Business Calc Homework w answers_Part_25

Business Calc Homework w answers_Part_25 - 121 Chapter 3...

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Unformatted text preview: 121 Chapter 3 Review 38. Use implicit differentiation. y y3 41. y 2 cos x x 2 x1 d x dx x 1 (x 1)(1) (x)(1) (x 1)2 1 2y(x 1)2 d2 y dx dy 2y dx dy dx 39. x d3 (x ) dx 3x 2 3 y 3 d3 (y ) dx d (y) dx 3y 2y (3y 2 y 2 sin x 1)y 2 sin x 1 d3 (y ) dx 3y 2y 0 y x2 y2 2 sin x 3y 2 1 d dx (3y 2 1)(2 cos x) (2 sin x)(6yy ) (3y 2 1)2 (3y 2 1)(2 cos x) 2 x y2 d dx y 2 sin x 3y 2 1 y d (1) dx y d (2 cos x) dx (y 2)(2x) (x 2)(2y)(y ) y4 (y 2)(2x) (x 2)(2y) (3y 2 2xy 2x x y2 2 x 1/3 42. 4 5 y 2x(x 3 y 3) y5 2x y5 since x 3 y3 (3y 2 d 1/3 (x ) dx 1 2/3 x 3 y 1/3 d 1/3 (y ) dx 1 2/3 y y 3 y 1 y 40. y2 d2 (y ) dx 2yy y y 1 2 x d d2 (1) dx dx x 2 x2 2 1 x 2(2y) x 2y d1 dx x 2y 1 d2 (x y) (x 2y)2 dx 1 [(x 2)(y ) (x 2y)2 1 1 (x 2) 2 (x 2y)2 xy 11 2xy x 4y 2 y 1 2xy 2 x 4y 3 1)2 1)2 cos x 12y sin 2 x (3y 2 1)3 4 d (4) dx 0 x y y 2/3 x 2/3 2/3 d dx y 2/3 x 2y 3x 1/3 2y 3x 1/3 xy (x) (y)(1) x2 y 2/3 x 43. y 2x 3 3x 1, y 6x 2 3, y 12x, y (4) 12, and the rest are all zero. 2xy x4 , 24 3 x , 6 2 x , 2 44. y y y y (4) y (5) x, 1, and the rest are all zero. y x2 2 1/3 1/3 xy ( x 5/3y 2/3 3 2 4/3 1/3 2 5/3 2/3 x y x y 3 3 (y)(2x)] 2 sin x 3y2 1 2 y4 3 (12y sin x) x 2 y) 122 Chapter 3 Review 45. dy d dx dx x2 (2x x2 32 2(3) 3 1 2 32 2(3) x 2) 2x 3, we have y At x 2 dy dx and 1 2x (x 3 3 or y 46. dy dx 2 d (4 dx 3) 2 3 or y csc 2 x At x y 2 4 3 3) dy dx At t 3 y 2 csc x) 2 csc x cot x csc 2 2 csc 2 y 2 (a) Tangent: y (b) Normal: y 4 2 2 csc cot 2 1x 0 2 and 1 2(1)(0) 2 or y 2 or y 2 50. 2 2 2 1x x x 1. 2 2 dy dx At t y 2y 2 d (2y 2) dx dy 2x 4y dx dy dx Slope at (1, 2): y d (9) dx dy 51. dx 0 2x 4y x 2y (a) Tangent: y At t 1 4 1 2(2) 1 (x 4 11 5 1) dy dt dx dt 2 sin t 2 cos t tan t 3 , we have x 4 2 cos 3 4 3 4 2 sin 2, and 2, dy dx tan 3 4 1. 1 x 4 2 or y 1(x 2) 2), or y 4 cos t 3 sin t dy dt dx dt ( 4 cot t 3 3 , we have x 4 4 sin 3 4 2 3 cos 2, and x 2 3 4 2. 3 2 2 dy dx , 4 3 cot 3 4 4 . 3 The equation of the tangent line is 9 d2 (x ) dx 4 x 5 1 or y 6 2 2 47. Use implicit differentiation. x2 4) 5 x 4 1 or y The equation of the tangent line is , we have cot dy dx x 49. 3 cot x 4) . 3 (x 2 53 2 x 4 (x 5 (b) Normal: y 3 (b) Normal: y 5 (x 4 (a) Tangent: y 2x 3 2 (a) Tangent: y 1 x2 9 4 y 4 x 3 32 2 dy dt dx dt 6 2 5 sec 2 t 3 sec t tan t , we have x 5 tan 5 sec t 3 tan t 3 sec 4 2. 5 3 sin t 2 6 53 dy , and dx 3 6 4 x 3 2, or y 3, 10 . 3 5 3 sin 6 The equation of the tangent line is (b) Normal: y 4(x 1) 2 or y 4x 2 y 48. Use implicit differentiation. xy x d (x) dx 1 1 2 (x) xy d ( dx dy dx xy) (y)(1) x 2 dy xy dx dy dx 6 d (6) dx 0 y 1 2 xy x 2 1 4 1 4 2 2 y 1 2 y 2 x Slope at (4, 1): xy 2 1 4 y x 5 4 xy 10 (x 3 2 3) 53 , or y 3 10 x 3 5 3. Chapter 3 Review (c) Nowhere in its domain dy dt dx dt dy 52. dx At t 1 cos t sin t 59. y 2 4 , we have x 2 cos 4 2 y = f (x ) , –3 y dy dx 2 sin 4 1 123 4 cos 4 4 sin 2 2 2 2 4 y 1. y = f (x ) The equation of the tangent line is y (2 y (1 2 1) x 2)x 2 4 2 2 1 This is approximately y 2 4 x –2 , and 60. 2 2 1 3 , or x . 2.414x 3.200. 61. (a) iii (b) i (c) ii 53. (a) 62. The graph passes through (0, 5) and has slope x 2 and slope 0.5 for x 2. 2 for y 7 [ 1, 3] by 5 3 1, (b) Yes, because both of the one-sided limits as x → 1 are equal to f (1) 1. (c) No, because the left-hand derivative at x the right-hand derivative at x 1 is 1. 1 is 1 and 54. (a) The function is continuous for all values of m, because the right-hand limit as x → 0 is equal to f (0) 0 for any value of m. (b) The left-hand derivative at x 0 is 2 cos (2 0) 2, and the right-hand derivative at x 0 is m, so in order for the function to be differentiable at x 0, m must be 2. 55. (a) For all x 0 (b) At x y = f (x ) 7 x 63. The graph passes through ( 1, 2) and has slope 2 for x 1, slope 1 for 1 x 4, and slope 1 for 4 x 6. y 7 0 (c) Nowhere 56. (a) For all x y = f (x ) (b) Nowhere (c) Nowhere 7 57. Note that lim f (x) lim (2x x→0 lim f (x) x→0 x→0 lim (x 3) x→0 3) x 3 and 3. Since these values agree with f (0), the function is continuous at x 0. On the other 64. i. If f (x) f (x) 9 7/3 x 28 9, then f (x) 3 4/3 x and 4 x 1/3, which matches the given equation. hand, 2, 1 x 0 f (x) , so the derivative is undefined at 1, 0 x x 4 (0, 4] (b) At x 0 58. Note that the function is undefined at x 0. (c) Nowhere in its domain (a) [ 2, 0) (0, 2] 9 7/3 x 28 3 4/3 x , which 4 2, then f (x) contradicts the given equation f (x) 0. (a) [ 1, 0) ii. If f (x) (b) Nowhere iii. If f (x) 3 4/3 x 4 6, then f (x) the given equation. x 1/3. x 1/3, which matches 124 Chapter 3 Review 64. continued (d) 3 4/3 x 4 iv. If f (x) 4, then f (x) x 1/3 and f (x) 5 tan 0)( 5 sec 2 0) (e) however that i and iii could not simultaneously be true. (2 (f) [ 1, 5] by [ 10, 80] (b) t interval avg. vel. 38 0.5 58 1 70 1.5 74 2 70 2.5 58 3 38 3.5 10 4 [0, 0.5] [0.5, 1] [1, 1.5] [1.5, 2] [2, 2.5] [2.5, 3] [3, 3.5] [3.5, 4] 10 0 38 0.5 58 1 70 1.5 74 2 70 2.5 58 3 38 3.5 1 2 67. (a) 8 f (1) 1 At x 2 d f( dx (b) 56 f(x) At x d dx g (x) g ( 1) 3(2) 1 5. d2 [f (x)g 3(x)] dx (c) d g(f (x)) dx At x f (x) g (f (x))f (x) 1, the derivative is x 1 ( 3) 2 13 . 10 (d) d f (g(x)) dx g (0)f ( 1) (4)(2) 8. f (g(x))g (x) 1, the derivative is f (x) f (x) f ( x) 3f (x) 2 5 ( 3)2(0) f 2(x) 3g 2(x)g (x) g 3(x) 2f (x)f (x) f (x)g 2(x)[3f (x)g (x) 2g(x)f (x)] At x 0, the derivative is f (0)g 2(0)[3f (0)g (0) 2g(0)f (0)] ( 1)( 3)2[3( 1)(4) 2( 3)( 2)] 9[ 12 12] 0. 1 5 x f (g( 1))g ( 1) 2 f (x) f (0) 0, the derivative is x) 5 f 2(1) cos 1 (1) 5 g (f ( 1))f ( 1) 2 (c) 2 1, the derivative is 3f ( 1) 40 At x 1 f(x) 2 x f (x) cos 2 2 g(x)] d [3f (x) dx At x 24 1 1 x 2 12. 1, the derivative is 1 f (1) 10f 2(x) cos 8 (d) Average velocity is a good approximation to velocity. 2 2 . 3 1, the derivative is 24 [ 1, 5] by [ 80, 80] x f (x) x2 f (x)] 2 x 10 sin (2f (x)f (x)) 2 x 20 f (x)f (x) sin 5 2 3f (0) 32 d [10 sin dx 20( 3) (c) x f (x)] 0, the derivative is 20 f (1) f (1) sin 40 cos x)( f (x)) ( f (x))( sin x) (2 cos x)2 cos 0)(f (0)) (f (0))( sin 0) (2 cos 0)2 At x 56 (2 d f (x) dx 2 cos x At x 65. (a) d dx 1 ( 5) 5 f (1)( 5) 1. Answer is D: i and iii only could be true. Note, (b) 5 tan x)( 5 sec 2 x) 0, the derivative is x 1/3. At x f (1 f (1 1 2/3 x , which contradicts the given equation 3 d [ dx 5 tan x) At x f (x) 66. (a) d f (1 dx 2 f (0) 29 f( 1 . 3 d f (x) dx g(x) 2 (g(x) 1 5 2 2. 0, the derivative is (g(0) 1 . 10 (2)(1) 2)f (x) f (x) g (x) (g(x) 2)2 At x x) 2x f ( 1) f (1) 1, the derivative is 2 21 (e) f ( 1)g ( 1) 2)f (0) f (0)g (0) (g(0) 2)2 6. (3 2)( 2) ( 1)(4) ( 3 2)2 Chapter 3 Review d g(x dx (f) f (x)) g (x f (x))(1 At x dw ds f (x)) (x 1)[1 dw dr dr ds ( 2)] d [sin ( dr 1 cos ( r f (0))[1 (1)( 1) r 2) 0, we have r dw ds cos ( 4 cos 0 4 2)] 1 4 t 0, the particle first reaches the origin at 2 . The velocity is given by v(t) 1 10 sin t 4 is 10 sin speed at t 6 t 4 4 , so the velocity at 10, and the 2 is 4 by a(t) 6 10 10. The acceleration is given 10 cos t is 10 cos , so the acceleration at 4 0. 2 4 and so 6 71. (a) 8 cos 0 6 24 183 4 2 6 f (0)] d 8 sin s ds 8 cos s 8 sin 0 2) 8 cos (d) Since cos t 2r At s f (x)) f (x)) 0, the derivative is g (0 g (0 68. d dx g (x 3 ds dt d 2s dt 2 d (64t dt d (64 dt 16t 2) 64 32t) 32t 32 (b) The maximum height is reached when 69. Solving 2 t 1 for t, we have 1 t 2 1 2 d2 ( d 12 ( 3 7) 2(2 3( 2 dr dt 7) 2 4( 2 3( ) At t and so dr dt 2 0, we may solve 2(1)4(12 3(1 2 t 7) 2) 2 (b) s(0) 10 cos t 10 cos 4 5 4 2/3 2(8) 2/3 3 1, (b) Since v(t) 1 . 6 (980) , y(t) 12.3 sec. 393.8 ft. 4 sec. 7 280 cm/sec. 980t, the velocity is 4 = 560 cm/sec. Since a(t) 7 dV dx 73. 2 d dx 10 (20x 1 dv dt 980, the 1, we have s(t) 10. 1, we have s(t) 10. 3 x2 x 3 x2 x 40 d dx 13 x 3 x 2) 9x 10x 2 32 x 20 13 x 1600 (b) The marginal revenue is r (x) 4 ds dt 5.2t, the 64 5.2 acceleration is 980 cm/sec 2. Farthest right: When cos t 64 5.2 4 7 74. (a) r(x) 4 64 The average velocity is 160 (c) Farthest left: When cos t 2.6t 2) 490t 2, it takes 72. (a) Solving 160 7) 2) 0, which 64, so the velocity is 64 ft/sec. d (64t dt ) 70. (a) One possible answer: x(t) ds dt ds dt The maximum height is s 1 to obtain 2/3 0, ) 3 ds dt 2 sec. maximum height is reached at t 1 dr (2 dt 2/3 2 (d) Since dr d ( dt d (2 ) 3 (c) When t dr dt dt d 7)1/3 2/3 occurs at t , and we may write: dr d 125 9 3 (x 2 1600 3 (x 1600 3 x 10 160x 40)(x 32 x 1600 4800) 120), which is zero when x 40 or x 120. Since the bus holds only 60 people, we require 0 x 60. The marginal revenue is 0 when there are 40 people, and the 40 2 corresponding fare is p(40) 3 $4.00. 40 ...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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