Business Calc Homework w answers_Part_89

Business Calc Homework w answers_Part_89 - Cumulative...

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Unformatted text preview: Cumulative Review 74. (a) r(t) x(t) y(t) (b) ymax tmax (155 cos 18 11.7)ti (4 155 sin 18 t (155 cos 18 11.7)t 4 155 sin 18 t 16t 2 441 16t 2)j Cumulative Review Exercises (pp. 573576) 1. Since the function has no discontinuity at x 1 1 2(1) 12 1 12 2 1, the limit is (155 sin 18 )2 4 39.847 feet, reached at 2(32) 155 sin 18 1.497 sec 32 0. sin 3x 4x 1 x 1 (c) y(t) 0 when t 3.075 sec (found using the quadratic formula), and then x (155 cos 18 11.7)(3.075) 417.307 ft. (d) Solve y(t) t 25 using the quadratic formula: 1552 sin2 18 32 4(16)(21) 2. By l'Hpital's Rule, lim x0 lim x0 3 cos 3x 4 1 3 . 4 1)2 1 3. By l'Hpital's Rule, lim x0 x ex ex lim x0 (x 1 ex ex 1. 4. By l'Hpital's Rule, lim lim x 155 sin 18 0.534 and 2.460 seconds. At those times, x 72.406 and (155 cos 18 11.7)t equals ex ex x x x 1 x 1 lim 1. t(1 t cos t) sin t t cos t 2 sin t lim sin t t0 5. By l'Hpital's Rule, lim t0 333.867 feet from home plate. (e) Yes, the batter has hit a home run. When the ball is 380 feet from home plate (at t 2.800 seconds), it is approximately 12.673 feet off the ground and therefore clears the fence by at least two feet. 75. (a) r(t) (155 cos 18 4 32 (1 0.092 t sin t (1 cos t) 1 cos t t sin t 3 cos t lim 3 cos t t0 lim t0 ex ln (e x 1) 6. By l'Hpital's Rule, lim ln x x0 x0 e lim x0 (e x 1 x 1) 11.7) 1 (1 0.09 e 0.09t ) i lim xe x x 1 lim x0 xe x e x ex 1 ln (e x x x) 155 sin 18 (1 0.09 e 0.09t 0.09t ) 7. Use f (x) (e x x) x)1/x. Then ln f (x) lim x0 , and 0.09t e ) j e ) ) 0.09t x(t) y(t) (155 cos 18 4 1 (1 11.7) 0.09 ) ln (e x x x0 lim (e x 1)/(e x 1 x) 2. 155 sin 18 (1 0.09 32 (1 0.09t 0.092 e e 0.09t So lim (e x x0 x)1/x 1 1 sin x 1) cos x x cos x lim e ln f(x) x0 e 2. 1) sin x x sin x x 0.09t 8. lim x0 3x x (3x lim x0 (3x 1 (b) Plot y(t) and use the maximum function to find y 36.921 feet at t 1.404 seconds. (c) Plot y(t) and find that y (t) 0 at t 2.959, then plug this into the expression for x(t) to find x(2.959) 352.520 ft. (d) Plot y(t) and find that y (t) 30 at t 0.753 and 2.068 seconds. At those times, x 98.799 and 256.138 feet (from home plate). (e) No, the batter has not hit a home run. If the drag coefficient k is less than 0.011, the hit will be a home run. (This result can be found by trying different k-values until the parametrically plotted curve has y 10 for x 380.) 76. (a) BD E (b) AP E (c) AC E AD E AB E AB E AB E 1 1 BD = AB E E 2 2 1 AD E 2 1 AC. E 2 lim x0 3 sin x sin x lim x0 (3x 1) sin x 6 cos x x sin x 2 cos x 3 AD, so by part (b), AP E E 77. The widths between the successive turns are constant and are given by 2 a. 442 Cumulative Review 12 1 1 20. y f (1) 1 21. y 2(1 2 h2 h h) 2h 1 h (1 h h)2 2h h2 1 1 9. (a) 2(1) (b) 2 1 19. y 2x tan 3x 1 x x2 1 x2 3 (c) 1 [from (a) and (b)] (d) Yes, since lim f (x) (e) No. Left-hand derivative: lim h0 x1 4 x e e xx (x 3 3x 4 )e x csc x sin x 3 2 (1 csc x 1 cos x cos x)( csc x cot x) (1 cos x)2 f (1 h) h f (1) lim h0 3 csc 2 x (1 (1 cos x)4 (1 (1 3 csc2 x (1 cos x)4 3 csc2 x (1 cos x)4 csc x cot x cot x csc x csc 2 x cos x csc x cot x) cot 2 x) cot x csc x 2 cot 2 x) cot 2 x) lim h0 lim h0 csc 2 x lim h0 h 0 3 csc2 x (csc 2 x (1 cos x)4 3 (sin x)(1 cos x)4 2 cot x csc x 1 (1 Right-hand derivative: lim h0 cos x 2 cos2 x sin2 x cos x)(1 2 cos x) sin 2 x f (1 h) h f (1) lim h0 2 h h (1 h h) 1 3 (sin 2 x)(1 cos x)4 3(1 2 cos x) (sin 4 x)(1 cos x)3 lim h0 1 Since the left- and right-hand derivatives are not equal, f is not differentiable at x 1. 10. Solve 4 x2 0: all x 2 and x , 2x 2 22. y d dx 2 1 1 sin x 2 1 x 1 x2 d dx 2 tan 1 x 2. x 0, x 1 (2 x) 1 11. Horizontal: since as x 1 cos x Vertical: solve 2x 2 x while 0. 1 . 2 1, the end behavior at both ends is y 0 to find x 3 23. d [cos (xy) dx y2 y) ln x] 2yy d (0) dx 1 x ,x x 2 2 y 24. y sin (xy)(xy 1 x 0 12. One possible function is y 3 8 , x y sin (xy) 2y 1 2 x sin (xy) 1 d x 1/2 x 2 dx dy/dt dx/dt 1 xy sin (xy) 2xy x 2 sin (xy) x x x 1 x 2x 1 y x x [ 10, 10] by [ 4, 4] 25. 4 1 5 3 (x 2)2 1/2 dy dx cos t sin t cot t 13. f (5) 5 f (0) 0 (x 2)(1) (x 9 5 (x 2)2 26. ln y ln y 1 dy y dx dy dx ln[(cos x)x] x ln (cos x) x y 1 ( sin x) cos x 14. y 1)(1) 15. y 3 sin 2 sin ( 1 1 1 3x 3x 2 1 3x) (1 2 ln cos x 3x) ( 3) x sin x cos x x sin x x (cos x) ln (cos x) cos x ln (cos x) 16. y sin x sec x (sin x)(1 cos 2 x) cos 2 x tan x cos x sin x cos 2 x (cos x)x sin x 1 [cos x ln (cos x) x sin x] 27. By the Fundamental theorem of Calculus, y 1 x 3. x2 17. y 18. y 1 x 2 2 1 x (2x) 1) 2x x 2 1 28. y 1)e x 2 cos t 2x cos (x 2) 2 sin (2x) cos (2x); (e x )(2x (2x x y 2x sin (x 2) Cumulative Review d 2 (y dx d (sec x) dx 3 2 3 443 29. 2y) 2y 35. At t dy dx 3 :x 1, y , and 3 2 2yy y y sec x tan x, dy/dt dx/dt 3 3 2 3 2 3 3 2 2 3 3 cos ( /3) 2 sin ( /3) 3 2 . sec x tan x , 2y 2 (2y 2)(sec 3 x (2y sec x tan 2 x) 2y sec x tan x (2y 2)2 2 3 2) (sec x sec x tan 2 x) 2 sec 2 x tan 2 x (2y 2)3 (1 3)( 1) (2)(3) (1 3)2 (a) y y (x 3 1) or 0.866x 3.464 x 2 2 3 5 2 3 30. (1 v)u uv x 0 (1 v)2 dx dt dv dt 1 (b) y y (x 1) or 1.155x 1.443 x 31. (a) v a 3t 2 6t 12t 12 9 36. At t 1)(t 3) t 0; t 5 1 or r dy dx 4 :r 4 sec tan 2. 4 4 i i tan sec 2 4 4 j j 2i 2i j and 2j, so that (b) Solve v t 3. 0 for t: 3(t sec 2 2 (c) Right: v 0 for 0 t 1, 3 left: v 0 for 1 t 3 (d) a v 32. For x dy dx 0 at t 3(2)2 1, y 6(1) 1 1 ,y sin 3 3 6 3 6 2 2, and at that instant 12(2) 9 3 m/sec 1 and 12(1) 2(x 4 1) or y 1) or y cos cos 1 x 2 (a) y (b) y y 1 1 1 2 2(x 1 2 2) or y (x 2) or 0.707x C1, x C2, x C2 1. 2 2x 1 1.414x 1 x x 2x 2(3) 2 2 2x 3 2 (a) y (b) y 33. For x dy dx 1 37. With f (x) 3 f (x) x 2x C1 1 (x 2 3 , choose C1, C2 so that 3 3 3 3 3 3 3 6 and 3 6 1 2 3 6 3 4, x 5, x 3 . 3 . (a) y y 6 3 x 3 6 or 0.407x 0.950 [ 3, 6] by [ 1, 5] x 3 (b) y y 6 6 6 3 3 3 3 38. (a) x x x 3 6 x 2 2y 9 0, 2 0 2 0: 2, 2, 3 or 2.458x 2.050 (b) x (c) x 3 (d) Absolute maximum of 2 at x absolute minimum of 0 at x 1 34. (2x) 4 1 (2yy ) 9 0; y 3 2 9x . 4y 3 2 39. According to the Mean Value Theorem the driver's speed at some time was . 111 1.5 At x (a) y y (b) y 1, y 3 74 mph. , the slope is y 3 2 3 3 2 3 x 2 3 3 2 2 3 (x 3 1) or 0.866x 1) 1.155x 1.443 3.464 2 2 3 (x 5 3 2 or y x 444 Cumulative Review (e) Local (and absolute) maximum of approximately 3.079 at x 2 3 6 40. (a) Increasing in [ 0.7, 2] (where f 0), decreasing in [ 2, 0.7] (where f 0), and has a local minimum at x 0.7. (b) y 2x 2 3x 3 and x 2 3 6 ; 0 and local (and absolute) minimum of 0 at x at x (f) [ 3, 3] by [ 15, 10] 2 ( 1.042, 1.853) 1 and an absolute 2. 3, 3 , 2 (c) f (x) f (0) 41. f (x) f (x) 42. x2 x2 2 3 x 3 1: f (x) 3x 3x 3 2 x 3x C; choose C so that 2 2 3 3 2 x x 3x 1. 3 2 43. (a) f has an absolute maximum at x minimum at x 3. (b) f has a point of inflection at x 2: (c) The function f (x) cos x cos x C; choose C so that f (0) 1. 1 (x 1)2 2 7 x 2 2 1 2 x x 2 3 is one example of a function with the given properties. [ 2.35, 2.35] by [ 0.5, 3.5] f (x) is defined on [ 2, 2]. f (x) f (x) 2x 4 x 2 x3 4 x2 8x 4 3x 3 x2 [ 3.7, 5.7] by [ 3, 5] ; solve 44. y = 2 1 A(x) A (x) A (x) 4x 0 for x to find x 0, x 2 6 . 3 The graph of y f (x) is shown. 16 x2 , and the area of the rectangle for x 16 x2 1 x 16 x 2. 16 x2 2(8 x 2) 2 0 is x 16 x2 16 x2 , and so 0 when x = 2 2) 2 and y 2. The maximum possible area is A(2 [ 2.35, 2.35] by [ 10, 10] 8, with dimensions 4 45. f 2 by 2. 2 and f 2 4 (a) (b) 2, 2 6 2 6 , 0, 3 3 4 4 sec 2 x 2 4 4 tan or 4 2. The 2 6 2 6 ,0 , ,2 3 3 equation is y 0 for y 2 x Use NDER to plot f (x) and find that f (x) x 1.042. 1.414x 0.303 46. V s 3 dV 3s 2 ds Since ds 0.01s, the error of the volume calculation is approximately dV 3s 2(0.01s) 0.03s 3 0.03V, or 3%. [ 2.35, 2.35] by [ 15, 5] (c) Approximately ( 1.042, 1.042) (d) Approximately ( 2, 1.042), (1.042, 2) Cumulative Review 47. Let s be the rope length remaining and x be the horizontal distance from the dock. (a) x s2 52, ds dt 445 4 54. 2 x x 4 1 dx 1 2 x 4 1 dx 4 x x 1 12 5 1.5, and 8 ft, 8 64 25 5 5 8 64 s 25 s2 dx dt s s2 ds , 25 dt 7 1 dx. x 2 which means that for s speed (b) s dx dt 55. Let u 1.9 ft/sec Then ds , which for 25 dt ln x, so du dx x(ln x)2 2e e ( 1.5) u dx x(ln x)2 1 du 1 u C 1 ln x C. d s sec 1 , so dt 5 Therefore, 1 ln x 2e e 8 ft becomes ( 1.5) 0.15 rad/sec. 1 ln 2 1 1 1 ln 2 ln 2 0.409 48. (a) Let h be the level of the coffee in the pot, and let V be 56. the volume of the coffee in the pot. h dh V , so dt 16 dV/dt 16 9 16 (3 (3t 2t)i t 2)i ex 1 j dt t 3 0.179 in./min. (ln t)j 1 (ln 3)j e x dx. csc2 u cot u du (csc2 u cot u cot (e x 2 2i 2i (ln 3)j (b) Now let h be the level of the coffee in the cone, and let V be the volume of the coffee in the cone. V 1 h 2 h 3 2 dV 4 h 2 dt dV h , so dt 12 4 ( 9) 25 3 57. Let u 1, so du Use the identity cot2 u e cot (e x 2 x 1. 1) dx 4 36 25 h 2 dh dh and dt dt 1) du u 1) C (e x 1) C 0.458 in./min. Since dh is negative, the level in the cone is falling at dt Since 1 C is an arbitrary constant, we may redefine C cot (e x 1) ex C. the rate of about 0.458 in./min. 49. (a) (1)(0 (b) (1)(1.8 1 0 and write the solution as 58. Let u 1.8 6.4 6.4 12.6 ... ... 1 16.2) 0) 165 in. 165 in. 1 2 x 2 0 2 1 0 50. 2 x dx 2 x dx 0 x dx 1 2 x 2 2 1 2 ds s , so du . 2 2 ds 1 ds ds 2 2[(s/2)2 s2 4 4(s/2)2 4 1 1 s tan 1 u C tan 1 C 2 2 2 1] 1 2 du u2 1 2.5 59. Let u 2, 2 2 cos (x 3) dx 3) 3) 3) , so du ( u C 3 sin (x 1 2 u 2 3) dx. C 51. Using Number 29 in the Table of Integrals, with a 2 2 4 ( x dx ) 2 . 2 x 2 4 x 2 2 sin 1 x 2 sin (x cos3 (x ) du 1 2 cos2 (x Alternately, observe that the region under the curve and above the x-axis is a semicircle of radius 2, so the area is 1 (2)2 2 3 2 . 1 1 3 dx x x 3 26 9.765 3 3 52. 1 x2 ln 3 /4 ln x 1 9 ln 3 1 3 53. 0 sec2 x dx /4 tan x 0 1 ...
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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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