to Even-Numbered Exercises
2. A
2x 5 x 2 2x 1 dx
4. A
x 2 x 3 dx
2
2
2
2
x 2 4 dx
1
1
0
0
1
3
1
8.
x 1 3 x 1 dx
6. A 2
1 x 2 x 2 1 dx
10.
2
4
x3 x x dx 3 3
12.
sec2 x cos x dx
4
y
y
y
7 2
3
(3, 6)
6
−π,2 4 2
(
5 4
)
( 4π , 2)
3 x
−2
2
2 1
(2, 23 )
(− 4π , 22 ) (π4 , 22 )
(3, 1) x
−1
−2
2
4
5
6
−π 4
7
8
1 14. f x 2 2 x
16. A
2
gx 2 x
8
A 1
2
Matches (a)
y
1 3 10 x xx 8 2 8
dx
3 2 7 x x 10 dx 8 2
x8 7x4 3
2
8
10x
2
64 112 80 1 7 20 18
4
y 3
(2, 9) (0, 2)
8
(8, 6)
6 1
(4, 0)
4 x
1
2
3
(2, 92 )
2
(8, 0) x 2
264
π 4
4
6
10
x
Section 6.1 18. The points of intersection are given by x2
Area of a Region Between Two Curves
20. The points of intersection are given by:
4x 1 x 1
x 2 4x 2 x 2
x2 3x 0
x3 x 0 when x 0, 3
x2 3x when x 0, 3
3
A
x2 4x 1 x 1 dx
A
3
3
x2 3x dx
3
2 3
x 3x 3 2
x 2 4x 2 x 2 dx
0
0 3
f x gx dx
0
0
3
x 2 3x dx
0
0
y
27 9 9 2 2
6
(3, 5)
5
y
4
6
3
5
(0, 2)
(3, 4)
4 3 2
x
−1
(0, 1)
1
2
3
5
4
x 1
5
22. A
3
2
4
5
6
1 1 0 dx x2 x
1
5 1
4 5
y
(1, 1)
1.0 0.8 0.6 0.4 0.2
(5, 0.04) x 1
2
3
4
5
24. The points of intersection are given by 3 x 1 x 1
y
x 1 x 13 x3 3x2 3x 1 3x 2x 0
x3
1
(2, 1)
2
(1, 0)
xx2 3x 2 0 xx 2x 1 0 ⇒ x 0, 1, 2
1
A2
0
3 x 1 dx x 1
2 x 4x 1
2
x2
3
1
4 3
0
12 1 0 43 21
2
x 2
(0, −1)
x3
3
3 2 x 2
3 0
9 2
265
266
Chapter 6
Applications of Integration
26. The points of intersection are given by: 2y y 2 y
y
(−3, 3)
y y 3 0 when y 0, 3
3
3
A
f y g y dy
1
0
(0, 0)
3
−3
2y y 2 y dy
x
−2
1 −1
0 3
3y y 2 dy
0
32 y
2
1 3 y 3
3 0
9 2
3
28. A
y
f y g y dy 4
0
3
0
y
16 y 2
(
3
3 ,3 7
)
2
3
1 2
16 y 21 22y dy
1
0
16 y 2
1
30. A
0 dy
4
0
3 0
4 7 1.354
x
−1
1
4 dx 2x
y
(1, 4)
0
1
4x 4 ln 2 x
3
2
3
(0, 2)
4 4 ln 2
1
1.227
x
−1
34. The points of intersection are given by:
32. The point of intersection is given by:
x 4 2x 2 2x 2
x 3 2x 1 2x
1
x 2x 2 4 0 when x 0, ± 2
x 3 1 0 when x 1
A2
f x gx dx
1
2x 2 x 4 2x 2 dx
0 2
1
2
1
A
3
1
2
x 3 2x 1 2x dx
1
x4 x x 1 dx 4 1 3
Numerical Approximation: 2.0
1 1
4x 2 x 4 dx
0
2
2
4x3
3
x5 5
2 0
128 15
Numerical Approximation: 8.533 10
y
(−2, 8)
(2, 8)
(−1, 2) −4
(1, 0) −2
x
−1
2 −1 −2
(1, −2)
4 −2
(0, 0)
Section 6.1
Area of a Region Between Two Curves
36. f x x 4 4x 2, gx x 3 4x
y
g 4
The points of intersection are given by:
3
x 4 4x 2 x 3 4x −4 −3
x 4 x 3 4x 2 4x 0 xx 1x 2x 2 0 when x 2, 0, 1, 2
0
A
2
1
3
4
−3 −4
1
x 3 4x x 4 4x 2 dx
x
−1
f
2
x 4 4x 2 x 3 4x dx
0
x3 4x x4 4x2 dx
1
248 37 53 293 30 60 60 30
Numerical Approximation: 8.267 0.617 0.883 9.767
3
38. A
0
x2
3 ln x 2 1
40. A
3
( 3, )
3
2
0
−1
5
(0, 0) −1
−1
6.908
5
−1
Numerical Approximation: 6.908
6
44 xx dx 3.434
x
0
9 5
3 ln 10
42. A
4
6x 0 dx 1
1
44. A
cos 2x sin x dx
2
0
6
43 23 0 3 4 3 1.299
1 sin 2x cos x 2
2
y
(
π, 1 6 2
2 4 x 4 sec
)
2 4
2
x2 4x
2 4
2
2
2
x x tan dx 4 4
4 x sec 4
1
0
4 4 4 2
2
4 1 2 2.1797
y
−π 2
π 6
(− π2 , −1)
4
x
3
−1
2
(1,
2)
1 x 1
46. From the graph we see that f and g intersect twice at x 0 and x 1.
1
A
48. A
2
(0, 1)
2x 1 3 x dx
2
0
1 x 3 x2 x ln 3
1 sin 2x 2
(1, 3)
1
2 cos x
3
0
2 sin x cos 2x 0 dx
0
y
gx f x dx
0
1 2 1 0.180 ln 3
−1
5 4
− 4
(0, 1)
1
x 1
2
(π, 1)
−2
0
4
267
268
Chapter 6
5
50. A
Applications of Integration
4 ln x 0 dx x
1
2ln x 2
52. (a) y x e x, y 0, x 0, x 1
1
5
(b) A
2ln 5 2 5.181
1
x e x dx.
0
No, it cannot be evaluated by hand.
2
(5, 1.29) 0
(c) 1.2556
y = xe x
4
6
(1, 0)
−2
−1
2 −1
x
54. Fx
0
12 1 t 2 dt t3 2t 2 6
x
x3 2x 6
0
(b) F4
(a) F0 0
43 56 24 6 3
(c) F6 36 12 48 y
y y 20
20 20
16
16 16
12
12 12
8
8 8
4
4 4
x
x 1
2
3
4
5
x
6 1
y
56. F y
1
4ex 2 dx 8ex 2
1
(b) F0 8 8e1 2 3.1478
30
25
25
20
20
20
15
15
15
10
10
10
5
5
2
4
2
3
2
74 x
2
6
4
7
2
y = 29 x − 12 (4, 6) y = − 25 x + 16
2
(6, 1) x 6
2
8
−2
(2, −3)
y=x−5
10
6
4
2
21x
6 4
1
2
3
4
5 x 16 x 5 dx 2
7 x 21 dx 2
4
4 x
7x
4
−4
9 x 12 x 5 dx 2
7 x 7 dx 2
4
5
6
5 x
−1
4
3
y
30
1
2
(c) F4 8e2 8e1 2 54.2602
25
x
6
1
6
30
4
y
5
y
−1
4
8e y 2 8e1 2
y
3
y
(a) F1 0
58. A
2
7 7 14
−1
x 1
2
3
4
Section 6.1
Area of a Region Between Two Curves
1 x2 1
60. f x
f x
y
(0, 1)
2x x 2 1 2
(1, 21 )
1 2
1 1 At 1, , f 1 . 2 2 Tangent line:
1 4
y=− 1x+1 2
x 1 2
1 1 1 x 1 or y x 1 2 2 2
y
1
0
1 1 x1 1 2
x2
3 2
1
1 at x 0. x2 1
The tangent line intersects f x A
1 f ( x) = 2 x +1
3 4
dx arctan x x4 x
2
1 0
3 0.0354 4
62. Answers will vary. See page 417. 64. x 3 ≥ x on 1, 0
y
x 3 ≤ x on 0, 1
(1, 1)
1
Both functions symmetric to origin
0
(0, 0)
1
1
x 3 x dx
x 3 x dx.
1
0
1
Thus,
x
−1
1
−1
x 3 x dx 0.
1
A2
x x 3 dx 2
0
(− 1, − 1)
x2 x4
4 1
2
0
1 2
66. Proposal 2 is better, since the cummulative deficit (the area under the curve) is less.
9
68. A 2
9 x dx 2 9x
0
9b
2
0
x2 2
9 0
9 b x dx
0
12 9
81 9 x b dx 2
9b
2
y
81
x2 2
9b
81 2
81 2
9 b9 b
81 2
2 9 bx
0
9b b9
9
2
9
2
2.636
6
(− (9 −b), b)
−6
(9 −b, b) x
−3
3 −3 −6
6
2
269
270
Chapter 6 n
70. lim
→0 i1
5
4i 4 and x is the same as n n
2
2
y
4 x i2 x
where xi 2
Applications of Integration
4 x 2 dx 4x
x3 3
2 2
f ( x) = 4 − x 2
3 2
32 . 3
1
(− 2, 0) −3
(2, 0) x
−1
1
3
−1
5
72.
5
7.21 0.26t 0.02t 2 7.21 0.1t 0.01t 2 dt
0
0.01t 2 0.16t dt
0
0.01t 3
29 billion $ 2.417 billion 12
3
0.16t 2 2
5 0
74. 5% : P1 893,000 e 0.05t 312 %: P2 893,000 e 0.035t Difference in profits over 5 years:
5
893,000e 0.05t 893,000e 0.035t dt 893,000
0
e0.05 e0.035
0.05t
0.035t 5 0
893,00025.6805 34.0356 20 28.5714 893,0000.2163 $ 193,156 Note: Using a graphing utility you obtain $193,183. 76. The curves intersect at the point where the slope of y2 equals that of y1, 1. y2 0.08x2 k ⇒ y2 0.16x 1 ⇒ x
1 6.25 .16
6.25
(b) Area 2
(a) The value of k is given by
y2 y1 dx
0
y1 y 2
6.25
2
6.25 0.086.25 2 k
0.08x 2 3.125 x dx
0
k 3.125.
2
0.08x 3
3
3.125x
x2 2
6.25 0
26.510417 13.02083
16 2 8 5.908
78. (a) A 6.031 2
1
2
(b) V 2A 25.908 11.816 m 3 80. True
1
2
(c) 5000V 500011.816 59,082 pounds
Section 6.2
Section 6.2
3
2
4 x 2 2 dx
0
4. V
x 4 8x 2 16 dx
0
9 x2
0
3
2 dx
9 x2 dx
5 x5
8x 3 16x 3
2
x3 3
3 0
256 15
0
V
x2
8
4
16 y 2
0
y3 16y 3
4 0
2 dy
16 y 2 dy
128 3
x4
80
2
x5
8
V
4
0
8 0
y 4 8y 3 16y 2 dy
1
y5 2y 5
4
16y 3 3
2
4x 4 dx
0
y
4 1
−4
8
8
6
6
4
4
2
−4
4
(c) Rx 8, rx 8 2x 2
2
32x 2
dx 4
4x 4
8
8x 2
x4
dx
0
0
8 3 1 5 x x 3 5
2 0
896 15
2y dy y y 4 4 dy 2 2 2
2
0
2
0
4y
y
42 3 2 y 2 y 3 4
y 8
−4
−2
128 5
4
8
2
x
−2
V
64 64 32x 2 4 x 4 dx
0
4
0
(d) R y 2 y 2, r y 0
2
5 2
2 x
−2
V
4x5
y
2
6
6
4
4
2
2 x 2
4
−4
−2
0
4
y 2 4y 2 dy
22
4482 132.69 15
459 153 15 5
V
16
12 dx
(b) Rx 2x 2, rx 0
y y2 dy 4y 2 4
2
2x3 12x 3
12. y 2x 2, y 0, x 2 (a) R y 2, r y y 2
22 dx
1282 322 242 80 3
1
0
16 2x
2
4
4
2
22
0
10. V
x2
2
x ± 22
8. y 16 x 2 ⇒ x 16 y 2
4 4
22
8 16 x2
18
22
x2 4
6. 2 4
0
9x
V
271
Volume: The Disk Method
2
2. V
Volume: The Disk Method
x 4
8
0
16 3
272
Chapter 6
Applications of Integration
14. y 6 2x x 2, y x 6 intersect at 3, 3 and 0, 6. (a) Rx 6 2x x 2, rx x 6
(b) Rx 6 2x x2 3, rx x 6 3
0
V
3
V
0
0
6 2x x 2 2 x 6 2 dx
3 2x x 2 2 x 3 2 dx
3 0
x 4 4x 3 9x 2 36x dx
3
5 x 1
5
x 4 3x 3 18x 2
0 3
243 5
x 4 4x 3 3x 2 18x dx
3
5x 1
x 4 x 3 9x 2
5
y
−6
−4
2
V
0
8
8
4
4
2
2 x
x3 , rx 0 2
dx
x6 4
x7 28
32 16
x
−2
2
3
4 2 4 sec x 2 dx
dx
3
8 sec x sec 2 x dx
0
128 144 28 7
5
3 3 2
1 x 1
2
3
1
4
−1
π 9
20. R y 6, r y 6 6 y y
4
V
x
π 3
2π 9
4π 9
5π 9
y 5
6 2 y 2 dy
4
0
36y
y3 3
4 0
368 3
0
8 ln 2 3 3 27.66 y
2
3
8 ln 2 3 3 8 ln 1 0 0
0
(2, 4)
8 ln sec x tan x tan x
2
y
−1
108 5
0
0
16x
−4
V
16 4x3 x4
18. Rx 4, rx 4 sec x
2
2
−6
2
x3 4 2
3
y
−2
16. Rx 4
0
3 2 1 x 1 −1
2
3
4
5
Section 6.2 6 22. R y 6 , r y 0 y
6
V
6
2
6
V 2
2
dy
2
2
1 y
35
rx 0
x4 x 2 dx 2
2
36 y 2 ln y
6
0
2 1 1 2 dy y y
36
36
24. Rx x4 x 2, 2
6 y
Volume: The Disk Method
2
6 2
32 2 ln 2
2 ln 6
4x 2 x 4 dx
0
4x3
3
x5 5
2 0
128 15 y
13 1 36 2 ln 1213 6 ln 3 241.59 3 3
3 2
y
1 6 −3
5
x
−1
1
2
3
4 −2 3 −3 2 1 x 1
2
3
5
3 , rx 0 x1
26. Rx
8
V
4
0
3 x1
28. Rx e x 2, rx 0
dx
4
x 1
2
dx
8 0
e x dx
0
0
1 9 x1
e x 2 2 dx
0
8
9
4
V
2
e x
8
4 0
e 4 1 168.38
y
y
4 8 3 6 2 4 1 2 x 2
4
6
8 −2
4
30. V
0
4
0
1 4 x 2
2
x
56 88 48 3 3
4 0
8
dx
x2 5x 16 dx 4
x3 5x2 16x 12 2
2
4
8
4
x
x 2
4
2 4 21 x dx
6
2
y 4
x2 5x 16 dx 4
x3 5x2 16x 12 2
2
8 4
(4, 2)
3
1 x −2
2 −1
4
6
8
10
273
274
Chapter 6
Applications of Integration
32. y 9 x 2, y 0, x 2, x 3
y 9 8 7 6 5 4 3 2 1
x 9 y
5
V
9 y
0
2 22 dy
5
5 y dy
(2, 5)
x
0
1 2 3 4 5 6 7 8 9
y2 2
25 25 2 2
5y 25
34. V
2
5 0
3
cos x 2 dx 2.4674
36. V
0
ln x 2 dx 3.2332
1
y
2
1
x 2
1
5
38. V
2 arctan 0.2x 2 dx 15.4115
0
b
3 40. A 4
42. V
y
d
or
V
a
1
Matches (b)
Ax dx
Ay dy
c
3 4 1 2 1 4
x 1 4
44. (a)
1 2
3 4
z
1
(b) 2
4
2 4 x
8
y
x
8
8
−4
h
0
r2 2 x dx h2
r3h x 2
y
r
y= r x h
(h, r)
h
3
2
2
y
8 x
a < c < b.
r 46. Rx x, rx 0 h
z 4
4
2
V
(c)
z
4
0
1 r 2 h 3 r 2h 3h 3
x h
16
y
Section 6.2
Volume: The Disk Method
4
50. (a) V
48. x r 2 y 2 , R y r 2 y 2 , r y 0
0
r
V
r 2 y 2
h
2 dy
c
r y dy 2
2
r3
r3 h3 r 2h 3 3
2r3
h3 3
3
c2
y3 3
r2y
x dx
0
h
2 d x
4
x dx
0
2x
2 4 0
8
Let 0 < c < 4 and set
r
x
275
r
x2
2
c 0
c 2 4. 2
8
c 8 22
h
r 2h
Thus, when x 22, the solid is divided into two parts of equal volume.
c
(b) Set
x dx
0
2r 3 3r 2h h 3 3
8 (one third of the volume). Then 3
43 c 2 8 2 16 4 , c , c . 2 3 3 3 3
d
y
To find the other value, set of the volume). Then
r
h
x dx
0
16 (two thirds 3
32 46 d 2 16 2 32 , d , d . 2 3 3 3 3
x
The x-values that divide the solid into three parts of equal volume are x 43 3 and x 46 3.
52. y
2.95,0.1x
11.5
V
0
3
2.2x 2 10.9x 22.2, 0 ≤ x ≤ 11.5 11.5 < x ≤ 15
0.1x 3 2.2x 2 10.9x 22.2 2 dx
0.1x 4 2.2x 3 10.9x 2 22.2x 4 3 2
11.5 0
y 8
15
6
2.95 2 dx
11.5
2.95 2x
4
15 2
11.5 x
1031.9016 cubic centimeters
4
8
12
54. (a) First find where y b intersects the parabola: b4
x2 4 z
x 2 16 4b 44 b 5
x 24 b
24b
V
0
4
4
4
0
4
0
x2 b 4
x2 b 4
2
4
dx
24b
b4
x2 4
2
dx
2
2
dx
x
x4 bx 2 2x 2 b 2 8b 16 dx 16 2
80
2x 3 bx 3 b 2x 8bx 16x 3 6
5
128 32 64 512 b 4b 2 32b 64 4b 2 b 3 3 3 15
x5
64
—CONTINUED—
4 0
2
4
y
16
276
Chapter 6
Applications of Integration
54. —CONTINUED—
(b) graph of Vb 4b 2
64 512 b 3 15
(c) Vb 8b
120
64 64 3 8 223 0 ⇒ b 3 8 3
V b 8 > 0 ⇒ b
0
8 is a relative minimum. 3
4 0
Minimum Volume is 17.87 for b 2.67
10
56. (a) V
f x 2 dx
0
Simpson’s Rule: b a 10 0 10, V
n 10
2.1 2 41.9 2 22.1 2 42.35 2 22.6 2 42.85 2 22.9 2 42.7 2 22.45 2 4 2.2 2 2.3 2 3 178.405 186.83 cm 3 3
(b) f x 0.00249x 4 0.0529x 3 0.3314x 2 0.4999x 2.112 6
0
10 0
10
(c) V
f x 2 dx 186.35 cm 3
0
1 58. V 1023 30 m3 2
60.
1 1 (b) Ax bh 24 x 2 34 x 2 2 2
y 3
3 4 x 2
1
2
V 3
x
−3
3
1 −1
2
4 x 2 dx
3 4x
−3
x3 3
2 2
Base of Cross Section 24 x 2 (a) Ax b 2 24 x 2
2 4 − x2
2
2 4 − x2
2
V
2
44 x 2 dx
4 4x
3 2
x 3
2
2 4 − x2
128 3 2 4 − x2
—CONTINUED—
2 4 − x2
323 3
Section 6.2
Volume: The Disk Method
60. —CONTINUED— 1 (d) Ax bh 2
1 (c) Ax r 2 2
2
V
4 x 2 2 2 4 x 2 2
2
2
4 x 2 dx
x3 4x 2 3
2 2
1 24 x 2 4 x 2 4 x 2 2
2
16 3
V
2
4 x 2 dx 4x
x3 3
2 2
32 3
4 − x2 2 4 − x2 2 4 − x2
62. The cross sections are squares. By symmetry, we can set up an integral for an eighth of the volume and multiply by 8. A y b 2 r 2 y 2
2 y
r
V8
x
r 2 y 2 dy
0
1 3 y 3
8 r2y
r 0
16 3 r 3
64. V
R 2 r 2
R 2 r 2 R 2 r 2
2
R 2 x 2 2 r 2 dx R
R 2 r 2 x 2 dx
0
2 R 2 r 2 x
2
R2
r 2 3 2
x3 3
r
R 2 r 2
0
R 2 r 23 2 3
R2 − r 2
R
4 R 2 r 2 3 2 3
When a 2: x2 y2 1 represents a circle. (b) y 1 xa1 a
66. (a) When a 1: x y 1 represents a square.
1
A2
1
1 xa1 a dx 4
1
1 x a1 a dx
y
1
a=1
−1
1
0
To approximate the volume of the solid, form n slices, each of whose area is approximated by the integral above. Then sum the volumes of these n slices.
a=2
−1
x
277
278
Chapter 6
Applications of Integration
Section 6.3
Volume: The Shell Method
2. px x
4. px x
hx 1 x
hx 8 x 2 4 4 x 2
1
V 2
2
x1 x dx
V 2
0
x4 x 2 dx
0
1
x x2 dx 2
2
0
x2 3 2
x3
1 0
3
2
2
4x x 3 dx
0
2 2x 2
6. px x
8. px x
6
0
0
8
hx 4 2x
2
2
V 2
1 3 x dx 2
4 6
x 4
2
10. px x
hx 4 x
1 hx x2 2 V 2
x4 4
0
V 2
0
2 0
2
8
2
4x 2x 2 dx
0
y
2 2 2x 2 x 3 3
y 18
x4 2x dx
0
1 2 2x 2 x 4 4
324
2
4x x 3 dx
2 0
16 3
y
3
15 4
2
12 9
3
1
6 3 −1 −3
−2
x 1
2
3
4
5
6
2
x
−1
1
2 1 x
−1
12. px x hx
0
2
x
2
sinx x dx
1
sin x dx 2 cos x
0
14. p y y p y ≥ 0 on 2, 0 h y 4 2 y 2 y
2
y2 y dy
2
π 2
x
π
3π 4
−1
16. p y y h y 16 y2
y 4
V 2
y16 y2 dy
0
3 2 1
4
0
2
4
π 4
4
0
2
3
3
0
V 2
2
y
sin x x
V 2
1
2y y 2 dy
y 2
y3 3
0 2
8 3
2
16y y3 dy
0
y4 2 8y2 4
4 0
2 128 64 128
x −1 −2 −3 −4
4
8
12
Section 6.3 20. px 6 x
18. px 2 x hx 4x x x 4x 2x 2
2
hx x
2
2 x4x 2x 2 dx
V 2
6 x x dx
0
0
4
2
2
4
2
V 2
Volume: The Shell Method
8x 8x 2 2x 3 dx
2
6x 1 2 x 3 2 dx
0
0
8 1 2 4x 2 x 3 x 4 3 2
2 0
16 3
2 2 4x3 2 x5 2 5
y
4 0
192 5
y 4
4
3 3 2 2
1 x
1
1
2
3
4
5
−1 x
−1
1
−2
3
22. (a) Disk
(b) Shell
5
V
y
10 , rx 0 x2
Rx
10 x2
1
8
2
dx
5
2 −1
1
x3 100 3
20
x 1
−2
2
3
4
5
1
x
1
4
x4 dx
5
V 2
6
5
100
Rx x, rx 0
10
10 dx x2
1 dx x
5
20 ln x
1
5 1
20 ln 5
100 1 496 1 3 125 15
(c) Disk 10 x2
Rx 10, rx 10
5
V
102 10
1
200 100 3x x
5
3
1
10 x2
dx 2
1904 15 y
24. (a) Disk
(b) Same as part a by symmetry
Rx a 2 3 x 2 3 3 2
y
(0, a)
rx 0
(0, a)
a
V
a
(a, 0)
a 2 3 x 2 3 3 dx
(−a, 0)
(a, 0)
x x
a
2
a 2 3a 4 3x 2 3 3a 2 3x 4 3 x 2 dx
0
9 4 3 5 3 9 2 3 7 3 1 3 a x a x x 5 7 3
9 3 9 3 1 3 32 a 3 a a a 5 7 3 105
2 a 2x 2 a 3
a 0
(0, −a)
279
280
Chapter 6
Applications of Integration
z
26. (a)
(b)
z
(c)
z
2
2
2
5
x
x
x
5
5
a < c < b
−2
4
28. 2
y
5
2
x
0
2x dx
y
30. (a) 1
represents the volume of the solid generated by revolving the region bounded by y x 2, y 0, and x 4 about the y-axis by using the Shell Method.
2
10
y
3 4 1 2 1 4
2
16 2y dy 2
0
4 2y dy 2
2
x 1 4
0
1 2
represents this same volume by using the Disk Method.
3 4
1
1
y
(b) V 2
4
x 1 x 3 dx 2.3222
0
3 2 1 x 1
2
3
4
5
−1
Disk Method
32. (a)
34. y tan x, y 0, x 0, x
y 4
4
Volume 1
3
Matches (e)
2
y
1 x 1
2
3
3
(b) V 2
1
2
4
2x dx 19.0162 1 e 1 x
1
π 4
x
π 2
36. Total volume of the hemisphere is 12 43 r 3 23 3 3 18. By the Shell Method, px x, hx 9 x 2. Find x0 such that
x0
6 2
x 9 x 2 dx
y
0
6
x0
9 x21 2 2x dx
0
2 9 x 2 3 2 3
1 x
x0 0
2 18 9 x 02 3 2 3
9 x 02 3 2 18 x0 9 18 2 3 1.460. Diameter: 2 9 18 2 3 2.920
2
−3 −2 −1 −1 −2 −3
1
2
3
y
Section 6.3
Volume: The Shell Method
r
38. V 4
r
R x r 2 x 2 dx
4 R
4 R
2r 2 23 r
r
r
r2 x2 dx 4
r
2
2
x r 2 x 2 dx
r
x 2 3 2
r r
2 2 r 2R
abn axn dx
xn1 xa n1
b
40. (a) Area region
(c) Disk Method:
0
abn1
a
abn1
V 2
b
R1n
0
b
2a
bn1 n1
xbn xn1 dx
0
1 n 1 abn1 n1 n1
b2 x
b 2
n
2a
2
2a
R2n
n 1 n→ n 1
(b) lim R1n lim n→
xabn axn dx
0
n n1 n abnb n1
abn1
b
abn
n2
abn2
n→
b 0
b2abn n→
n bn2 abn2 n2 n2
n n 2
(d) lim R2n lim
lim abnb
n→
xn2 n2
n n 2
n n 2 1
lim b2abn
n→
(e) As n → , the graph approaches the line x 1.
4
42. (a) V 2
x f x dx
0
2 40 0 41045 22040 43020 0 34
20 5800 121,475 cubic feet 3
(b) Top line: y 50
1 1 40 50 x 0 x ⇒ y x 50 20 0 2 2
Bottom line: y 40
20
V 2
0 20
2
0
2 2
0 40 x 20 2x 20 ⇒ y 2x 80 40 20
1 x x 50 dx 2 2
21 x
2
40
40
50x dx 2
x3 25x 2 6
20 0
x 2x 80 dx
20
2x 2 80x dx
20
2
2x 3 40x 2 3
40 20
32,000 2 26,000 3 3
121,475 cubic feet (Note that Simpson’s Rule is exact for this problem.)
281
282
Chapter 6
Applications of Integration
Section 6.4
Arc Length and Surfaces of Revolution
2. 1, 2 , 7, 10
4. y 2x3 2 3
(a) d 7 1 2 10 2 2 10
y
9
2 4 x 3 3
(b) y
s
4 3
272 1 9x
9
3 2
0
1
1
43
2
dx
53 x
7 1
10
3 6. y x2 3 4 2
2 823 2 1 54.929 27
8.
y x1 3, 1, 27
27
s
1
1
27
1
x1
32 23 x
1 4 1 x 4 2 2x
s
27
2
1
0
14. (a) y
1 x e ex 2
101 x
2 1
779 3.246 240
−5
dx
(b)
y 2x 1 1 y 2 1 4x 2 4x 1
1
2 0
L
1 1 e 2 2 3.627 2 e
(b)
y
1 1 x 2
1 y 2 1
y= 1 x+1
2 4x 4x 2 dx
L
1
0
(c) L 1.132
1 1 x 4
1
2
2
(c) L 5.653
5
−2
1 6x 3
2
0
1 ,0 ≤ x ≤ 1 1x
−1
5
2
e x ex dx
−3
1 x e ex 2
2
5
2
1 2
dx
12. (a) y x 2 x 2, 2 ≤ x ≤ 1
2 1 x e ex , 0, 2 1 y 2
1 4 1 x 4 2 2x
1 4 1 x 4 dx 2 2x
1
1 y e x ex , 0, 2 2
2
1 2 , 1, 2 2x 4
1 y 2 dx
1
1 y e x ex 2
s
2
2
4
a
103 2 23 2 28.794
10.
12 x b
1 3 2
2 3
y 1 y 2
x2 3
x5 1 10 6x 3
dx
1 3
2 1 1 3 dx 3x
1
y
2
x2 3 1 dx x2 3
27
3 2
1 9x dx
0
7
s
y 3x1 2, 0, 9
1 dx 1 x 4
Section 6.4
16. (a) y cos x,
≤ x ≤ 2 2
Arc Length and Surfaces of Revolution
y sin x
(b)
(c) 3.820
1 y 2 1 sin 2 x
2
L
y
1 x
y = cos x −2
2
2
2
1 sin 2 x dx
−2
18. (a) y ln x, 1 ≤ x ≤ 5
(b)
(c) L 4.367
2 −1
1 y 2 1
6
1 x2
5
L
1
1
−6
20. (a) x 36 y 2 , 0 ≤ y ≤ 3
(b)
dx 1 36 y 2 1 2 2y dy 2
y 36 x 2, 33 ≤ x ≤ 6
10
y
1
0
10
3
−5
(c) L 3.142 !
36 y 2 3
L −10
1 dx x2
0
y2 dy 36 y 2
6
36 y 2
dy
Alternatively, you can convert to a function of x. y 36 x 2 y
dy x dx 36 x 2
6
L
33
1 36 x x dx
6
2
2
33
6 36 x 2
dx
Although this integral is undefined at x 0, a graphing utility still gives L 3.142.
4
22.
0
1 dxd tan x
y
2
dx 2
s 1
( π4 , 1(
y = tan x 1
Matches (e)
(0, 0)
π 4
x 3π 8
24. f x x 2 4 2, 0, 4 (a) d 4 0 2 144 16 2 128.062 (b) d 1 0 2 9 16 2 2 1 2 0 9 2 3 2 2 25 0 2 4 3 2 144 25 2 160.151
4
(c) s
1 4x x 2 4 2 dx 159.087
0
(d) 160.287
283
284
Chapter 6
Applications of Integration
26. Let y ln x, 1 ≤ x ≤ e, y
1 and L1 x
Equivalently, x e y, 0 ≤ y ≤ 1,
e
1
1
dx e y, and L 2 dy
1 dx. x2
1
1
1 e 2y dy
0
1 e 2x dx.
0
Numerically, both integrals yield L 2.0035 y 31 10e x 20 ex 20
28.
1 y e x 20 ex 20 2
s
1 2
20
20
1 x 10 1 x 20 e 2 ex 10 e ex 20 4 2
1 y 2 1
12 e
x 20
ex 20
20
20
2
2
dx
e x 20 ex 20 dx 10e x 20 ex 20
20 20
20 e
1 47 ft e
Thus, there are 10047 4700 square feet of roofing on the barn. 30. y 693.8597 68.7672 cosh 0.0100333x y 0.6899619478 sinh 0.0100333x
299.2239
s
1 0.6899619478 sinh 0.0100333x 2 dx 1480
299.2239
(Use Simpson’s Rule with n 100 or a graphing utility.) 32.
34. y 2x
y 25 x 2 y 1 y 2
x
3 4
3
S 2
1 1x dx
2x
9
25 dx 25 x 2
5 arcsin 5 arcsin
, 4, 9
4
4
x 5
8 x 1 3 2 3
4
3
7.8540
4 3 arcsin 5 5
1 2 5 7.8540 s 4
x 1 dx
4
5 dx 25 x 2
1 x
9
25 25 x 2 4
s
y
25 x 2
9 4
8 3 2 10 53 2 171.258 3
Section 6.4
36.
y
x 2
y
1 2
y 2x
3
S 2
x 2
0
40.
1 y 2
x1 4x2 dx
0
54 dx
25 2 x 8
6 0
4
3
1 4x2 1 2 8x dx
0
6 1 4x
3
2 3 2
0
95
y ln x y
S 2
6
3 2 37 1 117.319 6
42. The precalculus formula is the distance formula between two points. The representative element is
1 x
xi 2 yi 2
y 1 x xi. i
2
i
x2 1 , 1, e x2
x x 1 dx
e
S 2
2
x
1
2
e
2
x 2 1 dx 22.943
1
44. The surface of revolution given by f1 will be larger. rx is larger for f1. 46. y r 2 x 2 y 1 y 2
285
38. y 9 x2, 0, 3
5 1 y 2 , 0, 6 4
Arc Length and Surfaces of Revolution
48. From Exercise 47 we have: S 2
r
2
r
a
r
0
r
r
r
rx
r 2 x 2
r2 2
r dx 2 rx
r r
x2
dx
4 r2
dx
r 2 x 2
0
r2 r2 x2
S 2
a
x r 2 x 2
2x dx r 2 x 2
2rr 2 x 2
2r 2
2rr 2
a 0
a2
2r r r 2 a 2 2 rh where h is the height of the zone y
r
h r−h
x
x2 + y2= r2
286
Chapter 6
Applications of Integration
50. (a) We approximate the volume by summing 6 disks of thickness 3 and circumference Ci equal to the average of the given circumferences: 6
r
V
i
2 3 4 C 6
3
2
i1
Ci
2
6
3
i
i1
2
i1
50 2 65.5 65.5 2 70 70 2 66 66 2 58 58 2 51 51 2 48 2
2
2
2
3 4
3 57.75 2 67.75 2 68 2 62 2 54.5 2 49.5 2 4
2
2
3 21813.625 5207.62 cubic inches 4 (b) The lateral surface area of a frustum of a right circular cone is sR r . For the first frustum,
S1 32
65.52 50 250 65.5 2 2 1 2
50 2 65.59 65.52 50
r s
2 1 2
h
.
Adding the six frustums together, S
R
50 2 65.59 15.5 2
2 1 2
65.5 2 709 4.5 2
2 1 2
70 2 669 24
66 2 589 28
58 2 519 27
51 2 489 23
2 1 2
2 1 2
2 1 2
2 1 2
224.30 208.96 208.54 202.06 174.41 150.37 1168.64
18
(c) r 0.00401 y 3 0.1416 y 2 1.232 y 7.943
(d) V
r 2 dy 5275.9 cubic inches
0
20
18
S
2 r y 1 r y 2 dy
0
1179.5 square inches −1
19 −1
52. Individual project, see Exercise 50, 51.
54. (a)
1 x9 , 0 ≤ x ≤ 3
x2 y2 1 9 4
1 x9
2
Ellipse: y1 2 y2 2
2
(b) y 2
1 x9
2
y 2
2x
1 9
x2
9
4
−5
2 1 2
121 x9 2x 9
5
L
3
0
1
2x 39 x2
4x2 dx 81 9x2
2
x2 + y = 1 9 4 −4
(c) You cannot evaluate this definite integral, since the integrand is not defined at x 3. Simpson’s Rule will not work for the same reason. Also, the integrand does not have an elementary antiderivative. 56. Essay
Section 6.5
Section 6.5
Work 4. W Fd 92000 2 5280 47,520,000 ft lb
2. W Fd 28004 11,200 ft lb
Work
1
b
6. W
Fx dx is the work done by a force F moving an object along a straight line from x a to x b.
a
9
8. (a) W
0
7
(b) W
10. W
0
9
20 dx
0
10
6 dx 54 ft lbs
5 5 2 x dx x 4 8
6
lb 3.33 ft lb
40 in
10x 90 dx 140 20
10
7
160 ft lbs
9
(c) W
0
1 2 x3 x dx 27 81
9
(d) W
9
2 32 x 3
x dx
0
9 ft lbs
0
9
0
2 27 18 ft lbs 3
12. Fx kx
14. Fx kx
800 k70 ⇒ k
70
W
80 7
70
Fx dx
0
0
16
kx dx
0
kx 2 2
524
W
540x dx 270x 2
16
4
80 40x 2 x dx 7 7
W2
70
16 0
524 16
15x dx 15x 2
0
0
4 0
240 ft lb
cm 280 Nm
28000 n
16. W 7.5
15 k1 k
k ⇒ k 540 72
h
18. W
80,000,000 80,000,000 dx 2 x x 4000
4.21875 ft lbs
h 4000
80,000,000 20,000 h
lim W 20,000 miton 2.1 1011 ft lb
h→
20. Weight on surface of moon: 16 12 2 tons Weight varies inversely as the square of distance from the center of the moon. Therefore, Fx 2
k x2 k 1100 2
k 2.42 10 6
1150
W
1100
2.42 10 6 2.42 10 6 dx x2 x
1150 1100
2.42 10 6
1 1 1100 1150
95.652 mi ton 1.01 10 9 ft lb
287
288
Chapter 6
Applications of Integration
22. The bottom half had to be pumped a greater distance then the top half. 24. Volume of disk: 4 y Weight of disk: 98004 y Distance the disk of water is moved: y W
12
y98004 dy 39,200
10
2
12
y2
10
39,200 22 862,400 newton–meters
26. Volume of disk:
23 y y 2
Weight of disk: 62.4
y 7
23 y y 2
5 4 3 2
Distance: y 4 (a) W 62.4 9 (b) W
4 62.4 9
x
2
y3
0
28. Volume of each layer:
4 1 dy 62.4 y 4 9 4
6
4
y 3 dy
2 0
49 62.4 14 y
6
4
4
110.9 ft lb
y3 3 y y 3 y 3
Distance: 6 y
−4 −3 −2 −1
2
4
3
y 4
(2, 3) 3
3
1
7210.7 ft lb
Weight of each layer: 55.6y 3 y
W
y
2
3
55.66 yy 3 dy 55.6
0
18 3y y2 dy
1
y = 3x − 3
0
x
55.6 18y
3y2 2
1
2
3
4
y3 3 3
0
3252.6 ft lb 30. Volume of layer: V 122254 y 2 y
y
Weight of layer: W 4224254 y 2 y
Ground
12
level
Distance:
6
19 y 2
19 25 1008 y 2 4 2.5
−y x
2.5
W
19 2
3 −9 −6 −3 −3
25 19 y2 y dy 4 2
4224
2.5
2.5
2.5
2
dy
2.5
3
6
9
−6
254 y y dy 2
The second integral is zero since the integrand is odd and the limits of integration are symmetric to the origin. The first integral represents the area of a semicircle of radius 52 . Thus, the work is W 1008
192 52 12 29,925 ft lb 94,012.16 ft lb. 2
Section 6.5
Work
32. The lower 10 feet of chain are raised 5 feet with a constant force. W1 3105 150 ft lb The top 5 feet will be raised with variable force. Weight of section: 3 y Distance: 5 y
5
W2 3
0
3 5 y dy 5 y 2 2
W W1 W2 150
5 0
75 ft lb 2
75 375 ft lb 2 2
34. The work required to lift the chain is 337.5 ft lb (from Exercise 31). The work required to lift the 500-pound load is W 50015 7500. The work required to lift the chain with a 100-pound load attached is W 337.5 7500 7837.5 ft lbs
6
36. W 3
0
3 12 2y dy 12 2y 2 4
6 0
3 12 2 108 ft lb 4
38. Work to pull up the ball: W1 50040 20,000 ft lb
40.
p
Work to pull up the cable: force is variable 2500
Weight per section: 1 y
40
0
k ⇒ k 2500 1
3
Distance: 40 x W2
k V
W
1 40 x dx 40 x 2 2
1
40
2500 dV 2500 ln V V
3 1
2500 ln 3 2746.53 ft lb
0
800 ft lb W W1 W2 20,000 800 20,800 ft lb 42. (a) W FD 80002 16,000 ft lbs (b) W
20 0 420,000 222,000 415,000 210,000 45000 0 36
24,88.889 ft lb (c) Fx 16,261.36x 4 85,295.45x 3 157,738.64x 2 104,386.36x 32.4675 25,000
0
2 0
(d) Fx 0 when x 0.524 feet. F x is a maximum when x 0.524 feet.
2
(e) W
0
4
44. W
0
Fx dx 25,180.5 ft lbs
ex 1 dx 11,494 ft lb 100 2
2
46. W
0
1000 sinh x dx 2762.2 ft lb
289
290
Chapter 6
Section 6.6 2. x
Applications of Integration
Moments, Centers of Mass, and Centroids
73 42 35 86 17 7438 11
4. x
126 14 62 30 118 0 12 1 6 3 11
8. 200x 5505 x (Person on left)
6. The center of mass is translated k units as well.
200x 2750 550x 750x 2750 x 323 feet
10.
101 25 54 0 x 10 2 5 y
12.
x
101 25 50 0 10 2 5
x, y 0, 0
y
y
15 6 152 2 93 62 15 40.5 27 15 12 6 2
122 61
15 8 152 2 96 64 15 40.5 27 15 12 6 2
123 65
8 6
m2 (5, 5)
4
m3 (− 4, 0)
x, y
2
64 , 62 27 27
y
−4
m3 (6, 8)
8 x
−2
m1 4 (1, −1)
−2
6
6
m2 (− 1, 5)
m1 (2, 3)
2
x −2
2 −2
2
14.
m
0
1 1 2 x 2 2
0
2
x
0
x
x, y
0
4 3
y
2
2
4 5 Mx 3 y m 4 5 3 My
2
12x dx 8 x
2
Mx
1 2 x3 x dx 2 6
1 1 2 x dx 2 2
2 My 3 m 4 2 3
32, 35
4
dx
40 x
2
5
0
0
32 4 40 5
2
1
( 32 , 35 ) x 1
2
0
x3 dx
8x
2
4
0
2
2
m4 (2, − 2)
6
8
Section 6.6
1
m
16.
x x dx
0
1
Mx
23 x
x x dx
2
2
1
0
1 0
6
y 1
x2 x3 dx 2 2 3
1
x
x2
0
12 0 1
3 4 1 2
Mx 1 6 m 12 2
My
x2 2
x x
0
y
3 2
Moments, Centers of Mass, and Centroids
( x, y )
1 4
1
x x x dx
x 3 2 x 2 dx
0
25 x
5 2
x3 3
1 0
15
x 1 4
My 2 6 x m 15 5
25 , 12
x, y
9
18. m
0
23x
3 2
9 x
Mx
0
2
9
0
x2 9 6
13x 1 dx x 31x dx 9
x 1
0
18
0
1 1 x1 1 3 x 1 x 1 dx 2 3 2
9 0
9
0
1 x x 1 x 1 dx 3
9
0
81 81 486 5 5
185, 25
y 5
(9, 4)
4 3
(185, 52 )
2 1 −2
(0, 1) x
−1
2
4
6
8
10
9
1 3
x x 2
0
9
0
x 31x dx
1 1 x x2 2 x dx 3 9
45 27 27 36 2 2 3
81 45 5 4 My 18 Mx 5 x ;y m 9 5 m 9 2 2 2
x, y
1 1 1 2 x x3 2 x3 2 x2 2 x x dx 3 3 9 3 2
x2 x3 4 x3 2 2 6 27 3
My
27 9 2 2
1 2 1 x3 2 x2 dx x5 2 x3 3 5 9
9 0
1 2
3 4
1
291
292
Chapter 6
20.
m 2
Applications of Integration
8
3 4 x 2 3 dx 2 4x x 5 3 5
0
8
0
128 5
y 12
By symmetry, My and x 0.
8
Mx 2
0
8
4 x 2 3 3 4 x 2 3 dx 16x x 7 3 2 7
8 0
512 5 20 y 7 128 7
20 7
x, y 0 ,
−8
m
2y y 2 dy y 2
2
0
y3 3
2
0
4 3
2y y 2 4y 3 y5 2y y 2 dy y4 2 2 3 5
2 0
8 15
2
( x, y ) 1
2
y 2y y 2 dy
0
2y 3 y 4 3 4
0
x
4 3
1
25 , 1
m
1
y 2 y 2 dy
y2 2y y3
3 2
2
1
9 2
2
y 2 y 2 y 2 y 2 dy 2 1 2
1
y 2 2 y 4 dy
2
y 2 3 y 5 2 3 5
2 1
36 5
My 36 2 8 m 5 9 5
x
1
y y 2 y 2 dy
2
85 , 12
4
A
1
Mx
1
4
My
x
1
1 dx ln x x 4
1 2
y3 y4 3 4
Mx 9 2 1 m 4 9 2
y
26.
1
2y y 2 y 3 dy y 2
4 1
ln 4
8 2 8
1 1 1 dx x2 2 x
1 dx x x
4 1
4 1
3
1
1
3
2 1
9 4
( x, y )
1
x 1 −1
2
Mx
y 3
2
My
x, y
2
2
24.
2
Mx 4 3 1 m 3 4
y
x, y
8
y
My 8 3 2 x m 15 4 5 Mx
4
0
My
x
−4 −4
2
22.
( x, y )
512 7
2
3
4
Section 6.6
2
28.
A Mx
2
2
x 2 4 dx 2
0
2
1 2
2
x 2 44 x 2 dx
2
1 x 5 8x 3 16x 2 5 3
4 x 2 dx 8x
2
1 2
2x 3 3
2 0
Moments, Centers of Mass, and Centroids
16
16 32 3 3
2
x 4 8x 2 16 dx
2
5
32
64 256 32 3 15
My 0 by symmetry.
4
30. m
3
xex 2 dx 2.3760
0
4
Mx
0
xex 2 xex 2 dx 2 2
( x, y )
4
dx 0.7619
x 2ex
−1 −1
4
My
5
0
x 2ex 2 dx 5.1732
0
x
My 2.2 m
y
Mx 0.3 m
Therefore, the centroid is 2.2, 0.3.
2
32. m
2
8 dx 6.2832 x2 4
2
Mx
1 8 2 2 2 x 4
x
3
8 dx 32 4
2
2
2
1 dx 5.14149 x 2 4 2
−3
3
Mx y 0.8 m
−1
x 0 by symmetry. Therefore, the centroid is 0, 0.8. 34.
A bh ac 1 1 A ac x
y
x, y
c
1 1 ac 2
1 2ac
y
0 c
0
y= c x b
b ya c
0
y
c 0
dy
( x, y ) y = c (x − a ) b x
(0, 0)
1 1 abc a 2c b a 2ac 2
b b ya y c c
(a + b, c)
(b, c)
2
c
b y c
2ab y a 2 dy c
1 ab 2 y a 2y 2ac c 1 ac
2
dy
1 y2 c 2
b 2 a , 2c
This is the point of intersection of the diagonals.
c 0
c 2
(a, 0)
293
294
Chapter 6
Applications of Integration
36. x 0 by symmetry
y
1 A r2 2
r
2 1 A r2 y
r
2 1 r2 2
r
1 x3 2 2 r x r 3
4r 3
x, y 0,
r r
1 4r 3 4r r2 3 3
1
38.
A
1 3
1 2x x 2 dx
0
1 3 A
1
x3
1 2x x 2 3 1 2x x 2 dx 2 2
0
x, y
3 2
x 2x 2 x 3 dx 3
0
1
1
x 1 2x x 2 dx 3
0
y3
x
r
r 2 x 2 2 dx
1
1 4x 2 4x 3 x 4 dx
0
x2 32 x 2
3
x4 4
1 0
1 4
1
1 2x x 2 2 dx
0
3 4 x5 x x3 x4 2 3 5
1 0
7 10
14, 107
40. (a) M y 0 by symmetry My
b b because there is more area above y 2 2 than below.
(b) y >
2n
b
2n b
x b x 2n dx 0
because bx x 2n1 is an odd function.
b x2nb x2n dx (c) M x 2 2n b
2n
b
1 2 x 4n1 b x 2 4n 1
b 2 b 1 2n
(d)
b
1 2 b x 4n dx 2 2n b
y
b 1 2n
2
3
4
y
5 9b
7 13 b
9 17 b
2n
b
(e) lim y lim
2n b
n→
x 2n1 A b x 2n dx 2 bx 2n 1 2n b 2 b
1 3 5b
b
b 4n1 2n 4n b 4n1 2n 4n 1 4n 1
2n
n
2n
n→
2n 1 1 b b 4n 1 2
(f) As n → , the figure gets narrower. y
2n
y = x 2n
b
0
b 2n1 2n 4n b 2n1 2n 2n 1 2n 1
Mx 4n b 4n1 2n 4n 1 2n 1 b A 4n b 24n1 2n 2n 1 4n 1
y=b x
−
2n
b
2n
b
Section 6.6
Moments, Centers of Mass, and Centroids
295
42. Let f x be the top curve, given by l d. The bottom curve is dx. x
0
0.5
1.0
1.5
2.0
f
2.0
1.93
1.73
1.32
0
d
0.50
0.48
0.43
0.33
0
2
(a) Area 2
(b) f x 0.1061x 4 0.06126x 2 1.9527
f x dx dx
0
2
dx 0.02648x 4 0.01497x 2 .4862
2 1.50 4 1.45 21.30 4 .99 0 34
1 13.86 4.62 3
y
(c)
Mx 4.9133 1.068 A 4.59998
x, y 0, 1.068
f x dx f x dx dx 2 2 2
Mx
2
f x 2 dx 2 dx
0
2 3.75 4 3.4945 22.808 4 1.6335 0 34
3
f
1 29.878 4.9797 6 y
d −2
Mx 4.9797 1.078 A 4.62
2 0
x, y 0, 1.078 44. Centroids of the given regions:
12, 32, 2, 12, and 72, 1
Area: A 3 2 2 7
x, y
33 2 21 2 21 15 2 15 7 7 14
2
m1 m3
m2 1
x 1
2
3
y
7 7 287 55 6 , P 0, 8 8 64 2 16
5 4 3
By symmetry, x 0. y
4
15 , 25 14 14
7 7 7 46. m1 2 , P1 0, 8 4 16 m2
4 3
31 2 22 27 2 25 2 25 x 7 7 14 y
y
2
7 47 16 287 6455 16 16,569 5523 7 4 287 64 6384 2128
x, y 0,
m2
5523 0, 2.595 2128
m1 − 23
−1
− 21
x 1 2
1
3 2
296
Chapter 6
Applications of Integration
48. Centroids of the given regions: 3, 0 and 1, 0
50. V 2 rA 2 34 24 2
Mass: 8 y0 x
x, y
81 3 8 3 8 8
883 , 0 1.56, 0
6
52.
A
2
6
My
4 2 x 2 dx x 23 2 3
6
2
32 3
y
6
6
x2 x 2 dx 2
2
x x 2 dx
Let u x 2, x u 2, du dx:
4
My 2
4
u 2 u du 2
0
645 323
2 x
2
5u
u3 2 2u1 2 du 2
0
(6, 4)
4
2
2
5 2
4 u3 2 3
( x, y )
4 0
x 2
4
6
704 15
My 704 15 22 A 32 3 5
rx
22 5
V 2rA 2
294.89 225323 1408 15
54. A planar lamina is a thin flat plate of constant density. The center of mass x, y is the balancing point on the lamina.
56. Let R be a region in a plane and let L be a line such that L does not intersect the interior of R. If r is the distance between the centroid of R and L, then the volume V of the solid of revolution formed by revolving R about L is V 2rA where A is the area of R. 58. The centroid of the circle is 1, 0. The distance traveled by the centroid is 2. The arc length of the circle is also 2. Therefore, S 22 4 2. y 2
C
1
d x
−1
1 −1 −2
3
Section 6.7
Section 6.7
Fluid Pressure and Fluid Force
Fluid Pressure and Fluid Force
2. F PA 62.4516 4992 lb
4. F 62.4h 448 62.4h48 62.4448 11,980.8 lb
6. h y 3 y
8. h y y L y 2 4 y 2
4 L y y 3 F 62.4
0
3
0
4 62.4 3
F 62.4
4 3 y y dy 3
62.4
3
3y y 2 dy
0
y2 4 y 2 dy
2
23 4 y
0
2 3 2
2
332.8 lb
y
y3 3
3y 2
4 62.4 3 2 3
374.4 lb
0
1 x −1
Force is one-third that of Exercise 5.
1
y −3
4
2 1 x −2
−1
1
2
10. h y y
y x
4 L y 9 y 2 3
−1
1 −1
0
F 62.4
−2
4 y 9 y 2 dy 3 3
62.4
0
2 3
3
−4
9 y 2 1 22y dy
4 62.4 9 y 2 3 2 9
0 3
748.8 lb
12. hy 1 3 2 y
y
1+ 3 2
L1y 2y lower part L2y 23 2 y upper part
3 2 2
F 29800
0
19,600
19,600
y2 2
1 3 2 yy dy
3 2y
y3 3 2 2 3
3 2 2
3
1 3 2 y3 2 y dy
3 2y 18y
0
9 2 2 1 9 2 1 4 4
44,1003 2 2 Newtons
3 2
3 2 3 2 2
y3 3
6 2 1 y 2
3 2
3 2 2
3 −3
−2
−1
x 1
2
3
297
298
Chapter 6
Applications of Integration 16. h y y
14. h y 6 y L y 1
L y 2
5
F 9800
16 y dy
y 9800 6y 2
0
2
0
0 2 5
43 9 y
F 140.7 171,500 Newtons
y
3
y2
43 9 y dy 2
140.74 3
140.73 4 23 9 y
6
0
3
9 y 2 2y dy 0
2 3 2
3
3376.8 lb
y 2 x −3 −2 −1
1
2
x
3
−2
2
−4 −6
18. h y y
y 1
5 L y 5 y 3
0
F 140.7
140.7
x −1
5 y 5 y dy 3 3
−2
0
−5
3
20. h y
4
6
(5, −3)
0 3
452 15 1055.25 lb
3 y 2
L y 2
12 9 4y
3 2
F 42
3
−4
5 5y y2 dy 3
5 5 140.7 y2 y3 2 9 140.7
−3
2
3 2
2
32 y 9 4y
3 2
2
dy 63
3 2
9 4y 2 dy
21 4
3 2
3 2
9 4y 2 8y dy
The second integral is zero since it is an odd function and the limits of integration are symmetric to the origin. 3 The first integral is twice the area of a semicircle of radius 2 .
9 4y 2 2 9 4 y 2 9 Thus, the force is 63 4 141.75 445.32 lb. 22. (a) F wk r2 62.47 22 1747.2 lbs (b) F wk r2 62.4532 2808 lbs
24. (a) F wkhb 62.4
112 35 5148 lbs
(b) F wkhb 62.4
175 510 10,608 lbs
Review Exercises for Chapter 6 26. From Exercise 21: F 64 15
12 753.98 lb 2
28. h y 3 y
30. h y 12 y
Solving y 5x 2 x 2 4 for x, you obtain x 4y 5 y.
L y 2
L y 2
716 y 2
2
716 y 2
4
F 62.4
4y 5y
3
F 62.4 2
y 3 y dy 546.265 lb 5y
3 y
0
3
2124.8
0
12 y716 y 2 dy
0
4y dy 5y
4
62.47
12 y16 y 2 dy 21373.7 lb
0
y
10 8
y
6 5 4 x −6 −4 −2
2
2
4
6
1 x −3 −2 −1 −1
1
2
3
32. Fluid pressure is the force per unit of area exerted by a fluid over the surface of a body.
34. The left window experiences the greater fluid force because its centroid is lower.
Review Exercises for Chapter 6
5
2. A
1 2
4x
1 x
5 1 2
1
1 dx x2
4
4. A
y 2 2y 1 dy
0 1
81 5
y 2 2y 1 dy
0
y
1
6
y 1 2 dy
0
5
( 21 , 4 )
(5, 4)
3
y 13 3
1 0
1 3
2
y
1
( 5, 251 ) 1
2
3
4
x
6 3 2
(−1, 1)
1 1 2
−2
− 23
− 21
x
299
300
Chapter 6
Applications of Integration
2
6. A
y 3 y 2 1 dy
1
8. A 2
2
1 1 2y y 2 y 3 2 3
2 1
2
6
2 csc x dx
2 y y 2 dy
1
2
6
2 2x ln csc x cot x
9 2
2 0
y
2
3
3 ln 2 3
23 ln 2 3 1.555
(5, 2) 2
y 3
1
( π6 , 2 )
x 2
3
4
5
( 5π6 , 2)
−1
(2, −1)
−2
1
10. A
5 3
3
12 cos y dy cos y 21 dy
x 2π 3
5π 6
7 3
5 3
y sin y 2
5 3
3
y sin y 2
7 3 5 3
23 3
12. Point of intersection is given by: x3 x2 4x 3 0 ⇒ x 0.783.
3 4x x 2 x 3 dx
0
1 1 3x 2x 2 x 3 x 4 3 4
3
0.783 0
1.189
( 21 , 7π3 ) ( 21 , 5π3 )
2
0.783
A
y
4
( 21 , π3 ) −2
π 2
π 3
π 6
−1
3
x
−1
−2 (.7832, .4804)
2
2
14. A 2
2x 2 x 4 2x 2 dx
16. y x 1 ⇒ x y 2 1
0 2
2
4x 2 x 4 dx
y
0
x1 ⇒ x 2y 1 2
2
4 1 2 x3 x5 3 5
2 0
128 8.5333 15
A
5
10
(−2, 8)
x 1
1
(2, 8)
−4
2y 1 y 2 1 dy
0
23 x 1
x1 dx 2
1 x 1 2 4
3 2
4 y −2
(0, 0) 3
(5, 2) 2 1 x
(1, 0) 2 −1 −2
3
4
5
5 1
4 3
Review Exercises for Chapter 6
1
18. A
5
2 dx
0
1
2 x 1 dx
301
y 5
x y2 1
4
2
A
3
y 2 1 dy
(5, 2)
0
3 y 1
3
y
2 0
14 3
1
(1, 0) x 2
20. (a) R1t 5.28341.2701t 5.2834 e0.2391t
3
4
5
(b) R2t 10 5.28 e0.2t
15
40
Difference
R1t R2t dt 171.25 billion dollars
10
0
10 0
22. (a) Shell
(b) Shell
2
V 2
y 3 dy
0
4 y 2
2 0
2
8
V 2
2 yy 2 dy
0 2
2
y
2y 2 y 3 dy
0
4
2
3
23 y
3
1 y4 4
2 0
8 3
2 y 1 4 x 1
2
3
4
3
1 x 1
(c) Disk
(d) Disk
2
V
y 4 dy
0
5 y 5
2 0
32 5
3
4
2
V
y 2 1 2 12 dy
0 2
y
2
y 4 2y 2 dy
0
4
3
15 y
5
2 y3 3
2 0
176 15
2 y 1
5 x 1
2
3
4
4 3 2 1 x 1
2
3
4
302
Chapter 6
24. (a) Shell
Applications of Integration (b) Disk
a
V 4
b x a 2 x 2 dx a
0
2 b a
a 2 x 2 1 22x dx
2 b 2 2 1 a x x3 a2 3
0
43ab a
2
x 2 3 2
0
b2 2 a x 2 dx a2
0
a
a
a
V 2
4 a 2b 3
V 2
1 1 x 2
0
2 arctan x 2
2 x2 + y =1 a2 b2
(0, b)
x
x
(a, 0)
(a, 0)
28. Disk
0
y 2 x2 + y =1 a2 b2
(0, b)
1
a
4 ab 2 3
y
26. Disk
1
2
dx
V
ex 2 dx
0 1
1
0
0
4 0
e2x dx e 2x 2
1 0
2e 2 2 1 e1 2
2 2
2
y
1
y 3 2
x 1
−2
x
−1
1
2
−1
30. (a) Disk
(b) Shell
0
V
1 0 1
x 2x 1 dx
u x 1
y
x u2 1
x3
x4 x3 4 3
dx
x2
12 1 0
x
−1
dx 2u du 0
V 2
1
x 2x 1 dx
−1
1
y
4
u 2 1 2 u 2du
0
x
−1
1
4
u 6 2u 4 u 2 du
0
−1
4
17 u
7
2 1 u5 u3 5 3
1 0
32 105
Review Exercises for Chapter 6 1 1 32. Ax bh 2a 2 x 2 3a 2 x 2 2 2
34.
3 a 2 x 2
3
a
1 x3 6 2x
1 1 y x 2 2 2 2x
a
V 3
y
303
a 2 x 2 dx 3 a 2x
x3 3
a a
4a3
1 y 2
3
12 x
3
s
1
2
1 2x 2
2
1 1 1 2 1 x 2 dx x 3 2 2x 6 2x
3 1
14 3
Since 43 a 3 3 10, we have a 3 53 2. Thus, a
5 2 3 1.630 meters. 3
2 a 2− x 2
2 a 2− x 2
2 a 2− x 2
36. Since f x tan x has f x sec 2 x, this integral represents the length of the graph of tan x from x 0 to x 4. This length is a little over 1 unit. Answers (b).
38. y 2x y
1 x
1 y 2 1
1 x1 x x
x x 1 d x 4
3
S 2
3
2x
0
4
40. F kx
0
3
3 2
0
56 3
8 dV 4 gal min 12 gal min ft3 min dt 7.481 gal ft3 7.481
50 9
50 x 9 9
W
23x 1
42. We know that
50 k9 ⇒ k F
x 1 dx
0
V r 2h
50 25 2 x dx x 9 9
225 in
9 0
lb 18.75 ft lb
19 h
dV dh dt 9 dt
dh 9 dV 9 8 3.064 ft min. dt dt 7.481 Depth of water: 3.064t 150 150 49 minutes 3.064 4912 588 gallons pumped
Time to drain well: t
Volume of water pumped in Exercise 41: 391.7 gallons 391.7 588 52 x x
58852 78 391.7
Work 78 ft ton
304
Chapter 6
Applications of Integration
44. (a) Weight of section of cable: 4 x Distance: 200 x
200
W4
200 x dx 2200 x 2
0
200 0
80,000 ft lb 40 ft ton
(b) Work to move 300 pounds 200 feet vertically: 200300 60,000 ft lb 30 ft ton Total work work for drawing up the cable work of lifting the load 40 ft ton 30 ft ton 70 ft ton
b
46.
W
Fx dx
a
Fx
x 6, 2 9 4 3x 16,
9
W
0 ≤ x ≤ 9 9 ≤ x ≤ 12
12
2 x 6 dx 9
0
9
32 x
1 x 2 6x 9
9
2
0
34 x 16 dx
16x
12 9
9 54 96 192 54 144 51 ft lbs
3
48.
A
3 1 A 32 x
1 2x 3 x 2 dx x 2 3x x 3 3 1
3 32
3
1
x2x 3 x 2 dx
323 12
3 32
1
17 x, y 1, 5
3 1
32 3
3x 2x 2 x 3 dx
1
2x 3 2 x 4 dx
3 4 1 9x 6x 2 x 3 x 5 64 3 5
1
3
3
y
3
3 64
3 3 2 2 3 1 4 x x x 32 2 3 4
3
1
1
3
1
y
9 12x 4x 2 x 4 dx 9
17 5
6
( x, y )
3
x
−3
8
50.
A
0
5 1 A 16 x
5 16
8
16 5
y 6
0
( x, y )
1 x x 2 3 x dx 2
165 12 x 8
4 3
0
x, y
0
2
8
6
4
5 3 8 3 1 3 x x 16 8 6 y
3 1 1 x 2 3 x dx x 5 3 x 2 2 5 4
3
37 x
1 5 2 16
40 103 , 21
7 3
8
x 2 −2
10 3
0
1 x 2 dx 4
1 3 x 12
8 0
40 21
4
6
8
Review Exercises for Chapter 6 26. From Exercise 21: F 64 15
12 753.98 lb 2
28. h y 3 y
30. h y 12 y
Solving y 5x 2 x 2 4 for x, you obtain x 4y 5 y.
L y 2
L y 2
716 y 2
2
716 y 2
4
F 62.4
4y 5y
3
F 62.4 2
y 3 y dy 546.265 lb 5y
3 y
0
3
2124.8
0
12 y716 y 2 dy
0
4y dy 5y
4
62.47
12 y16 y 2 dy 21373.7 lb
0
y
10 8
y
6 5 4 x −6 −4 −2
2
2
4
6
1 x −3 −2 −1 −1
1
2
3
32. Fluid pressure is the force per unit of area exerted by a fluid over the surface of a body.
34. The left window experiences the greater fluid force because its centroid is lower.
Review Exercises for Chapter 6
5
2. A
1 2
4x
1 x
5 1 2
1
1 dx x2
4
4. A
y 2 2y 1 dy
0 1
81 5
y 2 2y 1 dy
0
y
1
6
y 1 2 dy
0
5
( 21 , 4 )
(5, 4)
3
y 13 3
1 0
1 3
2
y
1
( 5, 251 ) 1
2
3
4
x
6 3 2
(−1, 1)
1 1 2
−2
− 23
− 21
x
299
300
Chapter 6
Applications of Integration
2
6. A
y 3 y 2 1 dy
1
8. A 2
2
1 1 2y y 2 y 3 2 3
2 1
2
6
2 csc x dx
2 y y 2 dy
1
2
6
2 2x ln csc x cot x
9 2
2 0
y
2
3
3 ln 2 3
23 ln 2 3 1.555
(5, 2) 2
y 3
1
( π6 , 2 )
x 2
3
4
5
( 5π6 , 2)
−1
(2, −1)
−2
1
10. A
5 3
3
12 cos y dy cos y 21 dy
x 2π 3
5π 6
7 3
5 3
y sin y 2
5 3
3
y sin y 2
7 3 5 3
23 3
12. Point of intersection is given by: x3 x2 4x 3 0 ⇒ x 0.783.
3 4x x 2 x 3 dx
0
1 1 3x 2x 2 x 3 x 4 3 4
3
0.783 0
1.189
( 21 , 7π3 ) ( 21 , 5π3 )
2
0.783
A
y
4
( 21 , π3 ) −2
π 2
π 3
π 6
−1
3
x
−1
−2 (.7832, .4804)
2
2
14. A 2
2x 2 x 4 2x 2 dx
16. y x 1 ⇒ x y 2 1
0 2
2
4x 2 x 4 dx
y
0
x1 ⇒ x 2y 1 2
2
4 1 2 x3 x5 3 5
2 0
128 8.5333 15
A
5
10
(−2, 8)
x 1
1
(2, 8)
−4
2y 1 y 2 1 dy
0
23 x 1
x1 dx 2
1 x 1 2 4
3 2
4 y −2
(0, 0) 3
(5, 2) 2 1 x
(1, 0) 2 −1 −2
3
4
5
5 1
4 3
Review Exercises for Chapter 6
1
18. A
5
2 dx
0
1
2 x 1 dx
301
y 5
x y2 1
4
2
A
3
y 2 1 dy
(5, 2)
0
3 y 1
3
y
2 0
14 3
1
(1, 0) x 2
20. (a) R1t 5.28341.2701t 5.2834 e0.2391t
3
4
5
(b) R2t 10 5.28 e0.2t
15
40
Difference
R1t R2t dt 171.25 billion dollars
10
0
10 0
22. (a) Shell
(b) Shell
2
V 2
y 3 dy
0
4 y 2
2 0
2
8
V 2
2 yy 2 dy
0 2
2
y
2y 2 y 3 dy
0
4
2
3
23 y
3
1 y4 4
2 0
8 3
2 y 1 4 x 1
2
3
4
3
1 x 1
(c) Disk
(d) Disk
2
V
y 4 dy
0
5 y 5
2 0
32 5
3
4
2
V
y 2 1 2 12 dy
0 2
y
2
y 4 2y 2 dy
0
4
3
15 y
5
2 y3 3
2 0
176 15
2 y 1
5 x 1
2
3
4
4 3 2 1 x 1
2
3
4
302
Chapter 6
24. (a) Shell
Applications of Integration (b) Disk
a
V 4
b x a 2 x 2 dx a
0
2 b a
a 2 x 2 1 22x dx
2 b 2 2 1 a x x3 a2 3
0
43ab a
2
x 2 3 2
0
b2 2 a x 2 dx a2
0
a
a
a
V 2
4 a 2b 3
V 2
1 1 x 2
0
2 arctan x 2
2 x2 + y =1 a2 b2
(0, b)
x
x
(a, 0)
(a, 0)
28. Disk
0
y 2 x2 + y =1 a2 b2
(0, b)
1
a
4 ab 2 3
y
26. Disk
1
2
dx
V
ex 2 dx
0 1
1
0
0
4 0
e2x dx e 2x 2
1 0
2e 2 2 1 e1 2
2 2
2
y
1
y 3 2
x 1
−2
x
−1
1
2
−1
30. (a) Disk
(b) Shell
0
V
1 0 1
x 2x 1 dx
u x 1
y
x u2 1
x3
x4 x3 4 3
dx
x2
12 1 0
x
−1
dx 2u du 0
V 2
1
x 2x 1 dx
−1
1
y
4
u 2 1 2 u 2du
0
x
−1
1
4
u 6 2u 4 u 2 du
0
−1
4
17 u
7
2 1 u5 u3 5 3
1 0
32 105
Review Exercises for Chapter 6 1 1 32. Ax bh 2a 2 x 2 3a 2 x 2 2 2
34.
3 a 2 x 2
3
a
1 x3 6 2x
1 1 y x 2 2 2 2x
a
V 3
y
303
a 2 x 2 dx 3 a 2x
x3 3
a a
4a3
1 y 2
3
12 x
3
s
1
2
1 2x 2
2
1 1 1 2 1 x 2 dx x 3 2 2x 6 2x
3 1
14 3
Since 43 a 3 3 10, we have a 3 53 2. Thus, a
5 2 3 1.630 meters. 3
2 a 2− x 2
2 a 2− x 2
2 a 2− x 2
36. Since f x tan x has f x sec 2 x, this integral represents the length of the graph of tan x from x 0 to x 4. This length is a little over 1 unit. Answers (b).
38. y 2x y
1 x
1 y 2 1
1 x1 x x
x x 1 d x 4
3
S 2
3
2x
0
4
40. F kx
0
3
3 2
0
56 3
8 dV 4 gal min 12 gal min ft3 min dt 7.481 gal ft3 7.481
50 9
50 x 9 9
W
23x 1
42. We know that
50 k9 ⇒ k F
x 1 dx
0
V r 2h
50 25 2 x dx x 9 9
225 in
9 0
lb 18.75 ft lb
19 h
dV dh dt 9 dt
dh 9 dV 9 8 3.064 ft min. dt dt 7.481 Depth of water: 3.064t 150 150 49 minutes 3.064 4912 588 gallons pumped
Time to drain well: t
Volume of water pumped in Exercise 41: 391.7 gallons 391.7 588 52 x x
58852 78 391.7
Work 78 ft ton
304
Chapter 6
Applications of Integration
44. (a) Weight of section of cable: 4 x Distance: 200 x
200
W4
200 x dx 2200 x 2
0
200 0
80,000 ft lb 40 ft ton
(b) Work to move 300 pounds 200 feet vertically: 200300 60,000 ft lb 30 ft ton Total work work for drawing up the cable work of lifting the load 40 ft ton 30 ft ton 70 ft ton
b
46.
W
Fx dx
a
Fx
x 6, 2 9 4 3x 16,
9
W
0 ≤ x ≤ 9 9 ≤ x ≤ 12
12
2 x 6 dx 9
0
9
32 x
1 x 2 6x 9
9
2
0
34 x 16 dx
16x
12 9
9 54 96 192 54 144 51 ft lbs
3
48.
A
3 1 A 32 x
1 2x 3 x 2 dx x 2 3x x 3 3 1
3 32
3
1
x2x 3 x 2 dx
323 12
3 32
1
17 x, y 1, 5
3 1
32 3
3x 2x 2 x 3 dx
1
2x 3 2 x 4 dx
3 4 1 9x 6x 2 x 3 x 5 64 3 5
1
3
3
y
3
3 64
3 3 2 2 3 1 4 x x x 32 2 3 4
3
1
1
3
1
y
9 12x 4x 2 x 4 dx 9
17 5
6
( x, y )
3
x
−3
8
50.
A
0
5 1 A 16 x
5 16
8
16 5
y 6
0
( x, y )
1 x x 2 3 x dx 2
165 12 x 8
4 3
0
x, y
0
2
8
6
4
5 3 8 3 1 3 x x 16 8 6 y
3 1 1 x 2 3 x dx x 5 3 x 2 2 5 4
3
37 x
1 5 2 16
40 103 , 21
7 3
8
x 2 −2
10 3
0
1 x 2 dx 4
1 3 x 12
8 0
40 21
4
6
8
Problem Solving for Chapter 6 54. F 62.4165 4992 lb
52. Wall at shallow end:
5
F 62.4
y20 dy 1248
0
y2 2
5
15,600 lb
0
Wall at deep end:
10
F 62.4
y20 dy 624 y 2
0
10
62,400 lb
0
Side wall:
5
F1 62.4
y 40 dy 1248 y 2
0 5
F2 62.4
5
31,200 lb
0
5
10 y 8y dy 62.4
0
80y 8y 2 dy
0
F F1 F2 72,800 lb y 20 15 10 5 x
−5
5 10 15 20 25 30 35 40 45
Problem Solving for Chapter 6
1
2. R
x1 x dx
0
x2 x3 2
3 1
0
1 1 1 2 3 6
Let c, mc be the intersection of the line and the parabola.
4. 8y2 x21 x2 y±
x 1 x2 2 2
Then, mc c1 c ⇒ m 1 c or c 1 m.
1 1 2 6
y
1m
x x2 mx dx
0.5
0
0.25
1 x2 x3 x2 m 12 2 3 2
1m 0
−1.5
x
− 0.5
0.5
1.5
− 0.25
1 m2 1 m3 1 m2 m 2 3 2
− 0.5
1 61 m2 41 m3 6m1 m2 1 m26 41 m 6m 1 m22 2m
13
12
S 22
13
x
0
1m
m1
1 2x2 2 2 1 x2
1
1 1 m3 2
12
For x > 0, y
0.2063
5 2 3
1
1 2x2 2 2 1 x2
dx 2
305
306
Chapter 6
Applications of Integration 8. fx2 ex
6. By the Theorem of Pappus,
fx e x2
V 2 r A
f x 2e x2 C
1 2 d w2 l2 lw 2
f 0 0 ⇒ C 2 f x 2e x2 2
10. Let f be the density of the fluid and 0 the density of the iceberg. The buoyant force is
0
F f g
h
Ay dy
where Ay is a typical cross section and g is the acceleration due to gravity. The weight of the object is
Lh
W 0 g
h
Ay dy.
FW
0
f g
h
Lh
Ay dy 0 g
h
Ay dy
0 submerged volume 0.92 103 0.893 or 89.3% f total volume 1.03 103 12. (a) y 0 by symmetry
6
My 2
x
1 6
m2
1
y
6
1 dx 2 x4
1
b
1 b
m2
1
2
−1
lim x
x −1 −2
63 x, y ,0 43
−3
2
3
4
5
y = − 14 x
b2 1 1 dx 3 x b2 1 2b3 1 dx 4 x 3b3
b2 1b2 3bb 1 x 2b3 13b3 2b2 b 1 b→
y = 14 x
1
215 1 dx x4 324
3536 63 x 215324 43 (b) My 2
3
1 35 dx x3 36
3 2
x, y
14. (a) Trapezoidal: Area (b) Simpson’s: Area
x, y
2b3bb b 11, 0 2
32, 0
160
0 250 254 282 282 273 275 280 0 9920 sq ft 28
160
0 450 254 482 282 473 275 480 0 10,41313 sq ft 38
16. Point of equilibrium: 1000 0.4x2 42x x 20, p 840
P0, x0 840, 20
20
1000 0.4x2 840 dx 2133.33
Consumer surplus
0
20
Producer surplus
0
840 42x dx 8400
PA R T
I C H A P T E R 6 Applications of Integration Section 6.1
Area of a Region Between Two Curves
. . . . . . . . . . .2
Section 6.2
Volume: The Disk Method . . . . . . . . . . . . . . . . . . 9
Section 6.3
Volume: The Shell Method . . . . . . . . . . . . . . . . . 17
Section 6.4
Arc Length and Surfaces of Revolution . . . . . . . . . . . 22
Section 6.5
Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
Section 6.6
Moments, Centers of Mass, and Centroids . . . . . . . . . 30
Section 6.7
Fluid Pressure and Fluid Force
. . . . . . . . . . . . . . . 37
Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
C H A P T E R 6 Applications of Integration Section 6.1
Area of a Region Between Two Curves
Solutions to Odd-Numbered Exercises
0 x 2 6x dx
x 2 2x 3 x 2 4x 3 dx
6
1. A
6
0
x 2 6x dx
0
3
3. A
3
0
0
1
0
3x 3 x dx 6
4
7.
2x 2 6x dx
0
5. A 2
0
6
or
x 3 x dx
0
6
x dx 2
x 1
1
1
x 3 x dx
9.
4 2x 3
0
y
3
x dx 6
11.
3
2 sec x dx y
y 6
5
3
5
4 3
3 2
2
1
2π 3
1
x 1
2
3
4
x
5
1
2
3
4
5
6
2
13. f x x 1 gx x 1
15. A
0
2
2
A 4
0
Matches (d)
1 3 x 2 x 1 dx 2
1 3 x x 1 dx 2
x8 x2 x 4
2
2
0
y
(3, 4)
3
168 24 2 0 2
2
y
(2, 6)
6 5
(0, 1) 4 3
x 1
2
1 −2
2
(2, 3)
(0, 2)
3
(0, 1) x 1
3
4
π
π
3
3
−1
2π 3
x
Section 6.1 17. The points of intersection are given by:
Area of a Region Between Two Curves
19. The points of intersection are given by:
x 2 4x 0
x 2 2x 1 3x 3
xx 4 0 when x 0, 4
x 2x 1 0 when x 1, 2
4
A
0
2
gx f x dx
A
1
4
2
x 2 4x dx
1
0
gx f x dx
x3 2x 3
3x 3 x 2 2x 1 dx
2
4
2
0
1
2 x x 2 dx
32 3
x2 x3 2 3
2x
2 1
9 2
y
(0, 0)
y
(4, 0)
1
2
3
x
5
10
−1
(2, 9)
8
−2
6
−3
4
( 1, 0)
−4
x
−5
4
21. The points of intersection are given by:
3
x0
1
A
2 y y dy 2y y 2
0
2
3x 1 x 1
x2
1
23. The points of intersection are given by:
x 2 x and x 0 and 2 x 0 x1
2
1 0
3x x when x 0, 3
3
1
A
f x gx dx
0
3
Note that if we integrate with respect to x, we need two integrals. Also, note that the region is a triangle.
3x 1 x 1 dx
0
3
y
3x1 2 x dx
0
3
2
29 3x
(1, 1)
1
3 2
x2 2
3 0
y
(2, 0) x
(0, 0)
2
5
3
4
(3, 4)
3 2
(0, 1) x
−2
25. The points of intersection are given by: y2 y 2
1
1
(4, 2)
1
x 1
g y f y dy
1
(1,
y 2 y 2 dy
2y
y2 y3 2 3
2 1
2
1
2
4
3
2
A
3
y
2
y 2 y 1 0 when y 1, 2
2
3
9 2
1)
3
4
5
3 2
3
4
Chapter 6
Applications of Integration
2
27. A
f y g y dy
1
29. y
10 10 ⇒ x x y
10 dy y
10
2
A
y 2 1 0 dy
1
2
y3 y 3
2 1
6
2
10ln 10 ln 2
y
10 ln 5 16.0944
3
(0, 2)
10
10 ln y
(5, 2)
y 1 12
x
2
3
4
5
(0, −1)
(0, 10)
6
(2, −1)
(1, 10)
8
2
6 4
(0, 2) −4
(5, 2) x
−2
2
4
6
8
31. The points of intersection are given by: x 3 3x 2 3x x 2
11
xx 1x 3 0 when x 0, 1, 3
1
A
3
f x gx dx
0
gx f x dx
1
1
1
(0, 0)
(1, 1) −1
x 2 x 3 3x 2 3x dx
1
3
x 3 4x 2 3x dx
0
−6
3
x 3 3x 2 3x x 2 dx
0
(3, 9)
x 3 4x 2 3x dx
1
x4 34 x 4
3
x4
3 2 x 2
1
4
0
3 4 x3 x2 3 2
3 1
37 12
Numerical Approximation: 0.417 2.667 3.083 33. The points of intersection are given by: x 4x 3 3 4x x 2
2xx 4 0 when x 0, 4
4
A
3 4x x 2 x 2 4x 3 dx
0 4
2x 2 8x dx
0
2x 3 4x 2 3
4 0
9
2
64 3
Numerical Approximation: 21.333
(0, 3) −6
(4, 3) 12
−3
12
Section 6.1
Area of a Region Between Two Curves
35. f x x 4 4x 2, gx x 2 4
2
The points of intersection are given by:
−4
x 4 4x 2 x 2 4
(− 2, 0)
x 4 5x 2 4 0
4
(2, 0)
(−1, − 3)
(1, − 3) −5
x 2 4x 2 1 0 when x ± 2, ± 1 By symmetry,
1
A2
0 1
2
2
x 4 4x 2 x 2 4 dx 2
x 2 4 x 4 4x 2 dx
1
2
x 4 5x 2 4 dx 2
0
x 4 5x 2 4 dx
1
2
5
2
5 3 4 2 5
x5 1
1
5x 3 4x 3
0
2
5
32
x 5 5x 3 4x 5 3
2 1
40 1 5 8 4 3 5 3
8.
Numerical Approximation: 5.067 2.933 8.0
39. 1 x 3 ≤
37. The points of intersection are given by: 1 x2 1 x2 2
x2
2
x2
Numerical approximation: 1.759
3
1 0
A
0
−3
5
x ±1
−1
0
−1
1 x2 dx 1 x2 2
2 arctan x 2
(0, 1)
−4
0
2
x3 6
1 0
4 16 2 13 1.237
Numerical Approximation: 1.237
41. A 2
3
f x gx dx
y
0
2
3
g 4
2 sin x tan x dx
π , 3
3
0
(2, 3)
(0, 2)
f x gx dx
1
1 x 2 1 x 3 dx 1.759 2
3
1
A2
2
(−1, ( ( 1, 12 ( 1 2
x4 x2 2 0
1 x 2 on 0, 2 2
2
3
0
2 2 cos x ln cos x
3 f
1
π
(0, 0)
2
x
π 2
21 ln 2 0.614 3 4
π , 3
3
5
5
6
Chapter 6
43. A
2
Applications of Integration
45. A
xex 0 dx 2
0
0
2
2
1
2 cos x cos x dx
0
2
2 x sin x
1 2 ex 2
1 cos x dx 4 12.566
1 0
1 1 1
0.316 2 e
y
0
y
1
3
(1, e1 )
(0, 1) 2
g
(2π, 1)
(0, 0)
x 1
f π 2
−1
47. A
π
x
2π
2 sin x sin 2x 0 dx
1 cos 2x 2
1
0
2 cos x
0
1 1 x e 0 dx x2
e 1 x
4.0
3 1
e e 1 3 1.323
4
3
( π, 0)
(0, 0)
0
3
49. A
0
4 x x , y 0, x (b) A dx, 4x 3
51. (a) y
3
6 0
0
x3
3
0
No, it cannot be evaluated by hand. (c) 4.7721
6
−1
4 −1
x
53. Fx
0
1 t2 t 1 dt t 2 4
(a) F0 0
x 0
x2 x 4 (b) F2
22 23 4
(c) F6
62 6 15 4
y y
y 6 5 4 3 2
−1 −1
6
6
5
5
4
4
3
3
2
2
t 1
2
3
4
5
6
−1 −1
t 1
2
3
4
5
6
−1 −1
t 1
2
3
4
5
6
Section 6.1
55. F
cos
1
2 d sin 2 2
1
Area of a Region Between Two Curves
2 2 sin 2 (b) F0
(a) F1 0
2
0.6366
(c) F
12 2
2
1.0868
y y 3 2
1 2
− 21
c
57. A
0
c
1 2
1 2
− 21
− 21
1 2
1
θ − 21
ba b y a y dy c c
a 2 y ay 2c
(b, c)
y= cx b
y=
c b − a (x − a )
c 0
ac ac ac 2 2
x
1 baseheight 2
(0, 0)
(a, 0)
59. f x x 3
y 8
f x 3x 2
6
2
Tangent line:
2
x 3 3x 2 dx
1
2
4
3
f (x) = x3 −6
(− 2, − 8)
The tangent line intersects f x x 3 at x 2. 1
(1, 1) x
−4 −3 −2
y 1 3x 1 or y 3x 2
y = 3x − 2
4
At 1, 1, f 1 3.
A
− 21
y
a y a dy c
0
3 2
θ
1
1 2
− 21
y
3 2
x4 3x2 4
2
2x
1 2
−8
27 4
61. The variable is y. 63. x 4 2x 2 1 ≤ 1 x 2 on 1, 1
y
1
A
1
1 x 2 x 4 2x 2 1 dx
2
(0, 1)
1
1
x2 x 4 dx
x3 x5 3 5
x
1 1
4 15
You can use a single integral because x 4 2x 2 1 ≤ 1 x 2 on 1, 1. 65. Offer 2 is better because the accumulated salary (area under the curve) is larger.
( 1, 0)
(1, 0)
1 2
1
θ
7
8
Chapter 6
Applications of Integration
3
A
67.
3
9 x 2 dx 36
y 10
9b
9b
9b
9 x 2 b dx 18
6 4
9 b x 2 dx 9
2
x
0
6
9 bx 3 x3
9b
0
(
9
2
9
2
b, b)
(
x
x2
6
b, b)
9
2 9 b 3 2 9 3
9 b 3 2 9b b9
n
x
69. lim
→0 i1
i
9 3
4
27 2 9 3
4
3.330
x i2 x
y
0.6
i 1 where xi and x is the same as n n
1
x
0
x2 x3 dx 2 3
x2
(1, 0) x
1 . 6 0
0.2 0.4
0.13t dt
0
73. (a) y1 275.06751.0537t 275.0675e0.0523t] 460
10
15
10
0.13t 2
2 5 0
$1.625 billion
(b) y2 239.94071.0417t 239.9407e0.0408t 460
0 240
0.6 0.8 1.0
(0, 0)
5
7.21 0.58t 7.21 0.45t dt
0
(c)
f (x)
0.2
1
5
71.
0.4
y1 y2 dt 649.5 billion dollars
0 240
10
(d) No, model y1 > y2 forever because 1.0537 > 1.0417. No, these models are not accurate. According to news reports, E > R eventually.
Section 6.2
Volume: The Disk Method
75. The total area is 8 times the area of the shaded region to the right. A point x, y is on the upper boundary of the region if
9
y
2
y=x
x 2 y 2 2 y
x 2 y 2 4 4y y 2
1
( x, y )
x 2 4 4y
x 1
4y 4 x 2 y1
2
x2 . 4
We now determine where this curve intersects the line y x. x1
x2 4
x 2 4x 4 0 x Total area 8
4 ± 16 16 2 ± 22 ⇒ x 2 22 2
222
1 x4 x dx 2
0
8 x
x3 x2 12 2
222 0
16 42 5 80.4379 3.503 3
1 31 5 x dx 5
77. (a) A 2
5.5
0
1 0 dx
5
x 92 5 x x 2 5 109 5 5.5 5 6.031 m
2
5
5.5
0
5
3 2
(b) V 2A 26.031 12.062 m 3
2
(c) 5000 V 500012.062 60,310 pounds
79. True 81. False. Let f x x and gx 2x x2. f and g intersect at 1, 1, the midpoint of 0, 2 . But
b
2
f x gx dx
a
x 2x x2 dx
0
Section 6.2
Volume: The Disk Method
1
1. V
1
x 1 2 dx
0
x
1
2 dx
4
x dx
1
1
5. V
0
7. y x 2 ⇒ x y
4
y
0
2 y2
2 x2
4 1
4 0
2 dy
8
x3 x
3
x
2
1 0
3
15 2
1
x 2 2 x 3 2 dx
0
V
x 2 2x 1 dx
0
4
3. V
2 0. 3
x 4 x 6 dx
x5 x7 5
7 1 0
2 35
9. y x 2 3 ⇒ x y 3 2
4
0
1
y dy
V
0
1
y 3 2 2 dy
0
y 3 dy
4 y4
1 0
4
10
Chapter 6
Applications of Integration
11. y x, y 0, x 4 (b) R y 4, r y y 2
(a) Rx x, rx 0
4
V
x
0
4
x dx
0
2
2 dx
V
16 y 4 dy
0
2 x
4
0
1 16y y 5 5
8
2
y
2 0
128 5
y
3
3
2
2
1
1 x 1
2
3
x
4
1
−1
2
3
4
−1
(c) R y 4 y 2, r y 0
(d) R y 6 y 2, r y 2
2
V
4 y 2 2 dy
6 y 2 2 4 dy
0
2
2
V
0
16 8y 2 y 4 dy
2
0
8 1 16y y 3 y 5 3 5
2 0
32 12y 2 y 4 dy
0
256 15
1 32y 4y 3 y 5 5
2 0
192 5
y y 3 4 3
2
2 1 1 x 1
2
x
3
1
2
3
4
5
−1
−1
−2
13. y x2, y 4x x 2 intersect at 0, 0 and 2, 4. (a) Rx 4x x 2 rx x 2
(b) Rx 6 x2, rx 6 4x x 2
2
V
V
0
2
16x 2 8x 3 dx
8
0
6 x 2 2 6 4x x 2 2 dx
0
2
2
4x x 2 2 x 4 dx
163 x
x 3 5x 2 6x dx
0
3
2x 4
2 0
32 3
8
x4 3 x
y
4
5
3
3x 2
y
4 5 3 4 3
2
2 1 1 −1
x 1
2
3
−2
−1
x 1
2
3
4
2 0
64 3
Section 6.2
17. Rx 4, rx 4
15. Rx 4 x, rx 1
3
V
4 x 2 1 2 dx
0
3
42 4
0
x 2 8x 15 dx
3
0
V
3
3 4x x3
2
Volume: The Disk Method
3
15x
0
0
18
1 1x 1 1x
dx 2
8 1 dx 1 x 1 x2
8 ln1 x
1 1x
y
5
8 ln 4
3
8 ln 4
2
1 1 4
3 32.485 4
y 1 x
−1
1
2
3
4
3 2 1 x
−1
1
2
3
4
2
3
4
5
−1
19. R y 6 y, r y 0
y 5
4
V
6 y 2 dy
4 3
0 4
y 2 12y 36 dy
2 1
0
x
y3 6y 2 36y 3
1
4
−1
0
208 3
21. R y 6 y 2, r y 2
23. Rx
2
V
6 y 2 2 2 2 dy
2
3
y4 12y 2 32 dy
0
0
y5 4y 3 32y 2 5
3
V
0
2
2
1 x 1
384 5
2
, rx 0
1 x 1
2
1 dx x1
0 ln 4
ln x 1
0
dx 3
y
2
1
x 1 −1
2
3
3 0
11
12
Chapter 6
Applications of Integration
1 25. Rx , rx 0 x
4
V
1 x
1
1 x
27. Rx ex, rx 0
1
V
2
dx
ex 2 dx
0 1
4
e2x dx
0
1
e2x 2
3 4
y
2
1 0
1 e2 1.358 2
y 1
2 x 1
2
3 1
−1
x 1
2
0 2
6
0
9
1 6 y 3
1 6 y 3
2
3 2
2
−1
33. V
sin x 2 dx 4.9348
0
y
dy 3
6
36 12y y 2 dy
2
0
6
8 y 6 5 4 3 2 1 x 3
4
5
6
1
0
216 216 216 9 3
(2, 5)
6
8 x3 x3 10x2 24x 3 0
2
y3 36y 6y 2 9 3
1
8
4x3 8x2 20x 24 dx
152 125 277 3 3 3
31. y 6 3x ⇒ x
y 10
2
8 x4 x3 10x2 24x 3
V
x2 12 5 2x x22 dx
2
3
4x3 8x2 20x 24 dx
0
3
5 2x x22 x2 12 dx
29. V
2
x 1
2
3
x −2
1
2
3
4
Section 6.2
2
35. V
Volume: The Disk Method
2
ex 2 dx 1.9686 2
37. V
0
39. A 3
1
e x 2 ex 2 2 dx 49.0218
41. Disk Method:
Matches (a)
b
V
y
d
Rx 2 dx or V
a
Ry 2 dy
c
Washer Method:
2
b
V
1
Rx 2 rx 2 dx or
a
d
V
x 1
2
Ry 2 ry 2 dy
c
43.
y
y 4
−2
3
3
2
2
1
1 x
−1
x
2
1
1
2
3
4
The volumes are the same because the solid has been translated horizontally. 1 45. Rx x, rx 0 2
r
6
V
47. Rx r 2 x 2, rx 0
0
V
1 2 x dx 4
r
r 2 x 2 dx
r
3 x 12
6 0
2
18
r 2 x 2 dx
0
1 2 r 2x x 3 3
1 Note: V r 2h 3
r 0
1 4 2 r 3 r 3 r 3 3 3
1 326 3
y
18
y = r2 − x2
y 4 3 2
(−r, 0)
1 x 1 −1 −2
2
3
4
5
6
x
(r, 0)
13
14
Chapter 6
Applications of Integration
h
V
r y y yr 1 , R y r 1 , r y 0 H H H
49. x r
r 1
0
y H
2
h
dy r 2
1
0
h2 h3 H 3H 2
r 2h 1
2
51. V
0
1 2 x 2 x 8
53. (a) Rx V
2
dx
1 2 1 3 y y H 3H 2
r2 h
64
H
2 1 y 2 y 2 dy H H
r2 y
y
h
h 0
−r
h h2 H 3H 2
2
x 4 2 x dx
0
2x 5 x 6 64 5 6
3 25 x 2 , rx 0 5
9 25
2 0
18 25
25 x 2 dx
V
5
25 x 2 dx
0
18 x3 25x 25 3
5 0
30
(b) R y
5
5
x
r
5 9 y 2, r y 0, x ≥ 0 3
25 9
3
9 y 2 dy
0
25 y3 9y 9 3
3 0
50
y
60 6
y 4 8 6
2
4 x 2 −6
−4
2
6
4
x
−2
2
4
6
−2 −4
55. Total volume: V
4 503 500,000 3 ft 3 3
y 60 40
Volume of water in the tank:
y0
2500
2 y2
50
dy
2500 y 2 dy
y3 3
20
y0
−60
50
2500y
2500y0
y03 3
−40
y0
−60
50
250,000 3
When the tank is one-fourth of its capacity: y 3 250,00 1 500,000 2500y0 0 4 3 3 3 125,000 7500y0 y03 250,000
−20
y03 7500y0 125,000 0 y0 17.36 Depth: 17.36 50 32.64 feet When the tank is three-fourths of its capacity the depth is 100 32.64 67.36 feet.
x 20
40
60
Section 6.2
b
h
57. (a)
(b)
r 2 dx (ii)
0
a
b
1
x2 b2
Volume: The Disk Method
2
is the volume of an ellipsoid with axes 2a and 2b.
is the volume of a right circular cylinder with radius r and height h.
r
(c)
dx (iv)
r
r 2 x 2
y=r
(0, a)
2 dx (iii)
is the volume of a sphere with radius r.
y
y
15
y 2 y=a 1− x b2
y=
r 2 − x2
(h, r)
x
(−r, 0)
x
(−b, 0)
(r, 0)
(b, 0)
x
h
(d)
0
rx h
r
2
(e)
dx (i)
is the volume of a right circular cone with the radius of the base as r and height h. y
R r 2 x 2 R r 2 x 2 dx (v) 2
r
2
is the volume of a torus with the radius of its circular cross section as r and the distance from the axis of the torus to the center of its cross section as R. y
(h, r) r 2 − x2
R+
y= r x h
R x
R−
r2− x2 −r
59.
x
r
y 4 3 2
x 2
3
4
Base of Cross Section x 1 x 2 1 2 x x 2 (b) Ax bh 2 x x 21
(a) Ax b 2 2 x x 2 2
V
2
V
1
4 4x 3x 2 2x 3 x 4 dx
1 1 4x 2x 3 x 3 x 4 x 5 2 5
2 + x − x2
2 + x − x2
2
4 4x 3x 2 2x 3 x 4
2 1
81 10
1
2 x x 2 dx 2x
1 2 + x − x2
x2 x3 2 3
2 1
9 2
16
Chapter 6
61.
y
Applications of Integration 3 y (a) A y b 2 1
1
1
V
3 4
0
1 2
2
3 y 1 dy 2
1−
1
1 4
1 2y 1 3 y 2 3 dy
1−
0
3 3 y y 4 3 y 5 3 2 5
x 1 4
1 2
3 4
1
1 0
1 10
3 Base of Cross Section 1 y
3 y 1 1 1 (b) A y r 2 2 2 2
V
1 8
1
0
3 y 1 dy 2
81 1 2
1
4
1
1
3
V
4
0
2
1 8 10 80
1 1 3 y (c) A y bh 1 23 2 2 3
3 y
3 y
1−
3
y
1−
3
1−
y
3
y
3 y 2
3 y 1 dy 2
3
4
101 403
1−
3
y
3 1 3 y 1 (d) A y ab 2 1 2 y 2 2
3 y 2 1 2
2
V
1
0
a
3 y 1 dy 2
1 2 10 20
b 1−
63. Let A1x and A 2x equal the areas of the cross sections of the two solids for a ≤ x ≤ b. Since A1x A 2x, we have
b
V1
65.
A1 x dx
y
1 4 4 25 r 2 3 2 125 3 2 3
25 r 2 3 2
b
a
3
A2 x dx V2
a
25 r 2
Thus, the volumes are the same.
25
125 2
125 2
2 3
25 r2 2 2 3
251 22 3 r 2 r 51 22 3 3.0415 67. (a) Since the cross sections are isosceles right triangles: 1 1 1 Ax bh r 2 y 2 r 2 y 2 r 2 y 2 2 2 2 V (b) Ax V
1 2
r
r
r
r 2 y 2 dy
r 2 y 2 dy r 2y
0
y3 3
r 0
2 r3 3
x
y
1 1 tan 2 bh r 2 y 2r 2 y 2 tan r y 2 2 2 2
tan 2
r
r
r
r 2 y 2 dy tan
As → 90 , V → .
0
r 2 y 2 dy tan r 2 y
y3 3
r 0
2 3 r tan 3
3
y
3
y
Section 6.3
Section 6.3
3. px x
hx x
hx x
2
xx dx
0
23x
3 2 0
2 x 3 3
2
0
16 3
4
V 2
16 3
4
2
45 x
4
52
0
128 5
7. px x
hx x
hx 4x x 2 x 2 4x 2x 2
2
2
2
x 3 dx
V 2
y
0
x4x 2x 2 dx
4
2 0
4
2
8
4
3
2x 2
dx
x3
4
1
23 x
3
2
1 x4 4
0
16 3
2 1
x
−1
1
9. px x
2
x
−1
3
1
2
11. px x
hx 4 4x x 2 x 2 4x 4
hx
1
ex 2 2
2
2
1
x 3 4x 2 4x dx
V 2
0
x
0
x4 34 x 4
3
2x 2
2 0
8 3
1 2
ex
dx
22
1
2
ex 2 x dx 2
0
y
2 ex 2
4
2
1 0
2 1
y
3 1
2
3 4
1 x
−1
1
2
1 2
3 1 4
x 1 4
13. p y y h y 2 y
2
y 2 y dy
0
2
2
3
0
2
V 2
y
0
4 x 2
2
x 32 dx
0
5. px x
V 2
xx dx
0
V 2
17
Volume: The Shell Method
1. px x
V 2
Volume: The Shell Method
2y y 2 dy
0
2 y 2
y3 3
2 0
8 3
1 2
3 4
1
1 e
0.986
3
18
Chapter 6
Applications of Integration
15. p y y and h y 1 if 0 ≤ y <
1 . 2
y 1
1 1 p y y and h y 1 if ≤ y ≤ 1. y 2
12
V 2
1 2
1
y dy 2
0
1 y dy
1 4
12
2 y2
2
3 4
12
y2 2
2 y
0
1 12
4 4 2
x 1 2
17. px 4 x
hx 4x x 2
2
V 2
V 2
4
2
x 3 6x 2 8x dx
x4 2x 4
3
4x 2
2
2
16
0
x 3 9x 2 20x dx
0
0
4
5 x4x x 2 dx
0
2
2 2
4
4 x4x 2x 2 dx
0
2
19. px 5 x
hx 4x x 2 x 2 4x 2x 2
3 2
1
y
x4 3x 4
3
10x 2
4 0
64
y
4
4 3
3
2
2
1 1 x 1
x 1
2
2
3
4
−1
3
21. (a) Disk
(b) Shell
Rx
px x
x3
rx 0
h x x 3
2
V
x 6 dx
0
x7
7 2 0
128 7
y
x5
5 2
y 8
6
6
4
4
2
2 x
−1
1
2
−1
3
(c) Shell px 4 x
y
h x x 3
8
2
4 xx 3 dx
6
0 4
2
2
x 4 dx 2
0
8
V 2
2
V 2
4x 3
dx
x4
0
1 2 x 4 x 5 5
2 0
96 5
2 x 1
2
3
4
x 1
2
3
0
64 5
Section 6.3 23. (a) Shell
Volume: The Shell Method
(c) Shell
p y y
px a x
h y a 12 y 12 2
hx a 12 x 12 2
a
V 2
V 2
a xa 12 x 12 2 dx
0
a
2
a
y a 2a 12 y 12 y dy
0
a
ay 2a 12 y32 y 2 dy
2
0
a2 2a 32 x 12 2a 12 x 32 x 2 dx
0
2
a2 y
2
a2 4a5
2
4a 12 52 y 3 y 5 3
3
3
3
a 3
a
2 a 2x
0
15a
3
4 32 32 4 12 52 1 3 a x a x x 3 5 3
a 0
4 a 3 15
y
(0, a)
y
(0, a)
(a, 0) x
(a, 0) x
(b) Same as part (a) by symmetry
d
25. V 2
b
p yh y dy
or V 2
pxhx dx
a
x
5
27.
5
x 1 dx
1
x 1
1
2 dx
This integral represents the volume of the solid generated by revolving the region bounded by y x 1, y 0, and x 5 about the x-axis by using the Disk Method.
29. (a)
1.5
y = (1 − x 4/3) 3/4
−0.25
1.5 −0.25
2
2
y 5 y 2 1 dy
0
represents this same volume by using the Shell Method.
(b) x 43 y 43 1, x 0, y 0 y 1 x 43 34
y
1
V 2
4
0
3 2 1 x 1
2
19
3
−1
Disk Method
4
5
x1 x 43 34 dx 1.5056
20
Chapter 6
31. (a)
Applications of Integration 33. y 2ex, y 0, x 0, x 2
7
y=
(x −
3
−
2) 2 (x
6) 2
Volume 7.5 Matches (d)
−1
7 y
−1
6
(b) V 2
3 x 22x 62 dx 187.249 x
2
2 1
x 1
2
35. px x 1 2 x 2
hx 2
2
V 2
x 2
0
2
1 2 x dx 2 2
Now find x0 such that
2
x0
0
2x
0
12
x2
1 3 x dx 2
2 0
4 total volume
y
2
1 x4 8
1 2x 02
1 3 1 x dx 2 x 2 x 4 2 8
2x
x0 0
1
1 4 x 4 0
x 2
1
x 04 8x 02 4 0 x 02 4 ± 23
(Quadratic Formula)
Take x0 4 23 since the other root is too large. Diameter: 24 23 1.464
1
37. V 4
1
2 x1 x 2 dx
1
8
1
39. Disk Method
R y r 2 y 2
1
1
x2
2 2
dx 4
1
x1
x2
dx
r y 0
1
8
1
x 1 x 2 122 dx
2
r 2 y 2 dy
rh
3 1 x
4 2 2
r
V
1
2 32
1
4 2
r 2y
y3 3
r rh
y
r
−r
x r
1 h 23r h 3
Section 6.3
r
41. (a) 2
hx 1
0
x dx (ii) r
is the volume of a right circular cone with the radius of the base as r and height h. y
r
(b) 2
r
Volume: The Shell Method
R x 2r 2 x 2 dx (v)
is the volume of a torus with the radius of its circular cross section as r and the distance from the axis of the torus to the center of its cross section as R. y
x=R y=h 1− x r
(
(0, h)
(
y=
r2 − x2
x
(r, 0) (r, 0)
(−r, 0)
x
y=−
r2 − x2
r
r
(c) 2
2xr 2 x 2 dx (iii) is the
(d) 2
hx dx (i) is the volume of a
0
0
volume of a sphere with radius r. y
r2 − x2
y=
right circular cylinder with a radius of r and a height of h. y
(r, h) (r, 0)
y=−
x
r2 − x2 x
b
(e) 2
2ax1 x 2b 2 dx (iv)
0
is the volume of an ellipsoid with axes 2a and 2b. y
y =a (0, a)
2 1 − x2 b
(b, 0)
(0, −a)
y = −a
x
2 1− x b2
200
43. (a) V 2
x f x dx
0
2 200 0 42519 25019 47517 210015 412514 215010 41756 0 38
1,366,593 cubic feet (b) d 0.000561x 2 0.0189x 19.39 24
−20
225 −6
200
(c) V 2
xd x dx 2 213,800 1,343,345 cubic feet
0
(d) Number gallons V 7.48 10,048,221 gallons
21
22
Chapter 6
Applications of Integration
Section 6.4
Arc Length and Surfaces of Revolution
1. 0, 0 , 5, 12
3. y
(a) d 5 0 2 12 0 2 13
y x 1 2, 0, 1
12 x 5
(b) y
1
0
y
1
13 x 5
5 0
13
2 8 1 1.219 3
2
y
x1 dx 2
1 y 2
1 3
8
3 2
x 2 3 1
1
0
1 x4 2 8 4x 1 3 1 x 3, 1, 2 2 2x
12 x
3
1 2 , 1, 2 2x 3
b
s
1 y 2 dx
a
2
2 dx 3x 1 3
3 2 2 3 x 1 3 2 2 3
1
3 2
y
x 2 3 1 dx x 2 3
1
dx
3 1 x
1 , 1, 8 x 1 3
8
2
7.
1
12 5
3 2 3 x 2
8
s
1 x dx
0
5
5. y
1
s
12 y 5 s
2 3 2 x 1 3
1
8 1
18 x
1 3 1 x 3 dx 2 2x
4
1 4x 2
2 1
33
2.063 16
55 22 8.352
y lnsin x ,
9.
y
3
4, 4
1 cos x cot x sin x
1 y 2 1 cot2 x csc2 x s
3 4
csc x dx
4
3 4
4
ln csc x cot x
ln2 1 ln2 1 1.763 11. (a) y 4 x 2, 0 ≤ x ≤ 2
y 2x
(b) 1
5
y 2
1 4x
(c) L 4.647 2
2
L
0
−1
3 −1
1 4x 2 dx
Section 6.4 1 13. (a) y , 1 ≤ x ≤ 3 x
1 x2
y
(b)
2
1 y 2 1 −1
1 x4
L
1
1
−1
15. (a) y sin x, 0 ≤ x ≤
1 dx x4
y cos x
(b)
(c) L 3.820
1 y 2 1 cos 2 x
2
L −
(c) L 2.147
3
4
Arc Length and Surfaces of Revolution
2
1 cos 2x dx
0
3 2 −0.5
17. (a) x ey, 0 ≤ y ≤ 2
y
(b)
y ln x
1 x
1 y 2 1
1 ≥ x ≥ e2 0.135
1
L
3
e
−1
2
(c) L 2.221 1 x2
1 x1 dx 2
3 −1
Alternatively, you can do all the computations with respect to y. (a) x e y 0 ≤ y ≤ 2
dx ey dy
(b) 1
(c) L 2.221
1e dx dy 2
2y
2
L
1 e2y dy
0
19. (a) y 2 arctan x, 0 ≤ x ≤ 1
(b) y
2 1 x2
1
3
L
1
0
−0.5
1.5
−3
(c) L 1.871 4 dx 1 x 2 2
23
24
Chapter 6
2
21.
Applications of Integration
dxd x
1
0
5 1
2
2
y
dx 5
(0, 5) y = 25 x +1
4
s 5
3
Matches (b)
2
(2, 1)
1
x −1
1
2
3
4
23. y x 3, 0, 4 (a) d 4 0 2 64 0 2 64.125 (b) d 1 0 2 1 0 2 2 1 2 8 1 2 3 2 2 27 8 2 4 3 2 64 27 2
64.525
4
(c) s
4
1 3x 2 2 dx
0
1 9x 4 dx 64.666
0
(d) 64.672
4
(c) y1 1, L1
y
25. (a) 4
y1
y4
1 −1
−1
y2
y2
3 2
2 dx 5.657
0
5
y3 1
2
y3
x
3
4
4
3 1 2 x , L2 4 1 x, L 3 2
5
1
0
4
1
0
5 3 2 x , L4 y4 16
1
0
1 3 2 x 3x 1 2 2 3 2
2
When x 0, y 3 . Thus, the fleeting object has traveled 3 units when it is caught. y
1 3 1 2 3 1 2 1 x1 x x 3 2 2 2 x 1 2
1 y 2 1
1
s
0
x 1 2 x 1 2 4x 4x
x1 1 dx 2x 1 2 2
1
x 1 2 x1 2 dx
0
1 2 3 2 x 2x 1 2 2 3
1 0
4 2 2 3 3
The pursuer has traveled twice the distance that the fleeing object has traveled when it is caught. 29.
y 20 cosh y sinh
x , 20 ≤ x ≤ 20 20
x 20
1 y 2 1 sinh 2
20
L
20
cosh
x x cosh 2 20 20 x dx 2 20
20
0
40 sinh1 47.008 m.
cosh
x x dx 220 sinh 20 20
20 0
x2 dx 5.916 4
4
(b) y1, y2, y3, y4
27. y
9x dx 5.759 16
25 3 x dx 6.063 256
Section 6.4
33. y
y 9 x 2
31.
y
x
Arc Length and Surfaces of Revolution x3 3
y x 2, 0, 3
9 x 2
3
1 y 2
2
s
0
2
S 2
9 9 x2
0
0
9 dx 9 x2
3 dx 9 x 2
3 arcsin
3 arcsin
y
x3 1 6 2x
y
x2 1 2 2 2x
1 y 2
x 3
2
2
1
2
2
1
2
9 1 x
8282 1 258.85 9
0
3 x 2 37. y
y
1 9x1
x
1
x2 1 2 dx 2 2x
x5 x 1 dx 12 3 4x 3
x6 x2 1 2 72 6 8x
8
x3 1 6 2x
1 , 1, 8 3x 2 3
S 2
2 1
47 16
39. y sin x
2 3
18
4 3
dx
8
x 1 39x 4 3 1 dx
1 8
9x 4 3 1 1 212x 1 3 dx
1
27 9x
145145 1010 199.48 27
4 3
1 3 2
8 1
41. A rectifiable curve is one that has a finite arc length.
y cos x, 0, S 2
3
4 3 2
x2 1 2 2 , 1, 2 2 2x
S 2
1 x 4 1 24x 3 dx
0
0
2 arcsin 0 3
6
3
2
2.1892 3
3 arcsin
35.
x3 1 x 4 dx 3
sin x1 cos 2 x dx
0
14.4236 43. The precalculus formula is the surface area formula for the lateral surface of the frustum of a right circular cone. The representative element is y 1 x x .
2 f di xi2 yi2 2 f di
2
i
i
i
25
26
Chapter 6
Applications of Integration
y
hx r
y
h r
45.
y 9 x 2
47.
y
r2 h2 r2
1 y 2
r
S 2
x
0
r 0
3 9 x 2
2
r2 h2 dx r2
2r 2 h 2 x 2 r 2
1 y 2
x 9 x 2
S 2
0
3x 9 x 2
2
3
r r 2 h 2
0
dx
2x 9 x 2
69 x 2
dx
2 0
6 3 5 14.40 See figure in Exercise 48. 1 1 2 x x 3 2 3
y
49.
1 1 2 3 1 2 1 1 2 x x x 9x 1 2 6 2 6
y
1 1 1 x 18 81x x1 2 9x 1 2 2 36 36
1 y 2 1
1 3
S 2
0
3
361 x
1 1 2 x x 3 2 3
1 2
9 1 2 2 dx
1 1 2x 9x 2 dx x x 2 3x 3 3 3 3
1 3
0
Amount of glass needed: V
2 6
1 3 0
1 3
0
13 x
1 2
x 3 2 x1 2 9x 1 2 dx
2 ft 0.1164 ft2 16.8 in 2 27
0.015
0.00015 ft 3 0.25 in 3 27 12
51. (a) y f x 0.0000001953x4 0.0001804x3 0.0496x2 4.8323x 536.9270
400
(b) Area
f x dx 131,734.5 square feet
0
3.0 acres (Answers will vary.)
400
(c) L
1 fx 2 dx 794.9 feet
0
(Answers will vary.)
b
53. (a) V
1
1 dx x2 x
b 1
1
1 b
(b) S 2 2
y = 1x
b
1
y
2
b
1
2
1
b
1
x 1
—CONTINUED—
b
1 x 1 x
1 x1 dx 2
2
1 x1 dx 4
x 4 1
x3
dx
Section 6.5
Work
27
53. —CONTINUED—
(c) lim V lim 1 b→
b→
1 b
(d) Since x 4 1
x 4
>
x3
x3
we have
b
x 4 1
x3
1
b
dx >
1
1 > 0 on 1, b x
1 dx ln x x
b 1
ln b
and lim ln b → . Thus, b→
b
lim 2
b→
55. (a) Area of circle with radius L: A L 2 Area of sector with central angle (in radians) 1 S A
L 2 L 2 2 2 2
x 4 1
x3
1
dx .
(b) Let s be the arc length of the sector, which is the circumference of the base of the cone. Here, s L 2 r, and you have S
1 2 1 1 1 s L L2 Ls L 2 r rL 2 2 L 2 2
(c) The lateral surface area of the frustum is the difference of the large cone and the small one. S r2 L L 1 r1L 1 L
r2 L L 1 r2 r1
r2
L1
L L1 L1 ⇒ Lr1 L 1 r2 r1 By similar triangles, r2 r1
r1
Hence, S r2 L L 1 r2 r1 r2 L Lr1 L r1 r2 .
Section 6.5
Work
1. W Fd 100 10 1000 ft lb
3. W Fd 112 4 448 joules (newton-meters)
5. Work equals force times distance, W FD.
7. Since the work equals the area under the force function, you have c < d < a < b .
9. F x kx
11. F x kx
5 k 4 k
5 4
50
7
W
250 k 30 ⇒ k
0
W
5 2 5 x dx x 4 8
7 0
245 in lb 8 30.625 in lb 2.55 ft lb
20
25 3
50
F x dx
20
25 25x 2 x dx 3 6
8750 n cm 87.5 joules or Nm
50 20
28
Chapter 6
Applications of Integration
1 3
13. F x kx
15. W 18
20 k 9
0
W
0
1 3
7 12
324x dx 162x 2
1 3
12
kx 2 2
7 12
20 k 9 W
kx dx
20 10 2 x dx x 9 9
12 0
Note:
40 ft lb 3
0
1 3
k ⇒ k 324 18
37.125 ft lbs
1 4 inches 3 foot
17. Assume that Earth has a radius of 4000 miles. k x2
F x
4000
k
4000 2
s
k 80,000,000
4100
(a) W
4000
4100
487.8 mi
4000
tons
80,000,000 dx 1395.3 mi ton x2 1.47 1010 ft lb
80,000,000 x2
F x
5.15 109 ft lb
4300
(b) W
80,000,000 80,000,000 dx x2 x
19. Assume that the earth has a radius of 4000 miles. k x2
F x
15,000
(a) W
4000
160,000,000 160,000,000 dx x2 x
15,000 4000
29,333.333 mi ton
k
4000 2
10
2.93 10 4 mi ton
k 160,000,000 160,000,000 F x x2
10,666.667 40,000
3.10 1011 ft lb
26,000
(b) W
4000
160,000,000 160,000,000 dx x2 x
26,000 4000
6,153.846 40,000 33,846.154 mi ton 3.38 10 4 mi ton 3.57 1011 ft lb
21. Weight of each layer: 62.4 20 y
y 6
Distance: 4 y
4
(a) W
2
4
(b) W
0
5
62.4 20 4 y dy 4992y 624y 2
62.4 20 4 y dy 4992y 624y 2
4 2 4 0
4
2496 ft
lb lb
Weight of disk of water: 9800 4 y Distance the disk of water is moved: 5 y
4
0
5 y dy
y2 2
4
5 y 9800 4 dy 39,200
0
39,200 5y
2 1
9984 ft
23. Volume of disk: 2 2 y 4 y
W
4−y
3
4 0
39,200 12 470,400 newton–meters
x 1
2
3
4
5
6
Section 6.5
25. Volume of disk:
23 y y 2
Weight of disk: 62.4
y
23 y y 2
4 3 2
x
6
6 y
y 2 dy
0
4 1 62.4 2y 3 y 4 9 4
6 0
−4 −3 −2 −1
2995.2 ft lb
1
3
4
10
Weight of disk: 62.4 36 y 2 y
8
4
6
y
y 36 y 2 dy
x −6 −4 −2 −2
0 6
62.4
2
y
2
W 62.4
6−y
5
27. Volume of disk: 36 y 2 y
Distance: y
29
7
Distance: 6 y 4 62.4 W 9
Work
0
1
36y y 3 dy 62.4 18y 2 y 4 4
2
4
6
6 0
20,217.6 ft lb 29. Volume of layer: V lwh 4 2 9 4 y 2 y
y
Weight of layer: W 42 8 9 4 y 2 y
Tractor
6 4
13 y Distance: 2
1.5
336
−y x
42 8 9 4 y 2 13 2
13 2
2
1.5
W
8
1.5
1.5
−6 −4 −2 −2
13 y dy 2
1.5
9 4 y 2 dy
1.5
2
4
6
−4
9 4 y 2 y dy
The second integral is zero since the integrand is odd and the limits of integration are symmetric to the origin. The first integral represents the area of a semicircle of radius 32 . Thus, the work is W 336
132 32 12 2457 ft lb 2
31. Weight of section of chain: 3 y Distance: 15 y
15
W3
15 y dy
0
3 15 y 2 2 337.5 ft
lb
33. The lower 5 feet of chain are raised 10 feet with a constant force. W1 3 5 10 150 ft lb The top 10 feet of chain are raised with a variable force.
15
Weight per section: 3 y
0
Distance: 10 y
10
W2 3
0
3
10 y dy 10 y 2 2 150 ft
W W1 W2 300 ft lb
lb
10 0
30
Chapter 6
Applications of Integration 37. Work to pull up the ball: W1 50015 7500 ft lb
35. Weight of section of chain: 3 y
Work to wind up the top 15 feet of cable: force is variable
Distance: 15 2y W3
7.5
0
3 15 2y dy 15 2y 2 4
7.5
Weight per section: 1 y
0
Distance: 15 x
3 152 168.75 ft 4
15
lb
W2
0
1 15 x dx 15 x 2 2
15 0
112.5 ft lb Work to lift the lower 25 feet of cable with a constant force: W3 12515 375 ft lb W W1 W2 W3 7500 112.5 375 7987.5 ft lb
39.
p
k V
41. Fx
1000
k 2
W
k 2 x 2
k 2000
3
W
2
5
0
1. x
32 810.93 ft lb
(b) x
0
3. x
126 3 14 3 62 3 30 3 118 3 99 3 12 1 6 3 11 33
125x 750 x 6 feet
y
x, y
17 18 112 115 118 12 11111
7 5 8 5 12 5 15 5 18 5 17 12 5 5
50x 750 75x
x
100x 125 x 3 dx 10,330.3 ft lb
Moments, Centers of Mass, and Centroids
7. 50x 75L x 7510 x
9.
5
45. W
65 31 53 6 635 7
5. (a) x
3
1000 1.8 lnx 1 dx 3249.44 ft lb
Section 6.6
2
2000 dV 2000 ln V V
2000 ln
43. W
1 k k 1 dx k 1 2 2 x 2 4 2 2 x 3k units of work 4 1
52 13 31 10 513 9 52 11 34 1 513 9
109 , 91
y
m1 (2, 2)
2
m2 1 (− 3, 1)
x −3 −2 −1 −1
1
2
3
−2 −3 −4
m3 (1, − 4)
Section 6.6
11.
x
7 32 41 27 10 63 34216 8
y
7 33 40 21 10 60 34216 16
7 7 x, y , 8 16
m
x dx
0 4
Mx y
m5 (− 3, 0) −4 −2 −2
23 x
4
32
0
x2
x dx 2 4
4 0
16 3
My
0
x 4
6
8
4
4
3 2
2
y
Mx 3 3 4 m 16 4 4
m3 (7, 1)
m4 (0, 0)
m1 (−2, −3)
x
0
31
y
m2 6 (− 1, 0)
4
13.
Moments, Centers of Mass, and Centroids
1
2 x x dx x 52 5
4
0
( x, y )
64 5
x 1
2
3
4
My 64 3 12 x m 5 16 5
x, y
125 , 34
1
15.
m
x 2 x 3 dx
0 1
Mx
Mx 12 12 m 35 35
1
My
x, y
0
12
y
1
x4
x6
0
xx 2 x 3 dx
x 3 x 4 dx
4
5 1 0
20
x 1 4
1 2
3 4
1
35 , 12 35
x 2 4x 2 x 2 dx
0
2
3
x 2 5x 4x 2 3x dx
0
x5
11x 3
3 0
2
x3
3x 2
3 0
9 2
y 6
( x, y )
1 −1
3
x x 2 4x 2 x 2 dx
0
32 , 225
2
99 5
3
0
3
x 4 8x 3 11x 2 12x dx
0
My 27 2 3 x m 4 9 2
5 4
3
Mx 99 2 22 m 5 9 5
3 2
x 2 4x 2 x 2 x 2 4x 2 x 2 dx 2
2x 4 6x 2 2 5 3
My
x, y
x4 x5
Mx
y
3 4
( x, y )
0
0 3
My 12 3 m 20 5
m
1
35 0 1
1 4
3
17.
x5 x7 dx 2 5 7
1 2
1
0
x
4 1
3
x 2 x 3 2 x x 3 dx 2 2
0
y
x3 x4
x 3 3x 2 dx
x4 x3 4
3 0
27 4
x 1
2
3
4
5
32
Chapter 6
Applications of Integration
8
19.
m
x 23 dx
0 8
Mx
35 x
96 5
0
x 23 23 3 73 x dx x 2 2 7
0
8
y
0
6
192 7
4
Mx 192 5 10 m 7 96 7
y
8
My
2
( x, y ) x
3 xx 23 dx x 83 8
0
2
8 0
96
4
6
8
−2
My 5 5 96 m 96
x
10 7
x, y 5 ,
2
21.
8
53
m 2
4 y 2 dy 2 4y
0
2
My 2
0
y3 3
2 0
32 3
y
4 y2 8 y5 4 y 2 dy 16y y 3 2 3 5
2 0
2
256 15
1
( x, y ) x
My 256 3 8 x m 15 32 5
1
−2
85 , 0
3
23.
m
2y y 2 y dy
0 3
My
0
2
3y2
3
y 4 4y 3 3y 2 dy
0
3 0
0
1
x 2 x 4 dx
0
1
x 2 x 3 dx
2
x3 3
1 0
1 x3 x5 2 3 5
1 6
1 0
1 1 1 1 2 3 5 15
x3 4 13 4 12 3
x4
y y 23y y 2 dy
( x, y )
0
27 10
1
3
3y 2 y 3 dy y 3
0
2 x
3
x x 2 dx
y 3
1
−3
−2
x
−1
1 −1
y 2y y 2 y dy
1
0
9 2
My
0
3 3 x, y , 5 2
1 2
3
Mx 27 2 3 m 4 9 2
y
Mx
y3 3
0
A
y5 y4 y3 2 5
3
Mx
2
2y y 2 y 2y y 2 y dy 2 2
My 27 2 3 m 10 9 5
x
25.
3
−1
By symmetry, M x and y 0.
x, y
2
1 0
1
1
y4 4
3 0
27 4
Section 6.6
3
27.
A
0
Mx
3
1 2
3 0
2
0
2x 2 8x 8 dx
0
2x 2 4x dx
0
3
2x 3
33
9 12 21
3
2x 4 dx
3
My
2x 4 dx x 2 4x
Moments, Centers of Mass, and Centroids
2x 2
3
3
2x 3
4x 2 8x
3 0
18 36 24 78
18 18 36
0
5
29. m
10x 125 x 3 dx 1033.0
0
5
Mx
10x 125 x 3 2
0
10x 125 x 3 dx
5
My
10x 2 125 x 3 dx
0
x
My 3.0 m
y
Mx 126.0 m
10 3
3,124,375 130,208 24
5
50
x 2 125 x 3 dx
0
5
125 x 3 3x 2 dx
0
12,500 5 3105.6 9
400
−1
Therefore, the centroid is 3.0, 126.0.
20
31. m
3 5 400 x 2 dx 1239.76
20 20
Mx
3 400 x 2 5 3 400 x 2 dx 5 2 20
25 2 y
20
20
400 x
2 23
dx 20064.27
6 −50
33.
1 A 2ac ac 2 1 1 A ac x
x 0 by symmetry. Therefore, the centroid is 0, 16.2.
y
50
−25
1 2ac
x, y
c
2
0
4ab 4ab y 2 y 2 dy c c
1 2ab 2 4ab 3 y 2y 2ac c 3c 1 ac 1 ac
25 −5
2
c
0
Mx 16.18 m
ac1 12 b c a y a b c a y a dy
c
y
0 c
0
1 2 b abc 2ac 3 3
b c a y a b c a y a dy
y
0
c
2a 2 y 2a dy c c
2 y2 y3 c 2 3c
c 0
c
0
y
y2 dy c
c 3
b3 , 3c
In Exercise 566 of Section P.2, you found that b3, c3 is the point of intersection of the medians. y
(b, c) y = c (x + a) b+a
y=
c b − a (x − a )
( x, y ) x
(−a, 0)
(a, 0)
34
Chapter 6
Applications of Integration
c 35. A a b 2 2 1 A ca b x y
2 c a b
c
b c a x a dx ca 2 b b c a x c
x
0
c2
2 1 ca b 2
c
0
ac 2 2
ca b 2
2 ba 1 x a dx c ca b
ba c
2bc 2
c
0
x3 2abc a x2 a x 2
3
2ac 2
2
3ac 2
6 ba c
c
x 2
2
c2b a
c 0
a 2bc
1 b a2c acb a a 2c ca b 3
1 b 2 2ab a 2c 3acb a 3a 2c 3ca b
a 2 ab b 2 1 b 2 2ab a 2 3ab 3a 2 3a 2 3a b 3a b
0
2ab a x a 2 dx c
1 ca b
3a b 3a b
2
2 b a x 3 ax 2 ca b c 3 2
ax dx
0
b a 2 ca b 3
2
y
a 2bc a 2 ab b 2 , . Thus, x, y 3a b 3a b
y = b −c a x + a
The one line passes through 0, a2 and c, b2. It’s equation is y
ba a x . 2c 2
The other line passes through 0, b and c, a b. It’s equation is y
a
(0, a) ( x, y )
( c, b)
a 2b x b. c
(0, 0) x
(c, 0)
b
x, y is the point of intersection of these two lines. 37. x 0 by symmetry
y
1 A ab 2 b
2 1 A ab 2 1 y ab 2
a
a
−a
b
a 2 x 2 a
a x x3
1 b2 ab a 2
4b 3
x, y 0,
39. (a)
3 a
2
a
a
dx
b 4a 3 4b a 3 3 3
(e) Mx
y
y=b
b x 2b x 2 dx 2 b
b
b2 x4 1 x5 d x b 2x 2 2 5 b
b
b 2 b −5 −4 −3 −2 −1
x 1 2 3 4 5
A (b) x 0 by symmetry (c) M y (d) y >
b
b
x
2
xb x 2 dx 0 because bx x 3 is odd
b b since there is more area above y than below 2 2
5
b b b b 24 3 3
b
b
5
b x 2 dx bx
b
b
4b 2 b
b b y
b 2 b
Mx 4b 2 b5 3 b. A 5 4b b3
x3 3
b
b
Section 6.6
Moments, Centers of Mass, and Centroids
41. (a) x 0 by symmetry
40
A2
f x dx
0
240 20 5560 30 429 226 420 0 278 34 3 3
f x 2 40 10 72160 dx 30 2 4 29 2 226 2 4 20 2 0 7216 2 34 3 3 40 40
Mx y
Mx 721603 72160 12.98 A 55603 5560
x, y 0, 12.98 (b) y 1.02 105 x 4 0.0019x 2 29.28 (c) y
Mx 23697.68 12.85 A 1843.54
x, y 0, 12.85 43. Centroids of the given regions: 1, 0 and 3, 0 Area: A 4
2
4 1 3 4 3 4 4
x
y
4 0 0 y 0 4
1 x 1
3
−1 −2
443 , 0 1.88, 0
x, y
32, 0, 5, and 0, 152
y
45. Centroids of the given regions: 0,
7
Area: A 15 12 7 34
6 5
150 120 70 0 x 34 y
1532 125 7152 135 34 34
x, y 0,
135 34
Mass: 4 2 4 1 2 3 2 3 4 2 2
y0
x, y
3 2 1 −4 −3 −2 −1
x 1
2
3
4
47. Centroids of the given regions: 1, 0 and 3, 0
x
4
223 , 0 2.22, 0
49. V 2 rA 2 516 160 2 1579.14
35
36
Chapter 6
Applications of Integration
1 51. A 44 8 2
1 1 y 8 2 ry
4
0
y 4
x3 1 4 x4 x dx 16x 16 3
4 0
8 3
2
8 3
1
V 2 rA 2
838 1283 134.04
53. m m1 . . . mn
x 1
2
55. (a) Yes. x, y
My m1x1 . . . mn xn Mx m1y1 . . . mn yn x
( x, y )
3
My Mx ,y m m
3
4
56, 185 2 56, 41 18
(b) Yes. x, y
56 2, 185 176, 185
(c) Yes. x, y
56, 185
(d) No. 57. The surface area of the sphere is S 4 r 2. The arc length of C is s r. The distance traveled by the centroid is
y
S 4 r 2 4r. s r
d
r
This distance is also the circumference of the circle of radius y.
(0, y) −r
x
r
d 2 y Thus, 2 y 4r and we have y 2r. Therefore, the centroid of the semicircle y r 2 x 2 is 0, 2r.
1
59.
A
x n dx
0
m A Mx
2
nx 1
1 0
1 n1
n1
1
x n 2 dx
0 1
My
n1
y
2 2nx 1 2n1
xx n dx
0
0
22n 1
1
(1, 1)
y=xn
n 2 0 n 2 x n2
x
My n 1 m n2
y
Mx n1 n1 m 22n 1 4n 2
Centroid:
1
1
nn 12, 4nn 12
As n → , x, y → 1, 14 . The graph approaches the x-axis and the line x 1 as n → .
x 1
Section 6.7
Section 6.7
Fluid Pressure and Fluid Force
Fluid Pressure and Fluid Force
1. F PA 62.453 936 lb
3. F 62.4h 26 62.4h6 62.426 748.8 lb
5. h y 3 y L y 4
7. h y 3 y L y 2
3
F 62.4
3 y4 dy
3y 1
3
3
3
124.8
0
y2 249.6 3y 2
3 0
3 y
0
3 y dy
249.6
F 262.4
0
y2 dy 3
3
0
1123.2 lb
124.8 3y
y
y3 9
3y 1 dy
3 0
748.8 lb
y 4 4
3 2
2
1
1
x 1
2
3
4 x −2
9. h y 4 y
−1
1
11. h y 4 y
L y 2 y
L y 2
4
F 262.4
4 y y dy
0
4y 1 2 y 3 2 dy
8y3
3 2
24 y dy
0
0
124.8
2
F 9800
4
124.8
2
2y 5 2 5
4 0
9800 8y y 2
1064.96 lb
2 0
117,600 Newtons
y
y 3
3
x −2
1 x −2
−1
1
2
−1
1
2
37
38
Chapter 6
Applications of Integration
13. h y 12 y
L y 10
2y 3
L y 6
9
F 9800
15. h y 2 y
0
2
2y 3 9800 72y 7y 2 9
9 0
2 y10 dy
F 140.7
2y 12 y 6 dy 3
0 2
1407
2,381,400 Newtons
2 y dy
0
1407 2y
y
y2 2
2
2814 lb
0
y 9 4 6
3
3 x −3
3
6
x
9
−6 −4 −2 −1
2
4
6
−2
17. h y 4 y L y 6
19. h y y L y 2
4
F 140.7
4 y6 dy
4
4 y dy
0
844.2 4y
y2 4 2
0
2
0
F 42
0
844.2
12 9 4y
6753.6 lb
3 2
y 9 4y 2 dy
0
42 8
214 23 9 4y
y
3 2
9 4y 2 1 2 8y dy 0
2 3 2
3 2
94.5 lb
y
5 2 3 1
x
1
−2
x −3 −2 −1 −1
1
2
−1
1
2
−1
3
−2
21. h y k y
y
water level
L y 2 r 2 y 2
r
r
Fw
r
k y r 2 y 2 2 dy
r
w 2k
r
−r
r
r 2 y 2 dy
r
r 2 y 2 2y dy
r
x
−r
The second integral is zero since its integrand is odd and the limits of integration are symmetric to the origin. The first integral is the area of a semicircle with radius r.
F w 2k
r2 0 wk r 2 2
Section 6.7 23. h y k y
Fluid Pressure and Fluid Force
39
25. From Exercise 22:
L y b
F 641511 960 lb
h 2
Fw
k yb dy
h 2
y2 2
wb ky
h 2 h 2
wbhk wkhb
y
water level h 2
−b 2
x
b 2 −h 2
27. h y 4 y
4
F 62.4
4 y L y dy
0
Using Simpson’s Rule with n 8 we have:
4380 0 43.53 235 42.58 229 41.510 2110.25 40.510.5 0
F 62.4
3010.8 lb 29. h y 12 y L y 2 4 2 3 y 2 33 2
4
F 62.4
212 y4 2 3 y 2 33 2 dy
0
31. (a) If the fluid force is one half of 1123.2 lb, and the height of the water is b, then h y b y L y 4
b
6448.73 lb
F 62.4
0
y
1 b y4 dy 1123.2 2
b
10 8
by 2 y2
6 4
x −6 −4 −2
b y dy 2.25
0
2
4
6
b2
b 0
2.25
b2 2.25 2 b2 4.5 ⇒ b 2.12 ft.
(b) The pressure increases with increasing depth.
d
33. F Fw w
c
hyLy dy, see page 471.
40
Chapter 6
Applications of Integration
Review Exercises for Chapter 6
5
1. A
1
1 1 dx x2 x
5 1
1 dx x2 1
1
4 5
3. A
y
1
arctan x (1, 1)
1
1 1
4 4 2
1 5,, 25
y
x 2
3
(5, 0)
4
(1, 0)
2
1,,
1
1 2
1,
1 (−1, 0)
1
5. A 2
1 (1, 0)
x
2
x x 3 dx
7. A
0
e 2 e x dx
0
12 x
2
1 2
2
1 x4 4
1
xe 2 e x
0
2 0
e2 1
1 2
y y
(0, e 2) (2, e 2)
6 1
(1, 1)
4
x
(0, 0) 1
1
(0, 1) x
1
1
2
3
1
( 1,
9. A
1)
5 4
4
11. A
3 8x x 2 x 2 8x 3 dx
0
cos x sin x
5 4
8
4
12 12 12 12
4 2 2 2
16x 2x 2 dx
0
8
sin x cos x dx
2 8x 2 x 3 3
8 0
512 170.667 3
20
(8, 3)
−4
(0, 3)
− 16
10
Review Exercises for Chapter 6 13. y 1 x
2
1
A
0
2
(0, 1)
1 x dx 2
−1
2
(1, 0)
1
1 2x 1 2 x dx
−1
0
4 1 x x3 2 x 2 3 2
1 0
1 0.1667 6
15. x y 2 2y ⇒ x 1 y 12 ⇒ y 1 ± x 1
0
A
1
1 x 1 1 x 1 dx
2
A
2
0 y 2 2y dy
0
0
y 3
0
1
2 x 1 dx
1 2y y 2 dy y 2 y 3 3
(0, 2)
2 0
1
4 3
(0, 0) −2
x
−1
2 −1
2
17. A
x 2
1 1
0
2
0
x dx 2
y1
dx
3
y
1 x 2 dx 3
2
3
3 x dx
2
2
(2, 0) x
y x 2 ⇒ x y 2, y 1
(3, 1)
(0, 1)
x ⇒ x 2 2y 2 1
3
1
A
y 2 2 2y dy
0 1
3y dy
0
32 y
1
2
0
3 2
19. Job 1 is better. The salary for Job 1 is greater than the salary for Job 2 for all the years except the first and 10th years. 21. (a) Disk
(b) Shell
4
V
x 2 dx
0
3x
3 4 0
64 3
x 2 dx
23 x
2
4
4
V 2
0
y
y
4
4
3
3
2
2
1
1 x 1
2
3
—CONTINUED—
4
x 1
3
4
3
0
128 3
41
42
Chapter 6
Applications of Integration
21. —CONTINUED— (c) Shell
(d) Shell
4
V 2
V 2
0 4
2
4
4 xx dx
6 xx dx
0 4
4x x 2 dx
2
0
6x x 2 dx
0
2 2x 2
x3 3
4 0
64 3
1 2 3x 2 x 3 3
y
4 0
160 3
y 5
4
4 3 3 2
2 1
1
x 1
x 1
23. (a) Shell
2
(b) Disk
4
V 4
x
0
3
212316 x
4
3 16 x 2 dx 4
3
2
V 2
0
4
2 3 2
0
64
3 16 x 2 4
4 0
4
2
2 1 x
−3 −2 −1
2
1
3
−3 −2 −1
−2
−2
−4
−4
x 1
2
3
y
1
V 2
0
1
0
x dx x4 1
1
2x dx x 2 2 1 x
arctan x
2
1
1
0
4 0 4
2
2
dx
48
y
4
1
5
9 x3 16x 8 3
y
25. Shell
4
−1
3
Review Exercises for Chapter 6 27. Shell
y 3
u x 2 x
43
2
2
u2
1 x
dx 2u du
1
6
V 2
2 2
4
0
1 x 2
1
dx 4
u 3 2u du 4 1u
3 u
4
3
2
x
0
u 2 2u du 1u
2
u2 u 3
0
1 u 2 3u 3 ln 1 u 2
2
3
4
5
6
−1
2 0
−2 −3
3 du 1u
4 20 9 ln 3 42.359 3
0
29. Since y ≤ 0, A
31. From Exercise 23(a) we have: V 64 ft 3
x x 1 dx.
1
1 V 16 4
ux1 xu1 dx du
1
A
1
u 1 u du
0
Disk:
1 0
16 9 y 2 dy 16 9
y0
3
9 y 2 dy 1
9y 31 y
4 15
y0
3
9y
y
0
3
9
1 y03 27 9 9 3 y03 27y0 27 0
x
−1
3
1 9
u 3 2 u 1 2 du
0
2 5 2 2 3 2 u u 5 3
y0
By Newton’s Method, y0 1.042 and the depth of the gasoline is 3 1.042 1.958 ft.
−1
33.
4 f x x 5 4 5
35.
f x x 1 4
y 300 cosh y
1 f x 2 1 x u 1 x x u 1 2 dx 2u 1 du
4
s
0
u u 1 du
1
3
u 3 2 u 1 2 du
2
1
2
25 u
5 2
2 u 3 2 3
3 1
4 3 2 u 3u 5 15
8 1 6 3 6.076 15
3 x sinh 20 2000
s
1 20
2000
2000
x 1 203 sinh 2000
2000
2000
2
dx
x 400 9 sinh 2000 dx 2
4018.2 ft (by Simpson’s Rule or graphing utility)
3
1 x dx 2
x 2000 280, 2000 ≤ x ≤ 2000
3 1
44
Chapter 6
Applications of Integration
3 37. y x 4
39. F kx y
1 y 2
4 k1
3 4
F 4x
25 16
W
5
4
S 2
0
4x dx 2x 2
0
3 x 4
15 dx 25 16 8 2
x2 4 0
50 in
15
5 0
lb 4.167 ft lb
y 4
(4, 3) 3
2 1 x 1
2
3
4
13 y 2
41. Volume of disk:
Weight of disk: 62.4
43. Weight of section of chain: 5 x Distance moved: 10 x
13 y 2
Distance: 175 y 62.4 9
W
10
W5
0
150
175 y dy
0
62.4 y2 175y 9 2
5 10 x dx 10 x 2 2 250 ft lb
150 0
104,000 ft lb 163.4 ft ton
b
45. W
Fx dx
a 4
80
ax 2dx
0
ax 3 3
4 0
64 a 3
380 15 a 3.75 64 4
a
47. A
a x
0
6 1 A a2 x
6 a2
a
0
a6 12
x, y
2
3 a2
a
0
0
4 1 a 2 a x 1 2 x dx ax a x 3 2 x 2 3 2
x a x 2 dx a
y
2 dx
6 a2
a 0
a2 6
a
ax 2 a x 3 2 x 2 dx
y
0
a x 4 dx
a
( x, y )
a
a 2 4a 3 2x 1 2 6ax 4a 1 2x 3 2 x 2 dx
0
3 2 8 8 1 a x a 3 2x 3 2 3ax 2 a 1 2x 5 2 x 3 a2 3 5 3
a5 , 5a
a
a 0
a 5
x
10 0
Review Exercises for Chapter 6 49. By symmetry, x 0.
y
1
A2
a 2 x 2 dx 2 a 2x
0
x3 3
a
4a 3 3
0
a2
1 3 A 4a 3
( x, y )
4a3 12
a
y
3
a
a 2 x 2 2 dx
−a
a 4 2a 2x 2 x 4 dx
0
a
2a 2 3 1 5 6 4 x x 3 a x 8a 3 5
6 2 5 1 5 2a 2 5 3 a a a 8a 3 5 5
x
a
6 8a 3
x, y 0,
a
0
2a 2 5
51. y 0 by symmetry
y 4
For the trapezoid:
6
x
0
6
1
1 1 x1 x1 6 6
x
x3
−1
dx
1 2 x 2x dx x2 3 9
0
6 0
−2 −3 −4
60
For the semicircle: m
122 2 2
8
My
x 4 x 6 2 4 x 6 2 dx 2
6
8
x 4 x 6 2 dx
6
Let u x 6, then x u 6 and dx du. When x 6, u 0. When x 8, u 2.
2
My 2
2
u 6 4 u 2 du 2
0
234 u 1
2
2
2 3 2
0
12
2
u 4 u 2 du 12
0
2
x
2 2 16 4 4 9 12 4 3 3
4 4 9 3
180 4 4 9 3
The centroid of the blade is
1
2 9
, 0. 239 49 9
4 u 2 du
0
Thus, we have: x 18 2 60
(6, 2)
2
m 46 16 18 My
y = 61 x + 1
3
29 49 3 9
1
2
3
4
5
7
(6, −2)
45
46
Chapter 6
Applications of Integration
53. Let D surface of liquid; weight per cubic volume.
d
F
D y f y g y dy
c
d
D f y g y dy
c
d
c
y f y g y dy
d
f y g y dy D
D
d
c
y
c
y f y g y dy
d
f y g y dy
c
d f
c
x
AreaD y Areadepth of centroid
Problem Solving for Chapter 6 1 1 1. T c c2 c3 2 2
c
R
cx x2 dx
0
cx2
2
x3 3
c
0
c3 c3 c3 2 3 6
1 3 c T 2 lim lim 3 c→0 R c→0 1 c3 6
3. (a)
1 V 2
1
2 1 y2 2 1 y2 dy
0
1
2
2
4 41 y2 1 y2 4 41 y2 1 y2 dy
0
1
8
1 y2 dy
Integral represents 1 4 area of circle
0
8
4 2
⇒ V 42
2
(b) x R2 y2 r2 ⇒ x R ± r2 y2 1 V 2
r
R r2 y2 R r 2 y2 dy
0
2
2
r
4Rr2 y2 dy
0
1 4R r2 2 r2 R 4 V 2 2 r 2 R
r
5. V 22 2
r2 h2 4
23r
2
xr2 x2 dx
x23 2
x2 + y2 = r2
r
r
r2 h2 4
4 h3 h3 which does not depend on r! 3 8 6
( x, y )
g
2 r2 − h 4 r
h 2
Problem Solving for Chapter 6 (b) Tangent at Aa, a3: y a3 3a2x a
7. (a) Tangent at A: y x3, y 3x2 y 1 3x 1
y 3a2x 2a 3
y 3x 2
x3 3a2x 2a3 0
To find point B:
x a2x 2a 0 ⇒
x3 3x 2
To find point B:
B 2a, 8a3
x3 3x 2 0
x 12x 2 0 ⇒ B 2, 8
y 8a3 12a2x 2a
Tangent at B:
y x3, y 3x2
Tangent at B:
y 12a2x 16a3
y 8 12x 2
x3 12a2x 16a3 0
To find point C:
x 2a2x 4a 0 ⇒
y 12x 16 x3 12x 16
To find point C:
x3 12x 16 0
x 2 x 4 0 ⇒ C 4, 64
Area of R
2
Area of S
27 4
x3 3x 2 dx
4
2
x3 3a2x 2a3 dx
27 4 a 4
2a
12a2x 16a3 x3 dx 108a4
Area of S 16area of R
12x 16 x3 dx 108 area S 16 area R
Area of S 16area of R
2a
4a
1
Area of S
C 4a, 64a3
a
2
Area of R
47
x
9. sx
1 ft2 dt
ds 1 fx2 dx
(a) sx
ds 1 fx2 dx
(b)
dydx dx 2
ds2 1 fx2dx2 1
x
(c) sx
1
1
2
(d) s2
32t
2
1 2
2
x
dt
9 1 t dt 4
1
9 8 9 1 t dt 1 t 4 27 4
1
dx2 dy2
This is the length of the curve y x
3 2
3 2 2 1
22 13 22 13 2.0858 27 27
from x 1 to x 2.
dx
6
My
1
6
1
x
5 3 12 35 36 7
2 1
x
−2 −3
3
4
5
6 1
6
(b) m 2
2 1 dx 2 x2 x
6 1
1
5 3
6
My 2
1
35 36
x, y
3
2
1
1 1 dx 2 x3 x
y
−1
6
1 1 x 3 3 x x
m2
b
11. (a) y 0 by symmetry
127, 0
x
1 b2 1 dx 3 x b2
1 2b 1 dx x2 b
2b 1 b 2b b2 1 b2 b 1
(c) lim x lim b→
b→
2b 2 b1
x, y
b 2b 1, 0
x, y 2, 0
48
Chapter 6
Applications of Integration
13. (a) W area 2 4 6 12 (b) W area 3 1 1 2
15. Point of equilibrium: 50 0.5x 0.125x x 80, p 10
1 1 7 2 2
P0, x0 10, 80
80
50 0.5x 10 dx 1600
Consumer surplus
0
80
Producer surplus
10 0.125x dx 400
0
17. (a) Wall at shallow end From Exercise 22: F 62.42420 9984 lb (b) Wall at deep end From Exercise 22: F 62.44820 39,936 lb 20
(c) Side wall From Exercise 22: F1 62.42440 19,968 lb
F2 62.4
15
y=8 10
4
8 y10y dy
4
8y y 2 dy 624 4y 2
0
26,624 lb Total force: F1 F2 46,592 lb
x = 40
5
0
624
y
y3 3
4
x 5 10 15 20 25
0
40 45 1x y = 10
C H A P T E R 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals Section 7.1
Basic Integration Rules . . . . . . . . . . . . . . . . . . . 308
Section 7.2
Integration by Parts . . . . . . . . . . . . . . . . . . . . . 312
Section 7.3
Trigonometric Integrals . . . . . . . . . . . . . . . . . . . 321
Section 7.4
Trigonometric Substitution . . . . . . . . . . . . . . . . . 328
Section 7.5
Partial Fractions
Section 7.6
Integration by Tables and Other Integration Techniques . . 343
Section 7.7
Indeterminate Forms and L’Hôpital’s Rule
Section 7.8
Improper Integrals
Review Exercises
. . . . . . . . . . . . . . . . . . . . . . 336
. . . . . . . . 348
. . . . . . . . . . . . . . . . . . . . . 353
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363
C H A P T E R 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals Section 7.1
Basic Integration Rules
Solutions to Even-Numbered Exercises
2. (a)
d x 2 ln x2 1 C 21 x2 2x dx 1 x 1
(b)
d 2x x2 122 2x2x2 12x_ 21 3x2 C 2 dx x2 12 x2 14 x 13
(c)
1 d arctan x C dx 1 x2
(d)
2x d lnx2 1 C 2 dx x 1
x dx matches (a). x2 1
4. (a)
d 2x sinx2 1 C 2xcosx2 12x 2 sinx2 1 22x2 cosx2 1 sinx2 1 dx
(b)
d 1 1 sinx2 1 C cosx2 12x x cosx2 1 dx 2 2
(c)
d 1 1 sinx2 1 C cosx2 12x x cosx2 1 dx 2 2
(d)
d 2x sinx2 1 C 2xcosx2 12x 2 sinx2 1 22x2 cosx2 1 sinx2 1 dx
x cosx2 1 dx matches (c).
6.
2t 1 dt t2 t 2
u t 2 t 2, du 2t 1 dt Use
12.
du . u
sec 3x tan 3x dx
u 3x, du 3 dx Use
308
sec u tan u du.
8.
2 dt 2t 12 4
10.
u 2t 1, du 2dt, a 2 Use
2x x2 4
u x2 4, du 2x dx, n
du . u2 a2
Use
14.
1 dx x x2 4
u x, du dx, a 2 Use
dx
du . u u2 a2
un du.
1 2
Section 7.1
309
18. Let u t 9, du dt.
16. Let u x 4, du dx.
Basic Integration Rules
6x 45 dx 6 x 45 dx 6
x 46 C 6
2 2 dt 2 t 92 dt C t 92 t9
x 46 C
20. Let u 4 2x2, du 4x dx.
x 4 2x2 dx
22.
x
3 dx 2x 32
1 4 2x21 24xdx 4
1 4 2x23 2 C 6
x dx
3 2
x2 3 2x 31 C 2 2 1
3 x2 C 2 22x 3
24. Let u x2 2x 4, du 2x 1 dx.
x1 x2 2x 4
dx
1 x2 2x 41 22x 1 dx 2
x2 2x 4 C
26.
28.
2x dx x4
2 dx
8 dx 2x 8 ln x 4 C x4
1 1 1 1 1 1 dx 3 dx 3 dx 3x 1 3x 1 3 3x 1 3 3x 1
30.
32.
x 1
1 x
3
sec 4x dx
1 4
x 1
1 1 1 3x 1 ln 3x 1 ln 3x 1 C ln C 3 3 3 3x 1
3 1 3 2 3 dx x x x
x3
3 1 1 1 2 dx x2 3x 3 ln x C x x 2 x
sec4x4 dx
34. Let u cos x, du sin x dx.
1 ln sec 4x tan 4x C 4
sin x cos x
dx cos x1 2sin x dx 2 cos x C
2
36. Let u cot x, du csc x dx.
csc2 xecot x dx ecot xcsc2 x dx ecot x C
38.
3ex
5 dx 5 2 5
3ex
1 2
x
ee dx x
ex dx 3 2ex
5 1 2ex dx 2 3 2ex
5 ln 3 2ex C 2
2
2x 3 2 dx
310
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals sin x dx cos x
40. Let u lncos x, du
tan xln cos x dx
1 cos d sin
csc d
cot d
ln cos xtan x dx
2 1 2 dx 3sec x 1 3 sec x 1
sec x 1
sec x 1 dx
46.
2 sec x 1 dx 3 tan2 x
2 sec x 2 dx cot2 x dx 3 tan2 x 3
2 cos x 2 dx csc2 x 1 dx 3 sin2 x 3
2 1 2 2 cot x x C 3 sin x 3 3
3 dt 3 arctan t C t2 1
2 csc x cot x x C 3 1 1 50. Let u , du 2 dt. t t
48. Let u 3 x, du 3 dx.
3 1 1 dx dx 4 3x2 3 4 3x 2
3 x 1 arctan C 2 2 3
52.
54.
56.
dy tan22x, 0, 0 dx
1
x 1 4x2 8x 3 1 1 4x x2
dx
dx
2
2x 1 2x 12 1 1
5 x2 4x 4
y
(a)
(b)
1
dx
tan22xdx
e1 t 1 dt e1 t 2 dt e1 t C t2 t
dx arcsec 2x 1 C
1 5 x 22
dx arcsin
sec22x 1 dx
x 52 C
a 5
1 tan2x x C 2 1.2
0, 0: 0 C x
−0.6
0.6
y
1 tan2x x 2
− 1.2
1.2
−1
58.
− 1.2
60. r
(0, 1) 5
−2
2 −2
ln csc cot ln sin C
lncos x2 C 2
tan x dx
44.
42.
1 et2 dt et
1 2et e2t dt et
et 2 et dt et 2t et C
Section 7.1 62. Let u 2x, du 2 dx. y
1 dx x 4x2 1
Basic Integration Rules
64. Let u sin t, du cos t dt.
2 dx 2x 2x2 1
sin2 t cos t dt
0
13 sin t 3
0
0
arcsec 2x C
66. Let u 1 ln x, du
e
1
1 ln x dx x
68.
1
e
1 ln x
1
4
0
72.
1 25 x2
dx arcsin
x 5
4 0
x2 dx x
e
1
arcsin
2
1
1
2 dx x
1x dx
1 1 ln x2 2
70.
2
1 dx. x
2
x 2 ln x
1
1 ln 4 0.386
1 2
4 0.927 5 6
4 x2 1 x2 dx lnx2 4x 13 arctan C x2 4x 13 2 3 3 `
−10
10
The antiderivatives are vertical translations of each other. −6
74.
ex ex 2
3
dx
1 3x e 9ex 9ex e3x C 24
sec u tan u du sec u C
76.
The antiderivatives are vertical translations of each other. 5
−5
5
−5
78. Arctan Rule:
du 4 1 arctan C a2 u2 a a
1 82. f x x3 7x2 10x 5
5
f x dx < 0 because
80. They differ by a constant: sec2 x C1 tan2 x 1 C1 tan2 x C.
2
84.
0
4 dx 4 x2 1
86. A
sin 2x dx
0
Matches (d).
0
2
y
more area is below the x-axis than above.
2
1 cos 2x 2
0
1
y
3
1
5 2 1 2
1 0
5 x 1
−5
2
3
4
x π 4
π 2
311
312
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
2
88.
2 x2 dx
0
2
A
(c) Let f x x over the interval 0, 2. Revolve this region about the y-axis.
(b) Let f x 2 x over the interval 0, 2. Revolve this region about the x-axis.
(a) Let f x 2 x2 over the interval 0, 2.
2 x2 dx
V
V 2
2
2 x dx
0
y
f ( x) =
2π x 2
20
xx dx
0
2
2
25
2
2
0
2 x2
dx
2 x2 dx
0
0
y 15
y
10 3
f ( x) =
2x
2
f ( x) = x
5 x 1
2
2
1
1 x 1
x 1
90. (a) (b)
1 n 0
n
sinnx dx
0 3
1
n
sinnxn dx
0
1 1 dx arctan x 3 3 31 x2 6 1
3 3
y
n
1 cosnx
94.
1
0
2
y x 2 3 y
x
1 y 2 1
1 x1 x x 2 x
0
2 3x1 3
1 y 2 1
x x 1 dx
9
S 2
3
1 arctan 3 3
y 2 x
92.
2
2
4 9x 2 3
8
s
1
1
4 dx 7.6337 9x2 3
9
2
2 x 1 dx
0
3 x 1
4
2
3 2
9 0
8 10 10 1 256.545 3
y 12 9 6 3 x 3
6
Section 7.2 2.
9
12
Integration by Parts
d 2 x sin x 2x cos x 2 sin x x2 cos x 2x sin x 2x sin x 2 cos x 2 cos x x2 cos x. Matches (d) dx
Section 7.2
Integration by Parts
4.
1 d x x ln x 1 x ln x ln x. Matches (a) dx x
6.
x2 e2x dx
8.
u x2, dv e2x dx
12. dv ex dx ⇒
ln 3x dx
10.
u ln 3x, dv dx
v
ex dx ex
ux
⇒ du dx
2
x dx 2 xex dx ex
14.
x2 cos x dx
u x2, dv cos x dx
e1 t 1 dt e1 t 2 dt e1 t C t2 t
ex dx 2xex ex C
2 xex
2xex 2ex C
16. dv x 4 dx ⇒
⇒ du
u ln x
x 4 ln x dx
20. dv
v
⇒ du
2
1 dx x ln x 3
x5 x5 1 1 4 dx ln x x dx 5 x 5 5
x5
5 ln x 1 C 25
1 1 dx x2 x
1 dx x
1 ln x 1 C dx x2 x x
v
x2 1 2 x dx
1 2 x2 1
⇒ du 2x3ex 2xex dx 2xex x2 1 dx 2
2
2
x3ex x2ex dx 2 2
1 2 x 1 x2
24. dv
1 dx ⇒ v x2
u ln 2x ⇒ du
x dx ⇒
x2 1 2
u x2ex
x5 1 ln x x5 C 5 25
ln x ln x dx x2 x
22. dv
18. Let u ln x, du
1 dx x
x5 ln x 5
1 dx ⇒ x2
u ln x
x5 5
v
2
2
2
xex dx 2
2
1 1 dx x2 x
1 dx x
ln 2x ln 2x dx x2 x
2
x2ex ex ex C C 2 2 2 x 1 2 2 x 1
ln 2x 1 1 ln 2x 1 dx C C x2 x x x
1 dx. x
ln x 3
1x dx 2 1 ln x
2
C
313
314
Chapter 7
26. dv
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
1 2 3x
dx ⇒
2
2 3x 1 2 dx 2 3x 3
⇒ du dx
ux
v
x 2 3x
2x 2 3x 2 2 3x dx 3 3
dx
2x 2 3x 4 2 2 3x 2 2 3x 2 3x 3 2 C 9x 2 2 3x C
3x 4 C 3 27 27 27
28. dv sin x dx ⇒
v cos x
⇒ du dx
ux
x sin dx x cos x
cos x dx x cos x sin x C
30. Use integration by parts twice. (2) u x, du dx, dv sin x dx, v cos x
(1) u x2, du 2x dx, dv cos x dx, v sin x
x2 cos x dx x2 sin x 2 x cos x
x2 cos x dx x2 sin x 2 x sin x dx
x2 sin x 2x cos x 2 sin x C
32. dv sec tan d ⇒
v
sec tan d sec
34. dv dx
⇒ du d
u
sec tan d sec
v
sec d
v
(2) dv e x dx ⇒
e x dx e x
v
ex cos 2x dx ex cos 2x 2 ex sin 2x dx ex cos 2x 2 ex sin 2x 2 ex cos 2x dx
5 e x cos 2x dx ex cos 2x 2ex sin 2x e x cos 2x dx
38. dv dx
⇒
ex
cos 2x 2 sin 2x C 5
vx
u ln x ⇒ du
1 dx x
y ln x y
ln x dx x ln x
x
x 1 x2
e x dx e x
u sin 2x ⇒ du 2 cos 2x dx
u cos 2x ⇒ du 2 sin 2x dx
dx
dx
4 x arccos x 1 x2 C
36. Use integration by parts twice. ⇒
1 1 x2
4 arccos x dx 4 x arccos x
sec ln sec tan C
(1) dv e x dx
dx x
u arccos x ⇒ du
⇒
1 dx x ln x x C x 1 ln x C x
cos x dx
Section 7.2 40. Use integration by parts twice. (1) dv x 1 dx ⇒
v
315
2
x 1 1 2 dx x 1 3 2 3
⇒ du 2x dx
u x2
(2) dv x 1 3 2dx ⇒
v
2
x 1 3 2 dx x 1 5 2 5
⇒ du dx
ux y
Integration by Parts
x 2 x 1 dx
2 4 4 2 2 2 x2 x 1 3 2 x x 1 3 2 dx x 2 x 1 3 2 x x 1 5 2 3 3 3 3 5 5
x 1 5 2 dx
2 x 1 3 2 2 8 16 x 2 x 1 3 2 x x 1 5 2
x 1 7 2 C
15x2 12x 8 C 3 15 105 105
⇒
42. dv dx u arctan y
v
dx x
1 2 1 x ⇒ du dx dx 2 1 x 2 2 2 4 x2
arctan
x x dx x arctan 2 2
y
44. (a)
(b)
4
x 2x dx x arctan ln 4 x 2 C 4 x2 2
y x
−6
dy 18 ex 3 sin 2x, 0, dx 37
ex 3 sin 2x dx
Use integration by parts twice.
4
(1) u sin 2x, du 2 cos 2x −4
dv ex 3 dx, v 3ex 3
ex 3 sin 2x dx 3ex 3 sin 2x
6ex 3 cos 2x dx
(2) u cos 2x, du 2 sin 2x dv ex 3 dx, v 3ex 3
ex 3 sin 2x dx 3ex 3 sin 2x 6 3ex 3 cos 2x
37 ex 3 sin 2x dx 3ex 3 sin 2x 18ex 3 cos 2x C y
ex 3 sin 2x dx
1 3ex 3 sin 2x 18ex 3 cos 2x C 37
18 1 : 0 18 C ⇒ C 0 0, 18 37 37 37 y
1 x 3 3e sin 2x 18ex 3 cos 2x 37
6ex 3 sin 2x dx C
316
46.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
dy x sin x, y 0 4 dx y
48. See Exercise 3.
(0, 4)
8
1
0
−5
x2e x dx x2e x 2xe x 2e x
1 0
e 2 0.718
10 −2
50. dv sin 2x dx ⇒
1 sin 2x dx cos 2x 2
v
⇒ du dx
ux
1 1 x sin 2x dx x cos 2x 2 2
52. dv x dx
Thus,
cos 2x dx
x arcsin x2 dx
1 1 x cos 2x sin 2x C 2 4
x sin 2x dx
0
v
u arcsin x2 ⇒ du
1 sin 2x 2x cos 2x C 4
⇒
14 sin 2x 2x cos 2x
0
. 2
2x
dx
1 x4
x3 dx 1 x 4
1 x2 arcsin x2 2 1 x 4 1 2 C 2 4
1 2 x arcsin x2 1 x 4 C 2
x arcsin x2dx
0
x2 2
x dx
x2 arcsin x2 2
1
Thus,
1 2 . 4 54. Use integration by parts twice. (1) dv ex, v ex, u cos x, du sin x dx
ex cos x dx ex cos x
ex sin x dx
(2) dv ex dx, v ex, u sin x, du cos x dx
ex cos x dx ex cos x ex sin x
ex cos x dx ⇒ 2 ex cos x dx ex sin x ex cos x
Thus,
2
ex cos x dx
0
⇒
56. dv dx
x
e
e 2
sin x ex cos x 2
2
v
u ln 1 x2 ⇒ du
1
0
0
1 2
dx x
2x dx 1 x2
x ln 1 x2 2
2
ln 1 x2 dx x ln 1 x2
Thus,
sin 2 cos 2
2x2 dx 1 x2 1
1 dx x ln 1 x2 2x 2 arctan x C 1 x2
ln 1 x2 dx x ln 1 x2 2x 2 arctan x
1 0
ln 2 2
. 2
1 2 x arcsin x2 1 x 4 2
1 0
Section 7.2 58. u x, du dx, dv sec2 x dx, v tan x
x sec2 x dx x tan x
60.
4
tan x dx
2
4 ln 22 0
1 ln 2 4 2
x3 cos 2x dx x3
u and its derivatives
x3
e2x
3x2
2e2x
6x
1 2x 4e
6
8e2x
0
1 2x 16 e
2 8 16 x2 x 2 3 2dx x2 x 2 5 2 x x 2 7 2
x 2 9 2 C 5 35 315
76.
5
0
68. Yes. u ln x, dv x dx
v and its antiderivatives
u and its derivatives
x3
3x2
1 2
6x
14
cos 2x
6
18
sin 2x
0
1 16
Alternate signs
u and its derivatives
x2
2x
2 5
x 2 5 2
2
4 35
x 2 7 2
0
8 315
x 2 9 2
2
x 2 5 2 35x2 40x 32 C 315
66. Answers will vary. See pages 488, 493.
4 sin d
1
Alternate signs
1
2
1 4x3 sin 2x 6x2 cos 2x 6x sin 2x 3 cos 2x C 8
74.
v and its antiderivatives
Alternate signs
12 sin 2x 3x 41 cos 2x 6x 81 sin 2x 6161 cos 2x C
3 3 3 1 x3 sin 2x x2 cos 2x x sin 2x cos 2x C 2 4 4 8
64.
4
0
1 1 2x 1 1 x 3e2x dx x 3 e2x 3x2 e2x 6x e2x 6 e C 2 4 8 16 1 e2x 4x3 6x2 6x 3 C 8
62.
x sec x dx x tan x ln cos x
0
317
Hence,
Integration by Parts
70. No. Substitution.
cos 2x sin 2x
cos 2x
v and its antiderivatives
x 2 3 2
72. No. Substitution.
1 4 cos 4 3 sin 12 2 cos 24 sin 24 cos C 5
x 4 25 x2 3 2 dx
1,171,875 arcsin x 5 x 2x2 25 25 x2 5 2 625x 25 x2 3 2 46,875x 25 x2 128 16 64 128
14,381.0699
5 0
318
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
78. (a) dv 4 x dx ⇒
v
2
4 x 1 2 dx 4 x 3 2 3
⇒ du dx
ux
x
2 2 4 x dx x 4 x 3 2 3 3
4 x 3 2 dx
4 2 2 x 4 x 3 2 4 x 5 2 C 4 x 3 2 3x 8 C 3 15 15 (b) u 4 x ⇒ x u 4 and dx du
x 4 x dx
u 4 u1 2du
u3 2 4u1 2 du
8 2 2 u5 2 u3 2 C u3 2 3u 20 C 5 3 15
2 2
4 x 3 23 4 x 20 C 4 x 3 2 3x 8 C 15 15
80. (a) dv 4 x dx ⇒
v
4 x 1 2 dx
82. n 0:
2 4 x 3 2 3 ⇒ du dx
ux
2 2 x 4 x dx x 4 x 3 2 3 3
n 1: n 2:
4 x
3 2 dx
4 2 x 4 x 3 2 4 x 5 2 C 3 15
(b) u 4 x ⇒ x 4 u and dx du
4u1 2 u3 2 du 2 8 u3 2 u5 2 C 3 5
u xn
xex dx xex ex C xex
ex dx
x2ex dx x2ex 2xex 2ex C
n 3:
x3ex dx x3ex 3x2ex 6xex 6ex C
x3ex 3 x2e x dx
n 4: x 4ex dx x 4ex 4x3ex 12x2ex 24xex 24ex C
x 4 x dx 4 u u du
84. dv cos x dx ⇒
ex dx ex C
x2ex 2 xex dx
2 4 x 3 25x 2 4 x C 15 2 4 x 3 2 3x 8 C 15
2 3 2 u 20 3u C 15
2
4 x 3 220 3 4 x C 15
2
4 x 3 2 3x 8 C 15
v sin x
⇒ du n x n1 dx
x n cos x dx xn sin x n xn1 sin x dx
x 4ex 4 x3ex dx In general, x nex dx xnex nx n1ex dx. (See Exercise 86)
1 86. dv eax dx ⇒ v eax a u x n ⇒ du nx n1 dx
x neax dx
x neax n a a
x n1eax dx
Section 7.2
Integration by Parts
88. Use integration by parts twice. 1 v eax a
(1) dv eax dx ⇒
u cos bx ⇒ du b sin bx
eax cos bx dx
Therefore, 1
eax cos bx b a a
u sin bx ⇒ du b cos bx
eax sin bx dx
eax cos bx beax sin bx b2 2 a a2 a b2 a2
e
ax
1 v eax a
(2) dv eax dx ⇒
eax cos bx b eax sin bx b a a a a
eax cos bx dx
eax cos bx dx
eax a cos bx b sin bx a2
cos bx dx
eax a cos bx b sin bx C. a2 b2
eax cos bx dx
90. n 2 (Use formula in Exercise 84.)
x2 cos x dx x2 sin x 2 x sin x dx (Use formula in Exercise 83.) n 1
x2 sin x 2 x cos x
cos x dx x2 sin x 2x cos x 2 sin x C
92. n 3, a 2 (Use formula in Exercise 86 three times.)
x3e2x dx
x3e2x 3 2 2
1 A 9
1 9
x2e2x dx n 3, a 2
x3e2x 3 x2e2x 2 2 2
x3e2x 3x2e2x 3 xe2x 1 2 4 2 2 2
e2x 3
4x 6x2 6x 3 C 8
xe2x dx n 2, a 2
94. dv ex 3 dx ⇒ ux
v 3ex 3
e2x dx
x3e2x 3x2e2x 3xe2x 3e2x C n 1, a 2 2 4 4 8
96. A
x 3
xe
0
⇒ du dx
3
See Exercise 83.
dx
3
0
3xe 3 e x 3
3
3
x 3
0
0
1 9 9ex 3 9 e
dx
0
1 1 2 1 1 0.264 e e e 0.4
−1
4
− 0.1
−1
4
3
−1
x sin x dx x cos x sin x
0
319
320
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
98. In Example 6, we showed that the centroid of an equivalent region was
1, 8 . By symmetry, the centroid of this region is 8, 1 . π 2
You can also solve this problem directly.
1
A
0
arcsin x dx x x arcsin x 1 x2 2 2
My A
x
Mx A
y
1
x
0
1
0
y
1 0
Example 3
x 1
2 2 0 1 1
arcsin x dx 2 8
2 arcsin x arcsin x dx 1 2 2
2
100. (a) Average
1 4
(b) Average
2
1.6t ln t 1 dt 0.8t2 ln t 0.4t2 t
1
1.6t ln t 1 dt 0.8t2 ln t 0.4t2 t
3
4 3
3.2 ln 2 0.2 2.018 12.8 ln 4 7.2 ln 3 1.8 8.035
102. c t 30,000 500t, r 7%, t1 5
5
30,000 500t e0.07t dt 500
P
0
5
60 t e0.07t dt
0
Let u 60 t, dv e0.07t dt, du dt, v
60 t 1007 e
P 500
0.07t
5 0
100 0.07t e . 7
5
100 7
e0.07t dt
0
e $131,528.68 60 t 1007 e 10,000 49
500
104.
0.07t
5
0.07t
0
0
x 2 cos nx dx
5
x2 2 2x sin nx 2 cos nx 3 sin nx n n n
2 2 cos n 2 cos n n2 n
4 cos n n2
44 n n ,, ifif nn isis even odd
1 n 4 n2
2 2
106. For any integrable function, f x dx C f x dx, but this cannot be used to imply that C 0.
2 , sin x ≤ 1 ⇒ x sin x ≤ x ⇒
108. On 0,
2
0
x sin x dx ≤
2
0
x dx.
Section 7.3
Trigonometric Integrals
321
110. fx cosx, f 0 2 (a) It cannot be solved by integration.
Section 7.3
(b) You obtain the points 4
n
xn
yn
0
0
2
1
0.05
2.05
2
0.10
2.098755
3
0.15
2.146276
80
2.8403565
4.0
0
4 0
Trigonometric Integrals
2. (a) y sec x ⇒ y sec x tan x sin x sec2 x.
(b) y cos x sec x ⇒ y sin x sec x tan x sin x sec2 x sin x
Matches (iii)
sin x1 sec2 x sin x tan2 x Matches i (c) y x tan x
1 3 tan x ⇒ y 1 sec2 x tan2 x sec2 x 3 tan2 x tan2 x1 tan2 x tan4 x Matches iv
(d) y 3x 2 sin x cos3 x 3 sin x cos x ⇒ y 3 2 cos xcos3 x 6 sin x cos2 xsin x 3 cos2 x 3 sin2 x 3 2 cos4 x 6 cos2 x1 cos2 x 3 cos2 x 31 cos2 x 8 cos4 x Matches ii
4.
cos 3 x sin4 x dx
cos x1 sin2 xsin4 x dx
sin4 x sin6 xcos x dx
sin5 x sin7 x C 5 7
6. Let u cos x, du sin x dx.
sin3 x dx
1 x x 8. Let u sin , du cos dx. 3 3 3
cos3
x dx 3
cos
3
x 3
1 sin 3x dx
1 sin2
2
x 3
13 cos 3x dx
3 sin
x 1 x sin3 C 3 3 3
3 sin
x x sin3 C 3 3
sin x1 cos2 x dx
cos2 xsin x dx
1 cos3 x cos x C 3
sin x dx
322
10.
Chapter 7
sin5 t dt cos t
12.
sin2 2x dx
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
sin t1 cos2 t2cos t12 dt
sin t1 2 cos2 t cos4 tcos t12 dt 4 2 cos t12 2cos t32 cos t72 sin t dt 2cos t12 cos t52 cos t92 C 5 9
1 cos 4x 1 1 dx x sin 4x C 2 2 4
14.
sin4 2 d
1 4x sin 4x C 8
1 cos 4 2
1 cos 4 d 2
1 1 2 cos 4 cos2 4 d 4 1 cos 8 d 2
1 4
1 2 cos 4
1 4
3 1 2 cos 4 cos 8 d 2 2
1 3 1 1 sin 4 sin 8 C 4 2 2 16
1 1 3 sin 8 C sin 4 8 8 64 16. Use integration by parts twice. dv sin2 x dx
1 cos 2x x sin 2x 1 2x sin 2x ⇒ v 2 2 4 4
u x2 ⇒ du 2x dx dv sin 2x dx ⇒
1 v cos 2x 2
⇒ du dx
ux
1 1 x2 sin2 x dx x22x sin 2x 4 2
2x2 x sin 2x dx
1 1 1 1 x3 x2 sin 2x x3 2 4 3 2
x sin 2x dx
1 1 1 1 1 x3 x2 sin 2x x cos 2x 6 4 2 2 2
cos 2x dx
1 1 1 1 x3 x2 sin 2x x cos 2x sin 2x C 6 4 4 8
1 4x3 6x2 sin 2x 6x cos 2x 3 sin 2x C 24
2
18. Let u sin x, du cos x dx.
2
cos5 x dx
0
20.
0
2
1 sin2 x2 cos x dx
0
2
1 2 sin2 x sin4 x cos x dx
0
sin x
8 15
sin2 x dx
2
2 3 1 sin x sin5 x 3 5
0
1 2
2
1 cos 2x dx
0
2
1 1 x sin 2x 2 2
0
4
Section 7.3
22.
sec22x 1 dx
1 tan2x 1 C 2
24.
sec6 3x dx
28.
30. Let u sec 2t, du 2 sec 2t tan 2t
32.
36.
26.
tan2 x dx
sec2 x 1dx tan x x C
tan3 2t sec3 2t dt
34.
sec2
42. y
sin2 x cos2 x
2 1 1 tan 3x tan3 3x tan5 3x C 3 9 15
tan5 2x sec2 2x dx
tan3 3x dx
x x x 1 x tan dx 2 tan sec2 dx 2 2 2 2 2
tan2 x sec5 x
1 2 tan2 3x tan4 3xsec2 3x dx
1 x x x sec2 dx C tan4 2 2 2 2
x C 2
tan2
38.
1 tan2 3x2 sec2 3x dx
1 tan6 2x C 12
sec5 2t sec3 2t C 10 6
sec4 2t sec2 2tsec 2t tan 2t dt
sec2 or
323
sec2 2t 1sec3 2t tan 2t dt
x x x 1 x x tan dx 2 sec sec tan dx 2 2 2 2 2 2
sec2
tan3
Trigonometric Integrals
sec2 3x 1tan 3x dx
1 3 sin 3x 1 tan 3x3 sec2 3x dx dx 3 3 cos 3x
1 2 1 tan 3x ln cos 3x C 6 3
x C 2
cos2 d 2 2
cos5 x dx
40. s
sin2 x cos3 x dx
sin2 x1 sin2 xcos x dx
sin2 x sin4 xcos x dx
sin 2 1 C 8 2
1 2 sin 2 C 16
sin3 x sin5 x C 3 5
tan x sec4 x dx
tan12 xtan2 x 1sec2 x dx
tan52 x tan12 x sec2 x dx
2 2 72 tan x tan32 x C 7 3
sin2
1 cos 2
1 4
1 2cos d 1 cos 4
sin2 d
1 8
1 cos 2 d
2
d
324
Chapter 7
44. (a)
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
y
(b)
1
y
y
tan3 x C 3
x
−1
1
( ) 1
0, 4
−1
1 dy sec2 x tan2 x, 0, dx 4
sec2 x tan2 x dx u tan x, du sec2 x dx
0, 41: 41 C ⇒ y 31 tan
3
46.
dy 3y tan2 x, y0 3 dx 8
48.
cos 4 cos3 d
(0, 3)
−1
1
x
1 4
cos 4 cos 3 d
1 cos 7 cos d 2
sin 7 sin C 14 2
−2
50.
x 1 x 52. Let u tan , du sec2 dx. 2 2 2
sin4x cos 3x dx sin 4x cos 3x dx
1 2
1 1 cos x cos 7x C 2 7
sin x sin 7xdx
tan4
x x sec4 dx 2 2
csc2 3x cot 3x dx
1 3
56.
tan4
2
1 7 cos x cos 7x C 14
54. u cot 3x, du 3 csc2 3x dx
cot3 t dt csc t
cot 3x3 csc2 3x dx
1 cot2 3x C 6
x x x tan2 1 sec2 dx 2 2 2
tan6
x x tan4 2 2
12 sec 2x dx 2
x 2 2 7x tan tan5 C 7 2 5 2
cos3 t dt sin2 t
cos t dt sin2 t
1 sin2 tcos t dt sin2 t
cos t dt
1 sin t C sin t
csc t sin t C
58.
60.
sin2 x cos2 x dx cos x
1 sec t dt cos t 1
1 2 cos 2 x dx cos x
sec x 2 cos x dx ln sec x tan x 2 sin x C
cos t 1 dt cos t 1 cos t
62.
0
3
0
sec2 ttan t dt
2
3 tan t 32
0
2 3
3
sec2 x 1 dx
0
3
tan x x
66. 4
tan2 x dx
sec t dt ln sec t tan t C
64. Let u tan t, du sec2 t dt. 4
0
sin 3 cos d
1 2
3
3
sin 4 sin 2 d
1 1 1 cos 4 cos 2 2 4 2
0
Section 7.3
2
68.
sin2 x 1 dx
2
70.
sin2 x cos2 x dx
2
2
2
2
Trigonometric Integrals
325
2x 1 dx 1 cos 2 2
32 21 cos 2x dx 32 x 41 sin 2x
1 4x sin 4x C 32
72.
2
3 2
tan31 x dx
tan21 x ln cos1 x C 2
2 2
−6
6
−3
3
−2
74.
−2
sec41 x tan1 x dx
2
sec41 x C 4
76.
1 cos 2 d
0
2
− 3.5
3
1
3 2 4
3.5
−2
2
78.
sin6 x dx
0
2
1 5x 3 1 2 sin 2x sin 4x sin3 2x 8 2 8 6
0
5 32
80. See guidelines on page 500. 82. (a) Let u tan x, du sec2 x dx.
sec2 x tan x dx
(b)
8
1 2 tan x C1 2 −4
Or let u sec x, du sec x tan x dx.
(c)
1 1 1 1 1 sec2 x C tan2 x 1 C tan2 x C tan2 x C2 2 2 2 2 2
84. Disks
y
Rx tan x rx 0 V 2
1 1 2
4 2
tan x dx
− 12
0
2
4 −2
1 sec xsec x tan x dx sec2 x C 2
−1
4
sec x 1 dx 2
0
2 1
4
2 tan x x
0
1.348 4
π 8
π 4
x
2
2 2 sin 4 sin 2
0
326
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
2
86. (a) V
cos2 x dx
0
(b) A
2
2
2
0
2
cos x dx sin x
0
1 cos 2x dx
0
1 x sin 2x 2 2
2
0
2 4
1
Let u x, dv cos x dx, du dx, v sin x. x
2
0
y
1 2 1 4
0
2
2
sin x dx x sin x cos x
0
0
2 1 2 2
2
cos2 x dx
y
0
2
1
1 cos 2x dx
0
2
1 1 x sin 2x 4 2
x, y
2
x cos x dx x sin x
0
( π 2− 2 , π8 (
1 2
8
π 4
2 2, 8
88. dv cos x dx ⇒
π 2
x
v sin x
u cosn1 x ⇒ du n 1cosn2 x sin x dx
cosn x dx cosn1 x sin x n 1 cosn2 x sin2 x dx cosn1 x sin x n 1 cosn2 x1 cos2 x dx
cosn1 x sin x n 1 cosn2 x dx n 1 cosn x dx
Therefore, n cosn x dx cosn1 x sin x n 1 cosn2 x dx cosn x dx
cosn1 x sin x n 1 n n
cosn2 x dx.
90. Let u secn2 x, du n 2secn2 x tan x dx, dv sec2 x dx, v tan x.
secn x dx secn2 x tan x
n 2 secn2 x tan2 x dx
secn2 x tan x n 2 secn2 xsec2 x 1 dx secn2 x tan x n 2
secn x dx
secn2 x dx
n 1 secn x dx secn2 x tan x n 2 secn2 x dx secn x dx
92.
cos4 x dx
1 n2 secn2 x tan x n1 n1
cos3 x sin x 3 4 4
cos2 x dx
secn2 x dx
cos3 x sin x 3 cos x sin x 1 4 4 2 2
dx
1 3 3 1 cos3 x sin x cos x sin x x C 2 cos3 x sin x 3 cos x sin x 3x C 4 8 8 8
Section 7.3
94.
sin4 x cos2 x dx
cos3 x sin3 x 1 6 2
cos2 x sin2 x dx
cos3 x sin x 1 cos3 x sin3 x 1 6 2 4 4
cos2 x dx
x 1 1 cos x sin x 1 C cos3 x sin3 x cos3 x sin x 6 8 8 2 2
1 8 cos3 x sin3 x 6 cos3 x sin x 3 cos x sin x 3x C 48
96. (a) n is odd and n ≥ 3.
2
cosn x dx
0
cosn1 x sin x n
2
0
2
n1 n
n1 n
n 2
n1 n
n2n4
n1 n
n2n4...
cosn3 x sin x n2
n3 n3 n3
2
0
n3 n2
cosn5 x sin x n4
n5
cosn2 x dx
0
2
0
0
n5
2
cos x dx
0
n3
n5
n2n4...1
0
(Reverse the order)
234567. . .n n 1
234567. . .n n 1 n1 n
n3
n5
n2n4...
n1 n
n3
2
cos2 x dx
(From part (a).)
0
n5
2
n 2 n 4 . . . 2 4 sin 2x 0
n1 n
2 123456. . .n n 1
123456. . .n n 12
n3
n5
x
n2n4... 4
1
(Reverse the order)
cosn6 x dx
0
2
(b) n is even and n ≥ 2.
0
n5 n4
2
n1 n
cosn x dx
cosn4 x dx
0
cosn6 x dx
2
2
2
n n 1 nn 32 nn 54 . . . sin x
1
n1 n
Trigonometric Integrals
327
328
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
Section 7.4 2.
Trigonometric Substitution
1 x 1 d x x2 16 1 1 x 8 ln x2 16 x xx2 16 C 8 x2 16 2 2 dx 2 2 2 x 16 x x 16
4.
x2 x2 16
x2 16 8 x x2 16 x2 2 2 2 x 16 x 16 x 2x 16
16 x2 x2 16 2x2 16
Indefinite integral:
2
x2 x2 16
Matches (d)
x 3 x 37 6x x2 1 d 8 arcsin C 8 dx 4 2 1 x 34 2
8 16 x 3
2
4 2 x 3 1
3x
1
7 6x x2
16 x 32 x 32 2 2 216 x 3
16 x2 6x 9 16 x2 6x 9 216 x 32
2 16 x 32 216 x 32
16 x 32
Indefinite integral:
7 6x x2
7 6x x2 dx
Matches (c)
6. Same substitution as in Exercise 5.
10 x225
x2
dx 10
5 cos d
25
sin2
5 cos
2 2 225 x2 csc2 d cot C C 5 5 5x
8. Same substitution as in Exercise 5
x2 dx 25 x2
25 sin2 25 5 cos d 5 cos 2
1 25 25 sin 2 C sin cos C 2 2 2
25 x x arcsin 2 5 5
25 x2
5
10. Same substitution as in Exercise 9
x2 4
x
1 cos 2 d
dx
2 tan 2 sec tan d 2 2 sec
2tan C 2
x2 4
2
C 21 25 arcsin 5x x
25 x2 C
tan2 d 2 sec2 1 d arcsec
1 7 6x x2 2
2x C x
2
4 2 arcsec
2x C
Section 7.4 12. Same substitution as in Exercise 9
x3 dx 4
x2
8 sec3 2 sec tan d 8 sec4 d 2 tan
8 1 tan2 sec2 d 8 tan
8 x2 4 3 2
3 x 4 4 C 31 2
14. Same substitution as in Exercise 13.
Trigonometric Substitution
tan3 8 C tan 3 tan2 C 3 3
1 3
x2 4 12 x2 4 C x2 4 x2 8 C
9x3 tan3 sec3 dx 9 sec2 d 9 sec2 1sec tan d 9 sec C 2 1 x sec 3
3 sec sec2 3 C 31 x2 1 x2 3 C 31 x2x2 2 C 16. Same substitution as in Exercise 13
x2 dx 1 x22
x2 dx 1 x2 4
1 2
tan2 sec2 d sec4
1 cos 2 d
sin2 d
1 sin 2 1 sin cos C 2 2 2
x 1 arctan x 2 1 x2
1 1 x2
C 21 arctan x 1 x x C 2
18. Let u x, a 1, and du dx.
1 x2 dx
20.
1 x1 x2 ln x 1 x2 C 2
x 1 dx 9 x21 22x dx 2 9 x 2 9 x21 2 C
22.
1 25 x2
dx arcsin
x C 5
(Power Rule)
24. Let u 16 4x2, du 8x dx.
x16 4x2 dx
1 8
16 4x21 28x dx
1 2 16 4x23 2 C 4 x23 2 C 12 3
26. Let u 1 t 2, du 2t dt.
t 1 dt 1 t 23 2 2
28. Let 2x 3 tan , dx
4x2 9
x4
1 t 23 22t dt
C
3 sec2 d, 4x2 9 3 sec . 2
dx
8 9
8 C 27 sin3
3 sec 3 2 sec2 d 3 24 tan4
1 1 t 2
cos d sin4
8 csc3 C 27
4x2 93 2 C 27x3
4x 2 + 9 2x
θ 3
329
330
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
30. Let 2x 4 tan , dx 2 sec2 d, 4x2 16 4 sec .
1 dx x4x2 16
1 4
2 sec2 d 2 tan 4 sec
sec 1 d tan 4
x2 + 4
x
csc d
θ
2
1 1 x2 4 2 ln csc cot C ln C 4 4 x
32. Let x 3 tan , dx 3 sec2 d, x2 3 3 sec2 .
x2
1 dx 33 2
1 3
3 sec2 d
33 sec3
x2 + 3
x
cos d
θ 3
1 sin C 3
x C 3x2 3
34. Let u x2 2x 2, du 2x 2 dx.
x 1x2 2x 2 dx
1 2
1 x2 2x 21 22x 2 dx x2 2x 23 2 C 3
36. Let x sin , x sin2 , dx 2 sin cos d, 1 x cos .
1 x x
dx
cos 2 sin cos d sin 1
2 cos2 d
x
θ 1− x
1 cos 2 d
sin cos C arcsinx x1 x C 38. Let x tan , dx sec2 d, x2 1 sec2 .
1 x3 x 1 dx x 4 2x2 1 4
x4
4x3 4x dx 2x2 1
1 lnx 4 2x2 1 4
1 1 lnx2 1 2 2
1 dx x2 12
d sec4
θ 1
1 cos 2 d
1 1 lnx2 1 sin cos C 2 2
1 x lnx2 1 arctan x 2 C 2 x 1
x2 + 1 x
sec2
Section 7.4
40. u arcsin x, ⇒ du
x arcsin x dx
1 x2 dx, dv x dx ⇒ v 2 2 1 x
x2 1 x2 arcsin x dx 2 2 1 x2
x sin , dx cos d, 1 x2 cos
x arcsin x dx
Trigonometric Substitution
x2 1 arcsin x 2 2
x2 1 sin2 cos d arcsin x cos 2 4
1 cos 2 d
1 1 1 x2 x2 arcsin x sin 2 C arcsin x sin cos C 2 4 2 2 4
1 1 x2 arcsin x arcsin x x1 x2 C 2x2 1 arcsin x x1 x2 C 2 4 4
42. Let x 1 sin , dx cos d, 1 x 12 2x x2 cos .
x2 2x x2
dx
x2
1 x 12
dx 1
1 sin 2cos d cos
1 2 sin
sin2
x−1
θ 1 − (x − 1)2
d
3 1 2 sin cos 2 d 2 2
1 3 2 cos sin 2 C 2 4 1 3 2 cos sin cos C 2 2
1 3 arcsinx 1 22x x2 x 12x x2 C 2 2
3 1 arcsinx 1 2x x2x 3 C 2 2
44. Let x 3 2 sec , dx 2 sec tan d, x 32 4 2 tan .
x dx x 6x 5 2
x dx x 32 4
2 sec 3 2 sec tan d 2 tan
2 sec2 3 sec d
2 tan 3 ln sec tan C1
2
x 33 4
2
3 lnx 2 3
x 32 4
2
C1
x2 6x 5 3 ln x 3 x2 6x 5 C
331
332
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
46. Same substitution as in Exercise 45 (a)
1 dt 1 t25 2
3 2
Thus,
0
cos d cos5
sec4 d
tan2 1 sec2 d
1 3 1 t tan tan C 3 3 1 t 2
t 1 t3 2 5 2 dt 1 t 31 t 23 2 1 t 2
3
t 1 t 2
C
3 2
0
3 2 33 8 3 3 23 3.464. 31 43 2 1 4
(b) When t 0, 0. When t 3 2, 3. Thus,
3 2
0
1 1 3 dt tan tan 1 t25 2 3
48. (a) Let 5x 3 sin , dx
1 3 3 3 23 3.464. 3
3 cos d, 9 25x2 3 cos . 5
9 1 cos 2 d 5 2
0
3 3 cos cos d 5
9 25x2 dx
9 1 sin 2 C 10 2
9 sin cos C 10
5x 5x 9 arcsin 10 3 3
3 5
Thus,
3
9 25x2 dx
0
9 25x2
3
C
9 5x 5x9 25x2 arcsin 10 3 9
3 5
0
9 9 . 10 2 20
3 (b) When x 0, 0. When x , . 5 2
3 5
Thus,
9 25x2 dx
2
109 sin cos
0
0
9 9 . 10 2 20
50. (a) Let x 3 sec , dx 3 sec tan d, x2 9 3 tan .
x2 9
x2
dx
3 tan 3 sec tan d 9sec2 tan2 d sec sin2 d cos 1 cos2 d cos
sec cos d
ln sec tan sin C ln —CONTINUED—
x 3
x2 9 9 C 3 x
x2
Section 7.4
Trigonometric Substitution
333
50. —CONTINUED—
6
Hence,
x2 9
x2
3
dx ln
. 3
(b) When x 3, 0; when x 6,
6
Hence,
x2 9
x2
3
52.
54.
x2 9 x2 9 x 3 3 x
dx ln sec tan sin
3
0
6
3
ln 2 3
3
ln 2 3
2
3
2
.
.
1 75 x2 2x 113 2 dx x 1x2 2x 26x2 2x 11 ln x2 2x 11 x 1 C 4 2
x2x2 4 dx
1 3 2 1 x x 4 xx2 4 2 ln x x2 4 C 4 2
56. (a) Substitution: u x2 1, du 2x dx (b) Trigonometric substitution: x sec 58. (a) x2 y k2 25
1 (b) Area square circle 4
y
Radius of circle 5 k 2 52 52 50
1 25 52 25 1 4 4
(0, k)
k 52
5
1 (c) Area r 2 r 2 r 2 1 4 4
5 5
x
60. (a) Place the center of the circle at 0, 1; x2 y 12 1. The depth d satisfies 0 ≤ d ≤ 2. The volume is
V32 6
d
1 y 12 dy
0
1 arcsiny 1 y 11 y 12 2
d
(Theorem 7.2 (1)) 0
3 arcsind 1 d 11 d 12 arcsin1 (b)
3 3 arcsind 1 3d 12d d 2. 2
d
10
V6
(d)
1 y 12 dy
0
dV dV dt dd 0
dd 61 d 12 dt
1
d t 4
2 0
(c) The full tank holds 3 9.4248 cubic meters. The horizontal lines y
⇒ d t (e)
1 241 d 12
0.3
9 3 3 , y , y 4 2 4
intersect the curve at d 0.596, 1. 0, 1.404. The dipstick would have these markings on it.
0
2 0
The minimum occurs at d 1, which is the widest part of the tank.
334
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
62. Let x h r sin , dx r cos d, r 2 x h2 r cos .
y
Shell Method:
hr
V 4
xr 2 x h2 dx
x
hr
4
2
2
h2
2
4 r 2
2
1 cos 2 d r
2
2 r 2h
h r sin r cos r cos d 4 r 2
1 sin 2 2
2
2
h−r
2
4 r 3
h r sin cos2 d
sin cos2 d
3
h+r
2
2
cos3
h
2
r
2 2 r 2h
2
x−h
θ r 2 − (x − h)2
1 64. y x2, y x, 1 y 2 1 x2 2
4
s
1 x2 dx
0
12xx
2
4
1 ln x x2 1
(Theorem 7.2)
0
1 417 ln4 17 9.2936 2
66. (a) Along line: d1 a2 a4 a1 a2
y
Along parabola: y x2, y 2x
(a, a 2)
a
d2
1 4x2 dx
0
y = x2
a
1 2x4x2 1 ln 2x 4x2 1 4
(Theorem 7.2)
0
x
(0, 0)
1 2a4a2 1 ln2a 4a2 1 4
(b) For a 1, d1 2 and d2
5
2
1 ln2 5 1.4789. 4
For a 10, d1 10101 100.4988 d2 101.0473. (c) As a increases, d2 d1 → 0. 68. (a)
(b) y 0 for x 72
25
− 10
80 −5
(c) y x
x2 x x , y 1 , 1 y 2 1 1 72 36 36
72
s
1 1
0
36 2
x 36
2
72
dx 36
18
x 36
0
1 1
x 36
2
x 36
1 1
2 ln 1 2
1
2
ln 1
361 dx 2
x 36
1 1
36
2 ln 1 2
x 36
2 18 ln
2
72 0
2 1 2 1
82.641
Section 7.4
Trigonometric Substitution y
70. First find where the curves intersect. y 2 16 x 42
6
1 4 x 16
(4, 4) 4 2
16 2 16x 4 2 x4
x −2 −2
16 2 16x 2 128x 162 x4
x4
16x2
2
4
0
−6
128x 0
1 1 1 2 x dx 42 x3 4 4 12
4
My
x
0
x4 16
1 2 x dx 4
4 0
4
y=
16 4 3
8
x16 x 42 dx
4
8
4
0
8
x 416 x 42 dx
4
16
416 x 42 dx
4
13 16 x 4
8
2 3 2
4
2 16 arcsin
x4 x 416 x 42 4
1 16 163 2 2 16 3 2
16 16 643 16 112 3
4
Mx
0
dx
8
2
1 1 2 x 2 4
4
1 16 x 42 dx 2
321 x5 8x x 6 4
64 416 32 64 32 5 6 15
5 4
3 8
0
4
x
My 112 3 16 112 48 28 12 4.89 A 16 3 4 16 12 4 3
y
Mx 416 15 104 1.55 A 16 3 4 54 3
x, y 4.89, 1.55 72. Let r L tan , dr L sec2 d, r 2 L2 L2 sec2 . 1 R
R
0
b
2mL 2mL dr r 2 L23 2 R 2m RL
a
L sec2 d L3 sec3
b
cos d
RL sin
2m RL
b
a
r r 2 L2
2m LR 2 L2
r 2 + L2 r
θ L
a
2m
6
10
−4
xx 4x2 4x 32 ⇒ x 0, 4 A
4
R 0
8 4
16 − (x − 4)
2
335
336
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
0.8
74. (a) Finside 48
1
0.8 y21 y 2 dy
0.8
96 0.8
1
0.8
1 y2 dy
1
y1 y 2 dy
2 arcsin y y1 y 31 y 0.8
96
0.8
1
2
961.263 121.3 lbs
2 32
1
0.4
(b) Foutside 64
1
0.4 y21 y 2 dy
0.4
128 0.4 128
1
0.4
1 y 2 dy
y1 y 2 dy
1
0.4 arcsin y y1 y 2 31 1 y 232 2
76. S 1520.4 111.2t 15.8t2 (a)
0.4
1
92.98
78. False
60
x2 1
x
dx
0
7
tan sec tan d sec
tan2 d
0
1 (b) St 1520.4 111.2t 15.8t212111.2 31.6t 2 S5 2.71 (c) Average value
80. True
1
1
St dt 68.24
x21 x2 dx 2
1 2
12
10
1
x21 x2 dx 2
0
Section 7.5
sin2 cos cos d 2
0
2
sin2 cos2 d
0
Partial Fractions
2.
4x 2 3 A B C x 53 x 5 x 52 x 53
6.
2x 1 A Bx C Dx E 2 2 xx 2 12 x x 1 x 12
8.
1 1 A B 4x 2 9 2x 32x 3 2x 3 2x 3
4.
1 A2x 3 B2x 3 3 2,
1 1 dx 4x2 9 6
2x 1 3 dx 2x 1 3 dx
1 ln 2x 3 ln 2x 3 C 12
1 2x 3 ln C 12 2x 3
10.
x2 x2 A B x 2 4x 3 x 1x 3 x 1 x 3
x1 dx x2 4x 3
1 6.
When x 1 6A, A When x 32 , 1 6B, B 16 .
2
x 1 dx x 1x 3
1 dx ln x 3 C x3
Section 7.5
12.
Partial Fractions
337
5x2 12x 12 A B C xx 2x 2 x x2 x2 5x2 12x 12 Ax2 4 Bxx 2 Cxx 2 When x 0, 12 4A ⇒ A 3. When x 2, 16 8B ⇒ B 2. When x 2, 32 8C ⇒ C 4.
14.
5x2 12x 12 dx x3 4x
3 dx x
2 dx x2
4 dx 3 ln x 2 ln x 2 4 ln x 2 C x2
x3 x 3 2x 1 A B x1 x1 x2 x 2 x 2x 1 x2 x1 2x 1 Ax 1 Bx 2 When x 2, 3 3A, A 1. When x 1, 3 3B, B 1.
x3 x 3 dx x2 x 2
16.
x1
1 1 dx x2 x1
x2 x2 x ln x 2 ln x 1 C x ln x 2 x 2 C 2 2
x2 A B xx 4 x 4 x
18.
2x 3 A B x 12 x 1 x 12
x 2 Ax Bx 4
2x 3 Ax 1 B
When x 4, 6 4A, A 32 . When x 0, 2 4B, B 12 .
x2 dx x 2 4x
When x 1, B 1. When x 0, A 2.
32 12 dx x4 x
2x 3 dx x 12
2 1 dx x 1 x 12
2 ln x 1
3 1 ln x 4 ln x C 2 2
20.
1 C x1
4x2 4x2 4x2 A B C x3 x2 x 1 x2x 1 x 1 x2 1x 1 x 1 x 1 x 12 4x2 Ax 12 Bx 1x 1 Cx 1 When x 1, 4 2C ⇒ C 2. When x 1, 4 4A ⇒ A 1. When x 0, 0 1 B 2 ⇒ B 3.
x3
4x2 dx x1 x2
1 dx x1
3 dx x1
ln x 1 3 ln x 1
22.
2 dx x 12
2 C x 1
6x A 6x Bx C x3 8 x 2x2 2x 4 x 2 x2 2x 4 6x Ax2 2x 4 Bx Cx 2 When x 2, 12 12A ⇒ A 1. When x 0, 0 4 2C ⇒ C 2. When x 1, 6 7 B 21 ⇒ B 1.
6x dx x3 8
1 dx x2
x 2 dx x2 2x 4
1 dx x2
3 arctan
x 1 dx x2 2x 4
1 3 x1 arctan C ln x2 2x 4 3 2 3
1 ln x2 2x 4 2
ln x 2 ln x 2
3x 1
3
C
3 dx x2 2x 1 3
338
24.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
x2 x 9 Ax B Cx D 2 2 x2 92 x 9 x 92 x2 x 9 Ax Bx2 9 Cx D Ax3 Bx2 9A Cx 9B D By equating coefficients of like terms, we have A 0, B 1, D 0, and C 1.
x2 x 9 x2 92
26.
1 dx x2 9
x dx x2 92
x 1 1 arctan C 3 3 2x2 9
x 2 4x 7 A Bx C x 1x 2 2x 3 x 1 x 2 2x 3 x 2 4x 7 Ax 2 2x 3 Bx Cx 1 When x 1, 12 6A. When x 0, 7 3A C. When x 1, 4 2A 2B 2C. Solving these equations we have A 2, B 1, C 1.
x 2 4x 7 dx 2 x3 x 2 x 3
1 dx x1
2 ln x 1
28.
x 1 dx x 2 2x 3
1 ln x 2 2x 3 C 2
x 2 x 3 Ax B Cx D 2 2 x 2 32 x 3 x 32 x 2 x 3 Ax Bx 2 3 Cx D Ax 3 Bx 2 3A Cx 3B D By equating coefficients of like terms, we have A 0, B 1, 3A C 1, 3B D 3. Solving these equations we have A 0, B 1, C 1, D 0.
x2 x 3 dx x4 6x 2 9
30.
1 x dx x 2 3 x 2 32
1 3
arctan
x 3
1 C 2x 2 3
x1 A B C 2 x 2x 1 x x x1 x 1 Axx 1 Bx 1 Cx 2 When x 0, B 1. When x 1, C 2. When x 1, 0 2A 2B C. Solving these equations we have A 2, B 1, C 2.
5
1
x1 dx 2 x 1
x2
5
1
1 dx x
5
1
5
1 dx 2 x2
1
1
1 2 ln x 2 ln x 1 x
2 ln 2 ln
x 1 x1 x
5 4 3 5
5 1
1 dx x1 5
Section 7.5
x2
x1 2 7 6x 2 1 1 dx 3 ln C x 2x 13 x x x 1 2x 12
1
32.
0
34.
x2 x dx x1
1
0
0
x2
2x 1 dx x ln x 2 x 1 x1
0 1 ln 3
1
2, 1: 3 ln
1
dx
Partial Fractions
4
1 1 2 7 1 C 1⇒ C 2 3 ln 2 2 1 2 2
(2, 1)
− 4.7
4.7
−4
36.
x2
3, 4:
x3 1 2 C dx ln x 2 4 2 4 2 2 x 4
8
2 22 1 1 ln 5 C 4 ⇒ C ln 5 2 5 5 2
(3, 4) −8
8 −2
38.
x2x 9 1 5 dx 2 ln x 2 C x 3 6x 2 12x 8 x 2 x 22
10
3, 2: 0 1 5 C 2 ⇒ C 4
(3, 2) − 10
10 −3
40.
x2 x 2 dx arctan x ln x 1 C x3 x2 x 1
10
2, 6: arctan 2 0 C 6 ⇒ C 6 arctan 2
(2, 6) −2
5 −2
42. Let u cos x, du sin x dx.
44.
1 A B uu 1 u u1
1 Au 1 Bu
1 Au 1 Bu When u 0, A 1. When u 1, B 1, u cos x. du sin dx.
sin x 1 dx du cos x cos2 x uu 1
1 A B , u tan x, du sec2 x dx uu 1 u u1
1 du u
ln u 1 ln u C
ln
u1 C u
ln
cos x 1 C cos x
sec2 x dx tan xtan x 1
1 du u1
When u 0, A 1. When u 1, 1 B ⇒ B 1.
ln 1 sec x C
1 du uu 1
1 1 du u u1
ln u ln u 1 C
ln
u C u1
ln
tan x C tan x 1
339
340
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
46. Let u ex, du ex dx. 1 A Bu C 2 u2 1u 1 u 1 u 1 1 Au2 1 Bu Cu 1 When u 1, A 12 . When u 0, 1 A C. 1 1 1 When u 1, 1 2A 2B 2C. Solving these equations we have A 2 , B 2 , C 2 , u e x, du e x dx.
48.
e
2x
ex dx 1e x 1
1 2
1 1 ln u 1 ln u2 1 arctan u C 2 2
1 2 ln e x 1 ln e 2x 1 2 arctan e x C 4
1 du u2 1u 1
u 1 1 du uu 11 du 2
1 A B a2 x2 a x a x
1 A C B 2 x2a bx x x a bx
50.
1 Axa bx Ba bx Cx 2
1 Aa x Ba x
When x 0, 1 Ba ⇒ B 1a. When x ab, 1 Ca2b2 ⇒ C b2a2. When x 1, 1 a bA a bB C ⇒
When x a, A 12a. When x a, B 12a.
1 1 dx a2 x2 2a
1 1 dx ax ax
1 ln a x ln a x C 2a
A ba 2. 1 dx x a bx 2
1 ax ln C 2a a x
52.
dy 4 , y0 5 dx x2 2x 3
54. (a)
8
(b)
−1
56. A 2
0
y
3
2
3
dx 14
0
2x 6
0
5 2
2 3 2
1 dx 16 x2
14 4 x ln 8 4x
7 ln 7 2.595 4
3 0
x
(From Exercise 46)
−3 −2 −1
1 b b ln x ln a bx C a2 ax a2
a bx b 1 ln C ax a2 x
1 x b ln C ax a2 a bx
Nx A1 B1x An Bn x . . . Dx ax2 bx c ax2 bx cn
−2
7 1 dx 16 x2
ba2 1a b2a2 2 dx x x a bx
Nx A1 A2 Am . . . Dx px q px q2 px qm
3
3
1
2
3
Section 7.5 58. (a)
Partial Fractions
341
(e) k 1, L 3
y 6
i y0 5: y
5 4
15 5 2e3t
1 32 3 ii y0 : y 2 12 52e3t 1 5e3t
3 2 1
5
t 1
2
3
4
5
6
(b) The slope is negative because the function is decreasing. (c) For y > 0, lim yt 3.
0
B A dy yL y y Ly
(d)
dy kyL y dt
(f )
dy dy d 2y k y L y 0 dt 2 dt dt
1 1 1 AL y By ⇒ A , B L L
dy yL y
1 L
1 dy y
5 0
t →
⇒ y
k dt
1 dy Ly
⇒
L 2
From the first derivative test, this is a maximum.
1 ln y ln L y kt C1 L
ln
y k L t LC1 Ly C2ekLt
When t 0,
y Ly
y y0 y0 C2 ⇒ ekLt. L y0 L y L y0 y0L . y0 L y0ekLt
Solving for y, you obtain y
3
60. (a) V
0
2x 2 dx 4 2 x 1
3
0
3
4
0
x2
x2 dx 12
1 1 dx x2 1 x2 12
(partial fractions)
1 x arctan x 2 2 x 1
x x2 1
4 arctan x 2 arctan x —CONTINUED—
dy dy L y dt dt y
k dt
3 0
3
(trigonometric substitution)
0
2 arctan 3
3 5.963 10
342
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
60. —CONTINUED—
3
(b) A
0
x
1 A
x2 3
0
2x2 1 dx 2 x 1 ln 10
2 ln 10
3
0
3
0
0
ln 10
3
2
0
3
1 2x 2 arctan x ln 10 1 1 y A 2
3
2x dx lnx2 1 1
0
y
2
2 dx x2 1
(1.521, 0.412) 1
2 3 arctan 3 1.521 ln 10
2 2x 2 dx x2 1 ln 10
(partial fractions)
1 x 2 arctan x arctan x 2 ln 10 2 x 1
2 1 x arctan x ln 10 2 2x2 1
3 0
3
x2 dx x2 12
1 1 dx x2 1 x2 12
2
−1
3
0
x 1
3
(trigonometric substitution)
0
1 x arctan x 2 ln 10 x 1
3 0
1 3 arctan 3 0.412 ln 10 10
x, y 1.521, 0.412 B 1 1 A 1 ,A ,B y0 xz0 x y0 x z0 x z0 y0 z0 y0
62. (a) 1 z0 y0
(Assume y0 z0)
1 1 dx kt C y0 x z0 x
1 z x ln 0 kt C, when t 0, x 0 z0 y0 y0 x C
z 1 ln 0 z0 y0 y0
kt
1 z x z ln 0 ln 0 z0 y0 y0 x y0
y0z0 x z0 y0k t 0 0 x
z y
ln
y0z0 x ez0 y0kt z0 y0 x x (b) (1) If y0 < z0, lim x y0. t →
(2) If y0 > z0, lim x z0. t →
y0z0 ez0 y0kt 1 z0ez0 y0kt y0 (c) If y0 z0, then the original equation is
1 dx y0 x 2
k dt
y0 x1 kt C1 x 0 when t 0 ⇒
1 C1 y0
kt y0 1 1 1 kt y0 x y0 y0 y0 x
y0 kt y0 1
x y0
y0 kt y0 1
As t → , x → y0 x0.
Section 7.6
Section 7.6
Integration by Tables and Other Integration Techniques
Integration by Tables and Other Integration Techniques
2. By Formula 13: b 2, a 5
2 2 1 4 1 5 4x x dx ln 3 x22x 52 3 25 x5 2x 5 2x 5
2 4x 5 x 8 ln C 375 2x 5 75 x2x 5
4. By Formula 29: a 3
C
6. By Formula 41:
1 1 x 9 x dx x2 9 arcsec C 3 x 3 3 2
8. By Formulas 51 and 47:
cos3 x dx 2
x
u x, du 10. By Formula 71:
1 1 dx 1 tan 5x 5
2
x
9 x 4
2 1 x dx
x sin x
3
1 dx 2 x
1 1 u ln cos u sin u C 5 2
1 5x ln cos 5x sin 5x C 10
u 5x, du 5 dx
12. By Formula 85:
a 21, b 2
ex 2 sin 2x dx
ex 2 1 sin 2x 2 cos 2x C 1 4 4 2
4 x 2 1 e sin 2x 2 cos 2x C 17 2
14. By Formulas 90 and 91:
ln x3 dx xln x3 3 ln x2 dx xln x3 3x 2 2 ln x ln x2 C x ln x3 3ln x2 6 ln x 6 C
16. (a) By Formula 89:
x 4 ln x dx
x5 x5 1 5 x ln x C 2 1 4 1 ln x C 5 25 5
(b) Integration by parts: u ln x, du
x 4 ln x dx
x5 ln x 5
x5 1 dx, dv x 4 dx, v x 5
x5 1 x5 x5 dx ln x C 5 x 5 25
1 2x dx 2 32 x 22 1 x2 arcsin C 2 3
2 1 2 cos x dx sin x cos 2 x 2 C 3 3 2 x
1 5 dx 1 tan 5x
dx
cos3 x
cos
2
343
344
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
18. (a) By Formula 24: a 75, x u, and
(b) Partial fractions:
1 1 x 75 dx ln C x 2 75 2 75 x 75
3
30
ln
A 1 B x 2 75 x 75 x 75 1 A x 75 B x 75
x 75 C x 75
x 75: 1 2A 75 ⇒ A
3 1 1 30 2 75 10 3
x 75: 1 2B 75 ⇒ B
1 dx x 2 75
20. By Formula 21:
22. By Formula 79:
x
1 x
30
3 30
3 30 dx x 75 x 75
3
30
3
ln
x 75 C x 75
2 dx 2 x 1 x C 3
arcsec 2x dx
1 2x arcsec 2x ln 2x 4x 2 1 C 2
u 2x, du 2 dx
24. By Formula 52:
26. By Formula 7:
28. By Formula 14:
x sin x dx sin x x cos x C
1 sin
3
1 2 2x 2 dx arctan C arctanx 1 C x 2 2x 2 2
4
30. By Formula 56:
2
C
x2 1 25 10 ln 3x 5 dx 3x 3x 52 27 3x 5
d
32. By Formula 71:
1 1 3 2 d 3 1 sin 3
1 dt t 1 ln t2
u ln t, du
u e x, du e x dx
1 tan 3 sec 3 C 3
34. By Formula 23:
ex 1 dx e x ln cos e x sin e x C 1 tan e x 2
1 1 dt arctanln t C 1 ln t2 t
1 dt t
36. By Formula 26:
1 x x 2 3 3 ln x x 2 3 C 2
38. By Formula 27:
1 3x2 2 2 3x2 3 dx 27
3 x 2 dx
x 2 2 3x2 dx
1 3x18x 2 2 2 9x 2 4 ln 3x 2 9x 2 C 827
Section 7.6
40. By Formula 77:
2 3
x arctanx3 2 dx
42. By Formula 45:
arctanx3 2
Integration by Tables and Other Integration Techniques
32 x dx
2 3 2 x arctanx3 2 ln 1 x3 C 3
ex ex dx C 2x 3 2 1 e
1 e 2x
u e x, du e x dx 44. By Formula 27:
2x 32 2x 32 4 dx
1 2x 32 2x 3)2 42 dx 2
1 2x 3 2x 32 2 2x 32 4 ln 2x 3 2x 32 4 C 8
u 2x 3, du 2 dx
46. By Formula 31:
cos x
sin2 x 1
dx ln sin x sin2 x 1 C
u sin x, du cos x dx
48.
3x dx 3x
3x
9 x 2
3
1 dx
9 x 2
3 arcsin
50. By Formula 67:
dx
un
a bu
du
x 9 x 2 C 3
du
a bu
2 , v a bu b
a bu 2n n1 2u n
a bu u a bu du b b
a bu
2n au n1 bu n 2u n
a bu du b b
a bu
2u n 2na u n1 un
a bu du 2n du b b a bu
a bu
54.
2u n 2n n1
a bu u a bu du b b
Therefore, 2n 1
tan3 d
x dx
9 x 2
52. Integration by parts: w u n, dw nu n1 du, dv
un
a bu
un
a bu
du
2 n un1 u a bu na du and b
a bu
2 un1 un a bu na du . 2n 1b
a bu
uncos u du un sin u n un1sin u du
w un, dv cos u du, dw nun1 du, v sin u
tan2 2
tan d
tan2 ln cos x C 2
345
346
56.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
ln un du uln un
nln un1
1uu du uln u
w ln un, dv du, dw nln un1
58.
x x 2 2x dx
0, 0:
n
n ln un1 du
1u du, v u
1 2x 2 2x3 2 3x 1 x 2 2x 3 ln x 1 x 2 2x C 6
15
1 3 ln 1 C 0 ⇒ C 0 6
−6
6
(0, 0)
−15
60.
2 2x x 2
x1
0, 2 :
dx 2 2x
x2
3 ln
3 2 2x x 2
x1
3
(0, 2 )
C
2 3 ln 3 2 C 2 ⇒ C 3 ln 3 2
−2.5
1
−6
62.
0, 1: C 1 ⇒ y
64.
sin 1 sin 1 sin d ln cos 1 sin 2 1 sin cos
1 sin 1 sin ln 2 1 sin cos
C
10
1
sin sin d d 1 cos2 1 cos 2
−8
8
(0, 1) −2
2
66.
0
1 d 3 2 cos
arctancos C
1
0
1
2
0
u tan
68.
cos d 1 cos
cos 1 cos d 1 cos 1 cos
70.
2
1 d sec tan
2u 1 u2 21 u2 3 1 u2
1 du 5u2 1
25 arctan 5 u
1 0
2
5
arctan 5
1 d 1 cos sin cos
cos cos2 d sin2
csc cot cot2 d
ln 1 sin C
csc cot csc2 1 d
csc cot C
cos d 1 sin
u 1 sin , du cos d
Section 7.6
2
72. A
0
1 2
0
1 2
2x dx 2 1 ex
1 4
2
1 2 2 x ln1 e x 2
1 1 4 ln1 e4 ln 2 2 2
347
y
x 2 dx 1 ex 2
Integration by Tables and Other Integration Techniques
x 1
2
0
0.337 square units
74. Log Rule:
1 du, u ex 1 u
78. Formula 16 with u e2x
76. Integration by parts
5
82. W
80. A reduction formula reduces an integral to the sum of a function and a simpler integral. For example, see Formula 50, 54.
0
500x
26 x 2
dx
5
250
26 x 21 22x dx
0
500 26 x 2 500 26 1 2049.51 ft lbs
84.
2
1 20
0
2500 5000 dt 1 e4.81.9t 1.9
2
0
1.9 dt 1 e4.81.9t
2500 4.8 1.9t ln1 e4.81.9t 1.9
2500 1 ln1 e 4.8 ln1 e4.8 1.9
2500 1e 3.8 ln 1.9 1 e4.8
86. (a)
401.4
20
k
6x 2ex 2
dx 50
0
By trial and error, k 5.51897.
0
5.51897
(b)
0
88. True
6x 2ex 2 dx
2
−1
5.52
0
5 0
348
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
Section 7.7
Indeterminate Forms and L’Hôpital’s Rule
1 ex 1 x →0 x
2. lim
1
−1
0.1
0.01
0.001
0.001
0.01
0.1
0.9516
0.9950
0.9995
1.00005
1.005
1.0517
x f x
6x
4. lim
x →
3x2 2x
3.4641
exact: 63
5
1
10
102
103
104
105
f x
6
3.5857
3.4757
3.4653
3.4642
3.4641
x →1
(b) lim
x →1
8. (a) lim
x →0
x →
(b) lim
x →
0
100 0
2x2 x 3 2x 3x 1 lim lim 2x 3 5 x →1 x →1 x1 x1 2x2 x 3 ddx 2x2 x 3
4x 1 lim 5 lim x →1 x →1 x1 ddx x 1
1
sin 4x sin 4x lim 2 21 2 x →0 2x 4x
10. (a) lim
−2
x
6. (a) lim
(b) lim
x →0
sin 4x 4 cos 4x ddx sin 4x
lim lim 2 x →0 x →0 2x ddx 2x
2
2x 1 2x 1x2 0 lim 0 2 x → 4x x 4 1x 4 2x 1 ddx 2x 1
2 lim lim 0 4x2 x x → ddx 4x2 x x → 8x 1
x2 x 2 2x 1 lim 3 x →1 x →1 x1 1
12. lim
14. lim x →2
4 x2
x2
lim x →2
lim x →2
16. lim x →0
ex 1 x ex 1 lim 3 x →0 x 3x2 lim x →0
ex 6x
sin ax a cos ax a lim x→0 sin bx x→0 b cos bx b
20. lim
x →
x1 1 lim 0 x2 2x 3 x → 2x 2
x2 2x 2 lim x lim x 0 x→ ex x→ e x→ e
28. lim
x4 x2 1 x 4 x2
2 18. lim ln x lim 2 ln x x→1 x2 1 x→1 x2 1
lim
2x 2x
lim
1 1 x2
x→1
x→1
24. lim
1
arctan x 4 11 x2 1 lim x →1 x →1 x1 1 2
22. lim
26. lim
x→
x3 3x2 lim x→ x1 1
Section 7.7
30. lim
x→
x2 x lim x→ 1 1x2 1
Indeterminate Forms and L’Hôpital’s Rule
32. lim
x2
x→
sin x 0 x
Note: Use the Squeeze Theorem for x > .
e x2 12e x2 lim x→ x x→ 1
ln x 4 4 ln x 4x lim lim x→ x3 x→ x→ 3x2 x3
36. lim
34. lim
lim
x→
4 0 3x3
38. (a) lim x3 cot x 0 (b) lim x3 cot x lim (c)
x→0
40. (a) lim x tan x→
x→0
x→0
1 sin x 1 ≤ ≤ x x x
x3 tan x
lim x→0
3x2 sec2 x
0
(b) lim x tan x→
1 0 x
1 tan1x lim x x→ 1x
1
lim
x→
0
1x2 sec21x 1x2
3
lim sec2 x→
−1
(c)
1x 1
2
1
10
−1
42. (a) lim e x x2x 1 x→0
(b) Let y lim e x x2x. x→0
ln y lim x→0
2 ln
ex
x
1 1 (b) Let y lim 1 . x x→
1 x
x
x
x→
x
ln y lim
2e x 1e x x lim 4 x→0 1
x→
x ln1 1x lim ln 1 1x1x
x→
1 1x 1x2
Thus, ln y 4 ⇒ y e 4 54.598. (c)
44. (a) lim 1
lim
x→
60
1x2
lim
x→
1 1 1 1x
Thus, ln y 1 ⇒ y e1 e. Therefore,
lim 1
x→ 0
1 x
x
e.
2 0
(c)
5
0 −1
10
349
350
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals 48. (a) lim 3x 4 x4 00
46. (a) lim 1 x1x 0 x →
x→4
(b) Let y lim 3x 4 x4.
(b) Let y lim 1 x
1x.
x→
ln y lim
x→
lim
x→
x→4
ln y lim x 4ln 3x 4
ln1 x x
x→4
111 x 0
ln 3x 4
1x 4
lim
1x 4 1x 42
x→4
x→4
Thus, ln y 0 ⇒ y e0 1. Therefore, lim 1 x1x 1.
lim x 4 0
x →
(c)
lim
x→4
Hence, lim 3x 4 x4 1.
5
x→4
(c)
2
10
0 −1
4
7 0
2 x
x
50. (a) lim cos x→0
00
2 x . x
(b) Let y lim cos x→0
x 1 4 x x41 1 x1 1 x1 lim lim x 4 x 4 x 4
52. (a) lim (b)
x →2
2
x →2
2
2
2
x →2
2 x
ln y lim x ln cos x→0
lim x →2
000
2 x
x
Hence, lim cos x→0
(c)
lim x →2
1. (c)
2
1 2x 1 2x 1 4xx 1
0.25
2 4
2
− 0.25
0
3 0
54. (a) lim x→0
(b) lim x→0
10x x3
(c)
2
10x x3 lim 10xx 3 2
x→0
10
0
5
2
−20
56. (a)
(b) Let y sin xx, then ln y x lnsin x.
2
−
lim
x →0
2
−1
lnsin x cos xsin x x2 2x lim lim lim 0 2 x →0 x →0 tan x x →0 sec2 x 1x 1x
Therefore, since ln y 0, y 1 and lim sin xx 1. x →0
1 8
Section 7.7
58. (a)
Indeterminate Forms and L’Hôpital’s Rule x3 3x2 6x 6 lim 2x lim 2x lim 2x 0 2x x→ e x→ 2e x→ 4e x→ 8e
(b) lim
1 −1
5
−3
62. Let f x x 25 and gx x.
60. See Theorem 7.4.
64. lim
x3 3x2 6x 6 lim lim 0 lim e2x x→ 2e2x x→ 4e2x x→ 8e2x
66. lim
ln x2 2 ln xx lim x→ x3 3x2
x→
x→
x→
xm mxm1 lim enx x→ nenx mm 1xm2 n2enx
lim
2 ln x 3x3
lim
lim
2x 2 lim 3 0 x→ 9x 9x2
. . . lim
x→
x→
70.
68. lim
x→
x→
m! 0 nmenx
x
1
5
10
20
30
40
50
100
ex x5
2.718
0.047
0.220
151.614
4.40 105
2.30 109
1.66 1013
2.69 1033
ln x x Horizontal asymptote: y 0 (See Exercise 29)
74. y
72. y x x, x > 0 lim x x and lim x x 1
x→
x→0
dy x1x ln x1 1 ln x 0 dx x2 x2
No horizontal asymptotes ln y x ln x
Critical number:
1 1 dy x ln x y dx x
0, e Sign of dydx: y f x: Increasing 1 Relative maximum: e, e
dy x x1 ln x 0 dx
e1, 0 Increasing 1 1 1 Relative minimum: e1, e1e , e e
e, Decreasing
x e1
Critical number:
xe
Intervals:
0, e1 Sign of dydx: y f x: Decreasing Intervals:
1 −1
4
(e, 1e (
1e
−4
4
(
−1
((
1
1, 1 e e e
( 4
−1
sin x 1 0 x→ x
76. lim
(Numerator is bounded)
Limit is not of the form 00 or . L’Hôpital’s Rule does not apply.
ex 0 0 x → 1 ex 10
78. lim
Limit is not of the form 00 or . L’Hôpital’s Rule does not apply.
351
352
Chapter 7
80. A P 1
r n
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals nt
r ln A ln P nt ln 1 ln P n
lim
n →
ln 1
r n
lim
r n
1 nt
1 r n2 1 rn 1 2 nt
n →
1 nt
ln 1
lim
n →
rt
1
1
r n
rt
Since lim ln A ln P rt, we have lim A eln Prt eln Pert Pert. Alternatively, n→
r lim A lim P 1 n → n → n
n→
nt
1 n
lim P n →
r
nr rt
Pert.
82. Let N be a fixed value for n. Then
N 1x N2 N 1N 2x N3 N 1! x N1 lim lim . . . lim 0. x x x→ e x→ x→ x→ e ex ex
lim
84. f x
xk 1 k
k 1,
k=1
f x x 1 1 10x 0.1 1 0.1
x 0.1
f x
k 0.01,
x 0.01 1 100x 0.01 1 f x 0.01
k →0
(See Exercise 68)
6
k 0.1,
lim
k = 0.1 k = 0.01 −2
10 −2
xk 1 xk ln x lim ln x k →0 k 1
1 86. f x , gx x2 4, 1, 2
x f c f 2 f 1 g2 g1 g c
88. f x ln x, gx x3, 1, 4
f 4 f 1 f c g4 g1 g c
12 1c2 3 2c
ln 4 1c 1 2 3 63 3c 3c 3c3 ln 4 63
1 1 3 6 2c
c3
2c3 6 3 3 c
90. False. If y exx2, then y
x2ex 2xex xex x 2 ex x 2 . x4 x4 x3
c
21 ln 4
ln214 2.474 3
92. False. Let f x x and gx x 1. Then lim
x →
x 1, but lim x x 1 1. x → x1
Section 7.8
e1x , 0,
94. gx
2
Improper Integrals
x0
y
x0
1.5
gx g0 e1x lim g0 lim x →0 x →0 x0 x
2
e1x e1x , then ln y ln Let y x x
2
2
0.5
1 1 x2 ln x 2 ln x . Since x x2
x −2
−1
1
2
− 0.5
ln x 1x x2 lim lim 0 lim x2 ln x lim 2 3 x →0 x →0 1x x →0 2x x →0 2
1 x x
2
we have lim
x →0
ln x
2
. Thus, lim y e
x →0
0 ⇒ g0 0.
Note: The graph appears to support this conclusion—the tangent line is horizontal at 0, 0. 98. lim x ln 21ln x
96. lim f xgx
x →0
x →a
Let y x ln 21ln x, then:
y f xgx ln y gx ln f x
ln y
lim gx ln f x
ln 2 1 ln x
x →a
As x → a, ln y ⇒
, and hence y . Thus,
lim ln y lim
x →0
lim f xgx .
x →0
ln 2ln x ln 2x lim x →0 1 ln x 1x
x→0
Thus, lim y eln 2 2. x →
Improper Integrals
2. Infinite discontinuity at x 3.
4
3
1 dx lim b →3 x 332
x 332 dx
2 b 3
4
b
lim 2x 312 b →3
lim 2 b →3
4 b
Diverges 4. Infinite discontinuity at x 1.
2
0
1 dx x 123
1
0
1 dx x 123
b
lim b→1
0
b→1
2
1
1 dx x 123
1 dx lim c→1 x 123
3 x 1 lim 3
Converges
ln 2ln x 1 ln x
lim ln 2 ln 2
x →a
Section 7.8
ln x
b 0
2
c
1 dx x 123
3 x 1 lim 3 c→1
2 c
0 3 3 0 6
353
354
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
6. Infinite limit of integration.
0
8.
0
e2x dx lim
b→
ex dx 0. You need to evaluate the limit.
0
lim
b→
0
1 2x e 2
lim
b→
b
b
e2x dx
b
e
ex dx lim b→
0
1 1 0 2 2
lim
b→
b
x
0
e
1 1
b
Converges
10.
5
1
b
dx lim b→ x3
1
b→
xex2 dx lim
x 1ex dx lim
eax sin bx dx lim
b→
0
b→
18.
e
x2
c→
0
x 1ex dx lim xex b→
0
ax
e
a sin bx b cos bx a2 b2
x2
0
x3 dx lim b→ 12
b
0
x2
b→
x dx lim b→ 1
1 lnx 2
lim
2
b
0
1
1
lim
b 0
b→
0 0 by L’Hôpital’s Rule. b e b
20.
1
ln x dx lim b→ x b→
b
0
1 2 1 x2
x
x2 1 2
dx
Diverges
b→
lim
3 4
0
b b 0 2 a b2 a 2 b2
22.
163x
lim eb22b 4 4 4
b
c
4x14 dx
1
2x 4
b
0
b→
b→
xex2 dx lim
b
lim
b
b→
dx lim
5 2
1
0
16.
b
4 4 x
1
2
14.
12.
25x
lim
5 dx x3
24.
0
b
1
ln x dx x
ln x2 2
b 1
Diverges
b
ex dx lim ln1 ex b→ 1 ex
ln 2
0
b
Diverges
0
1 2
Diverges
26.
6
30.
x x b dx lim 2 cos b→ 2 2 0 0 x Diverges since cos does not approach a limit as x → . 2 sin
0
4 dx lim b→6 6 x
0
4
b
8 dx lim 8 ln x b→0 x
4 b
32.
0
b
e
ln x2 dx lim b→0
2 ln x dx
0
lim 2x ln x 2x b→0
0
e b
80 8 6
lim 2e 2e 2b ln b 2b
8 6
0
sec d lim
Diverges
e
46 x12 dx
8 dx lim b→0 x
Diverges
0
2
34.
0
b
lim 86 x12 b→6
4
28.
b→ 2
b→0
ln sec tan
b 0
2
,
36.
0
1 x dx lim arcsin b→2 2 4 x2
b 0
2
Section 7.8
2
38.
0
1 dx lim b→2 4 x2
b
0
1 1 1 1 2x dx lim ln b→2 4 2x 2x 4 2x
Diverges
3
40.
1
2
2 dx x 283
b 0
Improper Integrals
355
0
3
2x 283 dx
1
2x 283 dx
2
b
lim b→2
1
2x 283 dx
3
2x 283 dx lim c→2
6 lim x 253 b→2 5
b
c
6 lim x 253 c→2 5
1
3 c
Diverges
42.
Thus,
1
1 dx x ln x
0
1 dx x ln x 1
lim lnln x b→1
44. If p 1,
e
e
lim
1
1
1 dx ln ln x C x ln x
c→
1 dx lim ln x a→0 x
a
lim ln a . a→0
Diverges. If p 1,
e
1
1 dx x ln x
0
lnln x
.
1 x1p dx lim a→0 xp 1p
1 a
lim a→0
1 1 p 1a p . 1p
1 if 1 p > 0 or p < 1. 1p
This converges to
e
Diverges
1
46. (a) Assume
gx dx L (converges).
a
Since 0 ≤ f x ≤ gx on a, , 0 ≤
f x dx ≤
a
gx dx diverges, because otherwise, by part (a), if
a
1
48.
0
3 x
56.
f x dx converges.
a
gx dx converges, then so does
dx
f x dx.
1 3 converges. 1 13 2
1 1 ≥ on 2, and x x 1
a
50.
x 4ex dx converges.
0
See Exercise 44, p 13 .
(See Exercise 45.)
1
2
x
1 1 ≤ 32 on 1, and x x1 x
54. Since
gx dx L and
a
1
52. Since
a
(b)
dx diverges by Exercise 43,
2
1
1 x32
1 x 1
dx diverges.
dx converges by Exercise 43,
1 1 ≥ since x ln x < x on 2, . Since x x ln x
1
1
2
x
1 x1 x
dx diverges by Exercise 43,
2
dx converges.
1 x ln x
dx diverges.
58. See the definitions, pages 540, 543. 60. Answers will vary.
(a)
1e
Converges
62. f t t
ex 2x
dx
(b)
x dx
(Example 4)
Diverges
Fs
st 1e
s
test dt lim b→
0
1
2
1 ,s > 0 s2
st
b 0
356
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
64. f t eat Fs
66. f t sin at
eat est dt
0
etas dt
0
1 e as
b
t as
0
1 1 ,s > a as sa
68. f t sinh at
est sinh at dt
0
est sin at dt
b→
2
est s sin at a cos at a2
a a ,s > 0 s2 a2 s2 a2
eat 1 dt 2 2
at
s
0
et sa et sa dt
0
b
1 1 1 et sa et sa b→ 2 s a s a
lim
0
e
est
0
lim
0
lim b→
Fs
Fs
0
0
1 1 1 2 s a s a
1 1 a 1 2 ,s > a 2 s a s a s a2
70. (a) A
1
1 1 dx x2 x
1
1
(b) Disk: V
1
(c) Shell: V 2
1
x
x1 dx lim 2 ln x
b
2
b→
1
1
dx lim 3 b→ x4 3x
b 1
Diverges 72. x 22 y2 1 2x 2 2yy 0 y
x 2 y
1 y 2 1
3
S 4
1
x dx 4
y
x 2 2y 2
3
1
1 (Assume y > 0. y
x dx 4
1 x 22
3
1
74. (a) Fx W (b)
2 1 x 22
dx
4
0 2 arcsin1 2 arcsin1 8
lim 4 1 x 22 2 arcsinx 2 a→1 b→3
x2 1 x 22 b
a
K K ,5 , K 80,000,000 x2 40002
80,000,000 80,000,000 dx lim b→ x2 x 4000
80,000,000 W 10,000 2 x
b 4000
80,000,000 10,000 b b 8000 Therefore, 4000 miles above the earth’s surface.
b 4000
20,000 mi-ton
80,000,000 20,000 b
2
3
b 0
Section 7.8
76. (a)
2 2t5 e dt 5 4
(b)
0
0
b 2 2t5 e dt lim e2t5 1 b→ 5 0
4
2 2t5 e dt e2t5 5
0
e85 1
0.7981 79.81%
(c)
Improper Integrals
25 e dt lim te 2t5
t
0
2t5
b→
5 e2t5 2
b 0
5 2
5
78. (a) C 650,000
25,0001 0.08te0.06t dt
0
650,000 25,000
$778,512.58
$905,718.14
1 0.06t t 0.06t 1 e 0.08 e e0.06t 0.06 0.06 0.062
5 0
10
(b) C 650,000
25,0001 0.08te0.06t dt
0
650,000 25,000 (c) C 650,000
1 0.06t t 0.06t 1 e e 0.08 e0.06t 0.06 0.06 0.062
10 0
25,0001 0.08te0.06t dt
0
t e 0.06
650,000 25,000 lim
0.06t
b→
80. (a)
1
1
xn
1
1 e0.06t 0.062
b 0
$1,622,222.22
(b) It would appear to converge. y
1
1.00
dx will converge if n > 1 and will diverge if n ≤ 1.
0.50 0.25
u
1
0.06t
0.75
(c) Let dv sin x dx ⇒
b
t e 0.06
1
1 1 dx lim b→ x2 x
1
1
b
1 dx lim ln x b→ x
0.08
1 x
v cos x
⇒ du
sin x cos x dx lim b→0 x x cos 1
1
b 1
x −5 − 0.25
1 dx x2
1
15
20
cos x dx x2
cos x dx x2
Converges 82. (a) Yes, the integral is not defined at x 2.
(b)
5
(c) As n → , the integral approaches 4 4 . (d) In
2
0
4 dx 1 tan xn
I2 3.14159 I4 3.14159 I8 3.14159 I12 3.14159
0 −2
2
357
358
Chapter 7
84. (a) f x
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
1 2 ex70 18 32
(b) P72 ≤ x <
0.2525
(c) 0.5 P70 ≤ x ≤ 72 0.5 0.2475 0.2525
90
f x dx 1.0
These are the same answers because by symmetry,
50
P70 ≤ x <
0.4
0.5
and 50
0.5 P70 ≤ x <
90
P70 ≤ x ≤ 72 P72 ≤ x <
− 0.2
86. False. This is equivalent to Exercise 85.
.
88. True
Review Exercises for Chapter 7 2.
2
xe x1 dx
2 1 x1 e 2x dx 2
4.
x 1 dx 1 x2122x dx 2 1 x 2
1 2 e x1 C 2
1 1 x212 C 2 12
1 x2 C
6.
u4 3u2 du
2x2x 3 dx
10.
8.
x x 4 2x2 x 1 1 2 x 4 2x2 1 x 12
22x 332 x 1 C 5
u 2x 3, x
u5 u3 C 5
x4 2x2 x 1 dx x2 12
u2 3 , dx u du 2
dx
x
1 2
2x dx x2 12
1 C 2x2 1
x2 1ex dx x2 1ex 2 xex dx x2 1ex 2xex 2 ex dx exx2 2x 1 1
(1) dv ex dx
⇒
v ex
u x2 1 ⇒ du 2x dx (2) dv ex dx ⇒ ux
⇒ du dx
12. u arctan 2x, du
v ex
2 dx, dv dx,v x 1 4x2
arctan 2x dx x arctan 2x x arctan 2x
14.
lnx2 1 dx
2x dx 1 4x2
1 ln1 4x2 C 4
1 2
lnx2 1 dx
1 x ln x2 1 2
1 x ln x2 1 2
x2
x2 dx 1
dx
1 dx x2 1
1 x1 1 C x ln x2 1 x ln 2 2 x1
dv dx
⇒
vx
u lnx2 1 ⇒ du
2x dx x2 1
Review Exercises for Chapter 7
16.
ex arctanex dx ex arctanex ex arctanex ⇒
dv ex dx
20.
x dx 2
sin2
ex dx 1 e2x
or
tan sec4 d
22.
tan3 tan sec2 d
sec3 sec tan d
cos 2sin cos 2 d
x2 9
x
dx
24.
1 ln1 e2x C 2
1 1 1 1 1 cos x dx x sin x C x sin x C 2 2 2
tan sec4 d
e2x dx 1 e2x
v ex
u arctan ex ⇒ du
18.
359
1 4 1 tan tan2 C1 4 2
1 sec4 C2 4
cos 2 sin2 sin cos 2 d
1 sin cos 3cos sin d sin cos4 C 4
3 tan 3 sec tan d 3 sec
x
3 tan2 d
x2 − 9
θ 3
3 sec2 1 d 3tan C x2 9 3 arcsec
3x C
x 3 sec , dx 3 sec tan d, x2 9 3 tan
26.
9 4x2 dx
1 9 2x2 2 dx 2
2 9 arcsin
1 2
9 2x x arcsin 9 4x2 C 4 3 2
1
28.
2x 2x9 4x2 C 3
sin 1 d 1 2 cos2 2
1 2
1 2 sin d 1 2 cos2
arctan 2 cos C
u 2 cos , du 2 sin d
360
Chapter 7
30. (a)
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
x4 x dx 64 tan3 sec3 d
(b)
x4 x dx 2 u4 4u2 du
64 sec4 sec2 sec tan d
64 sec3 3 sec3 5 C 15
24 x32 3x 8 C 15
2u3 2 3u 20 C 15
24 x32 3x 8 C 15
u2 4 x, dx 2u du
x 4 tan2 , dx 8 tan sec2 d, 4 x 2 sec
(c)
x4 x dx
u32 4u12 du
(d)
x4 x dx
4 x32 dx
2u32 3u 20 C 15
4 2x 4 x32 4 x52 C 3 15
24 x32 3x 8 C 15
24 x32 3x 8 C 15
dv 4 x dx ⇒
2 v 4 x32 3
⇒ du dx
ux 2x3 5x2 4x 4 4 3 2x 3 x2 x x x1
32.
34.
u 4 x, du dx
2x 2 4 x32 3 3
2x3 5x2 4x 4 dx x2 x
2x 3
3 4 dx x2 3x 4 ln x 3 ln x 1 C x x1
4x 2 A B 3x 12 x 1 x 12 4x 2 3Ax 1 3B Let x 1: 2 3B ⇒ B
2 3
Let x 2: 6 3A 3B ⇒ A
36.
4x 2 4 dx 3x 12 3
sec2 d tan tan 1
4 3
2 1 dx x1 3
1 du uu 1
1 du u1
ln u 1 ln u C ln u tan , du sec2 d 1 A B uu 1 u u1 1 Au 1 Bu Let u 0: 1 A ⇒ A 1 Let u 1: 1 B
4 1 2 2 1 dx lnx 1 C 2 lnx 1 C x 12 3 3x 1 3 x1
1 du u
tan 1 C ln 1 cot C tan
Review Exercises for Chapter 7
38.
x 2 3x
dx
24 3x 2 3x C (Formula 21) 27
40.
x 1 1 dx 2 dx 1 ex 2 1 eu
6x 8 2 3x C 27
u x2
1 u ln1 eu C 2
(Formula 84)
1 2 x2 ln1 e x C 2
42.
3 3 1 dx 3 dx 2 3x3x2 1 2x9x2 1
44.
(Formula 33)
tann x dx
50.
u 3x
1 1 1 dx dx 1 tan x 1 tan x
46.
3 arcsec 3x C 2
u x
11 x lncos x sin x C (Formula 71) 2
tann2xsec2 x 1 dx
tann2 x sec2 x dx
1 tann1 x n1
1 x dx
48.
4u4 4u2 du
4u5 4u3 4 C 1 x 323x 2 C 5 3 15
u 1 x, x u4 2u2 1, dx 4u3 4u du
52.
3x3 4x Cx D Ax B 2 2 x2 12 x 1 x 12 3x3 4x Ax Bx2 1 Cx D Ax3 Bx2 A Cx B D A 3, B 0, A C 4 ⇒ C 1, BD0 ⇒ D0
3x3 4x x dx dx 3 2 x2 12 x 1
54.
x dx x2 12
3 1 lnx2 1 C 2 2x2 1
sin cos 2 d
sin2 2 sin cos cos2 d
1 sin 2 d
1 dx u 2x, du 2x
tann2 x dx
u4u3 4u du
csc2x 1 dx 2 csc2x dx x 2x
2 ln csc2x cot2x C
tann2 x dx
1 1 cos 2 C 2 cos 2 C 2 2
361
362
Chapter 7
56. y
4 x2
2x
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
dx
2 cos 2 cos d 4 sin
csc sin d
ln csc cos cos C ln
4 x2 2 4 x2 C x 2
x 2 sin , dx 2 cos d, 4 x 2 2 cos
1 cos d
58. y
sin 1 cos
d 1 cos 12sin d 21 cos C
u 1 cos, du sin d
1
60.
x
x 2x 4
0
2
0
dx 2 ln x 4 ln x 2
1
62.
xe3x dx
0
e9 3x 1
2
3x
0
1 5e6 1 224.238 9
2 ln 3 2 ln 4 ln 2 ln
3
64.
0
9 0.118 8
x 22 x 1 x dx 3 1 x
3 0
4 4 8 3 3 3
1 dx 25 x2
1 x5 ln 10 x 5
4
66. A
0
68. By symmetry, y 0. A 4 5 1 44 x 4
17 3.4 5
70. s
y
4 0
1 1 1 ln ln 9 0.220 10 9 10
1 sin2 2x dx 3.82
0
3 2 1
(3.4, 0) x
−1
1
3
4
5
−2 −3
x, y 3.4, 0
72. lim
x →0
76.
sin x cos x 1 lim sin 2 x x →0 2 cos 2 x 2 2
74. lim xex lim 2
x →
x →
y lim x 1ln x x →1
ln y lim ln x lnx 1 x →1
lim x →1
lnx 1 lim x →1 1 ln x
lim 2xln x 0 x →1
Since ln y 0, y 1.
1 1 2 ln x x1 ln2 x x lim lim x →1 x →1 1 1 x1 1 x ln2 x x x2
x 1 2 lim 2 0 x → 2xe x ex
358
Chapter 7
84. (a) f x
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
1 2 ex70 18 32
(b) P72 ≤ x <
0.2525
(c) 0.5 P70 ≤ x ≤ 72 0.5 0.2475 0.2525
90
f x dx 1.0
These are the same answers because by symmetry,
50
P70 ≤ x <
0.4
0.5
and 50
0.5 P70 ≤ x <
90
P70 ≤ x ≤ 72 P72 ≤ x <
− 0.2
86. False. This is equivalent to Exercise 85.
.
88. True
Review Exercises for Chapter 7 2.
2
xe x1 dx
2 1 x1 e 2x dx 2
4.
x 1 dx 1 x2122x dx 2 1 x 2
1 2 e x1 C 2
1 1 x212 C 2 12
1 x2 C
6.
u4 3u2 du
2x2x 3 dx
10.
8.
x x 4 2x2 x 1 1 2 x 4 2x2 1 x 12
22x 332 x 1 C 5
u 2x 3, x
u5 u3 C 5
x4 2x2 x 1 dx x2 12
u2 3 , dx u du 2
dx
x
1 2
2x dx x2 12
1 C 2x2 1
x2 1ex dx x2 1ex 2 xex dx x2 1ex 2xex 2 ex dx exx2 2x 1 1
(1) dv ex dx
⇒
v ex
u x2 1 ⇒ du 2x dx (2) dv ex dx ⇒ ux
⇒ du dx
12. u arctan 2x, du
v ex
2 dx, dv dx,v x 1 4x2
arctan 2x dx x arctan 2x x arctan 2x
14.
lnx2 1 dx
2x dx 1 4x2
1 ln1 4x2 C 4
1 2
lnx2 1 dx
1 x ln x2 1 2
1 x ln x2 1 2
x2
x2 dx 1
dx
1 dx x2 1
1 x1 1 C x ln x2 1 x ln 2 2 x1
dv dx
⇒
vx
u lnx2 1 ⇒ du
2x dx x2 1
Review Exercises for Chapter 7
16.
ex arctanex dx ex arctanex ex arctanex ⇒
dv ex dx
20.
x dx 2
sin2
ex dx 1 e2x
or
tan sec4 d
22.
tan3 tan sec2 d
sec3 sec tan d
cos 2sin cos 2 d
x2 9
x
dx
24.
1 ln1 e2x C 2
1 1 1 1 1 cos x dx x sin x C x sin x C 2 2 2
tan sec4 d
e2x dx 1 e2x
v ex
u arctan ex ⇒ du
18.
359
1 4 1 tan tan2 C1 4 2
1 sec4 C2 4
cos 2 sin2 sin cos 2 d
1 sin cos 3cos sin d sin cos4 C 4
3 tan 3 sec tan d 3 sec
x
3 tan2 d
x2 − 9
θ 3
3 sec2 1 d 3tan C x2 9 3 arcsec
3x C
x 3 sec , dx 3 sec tan d, x2 9 3 tan
26.
9 4x2 dx
1 9 2x2 2 dx 2
2 9 arcsin
1 2
9 2x x arcsin 9 4x2 C 4 3 2
1
28.
2x 2x9 4x2 C 3
sin 1 d 1 2 cos2 2
1 2
1 2 sin d 1 2 cos2
arctan 2 cos C
u 2 cos , du 2 sin d
360
Chapter 7
30. (a)
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
x4 x dx 64 tan3 sec3 d
(b)
x4 x dx 2 u4 4u2 du
64 sec4 sec2 sec tan d
64 sec3 3 sec3 5 C 15
24 x32 3x 8 C 15
2u3 2 3u 20 C 15
24 x32 3x 8 C 15
u2 4 x, dx 2u du
x 4 tan2 , dx 8 tan sec2 d, 4 x 2 sec
(c)
x4 x dx
u32 4u12 du
(d)
x4 x dx
4 x32 dx
2u32 3u 20 C 15
4 2x 4 x32 4 x52 C 3 15
24 x32 3x 8 C 15
24 x32 3x 8 C 15
dv 4 x dx ⇒
2 v 4 x32 3
⇒ du dx
ux 2x3 5x2 4x 4 4 3 2x 3 x2 x x x1
32.
34.
u 4 x, du dx
2x 2 4 x32 3 3
2x3 5x2 4x 4 dx x2 x
2x 3
3 4 dx x2 3x 4 ln x 3 ln x 1 C x x1
4x 2 A B 3x 12 x 1 x 12 4x 2 3Ax 1 3B Let x 1: 2 3B ⇒ B
2 3
Let x 2: 6 3A 3B ⇒ A
36.
4x 2 4 dx 3x 12 3
sec2 d tan tan 1
4 3
2 1 dx x1 3
1 du uu 1
1 du u1
ln u 1 ln u C ln u tan , du sec2 d 1 A B uu 1 u u1 1 Au 1 Bu Let u 0: 1 A ⇒ A 1 Let u 1: 1 B
4 1 2 2 1 dx lnx 1 C 2 lnx 1 C x 12 3 3x 1 3 x1
1 du u
tan 1 C ln 1 cot C tan
Review Exercises for Chapter 7
38.
x 2 3x
dx
24 3x 2 3x C (Formula 21) 27
40.
x 1 1 dx 2 dx 1 ex 2 1 eu
6x 8 2 3x C 27
u x2
1 u ln1 eu C 2
(Formula 84)
1 2 x2 ln1 e x C 2
42.
3 3 1 dx 3 dx 2 3x3x2 1 2x9x2 1
44.
(Formula 33)
tann x dx
50.
u 3x
1 1 1 dx dx 1 tan x 1 tan x
46.
3 arcsec 3x C 2
u x
11 x lncos x sin x C (Formula 71) 2
tann2xsec2 x 1 dx
tann2 x sec2 x dx
1 tann1 x n1
1 x dx
48.
4u4 4u2 du
4u5 4u3 4 C 1 x 323x 2 C 5 3 15
u 1 x, x u4 2u2 1, dx 4u3 4u du
52.
3x3 4x Cx D Ax B 2 2 x2 12 x 1 x 12 3x3 4x Ax Bx2 1 Cx D Ax3 Bx2 A Cx B D A 3, B 0, A C 4 ⇒ C 1, BD0 ⇒ D0
3x3 4x x dx dx 3 2 x2 12 x 1
54.
x dx x2 12
3 1 lnx2 1 C 2 2x2 1
sin cos 2 d
sin2 2 sin cos cos2 d
1 sin 2 d
1 dx u 2x, du 2x
tann2 x dx
u4u3 4u du
csc2x 1 dx 2 csc2x dx x 2x
2 ln csc2x cot2x C
tann2 x dx
1 1 cos 2 C 2 cos 2 C 2 2
361
362
Chapter 7
56. y
4 x2
2x
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
dx
2 cos 2 cos d 4 sin
csc sin d
ln csc cos cos C ln
4 x2 2 4 x2 C x 2
x 2 sin , dx 2 cos d, 4 x 2 2 cos
1 cos d
58. y
sin 1 cos
d 1 cos 12sin d 21 cos C
u 1 cos, du sin d
1
60.
x
x 2x 4
0
2
0
dx 2 ln x 4 ln x 2
1
62.
xe3x dx
0
e9 3x 1
2
3x
0
1 5e6 1 224.238 9
2 ln 3 2 ln 4 ln 2 ln
3
64.
0
9 0.118 8
x 22 x 1 x dx 3 1 x
3 0
4 4 8 3 3 3
1 dx 25 x2
1 x5 ln 10 x 5
4
66. A
0
68. By symmetry, y 0. A 4 5 1 44 x 4
17 3.4 5
70. s
y
4 0
1 1 1 ln ln 9 0.220 10 9 10
1 sin2 2x dx 3.82
0
3 2 1
(3.4, 0) x
−1
1
3
4
5
−2 −3
x, y 3.4, 0
72. lim
x →0
76.
sin x cos x 1 lim sin 2 x x →0 2 cos 2 x 2 2
74. lim xex lim 2
x →
x →
y lim x 1ln x x →1
ln y lim ln x lnx 1 x →1
lim x →1
lnx 1 lim x →1 1 ln x
lim 2xln x 0 x →1
Since ln y 0, y 1.
1 1 2 ln x x1 ln2 x x lim lim x →1 x →1 1 1 x1 1 x ln2 x x x2
x 1 2 lim 2 0 x → 2xe x ex
Problem Solving for Chapter 7 78. lim x →1
ln2x x 2 1
lim
x →1
1
0
2 2x
x 11x ln x
lim
2x 2 2 lim 1 x 1 x ln x x→1 1 1 ln x
x→1
lim x →1
80.
2x 2 2 ln x ln xx 1
0 b
6 dx lim 6 ln x 1 b→1 x1
82.
e1x
x
0
2
a→0 b →
Diverges
84. V
xex 2dx
0
x2e2x dx
0
lim
b→
e4
2x
86.
2
b→
1 1 1 dx < x5 x10 x15
4x1
lim
1 1 9x9 14x14
4
b
2x2 2x 1
0
2
b
<
2
2
0.015846 <
2
4
1 dx < x5 1
x5
2
1 1 2 dx x5 x10 x15
1 1 1 1 dx < lim 4 9 14 b→ 1 4x 9x 7x
1 dx < 0.015851 x5 1
Problem Solving for Chapter 7
1
2. (a)
0
ln x dx lim x ln x b→0
1 b
1 lim b ln b b 1 b→0
Note: lim b ln b lim
b→0
b→0
1
0
ln b 1b lim 0 b1 b→0 1b2
ln x2 dx lim xln x2 2x ln x 2x b→0
1 b
2 lim bln b2 2b ln b 2b 2 b→0
(b) Note first that lim bln bn 0 (Mathematical induction). Also,
b→0
ln x
n1
dx xln xn1 n 1 ln xn dx.
1
Assume
ln xn dx 1n n!.
0
1
Then,
0
ln xn1 dx lim xln xn1 b→0
1 b
1
n 1
ln xn dx
0
0 n 11n n! 1n1n 1!.
dx lim e1x
b 2
b a
101
363
364
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
lim
4.
x→
xx cc
lim x ln
x→
lim
x→
x
1 4
xx cc ln 41
lnx c lnx c ln 4 1x 1 1 xc xc lim ln 4 x→ 1 2 x
lim
x→
2c x2 ln 4 x cx c lim
x→
2cx2 ln 4 c2
x2
2c ln 4 2x 2 ln 2 c ln 2 6. sin BD, cos OD 1 1 Area DAB DABD 1 cos sin 2 2 Shaded area R
1 1 1BD sin 2 2 2 2 DAB 121 cos sin Shaded area 12 sin
lim R lim
→0
→0
1 cos sin 1 cos cos sin2 lim →0 sin 1 cos
lim
1 cos sin cos sin 2 sin cos sin
lim
sin 4 cos sin 4 cos 1 lim 3 →0 sin 1
→0
→0
1 u2 x 1 u2 3 u2 8. u tan , cos x , 2 cos x 2 2 2 1u 1 u2 1 u2 dx
2
0
2 du 1 u2
1 dx 2 cos x
1
0
1
0
1 2 u du 2
2 du 3 u2
2 1 arctan 3 3
2
1 u2 3 u2
1 u arctan 3 3
2 3 6
1
0
3 0.6046 9
Problem Solving for Chapter 7 10. Let u cx, du c dx.
b
cb
ec
2 x2
dx
0
du 1 c c
eu
2
0
As b → , cb → . Hence,
cb
eu du 2
0
dx
ec
2 x2
0
1 c
ex dx. 2
0
x 0 by symmetry.
y
Mx m
ec2x2
2
0
dx
2
ec x dx
2
2 2
0
1 2
0
e2c x dx 2 2
0
ec x dx 2 2
1 2 ex dx 1 2c 0 2 1 x2 e dx c 0
1 2 2
Thus, x, y 0,
2
4
2
4
.
12. (a) Let y f 1x, f y x, dx fy dy.
f 1x dx
y fy dy
y f y
u y, du dy
dv fy dy, v f y
f y dy
x f 1x
f y dy
(b) f 1x arcsin x y, f x sin x
arcsin x dx x arcsin x
sin y dy 1
x arcsin x cos y C x arcsin x 1 x2 C
y 1 − x2
(c) f x ex, f 1x ln x y
e
1
ln x dx x ln x
1
e
1
e y dy
0
e ey
1 0
e e 1 1
x 1 ⇔ y 0; x e ⇔ y 1
x
365
366
Chapter 7
14. (a) Let x
I
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
u, dx du. 2
2
0
sin x dx cos x sin x
2
2
0
Hence, 2I
2
sin x dx cos x sin x
0
2
1 dx
0
0
(b) I
2
2
0
cos x dx sin x cos x
⇒ I . 2 4
2 u du cos u sin u 2 2 n
n
cosn u du sinn u cosn u
Thus, 2I
2
0
16.
cos u du sin u cos u
sinn
2
0
2 u du cos u sin u 2 2 sin
0
1 dx
⇒ I . 2 4
P1 P2 Pn Nx ... Dx x c1 x c2 x cn Nx P1x c2x c3. . .x cn P2x c1x c3. . .x cn . . . Pnx c1x c2. . .x cn1 Let x c1: Nc1 P1c1 c2c1 c3. . .c1 cn P1
Nc1 c1 c2c1 c3. . .c1 cn
Let x c2: Nc2 P2c2 c1c2 c3. . .c2 cn P2
Nc2 c2 c1c2 c3. . .c2 cn
Let x cn: Ncn Pncn c1cn c2. . .cn cn1 Pn
Ncn cn c1cn c2. . .cn cn1
If Dx x c1x c2x c3. . .x cn, then by the Product Rule Dx x c2x c3. . .x cn x c1x c3. . .x cn . . . x c1x c2x c3. . .x cn1 and Dc1 c1 c2c1 c3. . .c1 cn Dc2 c2 c1c2 c3. . .c2 cn
Dcn cn c1cn c2. . .cn cn1. Thus, Pk Nck Dck for k 1, 2, . . ., n.
Problem Solving for Chapter 7
18.
st
50,000 dt 50,000 400t
32t 12,000 ln
16t 2 12,000 ln 50,000 ln50,000 400t dt
16t 2 12,000t ln 50,000 12,000 t ln50,000 400t
400t dt 50,000 400t
16t 2 12,000t ln
50,000 12,000t 50,000 400t
16t 2 12,000t ln
50,000 12,000t 1,500,000 ln50,000 400t C 50,000 400t
1
50,000 dt 50,000 400t
s0 1,500,000 ln 50,000 C 0 C 1,500,000 ln 50,000
st 16t 2 12,000t 1 ln
50,000 50,000 400t 1,500,000 ln 50,000 400t 50,000
When t 100, s100 557,168.626 feet 20. Let u x ax b, du x a x b dx, dv f x dx, v fx.
b
a
b
x ax b dx x ax b fx
b
a
b
a
x a x b fx dx u 2x a b
dv fx dx
2x a b fx dx
2x a b f x 2
b
a
a
f x dx
b a
b
a
2 f x dx
367
C H A P T E R 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals Section 7.1
Basic Integration Rules
Section 7.2
Integration by Parts
Section 7.3
Trigonometric Integrals . . . . . . . . . . . . . . . . . . . 65
Section 7.4
Trigonometric Substitution
Section 7.5
Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . 84
Section 7.6
Integration by Tables and Other Integration Techniques
Section 7.7
Indeterminate Forms and L’Hôpital’s Rule . . . . . . . . . 96
Section 7.8
Improper Integrals
Review Exercises
. . . . . . . . . . . . . . . . . . . 50
. . . . . . . . . . . . . . . . . . . . . 55
. . . . . . . . . . . . . . . . . 74
. . 90
. . . . . . . . . . . . . . . . . . . . . 102
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
C H A P T E R 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals Section 7.1
Basic Integration Rules
Solutions to Odd-Numbered Exercises 1. (a) (b)
d x2 1 C 21x2 11 22x 2x dx x 1
(c)
1 1 2 x d 1 2 x 1 C x 11 22x dx 2 2 2 2x2 1
(d)
2x d lnx2 1 C 2 dx x 1
x x2 1
3. (a)
dx matches (b).
d 2x x2 122 2x2x2 12x 21 3x2 2 2 2 C dx x 1 x2 14 x 13
(c)
1 d arctan x C dx 1 x2
(d)
2x d lnx2 1 C 2 dx x 1
1 dx matches (c). x2 1
3x 24 dx
u 3x 2, du 3 dx, n 4 Use
11.
t sin t 2 dt
Use
7.
sin u du.
1
x 1 2x
dx
9.
u 1 2x, du
un du.
u t 2, du 2t dt
50
x d 2 lnx2 1 C 21 x2 2x dx 1 x 1
(b)
5.
d 2x2 1 C 2 12 x2 11 22x 2x dx x2 1
Use
1 x
dx
3 1 t2
u t, du dt, a 1 Use
du . u
13.
cos xesin x dx
u sin x, du cos x dx Use
eu du.
dt
du a 2 u 2
Section 7.1
17. Let u z 4, du dz
15. Let u 2x 5, du 2 dx.
2x 53 2 dx
1 2x 53 22 dx 2
5 z 44 C dz 5 z 45 dx 5 5 z 4 4
1 2x 55 2 C 5
19. Let u t3 1, du 3t2 dt.
3 t3 1 dt t2
1 3
Basic Integration Rules
21.
v
5 C 4z 44
1 dv 3v 13
t3 11 33t2 dt
1 t3 14 3 C 3 4 3
t3 14 3 C 4
v dv
1 3v 133dv 3
1 1 v2 C 2 63v 12
23. Let u t3 9t 1, du 3t2 9 dt 3t2 3 dt.
25.
t2 3 1 3t 2 3 1 dt dt ln t 3 9t 1 C t 3 9t 1 3 t 3 9t 1 3
x2 dx x1
29.
x 1 dx
1 dx x1
1 2 x x ln x 1 C 2
1 2x22 dx
27. Let u 1 ex, du ex dx.
ex dx ln1 ex C 1 ex
4 4 x 4x 4 4x2 1dx x5 x3 x C 12x 4 20x2 15 C 5 3 15
31. Let u 2 x2, du 4 x dx.
xcos 2 x2 dx
1 cos 2 x24 x dx 4 1 sin 2 x 2 C 4
33. Let u x, du dx.
cscx cot x dx
1 csc x cot x dx 1 cscx C
35. Let u 5x, du 5 dx.
e5x dx
1 5x 1 e 5 dx e5x C 5 5
37. Let u 1 e x, du e x dx.
2 dx 2 ex 1
2
1 ex 1
ee dx x x
ex dx 2 ln1 e x C 1 ex
51
52
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
39.
ln x2 1 ln x 2 dx 2 ln x dx 2 C ln x C x x 2
41.
1 sin x dx cos x
2
sec x tan x dx ln sec x tan x ln sec x C ln sec xsec x tan x C
1 1 cos 1 cos 1
43.
csc 1 d cos 1
cos 1
cos 1
cos 1 cos2 1
cos 1 sin2
cot csc2
csc cot csc2 d
csc cot C
45.
1 cos C sin sin
1 cos C sin
3z 2 3 2z dz dz dz 2 2 z2 9 2 z2 9 z 9
47. Let u 2t 1, du 2 dt.
3 2 z lnz2 9 arctan C 2 3 3
1 1 2t 12
dt
1 2 dt 2 1 2t 12
1 arcsin2t 1 C 2
49. Let u cos
2t , du 2 sint 2 t dt. 2
1 tan2 t 1 2 sin2 t dt dt t2 2 cos2 t t2
1 2 ln cos 2 t
51.
6x x2
53.
4 dx 4x2 4x 65
55.
ds t 1 , 0, dt 2 1 t 4
3
dx 3
C
1 9 x 32
dx 3 arcsin
(b) u t 2, du 2t dt
1
t
−1
x 3 3 C
1 1 1 x 1 2 2x 1 dx arctan C arctan C x 1 22 16 4 4 4 8
s
(a)
1
0,
t 1 t 4
dt
1 1 2t dt arcsin t 2 C 2 2 1 t22 0.8
1 1 1 1 : arcsin 0 C ⇒ C 2 2 2 2 −1.2
−1
1.2
1 1 s arcsin t 2 2 2 − 0.8
Section 7.1
57.
59. y
10
Basic Integration Rules
1 ex2 dx
53
e2x 2ex 1 dx
1 e2x 2ex x C 2
−10
10 −2
y 3e0.2x
61.
dy sec2 x dx 4 tan2 x
63. Let u 2x, du 2 dx.
4
Let u tan x, du sec2 x dx. y
0
sec2 x 1 tan x dx arctan C 4 tan2 x 2 2
1
2
x
xe
dx
0
1 2
1
0
1 2 2x dx ex 2
2
x
e
4
1
2 3
0
1 1 dx 4 9x2 3
2 3
0
4
cos 2x2 dx
0
4
2 sin 2x
0
0
2x dx x2 9
1 1 e1 0.316 2
69. Let u 3x, du 3 dx.
1 2
1
0
1 2
67. Let u x2 9, du 2x dx.
65. Let u x2, du 2x dx.
cos 2x dx
3 4 3x2
x2 91 22x dx
4
0
2x2 9
71.
1 3x arctan 6 2
dx
4 0
4
1 1 x2 dx arctan C x2 4x 13 3 3
The antiderivatives are vertical translations of each other. 1
2 3 0 −7
0.175 18
5
−1
73.
1 2 d tan sec C or 1 sin 1 tan 2
The antiderivatives are vertical translations of each other.
75. Power Rule:
un du
un1 C, n 1. n1
u x2 1, n 3
6
− 2
7 2
−6
77. Log Rule:
du ln u C, u x2 1. u
79. The are equivalent because exC1 e x eC1 Ce x, C eC1
54
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
81. sin x cos x a sinx b sin x cos x a sin x cos b a cos x sin b sin x cos x a cos b sin x a sin b cos x Equate coefficients of like terms to obtain the following. 1 a cos b
1 a sin b
and
Thus, a 1 cos b. Now, substitute for a in 1 a sin b.
cos1 b sin b
1
1 tan b ⇒ b Since b
2
83.
0
4
1 ,a 2. Thus, sin x cos x 2 sin x . 4 cos 4 4
dx sin x cos x
dx 2 sinx 4
1 2
csc x
1 dx ln csc x cot x 4 4 4 2
C
1 a
4x dx 3 x2 1
85. Let u 1 x2, du 2x dx.
0
12x
1 6a2
x1 x2 dx
A4
y
x ax2 dx
0
1
Matches (a).
87.
2
a x3 3
1
2
3
1 x21 22x dx
0
2
4 1 x23 2 3
1
1 0
Let
4 3
1 2 1 , 12a 2 3, a . 6a 2 3 2 y
( a1 , a1)
2
y
y=x
x 1
2
3 1
1
y = ax 2
1 2
x
x − 12
1
1 2
− 12
2
−1
89. (a) Shell Method: Let u x2, du 2x dx.
1
V 2
(b) Shell Method:
y
V 2
dx
1 2
ex
0 1
2
x
2x dx
e
0
xex dx 2
0
2
x
xe
b
1
x 1 2
2
b 0
1
1 eb 2
e
x2
1 0
1 e 1.986 1
eb 2
b
4 3
3 4 3
ln 33 4
0.743
1 a 0
Section 7.2
4
91. A
0
x
1 A
x 5 dx 5 arcsin 5
25 x2
4
x
0
4
y 4
3
25 x 4
2
2 12
2x dx
(2.157, y )
1
0
x
1 5 25 x212 5 arcsin45
4 5
5 dx
25 x2
1 5 5 arcsin45 2
5 arcsin
0
Integration by Parts
1
4
2
3
4
0
1 3 5 arcsin45
2 2.157 arcsin45 y tan x
93.
y sec2 x 1 y 2 1 2 sec4 x
14
s
1 2 sec4x dx
0
1.0320
Section 7.2
Integration by Parts
1.
d sin x x cos x cos x x sin x cos x x sin x. Matches (b) dx
3.
d 2 x x e 2xex 2ex x2ex 2xex 2xex 2ex 2ex x2ex. Matches (c) dx
5.
xe2x dx
7.
ux
ln x2 dx
u ln x 2, dv dx
u x, dv e2x dx
11. dv e2x dx ⇒
v
1 e2x dx e2x 2
⇒ du dx
1 xe2x dx xe2x 2
1 e2x dx 2
1 1 1 xe2x e2x C 2x 2x 1 C 2 4 4e
9.
x sec2 x dx
u x, dv sec2 x dx
55
56
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
13. Use integration by parts three times. (1) dv ex dx ⇒
v
(2) dv ex dx ⇒
ex dx ex
⇒ du 3x2 dx
u x3
u x2
v
(3) dv ex dx ⇒
ex dx ex
⇒ du 2x dx
v
⇒ du dx
ux
x3ex dx x3ex 3 x2ex dx x3ex 3x2ex 6 xex dx x3ex 3x2ex 6xex 6ex C exx3 3x2 6x 6 C
15.
x2 ex dx 3
17. dv t dt
1 3
1 3 3 ex 3x2dx ex C 3
⇒
v
t dt
t2 2
1 dt t1
u lnt 1 ⇒ du
t lnt 1 dt
t2 1 lnt 1 2 2
t2 dt t1
t2 1 lnt 1 2 2
t1
t2 2
lnt 1
1 dt t1
t2
1 t lnt 1 C 2 2
1 2t 2 1 ln t 1 t 2 2t C 4
19. Let u ln x, du
ln x2 dx x
1 dx. x
21. dv
ln x2
1x dx ln3x
3
1 dx ⇒ 2x 12
v
u x2
xe2x xe2x dx 2x 12 22x 1
⇒ du 2x dx
x2 1ex dx
x2ex dx
ux
e2x dx 2
e2x xe2x C 22x 1 4
e2x C 42x 1
v
ex dx ex
⇒ du dx
exdx x2ex 2 xex dx ex
x2ex 2 xex
(2) dv ex dx ⇒
ex dx ex
1 22x 1
e2x2x 1 dx
23. Use integration by parts twice. v
2x 12 dx
⇒ du 2xe2x e2x dx
u xe2x
(1) dv ex dx ⇒
C
ex dx ex x2ex 2xex ex C x 12ex C
ex dx ex
Section 7.2
25. dv x 1 dx ⇒
2 x 112dx x 132 3
v
27. dv cos x dx ⇒
⇒ du dx
ux
2 2 x x 1 dx xx 132 3 3
x 132 dx
cos x dx sin x
x cos x dx x sin x
sin x dx x sin x cos x C
4 2 xx 132 x 152 C 3 15
2x 132 3x 2 C 15
29. Use integration by parts three times. (1) u x3, du 3x2, dv sin x dx, v cos x
x3 sin dx x3 cos x 3 x2 cos x dx
(2) u x2, du 2x dx, dv cos x dx, v sin x
x3 sin x dx x3 cos x 3 x2 sin x 2 x sin x dx x3 cos x 3x2 sin x 6 x sin x dx
(3) u x, du dx, dv sin x dx, v cos x
x3 sin x dx x3 cos x 3x2 sin x 6 x cos x
cos x dx
x3 cos x 3x2 sin x 6x cos x 6 sin x C
t csc t cot t dt t csc t
csc t dt
⇒
33. dv dx
31. u t, du dt, dv csc t cot dt, v csc t
v
u arctan x ⇒ du
t csc t ln csc t cot t C
dx x
1 dx 1 x2
arctan x dx x arctan x x arctan x
35. Use integration by parts twice. (1) dv e2xdx ⇒
v
1 e2x dx e2x 2
1 1 e2x sin x dx e2x sin x 2 2
5 4
(2) dv e2x dx ⇒
1 1 1 2x 1 e2x cos x dx e2x sin x e cos x 2 2 2 2
1 1 e2x sin x dx e2x sin x e2x cos x 2 4 1 e2x sin x dx e2x2 sin x cos x C 5
v
x dx 1 x2
1 ln1 x2 C 2
1 e2x dx e2x 2
u cos x ⇒ du sin x dx
u sin x ⇒ du cos x dx
57
⇒ du dx
ux
v
Integration by Parts
e2x sin x dx
58
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
37. y xex
2
y
1 2 2 xe x dx ex C 2
39. Use integration by parts twice. (1) dv
1
2 3t
dt ⇒
v
(2) dv 2 3t dt ⇒
41.
v
2 2 3t12 dt 2 3t32 9
⇒ du dt
ut
2 2 3t12 dt 2 3t 3
⇒ du 2t dt
u t2
y
2t 2 2 3t 4 t2 dt 3 3
2 3t
t 2 3t dt
2 2t 2 2 3t 4 2t 2 3t32 3 3 9 9
8t 16 2t 2 2 3t 2 3t32 2 3t52 C 3 27 405
2 2 3t 27t 2 24t 32 C 405
2 3t32 dt
cos yy 2x
cos y dy
2x dx
sin y x 2 C 43. (a)
dy x y cos x, 0, 4 dx
(b)
y
8
6
dy
y
y12 dy
2
2
2
x cos x dx
4
u x, du dx, dv cos x dx, v sin x
x cos x dx
2y12 x sin x
x 4
sin x dx
6
x sin x cos x C
0, 4: 2412 0 1 C ⇒ C 3 2 y x sin x cos x 3
45.
dy x x8 e , y0 2 dx y
10
−10
10 −2
−6
6 −2
Section 7.2
Integration by Parts
47. u x, du dx, dv ex2 dx, v 2ex2
xex2 dx 2xex2
4
Thus,
2ex2 dx 2xex2 4ex2 C
xex2 dx 2xex2 4ex2
0
8e2 4e2 4
4 0
12e2 4 2.376. 49. See Exercise 27.
2
0
1
51. u arccos x, du
1 x2
0
1 2
dx, dv dx, v x
arccos x dx x arccos x
x
1 x2
dx x arccos x 1 x2 C
12
Thus,
2
x cos x dx x sin x cos x
arccos x x arccos x 1 x2
0
12 0
34 1
1 1 arccos 2 2
3 1 0.658. 6 2
53. Use integration by parts twice. (1) dv e x dx ⇒
v
(2) dv ex dx ⇒
e x dx e x
u sin x ⇒ du cos x dx
e x sin x dx e x sin x
e x cos x dx e x sin x e x cos x
v
ex dx ex
u cos x ⇒ du sin x dx
e x sin x dx
2 e x sin x dx e xsin x cos x e x sin x dx
ex sin x cos x C 2
1
Thus,
e x sin x dx
0
55. dv x2 dx, v
x2 ln x dx
x
1 0
e 1 esin 1 cos 1 1 sin 1 cos 1 0.909. 2 2 2
1 x3 , u ln x, du dx 3 x
x3 ln x 3
x3 1 dx 3 x
1 2 x3 ln x x dx 3 3
2
Hence,
e2 sin x cos x
x2 ln x dx
x3 ln x 9x 3
1
2
3
1
1
8 1 8 7 8 ln 2 ln 2 1.071. 3 9 9 3 2
59
60
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
57. dv x dx, v
x2 1 dx , u arcsec x, du
2 x x2 1
x2 x arcsec x dx arcsec x 2
Hence,
4
x22 dx x x2 1
2 arcsec x 2 x x2
x arcsec x dx
2
1
x2 2x 1 dx arcsec x 2 4 x2 1
8 arcsec 4
x2 1 arcsec x x2 1 C 2 2
8 arcsec 4
2
15
2
15
2
1
4 2
23 23
3
2
2 3
7.380.
59.
x2e2x dx x2
12 e 2x 14 e 2 18 e C 2x
2x
2x
u and its derivatives
x2
e2x
2x
1 2x 2e
2
1 2x 4e
0
1 2x 8e
1 1 1 x2e2x xe2x e2x C 2 2 4 1 e2x2x2 2x 1 C 4
61.
63.
u and its derivatives
x3 cos x 3x2 sin x 6x cos x 6 sin x C
x3
sin x
3x2 6 sin x x3 6x cos x C
3x2
cos x
6x
sin x
6
cos x
0
sin x
x sec2 x dx x tan x ln cos x C
71. Yes. Let u x and du
v and its antiderivatives
Alternate signs
u and its derivatives
x
sec2 x
1
tan x
0
ln cos x
67. No. Substitution.
1 , dx.
x 1
73.
t 3e4t dt
Substitution also works. Let u x 1
2
0
v and its antiderivatives
Alternate signs
x3 sin x dx x3cos x 3x2sin x 6x cos x 6 sin x C
65. Integration by parts is based on the product rule.
75.
v and its antiderivatives
Alternate signs
e2x sin 3x dx
e2x2 sin 3x 3 cos 3x 13
2
0
1 2e 3 0.2374 13
69. Yes. u x2, dv e2x dx
e4t 32t3 24t2 12t 3 C 128
Section 7.2 77. (a) dv 2x 3 dx ⇒
v
Integration by Parts
1 2x 312 dx 2x 332 3
⇒ du 2 dx
u 2x
2 2 2x 2x 3 dx x2x 332 3 3
2x 332 dx
2 2 x2x 332 2x 352 C 3 15
2 2 2x 3323x 3 C 2x 332x 1 C 15 5 1 u3 and dx du 2 2
(b) u 2x 3 ⇒ x
2x 2x 3 dx
2
1 u 3 12 1 u du 2 2 2
u32 3u12 du
1 2 52 u 2u32 C 2 5
1 1 2 u32u 5 C 2x 3322x 3 5 C 2x 332x 1 C 5 5 5 x
79. (a) dv
4 x2
dx ⇒ v
4 x212x dx 4 x2
u x2 ⇒ du 2x dx
x3 dx x2 4 x2 2 x 4 x2 dx
4 x2 2 1 x2 4 x2 4 x232 C 4 x2 x2 8 C 3 3
(b) u 4 x2 ⇒ x2 u 4 and 2x dx du ⇒ x dx
x3 dx
4 x2
1 2
x2 x dx
4 x2
1 du 2
u41 du
u 2
u12 4u12 du
1 2 32 u 8u12 C 2 3
1 1 1 u12u 12 C 4 x2 4 x2 12 C 4 x2 x2 8 C 3 3 3
81. n 0: n 1: n 2: n 3: n 4:
ln x dx xln x 1 C
x ln x dx
x2 2 ln x 1 C 4
x2 ln x dx
x3 3 ln x 1 C 9
x3 ln x dx
x4 4 ln x 1 C 16
x 4 ln x dx
x5 5 ln x 1 C 25
In general,
xn ln x dx
xn1 n 1ln x 1 C. (See Exercise 85.) n 12
61
62
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
83. dv sin x dx ⇒
⇒ du nx n1 dx
u xn
x n sin
85. dv x n dx ⇒
v cos x
x dx
x n
cos x n
u ln x
x n1
cos x dx
v
1 dx x
⇒ du
x n ln x dx
x n1 n1
x n1 ln x n1
xn dx n1
x n1 x n1 C ln x n1 n 12
x n1 n 1ln x 1 C n 12
87. Use integration by parts twice. 1 v eax a
(1) dv eax dx ⇒
u cos bx ⇒ du b sin bx dx
u sin bx ⇒ du b cos bx dx
eax cos bx dx
eax sin bx b eax cos bx b a a a a
Therefore, 1
eax sin bx b a a
eax sin bx dx
b2 a2
e
ax
sin bx dx
eax sin bx dx
eax sin bx dx
x3 ln x dx
eax sin bx b2 2 a a
eax sin bx dx
eaxa sin bx b cos bx a2 eaxa sin bx b cos bx C. a2 b2 91. a 2, b 3 (Use formula in Exercise 88.)
89. n 3 (Use formula in Exercise 85.)
1 v eax a
(2) dv eax dx ⇒
x4 4 ln x 1 C 16
e2x cos 3x dx
e2x2 cos 3x 3 sin 3x C 13
1
93. dv ex dx ⇒
4
A
xex dx xex
0
1
4
0
4
0
ex dx
4 ex e4
4
0
5 0.908 e4
exsin x cos x 1 2
1 1 1 1 2 e 1 2 e
0.395 See Exercise 87.
3
−1
ex sin x dx
0
⇒ du dx
ux
95. A
v ex
1
7 −1
0
1.5 0
1 0
Section 7.2
e
97. (a) A
e
ln x dx x x ln x
1
1
1 See Exercise 4.
y
(b) Rx ln x, rx 0
Integration by Parts
2
(e, 1)
e
V
ln x2 dx
1
1 e
xln x2 2x ln x 2x
1
x
Use integration by parts twice, see Exercise 7.
1
2
3
e 2 2.257 e
(c) px x, hx ln x
(d)
e
y
e 2 1 13.177 See Exercise 85. 2
x, y
x ln x dx 2
1
e
V 2
99. Average value
1
x
x2 1 2 ln x 4
1
4 sin 2t 2 cos 2t 1 4t 4 cos 2t 2 sin 2t e 5e4t 20 20
7 1 e4 0.223 10
10
100,000 4000te0.05t dt 4000
0
25 te0.05t dt
0
Let u 25 t, dv e0.05t dt, du dt, v
100 0.05t e 5
e dt
100 e e 4000 25 t 10,000 $931,265 5 25
25 t
100 0.05t e 5 0.05t
10 0
10
cos n cos n n n
2 cos n n
22nn, , ifif nn isis even odd
10
0.05t
0
10 0
x 1 x sin nx dx cos nx 2 sin nx n n
100 5
0.05t
0
103.
ln x2dx e 2 0.359 1 2
e
1 e2 , 2.097, 0.359 4 2
2
e4t cos 2t 5 sin 2t dt
10
P 4000
1 e 2 1
0
101. ct 100,000 4000t, r 5%, t1 10 P
1 x ln x dx e2 1 2.097 1 4
0
From Exercises 87 and 88
63
64
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
105. Let u x, dv sin
1
I1
x sin
0
n2 x dx, du dx, v n2 cos n2 x .
n 2x n x dx cos x 2 n 2
2
x 2 sin
1
2 n
0
2 n cos n 2
n 2 2 cos n 2 n
Let u x 2, dv sin I2
1
1
cos
0
n2 x dx
n2 sin n2 x
1
2
0
sin n2 2
n2 x dx, du dx, v n2 cos n2 x .
n 2x 2 n x dx cos x 2 n 2
2 n cos n 2
2 n 2 cos n 2 n
2 1
2 n
2
1
n2 sin n2 x
cos
n2 x dx
2
2
1
sin n2 2
n2 sin n2 n2 sin n2 n sin n2 2
hI1 I2 bn h
8h
2
2
107. Shell Method:
b
V 2
x f x dx
y
a
y = f ( x)
f (a )
dv x dx ⇒
x2 v 2
f (b )
u f x ⇒ du fx dx V 2
x2 f x x2 fxdx 2
b
2
x a
a
b
b2f b a2f a
b
x2 fx dx
a
Disk Method: V
f a
b2 a2 dy
0
f b
f a
b2 f 1 y 2 dy
b2 a2 f a b2 f b f a
b2f b a2f a
f b
f a
f b
f a
f 1y2 dy
f 1 y2 dy
Since x f 1 y, we have f x y and fxdx dy. When y f a, x a. When y f b, x b. Thus,
f b
f a
f 1 y 2 dy
b
x 2fx dx
a
and the volumes are the same.
Section 7.3
Trigonometric Integrals
109. fx xex (a) f x
xex dx xex ex C
(b)
1
Parts: u x, dv ex dx f 0 0 1 C ⇒ C 1 0
f x xex ex 1 (c) You obtain the points n
xn
4 0
(d) You obtain the points
yn
n
xn
yn
0
0
0
0
0
0
1
0.05
0
1
0.1
0
2
0.10
2.378 103
2
0.2
0.0090484
3
0.15
0.0069
3
0.3
0.025423
4
0.20
0.0134
4
0.4
0.047648
80
0.9064
40
0.9039
4.0
1
4.0
1
0
4
0
0
4 0
(e) f 4 0.9084 The approximations are tangent line approximations. The results in (c) are better because x is smaller.
Section 7.3
Trigonometric Integrals
1. f x sin4 x cos4 x (a) sin4 x cos4 x
2x 2x 1 cos
1 cos
2 2 2
2
1 1 2 cos 2x cos2 2x 1 2 cos 2x cos2 2x 4
1 1 cos 4x 22 4 2
1 3 cos 4x 4 (b) sin4 x cos4 x sin2 x2 cos4 x 1 cos2 x2 cos4 x 1 2 cos 2 x 2 cos4 x (c) sin4 x cos4 x sin4 x 2 sin2 x cos2 x cos4 x 2 sin2 x cos2 x sin2 x cos2 x2 2 sin2 x cos2 x 1 2 sin2 x cos2 x —CONTINUED—
65
66
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
1. —CONTINUED— (d) 1 2 sin2 x cos2 x 1 2 sin x cos xsin x cos x 1 sin 2x 1
12 sin 2x
1 2 sin 2x 2
(e) Four ways. There is often more than one way to rewrite a trigonometric expression. 3. Let u cos x, du sin x dx.
5. Let u sin 2x, du 2 cos 2x dx.
cos3 x sin x dx cos3 xsin x dx
sin5 2x cos 2x dx
1 cos4 x C 4
1 2
sin5 2x2 cos 2xdx
1 sin6 2x C 12
7. Let u cos x, du sin x dx.
sin x1 cos2 x2 cos2 x dx
sin5 x cos2 x dx
cos2 x 2 cos4 x cos6 xsin x dx
9.
cos3 sin d
cos 1 sin2 sin 1 2 d
sin 1 2 sin 5 2cos d
2 2 sin 3 2 sin 7 2 C 3 7
13.
sin2
cos2 d
1 cos 2 2
1 cos 2 d 2
1 1 cos2 2 d 4 1 4
1
1 cos 4 d 2
1 1 cos 4 d 8
1 1 sin 4 C 8 4
1 4 sin 4 C 32
11.
1 2 1 cos3 x cos5 x cos7 x C 3 5 7
cos2 3x dx
1 cos 6x dx 2
1 1 x sin 6x C 2 6
1 6x sin 6x C 12
Section 7.3
Trigonometric Integrals
15. Integration by parts. 1 cos 2x x sin 2x 1 ⇒ v 2x sin 2x 2 2 4 4
dv sin2 x dx u x ⇒ du dx
x sin2 x dx
1 1 x2x sin 2x 4 4
2x sin 2x dx
1 1 1 1 x2x sin 2x x2 cos 2x C 2x2 2x sin 2x cos 2x C 4 4 2 8 17. Let u sin x, du cos x dx.
2
2
cos3 x dx
0
1 sin2 x cos x dx
0
sin x
2
1 3 sin x 3
0
2 3
19. Let u sin x, du cos x dx.
2
2
cos7 x dx
0
1 sin2 x3 cos x dx
0
2
1 3 sin2 x 3 sin4 x sin6 x cos x dx
0
sin x sin3 x
21.
sec3x dx
1 ln sec 3x tan 3x C 3
25. dv sec2 x dx ⇒
sec3 x dx
sec3 x dx
27.
tan5
x dx 4
sec4 5x dx
0
16 35
1 tan2 5x sec2 5x dx
tan3 5x 1 tan 5x C 5 3
tan 5x 3 tan2 5x C 15
1 tan x
1 sec x tan x
2 sec3 x dx
⇒ du sec x tan x dx
u sec x
v
23.
2
3 5 1 sin x sin7 x 5 7
sec x tan2 x dx
1 sec x tan x
sec xsec2 x 1 dx
1 sec x tan x ln sec x tan x C1
1 sec x tan x ln sec x tan x C 2
sec2
tan3
x x 1 tan3 dx 4 4
x x sec2 dx 4 4
29. u tan x, du sec2 x dx
tan3
x dx 4
tan4
x 4
tan4
x x x 2 tan2 4 ln cos C 4 4 4
sec2
x x 1 tan dx 4 4
sec2 x tan x dx
1 2 tan x C 2
67
68
31.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
tan2 x sec2 x dx
tan3 x C 3
33.
sec6 4x tan 4x dx
35. Let u sec x, du sec x tan x dx.
sec3 x tan x dx
39. r
37.
sec2 xsec x tan x dx
1 4
sec5 4x4 sec 4x tan 4x dx
sec6 4x C 24
sec2 x 1 dx sec x
sec x cos x dx
ln sec x tan x sin x C
1 cos2 2 d
41. y
1 4
1 4
1 1 1 sin2 sin4 C 4 2 8
1 12 8 sin2 sin4 C 32
1 2 cos2 cos22 d 1 2 cos2
1 sec3 x C 3
sin4 d
1 cos4 d 2
43. (a)
tan2 x dx sec x
1 4
(b)
y 4
dy sin2 x, 0, 0 dx y
sin2 x dx
tan3 3x sec 3x dx
sec2 3x 1sec 3x tan 3x dx
1 1 sec2 3x3 sec 3x tan 3x dx 3 sec 3x tan 3x dx 3 3 1 1 sec3 3x sec 3x C 9 3
4
1 cos 2x dx 2
−6
6
x
sin 2x 1 C x 2 4
4
−4
45.
−4
1 sin 2x 0, 0: 0 C, y x 2 4
dy 3 sin x , y0 2 dx y
47.
sin 3x cos 2x dx
8
−9
9
1 2
sin 5x sin x dx
1 1 cos 5x cos x C 2 5
1 cos 5x 5 cos x C 10
−4
49.
sin sin 3 d
1 2
cos 2 cos 4 d
1 1 1 sin 2 sin 4 C 2 2 4
1 2 sin 2 sin 4 C 8
51.
cot3 2x dx
csc2 2x 1 cot 2x dx
1 2 cos 2x 1 cot 2x2csc2 2x dx dx 2 2 sin 2x
1 1 cot2 2x ln sin 2x C 4 2
1 ln csc2 2x cot2 2x C 4
Section 7.3
53. Let u cot , du csc2 d.
csc4 d
1 dx sec x tan x
cot2 t dt csc t
csc2 1 cot2 d
csc2 d
cot
57.
55.
Trigonometric Integrals
csc2 t 1 dt csc t
csc t sin tdt
ln csc t cot t cos t C
csc2 cot2 d
1 3 cot C 3
cos2 x dx sin x
1 sin2 x dx sin x
csc x sin x dx
ln csc x cot x cos x C
59.
tan4 t sec4 t dt
tan2 t sec2 ttan2 t sec2 t dt
tan2 t sec2 t 1
tan2 t sec2 t dt 2 sec2 t 1 dt 2 tan t t C
61.
sin2 x dx 2
0
x
1 sin 2x 2
0
4
1 cos 2x dx 2
63.
tan3 x dx
0
4
sec2 x 1 tan x dx
0
4
sec2 x tan x dx
0
4
0
sin x dx cos x
4
0
1 2 tan x ln cos x 2
1 1 ln 2 2 65. Let u 1 sin t, du cos t dt.
2
0
67. Let u sin x, du cos x dx. 2
0
cos t dt ln 1 sin t 1 sin t
2
ln 2
2
2
cos3 x dx 2
1 sin2 x cos x dx
0
2 sin x
69.
2
1 3 sin x 3
0
1 x cos4 dx 6x 8 sin x sin 2x C 2 16 6
−9
9
−6
71.
sec5 x dx
3
1 3 sec3 x tan x sec x tan x ln sec x tan x C 4 2
−3
3
−3
4 3
69
70
73.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
sec5 x tan x dx
4
1 sec5 x C 5
75.
sin 2 sin 3 d
0
1 1 sin sin 5 2 5
4
0
3 2 10
5
−2
2
−5
2
77.
sin4 x dx
0
2
1 3x 1 sin 2x sin 4x 4 2 8
79. (a) Save one sine factor and convert the remaining sine factors to cosine. Then expand and integrate.
0
3 16
(b) Save one cosine factor and convert the remaining cosine factors to sine. Then expand and integrate. (c) Make repeated use of the power reducing formula to convert the integrand to odd powers of the cosine.
81. (a) Let u tan 3x, du 3 sec2 3x dx.
(b)
sec4 3x tan3 3x dx
1 3
1 3
tan6 3x tan4 3x C1 18 12
sec2 3x tan3 3x sec2 3x dx
0.05
− 0.5
tan2 3x 1 tan3 3x3 sec2 3x dx
0.5
− 0.05
tan5 3x tan3 3x3 sec2 3x dx
Or let u sec 3x, du 3 sec 3x tan 3x dx.
(c)
sec4 3x tan3 3x dx
1 3
sec6 3x sec4 3x C 18 12
sec3 3x tan2 3x sec 3x tan 3x dx
sec3 3xsec2 3x 13 sec 3x tan 3x dx
sec6 3x sec4 3x 1 tan2 3x3 1 tan2 3x2 C C 18 12 18 12
1 1 1 1 1 1 1 tan6 3x tan4 3x tan2 3x tan4 3x tan2 3x C 18 6 6 18 12 6 12
tan6 3x tan4 3x 1 1 C 18 12 18 12
tan6 3x tan4 3x C2 18 12
1
83. A
sin2 x dx
y
0 1
0
1 cos2 x dx 2
x 1 sin2 x 2 4 1 2
1
1
1 2
0
x 1 2
1
Section 7.3
85. (a) V
2
sin2 x dx
0
(b) A
0
sin x dx cos x
0
1 x sin 2x 2 2
1 cos 2x dx
0
0
Trigonometric Integrals
2 2
112
Let u x, dv sin x dx, du dx, v cos x. x
1 A
x sin x dx
0
y
0
0
sin2 x dx
0
2
y
1
1 cos 2x dx
0
1 2
( π2 , π8 (
8 0
π 2
2 , 8
87. dv sin x dx ⇒
1 x cos x sin x 2
cos x dx
0
1 1 x sin 2x 8 2
x, y
x cos x
1 2A 1 8
1 2
x
π
v cos x
u sinn1 x ⇒ du n 1sinn2 x cos x dx
sinn x dx sinn1 x cos x n 1 sinn2 x cos2 x dx sinn1 x cos x n 1 sinn2 x1 sin2 x dx
sinn1 x cos x n 1 sinn2 x dx n 1 sinn x dx
Therefore, n sinn x dx sinn1 x cos x n 1 sinn2 x dx sinn x dx
sinn1 x cos x n 1 n n
sinn2 x dx.
89. Let u sinn1 x, du n 1sinn2 x cos x dx, dv cosm x sin x dx, v
mn m1
cosm1 x . m1
cosm x sinn x dx
sinn1 x cosm1 x n1 m1 m1
sinn2 x cosm2 x dx
n1 sinn1 x cosm1 x m1 m1
sinn2 x cosm x1 sin2 x dx
n1 sinn1 x cosm1 x m1 m1
sinn2 x cosm x dx
cosm x sinn x dx
sinn1 x cosm1 x n1 m1 m1
sinn2 x cosm x dx
cosm x sinn x dx
cosm1 x sinn1 n1 mn mn
cosm x sinn2 x dx
n1 m1
sinn x cosm x dx
71
72
91.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
sin5 x dx
sin4 x cos x 4 5 5
sin3 x dx
sin2 x cos x 2 sin4 x cos x 4 5 5 3 3
sin x dx
4 8 1 sin2 x cos x cos x C sin4 x cos x 5 15 15
93.
sec4
cos x 3 sin4 x 4 sin2 x 8 C 15
2 5 x dx 25 sec 2 5 x 25 dx 4
2 2 x 2 x 5 1 sec2 tan 2 3 5 5 3
2 x 2 x 2 x 5 sec2 tan 2 tan 6 5 5 5
2 x 5 tan 6 5
a1
1 b1 6 a0
sec2
2 5 x 25 dx
C
sec 2 5 x 2 C
2
t t b1 sin where: 6 6
12
1 12 1 6
95. (a) f t a0 a1 cos a0
f t dt
0
12
f t cos
0
12
f t sin
0
t dt 6 t dt 6
12 0 30.9 432.2 241.1 453.7 264.6 474.0 278.2 477.0 271.0 3122 460.1 247.1 435.7 30.9 55.46
a1
12 0 2 30.9 cos 0 4 32.2 cos 2 41.1 cos 4 53.7 cos 2 64.6 cos 6312 6 3 2 3
5 7 4 278.2 cos 4 77.0 cos 2 71.0 cos 6 6 3
3 5 11 2 47.1 cos 4 35.7 cos 30.9 cos 2 23.88 2 3 6
4 74.0 cos 4 60.1 cos b1
12 0 2 30.9 sin 0 4 32.2 sin 2 41.1 sin 4 53.7 sin 2 64.6 sin 6312 6 3 2 3
5 7 4 278.2 sin 4 77.0 sin 2 71.0 sin 6 6 3
3 5 11 2 47.1 sin 4 35.7 sin 30.9 sin 2 3.34 2 3 6
4 74.0 sin 4 60.1 sin Ht 55.46 23.88 cos —CONTINUED—
t t 3.34 sin 6 6
Section 7.3
Trigonometric Integrals
95. —CONTINUED— (b) a0
12 0 18.0 417.7 225.8 436.1 245.4 455.2 259.9 459.4 253.1 3122 443.2 234.3 424.2 18.0 39.34
a1
12 0 2 18.0 cos 0 4 17.7 cos 2 25.8 cos 4 36.1 cos 2 45.4 cos 6312 6 3 2 3
5 7 4 259.9 cos 4 59.4 cos 2 53.1 cos 6 6 3
3 5 11 2 34.3 cos 4 24.2 cos 18 cos 2 20.78 2 3 6
4 55.2 cos 4 43.2 cos b1
12 0 2 18.0 sin 0 4 17.7 sin 2 25.8 sin 4 36.1 sin 2 45.4 sin 6312 6 3 2 3
5 7 4 259.9 sin 4 59.4 sin 2 53.1 sin 6 6 3
3 5 11 2 34.3 sin 4 24.2 sin 18 sin 2 4.33 2 3 6
4 55.2 sin 4 43.2 sin Lt 39.34 20.78 cos
t t 4.33 sin 6 6
(c) The difference between the maximum and minimum temperatures is greatest in the summer. 85
H L
0
12 15
97.
cosmx cosnx dx
sinmx sinnx dx
sinmx cosnx dx
1 sinm nx sinm nx 2 mn mn
1 2
0, m n
sinm nx sinm nx dx
1 cosm nx cosm nx 2 mn mn
1 2
, m n
cosm n cosm n cosm n cosm n mn mn mn mn
0, since cos cos. sinmx cosmx dx
0, m n
cosm nx cosm nx dx
1 sinm nx sinm nx 2 mn mn 1 2
1 sin2mx m 2
0
73
74
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
Section 7.4 1.
Trigonometric Substitution
Indefinite integral:
x2 16
x
dx
2
x x2 16
2
4x
x2 16 x2 16 4
4x2
xx x 16 164 4x
4
3.
x2 16 4 d d 4 ln x2 16 C 4 ln x2 16 4 4 ln x x2 16 C dx x dx
4 x x x2 16
xx2 16 x2 16 4
4x2 4x2 16 16x2 16 x2x2 16 4x2 xx2 16 x2 16 4 x2 16x2 16 4x2 16
xx2 16 x2 16 4
x2 16 x2 16 x2 16 4 2 2 x x x 16 x 16 4
Matches (b)
x x16 x2 14 x1216 x2122x 16 x2 d C 8 8 arcsin dx 4 2 2 1 x42
16 x2 8 x2 2 2 2 16 x 216 x
16 x2 16 x2 x2 2 2 2 16 x2 216 x 216 x 216 x
Matches (a) 5. Let x 5 sin , dx 5 cos d, 25 x2 5 cos .
1 dx 25 x232
5 cos d 5 cos 3
1 25
5 x
sec d 2
θ 25 − x 2
1 tan C 25
x 2525 x2
7. Same substitution as in Exercise 5
25 x2
x
dx
C
25 cos2 d 1 sin2 5 d 5 csc sin d 5 sin sin
4x2 16 x2 16 4 x2 x2 16 4
5 ln csc cot cos C 5 ln
5 25 x2 25 x2 C x
Section 7.4
Trigonometric Substitution
9. Let x 2 sec , dx 2 sec tan d, x2 4 2 tan .
1 dx x2 4
2 sec tan d 2 tan
ln
x 2
x2 4
2
x
x2 − 4
sec d ln sec tan C1 θ 2
C1
ln x x 4 ln 2 C1 ln x x2 4 C 2
11. Same substitution as in Exercise 9
x3x2 4 dx
8 sec3 2 tan 2 sec tan d 32 tan2 sec4 d
32 tan2 1 tan2 sec2 d 32
tan3 tan5 C 3
5
32 x2 432 x2 4 32 3 tan 5 3 tan2 C 53 C 15 15 8 4
1 1 2 x 432 20 3x2 4 C x2 4323x2 8 C 15 15
13. Let x tan , dx sec2 d, 1 x2 sec .
x1 x2 dx
1 + x2
sec3 1 C 1 x232 C 3 3
tan sec sec2 d
x
θ
Note: This integral could have been evaluated with the Power Rule.
1
15. Same substitution as in Exercise 13
1 dx 1 x22
1 1 x2
4
dx 1 + x2 x
d sec4
sec2
θ
cos2 d
sin 2 1 2 2
1
sin cos C 2
x 1 arctan x 2 1 x2
1 x arctan x C 2 1 x2
1
1 1 cos 2 d 2
1 1 x2
C
17. Let u 3x, a 2, and du 3 dx.
4 9x2 dx
1 22 3x2 3 dx 3
1 1 3x4 9x2 4 ln 3x 4 9x2 C 3 2
2 1 x4 9x2 ln 3x 4 9x2 C 2 3
75
76
19.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
x 1 dx 9 2
x2
x2 9122x dx
x2
21.
1 16
x2
dx arcsin
4 C x
9C
(Power Rule) 23. Let x 2 sin , dx 2 cos d, 4 x2 2 cos .
16 4x2 dx 2 4 x2 dx
2 x
θ
2 2 cos 2 cos d
4 − x2
8 cos2 d 4 1 cos 2 d
4
1 sin 2 C 2
4 4 sin cos C 4 arcsin
2x x4 x
2
C
27. Let x sin , dx cos d, 1 x2 cos .
25. Let x 3 sec , dx 3 sec tan d, x2 9 3 tan .
1 x2 9
dx
3 sec tan d 3 tan
1 x2
x4
dx
sec d
cos cos d sin4
cot2 csc2 d
1 cot3 C 3
ln sec tan C1
x2 9 x ln C1 3 3
ln x x2 9 C
1
1 x232 C 3x3
x
θ
x
x2 − 9
1 − x2
θ 3
29. Same substitutions as in Exercise 28
1 dx x4x2 9
1 3
32 sec2 d 32 tan 3sec
1 1 4x2 9 3 csc d ln csc cot C ln C 3 3 2x
Section 7.4 31. Let x 5 tan , dx 5 sec2 d, x2 5 5 sec2 .
5x dx x2 532
55 tan 5 sec2 d 5 sec2 32
5
x2 + 5 x
tan d sec
θ 5
5 sin d 5 cos C 5
5
x2 5
C
5 C 5
x2
33. Let u 1 e2x, du 2e2x dx.
e2x1 e2x dx
1 2
1 1 e2x122e2x dx 1 e2x32 C 3
35. Let ex sin , ex dx cos d, 1 e2x cos .
ex1 e2x dx
cos2 d
1 2
1
1 cos 2 d
sin 2 1 2 2
ex
θ 1 − e 2x
1 1 sin cos C arcsin ex ex1 e2x C 2 2 37. Let x 2 tan , dx 2 sec2 d, x2 2 2 sec2 .
1 dx 4 4x2 x4
1 dx x2 22 2
2
4
sec2
4 sec4
cos2
x2 + 2 x
d
θ 2
d
2 1 cos 2 d
2 1
4 2
8 2
8
21 sin 2 C sin cos C
1 x 1 x arctan C 4 x2 2 2 2
Trigonometric Substitution
77
78
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
1 39. Since x > , 2 1 dx, dv dx ⇒ v x x4x2 1
u arcsec 2x, ⇒ du
arcsec 2x dx x arcsec 2x
2x sec , dx
arcsec 2x dx x arcsec 2x
1 4x x2
dx
1 sec tan d, 4x2 1 tan 2
x arcsec 2x
41.
1 4x2 1
dx
1 12 sec tan d x arcsec 2x tan 2
sec d
1 1 ln sec tan C x arcsec 2x ln 2x 4x2 1 C. 2 2
1 4 x 22
dx arcsin
x 2 2 C
43. Let x 2 2 tan , dx 2 sec2 d, x 22 4 2 sec .
x dx x 4x 8 2
x dx x 22 4
2 tan 22 sec2 d 2 sec
2 tan 1sec d
2 sec ln sec tan C1
2
x 22 4
2
ln
x 4x 8 2 ln
x 22 4
2
x2 2
C1
x 4x 8 x 2 C
x2 4x 8 2 ln x2 4x 8 x 2 ln 2 C1
2
2
45. Let t sin , dt cos d, 1 t 2 cos2 . (a)
t2 dt 1 t232
sin2 cos d cos3 1
tan2
d
t
θ 1 − t2
sec2 1 d
tan C
32
Thus,
0
t 1 t2
arcsin t C
32
t2 t dt arcsin t 1 t232 1 t 2
0
32 14
(b) When t 0, 0. When t 32, 3. Thus,
32
0
t2 dt tan 1 t232
3
0
3
0.685. 3
arcsin
3
2
3
0.685. 3
Section 7.4
Trigonometric Substitution
47. (a) Let x 3 tan , dx 3 sec2 d, x2 9 3 sec .
x3 x2 9
dx
27 tan3 3 sec2 d 3 sec
27 sec2 1 sec tan d
13 sec
27
3
9
3
Thus,
0
sec C 9 sec3 3 sec C
3
x2 9
3
3
C 31 x 9
x2 9
2
3
13 54
9x2 9 C
3
x3 1 dx x2 932 9x2 9 3 x2 9
32
0
9 27
2 272
18 92 9 2 2 5.272. (b) When x 0, 0. When x 3, 4. Thus,
3
0
x3 dx 9 sec3 3 sec 9
x2
4
0
9 22 32 91 3 9 2 2 5.272.
49. (a) Let x 3 sec , dx 3 sec tan d, x2 9 3 tan .
x2 x2 9
dx
9 sec2 3 sec tan d 3 tan
9
sec3
x
d
θ 3
12 sec tan 21sec d
9
(7.3 Exercise 90)
9 sec tan ln sec tan 2
9 x 2 3
x2 9
3
Hence,
6
4
ln
x2 9 x 3 3
x2 9 x2 9 xx2 9 x dx ln x2 9 2 9 3 3
9 2
27 627 ln 2 9 3
6
4
4 9 7 ln 43 37
93 27
6 27 4 7 9 ln ln 2 3 3
93 27
9 6 33 ln 2 4 7
93 27
9 4 7 2 3 12.644. ln 2 3
—CONTINUED—
x2 − 9
79
80
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
49. —CONTINUED— (b) When x 4, arcsec
43 .
When x 6, arcsec2
6
x2 9 dx sec tan ln sec tan 9 2
x2
4
. 3
9
2 2
x2
53.
x2
9 4 7 4 7 ln 3 3 3 3
3 ln 2 3 2
93 27
51.
3
arcsec43
9 6 33 12.644 ln 2 4 7
1 x2 dx x2 10x 9 x 15 33 ln x 5 x2 10x 9 C 2 10x 9
x2 1 dx xx2 1 ln x x2 1 C 2 1
a
57. A 4
0
55. (a) u a sin
b a2 x2 dx a
y
a
4b a
b y= a
(b) u a tan
a2 − x2
b
a2 x2 dx
0
4b 1 a 2
a2 arcsin
x xa2 x2 a
−a
a
x
a
0
−b
2b 2 a a 2
ab Note: See Theorem 7.2 for a2 x2 dx. 59. x2 y2 a2 x ± a2 y2
a
A2
a2 y2 dy a2 arcsin
h
ay ya
2
2 a
a2
2
arcsin
y2
a
(Theorem 7.2)
h
ha ha
2
h2
h a2 a2 arcsin ha2 h2 2 a
61. Let x 3 sin , dx cos d, 1 x 32 cos .
y
Shell Method:
2
4
1
V 4
x1 x 32 dx
x
2
4
1
2
−1
3 sin cos 2 d
2
−2
4
32
4
2 2 sin 2 3 cos
2
1 cos 2 d
2
3
1
1
2
cos2 sin d
2
2
3
2
6 2
3
(c) u a sec
Section 7.4
Trigonometric Substitution
1 1 x2 1 63. y ln x, y , 1 y 2 1 2 x x x2 Let x tan , dx sec2 d, x2 1 sec .
5
s
1
sec sec2 d tan
b
x2 1 dx x2
a
5
x2 1
x
1
b
a
x2 + 1 x
dx θ
sec 1 tan2 d tan
1
b
csc sec tan d
a
ln csc cot sec
b
a
x 26 1 ln
26 ln 2 1 2 5 5 2 1 26 1 26 2 4.367 or ln ln 26 1 5 2 1 x2 1
ln
1 x2 1 x
1
5
26 2
65. Length of one arch of sine curve: y sin x, y cos x L1
1 cos 2 x dx
0
Length of one arch of cosine curve: y cos x, y sin x L2
2
1 sin2 x dx
2
2
1 cos x 2 dx
ux
2
2
, du dx 2
0
1 cos 2 u du
1 cos 2 u du L1
0
67. (a)
(b) y 0 for x 200 (range)
60
−25
250 −10
(c) y x 0.005x 2, y 1 0.01x, 1 y2 1 1 0.01x 2 Let u 1 0.01x, du 0.01 dx, a 1. (See Theorem 7.2.)
200
s
200
1 1 0.01x2 dx 100
0
1 0.01x2 1 0.01 dx
0
50 1 0.01x1 0.01x2 1 ln 1 0.01x 1 0.01x2 1
50 2 ln 1 2 2 ln 1 2 1002 50 ln
2 1 2 1
229.559
200
0
81
82
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
69. Let x 3 tan , dx 3 sec2 d, x2 9 3 sec .
4
A2
0
4
3 dx 6 2 x 9
b
0
sec d 6 ln sec tan
6
a
b
dx 6 x2 9
a b
a
3 sec2 d 3 sec
6 ln
x 0 (by symmetry) y
4
1 1 2 A
4
3 9
x2
3
4
6 ln 3 0 y
dx 2
3 4 1 2
4
9 12 ln 3
3 1 x arctan 4 ln 3 3 3
1 dx 2 4 x 9
2 4 arctan 0.422 4 ln 3 3
x2 9 x
(0, 0.422) 1 4
x
4
−4
−2
2
4
4
1 4 arctan 0, 0.422 2 ln 3 3
x, y 0,
1 y 1 4x 2
y 2x,
71. y x2,
2x tan , dx
1 sec2 d, 1 4x2 sec 2
(For sec5 d and sec3 d, see Exercise 80 in Section 7.3)
2
S 2
a
0
4
b
x21 4x2 d x 2
b
sec3 tan2 d
a
4
tan 2 1 sec sec2 d 2 2
sec d sec d b
b
5
3
a
a
3 1 1 sec3 tan sec tan ln sec tan sec tan ln sec tan 4 4 2 2
1 1
1 4x 2322x 1 4x 2122x ln 1 4x2 2x 4 4 8
51 2 ln 3 2 2
32 102 4 4 8
b a
2
0
542 62 1 ln 3 22 4 4 6 8
2 ln 3 22
13.989
73. (a) Area of representative rectangle: 21 y2 y
y
Pressure: 262.43 y1 y 2 y
2
1
F 124.8
1
x=
3 y1 y d y
1
124.8 3
1
1
1 y 2 dy
x −2
y1 y 2 dy
1
32 arcsin y y1 y 21 23 1 y
1
124.8
2
2 32
1
62.43 arcsin 1 arcsin1 187.2 lb
1
(b) F 124.8
1
1 − y2
2
1
d y1 y 2 dy 124.8 d 124.8
1
1
1 y 2 dy 124.8
1
d2 arcsin y y1 y
y1 y 2 dy
1
2
1
124.80 62.4 d lb
2
Section 7.4
75. (a) m
dy y y 144 x2 dx x0
y
( 0, y +
Trigonometric Substitution
144 − x 2 (
12
144 x2
x
12
144 − x 2
( x, y ) x
y
x 2
(b) y
144 x2
x
4
6
8
10 12
dx 12
Let x 12 sin , dx 12 cos d, 144 x2 12 cos . y
x
θ
12 cos 1 sin2 12 cos d 12 d 12 sin sin
144 − x 2
12 csc sin d 12 ln csc cot 12 cos C
C
12 ln
144 x2 12 144 x2 12 x x 12
12 ln
12 144 x2 144 x2 C x
0
12
144 x2
x
144 x2.
12 144 x2 > 0 for 0 < x ≤ 12 x
(c) Vertical asymptote: x 0 (d) y 144 x2 12 ⇒ y 12 144 x2 Thus, 12 144 x2 12 ln
12
12
1 ln
144 x2
x
144 x2
x
144 x2
xe1 12 144 x2
xe1 122 144 x2
2
x2e2 24xe1 144 144 x2 x2e2 1 24xe1 0 x xe2 1 24e1 0 x 0 or x
24e1 7.77665. e2 1
Therefore,
12
s
7.77665 12
7.77665
1
144 x2
x
2
12
dx
7.77665
x
2
144 x2 dx x2
7.77665 12ln 12 ln 7.77665 5.2 meters.
12 dx 12 ln x x
12
12 0
When x 12, y 0 ⇒ C 0. Thus, y 12 ln Note:
30
83
84
Chapter 7
77. True
dx 1 x2
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
cos d cos
79. False
3
d
0
dx 1 x2 3
3
0
sec2 d sec3
3
cos d
0
81. Let u a sin , du a cos d, a 2 u 2 a cos .
a2 u2 du
a2 cos2 d a2
1 cos 2 d 2
1 a2 a2 sin 2 C sin cos C 2 2 2
a2 u u arcsin 2 a a
a2 u2
a
C 21 a arcsin ua u
a2 u2 C
2
Let u a sec , du a sec tan d, u2 a2 a tan .
u 2 a 2 du
a tan a sec tan d a 2 tan2 sec d
a 2 sec2 1 sec d a2 sec3 sec d
12 sec tan 21 sec d a sec d a 12 sec tan 21 lnsec tan u a a u ln 2 a a ua u a a C a2
2
2
2
2
2
2
2
1
1 uu2 a2 a2 ln u u2 a2 C 2
Let u a tan , du a sec2 d, u2 a2 a sec d.
u2 a2 du
a sec a sec2 d
12 sec tan 21 lnsec tan C
a2 sec3 d a2
Section 7.5
a2 u2 a2 2 a
u
1
a ln
u2 a2
a
u a
C1
1 uu2 a2 a2 ln u u2 a2 C 2
Partial Fractions
1.
5 A B 5 x 2 10x xx 10 x x 10
3.
2x 3 2x 3 A Bx C 2 x3 10x xx 2 10 x x 10
5.
16x 16x A C B 2 x3 10x 2 x 2x 10 x x x 10
7.
1 1 A B x 2 1 x 1x 1 x 1 x 1 1 Ax 1 Bx 1 When x 1, 1 2A, A 12 . When x 1, 1 2B, B 12 .
1 1 dx x2 1 2
1 1 dx x1 2
1 dx x1
1 1 ln x 1 ln x 1 C 2 2
1 x1 ln C 2 x1
Section 7.5
9.
3 3 A B x 2 x 2 x 1x 2 x 1 x 2
5 x Ax 1 B2x 1
When x 1, 3 3A, A 1. When x 2, 3 3B, B 1. 3 dx x2 x 2
1 9 3 When x 2 , 2 2 A, A 3.
1 dx x1
1 dx x2
When x 1, 6 3B, B 2.
ln x 1 ln x 2 C
ln
13.
5x dx 3 2x 2 x 1
x1 C x2
1 1 dx 2 dx 2x 1 x1
3 ln 2x 1 2 ln x 1 C 2
B C x 2 12x 12 A xx 2x 2 x x2 x2 x 2 12x 12 Ax 2x 2 Bxx 2 Cxx 2 When x 0, 12 4A, A 3. When x 2, 8 8B, B 1. When x 2, 40 8C, C 5.
x 2 12x 12 dx 5 x3 4x
1 dx x2
1 dx 3 x2
1 dx x
5 ln x 2 ln x 2 3 ln x C
15.
2x3 4x 2 15x 5 x5 A B 2x 2x x 2 2x 8 x 4x 2 x4 x2 x 5 Ax 2 Bx 4 When x 4, 9 6A, A 32 . When x 2, 3 6B, B 12 .
2x3 4x 2 15x 5 dx x2 2x 8
2x
x2
17.
12 32 dx x4 x2
3 1 ln x 4 ln x 2 C 2 2
B 4x 2 2x 1 A C 2 x 2x 1 x x x1 4x 2 2x 1 Axx 1 Bx 1 Cx 2 When x 0, B 1. When x 1, C 1. When x 1, A 3.
4x 2 2x 1 dx x3 x 2
1 3 1 1 2 dx 3 ln x ln x 1 C x x x1 x
1 ln x 4 x 3 C x
2 2 C B 19. x 3x 4 x 3x 4 A 3 2 x 4x 4x xx 22 x x 2 x 22
x2 3x 4 Ax 22 Bxx 2 Cx When x 0, 4 4A ⇒ A 1. When x 2, 6 2C ⇒ C 3. When x 1, 0 1 B 3 ⇒ B 2.
x2 3x 4 dx x3 4x2 4x
1 dx x
85
5x 5x A B 2x 2 x 1 2x 1x 1 2x 1 x 1
11.
3 x 2 Bx 1
Partial Fractions
2 dx x 2
ln x 2 ln x 2
3 dx x 22
3 C x 2
86
21.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
x2 1 A Bx C 2 xx 2 1 x x 1 x 2 1 Ax 2 1 Bx Cx When x 0, A 1. When x 1, 0 2 B C. When x 1, 0 2 B C. Solving these equations we have A 1, B 2, C 0.
x2 1 dx x3 x
1 dx x
2x dx x2 1
ln x 2 1 ln x C
ln
23.
x2
1 C x
A B Cx D x2 2 x 4 2x 2 8 x 2 x 2 x 2 x 2 Ax 2x 2 2 Bx 2x 2 2 Cx Dx 2x 2 When x 2, 4 24A. When x 2, 4 24B. When x 0, 0 4A 4B 4D, and when x 1, 1 9A 3B 3C 3D. Solving these equations we have A 16 , B 16 , C 0, D 13 .
x4
x 1 2 dx x 1 2 dx 2 x 1 x2 x ln 2 arctan C 6 x 2 2
x2 1 dx 2x 2 8 6
25.
2
1 dx 2
x A B Cx D 2x 12x 14x 2 1 2x 1 2x 1 4x 2 1 x A2x 14x 2 1 B2x 14x 2 1 Cx D2x 12x 1 When x 12 , 12 4A. When x 12 , 12 4B. When x 0, 0 A B D, and when x 1, 1 15A 5B 3C 3D. Solving these equations we have A 18 , B 18 , C 12 , D 0.
x 1 dx 16x 4 1 8
27.
2x 1 1 dx 2x 1 1 dx 4 4x x 1 dx 2
1 4x 2 1 ln C 16 4x 2 1
x2 5 A Bx C x 1x 2 2x 3 x 1 x 2 2x 3 x 2 5 Ax 2 2x 3 Bx Cx 1 A Bx 2 2A B Cx 3A C When x 1, A 1. By equating coefficients of like terms, we have A B 1, 2A B C 0, 3A C 5. Solving these equations we have A 1, B 0, C 2.
x3
x2 5 dx x2 x 3
1 dx 2 x1
1 dx x 12 2
ln x 1 2 arctan
x21 C
Section 7.5
29.
Partial Fractions
3 A B 2x 1x 2 2x 1 x 2 3 Ax 2 B2x 1 When x
1
0
1 2 ,
A 2. When x 2, B 1.
3 dx 2x 5x 2 2
1
0
2 dx 2x 1
1 dx x2
0
1
0
1
ln 2x 1 ln x 2 ln 2
31.
x1 A Bx C 2 xx 2 1 x x 1 x 1 Ax 2 1 x Cx When x 0, A 1. When x 1, 2 2A B C. When x 1, 0 2A B C. Solving these equations we have A 1, B 1, C 1.
2
1
x1 dx xx 2 1
2
1
1 dx x
ln x
2
1
x2
x dx 1
2
1
1 dx x2 1
1 lnx 2 1 arctan x 2
2 1
1 8 ln arctan 2 2 5 4
0.557
33.
3x dx 9 3 ln x 3 C x 2 6x 9 x3
4, 0: 3 ln4 3
35.
9 C0⇒C9 43
2 x2 x 2 x 1 arctan dx C x 2 22 2 2x 2 2 2
0, 1: 0
30
1 5 C1⇒C 4 4 3
(0, 1) −6
(4, 0)
−3
10
3 −1
−10
37.
2x 2 2x 3 1 2x 1 dx ln x 2 ln x 2 x 1 3 arctan C x2 x 2 2 3
x3
3, 10: 0
39.
20
(3, 10)
7 1 7 1 C 10 ⇒ C 10 ln 13 3 arctan ln 13 3 arctan 2 2 3 3
1 1 x2 dx ln C x2 4 4 x2
6, 4:
1 4 1 1 1 ln C 4 ⇒ C 4 ln 4 ln 2 4 8 4 2 4
10
(6, 4) − 10
10 −3
−2
6 −5
87
88
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
41. Let u cos x du sin x dx.
43.
3 cos x 1 dx 3 2 du sin2 x sin x 2 u u2
A B 1 uu 1 u u1
1 Au 1 Bu When u 0, A 1. When u 1, B 1, u cos x, du sin x dx.
u1 C u2
ln
1 sin x C 2 sin x
(From Exercise 9 with u sin x, du cos x dx)
sin x 1 dx du cos xcos x 1 uu 1
ln
1 du u
1 du u1
ln u ln u 1 C
u C ln u1 ln
cos x C cos x 1
45. Let u ex, du ex dx.
47.
1 A B u 1u 4 u 1 u 4
1 A B xa bx x a bx 1 Aa bx Bx When x 0, 1 a A ⇒ A 1a. When x ab, 1 abB ⇒ B ba.
1 Au 4 Bu 1
1 1 When u 1, A 5 . When u 4, B 5 , u ex, x dx. du e
ex dx x e 1ex 4
1 5
49.
1
u 1u 4
1 du u1
du
1 u1 ln C 5 u4
1 ex 1 ln C 5 ex 4
1 du u4
x A B a bx2 a bx a bx2
51.
−2
1b ab dx a bx a bx2
1 b
1 a dx a bx b
1 dx a bx2
a 1 1 ln a bx 2 C b2 b a bx
1 a ln a bx b2 a bx
1 x ln C a a bx
C
2 −4
1 b dx x a bx
1 ln x ln a bx C a
10
When x ab, B ab. When x 0, 0 aA B ⇒ A 1b.
6 dy , y0 3 dx 4 x2
x Aa bx B
x dx a bx2
1 1 dx xa bx a
Section 7.5 53. Dividing x3 by x 5.
Partial Fractions
89
55. (a) Substitution: u x2 2x 8 (b) Partial fractions (c) Trigonometric substitution (tan) or inverse tangent rule
57. Average Cost
1 80 75
1 5
80
75
80
75
124p dp 10 p100 p
124 1240 dp 10 p11 100 p11
1 124 1240 ln10 p ln100 p 5 11 11
80 75
1 24.51 4.9 5 Approximately $490,000.
3
59. A
1
10 dx 3 xx2 1
Matches (c) y 5 4 3 2 1 x 1
2
3
5
A B 1 1 ,AB x 1n x x 1 n x n1
61. 1 n1
4
1 1 dx kt C x1 nx
x1 1 ln kt C n1 nx When t 0, x 0, C
1 1 ln . n1 n
x1 1 1 1 ln kt ln n1 nx n1 n
x1 1 1 ln ln kt n1 nx n
ln
nx n n 1k t nx nx n en1kt nx x
nen1kt 1 n en1kt
Note: lim x n t →
90
63.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
Cx D x Ax B 1 x4 x2 2 x 1 x2 2 x 1 x Ax B x2 2 x 1 Cx D x2 2 x 1 A Cx3 B D 2 A 2 Cx2 A C 2 B 2 Dx B D 0 A C ⇒ C A 0 B D 2 A 2 C 1 A C 2 B 2 D
22 A 0 ⇒ A 0 and C 0 22 B 1 ⇒ B
2
and D
4
2
4
0 B D ⇒ D B Thus,
1
0
x dx 1 x4
1
0
1
2
4
2
4
2 4 2 4 2 dx x 2 x 1 x 2 x 1 2
0
1 1 dx 2 2 x 2 2 1 2 x 2 2 1 2
1 x 2 2 x 2 2 arctan arctan 1 2 1 2 1 2
1 0
1
1 arctan 2 x 1 arctan 2 x 1 2
1
arctan 2 1 arctan 2 1 arctan 1 arctan1 2
1 arctan 2 1 arctan 2 1 . 2 4 4
0
Since arctan x arctan y arctan x y 1 xy , we have:
1
0
1 x 2 1 2 1 1 1 2 dx arctan arctan 1 x4 2 1 2 1 2 1 2 2 2 2 2 4 2 8
Section 7.6 1. By Formula 6:
3. By Formula 26:
Integration by Tables and Other Integration Techniques
x x2 dx 2 x ln 1 x C 1x 2
e x1 e 2x dx
1 x
e e 2x 1 ln e x e 2x 1 C 2
u e x, du e x dx
5. By Formula 44:
1 x 2 1 dx C x x 21 x 2
Section 7.6
7. By Formulas 50 and 48:
sin42x dx
Integration by Tables and Other Integration Techniques
1 sin42x2 dx 2
1 sin 2x cos2x 3 2x sin 2x cos 2x C 2 4 8
1 sin32x cos2x 3 sin22x2 dx 2 4 4 3
9. By Formula 57:
1
x 1 cos x
u x, du
1 6x 3 sin 2x cos 2x 2 sin3 2x cos 2x C 16
dx 2
2 cot x csc x C
1 dx 2x
11. By Formula 84:
1 1 dx 1 cos x 2x
13. By Formula 89:
1 1 dx x ln1 e 2x C 1 e 2x 2
x 3 ln x dx
x4 4 ln x 1 C 16
x 2e x dx x 2e x 2 xe x dx
15. (a) By Formulas 83 and 82:
x 2e x 2 x 1e x C1 x 2e x 2xe x 2e x C (b) Integration by parts: u x 2, du 2x dx, dv e x dx, v e x
x 2e x dx x 2ex
2xe x dx
Parts again: u 2x, du 2 dx, dv e x dx, v e x
x 2e x dx x 2e x 2 xe x
2e x dx x 2e x 2xe x 2e x C
17. (a) By Formula: 12, a b 1, u x, and
1 1 1 1 x ln dx x 2x 1 1 x 1 1x
(b) Partial fractions:
C
1 x C ln x 1x
1 x1 ln C x x
1 A B C 2 x 2x 1 x x x1 1 Axx 1 Bx 1 Cx 2 x 0: 1 B x 1: 1 C x 1: 1 2A 2 1 ⇒ A 1
1 dx x 2x 1
1 1 1 2 dx x x x1
ln x
1 ln x 1 C x
1 x ln C x x1
91
92
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
19. By Formula 81:
21. By Formula 79:
1 2 2 xex e x C 2
x arcsecx 2 1 dx
1 arcsecx 2 1(2x dx 2 1 2
x 1 arcsecx 2 1 ln x 2 1 x 4 2x 2 C 2
u x 2 1, du 2x dx
23. By Formula 89:
27. By Formula 4:
x 2 ln x dx
x3 1 3 ln x C 9
25. By Formula 35:
31. By Formula 73:
e x arccos e x dx e x arccos e x 1 e 2x C
u e x, du e x dx
cos x dx arctansin x C 1 sin2 x
2 cos 1 sin d arctan C 3 2 sin sin2 2 2
u sin , du cos d
1 dx 3 x 22 9x 2
3
3x 2 3x2 2
2
32 9x 2 C 6x 2 9x 2
2x
C
39. By Formulas 54 and 55:
x 2 4
4x
C
x 1 2x dx dx 1 sec x 2 2 1 sec x 2 1 x 2 cot x 2 csc x 2 C 2
u sin x, du cos x dx
37. By Formula 35:
4
dx
35. By Formula 14:
1 x 2x 2
2x x 2 1 dx 2 dx ln1 3x C 1 3x2 1 3x2 9 1 3x
29. By Formula 76:
33. By Formula 23:
t3 cos t dt t3 sin t 3 t2 sin t dt
t3 sin t 3 t2 cos t 2 t cos t dt
t3 sin t 3t2 cos t 6 t sin t
sin t dt
t3 sin t 3t2 cos t 6t sin t 6 cos t C
dx
Section 7.6
41. By Formula 3:
Integration by Tables and Other Integration Techniques
ln x 1 dx 2 ln x 3 ln 3 2 ln x C x3 2 ln x 4
1 dx x
u ln x, du
43. By Formulas 1, 25, and 33:
x 1 2x 6 6 dx dx x 2 6x 102 2 x 2 6x 102
3 x3 1 arctanx 3 C 2x 2 6x 10 2 x 2 6x 10
45. By Formula 31:
x x 4 6x 2 5
3x 10 3 arctanx 3 C 2x 2 6x 10 2
1 2x dx 2 x 2 32 4
dx
1 1 x 2 6x 1022x 6 dx 3 dx 2
x 32 1 2
1 ln x 2 3 x 4 6x 2 5 C 2
u x 2 3, du 2x dx
47.
x3 dx 4 x 2
8 sin3 2 cos d 2 cos
x
8 1
cos2
sin d
θ 4 − x2
8 sin
cos 2
8 cos
8 cos3 C 3
2
sin d
4 x 2 2 x 8 C 3
x 2 sin , dx 2 cos d, 4 x 2 2 cos
49. By Formula 8:
e 3x dx 1 e x3
e x2 e x dx 1 e x3
2 1 ln 1 e x C 1 e x 21 e x 2
u e x, du e x dx
51.
1 2a bu a2 b2 1 A B u2 2 2 2 a bu b a bu2 b a bu a bu2
2a a2 u 2 Aa bu B aA B bAu b b
Equating the coefficients of like terms we have aA B a2 b2 and bA 2a b. Solving these equations we have A 2a b2 and B a2 b2.
a 1 bub du ba 1ba 1bu b du b1 u 2ab lna bu ba a 1 bu C
u2 1 2a 1 du 2 du 2 a bu2 b b b
2 2
C
a2 1 bu 2a ln a bu b3 a bu
2
2
2
3
3
93
94
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
53. When we have u2 a2:
When we have u2 a2:
u a tan du a
u2
a2
1 du u2 a23 2
sec2
a2
u a sec
d
sec2
du a sec tan d
a sec2 d a3 sec3
a2 a2 tan2
1 du u2 a23 2
a sec tan d a3 tan3
1 cos d a2
1 sin C a2
u C a2u2 a2
u
u
θ
1 cos d a2 sin2
u2 + a2
1 csc C a2
u C a2u2 a2
u2 + a2
θ a
a
55.
u2
arctan u du u arctan u u arctan u
1 2u du 2 1 u2 1 ln1 u2 C 2
u arctan u ln1 u2 C w arctan u, dv du, dw
57.
du ,vu 1 u2
21 x 1 dx C x x3 21 x
8
12, 5: 21 21 2 C 5 ⇒ C 7
( 12, 5) −0.5
1.5
21 x 7 y x 59.
1 1 x3 dx tan1x 3 2 C x2 6x 102 2 x 6x 10
3, 0: y
61.
−2
2
1 0 0 C0⇒C0 2 10
1 x3 tan1x 3 2 2 x 6x 10
−8
−2
1 d csc C sin tan
4 , 2: 22 C 2 ⇒ C 2 2 y csc 2 2
(3, 0)
10
(π4, 2) −
2
2 −2
8
Section 7.6
63.
1 d 2 3 sin
2
2 du 1 u2 2u 23 1 u2
Integration by Tables and Other Integration Techniques
2
65.
0
1 d 1 sin cos
2 du 1 u2 2u 1 u2 1 1 u2 1 u2
1 du 1u
0
1 du u 3u 1
0
ln 1 u
1
ln 2
1 du 3 2 5 u 2 4
u 23 25 u 23 25
u tan
1 ln 5
0
1
2 du 1 u2 6u 2
1
95
2
C
2u 3 5 C 2u 3 5 2 tan 3 5 2 1 ln C 5 2 tan 3 5 2
1
5
ln
u tan
67.
2
1 sin d 3 2 cos 2
2 sin d 3 2 cos
69.
cos 1 d 2 cos d 2
2 sin C
1 ln u C 2
u , du
1 ln3 2 cos C 2
1 d 2
u 3 2 cos , du 2 sin d
8
71. A
0
x dx x 1
y
73. Arctangent Formula, Formula 23,
4
22 x x 1 3
8
3
0
2
4 12 3
1 x 2
1 du, u ex u2 1
4
6
8
40 13.333 square units 3
75. Substitution: u x2, du 2x dx Then Formula 81.
77. Cannot be integrated.
79. Answers will vary. For example,
2xe2x dx
can be integrated by first letting u 2x and then using Formula 82.
96
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
5
81. W
2000xex dx
0
5
2000
xex dx
0
5
2000
xex1 dx
0
2000 xex ex
2000
6 1 e5
5 0
1919.145 ft lbs
3
83. (a) V 202
2
1 y 2
0
80 ln y 1 y 2
80 ln 3 10
W 148 80 ln 3 10
dy 3
11,840 ln 3 10
0
21,530.4 lb
145.5 cubic feet (b) By symmetry, x 0.
3
M 2
0
3
Mx 2
0
y
2
1 y 2
dy 4 ln y 1 y 2
2y dy 4 1 y 2
1 y 2
3 0
0 4 ln 3 10 3
4 10 1
Mx 4 10 1 1.19 M 4 ln 3 10
Centroid: x, y 0, 1.19
4
85. (a)
0
k
4
0
4
k dx 10 2 3x
(b)
0
15.417 dx 2 3x
8
10 10 1 0.6486 dx 2 3x
15.417
4
0 −1
ln307
87. False. You might need to convert your integral using substitution or algebra.
Section 7.7 1. lim
x →0
Indeterminate Forms and L’Hôpital’s Rule
sin 5x 5 2.5 exact: sin 2x 2
x f x
3
0.1
0.01
0.001
0.001
0.01
0.1
2.4132
2.4991
2.500
2.500
2.4991
2.4132
−1
1 −1
Section 7.7
Indeterminate Forms and L’Hôpital’s Rule
3. lim x5ex 100 0
10,000,000
x →
x f x
1
10
102
103
104
105
0.9901
90,484
3.7 109
4.5 1010
0
0 0
100 0
5. (a) lim
x →3
(b) lim
x →3
7. (a) lim
1 2x 3 2x 3 2 lim lim x →3 x 3x 3 x →3 x 3 x2 9 3 2 1 2x 3 2 d dx 2x 3 lim lim x →3 d dx x2 9 x →3 2x x2 9 6 3
x 1 2
x3
x →3
(b) lim
x 1 2
x →3
x3
lim
x 1 2
x3
x →3
lim
x →3
x 1 2
x 1 2
lim
x →3
1 x 1 4 1 lim x 3 x 1 2 x →3 x 1 2 4
1 2 x 1 1 d dx x 1 2 lim x →3 d dx x 3 1 4
5x2 3x 1 5 3 x 1 x2 5 lim 2 x → x → 3x 5 3 5 x2 3
9. (a) lim
5x2 3x 1 d dx 5x2 3x 1 10x 3 d dx 10x 3 10 5 lim lim lim lim 2 x → x → x → x → x → 6 3x 5 d dx 3x2 5 6x d dx 6x 3
(b) lim
x2 x 2 2x 1 lim 3 x →2 x →2 x2 1
11. lim
15. lim
x →0
13. lim
4 x2 2
x →0
x
x 4 x2 0 x →0 1
lim
ex 1 x ex 1 lim 2 x →0 x 1
17. Case 1: n 1 lim
x →0
ex 1 x ex 1 lim 0 x →0 x 1
Case 2: n 2 lim
x →0
ex 1 x ex 1 ex 1 lim lim 2 x →0 x →0 2 x 2x 2
Case 3: n ≥ 3 lim
x →0
ex 1 x ex 1 ex lim n1 lim n x →0 nx x →0 nn 1xn2 x
sin 2x 2 cos 2x 2 lim x→0 sin 3x x→0 3 cos 3x 3
19. lim
arcsin x 1 1 x2 lim 1 x →0 x →0 x 1
3x2 2x 1 6x 2 lim x→ x→ 2x2 3 4x
21. lim
23. lim
lim
x→
x2 2x 3 2x 2 lim x→ x → x1 1
25. lim
6 3 4 2
x3 3x2 lim x 2 x→ e x→ 1 2e x 2
27. lim
lim
x→
6x 6 lim 0 1 4e x 2 x→ 1 8e x 2
97
98
Chapter 7 x
29. lim
x→
x2 1
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
lim
1
x→
1 1 x2
1
31. lim
x→
cosx x ≤ 1x
Note: L’Hôpital’s Rule does not work on this limit. See Exercise 79. ln x 1 x 1 lim lim 2 0 x→ x2 x→ 2x x→ 2x
cos x 0 by Squeeze Theorem x
ex ex ex lim lim 2 x→ x x→ 2x x→ 2
33. lim
35. lim
37. (a) lim x ln x 0 0
39. (a) lim x sin
x →0
ln x (b) lim x ln x lim x →0 x →0 1 x lim x →0
x→
(b) lim x sin x→
1 x 1 x2
1 sin1 x lim x x→ 1 x
1 x2 cos1 x x→ 1 x2
lim
lim x 0 x →0
(c)
1 0 x
lim cos x→
2 0
(c)
10
1.5
−1
−12
1x 1
1 − 0.5
41. (a) lim x1 x 0 0, not indeterminant x →0
43. (a) lim x1 x 0 x →
(b) Let y lim x1 x.
(See Exercise 95)
x→
y x1 x
(b) Let
ln y ln
x1 x
ln y lim
1 ln x. x
x→
Thus, ln y 0 ⇒ y e0 1. Therefore,
1 ln x → . Hence, x
Since x → 0 ,
ln y → ⇒ y → 0.
ln x 1 x lim 0 x→ x 1
lim x1 x 1.
x →
(c)
2
Therefore, lim x1 x 0. x→0
(c)
−5
2
20 −0.5
− 0.5
2 − 0.5
45. (a) lim 1 x1 x 1
(c)
x →0
6
(b) Let y lim 1 x1 x. x →0
ln1 x x 1 1 x 1 lim x →0 1
ln y lim x →0
Thus, ln y 1 ⇒ y e1 e. Therefore, lim 1 x1 x e. x →0
−1
4 −1
Section 7.7 47. (a) lim 3xx 2 00
(c)
x→0
Indeterminate Forms and L’Hôpital’s Rule
99
7
(b) Let y lim 3xx 2. x→0
x ln x 2
ln x 2 x
ln y lim ln 3 x→0
lim ln 3 x→0
−6
lim ln 3 lim
1 x 2 x2
lim ln 3 lim
x 2
x→0
x→0
x→0
x→0
6 −1
ln 3 Hence, lim 3xx 2 3. x→0
49. (a) lim ln xx1 00
ln x (b) Let y lim
51. (a) lim
x
(b) lim
x
2
x →2
x→1
x1
x1
lim x 1ln x 0
2
x →2
x→1
x 8 4 x2 8 8 xx 2 x lim x →2 4 x2 x2 4
Hence, lim ln xx1 1 x→1
(c)
lim
2 x4 x x 2x 2
lim
x 4 3 x2 2
x →2
6
x →2
−4
8
(c)
4
−2 −7
5
−4
53. (a) lim x →1
(b) lim x →1
ln3x x 2 1 ln3x x 2 1
lim
x →1
lim x →1
55. (a)
3x 3 2 ln x x 1ln x
3
−1
7
3 2 x
x 1 x ln x
−1
(b) lim
(c)
x →3
8
x3 1 lim ln2x 5 x →3 2 2x 5 lim
x →3
−1
2x 5 1 2 2
4
−4
57. (a)
(b) lim x2 5x 2 x lim x2 5x 2 x
10
x→
x→
x2 5x 2 x x2 5x 2 x
x2 5x 2 x2 x→ x2 5x 2 x
lim −8
10 −2
lim
x→
lim
x→
5x 2
x2 5x 2 x
5 2 x
1 5 x 2 x2 1
5 2
100
59.
Chapter 7 0 , ,0 0
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
, 1, 00,
61. (a) Let f x x2 25 and gx x 5. (b) Let f x x 52 and gx x2 25. (c) Let f x x2 25 and gx x 53.
63. lim
x2 2x 2 lim lim 0 e5x x → 5e5x x → 25e5x
65. lim
ln x3 3ln x21 x lim x→ x 1
x →
x→
67. lim
x→
ln xn nln xn1 x lim x→ xm mxm1
3ln x2 x
lim
6ln x1 x x→ 1
lim
lim
x→
x→
nn 1ln xn2 x→ m2xm
lim
lim
x→
69.
x
ln x4 x
6ln x 6 lim 0 x→ x x
. . . lim
x→
10
102
104
106
108
1010
2.811
4.498
0.720
0.036
0.001
0.000
71. y x1 x, x > 0
lim
x →
1 ln x x
2x 2 lim x 0 x → e ex
Horizontal asymptote: y 0
1 1 dy 1 1 ln x 2 y dx x x x
1 dy x1 x 2 1 ln x x1 x 21 ln x 0 dx x Critical number:
n! 0 mnxm
73. y 2xex
Horizontal asymptote: y 1 (See Exercise 37) ln y
nln xn1 mxm
xe
0, e Sign of dy dx: y f x: Increasing Relative maximum: e, e1 e Intervals:
e, Decreasing
dy 2xex 2ex dx 2ex 1 x 0 Critical number:
x1
Intervals: , 1 Sign of dy dx:
y f x: Increasing Relative maximum:
4
1, Decreasing
1, 2e
3
(1, 2e ) −2
(e, e1/e) 0
6
−5
0
e2x 1 0 0 x →0 ex 1
75. lim
Limit is not of the form 0 0 or . L’Hôpital’s Rule does not apply.
10
77. lim x cos x →
1 1 x
Limit is not of the form 0 0 or . L’Hôpital’s Rule does not apply.
Section 7.7
79. (a) lim
x→
x
x2 1
x x
lim
x→
x→
lim
x→
(c)
(b) lim
x2 1 x
lim
Indeterminate Forms and L’Hôpital’s Rule
x→
x
x2 1
1
x 1 x2
1 x x2 1
x2 1
x
x→
1
lim
1 1 x2
1
x→
lim
2
1 0
lim
x→
lim
Applying L’Hôpital’s rule twice results in the original limit, so L’Hôpital’s rule fails.
1
−1.5
81. lim
v0kekt 32
k
k→0
321 ekt lim v0ekt k→0 k→0 k
lim lim
k→0
v0 320 tekt lim kt 32t v0 k→0 e 1
1 2x1 cos x x x cos x 2 Shaded area: Area of rectangle Area under curve
83. Area of triangle:
x
2x1 cos x 2
1 cos t dt 2x1 cos x 2 t sin t
0
x 0
2x1 cos x 2x sin x 2 sin x 2x cos x Ratio: lim
x →0
x x cos x 1 x sin x cos x lim 2 sin x 2x cos x x →0 2 cos x 2x sin x 2 cos x lim
1 x sin x cos x 2x sin x
lim
x cos x sin x sin x 2x cos x 2 sin x
lim
x cos x 2 sin x 2x cos x 2 sin x
lim
x 2 tan x 2x 2 tan x
x →0
x →0
x →0
x →0
1 cos x
1 cos x
1 2 sec2 x 3 x →0 2 2 sec2 x 4
lim
85. f x x3, gx x2 1, 0, 1 f c f b f a gb ga g c f 1 f 0 3c2 g1 g0 2c 1 3c 1 2 c
2 3
x x2 1 1
x
6
32 1 ekt
x→
x2 1
1.5
−6
101
2
87. f x sin x, gx cos x, 0, f c f 2 f 0 g 2 g0 g c 1 cos c 1 sin c 1 cot c c
4
102
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
89. False. L’Hôpital’s Rule does not apply since
91. True
lim x2 x 1 0.
x →0
x2 x 1 1 1 lim x 1 x →0 x →0 x x
lim
93. (a) sin BD cos DO ⇒ AD 1 cos 1 1 1 1 Area ABD bh 1 cos sin sin sin cos 2 2 2 2 (b) Area of sector:
1 1 1 1 1 Area OBD cos sin sin cos 2 2 2 2 2
Shaded area: (c) R
1 2
12 sin 12 sin cos sin sin cos 12 12 sin cos sin cos sin 12 sin 2 12 sin 2
(d) lim R lim →0
→0
cos cos 2 sin 2 sin 2 cos 4 cos 2 3 lim lim →0 →0 1 cos 2 2 sin 2 4 cos 2 4
lim
→0
95. lim f xgx x →a
y f xgx ln y gx ln f x lim gx ln f x
x →a
As x → a, ln y ⇒ , and hence y 0. Thus, lim f xgx 0.
x →a
b
97. f ab a
f tt b dt f ab a
a
f tt b
b a
b
f ab a f aa b f t dv f tdt ⇒
v f t
u t b ⇒ du dt
Section 7.8
Improper Integrals
1. Infinite discontinuity at x 0.
4
0
1 x
4
dx lim b →0
b
1 x
lim 2 x b →0
dx 4 b
lim 4 2 b 4 b →0
Converges
f t dt
a b a
f b f a
Section 7.8 3. Infinite discontinuity at x 1.
2
0
1 dx x 12
1
1 dx x 12
0
lim
lim
b
b→1
b→1
0
2
1
b
lim
0
103
1 dx x 12
1 dx lim c→1 x 12
1 x1
Improper Integrals
c→1
2
c
1 dx x 12
1 x1
2
1 1
c
Diverges
1
5. Infinite limit of integration.
7.
b
ex dx lim
b→
0
ex dx
because the integrand is not defined at x 0. Diverges
0
e
lim
x
b→
1 dx 2 x2
1
b 0
011
Converges
9.
1
1 dx lim b→ x2
b
1
1x
b
lim b→
1 dx x2
11.
3 x
1
dx lim
b→
b→
3x13 dx
1
9x 2
b
lim
1
1
b
3
23
1
Diverges
0
13.
0
xe2x dx lim
b→
0
1 2x 1e2x b→ 4
xe2x dx lim
b
b
lim
b→
1
1 2b 1e2b (Integration by parts) 4
Diverges
15.
b→
0
Since lim b→
b
x2ex dx lim
b→
0
b 0
lim b→
b2 2b 2 2 2 eb
ex cos x dx lim
0
x2 2x 2
x
b2 2b 2 0 by L’Hôpital’s Rule. eb
17.
e
x2ex dx lim
b→
1 x e cos x sin x 2
b 0
1 1 0 1 2 2
19.
4
1 dx lim b→ xln x3
1 ln x3 dx x 4
b
21.
2 2 dx 4 x
0
2 2 dx 4 x
0
2 ln x
lim
1 1 ln b2 ln 42 2 2
lim
1 1 1 2 2 ln 22 8ln 22
0 2
lim b→
1
2
b 4
b→
b→
b
0
2 dx 4 x2
2 dx lim c→ 4 x2
c
2 dx 4 x2 0
arctan2 lim arctan2 x
0 b
x
c→
0
2
c 0
104
Chapter 7
23.
0
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
1 1 dx lim b →0 x2 x
cos x dx lim
b→
0
sin x
b
1
0
Diverges since sin x does not approach a limit as x → .
arctane
0
0
25.
2 4 4
1
0
ex dx 1 e2x x
b→
b
b
lim
27.
1 dx lim b→ ex ex
1
1
b
Diverges
8
29.
1 dx lim 3 b→8 0 8 x
1
31.
0
x ln x dx lim b→0
b
0
tan d
0
x2 lnx 4 2
1
x2
b
2
33.
1 3 dx lim 8 x 23 b→8 2 8x
3
lnsec
b
lim
b→ 2
lim b→0
14
b 0
6
b2 ln b b2 1 since lim b2 ln b 0 by L’Hôpital’s Rule. b→0 2 4 4
4
35.
2
0
2 dx lim b→2 x x2 4
2 dx x x2 4
x 2
4
b
lim arcsec
Diverges
b→2
4 b
lim arcsec 2 arcsec b→2
4
37.
2
4
1 lim ln x x2 4 b→2 x 4 2
0 3 3
b
ln 4 2 3 ln 2 ln 2 3 1.317
2
39.
0
1 dx 3 x 1
1
0
1 dx 3 x 1
lim b→1
41.
0
4 dx xx 6
4 xx 6
Thus,
0
dx 4
xx 6
1
23
0
1
0
1 dx x1
3
b
3
2
2 x 1
lim c→1
4 dx xx 6
Let u x, u2 x, 2u du dx.
2 x 1
2
3
23
c
4 xx 6
1
3 3 0 2 2
dx
C
x 42u du du 4 8 8 8 2 arctan C arctan 2 uu 6 u 6 6 6 6 6
dx lim b→0
8 6
6
arctan
x
86 arctan 16
2 6 8
. 2 6 3
1
lim
b
8 6
c→
86 2
0
c
x 8 arctan 6 6
8 6
1
16
arctan
b2
Section 7.8
43. If p 1,
1
1 dx lim b→ x
b
1
Improper Integrals
105
b 1 dx lim ln x . b→ x 1
Diverges. For p 1,
1
1 x1p lim p dx b→ x 1p
This converges to
b 1
b 1 p. 1p 1p
lim b→
1
1 if 1 p < 0 or p > 1. p1
45. For n 1 we have
b
xex dx lim
b→
0
lim
b→
xex dx
0
e
xx
b
ex
Parts: u x, dv ex dx
0
lim ebb eb 1 b→
lim
b→
b
e
b
1 1 1 (L’Hôpital’s Rule) eb
Assume that
x nex dx converges. Then for n 1 we have
0
x n1ex dx x n1ex n 1 x nex dx
by parts u xn1, du n 1xn dx, dv ex dx, v ex. Thus,
b→
0
1
47.
0
x
xn1ex dx lim
n1ex
b 0
n 1
0
1 dx diverges. x3
53. Since
x2
49.
1
x3
1
1
x2
1
dx converges by Exercise 43,
1 1 ≥ 3 2 on 2, and 3 x xx 1
2
1 1 converges. 31 2
dx
(See Exercise 43, p 3.
1 1 ≤ 2 on 1, and 5 x
55. Since ex ≤ ex on 1, and
xnex dx, which converges.
0
(See Exercise 44, p 3 1.
51. Since
xnex dx 0 n 1
1
2
1 dx converges. x2 5
1 dx diverges by Exercise 43, 3 2 x
ex dx converges (see Exercise 5),
0
2
1 3 xx 1
dx diverges.
ex dx converges. 2
0
1
57. Answers will vary. See pages 540, 543.
59.
1 3 dx x 1
0
1 3 dx x 1
1
0
1 dx x3
These two integrals diverge by Exercise 44. 61. f t 1 Fs
63. f t t2
0
s e
est dx lim b→
1
st
b
1 ,s > 0 s 0
Fs
s s
t 2est dx lim b→
0
1
3
2 ,s > 0 s3
2 t2
2st 2est
b 0
106
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
65. f t cos at Fs
est cos at dt
0 st
e s cos at a sin at s a
lim b→
b
2
2
0
s s ,s > 0 s2 a2 s2 a2
0
67. f t cosh at Fs
est cosh at dt
0
est
0
e
at
etsa etsa dt
0
1 1 1 et sa et sa b→ 2 s a s a
lim
eat 1 dt 2 2
b 0
0
1 1 1 2 s a s a
1 1 s 1 2 ,s > a 2 s a s a s a2
69. (a) A
ex dx
(b) Disk:
0
lim
b→
ex
b
V
0 1 1
ex 2 dx
0
0
1 lim e2x b→ 2 (c) Shell: V 2
xex dx
0
2
lim 2 exx 1 b→
b 0
x23 y23 4
71.
2 13 2 13 x y y 0 3 3 y 1 y 2
y13 x13
1 xy
8
s4
0
y
8
2
(− 8, 0) −8
(0, 8)
−2
−8
(8, 0) x
2
(0, −8)
8
2 13
x
23 23
x
y23 x 23
2 x 23b 48
23
dx lim 8 b→0
3
8
x 4
23
2 x1/3
b 0
2
Section 7.8
Improper Integrals
73. n
xn1ex dx
0
(a) 1
b→
0
2
1
e
x 1
b→
0
x
(b) n 1
b→
x e
x 2ex dx lim
0
b
x
xex dx lim
0
3
e
ex dx lim
b→
1
x e
xnex dx lim
0
0
b
2xex 2ex
2 x
b
n x
b 0
0
2
b
lim n b→
xn1ex dx 0 n n
u xn, dv ex dx
0
(c) n n 1!
1 t7 e dt 7
75. (a)
0
1 t7 e dt lim et7 b→ 7
b 0
4
1
(b)
0
1 t7 e dt et7 7
4 0
et7 1
0.4353 43.53%
(c)
17 e dt lim te t7
t
0
t7
b→
7et7
b 0
077
5
77. (a) C 650,000
25,000 e0.06t dt 650,000
0
e 25,000 0.06
5
0.06t
0
$757,992.41
10
(b) C 650,000
25,000e0.06t dt $837,995.15
0
(c) C 650,000
25,000e0.06t dt 650,000 lim
b→
0
e 25,000 0.06
0.06t
b 0
$1,066,666.67
79. Let x a tan , dx a sec2 d, a2 x2 a sec .
1 dx a2 x232
a sec2 d 1 2 cos d a3 sec3 a
Pk
1
81.
x 1 k dx 2 lim a x232 a b→ a2 x2 2
x
θ a
1 1 x 2 sin 2 a a a2 x2 Hence,
a2 + x 2
b 1
k 1 k a2 1 1 1 . a2 a2 1 a2 a2 1
10 10 ⇒ x 0, 2. x2 2x xx 2 You must analyze three improper integrals, and each must converge in order for the original integral to converge.
3
0
1
f x dx
0
2
f x dx
1
3
f x dx
2
f x dx
107
108
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
83. For n 1, I1
x2
0
x 1 dx lim b→ 2 14
b
0
1 x2 142x dx lim 1 b→ 6 x2 13
For n > 1,
In
x2n1 x2n2 lim n3 dx b→ 2 n 2 x 1 2n 2x 1
2
0
0
(b)
0
(c)
0
1 x dx lim 2 b→ x2 14 6x 13
3
1 x dx x2 15 4
2 x5 x 16 5 2
0
0
0
n1 n2
0
b 0
1 . 6
n1 x2n3 dx 0 I x 1n2 n 2 n1 2
x 1 dx, v x2 1n3 2n 2x2 1n2
u x 2n2, du 2n 2x 2n3 dx, dv (a)
b
b 0
1 6
1 1 1 x dx x2 14 4 6 24
2 1 1 x3 dx x 15 5 24 60 2
85. False. f x 1 x 1 is continuous on 0, , lim 1 x 1 0, but x →
0
0 .
1 dx lim ln x 1 b→ x1
Diverges 87. True
Review Exercises for Chapter 7 1.
x x2 1 dx
1 x2 11 22x dx 2
3.
x 1 2x dx dx x2 1 2 x2 1
1 x2 13 2 C 2 3 2
1 ln x2 1 C 2
1 x2 13 2 C 3
5.
ln2x ln 2x2 dx C x 2
1 2 e2x sin 3x dx e2x cos 3x 3 3
9.
13 9
7.
16 16 x2
dx 16 arcsin
e2x cos 3x dx
2 1 2x 2 1 e sin 3x e2x cos 3x 3 3 3 3
e2x sin 3x dx
1 2 e2x sin 3x dx e2x cos 3x e2x sin 3x 3 9 e2x sin 3x dx
e2x 2 sin 3x 3 cos 3x C 13
(1) dv sin 3x dx ⇒ u e2x
1 v cos 3x 3
⇒ du 2e2x dx
(2) dv cos 3x dx ⇒ u e2x
v
1 sin 3x 3
⇒ du 2e2x dx
4x C
b
Review Exercises for Chapter 7 2 11. u x, du dx, dv x 51 2 dx, v x 53 2 3
2 x x 5 dx xx 53 2 3
1 x2 sin 2x dx x2 cos 2x 2
13.
2 x 53 2 dx 3
4
(1) dv sin 2x dx ⇒
156 x 34 C
u x2
x 53 2
x2 arcsin 2x 2
x arcsin 2x dx
19.
21.
dx
22x2 dx 1 2x2
1 1 x2 arcsin 2x
2x 1 4x2 arcsin 2x C (by Formula 43 of Integration Tables) 2 8 2
1
8x2 1 arcsin 2x 2x 1 4x2 C 16
v
cos3 x 1 dx
sec4
⇒ du dx
1 sin 2x 2
1 x2 arcsin 2x 2 8
u arcsin 2x ⇒ du
17.
x2
1 4x2
v
⇒
dv x dx
ux
1 v cos 2x 2
⇒ du 2x dx
(2) dv cos 2x dx ⇒
2 x 53 2 3x 10 C 15
x2 2 2 dx 1 4x2
1 sin2 x 1 cos x 1 dx
1 1 sin x 1 sin3 x 1 C 3
1 sin x 1 3 sin2 x 1 C 3
1 sin x 1 3 1 cos 2 x 1 C 3
1 sin x 1 2 cos 2 x 1 C 3
2x dx tan 2x 1 sec 2x dx 2
2
x x x 2 3 x 2 tan 2 tan C tan3 3 tan 3 2 2 3 2 2
tan2
1 d 1 sin
2x sec 2x dx sec 2x dx 2
2
1 sin d cos2
sin 2x dx
1 x 1 x2 cos 2x sin 2x cos 2x C 2 2 4
23x 15 x 5 C
15.
x cos 2x dx
1 1 1 x2 cos 2x x sin 2x 2 2 2
2 4 xx 53 2 x 55 2 C 3 15 x 53 2
109
C
sec2 sec tan d tan sec C
110
23.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
12 dx x2 4 x2
24 cos d 4 sin2 2 cos
x
csc2
3
2
d
θ 4 − x2
3 cot C 3 4 x2 C x
x 2 sin , dx 2 cos d, 4 x2 2 cos x 2 tan
25.
x2 + 4
dx 2 sec2 d
x
θ
4 x2 4 sec2
x3 dx 4 x2
2
8 tan3 2 sec2 d 2 sec
8 tan3 sec d 8 sec2 1tan sec d
sec3 sec C 3
8
x2 43 2 x2 4 C 24 2
8
x2 4
13 x
2
4 4 C
1 8 x2 x2 4 x2 4 C 3 3 1 x2 41 2x2 8 C 3
27.
4 x2 dx
2 cos 2 cos d 2 x
2 1 cos 2 d
θ 4 − x2
1 2 sin 2 C 2 2 sin cos C 2 arcsin
2x 2x
4 x2
2
C
1 x 4 arcsin x 4 x2 C 2 2
x 2 sin , dx 2 cos d, 4 x2 2 cos
Review Exercises for Chapter 7
29. (a)
x3 4 x2
dx 8
sin3 d cos4
(b)
x3 4 x2
8 sec sec2 3 C 3 4 x2
3
(c)
x3 dx x2 4 x2 4 x2
x dx ⇒ 4 x2
2x 4 x2 dx
31.
v 4 x2
⇒ du 2x dx
u x2
x 28 A B x2 x 6 x 3 x 2 x 28 Ax 2 Bx 3 x 2 ⇒ 30 B5 ⇒ B 6 ⇒ 25 A5
x3
33.
x 28 dx x2 6 6
⇒ A 5
5 6 dx 5 ln x 3 6 ln x 2 C x3 x2
x2 2x A Bx C 2 x 1x2 1 x 1 x 1 x2 2x Ax2 1 Bx Cx 1 Let x 1: 3 2A ⇒ A
3 2 3 2
Let x 0: 0 A C ⇒ C
Let x 2: 8 5A 2B C ⇒ B
x2 2x 3 dx x x2 x 1 2 3
1 2
1 1 dx x1 2 1 1 dx x1 4
x3 dx x2 1 2x 3 dx x2 1 2
1 dx x2 1
3 2
1 3 3 ln x 1 ln x2 1 arctan x C 2 4 2
1 6 ln x 1 lnx2 1 6 arctan x C 4
4 x2
3
x2 8 C
u2 4 x2, 2u du 2x dx
4 x2 2 2 x2 4 x2 4 x23 2 C x 8 C 3 3
dv
u2 4 du
u u2 12 C 3
x2 8 C
x 2 tan , dx 2 sec2 d
1 u3 4u C 3
8 cos4 cos2 sin d
dx
111
112
35.
Chapter 7
x2
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
15 2x x2 1 2 2x 15 x 2x 15
B 15 2x A x 3x 5 x 3 x 5 15 2x Ax 5 Bx 3 Let x 3:
9 8A ⇒ A
9 8
Let x 5: 25 8B ⇒ B
x2 dx x2 2x 15
dx
x
37.
9 8
25 8
25 1 dx x3 8
1 dx x5
9 25 ln x 3 ln x 5 C 8 8
x 1 2 dx ln 2 3x C 2 3x2 9 2 3x
39.
x 1 1 dx du 1 sin x2 2 1 sin u 1 tan u sec u C 2
(Formula 4)
1 tan x2 sec x2 C 2
41.
x 1 1 dx ln x2 4x 8 4 2 dx x2 4x 8 2 x 4x 8
(Formula 15)
1 2 2x 4 ln x2 4x 8 2 arctan 32 16 32 16 2
43.
C
(Formula 14)
1 x ln x2 4x 8 arctan 1 C 2 2
1 1 1 dx dx sin x cos x sin x cos x
1 ln tan x C
u x
(Formula 58)
45. dv dx
⇒
vx
1 u ln xn ⇒ du nln xn1 dx x
ln xn dx xln xn n ln xn1 d x
47.
sin cos d
1 2
sin 2 d
1 1 cos 2 4 4 dv sin 2 d ⇒ u
1 v cos 2 2
⇒ du d
1 1 1 cos 2 d cos 2 sin 2 C sin 2 2 cos 2 C 4 8 8
u x2 (Formula 56)
Review Exercises for Chapter 7
49.
x1 4 uu3 dx 4 du 1 x1 2 1 u2 4 4
u2 1
13 u
3
51.
1 cos x dx
1 du u2 1
sin x 1 cos x
113
dx
1 cos x1 2sin x dx
2 1 cos x C
u arctan u C
u 1 cos x, du sin x dx
4 x3 4 3x1 4 3 arctanx1 4 C 3 4 x , x u4, dx 4u3 du y
53.
cos x lnsin x dx sin x lnsin x
3 x3 9 dx ln C x2 9 2 x3 (by Formula 24 of Integration Tables)
sin x lnsin x sin x C dv cos x dx ⇒
v sin x
u lnsin x ⇒ du
57. y
55. y
cos x dx
cos x dx sin x
lnx2 xdx x ln x2 x x ln x 2 x x ln x2 x
5
2x2 x dx x2 x
59.
xx2 43 2 dx
2
15 x
2
45 2
5
2
1 5
2x 1 dx x1
2dx
1 dx x1
x ln x2 x 2x ln x 1 C ⇒
dv dx
vx
u lnx2 x ⇒ du
4
61.
1
ln x 1 dx ln x2 x 2
4
65. A
4 1
2x 1 dx x2 x
1 ln 42 2ln 22 0.961 2
0
x 4 x dx
0
4 u2u2u du
2
63.
0
2u4 4u2 du
y
2
u5 4u3
2
5
3
0 2
128 15
1 1 x x3 3
0
1 67. By symmetry, x 0, A . 2
0
x sin x dx x cos x sin x
1
2 1 2
1
x, y 0,
4 3
1 x22 dx
u 4 x, x 4 u2, dx 2u du
69. s
1 cos2 x dx 3.82
0
e2x 2e2x 4e2x lim lim 2 x → x x → 2x x → 2
73. lim
71. lim
x →1
ln x2
x 1 lim x →1
21 xln x 0 1
75. y lim ln x2 x x →
ln y lim
x →
2 x ln x 2 lnln x 0 lim x → x 1
Since ln y 0, y 1.
1 1
4 3
108
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
83. For n 1, I1
x2
0
x 1 dx lim b→ 2 14
b
0
1 x2 142x dx lim 1 b→ 6 x2 13
For n > 1,
In
x2n1 x2n2 lim n3 dx b→ 2 n 2 x 1 2n 2x 1
2
0
0
(b)
0
(c)
0
1 x dx lim 2 b→ x2 14 6x 13
3
1 x dx x2 15 4
2 x5 x 16 5 2
0
0
0
n1 n2
0
b 0
1 . 6
n1 x2n3 dx 0 I x 1n2 n 2 n1 2
x 1 dx, v x2 1n3 2n 2x2 1n2
u x 2n2, du 2n 2x 2n3 dx, dv (a)
b
b 0
1 6
1 1 1 x dx x2 14 4 6 24
2 1 1 x3 dx x 15 5 24 60 2
85. False. f x 1 x 1 is continuous on 0, , lim 1 x 1 0, but x →
0
0 .
1 dx lim ln x 1 b→ x1
Diverges 87. True
Review Exercises for Chapter 7 1.
x x2 1 dx
1 x2 11 22x dx 2
3.
x 1 2x dx dx x2 1 2 x2 1
1 x2 13 2 C 2 3 2
1 ln x2 1 C 2
1 x2 13 2 C 3
5.
ln2x ln 2x2 dx C x 2
1 2 e2x sin 3x dx e2x cos 3x 3 3
9.
13 9
7.
16 16 x2
dx 16 arcsin
e2x cos 3x dx
2 1 2x 2 1 e sin 3x e2x cos 3x 3 3 3 3
e2x sin 3x dx
1 2 e2x sin 3x dx e2x cos 3x e2x sin 3x 3 9 e2x sin 3x dx
e2x 2 sin 3x 3 cos 3x C 13
(1) dv sin 3x dx ⇒ u e2x
1 v cos 3x 3
⇒ du 2e2x dx
(2) dv cos 3x dx ⇒ u e2x
v
1 sin 3x 3
⇒ du 2e2x dx
4x C
b
Review Exercises for Chapter 7 2 11. u x, du dx, dv x 51 2 dx, v x 53 2 3
2 x x 5 dx xx 53 2 3
1 x2 sin 2x dx x2 cos 2x 2
13.
2 x 53 2 dx 3
4
(1) dv sin 2x dx ⇒
156 x 34 C
u x2
x 53 2
x2 arcsin 2x 2
x arcsin 2x dx
19.
21.
dx
22x2 dx 1 2x2
1 1 x2 arcsin 2x
2x 1 4x2 arcsin 2x C (by Formula 43 of Integration Tables) 2 8 2
1
8x2 1 arcsin 2x 2x 1 4x2 C 16
v
cos3 x 1 dx
sec4
⇒ du dx
1 sin 2x 2
1 x2 arcsin 2x 2 8
u arcsin 2x ⇒ du
17.
x2
1 4x2
v
⇒
dv x dx
ux
1 v cos 2x 2
⇒ du 2x dx
(2) dv cos 2x dx ⇒
2 x 53 2 3x 10 C 15
x2 2 2 dx 1 4x2
1 sin2 x 1 cos x 1 dx
1 1 sin x 1 sin3 x 1 C 3
1 sin x 1 3 sin2 x 1 C 3
1 sin x 1 3 1 cos 2 x 1 C 3
1 sin x 1 2 cos 2 x 1 C 3
2x dx tan 2x 1 sec 2x dx 2
2
x x x 2 3 x 2 tan 2 tan C tan3 3 tan 3 2 2 3 2 2
tan2
1 d 1 sin
2x sec 2x dx sec 2x dx 2
2
1 sin d cos2
sin 2x dx
1 x 1 x2 cos 2x sin 2x cos 2x C 2 2 4
23x 15 x 5 C
15.
x cos 2x dx
1 1 1 x2 cos 2x x sin 2x 2 2 2
2 4 xx 53 2 x 55 2 C 3 15 x 53 2
109
C
sec2 sec tan d tan sec C
110
23.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
12 dx x2 4 x2
24 cos d 4 sin2 2 cos
x
csc2
3
2
d
θ 4 − x2
3 cot C 3 4 x2 C x
x 2 sin , dx 2 cos d, 4 x2 2 cos x 2 tan
25.
x2 + 4
dx 2 sec2 d
x
θ
4 x2 4 sec2
x3 dx 4 x2
2
8 tan3 2 sec2 d 2 sec
8 tan3 sec d 8 sec2 1tan sec d
sec3 sec C 3
8
x2 43 2 x2 4 C 24 2
8
x2 4
13 x
2
4 4 C
1 8 x2 x2 4 x2 4 C 3 3 1 x2 41 2x2 8 C 3
27.
4 x2 dx
2 cos 2 cos d 2 x
2 1 cos 2 d
θ 4 − x2
1 2 sin 2 C 2 2 sin cos C 2 arcsin
2x 2x
4 x2
2
C
1 x 4 arcsin x 4 x2 C 2 2
x 2 sin , dx 2 cos d, 4 x2 2 cos
Review Exercises for Chapter 7
29. (a)
x3 4 x2
dx 8
sin3 d cos4
(b)
x3 4 x2
8 sec sec2 3 C 3 4 x2
3
(c)
x3 dx x2 4 x2 4 x2
x dx ⇒ 4 x2
2x 4 x2 dx
31.
v 4 x2
⇒ du 2x dx
u x2
x 28 A B x2 x 6 x 3 x 2 x 28 Ax 2 Bx 3 x 2 ⇒ 30 B5 ⇒ B 6 ⇒ 25 A5
x3
33.
x 28 dx x2 6 6
⇒ A 5
5 6 dx 5 ln x 3 6 ln x 2 C x3 x2
x2 2x A Bx C 2 x 1x2 1 x 1 x 1 x2 2x Ax2 1 Bx Cx 1 Let x 1: 3 2A ⇒ A
3 2 3 2
Let x 0: 0 A C ⇒ C
Let x 2: 8 5A 2B C ⇒ B
x2 2x 3 dx x x2 x 1 2 3
1 2
1 1 dx x1 2 1 1 dx x1 4
x3 dx x2 1 2x 3 dx x2 1 2
1 dx x2 1
3 2
1 3 3 ln x 1 ln x2 1 arctan x C 2 4 2
1 6 ln x 1 lnx2 1 6 arctan x C 4
4 x2
3
x2 8 C
u2 4 x2, 2u du 2x dx
4 x2 2 2 x2 4 x2 4 x23 2 C x 8 C 3 3
dv
u2 4 du
u u2 12 C 3
x2 8 C
x 2 tan , dx 2 sec2 d
1 u3 4u C 3
8 cos4 cos2 sin d
dx
111
112
35.
Chapter 7
x2
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
15 2x x2 1 2 2x 15 x 2x 15
B 15 2x A x 3x 5 x 3 x 5 15 2x Ax 5 Bx 3 Let x 3:
9 8A ⇒ A
9 8
Let x 5: 25 8B ⇒ B
x2 dx x2 2x 15
dx
x
37.
9 8
25 8
25 1 dx x3 8
1 dx x5
9 25 ln x 3 ln x 5 C 8 8
x 1 2 dx ln 2 3x C 2 3x2 9 2 3x
39.
x 1 1 dx du 1 sin x2 2 1 sin u 1 tan u sec u C 2
(Formula 4)
1 tan x2 sec x2 C 2
41.
x 1 1 dx ln x2 4x 8 4 2 dx x2 4x 8 2 x 4x 8
(Formula 15)
1 2 2x 4 ln x2 4x 8 2 arctan 32 16 32 16 2
43.
C
(Formula 14)
1 x ln x2 4x 8 arctan 1 C 2 2
1 1 1 dx dx sin x cos x sin x cos x
1 ln tan x C
u x
(Formula 58)
45. dv dx
⇒
vx
1 u ln xn ⇒ du nln xn1 dx x
ln xn dx xln xn n ln xn1 d x
47.
sin cos d
1 2
sin 2 d
1 1 cos 2 4 4 dv sin 2 d ⇒ u
1 v cos 2 2
⇒ du d
1 1 1 cos 2 d cos 2 sin 2 C sin 2 2 cos 2 C 4 8 8
u x2 (Formula 56)
Review Exercises for Chapter 7
49.
x1 4 uu3 dx 4 du 1 x1 2 1 u2 4 4
u2 1
13 u
3
51.
1 cos x dx
1 du u2 1
sin x 1 cos x
113
dx
1 cos x1 2sin x dx
2 1 cos x C
u arctan u C
u 1 cos x, du sin x dx
4 x3 4 3x1 4 3 arctanx1 4 C 3 4 x , x u4, dx 4u3 du y
53.
cos x lnsin x dx sin x lnsin x
3 x3 9 dx ln C x2 9 2 x3 (by Formula 24 of Integration Tables)
sin x lnsin x sin x C dv cos x dx ⇒
v sin x
u lnsin x ⇒ du
57. y
55. y
cos x dx
cos x dx sin x
lnx2 xdx x ln x2 x x ln x 2 x x ln x2 x
5
2x2 x dx x2 x
59.
xx2 43 2 dx
2
15 x
2
45 2
5
2
1 5
2x 1 dx x1
2dx
1 dx x1
x ln x2 x 2x ln x 1 C ⇒
dv dx
vx
u lnx2 x ⇒ du
4
61.
1
ln x 1 dx ln x2 x 2
4
65. A
4 1
2x 1 dx x2 x
1 ln 42 2ln 22 0.961 2
0
x 4 x dx
0
4 u2u2u du
2
63.
0
2u4 4u2 du
y
2
u5 4u3
2
5
3
0 2
128 15
1 1 x x3 3
0
1 67. By symmetry, x 0, A . 2
0
x sin x dx x cos x sin x
1
2 1 2
1
x, y 0,
4 3
1 x22 dx
u 4 x, x 4 u2, dx 2u du
69. s
1 cos2 x dx 3.82
0
e2x 2e2x 4e2x lim lim 2 x → x x → 2x x → 2
73. lim
71. lim
x →1
ln x2
x 1 lim x →1
21 xln x 0 1
75. y lim ln x2 x x →
ln y lim
x →
2 x ln x 2 lnln x 0 lim x → x 1
Since ln y 0, y 1.
1 1
4 3
114
Chapter 7
77. lim 1000 1 n →
Let y lim
n →
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals 0.09 n
n
1000 lim
n →
1 0.09 n
n
. 1 0.09 n n
0.09 ln y lim n ln 1 lim n → n → n
ln 1
0.09 n
n →
16
0
1
dx lim
4 x
b→0
3 x 4
16
34
b
lim
n →
1 n
Thus, ln y 0.09 ⇒ y e0.09 and lim 1000 1
79.
0.09 n
n
t0
n →
81.
x2 ln x dx lim
b→
1
0.09 0.09 0.09 1 n
9 1 3 ln x
Diverges
Converges
83.
lim
1000e0.09 1094.17.
32 3
0.09n2 1 0.09n 1 2 n
500,000e0.05t dt
0
e
500,000 0.05
0.05t
t0 0
500,000 0.05t0 e 1 0.05
10,000,0001 e0.05t0 (a) t0 20: $6,321,205.59 (b) t0 → : $10,000,000
1 0.95 2
(b) P15 ≤ x < 20
1 0.95 2
85. (a) P13 ≤ x <
ex12.9 20.95 dx 0.4581 2
2
13
ex12.9 20.95 dx 0.0135 2
2
15
Problem Solving for Chapter 7
1
1. (a)
1
1
1
x3 3
1 x2 dx x
1 1
2 1
1 4 3 3
1
1 x22 dx
1
1 2x2 x4 dx x
2x3 x5 3 5
1 1
2 1
(b) Let x sin u, dx cos u du, 1 x2 1 sin2 u cos2 u.
1
1
1 x2n dx
2
cos2 un cos u du
2
2
cos2n1 u du
2
2n
3 5 7 . . . 2n 1
2
2
4
6
2 3 4 5. . . 2n2n 1
2
22
42
62 . . . 2n2
22n1n!2 222nn!2 2n 1! 2n 1!
2 1 16 3 5 15
(Wallis’s Formula)
x3
b 1
Problem Solving for Chapter 7
lim
3.
x→
xx cc
lim x ln
x→
lim
x→
x
9
xx cc ln 9
lnx c lnx c ln 9 1x 1 1 xc xc lim ln 9 x→ 1 2 x
lim
x→
2c x2 ln 9 x cx c
x 2cx c ln 9 2
lim
x→
2
2
2c ln 9 2c 2 ln 3 c ln 3
5. sin
PB PB, cos OB OP
AQ AP BR OR OB OR cos The triangles AQR and BPR are similar: AR OR 1 OR cos BR ⇒ AQ BP sin sin OR sin OR cos OR lim OR lim
→0
→0
cos sin sin
sin cos cos cos 1
lim
sin cos 1
lim
sin cos sin
→0
→0
lim
→0
2
Q
cos sin sin
lim →0
y
cos cos sin cos
P
θ
R
O
B
A (1, 0)
x
115
116
Chapter 7
7. (a)
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals Area 0.2986
0.2
0
4 0
(b) Let x 3 tan , dx 3 sec2 d, x2 9 9 sec2 .
x2 dx 9 x2
Area
9 tan2 3 sec2 d 9 sec2 32
x2 + 9 x
tan2 d sec
θ 3
sin2 d cos 1 cos2 d cos
ln sec tan sin C
4
0
x2
x2
dx ln sec tan sin
9
32
ln
tan14 3
0
9 x x x2 9 3 3
x2
4
0
4 5 4 4 ln ln 3 3 3 5 5 (c) x 3 sinh u, dx 3 cosh u du, x2 9 9 sinh2 u 9 9 cosh2 u
4
A
0
x2 dx 2 x 932
sin14 3
0
9 sinh2 u 3 cosh u du 9 cosh2 u32
sinh14 3
tanh2 u du
0
sinh14 3
1 sech2 u du
0
sinh14 3
u tanh u
0
sinh1
4 4 tanh sinh1 3 3
43 169 1 tanh ln43 169 1
ln ln
43 35 tanhln43 35
ln 3 tanhln 3 ln 3
3 13 3 13
ln 3
4 5
Problem Solving for Chapter 7 9. y ln1 x2, y
1 x2 4x2 1 2x2 x4 4x2 1 x22 1 x22 1 x2
1 y 2 1
2x 1 x2
2
12
Arc length
1 y 2 dx
0 12
2 2
0 12
11 xx dx 1 1 2 x dx 2
0 12
1 x 1 1 1 1 x dx x ln1 x ln1 x
0
12 0
3 1 1 ln ln 2 2 2
1 ln 3 ln 2 ln 2 2 ln 3
11. Consider
1 dx. ln x
Let u ln x, du
If
1 0.5986 2
1 dx, x eu. Then dx
1 dx were elementary, then ln x
Hence,
1 dx ln x
1 u e du u
eu du. u
eu du would be too, which is false. u
1 dx is not elementary. ln x
13. x4 1 x2 ax bx2 cx d x4 a cx3 ac b dx2 ad bcx bd a c, b d 1, a 2
1
0
x4 1 x2 2 x 1x2 2x 1 Ax B dx x2 2x 1
1 2 x 1 2 4 dx 2 0 x 2 x 1
1 dx x4 1
1
0
2
4 2
4
1
0
Cx D dx x2 2 x 1
1
0
1 2 x 2 4 dx x2 2 x 1
arctan 2x 1 arctan 2x 1
1
0
arctan 2 1 arctan 2 1
0.5554 0.3116 0.8670
2
8
2
8
lnx
2
2x 1 lnx2 2x 1
ln2 2 ln2 2
2
4
1 0
4 4
2
8
0
117
118
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
15. Using a graphing utility,
1 x
1 0 x
1 x
(a) lim cot x x→0
(b) lim cot x x→0
(c) lim cot x x→0
cot x 1x 32.
Analytically,
1 x
1 x cot x 1 x cos x sin x lim lim x→0 x→0 x x x sin x
(a) lim cot x x→0
(b) lim cot x x→0
lim
cos x x sin x cos x x sin x lim x→0 sin x x cos x sin x x cos x
lim
sin x x cos x 0. cos x cos x x sin x
x→0
x→0
(c)
cot x 1x cot x 1x cot
2
lim
x→0
x
x2 cot2 x 1 x2
x2 cot2 x 1 2x cot2 x 2x2 cot x csc2 x lim 2 x→0 x 2x lim
cot2 x x cot x csc2 x 1
lim
cos2 x sin x x cos x sin3 x
lim
1 sin2 xsin x x cos x sin3 x
lim
sin x x cos x 1 sin3 x
x→0
x→0
x→0
x→0
Now, lim x→0
sin x x cos x cos x cos x x sin x lim x→0 sin3 x 3 sin2 x cos x lim
x 3 sin x cos x
lim
sinx x3 cos1 x 31.
x→0
x→0
Thus, lim cot x x→0
1 x2
1 x
cot x 1x 31 1 32.
The form 0 is indeterminant.
Problem Solving for Chapter 7
17.
P1 P2 P3 P4 x3 3x2 1 ⇒ c1 0, c2 1, c3 4, c4 3 2 x 13x 12x x x1 x4 x3 4
Nx x3 3x2 1 D x 4x3 26x 12 P1
N0 1 D 0 12
P2
N1 1 1 D 1 10 10
P3
N4 111 111 D 4 140 140
P4
N3 1 D 3 42
x3 3x2 1 112 110 111140 142 . x 13x2 12x x x1 x4 x3
Thus,
4
19. By parts,
b
f xg x dx f xg x
a
b
a
b
f xg x dx
a
f xgx
b
a
f xg x dx
b
a
f xgx dx.
b
a
gx f x dx
119
C H A P T E R Infinite Series
8
Section 8.1
Sequences . . . . . . . . . . . . . . . . . . . . . 369
Section 8.2
Series and Convergence . . . . . . . . . . . . . . 373
Section 8.3
The Integral Test and p-Series
Section 8.4
Comparisons of Series
Section 8.5
Alternating Series . . . . . . . . . . . . . . . . . 385
Section 8.6
The Ratio and Root Tests . . . . . . . . . . . . . 389
Section 8.7
Taylor Polynomials and Approximations . . . . . 393
Section 8.8
Power Series . . . . . . . . . . . . . . . . . . . . 398
Section 8.9
Representation of Functions by Power Series
. . . . . . . . . . 378
. . . . . . . . . . . . . . 381
Section 8.10 Taylor and Maclaurin Series
. . 403
. . . . . . . . . . . 408
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 414 Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . 421
C H A P T E R Infinite Series Section 8.1
8
Sequences
Solutions to Even-Numbered Exercises
32
2. an
2n n3
a1
2 1 4 2
a1
a2
4 5
a2
6 1 6
8 a3 27
a4
8 7
16 a4 81
a5
10 5 8 4
a5
a3
8. an 1n1 a1
4. an
2n
a3
2 1 a4 4 2 2 a5 5
k 2 1a
k
n 2
0 2
a2 cos 1
4 9
a3 cos
3 0 2
a4 cos 2 1 a5 cos
5 0 2
32 243 6 2 n n2
12. an
3n! 3n n 1!
a1 10 2 6 18
a1 31 3
3 25 2 2
a2 32 6
2 2 34 a3 10 3 3 3
2 3
14. a1 4, ak1
a1 cos
a2 10 1
2 a2 1 2
6. an cos
2 3
10. an 10
2 2 1
n
a4 10
1 3 87 2 8 8
a5 10
2 6 266 5 25 25
a3 33 9 a4 34 12 a5 35 15
1 16. a1 6, ak1 ak2 3
a2
1 2 1a
4
1 1 a2 a12 62 12 3 3
a3
2 2 1a
6
1 1 a3 a22 122 48 3 3
a4
3 2 1a
12
1 1 a4 a32 482 768 3 3
a5
4 2 1a
30
1 1 a5 a42 7682 196,608 3 3
1
2
3
4
369
370
Chapter 8
Infinite Series
18. Because the sequence tends to 8 as n tends to infinity, it matches (a). 22.
24.
4
−1
20. This sequence increases for a few terms, then decreases a2 16 2 8. Matches (b). 26.
10
12 −1
−1
12
4 an 2 , n 1, . . . , 10 n 28. an
n6 2
an 80.75n1, n 1, 2, . . . , 10
5 6 11 2 2
a6
66 6 2
32.
a5 240 80
36.
2425 600
2n 2! 2n!2n 12n 2 2n! 2n!
n 1n 2
n→
1 505 n2
2n 12n 2
40. lim
n→
5n 5 lim n→ 1 4n2 n2 4
44.
12
42. lim cos n→
2n 1
5 5 1 46.
2
−1
3n2 , n 1, . . . , 10 n2 1
25! 23!2425 23! 23!
a6 280 160
n 2! n!n 1n 2 n! n!
38. lim 5
an
30. an1 2an, a1 5
a5
12 −1
−1
−3
34.
4
4
−1
12 −1
−1
The graph seems to indicate that the sequence converges to 0. Analytically, lim an lim
n→
n→
The graph seems to indicate that the sequence converges to 3. Analytically,
1 1 lim 0. n32 x→ x32
48. lim 1 1n n→
lim an lim 3
n→
n→
3 n
50. lim
3 n n→
1
1 3 0 3. 2n
1, converges
does not exist, (alternates between 0 and 2), diverges.
52. lim
n→
1 1n 0, converges n2
54. lim
n→
lnn 12 ln n lim n→ n n
lim
n→
1 0, converges 2n
(L’Hôpital’s Rule) 56. lim 0.5n 0, converges n→
58. lim
n→
n 2! 1 lim 0, converges n→ nn 1 n!
Section 8.1
Sequences
2nn 1 2nn 1 lim 4n2n 1 21, converges 2
60. lim
n→
2
2
n→
2
1 n
62. an n sin
1 Let f x x sin . x lim x sin
x→
1 sin1x 1x 2 cos1x 1 lim lim lim cos cos 0 1 (L’Hôpital’s Rule) x→ x→ x x→ 1x 1x 2 x
or, lim
x→
sin1x sin y 1 lim 1. Therefore lim n sin 1. y→0 n→ 1x y n
64. lim 21n 20 1, converges n→
70. an
1n1 n2
76. an 1
2n 1 2n
66. lim
cos n 0, converges n2
68. an 4n 1
72. an
n2 3n 1
74. an 1n
78. an
1 n!
80. an
n→
3n2 2n1
xn1 n 1!
2n1 1 2n
82. Let f x
3x 6 . Then fx . x2 x 22
Thus, f is increasing which implies an is increasing.
an < 3, bounded
n2 84. an ne
a1 0.6065 a2 0.7358 a3 0.6694
Not monotonic; an ≤ 0.7358, bounded 86. an 23
88. an 32 <
n
n
Monotonic; lim an , not bounded
a1 23 a2
n→
4 9
8 a3 27
2 Not monotonic; an ≤ 3 , bounded
90. an
cos n n
a1 0.5403 a2 0.2081 a3 0.3230 a4 0.1634
32 n1 an1
Not monotonic; an ≤ 1, bounded
371
372
Chapter 8
Infinite Series
92. (a) an 4
4
3 n
(b)
3 < 4 ⇒ bounded n −1
3 4 < 3 an1 ⇒ monotonic n n1
an 4
Therefore, an converges.
94. (a) an 4
4
8
1 2n
12 0
lim 4
n→
(b)
3 4 n
6
1 ≤ 4.5 ⇒ an bounded 2n −1
1 1 an 4 n > 4 n1 2 2 an1 ⇒ an monotonic
(a) A1 $101.00 A2 $203.01
(b) A60 $8248.64
lim 4
n→
Therefore, an converges. 96. An 1001011.01n 1
12 −1
1 4 2n
98. The first sequence because every other point is below the x-axis.
(c) A240 $99,914.79
A3 $306.04 A4 $410.10 A5 $515.20 A6 $621.35 100. Impossible. The sequence converges by Theorem 8.5.
102. Impossible. An unbounded sequence diverges.
104. Pn 16,0001.045n
106. (a) an 410.9212n 6003.8545
P1 $16,720.00
12,000
P2 $17,472.40 P3 $18,258.66 P4 $19,080.30 P5 $19,938.91
108. an 1
1 n
lim sn L > 0,
n→
there exists for each > 0, an integer N such that sn L < for every n > N. Let L > 0 and we have,
a3 2.3704 a5 2.4883
< L, L < sn L < L, or 0 < sn < 2L
for each n > N.
a6 2.5216 n→
sn L
a4 2.4414
1 n
(b) For 2004, n 14 and an 11,757, or $11,757,000,000. 110. Since
a2 2.2500
10 0
n
a1 2.0000
lim 1
0
n
e
Section 8.2
Series and Convergence
373
112. If an is bounded, monotonic and nonincreasing, then a1 ≥ a2 ≥ a3 ≥ . . . ≥ an ≥ . . . . Then a1 ≤ a2 ≤ a3 ≤ . . . ≤ an ≤ . . . is a bounded, monotonic, nondecreasing sequence which converges by the first half of the theorem. Since an converges, then so does an. 114. True
116. True 1 1 , n 1, 2, . . . x 2 n1 xn1
118. x0 1, xn x1 1.5
x6 1.414214
x2 1.41667
x7 1.414214
x3 1.414216
x8 1.414114
x4 1.414214
x9 1.414214
x5 1.414214 x10 1.414214 The limit of the sequence appears to be 2. In fact, this sequence is Newton’s Method applied to f x x 2 2.
Section 8.2
Series and Convergence
2. S1 16 0.1667 S2
1 6
1 6
4. S1 1
0.3333
S2 1 13 1.3333
3 S3 16 16 20 0.4833
S3 1 13 15 1.5333
3 2 S4 16 16 20 15 0.6167
S4 1 13 15 19 1.6444
1 1 3 2 5 S5 6 6 20 15 42 0.7357
1 S5 1 13 15 19 11 1.7354
6. S1 1
8.
S2 1
0.5
S3 1
1 2
16 0.6667
S5 1
10.
21.03
n
1 24
1 120
Geometric series
Diverges by Theorem 8.6
0.6333
12.
Geometric series
n
2n 3
n1
n0
r 1.03
n
4 > 1 3
r
1 S4 1 12 16 24 0.6250 1 6
4
n0
1 2
1 2
3
> 1
lim
n→
Diverges by Theorem 8.6
n 1 0 2n 3 2
Diverges by Theorem 8.9
14.
n
n1
lim
n→
n 1
16.
2
n n2 1
n!
2
n1
lim
n→
1 1 1 n2
Diverges by Theorem 8.9
10
lim
n→
n
n! 2n
Diverges by Theorem 8.9
374
18.
Chapter 8
3
2
n
Infinite Series
1
n0
3 9
17
8
n
n0
17 8 64 . . . 1 3 9 81
S0
Matches graph (b).
Matches (d).
Analytically, the series is geometric:
Analytically, the series is geometric:
2
n
n0
1 1 3 1 2 3 1 3
17 , S 0.63, S3 5.1, . . . 3 1
17 8 9 n0 3
n
17 3 17 3 3 1 8 9 17 9
nn 2 2n 2n 2 2 6 4 8 6 10 8 12 10 14 . . . 1
n1
1
1
1
1
1
1
1
1
1
1
1
1
n1
1
nn 2
n1
24.
20.
5 S0 1, S1 , S2 2.11, . . . . 3
3
22.
2 4 . . . 3 9
2 2
1
lim Sn lim
n→
n→
12 41 2n 1 1 2n 2 2 41 43 1
1
n
26.
n0
n
n0
0.6
1 Geometric series with r 2 < 1.
Geometric series with r 0.6 < 1.
Converges by Theorem 8.6
Converges by Theorem 8.6
nn 4 n n 4
28. (a)
4
n1
1 1 (b)
(c)
1
1
n1
1 1 1 1 1 1 1 1 1 1 1 . . . 5 2 6 3 7 4 8 5 9 6 10
1 1 1 25 2.0833 2 3 4 12
n
5
10
20
50
100
Sn
1.5377
1.7607
1.9051
2.0071
2.0443
4
0
11 0
(d) The terms of the series decrease in magnitude slowly. Thus, the sequence of partial sums approaches the sum slowly.
30. (a)
30.85
n1
n1
(b)
(c)
3 20 1 0.85
(Geometric series)
n
5
10
20
50
100
Sn
11.1259
16.0625
19.2248
19.9941
19.999998
24
0
11 0
Section 8.2
5 3
32. (a)
1
n1
n1
(b)
5 15 3.75 1 1 3 4
(c)
n
5
10
20
50
100
Sn
3.7654
3.7499
3.7500
3.7500
3.7500
Series and Convergence
7
0
11 0
(d) The terms of the series decrease in magnitude rapidly. Thus, the sequence of partial sums approaches the sum rapidly.
34.
nn 2 2 n n 2 2 1 3 2 4 3 5 . . . 2 1 2 3 4
n1
36.
1
1
1
1
1
1
6 5
n
n
4
1
3 4
8
0.7
n
6 30 1 4 5
1
1
1
(Geometric)
1
1
1
1
0.9n
44.
81
1
n
n
4 2
7
n
10
9
n
2
n0
1
1 1
2 6 1 2 3 5
1
4 8 1 1 2 3
1 1 10 34 10 2 2 1 7 10 1 9 10 3 3
n
50. 0.21515
n0
81 1 Geometric series with a 100 and r 100
81 81 100 9 a 1 r 1 1 100 99 11
S
2
n0
10
100 100
2 3
n0
n0
48. 0.8181
40.
8 32 1 3 4
n1
52.
1
n1
n0
46.
1
2n 12n 3 2 2n 1 2n 3 2 3 5 5 7 7 9 . . . 2 3 6
n0
42.
1
n1
n1
38.
1
3 1 1 5 n0 200 100
3 1 Geometric series with a 200 and r 100
S
a 1 3 200 71 1 5 1 r 5 99 100 330
n1
2n 1
n1
n1 1 0 2n 1 2
lim
n→
Diverges by Theorem 8.9
54.
1
n1
56.
nn 3 3 n n 3
1
1
1
n1
1 41 12 51 13 61 14 71 15 81 16 91 . . .
1 3
1 1 1 11 1 , converges 3 2 3 18
3n
n
n1
3
ln 23n ln 22 3n ln n3 3n 3n lim lim lim 3 2 n→ n n→ n→ n→ 3n 6n 6 lim
(by L’Hôpital’s Rule) Diverges by Theorem 8.9
n
1
375
376
58.
Chapter 8
Infinite Series
1
4
60.
n
n0
2n
100
62.
n1
Geometric series with r
1 4
k
n
n1
Geometric series with r 2
Converges by Theorem 8.6
1 n
lim 1
n→
Diverges by Theorem 8.6
k n
n
ek 0
Diverges by Theorem 8.9
64. lim an 5 means that the limit of the sequence an is 5. n→
a
n
66. If lim an 0, then n→
a
n
diverges.
n1
a1 a2 . . . 5 means that the limit of the
n1
partial sums is 5. 68. (a) x 2 is the common ratio. (c) y1
(b) 1
2 2x
−5
Geometric series:
12
1 1 x 2
2 , x < 2 2x
1 < 0.0001 2n
72.
y = 10
Horizontal asymptote: y 10 4
x x a 1, r , < 1 ⇒ x < 2 2 2
1 0.8x
1 0.8
n
5
−5
2 5
5
x y2 1 2
70. f x 2
x2 x3 x x 2 4 8 2 n0
10,000 < 2n This inequality is true when n 14.
n −2
n0
16
0.01n < 0.0001
−2
2 S 10 1 4 5
10,000 < 10n This inequality is true when n 5. This series converges at a faster rate.
The horizontal asymptote is the sum of the series. f n is the nth partial sum. 74. Vt 225,0001 0.3n 0.7n225,000
n1
76.
1000.60 i
i0
V5 0.75225,000 $37,815.75
1001 0.6n 1 0.6
2501 0.6n Sum 250 million dollars
78. The ball in Exercise 77 takes the following times for each fall. s1 16t 2 16
s1 0 if t 1
s2 16t 2 160.81
s2 0 if t 0.9
s3 16t 160.81
s3 0 if t 0.92
2
2
sn 16t 160.81 2
n1
sn 0 if t 0.9n1
Beginning with s2, the ball takes the same amount of time to bounce up as it takes to fall. The total elapsed time before the ball comes to rest is t12
0.9
n1
1
n
1 2
0.9
n0
2 19 seconds. 1 0.9
n
80. Pn
1 2 3 3
P2
1 2 3 3
3 3
n0
1 2
n
n
2
4 27
1 3 1 1 2 3
million dollars
Section 8.2
Series and Convergence
377
82. (a) 64 32 16 8 4 2 126 in.2
64 2
(b)
1
n
n0
64 128 in.2 1 1 2 16 in.
Note: This is one-half of the area of the original square! 16 in.
13 9 2
84. Surface area 4 12 9 4
P 1 12
12t1
86.
r
n
12 P r P
Per 12n
n0
4 9
1
2
. . . 4 . . .
12r r 1 1
12 12t
P 1 1
n0
12t1
2
88. P 75, r 0.05, t 25
12t
r 1 1 12
12r
1 12r
12t
(a) A 75 (b) A
1
(b) A
1250
1 $75,743.82
75e0.0525 1 $44,732.85 e0.05 12 1
39
40,0001.04
n
n0
20e0.0650 1 $76,151.45 e0.06 12 1
40,000
111.04 1.04
40
$3,801,020 94. x 0.a1a2a3 . . . aka1a2a3 . . . ak
0.a1a2a3 . . . ak 1
101
1 1 10k 10k
10
1
n0
0.a1a2a3 . . . ak
2
k
3
. . .
n
k
1 11 10 a rational number k
96. Let Sn be the sequence of partial sums for the convergent series lim Sn L and since Rn
n→
a
n
L. Then
n1
ak L Sn,
kn1
we have lim Rn lim L Sn lim L lim Sn L L 0.
n→
1 $44,663.23
92. T 40,000 40,0001.04 . . . 40,0001.0439
12 0.06
1 0.06 12
0.a1a2a3 . . . ak
1225
P1 er 1212t Pert 1 r 12 1 er 12 e 1
90. P 20, r 0.06, t 50 (a) A 20
12 0.05
1 0.05 12
n→
n→
n→
98. If an bn converged, then an bn an bn would converge, which is a contradiction. Thus, an bn diverges.
378
Chapter 8
Infinite Series
100. True
104.
102. True;
1 1 1 1 1 2 3. . . r r r n0 r r
n
1r 1 1 1r r 1
This is a geometric series which converges if
Section 8.3 2.
since
1 < 1 r
4.
n2
Let f x xex2.
2 . 3x 5
f is positive, continuous, and decreasing for x ≥ 3.
f is positive, continuous, and decreasing for x ≥ 1.
1
ne
n1
n1
The Integral Test and p-Series
2
n 1 0 1000n 1 1000
1 < 1 ⇔ r > 1. r
3n 5
Let f x
lim
n→
2 2 dx ln3x 5 3x 5 3
1
Since fx
2x < 0 for x ≥ 3. 2e x2
Diverges by Theorem 8.10
xex2 dx 2x 2ex2
3
10e32
3
Converges by Theorem 8.10
6.
1
2n 1
8.
1 . 2x 1
1
n 3
Let f x
1 dx ln 2x 1 2x 1
1
x . x2 3
f x is positive, continuous, and decreasing for x ≥ 2 since
f is positive, continuous, and decreasing for x ≥ 1.
2
n1
n1
Let f x
n
3 x2 < 0 for x ≥ 2. x 2 3
fx
Diverges by Theorem 8.10
1
x dx ln x2 3 x2 3
1
Diverges by Theorem 8.10
10.
ne
k n
12.
1
n
13
n1
n1
xk Let f x x . e
Let f x
f is positive, continuous, and decreasing for x > k since
f is positive, continuous, and decreasing for x ≥ 1.
fx
1
for x > k. We use integration by parts.
1
xk ex dx xk ex
1
k
1 x13
dx
xk1 ex dx
1
2 x 3
23
Diverges by Theorem 8.10
1 k kk 1 . . . k! e e e e
Converges by Theorem 8.10
xk1k x < 0 ex
1 . x13
1
Section 8.3
14.
3
n
n1
2
n1
Divergent p-series with p 23 < 1
22.
1
n
20.
23
n1
2
Convergent p-series with p 2 > 1
5 > 1 3
1
n
1
n1
Convergent p-series with p
18.
n
16.
53
2
2
The Integral Test and p-Series
Convergent p-series with p > 1
2
n123. . .
2
n
24.
2
2
2 2 . . . 22 32
n1
n1
S1 2
S1 2
S2 3
S2 2.5
S3 3.67
S3 2.722
Matches (d)
Matches (c)
Diverges—harmonic series
Converges—p-series with p 2 > 1.
26. (a)
8
n
5
10
20
50
100
Sn
3.7488
3.75
3.75
3.75
3.75 0
11 0
The partial sums approach the sum 3.75 very rapidly. (b)
5
n
5
10
20
50
100
Sn
1.4636
1.5498
1.5962
1.6251
1.635 0
11 0
The partial sums approach the sum
28. x
n
x
n1
2
1.6449 slower than the series in part (a). 6
1
n
x
n1
Converges for x > 1 by Theorem 8.11.
30.
ln n p n2 n
If p 1, then the series diverges by the Integral Test. If p 1,
2
ln x dx xp
xp ln x dx
2
xp1
p 1
2
1 p 1 ln x
. (Use Integration by Parts.)
2
Converges for p 1 < 0 or p > 1.
32. A series of the form
n1
1 is a p-series, p > 0. np
The p-series converges if p > 1 and diverges if 0 < p ≤ 1.
34. The harmonic series
1
n.
n1
379
380
Chapter 8
Infinite Series
38. S4 1
36. From Exercise 35, we have: 0 ≤ S SN ≤
f x dx
R4 ≤
N
SN ≤ S ≤ SN N
≤ S ≤
n
n1
N
a
n
f x dx
n1
1 1 dx 4 x5 4x
1
n
1.0363 ≤
0.0010
4
≤ 1.0363 0.0010 1.0373
5
n1
f x dx
N
1 1 1 1 . . .
1.9821 2ln 23 3ln 33 4ln 43 11ln 113
40. S10
R10 ≤
10
1 1 dx x 1lnx 13 2lnx 12
1
n 1lnn 1
1.9821 ≤
10
1 1 1 1 2 3 4 0.5713 e e e e
R4 ≤
ex dx ex
4
e
0.5713 ≤
n
4
1
0.0870 2ln 113
≤ 1.9821 0.0870 2.0691
3
n1
42. S4
4
N
a
1 1 1 1.0363 25 35 45
44. 0 ≤ RN ≤
1 x32
N
dx
2 x12
N
2 N
< 0.001
N12 < 0.0005
0.0183
N > 2000
N ≥ 4,000,000
≤ 0.5713 0.0183 0.5896
n0
46.
RN ≤
ex2 dx 2ex2
N
2 e N2 e N2
N
2 e N2
< 0.001
48. Rn ≤
N
< 0.001
2 N arctan < 0.001 5 2 5
2 1 x dx 2 arctan x2 5 5 5
N
N arctan < 0.001118 2 5
> 2000
N > ln 2000 2
1.56968 < arctan
N > 2 ln 2000 15.2 N
N ≥ 16
5
N5
> tan 1.56968
N ≥ 2004
50. (a)
10
1 xp1 dx p x p 1
(b) f x
10
1 .p > 1 p 110 p1
1 xp
R10p
1 under the graph of f over the interval 10, p ≤ Area n n11
(c) The horizontal asymptote is y 0. As n increases, the error decreases.
52.
ln1 n ln
n2
1
2
n2
n2 1 n 1(n 1 ln lnn 1 lnn 1 2 ln n 2 2 n n n2 n2
ln 3 ln 1 2 ln 2 ln 4 ln 2 2 ln 3 ln 5 ln 3 2 ln 4 ln 6 ln 4 2 ln 5 ln 7 ln 5 2 ln 6 ln 8 ln 6 2 ln 7 ln 9 ln 7 2 ln 8 . . . ln 2
Section 8.4
54.
1
nn
2
n2
Let f x
56. 3
1
1
n
n1
Comparisons of Series
0.95
p-series with p 0.95
1 . xx 2 1
Diverges by Theorem 8.11
f is positive, continuous, and decreasing for x ≥ 2.
2
1 dx arcsec x xx 2 1
2
2 3
Converges by Theorem 8.10
58.
1.075
n
60.
n0
1
n1
Geometric series with r 1.075
2
1 1 1 3 2 3 n n1 n n1 n
62.
Since these are both convergent p-series, the difference is convergent.
Diverges by Theorem 8.6
64.
n
lnn
n2
lim lnn
n→
Diverges by Theorem 8.9
ln n 3 n2 n
Let f x
ln x . x3 1 3 ln x < 0 for x ≥ 2. x4
f is positive, continuous, and decreasing for x ≥ 2 since fx
2
ln x ln x dx 2 x3 2x
2
1 2
2
1 ln 2 1 dx 2 x3 8 4x
2
ln 2 1 (Use Integration by Parts.) 8 16
Converges by Theorem 8.10. See Exercise 14.
Section 8.4 2. (a)
2
Comparisons of Series 2
n 2 2 . . .
S1 2
an
an =
n1
4
2 2 2 . . . S1 4 0.5 2 0.5 n1 n 0.5
3
2
4 4 4 . . . S1 3.3 n 0.5 1.5 2.5 n1
(b) The first series is a p-series. It diverges p
2 n − 0.5
1 2
4 n + 0.5
an =
an = 2 n
1
< 1.
n 2
4
6
8
10
Σ
4 n + 0.5
(c) The magnitude of the terms of the other two series are greater than the corresponding terms of the divergent p-series. Hence, the other two series diverge. (d) The larger the magnitude of the terms, the larger the magnitude of the terms of the sequence of partial sums.
Sn
20
Σ
16
2 n − 0.5
12 8 4
Σ
2 n
8
10
n 2
4
6
381
382
Chapter 8
Infinite Series
4.
1 1 < 3n2 2 3n2
6.
Therefore,
1
2
n1
1 2
diverges by comparison with the divergent p-series
3n 3 < 4 5 4 n
n
10.
1
converges by comparison with the convergent p-series
3 n . 4
1 1 > 3 n 1 4 n 4 4 Therefore,
14.
3n
4n 4n > n 1 3
Therefore,
1
3
diverges by comparison with the divergent p-series
n1
2 3n 5 2 3n 2 lim n n n→ n→ 3 5 1 3
Therefore,
2 5
3 . n
n1
5n 3 n2 2n 5 5n2 3n 20. lim lim 2 5 n→ n→ n 2n 5 1 n Therefore,
n
2
n1
1
n.
n1
n1
3 n2 4 3n 3 lim n→ n→ n2 4 1 n
18. lim
3 4
n
2
n3
diverges by a limit comparison with the divergent harmonic series
1
n.
n3
1 nn2 1 n3 22. lim lim 3 1 n→ n→ n n 1 n3 Therefore,
5n 3 2n 5
diverges by a limit comparison with the divergent p-series
n
4
Therefore,
converges by a limit comparison with the convergent geometric series
4n 1
3 .
16. lim
3n
diverges by comparison with the divergent geometric series
1 1 . 4 n 4n1
1
3 2 .
n1
4 n 1
1
n
n1
1 1
3
n1
converges by comparison with the convergent geometric series
n
n
1 n3 2
Therefore,
3n n 4 5 n0
3
<
n3 1
n1
1
n.
n2
Therefore,
12.
1
n2
1 1 . 3 n1 n2
n0
for n ≥ 2.
n 1
converges by comparison with the convergent p-series
n
Therefore,
3n
8.
1
>
n 1
nn
2
n1
1 1
converges by a limit comparison with the convergent p-series
1
n.
n1
3
Section 8.4
24. lim
n→
n n 12n1 n lim 1 n→ n 1 1 2n1
26. lim
n→
Therefore,
2 1
5
n n
n1
2
n1
4
diverges by a limit comparison with the divergent harmonic series
converges by a limit comparison with the convergent geometric series
n1
1
n.
.
n1
n1
n→
5 n n2 4 5n 5 lim n→ n n2 4 2 1 n
n
n1
28. lim
383
Therefore,
n 12
Comparisons of Series
1 n2 sec21 n 1 tan1 n lim lim sec2 1 n→ n→ 1 n 1 n2 n
Therefore,
tan n 1
n1
diverges by a limit comparison with the divergent p-series
1
n.
n1
30.
5 5 1
n
32.
1 2n 15
Converges
Converges Geometric series with r
2
n4
n0
34.
3n
15
Limit comparison with
1
n
n1
2
n 1 n 2 2 3 3 4 4 5 . . . 2 1
1
1
1
1
1
1
1
1
n1
Converges; telescoping series
36.
3
nn 3
n1
Converges; telescoping series
n n 3 1
1
n1
38. If j < k 1, then k j > 1. The p-series with p k j converges and since lim
n→
Pn Qn L > 0, 1 nkj
the series
Pn
Qn converges by the limit comparison test. Similarly, if j ≥ k 1, then k j ≤ 1 which implies that
n1
Pn Q n n1
diverges by the limit comparison test.
40.
1 1 1 1 1 1 . . . , 2 3 8 15 24 35 n 1 n2
which converges since the degree of the numerator is two less than the degree of the denominator.
42.
n
n1
3
n2 1
diverges since the degree of the numerator is only one less than the degree of the denominator.
384
Chapter 8
Infinite Series
n 1 lim lim n 0 ln n n→ 1 n n→
44. lim
n→
Therefore,
1
ln n diverges.
n2
48. This is not correct. The beginning terms do not affect the convergence or divergence of a series.
46. See Theorem 8.13, page 585. One example is
1
n 1 diverges because lim n→
n2
1 n 1 1 1 n
In fact, 1 1 1 . . . diverges (harmonic) 1000 1001 n1000 n
and
1
50.
and
diverges (p-series).
n2 n
1
1 1 1 1 . . . , 200 210 220 n0 200 10n
52.
diverges
54. (a)
1 1 1 . . . 2 converges (p-series). 4 9 n1 n
1 1 1 1 1 . . . 3, 201 208 227 264 n1 200 n
converges
1
2n 1
2
n1
1 4n 1
4n
2
n1
converges since the degree of the numerator is two less than the degree of the denominator. (See Exercise 38.) (b)
(c)
n
5
10
20
50
100
Sn
1.1839
1.02087
1.2212
1.2287
1.2312
1
2n 1
2
n3
(d)
2 S2 0.1226 8
1 2 S9 0.0277 2 2n 1 8 n10
56. True 58. False. Let an 1 n, bn 1 n, cn 1 n2. Then, an ≤ bn cn, but
c
converges.
n
n1
60. Since
a
n
a
converges, then
n1
n an
n1
a
2
n
1
62.
n
(b)
a
n1
2
converge, and hence so does
n 1
2
2
1
n . 4
converges by Exercise 59.
64. (a)
a
n
1
n
3
and
b
n
1
n . Since 2
an 1 n3 1 lim lim 0 n→ bn n→ 1 n2 n→ n lim
converges, so does
1
n . 3
and
1
n
2
lim
n
n→
1
n and b
n
1
n. Since
an 1 n lim n lim n→ bn n→ 1 n
diverges, so does
1
n.
and
1
n
Section 8.5
Section 8.5 2.
6.
1n1 6 6 6 6 . . . n2 1 4 9 n1
4.
1n1 10 10 10 . . . n2n 2 8 n1
S1 6, S2 4.5
S1 5, S2 3.75
Matches (d)
Matches (a)
1n1
385
Alternating Series
Alternating Series
1
n 1! e 0.3679
n1
(a)
(b)
n
1
2
3
4
5
Sn
1
0
0.5
0.3333
0.375
6 0.3667
7
8
9
10
0.3681
0.3679
0.3679
0.3679
2
0
11 0
(c) The points alternate sides of the horizontal line that represents the sum of the series. The distance between successive points and the line decreases. (d) The distance in part (c) is always less than the magnitude of the next series.
8.
1n1
2n 1! sin1 0.8415
n1
(a)
(b)
n
1
2
3
4
5
6
7
8
9
10
Sn
1
0.8333
0.8417
0.8415
0.8415
0.8415
0.8415
0.8415
0.8415
0.8415
(c) The points alternate sides of the horizontal line that represents the sum of the series. The distance between successive points and the line decreases.
2
0
11 0
10.
1n1 n n1 2n 1
lim
n→
(d) The distance in part (c) is always less than the magnitude of the next series.
12.
1n
lnn 1
n1
n 1 2n 1 2
an1
Diverges by the nth Term Test. lim
n→
1 1 < an lnn 2 lnn 1
1 0 lnn 1
Converges by Theorem 8.14
14.
1n1n 2 n1 n 1
an1 lim
n→
n1 n < 2 an n 12 1 n 1
n 0 n2 1
Converges by Theorem 8.14
16.
1n1n2 2 n1 n 5
lim
n→
n2 1 n2 5
Diverges by nth Term Test
386
18.
Chapter 8
Infinite Series
1n1 lnn 1 n1 n1
lim
n sin 1
n1
lnn 1 lnn 1 1 < for n ≥ 2 n 1 1 n1
an1
n→
20.
1n1 2n 1 2 n n1
Converges; (see Exercise 9)
lnn 1 1n 1 lim 0 n→ n1 1
Converges by Theorem 8.14
22.
1n n n1
1
n cos n
n1
24.
1n
2n 1!
n0
Converges; (see Exercise 9)
1 1 an < 2n 3! 2n 1!
an1 lim
n→
1 0 2n 1!
Converges by Theorem 8.14
26.
1n1 n 3 n n1
28.
n12 lim n16 n→ n13 n→
1n12en 21n1 n en e e2n 1 n1 n1
Let f x
lim
Diverges by the nth Term Test
fx
2ex . Then 1
e2x
2e2x 1 e2x < 0 for x > 0. e2x 12
Thus, f x is decreasing for x > 0 which implies an1 < an. lim
n→
2en 2en 1 lim 2n lim n 0 n→ e 1 n→ 2e
e2n
The series converges by Theorem 8.14.
30. S6
41n1
6
lnn 1 2.7067
n1
4
R6 S S6 ≤ a7 ln 8 1.9236; 0.7831 ≤ S ≤ 4.6303 32. S6
1n1n 0.1875 2n n1 6
7
R6 S S6 ≤ a7 27 0.05469; 0.1328 ≤ S ≤ 0.2422 34.
1n n n0 2 n!
36.
(a) By Theorem 8.15,
Rn
≤ aN1
(a) By Theorem 8.15, 1 < 0.001. 2N1N 1!
This inequality is valid when N 4. (b) We may approximate the series by
1 1 1 1 1 n n! 1 2 8 48 348 0.607. 2 n0 4
1n n0 2n!
n
(5 terms. Note that the sum begins with n 0.)
RN
≤ aN1
1 < 0.001. 2N 2!
This inequality is valid when N 3. (b) We may approximate the series by
1n 1 1 1 1 0.540. 2n ! 2 24 720 n0 3
(4 terms. Note that the sum begins with n 0.)
Section 8.5
38.
387
1n1 4n n n1
(b) We may approximate the series by
(a) By Theorem 8.15,
RN
≤ aN1
1n1 1 1 1 0.224. 4n n 4 32 192 n1 3
1 < 0.001. N 1
4N1
This inequality is valid when N 3.
40.
Alternating Series
1n1 n4 n1
(3 terms)
42.
By Theorem 8.15, RN ≤ aN1
1 < 0.001. N 14
1n1 n1 n 1
The given series converges by the Alternating Series Test, but does not converge absolutely since the series
This inequality is valid when N 5.
1
n1
n1
diverges by the Integral Test. Therefore, the series converge conditionally.
44.
1n1 n n n1
46.
1
1
n n n
n1
n1
32
1n n2 n0 e
n→
50.
1n1 n1.5 n1
1
1
n
n2
n1
converges by a comparison to the convergent geometric series
2n 3 2 Therefore, the series diverges by the n 10
nth Term Test.
e
n0
lim
which is a convergent p-series.
Therefore, the given series converges absolutely.
48.
1n12n 3 n 10 n1
1.5
is a convergent p-series.
Therefore, the given series converge absolutely.
e . n
1
n0
Therefore, the given series converges absolutely.
52.
1n
n 4
n0
The given series converges by the Alternating Series Test, but
1 4
n0 n
diverges by a limit comparison to the divergent p-series
1
n.
n1
Therefore, the given series converges conditionally.
54.
1
n1
arctan n
n1
lim arctan n
n→
0 Therefore, the series diverges by 2
the nth Term Test.
388
56.
Chapter 8
Infinite Series
1n1 sin2n 12 n n n1 n1
58. S Sn Rn ≤ an1 (Theorem 8.15)
The given series converges by the Alternating Series Test, but
n1
1 sin2n 12 n n1 n
is a divergent p-series. Therefore, the series converges conditionally.
60.
1 1 1 1n1 1 . . . (Alternating Harmonic Series) n 2 3 4 n1
62.
1n np n1
64.
1 1n1 converges, but diverges n n1 n1 n
If p 0, then lim
n→
1 1 np
and the series diverges. If p > 0, then lim
n→
1 1 1 0 and < p. np n 1p n
Therefore, the series converge by the Alternating Series Test.
66. (a)
xn n1 n
(b) When x 1, we have the convergent alternating series
1n . n n1
converges absolutely (by comparison) for 1 < x < 1,
When x 1, we have the divergent harmonic series
since
xn < xn and n
1 . n
x
n
Therefore,
is a convergent geometric series for 1 < x < 1.
xn n1 n
converges conditionally for x 1.
68. True, equivalent to Theorem 8.16
70.
n
n1
2
3 5
converges by limit comparison to convergent p-series 1
n . 2
3 2
72. Converges by limit comparison to convergent geometric 1 . series 2n
74. Diverges by nth Term Test. lim an
76. Converges (conditionally) by Alternating Series Test.
78. Diverges by comparison to Divergent Harmonic Series:
n→
ln n 1 > for n ≥ 3. n n
Section 8.6
Section 8.6 2.
The Ratio and Root Tests
389
The Ratio and Root Tests
2k 2! 2k 2! 1 2k! 2k2k 12k 2! 2k2k 1
4. Use the Principle of Mathematical Induction. When k 3, the formula is valid since
1 233!35 1. Assume that 1 6!
1 2n n!2n 32n 1 . . . 135 2n 5 2n! and show that 1 2n1n 1!2n 12n 1 . . . . 135 2n 52n 3 2n 2! To do this, note that:
35.
1
1 . . 2n 52n 3 1
1
1 . . 2n 5
2n 3
2n n!2n 32n 1 2n!
2n 3
2n n!2n 1 2n!
2n 2n 1n!2n 12n 1 2n!2n 12n 2
2n1n 1!2n 12n 1 2n 2!
35.
1
2n 12n 2
2n 12n 2
The formula is valid for all n ≥ 3.
6.
4 n! 4 16 2 . . . 3
n
3
1
9
1
8.
n1
1n1 4 4 4 . . . (2n! 2 24 n1
S1 2
3 S1 , S2 1.03 4
Matches (b)
Matches (c)
10.
4e
n
4
n0
4 . . . e
S1 4 Matches (e)
12. (a) Ratio Test: lim
n→
(b)
(c)
an1 an
n 12 1 n 1! n2 2n 2 lim lim 2 n→ n→ n 1 n2 1 n!
n
5
10
15
20
25
Sn
7.0917
7.1548
7.1548
7.1548
7.1548
n 1 1 0 < 1. Converges
(d) The sum is approximately 7.15485
10
(e) The more rapidly the terms of the series approach 0, the more rapidly the sequence of the partial sums approaches the sum of the series. 0
11 0
390
14.
Chapter 8
Infinite Series
3n
n!
n0
16.
n1
an1 3 lim n→ n 1! an
lim
n→
lim
n→
n!
3n
3 0 n1
lim
n→
lim
n→
3n 1 3 2n 2
1n1n 2 nn 1 n1
an 1 n 13 2n1 lim n→ an n3 2n n→
2n
n3n
Therefore, by the Ratio Test, the series diverges.
20.
lim
an1 n 13n1 lim n→ an 2n1 n→
n
n1
n
lim
n3
2
3
n1
Therefore, by the Ratio Test, the series converges.
18.
n 2
an1
n 13 1 2n3 2
lim
n→
Therefore, by the Ratio Test, the series converges.
n3 n2 ≤ an n 1n 2 nn 1
n2 0 nn 1
Therefore, by Theorem 8.14, the series converges. Note: The Ratio Test is inconclusive since lim
n→
The series converges conditionally.
22.
1n13 2n n2 n1
lim
n→
24.
an1 3 2n1 lim 2 n→ n 2n 1 an lim
n→
n2
3 2n
n1
n→
3n2 3 > 1 2n2 2n 1 2
nn
n!
n1
lim
n→
an1 n 1 lim n→ an n 1!
n1
n!
nn
n→
lim
n 1n 1n n! n 1n!nn
lim
n n 1
n→
n
lim
an1
n 1! 2 lim n→ 3n 3! an lim
n→
lim
n→
an1 24n4 lim n→ 2n 3! an
3n!
n!2
n 12 0 3n 33n 23n 1
Therefore, by the Ratio Test, the series converges.
2n 22n 1n5 n 15
n!2
n→
e > 1
1n 24n n0 2n 1!
3n!
Therefore, by the Ratio Test, the series diverges.
30.
n5
2n!
Therefore, by the Ratio Test, the series diverges.
n0
an1 2n 2! lim n→ n 15 an lim
28.
n→
2n! n5
lim
Therefore, by the Ratio Test, the series diverges.
26.
an1 1. an
2n 1! 24 lim 0 n→ 2n 32n 2 24n
Therefore, by the Ratio Test, the series converges.
Section 8.6
32.
1n 2 4 6 . . . 2n
2 5 8 . . . 3n 1
n1
lim
n→
an1 2 4 . . . 2n2n 2 lim n→ 2 5 . . . 3n 13n 2 an
2
5 . . . 3n 1 2 4 . . . 2n
Therefore, by the Ratio Test, the series converges. 2 24 24 Note: The first few terms of this series are 2 25 25
lim
n→
an1 1 lim n→ n 14 an
4
p
n4 n lim n→ n 1 1
np n lim n→ n 1 1
1
n lim
n→
an1 1 lim n→ n 1p an
n 1 2n
6. 8
. .
1
n
38.
3n
2n 1
n 2n 1
n
n
n →
lim
n→
n a n lim
2n 2 n→ n 1
n→
lim
lim
n→
2n3n 1 n
3n
2n3n 1 32 3
e
n
n0
n an lim lim
n→
n→
e1 1e n
n
Therefore, by the Root Test, the series converges.
42.
n1 3n n0
44.
n→
n 3 1 lim n
n→
n
n→
n n 1
3
x y lim x1
Let
n→
x ln y lim ln x 1 n→
lim
1 lnx 1 x
lim
lnx 1 1 0. x x1
n→
n→
Since ln y 0, y e0 1, so lim
n→
n 1
3
1 . 3
Therefore, by the Root Test, the series converges.
5
1
n 5 n
n1
n lim an lim
3
27 8
Therefore, by the Root Test, the series diverges.
Therefore, by the Root Test, the series diverges.
3n
n1
n a lim n lim
40.
2n 2 2 3n 2 3
1
n1
n→
n→
p
n1
lim
4
n1
(b)
1
n
34. (a)
36.
The Ratio and Root Tests
n1
This is the divergent harmonic series.
391
392
46.
Chapter 8
4
Infinite Series
n
48.
2n
2
n1
n1
Since 4 < 1,this is convergent geometric series.
n 1
n 2n2 1 n2 1 lim > 0 n→ 2n2 1 1 n 2
lim
n→
This series diverges by limit comparison to the divergent harmonic series
1
n.
n1
50.
10
3 n
52.
3
n1
lim
3 2
1 n
10 3
lim
n→
Therefore, the series converges by a limit comparison test with the p-series
1
n
n1
54.
3 2
.
1n n2 n lnn
lim
n→
56.
lnn 2 n1 n
1 lnn ≤ 3 2 n2 n
1 1 ≤ an n 1 lnn 1 n lnn
Therefore, the series converges by comparison with the p-series
1 0 n lnn
1n 3n n2n n1
1
n
Therefore, by the Alternating Series Test, the series converges.
58.
2n ln 22n ln 222n lim lim n→ n→ 1 8n 8
4n2
Therefore, the series diverges by the nth Term Test for Divergence.
an1
2n 1
2
n1
10 3n3 2
n→
4n
n1
3 2
.
lim
n→
an1 3n1 lim n→ n 12n1 an
n2n 3n 3 lim n→ 2n 1 3n 2
Therefore, by the Ratio Test, the series diverges. 60.
3
n1
lim
n→
57. 18n
. . 2n 1 2n 1n!
an1 3 5 7 . . . 2n 12n 3 lim n→ an 18n12n 12n 1n!
3
18n 2n 1n! 2n 32n 1 1 2 lim 5 7 . . . 2n 1 n→ 182n 12n 1 18 9
Therefore, by the Ratio Test, the series converge.
64. (a) and (b) are the same.
62. (b) and (c)
n 14
n0
3
n
n 4 3
n1
n1
12
34 334
1n
n 12
n2 2
4
34
3
. . .
n1
1 1 1 . . . 2 2 22 3 23
1n1 1 1 1 . . . n2n 2 2 22 3 23 n1
Section 8.7
66. Replace n with n 2.
68.
3k
3k 2k k!
1 3 5 . . . 2k 1 2k!2k 1
k0
2n2 2n n2 n 2! n0 n!
Taylor Polynomials and Approximations
k0
6k k!
2k 1!
k0
0.40967 (See Exercise 3 and use 10 terms, k 9.)
70. See Theorem 8.18.
72. One example is
100 n. 1
n1
74. Assume that
lim an1an L > 1 or that lim an1an .
n→
n→
Then there exists N > 0 such that an1an > 1 for all n > N. Therefore,
an1 > an,
n > N ⇒ lim an 0 ⇒ n→
a
n
diverges
76. The differentiation test states that if
U
n
n1
is an infinite series with real terms and f x is a real function such that f 1n Un for all positive integers n and d 2 fdx 2 exists at x 0, then
U
n
n1
converges absolutely if f 0 f0 0 and diverges otherwise. Below are some examples. Convergent Series 1
Divergent Series 1
n , f x x
n, f x x
1 cos n, f x 1 cos x
sin n, f x sin x
3
3
1
Section 8.7
1
Taylor Polynomials and Approximations 4. y e12 13 x 13 x 1 1
2. y 18 x 4 12 x 2 1
Cubic Matches (b)
y-axis symmetry Three relative extrema Matches (c)
6. f x
4 3 x
4x13
4 fx x43 3
f 8 2 f8
P1x f 8 f8x 8
1 2 x 8 12 P1x
1 8 x 12 3
8
1 12
(8, 2) −4
14
−4
393
394
Chapter 8
Infinite Series
8. f x tan x
1 4
f
fx sec2 x
f
12 x P1x 2x 1
− 2
2 4
P1 f f 4 4
3
4
x 4
−3
2
f
4 2
fx sec x tan x
f
4 2
f x sec3 x sec x tan2 x
f
10. f x sec x
P2x f
2
4
4 3 2
−3
4 f 4 x 4 f 24 x 4
P2x 2 2 x
3 2 x 4 2 4
3 0
2
2
0.585
0.685
4
0.885
0.985
1.785
f x
1.8270 1.1995
1.2913
1.4142
1.5791
1.8088
4.7043
P2x
15.5414
1.2936
1.4142
1.5761
1.7810
4.9475
2.15
x
1.2160
12. f x x 2ex, f 0 0 (a)
fx x 2 2xex f x
f0 0 ex
f 0 2
f x x 2 6x 6ex
f 0 6
x2
4x 2
f 4x x 2 8x 12ex
P4x x 2 x3
14.
12x 4 x4 x 2 x3 4! 2
f x ex
f 0 1
fx
ex
f0 1
f x
ex
f 0 1
f x
ex
f 0 1
f 4x
ex
f 5x ex
P2
f −3
3
P3 −1
(c)
6x3 x 2 x3 3!
2
P4
f 40 12
2x 2 P2x x2 2! P3x x 2
(b)
f 0 2 P2 0 f 0 6 P3 0 f 40 12 P440
(d) f n0 Pnn0
f 40 1 f 50 1
P5x f 0 f0x
f0 2 f 0 3 f 40 4 f 50 5 x 2 x3 x4 x5 x x x x 1x 2! 3! 4! 5! 2 6 24 120
Section 8.7 16.
f x e3x
f 0 1
fx
3e3x
f0 3
f x
9e3x
f 0 9
Taylor Polynomials and Approximations
f x 27e3x f 0 27 f 4x 81e3x f 40 81 P4x 1 3x
18.
9 9 2 27 3 81 4 9 27 4 x x x 1 3x x 2 x3 x 2! 3! 4! 2 2 8
f x sin x
f 0 0
20.
f x x 2ex
f 0 0
f0
fx
2xex
f x 2 sin x
f 0 0
f x
2ex
f x 3 cos x
f 0 3
f x 6ex 6xex x 2ex
fx cos x
P3x 0 x
4xex
f x
x x11 1 x 11 x1 x1
f 0 0 f0 1
f x 2x 13
f 0 2
fx 6x 14
f0 6
24x 1
f 40 24
5
2 6 24 P4x 0 1x x2 x3 x4 x x2 x3 x4 2 6 24 24. f x tan x
f 0 0
fx sec2 x
f0 1
f x 2
sec2
x tan x
fx 4
sec2
tan2
x
f 0 0
x2
sec4
f0 2
x
2 1 P3x 0 1x 0 x3 x x3 6 3
26.
1 2
f x 2x2
f 2
fx 4x3
f2
f x 12x4
f 2
1 2
3 4
3 2 15 f 4x 240x6 f 4x 4 1 1 3 1 5 P4x x 2 x 22 x 23 x 24 2 2 8 4 32 fx 48x5
fx
f 0 2 f 0 6 f 40 12
2 2 6 3 12 4 x x x 2! 3! 4! 1 x 2 x3 x 4 2
fx x 12
f 4x
x2ex
f 4x 12ex 8xex x 2ex
0 2 3 3 3 3 x x x x 2! 3! 6
P4x 0 0x
22.
f0 0
x 2ex
395
396
Chapter 8
Infinite Series
28. f x x13
f 8 2
1 fx x23 3
f8
2 f x x53 9
f 8
fx
10 83 x 27
P3x 2
32.
f x
f8
30. f x x 2 cos x fx cos x x 2 sin x
1 12
x2
P2x 2 2 x
1
cos x f 2 2
2 2 x 2 2
5
28 3456
1 1 5 x 8 x 82 x 83 12 288 20,736
1 x2 1
2
P4
2x fx 2 x 12
f −3
3
Q4 P2
23x 2 1 f x 2 x 13
−2
24x1 x 2 x 2 14
fx f 4x
f 2
f x 2 cos x 4x sin x
1 144
10 27
f 2
245x 4 10x 2 1 x 2 15 (b) n 4, c 0
(a) n 2, c 0 P2x 1 0x
2 2 x 1 x2 2!
P4x 1 0x
2 2 0 24 4 x x3 x 1 x2 x 4 2! 3! 4!
(c) n 4, c 1 Q4x
34.
1 1 1 1 12 0 3 1 1 x 1 x 12 x 13 x 14 x 1 x 12 x 14 2 2 2! 3! 4! 2 2 4 8
f x ln x P1x x 1 P4x x 1 12 x 12 13 x 13 14 x 14 (a)
x
1.00
1.25
1.50
1.75
2.00
ln x
0.0000
0.2231
0.4055
0.5596
0.6931
P1x
0.0000
0.2500
0.5000
0.7500
1.0000
P4x
0.0000
0.2230
0.4010
0.5303
0.5833
(b)
2
P1 f −1
5
P4 −2
(c) As the distance increases, the accuracy decreases. 36. (a)
f x arctan x P3x x
(b)
y
(c) π 2
x3 3
π 4
x
0.75
0.50
0.25
0
0.25
0.50
0.75
f x
0.6435
0.4636
0.2450
0
0.2450
0.4636
0.6435
P3x
0.6094
0.4583
0.2448
0
0.2448
0.4583
0.6094
x
−1
f
P3
−π 4 −π 2
1 2
1
Section 8.7
40. f x 4xex 4
38. f x arctan x y
P7
Taylor Polynomials and Approximations 2
P5 P1
y
2 y = 4xe(−x /4)
4
2
2
1
f (x) = arctan x x
−3 −2
1
x
−4
3
4
−2
P9 P13
P11 P3
f x x 2ex x 2 x3
42. f
1 4 x 2
15 0.0328 f x x 2 cos x 2 2 x
44. f
2
2 x 2 2
78 6.7954
46. f x ex; f 6x ex ⇒ Max on 0, 1 is e1. R5x ≤
e1 6 1 0.00378 3.78 103 6!
48. f x arctan x; f 4x
24xx 2 1 1 x 24
50.
f x ex f n1x ex
⇒Max on 0, 0.4 is f 40.4 22.3672.
Max on 0, 0.6 is e0.6 1.8221.
22.3672 0.44 0.0239 R3x ≤ 4!
Rn ≤
1.8221 0.6n1 < 0.001 n 1!
By trial and error, n 5.
52.
f x cos x2
54.
gx cos x 1
x2 x 4 x6 . . . 2! 4! 6!
f x g x 2
f 0.6 1
2x 4 4x8 6x12 . . . 2! 4! 6! 2 4 6 0.64 0.68 0.612 . . . 2! 4! 6!
Since this is an alternating series, Rn ≤ an1
R3x
2n 0.64n < 0.0001. 2n!
By trial and error, n 4. Using 4 terms f 0.6 0.4257.
x
x3 3!
x4 sin z 4 x ≤ < 0.001 4! 4!
x 4 < 0.024
x 22 x 24 x 26 . . . 1 2! 4! 6! 1
f x sin x x
< 0.3936
0.3936 < x < 0.3936
397
398
Chapter 8
Infinite Series
56. f c P2c, fc P2c, and f c P2 c
58. See Theorem 8.19, page 611.
60.
62. (a) P5x x
y 10
P2
f P1
8
P5x 1
6 4 2 −20
10
−2
x2 x 4 2! 4!
This is the Maclaurin polynomial of degree 4 for gx cos x. x6 x2 x 4 (b) Q6x 1 for cos x 2 4! 6!
x
P3
x5 x3 for f x sin x 3! 5!
20
−4
x5 x3 P5x 3! 5! x3 x4 x2 (c) Rx 1 x 2! 3! 4! x3 x2 Rx 1 x 2! 3! Q6x x
The first four terms are the same! 64. Let f be an odd function and Pn be the nth Maclaurin polynomial for f. Since f is odd, f is even: fx lim
h→0
f x h f x f x h f x f x h f x lim lim fx. h→0 h→0 h h h
Similarly, f is odd, f is even, etc. Therefore, f, f , f 4, etc. are all odd functions, which implies that f 0 f 0 . . . 0. Hence, in the formula Pnx f 0 f0x
f 0x2 . . . all the coefficients of the even power of x are zero. 2!
f ic . 66. Let Pn x a0 a1x c a2x c2 . . . an x cn where ai i! Pnc a0 f c For 1 ≤ k ≤ n, Pnkc an k!
Section 8.8
f k!ck! f k
Power Series 4. Centered at
2. Centered at 0
6.
2x
n
n0
L lim
n→
kc.
8.
un 1 2xn1 lim 2x n→ un 2xn
2x < 1 ⇒ R
1 2
1n xn 2n n0
L lim
n→
1 x 2
un1 1n1xn1 lim n→ un 2n1
1 x < 1⇒R2 2
2n
1n xn
Section 8.8
10.
2n!x2n n! n0
L lim
n→
un 1 2n 2 n 1! lim n→ un 2n!x2nn! lim
n→
!x2n2
2n 22n 1 n 1
x2
The series only converges at x 0. R 0.
14.
1
n0
lim
n→
n1
k
12.
un 1 1 n 2x lim n→ un 1nn 1xn lim
n→
n 2x x n1
Interval: 1 < x < 1
When x 1, the series
1
n
Since the series is geometric, it converges only if xk < 1 or k < x < k.
3xn
2n!
16. n1
n0
lim
n→
un1 3xn1 lim n→ 2n 1! un lim
n→
n 1 diverges.
n 1 diverges.
n0
Therefore, the interval of convergence is 1 < x < 1.
18.
1n xn
n 1n 2
n0
lim
n→
un1 1n1xn1 lim n→ n 2n 3 un
Interval: 1 < x < 1 When x 1, the alternating series
n 1n 2 n 1x lim x n→ 1n xn n3 1n
n 1n 2 converges.
n0
When x 1, the series
1
1
n 1n 2 converges by limit comparison to n .
n0
n1
2
Therefore, the interval of convergence is 1 ≤ x ≤ 1.
20.
1n n!x 4n 3n n0
lim
n→
3x 0 2n 22n 1
n0
When x 1, the series
2n!
3xn
Therefore, the interval of convergence is < x <
n1
un1 1n1n 1!x 4n1 lim n→ un 3n1
R0
Center: x 4 Therefore, the series converges only for x 4.
399
n0
n 1xn n2
x
Power Series
3n
1n n!x 4n
lim
n→
n 1x 4 3
.
400
22.
Chapter 8
Infinite Series
x 2n1
n 14
n1
n0
lim
n→
un1 x 2n2 lim n2 n→ un n 24
n 14n1
x 2n1
R4
lim
n→
1 x 2n 1 x 2 4n 2 4
Center: x 2 Interval: 4 < x 2 < 4 or 2 < x < 6 When x 2, the alternating series
1n1
n 1 converges.
n0
When x 6, the series
1
n 1 diverges.
n0
Therefore, the interval of convergence is 2 ≤ x < 6.
24.
1n1x cn ncn n1
lim
n→
un1 1n2x cn1 lim n→ un n 1cn1
ncn
1n1x cn
Rc
lim
n→
nx c 1 xc cn 1 c
Center: x c Interval: c < x c < c or 0 < x < 2c When x 0, the p-series
1 diverges. n1 n
When x 2c, the alternating series
26.
1n1 converges. Therefore, the interval of convergence is 0 < x ≤ 2c. n n1
1nx2n1 n0 2n 1
lim
n→
un1 1n1x2n3 lim n→ un 2n 3
2n 1
1nx2n1
R1
lim
n→
2n 1 2 x x2 2n 3
Interval: 1 < x < 1
1n
When x 1,
2n 1 converges.
n0
When x 1,
1n1 converges. n0 2n 1
Therefore, the interval of convergence is 1 ≤ x ≤ 1.
28.
1n x 2n n! n0
lim
n→
30.
un1 1 lim n→ un n 1!
n1x 2n2
lim
n →
x2 0 n1
n!
1n x 2n
n1
Therefore, the interval of convergence is < x <
n!xn
2n! lim
n→
un1 n 1!xn1 lim n→ un 2n 2! lim
n→
.
2n! n!xn
n 1x 0 2n 22n 1
Therefore, the interval of convergence is < x <
.
Section 8.8
32.
2
4 6.
. . 2n
3 5 7 . . . 2n 1 x
n1
lim
n→
2n1
Power Series
401
un1 2 4 . . . 2n2n 2x 2n3 lim n→ un 3 5 7 . . . 2n 12n 3
3
2
5 . . . 2n 1 4 . . . 2nx 2n1
R1
lim
n→
2n 2x 2 x 2 2n 3
When x ± 1, the series diverges by comparing it to
1
2n 1
n1
which diverges. Therefore, the interval of convergence is 1 < x < 1.
34.
n!x cn
1 3 5 . . . 2n 1
n1
lim
n→
un1 lim n→ 1 un
n 1!x cn1 1 3 5 . . . 2n 1 n 1x c 1 lim x c . . 2n 12n 1 n→ n!x c 2n 1 2
3 5.
R2
Interval: 2 < x c < 2 or c 2 < x < c 2 The series diverges at the endpoints. Therefore, the interval of convergence is c 2 < x < c 2.
1 3n! c5. .2 .2nc 1 1 3 5 .n!.2. 2n 1 1 23 45 .6. . 2n2n 1 > 1
n
36. (a) f x
1n1x 5n , 0 < x ≤ 10 n5n n1
(b) fx
1n1n 1x 5n2 , 0 < x < 10 5n n2
(c) f x
f x dx
40. g2
1n1x 2n ,1 < x ≤ 3 n n1
38. (a) f x
1n1x 5n1 , 0 < x < 10 5n n1
(c) f x
. . .
(b) fx
(d)
2
3 2
n
1n1x 5n1 , 0 ≤ x ≤ 10 nn 15n n1
1
n0
2 4 . . . 3 9
1
x 2n1, 1 < x < 3
n1
n1
(d)
1
n1
f x dx
42. g2
n 1x 2n2, 1 < x < 3
n2
1n1x 2n1 ,1 ≤ x ≤ 3 nn 1 n1
3 alternating. Matches (d) 2
n
n0
S1 1, S2 1.67. Matches (a) 44. The set of all values of x for which the power series converges is the interval of convergence. If the power series converges for all x, then the radius of convergence is R . If the power series converges at only c, then R 0. Otherwise, according to Theorem 8.20, there exists a real number R > 0 (radius of convergence) such that the series converges absolutely for x c < R and diverges for x c > R.
46. You differentiate and integrate the power series term by term. The radius of convergence remains the same. However, the interval of convergence might change.
48. (a) f x
xn
n!, < x <
(See Exercise 11)
n0
(b) fx
xn
x2
x3
x4
n! 1 x 2! 3! 4! . . .
n1
xn nx n1 x n1 f x n! n 1 ! n1 n1 n0 n!
(c) f x
f 0 1 (d) f x ex
402
Chapter 8
Infinite Series
y1
50.
n1
22n n!
1n 4nx 4n1 7 11 . . . 4n 1
2
y
2n n!
n1
2
y
2n
n1
y x 2y x 2
3
1n 4n4n 1 x 4n2 1n 4nx 4n2 x 2 2n . . . . . . 4n 5 n! 3 7 11 4n 1 n2 2 n! 3 7 11
2
2n n!
n2
n1
52. J1x x (a) lim
k→
22n2
1n 4nx 4n2 1n x 4n2 x2 2n . . . 4n 5 n1 2 n! 3 7 11 . . . 4n 1 3 7 11
1n1 4n 1x4n2 1n1 x 4n2 22n 1 0 2n . . . . . . n 1! 3 7 11 4n 1 n1 2 n! 3 7 11 4n 1 22n 1
1k x 2k 1k x 2k1 k 1! k0 22k1k!k 1!
2
k0
1n x 4n 3 7 11 . . . 4n 1
2k1k!
uk1 1k1 x 2k3 lim 2k3 k→ 2 uk k 1!k 2!
22k1k!k 1!
1k x 2k1
Therefore, the interval of convergence is < x < (b) J1x
1
2
k
k0
k→
1x 2 0 22 k 2k 1
x2k1
k 1!
2k1k!
J1x
1k 2k 1x2k 2k1k!k 1! k0 2
J1x
1k 2k 12kx 2k1 22k1k!k 1! k1
x 2J1 xJ1 x 2 1J1
1k2k 1x 2k1 1k 2k 12kx 2k1 2k1 2 k!k 1! 22k1k!k 1! k1 k0
2
k0
1k2k 1x 2k1 1k2k 12kx 2k1 x 2k1k!k 1! 2 2 22k1k!k 1! k1 k1
1kx2k1 1kx2k3 x 2k1k!k 1! 2 k1 22k1k!k 1! 2 k0
1kx 2k1 2k 12k 2k 1 1 1kx 2k3 2k1 2k1 2 k!k 1! k!k 1! k1 k0 2
1kx2k14kk 1 1kx2k3 2k1 2k1 2 k!k 1! k!k 1! k1 k0 2
2
1kx 2k1 1k x2k3 k 1!k! k0 22k1k!k 1!
2k1
1k1x 2k3 1kx 2k3 2k1 k!k 1! k0 22k1k!k 1!
2
k0
1kx 2k3 1 1 0 22k1k!k 1! k0
x 1 3 1 5 1 x x x7 2 16 384 18,432
(d)
J0x
1k1x 2k1 1k12k 1x 2k1 2k2 k 1!k 1! k0 22k1k!k 1! k0 2
J1x
4
6
−4
k1
−6
1kx 2k3 1k x 2k1 2k1 k 1! k0 2 k!k 1!
2k1k!
(c) P7x
.
lim
1k1x 2k1 1kx 2k1 k!k 1! k0 22k1k!k 1! k0 2
2k1
Note: J0x J1x
Section 8.9
54. f x
1
n
n0
x 2n1 sin x 2n 1!
Representation of Functions by Power Series
56. f x
1
n
n0
(See Exercise 47.)
2
−
−2.5
2.5
−2
x 2n1 arctan x, 1 ≤ x ≤ 1 2n 1
(See Exercise 38 in Section 8.7.)
2
58.
403
−
x 2n1
x 2n1
2n 1! 2n 1!
n0
2
60. True; if
n1
a
Replace n with n 1.
n
xn
n0
converges for x 2, then we know that it must converge on 2, 2. 62. True
1
1
f x dx
0
0
Section 8.9 2. (a) f x
an xn dx
n0
an xn1 n0 n 1
0
n0
n
Representation of Functions by Power Series
4 4 5 a 5 x 1 x 5 1 r
n a 1 1
4 x n0 5 5
n
4xn n1 n0 5
4. (a)
1 a 1 1 x 1 x 1 r
x
n0
This series converges on 5, 5. 4x3 4 4 4 2 x x . . . 5 25 125 625 (b) 5 x ) 4 4 4 x 5 4 x 5 4 4 x x2 5 25 4 2 x 25 4 2 4x3 x 25 125 4x3 125 4x3 4x 4 125 625
n
1
n
xn
n0
This series converges on 1, 1. 1 x x 2 x3 . . . (b) 1 x ) 1 1x x x x 2 x2 x 2 x3 x3 x3 x 4
404
Chapter 8
Infinite Series
6. Writing f x in the form
a , we have 1r
8. Writing f x in the form a 1 r, we have 3 3 1 a 2x 1 3 2x 2 1 2 3x 2 1 r
4 4 7 a 4 . 5 x 7 x 2 1 1 7x 2 1 r
which implies that a 1 and r 2 3x 2. Therefore, the power series for f x is given by
Therefore, the power series for f x is given by 4 1 4 ar n x 2 5 x n0 n0 7 7
n
n 2 3 ar n x 2 , 2x 1 n0 3 n0
4x 2 . 7n1 n0
x 2 < 7
or 5 < x < 9
x 2
10. Writing f x in the form a 1 r, we have
5 x 2
n
2nx 2n , 3n n0
<
3 1 7 or < x < . 2 2 2
4 4 1 2 a 3x 2 8 3x 2 1 3 8x 2 1 r
which implies that a 1 5 and r 2 5x. Therefore, the power series for f x is given by
which implies that a 1 2 and r 3 8x 2. Therefore, the power series for f x is given by 1 3 4 arn x 2 3x 2 n0 2 8 n0
2n xn
5
n0
n1 ,
5 5 5 x < or < x < . 2 2 2
x 2 14.
<
2 8 14 or < x < . 3 3 3
Writing f x as a sum of two geometric series, we have
2 x 1
n
22x
n
n0
n0
1 31n 1 1 2n1 xn, x < or < x < . 2n1 2 2 2
16. First finding the power series for 4 4 x, we have 1 1 x 1 1 4x n0 4
n
1n xn 4n n0
Now replace x with x2. 1n x 2n 4 . 4 x2 n0 4n
The interval of convergence is x 2 < 4 or 2 < x < 2 since lim
n→
18. hx
un1 1 x lim n→ un 4n1
n1 2n2
4n
1n x 2n
x2 x2 . 4 4
x 1 1 1 1 n 1nxn x x 1 21 x 21 x 2n0 2n0
2
1 1
1n 1xn 0 2x 0x2 2x3 0x4 2x5 . . . 2n0 2
1 2x2n1 x2n1, 1 < x < 1 2n0 n0
n
1 3n x 2n , 2 n0 8n
4x 7 3 2 3 2 3 2 2 2x2 3x 2 x 2 2x 1 2 x 1 2x 1 1 2x 1 2x 3 4x 7 2x 2 3x 2 n0 2
12. Writing f x in the form a 1 r, we have
1 1 1 5 a 2x 5 5 2x 1 2 5x 1 r
1 1 ar n 2x 5 n0 5 n0
n
Section 8.9
1 x
n n
n0
1 nx 1 nn 1x
d dx
n
n1
n1
n
n2
n2
22. By integrating, we have
d2 1 2 . Therefore, dx 2 x 1 x 13
20. By taking the second derivative, we have d2 2 2 3 x 1 dx
Representation of Functions by Power Series
1 dx ln1 x C1 and 1x
1 n 2n 1 x , 1 < x < 1. n
n
n0
1 dx ln1 x C2. 1x
f x ln1 x 2 ln1 x ln1 x. Therefore, ln1 x 2
1 dx 1x
1 dx 1x
1n x n dx
n0
x n dx C1
n0
x n1 1n x n1 C2 n1 n0 n0 n 1
2x 2n2 1x 2n2
1n 1 xn1 C C n1 n1 n0 n0 2n 2 n0
C
To solve for C, let x 0 and conclude that C 0. Therefore, ln1 x2
x2n2
n 1, 1 < x < 1
n0
24.
x2
2x 2x 1n x 2n (See Exercise 23.) 1 n0
1 2x n
2n1
n0
Since
2x d lnx 2 1 2 , we have dx x 1
lnx 2 1
1n 2x 2n1 dx C
n0
1n x 2n2 , 1 ≤ x ≤ 1. n1 n0
To solve for C, let x 0 and conclude that C 0. Therefore, lnx 2 1
26. Since
1n x 2n2 , 1 ≤ x ≤ 1. n1 n0
1 1 dx arctan2x, we can use the result of Exercise 25 to obtain 4x 2 1 2
arctan2x 2
1 dx 2 4x 2 1
n0
To solve for C, let x 0 and conclude that C 0. Therefore,
1n 4nx 2n1 1 1 , < x ≤ . 2n 1 2 2 n0
arctan2x 2
1n 4n x 2n1 1 1 , < x ≤ . 2n 1 2 2 n0
1n 4n x 2n dx C 2
405
406
Chapter 8
28. x
Infinite Series
x 2 x3 x 4 ≤ lnx 1 2 3 4
5
S5 f
x 2 x3 x 4 x5 ≤ x 2 3 4 5
−4
8
S4 −3
x x
2
x3
x4
x 2 3 4
lnx 1 x
x 2 x3 x 4 x5 2 3 4 5
0.0
0.2
0.4
0.6
0.8
1.0
0.0
0.18227
0.33493
0.45960
0.54827
0.58333
0.0
0.18232
0.33647
0.47000
0.58779
0.69315
0.0
0.18233
0.33698
0.47515
0.61380
0.78333
1
In Exercise 35–38, arctan x
n
n0
30. gx x
x 2n1 . 2n 1
x3 , cubic with 3 zeros. 3
32. gx x
Matches (d)
1
n0
arctan x 2
34. The approximations of degree 3, 7, 11, . . . 4n 1, n 1, 2, . . . have relative extrema.
Matches (b)
In Exercises 36 and 38, arctan x
36.
x3 x5 x7 , 3 5 7
n
x 4n2 2n 1
n
x 4n3 C, C 0 4n 32n 1
1
n0
3 4
arctan x 2 dx
0
1
3 44n3 4n 32n 1
34n3 4n 32n 144n3
n
n0
1n
n0
x2n1 . 2n 1
1
n0
arctan x 2 dx
n
27 2187 177,147 192 344,064 230,686,720
Since 177,147 230,686,720 < 0.001, we can approximate the series by its first two terms: 0.13427
x 2 arctan x
38.
1
n
x 2n3 2n 1
1
n
x 2n4 2n 42n 1
1 1 1 . . . 2n 42n 122n4 64 1152
n0
x 2 arctan x dx
n0
1 2
x 2 arctan x dx
0
Since
1
n0
n
1 < 0.001, we can approximate the series by its first term: 1152
1 2
0
x 2 arctan x dx 0.015625.
Section 8.9
In Exercises 40 and 42,
n1
n1
42. (a)
407
1 x n, x 1. 1 x n0
40. Replace n with n 1.
nx
Representation of Functions by Power Series
n 1x
n
(b)
n0
1 2 n 3 n1 3
n
9 1 n 10 n1 10
n
2 2 n 9 n1 3
n1
9 9 n 100 n1 10
9 100
2 1 2 9 1 2 32
n1
1
1 9 102 9
46. Integrate the series and multiply by 1.
44. Replace x with x2. 48. (a) From Exercise 47, we have
1 120 1 120 arctan arctan arctan 119 239 119 239
arctan
arctan
1 239 28,561 arctan arctan 1 1120 119 120 1191 239
28,561 4
10 1 1 21 5 5 1 arctan arctan arctan arctan arctan 5 5 5 24 12 1 1 52
(b) 2 arctan
4 arctan
120 1 1 5 5 25 12 1 2 arctan 2 arctan arctan arctan arctan arctan 5 5 5 12 12 119 1 5 122
4 arctan
1 1 120 1 arctan arctan arctan (see part (a).) 5 239 119 239 4
1 1 1 2 1 3 5 6 arctan arctan arctan 2 3 1 1 21 3 5 6 4
50. (a) arctan
(b) 4 arctan
1 1 arctan 2 3
12 1 23
3
4
1 25 1 27 1 1 33 1 35 1 37 4 5 7 3 3 5 7
40.4635 40.3217 3.14
54. From Example 5, we have arctan x
52. From Exercise 51, we have
1
n1
n1
13 1 ln 34 0.2877.
56. From Exercise 54, we have
1
n1
n1
1 1 1n 2n1 32n12n 1 n0 3 2n 1
1
n0
arctan
n
n0
1n11 3n 1 3n n n1 n
ln
1
n
1 32n1 2n 1
1 0.3218. 3
1
n0
n
1 1 1n 2n 1 n0 2n 1
arctan 1
2n1
0.7854 4
x 2n1 . 2n 1
408
Chapter 8
Infinite Series
58. From Example 5, we have arctan x
1n
n0
1n
x 2n1 . 2n 1
3 1n 2n 2n 1 3
3 2n 1 3
n0
n
n0
1n 13 2n1 2n 1 n0
3
3 arctan
6 2 3
3
Section 8.10
1 3
Taylor and Maclaurin Series
2. For c 0, we have f x e3x f nx 3ne 3x ⇒ f n0 3n e 3x 1 3x
3xn 9x2 27x3 . . . 2! 3! n0 n!
4. For c 4, we have: f x sin x
f
4
2
fx cos x
f
4
2
f x sin x
f
4 22
f x cos x
f
4 22
f 4
4
f 4x sin x
2 2
2
2
and so on. Therefore we have: sin x
f n4 x 4 n n! n0
2
2 2
2
1 x 4
1
n0
x 4 2 x 4 3 x 4 4 . . . 2! 3! 4!
x 4 n1 1 n 1!
nn12
6. For c 1, we have: f x ex f nx ex ⇒ f n1 e ex
x 1n f n1x 1n x 12 x 13 x 14 . . . e 1 x 1 e n! 2! 3! 4! n! n0 n0
Section 8.10 8. For c 0, we have: f x lnx 2 1
f 0 0
fx
2x x2 1
f0 0
f x
2 2x 2 x2 12
f 0 2
f x
4xx 2 3 x2 13
f 0 0
f 4x
12x 4 6x 2 1 x 2 14
f 40 12
f 5x
48xx 4 10x 2 5 x 2 15
f 50 0
f 6x
2405x6 15x 4 15x 2 1 x 2 16
f 60 240
and so on. Therefore, we have: lnx 2 1
f n0xn 2x 2 0x3 12x 4 0x5 240x6 . . . 0 0x n! 2! 3! 4! 5! 6! n0
x2
1n x 2n2 x 4 x6 . . . 2 3 n1 n0
10. For c 0, we have; f x tanx
f 0 0
fx sec2x
f0 1
f x 2 sec2xtanx
f 0 0
f x 2 sec4x 2 sec2xtan2x
f 0 2
f 4x 8 sec4xtanx sec2xtan3x
f 40 0
f 5x 8 2 sec6x 11 sec4xtan2x 2 sec2xtan4x
f 50 16
tanx
f n0xn 2x3 16x5 . . . x3 2 x x x5 . . . n! 3! 5! 3 15 n0
12. The Maclaurin Series for f x e2x is
2xn . n! n0
f n1x 2n1e2x. Hence, by Taylor’s Theorem,
0 ≤ Rnx Since lim
n→
f n1z n1 2n1e2z n1 x x . n 1! n 1!
2n1xn1 2xn1 lim 0, it follows that Rnx → 0 as n → . n→ n 1! n 1!
Hence, the Maclaurin Series for e2x converges to e2x for all x.
Taylor and Maclaurin Series
409
410
Chapter 8
Infinite Series kk 1x2 kk 1k 2x3 . . . , we have 2! 3! 1 1232x 2 123252x3 . . . 1 x 2 2! 3!
14. Since 1 xk 1 kx 12
1 x
1
13x 2 135x3 . . . x 2 222! 233!
1
35.
1
n1
kk 1x2 kk 1k 2x3 . . . , we have 2! 3!
16. Since 1 xk 1 kx
1 x13 1
x 13x 1323 2!
2
x 2x2 2 5x3 2 5 8x4 . . . 2 3 3 3 2! 3 3! 344!
1
1n12 x 3 n2
58.
1n1 1 x 2 n2
we have 1 x312 1
ex
132353x3 . . . 3!
1
18. Since 1 x12 1
20.
. . 2n 1xn 2n n!
xn
35.
x3
x4
.
. . 2n 3xn
2n n!
1n1 1 x3 2 n2
x2
. . 3n 4
3nn!
35.
(Exercise 14)
. . 2n 3x3n
2n n!
.
x5
n! 1 x 2! 3! 4! 5! . . .
n0
e3x
22.
1n 3n xn 3xn 9x 2 27x3 81x 4 243x5 . . . 1 3x n! n! 2! 3! 4! 5! n0 n0
cos x cos 4x
1nx2n x2 x4 x6 1 . . . 2n ! 2! 4! 6! n0
1n42nx2n 1n4x2n 2n! 2n! n0 n0
1
26.
16x2 256x4 . . . 2! 4!
ex 1 x
x2 x3 . . . 2! 3!
ex 1 x
x2 x3 . . . 2! 3!
ex ex 2
2x2 2x4 . . . 2! 4!
2 cos hx ex ex
x2n
2 2n!
n0
24.
sin x
1nx2n1 n0 2n 1!
1nx32n1 n0 2n 1!
2 sin x3 2
x9 x15 . . . 3! 5!
2x3
2x9 2x15 . . . 3! 5!
2 x3
Section 8.10
28. The formula for the binomial series gives 1 x12 1
1 x 212 1 ln x x 2 1
x
x
411
1n 1 3 5 . . . 2n 1xn , which implies that 2n n! n1
1n 1 3 5 . . . 2n 1x 2n 2n n! n1
1 dx 1
x 2
x
1n1 3 5 . . . 2n 12n1 2n2n 1n! n1
x3 2
3
x2 x4 . . . 2! 4!
30. x cos x x 1
Taylor and Maclaurin Series
1 3x5 1 3 5x7 . . .. 245 2467
32.
arcsin x 2n!x2n1 nn!22n 1 x 2 n0
x3 x5 . . . 2! 4!
1
x
2n!x2n
2 n! 2n 1, x 0
n0
n
2
1nx2n1 2n! n0
2x 2 2x 4 2x6 . . . (See Exercise 33.) 2! 4! 6!
34. eix eix 2
1n x 2n eix eix x2 x 4 x6 1 . . . cosx 2 2! 4! 6! 2n! n0
36. gx ex cos x
8
x2
1x
2
4
4
x x . . . 6 24
1 2 24x . . . x2
−6
4
6
g P4
x2 x2 x3 x3 x4 x4 x4 x3 x 4 . . . 1x . . .1x 2 2 6 2 24 4 24 3 6
−40
38. f x ex ln1 x
1x
x 2 x3 x4 . . . 2 6 24
x x2 x
3
−3
P5 3
−3
2
3
4
5
x2 x3 x3 x3 x4 x4 x4 x4 x5 x5 x5 x5 x5 . . . 2 3 2 2 4 3 4 6 5 4 6 12 24
x 2 x3 3x5 . . . 2 3 40 f
x x2 x3 x4 x5 . . .
412
Chapter 8
40. f x
Infinite Series
ex . Divide the series for ex by 1 x. 1x
x2 x3 2 3 x2 1x 1x 2 1x x2 0 2 x2 2 1
f x 1
3x 4 . . . 8 x3 x4 x5 . . . 6 24 120
x 2 x 3 3x 4 . . . 2 3 8 6
P4
−6
f
6 −6
x3 6 x3 2 x3 x4 3 24 x3 x4 3 3 3x 4 x5 8 120 3x4 3x5 8 8
42. y x
x5 x2 x 4 x3 x 1 x cos x. 2! 4! 2! 4!
44. y x 2 x3 x 4 x 21 x x 2 x 2 Matches (d)
Matches (b)
x
46.
x
1 t3 dt
0
1
0
1n11 t3 2 n2
35.
. . 2n 3t3n
2n n!
1n11 3 5 . . . 2n 3t3n1 t4 8 3n 12n n! n2
x
1n11 3 5 . . . 2n 3x3n1 x4 8 3n 12n n! n2
dt
x
0
1n x 2n1 x3 x5 x7 x . . . , we have 3! 5! 7! n0 2n 1!
48. Since sinx
1n
t
sin1
1 1 x.
1
1
1
x2
x3
x4
2n 1! 1 3! 5! 7! . . . 0.8415.
(4 terms)
n0
50. Since ex
xn
x5
n! 1 x 2! 3! 4! 5! . . . , we have e
1
n0
and
11
1 1 1 1 . . . 2! 3! 4! 5!
1n1 1 1 1 1 1 e1 1 e1 1 . . . 0.6321. (6 terms) e 2! 3! 4! 5! 7! n! n1
52. Since sin x
1n x 2n1 x3 x5 x7 x . . . 2n 1 ! 3! 5! 7! n0
1n x 2n x2 x 4 x6 sin x 1 . . . x 3! 5! 7! n0 2n 1!
we have lim
x→0
1n x 2n sin x lim 1. x→0 n0 2n 1! x
Section 8.10
12
54.
0
arctan x dx x
Since 1
12
0
3
x5 x7 . . . 2 2 5 7
12 0
< 0.0001, we have
x→0
1
1
cos x dx
0.5
1
arctan x 1. x
1 2!x 4!x 6!x 8!x . . . dx x 2x2! 3x4! 4x6! 5x8! . . . 2
0.5
1 210, 600
Since
3
4
2
3
4
5
1
< 0.0001, we have
cos x dx 1 0.5
0.5
1 4
14
x lnx 1 dx
0
x
2
0
1 1 1 1 1 0.52 1 0.53 1 0.54 1 0.55 0.3243. 4 72 2880 201,600
x3 x4 x5 . . . dx 2 3 4
x3 4 x 2 5 x 3 6 x 4 . . . 3
4
5
14
6
0
14 < 0.0001, 15 5
Since
143 144 0.00472. 3 8
14
x lnx 1 dx
0
60. From Exercise 19, we have 1 2
2
ex 2dx 2
1
1 2
2
1
1n x 2n 1 dx nn! 2 2 n0
1n x 2n1
2 n!2n 1 n0
2
n
1
1 1n 2n1 1 2 n0 2nn!2n 1
1 2
1 2 17 3 2
31
2
62. f x sin
8191
26
6! 13
2! 5
127
23
3! 7
511
24
4! 9
25
5! 11
x ln1 x 2
4
f −4
8
P5
The polynomial is a reasonable approximation on the interval 0.60, 0.73.
−4
3 x f x arctan x, c 1
3
P5x 0.7854 0.7618x 1 0.3412
x 2! 1 0.0424 x 3! 1
x 14 x 15 1.3025 5.5913 4! 5!
2047
32,767 131,071 524,287 0.1359. 27 7! 15 28 8! 17 29 9! 19
x 2 x3 7x 4 11x5 P5x 2 4 48 96
64.
0.5
0.55
1
6
2
Note: We are using lim
58.
4
1 arctan x 1 1 1 1 dx 0.4872. x 2 3223 5225 7227 9229
0
2
9229
12
56.
1 x3 x5 x7 . . . dx x 3x
Taylor and Maclaurin Series
2
The polynomial is a reasonable approximation on the interval 0.48, 1.75.
3
f P5 −2
4 −1
413
414
Chapter 8
Infinite Series
66. a2n1 0 (odd coefficients are zero)
70. 60, v0 64, k y 3x
1 , g 32 16
32x 2 11632x3 116232x 4 . . . 2642122 3643123 4644124
2264x 2 2
3x 32
2
3x 32
24x 4 23x3 . . . 3 364 16 4644162
2n xn
n64 16 n
n2
72. (a) f x
68. Answers will vary.
3x 32
n2
xn
n32 16
n2
n
n2
lnx 2 1 . x2 1.5
From Exercise 8, you obtain P
(b)
1n x 2n 1 1n x 2n2 2 x n0 n1 n0 n 1
0
x2 x4 x6 x8 P8 1 2 3 4 5
−0.5
lnt2 1 dt t2
x
(c) Fx
2
0
x
Gx
P8t d t
0
x
0.25
0.50
0.75
1.00
1.50
2.00
Fx
0.2475
0.4810
0.6920
0.8776
1.1798
1.4096
Gx
0.2475
0.4810
0.6920
0.8805
5.3064
652.21
(d) The curves are nearly identical for 0 < x < 1. Hence, the integrals nearly agree on that interval. 74. Assume e pq is rational. Let N > q and form the following.
e 11
1 1 1 1 . . . . . . 2! N! N 1! N 2!
Set a N! e 1 1 . . .
1 N!
, a positive integer. But,
N 1 1! N 1 2! . . . N 1 1 N 11 N 2 . . . < N 1 1 N 1 1
a N!
2
1 1 1 1 1 . . . N1 N 1 N 12 N1
1
1 1 N1
. . .
1 , a contradiction. N
Review Exercises for Chapter 8 2. an
n2
n 1
n 4. an 4 : 3.5, 3, . . . 2 Matches (c)
32
6. an 6
Matches (b)
n1
: 6, 4, . . .
Review Exercises for Chapter 8 n 2
8. an sin
10. lim
n→
1 n
0
Converges
2
0
12
−2
The sequence seems to diverge (oscillates). n : 1, 0, 1, 0, 1, 0, . . . 2
sin
n 1 lim n→ lnn n→ 1n
n→
1 2n
n
1 1
k
lim k→
k 12
e12
Converges; k 2n
Diverges 16. Let
14. lim 1
12. lim
y bn cn1n ln y
18. (a) Vn 120,0000.70n, n 1, 2, 3, 4, 5 (b) V5 120,0000.705 $20,168.40
lnbn cn n
lim ln y lim
n→
n→
1 bn ln b cn ln c bn cn
Assume b ≥ c and note that the terms bn ln b cn ln c bn ln b cn ln c n n n n n b c b c b cn converge as n → . Hence an converges. 20. (a)
(b)
k
5
10
15
20
25
Sk
0.3917
0.3228
0.3627
0.3344
0.3564
(c) The series converges by the Alternating Series Test.
1
0
12 0
22. (a)
(b)
k
5
10
15
20
25
Sk
0.8333
0.9091
0.9375
0.9524
0.9615
(c) The series converges, by the limit comparison test with
1
n .
2 2n2 n 4 3 n0 n0 3
n
0
2
12 0
2 26. Diverges. nth Term Test, lim an . n→ 3
24. Diverges. Geometric series, r 1.82 > 1.
28.
1
43 12
See Exercise 27.
30.
3
n0
2
n
2 1 n 1n 2 n0 3
n
n 1 n 2 1
1
n0
1 1 23
1 2 2 3 3 4 . . . 3 1 2 1
1
1
1
1
415
416
Chapter 8
Infinite Series
32. 0.923076 0.9230761 0.000001 0.0000012 . . .
0.9230760.000001
n
n0
39
32,0001.055
34. S
n
0.923076 923,076 1276,923 12 1 0.000001 999,999 1376,923 13
32,0001 1.05540 1 1.055
n0
36. See Exercise 86 in Section 8.2. AP
$4,371,379.65
12r 1 12r
100
12t
1
12
0.065 1 0.065 12
120
1
$16,840.32
38.
1
n 4
n1
3
n1
1 n34
40.
1 2
n1
Divergent p-series, p
42.
n
3 4
< 1
1 1 1 2 n 2n n n1 n1 2
The first series is a convergent p-series and the second series is a convergent geometric series. Therefore, their difference converges.
n1
nn 2
44. Since
n1
n1
1
3
n
converges,
3
n1
n
1 converges by the 5
Limit Comparison Test.
n 1nn 2 n1 lim 1 n→ n→ n 2 1n lim
1
n, the series diverges.
By a limit comparison test with
n1
46.
1nn n1 n 1
an1 lim
n→
48. Converges by the Alternating Series Test.
n 1
n2
≤
n
n1
an1
an
3 lnn 1 3 ln n 3 ln n < an, lim 0 n→ n1 n n
n 0 n1
By the Alternating Series Test, the series converges.
50.
n!
e
n
n1
lim
n→
an1 n 1! lim n→ an en1 lim
n→
en
n!
n1 e
By the Ratio Test, the series diverges. 52.
1 3
5.
. . 2n 1
2 5 8 . . . 3n 1
n1
lim
n→
an1 1 lim n→ 2 an
3. 5.
. . 2n 12n 1 . . 3n 13n 2
By the Ratio Test, the series converges.
2
1
5. 3.
. . 3n 1 2n 1 2 lim . . 2n 1 n→ 3n 2 3
Review Exercises for Chapter 8
417
54. (a) The series converges by the Alternating Series Test. (b)
(c)
x
5
10
15
20
25
Sn
0.0871
0.0669
0.0734
0.0702
0.0721
(d) The sum is approximately 0.0714.
0.3
0
12 0
56. No. Let an
3937.5 3937.5 is a convergent p-series. , then a75 0.7. The series 2 2 n n1 n
4 1
58. f x tan x
f
f x 2 sec2 x tan x
f
f
x 4 sec2 x tan2 x 2 sec4 x
f
4 4
62. e0.25 1 0.25
P6x 1 P10x 1
68.
4 16
2 x 4 4
2
8 x 3 4
66.
x2 x4 2! 4!
−10
x2 x4 x6 2! 4! 6!
lim
n
Geometric series which converges only if 2x < 1 or 1 1 2 < x < 2 .
10
f P6
2x
n0
P4
P10
−10
x2 x4 x6 x8 x10 2! 4! 6! 8! 10!
3nx 2n n n1 n→
3
10
0.252 0.253 0.254 0.779 2! 3! 4!
f x cos x P4x 1
0.752 0.754 0.756 0.7317 2! 4! 6!
4 2
f x sec2 x
P3x 1 2 x
64.
60. cos0.75 1
f
70.
un1 3n1x 2n1 lim n→ un n1
3x2
n
3nx 2n
1 3 Center: 2 R
5
Since the series converges at 3 and diverges at 73 , the interval of convergence is 53 ≤ x < 73 .
x2 x 2n n 2 2 n0 n0
n
Geometric series which converges only if
x2 < 1 or 2
0 < x < 4.
418
Chapter 8
Infinite Series
y
72.
y
3n12n 2x2n1 3n2nx2n1 n 2 n! 2n1n 1! n1 n0
y
3n12n 22n 1x2n 2n1n 1! n0
y 3xy 3y
74.
3nx 2n 2nn! n0
1n13n22n 2x2n2 1n3n1x2n 3n12n 22n 1x2n n1 n1 2 n 1! 2 n 1! 2nn! n0 n0 n0
1n13n2x2n2 1n3n1x2n 1n13n12n 2x 2n n n 2 n! 2 n! 2nn! n0 n0 n0
1n13n2x 2n2 1n3n1x2n 2n 1 1 n 2 n! 2nn! n0 n0
1n13n2x 2n2 1n3n1x2n 2n n 2 n! 2nn! n0 n0
1n3n1x 2n 2n 1n13n1x2n 2n n n1n 1! 2n 2 n! n1 n1 2
1n3n1x2n 2n 2n 0 2nn! n1
3 32 a 32 2 x 1 x2 1 x2 1 r
2 2 x
3
n
n0
76. Integral:
n1
n0
1n3xn n1 n0 2
1 1 x 3 78. 8 2x 3 x 32 x 33 . . . 8 2 8 4 n0
80.
1n3xn1
n 12
n
8 1 x 34
32 32 , 1 < x < 7 4 x 3 1 x
f x cos x f x sin x f x cos x f x sin x cos x
82.
2 f n 4x 4 n 2 2 x x n! 2 2 4 2 2! 4 n0
1nn12x 4 n1 n 1! n1
1 x 2 4
2
f x cscx f x cscx cotx f x csc3x cscx cot2x f x 5 csc3x cotx cscx cot3x f 4x 5 csc5x 15 csc3x cot2x cscx cot4x cscx
1 f n2x 2 n 1 x n! 2! 2 n0
2
5 x 4! 2
4
. . .
2
2
2
3!
x 4
3
2
2
4!
x 4
4
. . .
Review Exercises for Chapter 8 f x x12
84.
1 f x x12 2 f x
32
12 12 32 x
f x
52
f 4x x
12 12 x
12 12 32 52 x
72,
. . .
f n4x 4n n! n0
2
x 4 x 42 1 3x 43 1 3 5x 44 . . . 22 252! 283! 2114!
2
1n11 3 5 . . . 2n 3x 4n x 4 2 2 23n1n! n2
hx 1 x3
86.
h x 31 x4 h x 121 x5 h x 601 x6 h4x 3601 x7 h5x 25201 x8 1nn 2!x n 1nn 2n 1x 12x2 60x3 360x4 2520x5 . . . 1 1 3x 3 1 x 2! 3! 4! 5! 2n! 2 n0 n0
1
ln x
88.
n1
n1
ln
x 1n , n
0 < x ≤ 2 n
e23
n1
n1
1
n1
n1
92.
sin x
sin
n
x2n1 , 2n 1!
n
1 0.3272 32n12n 1!
13 1 n0
1 0.1823 5nn
1
n0
xn
< x <
x2
n! 1 x 2! . . .
94. ex
n0
x n1 x3 x x2 . . . 2! n0 n!
xex
1
xex dx xex ex
0
0
0
e e 0 1 1
x n1 x n2 dx n0 n! n0 n 2n!
1
1
ex
1 0
1
n 2n! 1
n0
xn
n!,
n0
65 1 65n 1
90.
< x <
2n 23n 1.9477 n n! n0 n0 3 n!
419
420
Chapter 8
Infinite Series
f x sin 2x
f 0 0
f x 2 cos 2x
f 0 2
f x 4 sin 2x
f 0 0
f x 8 cos 2x
f 0 8
96. (a)
f 4x 16 sin 2x
f 40 0
f 5x 32 cos 2x
f 50 32
f 6x 64 sin 2x
f 60 0
f 7x 128 cos 2x
f 70 128
4 0x2 8x3 0x4 32x5 0x6 128x7 . . . 4 8 7 . . . 2x x3 x5 x 2! 3! 4! 5! 6! 7! 3 15 315
sin 2x 0 2x (b) sin x sin 2x
1nx2n1 n0 2n 1!
1n2x2n1 2x3 2x5 2x7 . . . 2x 2n 1 ! 3! 5! 7! n0
2x
8x3 32x5 128x7 . . . 4 4 8 7 . . . 2x x3 x5 x 6 120 5040 3 15 315
(c) sin 2x 2 sin x cos x
x5 x7 x3 . . . 6 120 5040
4 x7 2x3 2x5 4 4 8 7 . . . . . . 2x x3 x5 x 3 15 315 3 15 315
2 x
2 x 2 x
cos t
98.
cos
x
cos
0
t
t
2
2
dt
4
1n t 2n 2n! n0
et
100.
2
et 1
1 2n!n 1
2n
2
et
x
0
x
1nx n1 2n!n 1
2n
x3 2
3
tn
tn
n!
n1 n t n1
2
n!
n0
n n
0
1 3x5 1 3 5x7 . . . 2 4 5 2 4 6 7
x2 1 3x4 1 3 5x6 arcsin x 1 . . . 2 3 2 4 5 2 4 6 7 x lim
1 t 2n 2n! n0
arcsin x x
x→0
6
x5 x5 x5 x7 x7 x7 x3 x3 x7 . . . 2 6 24 12 120 720 144 240 5040
n0
102.
2
n0
x . . . 1 x2 24x 720
arcsin x 1 x
1 arcsin x 1 x2 lim 1. By L’Hôpital’s Rule, lim x→0 x→0 x 1
tn1 1 t n1 n!
et 1 dt t
tn
x
n n! n1
0
xn
n n!
n1
414
Chapter 8
Infinite Series
66. a2n1 0 (odd coefficients are zero)
70. 60, v0 64, k y 3x
1 , g 32 16
32x 2 11632x3 116232x 4 . . . 2642122 3643123 4644124
2264x 2 2
3x 32
2
3x 32
24x 4 23x3 . . . 3 364 16 4644162
2n xn
n64 16 n
n2
72. (a) f x
68. Answers will vary.
3x 32
n2
xn
n32 16
n2
n
n2
lnx 2 1 . x2 1.5
From Exercise 8, you obtain P
(b)
1n x 2n 1 1n x 2n2 2 x n0 n1 n0 n 1
0
x2 x4 x6 x8 P8 1 2 3 4 5
−0.5
lnt2 1 dt t2
x
(c) Fx
2
0
x
Gx
P8t d t
0
x
0.25
0.50
0.75
1.00
1.50
2.00
Fx
0.2475
0.4810
0.6920
0.8776
1.1798
1.4096
Gx
0.2475
0.4810
0.6920
0.8805
5.3064
652.21
(d) The curves are nearly identical for 0 < x < 1. Hence, the integrals nearly agree on that interval. 74. Assume e pq is rational. Let N > q and form the following.
e 11
1 1 1 1 . . . . . . 2! N! N 1! N 2!
Set a N! e 1 1 . . .
1 N!
, a positive integer. But,
N 1 1! N 1 2! . . . N 1 1 N 11 N 2 . . . < N 1 1 N 1 1
a N!
2
1 1 1 1 1 . . . N1 N 1 N 12 N1
1
1 1 N1
. . .
1 , a contradiction. N
Review Exercises for Chapter 8 2. an
n2
n 1
n 4. an 4 : 3.5, 3, . . . 2 Matches (c)
32
6. an 6
Matches (b)
n1
: 6, 4, . . .
Review Exercises for Chapter 8 n 2
8. an sin
10. lim
n→
1 n
0
Converges
2
0
12
−2
The sequence seems to diverge (oscillates). n : 1, 0, 1, 0, 1, 0, . . . 2
sin
n 1 lim n→ lnn n→ 1n
n→
1 2n
n
1 1
k
lim k→
k 12
e12
Converges; k 2n
Diverges 16. Let
14. lim 1
12. lim
y bn cn1n ln y
18. (a) Vn 120,0000.70n, n 1, 2, 3, 4, 5 (b) V5 120,0000.705 $20,168.40
lnbn cn n
lim ln y lim
n→
n→
1 bn ln b cn ln c bn cn
Assume b ≥ c and note that the terms bn ln b cn ln c bn ln b cn ln c n n n n n b c b c b cn converge as n → . Hence an converges. 20. (a)
(b)
k
5
10
15
20
25
Sk
0.3917
0.3228
0.3627
0.3344
0.3564
(c) The series converges by the Alternating Series Test.
1
0
12 0
22. (a)
(b)
k
5
10
15
20
25
Sk
0.8333
0.9091
0.9375
0.9524
0.9615
(c) The series converges, by the limit comparison test with
1
n .
2 2n2 n 4 3 n0 n0 3
n
0
2
12 0
2 26. Diverges. nth Term Test, lim an . n→ 3
24. Diverges. Geometric series, r 1.82 > 1.
28.
1
43 12
See Exercise 27.
30.
3
n0
2
n
2 1 n 1n 2 n0 3
n
n 1 n 2 1
1
n0
1 1 23
1 2 2 3 3 4 . . . 3 1 2 1
1
1
1
1
415
416
Chapter 8
Infinite Series
32. 0.923076 0.9230761 0.000001 0.0000012 . . .
0.9230760.000001
n
n0
39
32,0001.055
34. S
n
0.923076 923,076 1276,923 12 1 0.000001 999,999 1376,923 13
32,0001 1.05540 1 1.055
n0
36. See Exercise 86 in Section 8.2. AP
$4,371,379.65
12r 1 12r
100
12t
1
12
0.065 1 0.065 12
120
1
$16,840.32
38.
1
n 4
n1
3
n1
1 n34
40.
1 2
n1
Divergent p-series, p
42.
n
3 4
< 1
1 1 1 2 n 2n n n1 n1 2
The first series is a convergent p-series and the second series is a convergent geometric series. Therefore, their difference converges.
n1
nn 2
44. Since
n1
n1
1
3
n
converges,
3
n1
n
1 converges by the 5
Limit Comparison Test.
n 1nn 2 n1 lim 1 n→ n→ n 2 1n lim
1
n, the series diverges.
By a limit comparison test with
n1
46.
1nn n1 n 1
an1 lim
n→
48. Converges by the Alternating Series Test.
n 1
n2
≤
n
n1
an1
an
3 lnn 1 3 ln n 3 ln n < an, lim 0 n→ n1 n n
n 0 n1
By the Alternating Series Test, the series converges.
50.
n!
e
n
n1
lim
n→
an1 n 1! lim n→ an en1 lim
n→
en
n!
n1 e
By the Ratio Test, the series diverges. 52.
1 3
5.
. . 2n 1
2 5 8 . . . 3n 1
n1
lim
n→
an1 1 lim n→ 2 an
3. 5.
. . 2n 12n 1 . . 3n 13n 2
By the Ratio Test, the series converges.
2
1
5. 3.
. . 3n 1 2n 1 2 lim . . 2n 1 n→ 3n 2 3
Review Exercises for Chapter 8
417
54. (a) The series converges by the Alternating Series Test. (b)
(c)
x
5
10
15
20
25
Sn
0.0871
0.0669
0.0734
0.0702
0.0721
(d) The sum is approximately 0.0714.
0.3
0
12 0
56. No. Let an
3937.5 3937.5 is a convergent p-series. , then a75 0.7. The series 2 2 n n1 n
4 1
58. f x tan x
f
f x 2 sec2 x tan x
f
f
x 4 sec2 x tan2 x 2 sec4 x
f
4 4
62. e0.25 1 0.25
P6x 1 P10x 1
68.
4 16
2 x 4 4
2
8 x 3 4
66.
x2 x4 2! 4!
−10
x2 x4 x6 2! 4! 6!
lim
n
Geometric series which converges only if 2x < 1 or 1 1 2 < x < 2 .
10
f P6
2x
n0
P4
P10
−10
x2 x4 x6 x8 x10 2! 4! 6! 8! 10!
3nx 2n n n1 n→
3
10
0.252 0.253 0.254 0.779 2! 3! 4!
f x cos x P4x 1
0.752 0.754 0.756 0.7317 2! 4! 6!
4 2
f x sec2 x
P3x 1 2 x
64.
60. cos0.75 1
f
70.
un1 3n1x 2n1 lim n→ un n1
3x2
n
3nx 2n
1 3 Center: 2 R
5
Since the series converges at 3 and diverges at 73 , the interval of convergence is 53 ≤ x < 73 .
x2 x 2n n 2 2 n0 n0
n
Geometric series which converges only if
x2 < 1 or 2
0 < x < 4.
418
Chapter 8
Infinite Series
y
72.
y
3n12n 2x2n1 3n2nx2n1 n 2 n! 2n1n 1! n1 n0
y
3n12n 22n 1x2n 2n1n 1! n0
y 3xy 3y
74.
3nx 2n 2nn! n0
1n13n22n 2x2n2 1n3n1x2n 3n12n 22n 1x2n n1 n1 2 n 1! 2 n 1! 2nn! n0 n0 n0
1n13n2x2n2 1n3n1x2n 1n13n12n 2x 2n n n 2 n! 2 n! 2nn! n0 n0 n0
1n13n2x 2n2 1n3n1x2n 2n 1 1 n 2 n! 2nn! n0 n0
1n13n2x 2n2 1n3n1x2n 2n n 2 n! 2nn! n0 n0
1n3n1x 2n 2n 1n13n1x2n 2n n n1n 1! 2n 2 n! n1 n1 2
1n3n1x2n 2n 2n 0 2nn! n1
3 32 a 32 2 x 1 x2 1 x2 1 r
2 2 x
3
n
n0
76. Integral:
n1
n0
1n3xn n1 n0 2
1 1 x 3 78. 8 2x 3 x 32 x 33 . . . 8 2 8 4 n0
80.
1n3xn1
n 12
n
8 1 x 34
32 32 , 1 < x < 7 4 x 3 1 x
f x cos x f x sin x f x cos x f x sin x cos x
82.
2 f n 4x 4 n 2 2 x x n! 2 2 4 2 2! 4 n0
1nn12x 4 n1 n 1! n1
1 x 2 4
2
f x cscx f x cscx cotx f x csc3x cscx cot2x f x 5 csc3x cotx cscx cot3x f 4x 5 csc5x 15 csc3x cot2x cscx cot4x cscx
1 f n2x 2 n 1 x n! 2! 2 n0
2
5 x 4! 2
4
. . .
2
2
2
3!
x 4
3
2
2
4!
x 4
4
. . .
Review Exercises for Chapter 8 f x x12
84.
1 f x x12 2 f x
32
12 12 32 x
f x
52
f 4x x
12 12 x
12 12 32 52 x
72,
. . .
f n4x 4n n! n0
2
x 4 x 42 1 3x 43 1 3 5x 44 . . . 22 252! 283! 2114!
2
1n11 3 5 . . . 2n 3x 4n x 4 2 2 23n1n! n2
hx 1 x3
86.
h x 31 x4 h x 121 x5 h x 601 x6 h4x 3601 x7 h5x 25201 x8 1nn 2!x n 1nn 2n 1x 12x2 60x3 360x4 2520x5 . . . 1 1 3x 3 1 x 2! 3! 4! 5! 2n! 2 n0 n0
1
ln x
88.
n1
n1
ln
x 1n , n
0 < x ≤ 2 n
e23
n1
n1
1
n1
n1
92.
sin x
sin
n
x2n1 , 2n 1!
n
1 0.3272 32n12n 1!
13 1 n0
1 0.1823 5nn
1
n0
xn
< x <
x2
n! 1 x 2! . . .
94. ex
n0
x n1 x3 x x2 . . . 2! n0 n!
xex
1
xex dx xex ex
0
0
0
e e 0 1 1
x n1 x n2 dx n0 n! n0 n 2n!
1
1
ex
1 0
1
n 2n! 1
n0
xn
n!,
n0
65 1 65n 1
90.
< x <
2n 23n 1.9477 n n! n0 n0 3 n!
419
420
Chapter 8
Infinite Series
f x sin 2x
f 0 0
f x 2 cos 2x
f 0 2
f x 4 sin 2x
f 0 0
f x 8 cos 2x
f 0 8
96. (a)
f 4x 16 sin 2x
f 40 0
f 5x 32 cos 2x
f 50 32
f 6x 64 sin 2x
f 60 0
f 7x 128 cos 2x
f 70 128
4 0x2 8x3 0x4 32x5 0x6 128x7 . . . 4 8 7 . . . 2x x3 x5 x 2! 3! 4! 5! 6! 7! 3 15 315
sin 2x 0 2x (b) sin x sin 2x
1nx2n1 n0 2n 1!
1n2x2n1 2x3 2x5 2x7 . . . 2x 2n 1 ! 3! 5! 7! n0
2x
8x3 32x5 128x7 . . . 4 4 8 7 . . . 2x x3 x5 x 6 120 5040 3 15 315
(c) sin 2x 2 sin x cos x
x5 x7 x3 . . . 6 120 5040
4 x7 2x3 2x5 4 4 8 7 . . . . . . 2x x3 x5 x 3 15 315 3 15 315
2 x
2 x 2 x
cos t
98.
cos
x
cos
0
t
t
2
2
dt
4
1n t 2n 2n! n0
et
100.
2
et 1
1 2n!n 1
2n
2
et
x
0
x
1nx n1 2n!n 1
2n
x3 2
3
tn
tn
n!
n1 n t n1
2
n!
n0
n n
0
1 3x5 1 3 5x7 . . . 2 4 5 2 4 6 7
x2 1 3x4 1 3 5x6 arcsin x 1 . . . 2 3 2 4 5 2 4 6 7 x lim
1 t 2n 2n! n0
arcsin x x
x→0
6
x5 x5 x5 x7 x7 x7 x3 x3 x7 . . . 2 6 24 12 120 720 144 240 5040
n0
102.
2
n0
x . . . 1 x2 24x 720
arcsin x 1 x
1 arcsin x 1 x2 lim 1. By L’Hôpital’s Rule, lim x→0 x→0 x 1
tn1 1 t n1 n!
et 1 dt t
tn
x
n n! n1
0
xn
n n!
n1
Problem Solving for Chapter 8
Problem Solving for Chapter 8 2. Let S
1
2n 1
2
n1
Then
1 1 1 2 2. . . 2 1 3 5
1 1 1 1 2 2 2 2 2. . . 6 1 2 3 4 S
1 1 . . . 22 42
S
1 1 1 1 2 2. . . 22 2 3
S
1 2 . 22 6
Thus, S
2 1 2 2 3 2 . 6 4 6 6 4 8
4. (a) Position the three blocks as indicated in the figure. The bottom block extends 1 6 over the edge of the table, the middle block extends 1 4 over the edge of the bottom block, and the top block extends 1 2 over the edge of the middle block. 1 6
The centers of gravity are located at bottom block:
1 1 1 6 2 3
middle block:
1 1 1 1 6 4 2 12
top block:
0 1 5 6 12
5 1 1 1 1 . 6 4 2 2 12
The center of gravity of the top 2 blocks is
121 125 2 61, which lies over the bottom block. The center of gravity of the 3 blocks is
31 121 125 3 0 which lies over the table. Hence, the far edge of the top block lies 1 1 1 11 6 4 2 12 beyond the edge of the table. n
(b) Yes. If there are n blocks, then the edge of the top block lies
1
2i from the
c1
edge of the table. Using 4 blocks, 4
1
1
1
1
1
25
2i 2 4 6 8 24
c1
which shows that the top block extends beyond the table. (c) The blocks can extend any distance beyond the table because the series diverges:
1
1
2i 2
c1
c1
1 . i
1 2 1 4
11 12
421
422
Chapter 8
6. a
Infinite Series
1n1a b a b b a b . . . 2 3 4 2n n1
If a b,
1n1 1n12a converges conditionally. a 2n n n1 n1
If a b,
ab 1n1a b diverges. 2n n1 n1 2n
No values of a and b give absolute convergence. a b implies conditional convergence. ex 1 x
8.
x2 . . . 2!
ex 1 x2 2
x4 . . . x12 . . . 2! 6!
f 120 12! 1 ⇒ f 120 665,280 12! 6! 6!
10. (a) If p 1,
2
If p > 1,
2
1 dx ln ln x x ln x
diverges.
2
1 ln b1p ln 21p converges. dx lim p b→ xln x 1p 1p
If p < 1, diverges. (b)
n4
1 1 1 diverges by part (a). n lnn2 2n4 n ln n
12. Let bn an r n.
bn
1 n
14. (a)
an
r n 1 n
r → Lr as n → .
an
1 n
1 1 0.01n 0.99 1 0.01 n0
1 0.01 0.012 . . .
1 Lr < r 1. r
1.010101 . . .
By the Root Test,
b
n
converges ⇒
a
n
rn
converges.
(b)
1 1 0.02n 0.98 1 0.02 n0
1 0.02 0.022 . . . 1 0.02 0.0004 . . . 1.0204081632 . . .
1
16. (a) Height 2 1 2
1
n
n1
2
(b) S 4 1
1 2
1 3
. . .
p-series, p 21 < 1
1 1 1 . . . 4 2 3 n1 n
4 1 1 (c) W 1 3 2 3 2 . . . 3 2 3 4 1 converges. 3 n1 n3 2
C H A P T E R Infinite Series
8
Section 8.1
Sequences . . . . . . . . . . . . . . . . . . . . . 121
Section 8.2
Series and Convergence . . . . . . . . . . . . . . 126
Section 8.3
The Integral Test and p-Series
Section 8.4
Comparisons of Series
Section 8.5
Alternating Series . . . . . . . . . . . . . . . . . 138
Section 8.6
The Ratio and Root Tests . . . . . . . . . . . . . 142
Section 8.7
Taylor Polynomials and Approximations . . . . . 147
Section 8.8
Power Series . . . . . . . . . . . . . . . . . . . . 152
Section 8.9
Representation of Functions by Power Series
. . . . . . . . . . 131
. . . . . . . . . . . . . . 135
Section 8.10 Taylor and Maclaurin Series
. . 157
. . . . . . . . . . . 160
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 167 Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . 172
C H A P T E R Infinite Series Section 8.1
8
Sequences
Solutions to Odd-Numbered Exercises 1. an 2n a1 21 2 a2 22 4 a3 23 8 a4 2 16 4
a5 2 32 5
21
3. an
1
21
2
3
1 8
4
1 16
21
5
a2
1 a3 2
a5
1nn12 n2
a1
11 1 12
1 13 a2 22 4
16 1 a3 32 9 1 110 a4 42 16 1 115 a5 52 25
13. a1 3, ak1 2ak 1 a2 2a1 1 23 1 4 a3 2a2 1 24 1 6 a4 2a3 1 26 1 10 a5 2a4 1
5. an sin
21
a1
1 a4 2
7. an
n
9. an 5
1 2
a1 sin
a3 sin
a3 5 a4 5 a5 5
3 1 2
a4 sin 2 0 a5 sin
5 1 2
1 32
1 1 n n2
11. an
a1 5 1 1 5 a2 5
1 2
a2 sin 0
1 4
n 2
1 1 19 2 4 4 1 1 43 3 9 9 1 1 77 4 16 16 1 1 121 5 25 25
3n n!
a1
3 3 1!
a2
32 9 2! 2
a3
33 27 3! 6
a4
34 81 4! 24
a5
35 243 5! 120
1 15. a1 32, ak1 ak 2 1 1 a2 a1 32 16 2 2 1 1 a3 a2 16 8 2 2 1 1 a4 a3 8 4 2 2 1 1 a5 a4 4 2 2 2
210 1 18
121
122
Chapter 8
Infinite Series 19. This sequence decreases and a1 4, a2 40.5 2. Matches (c).
17. Because a1 81 1 4 and a2 82 1 83 , the sequence matches graph (d). 21.
23.
8
25.
18
−1 −1
12
3
−1
12
12 −1
− 10
2 an n, n 1, . . . , 10 3
an 160.5
29. an
27. an 3n 1 a5 35 1 14 a6 36 1 17 Add 3 to preceeding term.
−1
n1,
n 1, . . . , 10
an
3 2n1
31.
2n , n 1, 2, . . . , 10 n1
10! 8!910 8! 8! 910 90
an
3 3 24 16
a6
3 3 25 32
Multiply the preceeding term by 12.
33.
n 1! n!n 1 n! n! n1
35.
2n 1! 2n 1! 2n 1! 2n 1!2n2n 1
5n2 5 2
37. lim
39. lim
n→
n2
n→
2n n2 1
lim
n→
43.
2
41. lim sin
1 1n2
n→
2
−1 −1
1n 0
2 2 1 45.
3
1 2n2n 1
12
12 −1
−2
The graph seems to indicate that the sequence converges to 1. Analytically, lim an lim
n→
n→
n1 x1 lim lim 1 1. x→ x→ n x
The graph seems to indicate that the sequence diverges. Analytically, the sequence is
an 0, 1, 0, 1, 0, 1, . . . . Hence, lim an does not exist. n→
47. lim 1n n→
n n 1
does not exist (oscillates between 1 and 1), diverges.
51. lim
n→
1 1n 0, converges n
49. lim
n→
3n2 n 4 3 , converges 2n2 1 2
lnn3 3 lnn lim n→ 2n n→ 2 n
53. lim
lim
n→
(L’Hôpital’s Rule)
3 1 0, converges 2 n
Section 8.1
34
55. lim
n→
59. lim
n→
n
0, converges
57. lim
n→
n n1 n 12 n2 lim n→ n n1 nn 1 n→
k n
63. an 1
61. lim
1 2n 0, converges n2 n
n
lim 1
n→
65. lim
n→
k n
n
n 1! lim n 1 , diverges n→ n!
np 0, converges n→ en p > 0, n ≥ 2
lim
Sequences
sin n 1 lim sin n 0, converges n→ n n
lim 1 u1u k ek u→0
k where u , converges n 69. an n2 2
67. an 3n 2
73. an
1n1 2n2
79. an
1
83.
75. an 1
1n1 1n1 2nn! . . 2n 1 2n!
35.
n ? n1 ≥ n1 2 2n2 2 ? n2 n3 2 n ≥ 2 n 1 ? 2n ≥ n 1 n ≥ 1 n ≥ 1
Hence,
2n ≥ n 1
1 n1 n n
71. an
n1 n2
77. an
n n 1n 2
1 1 < 4 an1, n n1 monotonic; an < 4 bounded.
81. an 4
85. an 1n
1n
a1 1 a2
1 2
a3
1 3
Not monotonic; an ≤ 1, bounded
2n3n ≥ 2n2n 1 n n1 ≥ n1 2 2n2 2 an ≥ an1
True; monotonic; an ≤ 18 , bounded 87. an 23 > n
23 n1 an1
Monotonic; an ≤ 23 , bounded
89. an sin
n6
a1 0.500 a2 0.8660 a3 1.000 a4 0.8660
Not monotonic; an ≤ 1, bounded
123
124
Chapter 8
Infinite Series
91. (a) an 5
5
1 n
1 ≤ 6 ⇒ an bounded n
an 5
1 1 1 n 3 3
1 1 > 5 n n1
n1
0.4
−1
12 −1
12 −0.1
lim 5
n→
< 131 3 1
Therefore, an converges. (b)
−1
1 ⇒ an bounded 3
an1 ⇒ an monotonic
Therefore, an converges. 7
<
1 1 1 n 3 3
an
an1 ⇒ an monotonic
(b)
1 1 1 n 3 3
93. (a) an
95. An P 1
1 5 n
r 12
lim
n→
1
n
n
97. (a) A sequence is a function whose domain is the set of positive integers.
(a) lim An , divergent. The amount will grow n→ arbitrarily large over time.
(b) An 9000 1
0.115 12
n
A1 $9086.25
A6 $9530.06
A2 $9173.33
A7 $9621.39
A3 $9261.24
A8 $9713.59
A4 $9349.99
A9 $9806.68
A5 $9439.60
A10 $9900.66
99. an 10
31 31 31
1 n
103. (a) An 0.8n 2.5 billion
(b) A sequence converges if it has a limit. (c) A bounded monotonic sequence is a sequence that has nondecreasing or nonincreasing terms, and an upper and lower bound.
101. an
3n 4n 1
105. (a) an 3.7262n2 75.9167n 684.25
(b) A1 $2 billion
1500
A2 $1.6 billion A3 $1.28 billion A4 $1.024 billion (c) lim 0.8n 2.5 0 n→
−1
8 0
(b) For 2004, n 14 and a14 1017, or $1017.
Section 8.1
107. an
Sequences
n n n1n 109. an
10n n!
a1 111 1
109 (a) a9 a10 9!
a2 2 1.4142
1,000,000,000 362,880
1,562,500 567
3 a3 3 1.4422 4 a4 4 1.4142 5 a5 5 1.3797 6 a6 6 1.3480
(b) Decreasing (c) Factorials increase more rapidly than exponentials.
Let y lim n1n. n→
ln y lim
n→
lim
n→
1n ln n ln n 1n lim 0 n→ 1 n
Since ln y 0, we have y e0 1. Therefore, n lim n 1. n→
111. an2 an an1 (a) a1 1
a7 8 5 13
a2 1
a8 13 8 21
a3 1 1 2
a9 21 13 34
a4 2 1 3
a10 34 21 55
a5 3 2 5
a11 55 34 89
a6 5 3 8
a12 89 55 144
(b) bn b1 b2 b3 b4 b5
an1 ,n ≥ 1 an 1 b6 1 1 2 b7 2 1 3 b8 2 5 b9 3 8 b10 5
(c) 1
1 1 1 bn1 anan1 1
an1 an
an an1 an1 bn an an
(d) If lim bn , then lim 1 n→
n→
1 . bn1
Since lim bn lim bn1 we have, n→
13 8 21 13 34 21 55 34 89 55
113. True
1 1 .
n→
1 2 0 2 1
1 ± 1 4 1 ± 5 2 2
Since an, and thus bn, is positive,
1 52 1.6180. 115. True
117. a1 2 1.4142 a2 2 2 1.8478 a3 2 2 2 1.9616 a4
2 2 2 2 1.9904
a5
2 2 2 2 2 1.9976
an is increasing and bounded by 2, and hence converges to L. Letting lim an L implies that 2 L L ⇒ L 2. n→ Hence, lim an 2. n→
125
126
Chapter 8
Infinite Series
Section 8.2
Series and Convergence
1. S1 1
3. S1 3
S2 1 14 1.2500
S2 3 92 1.5
S3 1 14 19 1.3611
S3 3 92 27 4 5.25
1 S4 1 14 19 16 1.4236
81 S4 3 92 27 4 8 4.875
1 1 1 1 S5 1 4 9 16 25 1.4636
9 27 81 243 S5 3 2 4 8 16 10.3125
5. S1 3 S2 3 32 4.5 S3 3 32 34 5.250 S4 3 32 34 38 5.625 3 S5 3 32 34 38 16 5.8125
7.
3 2 3
n
9.
Geometric series
n
11.
Geometric series
r 1.055 > 1
3 > 1 2
lim
n→
Diverges by Theorem 8.6
n
2
n1
lim
n→
n2 1
n2
15.
4 4 9 1
n
n0
Diverges by Theorem 8.9
5
9
15
1
n
n0
21
4 16, S2 4 16 2.95, .
S0
. .
15 1 1 . . . 1 4 4 16
15 45 , S , S 3.05, . . . 4 1 16 2
Matches graph (a).
Analytically, the series is geometric:
Analytically, the series is geometric:
44 9
1
n
n0
94 94 3 1 14 34
15 1 4 n0 4
n
154 154 3 1 14
54
n n 1 n n 1 1 2 2 3 3 4 4 5 . . . 1
n1
1
n1
1
1
1
1
1
1
1
1
1
n1
n n 1
23.
45
4 4
19.
Matches graph (c).
21.
2n 1 n1 n0 2
lim
1 9 1 . . . 1 4 4 16
9 9 S 0 , S1 4 4
n 10 n1
2n 1 1 2n 1 0 lim n1 n→ 2 n→ 2 2
n2 10 1
Diverges by Theorem 8.9
17.
n
Diverges by Theorem 8.9
Diverges by Theorem 8.6
13.
n1
n1
n0
n0
r
1000 1.055
2 4 3
lim Sn lim 1
n→
n→
1 1 n1
n
25.
0.9
n
n0
n0
Geometric series with r 34 < 1.
Geometric series with r 0.9 < 1.
Converges by Theorem 8.6
Converges by Theorem 8.6
Section 8.2
27. (a)
Series and Convergence
n n 3 2 n n 3 6
n1
1
1
n1
1 4 2 5 3 6 4 7 . . . 1
2
2 1 (b)
(c)
1
1
1
1
1
1
1 1 11 3.667 2 3 3
n
5
10
20
50
100
Sn
2.7976
3.1643
3.3936
3.5513
3.6078
5
0
11 0
(d) The terms of the series decrease in magnitude slowly. Thus, the sequence of partial sums approaches the sum slowly.
29. (a)
2 0.9
n1
n1
(b)
2 0.9
n
n0
2 20 1 0.9
(c)
n
5
10
20
50
100
Sn
8.1902
13.0264
17.5685
19.8969
19.9995
22
0
11 0
(d) The terms of the series decrease in magnitude slowly. Thus, the sequence of partial sums approaches the sum slowly.
31. (a)
10 0.25
n1
n1
(b)
10 40 13.3333 1 0.25 3
(c)
n
5
10
20
50
100
Sn
13.3203
13.3333
13.3333
13.3333
13.3333
15
11
0 0
(d) The terms of the series decrease in magnitude rapidly. Thus, the sequence of partial sums approaches the sum rapidly.
33.
n
n2
35.
2
12 12 1 1 1 1 1 n2 n 1 n 1 2 n2 n 1 n 1
1 2
1 3 1 1 2 2 4
n 1 n 2 8 n 1 n 2 8 2 3 3 4 4 5 . . . 82 4 8
2
1
n
10
n0
1
1
1
1
1
1
1
39.
2
1
1
n1
n0
41.
1 31 12 41 13 51 14 61 . . .
n1
37.
1
n
1 2 1 12
1 10 1 110
9
1
n
n0
43.
3 3
n0
1
n
1 2 1 12 3
3 9 1 13 4
127
128
45.
Chapter 8
Infinite Series
1 1 1 2 3 2 n
n0
n
n
n0
1 3
n
47. 0.4
n0
S
3 1 2 2 2
40100 3
1
n
51.
53.
3 40
and r
1 100
n 10
lim
1
1
n 10 1 0 10n 1 10
Diverges by Theorem 8.9
n n 2 1 3 2 4 3 5 4 6 . . . 1
1
1
1
1
1
1
1
n1
55.
3n 1
a 410 4 1 r 1 110 9
10n 1
n→
340 5 a 1 r 99100 66
S
n
n1
n0
Geometric series with a
4 1 10 10 n0
4 1 Geometric series with a 10 and r 10
1 1 1 12 1 13
49. 0.07575
2n 1
57.
4
2
n
4
2 1
1 3 , converges 2 2
n
59.
1.075
n
n1
n0
3n 1 3 0 n→ 2n 1 2
Geometric series with r 12
Geometric series with r 1.075
Converges by Theorem 8.6
Diverges by Theorem 8.6
lim
n0
n0
Diverges by Theorem 8.9
61.
n
ln n
63. See definition, page 567.
n2
lim
n→
n 1 lim ln n n→ 1n
(by L’Hôpital’s Rule) Diverges by Theorem 8.9 65. The series given by
67. (a) x is the common ratio. (b) 1 x x2 . . .
ar n a ar ar 2 . . . ar n . . . , a 0
n0
n0
x
n
1 , x < 1 1x
Geometric series: a 1, r x, x < 1
is a geometric series with ratio r. When 0 < r < 1, the a series converges to . The series diverges if r ≥ 1. 1r
(c) y1
3
1 1x
y2 1 x
−1.5
1.5 −1
11 0.5 0.5 x
69. f x 3
7
y=6
Horizontal asymptote: y 6
3 2 1
n
n0
S
3 6 1 12
−2
10 −1
The horizontal asymptote is the sum of the series. f n is the nth partial sum.
Section 8.2
71.
Series and Convergence
1 < 0.001 n n 1
10,000 < n2 n 0 < n2 n 10,000 n
1 ± 12 4 1 10,000
2
Choosing the positive value for n we have n 99.5012. The first term that is less than 0.001 is n 100.
18
n
< 0.001
10,000 < 8n This inequality is true when n 5. This series converges at a faster rate. n1
73.
8000 0.9 i
i0
8000 1 0.9 n1 1 1 0.9
80,000 1
n1
75.
100 0.75 i
i0
, n > 0
0.9n
100 1 0.75 n1 1 1 0.75
400 1 0.75n million dollars. Sum 400 million dollars
77. D1 16 D2 0.81 16 0.81 16 32 0.81
down
up
D3 16 0.81 2 16 0.81 2 32 0.81 2
D 16 32 0.81 32 0.81 2 . . . 16
32 0.81
n
16
n0
79. P n P 2
1 1 2 2
1 1 n0 2 2
1 1 2 2 n
2
n
2
32 152.42 ft 1 0.81
81. (a)
1
n
n1
n0
1 1
n
1 1 1 2 1 12
(b) No, the series is not geometric.
1 8
(c)
12 1 1 12
83. Present Value
2 2
19
50,000
n1
1 1.06
50,000 1 1.06 n0 1.06
1
n
2
n1
n
18
n 2 n1
0.01 2 i
i0 n
50,000 1 1.0619 1.06 1 1.061
85. w 1 , r 1.06
$557,905.82 The present value is less than $1,000,000. After accruing interest over 20 years, it attains its full value.
0.01 1 2n
0.01 2n 1
12
(a) When n 29: w $5,368,709.11 (b) When n 30: w $10,737,418.23 (c) When n 31: w $21,474,836.47
129
130
Chapter 8
Infinite Series
87. P 50, r 0.03, t 20 (a) A 50 (b) A
89. P 100, r 0.04, t 40
12 0.03 1 0.03 12
12 20
1 $16,415.10
(a) A 100
50 e0.03 20 1
$16,421.83 e0.0312 1
(b) A
91. (a) an 6110.1832 1.0544 x 6110.1832e0.05297n
12 0.04 1 0.04 12
12 40
1 $118,196.13
100 e0.04 40 1
$118,393.43 e0.0412 1
93. x 0.749999 . . . 0.74
0.009 0.1
n
n0 10,000
0.74
0.009 1 0.1
0.74 0.01 0.75 0 6,000
10
(b) 78,530 or $78,530,000,000 9
a
(c) Total
n
78,449 or $78,449,000,000
n0
95. By letting S0 0, we have an
n
a
k
k1
a
n
n1
97. Let
a
n
S
n
Sn1
n1
n
a
k
Sn Sn1. Thus,
k1
Sn1 c c
n1
1 and b
n
n0
S
n1
1 .
c S
c Sn .
n1
n1
99. False. lim
n0
n→
1 1 0, but diverges. n n1 n
Both are divergent series.
a
n
bn
n0
n0
1 1 1 1 0
101. False
ar
n
n1
1 a r a
The formula requires that the geometric series begins with n 0. 103. Let H represent the half-life of the drug. If a patient receives n equal doses of P units each of this drug, administered at equal time interval of length t, the total amount of the drug in the patient’s system at the time the last dose is administered is given by Tn P Pekt Pe2kt . . . Pe n1 kt where k ln 2 H. One time interval after the last dose is administered is given by Tn1 Pekt Pe2kt Pe3kt . . . Penkt. Two time intervals after the last dose is administered is given by Tn2 Pe2kt Pe3kt Pe4kt . . . Pe n1 kt and so on. Since k < 0, Tns→0 as s → , where s is an integer.
Section 8.3
Section 8.3 1.
1
3.
n1
e
1
n
n1
Let f x ex.
1 . x1
f is positive, continuous, and decreasing for x ≥ 1.
f is positive, continuous and decreasing for x ≥ 1.
131
The Integral Test and p-Series
n1
Let f x
The Integral Test and p-Series
1 dx lnx 1 x1
1
ex dx ex
1
1
1 e
Converges by Theorem 8.10
Diverges by Theorem 8.10
5.
n
2
n1
1 1
7.
n1
1 . x2 1 f is positive, continuous, and decreasing for x ≥ 1. Let f x
1
1 dx arctan x x2 1
1
lnn 1 n1
Let f x
lnx 1 x1
f is positive, continuous, and decreasing for x ≥ 2 since
4
fx
Converges by Theorem 8.10
1
1 lnx 1 < 0 for x ≥ 2. x 12
lnx 1 ln2x 1 dx x1 2
1
Diverges by Theorem 8.10
9.
nk1 k c
n
n1
Let f x
11.
1 3
n1
x k1 . xk c
Let f x
ck 1 x k c2
x k2
xk
1 . x3
f is positive, continuous, and decreasing for x ≥ 1.
f is positive, continuous, and decreasing for k ck 1 since x > fx
n
1
< 0
1 1 dx 2 x3 2x
1
1 2
Converges by Theorem 8.10
k for x > ck 1.
1
x k1 1 dx lnx k c xk c k
1
Diverges by Theorem 8.10
13.
1
1
n n
n1
5
n1
Divergent p-series with p
17.
1
n
1 2
n1
1 5
1 Divergent p-series with p 2 < 1
< 1
1
n
n1
15.
1 5
3 2
3 Convergent p-series with p 2 > 1
19.
1
n
n1
1.04
Convergent p-series with p 1.04 > 1
132 21.
Chapter 8
2
n
n1
4
2 2 2 . . . 1 23 4 33 4
3
Infinite Series 23.
2
nn 2 2 2
3 2
2 33 2 . . .
n1
S1 2
S1 2
S2 3.189
S2 2.707
S3 4.067
S3 3.092 Matches (b)
Matches (a) 3 4
Diverges—p-series with p
Converges—p-series with p 3 2 > 1
< 1
25. No. Theorem 8.9 says that if the series converges, then the terms an tend to zero. Some of the series in Exercises 21-24 converge because the terms tend to 0 very rapidly. N
27.
1
1
1
1
1
n1234. . .N
> M
n1
(a)
29.
M
2
4
6
8
N
4
31
227
1674
(b) No. Since the terms are decreasing (approaching zero), more and more terms are required to increase the partial sum by 2.
1
nln n
p
n2
If p 1, then the series diverges by the Integral Test. If p 1,
1 dx xln xp
2
ln xp
2
1 ln xp1 dx x p 1
.
2
Converges for p 1 < 0 or p > 1. 31. Let f be positive, continuous, and decreasing for x ≥ 1 and an f n. Then,
33. Your friend is not correct. The series
a
n
and
n1
f x dx
n10,000
1
is the harmonic series, starting with the 10,000th term, and hence diverges.
either both converge or both diverge (Theorem 8.10). See Example 1, page 578.
35. Since f is positive, continuous, and decreasing for x ≥ 1 and an f n, we have, RN S SN
a
n
0 ≤ RN ≤
n
n1
n1
Also, RN S SN
N
a
an > 0.
nN1
an ≤ aN1
nN1
N1
f x dx.
N
37. S6 1 R6 ≤
6
1.0811 ≤
1 1 1 1 1 1.0811 24 34 44 54 64
1 1 dx 3 x4 3x
1
n
n1
4
6
0.0015
≤ 1.0811 0.0015 1.0826
f x dx ≤
1 1 1 . . . n 10,000 10,001
N
f x dx. Thus,
Section 8.3 1 1 1 1 1 1 1 1 1 1 0.9818 2 5 10 17 26 37 50 65 82 101
39. S10
R10 ≤
1
n
0.9818 ≤
10
arctan 10 0.0997 2
≤ 0.9818 0.0997 1.0815
5
n1
1 dx arctan x x 1 2
10
1 2 3 4 4 9 16 0.4049 e e e e
41. S4
R4 ≤
The Integral Test and p-Series
x2
xe
4
ne
0.4049 ≤
1 2 dx ex 2 n2
4
43. 0 ≤ RN ≤
N
1 1 dx 3 x4 3x
N
1 < 0.001 3N 3
1 < 0.003 N3
5.6 108
N3 > 333.33
≤ 0.4049 5.6 108
n1
N > 6.93 N ≥ 7
45. RN ≤
e5x
N
1 dx e5x 5
N
e5N < 0.001 5
47. RN ≤
N
x2
1 < 0.005 e5N
N
arctan N < 0.001 2
arctan N < 1.5698
e5N > 200
arctan N > 1.5698
5N > ln 200
N > tan 1.5698
ln 200 5
N >
1 dx arctan x 1
N ≥ 1004
N > 1.0597 N ≥ 2
49. (a)
1
n
1.1 .
n2
This is a convergent p-series with p 1.1 > 1.
1
n ln n is a divergent series. Use the Integral Test.
n2
1 dx ln ln x 2 2 x ln x 6 1 1 1 1 1 1 (b) 1.1 21.1 31.1 41.1 51.1 61.1 0.4665 0.2987 0.2176 0.1703 0.1393 n n2
6
1
1
1
1
1
1
n ln n 2 ln 2 3 ln 3 4 ln 4 5 ln 5 6 ln 6 0.7213 0.3034 0.1803 0.1243 0.0930
n2
The terms of the convergent series seem to be larger than those of the divergent series! (c)
1 1 < n1.1 n ln n n ln n < n1.1 ln n < n0.1 This inequality holds when n ≥ 3.5 1015. Or, n > e40. Then ln e40 40 < e400.1 e4 55.
133
134
Chapter 8
Infinite Series
51. (a) Let f x 1 x. f is positive, continuous, and decreasing on 1, .
n
Sn 1 ≤
1
y
1 dx x
1
Sn 1 ≤ ln n
1 2
Hence, Sn ≤ 1 ln n. Similarly,
n1
Sn ≥
1
1 dx lnn 1. x
1
2
3 ... n −1
x n
n+1
Thus, lnn 1 ≤ Sn ≤ 1 ln n. (b) Since lnn 1 ≤ Sn ≤ 1 ln n, we have lnn 1 ln n ≤ Sn ln n ≤ 1. Also, since ln x is an increasing function, lnn 1 ln n > 0 for n ≥ 1. Thus, 0 ≤ Sn ln n ≤ 1 and the sequence an is bounded.
n1
(c) an an1 Sn ln n Sn1 lnn 1
n
1 1 dx ≥ 0 x n1
Thus, an ≥ an1 and the sequence is decreasing. (d) Since the sequence is bounded and monotonic, it converges to a limit, . (e) a100 S100 ln 100 0.5822 (Actually 0.577216.)
53.
1
2n 1
n1
1 . 2x 1
Let f x
f is positive, continuous, and decreasing for x ≥ 1.
1
1 dx ln 2x 1 2x 1
1
Diverges by Theorem 8.10
55.
nn n
1
p-series with p
5 4
1 4
n1
n1
2
n
n0
Geometric series with r 23 Converges by Theorem 8.6
Converges by Theorem 8.11
59.
3
57.
5 4
n 2 1 n n1
1 1 n
61.
n
n1
n 1 lim 10 lim n→ n2 1 n→ 1 1 n2
n→
Diverges by Theorem 8.9
Fails nth Term Test
lim 1
1 n
n
e0
Diverges by Theorem 8.9
63.
1
nln n
3
n2
Let f x
1 . xln x3
f is positive, continuous and decreasing for x ≥ 2.
2
1 dx xln x3
2
ln x3
1 ln x2 dx x 2
Converges by Theorem 8.10. See Exercise 13.
2
1 2ln x2
2
1 2ln 22
Section 8.4
Section 8.4 1. (a)
6
n
n1
32
Comparisons of Series
Comparisons of Series
6 6 . . . S1 6 1 232
an
6 an = 3/2 n
6
6 6 6 3 . . . S1 3 4 232 3 2
n
5
32
n1
an =
4 3
6 6 6 6 . . . S1 4.9 2 0.5 1.5 11.5 24.5
nn
n1
135
6 n n 2 + 0.5
an = 3/26 n +3
2 1
n
(b) The first series is a p-series. It converges p 32 > 1. (c) The magnitude of the terms of the other two series are less than the corresponding terms at the convergent p-series. Hence, the other two series converge. (d) The smaller the magnitude of the terms, the smaller the magnitude of the terms of the sequence of partial sums.
Sn
n
Σ
k=1
12
6
4
2
8
10
6 k 3/2 n
Σ
10
k=1
6 k k 2+ 0.5
8 6 4
n
Σ
2
k=1
6 k 3/2 + 3 n
2
3.
1 1 < 2 n2 1 n
5.
Therefore,
n
2
n1
diverges by comparison with the divergent p-series
Therefore,
1 1
n1
diverges by comparison with the divergent series
1
n 1.
n0
ln n
n1
converges by comparison with the convergent geometric series
1 ln n > . n1 n1
9. For n ≥ 3,
Therefore,
n0
1
n.
n2
1 1 < n 3n 1 3
n
1
n2
2
10
n1
1
3
8
1 1 > for n ≥ 2 n1 n
1 1
n.
7.
6
Therefore,
converges by comparison with the convergent p-series
n1
4
1 n . 3
n1
Note:
1
n 1 diverges by the integral test.
n1
11. For n > 3,
1 1 > . n2 n!
Therefore,
1
n!
n0
converges by comparison with the convergent p-series
1
n.
n1
2
13.
1 1 2 ≤ en en Therefore,
1
e
n0
n2
converges by comparison with the convergent geometric series
e .
n0
1
n
136
Chapter 8
15. lim
n→
Infinite Series
nn2 1 n2 lim 2 1 n→ n 1 1n
17. lim
n→
Therefore,
Therefore,
n 1
n
2
n1
n
diverges by a limit comparison with the divergent p-series
1
n.
n1
n3 nn 2 n2 3n lim 2 1 21. lim n→ n→ n 2n 1n
2n2 1 2 2n 1 2n5 n3 lim 3 5 n→ 3n 2n 1 1n 3
3n5
n→
Therefore,
Therefore,
3n
nn 2
n1
diverges by a limit comparison with the divergent p-series
converges by a limit comparison with the convergent p-series
1 3. n1 n
n1
1nn2 1 n2 lim 1 n→ n→ nn2 1 1n2
nk1nk 1 nk lim k 1 n→ n→ n 1 1n
25. lim
23. lim
Therefore,
Therefore,
2
nk1 k 1
n
1
nn
n1
1
n.
n3
2n2 1 2n 1
5
n1
1
n.
n1
19. lim
1 1
2
n0
diverges by a limit comparison with the divergent p-series
1n2 1 n 1 lim n→ n2 1 1n
1
n1
diverges by a limit comparison with the divergent p-series
converges by a limit comparison with the convergent p-series
1
n.
1 . 2 n n1
n1
sin1n 1n2 cos1n lim n→ n→ 1n 1n2
27. lim
29.
n
n1
n
1
n
n1
Diverges
1 lim cos 1 n→ n
1
p-series with p 2
Therefore,
sinn 1
n1
diverges by a limit comparison with the divergent p-series
1
n.
n1
31.
3
n1
n
1 2
33.
Converges Direct comparison with
n1
1 3
n
2n 3
35.
n
2
n 12
n1
n1
Diverges; nth Term Test
Converges; integral test
n
lim
n→
n 1 0 2n 3 2
Section 8.4 an lim nan by given conditions lim nan is finite n→ 1n n→ n→ and nonzero.
39.
37. lim
Comparisons of Series
1 2 3 4 5 n . . . , 2 2 5 10 17 26 n1 n 1
which diverges since the degree of the numerator is only one less than the degree of the denominator.
Therefore,
a
n
n1
diverges by a limit comparison with the p-series
1
n.
n1
41.
n
3
n1
1 1
5n n 3 lim 5n n 3 51 0 3
43. lim n n→
converges since the degree of the numerator is three less than the degree of the denominator.
1 45. See Theorem 8.12, page 583. One example is 2 1 n n1 converges because
47.
1 1 < and 1 n2
1.0
4
n3 diverges. 3
4
Terms of ∞ Σ an
0.8
n=1
0.6
1
n
n1
n→
Therefore,
n1
n2
4
4
5n
Terms of ∞ 2 Σ an
0.4
2
n=1
0.2
converges ( p-series).
n 4
8
12
16
20
For 0 < an < 1, 0 < an2 < an < 1. Hence, the lower terms are those of an2.
49.
1 1 1 1 . . . , diverges 200 400 600 n1 200n
51.
1 1 1 1 1 , converges 201 204 209 216 n1 200 n2
55. False. Let an 1n3 and bn 1n2. 0 < an ≤ bn and both
53. Some series diverge or converge very slowly. You cannot decide convergence or divergence of a series by comparing the first few terms.
1
n
n1
3
and
n1
57. True
b
n
converges, lim bn 0. There exists N such that bn < 1 for n > N. Thus, n→
n1
anbn < an for n > N and
a
n bn
n1
converges by comparison to the convergent series
a
n.
i1
61.
1
n
2
and
1
n
3
both converge, and hence so does
n n n . 1
1
1
2
3
5
1
n
converge.
59. Since
137
2
138
Chapter 8
Infinite Series
63. (a) Suppose bn converges and an diverges. Then there exists N such that 0 < bn < an for n ≥ N. This means that 1 < an bn for n ≥ N. Therefore, lim an bn 0. Thus, an must also converge. n→
(b) Suppose bn diverges and an converges. Then there exists N such that 0 < an < bn for n ≥ N. This means that 0 < an bn < 1 for n ≥ N. Therefore, lim an bn . Thus, an must also diverge. n→
65. Start with one triangle whose sides have length 9. At the nth step, each side is replaced by four smaller line segments each having 13 the length of the original side.
3 3
#Sides
Length of sides
3
3
9
3
42
9
9
3 4n
9 13
34 3
1 3
1 2 3
n
At the nth step there are 3 4n sides, each of length 9 13 . At the next step, there are 3 4n new triangles of side 9 13 area of an equilateral triangle of side x is 14 3 x 2. Thus, the new triangles each have area n
1 4 3
3
9
n1
The area of the 3
2
3 1
4 32n
. The
.
4n new triangles is
4
3 4n
n1
3 1
32n
33 4 4 9
n .
The total area is the infinite sum 33 4 93 4 4 9 n0
n
93 33 1 93 33 9 183 . 4 4 1 49 4 4 5 5
The perimeter is infinite, since at step n there are 3 4n sides of length 9 13 . Thus, the perimeter at step n is 27 43 → . n
n
Section 8.5 1.
6
n
2
n1
5.
Alternating Series
6 6 6 . . . 1 4 9
3.
10
n2
n1
n
10 10 . . . 2 8
S1 6, S2 7.5
S1 5, S2 6.25
Matches (b)
Matches (c)
1n1 0.7854 4 n1 2n 1
(a)
(b)
n
1
2
3
4
5
6
7
8
9
10
Sn
1
0.6667
0.8667
0.7238
0.8349
0.7440
0.8209
0.7543
0.8131
0.7605
1.1
0
11 0.6
(c) The points alternate sides of the horizontal line that represents the sum of the series. The distance between successive points and the line decreases. (d) The distance in part (c) is always less than the magnitude of the next term of the series.
Section 8.5
7.
Alternating Series
1n1 2 0.8225 n2 12 n1
(a)
(b)
n
1
2
3
4
5
6
7
8
9
10
Sn
1
0.75
0.8611
0.7986
0.8386
0.8108
0.8312
0.8156
0.8280
0.8180
1.1
0
11 0.6
(c) The points alternate sides of the horizontal line that represents the sum of the series. The distance between successive points and the line decreases. (d) The distance in part (c) is always less than the magnitude of the next term in the series.
9.
1n1 n n1
an1 lim
n→
11.
1 1 < an n1 n
an1
1 0 n
lim
n→
Converges by Theorem 8.14.
13.
1n n2 2 n1 n 1
lim
n→
n2
1n1 n1 2n 1
1 1 < an 2n 1 1 2n 1
1 0 2n 1
Converges by Theorem 8.14
15.
n2 1 1
1n n1 n
an1
Diverges by the nth Term Test
lim
n→
1 n 1
1 n
<
1 n
an
0
Converges by Theorem 8.14
17.
1n1n 1 lnn 1 n1
lim
n→
19.
sin
n1
n1 1 lim lim n 1 lnn 1 n→ 1n 1 n→
2n 1 1n1 2 n1
Diverges by the nth Term Test
Diverges by the nth Term Test
21.
n1
n1
cos n 1
n
Diverges by the nth Term Test
23.
1n n! n0
an1 lim
n→
1 1 < an n 1! n!
1 0 n!
Converges by Theorem 8.14
139
140
25.
Chapter 8
Infinite Series
1n1n n2 n1
n 1 n < for n ≥ 2 n 1 2 n2
an1
n
lim
n→
27.
n2
1n12en 1n12 n n e e e2n 1 n1 n1
Let f x
0
2ex . Then e 1
f x
2x
2ex e2x 1 < 0. e2x 12
Thus, f x is decreasing. Therefore, an1 < an , and
Converges by Theorem 8.14
lim
n→
2en 2en 1 lim 0. lim e2n 1 n→ 2e2n n→ en
The series converges by Theorem 8.14.
29. S6
31n1 2.4325 n2 n1 6
3
R6 S S6 ≤ a7 49 0.0612; 2.3713 ≤ S ≤ 2.4937 31. S6
21n 0.7333 n! n0 5
2
R6 S S6 ≤ a7 6! 0.002778; 0.7305 ≤ S ≤ 0.7361 33.
1n n! n0
35.
1n
n0
(a) By Theorem 8.15,
RN
2n 1!
≤ aN1
(a) By Theorem 8.15, 1 < 0.001. N 1!
RN
This inequality is valid when N 6.
≤ aN1
1 < 0.001.
2N 1 1!
This inequality is valid when N 2.
(b) We may approximate the series by
(b) We may approximate the series by
1n 1 1 1 1 1 11 n! 2 6 24 120 720 n0 6
2
1n
1
1
2n 1! 1 6 120 0.842.
n0
0.368.
(3 terms. Note that the sum begins with n 0.)
(7 terms. Note that the sum begins with n 0.)
37.
1n1 n n1
39.
(a) By Theorem 8.15,
RN
1 ≤ aN1 < 0.001. N1
This inequality is valid when N 1000. (b) We may approximate the series by
1n1 1 1 1 1 1 . . . n 2 3 4 1000 n1 1000
0.693. (1000 terms)
1n1 3 n1 2n 1
By Theorem 8.15,
RN
≤ aN1
1 < 0.001. 2N 13 1
This inequality is valid when N 7.
Section 8.5
41.
1n1 2 n1 n 1
1
n 1
2
n1
43.
1n1 n n1
The given series converges by the Alternating Series Test, but does not converge absolutely since
converges by comparison to the p-series
1
n
1
2
n1
is a divergent p-series. Therefore, the series converges conditionally.
Therefore, the given series converge absolutely.
45.
141
n.
n1
Alternating Series
1n1 n2 2 n1 n 1
47.
n2 1 n→ n 12
1n n2 lnn
The given series converges by the Alternating Series Test, but does not converge absolutely since the series
lim
Therefore, the series diverges by the nth Term Test.
n 2
1 ln n
diverges by comparison to the harmonic series
1
n.
n1
Therefore, the series converges conditionally.
49.
1n n 3 n2 n 1
n
3
n2
51.
n0
n 1
n0
is convergent by comparison to the convergent geometric series
2
1
n. 2
n2
1
2n 1!
converges by a limit comparison to the convergent p-series
1n
2n 1!
1
n
n0
Therefore, the given series converges absolutely.
since 1 1 < n for n > 0. 2n 1! 2 Therefore, the given series converges absolutely.
53.
1n cos n n0 n 1 n0 n 1
55.
The given series converges by the Alternating Series Test, but
n 1 n 1
cos n
n0
1
n0
diverges by a limit comparison to the divergent harmonic series,
1
n.
n1
lim
n→
cos n n 1 1, therefore the series 1n
converges conditionally.
1n cos n 2 n n2 n1 n1
1
n
n1
2
is a convergent p-series. Therefore, the given
series converges absolutely.
142
Chapter 8
Infinite Series
57. An alternating series is a series whose terms alternate in sign. See Theorem 8.14.
59.
a converges. is conditionally convergent if a diverges, but a converges.
a a
n
is absolutely convergent if
n
n
n
n
61. (b). The partial sums alternate above and below the horizontal line representing the sum.
63. Since
a converges we have
65.
n
n1
n1
1
n
2
converges, hence so does
1
n.
n1
4
lim an 0.
n→
Thus, there must exist an N > 0 such that aN < 1 for all n > N and it follows that an2 ≤ an for all n > N. Hence, by the Comparison Test,
a
2
n
n1
converges. Let an 1n to see that the converse is false.
67. False Let an
69.
n1
1n . n
71. Diverges by nth Term Test. lim an n→
75. Convergent Geometric Series r
1 e
or Integral Test
10
n
32
10
1
n
n1
32
convergent p-series
73. Convergent Geometric Series r 78 < 1
77. Converges (absolutely) by Alternating Series Test
79. The first term of the series is zero, not one. You cannot regroup series terms arbitrarily.
Section 8.6 1.
The Ratio and Root Tests
n 1! n 1nn 1n 2! n 2! n 2! n 1nn 1
3. Use the Principle of Mathematical Induction. When k 1, the formula is valid since 1 13
5.
. . 2n 1 2n! 2n n!
and show that 13
5.
. . 2n 12n 1
—CONTINUED—
2n 2! . 2n1n 1!
21! . Assume that 21 1!
Section 8.6
The Ratio and Root Tests
143
3. —CONTINUED— To do this, note that:
35.
1
. . 2n 12n 1 1
35.
. . 2n 1 2n 1
2n! 2n 1 2n n!
2n!2n 1 2n 2 2n 1 2n n!
2n!2n 12n 2 2n1n!n 1
2n 2! 2n1n 1
The formula is valid for all n ≥ 1.
5.
n 4 3
n
1
n1
34 2169 . . .
7.
3 S1 , S2 1.875 4
33 3n1 9 . . . n! 2 n1
5n 3
9.
4n
n
n1
S1 9
S1 2
Matches (f)
Matches (a)
4 8 2 7
2
. . .
Matches (d)
11. (a) Ratio Test: lim
n→
(b)
(c)
an1 n 1258n1 n1 lim lim n→ n→ an n258n n
n
5
10
15
20
25
Sn
9.2104
16.7598
18.8016
19.1878
19.2491
2
5 5 < 1. Converges 8 8
20
0
12 0
(d) The sum is approximately 19.26. (e) The more rapidly the terms of the series approach 0, the more rapidly the sequence of the partial sums approaches the sum of the series.
13.
n!
3
n0
lim
n→
15.
n
an1 n 1! lim n→ an 3n1 lim
n→
3n
n!
n1 3
n
2
lim
n→
n1
lim
n→
an1 n1 lim n→ 2n1 an lim
n→
2n n
n1 1 2n 2
Therefore, by the Ratio Test, the series converges.
n
an 1 n 134n1 lim n→ an n34n lim
n→
3n 1 3 4n 4
Therefore, by the Ratio Test, the series converges.
19.
n
3
n1
Therefore, by the Ratio Test, the series diverges.
17.
n 4
2n
n
n1
lim
n→
2
an1 2n1 lim n→ n 12 an lim
n→
2n2
n 12
n2
2n
2
Therefore, by the Ratio Test, the series diverges.
144
21.
Chapter 8
Infinite Series
1n 2n n! n0
an1 2n1 lim n→ n 1! an
lim
n→
n!
2n
2 0 n→ n 1
n!
n3
23.
n
n1
lim
n→
an1 n 1! lim n→ n 13n1 an n→
4n
n!
n0
lim
n→
an1 4n1 lim n→ n 1! an
n!
4n
4 0 n1
lim
n→
Therefore, by the Ratio Test, the series converges.
27.
3n
n 1
n
n0
lim
n→
an1 3n1 lim n→ n 2n1 an
n 1n 3n 1n 3 n1 lim lim n n→ n 2n1 n→ n 2 n 2 3
n→
n
0
1e 0
nn 12 , let y lim nn 12 . Then, n
To find lim
n→
ln y lim
n→
n
n→
ln y lim n ln
1n 2 0 nn 12 lim ln n 1n 0 n→
1n 1 1n 2 1 by L’Hôpital’s Rule 1n2
1 y e1 . e Therefore, by the Ratio Test, the series converges.
29.
3
4n 1
lim
an1 4n1 lim n1 n→ an 1 3
n0
n→
n
3n 1 43n 1 41 13n 4 lim n1 lim n n→ n→ 4 1 3 3 3 13n
Therefore, by the Ratio Test, the series diverges.
31.
1n1n!
1 3 5 . . . 2n 1
n0
lim
n→
an1 lim n→ 1 an
3
n 1! 5 . . . 2n 12n 3
1
35.
. . 2n 1 n!
Therefore, by the Ratio Test, the series converges. Note: The first few terms of this series are 1
n3n n!
Therefore, by the Ratio Test, the series diverges.
Therefore, by the Ratio Test, the series converges.
25.
n 3
lim
lim
1 1
3
2!
1
35
3!
1
lim
n→
357
n1 1 2n 3 2
. . .
Section 8.6
32
n1
lim
n→
32
12
n32 n lim n→ n 1 1
n12 n lim n→ n 1 1
1
1
12
n1
lim
n→
35.
an1 1 lim n→ n 132 an
n
(b)
an1 1 lim n→ n 112 an
2n 1 n
1
n
37.
2n n 1
n lim an lim
n
n
n→
n 1 2n 1 2
n→
n→
lim
n→
Therefore, by the Root Test, the series converges.
1n
n an lim lim
n
n→
lim
39.
ln n
n2
n1
n→
145
1
n
33. (a)
The Ratio and Root Tests
ln1n
n
n
1
ln n
n
0
Therefore, by the Root Test, the series converges.
2n 1
n
n
n1
n 2 n n 1 lim 2 n n 1n n→
n a lim n lim
n→
n→
n n, let y lim x x. Then To find lim n→
n→
ln y lim
n→
x x ln
1 ln x 1x lim ln x lim lim 0. n→ n→ x n→ 1 x
n n 1 21 1 3. Therefore, by the Root Test, the series diverges. Thus, ln y 0, so y e0 1 and lim 2 n→
41.
1
ln n
43.
n
n3
ln1n
n a lim n lim
n→
n
n→
n
lim
n→
1 0 ln n
1n1 5 n n1
an1
Therefore, by the Root Test, the series converges.
5 5 < an n1 n
lim
n→
5 0 n
Therefore, by the Alternating Series Test, the series converges (conditional convergence).
45.
3
1
nn 3 n
n1
n1
47.
32
2n
n1
n1
This is convergent p-series.
lim
n→
2n 20 n1
This diverges by the nth Term Test for Divergence.
49.
1n 3n 32 1 1n 3n2 3 n n 2 2 9 2 n1 n1 n1
n
3 Since r 2 > 1, this is a divergent geometric series.
51.
10n 3 n2n n1
lim
n→
10n 3n2n 10n 3 10 lim n→ 12n n
Therefore, the series converges by a limit comparison test with the geometric series
2 .
n0
1
n
146
53.
Chapter 8
Infinite Series
cosn 2n n1
55.
cosn 1 ≤ n 2n 2
n1
n7n n1 n! lim
n→
Therefore, the series
an1 n 17n1 lim n→ an n 1!
n!
n7n
lim
n→
7 0 n
Therefore, by the Ratio Test, the series converges.
cosn 2n
converges by comparison with the geometric series
2 . 1
n
n0
57.
1n 3n1 n! n1
lim
n→
an1 3n lim n→ an n 1!
n!
3n1
lim
n→
3 0 n1
Therefore, by the Ratio Test, the series converges.
59.
3n
3 5 7 . . . 2n 1
n1
lim
n→
an1 lim n→ 3 an
3n1 . . 5 7 . 2n 12n 3
3
57.
. . 2n 1 3 0 lim n→ 2n 3 3n
Therefore, by the Ratio Test, the series converges.
63. (a) and (b) are the same.
61. (a) and (c) n 15n1 n5n n 1! n1 n! n0
5
252 353 454 . . . 2! 3! 4!
65. Replace n with n 1.
n
4
n1
n
n1 n1 n0 4
67. Since
69. See Theorem 8.17, page 597.
310 1.59 105, 210 10! use 9 terms.
3k 0.7769 k k1 2 k! 9
71. No. Let an The series
1 . n 10,000
1
n 10,000 diverges.
n1
73. The series converges absolutely. See Theorem 8.17.
Section 8.7 75. First, let
147
Second, let
n a lim n r < 1
n→
n a lim n r > R > 1.
n→
and choose R such that 0 ≤ r < R < 1. There must exist n a some N > 0 such that n < R for all n > N. Thus, for n > N, we an < Rn and since the geometric series
Taylor Polynomials and Approximations
n a Then there must exist some M > 0 such that n > R for all n > M. Thus, for n > M, we have an > Rn > 1 which implies that lim an 0 which in turn implies that
n→
R
a
n
n
n0
diverges.
n1
converges, we can apply the Comparison Test to conclude that
a n
n1
converges which in turn implies that
a
n
converges.
n1
Section 8.7
Taylor Polynomials and Approximations 3. y e12x 1 1
1
1. y 2 x 2 1 Parabola
Linear
Matches (d)
Matches (a)
5. f x
4 x
fx 2x32
f 1 2
fx sec x tan x
P1x f 1 f1x 1 P1x f
4 2x 1 P1x 2x 6
4 2
f
4 2
4 f 4 x 4
P1x 2 2 x
10
P1
f
7. f x sec x
f 1 4
4x12
4
5
(1, 4)
f
f −2
( π4 , 2)
6 −2
−
P1
4
2 −1
9. f x
4 x
f 1 4
4x12
x
0
0.8
0.9
1.0
1.1
1.2
2
fx 2x
f1 2
f x
Error
4.4721
4.2164
4.0
3.8139
3.6515
2.8284
f x 3x
f 1 3
P2x
7.5
4.46
4.215
4.0
3.815
3.66
3.5
32
52
f 1 P2 f 1 f1x 1 x 12 2 3 4 2x 1 x 12 2 10
P2 (1, 4) f −2
6 −2
148 11.
Chapter 8
Infinite Series
f x cos x
(b)
P2x 1
1 2 2x
P4x 1
1 2 2x
P6x 1
1 2 2x
(a)
1 4 24 x
1 4 24 x
P2x x
f x cos x
P2 x 1
f 0 P2 0 1 1 6 720 x
f x sin x
P4 x x
f 4x cos x
2
P6
−3
fx sin x
f 40 1 P440 f 5x sin x P65x x
P4 3
f
P44x 1
P2
f 6x cos x
−2
P6x 1
f 60 1 P660 (c) In general, f n0 Pnn0 for all n. 13.
f x ex
f 0 1
fx ex
f x e2x
f 0 1
f0 1
fx 2e2x
f0 2
f x ex
f 0 1
f x 4e2x
f 0 4
fx ex
f0 1
fx 8e2x
f0 8
f 4x 162x
f 40 16
P3x f 0 f0x 1x
17.
2
f 0 2 f0 3 x x 2! 3!
6
1 2x 2x 2
f 0 0
fx cos x
f0 1
f x sin x fx cos x
f 0 0
f x
xex
f0 1
fx
xex xex
0
f 50 1
P5x 0 1x
0 2 1 3 0 1 x x x 4 x5 2! 3! 4! 5!
f x
f 4x
f 0 1
1 x 12
f 0 0
f0 1
2ex
f 0 2
3ex
f0 3
4ex
f 40 4
P4x 0 x
2 2 3 4 x x3 x 4 2! 3! 4!
x x2
fx sec x tan x
P2x 1 0x
f 0 2
fx
6 x 14
f0 6
f 4x
24 x 15
P4x 1 x
f 0 1
f x
2 x 12
f 40 24 2 2 6 3 24 4 x x x 2! 3! 4!
1 x x 2 x3 x 4
1 3 1 4 x x 2 6
23. f x sec x
f0 1
f x
4 3 2 4 x x 3 3
ex
1 5 1 3 x x 6 120
1 x1
fx
f x xex fx
f 5x cos x
x
19.
xex
f 40
sin x
4 2 8 16 4 x x3 x 2! 3! 4!
P4x 1 2x
x3
f x sin x
f 4x
21.
x2
15.
sec3 x
f0 0
sec x
tan2
x
f 0 1
1 2 1 x 1 x2 2! 2
Section 8.7 f x
25.
1 x
fx f x
f 4x
149
f 1 1 1 x2
f1 1
2 x3
fx
Taylor Polynomials and Approximations
f 1 2 6 x4
f1 6
24 x5
f 41 24
P4x 1 x 1
2 6 24 x 12 x 13 x 14 2! 3! 4!
1 x 1 x 12 x 13 x 14 f x x
27.
fx
f 1 1
1
f1
2x
1 f x 4xx fx
1 2
3
f1
15 16x3x
3 8
f 41
15 16
fx
fx 4 sec2 x tan2 x 2 sec4 x f 4x 8 sec2 x tan3 x 16 sec4 x tan x f 5x 16 sec2 x tan4 x 88 sec4 x tan2 x 16 sec6 x 6
2
f
Q3 −6
f 1 1 f1 2
6 x4
f 41 6
1 P4x 0 x 1 x 12 2 1 1 x 13 x 14 3 4
P3x 0 x
f x 2 sec2 x tan x
2 P3 P5
1 x2
(a) n 3, c 0
sec2 x
−
f1 1
2 x3
f 4x
5 1 x 13 x 14 16 128
f x tan x
31.
f 1 0
1 x
f x fx
1 1 P4x 1 x 1 x 12 2 8
f x ln x fx
1 f 1 4
8x 2x
f 4x
29.
0 2 2 1 x x3 x x3 2! 3! 3
(b) n 5, c 0 0 2 2 0 16 5 x x3 x 4 x 2! 3! 4! 5! 2 5 1 x x x3 3 15
P5x 0 x
(c) n 3, c
4
4 x 4 2! 4
2 x 4 4
2
Q3x 1 2 x 12 x
2
16 x 3! 4
8 x 3 4
3
3
150
Chapter 8
Infinite Series
f x sin x
33.
P1x x P3x x 16 x3 1 5 P5x x 16 x3 120 x 1 1 1 P7x x 6 x3 120 x5 5040 x7
(a)
(b)
x
0.00
0.25
0.50
0.75
1.00
sin x
0.0000
0.2474
0.4794
0.6816
0.8415
P1x
0.0000
0.2500
0.5000
0.7500
1.0000
P3x
0.0000
0.2474
0.4792
0.6797
0.8333
P5x
0.0000
0.2474
0.4794
0.6817
0.8417
P7x
0.0000
0.2474
0.4794
0.6816
0.8415
3
P1
P3
P7 f − 2
2
P5 −3
(c) As the distance increases, the accuracy decreases 35. f x arcsin x (a) P3x x (b)
x3 6
(c)
y
π 2
x
0.75
0.50
0.25
0
0.25
0.50
0.75
f x
0.848
0.524
0.253
0
0.253
0.524
0.848
P3x
0.820
0.521
0.253
0
0.253
0.521
0.820
x
−1
1
P3 −
P8
y
y
P4 4
6
x
−4 −3 −2
8
2
−4 −3
−6
41. f x ex 1 x
x 2 x3 2 6
12 0.6042 f x ln x x 1 12 x 12 13 x 13 14 x 14
f 1.2 0.1823 45. f x cos x; f 5x sin x ⇒ Max on 0, 0.3 is 1. R4x ≤
f(x) = ln (x 2 + 1)
1 x
P6 P2
43.
P2
2
f(x) = cos x
2
f
P6
3
6
−6
π 2
39. f x) lnx2 1
37. f x cos x
1 0.35 2.025 105 5!
P8
P4
3
4
f
Section 8.7
47. f x arcsin x; f 4x
151
x6x 2 9 ⇒ Max on 0, 0.4 is f 40.4 7.3340. 1 x 272
7.3340 0.44 0.00782 7.82 103 4!
R3x ≤
49. gx sin x gn1x Rnx ≤
Taylor Polynomials and Approximations
51. f x lnx 1
≤ 1 for all x
f n1x
1 0.3n1 < 0.001 n 1!
By trial and error, n 3.
Rn ≤
1n1n! ⇒ Max on 0, 0.5 is n!. x 1n1
0.5n1 n! < 0.0001 0.5n1 n 1! n1
By trial and error, n 9. (See Example 9.) Using 9 terms, ln1.5 0.4055.
53.
f x ex 1 x
x 2 x3 , x < 0 2 6
ez 4 x < 0.001 4!
R3x
ez x 4 < 0.024
55. The graph of the approximating polynomial P and the elementary function f both pass through the point c, f c and the slopes of P and f agree at c, f c. Depending on the degree of P, the nth derivatives of P and f agree at c, f c.
xe z4 < 0.3936 x <
0.3936 < 0.3936, z < 0 e z4
0.3936 < x < 0 57. See definition on page 607.
59. The accuracy increases as the degree increases (for values within the interval of convergence).
61. (a) f x e x
63. (a) Q2x 1
2x 22 32
(b) R2x 1
2x 62 32
P4x 1 x
1 2 1 3 1 4 x x x 2 6 24
gx xe x Q5x x
x2
1 1 1 5 x x3 x 4 2 6 24
(c) No. The polynomial will be linear. Translations are possible at x 2 8n.
Q5x x P4x (b) f x sin x P5x x
x5 x3 3! 5!
gx x sin x Q6x x P5x x 2 (c) gx
x6 x4 3! 5!
sin x 1 x2 x 4 P5x 1 x x 3! 5!
65. Let f be an even function and Pn be the nth Maclaurin polynomial for f. Since f is even, f is odd, f is even, f is odd, etc. (see Exercise 45). All of the odd derivatives of f are odd and thus, all of the odd powers of x will have coefficients of zero. Pn will only have terms with even powers of x. 67. As you move away from x c, the Taylor Polynomial becomes less and less accurate.
152
Chapter 8
Infinite Series
Section 8.8
Power Series
1. Centered at 0
5.
xn n1
1
n
n0
n→
n1
1n xn
n1 x x n2
lim
n→
9.
7.
un1 1n1xn1 lim n→ un n2
L lim
x
3. Centered at 2
L lim
n→
lim
n→
< 1⇒R1
2x2n n0 2n!
n→
11.
un 1 2x 2n 2! lim n→ un 2x2n2n! lim
n→
2n2
1n x n n n1 n→
n 12
un1 1 lim n→ un n1
n1xn1
n→
n
1n xn
nx x n1
Interval: 1 < x < 1
n
xn
lim
n→
un1 x n1 lim n→ un n 1! lim
n→
n!
n0
1
n diverges.
Therefore, the interval of convergence is 1 < x ≤ 1.
2n!2
n0
lim
n→
x
n
un1 2n 2!x n1 lim n→ un 2n1
2n
2n!x n
lim
n→
Therefore, the series converges only for x 0.
19.
2n 22n 1 x 2
1n1x n 4n n1
n!
xn
x 0 n1
The series converges for all x. Therefore, the interval of convergence is < x < .
n1
17.
1 2
1n When x 1, the alternating series converges. n n1 When x 1, the p-series
x
2x
Since the series is geometric, it converges only if x2 < 1 or 2 < x < 2.
15.
lim
2
n0
lim
2n2x
n2
2xn
2x 0 2n 22n 1 2
Thus, the series converges for all x. R .
un1 2xn1 lim n→ n 12 un
2 x < 1⇒R
L lim
13.
2xn 2 n1 n
Since the series is geometric, it converges only if x4 < 1 or 4 < x < 4.
Section 8.8
21.
1n1x 5n n5n n1
Power Series
153
lim
n→
un1 1n2x 5n1 lim n→ un n 15n1
n5n
1n1x 5n
R5
lim
n→
nx 5 1 x5 5n 1 5
Center: x 5 Interval: 5 < x 5 < 5 or 0 < x < 10 When x 0, the p-series
n1
1 diverges. n
When x 10, the alternating series
1n1 converges. n n1
Therefore, the interval of convergence is 0 < x ≤ 10.
23.
1n1x 1n1 n1 n0
lim
n→
un1 1n2x 1n2 lim n→ un n2
n1
1n1x 1n1
R1
lim
n→
n 1x 1 x 1 n2
Center: x 1 Interval: 1 < x 1 < 1 or 0 < x < 2
1
n 1 diverges by the integral test.
When x 0, the series
n0
When x 2, the alternating series
1n1 converges. n0 n 1
Therefore, the interval of convergence is 0 < x ≤ 2.
25.
x cn1 cn1 n1
lim
n→
27.
un1 x cn lim n→ un cn
cn1
x cn1
Rc
n
n 1 2x
n1
n1
1 xc c
lim
n→
un1 n 12xn lim n→ un n2 lim
Center: x c
n→
Interval: c < x c < c or 0 < x < 2c
When x 0, the series
1
n1
R diverges.
n1
When x 2c, the series
1 diverges.
n1
Therefore, the interval of convergence is 0 < x < 2c.
2xn 12 2x nn 2
1 2
Interval:
n1
n2xn1
1 1 < x < 2 2
1 n When x , the series diverges 2 n 1 n1 by the nth Term Test.
1n1n 1 When x , the alternating series diverges. 2 n1 n1
Therefore, the interval of convergence is
1 1 < x < . 2 2
154
29.
Chapter 8
Infinite Series
x 2n1
2n 1!
n0
lim
n→
un1 x 2n3 lim n→ 2n 3! un lim
n→
2n 1! x 2n1
x2
2n 22n 3
0
Therefore, the interval of convergence is < x < 31.
.
kk 1 . . . k n 1x n n! n1
lim
n→
un1 kk 1 . . . k n 1k nx n1 lim n→ un n 1!
kk 1 .
n! k nx lim x . . k n 1xn n→ n1
R1
When x ± 1, the series diverges and the interval of convergence is 1 < x < 1.
kk 11 2 . .k .n n 1 ≥ 1
. . .
33.
1n1 3 7 11 . . . 4n 1x 3n 4n n1
lim
n→
un1 1n2 3 lim n→ un lim
n→
7 11 .
. . 4n 14n 3x 3n1 4n1
1n1
4n 3 7 11 . . . 4n 1x 3n
4n 3x 3 4
R0
Center: x 3 Therefore, the series converges only for x 3.
35. (a) f x
2 , 2 < x < 2 n
x
(Geometric)
37. (a) f x
n0
(b) fx
22 n
x
n1
, 2 < x < 2
1n1x 1n1 ,0 < x ≤ 2 n1 n0
(d)
2 n
n2
39. g1
n1 2
2x
2
x
n2
, 2 < x < 2
1
n
1
n0
(c) f x
n1
, 2 ≤ x < 2
n0
3
1
n1
x 1n, 0 < x < 2
n0
n 1 2
f x dx
(b) fx
n1
(c) f x
1 1 . . . 3 9
(d)
1
n1
nx 1n1, 0 < x < 2
n1
f x dx
41. g3.1
1n1x 1n2 ,0 ≤ x ≤ 2 n 1n 2 n1
3 3.1
n
diverges. Matches (b)
n0
S1 1, S2 1.33. Matches (c) 43. A series of the form
a x c n
n
n0
is called a power series centered at c.
45. A single point, an interval, or the entire real line.
Section 8.8
1n x 2n1 , < x < n0 2n 1!
47. (a) f x
Power Series
(See Exercise 29.)
1n x 2n , < x < 2n! n0
gx
1n x 2n gx 2n! n0
(b) fx
(c) gx
1n1x 2n1 1n x2n1 1n x 2n1 f x 2n 1! n1 2n 1! n0 n0 2n 1!
(d) f x sin x and gx cos x
y
49.
x 2n n n!
2
n0
y
2nx 2n1 n n1 2 n!
y
2n2n 1x 2n2 2n n! n1
y xy y
2nx 2n x 2n 2n2n 1x 2n2 n n n 2 n! n1 n1 2 n! n0 2 n!
2n 1x 2n 2n2n 1x 2n2 n n! 2 2n n! n1 n0
(a) lim
n0
51. J0x
2n 22n 1x2n 2n 1x2n 2n 1 2n 1 2n1n 1! 2n n!
2n 1x 2n 2n 1 2n 1 0 2n1n 1! n0
1k x 2k 2k 2 k0 2 k!
k→
uk1 1k1 x 2k2 lim 2k2 k→ 2 uk k 1! 2
22k k!2
1k x 2k
lim
Therefore, the interval of convergence is < x < J0
(b)
1
k
k0
J0
4k k!2
2k2k 1x 2k2 2k 22k 1x 2k 1k1 4k k!2 4k1 k 1! 2 k0
k
1
k
1k1
k0
—CONTINUED—
x 2k 2kx 2k1 2k 2 x 2k1 1k1 k1 k 2 4 k! 4 k 1! 2 k0
k1
x 2J0 xJ0 x 2J0
.
1x 2 0 22k 12
1
k1
J0
k→
22k 1 x 2k2 2x 2k2 x 2k2 1k1 k1 1k k 2 k1 4 k 1!k! 4 k 1!k! k0 4 k! k0
1k x2k2 22k 1 2 1 1 1 4k k!2 4k 1 4k 1 k0
1k x 2k2 4k 2 2 4k 4 0 4k k!2 4k 4 4k 4 4k 4 k0
155
156
Chapter 8
Infinite Series
51. —CONTINUED— x2 x4 x6 4 64 2304
(c) P6x 1
1
(d)
0
0
3
−6
1
J0dx
1k x 2k k 2 dx k0 4 k!
x 4 1 k! 2k 1
k
k
k0
6
1
2k1
2
0
1k k 4 k! 22k 1 k0
−5
1
1 1 0.92 12 320
(exact integral is 0.9197304101)
1
53. f x
n
n0
x 2n cos x 2n!
2
x
n
n0
1 1 for 1 < x < 1 1 x 1 x 3
−2
2
−1
−2
n n
n0
(See Exercise 47.)
57.
1 x
55. f x
2 x
1 0
n
n0
(a)
2 34
n
n0
8 3
n
(b)
n0
n0
1 8 1.6 1 38 5
n
8 3
n
n0
1.80
−1
34 2
1 8 0.7272 1 38 11
1.10
−1
6 0
2 N
(c) The alternating series converges more rapidly. The partial sums of the series of positive terms approach the sum from below. The partial sums of the alternating series alternate sides of the horizontal line representing the sum.
(d)
1nx n n2n n0
converges for x 2 but diverges for x 2.
3
n
> M
n0
M 10 N
59. False;
6 0.60
4
100
1000
10,000
9
15
21
61. True; the radius of convergence is R 1 for both series.
Section 8.9
Section 8.9 1. (a)
Representation of Functions by Power Series
Representation of Functions by Power Series
1 12 a 2 x 1 x2 1 r
2 2
1 x
n
3. (a)
1 12 a 2 x 1 x2 1 r
xn
2
n0
n1
n0
2 2 1
x
n
n0
This series converges on 2, 2.
1n xn 2n1 n0
This series converges on 2, 2.
1 x x2 x3 . . . 2 4 8 16 (b) 2 x ) 1 x 1 2 x 2 x x2 2 4 x2 4 x 2 x3 4 8 x3 8 x3 x4 8 16
1 x x2 x3 . . . 2 4 8 16 (b) 2 x ) 1 x 1 2 x 2 x x2 2 4 x2 4 x 2 x3 4 8 x3 8 x3 x4 8 16
5. Writing f x in the form a1 r, we have
7. Writing f x in the form a1 r, we have
1 13 1 2 x 3 x 5 1 13x 5
3 3 a 2x 1 1 2x 1 r
which implies that a 13 and r 13x 5.
which implies that a 3 and r 2x.
Therefore, the power series for f x is given by
Therefore, the power series for f x is given by
1 1 1 ar n x 5 2 x n0 3 3 n0
x 5n
3
n0
n1
3 ar n 32xn 2x 1 n0 n0
n
<
1
n
< x <
1 . 2
3 3 32 a x 2 2 x 1 12x 1 r which implies that a 32 and r 12x. Therefore, the power series for f x is given by
which implies that a 111 and r 211x 3. Therefore, the power series for f x is given by
2x , 2x < 1 or 2
11. Writing f x in the form a1 r, we have
111 a 1 211x 3 1 r
1 1 ar n 2x 5 n0 11 n0
n0
1 1 2x 5 11 2x 3
3
, x 5 < 3 or 2 < x < 8.
9. Writing f x in the form a1 r, we have
x 3
157
3 1 3 ar n x x 2 n0 2 2 n0
2nx 3n , 11n1 n0
17 11 5 or < x < . 2 2 2
n
1n x n 3 x n , n1 2 2 2 n0 n0
112 x 3
n
3
x
< 2 or 2 < x < 2.
158
13.
Chapter 8
Infinite Series
3x 2 1 2 1 1 1 x 2 x 2 x 2 x 1 2 x 1 x 1 12x 1 x Writing f x as a sum of two geometric series, we have 3x 1 x x x 2 n0 2
2
n
1x
n
2
n0
1
n
n0
The interval of convergence is 1 < x < 1 since lim
n→
15.
un1 1 2n1x n1 lim n→ un 2n1
2n
1 2n xn
1 x n.
lim
n→
1 2n1x x. 2 2n1
1 1 2 1 x2 1 x 1 x Writing f x as a sum of two geometric series, we have 2 xn xn 1 1n x n 2x 2n. 1 x 2 n0 n0 n0 n0
The interval of convergence is x 2 < 1 or 1 < x < 1 since lim
17.
n→
un1 2x 2n2 lim x2 . n→ un 2x 2
1 1n x n 1 x n0
1 1n xn 12n x n xn 1 x n0 n0 n0
hx
2 1 1 1n x n xn 1n 1 x n x 2 1 1 x 1 x n0 n0 n0
2 0x 2x2 0x3 2x4 0x5 2x6 . . .
2x
2n,
1 < x < 1 (See Exercise 15.)
n0
19. By taking the first derivative, we have d 1 x 12 dx
d 1 1 . Therefore, dx x 1 x 12
1 x 1 nx n
n
n0
n
n1
n1
1
n1
n 1 x n, 1 < x < 1.
n0
21. By integrating, we have lnx 1
1 dx lnx 1. Therefore, x1
1n x n dx C
n0
1n x n1 , 1 < x ≤ 1. n1 n0
To solve for C, let x 0 and conclude that C 0. Therefore, lnx 1
23.
1n x n1 , 1 < x ≤ 1. n1 n0
1 1n x 2n 1n x 2n, 1 < x < 1 x 2 1 n0 n0
25. Since,
1 1 1 1 1n x n, we have 2 1n 4x 2n 1n 4n x 2n 1n 2x2n, < x < . x 1 n0 4x 1 n0 2 2 n0 n0
Section 8.9 x2 x 2 x3 ≤ lnx 1 ≤ x 2 2 3
27. x
x x
5
S3 f −4
x2 2
lnx 1 8
x
S2 −3
29. gx x, line, Matches (c)
31. gx x
1
In Exercises 35 and 37, arctan x
n
n0
35. arctan
Representation of Functions by Power Series
x2 2
x3 3
0.0
0.2
0.4
0.6
0.8
1.0
0.000
0.180
0.320
0.420
0.480
0.500
0.000
0.180
0.336
0.470
0.588
0.693
0.000
0.183
0.341
0.492
0.651
0.833
x3 x5 , Matches (a) 3 5
33. f x arctan x is an odd function (symmetric to the origin)
x2n1 . 2n 1
1 1 1 142n1 1n 1 . . . 1n 2n1 4 4 n0 2n 1 2n 1 4 192 5120 n0
1 1 Since 5120 < 0.001, we can approximate the series by its first two terms: arctan 14 14 192 0.245.
arctan x 2 x 4n1 1n x 2n 1 n0
37.
12
0
arctan x 2 x 4n2 dx 1n x 4n 22n 1 n0
arctan x 2 1 1 1 . . . dx 1n 4n2 8 1152 x 4n 2 2n 1 2 n0
Since
1 < 0.001, we can approximate the series by its first term: 1152
In Exercises 39 and 41, use
39. (a)
12
0
arctan x 2 dx 0.125 x
1 x n, x < 1. 1 x n0
d d 1 1 1 x2 dx 1 x dx
x nx , x < 1 n
n0
n1
n1
(b)
x x nx n1 nx n, x < 1 2 1 x n1 n1
(c)
1x 1 x nx n1 x n, x < 1 1 x2 1 x2 1 x2 n1
2n 1x , x < 1 n
n0
(d)
x1 x x 2n 1x n 2n 1x n1, x < 1 1 x2 n0 n0
41. Pn En
12
n1
n
nPn
1 2 n
n1
n
1 1 n 2 n1 2
159
n1
1 1 2 2 1 12 2 Since the probability of obtaining a head on a single toss is 12 , it is expected that, on average, a head will be obtained in two tosses.
160
Chapter 8
Infinite Series
43. Replace x with x.
45. Replace x with x and multiply the series by 5.
47. Let arctan x arctan y . Then, tanarctan x arctan y tan tanarctan x tanarctan y tan 1 tanarctan x tanarctan y xy tan 1 xy arctan
49. (a) 2 arctan
1x xyy . Therefore, arctan x arctan y arctan1x xyy for xy 1.
21 2 4 1 1 1 arctan arctan arctan arctan 3 2 2 2 1 1 22
(b) 8 arctan
1 1 1 0.5 0.5 0.5 4 arctan 8 2 7 2 3 5 7
3
5
lnx 1
1
417 1 73
3
1n1xn 1n xn1 n 1 n n0 n1
1
n1
1 75 1 77 3.14 5 7
n1
1n12 5n 2n 5n n n1 n
1n1xn . n n1
n1
n1
7
53. From Exercise 51, we have
51. From Exercise 21, we have
Thus,
1 1 4 1 4 3 1 7 25 arctan arctan arctan arctan arctan 1 arctan 2 7 3 7 1 4 31 7 25 4
2 arctan
ln
25 1 ln 57 0.3365.
1n11 2n 1 n 2 n n1 n
ln
12 1 ln 23 0.4055
55. From Exercise 54, we have
1
n
n0
1 1 22n1 1 1n arctan 0.4636. 22n12n 1 n0 2n 1 2
57. The series in Exercise 54 converges to its sum at a slower rate because its terms approach 0 at a much slower rate.
59. f x
1
n1
n1
f 0.5
n1
n1
Section 8.10
Taylor and Maclaurin Series
1. For c 0, we have: f x e2x f nx 2n e2x ⇒ f n0 2n e2x 1 2x
2xn 4x 2 8x3 16x 4 . . . 2! 3! 4! n0 n!
x 1n , 0 < x ≤ 2 n
1n1
0.5n 1 2n n n n1
1 2n 0.6931 n
Section 8.10
Taylor and Maclaurin Series
3. For c 4, we have: f x cosx
f
4
f x sinx
f
4 22
f x cosx
f
4 22
f x sinx
f
f 4x cosx
f 4
2
2
4
2
4
2
2 2
and so on. Therefore, we have: cos x
f n 4 x 4 n n! n0
2
2
1 x 4 x 2! 4
2
x 4 3 x 4 4 . . . 3! 4!
1nn1 2 x 4 n . n! n0
2
2
[Note: 1nn1 2 1, 1, 1, 1, 1, 1, 1, 1, . . .] 5. For c 1, we have, f x ln x f x
1 x
f 1 1 1 x2
f x
f 1 1
2 x3
f x
f 4x f 5x
f 1 0
f 1 2
6 x4
f 41 6
24 x5
f 51 24
and so on. Therefore, we have: ln x
f n1x 1n n! n0
0 x 1 x 1
1
n0
n
x 12 2x 13 6x 14 24x 15 . . . 2! 3! 4! 5!
x 12 x 13 x 14 x 15 . . . 2 3 4 5 x 1n1 n1
161
162
Chapter 8
Infinite Series
7. For c 0, we have: f x sin 2x
f 0 0
f x 2 cos 2x
f 0 2
f x 4 sin 2x
f 0 0
f x 8 cos 2x
f 0 8
f 4x
f 40 0
16 sin 2x
f 5x 32 cos 2x
f 50 32
f 6x 64 sin 2x
f 60 0
f 7x 128 cos 2x
f 70 128
and so on. Therefore, we have: sin 2x
f n0xn 0x 2 8x3 0x 4 32x5 0x6 128x7 . . . 0 2x n! 2! 3! 4! 5! 6! 7! n0
1n2x2n1 8x3 32x5 128x7 . . . 3! 5! 7! 2n 1! n0
2x
9. For c 0, we have: f x secx
f 0 1
f x secxtanx
f 0 0
f x sec3x secxtan2x f x 5 f 4x
5
secx
f 0 1
xtanx secx
sec3
x
f 0 0
tan3
x 18
sec5
sec3
x
x secx
tan2
x
tan4
f 40 5
f n0xn x 2 5x 4 . . . 1 n! 2! 4! n0
11. The Maclaurin series for f x cos x is
1x2n . n0 2n!
Because f n1x ± sin x or ± cos x, we have f n1z ≤ 1 for all z. Hence by Taylor’s Theorem,
0 ≤ Rnx Since lim
n→
f n1z
n 1!
xn1 ≤
x . n 1! n1
xn1 0, it follows that Rnx → 0 as n → . Hence, the Maclaurin series for cos x converges to cos x for all x. n 1!
kk 1x 2 kk 1k 2x3 . . . , we have 2! 3! 23x 2 234x3 2345x 4 . . . 1 x2 1 2x 1 2x 3x 2 4x3 5x 4 . . . 2! 3! 5!
13. Since 1 xk 1 kx
1 n 1x .
n0
n
n
Section 8.10
15.
1
2 1 2
121 2x
4 x 2
and since 1 x1 2 1
1n 1 1 1 1 2 2 4 x n1
17. Since 1 x1 2 1
ex
xn
3 5.
x3
x4
. . 2n 3xn
2n n!
1n1 1 x2 2 n2
x2
1n 1 3 5 . . . 2n 1xn , we have 2n n! n1
. . 2n 1x 22n 1n 1 1 2 n! 2 n1 n
1n1 1 x 2 n2
we have 1 x 21 2 1
19.
3 5.
Taylor and Maclaurin Series
3 5.
3 5.
. . 2n 1x2n . n!
3n1
2
(Exercise 14)
. . 2n 3x 2n
2n n!
.
x5
n! 1 x 2! 3! 4! 5! . . .
n0
ex 2 2
x 2n x 2 2n x2 x4 x6 x8 1 2 3 4 . . . n n! 2 2 2! 2 3! 2 4! n0 n0 2 n!
21. sin x sin 2x
23.
1n x 2n1 x3 x5 x7 x . . . 3! 5! 7! n0 2n 1!
1n 22n1x2n1 1n2x2n1 8x3 32x5 128x7 . . . 2x 2n 1! 2n 1! 3! 5! 7! n0 n0
1n x 2n x2 x 4 . . . 1 2n! 2! 4! n0
cos x
1n x3n 1n x3 22n x3 x6 1 . . . 2n! 2n! 2! 4! n0 n0
cos x3 2
25.
ex 1 x
x2 x3 x4 x5 . . . 2! 3! 4! 5!
ex 1 x
x2 x3 x4 x5 . . . 2! 3! 4! 5!
e x ex 2x
2x3 2x5 2x7 . . . 3! 5! 7!
1 x x3 x2n1 x5 x7 e ex x . . . 2 3! 5! 7! n0 2n 1!
sinhx
1 27. cos2x 1 cos2x 2
31.
29. x sin x x x
1 2x2 2x4 2x6 . . . 11 2 2! 4! 6!
x2
1n2x2n 1 1 2 2n! n0
sin x x x3 3! x5 5! . . . x x 1
x2 x4 . . . 2! 4!
1nx2n
2n 1!, x 0
n0
x3 x5 . . . 3! 5!
x6 x4 . . . 3! 5!
1nx2n2 n0 2n 1!
163
164
Chapter 8
Infinite Series
eix 1 ix
33.
eix 1 ix
ix2 ix3 ix4 . . . x 2 ix3 x 4 ix5 x6 1 ix . . . 2! 3! 4! 2! 3! 4! 5! 6! ix2 ix3 ix4 . . . x 2 ix3 x 4 ix5 x6 1 ix . . . 2! 3! 4! 2! 3! 4! 5! 6!
2ix3 2ix5 2ix7 . . . 3! 5! 7!
eix eix 2ix
1n x2n1 x3 x5 x7 eix eix x . . . sinx 2i 3! 5! 7! n0 2n 1!
35. f x ex sin x
14
1x
x2 2
x3 6
x4
. . .
24
x x2
x x2
x3 x5 . . . 3 30
x 6 120 . . . x3
x5
P5 f
x5 x5 x3 x3 x4 x4 x5 . . . 2 6 6 6 120 12 24
−6
37. hx cos x ln1 x
1
x2 2
x4 24
4
. . .
x
x2 2
x3 3
x4 4
x5 5
. . .
P5
−3
x2 x3 x3 x4 x4 x5 x5 x5 . . . 2 3 2 4 4 5 6 24
x
x2
39. gx
2
6
3x5 40
9
h
x
x3
6 −2
−4
. . .
sin x . Divide the series for sin x by 1 x. 1x
5x2 5x4 6 6 x3 x5 0x4 . . . 1 x x 0x2 6 120 x x2 x3 x2 6 x2 x3 5x3 0x4 6 5x4 5x3 6 6 x5 5x4 6 120 5x4 5x5 6 6
gx x x2 4
x x2
5x 3 5x 4 . . . 6 6
g −6
6
P4 −4
41. y x 2
x4 x3 x x x sin x. 3! 3!
Matches (a)
43. y x x 2 Matches (c)
x3 x2 x 1x xex. 2! 2!
Section 8.10
x
45.
0
0
1n1t2n2 dt n 1! n0
x
0
1n1 t2n3
2n 3n 1!
x
0
n0
1n1x 2n3
2n 3n 1!
n0
1n x 1n1 x 12 x 13 x 14 . . . x 1 n1 2 3 4 n0
47. Since ln x
1 1 1 . . . 1 1n1 0.6931. 2 3 4 n n1
we have ln 2 1
49. Since ex
1n t2n 1 dt n! n0
x
et 1 dt 2
Taylor and Maclaurin Series
xn
x2
(10,001 terms)
x3
n! 1 x 2! 3! . . . ,
n0
we have e2 1 2
2n 22 23 . . . 7.3891. 2! 3! n0 n!
(12 terms)
51. Since cos x
1n x 2n x2 x 4 x6 x8 1 . . . 2n ! 2! 4! 6! 8! n0
1 cos x
1n x 2n2 x6 x8 x2 x 4 . . . 2! 4! 6! 8! n0 2n 2!
1n x 2n1 x x3 x5 x7 1 cos . . . x 2! 4! 6! 8! n0 2n 2!
we have lim
x→0
1
53.
0
1x 2n1 1 cos x lim 0. x→0 n0 2n 2! x
sin x dx x
1
0
1nx 2n dx n0 2n 1!
1n x 2n1
2n 12n 1! n0
1
0
1n
2n 12n 1!
n0
Since 1 7 7! < 0.0001, we have
1
0
sin x 1 1 dx 1 . . . 0.9461. x 3 3! 5 5!
Note: We are using lim x→0
2
55.
sin x 1. x
2
x cos x dx
0
0
1nx4n1 2 dx 2n! n0
Since 219 2 766,080 < 0.0001, we have
1
x cos x dx 2
0
0.3
57.
0.3
1 x3 dx
0.1
0.1
Since
1 7 56 0.3
n0
1n x4n3 2 4n 3 2n! 2
1n 2x4n3 2 n0 4n 32n!
0
23 2 27 2 211 2 215 2 219 2 0.7040. 3 14 264 10,800 766,080
0.3
x 5x . . . 1 x2 x8 16x 5x128 . . . dx x x8 56x 160 1664 3
6
9
12
4
7
< 0.0001, we have
1 x3 dx
2
0
0.17
0.3
0.1
2
0.3 0.1 80.3 1
4
0.14
1 0.37 0.17 0.2010. 56
10
13
0.1
165
166
Chapter 8
Infinite Series
59. From Exercise 19, we have 1 2
1
ex 2 dx 2
0
1 2
1
0
1 1n x 2n dx n n! 2 2 n0
61.
f x x cos 2x P5x x
2x3
1 2
x 21 n!2n 1
n
2n1
1
n
n0
0
1 2 1 3 2 1
1n 4n x2n1 2n! n0
63.
1n 1 n 2 n0 2 n!2n 1
1
2! 5
2
1
0.3414.
f x x ln x, c 1
x 13 x 14 71x 15 24 24 1920
P5x x 1
2x5 3
3! 7
23
3
2
g
P5
f −3
−2
3
4
P5 −2
−2
The polynomial is a reasonable approximation on the interval 14 , 2 .
The polynomial is a reasonable approximation on the interval 34 , 34 .
67. (a) Replace x with x.
65. See Guidelines, page 636.
(b) Replace x with 3x. (d) Replace x with 2x, then replace x with 2x, and add the two together.
(c) Multiply series by x.
69. y tan
g g kx x 2 ln 1 kv0 cos k v0 cos
2
3
gx g kx 1 kx kv0 cos k2 v0 cos 2 v0 cos
tan x
gx gx gx 2 gkx3 gk2x 4 . . . kv0 cos kv0 cos 2v02 cos2 3v03 cos3 4v04 cos4
tan x
gx2 kgx3 k2gx 4 . . . 2 3 3 2v0 cos 3v0 cos 4v04 cos4
71. f x
e0,
1 kx 3 v0 cos
tan x
1 kx 4 vo cos
4
. . .
2
x0 x0
1 x2
,
f x f 0 e1 x 0 lim x→0 x0 x 2
(a)
(b) f 0 lim
y
x→0
2
e1 x . Then x→0 x 2
Let y lim
1 −3 −2 −1
x 1
2
3
ln y lim ln x→0
e x lim x1 ln x lim 1 x x 1 x2
x→0
2
x→0
Thus, y e 0 and we have f 0 0. (c)
f n0 n f 0x f 0x 2 . . . x f 0 0 f x n! 1! 2! n0
This series converges to f at x 0 only.
73. By the Ratio Test: lim
n →
xn1 n! x 0 which shows that xn converges for all x. nlim → n 1 n 1! xn n0 n!
2
2
ln x
.
Review Exercises for Chapter 8
Review Exercises for Chapter 8 1. an
7. an
1 n!
2 3. an 4 : 6, 5, 4.67, . . . n Matches (a)
5n 2 n
9. lim
n→
8
5. an 100.3n1: 10, 3, . . . Matches (d)
n1 0 n2
11. lim
n→
n3 n 1 2
Converges
0
12 0
The sequence seems to converge to 5. lim an n→ lim
n→
5n 2 n
2 5 n
n→ lim 5
13. lim n 1 n lim n 1 n n→
n→
n 1 n n 1 n
sinn 0 n Converges
lim
n→
1 n 1 n
17. An 5000 1
15. lim
n→
n 1, 2, 3
0
0.05 4
n
Converges
50001.0125n
(a) A1 5062.50
A5 5320.41
A2 5125.78
A6 5386.92
A3 5189.85
A7 5454.25
A4 5254.73
A85522.43
(b) A40 8218.10 19. (a)
(b)
k
5
10
15
20
25
Sk
13.2
113.3
873.8
6448.5
50,500.3
(c) The series diverges geometric r 32 > 1
21. (a)
0
12
− 10
k
5
10
15
20
25
Sk
0.4597
0.4597
0.4597
0.4597
0.4597
(c) The series converges by the Alternating Series Test.
120
23. Converges. Geometric series, r 0.82, r < 1.
(b)
1
0
12
− 0.1
25. Diverges. nth Term Test. lim an 0. n→
167
168
27.
Chapter 8
3 2
Infinite Series
n
29.
n0
2
1 n
n0
2 Geometric series with a 1 and r 3 .
S
1 n 1 n 1 n 3 n0 2 n0 3 1 1 3 1 2 1 1 2 1 1 3 2 2
1 1 a 3 1 r 1 2 3 1 3
0.090.01
31. 0.09 0.09 0.0009 0.000009 . . . 0.091 0.01 0.0001 . . .
n
n0
33. D1 8
35. See Exercise 86 in Section 8.2.
D2 0.78 0.78 160.7
A
D 8 160.7 160.72 . . . 160.7n . . . 8
160.7
n
8
n0
37.
b→
16 4513 meters 1 0.7
ln x 1 3x 9x 3
3
39.
1
1 3 2n n n1
43.
1 n3 2n n3 2 lim 1 n→ n→ n3 2n 1 n3 2
n1
1
3 2 ,
2
1 1 1 2 n n n1 n1 n
3 5 . . . 2n 1 2 4 6 . . . 2n
1
1 3 5 . . . 2n 1 2 4 6 . . . 2n
an
1 1 1 > 32 54 . . . 2n 2n 2 2n 2n
By a limit comparison test with the convergent p-series
1
n1
lim
n
200e0.062 1 e0.06 12 1
Since the second series is a divergent p-series while the first series is a convergent p-series, the difference diverges.
By the Integral Test, the series converges.
n
n1
1 1 9 9
0
Pe rt 1 er 12 1
$5087.14
b
x4 lnx dx lim
1
41.
0.09 1 1 0.01 11
1
1
1
2n 2 n diverges (harmonic series),
Since
the series converges.
n1
n1
so does the original series.
45. Converges by the Alternating Series Test (Conditional convergence)
49.
n
e
n1
lim
n→
47. Diverges by the nth Term Test
51.
n2
an1 n1 lim n12 n→ e an lim
n→
lim
n→
n 1 2n1n
2 en
2
en
2
en n
e 1 n n 1 2n1
01 0 < 1 By the Ratio Test, the series converges.
2n
n
n1
lim
n→
3
an1 2n1 lim n→ n 13 an lim
n→
3
n3
2n
2n 2 n 13
Therefore, by the Ratio Test, the series diverges.
Review Exercises for Chapter 8
53. (a) Ratio Test: lim
n→
an1 n 13 5n1 lim n→ an n3 5n
n n 135 53 < 1
lim
n→
Converges (b)
(c)
x
5
10
15
20
25
Sn
2.8752
3.6366
3.7377
3.7488
3.7499
(d) The sum is approximately 3.75.
4
0
12
−1
55. (a)
1
N
x2
1x
dx
N
1 N
N
5
10
20
30
40
n
1.4636
1.5498
1.5962
1.6122
1.6202
0.2000
0.1000
0.0500
0.0333
0.0250
5
10
20
30
40
n
1.0367
1.0369
1.0369
1.0369
1.0369
0.0004
0.0000
0.0000
0.0000
0.0000
N
1 2
n1
N
(b)
N
1 1 dx 4 x5 4x
N
1 4N4
1 dx x2
N N
1 5
n1
N
1 dx x5
The series in part (b) converges more rapidly. The integral values represent the remainders of the partial sums. 57. f x ex 2
f 0 1
1 fx ex 2 2 1 f x ex 2 4 1 fx ex 2 8
f0 f 0
1 4
f0
P3x f 0 f0x f 0
1 2
1 8
x2 x3 f 0 2! 3!
1 1 x2 1 x3 1 x 2 4 2! 8 3! 1 1 1 1 x x2 x3 2 8 48
59. sin95 sin
95 95 95 95 0.996 95180 95180 180 3! 180 5! 180 7! 180 9!
61. ln1.75 0.75
3
3
5
5
7
7
9
9
0.752 0.753 0.754 0.755 0.756 . . . 0.7515 0.560 2 3 4 5 6 15
169
170
Chapter 8
Infinite Series
63. f x cos x, c 0 Rnx
f n1z n1 x n 1!
f n1z
x n1 n 1!
≤ 1 ⇒ Rnx ≤
(a) Rnx ≤
0.5n1 < 0.001 n 1!
(b) Rnx ≤
This inequality is true for n 6.
This inequality is true for n 4. (c) Rnx ≤
0.5n1 < 0.0001 n 1!
(d) Rnx ≤
10 x
2n1 < 0.0001 n 1!
This inequality is true for n 10.
This inequality is true for n 5.
65.
1n1 < 0.001 n 1!
n
n0
Geometric series which converges only if x 10 < 1 or 10 < x < 10.
67.
1nx 2n n 12 n0
lim
n→
69.
un1 1n1x 2n1 lim n→ un n 22
n 12
1nx 2n
x2
n!x 2
n
n0
lim
n→
which implies that the series converges only at the center x 2.
R1 Center: 2 Since the series converges when x 1 and when x 3, the interval of convergence is 1 ≤ x ≤ 3.
y
71.
x2n n!2
1
n
4n
n0
y
1n12n 2x2n1 1n2nx2n1 n 2 4 n! 4n1n 1! 2 n1 n0
y
1n12n 22n 1x2n 4n1n 1! 2 n0
x2y xy x2y
1n12n 2x2n2 1n12n 22n 1x2n2 x2n1 1n n 2 n1n 1! 2 n1n 1! 2 4 4 4 n! n0 n0 n0
1
n1
n0
1n14n 12 1 1n n x 2n2 4n1n 1! 2 4 n! 2
1n11 1 1n n x 2n2 0 4nn! 2 4 n! 2
n0
73.
2 2 3 a 3 x 1 x 3 1 r
3 3
n0
2 x
n
n0
n0
2n 22n 1 1n12n 2 1n n1 n 2 x 2n2 n1 2 2 4 n 1! 4 n 1! 4 n!
1n12n 22n 1 1 1 1n n x 2n2 4n1n 1! 2 4 n! 2
n0
2x n 3n1
un1 n 1!x 2n1 lim n→ un n!x 2n
75. Derivative:
2nx n1 n1 n1 3
Review Exercises for Chapter 8 2x 4 8 2 77. 1 x x2 x3 . . . 3 9 27 n0 3
n
1 3 , 1 2x 3 3 2x
3 3 < x < 2 2
f x sinx
79.
fx cosx f x sinx f x cosx, . . . sinx
f nxx 3 4 n n! n0
2
2
2
2
x 34 2 22!x 34
81. 3x eln3x ex ln3 and since ex
2
2 1nn1 2x 3 4 n . . . 2 n0 n!
xn
n!, we have
f x
83.
n0
3x
x ln 3n n! n0
1 x ln 3
1 x
fx x2 ln2 3 x3 ln3 3 x 4 ln4 3 . . . . 2! 3! 4!
f x
1 x2
2 x3
6 fx 4, . . . x f n1x 1 n 1 x n0 n!
85. 1 xk 1 kx
n!x 1n x 1n n! n0 n0
kk 1x2 kk 1k 2x3 . . . 2! 3!
x 1 54 5x2 1 54 59 5x3 . . . 5 2! 3!
1 x1 5 1
1 1 4x2 1 4 9x3 . . . 1 x 2 5 5 2! 533!
ln x
87.
1n14 x 5 n2
1
x 2 6 3 . . . x2 x 5 25 125
1
n1
n1
ln
x 1n , n
54 1 5 4n 1
n1
n1
89.
ex
xn
n!,
n0 n
n1
1
. . 5n 6xn
5nn!
0 < x ≤ 2
n1
9 14 .
1
1 0.2231 4nn
e1 2
1
< x <
2 n! 1.6487
n0
n
171
172
Chapter 8
cos x
91.
Infinite Series
1
n
x2n , < x < 2n!
n
22n 0.7859 32n2n!
n0
cos
23 1 n0
95. (a) f x e2x
f 0 1
(b) ex
f0 2
f x 4e2x
f 0 4
f x 8e2x
f 0 8
e2x 1 2x
4x2 8x3 . . . 2! 3!
1 2x 2x2
x
0
e2x
1 x x2 x6 . . . 2
3
x2 x2 4 x3 x3 x3 x3 . . . 1 2x 2x 2 x 3 . . . 2 2 6 6 2 2 3
1nt 2n1 n0 2n 1!
1 1n t n 1 t n0
99.
1 sin t t n0 2n 1!
sin t dt t
n t 2n
ln1 t
1n t 2n1
x
2n 12n 1!
0
n0
1nx 2n1
2n 12n 1!
x
n0
arctan x x
101.
0
x3 x5 x7 x9 . . . 3 5 7 9
x 52 x 92 x132 x172 . . . arctan x x 3 5 7 9 x lim
x→0
arctan x 0 x
1 arctan x 2x 1 x2 lim lim 0. By L’Hôpital’s Rule, lim x→0 x→0 x→0 1 x2 1 x 2x
Problem Solving for Chapter 8 1. (a) 1
3 29 427 . . . 3 3 1
1
1
1 2
n0
1 2 (b) 0, , , 1, etc. 3 3 (c) lim Cn 1 n→
3 3
n0
2xn 4x 2 8x 3 . . . 1 2x n! 2! 3! n0
4 1 2x 2x 2 x 3 . . . 3
x2 x3 . . . 2 6
1 x x x2
xn
4 3 . . . x 3
(c) e2x ex ex 1 x
sin t
n!
n0
fx 2e2x
97.
93. The series for Exercise 41 converges very slowly because the terms approach 0 at a slow rate.
1 2
n
110
n
13 1 1 23
1n t n1 1 dt 1t n1 n0
1n t n lnt 1 t n0 n 1
lnt 1 dt t
1n t n1 2 n0 n 1
x
0
1n x n1 2 n0 n 1
Problem Solving for Chapter 8 nn 1 . 2
3. If there are n rows, then an For one circle,
3 2
1
a1 1 and r1
1 3 3 2
3
6
1
r1
23 1 2
For three circles, a2 3 and 1 23r2 2r2 r2
1 2 23
2r2 1
r2
3 r2
For six circles, a3 6 and 1 23r3 4r3 r3
1 23 4
2r3 1
Continuing this pattern, rn Total Area rn2an An lim An
n→
5. (a)
a x n
n
2
1
4
r3
3 r3
1 . 23 2n 1
23 12n 1
2
nn 1 2
nn 1 2 2 3 2n 12
8
1 2x 3x2 x3 2x4 3x5 . . . 1 x3 x6 . . . 2x x4 x7 . . . 3x2 x5 x8 . . . 1 x3 x6 . . . 1 2x 3x2 1 2x 3x2
1 1 x3
R 1 because each series in the second line has R 1. (b)
a x n
n
a0 a1x . . . ap1x p1 a0 x p a1x p1 . . . . . . a01 x p . . . a1 x1 x p . . . . . . a p1 x p11 x p . . . a0 a1 x . . . a p1x p11 x p . . . a0 a1 x . . . ap1 x p1
R1
1 . 1 xp
173
174
Chapter 8
Infinite Series
ex 1 x
7.
xex x x2
x2 . . . 2! x n1 x3 . . . 2! n0 n!
xex dx xex ex C
x n2
n 2n!
n0
Letting x 0, C 1. Letting x 1,
1
1
1
n 2n! 2 n 2n!.
1
n0
n1
1
1
n 2n! 2.
Thus,
n1
9. Let a1
0
sin x dx, a2 x
2
sin x dx, a3 x
3
2
sin x dx, etc. x
Then,
0
sin x dx a1 a2 a3 a4 . . . . x
Since lim an 0 and an1 < an, this series converges. n→
11. (a) a1 3.0 a2 1.73205 a3 2.17533 a4 2.27493 a5 2.29672 a6 2.30146 lim an
n→
1 13 [See part (b) for proof.] 2
(b) Use mathematical induction to show the sequence is increasing. Clearly, a2 a a1 aa > a a1. Now assume an > an1. Then an a > an1 a an a > an1 a
an1 > an. Use mathematical induction to show that the sequence is bounded above by a. Clearly, a1 a < a. Now assume an < a. Then a > an and a 1 > 1 implies aa 1 > an1 a2 a > an a2 > an a a > an a an1. Hence, the sequence converges to some number L. To find L, assume an1 an L: L a L ⇒ L2 a L ⇒ L2 L a 0 L
1 ± 1 4a . 2
Hence, L
1 1 4a . 2
Problem Solving for Chapter 8
13. (a)
1
2
n1
n 1n
S1
1 1 1 1 1 . . . 211 221 231 241 251 1 1 20
S1 1
1 9 8 8
S3
9 1 11 8 4 8
S4
11 1 45 8 32 32
S5
1 47 45 32 16 32
2n 1 21 an1 n1 1n 1 1 1n 1 an 2 2 n
(b)
n
This sequence is 18, 2, 18, 2, . . . which diverges. (c)
2 n
1 n 1n
2 2
21 1
n
1n
n
1 1 n 1 1 n 12 → 1 and n 2 → 1. → < 1 converges because 21 2, 2, 2, 2, . . . and n 1 2 2 n
15. S6 130 70 40 240 S7 240 130 70 440 S8 440 240 130 810 S9 810 440 240 1490 S10 1490 810 440 2740
175
C H A P T E R 9 Conics, Parametric Equations, and Polar Coordinates Section 9.1
Conics and Calculus . . . . . . . . . . . . . . . . . . . . 424
Section 9.2
Plane Curves and Parametric Equations . . . . . . . . . . 434
Section 9.3
Parametric Equations and Calculus
Section 9.4
Polar Coordinates and Polar Graphs . . . . . . . . . . . . 444
Section 9.5
Area and Arc Length in Polar Coordinates . . . . . . . . .452
Section 9.6
Polar Equations of Conics and Kepler’s Laws . . . . . . . 458
Review Exercises
. . . . . . . . . . . . 439
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 469
C H A P T E R 9 Conics, Parametric Equations, and Polar Coordinates Section 9.1
Conics and Calculus
Solutions to Even-Numbered Exercises 2. x2 8y
4.
Vertex: 0, 0
Center: 2, 1 Ellipse Matches (b)
p2 > 0 Opens upward Matches graph (a).
6.
x 22 y 12 1 16 4
x2 y2 1 9 9
8.
Circle radius 3. Matches (g)
x 22 y2 1 9 4 Hyperbola Center: 2, 0 Horizontal transverse axis. Matches (d)
10. x2 8y 0
12. x 12 8 y 2 0
x2 42y Vertex: 0, 0 Focus: 0, 2
x 12 42 y 2
y
(0, 0) −8
x
−4
8
4
y
Focus: 1, 4
−4
Directrix: y 2
Vertex: 1, 2 Directrix: y 0
−8
−8
x
−4
8
4 −4
(1, − 2) −12
−8 −12
14. y 2 6y 8x 25 0
16. y 2 4y 8x 12 0
y 2 6y 9 8x 25 9
y 2 4y 4 8x 12 4
y 32 42x 2
y 22 42x 2
Vertex: 2, 3 Focus: 4, 3
8
Focus: 0, 2
Directrix: x 0
4
Directrix: x 4
−20 −16 −12
−8
x
−4
(− 2, − 3)
y 4 2 −6
−4
−2
−12
x 4
(2, − 2) −4
−8
424
Vertex: 2, 2
y
−6 −8
6
Section 9.1 y 16x2 8x 6 16x2 8x 16 10
18.
x2 2x 1 8y 9 1
x 12 42 y 1
6y 10 x 42
x 42 6y 53 x 42 4
425
20. x2 2x 8y 9 0
6y x 42 10
32
Conics and Calculus
Vertex: 1, 1
2
Focus: 1, 3
y 5 3
Vertex: 4, 53
−8
10
Directrix: y 1
4
Focus: 4, 16
−10
Directrix: y
−2
19 6
10
−4
x 12 42 y 2
22.
24. Vertex: 0, 2
y 22 42x 0
x2 2x 8y 15 0
y 2 8x 4y 4 0 y 4 x 22 4x x2
26.
28. From Example 2: 4p 8 or p 2
x 4x y 0
Vertex: 4, 0
2
x 42 8 y 0 x2 8x 8y 16 0 30. 5x2 7y 2 70
y 6
y2 x2 1 14 10 a2
14,
b2
32.
1 3 a2 1, b2 , c2 4 4
4 2
10,
c2
4
−6
x
−2
2
Center: 0, 0
−4
Foci: ± 2, 0
−6
x 22 y 42 1 1 14
4
Center: 2, 4
6
Foci:
Vertices: ± 14, 0
2
, 4
3
e
2
16x2 25y2 64x 150y 279 0 16x2 4x 4 25y2 6y 0 279 64 225 10
x 22 y 32 1 58 25
y 1 x
−1
1
2
3
4
−1
5 2 9 a2, , b2 , c2 a2 b2 8 5 40
−2
Center: 2, 3
−4
Foci:
2
Vertices: e
±
2
c 3 a 5
310 , 3 20 ±
10
4
, 3
−3
(2, −3)
−5
−4
−3
−2
x
−1 −1
(− 2, − 4)
−2 −3 −4
Vertices: 1, 4, 3, 4
14 2 e 7 14
34.
2 ±
3
y
−5
426
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
36x2 9y 2 48x 36y 43 0
36.
4 4 9 y 2 4y 4 43 16 36 36 x2 x 3 9 9
x 23 y 2 1 14 1 2
2
1 3 a2 1, b2 , c2 4 4 Center: Foci:
32, 2
32, 2 ± 23
32, 3, 32, 1
Vertices:
3
Solve for y: 9 y 2 4y 4 36x2 48x 43 36
y 22
−4
2
36x2 48x 7 9
−1
1 y 2 ± 36x2 48x 7 3
(Graph each of these separately.)
2x2 y 2 4.8x 6.4y 3.12 0
38.
50x2 25y 2 120x 160y 78 0
12 32 36 256 x 25 y 2 y 78 72 256 250 5 25 5 25
50 x2
x 65 2 y 165 2 1 5 10 a2 10, b2 5, c2 5 Center: Foci:
56, 165
56, 165 ± 5
Vertices:
7
56, 165 ± 10
Solve for y: y2 6.4y 10.24 2x2 4.8x 3.12 10.24
−7
5 −1
y 3.22 7.12 4x 2x2 y 3.2 ± 7.12 4x 2x2 40. Vertices: 0, 2, 4, 2
(Graph each of these separately.) 42 Foci: 0, ± 5
1 2 Horizontal major axis
Major axis length: 14
Center: 2, 2
Center: 0, 0
a 2, c 1 ⇒ b 3
c 5, a 7 ⇒ b 24
x 22 y 22 1 4 3
y2 x2 1 24 49
Eccentricity:
Vertical major axis
Section 9.1
44. Center: 1, 2
46.
Vertical major axis
Conics and Calculus
y2 x2 1 25 9 a 5, b 3, c a2 b2 34
Points on ellipse: 1, 6, 3, 2
Center: 0, 0
From the sketch, we can see that h 1, k 2, a 4, b 2
y
y
Foci: ± 34, 0 (1, 2) x
−2
−8
3 Asymptotes: y ± x 5
(3, 2) −4
10 8 6 4 2
Vertices: ± 5, 0
(1, 6) 6
x 12 y 22 1. 4 16
−4 −2 −4 −6 −8 −10
4 −2
48.
427
y 1 2 x 4 2 1 122 52
x 4
8 10
50. y 2 9x2 36x 72 0 y 2 9x2 4x 4 72 36 36
a 12, b 5, c a 2 b2 13
x 22 y2 1 36 4
Center: 4, 1 Vertices: 4, 11, 4, 13 Foci: 4, 14, 4, 12 12 Asymptotes: y 1 ± x 4 5
a 6, b 2, c a2 b2 210 Center: 2, 0 Vertices: 2, 6, 2, 6
Foci: 2, 210, 2, 210
y 20
Asymptotes: y ± 3x 2 y
5 x −5
1
2
6
7
8 5 x
−2 −1
− 20
4
5
−5
52. 9x2 6x 9 4 y2 2y 1 78 81 4 1
54.
9x 2 y 2 54x 10y 55 0 9x 2 6x 9 y 2 10y 25 55 81 25
9x 32 4 y 12 1
y 12 x 32 1 14 19
1
13 1 1 a ,b ,c 2 3 6
Foci:
3, 21, 3, 23
1 3, 1 ± 13 6
2
10 1 a , b 1, c 3 3
1
Center: 3, 5
3
Center: 3, 1 Vertices:
x 32 y 52 1 19 1
y
3 Asymptotes: y 1 ± x 3 2
−3
x
−1 −1
3 ± 13, 5 10 , 5 Foci: 3 ± 3
10
Vertices:
−8
Solve for y: y2 10y 25 9x2 54x 55 25
y 52 9x2 54x 80 y 5 ± 9x2 54x 80 (Graph each curve separately.)
2 0
428
Chapter 9
Conics, Parametric Equations, and Polar Coordinates 58. Vertices: 0, ± 3
3y 2 x 2 6x 12y 0
56.
Asymptotes: y ± 3x
3 y 2 4y 4 x 2 6x 9 0 12 9 3
y 2 x 3 1 1 3 2
2
a 1, b 3, c 2
Vertical transverse axis a3 a Slopes of asymptotes: ± ± 3 b Thus, b 1. Therefore,
6
Center: 3, 2 Vertices: 3, 1, 3, 3
−4
10
Foci: 3, 0, 3, 4
y 2 x2 1. 9 1
−4
Solve for y: 3 y 2 4y 4 x2 6x 12
y 22
x2 6x 12 3
y2 ±
x
2
6x 12 3
(Graph each curve separately.) 62. Center: 0, 0
60. Vertices: 2, ± 3 Foci: 2, ± 5
Vertex: 3, 0
Vertical transverse axis
Focus: 5, 0
Center: 2, 0
Horizontal transverse axis
a 3, c 5, b2 c2 a2 16
a 3, c 5, b2 c2 a2 16
y2 x 22 Therefore, 1. 9 16
Therefore,
64. Focus: 10, 0
66. (a)
3 Asymptotes: y ± x 4
y2 x2 1. 9 16
y 2 x2 1, y 2 2x2 4, 2yy 4x 0, 4 2 y
4x 2x 2y y
Horizontal transverse axis Center: 0, 0 since asymptotes intersect at the origin. c 10
At x 4: y ± 6, y
± 24
6
±
4 3
4 At 4, 6: y 6 x 4 or 4x 3y 2 0 3
b 3 3 Slopes of asymptotes: ± ± and b a a 4 4 c2 a2 b2 100 Solving these equations, we have a2 64 and b2 36. Therefore, the equation is x2 y2 1. 64 36
4 At 4, 6: y 6 x 4 or 4x 3y 2 0 3 (b) From part (a) we know that the slopes of the normal lines must be 34. 3 At 4, 6: y 6 x 4 or 3x 4y 36 0 4 3 At 4, 6: y 6 x 4 or 3x 4y 36 0 4
68. 4x2 y 2 4x 3 0
70. 25x2 10x 200y 119 0
72. y2 x 4y 5 0
A 4, C 1
A 25, C 0
A 0, C 1
AC < 0
Parabola
Parabola
Hyperbola
Section 9.1 2x2 2xy 3y y 2 2xy
74.
Conics and Calculus
9x2 54x 81 36 4 y2 4y 4
76.
2x2 y 2 3y 0
9x2 4y2 54x 16y 61 0
A 2, C 1, AC > 0
A 9, C 4, AC > 0
Ellipse
Ellipse
78. (a) An ellipse is the set of all points x, y, the sum of whose distance from two distinct fixed points (foci) is constant.
x h y k x h y k 1 or 1 a2 b2 b2 a2 2
(b)
2
2
2
82. Assume that the vertex is at the origin. x2 4py
(a)
82 4p
c 80. e , c a2 b2 a
For e 1, the ellipse is elongated.
84. (a) Without loss of generality, place the coordinate system so that the equation of the parabola is x2 4py and, hence,
3 100
y
6400 y y 1600 3 3
2 ⇒ x± 100
(b)
1283 ± 6.53 meters.
dy 1 x1 dx 2
x 2x 9
4 100
)
( 0, 1003 )
( 8, 1003 )
3x 9
2 100
x3
1 100
−8
dy 0 dx
At 0, 0, the slope is 1: y x. At 6, 3, the slope is 2: y 2x 9. Solving for x,
y
(
x2 4x 4y 0 2x 4 4
5 100
3 −8, 100
2p1 x.
Therefore, for distinct tangent lines, the slopes are unequal and the lines intersect.
(b) The deflection is 1 cm when y
0 < e < 1
For e 0, the ellipse is nearly circular.
1600 p 3 x2 4
y 3.
x
−4
4
8
Point of intersection: 3, 3
86. The focus of x2 8y 42y is 0, 2. The distance from a point on the parabola, x, x28, and the focus, 0, 2, is d
x 0 x8 2 . 2
2
2
Since d is minimized when d 2 is minimized, it is sufficient to minimize the function f x x2
4
x2 2 8
3
16
xx
x 2 = 8y
(0, 2)
x x3 x. 4 16
fx 0 implies that x3
y
2 x2 2 . 8
fx 2x 2
429
2
x, x 8
( )
1 −3
−2
x
−1
1 −2
16x 1 0 ⇒ x 0. 2
This is a minimum by the First Derivative Test. Hence, the closest point to the focus is the vertex, 0, 0.
2
3
430
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
88. (a) C 0.0853t2 0.2917t 263.3559 (b)
1 x y2 4
90.
320
1 x y 2
0
1 x2 1
18 0
s
dC 0.1706t 0.2971 (c) dt
4
1
0
y
3
y4 dy 21 4 y 2
2
x −1
2
3
4
1 y4 y2 4 ln y 4 y 2 4
1 420 4 ln 4 20 4 ln 2 4
4 0
25 ln2 5 5.916
1 1
dy
0
2
−1
4
5 4
y2 4
5
The consumption of fruits is increasing at a rate of 0.1706 pounds/year.
h
92. x2 20y y y
94. A 2
x 20 x 10
y12 dy
0
x
0
1
x 10
2
r
dx 2
0
10 23100 x
r
h
4p
r
S 2
4py dy
0
2
2 32
0
4p x100 x2 dx 10
(b) The thumbtacks are located at the foci and the length of string is the constant sum of the distances from the foci.
98.
e 0.0167
Focus Vertex
h
32
0
8 ph32 3
100 r 232 1000 15
96. (a) At the vertices we notice that the string is horizontal and has a length of 2a.
23y
c a c 149,570, 000
c 2,497,819
Focus Vertex
Least distance: a c 147,072,181 km Greatest distance: a c 152,067,819 km
100. e
AP AP
122,000 4000 119 4000 122,000 4000 119 4000
121,881
0.9367 130,119
102.
y2 x2 1 a2 b2 y2 x2 2 2 2 1 2 a a b a y2 x2 2 2 1 2 a a a c2a2 x2 y2 1 a2 a21 e2 As e → 0, 1 e2 → 1 and we have x2 y2 2 1 or the circle x2 y 2 a2. 2 a a
Section 9.1
104.
x2 y2 1 2 4.5 2.52
Conics and Calculus
y
y2 2.5 2
x2 4.52 1
5 ft x
9 x ± 2.52 y 2 5
3 ft
9 ft
V Area of bottomLength Area of topLength V
4.52.5 16 16 2
90
144 5
0.5
9 2.52 y 2 dy (Recall: Area of ellipse is ab.) 5
0
2 y2.52 y 2 2.52 arcsin 2.5 0
0.5
y
1
90
72 1 0.56 2.52 arcsin
318.5 ft3 5 5
106. 9x2 4y 2 36x 24y 36 0 18x 8yy 36 24y 0
8y 24y 18x 36 y
18x 36 8y 24
y 0 when x 2. y undefined when y 3. At x 2, y 0 or 6. Endpoints of major axis: 2, 0, 2, 6 At y 3, x 0 or 4. Endpoints of minor axis: 0, 3, 4, 3 Note: Equation of ellipse is
4
108. (a) A 4
0
x 22 y 32 1 4 9
3 x 3 16 x2 dx x16 x2 16 arcsin 4 2 4
4
V 2
(b) Disk:
0
9 9 16 x2 dx 16 8
4 0
12
16x 31x
4
3
0
48
3 y 16 x2 4 y 1 y2
4
S 22
0
3x 416 x2
1 16169x x 2
2
16161616 x x 9x dx 4 43
3 16 x2 4
2
7x 3 7x256 7x2 256 arcsin 16 87
—CONTINUED—
4
2
2
0
4 0
16 x2
256 7x2
416 x2
dx
3 4
7 3 487 256 arcsin
138.93 4 87
4
0
256 7x2 dx
431
432
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
108. —CONTINUED—
4
V 4
(c) Shell:
0
416 x dx 3 2316 x
x
3
1
2
4
2
2 32
0
64
4 x 9 y 2 3 4y x 3 9 y2 1 x 2
S 22 4
3
0
110. (a)
3
0
1 9916y y 2
2
4 9 y 2 3
99 99y y 16y 2
2
2
dy
4 81 7y 2 dy 9
16 9 27
8 3712 81 ln37 12 81 ln 9 168.53 97
7y81 7y
2
81 ln 7y 81 7y 2
x2 y2 21 2 a b
At P, y
b2 a2
Slope of line through c, 0 and x0, y0: m2
y0 x0 c
x
y0 m. 0
y0 b2x0 2 x0 c a y0 a2y 2 b2x0x0 c 2 0 2 y0 bx a y0x0 c b2x0 y0 1 2 0 x0 c a y0
a2y02 b2x02 b2x0c a2b2 b2x0c b2a2 x0c b2 2 2 2 2 2 2 x0y0a b a y0c x0y0c a y0c y0cx0c a y0c
arctan
b2 b2 arctan y0c y0c
m m tan 1 1 m1m
y0 x0 c
xb2 ya2
m m (c) tan 2 1 m2m
0
(b) Slope of line through c, 0 and x0, y0: m1
2x 2yy 2 0 a2 b y
3
y0 b2x 2 0 x0 c a y0 a2y 2 b2x0x0 c 2 0 2 y0 bx a y0x0 c b2x0y0 1 2 0 x0 c a y0
a2y02 b2x02 b2x0c a2b2 b2x0c b2a2 x0c b2 2 2 2 2 2 2 2 a x0y0 a cy0 b x0 y0 x0y0a b a cy0 y0cx0c a y0c
arctan
yb c 2
0
Since , the tangent line to an ellipse at a point P makes equal angles with the lines through P and the foci.
Section 9.1
112. (a) e
(b)
Conics and Calculus
114. The transverse axis is vertical since 3, 0 and 3, 3 are the foci.
a2 b2 c ⇒ ea2 a2 b2. Hence, a a
3, 23
x h2 y k2 1 a2 b2
Center:
x h2 y k2 2 1. 2 a a 1 e2
3 5 c , 2a 2, b2 c2 a2 2 4 Therefore, the equation is
x 22 y 32 1 4 41 e2
y 32 2 x 32 1. 1 54
7
−3
433
9 −1
(c) As e approaches 0, the ellipse approaches a circle. 116. Center: 0, 0 Horizontal transverse axis Foci: ± c, 0 Vertices: ± a, 0 The difference of the distances from any point on the hyperbola is constant. At a vertex, this constant difference is
a c c a 2a. Now, for any point x, y on the hyperbola, the difference of the distances between x, y and the two foci must also be 2a. x c2 y 02 x c2 y 02 2a x c2 y 2 2a x c2 y 2
x c2 y 2 4a2 4ax c2 y 2 x c2 y 2 4xc 4a2 4ax c2 y 2 xc a2 ax c2 y 2
y
x2c2 2a2cx a4 a2x2 2cx c2 y 2
x2
c2
x2 a2
a2
a2y2
a2
c2
( x, y )
a2
(c, 0) (− c, 0) (− a, 0) (a, 0)
2
y 1 c2 a2
Since a2 b2 c2, we have x2a2 y 2b2 1.
118. c 150, 2a 0.001186,000, a 93, b 1502 932 13,851 x2 y2 1 2 93 13,851
752 13,851
x 110.3 miles.
y2 x2 21 2 a b 2x 2yy b2x 2 0 or y 2 2 a b ay y y0
When y 75, we have x2 932 1
120.
b2x0 x x0 a2y0
a2y0y a2y02 b2x0x b2x02 b2x02 a2y02 b2x0x a2y0 y a2b2 b2x 0 x a2y 0 y x0x y0 y 2 1 a2 b
x
434
Chapter 9
Conics, Parametric Equations, and Polar Coordinates Ax2 Cy2 Dx Ey F 0
122.
A x2
(Assume A 0 and C 0; see (b) below)
D E x C y 2 y F A C
D D2 E E2 D2 E2 x 2 C y2 y 2 F R A 4A C 4C 4A 4C
A x2
x 2AD y 2CE 2
2
C
A
(a) If A C, we have
R AC (b) If C 0, we have
x 2AD y 2CE 2
2
R A
A x
(c) If AC > 0, we have
C y
2
D
E
R A
2
F Ey
D2 . 4A
E 2C
2
F Dx
E2 . 4C
2
These are the equations of parabolas.
If A 0, we have
which is the standard equation of a circle.
x 2A y 2C
D 2A
1
R C
(d) If AC < 0, we have
x 2AD y 2CE 2
which is the equation of an ellipse.
R A
2
R C
±1
which is the equation of a hyperbola. 124. True
126. False. The y4 term should be y2.
Section 9.2
Plane Curves and Parametric Equations
2. x 4 cos2
y 2 sin
0 ≤ x ≤ 4
2 ≤ y ≤ 2
(a)
2
x
0
y
2
0
4
2
2
4
2
0
2
0
2
2
4
(c)
3
−1
(b)
(d) 3 2 1 x 1
2
3
5
−3
y
−2
128. True
x cos2 4 y2 sin2 4
5
y2 x 1 4 4 x 4 y2, 2 ≤ y ≤ 2 (e) The graph would be oriented in the opposite direction.
Section 9.2 4. x 3 2t
435
6. x 2t2
y 2 3t y23
Plane Curves and Parametric Equations
y t4 1
3 2 x
2x
y
2y 3x 13 0
2
x2 1, x ≥ 0 4
1
For t < 0, the orientation is right to left.
y
For t > 0, the orientation is left to right. y
6
6 5
4
4 3
2
2 x 2
6
4
1
8
x
−1 −1
1
2
3
4
5
6
8. x t 2 t, y t 2 t Subtracting the second equation from the first, we have x y 2t or t
xy 2
2
t
1
0
1
y
2 4
x y 2 x y y 4 2
x
2
0
0
2
6
y
6
2
0
0
2
3 2
Since the discriminant is
x
−1
B2 4AC 2 2 41 1 0,
3
2
4
−1
the graph is a rotated parabola. 4 t, t ≥ 0 10. x
12. x 1
y3t y3
x 4,
x ≥ 0
x1
3
y
2 1 −2
−1
1
2
yt2
1 1 implies t t x1
1 1 x1
x −1
14. x t 1
yt1
y
−3
1 t
x y 2 1 y 3 y 5 4 3
y
3 2 1
−2
1
−3
x
−2
x
2 1
2
3
4
5
−3
16. x et, x > 0 y e2t 1 y x2 1
18.
y 3 2
1 1, x > 0 x2
1 −3
−2
−1
x −1 −2 −3
1
3
x tan2
y
y sec2
4
sec2 tan2 1
3
yx1
2
x ≥ 0
1 x 1
2
3
4
436
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
x 2 cos
20.
x cos
22.
y 6 sin
2x 6y 2
2
x 4 2 cos
24.
y 2 sin 2
y 1 2 sin
y 4 sin cos
cos2 sin2 1
x 4 2 4 cos2
1 x2 sin2
y2 x2 1 ellipse 4 36
y 1 2 4 sin2
y ± 4x1 x2
x 4 2 y 1 2 4 3
3
y
−1
−2
4
7
2
2 −6 −4
4
−2
6
x
26. x sec
28.
2
y tan
x cos3
1.5
y sin3 −3
x2 sec2
3
y2 tan2
−2
x2 3 cos2
2
y2 3 sin2
−2
30. x ln 2t
−1.5
32. x e2t
3
4
y et
y t2 t
−5
−3
ex 2
−2
3 −1
e 2x 1 2x e y r 4
y2 x
−1
y > 0
3 −1
y x, x > 0
34. By eliminating the parameters in (a) – (d), we get x2 y 2 4. They differ from each other in orientation and in restricted domains. These curves are all smooth. 4t 2 1 1 1 4 2 y (b) x (a) x 2 cos , y 2 sin t t t
y
x ≥ 0, x 2
3
y0
y 1 −3
2 x
−1
1
3
1 x
−1
−3
1
3
−1 −2
(c) x t
y 4 t
x ≥ 0
y ≥ 0
(d) x 4 e2t
y et
2 < x ≤ 0
y > 0
y
y
3
3
2
1
1
x 1
2
3
−3
−2
−1
x
Section 9.2
Plane Curves and Parametric Equations
36. The orientations are reversed. The graphs are the same. They are both smooth. 38. The set of points x, y corresponding to the rectangular equation of a set of parametric equations does not show the orientation of the curve nor any restriction on the domain of the original parametric equations.
x h r cos
40.
x h a sec
42.
y k r sin
y k b tan
cos
xh r
xh sec a
sin
yk r
yk tan b
cos2 sin2
x h 2 y k 2 1 r2 r2
x h 2 y k 2 1 a2 b2
x h 2 y k 2 r 2 44. From Exercise 39 we have
46. From Exercise 40 we have
x 1 4t
x 3 3 cos
y 4 6t.
y 1 3 sin .
Solution not unique
48. From Exercise 41 we have a 5, c 3 ⇒ b 4 x 4 5 cos y 2 4 sin .
Solution not unique
Center: 4, 2 Solution not unique
50. From Exercise 42 we have a 1, c 2 ⇒ b 3 x 3 tan
52. y
Center: 0, 0 Solution not unique The transverse axis is vertical, therefore, x and y are interchanged. 56. x sin
54. y x2 Example
Example x t, y
y sec .
2 t1
x t, y
y t2
x t 3,
y t6
60. x 2 sin
y 2 4 cos
6
x t,
2 t 1
58. x 2 4 sin
y 1 cos
−6
2 x1
y 2 cos
9
6 −2
Not smooth at x 2n 1
−9
4
9 −3
−
0
Smooth everywhere
5
437
438
Chapter 9
62. x
3t 1 t3
y
3t 2 1 t3
Conics, Parametric Equations, and Polar Coordinates
64. Each point x, y in the plane is determined by the plane curve x f t , y gt . For each t, plot x, y . As t increases, the curve is traced out in a specific direction called the orientation of the curve.
2
−3
3
−2
Smooth everywhere 66. (a) Matches (ii) because 1 ≤ x ≤ 0 and 1 ≤ y ≤ 2.
(b) Matches (i) because x y 2 2 1 for all y.
68. x cos3
70. x cot
y 2 sin
y 4 sin cos
Matches (a)
Matches (c)
2
72. Let the circle of radius 1 be centered at C. A is the point of tangency on the line OC. OA 2, AC 1, OC 3. P x, y is the point on the curve being traced out as the angle changes AP . AB 2 and AP ⇒ 2. Form the right triangle CDP. The angle AB and OCE 2 DCP
3 . 2 2 2
x OE Ex 3 sin
y EC CD 3 sin cos 3
3
C 2
A
1
1
3 sin sin 3 2
Hence, x 3 cos cos 3, y 3 sin sin 3. 74. False. Let x t 2 and y t. Then x y 2 and y is not a function of x. 76. (a) x v0 cos t y h v0 sin t 16t 2 t
x x x ⇒ y h v0 sin 16 v0 cos v0 cos v0 cos y h tan x
(b) y 5 x 0.005x 2 h tan x h 5, tan 1 ⇒ 0.005 v02
2
16 sec2 2 x v02
16 sec2 2 x v02
(c)
80
, and 4
16 16 sec2 4 22 v02 v0 32 6400 ⇒ v0 80. 0.005
Hence, x 80 cos45 t y 5 80 sin45 t 16t 2.
0
250
−5
(d) Maximum height: y 55 at x 100 Range: 204.88
α D P = ( x, y )
θ
sin 3 3 cos cos 3 2 2
y
x
B E
x
Section 9.3
Section 9.3 2.
Parametric Equations and Calculus
Parametric Equations and Calculus
dy dydt 1 3t23 dx dxdt 13t23
4.
dy dyd 12e2 1 1 e32 32 dx dxd 2e 4 4e
8. x t2 3t 2, y 2t
6. x t, y 3t 1
dy 2 2 when t 0. dx 2t 3 3
dy 3 6t 6 when t 1. dx 12t
4 4 d 2y 222t 3 when t 0. dx2 2t 3 2t 32 9
3t d 2y 6 concave upwards dx2 12t
concave downward 10. x cos , y 3 sin dy 3 cos 3 cot dx sin
dx is undefined when 0.
d 2y 3 csc2 3 3 2 dx sin sin
d 2y is undefined when 0. dx2
dy
14. x sin , y 1 cos
12. x t, y t 1 dy 1 2t 1 dx 1 2t
t t 1
dy sin 0 when . dx 1 cos
1 cos cos sin2 1 cos 2 d 2 dx 1 cos
2 when t 2.
2y
d 2y t 1 2t t 12t 1 t 1 dx2 1 2t
1 1 when t 2. t 132
1 1 when . 1 cos 2 4
concave downward
concave downward 16. x 2 3 cos , y 3 2 sin
18. x t 1, y
2 cos 2 dy cot dx 3 sin 3 At 1, 3, 0, and
(a)
dy is undefined. dx
−3
(b) At t 1, x, y 0, 2, and dy dx dy 1, 1, 1 dt dt dx
Tangent line: y 5 3
5
−4
dy 0. At 2, 5, , and 2 dx
4 23
1 1, t 1 t
4
Tangent line: x 1
At
439
,2 ,
7 dy 23 , and . 6 dx 3
(c)
Tangent line: 23 4 33 y2 x 3 2
dy 1. At 0, 2, y 2 1x 0 dx y x 2
(d)
4
−3
5
23x 3y 43 3 0 −4
440
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
20. x 4 cos , y 3 sin , (a)
3 4 (b) At
4
−6
(d)
dy 3 4 3 . At , , dx 4 2 2
dx dy 32 dy 3 22, , dt dt 2 dx 4
6
−4
(c)
3 4 3 , x, y , , and 4 2 2
y
3 2
3 4 x 4 2
4
−6
6
3 y x 32 4
−4
22. x t2 t, y t3 3t 1 crosses itself at the point x, y 2, 1. At this point, t 1 or t 2. dy 3t2 3 dx 2t 1 At t 1, At t 2,
dy 0 and y 1. Tangent Line dx
dy 9 3 and y 1 3x 2 or y 3x 5. Tangent Line dt 3
24. x 2, y 21 cos Horizontal tangents:
dy 2 sin 0 when 0, ± , ± 2, . . . . d
Points: 4n, 0, 22n 1, 4 where n is an integer. Points shown: 0, 0, 2, 4, 4, 0 Vertical tangents:
dx 2 0; none d
26. x t 1, y t 2 3t Horizontal tangents: Point:
28. x t2 t 2, y t 3 3t
dy 3 2t 3 0 when t . dt 2
Vertical tangents:
dx 1 0; none dt
Point:
30. x cos , y 2 sin 2 Horizontal tangents: Points:
dy 3 5 7 4 cos 2 0 when , , , . d 4 4 4 4
22, 2 , 22, 2 , 22, 2 , 22, 2
Vertical tangents:
dy 3t 2 3 0 when t ± 1. dt
Points: 2, 2, 4, 2
21, 49
Vertical tangents:
Horizontal tangents:
dx sin 0 when 0, . d
Points: 1, 0, 1, 0
74, 118
dx 1 2t 1 0 when t . dt 2
Section 9.3 32. x 4 cos2 , y 2 sin
Since dxd 0 at 2 and 32, exclude them. dx 8 cos sin 0 when d
Vertical tangents:
441
34. x cos2 , y cos
dy 3 2 cos 0 when , . d 2 2
Horizontal tangents:
Parametric Equations and Calculus
Horizontal tangents:
dy sin 0 when x 0, . d
Since dxd 0 at these values, exclude them. Vertical tangents:
0, .
dx 2 cos sin 0 when d
3 , . 2 2
Point: 4, 0
Exclude 0, . Point: 0, 0
36. x t2 1, y 4t3 3, 1 ≤ t ≤ 0
dydt 2
dx dy dx 2t, 12t2, dt dt dt
0
s
1
2
38. x arcsin t, y ln1 t 2, 0 ≤ t ≤
4t2 144t4
0
4t2 144t4 dt
1 36t232 54
1
2t1 36t2 dt
1 1 3732 4.149 54
0 1
1 2
dx 1 t dy 1 2t , dt 2 1 t2 1 t2 1 t 2 dt
12
s
0 12
dxdt dydt dt 1 1 t dt 2
2
12
2 2
0
1 t1 ln 2 t1
0
1 dt 1 t2
12 0
1 1 1 ln3 0.549 ln 2 3 2 t5 1 dx 1 dy t 4 3, 1, 4 10 6t dt dt 2 2t
40. x t, y
2
S
1
1
2
1
2
1
t2 2t1 dt 4
t4 1 4 2 2t
4a
2
2 1
779 240
44. x cos sin , y sin cos , dy sin d S
2
2 cos2 2 sin2 d
0
2
0
d
2 2
2
0
2 2
a2 sin2 a2 cos2 d
2
0
dt
3
2
dx dy a sin , a cos d d
0
t4 1 4 dt 2 2t 1
S4
4
10t 6t 5
2
42. x a cos , y a sin ,
dx cos d
2
d 4a
0
2a
442
Chapter 9
46. x
Conics, Parametric Equations, and Polar Coordinates
4t 4t 2 , y 1 t3 1 t3
(a) x 3 y 3 4xy
(b)
dy 1 t 38t 4t 23t2 dt 1 t 32
4
−6
4t2 t3 3 2. 0 when t 0 or t 1 t32
6
Points: 0, 0,
4 3 2, 4 3 4 1.6799, 2.1165 3
3
−4
1
(c) s 2
0
1
8
41 2t 3 1 t 32
2
4t2 t 3 1 t 32
2
t 8 4t 6 4t 5 4t 3 4t 2 1
1 t 32
0
1
dt 2
0
16 t 8 4t 6 4t 5 4t 3 4t 2 1 dt 1 t 34
dt 6.557
48. x 3 cos , y 4 sin
4
dy dx 3 sin , 4 cos d d s
2
−6
6
9 sin2 16 cos2 d 22.1
−4
0
50. x t, y 4 2t,
1 52. x t3, y t 1, 1 ≤ t ≤ 2, y-axis 3
dx dy 1, 2 dt dt
2
(a) S 2
dx dy t2, 1 dt dt
4 2t1 4 dt
0
254t t 2
2 0
2
(b) S 2
(a) S 4
1
t1 4 dt 5 t 2
0
54. x a cos , y b sin ,
2
2 0
45
2
ab sin
1
a2 b2 4ab cos2 d a2 e
2ab e cos 1 e2 cos2 arcsine cos e
2b2
e
32 17 232 23.48 9
b sin a2 sin2 b2 cos2 d
0
13 4 4 t t 1 dt x 132 3 9
dx dy a sin , b cos d d
0
4
2
S 2
85
2aabb arcsin
a2 b2
a
—CONTINUED—
2
2
2
a2 b2
a
2
c : eccentricity a
2b
2
0
2
2
e sin 1 e2 cos2 d
0
ab e1 e2 arcsine e
abe arcsine
2 1
Section 9.3
Parametric Equations and Calculus
443
54. —CONTINUED— (b) S 4
2
a cos a2 sin2 b2 cos2 d
0
4
2
4a c
a cos b2 c2 sin2 d
0
2
c cos b2 c 2 sin2 d
0
2a c sin b2 c2 sin2 b2 ln c sin b2 c 2 sin2 c
2a c b2 c2 b2 ln c b2 c 2 b2 ln b c
2
0
2ab2 a a2 b2 b2 1e ln 2a 2 ln 2 2 b e 1e a b
2a 2
56. (a) 0
58. One possible answer is the graph given by x t, y t.
(b) 4
y
4 3 2 1 x
−4 −3 −2
1
2
3
4
−2 −3
−4
b
60. (a) S 2
dxdt dydt dt dx dy f t dt dt dt
gt
a
b
(b) S 2
a
2
2
2
2
62. Let y be a continuous function of x on a ≤ x ≤ b. Suppose that x f t, y gt, and f t1 a, f t 2 b. Then using integration by substitution, dx f t dt and
b
y dx
a
64. x 4 t, y t,
0
A
t
4
t2
gtf t dt.
t1
dx 1 , 0 ≤ t ≤ 4 dt 24 t
1 dt 24 t
2
4 u2 du
0
1 u u4 u2 4 arcsin 2 2
2 0
Let u 4 t, then du 1 24 t dt and t 4 u2. x y
x, y
1
0
4 tt
4
1 2
0
t
4
2
1 1 dt 2 24 t
1 1 dt 4 24 t
0
4
0
t dt
4
1 2 32 t 2 3
dx sin d
0
V 2
2
3 sin 2sin d
0
18
sin3 d 18 cos
2
0 4
8 3
t 1 28 t 4 t dt 4 3 4 t
38, 38
66. x cos , y 3 sin ,
cos3 3
0
2
12
0 4
8 3
444
Chapter 9
Conics, Parametric Equations, and Polar Coordinates dx 2 csc2 d
68. x 2 cot , y 2 sin2 ,
0
A2
0
2 sin2 2 csc2 d 8
2
12
144 x2
x
0
2
4
72. 2a2 is area of deltoid (c).
3 70. 8 a2 is area of asteroid (b).
76. (a) y 12 ln
2
d 8
144 x
74. 2ab is area of teardrop (e).
(b) x 12 sech
2
0 < x ≤ 12
t t , y t 12 tanh , 0 ≤ t 12 12
60
60
0
12 0
0
12
Same as the graph in (a), but has the advantage of showing the position of the object and any given time t.
0
(c)
dy 1 sech2t12 t sinh dx secht12 tant12 12
y 24
t t t Tangent line: y t0 12 tanh 0 sinh 0 x 12 sech 0 12 12 12
y t0 sinh
(0, y0)
16 12
t0 x 12
8
( x, y )
4
y-intercept: 0, t0
x 2
Distance between 0, t0 and x, y: d
12 sech 12t 12 tanh 12t
0
2
0
2
4
6
10
12
12
d 12 for any t ≥ 0. 78. False. Both dxdt and dydt are zero when t 0. By eliminating the parameter, we have y x 23 which does not have a horizontal tangent at the origin.
Section 9.4 2.
Polar Coordinates and Polar Graphs
7
2, 4
4.
7 0 6
7 0 6
74 2
y 0 sin
x 3 cos1.57 0.0024
y 3 sin1.57 3
x, y 0.0024, 3
π 2
x, y 0, 0
(− 0.0024, 3)
π 2
π
2
2,
x, y 2, 2 (−
6. 3, 1.57
x 0 cos
x 2 cos y 2 sin
7 2 4
0, 76
2) (0, 0) 0 1
2
0 1
0 1
2
Section 9.4
8. r, 2,
11 6
Polar Coordinates and Polar Graphs
10. r, 8.25, 1.3
12. x, y 0, 5 r ±5
x, y 2.2069, 7.9494
x, y 1.7321, 1
tan undefined
y
3 3 , , 5, , 5, 2 2 2 2
y 8
2
6
(−1.7321, 1)
−2
(2.2069, 7.9494)
4
1
y
2 −2
x
−1
1
x
−1
1
−2
2
2
−2
x
−1
1
−2
2
3
−1
−4
−1
−6
−2
−8
−3 −4
(0, −5)
−5
14. x, y 4, 2
y
r ± 16 4 ± 2 5
x 1
2
3
4
5
−1
2 1 tan 4 2
−2
0.464
−4
(4, −2)
−3
2 5, 0.464, 2 5, 2.678
−5
16. x, y 3 2, 3 2
18. x, y 0, 5
r, 5, 1.571
r, 6, 0.785
20. (a) Moving horizontally, the x-coordinate changes. Moving vertically, the y-coordinate changes. (b) Both r and values change. (c) In polar mode, horizontal (or vertical) changes result in changes in both r and . 22. x2 y 2 2ax 0
π 2
24.
r 2 2ar cos 0 rr 2a cos 0
π 2
r 10 sec
xy 4
26.
x 10 r cos 10
r 2a cos
0
a
0
2a
2
28.
8
r 2 r 2 9cos 2 0 r 2 9 cos 2
8 csc 2 π 2
π 2
0
0 4
6
x2 y 22 9x2 y 2 0
r 2 4 sec csc
2
4
r 22 9r 2 cos2 r 2 sin2 0
r cos r sin 4
445
1
2
12
446
Chapter 9
30.
r 2
Conics, Parametric Equations, and Polar Coordinates
r 5 cos
32.
r2 4
r 2 5r cos
x2 y 2 4
x2 y2 5x
x 25
1
2
y2
52
5 6
3 y x 3 2
y
3
3
x
−1 −1
x
y
y
1
5 6
tan tan
25 25 y2 4 4
x2 5x
y
34.
4
2
3 2
1
1 −2 −1
x 1
2
3
4
6
−2
x
−1
1
−2
−1
−3 −4
−2
38. r 51 2 sin
r 2 csc
36.
40. r 4 3 cos
0 ≤ < 2
r sin 2 y2
0 ≤ < 2
3 −10
y20
2
6
10 −4
y
3
−18
10
−6
1
x
−1
42. r
1
2
2 4 3 sin
44. r 3 sin
Traced out once on 0 ≤ ≤ 2
1 46. r2 .
5 2
0 ≤ < 4
Graph as
4
r1
3
−6
6
1
, r2
1
It is traced out once on 0, . 1.5
−3
3 −4 −1
.
−2
2
−1.5
Section 9.4
Polar Coordinates and Polar Graphs
48. (a) The rectangular coordinates of r1, 1 are r1 cos 1, r1 sin 1. The rectangular coordinates of r2, 2 are r2 cos 2, r2 sin 2. d 2 x2 x12 y2 y12 r2 cos 2 r1 cos 12 r2 sin 2 r1 sin 12 r22 cos2 2 2r1r2 cos 1 cos 2 r12 cos2 1 r22 sin2 22 2r1r2 sin 1 sin 2 r12 sin2 1 r22 cos2 2 sin2 2 r12 cos2 1 sin2 1 2 r1r2cos 1 cos 2 sin 1 sin 2 r12 r22 2r1r2 cos1 2 d r12 r22 2r1r2 cos1 2 (b) If 1 2, the points lie on the same line passing through the origin. In this case, d r12 r22 2r1r2 cos0
r1 r22 r1 r2
(c) If 1 2 90, then cos1 2 0 and d r12 r22, the Pythagorean Theorem! (d) Many answers are possible. For example, consider the two points r1, 1 1, 0 and r2, 2 2, 2. d
1 2
2
212 cos 0
5 2
Using r1, 1 1, and r2, 2 2, 52 , d
1
2
22 212 cos
5 5. 2
You always obtain the same distance.
50.
10, 76 , 3, d
102
52. 4, 2.5, 12, 1
32
7 2103 cos 6
d 42 122 2412 cos2.5 1 160 96 cos 1.5 12.3
109 60 cos 6 109 30
3 7.6
54. r 21 sin
56. (a), (b) r 3 2 cos
dy 2 cos sin 2 cos 1 sin dx 2 cos cos 2 sin 1 sin At 2, 0,
At 3,
dy 1. dx
7 dy , is undefined. 6 dx
3 dy At 4, , 0. 2 dx
4
−8
4
−4
r, 1, 0 ⇒ x, y 1, 0 Tangent line: x 1 (c) At 0,
dy does not exist (vertical tangent). dx
447
448
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
58. (a), (b) r 4
r a sin
60.
dy a sin cos a cos sin d
6
−8
2a sin cos 0
8
0,
−6
4 ⇒ x, y 2 2, 2 2
dx a sin2 a cos2 a1 2 sin2 0 d
at r, 4,
Tangent line: y 2 2 1 x 2 2
sin ±
y x 4 2
,
3 5 7 , , , 4 4 4 4
Horizontal: 0, 0, a,
a 2 2, 4 , a 2 2, 34
Vertical:
62.
1 2
2
dy , 1. 4 dx
(c) At
3 , , 2 2
r a sin cos2
64. r 3 cos 2 sec
dy a sin cos3 2a sin2 cos a cos3 sin d
2
2a sin cos3 sin3 cos
−2
4
2a sin cos cos2 sin2 0 −2
3 0, tan2 1, , 4 4 Horizontal:
2a
,
4
4
,
Horizontal tangents: 2.133, ± 0.4352
2a 3
4
,
4
, 0, 0
66. r 2 cos3 2
r 3 cos
68.
π 2
r 2 3r cos
2
x2 y2 3x −3
3
3 x 2
2
0
−2
Circle: r
Horizontal tangents:
1.894, 0.776, 1.755, 2.594, 1.998, 1.442 Center:
y2 3 2
32, 0
Tangent at pole:
70. r 31 cos
1
9 4
2
π 2
Cardioid Symmetric to polar axis since r is a function of cos . 0 1
0
3
2
2 3
r
0
3 2
3
9 2
6
2
4
Section 9.4 72. r sin5
Polar Coordinates and Polar Graphs
74. r 3 cos 2
Rose curve with five petals
Rose curve with four petals
Symmetric to 2
Symmetric to the polar axis,
Relative extrema occur when
Relative extrema: 3, 0, 3,
dr 3 5 7 9 5 cos5 0 at , , , , . d 10 10 10 10 10
Tangents at the pole:
2 3 4 Tangents at the pole: 0, , , , 5 5 5 5
0
3 , 4 4
0 2
76. r 2
80. r 5 4 sin
78. r 1 sin
Circle radius: 2
3 , 3, , 3, 2 2
π 2
1
y2
, and pole 2
7 5 and given the same tangents. 4 4
π 2
x2
449
Limaçon
Cardioid
4
Symmetric to
π 2
2
π 2
0
0
1
1
2
r
9
6
0
6
2
7
5
3
1
2 π 2
0 2
r
82.
6 2 sin 3 cos
2r sin 3r cos 6 2y 3x 6 Line π 2
84. r
1
Hyperbolic spiral
4
2
3 4
5 4
3 2
r
4
2
4 3
1
4 5
2 3
π 2
0 1
0 1
4
450
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
86. r2 4 sin
π 2
Lemniscate Symmetric to the polar axis,
Relative extrema: ± 2,
2
, and pole 2 0
2
0
6
2
5 6
r
0
± 2
±2
± 2
0
Tangent at the pole: 0 90. r 2 cos 2 sec
88. Since r 2 csc 2
1 , sin
the graphs has symmetry with respect to 2. Furthermore, r ⇒ as ⇒ 0
Strophoid r ⇒ as ⇒ r ⇒ as ⇒
r cos 4 cos2 2
r r 1 2 2 sin sin y
x 4 cos2 2 lim 4 cos2 2 2
ry 2y r r
2
r 2 cos 2 sec 22 cos2 1 sec
r ⇒ as ⇒ . Also, r 2
2
→ ±2
2y . y1
x = −2
Thus, r ⇒ ± as y ⇒ 1.
2
−3
3
4
−2 −4
4
y=1 −2
92. x r cos , y r sin x2 y2 r 2, tan
y x
94. Slope of tangent line to graph of r f at r, is dy f cos f sin . dx f sin f cos If f 0 and f 0, then is tangent at the pole.
96. r 4 cos 2
98. r 2 sec
Rose curve
Line
Matches (b)
Matches (d)
Section 9.4 100. r 6 1 cos (a) 0, r 6 1 cos
(c)
2
2 6 1 cos cos sin sin 2 2
9
r 6 1 cos
−9
15
6 1 sin
−9
(b)
Polar Coordinates and Polar Graphs
, r 6 1 cos 4 4
15
12 −12
12 −3
−9
The graph of r 6 1 cos is rotated through the angle 2.
15
−6
The graph of r 6 1 cos is rotated through the angle 4.
102. (a) sin
sin cos cos sin 2 2 2
(b) sin sin cos cos sin sin
cos
r f sin
2
r f sin
f sin
f cos
(c) sin
3 3 3 sin cos cos sin 2 2 2
cos
r f sin
3 2
f cos
104. r 2 sin 2 4 sin cos
(a) r 4 sin
cos 6 6
(b) r 4 sin
2
−3
−3
−2
3
−2
2 2 cos 3 3
(d) r 4 sin cos 4 sin cos 2
2 −3 −3
3
3 −2 −2
2
3
(c) r 4 sin
cos 4 sin cos 2 2
451
452
Chapter 9
Conics, Parametric Equations, and Polar Coordinates π 2
106. By Theorem 9.11, the slope of the tangent line through A and P is f cos f sin f sin f cos
Radial line
Polar curve r = f (θ)
P = (r, θ)
This is equal to tan
ψ
sin cos tan tan tan . 1 tan tan cos sin tan
Tangent line
θ
0
A
Equating the expressions and cross-multiplying, you obtain
f cos f sin cos sin tan sin cos tan f sin f cos f cos2 f cos sin tan f sin cos f sin2 tan f sin2 f sin cos tan f sin cos f cos2 tan f cos2 sin2 f tan cos2 sin2 tan
108. tan At
r 31 cos drd 3 sin
110. tan
3 1 22 2 2 , tan . 4 2 2
arctan
f r . f drd
At
2 2 2 1.041 59.64
r 4 sin 2 drd 8 cos 2
3 sin 3 , tan . 6 2 cos 3 2
arctan
−6
2
6
−4
−5
112. tan
4
5
−8
23 0.7137 40.89
r 5 undefined ⇒ . drd 0 2 6
−9
9
−6
114. True
Section 9.5 2. (a) r 3 cos
116. True
Area and Arc Length in Polar Coordinates (b) A 2
π 2
12 3 cos
9
2
2
d
0
2
cos2 d
0
0 2
A
32
4
2
9 4
9 2
2
1 cos 2 d
0
9 sin 2 2 2
2
0
9 4
Section 9.5
4
6 sin 22 d 36
0
sin2 2 d
6. A 2
0
4
36
4
1 2
4. A 2
Area and Arc Length in Polar Coordinates
0
1 cos 4 d 2
18
sin 4 4
1 2
10
cos 5 d 2
0
10
1 1 sin10 2 10
453
0
20
4
0
4 92
18
2
1 2
8. A 2
1 sin d 2
0
3 1 2 cos sin 2 2 4
10. A 2 2
0
3 8 4
1 2
2
4 6 sin 2 d
arcsin23
2
16 48 sin 36 sin2 d
arcsin23
2
2 16 48 sin 361 cos d 2
arcsin23
34 48 cos 9 sin 2
2
arcsin23
1.7635
2 −8
8
−12
12. Four times the area in Exercise 11, A 4 33 . More specifically, we see that the area inside the outer loop is
12
2
2
6
21 2 sin 2 d
2
6
4 16 sin 16 sin2 d 8 63.
6
The area inside the inner loop is 2
1 2
3 2
7 6
21 2 sin 2 d 4 63.
−4
4 −1
Thus, the area between the loops is 8 63 4 63 4 123. 14. r 31 sin
16. r 2 3 cos
r 31 sin
r cos
Solving simultaneously,
Solving simultaneously,
31 sin 31 sin 2 sin 0
0, . Replacing r by r and by in the first equation and solving, 31 sin 31 sin , sin 1, 2. Both curves pass through the pole, 0, 3 2, and 0, 2, respectively. Points of intersection: 3, 0, 3, , 0, 0
2 3 cos cos 1 2 5 , . 3 3 Both curves pass through the pole, (0, arccos 23), and 0, 2, respectively. cos
Points of intersection:
12, 3 , 12, 53 , 0, 0
454
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
20.
18. r 1 cos r 3 cos
4
π 2
r2
Solving simultaneously,
Line of slope 1 passing through the pole and a circle of radius 2 centered at the pole.
1 cos 3 cos cos
1 2
0 1
3
Points of intersection:
2, 4 , 2, 4
5 , . 3 3 Both curves pass through the pole, 0, , and 0, 2, respectively. Points of intersection:
32, 3 , 32, 53 , 0, 0
22. r 3 sin
π 2
Points of intersection:
r 2 csc
0 1
2
17 3
2
17 3
2
, arcsin
17 3
, arcsin
2
,
17 3
2
,
3.56, 0.596, 3.56, 2.545 The graph of r 3 sin is a limaçon symmetric to 2, and the graph of r 2 csc is the horizontal line y 2. Therefore, there are two points of intersection. Solving simultaneously, 3 sin 2 csc sin2 3 sin 2 0 sin
3 ± 17 2
arcsin
17 3
2
0.596.
24. r 31 cos r
r=
6 1 cos
The graph of r 31 cos is a cardioid with polar axis symmetry. The graph of
6 1 − cos θ
−10
5
r 61 cos is a parabola with focus at the pole, vertex3, , and polar axis symmetry. Therefore, there are two points of intersection. Solving simultaneously, 31 cos
5
r = 3(1 − cos θ )
6 1 cos
1 cos 2 2 cos 1 ± 2
arccos 1 2 . Points of intersection: 32, arccos 1 2 4.243, 1.998, 32, 2 arccos 1 2 4.243, 4.285
−5
Section 9.5
Area and Arc Length in Polar Coordinates
26. r 4 sin r 21 sin
2
1 2
28. A 4
2
18
Points of intersection: 0, 0, 4, 2
91 sin d 2
0
455
1 sin 2 d
0
9 3 8 2
(from Exercise 14)
7
The graphs reach the pole at different times ( values). 6
−7
7
r = 4 sin θ −7 −6
6 −2
r = 2 (1 + sin θ)
32. A 2
30. r 5 3 sin and r 5 3 cos intersect at 4 and 5 4.
1 A2 2
5 4
5 3 sin
4
2 d
59 9 30 cos sin 2 2 4
6
2
6
3 sin 2 d
4
4
−4
59 302 50.251 2
4
−4
8
−12
12
−8
34. Area Area of r 2a cos Area of sector twice area between r 2a cos and the lines , . 3 2
2 1 A a 2 a 2 3 2
2
2
3
2 a 2 2a 2 3
sin 2 2 a 2 2a 2 3 2
3
2 a 2 3 π 2
2a 2
tan 1, 4
2a cos 2 d
a2
3
6
33a 2
2a
4
1 cos 2 d 2
1 1 a2 2 4 2
0
π 2
r = a sin θ
0
0
a π θ =−3
4
1 1 a2 a2 4 8 a
a
a cos 2 d
0
2
2
1 sin 2 a2 2 2
2 a 2
1 2
0
1 cos 2 d
3 2 3 4
π θ = 3
36. r a cos , r a sin
A2
1 2
2
6
2 sin 2 d
4 cos 2 4 sin d
2 sin2 4 cos
5 4
2 2 59 5 9 59 9 30 30 2 4 2 4 2 4 2 4
2
1 2
r = a cos θ
2
6
33
456
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
38. By symmetry, A1 A2 and A3 A4.
6
1 2
A1 A2
2a cos 2 a2 d
3
6
a2 2
a2 sin 2 2
3
4 cos2 1 d 2a2
6
3
1 5 2 1 a 2 2 6 2
5 a 2 2a 2 12
A6 2
5 6
12
5 6
cos 2 d
3 a2 3 a 2 1 a2 1 2 2 2 4
5 6
0
6
6
4
6
4
6
r=2
6
0
2a
4 6
2a sin 2 a2 d
4 sin2 1 d a 2 sin 2
4
6
12 1 23
a2
[Note: A1 A6 A7 A4 a2 area of circle of radius a]
r sec 2 cos ,
40.
< < 2 2
y
1
r cos 1 2 cos2 x12
r 2 cos2 x2 12 2 2 r x y2
x 2 y 2x x 2 y 2 2x 2
x
−1
y 2x 1 x 2 x3 y2
1 2
A2
4
0
x 21 x 1x
4
sec 2 cos 2 d
0
sec2 4 4 cos2 d
4
0
sec2 4 21 cos 2 d tan 2 sin 2
0
r = 2a cos θ
π θ = −3
a2 d
a 2 3 a 2 5 3 a2 a2 12 3 2 12 12 2
π θ= 6
A1 a
4
A6 A4
π θ= 4 A7
a
A5
12
1 cos 2 d a 2
12
A2
5π θ= 6
3 5 a 2 3 a2 a2 12 3 2 12 2
π θ= 3
2a
r = 2a sin θ
1 cos 2 d
2a sin 2 d 2
a 2 2 sin 2
a2
6
2a cos 2 2a sin 2 d
2a sin 2 d
0
A7 2
4
6
A3
6
2a 2
6
4
π 2
5 a 2 a 2 2 sin 2 12
4
1 2 a 2 a 2 2 4
A3 A4 A5
a 2 sin 2
1 2
4
0
2
2
Section 9.5 42. r 2a cos
s
2
2
r 8 sin
2a cos 2 2a sin 2 d
2
457
44. r 81 cos , 0 ≤ ≤ 2
r 2a sin 2
Area and Arc Length in Polar Coordinates
s2
81 cos 2 8 sin 2 d
0
2a d 2
2
2
2 a
16
1 2 cos cos2 sin2 d
0
162
1 cos d
0
162
1 cos
0
162
0
sin 1 cos
3221 cos
1 cos 1 cos
d
d
0
64
46. r sec , 0 ≤ ≤
3
48. r e, 0 ≤ ≤
50. r 2 sin2 cos , 0 ≤ ≤
10
2
3
−2
−3
4
−25
−2
−5
−3
Length 7.78
Length 31.31
Length 1.73 exact 3
4
5
52. r a cos r a sin S 2
2
0
2 a2
a cos cos a2 cos a2 sin2 d
2
cos2 d a2
0
a2
2
1 cos 2 d
0
sin 2 2
2
0
2a2 2
54. r a1 cos r a sin S 2
a1 cos sin a21 cos 2 a2 sin2 d 2 a2
0
22 a2
1 cos 32sin d
sin 1 cos 2 2 cos d
42 a2 1 cos 52 5
0
32 a2 5
58. The curves might intersect for different values of :
56. r
S 2
0
0
r 1
0
See page 696.
sin 2 1 d 42.32
458
Chapter 9
60. (a) S 2
(b) S 2
Conics, Parametric Equations, and Polar Coordinates
f sin f 2 f2 d
f cos f 2 f2 d
62. r 8 cos , 0 ≤ ≤ (a) A
1 2
r 2 d
0
1 2
64 cos2 d 32
0
0
1 cos 2 sin 2 d 16 2 2
16
0
(Area circle r2 42 16) (b)
0.2
0.4
0.6
0.8
1.0
1.2
1.4
A
6.32
12.14
17.06
20.80
23.27
24.60
25.08
1 (c), (d) For 4 of area 4 12.57: 0.42 1 For 2 of area 8 25.13: 1.57 2 3 For 4 of area 12 37.70: 2.73
(e) No, it does not depend on the radius. 64. False. f 0 and g sin 2 have only one point of intersection.
Section 9.6 2. r
Polar Equations of Conics and Kepler’s Laws
2e 1 e cos
(a) e 1, r
4. r
2 , parabola 1 cos
2e 1 e sin
(a) e 1, r
2 , parabola 1 sin
(b) e 0.5, r
1 2 , ellipse 1 0.5 cos 2 cos
(b) e 0.5, r
1 2 , ellipse 1 0.5 sin 2 sin
(c) e 1.5, r
3 6 , hyperbola 1 1.5 cos 2 3 cos
(c) e 1.5, r
3 6 , hyperbola 1 1.5 sin 2 3 sin
4
e = 1.5
9
e=1
−9
e = 1.5
3
e=1 −9
9
e = 0.5 −4
6. r
e = 0.5
−3
4 1 0.4 cos (b) r
(a) Because e 0.4 < 1, the conic is an ellipse with vertical directrix to the left of the pole. (c)
9
7
−10
10 −8
8
4 1 0.4 cos
The ellipse is shifted to the left. The vertical directrix is to the right of the pole 4 r . 1 0.4 sin The ellipse has a horizontal directrix below the pole.
−7
8. Ellipse; Matches (f)
−5
10. Parabola; Matches (e)
12. Hyperbola; Matches (d)
Section 9.6
14. r
6 1 cos
16. r
Parabola since e 1
Polar Equations of Conics and Kepler’s Laws
1 5 5 3 sin 1 3 5sin
18. r3 2 cos 6 r
3 Ellipse since e < 1 5
Vertex: 3, 0 π 2
Vertices:
5 5 3 , , , 8 2 2 2
2 1 2 3 cos
Ellipse since e
π 2 0 4
6 3 2 cos
Vertices: 6, 0,
8
0 1
2
2 < 1 3
65,
π 2
0 1
20. r
6 2 3 7 sin 1 7 3sin
Hyperbola since e Vertices:
22. r
43, 0 , 4,
π 2
0
0
2
2
Hyperbola
4
−6
6
26.
Hyperbola
4
−6
−4
28. r
6
−4
6
1 cos 3
30. r
6 Rotate the graph of r 1 cos
counterclockwise through the angle . 3
6 3 7 sin 2 3
Rotate graph of r
6 . 3 7 sin
Clockwise through angle of 2 3. 4
6
−6 −14
6
10
−4 −10
4
4 1 2 cos
Vertices:
π 2
24.
3
Hyperbola since e 2 > 1
7 > 1. 3
3 3 3 , , , 5 2 2 2
2
5
459
460
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
32. Change to
:r 6
2
1 sin
6
34. Parabola
e 1, y 1, d 1 r
36. Ellipse
38. Hyperbola
3 e , y 2, d 2 4 r
40. Parabola Vertex: 5,
3 e , x 1, d 1 2
ed 1 e sin
r
e 1, d 10
ed 1 e cos
r
23 4 1 3 4 sin
3 2 1 3 2 cos
6 4 3 sin
3 2 3 cos
42. Ellipse
10 ed 1 e cos 1 cos
44. Hyperbola
3 Vertices: 2, , 4, 2 2
Vertices: 2, 0, 10, 0
3 10 e ,d 2 3
1 e ,d8 3
r
ed r 1 e sin 8 3 1 1 3 sin 8 3 sin
46. r
1 ed 1 e sin 1 sin
ed 1 e cos
5 1 3 2 cos
10 2 3 cos
4 is a parabola with horizontal directrix above the pole. 1 sin
(a) Parabola with vertical directrix to left pole.
(b) Parabola with horizontal directrix below pole.
(c) Parabola with vertical directrix to right of pole.
(d) Parabola (b) rotated counterclockwise 4.
x2 y2 1 a2 b2
48. (a)
x 2b 2 y 2a 2 a 2b 2 b 2 r 2 cos 2 a2r 2 sin2 a 2b 2 r 2 b 2 cos 2 a 21 cos 2 a 2b 2 r 2 a 2 cos 2 b 2 a 2 a 2b 2 r2
a2
a 2b 2 a 2b 2 2 2 2 a cos a c 2 cos 2 b2
b2 b2 1 c a 2 cos 2 1 e 2 cos 2
y2 x2 21 2 a b
(b)
x 2b 2 y 2a 2 a 2b 2 b 2r 2 cos 2 a 2r 2 sin 2 a 2b 2 r 2 b 2 cos 2 a 2 1 cos 2 a 2b2 r 2 a 2 cos 2 a 2 b 2 a 2b 2 r2
a 2b 2 b2 a 2 c 2 cos 2 1 c 2 a 2 cos 2 b 2 1 e 2 cos 2
Review Exercises for Chapter 9
50. a 4, c 5, b 3, e r2
5 4
52. a 2, b 1, c 3, e
9 1 2516 cos 2
54. A 2
1 2
2
r2
2
1 1 34 cos 2
2
1 d 3.37 3 2 sin 2 2
ed 1 e cos
56. (a) r
3
3 22 sin d 4 2
2
(b) The perihelion distance is a c a ea a1 e .
1 e2 a a1 e . 1e
When 0, r c a ea a a1 e .
When , r
Therefore,
The aphelion distance is a c a ea a1 e . ed a1 e 1e
When 0, r
a1 e 1 e ed
1 e2 a a1 e . 1e
a1 e 2 ed. Thus, r
1 e2 a . 1 e cos
58. a 1.427 109 km
60. a 36.0 10 6 mi, e 0.206
e 0.0543 r
r
1.422792505 109 1 e 2 a 1 e cos 1 0.0543 cos
34.472 10 6 1 e 2 a
1 e cos 1 0.206 cos
Perihelion distance: a1 e 28.582 10 6 mi Aphelion distance: a1 e 43.416 10 6 mi
Perihelion distance: a1 e 1.3495139 109 km Aphelion distance: a1 e 1.5044861 109 km r a sin b cos
62.
461
r 2 ar sin br cos x 2 y 2 ay bx x 2 y 2 bx ay 0 represents a circle.
Review Exercises for Chapter 9 2. Matches (b) - hyperbola
4. Matches (c) - hyperbola
6. y 2 12y 8x 20 0
y
y 2 12y 36 8x 20 36
16
y 6 2 42 x 2
12
Parabola Vertex: 2, 6 −4
x 8
12
462
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
8. 4x 2 y 2 16x 15 0
y
4x 2 4x 4 y 2 15 16
1
(2, 0)
x 2 2 y 2 1 14 1
x 1
2
3
−1
Ellipse Center: 2, 0
−2
Vertices: 2, ± 1 4x 2 4y 2 4x 8y 11 0
10.
y 4
1 4 y 2 2y 1 11 1 4 4
4 x2 x
x 12 2 y 1 2 1 2 2
3
1 x
−2
1
Hyperbola
−1
12, 1 1 Vertices: ± 2, 1 2 1 Asymptotes: y 1 ± x 2
−2
3
4
Center:
12. Vertex: 4, 2 Focus: 4, 0 Parabola opens downward p 2
x 4 2 42 y 2 x 2 8x 8y 0 14. Center: 0, 0
16. Foci: 0, ± 8
Solution points: 1, 2 , 2, 0
Asymptotes: y ± 4x
Substituting the values of the coordinates of the given points into
Center: 0, 0
x2 y2 2 1, b2 a
c8 a y x 4x asymptote → a 4b b
we obtain the system
b1 a4 1, 4b 2
2
2
1.
18.
16 x2 3y 2 1. and b 2 4, 3 4 16
21 y2 x2 1, a 5, b 2, c 21, e 4 25 5
By Example 5 of Section 9.1, C 20
2
0
b2 c2 a2 64 4b 2 ⇒ 17b2 64 ⇒ b2
Solving the system, we have a2
Vertical transverse axis
1 2521 sin d 23.01. 2
64 1024 ⇒ a2 17 17
x2 y2 1 102417 6417
Review Exercises for Chapter 9
463
1 2 x 200
20. y
(a) x 2 200y
y
1 2 x 200
y
1 x 100
(b)
x2 450 y Focus: 0, 50
x 1 10,000 x dx 38,294.49 S 2 x 1 10,000 2
1 y 2
100
2
0
a
22. (a) A 4
0
b 4b 1 a 2 x 2 dx a a 2
b
(b) Disk: V 2
0
b
S 4
0
4 a b2
b
b 2 y 2 dy
0
a 0
ab
2 a 2 2 1 b y y3 b2 3
b
4 a 2b 3
0
b
b 4 c 2y 2 dy
0
2 a cy b 4 c 2y 2 b 4 ln cy b 4 c 2y 2 b 2c
0
b
2 a 2 b c b 2 c 2 b 4 ln cb b b 2 c 2 b 4 lnb 2 b 2c
ab 2 ca ln c e
a
(c) Disk: V 2
0 a
S 22
0
a2 2 2 a 2 b y 2 dy 2 b2 b
x a
b 4 a 2 b 2 y 2 a b 2 y 2 dy b b b 2 y 2
2 a 2
x a 2 x 2 a 2 arcsin
4 b a2
2
2 a 2
b2 2 2 b 2 a x 2 dx 2 a2 a
eb ln11 ee 2
a
a 2 x 2 dx
0
a 4 a 2 b 2 x 2 b a 2 x 2 a a a 2 x 2
a
a 4 c 2x 2 dx
0
2 b
t x 4 ⇒ y x 4 2 Parabola
0
4 ab 2 3
2
2
a 0
abe arcsine
26. x 3 3 cos , y 2 5 sin
24. x t 4, y t2
a
dx
2 b cx cx a 4 c 2x 2 a 4 arcsin 2 a 2c a
c a b 2 a c a 2 c 2 a 4 arcsin a 2c a
2 b 2 2 1 a x x3 a2 3
x 3 3 y 5 2 2
2
1
28. x 5 sin3 , y 5 cos3
5x
23
x 3 2 y 2 2 1 9 25
y 7 6
23
1
y 6 4
4
y
3
2
2 1 −1
5y
x23 y23 523
Ellipse
5
x 1
2
3
4
5
6
7
7 6 5 4 3 2 1 −2 −1 −2 −3
−6
x
−4
2 −4 −6
x 1 2 3 4 5 6 7 8
4
6
464
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
30. x h 2 y k 2 r 2
y2 x2 1 16 9
32. a 4, c 5, b2 c2 a2 9,
x 5 2 y 3 2 2 2 4 Let
x2 y2 sec 2 and tan 2 . 16 9
Then x 3 tan and y 4 sec . 34. x a b cos t b cos y a b sin t b sin
a b b t
a b b t (b) a 3, b 1
(a) a 2, b 1
(c) a 4, b 1
x cos t cos t 2 cos t
x 2 cos t cos 2t
x 3 cos t cos 3t
y sin t sin t 0
y 2 sin t sin 2t
y 3 sin t sin 3t
4
2
4
y=0 −2 ≤ x ≤ 2 −3
−6
3
−6
6
6
−4
−2
−4
(e) a 3, b 2
(d) a 10, b 1 x 9 cos t cos 9t
(f) a 4, b 3
x cos t 2 cos
y 9 sin t sin 9t
y sin t 2 sin
10
−15
x cos t 3 cos
t 2
y sin t 3 sin
4
15
−6
−10
t 2
rcos sin
−6
6
6
−4
38. x t 4 y t2
y v w r sin r cos
(a)
r sin cos
dy 2t 2t 0 when t 0. dx 1 Point of horizontal tangency: 4, 0
y
(b) t x 4 y x 4 2
θ t
v
(c)
rθ
θ w
r
u
y 6 5
( x, y ) x
t 3
4
−4
36. x t u r cos r sin
t 3
4 3 2 1 x 1
2
3
4
5
6
Review Exercises for Chapter 9 40. x
1 t
42. x 2t 1
y t2 (a)
dy 2t 2t 3 dx 1t 2 No horizontal tangents
y
1 t 2 2t
(a)
dy t 2 2t 2 2t 2 dx 2
t 0
1 (b) t x
1t 0 when t 1. t 2t 2 2
Point of horizontal tangency: 1, 1
1 y 2 x
x1 2
(b) t
(c)
y
1 4 x 1 2 2 2 x 1 2 x 3 x 1
y
4 3
(c)
y
2
2
1 x
−2 −2
2
4
x
−1
1
2
44. x 6 cos
46. x e t
y 6 sin (a)
465
y et
dy 6 cos 3 cot 0 when , . dx 6 sin 2 2
(a)
Points of horizontal tangency: 0, 6 , 0, 6 (b)
6x 6y
(c)
y
2
2
1 1 dy et 2t 2 dx et e x No horizontal tangents
(b) t ln x
1
1 y eln x e ln1x , x > 0 x (c)
4
y
3
2 −4
x
−2
2
2
4
−2 1
−4
x 1
48. x 2 sin
50.
3
x 6 cos y 6 sin
y 2 cos (a), (c)
2
dx 6 sin d
8
−8
dy 6 cos d
8
−4
3 dx (b) At ,
1.134, 2 , 6 d 2
dy dy 0.5, and
0.441 dt dx
s
36 sin 2 36 cos 2 d 6
0
(one-half circumference of circle)
0
6
466
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
52. x, y 1, 3
y
(−1, 3)
r 1 2 3 2 10
2
arctan 3 1.89 108.43
1
r, 10, 1.89 , 10, 5.03
−3
−2
x
−1
1
2
3
−1 −2 −3
r 10
54.
r
56.
r 2 100
1 2 cos
2r r cos 1
x 2 y 2 100
2 ± x 2 y 2 x 1 4x 2 y 2 x 1 2 3x 2 4y 2 2x 1 0
58.
r 4 sec
4 3 cos 3
60.
3 4
tan 1
4 12 cos 32 sin
r cos 3 sin 8
y 1 x
x 3 y 8
y x
64. x 2 y 2 arctan
62. x 2 y 2 4x 0 r 2 4r cos 0
12
2
a2
r 2 2 a 2
r 4 cos
66.
y x
π 2
68. r 3 csc , r sin 3, y 3
π 2
Horizontal line
Line 0 1
2 0 1
70. r 3 4 cos Limaçon Symmetric to polar axis π 2
2
0
3
2
2 3
r
1
1
3
5
7
0
2
3
4
Review Exercises for Chapter 9
467
72. r 2 Spiral Symmetric to 2 π 2
0 2
4
8
0
r
0
4 5
2
3 4 3 2
5 4 5 2
2
3 2 3
74. r cos 5
π 2
Rose curve with five petals Symmetric to polar axis
2 3 4 , 1, , 1, , 1, 5 5 5 5 3 7 9 Tangents at the pole: , , , , 10 10 2 10 10
Relative extrema: 1, 0 , 1,
76. r 2 cos2 Lemniscate Symmetric to the polar axis Relative extrema: ± 1, 0
0
r
±1
6 ±
0 1
π 2
4
2
2
0 0 1 2
3 Tangents at the pole: , 4 4
78. r 2 sin cos 2
80. r 4sec cos
0.75
Bifolium
3
Semicubical parabola
Symmetric to 2 −1
1 −0.25
Symmetric to the polar axis r ⇒ as ⇒ 2 r ⇒ as ⇒ 2
−1
5
−3
82. r 2 4 sin2 (a) 2r
ddr 8 cos2
(b)
4 cos2 dr d r
Tangents at the pole: 0, (c)
2
−3
dy r cos 4 cos 2 sin r dx r sin 4 cos 2 cos r
3
2
cos2 sin sin2 cos cos2 cos sin2 sin
Horizontal tangents: dy 0 when cos2 sin sin 2 cos 0, dx
tan tan 2 , 0, , 0, 0 , ± 2 3, 3 3
Vertical tangents when cos 2 cos sin 2 sin 0: −2
tan 2 tan 1, 0, , 0, 0 , ± 2 3, 6 6
468
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
84. False. There are an infinite number of polar coordinate representations of a point. For example, the point x, y 1, 0 has polar representations r, 1, 0 , 1, 2 , 1, , etc. 86. r a sin , r a cos The points of intersection are a 2, 4 and 0, 0 . For r a sin , dy a cos sin a sin cos 2 sin cos . dx a cos 2 a sin 2 cos 2
m1
At a 2, 4 , m1 is undefined and at 0, 0 , m1 0. For r a cos , dy a sin 2 a cos 2 cos 2 . dx a sin cos a cos sin 2 sin cos
m2
At a 2, 4 , m2 0 and at 0, 0 , m2 is undefined. Therefore, the graphs are orthogonal at a 2, 4 and 0, 0 . 90. r 4 sin 3
88. r 51 sin
1 A2 2
32
2
51 sin
d 117.81 75 2
2
A3
1 2
3
4 sin 3 2 d
0
12.57 4
4 −8
4
8
−6
6
−12 −4
92. r 3, r 2 18 sin 2
4
9 r 2 18 sin 2 −6
sin 2
−4
12 A2
1 2
6
1 2
12
18 sin 2 d
0
1 2
512
12
9 d
1 2
2
512
18 sin 2 d
1.2058 9.4248 1.2058 11.84
94. r e, 0 ≤ ≤ A
1 2
96. r a cos 2,
e 2 d 133.62
0
s8
4
dr 2a sin 2 d
a 2 cos 2 2 4a 2 sin 2 2 d
0
10
8a
4
1 3 sin 2 2 d (Simpson’s Rule: n 4)
0
−25
5
−5
a 1 41.1997 21.5811 41.8870 2 6
9.69a
Review Exercises for Chapter 9
50. a 4, c 5, b 3, e r2
5 4
52. a 2, b 1, c 3, e
9 1 2516 cos 2
54. A 2
1 2
2
r2
2
1 1 34 cos 2
2
1 d 3.37 3 2 sin 2 2
ed 1 e cos
56. (a) r
3
3 22 sin d 4 2
2
(b) The perihelion distance is a c a ea a1 e .
1 e2 a a1 e . 1e
When 0, r c a ea a a1 e .
When , r
Therefore,
The aphelion distance is a c a ea a1 e . ed a1 e 1e
When 0, r
a1 e 1 e ed
1 e2 a a1 e . 1e
a1 e 2 ed. Thus, r
1 e2 a . 1 e cos
58. a 1.427 109 km
60. a 36.0 10 6 mi, e 0.206
e 0.0543 r
r
1.422792505 109 1 e 2 a 1 e cos 1 0.0543 cos
34.472 10 6 1 e 2 a
1 e cos 1 0.206 cos
Perihelion distance: a1 e 28.582 10 6 mi Aphelion distance: a1 e 43.416 10 6 mi
Perihelion distance: a1 e 1.3495139 109 km Aphelion distance: a1 e 1.5044861 109 km r a sin b cos
62.
461
r 2 ar sin br cos x 2 y 2 ay bx x 2 y 2 bx ay 0 represents a circle.
Review Exercises for Chapter 9 2. Matches (b) - hyperbola
4. Matches (c) - hyperbola
6. y 2 12y 8x 20 0
y
y 2 12y 36 8x 20 36
16
y 6 2 42 x 2
12
Parabola Vertex: 2, 6 −4
x 8
12
462
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
8. 4x 2 y 2 16x 15 0
y
4x 2 4x 4 y 2 15 16
1
(2, 0)
x 2 2 y 2 1 14 1
x 1
2
3
−1
Ellipse Center: 2, 0
−2
Vertices: 2, ± 1 4x 2 4y 2 4x 8y 11 0
10.
y 4
1 4 y 2 2y 1 11 1 4 4
4 x2 x
x 12 2 y 1 2 1 2 2
3
1 x
−2
1
Hyperbola
−1
12, 1 1 Vertices: ± 2, 1 2 1 Asymptotes: y 1 ± x 2
−2
3
4
Center:
12. Vertex: 4, 2 Focus: 4, 0 Parabola opens downward p 2
x 4 2 42 y 2 x 2 8x 8y 0 14. Center: 0, 0
16. Foci: 0, ± 8
Solution points: 1, 2 , 2, 0
Asymptotes: y ± 4x
Substituting the values of the coordinates of the given points into
Center: 0, 0
x2 y2 2 1, b2 a
c8 a y x 4x asymptote → a 4b b
we obtain the system
b1 a4 1, 4b 2
2
2
1.
18.
16 x2 3y 2 1. and b 2 4, 3 4 16
21 y2 x2 1, a 5, b 2, c 21, e 4 25 5
By Example 5 of Section 9.1, C 20
2
0
b2 c2 a2 64 4b 2 ⇒ 17b2 64 ⇒ b2
Solving the system, we have a2
Vertical transverse axis
1 2521 sin d 23.01. 2
64 1024 ⇒ a2 17 17
x2 y2 1 102417 6417
Review Exercises for Chapter 9
463
1 2 x 200
20. y
(a) x 2 200y
y
1 2 x 200
y
1 x 100
(b)
x2 450 y Focus: 0, 50
x 1 10,000 x dx 38,294.49 S 2 x 1 10,000 2
1 y 2
100
2
0
a
22. (a) A 4
0
b 4b 1 a 2 x 2 dx a a 2
b
(b) Disk: V 2
0
b
S 4
0
4 a b2
b
b 2 y 2 dy
0
a 0
ab
2 a 2 2 1 b y y3 b2 3
b
4 a 2b 3
0
b
b 4 c 2y 2 dy
0
2 a cy b 4 c 2y 2 b 4 ln cy b 4 c 2y 2 b 2c
0
b
2 a 2 b c b 2 c 2 b 4 ln cb b b 2 c 2 b 4 lnb 2 b 2c
ab 2 ca ln c e
a
(c) Disk: V 2
0 a
S 22
0
a2 2 2 a 2 b y 2 dy 2 b2 b
x a
b 4 a 2 b 2 y 2 a b 2 y 2 dy b b b 2 y 2
2 a 2
x a 2 x 2 a 2 arcsin
4 b a2
2
2 a 2
b2 2 2 b 2 a x 2 dx 2 a2 a
eb ln11 ee 2
a
a 2 x 2 dx
0
a 4 a 2 b 2 x 2 b a 2 x 2 a a a 2 x 2
a
a 4 c 2x 2 dx
0
2 b
t x 4 ⇒ y x 4 2 Parabola
0
4 ab 2 3
2
2
a 0
abe arcsine
26. x 3 3 cos , y 2 5 sin
24. x t 4, y t2
a
dx
2 b cx cx a 4 c 2x 2 a 4 arcsin 2 a 2c a
c a b 2 a c a 2 c 2 a 4 arcsin a 2c a
2 b 2 2 1 a x x3 a2 3
x 3 3 y 5 2 2
2
1
28. x 5 sin3 , y 5 cos3
5x
23
x 3 2 y 2 2 1 9 25
y 7 6
23
1
y 6 4
4
y
3
2
2 1 −1
5y
x23 y23 523
Ellipse
5
x 1
2
3
4
5
6
7
7 6 5 4 3 2 1 −2 −1 −2 −3
−6
x
−4
2 −4 −6
x 1 2 3 4 5 6 7 8
4
6
464
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
30. x h 2 y k 2 r 2
y2 x2 1 16 9
32. a 4, c 5, b2 c2 a2 9,
x 5 2 y 3 2 2 2 4 Let
x2 y2 sec 2 and tan 2 . 16 9
Then x 3 tan and y 4 sec . 34. x a b cos t b cos y a b sin t b sin
a b b t
a b b t (b) a 3, b 1
(a) a 2, b 1
(c) a 4, b 1
x cos t cos t 2 cos t
x 2 cos t cos 2t
x 3 cos t cos 3t
y sin t sin t 0
y 2 sin t sin 2t
y 3 sin t sin 3t
4
2
4
y=0 −2 ≤ x ≤ 2 −3
−6
3
−6
6
6
−4
−2
−4
(e) a 3, b 2
(d) a 10, b 1 x 9 cos t cos 9t
(f) a 4, b 3
x cos t 2 cos
y 9 sin t sin 9t
y sin t 2 sin
10
−15
x cos t 3 cos
t 2
y sin t 3 sin
4
15
−6
−10
t 2
rcos sin
−6
6
6
−4
38. x t 4 y t2
y v w r sin r cos
(a)
r sin cos
dy 2t 2t 0 when t 0. dx 1 Point of horizontal tangency: 4, 0
y
(b) t x 4 y x 4 2
θ t
v
(c)
rθ
θ w
r
u
y 6 5
( x, y ) x
t 3
4
−4
36. x t u r cos r sin
t 3
4 3 2 1 x 1
2
3
4
5
6
Review Exercises for Chapter 9 40. x
1 t
42. x 2t 1
y t2 (a)
dy 2t 2t 3 dx 1t 2 No horizontal tangents
y
1 t 2 2t
(a)
dy t 2 2t 2 2t 2 dx 2
t 0
1 (b) t x
1t 0 when t 1. t 2t 2 2
Point of horizontal tangency: 1, 1
1 y 2 x
x1 2
(b) t
(c)
y
1 4 x 1 2 2 2 x 1 2 x 3 x 1
y
4 3
(c)
y
2
2
1 x
−2 −2
2
4
x
−1
1
2
44. x 6 cos
46. x e t
y 6 sin (a)
465
y et
dy 6 cos 3 cot 0 when , . dx 6 sin 2 2
(a)
Points of horizontal tangency: 0, 6 , 0, 6 (b)
6x 6y
(c)
y
2
2
1 1 dy et 2t 2 dx et e x No horizontal tangents
(b) t ln x
1
1 y eln x e ln1x , x > 0 x (c)
4
y
3
2 −4
x
−2
2
2
4
−2 1
−4
x 1
48. x 2 sin
50.
3
x 6 cos y 6 sin
y 2 cos (a), (c)
2
dx 6 sin d
8
−8
dy 6 cos d
8
−4
3 dx (b) At ,
1.134, 2 , 6 d 2
dy dy 0.5, and
0.441 dt dx
s
36 sin 2 36 cos 2 d 6
0
(one-half circumference of circle)
0
6
466
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
52. x, y 1, 3
y
(−1, 3)
r 1 2 3 2 10
2
arctan 3 1.89 108.43
1
r, 10, 1.89 , 10, 5.03
−3
−2
x
−1
1
2
3
−1 −2 −3
r 10
54.
r
56.
r 2 100
1 2 cos
2r r cos 1
x 2 y 2 100
2 ± x 2 y 2 x 1 4x 2 y 2 x 1 2 3x 2 4y 2 2x 1 0
58.
r 4 sec
4 3 cos 3
60.
3 4
tan 1
4 12 cos 32 sin
r cos 3 sin 8
y 1 x
x 3 y 8
y x
64. x 2 y 2 arctan
62. x 2 y 2 4x 0 r 2 4r cos 0
12
2
a2
r 2 2 a 2
r 4 cos
66.
y x
π 2
68. r 3 csc , r sin 3, y 3
π 2
Horizontal line
Line 0 1
2 0 1
70. r 3 4 cos Limaçon Symmetric to polar axis π 2
2
0
3
2
2 3
r
1
1
3
5
7
0
2
3
4
Review Exercises for Chapter 9
467
72. r 2 Spiral Symmetric to 2 π 2
0 2
4
8
0
r
0
4 5
2
3 4 3 2
5 4 5 2
2
3 2 3
74. r cos 5
π 2
Rose curve with five petals Symmetric to polar axis
2 3 4 , 1, , 1, , 1, 5 5 5 5 3 7 9 Tangents at the pole: , , , , 10 10 2 10 10
Relative extrema: 1, 0 , 1,
76. r 2 cos2 Lemniscate Symmetric to the polar axis Relative extrema: ± 1, 0
0
r
±1
6 ±
0 1
π 2
4
2
2
0 0 1 2
3 Tangents at the pole: , 4 4
78. r 2 sin cos 2
80. r 4sec cos
0.75
Bifolium
3
Semicubical parabola
Symmetric to 2 −1
1 −0.25
Symmetric to the polar axis r ⇒ as ⇒ 2 r ⇒ as ⇒ 2
−1
5
−3
82. r 2 4 sin2 (a) 2r
ddr 8 cos2
(b)
4 cos2 dr d r
Tangents at the pole: 0, (c)
2
−3
dy r cos 4 cos 2 sin r dx r sin 4 cos 2 cos r
3
2
cos2 sin sin2 cos cos2 cos sin2 sin
Horizontal tangents: dy 0 when cos2 sin sin 2 cos 0, dx
tan tan 2 , 0, , 0, 0 , ± 2 3, 3 3
Vertical tangents when cos 2 cos sin 2 sin 0: −2
tan 2 tan 1, 0, , 0, 0 , ± 2 3, 6 6
468
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
84. False. There are an infinite number of polar coordinate representations of a point. For example, the point x, y 1, 0 has polar representations r, 1, 0 , 1, 2 , 1, , etc. 86. r a sin , r a cos The points of intersection are a 2, 4 and 0, 0 . For r a sin , dy a cos sin a sin cos 2 sin cos . dx a cos 2 a sin 2 cos 2
m1
At a 2, 4 , m1 is undefined and at 0, 0 , m1 0. For r a cos , dy a sin 2 a cos 2 cos 2 . dx a sin cos a cos sin 2 sin cos
m2
At a 2, 4 , m2 0 and at 0, 0 , m2 is undefined. Therefore, the graphs are orthogonal at a 2, 4 and 0, 0 . 90. r 4 sin 3
88. r 51 sin
1 A2 2
32
2
51 sin
d 117.81 75 2
2
A3
1 2
3
4 sin 3 2 d
0
12.57 4
4 −8
4
8
−6
6
−12 −4
92. r 3, r 2 18 sin 2
4
9 r 2 18 sin 2 −6
sin 2
−4
12 A2
1 2
6
1 2
12
18 sin 2 d
0
1 2
512
12
9 d
1 2
2
512
18 sin 2 d
1.2058 9.4248 1.2058 11.84
94. r e, 0 ≤ ≤ A
1 2
96. r a cos 2,
e 2 d 133.62
0
s8
4
dr 2a sin 2 d
a 2 cos 2 2 4a 2 sin 2 2 d
0
10
8a
4
1 3 sin 2 2 d (Simpson’s Rule: n 4)
0
−25
5
−5
a 1 41.1997 21.5811 41.8870 2 6
9.69a
Problem Solving for Chapter 9
98. r
2 ,e1 1 cos
100. r
4 45 3 ,e 5 3 sin 1 35sin 5
Ellipse
Parabola
π 2
π 2
0 2 0 1
102. r
8 4 5 ,e 2 5 cos 1 52cos 2
Hyperbola
2
104. Line Slope: 3 Solution point: 0, 0
π 2
y 3 x, r sin 3 r cos , tan 3,
3
0 1
2
106. Parabola
Vertex: 2, 2 Focus: 0, 0 e 1, d 4 4 r 1 sin
108. Hyperbola Vertices: 1, 0, 7, 0 Focus: 0, 0 4 7 a 3, c 4, e , d 3 4
4374 7 r 4 3 4 cos 1 cos 3
Problem Solving for Chapter 9 2. Assume p > 0. Let y mx p be the equation of the focal chord.
y
x2
= 4py (0, p)
First find x-coordinates of focal chord endpoints: x
x2 4py 4pmx p x2 4pmx 4p2 0
y = −p
4pm ± 16p2m2 16p2 2pm ± 2pm2 1 2 x x2 4py, 2x 4py ⇒ y . 2p
x
(a) The slopes of the tangent lines at the endpoints are perpendicular because 1 1 2pm 2pm2 1 2p 2pm 2pm2 1 4p1 2 4p2m2 4p2m2 1 4p1 2 4p2 1 2p —CONTINUED—
469
470
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
2. —CONTINUED— (b) Finally, we show that the tangent lines intersect at a point on the directrix y p. Let b 2pm 2pm2 1 and c 2pm 2pm2 1. b2 8p2m2 4p2 8p2mm2 1 c2 8p2m2 4p2 8p2mm2 1 b2 2pm2 p 2pmm2 1 4p c2 2pm2 p 2pmm2 1 4p Tangent line at x b: y
b2 b bx b2 x b ⇒ y 4p 2p 2p 4p
Tangent line at x c: y
c2 c c2 cx x c ⇒ y 4p 2p 2p 4p bx b2 cx c2 2p 4p 2p 4p
Intersection of tangent lines:
2bx b2 2cx c2 2xb c b2 c2 2x4pm2 1 16p2mm2 1 x 2pm Finally, the corresponding y-value is y p, which shows that the intersection point lies on the directrix.
4.
y2 x2 2 1, a2 b2 c2, MF2 Mf1 2a 2 a b y
y
β
b2x a2y
α
F2(−c, 0)
M (x 0 , y 0 ) x
b2x y y0 2 0 x x0 a y0
Tangent line at Mx0, y0:
β
F1(c, 0) Q
yy0 y02 x0 x x02 b2 a2 x0 x y0 y x02 y02 2 2 2 a2 b a b x0 x y0 y 2 1 a2 b
QF2 QF1 MQ
b2 b2 ⇒ Q 0, . y0 y0
At x 0, y
x
2
0
c2
b4 2d y0
y0
b2 y0
2
f
By the Law of Cosines,
F2Q2 MF2 2 MQ2 2MF2 MQcos d 2 MF2 2 f 2 2 f MF2 cos
F1Q2 MF12 f 2 2 f MF1cos d 2 MF12 f 2 2 f MF1cos . cos
F2 2 f 2 d 2 MF12 f 2 d2 , cos 2 f MF2 2 f MF1
MF2 MF1 2a. Let z MF1. Slopes: MF1: —CONTINUED—
y0 b2 b2 ; QF1: ; QF2: x0 c y0c y0c
Problem Solving for Chapter 9
471
4. —CONTINUED— To show , consider
MF2 2 f 2 d2 2 f MF1 MF12 f 2 d 2 2 f MF2 ⇔
z 2a2 f 2 d 2 z z2 f 2 d 2 z 2a
⇔
z2 2az f 2 d 2
⇔
x0 c2 y02 2az x02 y0
⇔
b2 y0
c 2
2
b4 y02
az x0c a2 0
⇔
ax0 c2 y02 x0c a2
⇔
x02b2 a2y02 a2b2 x02 y02 2 1. a2 b
⇔
Thus, and the reflective property is verified.
6. (a) y2
t21 t22 2 1 t22 ,x 1 t22 1 t22
11 tt 2t 1x 1x 1t 2 1 1t 2 2
2
r cos sin2 sin2 cos2 r cos3
t2
r cos sin2 cos2 cos2 sin2
2
r cos cos 2
11 xx.
r cos 2
(d) r 0 for
π 2
(c)
11 rr cos cos
sin2 1 r cos cos2 1 r cos
2
1
Thus, y2 x2
r 2 sin2 r 2 cos2
(b)
sec
3 , . 4 4
Thus, y x and y x are tangent lines to curve at the origin. 0 1
(e) yt
2
1 t21 3t2 t t32t 1 4t2 t4 0 1 t22 1 t22
t4 4t2 1 0 ⇒ t2 2 ± 5 ⇒ x
5 1
2
,±
5 1
2
2 5
1 2 ± 5 3 5 1 2 ± 5 1 ± 5 5 1 3 5 1 5 2
472
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
10. r
8. (a)
ab ,0 ≤ ≤ a sin b cos 2
r a sin b cos ab ay bx ab y x 1 b a Line segment
Generated by Mathematica
u2 du, (b) x, y cos 2 0 t
u2 sin du is on 2 0 t
1 Area ab 2
the curve whenever x, y is on the curve. (c) xt cos
t2 t2 , yt sin , xt2 yt2 1 2 2
a
Thus, s
dt a.
0
On , , s 2. 12. Let r, be on the graph. r 2 1 2r cos r 2 1 2r cos 1
r 2 12 4r 2 cos2 1 r 4 2r 2 1 4r 2 cos2 1 r 2r 2 4 cos2 2 0 r 2 4 cos2 2 r 2 22 cos2 1 r 2 2 cos 2 14. (a) r 2
(c) r 2 2 cos
4
4
Cardioid
Circle radius 2 −6
−6
6
−4
−4
(b) r 2 cos
(d) r 2 3 cos
4
Convex limaçon −6
6
−4
16. The curve is produced over the interval 0 ≤ ≤ 9.
6
Limaçon with inner loop
4
−6
6
−4
C H A P T E R 9 Conics, Parametric Equations, and Polar Coordinates Section 9.1
Conics and Calculus . . . . . . . . . . . . . . . . . . . . 177
Section 9.2
Plane Curves and Parametric Equations . . . . . . . . . . 188
Section 9.3
Parametric Equations and Calculus
Section 9.4
Polar Coordinates and Polar Graphs . . . . . . . . . . . . 198
Section 9.5
Area and Arc Length in Polar Coordinates
Section 9.6
Polar Equations of Conics and Kepler’s Laws . . . . . . . 210
Review Exercises
. . . . . . . . . . . . 192
. . . . . . . . 205
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222
C H A P T E R 9 Conics, Parametric Equations, and Polar Coordinates Section 9.1
Conics and Calculus
Solutions to Odd-Numbered Exercises 3. x 32 2 y 2
1. y 2 4x
5.
Vertex: 0, 0
Vertex: 3, 2
p1 > 0
p 12 < 0
Opens to the right Matches graph (h).
Opens downward Matches graph (e).
x2 y 2 1 9 4
7.
Center: 0, 0 Ellipse Matches (f)
Hyperbola Center: 0, 0 Vertical transverse axis. Matches (c)
9. y 2 6x 4 32 x
11. x 3 y 22 0
Vertex: 0, 0
y 22 4 14 x 3
y
3 Focus: 2 , 0
Directrix: x
y2 x2 1 16 1
8
3 2
Vertex: 3, 2 (0, 0)
12
8
y
Focus: 3.25, 2
4 x
4
4
Directrix: x 2.75
4
(− 3, 2) −8
−6
2
−4
x
−2 −2
8
−4
13. y2 4y 4x 0
15. x2 4x 4y 4 0
y 2 4y 4 4x 4
x 2 4x 4 4y 4 4
y 22 41x 1
x 22 41 y 2
Vertex: 1, 2
Vertex: 2, 2
y
Focus: 0, 2
6
Directrix: x 2
4
y
Focus: 2, 1
4
(− 2, 2)
Directrix: y 3 (−1, 2) −6
x
2
2 2
4
−4
x
−2
2 −2
6 −4
177
178
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
17. y 2 x y 0
19. y 2 4x 4 0
y2 y 14 x 14
y 12 2 4 14 x 14 1 1 Vertex: 4 , 2 1 Focus: 0, 2
y 2 4x 4 41x 1 Vertex: 1, 0
2
4
Focus: 0, 0
−5
2
Directrix: x 2
1
Directrix: x 2
−6
6
−3
−4
y 22 42x 3
21.
23.
x h2 4p y k x2 46 y 4
y 2 4y 8x 20 0
x2 24y 96 0 y 4 x2
25.
x2 y 4 0 27. Since the axis of the parabola is vertical, the form of the equation is y ax2 bx c. Now, substituting the values of the given coordinates into this equation, we obtain
29. x2 4y 2 4 x2 y 2 1 4 1
3 c, 4 9a 3b c, 11 16a 4b c. Solving this system, we have a Therefore,
5 3, b
14 3,
c 3.
Foci: ± 3, 0
−2
3
2
y
33.
12
(1, 5)
Center: 1, 5
9x2 4y 2 36x 24y 36 0 9x2 4x 4 4 y2 6y 9 36 36 36 36
4
x 22 y 32 1 4 9
x 8
4
4
8
Vertices: 1, 10, 1, 0 4 e 5
x
1
Vertices: ± 2, 0
a2 25, b2 9, c2 16 Foci: 1, 9, 1, 1
(0, 0) −1
Center: 0, 0
e
x 12 y 52 1 9 25
2
a2 4, b2 1, c2 3
2 y 53 x2 14 3 x 3 or 5x 14x 3y 9 0.
31.
y
a2 9, b2 4, c2 5 y
Center: 2, 3 Foci: 2, 3 ± 5
6
Vertices: 2, 6, 2, 0 e
(− 2, 3) 2
5
3
x 6
4
2
2
Section 9.1 12x2 20y 2 12x 40y 37 0
35.
12 x2 x
x
2
3x
a2 4, b2 2, c 2 2
a2 5, b2 3, c2 2 Center: Foci:
Center:
12, 1
Foci:
12 ± 2, 1
Vertices:
9 1 9 2 y 2 2y 1 2 4 4 4 4
x 32 2 y 12 1 4 2
60
x 12 2 y 12 1 5 3
32, 1
32 ± 2, 1 21, 1, 72, 1
Vertices:
12 ± 5, 1
Solve for y: 2 y 2 2y 1 x2 3x
Solve for y:
y 12
20 y 2 2y 1 12x2 12x 37 20
y 12
57 12x 12x2 20
y 1 ±
57 12x20 12x
y 1 ±
2
1 −2
−3
4
3
−3 −3
41. Vertices: 3, 1, 3, 9 Minor axis length: 6 Vertical major axis Center: 3, 5
39. Center: 0, 0 Focus: 2, 0 Vertex: 3, 0 Horizontal major axis a 3, c 2 ⇒ b 5
a 4, b 3
x2
x 32 y 52 1 9 16
9
y2 5
1
43. Center: 0, 0 Horizontal major axis Points on ellipse: 3, 1, 4, 0 Since the major axis is horizontal,
ax by 1. 2
2
2
2
Substituting the values of the coordinates of the given points into this equation, we have
a9 b1 1, and 16a 1. 2
2
2
The solution to this system is a2 16, b2 167. Therefore, x2 y2 x2 7y 2 1, 1. 16 167 16 16
1 2 4
1 7 3x x2 2 4
7 12x8 4x
(Graph each of these separately.)
(Graph each of these separately.) 1
179
x2 2y 2 3x 4y 0.25 0
37.
1 20y 2 2y 1 37 3 20 4
Conics and Calculus
2
180
45.
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
y 2 x2 1 1 4
47.
x 12 y 22 1 4 1
a 1, b 2, c 5
a 2, b 1, c 5
Center: 0, 0
Center: 1, 2
Vertices: 0, ± 1
Vertices: 1, 2, 3, 2
1 Asymptotes: y ± x 2
1 Asymptotes: y 2 ± x 1 2
Foci: 1 ± 5, 2
Foci: 0, ± 5
y
y
4
1 x
2
1 x
4
2
2
1
2
3
2
4
2 4 5
4
9x2 y 2 36x 6y 18 0
49.
x2 9y 2 2x 54y 80 0
51.
9x2 4x 4 y 2 6y 9 18 36 9
x2 2x 1 9 y 2 6y 9 80 1 81 0 x 12 9 y 32 0
x 22 y 32 1 1 9
1 y 3 ± x 1 3
a 1, b 3, c 10 Center: 2, 3
Degenerate hyperbola is two lines intersecting at 1, 3.
Vertices: 1, 3, 3, 3 Foci: 2 ± 10, 3
y
Asymptotes: y 3 ± 3x 2
x
4
y
2
2 2
x
2
2
4
4
6
2
6
4 6
53.
9y 2 x2 2x 54y 62 0
55.
9 y 2 6y 9 x2 2x 1 62 1 81 18
3x2 2y 2 6x 12y 27 0 3x2 2x 1 2 y 2 6y 9 27 3 18 12
y 32 x 12 1 2 18 a 2, b 32, c 25 Center: 1, 3
x 12 y 32 1 4 6
−5
7
Vertices: 1, 3 ± 2
9 y2 6y 9 x2 2x 62 81 x2 2x 19 9
1 y 3 ± x2 2x 19 3 (Graph each curve separately.)
1 −5
7
Foci: 1 ± 10, 3 −7
Solve for y:
Center: 1, 3 Vertices: 1, 3, 3, 3
Foci: 1, 3 ± 25
y 32
a 2, b 6, c 10
1
−7
Solve for y:
2 y 2 6y 9 3x2 6x 27 18
y 32
3x2 6x 9 2
y 3 ±
3x
(Graph each curve separately.)
2
2x 3 2
Section 9.1 57. Vertices: ± 1, 0 Asymptotes: y ± 3x Horizontal transverse axis Center: 0, 0
181
59. Vertices: 2, ± 3 Point on graph: 0, 5 Vertical transverse axis Center: 2, 0 a3
b b a 1, ± ± ± 3 ⇒ b 3 a 1 Therefore,
Conics and Calculus
Therefore, the equation is of the form y2 x 22 1. 9 b2
x2 y 2 1. 1 9
Substituting the coordinates of the point 0, 5, we have 25 4 21 9 b
9 or b2 . 4
Therefore, the equation is
61. Center: 0, 0 Vertex: 0, 2 Focus: 0, 4 Vertical transverse axis
63. Vertices: 0, 2, 6, 2 2 2 Asymptotes: y x, y 4 x 3 3 Horizontal transverse axis
a 2, c 4, b2 c2 a2 12 Therefore,
y2 x 22 1. 9 94
Center: 3, 2 a3
y2 x2 1. 4 12
b 2 Slopes of asymptotes: ± ± a 3 Thus, b 2. Therefore,
x 32 y 22 1. 9 4
65. (a)
x2 2x x y 2 1, 2yy 0, y 9 9 9y At x 6: y ± 3, y At 6, 3 : y 3
±6
93
(b) From part (a) we know that the slopes of the normal lines must be 9 23 .
± 23
At 6, 3 : y 3
9
23 x 6 9
or 9x 23y 60 0 At 6, 3 : y 3
or 2x 33y 3 0 At 6, 3 : y 3
9 x 6 23
23 x 6 9
9 23
x 6
or 9x 23y 60 0
or 2x 33y 3 0 67. x2 4y 2 6x 16y 21 0
69. y2 4y 4x 0
71. 4x2 4y 2 16y 15 0
A 1, C 4
A 0, C 1
AC4
AC 4 > 0
Parabola
Circle
Ellipse 73. 9x2 9y 2 36x 6y 34 0
3x2 6x 3 6 2y2 4y 2
75.
AC9
3x2 2y2 6x 4y 5 0
Circle
A 3, C 2, AC < 0 Hyperbola
182
Chapter 9
Conics, Parametric Equations, and Polar Coordinates 79. (a) A hyperbola is the set of all points x, y for which the absolute value of the difference between the distances from two distance fixed points (foci) is constant.
77. (a) A parabola is the set of all points x, y that are equidistant from a fixed line (directrix) and a fixed point (focus) not on the line. (b) x h2 4py k or y k2 4px h
(b)
(c) See Theorem 9.2.
y k2 x h2 x h2 y k2 1 or 1 2 2 a b a2 b2
b a (c) y k ± x h or y k ± x h a b 83. y ax2
81. Assume that the vertex is at the origin.
y 2ax
x2 4py
3 4p1 2
The equation of the tangent line is y ax02 2ax0x x0 or y 2ax 0 x ax 02.
9 p 4
Let y 0. Then:
The pipe is located 94 meters from the vertex.
ax02 2ax0x 2ax02
y
ax02 2ax0x
3
Focus
Therefore,
2
(− 3, 1)
x0 x is the x-intercept. 2
y
(3, 1) 1
−3
−2
−1
x 2
1
3
(x0, ax02 )
y = ax 2
x
( x2 , 0) 0
85. (a) Consider the parabola x2 4py. Let m0 be the slope of the one tangent line at x1, y1 and therefore, 1m0 is the slope of the second at x2, y2. From the derivative given in Exercise 32 we have: m0
1 x or x1 2pm0 2p 1
1 1 2p x2 or x2 m0 2p m0 Substituting these values of x into the equation x2 4py, we have the coordinates of the points of tangency 2pm0, pm02 and 2pm0, pm02 and the equations of the tangent lines are
y pm02 m0x 2pm0 and
2p x . y mp 1 m m 2
0
0
0
The point of intersection of these lines is
pm m 1, p and is on the directrix, y p. 2
0
y
x 2 = 4py
0
2p p − , 2 m0 m0
(
) (2pm0, pm02) x
y = −p
( p(mm− 1) , − p) 0
0
—CONTINUED—
Section 9.1
Conics and Calculus
85. —CONTINUED— (b) x2 4x 4y 8 0
x 22 4 y 1. Vertex 2, 1 2x 4 4
dy 0 dx dy 1 x1 dx 2
At 2, 5, dydx 2. At 3, 54 , dydx 12 . Tangent line at 2, 5: y 5 2x 2 ⇒ 2x y 1 0. Tangent line at 3, 54 : y 54 12 x 3 ⇒ 2x 4y 1 0. Since m1m2 2 12 1, the lines are perpendicular. 1 1 Point of intersection: 2x 1 x 2 4 5 5 x 2 4 x
1 2
y0 1 Directrix: y 0 and the point of intersection 2 , 0 lies on this line.
87.
y x x2 dy 1 2x dx At x1, y1 on the mountain, m 1 2x1. Also, m
y1 1 . x1 1
y1 1 1 2x1 x1 1
x1 x12 1 1 2x1x1 1 x12 x1 1 2x12 x1 1 x12 2x1 2 0 x1
2 ± 22 412 2 ± 23 1 ± 3 21 2
Choosing the positive value for x1, we have x1 1 3. m 1 2 1 3 3 23 m
01 1 x0 1 x0 1
1 3 23 Thus, x0 1 1 x0 1 3 23 3 23 1 x0 3 23 x0. 3 The closest the receiver can be to the hill is 233 1 0.155.
y 2
(− 1, 1) −2
1
−1
( x1 , y1 ) (x0, 0) 1
−1 −2
x
183
184
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
89. Parabola Vertex: 0, 4
Circle Center: 0, k Radius: 8
x2 4p y 4
x2 y k2 64
y
42 4p0 4
42 0 k2 64
p 1 −6
x2 4 y 4
−4
−2
4
6
k 43
8
−4
x2 y4 4
k2 48
x 2
−2
(Center is on the negative y-axis.)
x2 y 43 64
−6
2
−8
y 43 ± 64 x2 Since the y-value is positive when x 0, we have y 43 64 x2.
4
A2
4
0
x3 1 x 43x x64 x2 64 arcsin 12 2 8
64 1 163 248 32 arcsin 12 2
2 4x 2 16
x2 43 64 x2 dx 4
4 0
16 4 33 2
15.536 square feet 3
91. (a) Assume that y ax2.
y
(− 60, 20)
(60, 20) 20
2 1 1 2 x ⇒ y 20 a60 ⇒ a 360 180 180 2
(b) f x
1 2 1 x , fx x 180 90
60
1 x 1 90
S2
0
15 10
2
2 dx 90
5
60
902
dx
−60 −45 −30 −15
0
x2
60
2 1 x902 x2 902 ln x 902 x2 90 2
1 6011,700 902 ln 60 11,700 902 ln 90 90
1 180013 902 ln 60 3013 902 ln 90 90
2013 90 ln
60 9030
10 213 9 ln
13
(formula 26)
0
2 3 13 128.4 m
1 1 3 93. x2 4py, p , , 1, , 2 4 2 2
95.
y
As p increases, the graph becomes wider. 1
y
p=
1 4
p=2 p= p=
−8
3 2
1 2
x
8
3
p=1
24
−16
2
16
17
16
15
14
13
12
11
4
10
5
9
6
8
7
7
8
6
9
5
10
4
11
3
12
2
13
1
14
15
16
17
x
x 15 30 45 60
Section 9.1 5 97. a , b 2, c 2
52
2
22
Conics and Calculus
3 2
The tacks should be placed 1.5 feet from the center. The string should be 2a 5 feet long. c a
e
99.
y
A P 2a c
AP a 2
x
caP
103.
AP AP P 2 2
c A P2 A P a A P2 A P
e
101. e
(a, 0) P
A
35.34au 0.59au AP
0.9672 AP 35.34au 0.59au
y2 x2 1 102 52
105. 16x2 9y 2 96x 36y 36 0 32x 18yy 96 36y 0
2yy 2x 2 0 102 5 y
y18y 36 32x 96
52x x 102y 4y
At 8, 3: y
y
32x 96 18y 36
y 0 when x 3. y is undefined when y 2.
8 2 12 3
At x 3, y 2 or 6. Endpoints of major axis: 3, 2, 3, 6
The equation of the tangent line is y 3 3 x 8. It 2 25 will cross the y-axis when x 0 and y 3 8 3 3 . 2
At y 2, x 0 or 6. Endpoints of minor axis: 0, 2, 6, 2 Note: Equation of ellipse is
2
1 x 4 x2 dx x4 x2 4 arcsin 2 2
107. (a) A 4
0
2
V 2
(b) Disk:
0
2 0
or, A ab 21 2
2
1 1 1 4 x2 dx 4x x3 4 2 3
x 32 y 22 1 9 16
2 0
8 3
1 y 4 x2 2 y 1 y2
2
S 22
y
x 24 x2
1 16 x 4x 16 4y 3x
16 3x2
0
—CONTINUED—
4y
2
2
2
dx 2 3 3x16 3x
2
16 arcsin
3x
4
2 0
2 9 43 21.48 9
185
186
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
107. —CONTINUED—
2
V 2
(c) Shell:
2
x4 x2 dx
0
2x4 x212 dx
0
2 4 x232 3
2 0
16 3
x 21 y 2 x 1 x2
2y 1 y 2
1 1 4y y
1
S 22
21 y 2
2
1
0
3y2
1 y 2
2
1 3y 2 1 y 2
1
dy 8
1 3y 2 dy
0
436
8 3y1 3y 2 ln 3y 1 3y 2 23
0
1 e2 sin2 d
2100 10a ⇒ a 20
0
For
x2 25
y2 49
Hence, the length of the major axis is 2a 40.
1, we have
a 7, b 5, c 49 25 26, e
C 47
34.69
Area ellipse ab a10
2
C 4a
3 ln 2 3
111. Area circle r 2 100
109. From Example 5,
1
c 26 . a 7
2
0
1 4924 sin d 2
281.3558 37.9614 113. The transverse axis is horizontal since 2, 2 and 10, 2 are the foci (see definition of hyperbola).
115. 2a 10 ⇒ a 5 c 6 ⇒ b 11
Center: 6, 2 c 4, 2a 6, b2 c2 a2 7 Therefore, the equation is
x 62 y 22 1. 9 7
117. Time for sound of bullet hitting target to reach x, y: Time for sound of rifle to reach x, y:
2c x c2 y 2 vm vs
y
( x, y )
x c2 y 2
vs
2c x c2 y 2 x c2 y 2 Since the times are the same, we have: vm vs vs 4c2 4c x c2 y 2 x c2 y 2 x c2 y 2 2 vm vmvs vs2 vs2 x c2 y2
1 vv x 2 m 2 s
x2 c2vs2vm2
c2
vm2
2
y2
vm2x vs2c vsvm
vv
y2 1 vs2vm2
2 s 2 m
1 c2
(− c, 0) rifle
x
(c, 0) target
Section 9.1
Conics and Calculus
187
119. The point x, y lies on the line between 0, 10 and 10, 0. Thus, y 10 x. The point also lies on the hyperbola x236 y264 1. Using substitution, we have: x2 10 x2 1 36 64 16x2 910 x2 576 7x2 180x 1476 0 x
180 ± 1802 471476 27
180 ± 1922 90 ± 962 14 7
Choosing the positive value for x we have: x
90 962 160 962
6.538 and y
3.462 7 7 2y 2 x2 2y 2 x2 2 1 ⇒ 2 1 2, c 2 a2 b2 a2 b b a
121.
x2 2y 2 2y 2 x2 2 1 ⇒ 2 2 1 a2 b2 b b a b2 1
x2 x2 1 1 2 1 ⇒ 2 x2 2 2 2 a a b2 a a b2 x2
2aa2 b2 2ac 2a2a2 b2 ⇒ x± ± 2 2 2a b 2a2 b2 2a2 b2
2y2 1 2a2c 2 1 2 2 b a 2a2 b2 y2
⇒ 2yb
2
2
2a2
b2 b2
b4 b2 ⇒ y± 22a2 b2 22a2 b2
There are four points of intersection:
2ac 2a2
b2
,±
b2 , 22a2 b2
2ac 2a2
b2
,±
b2 22a2 b2
2y2 2x 4yy b2x x2 2 1 ⇒ 2 2 0 ⇒ ye 2 2 a b a b 2a y a2 At
2y 2 2x 4yy b 2x x2 2 1 ⇒ 2 2 0 ⇒ yh 2 2 b b c b 2c y 2ac
2a2
ye
, b2
b2 2a2
b2 , the slopes of the tangent lines are: 22a2 b2 2a b 2ac
2
22a
2
b2
2
b2
c a
b2 and
yh 2c 2
2ac 2a2 b2
22a
b2 2
b2
a c
Since the slopes are negative reciprocals, the tangent lines are perpendicular. Similarly, the curves are perpendicular at the other three points of intersection. 123. False. See the definition of a parabola.
125. True
127. False. y2 x2 2x 2y 0 yields two intersecting lines.
129. True
188
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
Section 9.2
Plane Curves and Parametric Equations
1. x t, y 1 t (b)
(a) t
0
1
2
3
4
x
0
1
2
3
2
y
1
y
1
1
0
2
x
−1
3
1
2
3
−1 −2
(c)
2
−3 −1
3
(d) x2 t y 1 x2, x ≥ 0
−4
5. x t 1
3. x 3t 1 y 2t 1 y2
x 3 1 1
7. x t 3
y t2
y 12 t 2
y x 12
x t 3 implies t x13 y 12 x 23
y
2x 3y 5 0
y
4 y
1 4 x
3 2
2
1
1
2
3
x
2
2
4
x
4
2 2
9. x t, t ≥ 0
11. x t 1
yt2
t t1
y t2
y
x1 x
y
4
x4 x 2 2 2 y
y
3
8
2
4 4
1
x −1
y
y x2 2, x ≥ 0 y
13. x 2t
2
3
4
5
2
6
x
4
−2
x
2
2
15. x et, x > 0
y 5
y e3t 1
4
y x3 1, x > 0
3 2 1 −2 −1
x −1
1
2
3
4
4
8
12
Section 9.2 17. x sec
19. x 3 cos , y 3 sin
y cos
x
y 2 cos 2 2
x sin2 2 16
x2 y 2 9.
xy 1 y
x 4 sin 2
21.
Squaring both equations and adding, we have
, < ≤ 2 2
0 ≤ <
Plane Curves and Parametric Equations
y
y2 cos2 2 4
4
1 x
2
≥ 1, y ≤ 1
y2 x2 1 16 4
x
4
2
2
4 4
2
y
4
3
−6
2
6
1 x
1
2
−4
3
2 3
x 4 2 cos
23.
x 4 2 cos
25.
y 1 sin
y 1 4 sin
x 4 cos2 4
x 42 cos2 4
y 12 sin2 1
y 12 sin2 16
2
x 42 y 12 1 4 1
x 42 y 12 1 4 16
2
3
−1
8
−2
−4
−5
x 4 sec
27.
10
29. x t 3
y 3 tan x2 sec2 16
x et
31.
y 3 ln t
y e3t
3 x ln x y 3 ln
et
2
y2 tan2 9
−1
x2 y2 1 16 9
1 x
3 y et 3 y
5
1 x
y
−2
6
1 x3
x > 0 y > 0
−9
9 3
−6 −1
5 −1
189
190
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
33. By eliminating the parameters in (a) – (d), we get y 2x 1. They differ from each other in orientation and in restricted domains. These curves are all smooth except for (b). (a) x t, y 2t 1
(b) x cos
3
1 ≤ y ≤ 3
dx dy 0 when 0, ± , ± 2, . . . . d d
2
y
1
−2
y 2 cos 1
1 ≤ x ≤ 1
y
x
−1
1
3
2
−1
2 1
−2
x
−1
1
2
−1
(c) x et
(d) x e t
y 2et 1
x > 0
y > 1
x > 0
y
y > 1
y
4
4
3
3
2
2
1
1 x
−1
y 2et 1
1
2
−1
3
x 1
2
3
35. The curves are identical on 0 < < . They are both smooth. Represent y 21 x2 37. (a)
4
(b) The orientation of the second curve is reversed.
4
(c) The orientation will be reversed. −6
−6
6
(d) Many answers possible. For example, x 1 t, y 1 2t, and x 1 t, x 1 2t.
−4
−4
39.
6
x x1 tx2 x1
x h a cos
41.
y y1 t y2 y1
y k b sin
x x1 t x2 x1 y y1
xh cos a
xx xx y 1
2
2
yk sin b
y1
1
x h2 y k2 1 a2 b2
y y1 y y1 2 x x1 x2 x1 y y1 mx x1 43. From Exercise 39 we have
45. From Exercise 40 we have
x 5t
x 2 4 cos
y 2t.
y 1 4 sin .
Solution not unique
Solution not unique
47. From Exercise 41 we have a 5, c 4 ⇒ b 3 x 5 cos y 3 sin . Center: 0, 0 Solution not unique
Section 9.2
Plane Curves and Parametric Equations
51. y 3x 2
49. From Exercise 42 we have a 4, c 5 ⇒ b 3
53. y x3
Example
Example
x 4 sec
x t,
y 3t 2
x t,
y t3
y 3 tan .
x t 3,
y 3t 11
3 t, x
yt
x tan t,
y tan3 t
Center: 0, 0 Solution not unique 55. x 2 sin
57. x 32 sin
59. x 3 cos3
y 21 cos
y 1 32 cos
y 3 sin3
5
5
4
−6 −2
−2
16
6
7 −1
−1
−4
Not smooth at 2n
61. x 2 cot
Not smooth at x, y ± 3, 0 and 0, ± 3, or 12 n. 63. See definition on page 665.
4
y 2 sin 2
−6
6
−4
Smooth everywhere 65. A plane curve C, represented by x f t, y gt, is smooth if f and g are continuous and not simultaneously 0. See page 670. 67. x 4 cos
69. x cos sin
y 2 sin 2
y sin cos
Matches (d)
Matches (b)
71. When the circle has rolled radians, we know that the center is at a, a. sin sin180
191
C BD b
AP cos cos180 b
b
or
or
BD b sin
AP b cos
Therefore, x a b sin and y a b cos . 73. False x t2 ⇒ x ≥ 0 x t2 ⇒ y ≥ 0 The graph of the parametric equations is only a portion of the line y x.
y
P b A
B
C θ a D
x
192
Chapter 9
75. (a) 100 mihr
Conics, Parametric Equations, and Polar Coordinates
1005280 440 ftsec 3600 3
x v0 cos t
(d) We need to find the angle (and time t) such that x
440 cos t 3
y3
y h v0 sin t 16t 2 3 (b)
sin t 16t 440 3
cos t 400 440 3 sin t 16t 440 3
10 3
1200 1200 sin 16 440 3 440 cos 440 cos
7 400 tan 16 400 0
It is not a home run—when x 400, y ≤ 20. (c)
10.
From the first equation t 1200440 cos . Substituting into the second equation,
2
30
0
2
400 tan 16
60
120 44
2
sec2
tan 120 44 2
2
1.
We now solve the quadratic for tan : 16
0
120 44
2
tan2 400 tan 7 16
120 44
2
tan 0.35185 ⇒ 19.4
400 0
Yes, it’s a home run when x 400, y > 10.
Section 9.3 1.
Parametric Equations and Calculus
dy dydt 4 2 dx dxdt 2t t
3.
dy dydt 2 cos t sin t 1 dx dxdt 2 sin t cos t
Note: x y 1 ⇒ y 1 x and dydt 1 5. x 2t, y 3t 1
7. x t 1, y t 2 3t
dy dydt 3 dx dxdt 2
dy 2t 3 1 when t 1. dx 1
d 2y 0 Line dx2
d 2y 2 concave upwards dx2
9. x 2 cos , y 2 sin dy 2 cos cot 1 when . dx 2 sin 4 d 2y csc2 csc3 2 when . 2 dx 2 sin 2 4 concave downward
11. x 2 sec , y 1 2 tan 2 sec2 dy dx sec tan
2 sec 2 csc 4 when . tan 6
d 2y 2 csc cot dx2 sec tan 2 cot3 6 3 when concave downward
. 6
0
2
Section 9.3 13. x cos3 , y sin3
15. x 2 cot , y 2 sin2
3 sin2 cos dy dx 3 cos2 sin
4 sin cos dy 2 sin3 cos dx 2 csc2
tan 1 when
. 4
At
sec2 1 d 2y dx2 3 cos 2 sin 3 cos4 sin
Parametric Equations and Calculus
2
,
3
3 2
, 23, and dydx 3 8 3.
y
Tangent line:
sec4 csc 4 2 when . 3 3 4
3 3 3 2 x 2 8
3
3 3x 8y 18 0 At 0, 2,
concave upward
dy , and 0. 2 dx
Tangent line: y 2 0
At 2 3,
3 1 dy , , and . 2 6 dx 8
y
Tangent line:
3 1 x 2 3 2 8
3x 8y 10 0
17. x 2t, y t2 1, t 2 (a)
19. x t 2 t 2, y t3 3t, t 1 (a)
10
−6
5
−1
6
−4
8
−3
(b) At t 2, x, y 4, 3, and
(b) At t 1, x, y 4, 2, and
dx dy dy 2, 4, 2 dt dt dx
dx dy dy 3, 0, 0 dt dt dx
(c)
dy 2. At 4, 3, y 3 2x 4 dx y 2x 5
(d)
(c)
dy 0. At 4, 2, y 2 0x 4 dx y2
(d)
10
5
(4, 2)
(4, 3) −5
5 −4
21. x 2 sin 2t, y 3 sin t crosses itself at the origin, x, y 0, 0. At this point, t 0 or t . dy 3 cos t dx 4 cos 2t At t 0:
3 dy 3 and y x. Tangent Line dx 4 4
At t ,
3 dy 3 and y x Tangent Line dx 4 4
−1
8
−3
193
194
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
23. x cos sin , y sin cos Horizontal tangents:
dy sin 0 when 0, , 2, 3, . . . . d
Points: 1, 2n 1 , 1, 2n where n is an integer. Points shown: 1, 0, 1, , 1, 2 dx 3 5 cos 0 when , , ,. . .. d 2 2 2
Vertical tangents:
1
2n 1 , 1n1 2
3 5 Points shown: , 1, , 1, , 1 2 2 2 n1
Points:
25. x 1 t, y t 2 Horizontal tangents:
27. x 1 t, y t 3 3t dy 2t 0 when t 0. dt
Horizontal tangents:
dy 3t 2 3 0 when t ± 1. dt Points: 0, 2, 2, 2
dx 1 0; none dt
Vertical tangents:
Point: 1, 0 Vertical tangents: 3
dx 1 0; none dt
3
(2, 2) −4 −2
5
4
(1, 0)
(0, − 2)
−1
−3
29. x 3 cos , y 3 sin Horizontal tangents:
dy 3 3 cos 0 when , . d 2 2
31. x 4 2 cos , y 1 sin
Points: 0, 3, 0, 3 dx 3 sin 0 when 0, . d
Vertical tangents:
dy 3 cos 0 when , . d 2 2
Horizontal tangents:
Points: 4, 0, 4, 2 dx 2 sin 0 when x 0, . d
Vertical tangents:
Points: 3, 0, 3, 0
Points: 6, 1, 2, 1 2
4
(0, 3) (4, 0) 0 −6
(− 3, 0)
(3, 0)
9
(2, − 1)
6
(6, − 1)
(4, − 2) (0, − 3) −4
−4
35. x t 2, y 2t, 0 ≤ t ≤ 2
33. x sec , y tan Horizontal tangents:
dy sec2 0; none d
Vertical tangents:
dx sec tan 0 when x 0, . d
Points: 1, 0, 1, 0 4
−6
(− 1, 0)
(1, 0)
−4
6
dydt
dx dy dx 2t, 2, dt dt dt
2
2
4t 2 4 4t 2 1
2
s2
t 2 1 dt
0
0
t t 2 1 ln t t2 1 2 5 ln 2 5 5.916
2
Section 9.3
37. x et cos t, y et sin t, 0 ≤ t ≤
2
2
0
dx dt
2
dy dt
2
0
0
2
dt
2e2t dt 2
2
et1 dt
2et
0
21 e2 1.12
1 1 9 dt 4t 2
dx 3a cos2 sin , d
dy 3a sin2 cos d
1 u2 du
0
1 ln 37 6 6 37 3.249 12
3
t
dt
dx dy a1 cos , a sin d d
S2
a2 1 cos 2 a2 sin2 d
sin cos cos2 sin2 d
2 2a
0
6a
0
0
2
6
0
dt
t
0
43. x a sin , y a1 cos ,
9a2 cos4 sin2 9a2 sin4 cos2 d
12a
1 36t
1 ln 1 u2 u u 1 u2 12
2
S4
1
6
1 6
u 6 t, du
41. x a cos3 , y a sin3 ,
0
2
1
S
195
dx 1 dy , 3 dt 2 t dt
39. x t, y 3t 1,
dx dy etsin t cos t, etcos t sin t dt dt s
Parametric Equations and Calculus
1 cos d
0
2
sin 2 d 3a cos 2
0
2
0
6a
2 2a
0
sin
1 cos
4 2a 1 cos
d
0
8a
45. x 90 cos 30t, y 90 sin 30t 16t 2 (a)
(d) y 0 ⇒ 90 sin t 16t2 ⇒ t
35
x 90 cos t 0
240 0
(b) Range: 219.2 ft (c)
dx dy 90 cos 30, 90 sin 30 32t. dt dt y 0 for t
45 . 16
90 cos 302 90 sin 30 32t2 dt
902 2 cos 2 0 ⇒ 45 32 By the First Derivative Test, 45 4 maximizes the range.
dx 90 cos , dt
90 dy 90 sin 32 90 sin 32 sin 90 sin dt 16 s
0
230.8 ft
902 902 cos sin sin 2 16 32
x
4516
s
90 sin 16
9016sin
90 cos 2 90 sin 2 dt
0
9016sin
0
0
9016sin
90 dt 90t
2
90 sin 16
ds 902 cos 0 ⇒ d 16 2 By the First Derivative Test, 90 maximizes the arc length.
196
Chapter 9
47. (a)
Conics, Parametric Equations, and Polar Coordinates
x t sin t
x 2t sin2t
y 1 cos t
y 1 cos2t
0 ≤ t ≤ 2
0 ≤ t ≤
3
The time required for the particle to traverse the same path is t 4.
−1
49. x t, y 2t,
dx dy 1, 2 dt dt
51. x 4 cos , y 4 sin ,
4
4
2t 1 4 dt 4 5
0
S 2
t dt
0
4
2 5 t 2
0
0
32
4
2
dx dy 3a cos2 sin , 3a sin2 cos d d
−4
x
−2
2
4
−2 −4
61. x r cos , y r sin
y
r sin
r 2 sin2 r 2 cos2 d
0
sin d
θ
0
2r 2 cos
2
0
1 cos
r 2
32
59. s
a
2
dx dt
dydt 2
0
12 2 a 5
2
dt
See Theorem 9.8, page 678.
x t, y t.
2
0
12a2 sin5 5
sin4 cos d
b
4
2
57. One possible answer is the graph given by
y
0
See Theorem 9.7, page 675.
2r 2
2
t dt
a sin3 9a2 cos4 sin2 9a2 sin4 cos2 d 12a2
dy dydt dx dxdt
cos d 32 sin
0
0
2
16 5
0
53. x a cos3 , y a sin3 ,
S 2
4 cos 4 sin 2 4 cos 2 d
0
5 t 2
55.
2
4
t 1 4 dt 2 5
dx dy 4 sin , 4 cos d d
0
32 5
4
S 4
1
3
−1
(b) S 2
y 1 cos 2 t
−
3
(c) x 12 t sin 12 t
3
−
(a) S 2
(b) The average speed of the particle on the second path is twice the average speed of a particle on the first path.
x
Section 9.3
Parametric Equations and Calculus
197
63. x t, y 4 t, 0 ≤ t ≤ 4
4
A
4 t
0
x y
3 16 3 32
1 2 t
dt
4
1 2
4
4t12 t12 dt
0
28 t 3t t 1
0 4
4 t 2
0
0
2 1 t dt 323 4 t dt 323 4t t2 4
4 t t
4
2
1 2 t
dt
3 64
2
0
0
4
4
16t12 8t12 t32 dt
0
16 3
3 4
4
3 16 2 32 t t t t 2 t 64 3 5
0
8 5
3 8 x, y , 4 5
65. x 3 cos , y 3 sin ,
dx 3 sin d
0
V 2
2
3 sin 23 sin d
0
54
sin3 d
2 0
54
1 cos2 sin d
2
54 cos
67.
cos3 3
0
2
36
x 2 sin2
y
y 2 sin2 tan
2
dx 4 sin cos d A
2
π 0≤θ< 2
2 sin2 tan 4 sin cos d 8
0
sin4 d
0
3 sin3 cos 3 8 sin cos 4 8 8
69. ab is area of ellipse (d).
75. (a) x
−2
2
2
0
1 x
−1
1 −1 −2
3 2
71. 6a2 is area of cardioid (f).
73. 83 ab is area of hourglass (a).
1 t2 2t , y , 20 ≤ t ≤ 20 1 t2 1 t2
2
The graph is the circle x 2 y 2 1, except the point 1, 0. Verify: x2 y 2
11 tt 1 2t t 2 2 2
2
2
−3
1 2t 2 t 4 4t 2 1 t 22 1 1 t 22 1 t 22
(b) As t increases from 20 to 0, the speed increases, and as t increases from 0 to 20, the speed decreases. 77. False d g t d 2y dt f t f tg t g t f t dx2 f t f t 3
3
−2
198
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
Section 9.4 1.
Polar Coordinates and Polar Graphs
4, 2
3.
5. 2, 2.36
0 2
x 4 cos
2 4
y 4 sin
x 4 cos y 4 sin
4, 3
x 2 cos2.36 1.004
2 3
y 2 sin2.36 0.996
x, y 1.004, 0.996
3 23
π 2
x, y 2, 23
x, y 0, 4 π 2
π 2
(4, 36π )
(
2, 2.36 )
0
(− 4, − π3 )
1
0 1
2
3
0 1
7. r, 5,
3 4
9. r, 3.5, 2.5
11. x, y 1, 1 r ± 2
x, y 2.804, 2.095
x, y 3.5355, 3.5355
tan 1
y
5 5 , , 2, , 2, 4 4 4 4
y
1 4
(−3.54, 3.54)
3
−1
1 −3
−2
−2 x
−1
1
2
3
y
−1
2
−4
x
2
(2.804, − 2.095)
−3
1
(1, 1)
−1
1
x 1
13. x, y 3, 4
y
r ± 9 16 ± 5 tan
2
5
(− 3, 4)
43
4 3
2.214, 5.356, 5, 2.214, 5, 5.356
2 1 −4
−3
−2
−1
x 1
5 4 17. x, y 2 , 3
15. x, y 3, 2
r, 3.606, 0.588
r, 2.833, 0.490 y
19. (a) x, y 4, 3.5
π 2
(b) r, 4, 3.5
4
(4, 3.5) 0
3
1
2
(4, 3.5) 1 x 1
2
3
4
Section 9.4 21. x2 y 2 a2
π 2
Polar Coordinates and Polar Graphs y4
23.
π 2
r sin 4
ra
r 4 csc 0 a 0 2
3x y 2 0
25.
y 2 9x
27.
3r cos r sin 2 0
r 2 sin2 9r cos
r3 cos sin 2 r
4
r
2 3 cos sin
9 cos sin2
r 9 csc2 cos
π 2
π 2
0 1
0 1
r3
29.
y2
3
4
5
r sin
31.
6
7
r
33.
r 2 r sin
r2 9 x2
2
2
y2
9
x2 y
x2
y
2
1 2
tan r tan
y 2
tanx2 y 2 1 4
x2 y 2 arctan
x2 y 2 y 0
1
y x y x
y x
2
1
1
y 2π
2
1
π
2
π
1 2
x
2π
−π x 1 2
35.
r 3 sec
37. r 3 4 cos
y
r cos 3
− 2π
1 2
6
0 ≤ < 2
3
− 12
x3
6
2
x30
−6
1
x 1
39. r 2 sin
2
41. r
4
0 ≤ < 2 −4
5
2 1 cos
Traced out once on < <
5
−10
5
−5 −2
199
200
Chapter 9
43. r 2 cos
Conics, Parametric Equations, and Polar Coordinates
32
45. r2 4 sin 2
2
0 ≤ < 4
−3
0 ≤ <
3
2
2
−3
3
−2
−2
r 2h cos k sin
47.
Radius: h2 k 2 Center: h, k
r 2 2rh cos k sin r 2 2hr cos kr sin x2 y 2 2hx ky x2 y 2 2hx 2ky 0
x2 2hx h2 y 2 2ky k2 0 h2 k 2 x h2 y k2 h2 k 2
49.
4, 23, 2, 6
51. 2, 0.5, 7, 1.2 d 22 72 227 cos0.5 1.2
20 16 cos 2 5 4.5 2
d
42
22
2 242 cos 3 6
53 28 cos0.7 5.6
53.
55. (a), (b) r 31 cos
r 2 3 sin dy 3 cos sin cos 2 3 sin dx 3 cos cos sin 2 3 sin
4
2 cos 3 sin 1 2 cos 3 sin 1 3 cos 2 2 sin 6 cos2 2 sin 3
2 ⇒ x, y 0, 3
r, 3,
Tangent line: y 3 1x 0
y x 3
dy (c) At , 1.0. 2 dx
57. (a), (b) r 3 sin
5
−4
5 −1
r,
3 2 3, 3 ⇒ x, y 3 4 3, 94
Tangent line: y
33 9 3 x 4 4 y 3x
(c) At
4
−4
dy , 0. 2 dx dy 2 At 2, , . dx 3 3 dy , 0. At 1, 2 dx At 5,
−8
9 2
dy , 3 1.732. 3 dx
Section 9.4 59.
r 1 sin
61.
dy 1 sin cos cos sin d
201
r 2 csc 3 dy 2 csc 3 cos 2 csc cot sin d
cos 1 2 sin 0
3 cos 0
1 3 5 cos 0, sin ⇒ , , , 2 2 2 6 6 Horizontal tangents:
Polar Coordinates and Polar Graphs
3 , 2 2
2, 32, 12, 6 , 12, 56
5, 2 , 1, 32
Horizontal:
dx 1 sin sin cos cos d sin sin2 sin2 1 2 sin2 sin 1 2 sin 1sin 1 0 sin 1, sin Vertical tangents:
1 7 11 ⇒ , , 2 2 6 6
32, 76, 32, 116 65. r 2 csc 5
63. r 4 sin cos2
10
2
−3
3
−12
12
−6
−2
Horizontal tangents:
2 , 3, 32
Horizontal tangents: 7,
0, 0, 1.4142, 0.7854, 1.4142, 2.3562 r 3 sin
67.
69. r 21 sin
π 2
r2 3r sin
Cardioid 0
x 2 y 2 3y
x2 y
3 2
Circle r
3 2
2
π 2
0
9 4
1
2
Symmetric to y-axis, 2
3
32
Center: 0,
Tangent at the pole: 0 71. r 2 cos3 π 2
Rose curve with three petals Symmetric to the polar axis
Relative extrema: 2, 0, 2,
2 , 2, 3 3
0 2
0
6
4
3
2
2 3
5 6
r
2
0
2
2
0
2
0
Tangents at the pole:
5 , , 6 2 6
2
1
2
3
202
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
73. r 3 sin 2 π 2
Rose curve with four petals
Symmetric to the polar axis, , and pole 2
Relative extrema: ± 3,
5 , ± 3, 4 4
Tangents at the pole: 0,
0 3
2
, 3 2 give the same tangents. 75. r 5
77. r 41 cos
π 2
π 2
Cardioid
Circle radius: 5 x2 y2 25
0
0 2
4
79. r 3 2 cos
2
6
r 3 csc
81.
π 2
r
1
10
π 2
y3
Symmetric to polar axis 0
6
r sin 3
Limaçon
4
0
3
2
2 3
2
3
4
5
Horizontal line
2
0 1
83. r 2
π 2
Spiral of Archimedes Symmetric to
2
0
4
2
3 4
5 4
3 2
r
0
2
3 2
2
5 2
3
0 1
Tangent at the pole: 0 85. r2 4 cos2 Lemniscate
Symmetric to the polar axis, , and pole 2
π 2
Relative extrema: ± 2, 0 0
0
6
r
±2
± 2
4 0
Tangents at the pole:
1
3 , 4 4
2
3
2
Section 9.4
Polar Coordinates and Polar Graphs 2
89. r
87. Since r 2 sec 2
1 , cos
Hyperbolic spiral r ⇒ as ⇒ 0
the graph has polar axis symmetry and the lengths at the pole are
, . 3 3
r
2 2 2 sin 2 sin ⇒ r r sin y
y
2 sin
x = −1 4
Furthermore,
−6
r ⇒ as ⇒ 2
6
lim
→0
2 sin 2 cos lim 2 →0 1
−4 3
r ⇒ as ⇒ . 2 Also, r 2
y=2
1 r r 2 2 cos r cos x
−3
3 −1
rx 2x r r
2x . 1x
Thus, r ⇒ ± as x ⇒ 1. 91. The rectangular coordinate system consists of all points of the form x, y where x is the directed distance from the y-axis to the point, and y is the directed distance from the x-axis to the point. Every point has a unique representation. The polar coordinate system uses r, to designate the location of a point. r is the directed distance to the origin and is the angle the point makes with the positive x-axis, measured clockwise. Point do not have a unique polar representation. 97. r 31 cos
95. r 2 sin circle
93. r a circle
b line
Cardioid
Matches (c)
Matches (a) 99. r 4 sin (a) 0 ≤ ≤
2
(b)
≤ ≤ 2
(c)
π 2
π 2
0 1
2
≤ ≤ 2 2 π 2
0 1
2
0 1
2
203
204
Chapter 9
Conics, Parametric Equations, and Polar Coordinates π 2
101. Let the curve r f be rotated by to form the curve r g. If r1, 1 is a point on r f , then r1, 1 is on r g. That is,
(r, θ +φ )
g1 r1 f 1. Letting 1 , or 1 , we see that
(r, θ )
g g1 f 1 f .
φ θ 0
103. r 2 sin
(a) r 2 sin
2 2 sin cos 4 2
(b) r 2 cos 2 cos 4
4
−6
−6
6
6
−4
−4
(d) r 2 cos
(c) r 2 sin 2 sin
4
4
−6
−6
6
6
−4
−4
105. (a) r 1 sin
(b) r 1 sin
π 2
4
π 2
Rotate the graph of r 1 sin
0 1
107. tan
2
r 21 cos dr d 2 sin
At , tan is undefined ⇒
109. tan
. 2
At
3
−6
r 2 cos 3 dr d 6 sin 3
, tan 0 ⇒ 0. 6 2
3
−3
0 1
through the angle 4.
−3
3
−2
2
Section 9.5
111.
r
Area and Arc Length in Polar Coordinates
6 dr 6 sin 61 cos 1 ⇒ 1 cos d 1 cos 2
7
ψ
6 r 1 cos 1 cos tan dr 6 sin sin d 1 cos 2 2 At , tan 3
21
1 3
θ −8
7
−3
3.
2
, 60 3 113. True
115. True
Section 9.5
Area and Arc Length in Polar Coordinates
1. (a) r 8 sin
(b) A 2
π 2
12 8 sin 2
64
2
d
0
2
sin2 d
0
32
2
1 cos 2 d
0
32
0 2
4
sin 2 2
2
16
0
A 42 16
3. A 2
1 2
6
1 sin 6 6
2 cos 32 d 2
0
6
0
3
5. A 2
7. A 2
2
1 2
2
1 sin 2 d
1 2
4
1 1 sin 4 2 4
9. A 2 2
32 2 cos 41 sin 2
2
3 2
cos 22 d
0
1 2
2 3
4
0
8
1 2 cos 2 d
3 4 sin sin 2
2 3
2 33 2
2
−1
4
−2
11. The area inside the outer loop is
12
2 3
2
2
1 2 cos 2 d 3 4 sin sin 2
0
2 3
0
4 33 . 2
−1
4
From the result of Exercise 9, the area between the loops is
4 33 2 33 33. A 2 2
−2
205
206
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
13. r 1 cos
15. r 1 cos
r 1 cos
r 1 sin
Solving simultaneously,
Solving simultaneously,
1 cos 1 cos
1 cos 1 sin
2 cos 0
cos sin tan 1
3 , . 2 2 Replacing r by r and by in the first equation and solving, 1 cos 1 cos , cos 1, 0. Both curves pass through the pole, 0, , and 0, 0, respectively.
2 , 1, 32, 0, 0
Points of intersection: 1,
3 7 , . 4 4
Replacing r by r and by in the first equation and solving, 1 cos 1 sin , sin cos 2, which has no solution. Both curves pass through the pole, 0, , and 0, 2, respectively. Points of intersection:
17. r 4 5 sin
19. r
r 3 sin
2 3
2 2
,
4
2 7
, 2 2
2
,
4
, 0, 0
π 2
r2
Solving simultaneously, 4 5 sin 3 sin sin
1 2
Solving simultaneously, we have 0
2 2, 4.
1
Points of intersection:
5 , . 6 6
2, 4, 2, 4
Both curves pass through the pole, 0, arcsin 4 5, and 0, 0, respectively. Points of intersection:
32, 6 , 32, 56, 0, 0
21. r 4 sin 2
π 2
r2 r 4 sin 2 is the equation of a rose curve with four petals and is symmetric to the polar axis, 2, and the pole. Also, r 2 is the equation of a circle of radius 2 centered at the pole. Solving simultaneously, 4 sin 2 2 2
5 , 6 6 5 , . 12 12
Therefore, the points of intersection for one petal are 2, 12 and 2, 5 12. By symmetry, the other points of intersection are 2, 7 12, 2, 11 12, 2, 13 12, 2, 17 12, 2, 19 12, and 2, 23 12.
0 1
3
Section 9.5 23. r 2 3 cos sec 2
r
25. r cos
r = sec θ 2
4
Area and Arc Length in Polar Coordinates
r 2 3 sin
−4
Points of intersection:
8
0, 0, 0.935, 0.363, 0.535, 1.006 −4
The graphs reach the pole at different times ( values).
r = 2 + 3 cos θ
The graph of r 2 3 cos is a limaçon with an inner loop b > a and is symmetric to the polar axis. The graph of r sec 2 is the vertical line x 1 2. Therefore, there are four points of intersection. Solving simultaneously, 2 3 cos
1
r = cos θ
−4
5
sec 2
−5
r = 2 − 3 sin θ
6 cos2 4 cos 1 0 cos
2 ± 10 6
arccos
2 6 10 1.376
arccos
2 6
10
2.6068.
Points of intersection: 0.581, ± 2.607, 2.581, ± 1.376 27. From Exercise 21, the points of intersection for one petal are 2, 12 and 2, 5 12. The area within one petal is
12
1 2
0
12
16
Total area 4
29. A 4
1 2
2
1 sin 4 4
12
0
12
5 12
sin22 d 2
12
0
8
5 12
1 2
4 sin 22 d
12
1 2
2
5 12
4 sin 22 d
4
d (by symmetry of the petal)
5 12
2
22 d
−6
4 3. 3
6
−4
43 3 163 43 34 4 33 3 2 sin 2 d
0
6
2
2 11 12 cos sin2
0
−9
9
11 24 −6
12
6
31. A 2
4 sin 2 d
0
6
12 41 sin2
16
0
1 2
2
6
8 2 23 4 33 3 3 5
−6
6
−3
2
4
22 d
6
33. A 2
a2
1 2
a 1 cos 2 d
0
32 2 sin sin42
3a 2 a 2 5a 2 2 4 4
0
a2 4
a 2 4
207
208
Chapter 9
35. A
Conics, Parametric Equations, and Polar Coordinates
a 2 1 8 2
a 2 a 2 8 2 a 2 8
a 2 8
a1 cos 2 d
2
2
π 2
32 2 cos cos22 d
sin 2 3 2 sin 2 2 4
a2
32 34 2 2 2
2
0
a2
a
2a
2
a2
37. (a) r a cos2
(b)
4
a=6
a=4
r3 ar2 cos2 −6
x 2 y 23 2 ax 2
6
−4
(c) A 4
12
2
6 cos2 2 4 cos2 2 d 40
0
10
2
cos4 d 10
0
2
2
1 cos 22 d
0
4 1 2 cos 2 1 cos d 10 32 sin 2 81 sin 4 2
2
0
0
15 2
39. r a cosn For n 1:
For n 2:
r a cos A
a 2
2
r a cos 2
a 2 4
A8
π 2
4
a cos 22 d
0
a 2 2
π 2
a
0
a
For n 3:
0
For n 4:
r a cos 3 A6
1 2
r a cos 4
12
6
a cos 32 d
0
π 2
a 2 4
A 16
12
8
a cos 42 d
0
a 2 2
π 2
a
0 a
0
In general, the area of the region enclosed by r a cosn for n 1, 2, 3, . . . is a 2 4 if n is odd and is a 2 2 if n is even.
Section 9.5
43. r 1 sin
41. r a r 0 s
Area and Arc Length in Polar Coordinates
r cos
2
a2 02 d a
0
2 0
2a
s2
(circumference of circle of radius a)
3 2
2
1 sin 2 cos 2 d
3 2
22
2
1 sin d
3 2
22
cos 1 sin
2
42 1 sin
d
3 2
2
42 2 0 8
2
45. r 2, 0 ≤ ≤
1 47. r , ≤ ≤ 2
49. r sin3 cos , 0 ≤ ≤ 1
4
0.5
−1
− 0.5 −1
2 −1
53. r ea r aea
r 6 sin
Length 4.39
Length 0.71
51. r 6 cos
S 2
−1
− 0.5
Length 4.16
2
S 2
6 cos sin 36 cos2 36 sin2 d
72
2
ea cos ea2 aea2 d
0
0
2
sin cos d
21 a2
0
2
e2a cos d
0
36 sin2
2
21 a2
0
36
21 4a2 1
a2
2a
2
ea 2a
r 8 sin 2 S 2
4
4 cos 2 sin 16 cos2 2 64 sin2 2 d
0
32
4
cos 2 sin cos2 2 4 sin2 2 d 21.87
0
Area Arc length
1 2
f 2d
1 2
59. (a) is correct: s 33.124.
r2 d
f 2 f 2 d
r2
ddr
2
d
2
4ae 1 2a cos sin
55. r 4 cos 2
57.
2
0.5
0
209
210
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
61. Revolve r a about the line r b sec where b > a > 0.`
π 2
f a f 0 S 2
2a a
2
b a cos a2 02 d
a
b
0
0
2a b a sin 2a2b
2 0
4 2ab
63. False. f 1 and g 1 have the same graphs. 65. In parametric form,
b
s
a
dx dt
dydt 2
2
dt.
Using instead of t, we have x r cos f cos and y r sin f sin . Thus, dx dy f cos f sin and f sin f cos . d d It follows that
ddx ddy 2
Therefore, s
Section 9.6 1. r
2
f 2 f2.
f 2 f2 d
Polar Equations of Conics and Kepler’s Laws
2e 1 e cos
(a) e 1, r
3. r
2 , parabola 1 cos
2e 1 e sin
(a) e 1, r
2 , parabola 1 sin
(b) e 0.5, r
1 2 , ellipse 1 0.5 cos 2 cos
(b) e 0.5, r
1 2 , ellipse 1 0.5 sin 2 sin
(c) e 1.5, r
3 6 , hyperbola 1 1.5 cos 2 3 cos
(c) e 1.5, r
3 6 , hyperbola 1 1.5 sin 2 3 sin
e = 1.0
4
e = 1.5
e = 0.5 −4
4
−9
9
8
e = 1.5 e = 1.0 −4
e = 0.5
−8
Section 9.6 5. r
Polar Equations of Conics and Kepler’s Laws
4 1 e sin
(a)
5
(b)
e = 0.1
− 30
5 − 30
30
30
e = 0.25 e = 0.5 e = 0.75 e = 0.9
− 40
− 40
The conic is a parabola.
The conic is an ellipse. As e → 1, the ellipse becomes more elliptical, and as e → 0 , it becomes more circular. (c)
e = 1.1
80
e=1 e=2 − 90
90
− 40
The conic is a hyperbola. As e → 1, the hyperbolas opens more slowly, and as e → , they open more rapidly. 7. Parabola; Matches (c)
13. r
9. Hyperbola; Matches (a)
1 1 sin
15. r
Parabola since e 1 1 3 Vertex: , 2 2
6 2 cos
17. r 2 sin 4 r
3 1 1 2 cos
Ellipse since e
π 2
11. Ellipse; Matches (b)
1 < 1 2
Vertices: 2, 0, 6, 0 1
2
4 2 sin 2 1 1 2 sin
Ellipse since e
π 2
Vertices:
1 < 1 2
43, 2 , 4, 32 π 2
0 1
3
0 1
19. r
5 5 1 2 cos 1 2 cos
Hyperbola since e 2 > 1
5 Vertices: 5, 0, , 3 π 2
21. r
3 3 2 2 6 sin 1 3 sin
Hyperbola since e 3 > 1 Vertices:
38, 2 , 43, 32 π 2
0 4
6
8
0 1
3
4
211
212
Chapter 9
23.
Conics, Parametric Equations, and Polar Coordinates Ellipse
1
−2
25.
Parabola
2
2 −2
1
−2
−2
27. r
1
1 sin 4
29. r
Rotate the graph of r
6
2 cos
6
Rotate the graph of
1 1 sin
r
counterclockwise through the angle . 4
6 2 cos
clockwise through the angle . 6 5
2 −2
10
−8
4
−6
31. Change to
−3
:r 4
5
5 3 cos
4
.
33. Parabola e 1, x 1, d 1 r
35. Ellipse 1 e , y 1, d 1 2 r
ed 1 e sin
1 2 1 1 2 sin
1 2 sin
41. Ellipse Vertices: 2, 0, 8, 3 16 e ,d 5 3 r
ed 1 e cos 16 5 1 3 5 cos 16 5 3 cos
ed 1 1 e cos 1 cos 39. Parabola
37. Hyperbola
2
e 2, x 1, d 1
Vertex: 1,
2 ed r 1 e cos 1 2 cos
e 1, d 2, r
43. Hyperbola
Vertices: 1,
3 3 , 9, 2 2
5 9 e ,d 4 5 r
ed 1 e sin
9 4 1 5 4 sin
9 4 5 sin
2 1 sin
45. Ellipse if 0 < e < 1, parabola if e 1, hyperbola if e > 1.
Section 9.6
1 < 1 2
213
4 49. a 5, c 4, e , b 3 5
47. (a) Hyperbola e 2 > 1 (b) Ellipse e
Polar Equations of Conics and Kepler’s Laws
r2
(c) Parabola e 1
9 1 16 25 cos 2
(d) Rotated hyperbola e 3 51. a 3, b 4, c 5, e
5 3
53. A 2
16 r2 1 25 9 cos 2
9
1 2
0
0
3 2 cos
d 2
1 d 10.88 2 cos 2
55. Vertices: 126,000, 0, 4119, a
126,000 4119 c 40,627 84,000 65,059.5, c 65,059.5 4119 60,940.5, e , d 4119 2 a 43,373 40,627
r
411984,000 43,373 345,996,000 ed 1 e cos 1 40,627 43,373 cos 43,373 40,627 cos
When 60 , r
345,996,000
15,004.49. 23,059.5
Distance between the surface of the earth and the satellite is r 4000 11,004.49 miles. 57. a 92.957 106 mi, e 0.0167 r
59. a 5.900 109 km, e 0.2481
92,931,075.2223 1 e2a 1 e cos 1 0.0167 cos
r
5.537 10 9 1 e 2 a
1 e cos 1 0.2481 cos
Perihelion distance: a1 e 91,404,618 mi
Perihelion distance: a1 e 4.436 10 9 km
Aphelion distance: a1 e 94,509,382 mi
Aphelion distance: a1 e 7.364 10 9 km
61. r
5.537 109 1 0.2481 cos
9
1 2
(a) A
0
1 2
9
9
1 2 248 1 2 (b)
10 1 5.537 0.2481 cos
0
2
0
2
d 9.341 1018 km2
10 d 1 5.537 0.2481 cos 5.537 10 1 0.2481 cos d 9
2
9
2
1 0.2481 cos 5.537 109
2
21.867 yr
d 9.341 1018
0.8995 rad In part (a) the ray swept through a smaller angle to generate the same area since the length of the ray is longer than in part (b). (c) r s
5.537 10 90.2481 sin 1 0.2481 cos 2
9
0
10 1.3737297 10 sin 1 5.537 0.2481 cos 1 0.2481 cos
9
2
9
2
2
d 2.559 109 km
2.559 109 km
1.17 108 km yr 21.867 yr s
0.899
10 1.3737297 10 sin d 4.119 1 5.537 0.2481 cos 1 0.2481 cos
9
4.119 109 km
1.88 108 km yr 21.867 yr
2
2
9
2
109 km
214
Chapter 9
63. r1
Conics, Parametric Equations, and Polar Coordinates
ed ed and r2 1 sin 1 sin
Points of intersection: ed, 0, ed, ed ed cos cos sin dy 1 sin 1 sin 2 r1: dx ed ed cos sin cos 1 sin 1 sin 2
At ed, 0,
dy dy 1. At ed, , 1. dx dx
ed ed cos cos sin dy 1 sin 1 sin 2 r2: dx ed ed cos sin cos 1 sin 1 sin 2
At ed, 0,
dy dy 1. At ed, , 1. dx dx
Therefore, at ed, 0 we have m1m2 11 1, and at ed, we have m1m2 11 1. The curves intersect at right angles.
Review Exercises for Chapter 9 1. Matches (d) - ellipse
3. Matches (a) - parabola
16x 2 16y 2 16x 24y 3 0
5.
x
2
x
y
1 3 9 3 1 9 y2 y 4 2 16 16 4 16
1 x
1 x 2
2
3 y 4
1 2
1
2
1 , 2
2
Circle Center:
1
3 4
12, 43
Radius: 1 3x 2 2y 2 24x 12y 24 0
7.
3x 2 2y 2 12x 12y 29 0
9.
3x 2 8x 16 2 y 2 6y 9 24 48 18
3x 2 4x 4 2 y 2 6y 9 29 12 18
x 4 2 y 3 2 1 2 3
x 2 2 y 3 2 1 13 12
Hyperbola
Ellipse
Center: 4, 3
Center: 2, 3
Vertices: 4 ± 2, 3 Asymptotes: y 3 ±
Vertices:
32 x 4
2, 3
±
2
2
y
x
1
y
1
2
1 6
2
4
3 4
2 x
6
4
2
(2, − 3)
3
Review Exercises for Chapter 9 13. Vertices: 3, 0, 7, 0
11. Vertex: 0, 2
15. Vertices: ± 4, 0
Directrix: x 3
Foci: 0, 0, 4, 0
Foci: ± 6, 0
Parabola opens to the right
Horizontal major axis
Center: 0, 0
p3
Center: 2, 0
Horizontal transverse axis
y 2 4 3x 0
a 5, c 2, b 21
a 4, c 6, b 36 16 25
y 2 4y 12x 4 0
x 22 y 2 1 25 21
x2 y2 1 16 20
2
17.
215
5 x2 y2 1, a 3, b 2, c 5, e 9 4 3
19. y x 2 has a slope of 1. The perpendicular slope is 1. y x 2 2x 2
By Example 5 of Section 9.1, C 12
2
0
dy 1 5 2x 2 1 when x and y . dx 2 4
1
5 sin2 d 15.87. 9
y
Perpendicular line:
5 1 1 x 4 2
4x 4y 7 0 21. (a) V abLength 12 16 192 ft 3
3
(b) F 262.4
8 4 3 y 9 y 2 dy 62.4 3 3 3 3
3
3
8 3 9 3 9 62.4 3 2 2 2 2
3
y9 y 2 dy
y 1 3 8 y 9 y 2 9 arcsin 9 y 2 32 62.4 3 2 3 3
3
9 y 2 dy 3 3
38 62.4272 7057.274
3
(c) You want 4 of the total area of 12 covered. Find h so that
h
2
x=
4 9 y 2 dy 3 3
0
h
9 y 2 dy
0
y 1 y 9 y 2 9 arcsin 2 3
h 0
y 4 3
9−
y2
4
h
9 8
Area of filled tank above x-axis is 3π.
2 x
−2
2 −2
9 8
−4
Area of filled tank below x-axis is 6π.
h3 94.
h9 h 2 9 arcsin
By Newton’s Method, h 1.212. Therefore, the total height of the water is 1.212 3 4.212 ft. (d) Area of ends 212 24 Area of sides PerimeterLength 16
2
1
0
256
167 sin d 16 from Example 5 of Section 9.1 2
122 1 167 sin 0 41 167 sin 8 21 167 sin 4 2
2
2
1 167 sin 38 1 167 sin 2 353.65
4
Total area 24 353.65 429.05
2
2
216
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
23. x 1 4t, y 2 3t t
y
x1 x1 ⇒ y23 4 4
4
2 1
11 3 y x 4 4
−1
4y 3x 11 0
x −1
1
2
3
5
−2
Line 25. x 6 cos , y 6 sin
6x 6y 2
x2
y2
2
27. x 2 sec , y 3 tan
y
1
x 22 sec2 1 tan2 1 y 32
4
x 22 y 32 1
2
36
−4 −2
Circle
y
x −2
2
4
8
Hyperbola
−4
4 2
x
−4
−2
2
4
8
−4
29. x 3 3 2t 3 5t
31.
x 3 2 y 4 2 1 16 9
y 2 2 6t 2 4t Let
(other answers possible)
x 3 2 y 4 2 cos 2 and sin 2 . 16 9
Then x 3 4 cos and y 4 3 sin . 35. (a) x 2 cot , y 4 sin cos , 0 < <
33. x cos 3 5 cos y sin 3 5 sin
4
5
−12 −7
12
8
−4
(b) 4 x2y 4 4 cot2 4 sin cos
−5
16 csc2 16
sin cos
cos sin
82 cot 8x 37. x 1 4t y 2 3t (a)
dy 3 dx 4
(b) t
x1 4
y
(c) 5
No horizontal tangents
3 3x 11 y 2 x 1 4 4
4
2 1 x
1
2
3
5
Review Exercises for Chapter 9
217
39. x 1 t y 2t 3 (a)
dy 2 2t 2 dx 1t 2
(b) t
No horizontal tangents t 0
1 x
y
(c)
y
6
2 3 x
4 2 x
4
41. x
1 2t 1
y
1 t 2t
2
y 2 5 sin (a)
2t 2 dy t 2 2t 2 t 12t 1 2 0 when t 1. (a) dx 2 t 2t 2 2 2t 1 2 Point of horizontal tangency: (b) 2t 1
13 , 1
1 1 1 1 ⇒ t x 2 x
dy 5 cos 3 2.5 cot 0 when , . dx 2 sin 2 2 Points of horizontal tangency: 3, 7, 3, 3
(b)
x 3 2 y 2 2 1 4 25 y
(c)
8
4
1 1 1x 2 x
12 1 x x 2
x
4
4x 2 4x 2 2 1 x 4x1 x 5x 1x 1
(c)
y 3 2
−2
x
−1
2
3
−1 −2
45. x cos 3 y 4 sin 3 (a)
dy 12 sin 2 cos 4 sin 4 tan 0 when 0, . dx 3 cos 2 sin cos But,
dy dx 0 at 0, . Hence no points of horizontal tangency. dt dt
(b) x 23
4y
23
1
y
(c) 4
x
4
4
43. x 3 2 cos
2
y
2
2
2
4
4
8 4
218
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
47. x cot
x r cos sin
49.
y rsin cos
y sin 2 2 sin cos (a), (c)
dx r cos d
2
−3
3
dy r sin d
−2
sr
dx 1 dy dy (b) At , 4, 1, and 6 d d dx 4
2 cos 2 2 sin 2 d
0
r
d
0
51. x, y 4, 4
1 2 r 2
x −1
1
2
3
4
5
−2
r, 42,
7 , 4
42,
3 4
−3
−4 −5
r 3 cos r2
(4, − 4)
r 2 2r 1 cos
3r cos
x 2 y 2 2 ± x 2 y 2 2x
x 2 y 2 2x 2 4x 2 y 2
x 2 y 2 3x 0 r 2 cos 2 cos 2 sin 2
r 21 cos
55.
x 2 y 2 3x
r4
0
1
7 4
57.
y
r 4 2 4 2 42
53.
r 2 2
r2
cos 2
59.
r 2 sin 2
r 4 cos 2 sec 4 2 cos 2 1
x 2 y 2 2 x 2 y 2
cos1
r cos 8 cos 2 4 x8
x
2
x2 4 y2
x 3 xy 2 4x 2 4y 2 y2 x2
61. x 2 y 2 2 ax 2y
63. x 2 y 2 a 2 arctan
r 4 a r 2 cos 2 r sin ra
cos 2
y x
2
r2 a22
sin
65. r 4
67. r sec
π 2
Circle of radius 4
1 cos
π 2
r cos 1, x 1
Centered at the pole Symmetric to polar axis,
2, and pole
44 xx
0 2
6
Vertical line
0 1
Review Exercises for Chapter 9 69. r 21 cos
71. r 4 3 cos
Cardioid
Limaçon
Symmetric to polar axis
Symmetric to polar axis
π 2
π 2
0
0
1
2
0
3
2
2 3
0
r
4
3
2
1
0
r
1
73. r 3 cos 2
4
3 5 2
2
2 3 11 2
4
π 2
Rose curve with four petals Symmetric to polar axis,
, and pole 2
0 4
3 Relative extrema: 3, 0, 3, , 3, , 3, 2 2
Tangents at the pole:
3 , 4 4
75. r 2 4 sin 2 2
π 2
r ± 2 sin 2 Rose curve with four petals
, and pole 2 3 Relative extrema: ± 2, , ± 2, 4 4 Tangents at the pole: 0, 2
0
Symmetric to the polar axis,
77. r
3 cos 4
79. r 4 cos 2 sec
Graph of r 3 sec rotated through an angle of 4 5
−1
2
Strophoid Symmetric to the polar axis r ⇒ as ⇒ 2 r ⇒ as ⇒ 2
8
4
−1
−6
6
−4
7
219
220
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
81. r 1 2 cos (a) The graph has polar symmetry and the tangents at the pole are
, . 3 3 (b)
2 sin 2 1 2 cos cos dy dx 2 sin cos 1 2 cos sin Horizontal tangents: 4 cos 2 cos 2 0, cos When cos
3 4
3 4
3 4
3 4
1 ± 33 1 33 ,r12 8 8
1 ± 1 32 1 ± 33 8 8
3 4 33,
1 8 33 0.686, 0.568 33 1 33 , arccos 0.686, 0.568 8 33 1 33 , arccos 2.186, 2.206 8 33 1 33 , arccos 2.186, 2.206. 8
33
, arccos
Vertical tangents: 1 sin 4 cos 1 0, sin 0, cos , 4
0, , ± arccos
14, 1, 0, 3,
12, ± arccos 41 0.5, ± 1.318 (c)
2.5
−5
1
−2.5
83. Circle: r 3 sin sin 2 dy 3 cos sin 3 sin cos dy tan 2 at , 3 dx 3 cos cos 3 sin sin cos 2 sin 2 6 dx Limaçon: r 4 5 sin dy 5 cos sin 4 5 sin cos dy 3 at , dx 5 cos cos 4 5 sin sin 6 dx 9 Let be the angle between the curves: tan
23.
3 39
1 13
Therefore, arctan
3
2 3 3 49.1 .
Review Exercises for Chapter 9 85. r 1 cos , r 1 cos The points 1, 2 and 1, 32 are the two points of intersection (other than the pole). The slope of the graph of r 1 cos is m1
r sin r cos dy sin 2 cos 1 cos . dx r cos r sin sin cos sin 1 cos
At 1, 2, m1 11 1 and at 1, 32, m1 11 1. The slope of the graph of r 1 cos is m2
sin 2 cos 1 cos dy . dx sin cos sin 1 cos
At 1, 2, m2 11 1 and at 1, 32, m 2 11 1. In both cases, m 1 1m 2 and we conclude that the graphs are orthogonal at 1, 2 and 1, 32. 87. r 2 cos A2
1 2
89. r sin
2 cos 2 d 14.14
0
9 2
A2
3
1 2
0.10
−3
cos 2 2
sin cos 2 2 d
0
32
6
0.5
−3 − 0.5
0.5 − 0.1
91. r 2 4 sin 2 A2
93. r 4 cos , r 2
12
2
4 sin 2 d 4
0
A2
12
−3
3
2
3
22 a
−3
1 cos d 22 a
0
0
2 ,e1 1 sin
Parabola
sin d 42 a1 cos 12 1 cos
99. r
0
8a
6 2 2 ,e 3 2 cos 1 23cos 3
Ellipse π 2
π 2
0
2
0
2
4 6
8
4 cos 2 d 4.91
a 21 cos 2 a 2 sin 2 d
0
97. r
6
−2
1 2
3
−3
4 d
0
2
95. s 2
3
221
214
Chapter 9
63. r1
Conics, Parametric Equations, and Polar Coordinates
ed ed and r2 1 sin 1 sin
Points of intersection: ed, 0, ed, ed ed cos cos sin dy 1 sin 1 sin 2 r1: dx ed ed cos sin cos 1 sin 1 sin 2
At ed, 0,
dy dy 1. At ed, , 1. dx dx
ed ed cos cos sin dy 1 sin 1 sin 2 r2: dx ed ed cos sin cos 1 sin 1 sin 2
At ed, 0,
dy dy 1. At ed, , 1. dx dx
Therefore, at ed, 0 we have m1m2 11 1, and at ed, we have m1m2 11 1. The curves intersect at right angles.
Review Exercises for Chapter 9 1. Matches (d) - ellipse
3. Matches (a) - parabola
16x 2 16y 2 16x 24y 3 0
5.
x
2
x
y
1 3 9 3 1 9 y2 y 4 2 16 16 4 16
1 x
1 x 2
2
3 y 4
1 2
1
2
1 , 2
2
Circle Center:
1
3 4
12, 43
Radius: 1 3x 2 2y 2 24x 12y 24 0
7.
3x 2 2y 2 12x 12y 29 0
9.
3x 2 8x 16 2 y 2 6y 9 24 48 18
3x 2 4x 4 2 y 2 6y 9 29 12 18
x 4 2 y 3 2 1 2 3
x 2 2 y 3 2 1 13 12
Hyperbola
Ellipse
Center: 4, 3
Center: 2, 3
Vertices: 4 ± 2, 3 Asymptotes: y 3 ±
Vertices:
32 x 4
2, 3
±
2
2
y
x
1
y
1
2
1 6
2
4
3 4
2 x
6
4
2
(2, − 3)
3
Review Exercises for Chapter 9 13. Vertices: 3, 0, 7, 0
11. Vertex: 0, 2
15. Vertices: ± 4, 0
Directrix: x 3
Foci: 0, 0, 4, 0
Foci: ± 6, 0
Parabola opens to the right
Horizontal major axis
Center: 0, 0
p3
Center: 2, 0
Horizontal transverse axis
y 2 4 3x 0
a 5, c 2, b 21
a 4, c 6, b 36 16 25
y 2 4y 12x 4 0
x 22 y 2 1 25 21
x2 y2 1 16 20
2
17.
215
5 x2 y2 1, a 3, b 2, c 5, e 9 4 3
19. y x 2 has a slope of 1. The perpendicular slope is 1. y x 2 2x 2
By Example 5 of Section 9.1, C 12
2
0
dy 1 5 2x 2 1 when x and y . dx 2 4
1
5 sin2 d 15.87. 9
y
Perpendicular line:
5 1 1 x 4 2
4x 4y 7 0 21. (a) V abLength 12 16 192 ft 3
3
(b) F 262.4
8 4 3 y 9 y 2 dy 62.4 3 3 3 3
3
3
8 3 9 3 9 62.4 3 2 2 2 2
3
y9 y 2 dy
y 1 3 8 y 9 y 2 9 arcsin 9 y 2 32 62.4 3 2 3 3
3
9 y 2 dy 3 3
38 62.4272 7057.274
3
(c) You want 4 of the total area of 12 covered. Find h so that
h
2
x=
4 9 y 2 dy 3 3
0
h
9 y 2 dy
0
y 1 y 9 y 2 9 arcsin 2 3
h 0
y 4 3
9−
y2
4
h
9 8
Area of filled tank above x-axis is 3π.
2 x
−2
2 −2
9 8
−4
Area of filled tank below x-axis is 6π.
h3 94.
h9 h 2 9 arcsin
By Newton’s Method, h 1.212. Therefore, the total height of the water is 1.212 3 4.212 ft. (d) Area of ends 212 24 Area of sides PerimeterLength 16
2
1
0
256
167 sin d 16 from Example 5 of Section 9.1 2
122 1 167 sin 0 41 167 sin 8 21 167 sin 4 2
2
2
1 167 sin 38 1 167 sin 2 353.65
4
Total area 24 353.65 429.05
2
2
216
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
23. x 1 4t, y 2 3t t
y
x1 x1 ⇒ y23 4 4
4
2 1
11 3 y x 4 4
−1
4y 3x 11 0
x −1
1
2
3
5
−2
Line 25. x 6 cos , y 6 sin
6x 6y 2
x2
y2
2
27. x 2 sec , y 3 tan
y
1
x 22 sec2 1 tan2 1 y 32
4
x 22 y 32 1
2
36
−4 −2
Circle
y
x −2
2
4
8
Hyperbola
−4
4 2
x
−4
−2
2
4
8
−4
29. x 3 3 2t 3 5t
31.
x 3 2 y 4 2 1 16 9
y 2 2 6t 2 4t Let
(other answers possible)
x 3 2 y 4 2 cos 2 and sin 2 . 16 9
Then x 3 4 cos and y 4 3 sin . 35. (a) x 2 cot , y 4 sin cos , 0 < <
33. x cos 3 5 cos y sin 3 5 sin
4
5
−12 −7
12
8
−4
(b) 4 x2y 4 4 cot2 4 sin cos
−5
16 csc2 16
sin cos
cos sin
82 cot 8x 37. x 1 4t y 2 3t (a)
dy 3 dx 4
(b) t
x1 4
y
(c) 5
No horizontal tangents
3 3x 11 y 2 x 1 4 4
4
2 1 x
1
2
3
5
Review Exercises for Chapter 9
217
39. x 1 t y 2t 3 (a)
dy 2 2t 2 dx 1t 2
(b) t
No horizontal tangents t 0
1 x
y
(c)
y
6
2 3 x
4 2 x
4
41. x
1 2t 1
y
1 t 2t
2
y 2 5 sin (a)
2t 2 dy t 2 2t 2 t 12t 1 2 0 when t 1. (a) dx 2 t 2t 2 2 2t 1 2 Point of horizontal tangency: (b) 2t 1
13 , 1
1 1 1 1 ⇒ t x 2 x
dy 5 cos 3 2.5 cot 0 when , . dx 2 sin 2 2 Points of horizontal tangency: 3, 7, 3, 3
(b)
x 3 2 y 2 2 1 4 25 y
(c)
8
4
1 1 1x 2 x
12 1 x x 2
x
4
4x 2 4x 2 2 1 x 4x1 x 5x 1x 1
(c)
y 3 2
−2
x
−1
2
3
−1 −2
45. x cos 3 y 4 sin 3 (a)
dy 12 sin 2 cos 4 sin 4 tan 0 when 0, . dx 3 cos 2 sin cos But,
dy dx 0 at 0, . Hence no points of horizontal tangency. dt dt
(b) x 23
4y
23
1
y
(c) 4
x
4
4
43. x 3 2 cos
2
y
2
2
2
4
4
8 4
218
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
47. x cot
x r cos sin
49.
y rsin cos
y sin 2 2 sin cos (a), (c)
dx r cos d
2
−3
3
dy r sin d
−2
sr
dx 1 dy dy (b) At , 4, 1, and 6 d d dx 4
2 cos 2 2 sin 2 d
0
r
d
0
51. x, y 4, 4
1 2 r 2
x −1
1
2
3
4
5
−2
r, 42,
7 , 4
42,
3 4
−3
−4 −5
r 3 cos r2
(4, − 4)
r 2 2r 1 cos
3r cos
x 2 y 2 2 ± x 2 y 2 2x
x 2 y 2 2x 2 4x 2 y 2
x 2 y 2 3x 0 r 2 cos 2 cos 2 sin 2
r 21 cos
55.
x 2 y 2 3x
r4
0
1
7 4
57.
y
r 4 2 4 2 42
53.
r 2 2
r2
cos 2
59.
r 2 sin 2
r 4 cos 2 sec 4 2 cos 2 1
x 2 y 2 2 x 2 y 2
cos1
r cos 8 cos 2 4 x8
x
2
x2 4 y2
x 3 xy 2 4x 2 4y 2 y2 x2
61. x 2 y 2 2 ax 2y
63. x 2 y 2 a 2 arctan
r 4 a r 2 cos 2 r sin ra
cos 2
y x
2
r2 a22
sin
65. r 4
67. r sec
π 2
Circle of radius 4
1 cos
π 2
r cos 1, x 1
Centered at the pole Symmetric to polar axis,
2, and pole
44 xx
0 2
6
Vertical line
0 1
Review Exercises for Chapter 9 69. r 21 cos
71. r 4 3 cos
Cardioid
Limaçon
Symmetric to polar axis
Symmetric to polar axis
π 2
π 2
0
0
1
2
0
3
2
2 3
0
r
4
3
2
1
0
r
1
73. r 3 cos 2
4
3 5 2
2
2 3 11 2
4
π 2
Rose curve with four petals Symmetric to polar axis,
, and pole 2
0 4
3 Relative extrema: 3, 0, 3, , 3, , 3, 2 2
Tangents at the pole:
3 , 4 4
75. r 2 4 sin 2 2
π 2
r ± 2 sin 2 Rose curve with four petals
, and pole 2 3 Relative extrema: ± 2, , ± 2, 4 4 Tangents at the pole: 0, 2
0
Symmetric to the polar axis,
77. r
3 cos 4
79. r 4 cos 2 sec
Graph of r 3 sec rotated through an angle of 4 5
−1
2
Strophoid Symmetric to the polar axis r ⇒ as ⇒ 2 r ⇒ as ⇒ 2
8
4
−1
−6
6
−4
7
219
220
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
81. r 1 2 cos (a) The graph has polar symmetry and the tangents at the pole are
, . 3 3 (b)
2 sin 2 1 2 cos cos dy dx 2 sin cos 1 2 cos sin Horizontal tangents: 4 cos 2 cos 2 0, cos When cos
3 4
3 4
3 4
3 4
1 ± 33 1 33 ,r12 8 8
1 ± 1 32 1 ± 33 8 8
3 4 33,
1 8 33 0.686, 0.568 33 1 33 , arccos 0.686, 0.568 8 33 1 33 , arccos 2.186, 2.206 8 33 1 33 , arccos 2.186, 2.206. 8
33
, arccos
Vertical tangents: 1 sin 4 cos 1 0, sin 0, cos , 4
0, , ± arccos
14, 1, 0, 3,
12, ± arccos 41 0.5, ± 1.318 (c)
2.5
−5
1
−2.5
83. Circle: r 3 sin sin 2 dy 3 cos sin 3 sin cos dy tan 2 at , 3 dx 3 cos cos 3 sin sin cos 2 sin 2 6 dx Limaçon: r 4 5 sin dy 5 cos sin 4 5 sin cos dy 3 at , dx 5 cos cos 4 5 sin sin 6 dx 9 Let be the angle between the curves: tan
23.
3 39
1 13
Therefore, arctan
3
2 3 3 49.1 .
Review Exercises for Chapter 9 85. r 1 cos , r 1 cos The points 1, 2 and 1, 32 are the two points of intersection (other than the pole). The slope of the graph of r 1 cos is m1
r sin r cos dy sin 2 cos 1 cos . dx r cos r sin sin cos sin 1 cos
At 1, 2, m1 11 1 and at 1, 32, m1 11 1. The slope of the graph of r 1 cos is m2
sin 2 cos 1 cos dy . dx sin cos sin 1 cos
At 1, 2, m2 11 1 and at 1, 32, m 2 11 1. In both cases, m 1 1m 2 and we conclude that the graphs are orthogonal at 1, 2 and 1, 32. 87. r 2 cos A2
1 2
89. r sin
2 cos 2 d 14.14
0
9 2
A2
3
1 2
0.10
−3
cos 2 2
sin cos 2 2 d
0
32
6
0.5
−3 − 0.5
0.5 − 0.1
91. r 2 4 sin 2 A2
93. r 4 cos , r 2
12
2
4 sin 2 d 4
0
A2
12
−3
3
2
3
22 a
−3
1 cos d 22 a
0
0
2 ,e1 1 sin
Parabola
sin d 42 a1 cos 12 1 cos
99. r
0
8a
6 2 2 ,e 3 2 cos 1 23cos 3
Ellipse π 2
π 2
0
2
0
2
4 6
8
4 cos 2 d 4.91
a 21 cos 2 a 2 sin 2 d
0
97. r
6
−2
1 2
3
−3
4 d
0
2
95. s 2
3
221
222
Chapter 9
101. r
Conics, Parametric Equations, and Polar Coordinates
4 2 3 ,e 2 3 sin 1 32sin 2
103. Circle Center:
Hyperbola
5, 2 0, 5 in rectangular coordinates
Solution point: 0, 0
π 2
x 2 y 5 5 25 x 2 y 2 10y 0
0
2
3
4
r 2 10r sin 0 r 10 sin
105. Parabola
107. Ellipse
Vertex: 2,
Vertices: 5, 0, 1,
Focus: 0, 0
Focus: 0, 0
e 1, d 4
2 5 a 3, c 2, e , d 3 2
4 r 1 cos
2352 5 r 2 3 2 cos 1 cos 3
Problem Solving for Chapter 9 y
1. (a) 10 8 6 4
(− 1, 14 )
(4, 4)
2
−6 −4 −2
x 2
−2
4
6
(b) x2 4y 2x 4y 1 y x 2 ⇒ y 2x 4
y 4 2x 4
1 1 1 1 y x 1 ⇒ y x 4 2 2 4
Tangent line at 4, 4
Tangent line at 1,
Tangent lines have slopes of 2 and 12 ⇒ perpendicular. (c) Intersection: 1 1 2x 4 x 2 4 8x 16 2x 1 10x 15 x
3 ⇒ 2
32, 1
Point of intersection, 32, 1, is on directrix y 1.
1 4
Problem Solving for Chapter 9 3. Consider x2 4py with focus 0, p.
y
A
B
Let pa, b be point on parabola. zx 4py ⇒ y yb
F
x 2p
P(a, b) x
Q
a x a Tangent line 2p
For x 0, y b
a 4pb a2 b b. a b 2p 2p 2p
Thus, Q 0, b. FQP is isosceles because
FQ p b FP a 02 b p2 a2 b2 2bp p2 4pb b2 2bp p2 b p2 b p. Thus, FQP BPA FPQ.
5. (a) In OCB, cos
2a ⇒ OB 2a sec . OB
r 2a tan sin
(c)
r cos 2a sin2
OA In OAC, cos ⇒ OA 2a cos . 2a
r 3 cos 2a r 2 sin2
x2 y2x 2ay2
r OP AB OB OA 2asec cos 2a
cos1 cos
2a
y2
x3 2a x
sin2 cos
2a tan sin (b) x r cos 2a tan sin cos 2a sin2 y r sin 2a tan sin sin 2a tan
sin2 , 2
2
< <
1 + t2
θ
Let t tan , < t < . t2 t2 t3 Then sin2 and x 2a , y 2a . 2 2 1t 1t 1 t2
7. y a1 cos ⇒ cos
arccos
t
1
ay a
a a y a
x a sin
a a y sinarccosa a y
a a y
a arccos
a arccos
x a arccos
2ay y2
a
a a y 2ay y , 0 ≤ y ≤ 2a 2
θ
a−y
2ay − y 2
223
224
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
3 5 7 , , , ,. . . 2 2 2 2
9. For t y
11. (a) Area
0
2 2 2 2 , , , ,. . . 3 5 7
Hence, the curve has length greater that
(b) tan
2 2 2 2 S . . . 3 5 7
2 1 1 1 1 . . . 3 5 7
2 1 1 1 1 . . . > 2 4 6 8
1 2
1 2 r d 2
x=1 r = sec θ
α
sec2 d
1
0
h 1 ⇒ Area 1tan
1 2 ⇒ tan
sec2 d
0
(c) Differentiating,
d tan sec2 . d
.
13. If a dog is located at r, , then its neighbor is at r,
: 2
x, y r cos , r sin and x, y r sin , r cos . The slope joining these points is r cos r sin sin cos slope of tangent line at r, . r sin r cos sin cos dr sin r cos sin cos d dr sin cos cos r sin d dr r d
⇒
dr d r ln r C1 r e C1 r Ce r
4 d2 ⇒ r Ce
4
Finally, r
d 2
d 2
⇒ C
d 2
e4
e4 .
15. (a) The first plane makes an angle of 70 with the positive x-axis, and is 150 miles from P:
(c)
280
x1 cos 70 150 375t y1 sin 70 150 375t
0
1 0
Similarly for the second plane, x2 cos 135 190 450t
The minimum distance is 7.59 miles when t 0.4145.
cos 45 190 450t y2 sin 135 190 450t sin 45 190 450t (b) d x2 x12 y2 y12
cos 45190 450t cos 70150 375t 2 sin 45190 450t sin 70150 375t 2 12
Problem Solving for Chapter 9 17.
4
−6
4
6
−6
4
4
6
−6
6
−6
6
−4
−4
−4
−4
4
4
4
4
−6
6
−6
6
−6
6
−4
−4
−4
4
4
4
−6
6
−4
−6
6
−4
−6
n 1, 2, 3, 4, 5 produce “bells”; n 1, 2, 3, 4, 5 produce “hearts”.
6
−4
6
−4
−6
225
PA R T
I C H A P T E R P Preparation for Calculus Section P.1
Graphs and Models . . . . . . . . . . . . . . . . . . . . . . 2
Section P.2
Linear Models and Rates of Change . . . . . . . . . . . . . 7
Section P.3
Functions and Their Graphs . . . . . . . . . . . . . . . . . 14
Section P.4
Fitting Models to Data . . . . . . . . . . . . . . . . . . . . 18
Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
C H A P T E R P Preparation for Calculus Section P.1
Graphs and Models
Solutions to Odd-Numbered Exercises
1 1. y 2 x 2
3. y 4 x2
x-intercept: 4, 0
x-intercepts: 2, 0, 2, 0
y-intercept: 0, 2
y-intercept: 0, 4
Matches graph (b)
Matches graph (a) 7. y 4 x2
5. y 32x 1 x
4
2
0
2
4
x
3
2
0
2
3
y
5
2
1
4
7
y
5
0
4
0
5
y
y 8
(4, 7)
6
2
(−2, 0)
(0, 1)
−8 −6 −4
2 −4
(0, 4)
(2, 4)
4
(−4, − 5)
6
x
4
6
−6
8
4
(− 3, − 5)
(3, − 5)
−4 −6
11. y x 4
9. y x 2 x
5
4
3
2
1
0
1
x
0
1
4
9
16
y
3
2
1
0
1
2
3
y
4
3
2
1
0
y
y 10 8 6 4 2
6 4
(−5, 3)
(−4, 2) 2 −6
−4
(1, 3) (0, 2)
−2
(−1, 1)
(−3, 1)
x
(− 2, 0) −2
2
6
−2
−8
(2, 0) x
−4
(− 2, −2)
−6
2
2
−4 −6 −8 − 10
(4, − 2)
(16, 0) x
2
(1, − 3) (0, − 4)
12 14 16 18
(9, − 1)
Section P.1 13.
15.
Xmin = -3 Xmax = 5 Xscl = 1 Ymin = -3 Ymax = 5 Yscl = 1
3
5
(− 4.00, 3) (2, 1.73) −6
6
−3
(a) 2, y 2, 1.73
Note that y 4 when x 0.
(b) x, 3 4, 3
y 02 0 2
y-intercept:
y 0225 02
y-intercept:
y 0; 0, 0
y 2; 0, 2 0 x2 x 2
x-intercepts:
y 5 2 3 1.73 3 5 4
19. y x225 x2
17. y x2 x 2
21. y
Graphs and Models
x-intercepts:
0 x225 x2
0 x 2x 1
0 x25 x5 x
x 2, 1; 2, 0, 1, 0
x 0, ± 5; 0, 0; ± 5, 0
32 x x
23. x2y x2 4y 0
y-intercept:
None. x cannot equal 0.
x-intercepts:
32 x 0 x
y-intercept: 02y 02 4y 0 y 0; 0, 0
0 2 x
x-intercept:
x 4; 4, 0
x20 x2 40 0 x 0; 0, 0
25. Symmetric with respect to the y-axis since
27. Symmetric with respect to the x-axis since
y x 2 x 2. 2
y2 y2 x3 4x.
2
31. y 4 x 3
29. Symmetric with respect to the origin since
xy xy 4.
No symmetry with respect to either axis or the origin.
x y x2 1 y
35. y x3 x is symmetric with respect to the y-axis
33. Symmetric with respect to the origin since
since y x3 x x3 x x3 x .
x . x2 1
37. y 3x 2
y
Intercepts:
23 , 0, 0, 2
2
(0, 2)
1
2 3,
Symmetry: none
0 x
1
2 1
3
4
Chapter P
39. y
Preparation for Calculus
x 4 2
43. y x 32
41. y 1 x2
Intercepts:
Intercepts:
Intercepts:
1, 0, 1, 0, 0, 1
8, 0, 0, 4
3, 0, 0, 9
Symmetry: y-axis
Symmetry: none
Symmetry: none
y y
y 12
2
2
(8, 0)
2
2
8
4
10
( 1, 0)
2
(0,
10
(0, 1)
x
x
−2
4)
2
6
1
8
2
2 − 10 − 8 − 6
10
47. y xx 2
45. y x3 2 3 2, 0 , 0, 2
−2
Symmetry: origin
Symmetry: none
y 4
Domain: x ≥ 2
y
3 2
5 y
4
2, 0)
(0, 2)
2
1
2
1
(− 2, 0) −4 −3
−1
3
4
−4
2
3
2
−3
3
1
1
−2
4
x
x
−4 −3 −2 −1
5
1 3
(0, 0)
6
3 3
4
Intercepts: 0, 0
0, 0, 2, 0
Symmetry: none
x 2
(− 3, 0)
49. x y3
Intercepts:
Intercepts:
(
(0, 9)
8
(1, 0)
(0, 0) x 1
2
3
4
−2
1 x Intercepts: none
51. y
53. y 6 x
y 3
y 8
Intercepts:
6
2
0, 6, 6, 0, 6, 0
1
Symmetry: origin
x 1
2
Symmetry: y-axis
3
(0, 6)
4 2
(− 6, 0) −8
−4 −2 −2
(6, 0) x 2
4
6
8
−4 −6 −8
57. x 3y2 6
55. y2 x 9 y2 x 9
3y2 6 x
y ± x 9
4
Intercepts:
0, 3, 0, 3, 9, 0 Symmetry: x-axis
y±
(0, 3) (−9, 0) − 11
1
(0, − 3) −4
2 3x
Intercepts:
3
(0, 2 ) (6, 0)
−1
8
6, 0, 0, 2, 0, 2 Symmetry: x-axis
(0, − 2 ) −3
Section P.1 59. y x 2x 4x 6 (other answers possible)
Graphs and Models
61. Some possible equations: yx y x3 y 3x3 x 3 x y
63.
xy2⇒y2x
xy7⇒y7x
65.
2x y 1 ⇒ y 2x 1
3x 2y 11 ⇒ y
2 x 2x 1 7x
3 3x 1x
3x 11 2
3x 11 2
14 2x 3x 11
The corresponding y-value is y 1.
5x 25
Point of intersection: 1, 1
x5 The corresponding y-value is y 2. Point of intersection: 5, 2
67. x2 y 6 ⇒ y 6 x2
69. x2 y 2 5 ⇒ y 2 5 x 2 xy1⇒yx1
xy4⇒y4x
5 x2 x 12
6 x2 4 x 0 x2 x 2
5 x2 x2 2x 1
0 x 2x 1
0 2x2 2x 4 2x 1x 2
x 2, 1
x 1 or x 2 The corresponding y-values are y 2 and y 1.
The corresponding y-values are y 2 (for x 2) and y 5 (for x 1).
Points of intersection: 1, 2, 2, 1
Points of intersection: 2, 2, 1, 5 71.
y x3
y x3 2x2 x 1
73.
yx
y x2 3x 1
x3 x
x3 2x2 x 1 x2 3x 1
x3 x 0 xx 1x 1 0
x3 x2 2x 0 xx 2x 1 0 x 1, 0, 2
x 0, x 1, or x 1 The corresponding y-values are y 0, y 1, and y 1.
1, 5, 0, 1, 2, 1 4
Points of intersection: 0, 0, 1, 1, 1, 1 −4
y = x 3 − 2x 2 + x − 1 (2, 1)
(0, −1)
6
(−1, −5) −8
y = −x 2 + 3 x − 1
5
6
Chapter P
Preparation for Calculus
75. 5.5x 10,000 3.29x
5.5x 2 3.29x 10,0002 30.25x 10.8241x2 65,800x 100,000,000 0 10.8241x2 65,830.25x 100,000,000
Use the Quadratic Formula.
x 3133 units The other root, x 2949, does not satisfy the equation R C. This problem can also be solved by using a graphing utility and finding the intersection of the graphs of C and R. 77. (a) Using a graphing utility, you obtain
(b)
250
y 0.0153t2 4.9971t 34.9405 (c) For the year 2004, t 34 and y 187.2 CPI.
−5
35 − 50
79.
400
0
100 0
If the diameter is doubled, the resistance is changed by approximately a factor of 14. For instance, y20 26.555 and y40 6.36125. 81. False; x-axis symmetry means that if 1, 2 is on the graph, then 1, 2 is also on the graph. 83. True; the x-intercepts are
b
± b2 4ac
2a
,0 .
85. Distance to the origin K Distance to 2, 0 x2 y2 Kx 22 y2, K 1
x2 y 2 K 2x2 4x 4 y2
1 K 2 x 2 1 K 2y 2 4K 2x 4K 2 0 Note: This is the equation of a circle!
Section P.2
Section P.2
Linear Models and Rates of Change
1. m 1
3. m 0
7.
9. m
y
m=1
5 4
2
1
5. m 12
2 4 53
11. m
6 3 2
y
m = − 32
m is undefined
1
−1
(2, 3)
3
Linear Models and Rates of Change
51 22 4 0
undefined
3
x 3
4
2
5
y
(5, 2)
1
m = −2 −1
6 x 1
2
3
5
6
4
−2
3
−3 −4
(2, 5)
5
7
2
(3, − 4)
(2, 1)
1
−5
−2 −1 −1
x 1
3
4
5
6
−2
13. m
23 16 12 34
y 3 2
12 2 14
(− 12 , 23 ) −3
−2
(− 34 , 16 ) x 1
2
3
−1 −2 −3
15. Since the slope is 0, the line is horizontal and its equation is y 1. Therefore, three additional points are 0, 1, 1, 1, and 3, 1. 17. The equation of this line is y 7 3x 1 y 3x 10 . Therefore, three additional points are 0, 10, 2, 4, and 3, 1. 19. Given a line L, you can use any two distinct points to calculate its slope. Since a line is straight, the ratio of the change in y-values to the change in x-values will always be the same. See Section P.2 Exercise 93 for a proof.
7
8
Chapter P
Population (in millions)
21. (a)
Preparation for Calculus (b) The slopes of the line segments are
270
255.0 252.1 2.9 21
260
257.7 255.0 2.7 32
250
260.3 257.7 2.6 43
1 2 3 4 5 6 7 8 9
Year (0 ↔ 1990)
262.8 260.3 2.5 54 265.2 262.8 2.4 65 267.7 265.2 2.5 76 270.3 267.7 2.6 87 The population increased most rapidly from 1991 to 1992.
m 2.9 23. x 5y 20 y
25. x 4
15 x
4
Therefore, the slope is m 0, 4. 27.
y 34 x 3
1 5
The line is vertical. Therefore, the slope is undefined and there is no y-intercept.
and the y-intercept is
y 23 x
29.
4y 3x 12
y 2 3x 3
31.
y 2 3x 9
3y 2x
0 3x 4y 12
2x 3y 0
y 3x 11 y 3x 11 0
y
y 5
4
4
3
y 3
(0, 3) 2
2
2
1
1
(0, 0) x
x
−4 −3 −2 −1
1
1
2
3
4
−2 −1 −1
x 1
2
4
5
6
(3, − 2)
−2
−1
3
−3 −4 −5
33. m
60 3 20
35. m
y 0 3x 0
1 3 2 20
y 1 2x 4
8
3y 8x 40 0
y
6
(2, 6)
4
2
y
(2, 1)
1
(0, 0) x 2
4
6
8
−2 −1 −1
x 2
−2
−8
40 8 y x 3 3
0 2x y 3
y
−8 −6 −4 −2
80 8 25 3 8 y 0 x 5 3
y 1 2x 2
y 3x
2
37. m
−3
−5
(0, −3)
3
4
5
9 8 7 6 5 4 3 2 1 −1 −2
(2, 8)
(5, 0) x 1 2 3 4
6 7 8 9
Section P.2 81 55
39. m
Undefined.
Vertical line x 5
41. m
Linear Models and Rates of Change
72 34 114 11 12 0 12 2 y
x3
43.
x30
3 11 x 0 4 2
y
y 9 8 7 6 5 4 3 2 1 −1
y
(5, 8)
11 3 x 2 4
2 1
22x 4y 3 0 −1
y
(5, 1) x 1 2 3 4
(3, 0) 1
4
6 7 8 9
3
−2
−2
( 12 , 72 )
2 1 −4 −3 −2 −1
y x 1 2 3
45.
( 0, 34 ) x 1
2
3
4
47.
3x 2y 6 0
y x 1 a a 1 2 1 a a 3 1 a a3⇒xy3 xy30
49.
51. y 2x 1
y 3 y30
y 3
y 2 1 x
−3 −2 −1
1
2
3
4
5
−2
−2
x
−1
1
2
−1
−4 −5 −6
y 2 32x 1
53.
y
3 2x
55. 2x y 3 0 y 2x 3
1 2
2y 3x 1 0
y 1
y
x
4
2
3
1
2 1
2
2
1 x
−4 −3 −2
1 −2 −3 −4
2
3
4
3
3
2
4
x
9
10
Chapter P
57.
Preparation for Calculus
10
10
− 10
− 15
10
− 10
15
− 10
The lines do not appear perpendicular.
The lines appear perpendicular.
The lines are perpendicular because their slopes 1 and 1 are negative reciprocals of each other. You must use a square setting in order for perpendicular lines to appear perpendicular. 61. 5x 3y 0
59. 4x 2y 3 y 2x 2
y 53x
m2
m 53
3
y 1 2x 2
(a)
(a)
24y 21 40x 30
y 1 2x 4
24y 40x 9 0
2x y 3 0 y1
(b)
y 78 53x 34
1 2 x
2
(b)
y 78 35x 34 40y 35 24x 18
2y 2 x 2
40y 24x 53 0
x 2y 4 0 63. (a) x 2 ⇒ x 2 0 (b) y 5 ⇒ y 5 0 65. The slope is 125. Hence, V 125t 1 2540 125t 2415 67. The slope is 2000. Hence, V 2000t 1 20,400 2000t 22,400 69.
5
(2, 4)
−3
(0, 0)
6
−1
You can use the graphing utility to determine that the points of intersection are 0, 0 and 2, 4. Analytically, x2 4x x2 2x2 4x 0 2xx 2 0 x 0 ⇒ y 0 ⇒ 0, 0 x 2 ⇒ y 4 ⇒ 2, 4. The slope of the line joining 0, 0 and 2, 4 is m 4 02 0 2. Hence, an equation of the line is y 0 2x 0 y 2x.
Section P.2
71. m1 m2
Linear Models and Rates of Change
10 1 2 1 2 0 2 2 1 3
m1 m2 The points are not collinear. y
73. Equations of perpendicular bisectors: y y
c ab ab x 2 c 2
c ba ab x 2 c 2
Letting x 0 in either equation gives the point of intersection:
0, a
2
(b, c)
( b −2 a , 2c )
( a +2 b , 2c )
(− a, 0)
x
(a, 0)
b2 c2 . 2c
This point lies on the third perpendicular bisector, x 0. 75. Equations of altitudes: y
y
ab x a c (b, c)
xb y
ab x a c
(a, 0) x
(− a, 0)
Solving simultaneously, the point of intersection is
b, a
2
b2 . c
77. Find the equation of the line through the points 0, 32 and 100, 212. 9 m 180 100 5 9 F 32 5 C 0
F 95 C 32 5F 9C 160 0 For F 72, C 22.2. 79. (a) W1 0.75x 12.50
(b)
50
W2 1.30x 9.20 (c) Both jobs pay $17 per hour if 6 units are produced. For someone who can produce more than 6 units per hour, the second offer would pay more. For a worker who produces less than 6 units per hour, the first offer pays more.
(6, 17) 0
30 0
Using a graphing utility, the point of intersection is approximately 6, 17. Analytically, 0.75x 12.50 1.30x 9.20 3.3 0.55x ⇒ x 6 y 0.756 12.50 17.
11
12
Chapter P
Preparation for Calculus
81. (a) Two points are 50, 580 and 47, 625. The slope is m
(b)
50
625 580 15. 47 50
p 580 15x 50
0
p 15x 750 580 15x 1330
1 If p 655, x 15 1330 655 45 units.
1 or x 15 1330 p
83. 4x 3y 10 0 ⇒ d
85. x y 2 0 ⇒ d
1500 0
1 (c) If p 595, x 15 1330 595 49 units.
40 30 10 10 2 42 32
5
12 11 2 1 1 2
2
5 5 2 2 2
87. A point on the line x y 1 is 0, 1. The distance from the point 0, 1 to x y 5 0 is d
10 11 5 1 5 12 12
2
4 2
2 2.
89. If A 0, then By C 0 is the horizontal line y CB. The distance to x1, y1 is
By C Ax By C. C B A B B
Ax C Ax By C. C A A B A
d y1
1
1
1
2
2
If B 0, then Ax C 0 is the vertical line x CA. The distance to x1, y1 is d x1
1
1
1
2
2
(Note that A and B cannot both be zero.) The slope of the line Ax By C 0 is AB. The equation of the line through x1, y1 perpendicular to Ax By C 0 is: y y1
B x x1 A
Ay Ay1 Bx Bx1 Bx1 Ay1 Bx Ay The point of intersection of these two lines is: Ax By C
⇒
Bx Ay Bx1 Ay1 ⇒
A2x ABy AC B
2x
ABy
B2x
1
(1) ABy1 (2)
A2 B2x AC B2x1 ABy1 (By adding equations (1) and (2)) x Ax By C
⇒
AC B2x1 ABy1 A2 B2
ABx B2y BC
Bx Ay Bx1 Ay1⇒ ABx
A2 y
ABx1
(3) A2 y1
(4)
A2 B2y BC ABx1 A2y1 (By adding equations (3) and (4)) y
—CONTINUED—
BC ABx1 A2y1 A2 B2
Section P.2
Linear Models and Rates of Change
89. —CONTINUED— Ay AC A Bx B ABy , BC A ABx point of intersection B 2
1
2
1
2
1
2
2
1
2
The distance between x1, y1 and this point gives us the distance between x1, y1 and the line Ax By C 0. Ay AC A Bx B ABy x BC A ABx B AC ABy A x By
BC A ABx A B B 2
d
1
2
1
2
2
1
2
1
2
1
1
2
2
2
2
1
2
2
2
1
1
2
2
2
2
1
2
1
2
2
1 2 2
1
y1
2
2
2
AC A ByB Ax BC A AxB By A B C Ax By A B
1
1
2
2
2
Ax1 By1 C A2 B2
91. For simplicity, let the vertices of the rhombus be 0, 0, a, 0, b, c, and a b, c, as shown in the figure. The slopes of the diagonals are then m1
y
(b, c)
(a + b , c )
c c . and m2 ab ba
x
(0, 0)
Since the sides of the Rhombus are equal, a2 b2 c2, and we have m1m2
c ab
c
c2
(a, 0)
c2
b a b2 a2 c2 1.
Therefore, the diagonals are perpendicular.
93. Consider the figure below in which the four points are collinear. Since the triangles are similar, the result immediately follows. y2 y1 y2 y1 x2 x1 x2 x1
95. True. a c a ax by c1 ⇒ y x 1 ⇒ m1 b b b b c bx ay c2 ⇒ y x 2 a a
y
m2 (x 2 , y2 )
(x *2 , y*2 )
(x1, y1 ) (x *1, y*1 )
x
1 m1
⇒ m2
b a
13
14
Chapter P
Preparation for Calculus
Section P.3
Functions and Their Graphs 3. (a) g0 3 02 3
1. (a) f 0 20 3 3 (b) f 3 23 3 9
(b) g3 3 3 3 3 0
(c) f b 2b 3
(c) g2 3 22 3 4 1
(d) f x 1 2x 1 3 2x 5
(d) gt 1 3 t 12 t2 2t 2
5. (a) f 0 cos20 cos 0 1 (c) f
2
4 cos2 4 cos 2 0
(b) f
3 cos23 cos23 12
7.
f x x f x x x3 x3 x3 3x2x 3xx2 x3 x3 3x2 3xx x2, x 0 x x x
9.
f x f 2 1x 1 1 x2 x2
1 x 1 1 x 1 2x 1 ,x2 x 2x 1 1 x 1 x 2x 11 x 1 x 11 x 1
11. hx x 3
13. f t sec
Domain: x 3 ≥ 0 ⇒ 3, Range: , 0
t 4
t 2k 1 ⇒ t 4k 2 4 2 Domain: all t 4k 2, k an integer Range: , 1, 1,
15. f x
1 x
Domain: , 0, 0, Range: , 0, 0,
17. f x
2x2x 1,2, xx ≤ 11 2
20. f x
2
x x54,, xx >≤ 55
2
(a) f 2 22 2 6
(a) f 3 3 4 1 1
(b) f 0 02 2 2
(b) f 0 0 4 2 (c) f 5 5 4 3
(c) f 1 12 2 3 (d) f
s2
2 2
s2
2 2
2s4
(d) f 10 10 52 25
10
8s2
(Note: s2 2 > 1 for all s)
Domain: 4,
Domain: ,
Range: 0,
Range: 2, 22. gx
4 x
1 24. f x 2 x3 2
y
Domain: ,
6
Domain: , 0, 0,
4
8 6
Range: ,
2
Range: , 0, 0,
y
4
x 2
4
6
x
−6 −4
26. f x x 4 x2 4
Domain: 2, 2 Range:
2, 22 2, 2.83
28. h 5 cos
y
(0, 2) 2, 0(
−4 −3 −2
y-intercept: 0, 2 x-intercept: 2, 0
x −1
1
2
3
6
5 4 3 2 1
Range: 5, 5 −2 π
4
4
y
Domain: ,
3
(−
2
2
2π
θ
−2 −3 −4
30. x2 4 y 0 ⇒ y x2 4 y is a function of x. Vertical lines intersect the graph at most once.
34. x2 y 4 ⇒ y 4 x2 y is a function of x since there is one value of y for each x.
−5
32. x2 y2 4 y ± 4 x2 y is not a function of x. Some vertical lines intersect the graph twice. x2 4 y is a function of x since there is one value of y for each x.
36. x2y x2 4y 0 ⇒ y
x2
294
Chapter P
Preparation for Calculus
38. p1x x3 x 1 has one zero. p2x x3 x has three zeros. Every cubic polynomial has at least one zero. Given px Ax3 Bx2 Cx D, we have p → as x → and p → as x → if A > 0. Furthermore, p → as x → and p → as x → if A < 0. Since the graph has no breaks, the graph must cross the x-axis at least one time.
2
P1 −3
−2
40. The function is f x cx. Since 1, 1 4 satisfies the equation, c 14. Thus, f x 14x.
3
P2
42. The function is hx c x . Since 1, 3 satisfies the equation, c 3. Thus, hx 3 x .
20 1 mimin during the first 40 2 4 minutes. The student is stationary for the following 62 2 minutes. Finally, the student travels 1 mimin 10 6 during the final 4 minutes.
44. The student travels
46. (a)
A 500 400 300 200 100 t 10
20
30
40
50
(b) A15 345 acresfarm 48. (a) gx f x 4
y 4
4
g6 f 2 1
Shift f left 2 units
3
3 2
2
g0 f 4 3 Shift f right 4 units
(b) gx f x 2
y
(6, 1)
1
x
−1
1
2
3
5
6
−7 −6 −5 −4 −3
7
(0, 1) x
−1
1
−2
−2
−3
(− 6, −3)
(0, −3)
−4
−4
(c) gx f x 4
(d) gx f x 1
y 6
Vertical shift upwards 4 units
y 2
(2, 5)
Vertical shift down 1 unit
5 4
1
(2, 0) x
−5 −4 −3 −2 −1
2
3
2
(− 4, 1)
−3
1
x
−5 −4 −3 −2 −1
1
2
(− 4, − 4)
3
(e) gx 2f x
1 (f) gx 2 f x
y
(2, 2) 2
g2 2 f 2 2 −5 −4 −3 −2 −1
−3
y 2
g2 12 f 2 12
1
(2, 12 )
1
x 1
2
3
g4 12 f 4 32
−5 −4 −3
(− 4, − 32 )
−4
(− 4, − 6 )
−5 −6
−2
g4 2f 4 6
−4
−5 −6
x −1 −2 −3 −4 −5 −6
50. (a) hx sinx 2 1 is a horizontal shift 2 units to the left, followed by a vertical shift 1 unit upwards. (b) hx sinx 1 is a horizontal shift 1 unit to the right followed by a reflection about the x-axis.
1
2
3
Section P.3 52. (a) f g1 f 0 0
Functions and Their Graphs
54. f x x2 1, gx cos x
(b) g f 1 g1 0
f gx f gx f cos x cos2 x 1
(c) g f 0 g0 1
Domain: ,
(d) f g4 f 15 15
g f x gx 2 1 cosx 2 1
(e) f gx f x2 1 x2 1
Domain: ,
(f) g f x g x x 1 x 1 x ≥ 0
No, f g g f.
2
56. f gx f x 2
1 x 2
Domain: 2,
g f x g
1x 1x 2 1 x 2x
You can find the domain of g f by determining the intervals where 1 2x and x are both positive, or both negative. + −2
+
+ − − +
−1 − 1 2
0
+
+ 1
+
x 2
Domain: , 12 , 0,
58. (a)
3 x 3 x f x 60. f x
25
Odd
100
0 0
(b) H1.6x 0.0021.6x2 0.0051.6x 0.029 0.00512x 2 0.008x 0.029 62. f x sin2x sinx sinx sin xsin x sin2x Even 64. (a) If f is even, then 4, 9 is on the graph. 66. f x a2nx2n a2n2x2n2 . . . a2x2 a0 a2nx2n a2n2x2n2 . . . a2x2 a0 f x Even 68. Let F x f xgx where f is even and g is odd. Then F x f xgx f x gx f xgx F x. Thus, F x is odd.
(b) If f is odd, then 4, 9 is on the graph.
295
296
Chapter P
Preparation for Calculus
70. (a) Let F x f x ± gx where f and g are even. Then, F x f x ± gx f x ± gx F x. Thus, F x is even. (b) Let F x f x ± gx where f and g are odd. Then, F x f x ± gx f x gx F x. Thus, F x is odd. (c) Let F x f x ± gx where f is odd and g is even. Then, F x f x ± gx f x ± gx. Thus, F x is neither odd nor even.
72. By equating slopes,
02 y2 03 x3 y2 y
L x2 y2
74. True
6 x3 6 2x 2 x3 x 3,
x x 2x 3 . 2
2
76. False; let f x x2. Then f 3x 3x2 9x2 and 3f x 3x2. Thus, 3f x f 3x.
Section P.4
Fitting Models to Data
2. Trigonometric function
4. No relationship
6. (a)
8. (a) s 9.7t 0.4
20
(b)
0
5
20 0 −1
No, the relationship does not appear to be linear. (b) Quiz scores are dependent on several variables such as study time, class attendance, etc. These variables may change from one quiz to the next. 10. (a) Linear model: H 0.3323t 612.9333 (b)
The model fits well. (c) If t 2.5, s 24.65 meterssecond. 12. (a) S 180.89x 2 205.79x 272 (b)
600
0
1300 0
45 −1
25000
14
0 0
The fit is very good. (c) When t 500, H 0.3323500 612.9333 446.78.
(c) When x 2, S 583.98 pounds.
Review Exercises for Chapter P 14. (a) t 0.00271s2 0.0529s 2.671 (b)
16. (a) T 2.9856 104 p3 0.0641 p2 5.2826p 143.1 (b)
21
100
20
350
0 150
0
110
(c) The curve levels off for s < 20.
(c) For T 300F, p 68.29 pounds per square inch.
(d) t 0.002s2 0.0346s 0.183
(d) The model is based on data up to 100 pounds per square inch.
21
100
0 0
The model is better for low speeds.
18. (a) Ht 84.4 4.28 sin
6t 3.86
(c)
100
One model is Ct 58 27 sin (b)
6t 4.1.
13
0 0
100
(d) The average in Honolulu is 84.4. The average in Chicago is 58. 0
(e) The period is 12 months (1 year).
13 0
(f) Chicago has greater variability 27 > 4.28.
20. Answers will vary.
Review Exercises for Chapter P 2. y x 1x 3 x 0 ⇒ y 0 10 3 3 ⇒ 0, 3
297
y-intercept
y 0 ⇒ 0 x 1x 3 ⇒ x 1, 3 ⇒ 1, 0, 3, 0 4. xy 4 x 0 and y 0 are both impossible. No intercepts.
x-intercepts 6. Symmetric with respect to y-axis since y x4 x2 3 y x4 x2 3.
298
Chapter P
Preparation for Calculus
8. 4x 2y 6 y 2x 3
2x 15y 25
y
2 y 15 x 53
Slope: 2
10 8
2 Slope: 15
y-intercept: 3
y-intercept:
y
6
5 3
4 2
1 −2
12. y x6 x
10. 0.02x 0.15y 0.25
y x
−1
2
−2
3
x 2
4
8
3
−1 −2
1
−3
x
−4
4
8
12
−1 −2
14. y x 4 4
3 x 6 16. y 8
18.
x 1 x2 7
y
Xmin = -40 Xmax = 40 Xscl = 10 Ymin = -40 Ymax = 40 Yscl = 10
x
−1 −1
2
3
4
5
6
−2 −3 −4
yx1 0 x2 x 6 No real solution No points of intersection The graphs of y x 1 and y x2 7 do not intersect.
−5 −6
20. y kx3 (a) 4 k13 ⇒ k 4 and y 4x3
(b) 1 k23 ⇒ k 18 and y 18 x3
(c) 0 k03 ⇒ any k will do!
(d) 1 k13 ⇒ k 1 ⇒ y x3
22.
24.
y 14
3 4 3 t 11
(7, 12)
12 10 8 6
44 9 3t
4 2 −2
36 3 1 3 t 3 8
1
2
3
4
5
6
(7, −1)
x
53 3t t
The line is vertical and has no slope.
53 3
Review Exercises for Chapter P 26. y 6 0x 2
28. m is undefined. Line is vertical. x5
y 6 Horizontal line
y
y 6
8
(−2, 6)
4
2 −4
−2
(5, 4)
2
4
−6
299
−4 x 2
−2
4
6
−2
x 2
4
6
8
−2 −4 −6
−4
2 y 3 x 1 3
30. (a)
32. (a) C 9.25t 13.50t 36,500 22.75t 36,500
3y 9 2x 2
(b) R 30t
2x 3y 11 0 (b) Slope of perpendicular line is 1.
(c)
30t 22.75t 36,500 7.25t 36,000
y 3 1x 1
t 5034.48 hours to break even.
yx2 0xy2 m
(c)
43 1 21
y 3 1x 1 yx2 0xy2 y3
(d)
y30 34. x2 y 0
36. x 9 y2
Function of x since there is one value for y for each x. y
Not a function of x since there are two values of y for some x. y
6 4
5 4
2
3
1
2
−12 −9 −6 −3 −1
1 −3
−2
−1
x 3
6
12
−2
x 1
2
3 −4
38. (a) f 4 42 2 18
f 1 1 2 1
(because 4 < 0)
40. f x 1 x2 and gx 2x 1
(b) f 0 0 2 2
(a) f x gx 1 x2 2x 1 x2 2x
(c)
(b) f xgx 1 x22x 1 2x3 x2 2x 1 (c) g f x g1 x2 21 x2 1 3 2x2
Review Exercises for Chapter P 15. (a) y 1.81x3 14.58x2 16.39x 10 (b)
17. (a) Yes, y is a function of t. At each time t, there is one and only one displacement y.
300
(b) The amplitude is approximately
2.35 1.652 0.35. 0
The period is approximately
7 0
20.375 0.125 0.5.
(c) If x 4.5, y 214 horsepower.
(c) One model is y 0.35 sin4 t 2. (d)
4
0.9
0 0
19. Answers will vary.
Review Exercises for Chapter P 1. y 2x 3 x 0 ⇒ y 20 3 3 ⇒ 0, 3 3 3 y 0 ⇒ 0 2x 3 ⇒ x 2 ⇒ 2 , 0
3. y
y-intercept x-intercept
x1 x2
5. Symmetric with respect to y-axis since
01 1 1 ⇒ 0, x0⇒y 02 2 2
y0⇒0
x2y x2 4y 0 y-intercept
x1 ⇒ x 1 ⇒ 1, 0 x2
7. y 12 x 32
x2y x2 4y 0.
x-intercept
11. y 7 6x x2
1 5 9. 3 x 6 y 1
25 x y 65
y
y
y 25 x 65
3 2
2 5
Slope:
1
y-intercept:
6 5
5
x 1
2
3
y x
1
10 3
2
2
x 3
2
1
1 1
5
5
19
20
Chapter P
Preparation for Calculus 17. 3x 4y 8
15. y 4x2 25
13. y 5 x
4x 4y 20
Domain: , 5
Xmin = -5 Xmax = 5 Xscl = 1 Ymin = -30 Ymax = 10 Yscl = 5
y 5 4 3
28
7x
x 4 y 1 Point: 4, 1
2 1
x 1
2
3
4
5
19. You need factors x 2 and x 2. Multiply by x to obtain origin symmetry y xx 2x 2. x3 4x.
21.
23.
y
15 1t 1 0 1 2
5
1t
4 5 2
( 5, )
3 2
t
1
( 32 , 1)
4 3
7 3
x 1
2
Slope
3
4
5
52 1 32 3 5 32 72 7
y 5 32x 0
25.
y 0 23x 3
27.
y 32x 5 2y 3x 10 0
y 23x 2 3y 2x 6 0 y
y 4
4
2
2 −4
−2
(−3, 0) x
−2 −4
−8
2
(0, −5)
4
6
8
−8
−6
−4
x −2 −4 −6 −8
2
4
Review Exercises for Chapter P
y4
29. (a)
7 x 2 16
5 (b) Slope of line is . 3
16y 64 7x 14
5 y 4 x 2 3
0 7x 16y 78
3y 12 5x 10
40 2 2 0
m
(c)
y 2x
0 5x 3y 22 x 2
(d)
2x y 0
21
x20
31. The slope is 850. V 850t 12,500. V3 8503 12,500 $9950 33. x y2 0
35. y x2 2x
y ± x
Function of x since there is one value of y for each x.
Not a function of x since there are two values of y for some x.
y 4 3
y 3 2
x −1
x
−2 −1
1 1
2
3
4
5
3
4
−2
6
−2 −3
37. f x
1 x
39. (a) Domain: 36 x2 ≥ 0 ⇒ 6 ≤ x ≤ 6 Range: 0, 6
(a) f 0 does not exist. 1 1 1 x 1 1 1 x f 1 x f 1 (b) x x 1 xx 1 , x 1, 0 1 x 41. (a) f x x3 c, c 2, 0, 2 y
c
Range: all y 0
or , 0, 0,
,
(c) Domain: all x or Range: all y or
,
(b) f x x c3, c 2, 0, 2 y
c
c
or , 5, 5,
(b) Domain: all x 5
0
3
1
c
0
2
1
2 x
3
2
2
3
x 2
2 2
c
2
—CONTINUED—
or 6, 6
3
c
2
3
22
Chapter P
Preparation for Calculus
41. —CONTINUED— (c) f x x 23 c, c 2, 0, 2
(d) f x cx3, c 2, 0, 2 y
y
2
c
3
2
c
2
2
c
1
0
1
c x 2
3
4
2
1
1
0 2
x 3
1 2
c
c
2
43. (a) Odd powers: f x x, gx x3, hx x5
Even powers: f x x2, gx x4, hx x6
g
2
2
g
4
h
h
−3
f
3
f −3 −2
3 0
The graphs of f, g, and h all rise to the left and to the right. As the degree increases, the graph rises more steeply. All three graphs pass through the points 0, 0, 1, 1, and 1, 1.
The graphs of f, g, and h all rise to the right and fall to the left. As the degree increases, the graph rises and falls more steeply. All three graphs pass through the points 0, 0, 1, 1, and 1, 1. (b) y x7 will look like hx x5, but rise and fall even more steeply. y x8 will look like hx x6, but rise even more steeply. 45. (a)
(b) Domain: 0 < x < 12
y
x
40
x
y
2x 2y 24
0
12 0
y 12 x A xy x12 x 12x x2
47. (a) 3 (cubic), negative leading coefficient (b) 4 (quartic), positive leading coefficient (c) 2 (quadratic), negative leading coefficient
(c) Maximum area is A 36. In general, the maximum area is attained when the rectangle is a square. In this case, x 6. 49. (a) Yes, y is a function of t. At each time t, there is one and only one displacement y. (b) The amplitude is approximately
0.25 0.252 0.25.
(d) 5, positive leading coefficient
The period is approximately 1.1. (c) One model is y (d)
1 2 1 cos t cos5.7t 4 1.1 4
0.5
0
−0.5
2.2
Problem Solving for Chapter P
23
Problem Solving for Chapter P x2 6x y2 8y 0
1. (a)
4 (b) Slope of line from 0, 0 to 3, 4 is Slope of tangent line 3 3 is Hence, 4
x2 6x 9 y2 8y 16 9 16 x 32 y 42 25 Center: 3, 4
3 3 y 0 x 0 ⇒ y x 4 4
Radius: 5
(c) Slope of line from 6, 0 to 3, 4 is
40 4 . 36 3
3 Slope of tangent line is . Hence, 4
3 3 9 (d) x x 4 4 2 3 9 x 2 2
3 9 3 y 0 x 6 ⇒ y x 4 4 2
Tangent line
x3 Intersection:
3. Hx
10
x ≥ 0 x < 0
Tangent line
3, 49
y 4 3 2 1 x
−4 −3 −2 −1 −1
1
2
3
4
−2 −3 −4
(a) Hx 2
(b) Hx 2
y
y
4
4
3
3
2
2
1
1 x
−4 −3 −2 −1 −1
1
2
3
x
−4 −3 −2 −1 −1
4
1
2
3
4
1
2
3
4
1
2
3
4
−2 −3
−3
−4
−4
(c) Hx
(d) Hx
y
y
4
4
3
3
2
2
1 x
−4 −3 −2 −1 −1
1
2
3
−2
−2
−3
−3
−4
−4
1 (e) 2Hx
(f ) Hx 2 2
y
x
−4 −3 −2 −1 −1
4
y
4
4
3
3
2 1 −4 −3 −2 −1 −1
1 x 1
2
3
4
−4 −3 −2 −1 −1
−2
−2
−3
−3
−4
−4
x
24
Chapter P
Preparation for Calculus
5. (a) x 2y 100 ⇒ y Ax xy x
100 x 2
7. The length of the trip in the water is 22 x2, and the length of the trip over land is 1 3 x2. Hence, the total time is
1002 x x2 50x 2
T
Domain: 0 < x < 100 (b)
4 x2
2
1 3 x2
1600
0
110 0
Maximum of 1250 m 2 at x 50 m, y 25 m. 1 (c) Ax x2 100x 2 1 x2 100x 2500 1250 2 1 x 502 1250 2 A50 1250 m 2 is the maximum. x 50 m, y 25 m. 9. (a) Slope
94 5. Slope of tangent line is less than 5. 32
(b) Slope
41 3. Slope of tangent line is greater than 3. 21
(c) Slope
4.41 4 4.1. Slope of tangent line is less than 4.1. 2.1 2
(d) Slope
f 2 h f 2 2 h 2
2 h2 4 h
4h h2 h
4 h, h 0 (e) Letting h get closer and closer to 0, the slope approaches 4. Hence, the slope at 2, 4 is 4. 11. (a) At x 1 and x 3 the sounds are equal. (b)
I x2 y2
3
2I x 32 y2
−6
3
x 32 y2 4x2 y2 3x2 3y2 6x 9 x2 2x y2 3
x 12 y2 4 Circle of radius 2 centered at 1, 0
−3
4
hours.
Problem Solving for Chapter P d1d2 1
13.
y
x 12 y2 x 12 y2 1 x 12x 12 y2x 12 x 12 y4 1 x2 12 y22x2 2 y4 1 x4 2x2 1 2x2y2 2y2 y4 1
x4 2x2y2 y4 2x2 2y2 0 x2 y22 2x2 y2 Let y 0. Then x4 2x2 ⇒ x 0
or
x2 2.
Thus, 0, 0, 2, 0 and 2, 0 are on the curve.
2
(− 2 , 0)
1
( 2 , 0) x
−2
2 −1 −2
(0, 0)
25
Review Exercises for Chapter P 14. (a) t 0.00271s2 0.0529s 2.671 (b)
16. (a) T 2.9856 104 p3 0.0641 p2 5.2826p 143.1 (b)
21
100
20
350
0 150
0
110
(c) The curve levels off for s < 20.
(c) For T 300F, p 68.29 pounds per square inch.
(d) t 0.002s2 0.0346s 0.183
(d) The model is based on data up to 100 pounds per square inch.
21
100
0 0
The model is better for low speeds.
18. (a) Ht 84.4 4.28 sin
6t 3.86
(c)
100
One model is Ct 58 27 sin (b)
6t 4.1.
13
0 0
100
(d) The average in Honolulu is 84.4. The average in Chicago is 58. 0
(e) The period is 12 months (1 year).
13 0
(f) Chicago has greater variability 27 > 4.28.
20. Answers will vary.
Review Exercises for Chapter P 2. y x 1x 3 x 0 ⇒ y 0 10 3 3 ⇒ 0, 3
297
y-intercept
y 0 ⇒ 0 x 1x 3 ⇒ x 1, 3 ⇒ 1, 0, 3, 0 4. xy 4 x 0 and y 0 are both impossible. No intercepts.
x-intercepts 6. Symmetric with respect to y-axis since y x4 x2 3 y x4 x2 3.
298
Chapter P
Preparation for Calculus
8. 4x 2y 6 y 2x 3
2x 15y 25
y
2 y 15 x 53
Slope: 2
10 8
2 Slope: 15
y-intercept: 3
y-intercept:
y
6
5 3
4 2
1 −2
12. y x6 x
10. 0.02x 0.15y 0.25
y x
−1
2
−2
3
x 2
4
8
3
−1 −2
1
−3
x
−4
4
8
12
−1 −2
14. y x 4 4
3 x 6 16. y 8
18.
x 1 x2 7
y
Xmin = -40 Xmax = 40 Xscl = 10 Ymin = -40 Ymax = 40 Yscl = 10
x
−1 −1
2
3
4
5
6
−2 −3 −4
yx1 0 x2 x 6 No real solution No points of intersection The graphs of y x 1 and y x2 7 do not intersect.
−5 −6
20. y kx3 (a) 4 k13 ⇒ k 4 and y 4x3
(b) 1 k23 ⇒ k 18 and y 18 x3
(c) 0 k03 ⇒ any k will do!
(d) 1 k13 ⇒ k 1 ⇒ y x3
22.
24.
y 14
3 4 3 t 11
(7, 12)
12 10 8 6
44 9 3t
4 2 −2
36 3 1 3 t 3 8
1
2
3
4
5
6
(7, −1)
x
53 3t t
The line is vertical and has no slope.
53 3
Review Exercises for Chapter P 26. y 6 0x 2
28. m is undefined. Line is vertical. x5
y 6 Horizontal line
y
y 6
8
(−2, 6)
4
2 −4
−2
(5, 4)
2
4
−6
299
−4 x 2
−2
4
6
−2
x 2
4
6
8
−2 −4 −6
−4
2 y 3 x 1 3
30. (a)
32. (a) C 9.25t 13.50t 36,500 22.75t 36,500
3y 9 2x 2
(b) R 30t
2x 3y 11 0 (b) Slope of perpendicular line is 1.
(c)
30t 22.75t 36,500 7.25t 36,000
y 3 1x 1
t 5034.48 hours to break even.
yx2 0xy2 m
(c)
43 1 21
y 3 1x 1 yx2 0xy2 y3
(d)
y30 34. x2 y 0
36. x 9 y2
Function of x since there is one value for y for each x. y
Not a function of x since there are two values of y for some x. y
6 4
5 4
2
3
1
2
−12 −9 −6 −3 −1
1 −3
−2
−1
x 3
6
12
−2
x 1
2
3 −4
38. (a) f 4 42 2 18
f 1 1 2 1
(because 4 < 0)
40. f x 1 x2 and gx 2x 1
(b) f 0 0 2 2
(a) f x gx 1 x2 2x 1 x2 2x
(c)
(b) f xgx 1 x22x 1 2x3 x2 2x 1 (c) g f x g1 x2 21 x2 1 3 2x2
300
Chapter P
Preparation for Calculus
42. f x x3 3x2
44. (a) f x x2x 62
6
−6
100
(0, 0) 6 −4
(2, − 4)
10
−6
−25
(a) The graph of g is obtained from f by a vertical shift down 1 unit, followed by a reflection in the x-axis:
(b) gx x3x 62 300
gx f x 1 x3 3x2 1 −2
(b) The graph of g is obtained from f by a vertical shift upwards of 1 and a horizontal shift of 2 to the right. gx f x 2 1
10 −100
(c) hx x3x 63
x 2 3x 2 1 3
2
200 −4
10
− 800
46. For company (a) the profit rose rapidly for the first year, and then leveled off. For the second company (b), the profit dropped, and then rose again later.
48. (a) y 1.204x 64.2667 (b)
70
0
33 0
(c) The data point 27, 44 is probably an error. Without this point, the new model is y 1.4344x 66.4387.
Problem Solving for Chapter P 2. Let y mx 1 be a tangent line to the circle from the point 0, 1. Then x2 y 12 1 x2 mx 1 12 1
m2 1x2 4mx 3 0 Setting the discriminant b2 4ac equal to zero, 16m2 4m2 13 0 16m2 12m2 12 4m2 12 m ± 3 Tangent lines: y 3x 1 and y 3x 1.
Problem Solving for Chapter P 4. (a) f x 1
(b) f x 1
y
y
4
−3
4
x
−1
1
x
−4
3
4
−2
−2
−4
−4
(c) 2f x
(d) f x
y
y
4
4 2
−4
x
−2
2
−4
4
−2
−2
−4
−4
(e) f x
(f) f x
y 4
4
2
4
y
2 x
−2
2
−4
4
x
−2
−2
−2
−4
−4
(g) f x
2
4
2
−4
x
−2
y 4 2
−4
x
−2
2
4
−2 −4
6. (a) 4y 3x 300 ⇒ y
300 3x 4
3x2 300x 300 3x Ax x2y x 2 2
y
(b) 4000 3500 3000 2500 2000
Domain: 0 < x < 100
1500 1000 500 x 25
50
75
100
Maximum of 3750 ft 2 at x 50 ft, y 37.5 ft. 3 (c) Ax x2 100x 2 3 x2 100x 2500 3750 2 3 x 502 3750 2 A50 3750 square feet is the maximum area, where x 50 ft and y 37.5 ft.
301
302
Chapter P
Preparation for Calculus
8. Let d be the distance from the starting point to the beach. Average velocity
distance time
2d d d 120 60
2 1 1 120 60
80 km hr y
10. 4 3 2
(4, 2)
1 x 1
2
3
4
5
−1
(a) Slope
1 32 1 . Slope of tangent line is greater than . 94 5 5
(b) Slope
21 1 1 . Slope of tangent line is less than . 41 3 3
(c) Slope
10 2.1 2 10 . Slope of tangent line is greater than . 41 4.1 4 41
(d) Slope
f 4 h f 4 4 h 4
(e)
4 h 2
4 h 2
h
h
4 h 2
h
4 h 2 4 h 2
4 h 4 h4 h 2 1 4 h 2
,h0
1 1 As h gets closer to 0, the slope gets closer to . The slope is at the point 4, 2. 4 4
Problem Solving for Chapter P I
12. (a)
kI
x2 y2
x 42 y2
x 42 y2 k2x2 y2 k2 1x2 k2 1y2 8x 16 If k 1, then x 2 is a vertical line. So, assume k2 1 0. Then x2 y2
8x 16 k2 1 k2 1
x k
4 1
2
x k
4 1
2
2
2
y2
16 16 k2 1 k2 12
y2
k 4k 1 , Circle
1 2
(b) If k 3, x
2
2
2
y2
32
2
(c) For large k, the center of the circle is near 0, 0, and the radius becomes smaller. y 2 1 x −2
(− 12 , 0)
1
2
−2
14. f x y
1 1x
(a) Domain: all x 1 Range: all y 0 (b) f f x f
1 1 x
1
1 1 1x
1 1x x1 1x1 x x 1x
Domain: all x 0, 1 (c) f f f x f
x x 1
1 1 x x1 1 1 x x
Domain: all x 0, 1 (d) The graph is not a line. It has holes at 0, 0 and 1, 1. y 2 1 x −2
1
−2
2
303
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