Selected Answers
A-15
CHAPTER 3
27. a. f ′1-32 = 8
Section 3.1 Exercises, pp. 134–136
1 b. y = -14x - 16 31. a. f ′a b = - 4 b. y = - 4x + 3 4 1 1 5 1 x 3 + b. y = x + 35. a. b. y = 33. a. 3 100 100 20 3 3
1. Given the point 1a, f 1a22 and any point 1x, f 1x22 near 1a, f 1a22, f 1x2 - f 1a2 the slope of the secant line joining these points is . The x - a limit of this quotient as x approaches a is the slope of the tangent line at the point. 3. The average rate of change over the interval 3a, x4 is f 1x2 - f 1a2 f 1x2 - f 1a2 . The limit lim is the slope of the tangent xSa x - a x - a line; it is also the limit of average rates of change, which is the instantaneous rate of change at x = a. 5. f ′1a2 is the slope of the tangent dy is line at 1a, f 1a22 or the instantaneous rate of change of f at a. 7. dx ∆y the limit of and is the rate of change of y with respect to x. ∆x y 9. a. 6 b. y = 6x - 14 c.
b. y = 8x
37. a. f ′1x2 = 6x + 2 y c.
b. y = 8x - 13
5
x
2
39. a. f ′1x2 = 10x - 6 y c.
b. y = 14x - 19
20
5
x
1
x
2
11. a. - 5
b. y = - 5x + 1
c.
y
1
x
1
13. a. - 1
b. y = -x - 2
c.
29. a. f ′1- 22 = -14
y
1 1 41. a. 2ax + b b. 8x - 3 c. 5 43. 45. 4 5 47. a. True b. False c. True 3 3x 13 49. a. f ′1x2 = b. y = + 10 5 213x + 1 -6 3x 5 51. a. f ′1x2 = b. y = 53. a. C, D 2 2 2 13x + 12 b. A, B, E c. A, B, E, D, C 55. a. Approximately 10 kW; approximately -5 kW b. t = 6 and t = 18 c. t = 12 1 1 ; a = 2; 59. f 1x2 = x4; a = 2; 32 57. f 1x2 = x + 1 9 61. No; f is not continuous at x = 2. 63. a = 4
Section 3.2 Exercises, pp. 141–144
1
1
x
1. The slope of a curve at a point is independent of the function value at that point. 3. Yes y 5. 7. a–C; b–C; c–A; d–B 2
15. a. 8.667, 13.333 11 b. 15, 15, 15 c. Millions of daily users per month d. Decreasing 17. a. 2 b. y = 2x + 1 19. a. 4 2 b. y = 4x - 8 21. a. 3 b. y = 3x - 2 23. a. 25 2 7 1 1 7 b. y = x + 25. a. b. y = x + 25 25 4 4 4
4
x
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Z01_BRIG5715_01_SE_ANS_Ch03.indd 15
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A-16
Selected Answers
9. a–D; b–C; c–B; d–A
11.
b.
y
Vertical tangent line x = - 1
y
2
y 5 f (x) y 5 f 9(x) x
21
x
c.
Vertical tangent line x = 4
y
y
13.
2
y 5 f 9(x) 1
22
1
1
21
x
2
x
4
21
d.
Vertical tangent line x = 0
y
22
3
15. a. x = 1
b. x = 1, x = 2
c.
y
3
2
4
x
x
1 -2>3 x and lim " f ′1x2 " = lim " f ′1x2 " = ∞ x S 0x S 0+ 3 31. b. 1 c. 1 d. f is not differentiable at 0 because it is not continuous at 0. 29. f ′1x2 =
17. a. True
b. True
19. Yes.
c. False 1 2 21. y = - x 3 3 x 3 23. y = + 2 2
y 3 2 1
22
1
21
x
2
21
25. b. f+′ 122 = 1, f-′122 = - 1 differentiable at x = 2. 27. a. y
c. f is continuous but not Vertical tangent line x = 2
Section 3.3 Exercises, pp. 151–154 1. Using the definition can be tedious. 3. f 1x2 = ex 5. Compute the product of the constant and the derivative of the function. 7. 5x4 9. 0 11. 1 13. 15x2 15. 8 17. 200t 19. 12x3 + 7 21. 40x3 - 32 23. 6w2 + 6w + 10 25. 18x2 + 6x + 4 27. 4x3 + 4x 29. 2w, for w ≠ 0 31. 1, for x ≠ 1 1 33. , for x ≠ a 35. a. y = -6x + 5 21x y b. 3
2
x
1
4
x
Copyright © 2014 Pearson Education, Inc.
Z01_BRIG5715_01_SE_ANS_Ch03.indd 16
30/08/13 7:24 PM
Selected Answers 37. a. y = 3x + 3 - 3 ln 3
b.
y
29. a. 2w, for w ≠ 0 b. y
33. a. y = -
31. 1
A-17
3x 17 + 2 2
2
1
2
x
39. a. x = 3 b. x = 4 41. a. 1- 1, 112, 12, -162 b. 1-3, -412, 14, 362 43. a. 14, 42 b. 116, 02 45. f ′1x2 = 20x3 + 30x2 + 3; f ″1x2 = 60x2 + 60x; f 1321x2 = 120x + 60 47. f ′1x2 = 1; f ″1x2 = f 1321x2 = 0 for x ≠ - 1 49. a. False b. True c. False d. False e. False 51. a. y = 7x - 1 b. y = - 2x + 5 c. y = 16x + 4 53. -10 1 55. 4 57. 7.5 59. a. f 1x2 = 1x; a = 9 b. f ′192 = 6 61. a. f 1x2 = x100; a = 1 b. f ′112 = 100 63. 3 65. 1 67. f 1x2 = ex; a = 0; f ′102 = 1 69. a. d′1t2 = 32 t; ft>s; the velocity of the stone b. 576 ft; approx. 131 mi>hr dD 71. a. = 0.10g + 35; mi>gal; the rate of change of the dg range of the car with respect to the capacity of the tank. b. 35 mi>gal 35.5 mi>gal 36 mi>gal; the gas mileage improves when driving longer distances. c. Approx. 427 mi y 79. b. c. Changing 0.01 to 0.001 1.0 does not change the graph. We see the graph 0.5 of the derivative of f.
x
1
35. a. y = 3x + 1
1
21
2
y 5 4 3 2 1
1
21
x
21
37. -27x -10 45. 45e3x
39. 6t -
42 t8
3 2 - 3 43. e7x17x + 12 t2 t 2 49. e x - e -x 3
41. -
47. e - 3x11 - 3x2
20 2 b b. p′152 ≈ 8.163 c. t = 0 t + 2 d. lim p′1t2 = 0; the population approaches a steady state. S 51. a. p′1t2 = a t
e.
∞ y
y ! p(t)
200 22
b.
x 150
20.5 100 21.0 50
81. b.
c. Changing 0.01 to 0.001 does not change the graph visibly. We see the graph of the derivative of f.
y
1.5
1
y 5 D(x)
1.5
0
0.5
1
1.5
2
2.5
3
3.5
x
Section 3.4 Exercises, pp. 161–163 d n d 1x 2 = nx n - 1, 3 f 1x2 # g1x24 = f ′1x2 g1x2 + f 1x2 g′1x2 3. dx dx for any integer n 5. y′ = ke kx, for any real number k 7. 36x5 - 12x3 9. ett 41t + 52 11. 4x3 13. ew1w3 + 3w2 - 12 15. a. 6x + 1 1 ex 17. a. 18y5 - 52y3 + 8y 19. 21. 1x + 122 1ex + 122 x 2 1x 2x - 12 e 1 23. e - x11 - x2 25. 27. 2 2 1x - 122 1t - 12 1.
y ! p"(t) 0
10
20
30
t
-0.0693t
53. a. Q′1t2 = - 1.386e b. -1.386 mg>hr; - 1.207 mg>hr c. lim Q1t2 = 0—eventually none of the drug remains in the S t
∞
Q′1t2 = 0—the rate at which the body excretes the bloodstream; lim tS ∞ 1 b. The line tangent to drug goes to zero over time. 55. a. x = 2 x 2 e 1x - x - 52 1 the graph of f at x = - is horizontal. 57. 2 1x - 222 x 2 e 1x + x + 12 59. 61. a. False b. False c. False d. True 1x + 122 63. f ′1x2 = x e3x 13x + 22 f ″1x2 = e3x 19x2 + 12x + 22 f ‴1x2 = 9e 3x 13x 2 + 6x + 22 x2 + 2x - 7 65. f ′1x2 = 1x + 122 16 f ″1x2 = 1x + 123
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A-18
Selected Answers
r - 61r - 1 2 69. 2 21r 1r + 122 15x + 12 71. 300x9 + 135x8 + 105x6 + 120x3 + 45x2 + 15 108 567 y 73. a. y = x + b. 169 169
67. x =
67. 8x -
69. a.
7p 11p + 2kp and x = + 2kp, where k is any integer 6 6 y b. v1t2 = 30 cos t 2
t
3
230
1
1
c.
x
1.8 * 1010 Qq 3 1 7 77. 79. 81. a. F ′1x2 = N>m 8 2 9 x3 b. - 1.8 * 1019 N>m c. " F ′1x2 " decreases as x increases. 83. One possible pair: f 1x2 = eax and g1x2 = ebx, a where b = , a ≠ 1. 87. f ″g + 2f ′g′ + fg″ a - 1 91. a. f ′gh + fg′h + fgh′ b. 2e2x1x2 + 3x - 22
30
75. -
2
Section 3.5 Exercises, pp. 170–172
sin x is undefined at x = 0. 3. The tangent and cotangent functions x are defined as ratios of the sine and cosine functions. 5. -1 7 1 7. 3 9. 11. 5 13. 7 15. 17. cos x - sin x 3 4 1 19. e-x1cos x - sin x2 21. sin x + x cos x 23. 1 + sin x 25. cos2 x - sin2 x = cos 2x 27. - 2 sin x cos x = -sin 2x csc x 33. sec x tan x - csc x cot x 35. e5x csc x15 - cot x2 37. 1 + csc x 39. cos2 z - sin2 z = cos 2z 41. 2 cos x - x sin x 43. 2ex cos x 45. 2 csc2 x cot x 47. 2 1sec2 x tan x + csc2 x cot x2 49. a. False a 3 53. 55. 0 b. False c. True d. True 51. 4 b 1 1 2 sin x 61. 57. x cos 2x + sin 2x 59. 2 2 sin x cos x - 1 11 + cos x22 p13 y 63. a. y = 13x + 2 b. 6 3
x
213p + 1 3
b.
y
t
3
p t = 12k + 12 , where k is any nonnegative 2 integer. The position is y = 0 if k is even or y = - 60 if k is odd. e. v1t2 is at a maximum at t = 2kp, where k is a nonnegative integer; the position is y = -30 y f. a1t2 = -30 sin t d. v1t2 = 0
1.
65. a. y = - 213x +
y
for
30
2
3
t
230
77. a = 0 79. a. 2 sin x cos x b. 3 sin2 x cos x c. 4 sin3 x cos x d. n sinn - 1 x cos x The conjecture is true for n = 1. If it holds for d n = k, then when n = k + 1, we have 1sink + 1 x2 = dx d 1sink x # sin x2 = sink x cos x + sin x # k sink - 1 x cos x = dx p 13 1k + 12 sink x cos x. 81. a. f 1x2 = sin x; a = b. 6 2 p 83. a. f 1x2 = cot x; a = b. -2 4 y 85. b. c. Changing 0.01 to 1.0 0.001 does not change y 5 D(x) the graph visibly. We 0.5 see the graph of the derivative of f. 22
1
21
2
x
20.5
21.0
1
x
Copyright © 2014 Pearson Education, Inc.
Z01_BRIG5715_01_SE_ANS_Ch03.indd 18
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Selected Answers y
87. b.
c. Changing 0.01 to 0.001 does not change the graph visibly. We see the graph of the derivative of f.
3
y 5 D(x) 2
1
b. v1t2 = 4t - 9; stationary at
A-19
y
9 9 t = , to the right on a , 3 d , to the 4 4 9 c 0, b left on 4
1
3
t
24
28
22
1
21
2
x
Section 3.6 Exercises, pp. 177–182 1. The average rate of change is
9 d. a a b = 4 ft>s2 4
c. v112 = - 5 ft>s; a112 = 4 ft>s2 13. a.
f 1x + ∆x2 - f 1x2
, whereas the ∆x instantaneous rate of change is the limit as ∆x goes to zero in this quotient. 3. Small 5. If the position of the object at time t is s1t2, d 2s then the acceleration at time t is a1t2 = 2 . 7. a. 40 mi>hr dt b. 40 mi>hr; yes c. - 60 mi>hr; - 60 mi>hr; south d. The police car drives away from the police station going north until about 10:08, when it turns around and heads south, toward the police station. It continues south until it passes the police station at about 11:02 and keeps going south until about 11:40, when it turns around and heads north until 12:00. y 9. a.
y
9 e. a , 3 d 4
40
0
1
2
3
4
5
t
6
b. v1t2 = 6t 2 - 42t + 60; stationary at t = 2 and t = 5, to the right on 30, 22 and 15, 64, to the left on 12, 52 y
60
2
2
30
t
4
1
b. v1t2 = 2t - 4; stationary at t = 2, to the right on 12, 54 , to the left on 30, 22
2
3
4
5
6
t
y
2
2
c. v112 = - 2 ft>s; a112 = 2 ft>s2 11. a. y
d. a122 = 2 ft>s2
12
4
e. 12, 5 4
t
c. v112 = 24 ft>s; a112 = - 30 ft>s2 d. a122 = - 18 ft>s; 7 a152 = 18 ft>s2 e. a2, b, 15, 64 15. a. v1t2 = -32t + 64 2 b. At t = 2 c. 96 ft d. At t = 2 + 16 e. - 3216 ft>s f. 1 2, 2 + 26 2 17. a. 98,300 people>yr b. 99,920 people>yr in 1997; 95,600 people>yr in 2005 c. p′1t2 = - 0.54t + 101; population increased, growth rate is positive but decreasing. 19. a. False b. True c. False d. True 21. 240 ft 23. 64 ft>s 25. a. t = 1, 2, 3 b. It is moving in the positive direction for t in 10, 12 and 12, 32; it is moving in the negative direction for t in 11, 22 and t 7 3. c. v
0
1
2
3
t
2 1
2
3
t
d. 10, 0.52, 11, 1.52, 12, 2.52, and 13, ∞ 2 Copyright © 2014 Pearson Education, Inc.
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A-20
Selected Answers
27. a. 1930, 1.1 million people>yr b. 1960, 2.9 million people>yr c. The population did not decrease. d. 11905, 19152, 11930, 19602, 11980, 19902 100 y 29. a. b. v = 1t + 122 100
33. a.
y
10
t
3 210
50
dx 3p + kp, where k is any = 10 cos t + 10 sin t c. t = dt 4 positive integer. d. The graph implies that the spring never stops oscillating. In reality, the weight would eventually come to rest. 35. a. Juan starts faster than Jean and opens up a big lead. Then, Juan slows down while Jean speeds up. Jean catches up, and the race finishes in a tie. b. Same average velocity c. Tie d. At t = 2, p u′122 = rad>min; u′142 = p = Jean>s greatest velocity e. At 2 p t = 2, w′122 = rad>min; w′102 = p = Juan’s greatest velocity 2 V 37. a. V102 = 4,000,000 m3
b. 0
c.
4
t
8
y 100
50
4 3 106 0
4
t
8
The marble moves fastest at the beginning and slows considerably over the first 5 s. It continues to slow but never actually stops. d. t = 4 s e. t = - 1 + 12 ≈ 0.414 s 31. a. R1p2 =
100p
2 3 106
y
p2 + 1
0
40
200
b. 200 hr V9 c.
40
d. The magnitude of the flow rate is greatest (most negative) at t = 0 and least (zero) at t = 200.
20 40 0
4
8
t
200
t
p V9(t) 5 200(t 2 200)
b. R′1p2 = y
10011 - p22 1p2 + 122
c. p = 1
100
0
8
p
240,000
39. a. v1t2 = - 15e-t 1sin t + cos t2; v112 ≈ - 7.6 m>s, v132 ≈ 0.63 m>s b. Down 10, 2.42 and 15.5, 8.62; up 12.4, 5.52 and 18.6, 102 c. Approximately 0.65 m>s 41. a. - T′112 = - 80, -T′132 = 80 b. -T′1x2 6 0 for 0 … x 6 2; - T′1x2 7 0 for 2 6 x … 4 c. Near x = 0, with x 7 0, - T′1x2 6 0, so heat flows toward the end of the rod. Similarly, near x = 4, with x 6 4, -T′1x2 7 0. 43. a. y 2
y!
x
0
3.5 4 4.5
x
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Z01_BRIG5715_01_SE_ANS_Ch03.indd 20
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Selected Answers b.
h
Approximation
79. a. h142 = 9, h′142 = -6 81. y = 6x + 3 - 3 ln 3
Error
0.1
0.25002
2.0 * 10-5
0.01
0.25
2.0 * 10-7
0.001
0.25
2.0 * 10-9
b. y = - 6x + 33 y
4
c. Values of x on both sides of 4 are used regardless of whether h is positive or negative. d. The centered difference approximations are more accurate than the forward and backward difference formulas. 45. a. 0.395, 0.415 b. 0.0204, 0.000343
2
1
Section 3.7 Exercises, pp. 189–192 dy dy du d # ; 1 f 1g1x222 = f ′1g1x22 # g′1x2 3. g1x2, x = dx du dx dx 5. The inner function is x 2 + 10 and the outer function is u -5 x 7. 30 13x + 729 9. 5 sin4 x cos x 11. 5e5x - 7 13. 2 2x + 1 15. 10x sec2 15x 22 17. ex sec ex tan ex 5 19. 1016x + 7213x 2 + 7x29 21. 210x + 1 315x2 23. 25. 3 sec 13x + 12 tan 13x + 12 27. ex sec2 ex 17x3 + 124 cos 121x2 29. 112x2 + 32 cos 14x3 + 3x + 12 31. 1x 33. 5 sec x 1sec x + tan x25 35. a. u = cos x, y = u3; dy dy = -3 cos2 x sin x b. u = x3, y = cos u; = -3x2 sin x3 dx dx 37. a. 100 b. - 100 c. - 16 d. 40 e. 40 39. - 1.004 hPa>min 41. y′ = 25112x5 - 9x2212x6 - 3x3 + 3224 cot x csc2 x 43. y′ = 30 11 + 2 tan x214 sec2 x 45. y′ = 21 + cot2 x 47. ex cos 1sin 1ex22 cos 1ex2 49. y′ = - 15 sin4 1cos 3x2 1sin 3x2 3cos 1cos 3x24 3e23x 1 1 51. y′ = sec2 1 e23x 2 53. y′ = a1 + b 213x 21x 23x + 1x 1.
5x4 1x + 126 x2 + 1 3 3 59. x e 12 sin x + 3 x cos x 2 61. 5u 2 sec 5u tan 5u + 2u sec 5u 2 x3 - sin 2x 63. 41x + 2231x2 + 12313x2 + 4x + 12 65. 2x4 + cos 2x 67. 2 1p + p21sin p2 + p1p + p2 cos p22 69. a. True b. True 55. y′ = f ′1g1x222g′1x22 2x
c. True
A-21
57.
83. a. -3p 87. a.
b. -5p
85. a.
d2 y dt
2
y
= -
x
y0 k k cos at b m Am
10
2
t
10
b. v1t2 = -5e-t>2 c
p pt pt sin a b + cos a b d 4 8 8 y
2
10
t
25
2p1t + 102 6p sin a b 365 365 c. 2.87 min>day; on March 1, the length of day is increasing at a rate of about 2.87 min>day. y d. 89. a. 10.88 hr
b. D′1t2 =
0.06
d. False 71. 2 cos x 2 - 4x 2 sin x 2 2
73. 4 e -2x 14x 2 - 12 77. y = -9x + 35
75. y′ = y
f ′1x2
100
22f 1x2
300
t
20.06
8
e. Most rapidly, approximately March 22 and September 22; least rapidly, approximately December 21 and June 21 pt 91. a. E′1t2 = 400 + 200 cos a b MW b. At noon; 12 E′102 = 600 MW c. At midnight; E′1122 = 200 MW
4
3
x
Copyright © 2014 Pearson Education, Inc.
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A-22 d.
Selected Answers y
55. a. y′ = -
8000
1 1 x + 2, y = - x + 2 2 2
b. y =
5 1 16x 57. a. a , b b. No 4 2 1x2 + 422 dy dy 1 = 0 on the y = 1 branch; 59. a. = on the other two dx dx 2y + 1 branches. - 1 + 14x - 3 - 1 - 14x - 3 b. f11x2 = 1, f21x2 = , f31x2 = 2 2 y c. c. -
y 5 E(t)
4000
y 5 E9(t) 5 P(t) 0
2xy x2 + 4
12
24
t
93. a. f ′1x2 = - 2 cos x sin x + 2 sin x cos x = 0 b. f 102 = cos2 0 + sin2 0 = 1; f 1x2 = 1 for all x; that is, cos2 x + sin2 x = 1 97. a. h1x2 = 1x2 - 325; a = 2 b. 20 p2 p b. p cos a b 99. a. h1x2 = sin 1x 22; a = 2 4
2
x
4
101. 10 f ′1252
Section 3.8 Exercises, pp. 199–201 1. There may be more than one expression for y or y′. 3. When dy is usually given in terms of both x and y. derived implicitly, dx 3 x 2 20x 3 5. a. - 3 b. 1 7. a. b. 1 9. a. y cos y y 1 - y cos 1xy2 1 b. - 1 13. b. - 20 11. a. sin y x cos 1xy2 - 1 1 1 15. 17. 1 + sin y 2y sin 1y 22 + e y 2 2 3x 1x - y2 + 2y 13y - 18x 2 19. 21. 2x 21y 2 - 13x 52x 4 + y 2 - 2x 3 23. 25. a. 22 + 2 # 1 + 12 = 7 y - 6y 2 2x 4 + y 2 5 7 p2 b. y = - x + 27. a. sin p + 5 a b = p2 4 2 5 p11 + p2 5 b. y = + x 1 + 2p 1 + 2p p p p x 29. a. cos a - b + sin = 12 b. y = 2 4 4 2 sin y 1 4e 2y 31. - 3 33. 35. 3 1cos y - 12 11 - 2e 2y23 4y 5 10 37. x 1>4 39. 4 315x + 121>3 3 2 41. - 7>4 3>4 43. 5>4 3 3 2>3 2 x 14x - 32 9x 21 + 1 x 24 1 45. 47. 49. - 5 51. a. False b. True c. False 13 4 d. False 53. a. y = x - 1 and y = -x + 2 y b.
61. a.
dy x - x3 = y dx
c.
y
b. f11x2 =
B
x2 -
x4 x4 ; f21x2 = - x2 2 B 2
1
1
x
4x 3 5 5
63. y =
y
1 1
65. y = y
x
1 + 2p 25 + p + 2p2 x + pa b 5 25
1 1 1
2
x
5
1
2
x
Copyright © 2014 Pearson Education, Inc.
Z01_BRIG5715_01_SE_ANS_Ch03.indd 22
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Selected Answers y
5p 4
67. y = - 2x +
9.
1 x
11.
2 x
13. cot x
69. a. Tangent line y = b.
9x 20 11x 2 + ; normal line y = 11 11 9 9
y
1
x
1
71. a. Tangent line y = b.
x 8 + ; normal line y = 3x - 4 3 3
y
2
x
2
dr h - 2r = ; -3 dh h dy dy y x = - . = m = ; for x2 + y2 = a2, 77. Note that for y = mx, x y dx dx dy y dy x = - . For x2 - y2 = b, = . Since 79. For xy = a, x y dx dx y x a - b # a b = - 1, the families of curves are orthogonal trajectories. x y 2y 215 + 8x1y2 7y 2 - 3x 2 - 4xy 2 - 4x 3 81. 83. 2y12x 2 + 2y 2 - 7x2 11 + 2x1y23 85. No horizontal tangent line; vertical tangent lines at 12, 12 and 1- 2, 12 87. No horizontal tangent line; 73. a.
dK K = dL 2L
b. - 4
75.
vertical tangent lines at 10, 02, a
313 313 , 13b, a , - 13b 2 2
2 1x 2 - 12
x2 + 1 1 + 2x ln x. 19. x x ln x 1 x 21. 23. 8 ln 8 25. 5 # 4x ln 4 x1ln x + 122 27. 3x # x 2 1x ln 3 + 32 29. A′ = 100011.04524t ln11.0452 46.512 s dT = - 2.74 # 2-0.274a ln 2 31. a. About 28.7 s b. c. da 1000 ft dT 0.4156 min At t = 8, = da 1000 ft 24.938 s . = 1000 ft If a plane is traveling at 30,000 feet and it increases its altitude by 1,000 feet, the time of useful consciousness would decrease by about 25 seconds. 33. a. About 67.19 hr b. Q′1122 = -9.815 mCi>hr Q′1242 = -5.201 mCi>hr Q′1482 = -1.461 mCi>hr The rate at which iodine-123 leaves the body decreases with time. 35. 2x ln 2 37. e yy e - 1 1y + e2 39. 2e 2u 1x 2x ln 2 41. 110x - 92 43. x 2 12 + 122 p 45. x cos x - 1 1cos x - x ln x sin x2; -ln a b 2 1x 2 + ln x 47. x a b; 412 + ln 42 21x 1sin x2ln x1ln 1sin x2 + x1ln x2 cot x2 ;0 49. x 51. y = x sin 1 + 1 - sin 1 53. y = e2>e and y = e-2>e 8x 57. -sin x 1ln 1cos2 x2 + 22 55. y′ = 2 1x - 12 ln 3 1x + 1210 8 ln 4 10 59. 61. c d x - 2 x ln2 x 12x - 428 x + 1 1ln x2 - 1 63. 2x ln x 1x + 123>2 1x - 425>2 # 65. 15x + 322>3 3 5 10 c + d 21x + 12 21x - 42 315x + 32 tan x 2 67. 1sin x2 11 + sec x ln 1sin x22 69. a. False b. False c. False d. False e. True 2 1 12 71. - 2 73. 75. 3x ln 3 77. x 3x + 1 x ln 10 2 3 8 1 79. 81. + + 2x - 1 x + 2 1 - 4x 2x y 83. y = 2 17.
x
15. -
A-23
Section 3.9 Exercises, pp. 209–212 1. x = e y 1 1 = e yy′1x2 1 y′1x2 =
1 1 = . x ey
d d d 1ln kx2 = 1ln k + ln x2 = 1ln x2 3. dx dx dx 1 1 5. f ′1x2 = . If b = e, then f ′1x2 = . x x ln b
1
7. f 1x2 = eh1x2 ln g1x2
x
Copyright © 2014 Pearson Education, Inc.
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A-24
Selected Answers
85. 10x10x11 + ln x2 cos x 87. x cos x a - ln x sin xb x x 1 1 1 89. a1 + b c ln a1 + b d x x x + 1 10 91. x9 + x 11 + 10 ln x2 y 93. a.
21. 25.
1 2
$ 2u + 1$ 2u + u 1
x$ ln x $ 21ln x22 - 1
2y
23.
1y2 + 122 + 1
27. -
ex sec2 1ex2
$ tan ex $ 2tan2 ex - 1
p 1 p es 4 2 31. y = x + 33. y = x + + 2s 4 2 3 16 13 1 + e 35. a. Approximately -0.00055 rad>m. The magnitude of the change du in angular size, ` ` , is greatest when the boat is at the skyscraper dx 1i.e., at x = 02. d b. 29. -
8000
4000
dx
0
x
200
t
20
b. t = 2 ln 12652 ≈ 11.2 years; about 14.5 years c. P′102 ≈ 25 fish>yr; P′152 ≈ 264 fish>yr y d. The population is growing fastest after about 10 years. 800
20.007
1 1 1 1 1 5 1 39. 41. 43. 4 45. 47. 49. 51. a. 2 3 5 12 4 4 2 2 3 c. Cannot be determined d. 53. a. True b. False b. 3 2 c. True d. True e. True y 55. a. 37.
400
0
4
t
95. b. r1112 ≈ 0.0133; r1212 ≈ 0.0118; the relative growth rate is decreasing. c. lim r1t2 = 0; as the population gets close to carrying tS ∞ capacity, the growth rate approaches zero. 97. a. A152 = +17,443 A1152 = +72,705 A1252 = +173,248 A1352 = +356,178 +5526.20>year, +10,054.30>year, +18,293>year b. A1402 = +497,873 dA c. = 600,000 ln 11.0052311.005212t 4 dt ≈ 12992.5211.005212t A increases at an increasing rate. 1 103. 2711 + ln 32 99. p = e1>e; 1e, e2 101. e
1
b. f ′1x2 = 2x sin-1 x +
x2 - 1 21 - x2 y
3 2
Section 3.10 Exercises, pp. 220–222
d 1 d 1 1sin-1 x2 = ; 1tan-1 x2 = 2 dx dx 1 + x2 21 - x d 1 2 1 1 1sec -1 x2 = 3. 5. 7. 2 dx 5 4 $ x$ 2x - 1 21 - 4x2 -2x 4w 2e 10 11. 13. 9. 2 2 -4x 100x + 1 21 - 4w 21 - e 4y 1 1 15. 17. 19. 21z 11 + z2 1 + 12y2 - 422 $ x $ 2x2 - 1
x
1
21
1
1.
21
0.5
20.5
1
x
21
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Selected Answers 57. a.
y
b.
1
11. - 40p ft2 >min
1 1 17. At the point a , b 2 4 21.
25.
x
5
31. b. f ′1x2 =
1 cm>s 2
-x
e - e-x tan-1 1x2 1 + x2
39.
y
45.
1
49. 53. 5
x
3 in.>min 80p
1 m>min; 2000 min 500 5 10 tan 20° km>hr ≈ 3.6 km>hr 23. ft>s 24 8 32 8 - ft>s, - ft>s 27. 2592p cm3 >s 29. ft>s 3 3 9p 2 9p ft3 >min 33. m2 >min 35. 57.89 ft>s 37. 4.66 in>s 5 315 ft>s 41. 720.3 mi>hr 43. 11.06 m>hr 2 du a. 187.5 ft>s b. 0.938 rad>s 47. = 0.543 rad>hr dt 1 1 du rad>s, rad>s 51. = 0 rad>s, for all t Ú 0 5 8 dt 13 -0.0201 rad>s 55. a. m>hr b. - 1 m2 >hr 10 19.
Chapter 3 Review Exercises, pp. 231–234 1. a. False
1 2′1x2 = 3
1 59. 1 f 61. 1 f 2′1x2 = 21x + 4 2 63. 1 f -12′1x2 = 2x 65. 1 f -12′1x2 = - 3 x 10 10 67. a. sin u = implies u = sin-1 . / / du 1 10 -2 # 1- 10/ 2 = Thus, = . 2 d/ 2 /2/ - 100 10 1 - a b / B b. du>d/ = -0.00408, - 0.0289, and - 0.198 du = - ∞ d. The length / is decreasing. c. lim + / S 10 d/ du 1 1 = b. 69. a. 2 2 dc D 2D - c p 73. Use the identity cot-1 x + tan-1 x = . 2 -1
-1
13.
A-25
3. a. 16
b. False
c. False
d. False
e. True
b. y = 16x - 10 y
10
1
5. a. -
3 4
b. y = -
3x 1 + 4 2
x
2
y
1
Section 3.11 Exercises, pp. 226–231 1. As the side length s of a cube changes, the surface area 6s2 changes as well. 3. The other two opposite sides decrease in length. 5. a. 40 m2 >s b. 80 m2 >s c. y 100
1
x
7. a. 2.70 million people>yr b. The slope of the secant line through the two points is approximately equal to the slope of that tangent line at t = 55. c. 2.217 million people>yr 9. a. 40 m>s b. 20>3 m>s c. 15 m>s d.
v 60
40 0
7. a. 4 m2 >s
30
b. 12 m2 >s
s
c. 212 m>s
20
9. a.
1 cm>s 4p
0
30
60
90
120
t
e. The skydiver deployed the parachute. Copyright © 2014 Pearson Education, Inc.
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A-26 13.
Selected Answers 4 b. - , 0 3
y
2
c.
2
x
5
d. - 1, 1
y 3
1
23
22
1
21
2
3 x
21 22
15. 2x2 + 2px + 7
17. 5t 2 cos t + 10t sin t 32u2 + 8u + 1 19. 18u + 122 sec2 1u 2 + 3u + 22 21. 18u + 122 sin x 2 sin x + 6x2 cos x - 2x cos x 9x 23. sec2 1sin u2 # cos u 25. 13x - 1 1 x2 - x ln 2 31. 27. 12 + ln x2 ln x 29. 12x - 12 2 ! x ! 2x2 - 1 dy y cos x p 33. 1 35. 13 + 37. = dx 1y - 1211 + sin x2 6 xy 4x 24 39. - 2 41. y = x 43. y = + 5 5 x + 2y 2 cos 1x 1x sin 1x + cos 1x 45. x = 4; x = 6 47. ,, 2 1x 4x 3>2 31x sin 1x + 13 - x2 cos 1x 49. x2 f ′1x2 + 2x f 1x2 8x 5>2 g1x21x f ′1x2 + f 1x22 - x f 1x2g′1x2 25 51. 53. a. 27 b. c. 294 27 g21x2 1 3p 55. f 1x2 = tan 1p13x - 112, a = 5; f ′152 = d. 1215 e. 9 4 6 3 1 -1 -1 57. 59. 1 f 2′1x2 = - 4 61. a. 1 f 2′ a b = 12 13 12 x 1 1 b. 1 c. 65. a. 6550 people>yr b. p′1402 = 63. a. 4 3 4800 people>yr 67. 50 mi>hr 69. - 5 sin 165°2 ft>s or -4.53 ft>s
71. - 0.166 rad>s
AP® Practice, Section 1, Parts A and B, pp. 243–244 l. B 2. C 3. C 4. E 5. E 6. A 7. B 8. D 9. C 11. C 12. C 13. C 14. C 15. D 16. C 17. B
AP® Practice, Section 2, Parts A and B, p. 245 1. a.
y 5 4
10. A
23
2. a. 0 b. (0, 0.087) c. (0.087, 0.405) d. t ≈ 0.405 e. t ≈ 0.405 3. a. A–c; B–d; C–a; D–b b. f9C; f ′9B; f ″9A y 4. 2
y 5 g(x)
x
2
22
22
5. The length of the hypotenuse is increasing at 3>5 m/s. 6. a. Note that lim -f 1x2 = lim f 1x2 = 1 which implies that xS0
xS0 +
lim f 1x2 = 1. Because f 102 is also equal to 1, f is continuous at 1. b. The slope of the line tangent to e - px at 0 is the same as the slope of the line tangent to 1 - sin px at 0 (both slopes are - p). xS0
c. f ′1x2 = e
-pe - px - p cos p x
if x … 0 ; x = 0.5 if x 7 0
d. y = - px + 1
x + y b. Horizontal tangents at 12, - 22 and 1-2, 22 x + 4y c. Vertical tangents at 14, -12 and 1-4, 12 7. a. -
CHAPTER 4 Section 4.1 Exercise, pp. 253–256 1. f has an absolute maximum at c in 3a, b4 if f 1x2 … f 1c2 for all x in 3a, b4. f has an absolute minimum at c in 3a, b4 if f 1x2 Ú f 1c2 for all x in 3a, b4. 3. The function must be continuous on a closed interval. y 5. y 7.
3 2 0
1
23
22
21
1
2
1
2
3
x
1
3 x 21
1
x
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