# Geodesic of a Surface in $\mathbb{R}^3$

I'm not familiar with geodesics. How can I show that a curve $c$ given by $c(t)=(t,f(t)\cos{\alpha},f(t)\sin{\alpha})$ for $\alpha$ constant is a geodesic on $M$ where $M=\left\{(x,y,z) \in \Bbb{R}^3 \mid f(x)=y^2+z^2\right\}$?

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Clairaut's relation. en.wikipedia.org/wiki/Clairaut%27s_relation –  Will Jagy Feb 12 '13 at 20:36
Also, meridians on a surface of revolution are always geodesics, and your notation is poor. –  Will Jagy Feb 12 '13 at 20:44
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## 1 Answer

One definition of a geodesic is that it's a curve whose geodesic curvature is zero. So, compute the geodesic curvature, and show that it's zero. Almost any decent differential geometry reference will tell you how to compute geodesic curvature; here's one: MathWorld.

Alternatively (and equivalently, using a different definition of geodesic), show that the principal normal of the curve coincides with the surface normal at every point.

Of course, these are dumb brute force techniques. A much smarter approach is to notice that your surface is a surface of revolution and your curve is a meridian, and use the ideas given in the two comments. The dumb approaches are valuable only because they will always work, and because they don't require any clever insights.

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