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.
But why must the magnitude of $\underline{r}'(t)$ be constant? It is written as a function of $t$, so as $t$ changes, $\underline{r}'(t)$ can change, and thus its magnitude can change.
You just need to realize how scalar multiplication works across cross products. The key is $$(k\,\mathbf{a})\times \mathbf{b}=\mathbf{a}\times(k\,\mathbf{b})=k\,(\mathbf{a}\times\mathbf{b}),$$ paying special attention to the last equality.
Then, using that last equality twice and the fact that $\underline{T}\times\underline{T}=\underline{0}$, we get \begin{align}\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)\\ &=s'(t)s'(t)(\underline{T}\times\underline{T'})+s's''(\underline{T}\times\underline{T})\\ &=[s'(t)]^2(\underline{T}\times\underline{T'}), \end{align} as desired.