Geodesics of the right circular cone

Question

I have to find teh geodesics of the cone:

$$C={(x,y,z)\in R^3:x^2+y^2=z^2,z>0}$$

My idea is use that the geodesic curvature must be 0 for be a geodesic.Then I use the parametrization:

$$X(u,v)=(ucosv,usinv,u)$$

but I get this conclusion:

$Xu=(cosv,sinv,1)$ $Xv=(-usinv,ucosv,0)$$Xu\times Xv=(-ucosv,-usinv,1)$

$norm(Xu\times Xv)=u\sqrt{2}$

$N=\frac{Xu\times Xv}{norm(Xu\times Xv)}=(-cosv/\sqrt{2},-sinv/\sqrt{2},1/\sqrt{2})$

$alpha'(v)=(-usinv,ucosv,0)$$alpha''(v)=(-ucosv,-usonv,0)$

$Kg=det\left(\begin{array}{c}alpha''(v)\\N\\alpha'(v)\end{array}\right)$

$$Kg=\frac{u^2}{\sqrt{2}} \Rightarrow Kg=0 \Leftrightarrow u=0$$

And it doesn't have any sense to me. So, which is the way to get the geodesics and why my idea doesn't work?.

Thanks in advance.

Answer 1

This won't help you with your computations, but I would just like to observe that when the cone is cut open and unrolled flat, the geodesics are straight line segments.

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          ^{Image from Mark Irons.}

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