Existence of Unit-speed parametrization. It seems obvious to me that for all regular smooth functions, there exists a unit-speed parametrization.  Is there a proof of that statement?
Andrew Pressley's Elem.Diff.Geometry tells me that "A parametrized curve has a unit-speed reparametrization if and only if it is regular". Does the proof of his statement prove existence?  In a follow-up example, he shows the twisted cubic, $\gamma(t)=(t,t^2,t^3),-\infty\le t \le +\infty$ and says that since it results in an elliptic integral when trying to find the arc length that we cannot express the unit-speed parametrization with elementary functions. That doesn't seem to exclude its existence.
 A: Consider the arc length function $L:t\mapsto s$ of the curve $C$. Its derivative is just the length element, which is non-zero since $C$ is regular. It follows by the inverse function theorem that $L^{-1}:s\mapsto t$ exists, and thus the arc length parametrisation $\alpha(s)=C(L^{-1}(s))$  exists.
A: The proof is given pretty much by just explicitly constructing the reparametrization:
Let $c:I \rightarrow \mathbb{R}^3$ be a regular parametrized curve. Let $t_0 \in I$ and define:
$$\psi(t) := \int_{t_0}^{t}||\dot{c}(\tau)||d\tau.$$
Set $J = \psi(I)$. Now $\psi$ is a diffeomorphism from $I$ to $J$: By the fundamental theorem of calculus it is differentiable, and since
$$\psi'(t) = ||\dot{c}(t)|| > 0,$$
it is monotonic increasing, hence bijective. Define $\phi(t) := \psi^{-1}(t)$. Now  $\phi(t)$ is the reparametrization of $c$ that makes it parametrized by arc-length. You can check this by computing the norm of the derivative of the composition.
Now this gives you the existence of the reparametrization, but as you may well know not all integrals are expressable in terms of elementary functions, so we have no way to know if we can actually 'do' the integral. Apparently, the twisted cubic you mentioned is such a case (which I haven't checked...).
