Business Calc Homework w answers_Part_65

Business Calc - Chapter 7 Review 82 36 The radius of a horizontal cross section is y 2 where 1 42 Use f(x 321 x 2/2 e 2 y 2 the y is distance below

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Unformatted text preview: Chapter 7 Review 82 36. The radius of a horizontal cross section is y 2, where 1 42. Use f (x) 321 x 2/2 e . 2 y 2), the y is distance below the rim. The area is (64 1 (a) y )( y) y. The total work is y ) dy 0.04 3 (64y 2 y ) dy 2 36 320y(2 y) dy 320 ( y 0 2 13 y 3 1280 3 23 ft, 3.5 in. 48 38. 5.75 in. f (x) dx 1 A shell has radius x and height 2x x 2 4y, and the [ 1, 3] by [ 1, 3] 2y) dy 2 y2 0 and 44. 0 320 0.9973 (99.73%) 43. Because f (x) 113.097 in.-lb. 2 f (x) dx 3 y)4y y. The total force is 2 0.9545 (95.45%) 3 2 37. The width of a thin horizontal strip is 2(2y) force against it is 80(2 f (x) dx (c) 8 14 y 4 32y 2 0.04 (b) 2 8 2 0.04 y(64 0.6827 (68.27%). 2 2 the rim is 0.04 (64 8 f (x) dx evaluates, using NINT, to 1 y 2) y, and the work to lift it over weight is 0.04 (64 1 volume is 2 (x) 0 0 3 x dx 2 1 x3 3 x. The total 2 . 0 45. 426.67 lb. 5 ft. 6 7 ft, and 10 in. 24 For the base, Force 57 23 48 [ 3, 3] by [ 3, 3] 5 6 7 24 6.6385 lb. 5/6 Force 0 2 5/6 399 1 2 y 24 2 7 57 y dy 24 5/6 57 0 23 y dy 48 39. A square’s height is y y2 ( 0 1/2 2 1 dx x 2 dx 2 2x 1/2 3. 1/2 46. 5/6 1311 1 2 y 48 2 9.4835 lb. 0 x)2, and its area is (6 x)4. The volume is 6 2 (x) 5.7726 lb. For the sides, Force 1 x A shell has radius x and height . The total volume is For the front and back, 6 ( 6 x)4 dx, 2 , 3 2 by [ 2, 2] 0 A shell has radius x and height sin x. The total volume is which using NINT evaluates to exactly 14.4. 2 (x)(sin x) dx 40. Choose 50 cm as a conveniently large upper limit. 50 20 1 e (x 17.2)2/(2 3.42) x cos x 2 2 . 0 47. dx, evaluates, 0.2051 (20.5%). [ 1, 4] by [ 4, 1] 41. Answers will vary. Find , then use the fact that 68% of the class is within of to find , and then choose a conveniently large number b and calculate 10 sin x 3.4 2 using NINT to b 2 0 1 2 e (x )2/(2 2 ) dx. The curves intersect at x x and height x 3 (x 2 1 and x 3. A shell has radius 3x) 2 x 4x 3. The total volume is 3 2 (x)( x 2 3 4x 3) dx ( x3 2 1 4x 2 3x) dx 14 x 4 43 x 3 32 x 2 1 2 16 . 3 3 1 322 Section 8.1 48. Use the intersect function on a graphing calculator to determine that the curves intersect at x has radius x and height 3 1x x 2 2 3 10 1.8933. A shell that l’Hôpital’s Rule does not say that lim . The volume, which lim 0 evaluates to 9.7717. 5 (x 4 49. (a) y . x2 3 dx, which using NINT 10 x2 2 (x) 31 x→0 y2 y1 x→0 y2 is calculated using the right half of the area, is 1.8933 x cos x sin x . The graphs of y3 and y5 clearly show x2 y1 3. y5 2)(x 2) 52 x 4 5 [ 3, 3] by [ 2, 2] Quick Review 8.1 (b) Revolve about the line x 4, using cylindrical shells. 1. x A shell has radius 4 52 x . The total 4 x and height 5 1 2 2 (4 2 10 2 320 3 1 100 x 12 x 2 4x 2 dL dx 1 x f (x) must equal 1 x 2 2 f (x) x 1 x2 2. 12 x 4 51. y 12 x 4 1 ln x 2 3 4 sec x, so the area is 2 (tan x) 2 2.7183 1 2.7183 2.7183 2.7183 0.00001 2.7183 As x→0 , x1/(ln x) approaches 2.7183. 2 (sec x) dx, 0 which using NINT evaluates to x1/(ln x) 0.0001 3 . The 4 2 ln x 3 . 4 x2 x 0.001 C, and the /4 2 0.1 x approaches 1.1052. x 0.01 1 ln x 2 requirement to pass through (1, 1) means that C function is f (x) 1.1052 0.1 ( f (x))2, and 1 . Then f (x) 2x 1.1052 As x→ , 1 ( f (x))2, 1 1.1052 10,000 2 1.1051 1,000,000 4 dx 335.1032 in3. ( f (x))2 f (x) x 2 13 x 3 14 x 16 10 50. Since 13 x 4 1.1046 1000 52 x dx 4 x) 5 1.1000 10 volume is 2 0.1 x x 1 3. x 3.84. 1 1x x 1 1 52. x and x y 2 1 2 1 y 1 0.5 0.1 1 , so the area is y2 12 dy, y2 0.78679 Chapter 8 L’Hôpital’s Rule, Improper Integrals, and Partial Fractions 0.99312 0.0001 5.02. 0.95490 0.99908 0.00001 which using NINT evaluates to 0.01 0.001 0.99988 0.000001 0.99999 1x approaches 1. x As x→0 , 1 4. 1x x x s Section 8.1 L’Hôpital’s Rule (pp. 417–425) Exploration 1 Exploring L’Hôpitals Rule Graphically 1. lim x→0 sin x x lim x→0 cos x 1 1 1.1 13.981 2. The two graphs suggest that lim y1 x→0 y2 lim y1 x→0 y2 . 105.77 1007.9 1.0001 1 1.01 1.001 10010 As x→ 1 , 1 1x goes to x . is equal to Section 8.1 5. 323 3. [0, 2] by [0, 3] t As t→1, 1 t [0, 3] by [0, 3] From the graph, the limit appears to be 1. The limit leads to the indeterminate form 0. approaches 2. 1 6. ln 1 ln 1 [0, 500] by [0, 3] lim As x→ , 1 x x ln 1 1 lim 1 x2 1 lim x→0 lim 1 x2 1/x x→0 7. 1 x 1 x x→0 x [ 5, 5] by [ 1, 4] 1 x 1 1 x 1 x x→0 4x 2 1 approaches 2. x1 1 x ln 1 1x x 0 1 Therefore, sin 3x As x→0, approaches 3. x 1x x lim 1 x→0 8. lim f (x) x→0 lim e ln f (x) e0 x→0 1. 4. [0, ] by [ 1, 2] tan →, 2 2 tan As [0, 1000] by [0, 1] approaches 1. From the graph, the limit appears to be about 0.714. 5x 2 2 x→ 7x lim 1 sin h h 9. y 10. y lim x→ x3 1 4x 3 x 3 x→1 h)1/h (1 3x 1 10x 3 14x lim x→ 3x 2 12x 2 1 x→1 5. lim lim 10 14 5 7 3 11 Section 8.1 Exercises 1. [0, 2] by [0, 1] [0, 2] by [0, 1] 1 4 The graph supports the answer. From the graph, the limit appears to be . x 2 x→2 x lim 2 4 1 x→2 2x lim 1 4 6. lim x→0 1 cos x x2 lim x→0 sin x 2x lim x→0 2. [ 2, 2] by [ 2, 6] [ 5, 5] by [ 1, 1] From the graph, the limit appears to be 5. lim x→0 sin 5x x lim x→0 5 cos 5x 1 5 The graph supports the answer. cos x 2 1 2 0.71429 324 Section 8.1 7. The limit leads to the indeterminate form 1 . ln(e x Let ln f (x) ln(e x x x→0 lim lim (e x x) ex ex lim 1 x x) . t ex x x→0 e 1 x lim 1 x→0 x)1/x x→0 ln(e x x x)1/x 2 1 lim e ln f (x) lim f (x) x→0 1 t 1 sin t 12. Let f (t) 2 e2 x→0 100 f (t) 0.1884 t sin t . t sin t 1 10 10 0.0167 2 10 0.0017 3 0.00017 Estimate the limit to be 0. 1 t 1 lim t→0 sin t lim t t→0 lim t→0 lim t→0 [0, 5] by [0, 10] 13. Let f (x) sin t t sin t 1 cos t t cos t sin t sin t t sin t cos t 0 cos t x)1/x. (1 The graph supports the answer. 8. lim x→ 3x 2x x3 x 1 3 1 4x 3x 2 lim x→ lim x→ 4 6x 0 102 10 x 2 103 104 105 f (x) 1.2710 1.0472 1.0069 1.0009 1.0001 Estimate the limit to be 1. ln (1 x) 1 x 1x ln (1 x) lim lim 1 x x→ x→ ln f (x) [ 5, 25] by [ 1, 2] The graph supports the answer. 9. (a) 102 10 x x)1/x lim (1 x→ 103 104 105 0 1 0 lim e ln f (x) e0 102 103 104 105 0.6525 0.6652 0.6665 0.6667 lim f (x) x→ x→ 1 x 2x2 . 3x2 5x 14. Let f (x) f (x) 1.1513 0.2303 0.0345 0.00461 0.00058 10 x Estimate the limit to be 0. ln x 5 x x→ 5 ln x x x→ 10. (a) 100 x 5/x 1 x→ lim (b) lim 1 10 10 2 f (x) 0 1 lim 10 3 10 4 f (x) 0.1585 0.1666 0.1667 0.1667 0.1667 1 6 Estimate the limit to be . (b) lim x→0 x sin x x3 0.5429 0 x 2x 2 2 5x x→ 3x lim 15. lim sin 2 2 cos 1 →0 x→0 lim x→0 1 16. lim → /2 1 sin cos 2 1 6 sin 3 . sin 4 17. lim 10 f( ) 0 10 1 10 3 4 sin 3 →0 sin 4 2 10 3 10 4 0.1865 0.7589 0.7501 0.7500 0.7500 Estimate the limit to be . lim lim 3 cos 3 →0 4 cos 4 4 6 (2)(0) cos (0)2 2 . 3 0 1 lim x→0 lim 11. Let f ( ) 2 lim →0 cos x 3x 2 sin x 6x cos x 6 2 . 3 1 4x lim lim 5 x→ x→ 6x Estimate the limit to be 3 4 cos t t t t→0 e 1 1 cos 2 sin 2 sin lim → /2 4 cos 2 sin /2 4 cos 1 4 lim → /2 lim t t→0 e sin t 1 lim t→0 c os t et 1 Section 8.1 18. lim t t→1 ln t 1 sin t 1 lim t→1 1 cos t t y→ /2 1 1 22. lim y tan y 2 ( 1) lim x→ 2 x ( 1)(1) (1) lim ln 2 x→0 x→0 ln 2 log3 (x 3) lim x→ lim (x lim lim x→0 tan 1 24. lim x tan x x→ lim x→ ln 3 ln 2 ln (y 2 2y) ln y y→0 21. lim x x→0 x ln 3 3 ln 3 x ln 2 lim x→ y→0 lim y→0 2 2y 1 1 sec2 x2 x 1 x2 lim sec2 1 x sec2 0 1 y(2y 2) y 2 2y 4y y→0 2y 4(0) 2(0) 1 x x→ 1 y 2y 2 lim 2 y→0 y lim 0 1 x lim 2y y2 1 x2 x lim 3) ln 3 x ln 2 ln 3 x→ ln 2 2 1 x lim lim ( 1) sin 2 1 x 1 x2 x→0 1 x ln 2 1 (x 3) ln 3 x→ x→ cos lim ln x 23. lim x ln x x→ x→ 2 sin 1 x ln 2 ( 1) sin y sin y 1 x ln 2 1 x→ x 20. lim y cos y y→ /2 lim log2 x 2 lim 1 y sin y cos y y→ /2 1 +1 ln (x 1) 19. lim log2 x x→ 2 lim 2 2 2y 2y 2 2 25. lim (csc x cot x x→0 2 2 1 lim x→0 lim x→0 lim x→0 1 sin x 1 cos x sin x cos x) cos x cos x cos x sin x sin x sin x cos x cos x cos x sin x sin x 1 2 325 ...
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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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