Partial Solutions to Brainfood #5

These are partial answers so that you can check to see if you are on target. We expect your actual answers to be "fleshed out" a bit more.
  1. (1/2, 5/2)
  2. Since the gradient is (1/2, 5/2), the norm of the gradient is sqrt(26)/2. Thus the gradient points in the direction u = (1/sqrt(26), 5/sqrt(26)). The directional derivative in the direction of the gradient is |grad(f)(1,1/2)| = sqrt(26)/2

    Therefore a tangent vector to the surface is (1/sqrt(26), 5/sqrt(26), sqrt(26)/2).

  3. (-5/2, 1/2) is a tangent vector to the level curve. Normalize it to get a direction.
  4. Clearly (0,0) is a minima. From the graph of the function, we suspect that there are maxima along the 45-degree diagonals in the (x,y)-plane. If we parametrize the diagonal by (t,t), then solve for the critical points along that line, we find that there are critical points at (sqrt(5)/2, sqrt(5)/2)) and (-sqrt(5)/2, -sqrt(5)/2)). The other diagonal is parametrized by (t, -t) and along this line the critical points are (sqrt(3)/2, sqrt(3)/2) and (-sqrt(3)/2, -sqrt(3)/2).

    Non-symmetric solutions (saddle points) occur somewhere near (0.259, -0.966), (-0.259, 0.966), (-0.966, 0.259), (0.966, -0.259).

  5. The gradients should all be orthogonal (perpendicular) to the level sets.
  6. The key observation is that to minimize walking uphill, you want to stay on level curves as much as possible. Thus one solution is to walk directly downhill to the level curve of the saddle points, then walk along this level curve until you are at the saddle point at (0.259, -0.966), then climb straight up the mountain, always following the gradient. This results in the minimum distance walked uphill.

Last modified: Mon Nov 4 14:31:25 1996