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Simple calculus - interpretation Euler-Lagrange equation

  1. Sep 2, 2006 #1
    This is not a homework question but one that is part of the course material and I can't really move on until I understand the basic calculus.

    I have a problem interpreting "d by dx of partial dF by dy' equals partial d by dy' of dF by dx" in the following question, which I set out and then highlight my problem.

    For F = (x^2 + y'^2)^1/2, find;

    partial dF/dx = x(x^2 + y'^2)^-1/2

    partial dF/dy = 0

    partial dF/dy' = y'(x^2 + y'^2)^-1/2

    dF/dx = p.dF/dx + p.dF/dy.y' + p.dF/dy'.y''

    Show that: d/dx(p.dF/dy') = p.d/dy'(dF/dx)

    Taking the RHS I believe is:

    p.d/dy'(dF/dx) = p.d/dy'[x(x^2 + y'^2)^-1/2 + p.dF/dy.y' +

    y'(x^2 + y'^2)^-1/2 y'']

    ditto = - xy'(x^2 + y'^2)^3/2 + p.d/dy' pdF/dy.y'

    - y'^2 y''(x^2 + y'^2)^3/2

    My problem is that I have forgotten how to interprete p.d/dy' pdF/dy.y' I know that the answer is y''(x^2 + y'^2)^-1/2 but I don't understand how one gets this answer when partial dF/dy = 0

    and I don't know how to interpret the LHS - d/dx(partial dF/dy') at all although I appreciate that it gives the same answer for this function as the RHS.
    I would appreciate some help on the fundamentals. Many thanks.
  2. jcsd
  3. Sep 2, 2006 #2


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    Homework Helper

    In this case [itex]\partial F/\partial y =0 [/itex], so that term drops out. In general you would write out that term like the others and do just what you did with them.

    The following:

    [tex]\frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right)[/tex]

    is computed the same way as [itex]dF/dx[/itex]. Remember that [itex]\partial F/\partial y' [/itex] is a function you know, just like F. Just write it out like your fifth line above.
  4. Sep 7, 2006 #3
    Thank you StatusX. I've got it at last.
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