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

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

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.