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solution-A5

# solution-A5 - MATHEMATICS 222 CALCULUS III ASSIGNMENT 5...

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MATHEMATICS 222 CALCULUS III ASSIGNMENT 5 ANSWERS 1. The function f is defined by 2 xy x 2 + y 2 ) for ( x, y ) 6 = (0 , 0) and f (0 , 0) = 0 (a) ∂f ∂x (0 , 0) = lim h 0 ˆ 2 h (0) h 2 + 0 2 - 0 ! /h = 0 Similarly ∂f ∂y (0 , 0) = 0 For any other point, simply apply the quotient rule to get ∂f ∂x ( x, y ) = 2 y ( y 2 - x 2 ) ( x 2 + y 2 ) 2 and ∂f ∂y ( x, y ) = 2 x ( x 2 - y 2 ) ( x 2 + y 2 ) 2 (b) In order to find the second partial derivatives at ( x, y ) 6 = (0 , 0) apply the quotient rule again to obtain f xx = 4 xy ( x 2 - 3 y 2 ( x 2 + y 2 ) 3 f yy = 4 xy ( y 2 - 3 x 2 ) ( x 2 + y 2 ) 3 f xy = 2(6 x 2 y 2 - x 4 - y 4 ) ( x 2 + y 2 ) 3 Substituting in to (recall the correction y 2 f yy ) x 2 f xx + 2 xyf xy + y 2 f yy = 4 xy ( x 2 + y 2 ) 3 h x 4 - 3 x 2 y 2 + y 4 - 3 x 2 y 2 + 6 x 2 y 2 - x 4 - y 4 i = 0 2. (a) if V = x 3 - 3 xy 2 then V xx = 6 x and V yy = - 6 x which add up to zero as required. (b) if V = e - y cos x then V xx = - e - y cos x and V yy = ( - 1) 2 e - y cos x which add up to zero. (c) If, for ( x, y ) 6 = (0 , 0), V = tan - 1 y x then V x = - y x 2 + y 2 and V y = x x 2 + y 2 . Hence V xx = - 2 xy ( x 2 + y 2 ) 2 and V yy = 2 xy ( x 2 + y 2 ) 2 ) which add up to zero as required.

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solution-A5 - MATHEMATICS 222 CALCULUS III ASSIGNMENT 5...

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