Steepest Ascent Theorem

Question

The temperature on the $Oxy$ plane is given by f$(x, y) = 100−x^2 −2y^2$.

Assuming that an ant follows the path of steepest ascent of temperature, find and sketch the path the ant follows if it starts at $(2, 4)$.

I calculated the directional derivative and used the steepest ascent theorem to obtain $4\sqrt{17}$.

I don't think this is right though. My other thought is finding the partial derivatives but then I'd just get $-2x$ and $-4y$. Can I have some help please?

Answer 1

The direction of the steepest ascent is $\nabla f=(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial x})$, which is a vector in the $(x,y)$-plane. Its norm $\|\nabla f\|$ is the ascent. The interpretation is: If you move by $\epsilon$ in the direction $\nabla f$, then $f$ will change by $\epsilon\|\nabla f\|$ (in a linear approximation).

In your example, the direction of the steepest ascent at the point $(2,4)$ is $(-4, -16)$, and the ascent is $4\sqrt{17}$.

If you need to draw the path the ant takes, I would do this in the $(x,y)$-plane: You start from $(2,4)$ and move 'a little' bit in the direction $(-4, -16)$. This gives you a new point $x_1$. There you compute $\nabla f(x_1)$, and get $x_2$, and you keep repeating till you are sufficiently close to $(0,0)$ (where the ant will stop).

Formally, the algorithm to draw the desired path would be: Fix a step-size $h$. Given a point $x_n$ in the $(x,y)$-plane:

1. Compute $\nabla f(x_n)$ and normalize it to length one. Call this vector $d_n$.
2. Compute $x_{n+1}=x_n + hd_n$.

Comment/further reading: The algorithm is a forward Euler scheme for the ODE $\dot x = \nabla f$.

Comment 2. Of course the actual path the ant takes is in 3D, on the graph of $f$. But you don't need to work much to get that: Once you have the 2D path $(x_0, x_1, ...)$, you can find the path on the graph by evaluating $f$ along the path, i.e. $(f(x_0), f(x_1), \ldots)$. But for visualisation purposes, the 2D path is certainly more convenient.