solutions for homework 4 - MODEL ANSWERS TO HWK#11(18.022...

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Z MODEL ANSWERS TO HWK #11 (18.022 FALL 2010) (1) (6.1.1) (a) x 0 = ( - 3 , 4) so || x 0 || = 5, hence Z Z 2 fds = 5 (2 - 3 t + 8 t - 2) dt = 50 . x 0 (b) x 0 = ( - sin t, cos t ) so || x 0 || = 1, hence Z Z fds = (cos t + 2 sin t ) = 4 . x 0 (2) (6.1.3) x 0 = (1 , 1 , 3 p t/ 2) so || x 0 || = p 2 + 9 t/ 4, hence Z Z 3 t + t 3 / 2 p fds = 2 + 9 t/ 4 dt . t + t 3 / 2 x 1 We perform the substitution t = 4 ( u - 2) and get that this equals 9 35 / 4 4 p u/ 9 du = 1 [17 3 / 2 - 35 3 / 2 ] . 27 17 / 4 (3) (6.1.7) x 0 = (cos t, sin t ), hence Z Z / 2 F ds = [( - cos t + 2) cos t + sin 2 t ] dt . · x 0 It easy to see by change of variable that R 0 / 2 sin 2 t = R 0 pi/ 2 cos 2 t and so the above intergal equals Z / 2 2 cos tdt = 2 . 0 (4) (6.1.11) x 0 = ( - 3 sin 3 t, 3 cos 3 t ), hence Z Z xdy - ydx = 3 cos 2 3 t + 3 sin 2 3 t = 3 . x 0 (5) (6.1.13) x 0 = (2 e 2 t cos 3 t - 3 e 2 t sin 3 t, 2 e 2 t sin 3 t + 3 e 2 t cos 3 t ), hence Z Z 2 4 t xdx + ydy 2 e = = 1 - e - 4 . ( x 2 + y 2 ) 3 / 2 e 6 t x 0 (6) (6.1.16) A parametrization of the curve (there is more than one) is x ( t ) = ( t, 5 - 4 t, 2 t - 1) for 1 t 2. We have x 0 = (1 , - 4 , 2), hence the work is Z 2 Z 2 1 [ t 2 (5 - 4 t ) - 4(2 t - 1) + 2(6 t - 5)] dt = [ - 4 t 3 + 5 t 2 + 4 t - 6] dt = - 3 .
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