How to tell whether a curve has a regular parametrization? A parametrization of a 1-dimensional curve is called regular if its velocity is always positive. For example, the following parametrization:
$$x(t)=t^3, y(t)=t^6$$
is not regular because its velocity is 0 in $t=0$.
But, this same curve can be re-parametrized as:
$$x(t)=t, y(t)=t^2$$
and this second parametrization is regular because its velocity anywhere is at least 1.
So, my question is: given a non-regular parametrization of a curve, is there an algorithm to tell whether the curve has a regular parametrization, and find it if it exists?
 A: $\newcommand{\Reals}{\mathbf{R}}$Let $I$ be a non-empty open interval of real numbers, and $\gamma:I \to \Reals^{n}$ a continuously-differentiable path. (This hypothesis is arguably substantial, but appears not out of line with the OP's intent.)
At each point where $\gamma$ is regular, define the unit tangent field
$$
T(t) = \frac{\gamma'(t)}{\|\gamma'(t)\|}.
$$
By continuity, the zero set of $\gamma'$ is a closed subset of $I$; that is, the domain of $T$ is an open subset of $I$. Without loss of generality, we may assume the domain of $T$ is the complement of a discrete set. (Loosely, if there are intervals on which $\gamma' = 0$, "excise them and push their endpoints together").
Claim: The following are equivalent:
(i) $\gamma$ has a regular $C^{1}$ parametrization.
(ii) $T$ has a continuous extension to $I$.
(i) implies (ii): The unit tangent field is independent of monotone reparametrization in the sense that if $\Gamma(t) = \gamma(\tau(t))$ for some differentiable function $\tau$ with positive derivative, then the unit tangent field of $\Gamma$ at $t$ is the unit tangent field of $\gamma$ at $\tau(t)$. If $\gamma$ has a regular $C^{1}$ reparametrization $\Gamma$, then the unit tangent field of $\Gamma$ continuously extends the unit tangent field of $T$.
(ii) implies (i). If $T$ has a continuous extension to $I$ (which we continue to denote $T$), then
$$
\Gamma(s) = \int_{0}^{s} T(t)\, dt
$$
is a regular parametrization of $\gamma$.
For the curve in question, we have
$$
\gamma(t) = (t^{3}, t^{6}),\qquad
\gamma'(t) = (3t^{2}, 6t^{5}),\qquad
T(t) = \frac{(3t^{2}, 6t^{5})}{\sqrt{9t^{4} + 36t^{10}}}
  = \frac{(1, 2t^{3})}{\sqrt{1 + 4t^{6}}},
$$
which has a continuous extension at $t = 0$ (given by the same formula).
For a cusp, say, we have
$$
\gamma(t) = (t^{2}, t^{3}),\qquad
\gamma'(t) = (2t, 3t^{2}),\qquad
T(t) = \frac{(2t, 3t^{2})}{\sqrt{4t^{2} + 9t^{6}}}
  = \frac{t(2, 3t)}{|t|\sqrt{4 + 9t^{2}}},
$$
which does not extend continuously to $0$.
