If u = Constant, then $S \equiv p(u,v)$ is a geodesic

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This has been my work so far:

I have a surface parameterized by $p(u,v) = p(u_0,v)$

To proof that it is a geodesic I used the following equations:

\begin{align*} \frac{d}{dt}\Big(E\dot{u} + F\dot{v}\Big) &= \frac{1}{2}\Big(\partial_uE\ \dot{u}^2 + 2\partial_uF\ \dot{u}\dot{v} + \partial_uG\ \dot{v}^2 \Big) \\ \frac{d}{dt}\Big(F\dot{u} + G\dot{v}\Big) &= \frac{1}{2}\Big(\partial_vE\ \dot{u}^2 + 2\partial_vF\ \dot{u}\dot{v} + \partial_vG\ \dot{v}^2 \Big) \end{align*}

Since $u=u_0$, I know that $$E = F = \partial_uE =\partial_uF = \partial_vE = \partial_vF= 0 $$ and $G = p_v\cdot p_v, \ \partial_vG=\partial_v (p_v\cdot p_v)\ $ and $\ \partial_u G = 0$

The first equation I came up with $0=0$, but I can´t figure it out how to solve the second equation.

Thank you!