ODDREV02 - 92 Chapter 2 Differentiation 51. x2 y2 25;...

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Unformatted text preview: 92 Chapter 2 Differentiation 51. x2 y2 25; acceleration of the top of the ladder dx dt x dx dt d 2x dt 2 dy dt dy dt dx dt d 2y dt 2 First derivative: 2x 2y y 0 0 dx dt y d 2y dt 2 dy dt dy dt d 2y dt 2 0 1 y x d 2x dt 2 dx dt 2 Second derivative: x dy dt 2 When x d 2y dt 2 7, y 1 24 24, 70 dy dt 2 2 7 dx , and 12 dt 7 12 2 2 (see Exercise 27). Since 1 24 4 49 144 1 24 29.10s dx d 2x is constant, 2 dt dt 625 144 206.2 0. 0.1808 ft sec2 53. (a) Using a graphing utility, you obtain m s (b) dm dt dm ds ds dt 1.762s 29.10 ds dt 1.2. 0.881s2 (c) If t Thus, s 1995 , then s dm dt 15.5 and ds dt 1.762 15.5 29.10 1.2 2.15 million. Review Exercises for Chapter 2 1. f x fx x2 lim x→0 2x fx x x2 3 x x x 2 fx 2x x x x 2 2 lim x x 2x 3 x2 2x x 2x 3 3 x2 2x 3 x→0 lim 2x x x→0 lim 3. f x fx x lim 2x x 2x x→0 lim 2x x→0 x 2 2x 2 1. 1 fx x x x x x x 1 x x x x x x x x x fx 1 x x x x 1 2x x x x 1 x x x x 5. f is differentiable for all x x→0 lim lim lim x→0 x→0 x→0 lim x→0 R eview Exercises for Chapter 2 93 7. f x 4 x 2 2. 2 because of the sharp turn 9. Using the limit definition, you obtain g x At x 1, g 1 4 3 1 6 3 2 (a) Continuous at x 4 x 3 1 . 6 (b) Not differentiable at x in the graph. y 7 6 5 4 3 2 x 123456 −1 −2 −3 11. (a) Using the limit defintion, f x At x y 2 y (b) −4 3x 2. 13. g 2 x→2 lim gx x x2 x x x3 x x g2 2 1 2 x2 2 2 x2 x x2 x 2 8 2 4 4 1, f 3x 3x 0 1 3. The tangent line is 1 1 x→2 2 x→2 lim lim (− 1, − 2) x→2 lim −4 x→2 lim x 2 15. 2 y 17. y f′ f 25 0 19. f x fx x8 8x7 y 1 x −1 1 21. h t ht 3t 4 12t 3 33 x x 23 23. f x fx 6x1 2 x3 3x2 2 t 3 2 3x2 6x 3x x 2 25. h x hx 6x 3x 12 3x1 x 3 27. g t gx 31. f f 3 3 1 x2 4 t 3 3 cos 3 4 3t3 sin 4 cos 4 29. f f 2 2 3 sin 3 cos 3 sin 94 Chapter 2 Differentiation 33. F Ft 200 T 100 T 4, F 4 9, F 9 50 vibrations/sec/lb. 331 vibrations/sec/lb. 3 35. st s 9.2 s0 16t2 16 9.2 1354.24 s0 2 s0 0 (a) When T (b) When T 37. (a) 15 y The building is approximately 1354 feet high (or 415 m). (c) Ball reaches maximum height when x (d) y y y0 x 20 40 60 25. x 1 1 0.6 0 0.02x2 0.04x 10 5 y 10 y 25 Total horizontal distance: 50 (b) 0 0 39. x t (a) v t at (c) v t x 3 2 x x1 t2 0.02x2 x implies x 50 2 2t 2 3 2. 3 2 y 30 y 50 50. 1 (e) y 25 0 0.2 1 3t xt vt 0 for t 2 t 3 2t 3 (b) v t < 0 for t < 2 . (d) x t v1 1 2 1 2 1 4 0 for t 21 22 1, 2. 3 3 1 1 1 v2 The speed is 1 when the position is 0. 3x2 3x2 2 45. f x fx 6x3 x 2x 7 x2 7 2x 9x2 2 41. f x fx 2x 2 16x 3 x2 7 2x 3 6x 43. h x hx x sin x 1 sin x 2x x2 x2 x2 x2 x2 1 x 1 1 2x 1 2 x1 2 sin x x cos x 2x 2 47. f x 21 1 x3 fx 1 1 x2 x2 12 x 1 2x 3 2 x3 1 x3 49. f x fx 4 4 3x 2 sec x 3x2 3x2 1 51. y 2 x2 cos x cos x 2x cos2 x tan x x sec2 x tan x x2 x sin x 2x cos x x 2 sin x cos2 x 6x 4 6x 3x2 2 y 55. y 53. y y 3x 2 sec x tan x 6x sec x y R eview Exercises for Chapter 2 95 57. y y x cos x x sin x sin x cos x cos x x sin x 59. g t gt gt t3 3t2 6t y y 3t 3 2 61. f f f 3 tan 3 sec2 6 sec sec tan 6 sec 2 63. 2 sin x 2 cos x 2 sin x 2 sin x 0 3 cos x 3 sin x 3 cos x 3 cos x 2 sin x 3 cos x tan y y y 65. f x fx 1 1 1 2 x3 12 67. h x 12 x x2 2 x x2 3 1 2 x3 3x2 hx 3 1 3 x2 x2 x2 1 3 11 x2 6x 1 x 12 3 2x 3x2 2 1 x3 s2 s2 s s2 52 52 2x 69. f s fs 1 1 s3 3s2 3s 5 s3 s2 1 3s 5 5 25 5 2 71. y s2 5 1 32 3 cos 3x 9 sin 3x 1 1 2s y 1 1 32 32 s3 s s2 8s3 73. y y 1 csc 2x 2 1 2 csc 2x cot 2x 2 csc 2x cot 2x 75. y y x 2 1 2 1 1 2 sin 2x 4 1 cos 2x 2 4 cos 2x sin2 x 77. y y 2 32 sin x 3 sin1 2 x cos x cos x cos3 x 2 72 sin x 7 sin5 2 x cos x sin2 x 79. y y sin x x2 x 2 cos x x 22 sin x sin x 1 sin x 1 1 4 5 81. f t ft t2 t tt 83. g x 2 gx 2x x x x 1 1 2 32 12 7t The zeros of f correspond to the points on the graph of f where the tangent line is horizontal. 0.1 g does not equal zero for any value of x in the domain. The graph of g has no horizontal tangent lines. 4 f′ − 0.1 1.3 f − 0.1 −2 g′ 7 g −2 96 Chapter 2 t 6t 1 5 1 Differentiation 12 85. f t ft t 1 13 t 1 56 87. y y tan 1 x 16 sec2 1 x 21 x f does not equal zero for any x in the domain. The graph of f has no horizontal tangent lines. 5 y does not equal zero for any x in the domain. The graph has no horizontal tangent lines. 5 f y − 20 7 2 f′ −2 −1 y′ −4 89. y y y 2x2 4x 4 sin 2x 2 cos 2x 4 sin 2x 91. f x fx f cot x csc2 x 2 csc x csc x cot x 93. f t ft ft t 1 t 1 2t 1 t 2 1 t3 2 t4 2 csc2 x cot x 95. g g g 97. T T tan 3 3 sec2 3 sin cos 1 1 sin 1 18 sec2 3 tan 3 700 t2 t2 4t 10 1 1400 t 2 4t 10 2 1, 1400 1 2 4 10 2 5, 1400 5 30 3xy 3y 3x 2 10 y3 3y2y y2 y y 2 (a) When t T 1 (b) When t 18.667 deg hr. T (d) When t 3.240 deg hr. T 9 3, 1400 3 2 12 10 2 10, 2 10 2 7.284 deg hr. (c) When t T 25 x2 2x 3xy 1400 10 100 40 0.747 deg hr. 99. 10 0 2x 2x 3x yx 3y 3y y2 xy y1 2 101. y 1 x 2 12 16 0 y y 2x x1 2y x 1 y 2 12 y x x y 2y 2 xy x y 2y y 2 xy y 2x 2 xy y 2x 2y 2 xy x 2y x 2x y yy xx R eview Exercises for Chapter 2 97 103. x cos y y y x cos y x sin y sin y cos x y y cos x y sin x y sin x y cos x sin y 105. 2x x2 y2 2yy y 20 0 −9 6 (2, 4) 9 y sin x sin y cos x x cos y x y −6 At 2, 4 : y Tangent line: y x Normal line: y 2x 1 2 4 2y 4 y 1 x 2 10 2x 0 0 2 2 107. y dy dt dy dt x 2 units sec dx 1 dx ⇒ dt 2 x dt 1 dx , 2 dt 1, 4, dx dt dx dt 2x dy dt 4x (a) When x (b) When x (c) When x 2 2 units/sec. 4 units/sec. 8 units/sec. 109. s h s dV dt 12 2 1 h 4 1 111. s t st s 35 60 9.8t 60 4.9t2 s (t) 4.9t2 25 5 4.9 st xt 3s t 3 9.8 5 4.9 30˚ x(t ) 4.9t2 t 1 h 4 5 8 4 4 2 hh h tan 30 1 3 xt dx dt 2 1 2 Width of water at depth h: w V dV dt dh dt When h 2 2s 5 2 2 5 4 2 4 2 h dh dt 2 h 2 h 3 ds dt 1 2 38.34 m sec 2 2 dV dt 54 h 1, dh dt 2 m min. 25 s h 2 98 Chapter 2 Differentiation Problem Solving for Chapter 2 1. (a) x 2 y r 2 r2 Circle y Parabola −3 3 x2 Substituting, y y2 y2 yy 2r y 2r y 2r r 2 3 −1 r2 r2 0 0 y y r2 y 1 Since you want only one solution, let 1 Graph y x 2 and x 2 y 12 2 1 4 2r 0⇒r 1 2 (b) Let x, y be a point of tangency: x 2 y x2 ⇒ y 2x 2b b Also, x 2 y y b y y y 2 y b 2 1 ⇒ 2x 2y by 0⇒y x b y circle . 2x (parabola). Equating, x b 1 1 ⇒b 2 b 2 y 3 y 1 and y y 1 2 x 2 imply y 1 2 1⇒y 1 2 1⇒y −3 −1 3 1⇒y 3 and b 4 5 . 4 Center: 0, Graph y 5 4 x 2 and x 2 y 5 4 a0 2 1 3. (a) fx f0 f0 P1 x cos x 1 0 1 P1 x P1 0 P 1 a1x 1 0 (b) fx f0 f0 f0 P2 x cos x 1 0 1 1 12 2x P2 x P2 0 P P 2 2 a0 a1x a2x 2 1 0 1 2 a0 ⇒ a0 a1 ⇒ a1 a0 ⇒ a0 a1 ⇒ a1 2a2 ⇒ a2 0 0 0 (c) x cos x P2 x 1.0 0.5403 0.5 0.1 0.9950 0.9950 0.001 1 1 0 1 1 0.001 1 1 0.1 0.9950 0.9950 1.0 0.5403 0.5 P2 x is a good approximation of f x (d) fx f0 f0 f0 f 0 x sin x 0 1 0 1 13 6x cos x when x is near 0. a0 a1x a2 x 2 0 1 0 1 6 P3 x P3 0 P P P 3 3 3 a3x3 a0 ⇒ a0 a1 ⇒ a1 2a2 ⇒ a2 6a3 ⇒ a3 0 0 0 P3 x P roblem Solving for Chapter 2 5. Let p x px Ax3 3Ax 2 B 2B A 3A Bx 2 2Bx C C B 2B C C 2D 2C 12 16 5 2C 12, you obtain 4A 8⇒A 2. Finally, C 1 2 99 Cx C D D At 1, 1 : A 3A At 1, 3: 1 14 D 3 2 2 Equation 1 Equation 2 Equation 3 Equation 4 Adding Equations 1 and 3: 2B Subtracting Equations 1 and 3: 2A Adding Equations 2 and 4: 6A 2C 4 Subtracting Equations 2 and 4: 4B Hence, B 4 and D 2C 4x 2 a 2 y2 x4 a2x 2 a a2x 2 a 2 1 2 2 2B Subtracting 2A Thus, p x 7. (a) x4 a2 y 2 y 2x3 a2 x 2 a2x 2 ± 4 and 6A 5. 4 2A 0 x4 x4 a2x 2 a x4 Graph: y1 (b) −3 and y2 a= 1 2 3 a=2 a=1 −2 ± a, 0 are the x-intercepts, along with 0, 0 . (c) Differentiating implicitly, 4x 3 y a2 2 2 2a2 x 2a2 y y x a2 a2y 2x 2 0 ⇒ 2x 2 a2 ⇒ x ±a 2a2x 4x3 2a2y a2 a2 2 a2y 2 a2y2 2 . a4 4 a2y 2 y2 y a4 2 a4 4 a2 4 ± a 2 aa ,, 22 a , 2 a , 2 aa ,, 22 a , 2 a 2 Four points: 100 9. (a) Chapter 2 y Differentiation Line determined by 0, 30 and 90, 6 : 30 (0, 30) y (90, 6) (100, 3) x 90 100 30 30 6 x 0 90 100, y 0 24 x 90 30 4 x⇒y 15 4 x 15 30 When x 4 100 15 10 > 3 ⇒ Shadow determined by man. 3 Not drawn to scale (b) 30 y Line determined by 0, 30 and 60, 6 : (0, 30) (60, 6) (70, 3) x 60 70 y When x 30 30 6 x 0 60 70, y 2 70 5 0 30 2 x⇒y 5 2 x 5 30 2 < 3 ⇒ Shadow determined by child. Not drawn to scale (c) Need 0, 30 , d, 6 , d 30 0 6 d 6 d d 3 10, 3 collinear. 10 ⇒ 24 d 3 ⇒d 10 80 feet dx dt 5. (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. y 30 y 6 x ⇒y dy 5 x and 4 dt 5 dx 4 dt 25 . 4 For x < 80, the shadow is determined by the child. y 30 y x 3 10 ⇒y 10 x 9 100 dy and 9 dt 10 dx 9 dt 50 . 9 Therefore, 25 4 50 9 x > 80 0 < x < 80 80. dy dt dy is not continuous at x dt 11. L x lim lim lim Lx Lx Lx x lim x→0 x x Lx x Lx Lx x→0 x→0 Also, L 0 But, L 0 Thus, L x Lx x L0 L0 0 L0 L 0 ⇒L 0 0. x→0 0 because L 0 L 0 , for all x. The graph of L is a line through the origin of slope L 0 . P roblem Solving for Chapter 2 13. (a) 101 z (degrees) sin z z sin z z z→0 0.1 0.0174524 0.01 0.0174533 0.0001 0.0174533 (b) lim z→0 0.0174533 sin z z lim z →0 In fact, lim (c) d sin z dz 180 sin z sin z z z cos z cos z z 180 2 sin z sin z z 1 cos z sin z sin z z lim z →0 lim z →0 sin z lim z →0 cos z sin z 0 (d) S 90 d Sz dz sin 90 cos z sin c 180 cos z cos 180 1 180 1; C 180 Cz 180 d sin cz dz cos cz 180 (e) The formulas for the derivatives are more complicated in degrees. 15. j t at (a) j t is the rate of change of the acceleration. (b) From Exercise 102 in Section 2.3, st vt at at 8.25t 2 16.5t 16.5 jt 0 66t 66 ...
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