Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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 ]( I picked up a lot of my useful knowledge from this book. – J. M. Oct 25 '10 at 15:30
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 $\frac1r \frac{dr}{ds}=\frac{1}{\tan b}$.

share|cite|improve this answer
@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 $\tan\alpha \approx (r d\theta)/dr$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.