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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.

Antoni Parellada
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2 Answers2

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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.

  • Thank you! Can you include the steps from $s$ to the final parameterization? The $\log(\cdot)$ indicates that you are using the inverse of $s$ to replace $t$ in the first set of equations, and since there is a $e^{\log(\cdot)}$ in the $a^{at}$ we end up with just the inverse ($e$ "cancels" $\log)$... Also, you are integrating from $-\infty$ whereas on my post it is from $0.$ In addition, you have left the tangent implied, but - just out of curiosity - considering all there is no contradiction between posts, is there? – Antoni Parellada Feb 24 '20 at 21:54
  • @Mathstunned: integration from $-\infty$ is for convenience, it simplifies the equations but is harmless. What do you mean by "tangent implied" ? –  Feb 24 '20 at 21:59
  • I mean that your final equations parametrize the curve by arc length, which is just perfect. Yet, the final step would be to get the derivatives of $x$ and $y$ to generate the tangent vector to the curve at each point, which - by construction - would have norm $1.$ I am planning on accepting your answer, but I would want to make sure - for my own curiosity - that there is no contradiction with what I wrote... – Antoni Parellada Feb 24 '20 at 22:03
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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:

enter image description here

Antoni Parellada
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