Let $\gamma:(0,1)\rightarrow \mathbb{R}^3$ be a smooth parametrized curve.

If $\|\gamma'(t)\|=1$ for all $t\in (0,1)$ then $\gamma$ is unit parametrized and its curvature (at point $t$ or say at $\gamma(t)$) is defined as $\|\gamma''(t)\|$.

Suppose $\gamma'(t)\neq 0$ for all $t$ but not necessarily of unit length for all $t$. So $\gamma$ is not unit parametrized.

Then curvature is given by formula $$\frac{\|\gamma''(t)\times \gamma'(t) \|}{\| \gamma'(t)\|^3}$$ I did not find any answer to following basic question. In the above definition of curvature, the curve $\gamma$ is taken to be regular parametrized curve, i.e. $\gamma'(t)\neq 0$ for all $t$. But if $\gamma'(t)=0$ for some $t$, do we say that curvature is not defined at this point $t$?

The above expression(s) give formulas for curvature; so if $\gamma'(t)=0$ for some $t$, then formula doesn't make sense; but my question is whether curvature is (can be) defined at such points?

For example one can take $\gamma(t)=(t^2,t^3)$. Is curvature definedat $t=0$?

  • 1
    $\begingroup$ Curvature is defined only for regular curves or equivalently for curves which can be parametrized with respect to arc length. $\endgroup$ Sep 3 '18 at 19:59
  • $\begingroup$ @Beginner Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$
    – user
    Sep 17 '18 at 20:23

We can consider

$$\lim_{t\to 0^+} \frac{\|\gamma''(t)\times \gamma'(t) \|}{\| \gamma'(t)\|^3}=\lim_{t\to 0^+} \frac{\|(2,6t)\times (2t,3t^2) \|}{\| (2t,3t^2)\|^3}=\lim_{t\to 0^+} \frac{6t^2}{t^3\sqrt{(4+9t^2)^3}}=\lim_{t\to 0^+} \frac{6}{t\sqrt{(4+9t^2)^3}}\to \infty$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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