The curvature of a curve (rate of change of the unit tangent vector with respect to arc length) is defined as $\kappa = \frac{|\underline{r}'(t) \times \underline{r}''(t)|}{|\underline{r}'(t)|^3}$ The proof of this is shown in my textbook, but I don't understand one step.
They say: $\underline{r}'(t) = \underline{T}(t) |\underline{r}'(t)| = \underline{T}(t) s'(t)$ and then $ \underline{r}''(t) = \underline{T}'(t) s'(t) + \underline{T}(t)s''(t).$
First question: Would $s''(t)$ not equal $0$? My reasoning being since $s'(t) = |\underline{r}'(t)|,$ a constant and then subsequently differentiate again to get $0$?
After the above step, they say $\underline{r}'(t) \times \underline{r}''(t) = [s'(t)]^2 \underline{T}(t) \times \underline{T}'(t)$. I am not sure how they get this. The expression I got when I simplified was:$ \underline{T}(t)s'(t) \times (\underline{T}'(t)s'(t) + \underline{T}(t)s''(t)) = \underline{T}(t)s'(t) \times \underline{T}'(t)s'(t) + \underline{T}(t)s'(t) \times \underline{T}(t)s''(t)$ And the latter term disappears because $\underline{T}(t) \times \underline{T}(t) = \underline{0}$ Many thanks.