Maximum steepness of hill

Question

A hill is given by $$ z = f(x,y) = \frac {32}{1 + x^2 + y^2}$$

where $z$ is the height of the hill in meters. At what height is the hill the steepest?

The standard way to do it, which is how I would do it, is to determine the gradient on vector form and then maximize $\| \text{grad} f(x,y)\|^2$ with respect to $x^2 + y^2 = r^2$ (I'll maximize the square since it is hairy to deal with square roots).

But what I would like to know is what other, perhaps more clever and sophisticated, ways there are to solve the same problem.

By rotational symmetry this can be treated as a one-dimensional problem. — Thomas, Feb 18 '14 at 17:02
@jibounet: That is certainly true but I am not sure what you are trying to get at. — thelionkingrafiki, Feb 18 '14 at 17:21
@Thomas: You mean we set $f(x,y) = g(r) = g(x^2 + y^2)$ and then maximize $g'(r)$? — thelionkingrafiki, Feb 18 '14 at 17:29

score 1 · Accepted Answer · answered Feb 18 '14 at 17:32

The graph you give has rotational symmetry about the $z$-axis.

To see this, put $x=r\cos\theta$ and $y=r\sin\theta$. We get $$\frac{32}{1+x^2+y^2} = \frac{32}{1+r^2}$$

This is independent of $\theta$. We might as well consider a planer cross section, e.g. the plane $y=0$.

This gives the curve $z=\frac{32}{1+x^2}$. For the steepest points you want to maximise or minimise the gradiet, i.e. solve $\frac{\mathrm{d}^2z}{\mathrm{d}x^2}=0$. Doing so gives $x=\pm\frac{1}{\sqrt{3}}$. Checking the sign of $\frac{\mathrm{d}^3z}{\mathrm{d}x^3}$ shows that $x=\frac{1}{\sqrt{3}}$ is a minimum and $x=-\frac{1}{\sqrt{3}}$ is a maximum.

These two points correspond to the circle $x^2+y^2 = \frac{1}{3}$.

Thanks. I however did not choose a planer cross section, instead I just wrote $x^2 + y^2 = r^2$ and then maximized $g'(r) = 0$. I guess it doesn't really matter whether we impose anything on $y$ or not? — thelionkingrafiki, Feb 18 '14 at 17:37

Maximum steepness of hill

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