Business Calc Homework w answers_Part_20

Business Calc Homework w answers_Part_20 - 96 Section 3.6...

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Unformatted text preview: 96 Section 3.6 38. f (u) u 2 u u 2 u g (x) du du u 12 1 (u 1 1 1 (u 1 1) ( f g) ( 1) d (u du 1du 1 du u 1) (u 1) 1 1 d (u du 1) 3 2x f (g( 1))g ( 1) dy dt dx dt 2 cos t 2 cos t 2 sin t cot t and has slope cot 2, or y x dx dt d (sin 2 t) dt (cos 2 t) (2 t) dy dt 42. 2) 6 sin u d (cos 2 t) dt dy dt dx dt 2 sin 2 t 2 cos 2 t and has slope tan 1) y 2 43. 6 sin 2u dy 40. (a) dx dx dt d (sec 2 t dt 2 , cos 31 , 2 2 6 2 3. Its equation is 6 1 , or y 2 3x 2. d dt (2 sec t) (sec t) (2 sec t)(sec t tan t) 2) dy dt 1)(1) 2x 2x cos (x 2 d tan t dt dy dx d 1) (x 2) dx 2x cos (u dy dt dx dt 1) sec 2 y 1) 1 cot t. 2 1, tan 4 1 cot 2 1 (x 2 1) 44. 2 sec2 t 2 sec2 t tan t slope (cos u)(2x) 2x cos u sec 2 t The line passes through 1) dy du du dx d d (sin u) (x 2 du dx 2x cos (x 6 2 sec 2 t tan t dy du du dx d sin (u du cos (u dy (b) dx 1) 2 sin 2 t tan 2 t 2 3 3x ( sin 2u)(2) (3) 2 cos 2 t d dt 2) dy du du dx d d (cos 2u) (3x du dx 6 sin (6x 2) 2. ( sin 2 t) (2 t) d dt The line passes through sin dy dx 2 (x dy dx dy du du dx d d (cos u) (6x du dx 6 sin (6x ( 2, 4 8 ( sin u)(6) (b) , 2 sin 1. Its equation is 4 2) 4 y ( 4)(2) dy dx d (2 sin t) dt 2 sin t The line passes through 2 cos f (0)g ( 1) 39. (a) d (2 cos t) dt dy dx 1) dx dt dy dt 41. (u 1)2 (u 1) 4(u 1) 1)2 (u 1)3 1) (u d (x 2 dx u u 2 dx dt dy dt d sec t dt d tan t dt dy dx dy dt dx dt 1, or y 2x 1 . 2 sec 2 t sec t tan t The line passes through sec or y 1 x 2 sec t tan t sec 2 t sec t tan t has slope csc 1) and has 1 . Its equation is 2 4 1) (1, 4 1 sin t 6 , tan csc t 2 6 2. Its equation is y 6 1 , 3 2x 2 1 3 3. and 3 , 3 Section 3.6 dx dt d t dt dy dt d dt dy dx 45. dy dt dx dt 1 t 1/(2 t) 1 1 2 y 1 4 (b) 11 , and has slope 42 1 4 1x dx dt dy dt dy dt dx dt 1 , or 2 (2t 4t 4t 3 Then dy dx dy dt dx dt 1 sin t) 1 cos t) 1 y 1)(sin t) 2 cos t (2t 1)2 du dx du , so dx dt dx du dt dx . Therefore, dt d dy dt dx (2t dx dt 1)(sin t) 2 cos t (2t 1)2 1)(sin t) 2 cos t (2t 1)3 cos t sin t (2t 1) (d) The expression in part (c). sin t cos t 50. Since the radius passes through (0, 0) and (2 cos t, 2 sin t), it has slope given by tan t. But the slope of the tangent is sin 3 ,1 cos dy dt dx dt dy dx 3 31 , and has slope 2 3 2 cos t 2 sin t cot t, which is the negative reciprocal of tan t. This means that the radius and the 2 tangent are perpendicular. (The preceding argument breaks 3 cos du dt d dy dx dx The line passes through sin 1) (2t d (t dt d (1 dt 1) 2 dy . dx (c) Let u t2 d dt (cos t) (2t 1)( sin t) (cos t)(2) (2t 1)2 (2t 3) 4t 3 4t dx dt dy dt 3 d cos t dt 2t 1 d (2t 1) (cos t) dt The line passes through (2( 1)2 3, ( 1)4) (5, 1) and has slope ( 1)2 = 1. Its equation is y 1(x 5) 1, or y x 4. 47. d dy dt dx (2t d (2t 2 dt d4 (t ) dt 1 cos t 2t 1 1 . 4 x dy dx 46. dy dt dx dt t 1. Its equation is y 1 4 d2 (t t) 2t dt d sin t cos t dt t 2 1 The line passes through , 4 1 2 dx dt dy dt dy dx 49. (a) 1 3. Its equation is down when t 3 1 , or 2 3 3x 3 2 k , where k is an integer. At these values, 2 either the radius is horizontal and the tangent is vertical or the radius is vertical and the tangent is horizontal, so the y 3x 2 . result still holds.) 3 48. 97 dx dt dy dt d cos t sin t dt d (1 sin t) cos t dt dy dx dy dt dx dt cos t sin t cot 2 ds dt ds d d dt ( sin ) When cot t The line passes through cos has slope 51. 2 ,1 sin 0. Its equation is y (0, 2) and 2 2. 52. dy dt d d (cos ) dt d d dt 3 d and 2 dt dy dx dx dt (2x When x 7) d2 (x dx 5, ds dt 7x 5) sin 3 (5) 2 5. dx dt dx dt 1 and dy dt 1 dy , 3 dt [2(1) 7] 1 3 3. 98 53. 54. Section 3.6 dy dx 1 x cos 2 2 1 x Since the range of the function f(x) cos is 2 2 dy 1 the largest possible value of is . dx 2 d x sin dx 2 dy dx xdx 2 dx 2 cos d (sin mx) dx d dx (cos mx) (mx) The desired line has slope y (0) (e) 11 ,, 22 through (0, 0), so its equation is y At x f (g(x))g (x) 2, the derivative is f (g(2))g (2) m cos mx m cos 0 d f (g(x)) dx (f) 1 ( 3) 3 f (2)g (2) 1 d dx f (x) f (x) d f (x) f (x) dx 2 At x f (x ) 2 m and passes mx. 2, the derivative is f (2) 55. dy dx x d 2 tan 4 dx x sec 2 2 4 y (1) 2 sec 2 2 sec 2 (g) 4 2 2)2 ( The tangent line has slope 1, 2 tan y . x (x 1) (h) 1 equation is y 12 2 d1 dx g 2(x) d [g(x)] 2 dx 2(5) ( 4)3 1) 2, or y x f 2(x) g 2(x) 2. g 2(x) 2, the derivative is 2 f (2) At x (c) g(x)] g (x) g(x)] f (x)g (x) g (3) 2 g(x) f (x) 3, the derivative is f (3)g (3) g(3) f (3) 5. 5 f (1) (b) 15 ( 4)(2 ) 8. 10 3 2 At x g(x) f (x) f (x)g (x) [g(x)]2 2, the derivative is g(2) f (2) f (2)g (2) [g(2)]2 74 3 4 37 . 6 (2) 1 3 (8)( 3) (2)2 22 3 17 5 f (x) g (x) 1, the derivative is g (1) d f (x)g 3(x) dx 1 3 5 8 3 d g 3(x) f (x) dx 3 f (x)g 2(x)g (x) 1. d dx f (x) g 3(x) d dx At x (2)( 3) 5 17 g(x)] g 3(x) f (x) g 3(x) f (x) 0, the derivative is 3 f (0)g 2(0)g (0) 3(1)(1)2 d f (x) (d) dx g(x) d dx 2g(x) g(x)] 82 f (x)[3g 2(x)] g(x) (3)(5) 1 3 g 2(2) d [5 f (x) dx At x 3, the derivative is f (3) d [ f (x) dx At x f (x) (8) g(2)g (2) 68 57. (a) d f (x) dx g 2(x)] g 2(x) 10 3 2 f (x) 2 . 3 d2 [ f (x) g 2(x) dx 2, the derivative is f 2(2) 1 2 3 2g (x) [g(x)]3 g(x) g(x)g (x) f 2(x) [ 4.7, 4.7] by [ 3.1, 3.1] d (b) [ f (x) dx [2f (x) 2 f (x) f (2) f (2) At x 2 f 2(x) 1 At x d [2 f (x)] dx dx 5 . 32 1 d dx 2 Graphical support: 56. (a) 10 64 f (x) f (x) (x 1 3d 2[g(x)] 3, the derivative is and passes through (1, 2). Its 1 . 2, or 2. 1 1 6(2 2) 2g (3) [g(3)]3 and passes through The normal line has slope 1 28 At x (1, 2). Its equation is y 4 1 3 2 f (2) xd x 4 dx 4 1. 1 3 (1)3(5) 6. g 3(0) f (0) 99 Section 3.6 [g(x) d d 1] f (x) f (x) [g(x) dx d f (x) (c) dx g(x) 1 1]2 [g(x) [g(x) 58. For y 1] d dx sin 2x, y (cos 2x) (2x) dx 2 cos 2x and the slope at the origin is 2. 1] f (x) f (x)g (x) [g(x) 1]2 For y x 2 sin , y cos 1 x cos and the 2 2 xdx 2 dx 2 1 . Since the slopes of the two 2 1 tangent lines are 2 and , the lines are perpendicular and 2 slope at the origin is At x 1, the derivative is [g(1) 1] f (1) f (1)g (1) [g(1) 1]2 (4 At x and y 1 3 1 3 8 (5) 3 g (1) f (0) f (x)] 2 2[g(x) f (x)] 59. Because the symbols 40 3 3d dx 2[g (1) f (1)] [g(1) f (1)]3 8 3 (4 1 3 values. [g(x) 6 1 3)3 d dx 6. 60. Velocity: s (t) 2 bA sin (2 bt) acceleration: s (t) 4 2b 2A cos (2 bt) 33 jerk: s (t) 8 b A sin (2 bt) The velocity, amplitude, and jerk are proportional to b, b 2, and b 3, respectively. If the frequency b is doubled, then the amplitude of the velocity is doubled, the amplitude of the acceleration is quadrupled, and the amplitude of the jerk is multiplied by 8. d 2 37 sin (x dt 365 61. (a) y (t) d (25) dt 101) f (x g(x))(1 g (x)) 37 cos 2 (x 365 101) d2 (x dx 365 At x 0, the derivative is 37 cos 2 (x 365 101) 2 365 f (0 g(0))(1 g (0)) f (0 1) 1 4 f (1) 3 14 33 g(x)] f (x)] g(x))] f (x g(x)) [x dy dy du , , and are not fractions. The dx du dx individual symbols dy, dx, and du do not have numerical 1, the derivative is 2 d [ f (x dx 1 x is shown. 2 [ 4.7, 4.7] by [ 3.1, 3.1] g ( f (x)) f (x) 2[g (x) f (x)] [g(x) f (x)]3 (g) 2x 1 . 9 0, the derivative is d [g(x) dx At x the curves are orthogonal. A graph of the two curves along with the tangents y 1. f (1)g (0) g ( f (0)) f (0) (f) 1)2 0, the derivative is d g( f (x)) dx At x 8 3 f (g(x))g (x) f (g(0))g (0) (e) (3) (4 9 9 d (d) f (g(x)) dx 1 3 1) 1 3 74 2 cos (x 365 365 101) Since cos u is greatest when u 4 . 9 y (t) is greatest when x 101) 2 (x 365 0, 101) 2 , and so on, 0, or 101. The temperature is increasing the fastest on day 101 (April 11). (b) The rate of increase is y (101) 74 365 0.637 degrees per day. 0 100 Section 3.6 d dt 62. Velocity: s (t) 1 1 4t 4 2 At t d (1 4t dt 21 66. dT du dT dL dL du 4t 1 1 2 m/sec 5 4(6) 1 2 4t 2 6, the velocity is 1 4t d dt (1 4t) (2) ( 21 1 d dt 1 4t 4t)2 1 d (1 4t dt 1 2 2 4t) 4t y 4t 1 4t 4 4t)3/2 (1 L g dL (kL) dL g 1 (kL) g k d f (g(x)) dx 6, the acceleration is f (g(x)) has a horizontal tangent at x f (g(1))g (1) dv dt 63. Acceleration k 2 (k 4 4(6)]3/2 [1 dv ds ds dt dv (v) ds 4 m/sec 2 125 d (k ds 0, so either g (1) s tangent at x 1, or the graph of y tangent at u 1, then 0 or f (g(1)) 0. This g(x) has a horizontal f (u) has a horizontal g(1). s) (k s) k2 , a constant. 2 s) kT 2 f (g(x))g (x). If the graph of means that either the graph of y At t L g 67. No, this does not contradict the Chain Rule. The Chain Rule states that if two functions are differentiable at the appropriate points, then their composite must also be differentiable. It does not say: If a composite is differentiable, then the functions which make up the composite must all be differentiable. 68. Yes. Note that 4 1 2 L g 2 d dt L (kL) g d 2 dL 2 1 Acceleration: s (t) 4t) 69. For h 1: 64. Note that this Exercise concerns itself with the slowing down caused by the earth’s atmosphere, not the acceleration [ 2, 3.5] by [ 3, 3] caused by gravity. For h Given: v 0.5: k s Acceleration dv dt dv ds ds dt k dv (v) ds (v) dv ds dk s ds s d [ 2, 3.5] by [ 3, 3] s (k) k ds k d ds For h 0.2: s ( s)2 s k (2 s) k s s k2 2s 2 [ 2, 3.5] by [ 3, 3] As h → 0, the second curve (the difference quotient) ,s 0 Thus, the acceleration is inversely proportional to s 2. 65. Acceleration dv dt df (x) dt df (x) dx dx dt f (x) f (x) approaches the first (y 2 cos 2x). This is because 2 cos 2x is the derivative of sin 2x, and the second curve is the difference quotient used to define the derivative of sin 2x. As h → 0, the difference quotient expression should be approaching the derivative. ...
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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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