# A curve is a geodesic on a surface proof

#### JoeG

##### New member
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 $S$.

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A curve on a surface is geodesic iff it's geodesic curvature is zero everywhere, so I understand that I've to show that $\displaystyle k_g = \ddot \gamma \cdot (n \times \dot \gamma) = 0$ (where 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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#### GJA

##### Well-known member
MHB Math Scholar
Hi JoeG ,

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 it's geodesic curvature is zero everywhere, so I understand that I've to show that $\displaystyle k_g = \ddot \gamma \cdot (n \times \dot \gamma) = 0$ (where 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.
Let's use $n$ to denote the normal to the surface and $T, N$, and $b$ to denote the Frenet frame for $\gamma.$ Doing so we have $\sigma_{u} = \dot{\gamma}-v\tau N = T-v\tau N$. From here $n=\sigma_{u}\times\sigma_{v}=T\times b - v\tau N\times b.$ Using the Frenet frame $T\times b = c_{1}N$ and $N\times b = c_{2}T,$ for some constants $c_{1}$ and $c_{2}.$ Hence $n=c_{1}N-v \tau c_{2} T.$ Using this, see if you can show the desired result. Feel free to ask any follow up questions.

#### JoeG

##### New member
Hi JoeG ,

Let's use $n$ to denote the normal to the surface and $T, N$, and $b$ to denote the Frenet frame for $\gamma.$ Doing so we have $\sigma_{u} = \dot{\gamma}-v\tau N = T-v\tau N$. From here $n=\sigma_{u}\times\sigma_{v}=T\times b - v\tau N\times b.$ Using the Frenet frame $T\times b = c_{1}N$ and $N\times b = c_{2}T,$ for some constants $c_{1}$ and $c_{2}.$ Hence $n=c_{1}N-v \tau c_{2} T.$ Using this, see if you can show the desired result. Feel free to ask any follow up questions.
Hi GJA,

This is probably wrong, but here goes

$k_g = \ddot \gamma \cdot (n \times \dot \gamma) = \ddot \gamma \cdot ((c_{1}N-v \tau c_{2} T) \times \dot \gamma)$

$= \ddot \gamma \cdot ((v \tau c_{2} T \times \dot \gamma- c_{1} N \times \dot \gamma)$

but $T = \dot \gamma$ by definition so $T \times \dot \gamma =0$

But also $\gamma$ is a unit speed curve so $N = \ddot \gamma$

So $k_g = N\cdot (-c_1 N \times T) = 0?$

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#### GJA

##### Well-known member
MHB Math Scholar
Looks good to me. Only small comment I would make is that $\ddot{\gamma}=\kappa N$ where $\kappa$ is the curvature, but this doesn't change the fact that the result is zero.

#### JoeG

##### New member
Looks good to me. Only small comment I would make is that $\ddot{\gamma}=\kappa N$ where $\kappa$ is the curvature, but this doesn't change the fact that the result is zero.
Thanks again.

In the solution to this problem, it's written

$\sigma_u \times \sigma_v (u, 0) = -n, k_g = \ddot \gamma \cdot \dot g \times n = 0$.

What did they do in the first equality?

Not where $g$ comes from either.

#### GJA

##### Well-known member
MHB Math Scholar
They used the equation we had for $\sigma_{u}\times\sigma_{v} = T\times b - v\tau N$ and set $v=0$. From there they used the Frenet frame relations to write $T\times b = -N;$ i.e., $n(u,0) = -N(u).$