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With help from https://en.wikipedia.org/wiki/Frenet–Serret_formulas

Let a curve be defined as $\vec{r}(t) : \mathbb{R} \rightarrow \mathbb{R}^n$
$\vec{r} = \{ r_i(t), 1 \le i \le n \}$

Velocity: $\vec{v} = \frac{\partial \vec{r}}{\partial t}$

Speed: $\sigma = |\vec{v}| = \sqrt{ \underset{j}{\Sigma} \left( \frac{\partial r_j}{\partial t} \right)^2 }$
$\sigma \ge 0$
$\frac{\partial \sigma}{\partial t} = \frac{1}{2 \sigma}$

Arclength: $s(t) := \int_{t_0}^t \sigma(t') dt'$
$\frac{\partial s}{\partial t} = \sigma$

$t = s^{-1}( s(t) )$
Notice the ambuguity that $(s)$ is a parameter (and can be replaced by any other letter) while $s^{-1}(\cdot)$ is the inverse of the arclength function.
Arclength inverse:
$t(s) = s^{-1}( s )$

Reparameterization by arclength: $r_i(s) := r_i(t(s))$
Notice that $r_i(t)$ and $r_i(s)$ are represented identically with the only difference being the parameter variable, however using a $t$ denotes the original function while using a $s$ denotes the arclength parameterization of the function.

Velocity wrt arclength:
$\frac{\partial \vec{r}}{\partial s} = \frac{\partial \vec{r}}{\partial t} \frac{\partial t}{\partial s} = \frac{1}{\sigma} \vec{v}$

un-normalized tangent:
$\vec{\bar{e}}_1 = \frac{\partial \vec{r}}{\partial s}$

un-normalized tangent length:
$|\vec{\bar{e}}_1| = |\frac{\partial \vec{r}}{\partial s}| = \frac{1}{\sigma} |\vec{v}| = \frac{\sigma}{\sigma} = 1$

tangent:
$\vec{e}_1 = unit(\vec{\bar{e}}_1) = \vec{\bar{e}}_1 = \frac{\partial \vec{r}}{\partial s} = \frac{1}{\sigma} \vec{v}$
$\vec{T} = \vec{e}_1$

tangent orthogonality to derivative:
$\vec{T} \cdot \vec{T} = 1$
$\frac{\partial}{\partial s} (\vec{T} \cdot \vec{T}) = 0$
$\frac{\partial \vec{T}}{\partial s} \cdot \vec{T} = 0$
$\frac{\partial^2 \vec{r}}{\partial s^2} \cdot \frac{\partial \vec{r}}{\partial s} = 0$

un-normalized normal:
$\vec{\bar{e}}_2 = \frac{\partial^2 \vec{r}}{\partial s^2} - (\frac{\partial^2 \vec{r}}{\partial s^2} \cdot \vec{e}_1) \vec{e}_1$
$= \frac{\partial^2 \vec{r}}{\partial s^2} - (\frac{\partial^2 \vec{r}}{\partial s^2} \cdot \frac{\partial \vec{r}}{\partial s}) \frac{\partial \vec{r}}{\partial s}$
$= \frac{\partial^2 \vec{r}}{\partial s^2}$

normal:
$\vec{N} = \vec{e}_2$
$= unit( \vec{\bar{e}}_2 )$
$= unit( \frac{\partial^2 \vec{r}}{\partial s^2} )$
$= \frac{ \frac{\partial^2 \vec{r}}{\partial s^2} }{ \left| \frac{\partial^2 \vec{r}}{\partial s^2} \right| }$

curvature:
$\chi_1 = \frac{1}{\left| \frac{\partial^2 \vec{r}}{\partial s^2} \right|}$
$\kappa = \chi_1$

So $\frac{\partial}{\partial s} \vec{e}_1 = \chi_1 \vec{e}_2$
i.e. $\frac{\partial}{\partial s} \vec{T} = \kappa \vec{N}$

Normal derivative:
$\frac{\partial}{\partial s} \vec{e}_2 = \frac{\partial}{\partial s} \left( \frac{ \vec{\bar{e}}_2 }{ \left| \vec{\bar{e}}_2 \right| } \right) $
$ = \frac{ \frac{\partial}{\partial s} \vec{\bar{e}}_2 \left| \vec{\bar{e}}_2 \right| - \vec{\bar{e}}_2 \frac{\partial}{\partial s} \left| \vec{\bar{e}}_2 \right| }{ \left| \vec{\bar{e}}_2 \right|^2 } $
$ = \chi_1^2 ( \frac{\partial^3 \vec{r}}{\partial s^3} \left| \vec{\bar{e}}_2 \right| - \frac{\partial^2 \vec{r}}{\partial s^2} \frac{\partial}{\partial s} \left| \vec{\bar{e}}_2 \right| )$

Frenet frame:
$\frac{\partial}{\partial s} \left[ \begin{matrix} \vec{T} \\ \vec{N} \\ \vec{B} \end{matrix} \right] = \left[ \begin{matrix} 0 & \chi_1 & 0 \\ -\chi_1 & 0 & \chi_2 \\ 0 & -\chi_2 & 0 \\ \end{matrix} \right] \left[ \begin{matrix} \vec{T} \\ \vec{N} \\ \vec{B} \end{matrix} \right]$

Arclength generalized by coordinate system / line element:
Start in our Cartesian basis:
$\frac{\partial s}{\partial t} = \sqrt{ \underset{i}{\Sigma} \left( \frac{\partial r^i}{\partial t} \right)^2 }$
Apply a change-of-basis:
Let $r^i = {e^i}_j x^j$
$\frac{\partial s}{\partial t} = \sqrt{ \underset{i}{\Sigma} \left( \frac{\partial}{\partial t}( \underset{j}{\Sigma} {e^i}_j x^j ) \right)^2 }$
Assuming the parameterization is independent of the change-of-basis:
$\frac{\partial s}{\partial t} = \sqrt{ \underset{i}{\Sigma} \left( \underset{j}{\Sigma} {e^i}_j \frac{\partial x^j}{\partial t} \right)^2 }$
$\frac{\partial s}{\partial t} = \sqrt{ \underset{i}{\Sigma} \left( \frac{\partial x^i}{\partial t} \underset{k}{\Sigma} \left( {e^k}_i \underset{j}{\Sigma} \left( {e^k}_j \frac{\partial x^j}{\partial t} \right) \right) \right) }$
$\frac{\partial s}{\partial t} = \sqrt{ \underset{i,j,k}{\Sigma} \left( {e^k}_i {e^k}_j \frac{\partial x^i}{\partial t} \frac{\partial x^j}{\partial t} \right) }$
Let $g_{ij} = \underset{k}{\Sigma} {e^k}_i {e^k}_j$:
$\frac{\partial s}{\partial t} = \sqrt{ \underset{i,j}{\Sigma} \left( g_{ij} \frac{\partial x^i}{\partial t} \frac{\partial x^j}{\partial t} \right) }$

As matrices:
$\frac{\partial s}{\partial t} = \sqrt{ | \vec{r} |^2 }$
$\frac{\partial s}{\partial t} = \sqrt{ \textbf{r}^T \textbf{r} }$
$\frac{\partial s}{\partial t} = \sqrt{ (\textbf{E} \textbf{x})^T \textbf{E} \textbf{x} }$
$\frac{\partial s}{\partial t} = \sqrt{ \textbf{x}^T \textbf{E}^T \textbf{E} \textbf{x} }$
Let $\textbf{G} = (g_{ij})$ be defined as $\textbf{G} = \textbf{E}^T \textbf{E}$, so $g_{ij} = \vec{e}_i \cdot \vec{e}_j$, for $\vec{e}_i$ the i'th column vector of $\textbf{E}$.
Now we are looking at a norm weighted by the metric:
$\frac{\partial s}{\partial t} = \sqrt{ \textbf{x}^T \textbf{G} \textbf{x} }$

A concise way of writing this is:
$ds = \sqrt{ \underset{i,j}{\Sigma} \left( g_{ij} \frac{\partial x^i}{\partial t} \frac{\partial x^j}{\partial t} \right) } dt$
$ds^2 = g_{ij} \frac{\partial x^i}{\partial t} \frac{\partial x^j}{\partial t} dt^2$
$ds^2 = g_{ij} dx^i dx^j$


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