Glueing smooth functions give a smooth function if reparametrized [closed]

Question

Given $\mathbf{r}:(t_0-\varepsilon, t_0+\varepsilon)\to\mathbb{R}^2$ be a $C^{1}$ application, with $\mathbf{r}'(t)=(0,0)\Longleftrightarrow\ t=t_0$, and

$$\mathbf{r}(t)=\begin{cases} \mathbf{r}_1 (t), \ t\in (t_0-\varepsilon, t_0] \\ \mathbf{r}_2 (t), \ t\in [t_0, t_0+\varepsilon)\end{cases},\ \mathbf{r}_1\in C^{\infty}((t_0-\varepsilon, t_0]),\ \mathbf{r}_2\in C^{\infty}([t_0, t_0+\varepsilon))$$

Is it true that we can always find $\mathbf{s}:(t_0-\varepsilon, t_0+\varepsilon)\to\mathbb{R}^2$, $\mathbf{s}\in C^{\infty}((t_0-\varepsilon, t_0+\varepsilon))$, such that $\mathbf{s}((t_0-\varepsilon, t_0+\varepsilon))=\mathbf{r}((t_0-\varepsilon, t_0+\varepsilon))$?

Answer 1

Yes, even if $\mathbf r$ is only $C^0$. Just reparameterize it in such a way that it stops infinitely flat at 0 (use $e^{-1/t^2}$).