Given a parametrized curve, we know that its arc length parametrization is its unit speed reparametrization. However, I wanted to know if there was any generic procedure to find any other reparametrizations of the same curve, which are not unit speed, and are non trivial?
[Math] Reparametrization of a curve
calculusdifferential-geometryvector analysis
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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 $C^1$.
Consider the spiral $\gamma: \Bbb R \to \Bbb R^2$ given by $$\gamma(t) = (e^t\cos t, e^t \sin t).$$ Then $$\gamma'(t) = (e^t(\cos t - \sin t),e^t (\cos t+\sin t)),$$ and so: $$\|\gamma'(t)\| = e^t\sqrt{(\cos t -\sin t)^2+(\cos t+\sin t)^2} = e^t\sqrt{2}.$$ With this: $$s(t) = \int_0^t e^\xi \sqrt{2}\,{\rm d}\xi = \sqrt{2}(e^t-1) \implies t = \ln\left(\frac{s}{\sqrt{2}}+1\right).$$ So we get a nice and non-trivial unit speed curve: $$\gamma(t(s)) \equiv \gamma(s) = \left(\left(\frac{s}{\sqrt{2}}+1\right)\cos\left(\ln\left(\frac{s}{\sqrt{2}}+1\right)\right),\left(\frac{s}{\sqrt{2}}+1\right)\sin\left(\ln\left(\frac{s}{\sqrt{2}}+1\right)\right)\right).$$
Best Answer
Assume you have a curve $\gamma : [a,b] \to \mathbb R^d$ and $\varphi : [a,b] \to [a,b]$ is a reparametrization, i.e., $\varphi'(t) > 0$. Then you can prescribe any speed function for your parametrization. Given a function $\sigma: [a,b] \to \mathbb R_{>0}$, define $\varphi$ via the ODE $$ \varphi'(t) = \frac{\sigma(t)}{|\gamma'(\varphi(t))|} \,. $$ Then the reparametrized curve $\tilde \gamma (t) = \gamma \circ \varphi(t)$ has speed $\sigma$, i.e., $$ |\tilde \gamma'(t)| = \sigma(t)\,. $$ The only restriction on $\sigma$ is that $\varphi(b) = b$.