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Where does this come from?

  1. Mar 26, 2005 #1
    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. jcsd
  3. Mar 26, 2005 #2
    The easiest way is to use polar coordinates:

    [tex]x = r \cos \theta[/tex]
    [tex]y = r \sin \theta[/tex]

    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 [tex]\Delta \theta_1[/tex] that arises from [tex]\Delta x[/tex], and then [tex]\Delta \theta_2[/tex] that arises from [tex]\Delta y[/tex]. The add them to get [tex]\Delta \theta[/tex]. Note that a positive delta x decreases the angle ([tex]\Delta \theta_1[/tex] is negative). But in the diagram delta x looks (slightly?) negative.
    Last edited: Mar 26, 2005
  4. Mar 26, 2005 #3
    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?
  5. Mar 26, 2005 #4
    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 [tex]r \Delta \theta[/tex] (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 [tex]r \Delta \theta[/tex] is equal in magnitude to [tex]\Delta x \ \sin \theta[/tex]. Looking at the picture, you can see that the sign of delta theta is negative, so we get

    [tex]\Delta \theta = -\frac{\Delta x \sin \theta}{r} = -\frac{y \Delta x}{r^2}[/tex]

    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
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