Calculating the shortest route around cylinder

Question

If I have the following situation, what would the path look like? Where would the path go and how would I calculate it?

Cylinder with diameter $10\,\hbox{cm}$ and height $5\,\hbox{cm}$. Use the shortest route from a point on the top edge ($A$) to a point on the bottom edge ($B$) diametrically opposite $A$.

I understand the following might help but how?

$\mathbf{r}(t) = a\cos(t)\mathbf{i}+ a\sin(t)\mathbf{j}+ ct\mathbf{k}$.

With the resultant formulae as follows: \begin{cases} x = r\cos t \\ y = r \sin t\\ z = \frac{h\cdot t}{2\cdot \pi\cdot n},& (0 ≤ z ≤ h) \end{cases}

Where:
$h$ = height
$r$ = radius
$n$ = the number of complete revolutions

Note: I know 2 methods of calculating this, but I don't know how to provide a value (or an approximation). Could a value and the process be provided?

Answer 1

If I understand the question properly, open up the cylindar to form a rectangle $5{\rm{ by 20}}\pi $

A is one corner of the rectangle and B is half way along the ${\rm{20}}\pi $ side.

Using pythagoras gives a distance around the cylindar of $\sqrt {100\mathop \pi \nolimits^2 + \mathop 5\nolimits^2 } $

Answer 2

Let's label the shortest path red and then flatten the shape:

So we need to minimize the path: $$\min\limits_{\theta}\left(\sqrt{h^2+(2r\theta)^2}+2r\cos\theta\right) = 10\,\hbox{cm} \cdot\min\limits_{\theta}\left(\sqrt{\frac{1}{4}+\theta^2}+\cos\theta\right)=15\,\hbox{cm},$$
according to wolframalpha.