{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–3. Sign up to view the full content.

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 0 x 6 + x 9 2 x 6 = lim x 0 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 = x 2 + y 2 and θ , we obtain F ( x, y ) = r 2 ( cos 4 θ - 2 cos 3 θ sinθ + sin 4 θ ), 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 0 , y 0 ) = A ( x - x 0 ) + B ( y - y 0 ) + η ( P - P 0 ), where η ( P ) = η x η y is a vector functon with lim P P 0 η = 0. Thus F ( x, y 0 ) - F ( x 0 , y 0 ) = A ( x - x 0 ) + B ( y 0 - y 0 ) + η x ( x - x 0 ) + η y ( y 0 - y 0 ), so F ( x,y 0 ) - F ( x 0 ,y 0 ) x - x 0 = A + η x . Then F x ( x 0 , y 0 ) = lim x x 0 F ( x,y 0 ) - F ( x 0 ,y 0 ) x - x 0 = lim x x 0 A + η x = A . Similarly F y ( x 0 , y 0 ) exists. 4. Show that any differentiable function is continuous. We have F ( P ) - F ( P 0 ) = J F ( P - P 0 ) + η F ( P - P 0 ), so that lim P P 0 F ( P ) = F ( P 0 ) + ( J F + η F )( P - P 0 ) = F ( P 0 ) + J F (0) = F ( P 0 ). 5. Let f ( x, y ) = xy/ ( x 2 + y 2 ) for ( x, y ) = (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 0 f ( h, 0) - f (0 , 0) h = lim h 0 0 = 0 and similarly f y (0 , 0) = 0. Substituting y = x , lim ( x,y ) (0 , 0) f ( x ) = lim x 0 x 2 2 x 2 = 1 2 = 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 ) = (0 , 0) and f (0 , 0) = 0. Show that f is differentiable at (0,0). f x (0 , 0) = lim h 0 f ( h, 0) - f (0 , 0) h = lim h 0 h = 0 and f y (0 , 0) = lim h 0 f (0 ,h ) - f (0 , 0) h = lim h 0 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 0 rcos 4 θ = 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 0 F ( P + hv ) - F ( P ) h . Given that F is differentiable, we have F ( P + hv ) = F ( P ) + F ( P )( hv ) + η ( P + hv )( hv ) where lim Q P η ( Q ) = 0. Thus D v ( F ) = lim h 0 F ( P ) v + η ( P + hv )( hv } /h ) = F ( P ) v , since hv /h = ± 1 is bounded. 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 8. Let F x y = x 2 + y 2 2 xy 2 x + y . Find F 1 3 and give the linear approximation L x y about the point. Find the directional derivative D v F at 1 3 in the direction v = 4 / 5 3 / 5 . F (1 , 3) = (10 , 6 , 5) and F = 2 x 2 y 2 y 2 x 2 1 = 2 6 6 2 2 1 So L ( x, y ) = F (1 , 3) + F (1 , 3) x - 1 y - 3 = 2 x + 6 y - 10 6 x + 2 y - 6 2 x + y . 9. Let G : R n R m and F : R m R k be differentiable functions and let H = F ( G ( P )). Show that H is differentiable and furthermore, J H = J F J G . Let Q = G ( P ) and Q 0 = G ( P 0 ). We have G ( P ) - G ( P 0 ) = J G ( P )( P - P 0 ) + η G ( P )( P - P 0 ) and F ( Q ) - F ( Q 0 ) = J F ( Q )( Q - Q 0 ) + η F ( Q )( Q - Q 0 ), where lim P P 0 η G ( P ) = 0 and lim Q Q 0 η F ( Q ) = 0.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 6

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online