Arc Length in Polar Coordinates

[planck42]

In the polar formula for arc length, ds^{2}=dr^{2}+r^{2}d{\theta}^{2}, what is the exact meaning of the r^2 term multiplying d{\theta}^2? Is it an initial distance from the origin? A final distance from the origin? The change in r from point a to point b? This baffles me to no end and nothing explains it.

[Jerbearrrrrr]

God I hate this stuff. But after you use it, you kind of realize how much effort it saves.

This is the meaning of that rather "disembodied" statement.

Let's draw a curve through space. And now let's parametrize that curve with a parameter t. So along this path, r=\tilda r(t), \ \ \theta = \tilda \theta (t).
Then the arclength satisfies the following 'differential' equation:

(\frac{ds}{dt})^2 = (\frac{d\ r (t)}{dt})^2 + r^2 (\frac{d\ \theta (t)}{dt})^2

I've included the (t) thing to make the method of calculation explicit. You literally differentiate the function r(t) wrt t. r=r(t) is a horrible abuse of notation that actually gets me confused from time to time but...it saves a lot of time too. Haha.

[l'Hôpital]

Do you mean a geometric reason, or an analytic reason?

To see how it got there, just consider the following:


dS = \sqrt{1 + (\frac{dy}{dx})^2}


Let y = r \sin \theta and x = r \cos \theta and work it out and you'll get your r^2 factor on the d\theta

[planck42]

I mean a geometric reason; the derivation is no problem.

[Jerbearrrrrr]

It gives the arclength gained by increasing r by dr and theta by dtheta.


(\frac{ds}{dt})^2 = (\frac{d\ r (t)}{dt})^2 + r^2 (\frac{d\ \theta (t)}{dt})^2

So yes, the r^2 corresponds to how far you are from the origin.
Think about small curves on a sphere. If your r is large, then your ds is going to be larger, for a given dtheta.

[planck42]

Let's take an example: suppose I am moving from the polar point (2, \frac{\pi}{4}) to (3, \frac{\pi}{2}). Would my distance traveled be 1+\frac{{\pi}^{2}}{4}?

[elibj123]

No, the relation you stated in the beginning is a differential relation, meaning it is good only for really close points. (It's like a taylor expansion)

Otherwise you need to develope a formula using trigonometry or using the definition of distance via cartesian coordinates then substituting with polar coordinates.

Back to your original question: dtheta denotes a change in the angle. But as you may know, the angle between two rays doesn't depend on their length. So the information of the angle itself doesn't give you a measure of distance. At small distances from the origin, taking a small angle difference will give you a small distance (the arc of a small circle). At large distances, taking a small angle step will give you a larger distance.
Therefore the r dependence comes in.

[Gib Z]

http://img716.imageshack.us/img716/3151/arcy.jpg (http://img716.imageshack.us/i/arcy.jpg/)

Differentiable curves exhibit local linearity, so if we zoom up to infinitesimal scales the curve is approximately a straight line. We wish to find the infinitesimal increments of the arc length of the curve between polar coordinates (r,\theta) and (r+dr, \theta + d\theta).

By the Pythagorean theorem, ds^2 = dr^2 + (ab)^2, but the length of the line segment ab can be approximated by the length of the arc that passes through a and b of the circle centered at the origin of radius r. The length of this arc is given by r d\theta.