I want to calculate the arc-length of the Log spiral given by the equation:

$\vec{x}(t)=\begin{pmatrix} e^{-\alpha t} \cos t \\ e^{-\alpha t} \sin t\end{pmatrix}$

$\alpha >0$ and $t \in [0,T]$

Our lecturer gave us the following formula to work with:


However, in this case I don't really have a traditional function to work with. How does this work with vectors?

I am also not sure how to get to the formula. All I know is that:

1) I have some path/image of a function on the intervall [a,b] and I want to find the length of it

2) I can apporximate the length of the path by using small polygonal chains.

$\implies L(x)=\sum ||x(t_a)-x(t_{a-1})||$

But how do I go from that to the integral formula?

  • $\begingroup$ I would try converting into polar coordinates with $r=e^{-a\theta}$. $\endgroup$ – Archaick Apr 18 '15 at 9:16
  • $\begingroup$ Use $$L(t_1,t_2)=\int_{t_1}^{t_2}\sqrt{x_1'(t)^2+x_2'(t)^2}dt$$ $\endgroup$ – String Apr 18 '15 at 9:17
  • $\begingroup$ @String. It is funny to see how simple the problem becomes ! $\endgroup$ – Claude Leibovici Apr 18 '15 at 9:18
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    $\begingroup$ @String. This is why this problem was set, I bet. $\endgroup$ – Claude Leibovici Apr 18 '15 at 9:20
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    $\begingroup$ There should be no $f(x)$ in the "lecturer's formula". $\endgroup$ – Yves Daoust Apr 18 '15 at 9:21

it should be easier to do in polar coordinates. the curve $$e^{-at}(\cos t, \sin t) \to r = e^{-a\theta}$$ the arc length formula is $$ ds = \sqrt{(r d \theta)^2 + (dr)^2} =\sqrt{a^2e^{-2a\theta}+e^{-2a\theta}}\, d\theta$$ therefore the arc length is $$\int_{t_1}^{t_2} e^{-a\theta}\sqrt{a^2+1}\, d\theta = \frac1a\sqrt{a^2 + 1}\left(e^{-at_1} - e^{-at_2}\right)$$

  • $\begingroup$ No, there is a wrong term. The arc length is $Ae^{-a\theta}$. $\endgroup$ – Yves Daoust Apr 18 '15 at 9:40

the characteristic feature of a logarithmic (or 'equi-angular') spiral is central self-similarity, expressed geometrically as a proportionality between the two elements of length in polar cordinates. we may write this: $$ dr = \alpha r d\theta \tag{1} $$ which gives, for the element of length $dl$: $$ dl = \left(dr^2 + (rd\theta)^2) \right)^{\frac12}=\beta dr $$ where $\beta = \sqrt{1+\alpha^{-2}}$

this integrates to: $$ L = \beta(r_2 - r_1) $$ which may be evaluated using (1) to obtain $$ r + const = e^{\alpha \theta} $$


Alright, so I tried to calculate it using what you guys have shown me. Here is what I got:

I want to find the lenth of the archimedean spiral for the parameter $\alpha>0$ and $t \in [0,T]$

Arclength: $s(0,T)=\int_0^T||\space \dot{x}(t)\space ||\space dt=\int_0^T\sqrt{\dot{x}_1^2+\dot{x}_2^2}\space dt$

Equation of my spiral: $\vec{x}=\begin{pmatrix}x_1 \\ x_2\end{pmatrix}=\begin{pmatrix}e^{-\alpha t} \cos t \\ e^{- \alpha t} \sin t\end{pmatrix}$

$\dot{x}_1=-\alpha e^{-\alpha t} \cos t-e^{-\alpha t} \sin t$

$\implies \dot{x}_1^2=\alpha^2e^{-2 \alpha t}\cos^2 t+2\alpha e^{-2\alpha t}\cos t \sin t+e^{-2 \alpha t} \sin^2 t$

$\dot{x}_2=-\alpha e^{-\alpha t} \sin t+e^{-\alpha t} \cos t$

$\implies \dot{x}_2^2=\alpha^2e^{-2 \alpha t}\sin^2 t-2\alpha e^{-2\alpha t}\cos t \sin t+e^{-2 \alpha t} \cos^2 t$


$\alpha^2e^{-2 \alpha t}\cos^2 t+2\alpha e^{-2\alpha t}\cos t \sin t+e^{-2 \alpha t} \sin^2 t+\alpha^2e^{-2 \alpha t}\sin^2 t-2\alpha e^{-2\alpha t}\cos t \sin t+e^{-2 \alpha t} \cos^2 t$

$=(\alpha^2+1) e^{-2 \alpha t}(\sin^2 t+\cos^2 t)=(\alpha^2+1)e^{-2 \alpha t}$

My integral becomes:

$\int_0^T \sqrt{(\alpha^2+1)e^{-2 \alpha t}}=\sqrt{(\alpha^2+1)}\int_0^T \sqrt{e^{-2 \alpha t}}=\sqrt{(\alpha^2+1)}\int_0^T e^{-\alpha t}$

Antiderivative of $e^{-\alpha t}=\frac{1}{-\alpha}e^{-\alpha t}$

$\implies \sqrt{\alpha^2+1}\frac{1}{-\alpha}e^{-\alpha T}-\sqrt{\alpha^2+1}\frac{1}{-\alpha}e^{0}=\sqrt{\alpha^2+1}\frac{1}{-\alpha}(e^{-\alpha T}+1)$

I hope this is correct.

Sign correction:

$\implies \sqrt{\alpha^2+1}\frac{1}{-\alpha}e^{-\alpha T}-\sqrt{\alpha^2+1}\frac{1}{-\alpha}e^{0}=\sqrt{\alpha^2+1}\frac{1}{-\alpha}(e^{-\alpha T}-1)$

$=\sqrt{\alpha^2+1}\frac{1}{-\alpha}e^{-\alpha T}-\sqrt{\alpha^2+1}\frac{1}{-\alpha}e^{0}=\frac{\sqrt{\alpha^2+1}}{\alpha}(1-e^{-\alpha T})$

  • $\begingroup$ Compare to Abel's solution and apply a sign correction. $\endgroup$ – Yves Daoust Apr 19 '15 at 8:46
  • $\begingroup$ @YvesDaoust Thanks, I fixed it. Thank you again for your help. $\endgroup$ – qmd Apr 19 '15 at 9:31

The Archimedean spiral is described by $$x=a\theta\cos(\theta),\\y=a\theta\sin(\theta)$$ or $$r=a\theta$$in polar coordinates.

Then $$ds=\sqrt{\dot r^2+r^2}\,d\theta=a\sqrt{1+\theta^2}\,d\theta,$$ and by integration,

$$a\int_{\theta=\theta_0}^{\theta_1}\sqrt{1+\theta^2}\,d\theta=\frac a2\left(\theta\sqrt{1+\theta^2}+\sinh^{-1}(\theta)\right)\Big|_{\theta_0}^{\theta_1}.$$


This would be so much simpler in the complex plane. Let


The arc length is given by

$$s=\int |\dot z(t)| dt$$


$$\dot z=(i-\alpha)e^{(i-\alpha)t}\ \ \ \text{and}\ \ \ |\dot z|= \sqrt{1+\alpha^2}e^{-\alpha t}$$

there follows

$$s=\int_0^T |\dot z(t)| dt=\sqrt{1+\alpha^2}\int_0^T e^{-\alpha t} dt=\frac{\sqrt{1+\alpha^2}}{-\alpha} e^{-\alpha t} \big|_0^T=\frac{\sqrt{1+\alpha^2}}{\alpha}(1-e^{-\alpha T})$$

The reference for this work is: Zwikker, C. (1968). The Advanced Geometry of Plane Curves and Their Applications, Dover Press.


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