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Help with chain rule

  1. Mar 28, 2012 #1
    I'm looking at one step in my thermodynamics book and they go from

    [tex] pV = \nu*R*T [/tex]

    to

    [tex] p*dV + V*dp = \nu * R * dT[/tex]

    I think there's an application of the chain rule in here but I don't see exactly how it's working. Could someone show me the steps in between?

    Thanks!
     
  2. jcsd
  3. Mar 28, 2012 #2
    It is a partial derivative. You are allowing P,V,and T to very.

    So in partial derivatives you can take the derivative with respect to one variable and then the derivative with respect to the next variable and so on. The sum of this being your answer. That is how you get that equation.

    P(n,V,T) Derivative with respect to P= dP*V, the derivative with respect to V= dV*P and derivative with respect to T= n*R*dT.

    So you end up with dP*V+dv*P=n*R*dT.
     
  4. Mar 29, 2012 #3
    In a sense, this can also be thought of as being an application of the product rule. :biggrin:
     
  5. Mar 29, 2012 #4
    thanks! I see it now
     
  6. Mar 29, 2012 #5
    Actually now I'm looking back at this and I'm not sure I follow exactly what you mean. What are you taking the derivative of? Each side independently and then summing them? As in...

    [tex] \frac {\partial} {\partial p} (pv) [/tex]

    and then
    [tex] \frac {\partial} {\partial v} (pv) [/tex]

    etc...and then summing the results, keeping everything on the original side?

    Then why do you get dp * v? I always learned that that would be

    [tex] \frac {\partial} {\partial p} (pv) = v [/tex]

    Where did the dp come from?

    Thanks!
     
  7. Mar 29, 2012 #6

    HallsofIvy

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    There is NO derivative with respect to "p" or "V". If pV= NRT and p, V, and T are functions (N and R are constants, of course) of some independent variable, x (whatever it is- I started to write t for time but this is true for any parameter) then differentiating both sides with respect to x,
    [tex]\frac{d(pV)}{dx}= \frac{d(NRT)}{dx}[/tex]
    then, using the product rule, not the chain rule,
    [tex]p\frac{dV}{dx}+ \frac{dp}{dx}V= NR\frac{dT}{dx}[/tex]
    which, in "differential notation", is
    [tex]pdV+ Vdp= NRdT[/tex]
     
  8. Mar 29, 2012 #7
    Oh! ok that makes sense. But then what exactly do you mean by 'differential notation'? I see that if you 'multiply' by dx then that's what you get, but I know that that isn't really very rigorous and not completely correct since a differential isn't a fraction. Does the "[tex]d[/tex]" mean in general a derivative with respect to some arbitrary independent variable? I always thought it was an infinitesimal.
     
  9. Mar 30, 2012 #8
    Technically, I think the concept of a differential instead of just a "fraction" of differentials has been rigourized in Infinitesimal Analysis. Yes, it's an infinitesimal, I think. Technically, without infinitesimal analysis, your first post is not rigorous, making this whole thread just pure gobbledygook. I've always thought of the differential as a term (technically an infinitesimal) since it's so helpful for some differential equations.
     
  10. Mar 31, 2012 #9

    HallsofIvy

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    On the other hand, to make "infinitesmal analysis" rigorous requires some rather deep symbolic logic. Most Calculus texts, after defining the derivative "dy/dx", which is NOT defined as a fraction but as a limit of fractions, then define the "differentials" dy and dx by taking "dx" as purely symbolic (avoiding any mention of "infinitesmal") and defining dy as (dy/dx) dx.
     
  11. Apr 1, 2012 #10
    Actually, there is no need for Non Standard Analysis here at all. The differential df of a function f is well defined without it; if you study differential geometry you will learn what the symbols dp, dV, dT in your equation actually mean. They are 1-forms, which are linear functions on tangent vectors; i.e they are elements of the dual space to the space of tangent vectors at a point.

    More specifically, the differential df of a function f is the 1-form which assigns to any vector field x the number x(f). Differentials are rarely rigorously defined in an introductory calculus sequence as they require a different way of thinking about geometry which would confuse new students, and frankly it is unnecessary to have them formally introduced at that stage. So don't worry about it too much, just treat them using the same rules as derivatives and you'll be fine
     
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