# Traveling between points on the surface $C(x,y)=e^{-2x^2-3y^2}$

We have a start point $(0,0)$ and an end point $(1,2)$, we would like to travel from the start to the end point. The surface is given by the equation $C(x,y)=e^{-2x^2-3y^2}$ and we would like to always travel in the direction of where the slope is the steepest (meaning that in each point we travel in the gradients direction).

This means we will get some path between the points, and this path can be seen as the graph of the function $y=\phi (x)$.

I want to determine what this function $\phi$ is. How do I approach this problem?

• How can you tell that travelling along the steepest descent you reach (1,2) from (0,0) ? – G Cab Aug 20 '17 at 17:43
• That I'm not sure of, but I guess it's assumed in my problem that there is such a path – osk Aug 20 '17 at 17:58

This is a bit misleading, since the gradient vanishes at $(0,0)$. But try following the negative of the gradient, starting at $(1,2)$. As a hint, show that the curve you want has slope $\dfrac{3y}{2x}$ at $(x,y)$. You can separate variables to find the curve passing through $(1,2)$. Unfortunately, you'll see that to follow the negative if the gradient will take you away from the origin, not toward it. So the original question is flawed.