Volcano-watching

It makes things simpler to perform the analysis a little more generally. Suppose that the equation of the surface of the volcano (in cylindrical polar coordinates $r,\theta,z$ ) is $z \; = \; R^2 - r^2 \hspace{2cm} \rho \le r \le R$ Then the vector equation of a path from $A$ to $B$ is given as $\mathbf{r} \; = \; \left(\begin{array}{c} r\cos\theta \\ r\sin\theta \\ R^2 - r^2 \end{array} \right)$ where $r = r(\theta)$ is a function such that $\rho \le r(\theta) \le R$ for all $0 \le \theta \le 2\pi$ , and $r(0) = \rho$ and $r(2\pi) = R$ . Then $\frac{d\mathbf{r}}{d\theta} \; = \; r'(\theta) \left(\begin{array}{c} \cos\theta \\ \sin\theta \\ -2r\end{array}\right) + r\left(\begin{array}{c} -\sin\theta \\ \cos\theta \\ 0 \end{array} \right)$ and so $\frac{ds}{d\theta} \; = \; \sqrt{(1 + 4r^2)(r')^2 + r^2}$ where $s$ is the arc-length. If $\mathbf{t}$ is the unit tangent for the path, then $\frac{ds}{d\theta}\,\mathbf{t} \; = \; \frac{d\mathbf{r}}{d\theta}$ and so the angle $\alpha$ between the unit tangent of the path and the horizontal tangent to the surface is given by the equation $\cos\alpha \; = \; \mathbf{t} \cdot \left(\begin{array}{c} -\sin\theta \\ \cos\theta \\ 0 \end{array}\right) \; = \; \frac{r}{\sqrt{(1 + 4r^2)(r')^2 + r^2}}$ Thus we want to find the function $r= r(\theta)$ which is suitably smooth and such that $r(0) = \rho$ , $r(2\pi) = R$ and such that $S[r] \; = \; \int_0^{2\pi} \sqrt{(1 + 4r^2)(r')^2 + r^2}\,d\theta \; = \; \int_0^{2\pi} F\,d\theta$ is minimized. Since $F \; = \; \sqrt{(1 + 4r^2)(r')^2 + r^2}$ does not depend explicity on $\theta$ , a standard result of the Calculus of Variations is that $c \; = \; F - r'\frac{\partial F}{\partial r'} \; = \; \frac{r^2}{\sqrt{(1 + 4r^2)(r')^2 + r^2}} \; = \; r\cos\alpha$ is constant (this relationship between $r$ and $\alpha$ is known as Clairault's Relation). At the point $A$ , we note that $c = \rho \cos\alpha_0$ , where $\alpha_0$ is the angle that we want to determine. Thus $-\rho < c < \rho$ .

Rearranging this equation, we deduce that $\frac{d\theta}{dr} \; = \; \frac{c}{r} \sqrt{\frac{4r^2+1}{r^2 - c^2}}$ and a similar set of calculations to those performed in the hint question show that $\theta \; = \; 2c \cosh^{-1}\left(\sqrt{\frac{1 + 4r^2}{1 + 4c^2}}\right) + \tan^{-1}\left(\frac{\sqrt{r^2 - c^2}}{c\sqrt{1 + 4r^2}}\right) + k$ for an arbitrary constant $k$ , and so we deduce that $\theta \; = \; 2c\left[\cosh^{-1}\left(\sqrt{\frac{1 + 4r^2}{1 + 4c^2}}\right) - \cosh^{-1}\left(\sqrt{\frac{1 + 4\rho^2}{1 + 4c^2}}\right)\right] + \tan^{-1}\left(\frac{\sqrt{r^2 - c^2}}{c\sqrt{1 + 4r^2}}\right) - \tan^{-1}\left(\frac{\sqrt{\rho^2 - c^2}}{c\sqrt{1 + 4\rho^2}}\right)$ where $c$ is to be chosen so that $A(c) = 2\pi$ , where $A(c) \; = \; 2c\left[\cosh^{-1}\left(\sqrt{\frac{1 + 4R^2}{1 + 4c^2}}\right) - \cosh^{-1}\left(\sqrt{\frac{1 + 4\rho^2}{1 + 4c^2}}\right)\right] + \tan^{-1}\left(\frac{\sqrt{R^2 - c^2}}{c\sqrt{1 + 4R^2}}\right) - \tan^{-1}\left(\frac{\sqrt{\rho^2 - c^2}}{c\sqrt{1 + 4\rho^2}}\right)$ Since $A'(c) \; = \; 2\left[\cosh^{-1}\left(\sqrt{\frac{1 + 4R^2}{1 + 4c^2}}\right) - \cosh^{-1}\left(\sqrt{\frac{1 + 4\rho^2}{1 + 4c^2}}\right)\right] + \sqrt{\frac{1 + 4\rho^2}{\rho^2 - c^2}} - \sqrt{\frac{1 + 4R^2}{R^2 - c^2}}$ it is easy to see that $A'(c) > 0$ for all $-\rho < c < \rho$ , and so the desired value of $c$ is unique.

With $\rho = 1$ and $R = 15$ , we calculate numerically that $c = 0.963915$ , and hence that $\alpha_0 = \cos^{-1}c = 15.4388^\circ$ . To $1$ decimal place, the answer is $\boxed{15.4^\circ}$ .

It is worth noting that, for the optimal curve, $\frac{ds}{d\theta} = \frac{r^2}{c}$ , and hence the length of the path $AB$ is $s \; =\; \int_0^{2\pi} \tfrac{r^2}{c}\,d\theta \; = \; \int_1^{15} r\sqrt{\tfrac{4r^2+1}{r^2-c^2}}\,dr \; =\; \tfrac12\int_1^{225} \sqrt{\tfrac{4u+1}{u-c^2}}\,du \; = \; 227.99$ which is shorter than the special case (which would not be handled by the Calculus of Variations) of running around the lip of the crater, then going straight down the hill; this special path has length $2\pi + \int_1^{15}\sqrt{1 + 4x^2}\,dx \; =\; 230.95$