# Show that the torsion satisfies $\tau(s)=\frac{-\alpha'(s)\wedge\alpha''(s)\cdot\alpha'''(s)}{|\kappa(s)|^2}$

Let $$\alpha:I\rightarrow\mathbb{R}^{3}$$ a curve parametrized by arc lenght $$s$$. Denote by $$t(s)=\alpha'(s)$$ the tangent vector, $$\kappa(s)=|\alpha'(s)|$$ the curvature, $$n(s)=\alpha''(s)/\kappa(s)$$ the normal vector, $$b(s)=t(s)\wedge n(s)$$ the binormal vector. The torsion is the number $$\tau(s)$$ such that $$b'(s)=\tau(s)n(s).$$

I need to show that $$\tau(s)=\frac{-\alpha'(s)\wedge\alpha''(s)\cdot\alpha'''(s)}{|\kappa(s)|^2}$$.

$$\frac{-\alpha'(s)\wedge\alpha''(s)\cdot\alpha'''(s)}{|\kappa(s)|^2}n(s)= \frac{-t(s)\wedge n(s)\cdot\alpha'''(s)}{\kappa(s)}n(s),$$

and, since $$b'(s)=t'(s)\wedge n(s)+t(s)\wedge n'(s),$$ I need to prove that

$$\frac{-t(s)\wedge n(s)\cdot\alpha'''(s)}{\kappa(s)}n(s)=t'(s)\wedge n(s)+t(s)\wedge n'(s).$$

How can I do that?

By definition, $$\alpha'(s) = t(s) ,$$ and differentiating twice, $$\alpha''(s) = t'(s) = \kappa(s) n(s) \\ \alpha'''(s) = \kappa'(s) n(s) + \kappa(s) n'(s) .$$ We can find what $$n'$$ is: \begin{align} n \cdot n &= 1 \implies 0 = (n \cdot n)' = n \cdot n' \\ n \cdot t &= 0 \implies 0 = (n \cdot t)' = n' \cdot t + n \cdot t' = n' \cdot t + \kappa n \cdot n = n' \cdot t + \kappa \\ n \cdot b &= 0 \implies 0 = (n \cdot b)' = n' \cdot b + n \cdot b' = n' \cdot b + \tau \end{align} so $$n' = -\kappa t - \tau b$$, one of the Frenet–Serret formulae (note that Wikipedia uses a different, more common convention that $$b' = -\tau n$$). Thus, $$\alpha'''(s) = \kappa'(s) n(s) - \kappa(s)^2 t(s) - \kappa(s)\tau(s)b(s)$$ If we now compute the triple product of the first three derivatives, we find $$\alpha'(s) \times \alpha''(s) = \kappa(s) b(s),$$ so $$(\alpha'(s) \times \alpha''(s)) \cdot \alpha'''(s) = -\kappa(s)^2 \tau(s),$$ whence the result. The key here is to work either completely in terms of $$\alpha$$, or completely in terms of $$t,n,b$$: mixing them tends to be confusing.
You can just calculate that $$\alpha'''=\left(\kappa N\right)'=\kappa N'+\kappa' N=\kappa \tau B+\cdots,$$ using the Frenet equations. We only care about the $$B$$ component due to the dot product we want to take. Hence, $$(\alpha'\times\alpha'')\cdot\alpha'''=\kappa^2\tau\implies \tau=\frac{(\alpha'\times\alpha'')\cdot\alpha'''}{\kappa^2}.$$
EDIT: Typically, one defines $$\tau$$ by $$B'=-\tau N.$$ I see now that you are using $$B'=\tau N.$$ In this case, the Frenet equations have some sign changes. Taking this into account, you get that $$\alpha'''=-\kappa\tau B+\cdots,$$ and taking the dot product gives $$(\alpha'\times\alpha'')\cdot\alpha'''=-\kappa^2\tau,$$ or $$\tau=-\frac{(\alpha'\times\alpha'')\cdot\alpha'''}{\kappa^2}.$$