Can I have an example of arc length parameterization of a curve?

Question

In the construction of the Frenet-Serret triad along a curve in $\mathbb R^3$ arc length parameterization of the tangent vector ensures orthogonal orientation of the curvature vector by making the norm of the tangent vector unitary. This is explained beautifully at this point of this lecture by Claudio Arezzo.

The example of a logarithmic exponential is elaborated on; however, the "trivial" ending is left out as mechanical - and it is. However, I wanted to ask for its completion, and then I figured I might as well ask myself. So I'm posting this as a Q&A for future reference, and in case it helps someone else.

Answer 1

With $$\begin{cases}x=e^{at}\cos t,\\y=e^{at}\sin t\end{cases}$$ the arc length is $$s=\int_{-\infty}^t e^{at}\sqrt{(a\cos t-\sin t)^2+(a\sin t+\cos t)^2}dt=\frac{\sqrt{a^2+1}\,e^{at}}{a}.$$

So,

$$\begin{cases}x=\dfrac{as}{\sqrt{a^2+1}}\cos\dfrac{\log\dfrac{as}{\sqrt{a^2+1}}}a,\\y=\dfrac{as}{\sqrt{a^2+1}}\sin\dfrac{\log\dfrac{as}{\sqrt{a^2+1}}}a\end{cases}$$

is the arc length parameterization.

Answer 2

The parameterized spiral used as an example is

$$ \begin{align} \alpha(t) &= \left (e^{-0.2t} \cos(t),e^{-0.2t}\sin(t)\right) \end{align} $$

and the function $s(t)$

$$\begin{align}I \to \mathbb R\\ s(t) =\int_{t_o}^t \vert \alpha'(u)\vert du \end{align}$$

corresponds to the length of the curve from $t_o$ to $t,$ which in the example under consideration is

$$ \begin{align}s(t) &=\int_0^t \vert \left (e^{-0.2t} \cos(t),e^{-0.2t}\sin(t)\right) \vert dt\\&=\small \int_0^t \sqrt{ (-0.2)^2e^{-0.2t} \cos^2(t)+e^{-0.2t} \sin^2(t)+(-0.2)^2e^{-0.2t}\sin^2(t)+e^{-0.2t} \cos^2(t)} dt\\ &=\frac{e^{-0.2t}-1}{-0.2}\sqrt{(-0.2)^2+1} \end{align}$$

Its inverse is

$$\varphi(g)=s^{-1}(g)=-5 \log(-0.196116 (g - 5.09902))$$

And the tangent to the curve parameterized by the arc length, i.e. $\beta(g)=\alpha\circ\varphi=\alpha(\varphi(g))$:

$$\begin{align} &\\\beta'(g)&=\alpha'\left( \varphi(g)\right)\varphi'(g)\\ &=\small{ (0.98058\sin(5\ln(-0.196116 (x - 5.09902))) - 0.196116\cos(5\ln(-0.196116 (g - 5.09902))),} \\ &\small{0.196116\sin(5\ln(-0.196116 (g - 5.09902))) + 0.98058\cos(5\ln(-0.196116 (g - 5.09902))))} \end{align}$$

yielding unitary tangent vectors to each point on the curve: