# Arc Length in Polar Coordinates

1. May 29, 2010

### 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.

2. May 29, 2010

### 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.

3. May 29, 2010

### 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 $$d\theta$$

4. May 29, 2010

### planck42

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

5. May 29, 2010

### 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.

6. Jun 7, 2010

### 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 $$1+\frac{{\pi}^{2}}{4}$$?

7. Jun 8, 2010

### 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.

8. Jun 10, 2010

### Gib Z

http://img716.imageshack.us/img716/3151/arcy.jpg [Broken]

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 $r d\theta$.

Last edited by a moderator: May 4, 2017