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.

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

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