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Let $r(t)$ and $s(t)$ ($t\in\mathbb{R}$) be two differentiable vector functions describing the motions of two particles $R$ and $S$ respectively travelling in the same direction along the same curve. We further assume that $r(0) = s(0)$.

Why is the following statement false?

If $r(t)$ is smooth (i.e. $r'(t)\neq <0,0,0>$ for all $t\in\mathbb{R}$), then $s(t)$ is smooth.

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Let $$r(t)=\begin{pmatrix} 0 \\ 0 \\ t\\ \end{pmatrix} \qquad s(t)=\begin{pmatrix} 0 \\ 0\\ e^{-\frac{1}{t^2}}\end{pmatrix}$$ both particales move along the $z$ axis, but the derivative of $s$ at the origin is 0.
Note that $\exp(-\frac{1}{t^2})$ has the continous differentiable extension $0$ in $0$.

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