Business Calc Homework w answers_Part_63

Business Calc Homework w answers_Part_63 - Section 7.4 /4...

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Unformatted text preview: Section 7.4 /4 sec2 y 8. (a) x 1, so Length sec y dy. /3 (e x 10. y e x) 2 3 , so Length ex 1 x2 e dx. 2 3 (b) sec2 y (b) x 1 tan y , ln (cos y) ln (cos y) y 3 so x 0 y 0 4 [ 3, 3] by [ 2, 12] (c) Length 12 (x 2 11. y 3 1 0 [ 2.4, 2.4] by (c) Length 2 , (x x 2 sec x tan x, so /3 1 2 sec x tan x dx. 3 2 3 2 1 0 , (c) Length 3 x 12. 4 dx 9x dx 4 9x 3/2 4 8 1 0 8 1 27 80 10 27 y2 by [ 1, 3] 3 1 3.139 2 1 4 0 . 3 1 , so the length is 4y 2 12 13 y2 dy y 3 4 y2 12 dy 4 y2 y2 1 1 1 4y 3 53 . 6 1 1 , so the length is 4y 3 y3 14. x 1 dx 0 2 x (b) 13. x 2x 2 x, so the length is /3 3 x4 3 13 x 3 1) dx 4 2 2, so the length is 3 0 (x 2 2 x x2 2)2 dx 0 2.198 Length 2)1/2(2x) 3 12. y 9. (a) y 20.036 2 12 dy 4 y3 y3 1 12 dy 4 y3 y3 1 14 y 4 1 8y 2 2 1 123 . 32 y2 2 15. x 1 , so the length is 2y 2 2 1 2 12 dy 2 y2 y2 2 1 1 13 y 6 x2 16. y 2x 1 1 2y 4 (x 4)2 (4x so the length is 2 1 (x 2 4(x 0 1 (x 3 1)3 2 4(x 1)2 2 1 1) dx 1)2 4(x 1 1)2 (x 2 1 1)2 0 0 12 dy 2y 2 y2 2 dx 53 . 6 2 1 1)2 17 . 12 1 4(x 1)2 311 312 Section 7.4 sec4 y 17. x /4 /4 (sec4 y 1 24. The area is 300 times the length of the arch. 1, so the length is 1) dy /4 y sec2 y dy /4 x , so the length is 50 2 x sin2 dx, which evaluates, using NINT, 2 50 sin 2 25 /4 tan y 1 2. 25 /4 to 3x 4 18. y 1 the cost (rounded to the nearest dollar): $38,422. 1 (3x 4 1 73.185. Multiply that by 300, then by $1.75 to obtain 1, so the length is 3x 2 dx 1) dx 2 2 1 1 3 3 x3 73 . 3 2 dy 2 dy 1 1 corresponds to here, so take as . Then dx dx 4x 2x 19. (a) y x C, and, since (1, 1) lies on the curve, C 0.2x y1 10 1 0. x. dx 2 dx 1 1 corresponds to 4 here, so take as 2 . dy dy y y 1 Then x C and, since (0, 1) lies on the curve, y 1 C 1. So y . 1x (b) Only one. We know the derivative of the function and the value of the function at one value of x. 21. y cos 2x, so the length is /4 5 22.3607, and . NINT fails to evaluate (y1 )2 dx because of the undefined slope at 22.36 (y1 )2 dx 1 2 0 52.548. Then, pretending the last little stretch at each end is a straight line, add 0.2(22.36)2 2 100 track 1 as 0.156 to get the total length of 52.704. Using a similar strategy, find the length of the right half of track 2 to be 2 cos x dx 0 2 sin x 32.274 + NINT Y2 /4 1. 32.274. Now 52.704 and 2 cos 2x dx 0 from the upper limit a little, and find enter Y1 /4 1 5 10 the limits, so use the track’s symmetry, and “back away” (b) Only one. We know the derivative of the function and the value of the function at one value of x. 20. (a) 0.2x 2 100 5 10 So y 0 at x 25. For track 1: y1 2 0.2t 1 150 0.2t 2 , t, x, 0 and graph in a [ 30, 0] by [0, 60] window to see the effect of 0 the x-coordinate of the lane-2 starting position on the length 22. y x 2/3)1/2x (1 1/3 , so the length is of lane 2. (Be patient!) Solve graphically to find the 1 8 1 x 2/3)x (1 2/3 intersection at x dx 2/4 coordinates ( 19.909, 8.410). 1 8 2/3 x dx 24 1 2/3 x 3 26. f (x) 1 8 x 1/3 dx 2 1 2/4 1 3 2 31 22 x 0. So, instead solve for x in terms of y using the quadratic formula. (x1/3)2 6. 23. Find the length of the curve y 20 0 to (f (x))2 dx because of the undefined slope at 2/4 1 x1/3 y 2 1/3 x , but NINT fails on 3 0 3 2 8 x 2/3 8 19.909, which leads to starting point 3 sin x for 0 20 x 20. 3 3 cos x, so the length is 20 20 2 3 3 1 cos x dx, which evaluates, using NINT, 20 20 21.07 inches. 1 (1 8 3 (1 8 x x 2 1 2 1/3 4y 4y 2/3 2 y 0, and . Using the positive values, 1)3. Then, 2 1)2 1 1 0 4y x1/3 (x )2 dy , and 4y 3.6142. 313 Section 7.5 (4x 2 27. f (x) 1) (8x 2 (4x 2 1)2 1 1/2 using NINT, to partition goes to zero will always be the length (b 4x 2 8x 1 2 dx which evaluates, (4x 2 1)2 1 length is 31. Because the limit of the sum Σ xk as the norm of the 4x 2 8x 1 , so the (4x 1)2 8x) a) of the interval (a, b). 2.1089. 32. No; the curve can be indefinitely long. Consider, for 28. There is a corner at x 0: 1 3 example, the curve sin 1 + 0.5 for 0 x x 1. 33. (a) The fin is the hypotenuse of a right triangle with leg lengths xk and [ 2, 2] by [ 1, 5] Σ k1 (b) lim n→ x3 x3 ( xk)2 2 0 x x n 0 and 1 Σ k1 lim n→ xk 1 ( f (x))2 dx b 2 3x 3x 2 y 5 5 x x 1 The length is 0 . 1 (y )2 dy 2 0 (3x 2 1 34. Yes. Any curve of the form y 5)2 dx a 1 (3x 2 5)2 dx, x)2, 0 (1 x (y )2 dx 1 c, where c is a 1, so that a 2 dx 0 0 which evaluates, using NINT for each part, to 29. y x constant, has constant slope 1 2 ( f (xk 1))2 a 2 0 1 f (xk 1) xk. ( f (xk 1) xk)2 1 5x 5x xk 1 n Break the function into two smooth segments: y df dx x xk a 2. 0 13.132. s Section 7.5 Applications from Science and Statistics 1 (pp. 401–411) Quick Review 7.5 1 1. (a) [ 0.5, 2.5] by [ 0.5, 1.5] x y 1 e into two equal segments by solving 1 . The total length is 2 4 which evaluates, using NINT, to x y 1 1 2. (a) 12 x 1/4 ex 1 e 1 1.718 /2 /2 dx, 1 e 1 0 (b) 3. (a) sin x dx x (b) 2 cos x /4 1.623. 2 /4 0.707 3 30. e x dx 0 1 with 1 0.632 1 0. So, split the curve x 0 (b) , but NINT may fail using y over the entire interval because y is undefined at x x: x dx 0 x y x e 4. (a) (x 2 13 x 3 2) dx 0 3 2x 15 0 (b) 15 2 5. (a) 1 x2 x 3 1 [0, 16] by [0, 2] 1 3/4 x , but NINT may fail using y over the entire 4 y interval, because y is undefined at x 4 x y ,0 2 1 y 2: x 0. So, use (b) 3 4y and the length is 6. 0.501 2 (x 2) sin x dx 0 0 7 16.647. 1 ln (x 3 1) 3 1 [ln 9 ln 2] 3 1 9 ln 3 2 7 (4y 3)2 dy, which evaluates, using NINT, to dx 7. (1 0 x 2)(2 x) dx 2 1 314 Section 7.5 7 cos2 x dx 8. 7 y2 (10 2 9. 0 7 3 10. 4 0 ks, so 150 k 1 and k 16 1 ,F 8 1 8 300 lb. 10. (a) F 0 s y) dy 2400 1/8 sin2 x dx 1/8 1200x 2 2400x dx (b) 2400 lb/in. Then for 18.75 in.-lb 0 0 Section 7.5 Exercises 5 1. xe x/3 dx 5 x/3 3e (3 0 0 3 2. x dx 4 x sin 0 44 x 4 sin 4 22 11. When the end of the rope is x m from its starting point, the 24 e 5/3 x) 9 4.4670 J (50 x 4 x cos F(x) 0 3. 3 2 (0.624)(50 x 2 dx x9 (esin x cos x 4. 9J (b) p1V11.4 10 esin x 19 19.5804 J 109,350 F(x) 490 1 490 20 W (490 24.5x) dx 12.25x 2 490x 0 0.4 V 4900 J. 20 4x/5 20 x 25 490 1 490 W (490 19.6x) dx 490x 9.8x 0 k1 F(x) ∑ 124.8yk 144 x/2 x 36 144 1 18 W (144 4x) dx 144x 2x (b) ks, so 800 k(14 10) and k 2 (b) F(x) 200x, and 100x 2 200x dx (c) F 9. (a) F ks, so 21,714 400 in.-lb. n Σ (62.4yk) 6 k1 7238 lb/in. (b) 374.4y 1 0 1/2 (b) F(x) 7238x. W 3619x 2 7238x dx 0 1/2 0 1 904.75 905 in.-lb, and W 7238x dx 1/2 3619x 2 1 2714.25 1/2 1, it follows that a thin y2 y, where y is 64 force is approximately 8 5) and k y2 82 distance from the top, and pressure 62.4y. The total 8 in. k(8 3 2 0 1600 200 200s, so s x2 32 horizontal rectangle has area 6 1 200 lb/in. 0 y 2)3/2 0 14. (a) From the equation 0 8. (a) F 41.6(9 1123.2 lb 1944 ft-lb 0 9 yk2) y yk2 y. 0 4x lb. 18 2 y 2 dy 3 Then 18 124.8y 5880 J. 144 9 k1 0 18 y 2 y, where y is n 20 7. When the bag is x ft off the ground, the sand weighs 32, it follows that a thin force is approximately ∑ (62.4yk)(2 9 n 2 y2 distance from the top, and pressure 62.4y. The total 19.6x N. Then 20 243 horizontal rectangle has area 2 9 0 490 109,350 and V1.4 32 V 6. When the bucket is x m off the ground, the water weighs F(x) p dV (p , V ) 11 dV 2.5V 13. (a) From the equation x 2 20 (p , V ) 22 pA dx 37,968.75 in.-lb 24.5x N. Then 780 J. 0 109,350, so p 5. When the bucket is x m off the ground, the water weighs x 20 50 (p , V ) 11 (p , V ) 2 2 109,350 V1.4 (p , V ) 11 Work 0 esin 10 (p , V ) 22 F(x) dx (50)(243)1.4 2x 0 20 x 490 20 12 x 2 0.624 50x (p , V ) 11 0 2) dx x) dx (p , V ) 22 3 x 2)3/2 10 x) N. The total work is 0 12. (a) Work 1 (9 3 0 (0.624)(50 50 3.8473 J 3 x) m of rope still to go weigh 3 2714 in.-lb. 1 yk2 64 y2 dy 64 n y Σ 374.4yk k1 7987.2 1 7987.2 lb yk2 1 y 2 3/2 64 64 8 0 y. 315 Section 7.5 3 y, it follows that a thin 8 3 horizontal rectangle has area y y, where y is the 4 18. The work needed to raise a thin disk is (10)2(51.2)y y, 15. (a) From the equation x distance from the top of the triangle, the pressure is 62.4(y ∑ 62.4(yk ∑ 46.8(yk y 2 3yk) y. k1 46.8(y 22 3y) dy 3494.4 3 ( 210.6) 15.6y 3 70.2y 2 8 5120 0 12 y 2 30 7,238,229 ft-lb 0 4 (62.4)( y 12 y 2 6 15y 0 84,687.3 ft-lb x2 , it follows that a thin 2 for the whole tank. Work to pump over the rim is y). The total force is approximately (2)2(62.4)(6 15) y for a thin disk and 6 0 n 4 (62.4)(21) dy 4 (62.4)(21)(6) 98,801.8 ft-lb for the whole tank. Through a hose attached to a valve in the yk)(2 2yk) y n 2(4 249.6 0 distance from the bottom, and pressure 62.4(4 Σ 124.8 k1 15) dy 3705 lb horizontal rectangle has area 2 2y y, where y is Σ 62.4(4 k1 15) y for a thin disk and 6 3 16. (a) From the equation y bottom is faster, because it takes more time for a pump with yk3/2) y. yk a given power output to do more work. 4 (b) 100 (51.2)y dy 19. Work to pump through the valve is (2)2(62.4)( y n 3 3) yk 4 k1 (b) 30 3). The total force is approximately n 8 where y is height up from the bottom. The total work is 3/2 124.8 2(4 y y ) dy 124.8 2 y3/2 0 8 3 1064.96 2 2 5/2 y 5 4 20. The work is the same as if the straw were initially an inch 0 long and just touched the surface, and lengthened as the 1506.1 lb 17. (a) Work to raise a thin slice 62.4(10 20 Total work 12)( y)y. 62.4 60y 2 62.4(120)y dy 0 liquid level dropped. For a thin disk, the volume is 20 0 1,497,600 ft-lb (b) (1,497,600 ft-lb) 100 min (250 ft-lb/sec) 5990.4 sec 10 62.4 60y 2 7 0 (c) Work to empty half the tank 17.5 2 14 y 17.5 2 14 y 17.5 14 y to y and the work to raise it is 4 (8 y) y. The total work is 9 24 (8 y) dy, which using NINT evaluates 9 91.3244 in.-oz. 62.4(120)y dy 0 10 374,400 ft-lb, and 21. The work is that already calculated (to pump the oil to the 0 374,400 250 1497.6 sec 25 min rim) plus the work needed to raise the entire amount 3 ft higher. The latter comes to (d) The weight per ft3 of water is a simple multiplicative factor in the answers. So divide by 62.4 and multiply 12 r h (57)(3) 3 total is 22,921.06 57 (4)2(8) 30,561.41 22,921.06 ft-lb, and the 53,482.5 ft-lb. by the appropriate weight-density For 62.26: 62.26 1,494,240 ft-lb and 62.4 62.26 5990.4 5976.96 sec 100 min. 62.4 1,497,600 For 62.5: 62.5 1,500,000 ft-lb and 62.4 62.5 5990.4 6000 sec 100 min. 62.4 1,497,600 22. The weight density is a simple multiplicative factor: Divide by 57 and multiply by 64.5. 30,561.41 64.5 57 34,582.65 ft-lb. ...
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