Suppose a region A is parametrized by $x(u,v)=(u\cos v,u\sin v,0)$ with $u\in[0,R]$ and $v\in [0,V]$, $V<2\pi$. Let $c=\frac{V}{2\pi}$ and consider a cone $C$ parametrized by $y(u,v)=(cu\cos \frac{v}{c},cu\sin \frac{v}{c},\sqrt{1-c^2}u)$, also with $u\in[0,R]$ and $v\in [0,V]$, $V<2\pi$. How can I prove that $y\circ x^{-1}: A\rightarrow C$ is a local isometry? Also, is there any geometrical way to describe this map?
I'm totally stuck at this question... Any help would be appreciated!
Local isometry iff $E,\ F, \ G$ are same : $$ |x_u|=|y_u|=1,\ |x_v|=|y_v|=|u|,\ x_u\cdot x_v=y_u\cdot y_v=0$$