Frenet-Serret formula: why is $T$'s magnitude unitary?

Question

Why is $T$'s tangent vector magnitude unitary?

$$T=\frac{dr}{ds}$$

Answer 1

I take it that our OP Francesco Venuti is asking for a proof that

$\left \Vert \dfrac{d {\mathbf r}(s)}{ds} \right \Vert = 1. \tag 0$

Let $\mathbf r(t)$ be a regular curve, that is

$\dot{\mathbf r}(t) \ne 0 \tag 1$

for every $t$ in the domain of $\mathbf r$; as is well-known the arc-length $s$ along $\mathbf r(t)$ satisfies

$\dfrac{ds}{dt} = \vert \dot{\mathbf r}(t) \vert; \tag 2$

we may arbitrarily assign the $s$-value $s_0$ to the point $\mathbf r(t_0)$ on the curve $\mathbf r(t)$; then in light of (2) we have

$s(t) - s(t_0) = \displaystyle \int_{t_0}^t \dfrac{ds}{dt} \; dt = \int_{t_0}^t \vert \dot{\mathbf r}(t) \vert \; dt, \tag 3$

whence

$s(t) = s(t_0) + \displaystyle \int_{t_0}^t \vert \dot{\mathbf r}(t) \vert dt; \tag 4$

this defines $s$ as the function $s(t)$ of $t$; we may thus re-parametrize $\mathbf r(t)$ by $s$, and so obtain

$\mathbf r(s(t)) = \mathbf r(t); \tag 5$

of course, $\dot{\mathbf r}(t)$ is tangent to the curve $\mathbf r(t)$; thus the unit tangent vector is given by

$\dfrac{\dot{\mathbf r}(t)}{\vert \dot{\mathbf r}(t) \vert} = \dfrac{\dot{\mathbf r}(t)}{\dfrac{ds}{dt}} = \dot{\mathbf r}(t)\dfrac{dt}{ds}; \tag 6$

now an application of the chain rule to the expression on the right yields

$T(s) = \dfrac{d {\mathbf r}(s)}{ds} = \dot{\mathbf r}(t) \dfrac{dt}{ds}, \tag 7$

from which it follows that

$\Vert T(t) \Vert = \left \Vert \dfrac{d {\mathbf r}(s)}{ds} \right \Vert = \left \Vert \dfrac{\dot{\mathbf r}(t)}{\vert \dot{\mathbf r}(t) \vert} \right \Vert = 1. \tag 8$