Finding torsion without computing binormal vector I've been asked to 'write down the torsion of [a space curve] without computing the binormal vector'.
Now, is this just a matter of using known vector identities with Tangent and Normal vectors to calculate the torsion, or am I forgetting a proposition?
 A: I'm going to use $\bf t$, $\bf p$, and $\bf b$ to refer to the tangent, normal, and binormal vectors, as well as $\kappa$ and $\tau$ to refer to the curvature and torsion, respectively. The function $\bf\dot{x}$ is the derivative with respect to arc length $s$ and $\bf x'$ is the derivative with respect to an aribtrary parameter $t$.
Depending on your perspective, this might not be possible. This is because the torsion of a space curve is defined using the binormal as $${\bf\dot{b}}=-\tau{\bf p}\tag{1}$$
by some authors. That being said, a few formulas can be constructed that sort of hide the fact that $\bf b$ is being computed. One example uses the relation $${\bf\dot{p}}=-\kappa{\bf t}+\tau{\bf b}\tag{2}$$
By taking the dot product with $\bf b$, we get $$\tau={\bf\dot{p}}\cdot{\bf b}={\bf\dot{p}}\cdot({\bf t}\times{\bf p})=det({\bf \dot{p}},{\bf t},{\bf p})\tag{3}$$
This may be a sufficient answer if you as keen to using arc length as your argument. Otherwise, we must convert to arbitrary parameterizations. The algebra is quite long and tedious, but it simplifies down to $$\tau=\frac{det({\bf x'},{\bf x''},{\bf x'''})}{|{\bf x'\times x''}|^2}=\frac{({\bf x'\times x''})\cdot{\bf x'''}}{|{\bf x'\times x''}|^2}\tag{4}$$
Whenever you see something like $\bf x'\times x''$, this is alluding to the need to calculate the binormal. I hope this helps.
