Prove that for any piecewise smooth curve it is possible to find the parametrisation Prove that for any piecewise smooth curve it is possible to find the parametrisation $\phi$ that is consistent with its length, ie.
length of a curve segment between $\phi(a)$ and $\phi(b)$ is equal to $|b-a|$.
It seems intuitive, but I don't have any idea how can I show this in the formal way.
 A: If the focus is on the existence of an arclength parametrization then, for a first step, it is convenient to assume that $\phi :[a,b]\rightarrow M$ is a differentiable curve with nonvanishing derivative. The length of that curve, restricted to $[a,t]$ is then given by
$$\ell(t) =\int_a^t |\phi^\prime(s)|\,ds $$
This is, by the assumption $\phi^\prime \neq0$, a strictly monotonic function. In fact, $\ell^\prime(t) = |\phi^\prime(t)|>0$. Therefore, by the inverse function theorem, there is a (differentiable) function $\rho: [0,\ell(b)]\rightarrow [a,b]$ such that $\ell\circ \rho (x) = x$. Also, $\rho^\prime = \frac{1}{|\phi^\prime|}$.
Now look at $\phi\circ\rho: [0,\ell(b)]\rightarrow M$. We have
$$\frac{d}{ds} \phi\circ\rho(s)=\frac{\phi^\prime}{|\phi^\prime|}$$
which is of length $1$. So with respect to this parametrization, $\int_0^c |(\phi\circ\rho)^\prime| ds = c $ is the length of $\phi\circ \rho|_{[0,c]}$
(I'm rather sure I answered exactly the same question during the last year, but cannot find that anymore. It probably got deleted by the crusaders who feel in charge of "cleaning up" this site).
