Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 1 down vote accepted

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 $0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.