Let $f:\mathbb{R}_{>0} \times (0,2\pi) \rightarrow \mathbb{R}^3$ $$f(t,\xi) := (r(t) \cos( \xi) , r(t) \sin(\xi),z(t))$$ be a surface of revolution, where we assume that $r>0$ and $r'^2+z'^2>0.$
Then the metric tensor of this immersion can be written as $(g_{ij})(t):=\begin{pmatrix} r'(t)^2+z'(t)^2 & 0 \\ 0 & r(t)^2 \end{pmatrix}.$
By exploiting the fact that there is an arc-length parametrization ($\psi(t) = \int_0^t r'(s)^2+z'(s)^2 ds$) for the underlying curve of the surface of revolution, the map $f$ can be reparametrized so that we get the metric tensor $\tilde{g_{ij}}(t) = \begin{pmatrix} 1 & 0 \\ 0 & r(\psi(t))^2 \end{pmatrix}.$
Now, I am asked in my exercise, if there is also a reparametrization $\phi: \hat{U} \rightarrow U$, such that we locally(!) (for some open set $U$ in the domain of $f$) get a map $\hat{f}: \hat{U} \rightarrow \mathbb{R}^3$ satisfying $\hat{f} = f \circ \phi$, where the new metric tensor $(\hat{g}_{ij})(p) = \lambda(p) \ Id$ for all $p \in \hat{U}$.
I know that a reparametrization would give rise to the following equation regarding the metric tensor: $$(\hat{g}_{ij}) = D \phi^T ( g_{ij}) D \phi$$.
I also got the hint that I shall look for an ODE and use the Picard-Lindelöf theorem to show the local existence, but I just don't see a differential equation showing up here anywhere.
If anything is unclear, please let me know.
We have to reparametrize the meridian curve $t\mapsto\bigl(r(t),z(t)\bigr)$ in such a way that the metric tensor obtains the desired form. Therefore we consider the representation $$g(\tau,\phi):=f\bigl(p(\tau),\phi\bigr)=\bigl(r(p(\tau))\cos\phi,r(p(\tau))\sin\phi,z(p(\tau))\bigr)$$ of the same surface, where $\tau\mapsto t=p(\tau)$ is a monotonically increasing function to be determined.
Computation gives $$g_\tau\cdot g_\tau=(r'^2+z'^2)\dot p^2, \quad g_\tau\cdot g_\phi=0,\quad g_\phi\cdot g_\phi=r^2\ ,$$ whereby the ${}'$ denotes differentiation with respect to $t$, and the dot denotes differentiation with respect to $\tau$. Since we want the metric tensor to be a multiple of the identity matrix we obtain the condition $$\dot p={r(t)\over\sqrt{r'^2(t)+z'^2(t)}}\ .$$ Here the left hand side is nothing else than ${dt\over d\tau}$, which means that we have derived the desired ODE. Obtaining the inverse function $$q:=p^{-1}:\quad t\mapsto \tau:=q(t)$$ reduces to a mere integration: $$q(t)=\int_{t_0}^t{\sqrt{r'^2(s)+z'^2(s)}\over r(s)}\ ds\ .$$