BRIEF ANSWERS FOR TEST 3

• T or F?
  • (If grad f(a,b) exists, then z = f(x,y) is differentiable at (a,b).) FALSE! The gradient of the function f(x,y) = (2xy)/(x^2+y^2) exists at (a,b) = (0,0), but the function is not differentiable there.
  • (If u is normal to the direction of maximum increase of = f(xx,y) at (x,y) = (a,b), then D_u f(a,b) = 0.) TRUE! Both conditions are equivalent to grad f(a,b) dot u = 0.
  • (Assume that z = f(x,y) is differentiable at x,yy)=(a,b). Then D_(-u) f(a,b) = -D_u f(a,b).) TRUE! This follows from D_u f(a,b) = grad f(a,b) dot u.
  • (If z = f(x,y) is continuous on the rectangle R = [a,b]x[c,d], then the double integral of f over R exists and equals the integral from c to d of the integral from a to b of f(x,y).) TRUE! This theorem was given in class.
• (Let F(x,y) = x + e^y. If x = r*cos(theta) and y = r*sin(theta), then find the r and theta partials of F at (1,Pi/4) in two ways. First use the Chain Rule. Then find these partials directly.) The chain rule is that F_r = F_x*x_r + F_y*y_r, and (for t = theta) F_t = F_x*x_t + F_y*y_t. Both calculations should lead to:
  • F_r = (1+e^(1/square_root(2)))/square_root(2)
  • F_t = (-1+e^(1/square_root(2)))/square_root(2)

• This is a problem to locate and classify the critical points of f(x,y) = 2xy+x+y+y^2 on the region R defined by 2x^2 + 2xy + y^2 less-or-equal to 2.
  • (Locate all critical points of f(x,y) in the interior of R.) Setting grad f = 0 we find that (x,y) = (0,-1/2) is the only critical point.
  • (Calculate the Hessian at each critical point found.) The Hessian is the 2x2 matrix [ab/cd] = [f_xx f_xy/f_yx f_yy]. In this problem the Hessian is [02/22].
  • (Use the method of Lagrange Multipliers to locate all extreme points of f(x,y) on the boundary of R.) We set grad f = lambda * grad g and solve the resulting equations, which are:
    • 2y+1 = lambda(4x+2y),
    • 2x+2y+1 = lambda(2x+2y),
    • g(x,y) = 0.

    If lambda = 0, then we are led to grad f = 0, which occurs at no boundary point. Therefore we may assume lambda is not zero. The difference of the first two equations is -2x = lambda(2x). If x = 0, we solve to find that y = +/- square_root(2). Thus (0,+/-square_root(2)) must be considered. Otherwise x is not zero and -2x = lambda(2x) implies that lambda = -1. This leads to 4x + 4y + 1 = 0, which together with 0 = g(x,y) = x^2 + (x+y)^2 - 2 leads to x^2 = 2 - 1/16 and y = -x - 1/4. Thus, (x,y) = (+/- square_root(31)/4, -1/4 -/+ square_root(31)/4) must be considered.
  • (Describe all critical and extreme points of f(x,y) in R.) By the second part, the critical point found is a saddle point. On the boundary we get that (0,square_root(2)) is where the absolute maximum occurs. The point (0,-square_root(2)) is where a local maximum occurs. The points (+/- square_root(31)/4, -1/4 -/+ square_root(31)/4) are where the absolute minimum occurs.
• (Find the centroid of the region in the xy-plane that is bounded above by y=0, below by y=x^2-x, on the left by x=0 and on the right by x=1.) M_x = -1/60, M_y = 1/12, M = 1/6, so the centroid = (bar(x),bar(y)) = (1/2,-1/10).