Hi just working on some past exam questions for an upcoming quiz for a class in Lagrangian dynamics, and I am a bit stuck on this question (which seems quite different from what we have done in class):
(i) Let $ds$ be the small distance between neighbouring points. Show that in cylindrical coordinates $ds^2=dr^2+r^2dθ^2+dz^2$
(ii) Find the expression for the path length on the surface of the cylinder with constant $r=a$
$$ I= \int_{z_0}^{z_1} \frac{ds}{dz}dz $$
(iii) Find the function $θ(z)$ that describes the path of least length between the points $(a,0,0)$ and $(a,π,H)$
I'm stuck on parts (ii) and (iii).
For (ii) I use the result to obtain:
$$\frac{ds}{dz} = \sqrt{r^2(\frac{d\theta}{dz})^2 + (\frac{dr}{dz})^2 + 1}$$ but then I'm not too sure how to integrate this?
I'm also assuming the answer to his question leads to (iii).
Any help would be greatly appreciated.
If the radius is fixed, then all the of the terms involving $ dr $ will vanish. So your result for (ii) should look like $$ I = \int_{z_0}^{z_1} \sqrt{ a^2 \left( \frac{d\theta}{dz}\right)^2 + 1} \; dz $$ You should not try to integrate this, but instead find the Euler-Lagrange equations in the usual way, and then try to solve them.
If you let $ x = a \theta $, then it is identical to finding the shortest line connecting two points on the plane. This is of course because the cylinder is flat - you can slice it along the length and spread it out into a rectangle, and then find the shortest length there.