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Math Help - Polar coordinate simple problem

  1. #1
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    Polar coordinate simple problem

    I am trying to convert dxdy in cartesian into polar plane.
    and have to prove: dxdy = rd \thetadr

    Here is my conversion proof:

    x = rcos \theta
    y = rsin \theta

    dx = -rsin \thetad \theta + cos \thetadr
    dy = rcos \thetad \theta + sin \thetadr

    ignoring 2nd order differentials, and keeping only drd \theta,
    we have:
    dxdy = -rsin \thetad \thetasin \thetadr
    + rcos \thetadrcos \thetad \theta
    = rd \thetadr(-sin ^2 \theta + cos ^2 \theta)

    I need to equate (-sin ^2 \theta + cos ^2 \theta) = 1,
    but i can't because of the -ve sign. Where did I go wrong?

    Thanks for your time.
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  2. #2
    Senior Member Spec's Avatar
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    The Jacobian is r, so dxdy=r\cdot drd\theta
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  3. #3
    Senior Member Pinkk's Avatar
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    Here's a geometric/visual proof:



    Consider the graph of a polar rectangle in the cartesian plane with area dxdy (a polar rectangle has height in radius and length in angle). Consider the polar rectangle as subtracting a "sliver" of a smaller circle (let's say with radius r_{2} from a "sliver" of a larger circle (radius r_{2}). Both have the same angle width of \theta_{2}-\theta_{1}; subtracting the smaller angle from the larger angle. So the area of the polar rectangle is dxdy=\frac{1}{2}(\theta_{2}-\theta_{1})r_{2}^{2}-\frac{1}{2}(\theta_{2}-\theta_{1})r_{1}^{2} Well, \theta_{2}-\theta_{1}=d\theta; it's the change in \theta. So let's rewrite this as:

    dxdy=\frac{1}{2}d\theta(r_{2}^{2}-r_{1}^{2})=\frac{1}{2}d\theta(r_{2}+r_{1})(r_{2}-r_{1})

    Again, we can make substitutions. \frac{1}{2}(r_{2}+r_{1}) can really be thought of as the average of radii, and as this polar rectangle gets infinitely smaller and smaller, the average will simply equal r. And as with d\theta, r_{2}-r_{1}=dr, or the change in r. Making all the substitutions, we finally arrive at:

    dxdy=rdrd\theta
    Last edited by Pinkk; May 16th 2009 at 10:01 AM.
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  4. #4
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    Hi Pinkk, that was pretty neat. Thanks for your rather rigorous geometrical proof. But I am still interested in my analytic approach, and I need to know how to get rid of the -ve sign.

    And I do not want to use the Jacobi method for the time being.

    Thanks again.
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  5. #5
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    Quote Originally Posted by chopet View Post
    I am trying to convert dxdy in cartesian into polar plane.
    and have to prove: dxdy = rd \thetadr

    Here is my conversion proof:

    x = rcos \theta
    y = rsin \theta

    dx = -rsin \thetad \theta + cos \thetadr
    dy = rcos \thetad \theta + sin \thetadr

    ignoring 2nd order differentials, and keeping only drd \theta,
    we have:
    dxdy = -rsin \thetad \thetasin \thetadr
    + rcos \thetadrcos \thetad \theta
    = rd \thetadr(-sin ^2 \theta + cos ^2 \theta)

    I need to equate (-sin ^2 \theta + cos ^2 \theta) = 1,
    but i can't because of the -ve sign. Where did I go wrong?

    Thanks for your time.

    Hello Chopet!

    What you are effectively trying to do is derive the jacobian from first principles.

    The reason your approach is going wrong is that you are forgetting that the \mathrm{d}x's, \mathrm{d}y's, \mathrm{d}r's and \mathrm{d}\theta's all have associated directions i.e. you should be treating them as vectors.

    Now you should know that the area of a parallelogram formed by two vectors is given by the magnitude of the cross product of those two vectors. So:

    <br />
\mathrm{d}x \, \mathrm{d}y = |\mathbf{dx} \times \mathbf{dy}|
    <br />
= |(-r\sin \theta \, \boldsymbol{\mathrm{d}\theta} + \cos \theta \mathbf{dr}) \times (r\cos \theta \, \boldsymbol{\mathrm{d}\theta} + \sin \theta \mathbf{dr})|<br />
    = |-r \sin^2 \theta \, \boldsymbol{\mathrm{d} \theta} \times \mathbf{dr} <br />
+r \cos^2 \theta \, \mathbf{dr} \times \boldsymbol{\mathrm{d}\theta}|
    = | r (\sin^2 \theta +\cos^2 \theta)  \, \mathbf{dr} \times \boldsymbol{\mathrm{d}\theta}|
    = r \, \mathrm{d}r \,\mathrm{d} \theta

    Now you'll notice about the above that the terms of order \mathrm{d}r^2 and \mathrm{d} \theta^2 drop out because their corresponding vector cross products are zero rather than because you arbitrarily decide they're smaller.

    Finally you should also note that had you written your two original equations in matrix form then taking the determinant of that matrix would be equivalent to the vector cross product that I've just shown and infact is just our old friend the Jacobian!

    If you need any more explanation please don't hesitate to ask.
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