Consider a surface with height $z(x,y) = 10 - x^2 - 2 y^2$. Find the path of steepest ascent starting at $(2, 1, 4)$. Express your answer as a curve in the $xy$ plane.
Solution
To find the path of steepest ascent, we need information on the direction of steepest ascent at any point.
The path of steepest ascent is the curve in $xy$ which is always tangent to the direction of steepest ascent of $z$.
For the curve $y(x)$ to be tangent to $\nabla z$, its slope must equal the rise-over-run of the 2d gradient vector: $$\frac{dy}{dx} = \frac{-4 y}{-2 x}.$$
Simplifying, the curve in $xy$ must solve $$\frac{dy}{dx}= 2 \frac{y}{x}.$$
This is an ordinary differential equation, which we can solve by separating the $x$'s and $y$'s and integrating.
Moving the $y$'s to the left-hand side and the $x$'s to the right hand side, we get $$\frac{dy}{y} = 2 \frac{dx}{x}.$$
Integrating gives \begin{align}\log y &= 2 \log x + c_1.\end{align}
Taking the exponential of both sides, and replacing the $e^{c_1}$ by $c_2$, we get $$y = c_2 x^2.$$
Because $x=2, y=1$ is on the curve, $c_2 = \frac{1}{4}.$
The curve of steepest ascent is $$y = \frac{1}{4} x^2.$$
