ps12sol - 18.02 Fall 2008 Problem Set 12, Part B Solutions...

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Unformatted text preview: 18.02 Fall 2008 Problem Set 12, Part B Solutions 1. a) x = cos t , y = sin t , z = t , so dx =- sin t dt , dy = cos t dt and dz = dt ; the range of integration is 0 t 2 . Hence Z ~ F d~ r = Z 2 ( a sin t + b cos t sin 2 t ) (- sin t ) dt + (2 cos 2 t sin t ) cos t dt + (cos t cos t- t 2 ) dt = Z 2 (- a sin 2 t- b sin 3 t cos t + 2 cos 3 t sin t + cos 2 t- t 2 ) dt =- a- b 1 4 sin 4 t 2 +- 1 2 cos 4 t 2 + - t 3 3 2 = (1- a ) - 8 3 / 3 . b) ~ F = ( R y- Q z ) +( P z- R x ) +( Q x- P y ) k = (0- 0) +( a cos z- cos z ) +(4 xy- 2 bxy ) k = h , ( a- 1) cos z, 2(2- b ) xy i . Therefore ~ F is conservative if and only if a = 1 and b = 2. c) By integrating f x = sin z + 2 xy 2 we obtain f = x sin z + x 2 y 2 + g ( y, z ). Comparing f y = 2 x 2 y + g y = x 2 y we obtain g y = 0, so in fact g = g ( z ). Finally f z = x cos z + g ( z ) = x cos z- z 2 gives g ( z ) =- z 2 , so g =- 1 3 z 3 + c . The potential function is therefore f ( x, y, z ) = x sin z + x 2 y 2- 1 3 z 3 + c ....
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This note was uploaded on 04/28/2009 for the course MATH 18.02 taught by Professor Auroux during the Fall '08 term at MIT.

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ps12sol - 18.02 Fall 2008 Problem Set 12, Part B Solutions...

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