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 $p\cdot p=1$ to obtain $\frac{dp}{ds}\cdot p=0$ so it takes on the form $\frac{dp}{ds}=at+cb$. We take the dot product of this with t to obtain $a=\frac{dp}{ds}\cdot t$. Taking the derivative of $p\cdot t=0$, we get $\frac{dp}{ds}\cdot t+p\cdot \frac{dt}{ds}=0$ or $\frac{dp}{ds}\cdot t=-p\cdot \frac{dt}{ds}=-\kappa p\cdot p=-\kappa$. Also, taking the dot product of $\frac{dp}{ds}=at+cb$ with b, we obtain $c=\frac{dp}{ds}\cdot p$. Taking the derivative of $p\cdot b=0$, we get $\frac{dp}{ds}\cdot b=-p\cdot \frac{db}{ds}=\tau$. 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.