Let $X:[a,b)\to\mathbb R^n\ (a<b)$ be a $C^1$ curve whose velocity does not vanish on $(a,b)$ (but $X'(a)$ could be $0$). Let $\varphi:[a,b)\to\mathbb R$ be defined by \begin{equation} \varphi(t) = \int_a^t |X'(u)|\ du\text{.} \end{equation} Then $\varphi$ is a bijective function onto its image, which is an interval. Define \begin{equation} Y=X\circ\varphi^{-1}\text{.} \end{equation} Then is it true that $Y$ is $C^1$, or at least differentiable on the whole domain?
I know that $Y$ is $C^1$ on the interval excluding $0$ (the endpoint), but I don't know what happens at $0$.
The answer is no.
Consider the example where $X\colon[0,b)\to\mathbb{R}^2$ is a smooth spiral $t\mapsto(\exp(-1/t^2)\cos(t^{-1}),\exp(-1/t^2)\sin(t^{-1}))$ for $t\neq 0$ and $X(0)=(0,0)$. This $X$ is $C^1$, but the reparametrised curve does not have a one-sided derivative at $0$ --- the spiral doesn't have a tangent direction at the limit point origin.