Business Calc Homework w answers_Part_59

Business Calc Homework w answers_Part_59 - 291 Section 7.1...

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Unformatted text preview: 291 Section 7.1 7. (a) Right when v(t) i.e., when 0 0, which is when cos t 3 or 2 2 t cos t 0, i.e., when cos t e sin t cos t dt e 0 [e 0 0 e] 16t /2 0 e sin t cos t dt e sin t cos t dt C1, where 6 ft/sec. 90t 0. Solve s(t) 90t when t 90 16t2 2t( 8t 45 8 0 or t 0 2 /2 esin t cos t dt 32(3) s(0) 2 32t 90. v(t) dt C2 0 esin t cos t dt (c) Distance 3 /2 (b) s(t) 0 2 v(0) 32 dt Then v(3) 2 sin t a(t) dt C1 3 . Stopped when 2 3 or . 2 2 0, i.e., when t (b) Displacement 11. (a) v(t) 2 . Left when t 2 2 t 0, C2, where 0: 45) 0 5.625 sec. The projectile hits the ground at 5.625 sec. 3 /2 (e 1) 1 e e 1 e 1 2 e 2e 4.7 (c) Since starting height 8. (a) Right when v(t ) 0, which is when 0 t 3. Left: never, since v(t ) is never negative. Stopped when t 0. 3 (b) Displacement 1 0 1 [ln (10) 2 t t 3 t 0 t 2) 0. 3 (d) Max. Height 0 1.15 ln 10 2 dt t2 1 1 ln (1 2 dt ln 10 2 ln (1)] (c) Distance 2 Displacement 16 s 5.625 2 2 Distance 1.15 ending height, 5.625 2 5.625 90 2 126.5625, and 2(Max. Height) 253.125 ft. c 9. (a) v(t) a(t) dt v(t) t 2t 3/2 C, and since v(0) 2t 3/2. Then v(9) t 9 2(27) 0, 63 mph. 12. Displacement v(t) dt 4 5 24 23 cm 0 c 13. Total distance v(t) dt 4 5 24 33 cm v(t) dt 15 4 11. v(t) dt 15 4 5 16. v(t) dt 15 4 5 24 0 a (b) First convert units: 14. At t t t 3/2 t 2t 3/2 mph mi/sec. Then 3600 1800 9 3/2 t t Distance dt 1800 0 3600 2 5/2 t t 9 27 9 0 0.06525 mi 7200 4500 0 800 500 344.52 ft. a, s s(0) 0 b At t b, s s(0) 0 c At t c, s s(0) 8. 0 15. At t a, where dv is at a maximum (the graph is steepest dt upward). 4 10. (a) Displacement (t 2) sin t dt 0 sin t 16. At t 4 t cos t 2 cos t 4 cos 4 2 cos 4) 2] 1.44952 m 17. Distance (b) Because the velocity is negative for 0 for 2 t , and negative for t t 2, positive 4, 2 Distance (t 0 2) sin t dt [(2 sin 2) ( ( 2 2 cos 4 (t 2) sin t dt 2 4 (t 2) sin t dt sin 2 2 cos 4 2 sin 2 dv is at a maximum (the graph is steepest dt upward). 0 [(sin 4 c, where sin 4 2)] 2 1.91411 m. 4 1 2 (a) Final position Initial position 2 4 6; ends at x 6. 12 4 Distance (b) 4 meters 18. (a) Positive and negative velocities cancel: the sum of signed areas is zero. Starts and ends at x 2. (b) Distance meters 2) sin 4 Area under curve Sum of positive areas 4(1 1) 4 292 Section 7.1 25. (Answers may vary.) Plot the speeds vs. time. Connect the points and find the area under the line graph. The definite integral also gives the area under the curve. 7 19. (a) Final position 2 v(t) dt 0 1 (1)(2) 2 1 (2)(2) 2 2 1 (1)(2) 2 1 (2)(1) 2 1(2) 26. (a) Sum of numbers in Sales column (b) Enter the table in a graphing calculator and use QuadReg: B(x) 1.6x 2 2.3x 5.0. 5; ends at x 5. 11 (1.6x 2 (b) 1 1 (1)(2) (1)(2) 2 2 1 (2)(1) 2 v(t) dt 0 v(t) dt 0 1 1 (2)(3) (1)(3) 2 2 1 (3)(3) 2 2 1 (1)(3) 2 (3)(3) (1.6x 2 1.6 3 x 3 10 v(t) dt 1 (1)(3) 2 3) 1 (1)(3) 2 3(3) 1 (3)(3) 2 21. 27.08 e t/25 b dt 27.08 25e 10 t/25 27.08[25e 0 0.4 25] 332.965 billion barrels 3.9 22. 0 93.6 t 2.4 sin dt 12 28.8 28.8 23. (a) Solve 10,000(2 (b) Width 3.9t r) 0: r 20,000 29. F(x) r2 20,000 80,000 3 13 r 3 20,000 4 r ) dr 8 3 30. F(x) 2 r: Area 3 (c) 8(10 (2 r) r in2 flow in 396 in3 sec x2 2x dx 0 kx; 10,000 2x. 12 kx 2 kx dx 9 81 N cm 0 k(1), so k (b) For total distance: W 16 (10r 3 16 0 1244.07 in3 sec 45 81 4 W in3 sec r 3) dr 0 14 r 4 2 and F(x) 9 0 2rr 3 r2)(2 r) dr 5r 2 2(110) 18N F(x) dx (a) W Cross section area 0 16 2(112) d 10,000. 12 kd 2 0 1 (10,000)(0.5)2 2 1 (10,000)(1)2 2 5000 For second half of distance: Inches per second in. sec 2(119) 1250 inch-pounds r, Length r 2) 2(115) 2(115) k(3), so k d 0 (b) Volume per second 8(10 2(120) 0 2 83,776 24. (a) Width 2(115) 2(9) (b) W (2r 0 kx; 6 (a) F(9) 0 2 2(110) 9 2 r)(2 r) dr f (xn) 121] Area 0 ∑ 2 f (xi) 1156.5 2rr Population density 10,000(2 798.97 thousand 0.5 i1 2(120) 0 2 miles. 2 r: Area 2 (d) 5.0x n1 f (x0) 24 t cos 12 93.6 kilowatt-hours r, Length (c) Population 28.8 a 2n 18 8 [120 2(10) 0 24 10.5 2.3 2 x 2 28. Treat 6 P.M. as 18 o’clock: 19.5 meters 10 5.0) dx (b) The answer in (a) corresponds to the area of midpoint rectangles. The curve now gives a better approximation since part of each rectangle is above the curve and part is below. 0 1 (2 2 2.3x 0.5 2.5. (b) Distance 0 10.5 27. (a) 2.5; ends at x 11 (d) The answer in (a) corresponds to the area of left hand rectangles. These rectangles lie under the curve B(x). The answer in (c) corresponds to the area under the curve. This area is greater than the area of rectangles. 10 2 5.0x 904.02 thousand 7 meters 20. (a) Final position 5.0) dx 2.3 2 x 2 0 1 (2)(2) 2 1(2) 2.3x 1.6 3 x 3 (c) 7 797.5 thousand 31. 5000 (12 0) [0.04 2(12) 1250 3750 inch-pounds 2(0.04) 2(0.05) 2(0.06) 2(0.05) 2(0.04) 2(0.04) 2(0.05) 2(0.04) 2(0.06) 2(0.06) 2(0.05) 0.05] 0 0.585 The overall rate, then, is 12 0.04875. 0.585 Section 7.2 32. (12 0) [3.6 2(12) 2(4.0) 2(3.1) 2(2.8) 2(2.8) 2(3.2) s Section 7.2 Areas in the Plane (pp. 374–382) 2(3.3) 2(3.1) 2(3.9) 4.0] 2(3.2) 2(3.4) 2(3.4) 40 thousandths or 0.040 Exploration 1 1. For k ∑mkxk . Taking dm ∑mk My 33. (a) x 293 M A Family of Butterflies 1: [(2 sin x) sin x] dx (2 0 dA as mk and letting 2 sin x) dx 0 2x 2 cos x 2 4 0 dA → 0, k → x dm yields . For k dm 2: /2 ∑mky . Taking dm ∑mk My (b) y M [(4 0 (4 dA as mk and letting (2 sin 2x)] dx 4 sin 2x) dx 0 /2 4x dA → 0, k → yields 2 sin 2x) /2 2 cos 2x y dm 2 4 0 . dm 2. It appears that the areas for k 2 4. 3 will continue to be /k 34. By symmetry, x 0. For y, use horizontal strips: 3. Ak [(2k k sin kx) k sin kx] dx 0 /k y dm y dA (2k y dA 2k sin kx) dx 0 y dm dA dA If we make the substitution u kx, then du k dx and the 4 y(2 y) dy u-limits become 0 to . Thus, 0 4 2 /k y dy 2 5 2k sin kx) dx 0 /k 4 0 0 2 3/2 4 y 3 0 0 2 y 5/2 2 (2k Ak 0 (2 (2 4. 2 12 5 2 sin kx)k dx 2 sin u) du. 4 5. Because the amplitudes of the sine curves are k, the kth butterfly stands 2k units tall. The vertical edges alone have lengths (2k) that increase without bound, so the perimeters are tending to infinity. 35. By symmetry, y 0. For x, use vertical strips: Quick Review 7.2 x dm x dA x dA 1. x dm dA sin x dx 0 x(2x) dx [1 1] 2 0 1 2 cos x 0 dA 2. 1 2x e 2 e2x dx 0 1 0 12 (e 2 1) 3.195 2 0 2x dx /4 3. sec2 x dx tan x /4 1 ( 1) 2 /4 2 32 x 30 x2 2 0 /4 2 4. x 3) dx (4x 0 3 4 3 5. 9 3 radius 3.) x 2 dx 2x 2 14 x 4 2 (8 4) 0 4 0 9 (This is half the area of a circle of 2 294 Section 7.2 6. Solve x 2 4x x 6. x 2 5x 6 0 (x 6)(x 1) 0 x 6 or x 1 y 6 6 12 or y 1 (6, 12) and ( 1, 5) 7. Solve e x 1 3. 1 6 [(12y 2 (x 1)] 0. Test: e0 e 0 x ( 12y 3 (x 10 3 1) is at a minimum. x2 2y)] dy 2y) dy 1 y2 0 4 3 1 1, i.e., when 5. Use the region’s symmetry: 1. So the solution is (0, 1). 2 2x 10y 2 10 3 y 3 3y 4 3 8. Inspection of the graphs shows two intersection points: (0, 0), and ( , 0). Check: 02 0 sin 0 0 and 2 2 sin 0. 9. Solve (2y 2 2 [2x 2 (x 4 2 2x 2)] dx 2 ( x4 0 2 15 x 5 43 x 3 2 32 5 2 2 x4 x4 x 1 2 (x 2 x2 x2 2 0 0 25 x 5 2 x3 1 2 0 1 3 2 5 2 or 1 1 Throw out the negative solution. 0 (2 y 4 3/2 y 3 y) dy 12 y 2 1 0 1 x3 y 4 3 1 (0, 0), ( 1, 1 2 5 6 0 8. Integrate with respect to y: 1) and (1, 1) 1 10. Use the intersect function on a graphing calculator: 0 [(2 y) y] dy 12 y 2 2y 2 3/2 y 3 1 1 2 2 0 2 3 0 9. Integrate in two parts: 0 [(2x 3 x2 [( x 2 3x) [ 2, 2] by [ 2, 2] ( 0.9286, 0.8008), (0, 0), and (0.9286, 0.8008) 0 (1 2 cos x) dx 0 1 x 2 1 sin 2x 4 /3 0 1 sec2 t 2 /3 (sec2 t 8 sin2 t) dt 0 /3 tan t 4t 2 sin 2t 0 3 4 3 4 3 3 0 x2 2 8x) dx 3x)] dx 5x)] dx ( 2x 3 8x) dx 0 14 x 2 2 0 [0 4 sin2 t dt (2x 3 (2x 3 2 2. Use symmetry: 2 0 ( x2 5x) 2 2 Section 7.2 Exercises 1. 128 15 22 15 7. Integrate with respect to y: 8 1 2 x 1 3 2x 4) dx 0 1 2 0 6. Use the region’s symmetry: x2 1 2 32 3 (0, 0) is a solution. Now divide by x. x2 4x 2) dx 0 x 3. 1 1 12 0 0 1, so that if they are 1 is zero when e x 1 12y 3) 1 14 y 4 0 1 always greater than or equal to x x 4. 5 1. From the graphs, it appears that e is ever equal, this is when e x 13 y 3 y 3) dy 0 x x dx [e dx ( y2 4x 2 (8 0 2 16)] 14 x 2 [( 8 4x 2 16) 2 0 0] 16 5 6 Section 7.2 10. Integrate in three parts: x 1 [( x 2) 2x 2 x2 4: x 2 2x 2) (x 2 4)] dx 13. Solve 7 1, so the curves intersect at 1. 2 (4 x )] dx 1 2 [(7 1 1 2 x 2) [(4 295 (x ( 3x 2 3) dx 1 2)] dx 1 1 3 x 2) dx (1 1 3 [( x 2) x 2)] dx (4 1 (x 2 2 x 2) dx (x 2 2 x 2) dx (x 2 x 2 3 3 1 3 13 x 3 3x 2 1 1 2 3 4 14. 2) dx 2 13 x 3 1 12 x 2 1 x3 3 2 1 2 2 1 x2 2 2x 1 [ 3, 3] by [ 1, 5] 3 1 x2 2 1 3 13 x 3 2x 2x The curves intersect at x 2 8 3 2 2 1 2 1 3 4 4x 2 9 2 49 6 8 2 1 2 (x 4 2 (x 4 4x 2 4)] dx 5x 2 2 2 ( x4 4) dx 1 5 1 2 5 5x 2 4) dx 1 53 x 3 2 x5 4 1 6 5 3 4 1 4x 15 x 5 2 0 32 5 2 40 3 2 53 x 3 4x 1 1 5 8 5 3 4 8 15. 11. Solve x 2 x 8 3 6 2 [x 2 1 0 9 2 x 2] dx 4) 0 1 2 2. Use the region’s symmetry: 4 2 [(x 4 8 3 1 and x 2 2: x 2 4, so the curves intersect at 2. 2 [2 2 (x 2 2)] dx x 2) dx (4 2 33 a, a by [ a 2, a 2] 22 2 4x 1 32 x 3 2 8 x2 3: x 2 12. Solve 2x 8 3 8 3 8 2x 3 (x 32 3 3)(x 10 1) 2 3 The curves intersect at x 0, 3 (2x x 2 3) dx x 1 and x 13 x 3 2 1 (9 9 32 3 10 9) 2 3 2 x a2 3. 3x 1 x 2 dx 2 0 3 1 20 1 3 3 a. Use the region’s symmetry: a so the curves intersect at x 0 and x 23 a 3 12 (a 3 x 2)3/2 a 0 13 a 3 ...
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