Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Where does this come from?

  1. Mar 26, 2005 #1
    For this diagram: http://ananth.ath.cx/coag.jpg

    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?
     
  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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Where does this come from?
  1. Where did pi come from? (Replies: 10)

Loading...