A fly sits on the outside surface of a cylindrical drinking glass . It must crawl to another point situated on inside of the glass . Find the shortest

Question

A fly sits on the outside surface of a cylindrical drinking glass . It must crawl to another point situated on inside of the glass . Find the shortest path possible(neglecting the thickness of the glass).

I am not quite getting wont it be equal to $2a$ where $a$ is the distance of the fly from the upper circular boundary of the cylinder.However, if that point is not situated exactly to its initial position, in that case i think the shortest path will be from the point of position to the upper corcular boundary($a$) of the cylinder and then from that upper circular boundary to the desired point i.e those two are connected by a straight line , so that it is $x$(say), so the total shortest path is $a+x$....

Answer 1

Using cylindrical coordinates $(r, \phi, z)$, let coordinates of $A$ be

$A = (R , \phi_1, z_1) $

and let the cylindrical coordinates of $B$ be

$B = (R, \phi_2, z_2) $

And let the height of the cylindrical glass be $H$. What you need to do is reflect point $B$ about the plane $z = H$, then its image would becomes

$B' = (R , \phi_2 , 2 H - z_2 ) $

Now the shortest distance between $A$ and $B'$ is by Pythagorean theorem equal to

$ D = \sqrt{ R^2 (\phi_2 - \phi_1)^2 + (2 H - z_1 - z_2)^2 } $

Answer 2

The easiest way to stretch your intuition for this problem is to take $2$ equal sized pieces of paper, and designate them as P1 and P2.

Roll P1 and P2 both into cylinders, where in both P1 and P2, the right edge of the paper just meets the left edge. So, P1 and P2 should now represent cylinders that have equal heights and equal radii.

On P1, mark any arbitrary outer-starting point, on the outside of the cylinder, with a pen. Also, mark any arbitrary inner-ending point, on the inside of the cylinder, with a pen. On P2, mark the exact same inner-ending point, on the inside of its cylinder, to correspond to the inner-ending point of P1.

Unwrap P1, so that it is now flat on a table, with the side containing the outer-starting point face up.

Unwrap P2, so that it is now flat on the same table, with the side containing the inner-ending point face up.

Place P2 directly above P1, but upside down, so that (in effect) the top edge of P1 is edge-to-edge with what represents the top edge of P2.

Draw the line from the (now visible) outer-starting point on P1 to the inner-ending point on P2. This represents the shortest distance possible between the arbitrarily selected outer-starting point, and inner-ending point.

For the portion of the path that is on P2, trace the same path on the underside of P1, where the underside contains the inner-ending point.

Now, take P1 and roll it back into the cylinder. The path that has been inked in on the outside and inside of P1 represent the path that must be followed to minimize the route going along the surface of the P1 cylinder, from the outer-starting point to the inner-ending point.