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Unformatted text preview: Math 237 Assignment 3 Due: Friday, Oct 10th 1. Determine all points where the function is differentiable. a) f ( x,y ) = x 1 / 3 Solution: Observe that f x ( x,y ) = 1 3 x 2 / 3 which does not exist for all points (0 ,y ), hence f is not differentiable at all points (0 ,y ), y ∈ R . We also have f y ( x,y ) = 0 and so since f x and f y are both continuous for all ( x,y ) ∈ R 2 , x 6 = 0 we have that f is differentiable for all points ( x,y ) ∈ R 2 , x 6 = 0. b) g ( x,y ) = ( x 2 y 2 x 2 + y 2 if ( x,y ) 6 = (0 , 0) if ( x,y ) = (0 , 0) . Solution: If ( x,y ) 6 = (0 , 0) then we have g x ( x,y ) = 2 xy 2 ( x 2 + y 2 ) 2 x ( x 2 y 2 ) ( x 2 + y 2 ) 2 which is continuous for ( x,y ) 6 = (0 , 0). By symmetry we have g y ( x,y ) is continuous for ( x,y ) 6 = (0 , 0) and so g is differentiable for all points ( x,y ) 6 = (0 , 0). For ( x,y ) = (0 , 0) we must use the definition of differentiability. Observe that g x (0 , 0) = lim h → f ( h, 0) f (0 , 0) h = 0 g y (0 , 0) = lim h → f (0 ,h ) f (0 , 0) h = 0 So, both partial derivatives exist at (0 , 0) and L (0 , 0) ( x,y ) = 0. Thus, R 1 , (0 , 0) ( x,y ) = x 2 y 2 x 2 + y 2 . Then we have  R 1 , (0 , 0) ( x,y )  p x 2 + y 2 0 = x 2 y 2 ( x 2 + y 2 ) 3 / 2 ≤ ( x 2 + y 2 )( x 2 + y 2 ) ( x 2 + y 2 ) 3 / 2 = p x 2 + y 2 Since lim ( x,y ) → (0 , 0) p x 2 + y 2 = 0 by continuity we have that lim ( x,y ) →...
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This note was uploaded on 04/18/2010 for the course MATH 237 taught by Professor Wolczuk during the Spring '08 term at Waterloo.
 Spring '08
 WOLCZUK
 Math

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