Serret-Frenet equations in ${ℝ}^2$

Given a plane curve, we may associate to each point on the curve anorthonormal basis consisting of the unit normal tangent vector andthe unit normal. In general, different points will have differentbases associated to them, so we may ask how the basis depends uponthe choice of point. The Serret-Frenet equations answer thisquestion by relating the rte of change of the basis vectors tothe curvature of the curve.

Suppose $I$ is an open interval and $c:I\to {ℝ}^2$ is a twicecontinuously differentiable curve such that $\parallel c^{\prime }\parallel =1$.Let us thendefine the tangent vector and normal vector as

$𝐓$$=$$c^{\prime },$

$𝐍$$=$$J\cdot 𝐓,$

where is therotational matrix that rotates the plane $90$ degrees counterclockwise.

Curvature

Differentiating $⟨c^{\prime },c^{\prime }⟩=1$ yields$⟨{𝐓}^{\prime },𝐓⟩=0$,so ${𝐓}^{\prime }$ is in the orthogonal complement of $𝐓$,which is $1$-dimensional. Since $J\cdot 𝐓$ is also inthe orthogonal complement,it follows that there exists a function $\kappa :I\to ℝ$ such that

$${𝐓}^{\prime }=\kappa J\cdot 𝐓.$$

Furthermore, $\kappa$ is uniquely determined by this equation.We define this unique $\kappa$ tobe the curvature of $c$. Explicitly,

$$\kappa =⟨{𝐓}^{\prime },J\cdot 𝐓⟩.$$

Serret-Frenet equations

By the definition of curvature

${𝐓}^{\prime }$$=$$\kappa J\cdot 𝐓=\kappa 𝐍$

and so

${𝐍}^{\prime }$$=$$J\cdot {𝐓}^{\prime }=\kappa J𝐍=-\kappa 𝐓$

since $J^2=-\mathrm{I}$. These are the Serret-Frenetequations