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Taking a derivative?

  1. Sep 7, 2009 #1
    1. The problem statement, all variables and given/known data

    I have a system that is defined by the equation:

    u = Av-2exp(s/R)

    I'm looking for the final temperature of the system knowing that as the system changes, pressure will be halved and (s/R) will be a constant.

    The equation which relates energy, temperature, and pressure is du = Tds - Pdv

    How do I take the derivative of Av-2exp(s/R) to get it in the form Tds - Pdv?

    2. Relevant equations

    [tex]\partial[/tex]u[tex]\partial[/tex]s=T
    [tex]\partial[/tex]u[tex]\partial[/tex]v=-P
    du = [tex]\partial[/tex]u[tex]\partial[/tex]s ds + [tex]\partial[/tex]u[tex]\partial[/tex]v dv

    3. The attempt at a solution

    Do I take the partial derivative of the whole thing for v, then add another term that is the partial derivative of the whole thing for s?

    -2Av-3exp(s/R) + partial derivative for exp(s/R) with respect to s = du?

    And, when you're taking the derivative of exp(s/R), does it come out something like: 1/Rexp(s/R)?
     
  2. jcsd
  3. Sep 7, 2009 #2

    Mark44

    Staff: Mentor

    I think what you're looking for is what's called the total differential, du.

    In your problem,
    [tex]du = \frac{\partial u}{\partial s}~ds~+~\frac{\partial u}{\partial v}~dv [/tex]

    Starting with u = Av-2es/R, take each partial derivative and substitute it into the preceding formula.

    It depends on which partial you're taking. If you take the partial of u w.r.t s, yes, that's right. If you're taking the partial of u wrt v, though, it doesn't. That partial is 0, since es/R is a constant as far as v is concerned.
     
  4. Sep 7, 2009 #3
    Thank you.
     
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