Why isn't torsion the magnitude of the derivative of the binormal vector? I am learning about the $TNB$, curvature and torsion and I found this weird derivation about torsion:
$$\tau=-B'(s)\cdot \hat{N}(s).$$
And the explaination I got basically is that since
 $$|\hat{B}(s)|=1,$$ then $$B'(s)\perp \hat{B}(s).$$
Also using some algebraic manipulation you get that
 $$\hat{T}(s)\perp B'(s).$$
Therefore $B'(s)$ must be parallel to the normal vector $\hat{N}(s)$ since it's perpendicular to both $B$ and $T$.
So you could say that  $B'(s)$ is some constant $\tau$ multiplied by $\hat{N}$, and that's how you get to the first formula.
So the question is, why is it done this way? Isn't it posible to say "torsion is the speed of change of the binormal vector", write it down as $\tau=|B'(t)|$? Am I missing something? Is there a historic or practical reason why it's defined this way? 
It just seemed very counterintuitive.
 A: The norm $\|B'(t)\|$ tells us only something about the size of the change of $B(s)$, it doesn't tell you in which directions the binormal changes. For this reason it is natural to have a look at the components of $B'(s)$ w.r.t. the Frenet frame:
$$
   B'(s) = \bigl( B'(s)\cdot T(s) \bigr) T(s) + \bigl( B'(s)\cdot N(s)\bigr) N(s) + \bigl(B'(s) \cdot B(s)\bigr) B(s). 
$$ 
Since we have shown that the 1st and 3rd component are zero, we have 
$$B'(s) = \bigl( B'(s)\cdot N(s)\bigr) N(s). \tag{$*$}$$ 
Since $B'(s)$ only has one component, your characterization $\tau(s) = \|B'(s)\|$ agrees with the definition, at least up to the sign. Also, $\tau = \|B'(s)\|$ doesn't specify a sign for $\tau$: a norm is just always non-negative, but $B'(s)\cdot N(s)$ can be negative.
Wikipedia and my old differential geometry notes claim that the sign is chosen historically. However, the sign intuitively makes sense. The torsion is a measure of how hard the curve is moving out of the $TN$-plane (osculating plane) at a point. If $\tau > 0$, it goes in the direction of $B$; if $\tau < 0$, it goes in the direction of $-B$.
