Computing the Distance Formula on a Metric Cone Given a Riemmanian $(n-1)-$Manifold $N$, with metric $\tilde{g}$, we have a so-called Metric cone with cross-section $N$, given by $\mathbf{R}_{>0}\times N$. I am trying to compute the distance between two points on the cone, $(r_1,x_1)$ and $(r_2,x_2)$. 

I am able to see that the product metric is of the form $g=dr^2+r^2\tilde{g}$, where $r$ denotes the coordinate on $\mathbf{R}_{>0}$. I understand that the distance should be the infimum over all lengths of paths between the two points. 

I am basically uncertain how to proceed, as this sort of problem is quite new to me. Any help would be appreciated. 
 A: I can't comment so I'll go ahead and provide my thoughts as an answer. If they are not helpful feel free to do whatever is necessary.
Let $\mathbb(R)_{>0}\times M:=C(M)$. Consider a curve $\ell \colon [0,1] \to C(M)$ starting at $(r_1,x_1)$ and ending at $(r_2,x_2)$. Now $\ell '$ can be written as $\frac{\partial}{\partial r} + \sum _i\frac{\partial}{\partial x_i}$, here $x_i$ are the local coordinates for M.
$$||\ell'||_g=g(\ell',\ell')^{\frac{1}{2}}$$
\begin{align*}
g(\ell',\ell')
& =(dr^2+r^2\tilde{g})\left(\frac{\partial}{\partial r} + \sum _i\frac{\partial}{\partial x_i},\frac{\partial}{\partial r} + \sum _i\frac{\partial}{\partial x_i}\right) \\
& =1+r^2\tilde{g}\left(\sum _i\frac{\partial}{\partial x_i},\sum _i\frac{\partial}{\partial x_i}\right) \\
& =1+r^2\tilde{g}(\tilde{\ell}',\tilde{\ell}') \\
& =1+r^2||\tilde{\ell}'||^2_{\tilde{g}},
\end{align*}
where $\tilde{\ell}' \colon [0,1] \to M$ with starting and ending points as $x_1,x_2$ respectively.
Now $r$ can be written in terms of $t$ since we know that we start at $r_1$ at $t=0$ and end at $r_2$ at $t=1$.
$$
\int_0^1 ||\ell'|| dt
=\int_0^1 (1+r(t)^2||\tilde{\ell}'||^2_{\tilde{g}})^{\frac{1}{2}} dt.
$$
Now taking $\inf$ over curves $\ell$ would be equivalent to taking $\inf$ over $\tilde{\ell}$ on the left side.
I think I have left one step but I hope this helps.
