exam2-v1-solutions

exam2-v1-solutions - Math233 Exam 2 - Solutions x2 + y 2...

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Math233 Exam 2 - Solutions Fall 2006 Problem 1 (15pts). Determine the set of points at which the function is continuous. f ( x,y ) = ( p x 2 + y 2 ln( x 2 + y 2 ) if ( x,y ) 6 = (0 , 0) 1 if ( x,y ) = (0 , 0) f is continuous on R 2 except maybe at (0 , 0) (as a product of continuous functions). To check continu- ity at (0 , 0) , we examine the limit of f ( x,y ) as ( x,y ) tends to (0 , 0) . For that we use polar coordinates (see very similar example done in class). Let x = r cos θ and y = r sin θ , then x 2 + y 2 = r 2 and ( x,y ) (0 , 0) is equivalent to r 0 . lim ( x,y ) (0 , 0) p x 2 + y 2 ln( x 2 + y 2 ) = lim r 0 r 2 ln( r 2 ) = lim r 0 2 r ln( r ) = lim r 0 2 ln( r ) 1 /r = lim r 0 2 1 /r - 1 /r 2 (L’Hopital’s Rule) = lim r 0 - 2 r = 0 The limit exists but it different from f (0 , 0) , therefore f is not continuous at (0 , 0) . Problem 2 (10pts). Give the definition of the partial derivatives by completing the following: f x ( a,b ) = lim h 0 f ( a + h,b
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exam2-v1-solutions - Math233 Exam 2 - Solutions x2 + y 2...

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