# Showing that a curve is a geodesic of a surface?

Let $$\gamma(u): I \to \mathbb{R}^3$$be a unit speed curve and let $$\vec{b}(u)$$ be its binormal vector. Consider the surface $$S$$ given by the surface patch $$\sigma(u,v) = \gamma(u)+v\vec{b}(u)$$. Show that $$\gamma$$ is a geodesic of $$S$$.

A curve on a surface is geodesic iff its geodesic curvature is zero everywhere, so I understand that I've to show that $$\displaystyle k_g = \ddot \gamma \cdot (\vec{n} \times \dot \gamma) = 0$$ (where $$\vec{n}$$ is the unit normal to $$\sigma$$), which I've trouble calculating.

I've $$\sigma_u =\dot \gamma +v b'(u)= \dot \gamma -v \tau \vec{n}$$ and $$\sigma_v = \vec{b}$$. But it doesn't seem clear what $$\sigma_u \times \sigma_v$$ would be.

How do I show that $$k_g = 0$$? Thanks.

• I think I'm missing something. why is $\sigma_u$ not $\dot{\gamma} + vb'(u)$? – Rylee Lyman Mar 6 at 1:56
• @RyleeLyman Yes, the sign was wrong. Thanks. – user651098 Mar 6 at 2:02
• The curve has unit speed, so $\upsilon = 1$ and $\sigma_u = \vec t - \tau\vec n$. Now take the cross product with $\vec b$? – Ted Shifrin Mar 6 at 2:23