June's Shortest Cylindrical Shell Path

Now, June is on a hollowed-out cylinder, also known as a “cylindrical shell.” The shell’s outer radius is $2$ meters and its inner radius is $1$ meter. The shell is $2$ meters tall. June is on the outer edge of one of the cylinder’s two flat faces. Her dinner is on the opposite face, and all the way around on the other end of that face.

Once again, your job is to help June find the shortest path along the surface of the shell so that she can chow down as quickly as possible. What’s the length of this shortest path?

In the classic problem, June was able to roughly ignore the rounded edge of the cylinder and use two straight lines to reach the food. Here, the hollowed out center will force her to take another approach. Let's again assume that June starts at $(2, 0, 2),$ her food is at $(-2, 0, 0),$ June's last stop on the upper face is at the point $(\cos \theta, \sin \theta, 2)$ and first stop on the lower face is at the point $(\cos \phi, \sin \phi, 0),$ for some $0 \leq \theta \leq \frac{\pi}{3}$ and $\frac{2\pi}{3} \leq \phi \leq \pi.$ Here, we note that we can constrain $\theta$ and $\phi$ a little bit more than in the Classic problem, since the distance from $(\cos \theta, \sin \theta, 2)$ to $(\cos \phi, \sin \phi, 0)$ is $$d_2(\theta, \phi) = \sqrt{4 + (\phi-\theta)^2},$$ which is always less than descending vertically down from the upper to lower faces and then traversing along the inner radius from angle $\theta$ to $\phi,$ or vice versa. This means that we can rule out any of the values for $\theta$ and $\phi$ in $[\pi/3, 2\pi/3],$ where the straight line path would cross within the hollowed out inner circle and require June to travel along the curved inner radius.

Therefore, taking the Euclidean distances along the upper and lower faces of the cylindrical shell, we have that the Extra Credit distance formula is given by $$d(\theta, \phi) = \sqrt{5 - 4 \cos \theta} + \sqrt{5 + 4 \cos \phi} + \sqrt{4 + (\phi - \theta)^2}.$$ so we need to find $$d^* = \inf \left\{ d(\theta, \phi) \mid 0 \leq \theta \leq \frac{\pi}{3}, \frac{2\pi}{3} \leq \phi \leq \pi \right\}.$$ From a symmetry perspective, let's assume that $\phi = \pi - \theta,$ which means that we can define \begin{align*}\tilde{d} (\theta) &= d(\theta, \pi-\theta) \\ &= \sqrt{5 - 4\cos \theta} + \sqrt{5 + 4 \cos (\pi - \theta)} + \sqrt{4 + (\pi - \theta - \theta)^2} \\ &= 2 \sqrt{5 - 4 \cos \theta} + \sqrt{4 + (\pi - 2\theta)^2}.\end{align*} Here we see that $$\tilde{d}^\prime (\theta) = \frac{4\sin \theta}{\sqrt{ 5 - 4 \cos \theta}} + \frac{ 2 (2\theta - \pi) }{\sqrt{ 4 + (\pi - 2\theta)^2} },$$ so setting this equal to zero we get $$\frac{4 \sin \theta}{\sqrt{5 - 4 \cos \theta}} = \frac{ 2 (\pi - 2\theta) }{\sqrt{4 + (\pi - 2\theta)^2}}.$$ Squaring both sides and using some trigonometry identities, we see that this is equivalent to the implicit function $$\theta + \frac{2\sin \theta}{2 \cos \theta - 1} = \frac{\pi}{2}.$$

Let's use Newton-Raphson method on the function $$f(\theta) = \theta + \frac{2\sin \theta}{2\cos \theta - 1} - \frac{\pi}{2},$$ where $$f^\prime(\theta) = 1 + \frac{4-2\cos \theta}{(2\cos \theta - 1)^2},$$ to figure out the proper root of this implicit function, we see that we can start with an example of $\theta_0 = \frac{1}{2},$ then $$\theta_{n+1} = \theta_n - \frac{f(\theta_n)}{f^\prime(\theta_n)},$$ for $n = 0, 1, 2, \dots.$ We see that after only a few steps we quickly settle into $$\theta^* = \lim_{n \to \infty} \theta_n = 0.457751785361\dots.$$ This translates into a minimal distance for June to get to her food on the cylindrical shell of $$\tilde{d}^* = \tilde{d}(\theta^*) \approx 5.368959019243\dots$$ meters.

$n$	$\theta_n$	$f(\theta_n)$	$f^\prime(\theta_n)$	$\tilde{d}(\theta_n)$
1	0.500000000000	0.198927402511	4.936412045223	5.371299778386
2	0.459702026353	0.008790035183	4.516178419967	5.368964120693
3	0.457755682704	0.000017530945	4.498196526681	5.368959019263
4	0.457751785377	0.000000000070	4.498160714274	5.368959019243
5	0.457751785361	0.000000000000	4.498160714132	5.368959019243