Vtyjkyuk

Let $gamma:(0,1)rightarrow mathbbR^3$ be a smooth parametrized curve.

If $|gamma'(t)|=1$ for all $tin (0,1)$ 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
$$frac$$
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 $t=0$?

We can consider

$$lim_tto 0^+ frac=lim_tto 0^+ frac(2,6t)times (2t,3t^2) ^3=lim_tto 0^+ frac6t^2t^3sqrt(4+9t^2)^3=lim_tto 0^+ frac6tsqrt(4+9t^2)^3to infty$$