Business Calc Homework w answers_Part_60

Business Calc Homework w answers_Part_60 - 296 Section 7.2...

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Unformatted text preview: 296 Section 7.2 18. Solve y 2 16. 2: y2 y y curves intersect at y 2 (y [ 2, 12] by [0, 3.5] 12 y 2 1 The curves intersect at three points: x 1, x 4 and x 9. Because of the absolute value sign, break the integral up at x 0 also: 0 x 4 6 5 1 x x dx 9 x x 6 12 x 2 2 ( x)3/2 3 5 6x 5 1 y 2 3/2 x 3 20 5 6x y 4 4 5 18 189 10 16 15 13 30 32 5 16 3 1 6 5 3 1 16 3 (y 2 1 3 4. y 4 4) 5)( y 0 y2 4 y3 12 32 5 y2 4 4 4 2 3 y 4 y2 4 0. 4: The curves intersect at y 5 11 10 y 4 1 1 2 1 2 1 and x 1 2 13 y 3 5 0, 4 and y 5. 0 4 0 y2 4 0, so the 4 9 12 x 2 2 3/2 x 3 y2 12 x 2 y2 4 4 1) 2. 8 3 4 9 2 19. Solve for x: x 2)(y 2y 2 Now solve 0 6x x dx dx 5 4 6 5 0 (y 1 and y y 2) dy 2 2 y 4 2 3 5 5y 4 25 8 20. Solve for x: x dy 5 dy y2 8 125 12 1 16 3 25 y 2 and x 2 243 8 20 30 3 8 2y 2. Now solve 3 17. y2 2y 2: y 2 3 1, so the curves intersect at y 1. Use the region’s symmetry: 1 [ 5, 5] by [ 1, 14] 2 0 2y 2 (3 1 y 2) dy 2 The curves intersect at x 0 and x 4. Because of the absolute value sign, break the integral up at x 2 also (where x 2 4 turns the corner). Use the graph’s symmetry: 2 2 0 x2 2 2 2 0 2 4 (4 4 2 2 4 2 3x dx 2 x3 2 20 2[4] x 2) dx 2 2 2 2 x3 6 32 3 x2 2 4 (x 2 3y 2) dy 0 (1 y 2) dy 13 y 3 6y 4) dx 1 0 1 3 61 0 4 8 dx 4 y 2 and x 21. Solve for x: x 8x 2 32 (3 1 6 2 x 2 0 4 3 16 64 3 21 1 3 y2 Now solve at y 0 3y 2 (2 1 4 3y 2: y 2 2 1, so the curves intersect 1. Use the region’s symmetry: 1 2 3y 2. 2 0 (1 y 2) dy y 2) dy 4y 1 2 0 2y 2) dy (2 13 y 3 1 41 0 1 3 0 8 3 Section 7.2 22. Solve for y: y 4x 2 Now solve 4 x4 4x 2 4x 2 and y 4 x4 (x 2 5 x4 297 27. 1. 1: 1)(x 2 5) 0. [ 1.5, 1.5] by [ 1.5, 1.5] The curves intersect at x = 1. The curves intersect at x 0 and x 1. Use the area’s Use the region’s symmetry: symmetry: 1 2 0 4x 2) [(4 1)] dx 1 0 ( x4 2 4 3 6 4 2 y and x y2 2 :y 4 2 0 2 2. 0 2 3 0 y 2 /4 2 y2 dy 4 2 y3 4 2(6 (2 sin x 0.273 [sec2 x (sec2 x 1)] dx /4 dx 2x 2 0 0 (tan2 y tan2 y) dy /4 4 /4 0 0 8 41 1 cos 2x 2 2 cos x 1 2 2 4 /2 3 sin y 30. cos y dy 2 (8 cos x 0 sec x) dx 2 8 sin x tan x 0.858 2 (cos y)3/2 3 3 2 3 30 /3 y 3 and x 31. Solve for x: x 0 4 4 0 1 2 0 2 y 0 25. Use the region’s symmetry: /3 tan2 y dy 0 4 tan y 0 2 tan2 x) dx 2 2) sin 2x) dx 2 3y 2 dy 4 3 0 2 3y 24. 0 29. Use the region’s symmetry: Use the region’s symmetry: 2 (sec2 x /4 4, so the curves intersect at y 0 /4 y2 . 4 2 y2 Now solve 3 1 28. Use the region’s symmetry, and simplify before integrating: 14 15 3 12 x 2 2 0 0 x 2 cos 1 2 2 5x 5 23. Solve for x: x 2 2 1 /4 104 15 x dx 5) dx 43 x 3 1 5 2 0 4x 2 15 x 5 x 2 sin 2 1 2 (x 4 /2 0 2 y. 0 2[(4 3 3) 0] 6 3 26. [ 1.5, 1.5] by [ 1.5, 1.5] The curves intersect at x [ 1.1, 1.1] by [ 0.1, 1.1] 1 The curves intersect at x cross at x 1 2 cos 2x 13 x 3 2 21 1 3 1 1, but they do not 0. x2 0 0 and x 2 x 2 sin 0 dx x 2 1 0 4 3 4 0.0601 symmetry: 2 ( y 0 y 3) dy 0 and x 1 2 2 y2 1. Use the area’s 14 y 4 1 0 1 2 298 Section 7.2 32. (b) The two areas in Quadrant I, where x c y, are equal: 4 y dy y dy 0 c 4 2 3/2 c 2 3/2 y y 3 3 0 c 2 3/2 2 3/2 2 3/2 c 4 c 3 3 3 [ 0.5, 2.5] by [ 0.5, 1.5] y 1 intersect at x x2 x and y 1 2 x dx 0 1 12 x 2 1 dx x2 1. Integrate in two parts: 1 1 x 0 1 2 1 2 8 c 3/2 1 ( 1) 33. The curves intersect when sin x 2c 3/2 2 1 4 42/3 c cos x, i.e., at x 4 . 24/3 (c) Divide the upper right section into a (4 c)-by- c /4 /4 (cos x sin x) dx sin x rectangle and a leftover portion: cos x 0 0 2 1 0.414 c 2 x 2) dx (c 34. (4 c) c c 13 x 3 cx c c [ 3, 3] by [ 2, 4] c c3/2 4c 2 2 0 1) dx 2 2 3/2 c 3 0 13 x 3 2 4x 8 3 (b) Solve y x 2 for x: x 3 y-intercepts are 3 y dy 16 3 4 c3/2 c 4 2 1 20 35. (a) 0 3 32 3 y. The c 42/3 5 y=4 2 (3 3 36. 3 y)3/2 1 16 3 32 3 y = x2 (2, 4) (–2, 4) [ 1, 5] by [ 1, 3] The key intersection points are at x 0, x 1 and x Integrate in two parts: 1 1 (–√c, c) 0 (√ c , c ) 3 ( x2 c, then x c, c) and ( c, c). x dx 4 x x If y 1 3/2 c 3 24/3 y y=c c 1 and 3. 3 2 1 3/2 c 3 4c 0 8 3 28 c 16 3 c3/2 2 4 4 3/2 c 3 x 2) dx (4 2 c 2. 2 x2 (3 4x 0 8 Use the region’s symmetry: 13 x 3 3/2 4 1 3/2 c 3 3/2 (a) The curves intersect at x x 2) dx (4 0 1 2 3 x c. So the points are x2 8 2 3/2 x 3 1 8 4 1 2 x dx 4 x 1 4 x2 8 x 0 (8 2) 4 4 1 1 8 11 3 4. 299 Section 7.2 37. First find the two areas. 43. First graph y 1 2 For the triangle, (2a)(a 2) a For the parabola, 2 (a 2 The ratio, then, is 2 a 2x 13 x 3 a 0 43 a 3 3 , which remains 4 a 43 a 3 a3 x 2) dx 0 3 x 2. cos x and y [ 1.5, 1.5] by [ 0.5, 1.5] constant as a The curves intersect at x 0.8241 approaches zero. 2 0.8241. Use NINT to find x 2) dx (cos x 1.0948. Multiplying both 0 b 38. functions by k will not change the x-value of any b [2f (x) f (x)] dx a f (x) dx, which we already know a intersection point, so the area condition to be met is equals 4. 0.8241 2 2 kx 2) dx (k cos x 0 39. Neither; both integrals come out as zero because the 1-to-0 and 0-to-1 portions of the integrals cancel each other. ⇒2 ⇒2 k(1.0948) 40. Sometimes true, namely when dA [ f (x) g(x)] dx is always nonnegative. This happens when f (x) g(x) over the entire interval. ⇒k 1.8269. 0.8241 k2 x 2) dx (cos x 0 44. (a) Solve for y: 41. x2 a2 y2 The curves intersect at x symmetry: 2 0 2x x2 1 x 3 b2 1 y [ 1.5, 1.5] by [ 1.5, 1.5] 1 y2 b2 dx 0 and x ln 4 14 x 4 1 1 2 2 ln 2 1 2 x2 a2 1. Use the area’s a 2 ln x x2 a2 b1 b1 1 x2 dx or 4 a2 a a 2 0 0.886 x2 a2 b1 (b) 2 1 a dx or a 0 x2 dx a2 b1 (c) Answers may vary. a (d, e) 2 42. x2 a2 b1 b1 a x2 dx a2 2b x 2 1 x2 a2 a 2b sin 1 (1) 2 a a x sin 1 2 a a a sin 1 ( 1) 2 ab [ 1.5, 1.5] by [ 1.5, 1.5] The curves intersect at x 0.9286 to find 2 (sin x 0 and x x 3) dx 0.9286. Use NINT 0.4303. 45. By hypothesis, f (x) g(x) is the same for each region, where f (x) and g(x) represent the upper and lower edges. 0 b But then Area [f (x) a each. g(x)] dx will be the same for 300 Section 7.3 Quick Review 7.3 1 : 2 46. The curves are shown for m 1. x 2 x 2. s x2 . 2 s2 , so Area 2 3. 12 x2 r or 2 2 4. 1 2 [ 1.5, 1.5] by [ 1, 1] In general, the intersection points are where which is where x x x2 mx, 1 (1/m) 1 x x2 0 1) 1 m 1 m m 1 1 0 2x m ln (m) 2x 1. s Section 7.3 Volumes x (pp. 383–394) b Exploration 1 x and h 1 bh 2 12 x 2 (2x)2 xk2. 3xk Area 2. Unrolling the cylinder, the circumference becomes one dimension of a rectangle, and the height becomes the other. The thickness x is the third dimension of a slab with dimensions 2 (xk 1) by 3xk xk2 by x. The volume is obtained by multiplying the dimensions together. 3 3. The limit is the definite integral 2 (x 1)(3x x 2) dx. 15 x, so 2 15 2 x. 4 Volume by Cylindrical Shells 1. Its height is f (xk) x2 . 4 8. 1 ln 1 bh 2 , so Area 32 x. 4 x2 . 2 1 bh 2 x, so Area x 1 bh 2 x, so Area 2 (1/m) 1 12 mx 2 h 2 1. Then, mx dx 1 1 2 ln (x 2 2 h 3 x and h 7. b 0 or else x 5. b 6. b 1 m because of symmetry, the area is 2 d2 x2 or 2 8 9. This is a 3-4-5 right triangle. b Area 1 bh 2 4x, h 3x, and 2 6x . 10. The hexagon contains six equilateral triangles with sides of 0 4. length x, so from Exercise 5, Area 45 2 Exploration 2 Surface Area b 1. 2y 1 (a) A r 2, where r [a, b]. (b) A sin x, so dy dx b a 2 sin x x, so (d) A cos2 x dx 1 dy dx 1 2x 12 4 2 0 s 2, where s w2 2 w , so A(x) 2 w2 w, so A(x) w , so A(x) 2 14.424. 0 3. y s 2, where s (c) A cos x and dy 2 dx dx 2y 1 x1 2 x 3 32 x. 2 1. In each case, the width of the cross section is w 2 1 x 2. The limit will exist if f and f are continuous on the interval 2. y 32 x 4 Section 7.3 Exercises dy 2 dx dx a 6 and dx 36.177. A(x) 4(1 w2 (1 x 2). 2(1 2 32 w (see Quick Review Exercise 5), so 4 3 4 (2 1 x 2)2 3(1 x 2). x 2). x 2). ...
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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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