I have to prove that for a regular parametrized curve there is essentially (up to sign and a constant) a unique reparametrization which makes it a unit-speed curve.
Let $x$ be a curve, $s(t) = \int_{t_0}^{t} \left \| \frac{\mathrm{d} x}{\mathrm{d} t} \right \| d\tau$ is an arc length parameter. We have the unit-speed reparametrization $y(s) = x(t(s))$, in fact
$$1=\left \| \dot{y}(s) \right \| = \left \| \frac{\mathrm{d} x}{\mathrm{d} t} \right \| \, \frac{\mathrm{d} t}{\mathrm{d} s}.$$
Suppose that u is a parameter that makes $y(u) = x(t(u))$ a unit-speed parametrization. Then
$$1=\left \| \dot{y}(u) \right \| = \left \| \frac{\mathrm{d} x}{\mathrm{d} t} \right \| \, \frac{\mathrm{d} t}{\mathrm{d} u}$$
and so
$$\frac{\mathrm{d} u}{\mathrm{d} t} = \pm \frac{\mathrm{d} s}{\mathrm{d} t}
$$
that finally yelds $u=\pm s + \mathrm{const}$.
I have no idea if this can be proved as I did, can you spot any errors? Thanks
Best Answer
Roughly, I would agree with you. The differential geometry of curves and surfaces course I took wasn't the absolutely most rigorous, but this is how we did it. The only small remark I would make, is that it would make things more clear to note that since $\displaystyle \pm 1=\dot{x} \frac{dt}{du}$ on a connected set, $\displaystyle \dot{x}\frac{dt}{du}$ is continuous, we know that $\displaystyle \dot{x}\frac{dt}{du}=1$ simultaneously for all values or $\displaystyle \dot{x}\frac{dt}{du}=-1$ for all values. This was probably what you meant, but I think it's important to note--it at least puts things on a more rigorous standing.
NB: I obviously assumed that our curve was at least $C^1$.