Mathematics 41, section 1 -- Analysis 3

Solutions -- Practice Exam 2

March 18, 1998

I.

• A) The derivative of alpha is alpha'(t)=(8e^4t - e^-t, -4e^4t - 2e^-t). On the other hand, F(alpha(t)) = (3(2e^4t + e^-t) - 2(-e^4t +2e^-t),-2(2e^4t + e^-t)). Simplifying, we get (8e^4t - e^-t, -4e^4t - 2e^-t) = alpha'(t). Hence, alpha(t) is a flow line of F(x,y).
• B) grad f(x,y) = (partial f/partial x, partial f/partial y) = (3x - 2y, -2x) = F(x,y).
• C) Since the gradient vector points in the direction of maximum rate of increase of f, the values will increase along a flow line of grad f.

II.

• A) lim_x->0 f(x,0) = lim_x->0 (2x⁵)/x⁴ = lim_x->0 2x = 0. Next, lim_y->0 f(0,y) = lim_y->0 (y⁴)/y⁴ = lim_y->0 1 = 1. Since these one-variable limits are not the same, lim_(x,y)->(0,0) f(x,y) does not exist.
• B) (partial f/partial x)(0,0) = lim_h->0 (f(h,0)-f(0,0))/h = lim_h->0 (2h⁵)/h⁵ = lim_h->02 = 2. (partial f/partial y)(0,0) = lim_h->0 (f(0,h)-f(0,0))/h = lim_h->0 (h⁴)/h⁵ = lim_h->01/h, which does not exist.
• The graph does not look locally linear, so f(x,y) is not differentiable.

III.

• A) Use D_u g(x,y) = grad g(x,y) dot u. grad g(x,y) = (cos(x-y),-cos(x-y)), so grad g(0,pi) = (-1,1), so D_u g(0,pi) = -u₁ + u₂.
• B) In the direction of grad g(0,pi): u = (-1/sqrt(2),1/sqrt(2)).

IV.

• A) Critical points: (1/sqrt(2),0), (-1/sqrt(2),0).
• B) The first is a sink for grad h, hence a local maximum of h; the second is a source for grad h, hence a local minimum of h.