MIT18_024S11_soln-pset7

MIT18_024S11_soln-ps - 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

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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 ∂F/∂y = ; = . ∂x ∂F/∂z ∂y 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 )( 2 )+ D 2 2 f ( u, v )( ) , = f ( u, v ) · ( ∂u/∂y, ∂v/∂y D 1 f ( u, v )(1 /x ) , = f ( u, v ) · ( ∂u/∂z, ∂v/∂z D 2 f ( u, v )(1 /x ) . ∂z An easy computation then gives yD 1 f ( u, v ) z D f ( u, v ) = + ; = 1 . xD 2 f ( u, v ) x D 2 f ( u, v ) Thus x + y = z = g ( x, y ) . 2. (a) First, DF has the form of a block matrix. I 0 ( 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 )) i i where here F i ( x , y x · e i . Then = δ ij for 1 i, j n and =0 j 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 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 . The new matrix ( a , b ) is lower triangular (everything above the main diagonal is ˜ zero). Recall in this case that det ( ( a , b )) = 1 and since row
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This note was uploaded on 01/18/2012 for the course MATH 18.024 taught by Professor Christinebreiner during the Spring '11 term at MIT.

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MIT18_024S11_soln-ps - 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

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