sol1 - Sample Problem Solutions for Exam One 1. Show that...

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Unformatted text preview: Sample Problem Solutions for Exam One 1. Show that lim ( x,y ) (0 , 0) ( x 3 y 2 + y 6 ) / ( x 6 + y 4 ) does not exist. We know that if the limit exists, then it is the same along every path approaching (0,0). First try y = 0, where the limit is clearly 0. Next try y = x 3 / 2 , where we obtain lim x x 6 + x 9 2 x 6 = lim x 1+ x 3 2 = 1 2 . Thus the limit does not exist. 2. Show that lim ( x,y ) (0 , 0) ( x 4- 2 x 3 y + y 4 ) / ( x 2 + y 2 ) = 0. Using the polar coordinates r = p x 2 + y 2 and , we obtain F ( x,y ) = r 2 ( cos 4 - 2 cos 3 sin + sin 4 ), so that F ( x,y ) 4 r 2 . Since lim ( x,y ) (0 , 0) r = 0, it follows that the limit is 0. 3. Show that if F ( x,y ) is differentiable, then the first partials exist. We have A and B such that F ( x,y )- F ( x ,y ) = A ( x- x ) + B ( y- y ) + ( P- P ), where ( P ) = x y is a vector functon with lim P P = 0. Thus F ( x,y )- F ( x ,y ) = A ( x- x ) + B ( y- y ) + x ( x- x ) + y ( y- y ), so F ( x,y )- F ( x ,y ) x- x = A + x . Then F x ( x ,y ) = lim x x F ( x,y )- F ( x ,y ) x- x = lim x x A + x = A . Similarly F y ( x ,y ) exists. 4. Show that any differentiable function is continuous. We have F ( P )- F ( P ) = J F ( P- P ) + F ( P- P ), so that lim P P F ( P ) = F ( P ) + ( J F + F )( P- P ) = F ( P ) + J F (0) = F ( P ). 5. Let f ( x,y ) = xy/ ( x 2 + y 2 ) for ( x,y ) 6 = (0 , 0) and f(0,0)=0. Show that f x and f y both exist at (0,0) but that f is not continuous there. f x (0 , 0) = lim h f ( h, 0)- f (0 , 0) h = lim h 0 = 0 and similarly f y (0 , 0) = 0. Substituting y = x , lim ( x,y ) (0 , 0) f ( x ) = lim x x 2 2 x 2 = 1 2 6 = f (0 , 0), so that f is not continuous at (0,0). 6. Let f ( x,y ) = x 4 x 2 + y 2 for ( x,y ) 6 = (0 , 0) and f (0 , 0) = 0. Show that f is differentiable at (0,0). f x (0 , 0) = lim h f ( h, 0)- f (0 , 0) h = lim h h = 0 and f y (0 , 0) = lim h f (0 ,h )- f (0 , 0) h = lim h 0 = 0. Thus the linear approximation L ( x,y ) = f (0 , 0) + Ax + By = 0 near (0,0). Now lim ( x,y ) (0 , 0) f ( x,y )- L ( x,y ) x 2 + y 2 = lim ( x,y ) (0 , 0) x 4 ( x 2 + y 2 ) 3 / 2 = lim r rcos 4 = 0, so that f is differentiable at (0,0). 7. Define the directional derivative D v F of a function F at a point P in direction v and show that if F is differentiable, then D v F = F ( P ) v . D v F = lim h F ( P + hv )- F ( P ) h . Given that F is differentiable, we have F ( P + hv ) = F ( P ) + F ( P )( hv ) + ( P + hv )( k hv k ) where lim Q P ( Q ) = 0. Thus D v ( F ) = lim h F ( P ) v + ( P + hv )( k hv } /h ) = F ( P ) v , since k hv k /h = 1 is bounded....
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This note was uploaded on 01/22/2012 for the course MAA 4103 taught by Professor Staff during the Fall '08 term at University of Florida.

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sol1 - Sample Problem Solutions for Exam One 1. Show that...

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