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MIT18_024S11_soln-pset7

MIT18_024S11_soln-pset7 - Solutions for PSet 7 1(9.8:7 Hint...

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Solutions for PSet 7 1. (9.8:7) Hint: It might help to define a scalar field F ( x, y, z ) = f ( u ( x, y, z ) , v ( x, y, z )) where u, v are as needed. We first assume that x = 0. Given g ( x, y ) = z , we know ∂g ∂F/∂x ∂g ∂F/∂y = ; = . ∂x ∂F/∂z ∂y ∂F/∂z Now, we need only use the chain rule to determine the result. First observe that u = ( y/x 2 , 1 /x, 0) , v = ( z/x 2 , 0 , 1 /x ). Now we compute ∂F = f ( u, v ) · ( ∂u/∂x, ∂v/∂x ) = D 1 f ( u, v )( y/x 2 ) + D 2 2 f ( u, v )( z/x ) , ∂x ∂F = f ( u, v ) · ( ∂u/∂y, ∂v/∂y ) = D 1 f ( u, v )(1 /x ) , ∂y ∂F = f ( u, v ) · ( ∂u/∂z, ∂v/∂z ) = D 2 f ( u, v )(1 /x ) . ∂z An easy computation then gives ∂g yD 1 f ( u, v ) z ∂g D f ( u, v ) = + ; = 1 . ∂x xD 2 f ( u, v ) x ∂y D 2 f ( u, v ) Thus ∂g ∂g x + y = z = g ( x, y ) . ∂x ∂y 2. (a) First, DF has the form of a block matrix. I 0 DF ( x , y ) = n | (1) D f x ( x , y ) | D f y ( x , y ) This comes when we consider F ( x , y ) = ( F 1 ( x , y ) , . . . , F n ( x , y ) , f 1 ( x , y ) , . . . , f m ( x , y )) ∂F i ∂F i where here F i ( x , y ) = x · e i . Then = δ ij for 1 i, j n and = 0 ∂x j ∂y k for 1 i n, 1 k m . The bottom portion of the matrix is exactly what we get based on our determination of D f x , D f y . 1

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(b) DF ( a , b ) = I n | 0 (2) D f x ( a , b ) | D f y ( a , b ) The invertibility of Df y gives that DF is invertible at ( a , b ). That is, recall that an invertible matrix has a row reduction that reduces it to the identity matrix. Using this particular row reduction, reduce the bottom ˜ m rows of DF . The new matrix DF ( a , b ) is
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