- (1/2, 5/2)
- 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).

- (-5/2, 1/2) is a tangent vector to the level curve. Normalize it to get a direction.
- 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).

- The gradients should all be orthogonal (perpendicular) to the level sets.
- 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