# Serret-Frenet for Non-unit “speed” space curves

The Serret-Frenet equations form a system of linear, often non-autonomous, ordinary differential equations that recover the local tangent, normal and binormal vectors of a unit "speed" space curve from the curve's curvature and torsion.

The equations are scaled for space curves of non unit "speed." (see sec. 5 of the previous Wikipedia article) This means that in order to find the local frame one must also supply data about the curve parameterization. Often, one wants to find the curve parameterization for known curvature/torsion variables.

Question 1: So, if one were to solve the SF equations

\begin{align} \dfrac{d\mathbf{T}}{ds} &= & \kappa \mathbf{N}, \\ \dfrac{d\mathbf{N}}{ds} &= - \kappa \mathbf{T} & & + \tau \mathbf{B},\\ \dfrac{d\mathbf{B}}{ds} &= & -\tau \mathbf{N}, \end{align}

for a given curvature and torsion, then would the associated space curve be identical, up to its "speed," to the one found by solving

$\frac{d}{ds} \begin{bmatrix} \mathbf{T}\\ \mathbf{N}\\ \mathbf{B} \end{bmatrix} = \|\mathbf{r}'(s)\| \begin{bmatrix} 0&\kappa&0\\ -\kappa&0&\tau\\ 0&-\tau&0 \end{bmatrix} \begin{bmatrix} \mathbf{T}\\ \mathbf{N}\\ \mathbf{B} \end{bmatrix}?$

It seems that this should not be the case since the $||\textbf{r}'(s)||$ prefactor could greatly complicate the system of equations.

Question 2: Assuming that the answer to question 1 is no, then what is the utility of the SF equations? Knowledge of $||\textbf{r}'(s)||$ likely comes from knowledge of $\textbf{r}'(s)$, which should allow one to find the tangent, normal and binormal vectors bypassing SF.

The Serret-Frenet formula in the case where the curve is not parametrised by arc-length are $$\frac{1}{V_r(t)}\frac{d}{dt}\begin{pmatrix} T\\ N\\ B \end{pmatrix} = \begin{pmatrix} 0 & \kappa & 0\\-\kappa & 0 & \tau \\ 0 &-\tau & 0\end{pmatrix} \begin{pmatrix} T\\N \\B \end{pmatrix}$$ where $V_r(t)=\Vert \dot{r}(t) \Vert$ is the speed of your curve. This is true because in the case of non-arc-legnth parametrisation, the definition of torsion and curvature changes, in order to take into account the speed. Namely, we have that for a $C^3$ biregular curve, the torsion is defined by $$\tau = \frac{1}{V_r} \langle B, \dot{N} \rangle$$ and the curvature $\kappa$ is defined as the norm of the curvature vector $$K = \frac{1}{V_r} \dot{T}$$
Assuming that all these quantities are well defined (which requires, as I mentionned earlier, that the curve be at least $C^3$ and biregular), then it holds that the "trace", or the image of the curve, is invariant under reparametrisation (actually we even need a lot less: assuming merely that the curve is $C^1$ and regular is enough, for instance).
Serret-Frenet formulas provide a cinematic interpretation of the curvature and the torsion. Locally, these quantities tell you how the curve is going to evolve. More generally, it is possible to show that (under the usual hypothesis $C^3$ and biregular) if two curves parametrised by arc-length possess same curvature and same torsion, then they are "equal" in the sense that there exists a deplacement of $\mathbb{R}^3$ from one onto the other. Therefore curvature and torsion control all the geometry of the curve, and thus solving Serret-Frenet completely determines the curve.
Observe, $$\frac{dT}{dt} = \frac{ds}{dt}\frac{dT}{ds} \ \& \ \frac{dN}{dt} = \frac{ds}{dt}\frac{dN}{ds} \ \& \ \frac{dB}{dt} = \frac{ds}{dt}\frac{dB}{ds}$$ but, the Serret-Frenet equations allow us to write the change in $T,N,B$ in terms of $\kappa, \tau$-weighted sum of $T,N,B$. That is, the equations you quote from Wikipedia. Except, note the distinction between $s$ (arclength) and $t$ (an arbitrary parameter). We define the curvature and torsion with respect to the change of $T,N,B$ in terms of arclength. That choice is important as without it the curvature and torsion would not be well-defined. As the equations above reveal, the magnitude of the rates of change in $T,N,B$ are different for different speed parametrizations of a given curve. That said, it is important to allow for non-unit-speed parametrizations in the theory because often the arclength parametrization is uglier than the $t$-parametrization. In fact, it may not even be possible to calculate the arclength explicitly ! So, we need the $s$-equations for logic, and we need the $t$-equations for computational flexibility. Why do we need the Frenet-Serret equations at all? I suppose it's nice to be able to show congruence of curves which defy highschool geometry contructive geometry. We just have to show the curvature and torsion match then we know the curves are the same curve up to some Euclidean motion. Moreover, this is merely the first chapter in a deeply interesting chapter of math known as Cartan's method of moving frames... much more to see here.