# 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.