# Definition of curvature and points of its definition

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 $t=0$?

• Curvature is defined only for regular curves or equivalently for curves which can be parametrized with respect to arc length. Sep 3 '18 at 19:59
• @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/…
– user
Sep 17 '18 at 20:23

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