Variational Mechanics for Geodesic Cylinder: Find Min Dist b/w 2 Points

[stunner5000pt]

for a geodesic right cylinder with radius R. Find the curve taht minimizes the distance between two points r1 and r2. where r = (R,phi, z) in cylindrical polar coordinates. Express your answer as z = z(phi)

pardon the sloppy math
not the most fun to get this type of question on a 50 minute test! Anyway,

x = r cos t
y = r sin t
z = z

$$\mbox{distance} = \int_{1}^{2} \sqrt{r^2 + \left( \frac{\partial z}{\partial \theta}\right)^2}$$
ist hat the distance between two points on a cylinder?
Or would this distane be represented by something of a helix? I mean a cylinder could be considered as helix... right?

 

[Gokul43201]

If your integration limits are angles $\phi_1~and ~\phi_2$ and you integrate w. r. to $\phi$, then you are good.

 

[stunner5000pt]

ok so do i plug this into the euler lagrange equation?

so $$f(z,\theta) = \sqrt{r^2 + \left( \frac{\partial z}{\partial \theta}\right)^2}$$

Euler Lagrange equation is
$$\frac{\partial f}{\partial \theta} - \frac{d}{d\theta} \frac{\partial f}{\partial z} = 0$$
is this the right way to go?