Finding a path where the function increases the most rapidly

Question

Consider the differentiable function $z = f(x,y)$. Construct a path $\gamma: \mathbb{R} \to \mathbb{R}^3$ on the graph of $z=f(x,y)$ of $\textbf{ steepest ascent}$ that starts at some point $(x_0,y_0)$ in the graph of $z=f(x,y)$

thoughts:

I know $\nabla f$ is a vector that always points in the direction of the steepest ascent of the function $z=f(x,y)$. Initially, at the point $(x_0,y_0)$, the direction of highest increase is $( f_x(x_0,y_0), f_y(x_0,y_0))$. This occurs at the level curve $f(x,y) = f(x_0,y_0)$

Let $\gamma(t) = (x,y)$ be such a curve. For this path we want its derivative to be in the direction of the gradient at each level curve. That is ,

$$ \dfrac{ \partial f}{\partial x} (a,b) = \dfrac{d x}{d t} $$

and

$$ \dfrac{ \partial f}{\partial y} (a,b) = \dfrac{d y}{d t} $$

and $\gamma(0) = (x_0,y_0)$. So, solving that system of ODE's will yield such a path.

Is this an efficient approach?

Answer 1

You don’t need to solve any ODEs.

Suppose $\nabla f (x_0, y_0) = (a,b)$. As you already know, this is the direction of steepest ascent of $f$.

The parametric equation $t \mapsto (x_0,y_0) + t(a,b)$ is a line in this direction passing through the point $(x_0,y_0)$. To get the desired curve, we just need to project this line vertically onto the surface $z=f(x,y)$. This projection sends the point $(x,y)$ to the point $(x,y, f(x,y))$.

Putting everything together, the desired curve is $$ \gamma(t) = \bigl(x_0 +ta, y_0 +tb, f(x_0+ta, y_0+tb) \bigr) $$ Note that when $t=0$ this curve “starts” at the point $ \bigl(x_0, y_0, f(x_0,y_0) \bigr)$, not at the point $(x_0, y_0)$. The latter point might not even lie on the surface $z=f(x,y)$, so the statement of the question is somewhat flawed.

Having read the question again, I now see that you might be asking for a curve that’s running in the direction of steepest ascent at every point along it’s length, not just at its start point. If so, then you probably do need to solve the ODEs you cited.