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!

It reads:

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.

Re: Meridians onf surface of revolution.

Consider a constant speed curve $\displaystyle \alpha : I \rightarrow \mathbb{R}^3$ along a meridian of the revolved surface S,

where $\displaystyle I$ is some open interval of $\displaystyle \mathbb{R}$.

To show $\displaystyle \alpha$ is a geodesic, you want to show that $\displaystyle \alpha''$ is orthogonal to S.

Since it has constant speed, you can say something about the relationship between $\displaystyle \alpha'$ and $\displaystyle \alpha''$.

Consider the plane P that defines the meridian by cutting through S.

Since $\displaystyle \alpha$ stays in P, so do all of its derivatives.

At any point along $\displaystyle \alpha$, P is spanned by $\displaystyle \alpha'$ and the normal to S at that point.

Put it all together, and you deduce something about $\displaystyle \alpha''$ that proves that $\displaystyle \alpha$ is a geodesic.