a4_soln

# a4_soln - Math 237 Assignment 4 Solutions 1 Determine all...

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Unformatted text preview: Math 237 Assignment 4 Solutions 1. Determine all points where the function is differentiable. a) f ( x, y ) = x 3 + y 3 x 2 + y 2 , if ( x, y ) 6 = (0 , 0) , if ( x, y ) = (0 , 0) . Solution: Observe that f x = x 4 +3 x 2 y 2- 2 xy 3 ( x 2 + y 2 ) 2 and f y = y 4 +3 x 2 y 2- 2 x 3 y ( x 2 + y 2 ) 2 are both continuous for ( x, y ) 6 = (0 , 0). Thus, f ( x, y ) is differentiable at all points ( x, y ) 6 = (0 , 0). At (0 , 0) we must use the definition of differentiability. We have f x (0 , 0) = lim h → f (0 + h, 0)- f (0 , 0) h = lim h → h 3 h 3 = 1 , so f y (0 , 0) = 1 by symmetry. Thus, R 1 , (0 , 0) ( x, y ) = f ( x, y )- L , ( x, y ) = x 3 + y 3 x 2 + y 2- x- y =- xy 2- yx 2 x 2 + y 2 . Then lim ( x,y ) → (0 , 0) | R 1 , (0 , 0) ( x, y ) | k ( x, y )- (0 , 0) k = lim ( x,y ) → (0 , 0) | - xy 2- yx 2 | ( x 2 + y 2 ) 3 / 2 Approaching the limit along y = x , we get lim ( x,y ) → (0 , 0) | - x 3- x 3 | ( x 2 + x 2 ) 3 / 2 = lim x → 2 | x 3 | 2 3 / 2 | x 3 | = 1 √ 2 6 = 0 Hence, f is not differentiable at (0 , 0)....
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## This note was uploaded on 11/30/2010 for the course MATH 235/237 taught by Professor Wilkie during the Spring '10 term at Waterloo.

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a4_soln - Math 237 Assignment 4 Solutions 1 Determine all...

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