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hw8_solutions - Solutions for Homework 8 Math 20E Winter...

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Solutions for Homework 8, Math 20E, Winter 2009 Page 559 Problem 5 Let P ( x, y ) = f ( x ) and Q ( x, y ) = f ( y ). Then ∂P ∂y = ∂y f ( x ) = 0 and ∂Q ∂x = ∂x f ( y ) = 0, since f ( x ) does not depend on y and f ( y ) does not de- pend on x . Thus by corollary 1 on page 557, there is a function g ( x, y ) such that F = g . Problem 6a Let r ( x, y, z ) = ( x, y, z ) and r ( x, y, z ) = ( x 2 + y 2 + z 2 ) 1 2 = r ( x, y, z ) . Then 1 r = ( x 2 + y 2 + z 2 ) - 1 2 , so 1 r = ( - 1 2 ( x 2 + y 2 + z 2 ) - 3 2 (2 x ) , - 1 2 ( x 2 + y 2 + z 2 ) - 3 2 (2 y, - 1 2 ( x 2 + y 2 + z 2 ) - 3 2 (2 z )) = ( - x r 3 , - y r 3 , - z r 3 ) = - r r 3 . Problem 6b There are many ways to describe what it means to move a particle ”to ”. Here is one: Since F = 1 r , we know that the work done in moving a particle from r 0 to another point q 0 is independent of the path along which the particle travels (theorem 7 on page 551), and in fact for any curve γ ( t ) connecting them (oriented to start at r 0 and end at q 0 ) we have Work = γ F · d s = 1 r ( q 0 ) - 1 r ( r 0 ) = 1 q 0 - 1 r 0 . For a fixed r 0 , define a curve c b ( t ) = (1 + t ) r 0 , where t [0 , b ]. Since we can
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