Business Calc Homework w answers_Part_62

Business Calc Homework w answers_Part_62 - 306 Section 7.3...

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Unformatted text preview: 306 Section 7.3 43. continued 45. (c) A shell has radius 1 2 0 8 5 8 5 y. The volume is y 12( y 2 1 24 y4 0 15 y 5 24 y 3) dy 2 y 0 1 24 0 24 The functions intersect at (2, 8). (a) Use washer cross sections: a washer has inner radius A(x) 2 12( y 2 y 3) dy 5 22 33 y4 y y dx 5 5 1 15 34 23 y y y 2. 5 20 15 0 y2 44. A shell has height 2 y2 y4 y2 . 2 4 2 2 ( y) y 2 0 (b) A shell has radius 2 2 2 (2 2 0 2 16 y 24 16 y 24 y4 y 3 2y 2 dy 2 14 23 15 y y y 4 3 10 2 0 2 0 2 y5 4 16 y 24 2 y 2 2 0 2 5 8 y5 4 16 y 24 512 . 21 0 y and 8 . 3 46. The functions intersect at (0, 0) and (1, 1). 2 8 . 5 0 y. The volume is 5y 4 4 15 y 4 y3 (a) Use cylindrical shells: a shell has radius x and height x2 1 2 (x)(x x x 2. The volume is x x 2) dx 2 0 53 y 3 13 x 3 y2 height 2x 2 8. x 2 1 2 (1 0 5 . The volume is 8 5y 4 32 x)(x x x 15 y 32 14 y 4 5y 2 dy 8 53 y 24 0 6 . x 2) dx 2 4. 0 x and x . The volume is 1 (x 3 2 2x 2 x) dx 23 x 3 12 x 2 0 6 dy y3 1 2 14 x 4 2 4 y 4 14 x 4 (b) Use cylindrical shells: a shell has radius 1 5y 2 dy 14 y 4 (d) A shell has radius y 0 2 y . The volume is 4 8 y 2 (8 y) y 1/3 dy 4 0 8 y2 2 8y 1/3 2y y 4/3 dy 4 0 8 3 7/3 832 13 2 6y 4/3 y 2 y y . 7 21 12 0 0 2 17 x 7 [ 0.5, 1.5] by [ 0.5, 1.5] y4 dy 4 y) y 2 0 2 16 3 x 3 height y 1/3 2x 2 (5 x 6). The volume is dy (c) A shell has radius 5 2 (16x 2 y. The volume is y 4 y) y 2 y5 4 r) 4 0 2 14 y 4 2 (16x 2 2 (b) Use cylindrical shells: a shell has a radius 8 y4 4 y4 dy 4 4x, and area x 6) dx (R 2 0 (a) A shell has radius y. The volume is 2 x 3, outer radius R r 2 . The volume is 5 (d) A shell has radius y 1 [ 1, 3] by [ 1.4, 9.1] 13 3 82 y y dy 5 5 1 13 4 83 y y 2. 20 15 0 . 1 0 307 Section 7.3 47. (a) Solve d dc 2 22 c 4c 2 2 0 2 c 4 0 [ 0.5, 2.5] by [ 0.5, 2.5] 1 1 , 1 , , 2 , and (1, 1). 4 4 The intersection points are and area (R 2 1 y4 1 1 y4 1 3y 3 2 r) 1 dy 16 1 (b) A shell has radius x and height 1. The volume is 2 0 2 c 1 1 , outer radius R , 4 y2 1 . The volume is 16 2 11 1 y . 48 16 1 (a) A washer has inner radius r 2 c This value of c gives a minimum for V because a 2V 2 dc 2 2 0. 2 22 Then the volume is 1 2 (x) 1/4 1 dx 2 3/2 x 3 2 x 48. (a) For 0 x x(sin x) x , x f (x) For x 0, x f (x) 01 x f (x) sin x for 0 1 1/4 2 anywhere else besides c 11 . 48 2 4 , the maximum must occur 2 at c sin x. sin 0 x 12 x 2 2 2 (b) Since the derivative with respect to c is not zero x 1 2 2 4 0 or c and for c sin x. So 1. The volume for c (3 1 it is 8) 0 is 2.238. c 2 2 4.935, 0 maximizes the volume. . (c) (b) Use cylindrical shells: a shell has radius x and height y. The volume is 2 xy dx, which from part (a) is 0 2 sin x dx 2 cos x 0 49. (a) A cross section has radius r r2 A(x) 6 144 0 (b) [0, 2] by [0, 6] 4. The volume gets large without limit. This makes sense, since the curve is sweeping out space in an ever-increasing radius. 0 144 (36x 2 (36x 2 x 4) dx x 12 36 x 2 and area x 4). The volume is 144 36 cm3 (8.5 g/cm3) 5 15 x 5 12x 3 6 0 51. (a) Using d 36 cm3. 5 C d2 2 , and A C2 yields the 4 following areas (in square inches, rounded to the nearest tenth): 2.3, 1.6, 1.5, 2.1, 3.2, 4.8, 7.0, 9.3, 10.7, 192.3 g 10.7, 9.3, 6.4, 3.2. 50. A cross section has radius r c sin x and area (b) If C( y) is the circumference as a function of y, then the r2 A(x) sin x)2 (c (c 2 2c sin x sin2 x). area of a cross section is The volume is C(y)/ 2 A(y) (c2 sin2 x) dx 2c sin x and the volume is 0 1 x 2 c2x 2c cos x 2 1 2 c 22 c 2c 1 sin 2x 4 2c 0 (c) 1 4 6 2 . 0 5.12 11.62 C 2( y) , 4 6 1 4 C 2( y) dy. 0 6 A( y) dy 2 4c 2 1 C 2( y) dy 40 16 0 [5.42 24 4 6.32 10.82 7.82 9.42 9.02) 2(4.52 10.82 6.32] 4.42 11.62 34.7 in.3 308 Section 7.3 52. (a) A cross section has radius r 5 r2 2 y. The volume is 2 y dy 0 dV dt dh so dt For h dh so dt 1 8 0 and x a. For revolution about the x-axis, a cross section has radius x 2 and area ax A(h). r2 A(x) dh , dt x 2)2 (ax (a 2x 2 2ax 3 x 4). The volume is a 8, units3 3 3 = sec 8 53. (a) 25 . 0 4, the area is 2 (4) 0: This is true at x 5 r dV A(h) dh, so dh dV dh A(h) dh dt dV 1 A(h) dt (b) V(h) y2 x2 55. Solve ax 2y and area (a 2x 2 2ax 3 123 ax 3 x 4) dx 0 units3 . sec 14 ax 2 15 x 5 a 0 1 a 5. 30 y For revolution about the y-axis, a cylindrical shell has 2 r x 2. The volume is radius x and height ax x a x 2) dx 2 (x)(ax 13 ax 3 2 0 The remaining solid is that swept out by the shaded region in revolution. Use cylindrical shells: a shell has r2 radius x and height 2 . x 2. The volume is 1 a5 30 14 1 a yields a 6 30 2 (x)(2 r 2 2 x 2) dx 22 1 , so a 6 5. 56. The slant height s of a tiny horizontal slice can be x2 written as s r2 0 14 a. 6 Setting the two volumes equal, –2 .r a 14 x 4 y2 ( g ( y))2 y. So the 1 surface area is approximated by the Riemann sum n 22 (r 3 4 ( 8) 3 Σ2 k1 r 2 3/2 x) r2 4 32 . 3 integral. dx dy 57. g ( y) (b) The answer is independent of r. (g ( y))2 y. The limit of that is the g( yk) 1 1 2 2 y1 0 , and 2y 1 2 2 dy 4y and measure the radius r (x) of the shadow region at these 6 points. Then use an approximation such as the trapezoidal (4y 13 3 b rule to estimate the integral a r 2(x) dx. dx dy 58. g ( y) 1 2 0 y3 3 3 2 1 dx dy y1/2 1 3 2 1 1)3/2 2 0 13.614 y 2, and 2 3 ( y 2)2 dy 9 59. g ( y) 1 dy 0 2y 54. Partition the appropriate interval in the axis of revolution y1/2 1 (1 6 (2 2 y 4)3/2 0 1) 1 1/2 y , and 2 1 3/2 3 1 3/2 3 1 1/2 2 y dy 2 1 1 1 dy. 4y Using NINT, this evaluates to 1 16.110 0.638. Section 7.3 309 66. Use washer cross sections: a washer has inner radius dx dy 2y 2y 11 60. g ( y) 1 , and 1 2 r 1 5/8 area (R2 2 1 dy 2y 1 2y dy 2 4 2 3/2 y 3 2 2 dy dx 61. f (x) 2 2 x2 a 1 a2 y 2)2 5/8 y 2)2] 4 b a2 a2 a y 2 dy 4b y 2 dy a a2 2 2.997. 22 2 ab 2x, and 67. (a) Put the bottom of the bowl at (0, 2 (2x)2 dx 1 2 x2 0 4x2 dx evaluates, 1 using NINT, to (a2 y 2) dy 13 y 3 a2y a 2 2 (3x x) 1 2 (3 2x) dx evaluates, using NINT, (52 0 to dh dt 44.877. dy dx 63. f (x) 1 2 x x2 2x 1.5 1 0.5 dx x2 2 1 dx 0.5 0.5 1 2 2 5 3/2 4 2 x 3 2 4 3 a 6.283 25 3/2 x Since A1 (a2 x 2) dx 49 3 x 2 and area ( a2 13 x 3 a2x x2 r2 1 1 x 2). The a a 13 a 3 a3 13 a 3 h 2 volume of cone y and area h r1 y2 h r2 1 2y h y2 . The h2 volume is 51.313 22 h) A1. A2 23 R. 3 a2 x 2)2 a A( y ) theorem. Thus, volume of hemisphere R3 a2 5 volume of cylinder m/sec. 43 a. 3 A2, the two volumes are equal by Cavalieri’s 13 R 3 ( dx Right circular cylinder with cone removed cross sectional h2 r2 1 65. Hemisphere cross sectional area: ( R area: R2 1 120 (b) A cross section has radius x 9 3/2 4 4 1 (0.2) 24 2 5 dx 4 x 1 1 dV A dt a3 1 11 5 . 24 . The rate of rise is 1 1 2 a h) 1 and the area of a cross section is , and x 5 x h2(3a 3 x 2 2 ha volume is 1.5 2 dy dx 12) A(x) 1.5 2 x 2x 64. f (x) 4, y 68. (a) A cross section has radius r , and x2 1 2x y 2). 2x, and (b) For h 3 ( a2 The volume for height h is 53.226. 3 y 2)2 horizontal cross section is ( a ha dy dx a). The area of a 2 0 62. f (x) y 2, and y 2. The volume is 4b 5 2 a2 b a2 (b a 5 16 1 3 a2 4b 5/8 y 2, outer radius R r 2) [(b 1 2 a2 b 0 r2 1 2y h y2 dy h2 r2 y 12 r h. 3 y2 h y3 3h2 h 0 310 Section 7.4 s Section 7.4 Lengths of Curves 3. (a) x (pp. 395–401) cos2 y dy. 1 cos y, so Length 0 (b) Quick Review 7.4 1. 1 1 x 1 2. x)2, which, since x (1 x2 4 1, equals x2 , which, since x 2 1 1 (tan x)2 equals sec x. [ 1, 2] by [ 1, 4] 2, equals (c) Length 2x x or . 2 2 1 3. x2 1. 2x x or x 4. (a) x (sec x)2, which, since 0 x 2 3.820 y 2) y(1 , 1/2 , so 1/2 Length y2 1 1/2 4. 12 x x 4 1 since x 5. 1 1 2 12 x 16 x2 0, equals 4 4x (x 2 1 4 1 x2 which, x 2 2 10 3 3 [ 1, 2] by [ 1, 1] (c) Length 6. f (x) has a corner at x d 7. (5x 2/3) dx (b) , 2 cos x. equals 8. d ( dx is undefined at x 2y 2x 2 2y 1 y 0. f (x) has a cusp x2 1 is undefined for x 5(x 3)4/5 10. d 1 dx 4x 3 1)2 (y 4 x 2 Length 3. 2x 2, and 1 1. Then y 2x 1 1 2x 1 2 , and 2 dx. (b) 2 has a corner at x 2. cos x is undefined for x 3(sin x)2/3 sin x 2x 7 f (x) has a vertical tangent there. 9. 1, so x y x 3) 1.047 5. (a) y 2 4. there. 5 dy. . 2 cos2 x, which, since cos 2x 4)2 x2 y2 1 k, [ 1, 7] by [ 2, 4] where k is any integer. f (x) has vertical tangents at these (c) Length 9.294 values of x. 6. (a) y Section 7.4 Exercises cos x x sin x Length cos x x sin x, so x 2 sin2 x dx. 1 0 1. (a) y 2x, so 2 Length 1 (b) 2 (2x)2 dx 4x 2 dx. 1 1 1 (b) [0, by [ 1, 4] (c) Length 7. (a) y [ 1, 2] by [ 1, 5] (c) Length 2. (a) y 6.126 2 4.698 /6 0 (b) y 0 sec x, so Length 1 tan x dx ln ( sec x ) 4 sec x dx. /3 (b) 0, 3 , 0 by [ 3, 1] (c) Length 2.057 1 tan x, so Length 6 (c) Length by [ 0.1, 0.2] 0.549 tan2 x dx. ...
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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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