Prove that the curves of the family $v^3/u^2=k$ are geodesics on a surface

Question

Prove that the curves of the family $v^3/u^2=k$ where $k$ is a constant are geodesics on a surface with the metric

$$v^2 \, du^2-2uv \, du+2u^2 \, dv^2$$ where $u,v \gt 0$.

Answer 1

if you want to prove a curve is geodesic, you can use the following equivalent condition :

$(r^{'},r^{''},n)=0$

your curve equation is :

$r(u,v)=v^{3}-ku^{2}=0$

since, $(r^{'},r^{''},n)=(r^{'}\times{r^{''}})\cdot{n}=\vert{r^{'}}\vert\cdot\vert{r^{''}}\vert\sin\theta{\cdot{n}}$

therefore, $sin\theta=0$$\Longrightarrow{cos\theta}=1$$\Longrightarrow$$\frac{r^{'}\cdot{r^{''}}}{\vert{r^{'}}\vert\vert{r^{''}}\vert}=1$

in which, $r^{'}=(-2ku,3v^{2})$ $r^{''}=(-2k,6v)$