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plugin-review-solution - Review excercises for Exam 2 Math...

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Math 255, Winter 2010 Review excercises for Exam 2 1. Let f ( x, y ) = ( ( x 2 + y 2 ) ln( x 2 + y 2 ) for ( x, y ) 6 = (0 , 0) 0 ( x, y ) = (0 , 0) . (a) Is f ( x, y ) continuous at (0 , 0)? (b) Is f ( x, y ) differentiable at (0 , 0)? Solution : (a) To see if f ( x, y ) is continuous at (0 , 0), you need to check if lim ( x,y ) (0 , 0) f ( x, y ) = f (0 , 0) . Since f (0 , 0) = 0, then you must check if lim ( x,y ) (0 , 0) f ( x, y ) = 0 . In polar coordinate, f ( x, y ) = 2 r 2 ln r for r 6 = 0. lim ( x,y ) (0 , 0) f ( x, y ) = lim r 0 2 r 2 ln r = lim r 0 2 ln r r - 2 = lim r 0 2 /r - 2 r - 3 = 0 . So f ( x, y ) is continuous at (0 , 0). (b) To see if f ( x, y ) is differentiable at (0 , 0), you need to check if ∂f ∂x and ∂f ∂y are continuous at (0 , 0. In other words you need to check if lim ( x,y ) (0 , 0) ∂f ∂x ( x, y ) = ∂f ∂x (0 , 0) lim ( x,y ) (0 , 0) ∂f ∂y ( x, y ) = ∂f ∂y (0 , 0) Step 1: Compute ∂f ∂x (0 , 0) and ∂f ∂y (0 , 0): the computation of the partial derivatives has to be divided in two cases since f ( x, y ) is a piecewise defined function around (0 , 0). For ( x, y ) 6 = (0 , 0): ∂f ( x,y ) ∂x = 2 x ln( x 2 + y 2 ) + 2 x and ∂f ( x,y ) ∂y = 2 y ln( x 2 + y 2 ) + 2 y At (0 , 0): ∂f ∂x (0 , 0) = lim h 0 f ( h, 0) - f (0 , 0) h = lim h 0 h 2 ln( h 2 ) h = lim h 0 h ln( h 2 ) = 0 (using L’Hospitals rule) (( x, y ) = ( h, 0) by fixing y = 0). ∂f ∂y (0 , 0) = lim h 0 f (0 ,h ) - f (0 , 0) h = lim h 0 h 2 ln( h 2 ) h = lim h 0 h ln( h 2 ) = 0 (( x, y ) = (0 , h ) by fixing x = 0). Hence ∂f ∂x (0 , 0) = 0 and ∂f ∂y (0 , 0) = 0. Step 2: Compute lim ( x,y ) (0 , 0) ∂f ∂x ( x, y ) and lim ( x,y ) (0 , 0) ∂f ∂y ( x, y ). lim ( x,y ) (0 , 0) ∂f ∂x ( x, y ) = lim ( x,y ) (0 , 0) 2 x ln( x 2 + y 2 ) + 2 x. 1
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Math 255, Winter 2010 The second term obviously goes to 0. We only have to show that the first term (2 x ln( x 2 + y 2 )) in the limit goes to zero. To show this, we use the squeeze theorem (the idea is to compare it with something that can be easily computed in polar coordinates whose limit is easy to compute) | 2 x ln( x 2 + y 2 ) | = | 2 x 2 ln( x 2 + y 2 ) | 6 | 2 p x 2 + y 2 ln( x 2 + y 2 ) | We will be done if we can show lim ( x,y ) (0 , 0) 2 p x 2 + y 2 ln( x 2 + y 2 ) = 0 . To show the above limit goes to zero, we use polar coordinates lim ( x,y ) (0 , 0) 2 p x 2 + y 2 ln( x 2 + y 2 ) = lim r 0 2 r ln r 2 = 0 (using L’Hospitals rule) . Hence we have shown that lim ( x,y ) (0 , 0) ∂f ∂x ( x, y ) = lim ( x,y ) (0 , 0) 2 x ln( x 2 + y 2 ) + 2 x = 0 = ∂f ∂x (0 , 0) . which implies the continuity of ∂f ∂x at (0 , 0). A similar argument can be used to show that lim ( x,y ) (0 , 0) ∂f ∂y ( x, y ) = lim ( x,y ) (0 , 0) 2 y ln( x 2 + y 2 ) + 2 y = 0 = ∂f ∂y (0 , 0) . You can see ∂f ∂x and ∂f ∂y exist near (0 , 0) (they are actually obtained through direct differentiation in Step 1) and continuous at (0 , 0). So by Theorem on Page 962, f ( x, y ) is differentiable at (0 , 0). 2. 2 sin( xyz ) = x 2 + y 2 + xz defines z = f ( x, y ) implicitly. a) f (1 , 0) =? b) Compute ∂z ∂x in terms of x , y and z . c) ∂z ∂x (1 , 0) =? Solution : (a) Plug x = 1 and y = 0 into the equation: 0 = 1 + z , and so z = - 1. f (1 , 0) = - 1.
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