## Problem on finding the path of steepest ascent 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. Recall that 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 $$y = \frac{1}{4} x^2.$$