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Partial differentiation question. Would all three methods work?

  1. Feb 12, 2012 #1

    s3a

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    I'm attaching the question and solution.

    I'm talking about the first part since the second part is the same just with different variables and stuff. I get what the solution is saying but:

    1) What if I computed a Jacobian, with
    F = x^2 + xy + y^2 - z = 0
    G = 2r + s - x = 0
    H = r - 2s - y = 0
    would it work? Whether it works or not, why does it work or not?

    2) Why am I getting it wrong when I substitute x and y with the r and s equivalents and then attempt taking ∂z/∂r? Am I just making a mistake whenever I compute it or is this attempt just bound to fail for some theoretical reason?

    I'm not necessarily concerned about the detailed algebra for (1) and (2) but rather just if the methods should work in theory and why or why they would or wouldn't work and if I was doing something wrong (sorry, the file which had my work was corrupted but I'm not concerned about the detailed algebra anyway apart from the part where you set things up prior to the tedious algebra/computation). Part of the reason why I am asking this is because, instinctively, I would have done method (2) and, for method (1), I'm just trying to see when Jacobians are better or worse to use. So, if Jacobians work for this problem, I'd be great if I can also be told how to recognize when Jacobians are the better or worse to use. Sorry, if I repeated myself a bit but I wanted to make my concerns very clear.

    Any input would be greatly appreciated!
    Thanks in advance!
     

    Attached Files:

    • P16.jpg
      P16.jpg
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  2. jcsd
  3. Feb 12, 2012 #2

    SammyS

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    attachment.php?attachmentid=43810&d=1329059321.jpg

    For your Question #1:
    How do you propose using that Jacobian to find ∂z/∂r and ∂z/∂s ?​

    For your Question #2:
    Yes, that should work. Without you showing your work, it's hard for us to tell what the problem is.​
     
  4. Feb 12, 2012 #3

    s3a

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    For (2), I think the problem was that I was tired and sick ;). As for (1), is what I'm uploading now correct?
     

    Attached Files:

  5. Feb 12, 2012 #4

    SammyS

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    I'll look over #1 when I get a chance. Hopefully that won't be a whole lot later or perhaps someone else will come along & look it over.

    attachment.php?attachmentid=43824&d=1329085725.jpg

    For #2, you can even show more generally that both methods give the same result.
    You have that [itex]\displaystyle \frac{\partial z}{\partial r}=14r-3s\,.[/itex]

    The attachment you had in your Original Post, has:

    [itex]\displaystyle \frac{\partial z}{\partial r}=5x+4y[/itex]

    Substituting for x & y gives:

    [itex]\displaystyle \frac{\partial z}{\partial r}=5(2s+r)+4(r-2s)[/itex]
    [itex]\displaystyle =14r-3s\,.[/itex]
     
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