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 $(1,2)$.

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 $$\frac{1}{2} y^2 = \frac{1}{2}x^2 + \frac{3}{2}.$$