Exam Solutions 5 - University of Toronto Scarborough...

This preview shows page 1 - 3 out of 4 pages.

University of Toronto Scarborough Department of Computer & Mathematical Sciences MAT B41H 2013/2014 Solutions #5 1. A normal vector to the plane 4 x 8 y z = 3 is (4 , 8 , 1) and a normal vector to the tangent plane to the graph of f ( x,y ) = x 2 + y 2 1 is (2 x, 2 y, 1). For the planes to be parallel we need (2 x, 2 y, 1) = k (4 , 8 , 1), for some k . From the third component we see that k = 1, so we have x = 2 and y = 4. The tangent plane passes through (2 , 4 ,f (2 , 4)) = (2 , 4 , 19) and has equation 4 x 8 y z = 21. 2 x -4 y 19 z 2. (a) We have f (0 , 0) = 0, so we can compute the partial derivatives from the defini- tion. ∂f ∂x (0 , 0) = lim h 0 f ( h, 0) f (0 , 0) h = lim h 0 0 0 h = 0 ∂f ∂y (0 , 0) = lim h 0 f (0 ,h ) f (0 , 0) h = lim h 0 0 0 h = 0 (b) Recall that for a function f to be differentiable at a point a , we need lim x a bardbl f ( x ) f ( a ) D f ( a ) ( x a ) bardbl bardbl x a bardbl = 0. Here we need to evaluate lim ( x,y ) (0 , 0) | x 1 / 3 y 1 / 3 0 (0 , 0) · ( x,y ) | bardbl ( x,y ) bardbl = lim ( x,y ) (0 , 0) | x 1 / 3 y 1 / 3 | radicalbig x 2 + y 2 . If we approach along the line y = x , with x > 0, this limit reduces to lim x 0 + x 2 / 3 2 x 2 = lim x 0 + x 2 / 3 x 2 = lim x 0 + 1 x 1 / 3 2 = . Since this limit is not 0, f is not differentiable at (0 , 0). 3. The rate of change in depth is the directional derivative. (a) The depth will increase most rapidly in the direction of the gradient; i.e., in direction D (1 , 2). Now D = ( 6 xy 2 , 6 x 2 y ) so the rubber duck swims in direction D (1 , 2) = ( 24 , 12). (b) The depth will stay constant if the duck stays on the level set which passes through (1 , 2). Since the gradient is orthogonal to level sets it can proceed in direction (1 , 2) or direction ( 1 , 2) since ± 1 (1 , 2) · D (1 , 2) = ± 1 (1 , 2) · ( 24 , 12) = 0.
MATB41H Solutions # 5 page 2 4. (a) Let x = u 2 v 2 and y = v 2 u 2 . Then g ( u,v ) = f ( x,y ) and the Chain Rule can be applied to give ∂g ∂u = ∂f ∂x ∂x ∂u + ∂f ∂y ∂y ∂u = ∂f ∂x ( 2 u ) + ∂f ∂y ( 2 u ) and ∂g ∂v = ∂f ∂x ∂x ∂v + ∂f ∂y ∂y

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture