Business Calc Homework w answers_Part_64

Business Calc Homework w answers_Part_64 - 316 Section 7.5...

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Unformatted text preview: 316 Section 7.5 (d) 0 if we assume a continuous distribution. Between 23. The work to raise a thin disk is r 2(56)h ( 102 56 (12 y 2)2(56)(10 2 59.5 in. and 60.5 in., the proportion is y) y 60.5 y 2) y. The total work is y)(100 f (x) dx 56 (12 y 2) dy, which evaluates using NINT y)(100 0 to 0.071 (7.1%) 59.5 10 1 28. Use f (x) (x 498)2/(2 1002) e 100 2 967,611 ft-lb. This will come to 500 (967,611)($0.005) (a) $4838, so yes, there’s enough money f (x) dx 0.34 (34%) 400 to hire the firm. (b) Take 1000 as a conveniently high upper limit: 1000 24. Pipe radius 1 ft; 6 700 360 12 (62.4)y dy 6 Work to fill pipe 0 385 0.217(300) 58,110,000 ft-lb. 30. The proportion of lightbulbs that last between 100 and 800 hours. 58,222,320 ft-lb, which will take 35,780,000 31. 58,222,320 1650 110,855 sec 6.5 people 29. Integration is a good approximation to the area (which represents the probability), since the area is a kind of Riemann sum. 360 Total work 0.217, which means about 112,320 ft-lb. (10)2(62.4)y dy Work to fill tank f (x) dx 6,370,000 31 hr. M 25. (a) The pressure at depth y is 62.4y, and the area of a thin 1000MG dr r2 1024, G 5.975 1 r 1000 MG 6.6726 10 35,780,000 , which for 6,370,000 11 evaluates to 1010 J. 5.1446 11 ft, so horizontal strip is 2 y. The depth of water is 6 32. (a) The distance goes from 2 m to 1 m. The work by an the total force on an end is 11/6 (62.4y)(2 dy) external force equals the work done by repulsion in 209.73 lb. 0 moving the electrons from a 1-m distance to a 2-m (b) On the sides, which are twice as long as the ends, the initial total force is doubled to 419.47 lb. When the tank is upended, the depth is doubled to 11/3 force on a side becomes 11 ft, and the 3 (62.4y)(2 dy) distance: 2 Work 23 dr 2 1 r 23 838.93 lb, 29 10 10 0 1.15 which means that the fluid force doubles. 26. 3.75 in. 5 ft, and 7.75 in. 16 31 ft. 48 31/48 Force on a side p dA (64.5y) 0 5 dy 16 4.2 lb. 27. (a) 0.5 (50%), since half of a normal distribution lies below the mean. 65 (b) Use NINT to find f (x) dx, where 63 f (x) 1 e 2 2 (x 63.4) /(2 3.2 ) . The result is 3.2 2 0.24 (24%). (c) 6 ft 72 in. Pick 82 in. as a conveniently high upper 82 limit and with NINT, find f (x) dx. The result is 72 0.0036 (0.36%). 28 10 2 1 r 29 1 J (b) Again, find the work done by the fixed electrons in pushing the third one away. The total work is the sum of the work by each fixed electron. The changes in distance are 4 m to 6 m and 2 m to 4 m, respectively. 6 Work 23 10 r2 4 23 10 7.6667 4 29 dr 29 10 1 r 29 6 4 J. 2 23 10 r2 4 1 r2 29 dr 317 Chapter 7 Review 33. F dv dt m x 2 W x 1 x 2 x 1 v 2 mv s Chapter 7 Review Exercises dv , so dx (pp. 413–415) F(x) dx 5 1. 0.2t 3) dt 0 13 t 3 mv dv 7 v 1 2. 1 mv 2 22 (t 2 0 dv dx dx mv 5 v (t) dt 0.001t 4) dt (4 0 1 mv 2 21 10.417 ft 0 7 c(t) dt 5 0.05t 4 0 0.0002t 5 4t 7 31.361 gal 0 34. Work Change in kinetic energy 12 mv . 2 100 3. 0.3125 lb 0.009765625 slug, and 32 ft/sec2 5280 ft 1 hr 90 mph 90 132 ft/sec, so 1 mi 3600 sec 1 Work change in kinetic energy (0.009765625)(132)2 2 0 1.6 oz 1 (0.003125)(280)2 2 Work 37. 2 oz 2 2 4. (x) dx (11 0 so Work 24 Work 39. 6.5 oz Work cos t 12 sin t 12 0 12 dt 24 14,400 0 6. 122.5 ft-lb. [ 1, 3] by [ 1, 2] The curves intersect at x 2 x dx 0.02832 slug, so 0 1 12 x 2 1 dx x2 1 2 109.7 ft-lb. 1 lb /(32 ft/sec2) 16 oz 1 (0.01270)(132)2 2 14 g 0.003125 slug, so 1 lb 14.5 oz /(32 ft/sec2) 16 oz 6.5 oz 2 0 300 2 0 1 lb 1 /(32 ft/sec2) slug, and 16 oz 256 5280 ft 1 hr 124 mph 181.867 ft/sec, 1 mi 3600 sec 11 (181.867)2 64.6 ft-lb. 2 256 1 (0.02832)(88)2 2 2x 2 11x 24 E(t) dt 5. 1 38. 14.5 oz 4x) dx 0 2 oz 124 mph 1464 0 300 2t 1 lb /(32 ft/sec2) 16 oz 100 33.333e0.03x 21x 85.1 ft-lb. 36. 1.6 oz e0.03x) dx (21 0 2 oz 1/8 lb 1 m slug, so 32 ft/sec2 32 ft/sec2 256 11 Work (160)2 50 ft-lb. 2 256 35. 0.3125 lb 100 B(x) dx 0.01270 slug, so 1. The area is 1 1 x 0 1 2 2 1 1 1. 7. 110.6 ft-lb. 1 1 lb slug. Compression energy of spring 8 256 12 1 12 ks (18) 0.5625 ft-lb, and final height is 2 2 4 0.5625 given by mgh 0.5625 ft-lb, so h 4.5 ft. (1/256)(32) 40. 2 oz [ 4, 4] by [ 4, 4] The curves intersect at x 1 [3 x2 2 and x 1 (x 1)] dx 2 ( x2 1. The area is x 2) dx 12 x 2 2x 2 13 x 3 1 3 9 . 2 1 2 2 1 2 8 3 2 4 318 Chapter 7 Review 8. x y 1 implies y x)2 (1 1 2 x x. 12. 2 [ 0.5, 2] by [ 0.5, 1] 1 The area is (1 2 x x) dx 4 3/2 x 3 x 0 12 x 2 1 . 6 by [ 3, 3] The area is 1 (2 sin x 0 sin 2x) dx 2 cos x 0 1 cos 2x 2 4. 0 13. x . 2 2y 2 implies y 9. x 3 2 , [ 5, 5] by [ 5, 5] The curves intersect at x [ 1, 19] by [ 1, 4] 2.1281 (4 x 2.1281 The curves intersect at x 18 x dx 2 3 0 3 or 0 y2 10. 4x 1 y 4 x 18 4 x 3/2 3x 23 y 3 2y 2 dy 18. The area is 32 2 2.1281. The area is cos x) dx, which using NINT evaluates to 18, 8.9023. 14. 0 3 18. 0 12 y 4 4 implies x 1, and 4x y 16 implies [ 4, 4] by [ 4, 4] 4. The curves intersect at x 0.8256 (3 0.8256. The area is sec2 x) dx, x 0.8256 which using NINT evaluates to 2.1043. [ 6, 6] by [ 6, 6] 15. Solve 1 The curves intersect at (3, cos x 2 cos x for the x-values at the two 4) and (5.25, 5). The area is ends of the region: x 5 1 y 4 4 5 12 y 4 4 12 y 4 4 12 y 8 425 24 38 3 3 , i.e., symmetry of the area: dy 7 /3 2 1 y 4 13 y 12 1 2 [(1 2 5 dy cos x) 2 cos x)] dx (2 cos x 1) dx 2 5 5y 2 2 sin x x 4 243 8 (2 7 /3 2 30.375. 2 3 3 7 /3 2 1.370. 5 /3 11. 16. [(2 /3 cos x) (1 5 /3 (1 2 cos x) dx /3 x [ 0.1, 1] by [ 0.1, 1] /4 The area is (x 0 sin x) dx 2 /4 12 x 2 cos x 2 2 32 0 2 0.0155. 1 2 sin x 3 4 3 5 /3 /3 7.653 cos x)] dx 5 7 or . Use the 3 3 Chapter 7 Review 17. Solve x 3 x x x x2 to find the intersection points at 1 (c) Use cylindrical shells. A shell has radius 4 height 2 x 21/4. Then use the area’s symmetry: 0 and x 2 (4 0 2 0 (x3 1 1 ln (x 2 2 2 x)(2 x x) dx (8 4x 2x 3/2 x 2) dx 2x 2 x 4 5/2 x 5 13 x 3 0 x x2 x. The total volume is 4 1/4 2 ln ( 2 x) dx 14 x 4 1) 1) 2 1 16 3/2 x 3 2 12 x 2 x and 4 the area is 2 21/4 4 64 . 5 0 (d) Use cylindrical shells. A shell has radius 4 0 319 1.2956. y and y2 . The total volume is 4 4 y2 2 (4 y) y dy 4 0 4 y3 2 4y 2y 2 dy 4 0 4 23 32 14 2 2y 2 y y . 3 3 16 0 height y 18. Use the intersect function on a graphing calculator to determine that the curves intersect at x 1.8933. The area is 1.8933 31 x2 3 dx, 10 x2 1.8933 which using NINT evaluates to 5.7312. 22. (a) Use disks. The volume is 2 19. Use the x- and y-axis symmetries of the area: 4 x sin x dx 4 sin x x cos x 0 4. 0 k (b) 0 0 A(x) r 1 V 1 k, 9 x. 9 x dx x 9 When k 1 y2 2y dy 2 4. 0 k2 0 8 8 k 2 dV (c) Since V 2 0 y2 2y dy 3x 4 and area 20. A cross section has radius r 2 2y)2 dy ( 1, 2. dt dk dt dk dt 1 dV 2 k dt 2k . 1 (2) 2 1 1 so the depth is increasing at the rate of 1 21. , unit per second. 23. The football is a solid of revolution about the x-axis. A [ 5, 5] by [ 5, 5] cross section has radius The graphs intersect at (0, 0) and (4, 4). r2 (a) Use cylindrical shells. A shell has radius y and height 2 y . The total volume is 4 4 4 y2 y3 2 ( y) y dy 2 y2 dy 4 4 0 0 4 13 14 2 y y 3 16 0 32 . 3 y 12 1 11/2 2 12 0 1 4x 2 and area 121 12 1 4x 2 . The volume is, given the symmetry, 121 11/2 4x 2 4x 2 dx 24 1 dx 121 121 0 11/2 21 2 24 x x3 3 11 0 11 11 24 2 6 88 276 in3. 24. The width of a cross section is 2 sin x, and the area is (b) Use cylindrical shells. A shell has radius x and height 2x x. The total volume is 4 4 2 (x)(2 x x) dx 2 0 (2x 3/2 2 x ) dx 0 2 4 5/2 x 5 128 . 15 13 x 3 4 0 12 1 r sin2 x. The volume is 2 2 1 1 1 sin2 x dx x sin 2x 22 4 02 2 0 4 . 320 Chapter 7 Review 29. 25. [ 1, 2] by [ 1, 2] [ 1, 3] by [ 3, 3] Use washer cross sections. A washer has inner radius r e x/2, and area (R2 outer radius R r 2) (e x 1, 1). 3x 2 y is at x 6x equals zero when x 0, the minimum at x = 2. The distance between 2 them along the curve is The volume is ln 3 (e x ex 1) dx using NINT evaluates to x 0 (3 ln 3 (2 4.5920 2 ln 3). 1) 26. Use cylindrical shells. Taking the hole to be vertical, a shell has radius x and height 2 22 3 2x 0 4 2 (4 3 2 x 2 dx x4 27. The curve crosses the x-axis at x 3 ( 2x)2 dx 1 3 1 2 x 4 dx 2 5 x 2 dx 2 2x, so the 13 x 3 5 39. 2 32. (a) (100 N)(40 m) = 4000 J (b) When the end has traveled a distance y, the weight of the remaining portion is (40 4x 2 dx, which 3 using NINT evaluates to 5 (F (x))2 dx 1 0 3. y 3 1, so 3 x2)3/2 4 (1 8) 3 28 29.3215 ft3. 3 length is 30. If (b) were true, then the curve y k sin x would have to get vanishingly short as k approached zero. Since in fact the curve’s length approaches 2 instead, (b) is false and (a) is true. 5 3 2 0 4.5920. The time taken is about 2.296 sec. 31. F (x) x 2) dx y)(0.8) 40 28. 0.8y) dy 0.4y 2 32y 0 The curves intersect at x 0 and x 1. Use the graphs’ 1 4 1 0 (3x 2 640 is (800)(8) 4750 x- and y-axis symmetry: 3x 40 640 J. 4640 J 33. The weight of the water at elevation x (starting from x [ 2, 2] by [ 2, 2] x) 0.8y. 0 (c) 4000 2 32 The total work to lift the rope is 19.4942. (32 d3 (x dx 6x)2 dx, which x 2. The volume of the piece cut out is 2 (x)(2 22 (3x 2 1 0 ln 3 0 0 or 2. The maximum is 0 4750 x/2 4750 128 4750 95 128 4750 95 1 x dx 2 1, and the total perimeter is 0) 1 x . The total work 2 128 4750x 95 12 x 4 4750 0 22,800,000 ft-lb. 1)2 dx, which using NINT evaluates to 34. F ks, so k 5.2454. 0.3 Work 0 F 800 80 N/m. Then s 3 0.3 0.3 800 800 2 x dx x 12 J. 3 6 0 To stretch the additional meter, 1.3 Work 0.3 800 x dx 3 800 2 x 6 1.3 213.3 J. 0.3 35. The work is positive going uphill, since the force pushes in the direction of travel. The work is negative going downhill. ...
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