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Showing that a partial derivative equation holds

  1. Mar 1, 2012 #1

    s3a

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    1. The problem statement, all variables and given/known data
    The question is attached as Question.jpg.


    2. Relevant equations
    Partial differentiation.


    3. The attempt at a solution
    This seems obvious to me but I don't know how to express myself mathematically. Basically, what I'd do is:

    [∂(u,v)/∂(x,y)] [∂(x,y)/∂(r,s)] = [∂(u,v)/∂(r,s)] [∂(x,y)/∂(x,y)] = ∂(u,v)/∂(r,s)

    as for ∂(u,v)/∂(x,y) = 1/[∂(u,v)/∂(x,y)], that's especially obvious to me! It's like saying 1/(1/2) = 1 * 2/1 = 2! Whenever I get an “It's obvious” feeling, that usually tells me that I'm not grasping the assumptions that are made when using these mathematical tools.

    Any assistance in solving this problem would be greatly appreciated!

    Edit: Forgot to attach the question! :P
     

    Attached Files:

    Last edited: Mar 1, 2012
  2. jcsd
  3. Mar 1, 2012 #2

    LCKurtz

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    Remember that a Jacobian is the determinant of a matrix. Write out the matrix for [∂(u,v)/∂(x,y)] and [∂(x,y)/∂(r,s)]. What do you get if you multiply those matrices? And see if you can use the hint that |AB|=|A||B| for the determinant of a product of square matrices.
     
  4. Mar 1, 2012 #3

    s3a

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    I attached the Jacobian stuff you asked for.

    For the hint, A and B are matrices, right? What do the || mean in this case?
     

    Attached Files:

  5. Mar 2, 2012 #4

    LCKurtz

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    |A| means the determinant of A. And in your last attachment, multiply the matrices themselves before calculating any determinants. I think you will find you don't have to actually evaluate any determinants to work the problem.
     
    Last edited: Mar 2, 2012
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