{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online