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How to find the arc length parametrization of the Archimedean spiral?

I know the curve is defined in the complex nubmer like this: $c(t)=e^{it} - ite^{it}$

and $|\frac{d}{dt}(c(t))|= te^{it}$ (is this correct?)

now I should find $\int te^{it} dt$ (the integration from 0 to t ,,,I am stuck here)

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2 Answers 2

up vote 2 down vote accepted

It is easier to use the polar formula for the Archimedean spiral: $r=a+b\theta$. The arc length in polar coordinates is $ds=\sqrt{dr^2 +r^2 d\theta^2}$ or $\frac {ds}{d\theta}=\sqrt{r^2+\frac {dr^2}{d\theta^2}}=\sqrt{(a+b\theta)^2+b^2}$ and integrate that.

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I don't see the necessity to involve complex plane here. And I think the post above mine was trying to find the total arc length, instead of arc length parameterisation of the path (Archimedean spiral).

You could use polar coordinate or Cartesian coordinate; either way you should find it quite easy to parameterise the path using angle, and you could find the following arc length - angle relationship:

$$ L(\theta)=\frac{P}{2\pi}[\frac{\theta}{2}\sqrt{1+\theta^{2}}+\frac{1}{2}\ln(\theta+\sqrt{1+\theta^{2}})] $$

where P is the pitch.

Now to find out arc length parameterisation we need to find $\theta(L)$, i.e. to invert the above relationship. However, I'm not sure if it's possible and this answer is the closest thing I've found:

How to place objects equidistantly on an Archimedean spiral?

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