I'm trying to find the Geodesic curvature of the circle $u = u_0$ for the circular cone with a vertex angle of $2ϕ$ that is parametrized by $ x(u,v)=(utanϕcosv,utanϕsinv,u)$ for $0≤u≤u0$ and $0≤v≤2π.$ The Equation for Geodesic curvature is $k_g = kN(n \times T)$ which I found to be $k_g = \frac{\cos \phi}{u_0}$. Now I want to show that this is the same as the curvature of the curve we get when we open up the cone. Like this:
And the textbook I'm using says the parametrization for the pacman region is $x(u,v) = (u\cos v, u\sin v, 0)$ for $0 \leq u \leq R, 0 \leq v \leq V$ where $V$ is less than $2 \pi$. Now I'm not sure how to parametrize the pacman curve. I thought you would take $u = u_0$ and then compute the curvature but I don't think that is right. I'm not sure how to compute the curvature of the curve that is identical to the circle at the bottom of the cone once opened.