## Problem on a path of steepest descent

Let $f(x,y) = xy$.
a) Sketch the level curves of $f$.
b) Sketch the path of steepest descent starting at $(1,2)$.
b) Find the path of steepest descent starting at $(1,2)$.
• ## Solution

#### (a) Level curves of $f$

Recall that
A sketch of these curves is:

#### (b) Sketch of path of steepest descent

Recall that
Now, we need to relate the direction of steepest descent to the level curves that we have already drawn.
Recall that
Hence, we seek a path that is always perpendicular to the level curves.
Starting at $(1,2)$ there are two perpendicular directions. We select the one that is pointing toward smaller values of $f$. The path begins as:
When the path cross the $c=1$ level curve, it must be perpendicular to that level curve:
Similarly, when the path cross the $c=0$ level curve, it must be pointing directly to the left:
We now draw the remaining directions at the $c=-1$ and $c=-2$ level curves. We complete the figure by connecting the arrows with a continuous path:

#### (c) Computation of path of steepest descent

Dividing the second equation by the first, we get
$$\frac{dy/dt}{dx/dt} = \frac{dy}{dx} = \frac{-g(t) x}{- g(t) y} = \frac{x}{y}.$$
Separating the variables, $$y dy = x dx.$$
Integrating, $$\frac{1}{2} y^2 = \frac{1}{2} x^2 + c.$$
Because $x=1, y=2$ is on the curve, we find that $c=3/2$.
Thus, the curve traced by the path of steepest descent is $$\frac{1}{2} y^2 = \frac{1}{2}x^2 + \frac{3}{2}.$$