Explain this passage on the proof that the curvature of the vector function $r(t)$ is $\frac{|r'(t) \times r''(t)|}{ |r'(t)|³}$? 
Explain this step on the proof of the curvature of the vector $r(t)$?
 A: Observe that $r$ and $T$ are functions of the variable $t$, so $r'$ is. But $r'$ is a product of functions, therefore, using Leibniz' rule, you can derive it, obtaining $r''$.
Now $T \times T$ is $0$ because, clearly, these vectors aren't linearly indipendent.
Finally, calculate the wedge product in $\mathbb{R}^3$ of the vectors $r'$ and $r''$. By distributive law, using the fact that $T \times T=0$, you obtain the result stated.
A: If your question is why $\mathbf{r'}\times \mathbf{r''}=(\frac{ds}{dt})^2(\mathbf{T} \times \mathbf{T'})$, here is how it worked:
$$\mathbf{r'}\times \mathbf{r''}=(\frac{ds}{dt}\mathbf{T})\times (\frac{d^2s}{dt^2}\mathbf{T}+\frac{ds}{dt}\mathbf{T'})\qquad\qquad \rightarrow (simple\ substitution)$$
$$ =(\frac{ds}{dt}\mathbf{T} \times \frac{d^2s}{dt^2}\mathbf{T}) + (\frac{ds}{dt}\mathbf{T} \times \frac{ds}{dt}\mathbf{T'}) \qquad\qquad\rightarrow(by\  distribution\  law) $$
$$=(\frac{ds}{dt}*\frac{d^2s}{dt^2})(\mathbf{T}\times \mathbf{T})+(\frac{ds}{dt})(\mathbf{T}\times\mathbf{T'}) \qquad\qquad\rightarrow (cross\ multiplication\ properties)$$
since $\mathbf{T}\times\mathbf{T}=0$ (the cross product of any vector by itself is $0$, try it out!)
then, the first summand will disappear and the final result is:
$$\mathbf{r'}\times \mathbf{r''}=(\frac{ds}{dt})^2(\mathbf{T} \times \mathbf{T'})$$
