Differential Geometry/Frenet-Serret Formulae

The derivatives of the vectors t, p, and b can be expressed as a linear combination of these vectors. The formulae for these expressions are called the Frenet-Serret Formulae. This is natural because t, p, and b form an orthogonal basis for a three-dimensional vector space.

Of course, we know already that $$\frac{dt}{ds}=\kappa p$$ and $$\frac{db}{ds}=-\tau p$$ so it remains to find $$\frac{dp}{ds}$$. First, we differentiate $$p \sdot p = 1$$ to obtain $$\frac{dp}{ds}\sdot p = 0$$ so it takes on the form $$\frac{dp}{ds}=at+cb$$. We take the dot product of this with t to obtain $$a=\frac{dp}{ds}\sdot t$$. Taking the derivative of $$p\sdot t=0$$, we get $$\frac{dp}{ds}\sdot t + p \sdot \frac{dt}{ds} = 0$$ or $$\frac{dp}{ds}\sdot t=-p \sdot \frac{dt}{ds} = -\kappa p \sdot p = -\kappa$$. Also, taking the dot product of $$\frac{dp}{ds}=at+cb$$ with b, we obtain $$c=\frac{dp}{ds}\sdot p$$. Taking the derivative of $$p\sdot b=0$$, we get $$\frac{dp}{ds}\sdot b=-p\sdot\frac{db}{ds}=\tau$$. Thus, we arrive at the following expression for $$\frac{dp}{dt}$$:

$$\frac{dp}{dt}=-\kappa t+\tau b$$.

This formula, combined with the previous two formulae, are together called the Frenet-Serret Formulae and they can be represented by a skew-symmetric matrix.