# Unit-speed reparametrizations

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

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If you have a curve $(f(t),g(t))$ with the corresponding arclength function $s(t)$, then a unit-speed parametrization would take the form $(f(t(s)),g(t(s)))$, where $t(s)$ is the inverse of the arclength function $s(t)$. How would you prove that $s(t)$ is an invertible function (e.g. via monotonicity, etc.)? – J. M. Nov 1 '11 at 17:53
It's not clear to me what are you exactly asking. However, it is not difficult to prove that $s$ is an invertible function, it follows clearly from regularity of the parametrization over a connected set (the function is strictly monotone increasing). – user14174 Nov 1 '11 at 18:12

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 $C^1$.