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Question about changing line element.

  1. Sep 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Say we have a function such that
    [itex] x = uv , y = (u^2 - v^2) /2 [/itex]
    Hence our line element in Cartesian coordinates is.
    [itex] ds^2 = dx^2 + dy^2 [/itex]

    Now I have two questions. I like to work on math problems algebraically if possible so I thought to convert our line element I could take the implicit derivative of both x and y.

    Hence we have

    [itex] dx = (du*v) + (u*dv), dy = (2u du - 2v dv) [/itex]
    However if i were to square these terms out I would then have some mixed components such as [itex] 4vu du dv[/itex]

    I tested this technique with polar coordinates and using [itex] x = r \cos \theta, y = r \sin\theta [/itex] and I managed to come out to the proper line element.

    Can someone please point me in the right direction as to what I need to do to get the proper line element?

    Also I had one more question. If i wanted to find the equation of the unit circle centered at the origin in our new coordinate system can I just take the cartesian equation [itex] x^2 + y^2 = 1 [/itex] , plug in our substitutions, and then solve?
     
    Last edited by a moderator: Sep 9, 2012
  2. jcsd
  3. Sep 9, 2012 #2

    vela

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    You forgot the factor of 1/2 when you calculated dy.
     
  4. Sep 9, 2012 #3
    thats true I forgot to write that down when I was making the post.. so this is the correct way to set up the line element?
     
  5. Sep 9, 2012 #4

    vela

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    Yup, and you should find with the correction the cross-term vanishes.

    And to answer your other question, yes, you can just substitute for x and y into the equation for a circle and get the equivalent equation in terms of u and v. It works the way you think it should.
     
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