A8_soln - Math 237 Assignment 8 Solutions 1. Consider the...

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Unformatted text preview: Math 237 Assignment 8 Solutions 1. Consider the maps F : R 2 R 2 defined by ( u, v ) = F ( x, y ) = ( e x + y , e x- y ). a) Show that F has an inverse map by finding F- 1 explicitly. Solution: We have u = e x + y and v = e x- y . Hence, we see that uv = e x + y e x- y = e 2 x thus x = 1 2 ln( uv ). Similarly, u v = e x + y e x- y = e 2 y , so y = 1 2 ln( u/v ). Hence the inverse mapping is ( x, y ) = F- 1 ( u, v ) = ( 1 2 ln( uv ) , 1 2 ln ( u v )) . b) Find the derivative matrices DF ( x, y ) and DF- 1 ( u, v ) and verify that DF ( x, y ) DF- 1 ( u, v ) = I . Solution: We have DF ( x, y ) = parenleftbigg e x + y e x + y e x- y- e x- y parenrightbigg and DF- 1 ( u, v ) = parenleftbigg 1 2 u 1 2 v 1 2 u- 1 2 v parenrightbigg . Hence DF ( x, y ) DF- 1 ( u, v ) = parenleftbigg e x + y e x + y e x- y- e x- y parenrightbiggparenleftbigg 1 2 u 1 2 v 1 2 u- 1 2 v parenrightbigg = parenleftbigg e x + y u e x- y v parenrightbigg = I, since u = e x + y and v = e x- y . c) Verify that the Jacobians satisfy ( x,y ) ( u,v ) = bracketleftBig ( u,v ) ( x,y ) bracketrightBig- 1 . Solution: We have ( x, y ) ( u, v ) = det parenleftbigg 1 2 u 1 2 v 1 2 u- 1 2 v parenrightbigg =- 1 2 uv and ( u, v ) ( x, y ) = det parenleftbigg e x + y e x + y e x- y- e x- y parenrightbigg =- 2 e 2 x . Hence, since uv = e 2 x we get ( x,y ) ( u,v ) =- 1 2 uv =- 1 2 e 2 x = bracketleftBig ( u,v ) ( x,y ) bracketrightBig- 1 2. Find the Jacobian for the following mappings. a) ( x, y ) = T ( r, ) = ( r cos , r sin ). Solution: ( x,y ) ( r, ) = det parenleftbigg cos - r sin sin r cos parenrightbigg = r cos 2 + r sin 2 = r b) ( x, y, z ) = T ( r, , z ) = ( r cos , r sin , z )....
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This note was uploaded on 10/21/2010 for the course MATHEMATIC stats taught by Professor Various during the Spring '10 term at Waterloo.

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A8_soln - Math 237 Assignment 8 Solutions 1. Consider the...

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