Where does this come from?

1. Mar 26, 2005

Moose352

For this diagram: http://ananth.ath.cx/coag.jpg [Broken]

The textbook gives the formula for dtheta as (x*dy-y*dx) / (x^2 + y^2) assuming that r' ~ r. How is this formula derived?

Last edited by a moderator: May 2, 2017
2. Mar 26, 2005

PBRMEASAP

The easiest way is to use polar coordinates:

$$x = r \cos \theta$$
$$y = r \sin \theta$$

From these you can express theta as a function of x and y, and then differentiate both sides of the equation. You can also find this formula directly from the diagram by first calculating the $$\Delta \theta_1$$ that arises from $$\Delta x$$, and then $$\Delta \theta_2$$ that arises from $$\Delta y$$. The add them to get $$\Delta \theta$$. Note that a positive delta x decreases the angle ($$\Delta \theta_1$$ is negative). But in the diagram delta x looks (slightly?) negative.

Last edited: Mar 26, 2005
3. Mar 26, 2005

Moose352

Thanks PBRMEASAP!

But I'm a bit confused by the second part. You write that i can simply add theta1 and theta2 together to find theta. How exactly do I find the theta1 and theta2 that arise from dx and dy?

4. Mar 26, 2005

PBRMEASAP

Well, it's hard to describe without a visual aid, but I don't know how to put up a picture that isn't too large a file (I tried to post something I drew in Paint one time and it wouldn't take it).

For the delta x part, draw a picture where r' only differs from r by a small delta x (delta y = 0). Draw the little segment of length delta x that connects the end of r to end of r'. Now draw another small segment from the end of r so that it intersects r' at 90 deg. You now have formed a little right triangle with delta x as the hypotenuse. And the length of that last little segment you drew is $$r \Delta \theta$$ (approximately).

Now here's the trick. Say that r makes an angle theta (not delta theta) with the x-axis. Then so does r', approximately, since it only differs from r by a small x displacement. Then by alternate interior angles, r' makes an angle theta with the little delta x segment (still with me? :-)). Now you can see that the segment of length $$r \Delta \theta$$ is equal in magnitude to $$\Delta x \ \sin \theta$$. Looking at the picture, you can see that the sign of delta theta is negative, so we get

$$\Delta \theta = -\frac{\Delta x \sin \theta}{r} = -\frac{y \Delta x}{r^2}$$

The contribution from delta y is found the same way. Were you able to get the answer from polar coordinates?

Last edited: Mar 26, 2005