A surface $S$ has first fundamental form $du^2 + G(u,v)dv^2$ and curvature $0$. Also the curve $u=0$ is a geodesic when parametrized by arclength.
Prove that $G(u,v) = 1$ i.e. that $S$ is isometric to the plane.
It seems like you should be able to do this using the formula for curvature $K=\frac{-\sqrt{G}_{uu}}{\sqrt{G}}$ along with the geodesic equations, but I can't get it to work.
Best Answer
Know a curve $γ(s)=(u(s), v(s))$ on a surface parametrized by arc length is a geodesic if and only if
$$\dfrac{d}{ds}(Eu'+Fv')=\dfrac{1}{2}(E_uu'^2+2F_uu'v'+G_uv'^2)$$ o $$\dfrac{d}{ds}(Fu'+Gv')=\dfrac{1}{2}(E_vu'^2+2F_vu'v'+G_vv'^2).$$
Then, use the form
$$du^2+G(u,v)dv^2$$
If $K=0$ then $\sqrt{G}_{uu}=0$ so $$G(u, v)=A(v)u+B(v).$$ But at $u=0$, $x_u$ and $x_v$ are unit so $B(v)=1$.
Also, the curve $u=0$ is a geodesic with $v$ arc length. So the geodesic equation
$$\dfrac{d}{ds}(Eu'+Fv')=\dfrac{1}{2}(E_uu'^2+2F_uu'v'+G_uv'^2)$$
gives $0=G_u(0, v)$ and this means in our case $A(v)=0$. The first fundamental form is therefore $$ds^2=du^2+dv^2$$ also is isometric to the plane.