HW5 - Math 191 6.5: 6 a. dx dy HW 5 Solutions = 1 + y -1/2...

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Unformatted text preview: Math 191 6.5: 6 a. dx dy HW 5 Solutions = 1 + y -1/2 dx dy 2 = 1 + y -1/2 2 S = 2 2 (y 1 + 2 y) 1 + (1 + y -1/2 )2 dx. 2 y+2 sqrt(y)=x 1.9 1.8 1.7 1.6 1.5 y 1.4 1.3 1.2 1.1 1 3 3.2 3.4 3.6 3.8 x 4 4.2 4.4 4.6 4.8 b. c. S 51.33. 6.5: 12 y = 2 2 2 d x = 2y-1 dx = 2; S = c 2x 1 + dx dy = 1 2(2y-1) 1 + 4dy = 2 5 1 (2y- dy dy 1)dy = 2 5; Geometry formula: r1 = 1, r2 = 3, slant height = (2 - 1)2 + (3 - 1)2 = 5 Frustum surface area = (1 + 3) 5 = 4 5 in agreement with the integral value. x 1 2+2 dy dx 28 3 1 = 2 x-1/2 dy dx 2 6.5: 14 = 1 4x S= 15/4 3/4 2 x 1+ 1 4x dx = 2 15/4 3/4 1 x + 4 dx = 2 2 3 x+ 15/4 1 3/2 4 3/4 = 6.5: 36 dx dt = a(1-cos t) and S= 6.5: 38 dx dt = [a(1 - cos t)]2 + (a sin t)2 = a 2 1 - cos t 2 2 2yds = 0 2a(1 - cos t) a 2 1 - cos t = 2 2a2 0 (1 - cos t)3/2 dt. dy dt = a sin t dx 2 dt + dy dt 2 2 1 dx 2 = h and dy = r + dy = h2 + r2 S = 2ds = 0 2rt h2 + r2 dt = dt dt dt 2 1 1 2r h2 + r2 0 tdt = 2rt h2 + r2 t2 = r h2 + r2 . Check: slant height is h2 + r2 Area 0 is r h2 + r2 . 6.6: 7 The force required to haul up the rope is equal to the rope's weight, which varies steadily and is proportional to x, the length of the rope still hanging: F (x) = 0.624x. The work done is: W = 50 50 F (x)dx = 0 0.624xdx = 780J. 0 1 6.6: 9 The force required to lift the cable is equal to the weight of the cable paid out: F (x) = (4.5)(180 - x) 180 180 where x is the position of the car off the first floor. The work done is: W = 0 F (x) = 4.5 0 (180 - x)dx = 72, 900ft lb. 2 ...
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This note was uploaded on 06/19/2008 for the course MATH 1910 taught by Professor Berman during the Spring '07 term at Cornell University (Engineering School).

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HW5 - Math 191 6.5: 6 a. dx dy HW 5 Solutions = 1 + y -1/2...

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