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Differential, why three members

  1. Jun 28, 2012 #1


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    I have this thermodynamical expression:
    [itex]dS=\sigma T^3 dV+4V\sigma T^2 dT+\frac{1}{3}\sigma T^3 dV=d(\frac{4}{3}\sigma T^3 V)[/itex]
    Basically saying:
    [itex]\frac{4}{3}\sigma T^3 V=S[/itex]

    Now, I do not get this.. d(expr) part, why are there three members to d(expr), with 2x dV and 1x dT.. nope.. :confused:
    I might add that [itex]\sigma[/itex] is a constant.
    Last edited: Jun 28, 2012
  2. jcsd
  3. Jun 28, 2012 #2


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    Where did you find that? With [itex]\sigma[/itex] constant, there should be two parts:
    [tex]d((4/3)\sigma T^3V)= 4\sigma T^2V dT+ (4/3)\sigma T^3 dV[/tex]

    Oh, wait, what they have done is just separate that last term:
    [tex](4/3)\sigma T^3dV= (1+ 1/3)\sigma T^3dV= \sigma T^3dV+ (1/3)\sigma T^3dV[/tex]
  4. Jun 29, 2012 #3

    I like Serena

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    Hi Uku! :smile:

    Can you clarify what you do not get?

    What you have is similar to:
    $$df(x,y)={\partial f \over \partial x}dx + {\partial f \over \partial y}dy$$
    This is how the derivative of a multi variable function is taken.
  5. Jun 29, 2012 #4


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    HallsofIvy nailed it, thanks!

  6. Jun 29, 2012 #5


    Staff: Mentor

    Fixed that for you:smile:
  7. Jun 29, 2012 #6

    I like Serena

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    Thanks. :wink:
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