June the ant is on a cylinder. More specifically, she is on the edge of one of the cylinder’s two circular faces. Her dinner is on the edge of the opposite circular face, and all the way around on the other side of that face. The radius of the cylinder is $2$ meters and its height is $2$ meters.
Your job is to help June find the shortest path along the surface of the cylinder so that she can chow down as quickly as possible. What’s the length of this shortest path?
Let's assume that June is located at the point $(2,0,2)$ while her food is located at $(-2,0,0),$ making the center of the cylinder aligned with the positive $z$-axis. Let's assume that June's path last leaves the top circular face at the point $(2\cos \theta, 2\sin \theta, 2),$ for some $0 \leq \theta \leq \pi.$ Similarly, let's assume that June's path first arrives on the bottom circular face at the point $(2 \cos \phi, 2\sin \phi, 0),$ for some $\theta \leq \phi \leq \pi.$ Though June's path could theoretically wander about anywhere, let's assume that she will walk in straight lines when on the two circular faces and on a straight line with respect to the curved outer edge. The path along the upper circular face will have a distance $$d_1(\theta) = \sqrt{ (2\cos \theta - 2)^2 + (2 \sin \theta)^2 + (2-2)^2 } = 2 \sqrt{ 1 - 2 \cos \theta } = 4 \sin \frac{\theta}{2}.$$ The path along the lower circular face will have a distance $$d_3(\phi) = \sqrt{ (2 \cos \phi + 2)^2 + (2 \sin \phi)^2 + (2-2)^2 } = 2 \sqrt{ 1 + 2 \cos \phi } = 4 \cos \frac{\phi}{2}.$$ The path along the curved outer edge is given by $d_2(\theta, \phi) = 2 \sqrt{1 + (\phi - \theta)^2},$ which we can see either by unrolling the curved outer edge into a rectangle and realizing that the path is the hypotenuse of a triangle with height $2$ and base $2 (\phi - \theta)$, or more formally by paramaterizing the path along the outer edge as the curve $x(t) = 2 \cos t,$ $y(t) = 2 \sin t,$ $z(t) = 2 \frac{\phi - t}{\phi - \theta},$ for $t \in [\theta, \phi]$ and finding \begin{align*}d_2(\theta, \phi) &= \int_\theta^\phi \sqrt{ \left(\frac{dx}{dt}\right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2 } \,dt \\ &= \int_\theta^\phi \sqrt{ (-2\sin t)^2 + (2 \cos t)^2 + \left(\frac{-2}{\phi - \theta}\right)^2 } \,dt \\ &= \int_\theta^\phi \sqrt{ 4 \sin^2 t + 4 \cos^2 t + \frac{4}{(\phi - \theta)^2} } \,dt \\ &= \frac{2 \sqrt{1 + (\phi - \theta)^2}}{\phi - \theta} \int_\theta^\phi \,dt \\ &= 2 \sqrt{ 1 + (\phi - \theta)^2 }.\end{align*} So the total distance is $$d(\theta, \phi) = d_1(\theta) + d_2(\theta, \phi) + d_3(\phi) = 4 \sin \frac{\theta}{2} + 4 \cos \frac{\phi}{2} + 2 \sqrt{ 1 + (\phi - \theta)^2 }.$$
So June would want to find $$d^* = \inf \{ d(\theta, \phi) \mid 0 \leq \theta \leq \phi \leq \pi \}.$$ Let's treat the case where $\theta = \phi$ and hence we have a univariate problem $$\tilde{d}(\theta) = 4 \sin \frac{\theta}{2} + 4 \cos \frac{\theta}{2} + 2.$$ We see that in this case, we have $$\frac{d}{d\theta} \tilde{d} = 2 \cos \frac{\theta}{2} - 2 \sin \frac{\theta}{2}$$ which has critical points whenever $\tan \frac{\theta}{2} = 1,$ that is $\frac{\theta}{2} = \frac{\pi}{4},$ or $\theta^* = \frac{\pi}{2}.$ Therefore, we see that the minimal value of $$ \tilde{d}^* = \inf_{0 \leq \theta \leq \pi} \tilde{d}(\theta) = \min \left\{ \tilde{d}(0), \tilde{d}(\pi), \tilde{d} \left( \frac{\pi}{2} \right) \right\} = \min \{ 6, 6, 4\sqrt{2}+2 \} = 6,$$ which occurs when either $\theta=0$ or $\theta=\pi.$ This amounts to either going straight down the curved side and then taking a straightline path across the diameter to the food, or vice versa.
Let's fix some $\theta \in [0,\pi)$ and then try to minimize with respect to $\phi.$ We see that $$\frac{\partial d}{\partial \phi} = -2 \sin \frac{\phi}{2} + \frac{ 2(\phi - \theta)}{\sqrt{1 + (\phi-\theta)^2}}.$$ Setting this equal to zero and doing a bunch of algebra and trigonometry identities we get the implicit function $$\phi - \tan \frac{\phi}{2} = \theta.$$ Letting $g(t) = t - \tan \frac{t}{2},$ we see that $g^\prime(t) = 1 - \frac{1}{2} \sec^2 \frac{t}{2},$ so the largest possible value of $g$ occurs when $\sec^2 \frac{t}{2} = 2$ or equivalently when $\cos \frac{t}{2} = \frac{\sqrt{2}}{2},$ that is when $t = \frac{\pi}{2},$ and in particular we see that $$g(t) \leq g^* = g( \frac{\pi}{2} ) = \frac{\pi}{2} - 1.$$ Therefore, we see that if $\theta \geq \frac{\pi}{2} - 1$ then we have $\frac{\partial d}{\partial \phi} \leq 0$ for all $\phi \in [\theta, \pi],$ so if $\theta \gt \frac{\pi}{2} - 1,$ then the shortest path that can be obtained by choosing $\phi = \pi$ is $$ \inf_{\phi \in [\theta, \pi]} d(\theta, \phi) = d(\theta, \pi) = 4 \sin \frac{\theta}{2} + 2 \sqrt{ 1 + (\pi - \theta)^2}.$$ We see that in this case if we set $$\delta(\theta) = 4 \sin \frac{\theta}{2} + 2 \sqrt{ 1 + (\pi - \theta)^2}$$ that $$ \delta^* = \inf_{\theta \in [\frac{\pi}{2} -1, \pi]} \delta(\theta) = \delta(\pi) = 6.$$
Ok, so we are now left with only the case where $\theta \in [0, \frac{\pi}{2} -1).$ In this case, we can approximate the implicit equation using a cubic function, since $\tan \frac{t}{2} \approx \frac{t}{2} + \frac{t^3}{24} + O(t^4)$ for $|t| \lt 1.$ This yields the new approximate implicit equation $$\phi - \left( \frac{\phi}{2} + \frac{\phi^3}{24} \right) = \theta,$$ or equivalently, $$\phi^3 - 12\phi + 24 \theta = 0.$$ This cubic has three real roots, and the middle one will yield a local minimum since it will represent the place where $\frac{\partial d}{\partial \phi} \lt 0$ to the left of the root and $\frac{\partial d}{\partial \phi} \gt 0$ to the right of the root. From the Viete formula, we see that this root will be at $$\phi^*(\theta) = 4 \cos \left( \frac{ \cos^{-1} \left( -\frac{3\theta}{2} \right) - 2\pi }{3} \right).$$ Importantly, we remember that we got into this mess because $\phi^*(\theta)$ roughly satisfies $$\frac{\partial d}{\partial \phi} (\theta, \phi^*(\theta)) \approx 0.$$ So let's define $$\Delta(\theta) = d(\theta, \phi^*(\theta)),$$ and look to find $\Delta^* = \inf \{\Delta(\theta) \mid \theta \in [0,\frac{\pi}{2}-1]\}.$ Thankfully, we see that \begin{align*}\Delta^\prime(\theta) &= 2 \cos \frac{\theta}{2} - 2 \sin \frac{\phi^*(\theta)}{2} (\phi^*)^\prime (\theta) + \frac{ 2(\phi^*(\theta) - \theta) }{ \sqrt{ 1 + (\phi^*(\theta) - \theta)^2 }} (\phi^*)^\prime (\theta) \\ &= 2 \cos \frac{ \theta}{2} + (\phi^*)^\prime (\theta) \left( - 2 \sin \frac{ \phi^*(\theta) }{2} + \frac{ 2( \phi^*(\theta) - \theta }{ \sqrt{ 1 + (\phi^* (\theta) - \theta)^2 }} \right) \\ & = 2 \cos \frac{ \theta}{2} + (\phi^*)^\prime (\theta) \frac{\partial d}{\partial \phi} (\theta, \phi^*(\theta)) \\ & \approx 2 \cos \frac{\theta}{2} \gt 0,\end{align*} for all $\theta \in [0, \frac{\pi}{2} - 1),$ therefore we see that $\Delta^* = \Delta(0) = 6.$
Therefore, since we went through all of the cases, we now anticlimactically confirm that in fact, June's shortest path to the food is 6 meters long.
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