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I am trying to understand a simple calculation in polar coordinates - and I am getting totally discombobulated (the original source can be found: here).

Please have a look at the following picture:

alt text

My questions

(1) I don't understand how the length labeled $dr \over d \theta$ can possibly be the change of $r$ with respect to a change in $\theta$.
(2) It is then said that $tan \alpha = {r \over {dr \over d \theta}}$. But how can you calculate the tan-function if you don't even have a right angled triangle - and anyway, the formular would suggest that you calculate the tan of the angle that is not labeled in the pic (because $tan={opposite \over adjecent}$)

Sorry, if this is too elementary - but it really bothers me right now... Thank you!

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If you're still trying to get intuition for the properties of the equiangular spiral, may I suggest taking a look at E.H. Lockwood's [ A Book of Curves ](amazon.com/dp/0521044448)? I picked up a lot of my useful knowledge from this book. –  J. M. Oct 25 '10 at 15:30

2 Answers 2

up vote 3 down vote accepted

Let me re-draw the picture

enter image description here

Instead of $\theta$ I wrote $s$. We are looking at a sweep from $r\to r′$ over an angle of $ds$. The arc formed by sweeping $r$ with no radial change has length $r~ds$. The change of radius is $r' - r = dr$. As you can see, when $ds$ is really small, the region formed by $dr$, the arc $r~ds$, the the un-labeled segment connecting the tip of $r$ to that of $r′$ approximates a right triangle

For the angle $b$ (which is your $\alpha$), the opposite side is the arc, of length $r~ds$, and the adjacent side is the segment $dr$. So you have $\frac1r \frac{dr}{ds}=\frac{1}{\tan b}$.

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@Willie: Is the formula in your second paragraph $\frac{r}{\frac{dr}{d\theta}}$? –  Jack Jul 18 '11 at 2:56
    
@Jack: Now that I have the rep to post images, I re-posted the image and re-wrote the paragraphs in terms of the image. This I hope looks better now. –  Willie Wong Jul 18 '11 at 10:19

(Willie beat me by a few seconds, but this is what I had written.)

No wonder you're confused. That's a horrible picture! You should draw an arc of a circle ($\approx$ a line) at right angles to the side labeled $r_0$; this will have length $r d\theta$ and split the line labeled $r$ into two pieces of lenghts $r_0$ and $dr$ (approximately, for small $d\theta$). Then you have a small right triangle such that $\tan\alpha \approx (r d\theta)/dr$.

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