m425b-sm2soln-s08

m425b-sm2soln-s08 - MATH 425b SAMPLE MIDTERM 2 SOLUTIONS...

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MATH 425b SAMPLE MIDTERM 2 SOLUTIONS SPRING 2008 Prof. Alexander (1) Let γ ( t ) = t y + (1 - t ) x = x + t ( y - x ) for 0 t 1, and g ( t ) = f ( γ ( t )). By the chain rule, g 0 ( t ) = f 0 ( γ ( t )) γ 0 ( t ) = f 0 ( γ ( t ))( y - x ). By the usual MVT for function on R , f ( y ) - f ( x ) = g (1) - g (0) = g 0 ( u )(1 - 0) = f 0 ( γ ( u ))( y - x ) for some u (0 , 1), so we take ξ = γ ( u ). We know ξ E because E is convex. (2)(a) Let f 1 ( x,y,z ) = x 2 - y 2 + z 2 - 1 , f 2 ( x,y,z ) = xy + xz , and f = ( f 1 ,f 2 ). Then Df ( x ) = ± 2 x - 2 y 2 z y + z x x ² , which at (1 , 1 , - 1) is ± 2 - 2 - 2 0 1 1 ² . Then A ( x,y ) = ± 2 - 2 0 1 ² is invertible so we can solve for ( x,y ) = h ( z ) in a neighborhood; the curve in this neigh- borhood then consists of all points ( h 1 ( z ) ,h 2 ( z ) ,z ). So the parametrization works for the variable z . Next, A ( x,z ) = ± 2 - 2 0 1 ² so similarly we can solve for ( x,z ) in terms of y , i.e. it works for the variable y . But A ( y,z ) = ± - 2 - 2 1 1 ² is not invertible so it might not work for
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m425b-sm2soln-s08 - MATH 425b SAMPLE MIDTERM 2 SOLUTIONS...

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