# Meridians onf surface of revolution.

• Sep 30th 2012, 10:29 PM
bluesboy91
Meridians onf surface of revolution.
Hi Guys,

I have a homework questions that I am stuck on, it's the last in the set and I can't do it!

Prove that the meridians are geodesics when the curve: (x(u),0,u) is revolved around the z-axis.

I have an idea of using the Geodesic Curvature formula etc but do not know where to start.

If anyone could give me a hand, please let me know!

Cheers.

Patrick.
• Oct 2nd 2012, 06:54 AM
johnsomeone
Re: Meridians onf surface of revolution.
Consider a constant speed curve $\alpha : I \rightarrow \mathbb{R}^3$ along a meridian of the revolved surface S,
where $I$ is some open interval of $\mathbb{R}$.
To show $\alpha$ is a geodesic, you want to show that $\alpha''$ is orthogonal to S.

Since it has constant speed, you can say something about the relationship between $\alpha'$ and $\alpha''$.
Consider the plane P that defines the meridian by cutting through S.
Since $\alpha$ stays in P, so do all of its derivatives.
At any point along $\alpha$, P is spanned by $\alpha'$ and the normal to S at that point.
Put it all together, and you deduce something about $\alpha''$ that proves that $\alpha$ is a geodesic.