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Show that (product of these three partial derivatives) = -1.

  1. Mar 20, 2012 #1

    s3a

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    1. The problem statement, all variables and given/known data
    The question is attached along with its solution.


    2. Relevant equations
    Partial differentiation and the implicit function theorem.


    3. The attempt at a solution
    My work is attached. I feel it's correct but is it incomplete?

    I have the following questions/confusions.:

    1) What's the purpose of the solution showing that T=T(P,V) and then F( P,V,T(P,V) )? Couldn't I just as easily have said V=V(P,T) or P=P(V,T) and modified F accordingly?

    2) Prior to making this forum thread, I was completely confused as to what the 0 = stuff meant before the implication arrow however, I know see that it is the implicit function theorem written in one line. Is this important? Why is it being done in the solution?

    Any input would be appreciated!
    Thanks in advance!
     

    Attached Files:

  2. jcsd
  3. Mar 20, 2012 #2
    1) yes, that is what the solution is doing.

    2) {(x,y,z) : F(x,y,z)=0} is a 3-1 hypersurface. the implicit function theorem states, by local differentiability i think, that z can be written as a function of x and y, z=z(x,y). substituting this back in F, we get an implicit map in x and y {(x,y, z(x,y)) : F(x,y, z(x,y))=0 }, so we can apply implicit partial differentiation to it (the meaning of the double quotient formulas), allowing "cancellation" of the F terms.
     
  4. Mar 21, 2012 #3

    s3a

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    Thanks for your answer but I still have some questions/concerns:

    1) You didn't tell me if what I did myself is correct and complete. So, is it?

    2) Generally speaking, you said by substituting z = f(x,y) into F(x,y,z), we get F(x,y,f(x,y)) and that we could use the implicit function theorem with this after but what is the point of doing this? For example, when I did the problem (my work is attached), I didn't even bother to set x, y, or z as a function of the other two variables and concluded with the same answer. Is this extra step necessary? Whether it is or not, what's the benefit of doing it?
     
  5. Mar 22, 2012 #4
    i would say not, since i would have started from the beginning with the solution set {(x,y, z(x,y)) : F(x,y,z(x,y))=0} then use knowledge of the implicit mapping theorem in calculus. but that depends how much you want to assume as given, i suppose.

    i think the purpose of this is to derive a kind of "cancellation law" similar to df/dy= df/dx / fy/dx . so if you are already assuming df/dx / df/dy = dy/dx, then yes, the problem looks somewhat academic.
     
  6. Mar 24, 2012 #5

    s3a

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    After some more extensive analysis of this problem, I'm thinking that the first partial derivative on the top left of MyWork.jpg is correct but still needs to be proved because of the interdependence of P, V and T.

    For instance, ∂T/∂P = (∂F/∂P) / (∂F/∂T) is not true in general for partial derivitives because of this interdependence of P, V, and T. I'm still confused as to how I prove that first partial derivative (on the top-left of the MyWork.jpg image) is correct though. Like, how do I do this specifically?
     
  7. Mar 24, 2012 #6
    i'm not sure what you're trying to get at. we've created a system of two equations which are not independent, so describe the same solution space. this means the solution space is not reduced when the second is substituted into the first.

    i think it is the continuity and restricted invertibility of the jacobian determinant over F near (x,y,z) that decides whether there is an explicit function z=-z(x,y) (or z(x,y)) for the implicit map F(x,y,z)=0. since there IS (and z(x,y) is still differentiable near (x,y,z(x,y)) ), we can apply implicit differentiation and the chain rule to F(x,y, -z(x,y))=0 to yield the result:

    0 = dF/dx = dF/dx * dx/dx + dF/dz * -dz/dx

    solving for dz/dx we get a "cancelation" of F.

    <a href=http://en.wikipedia.org/wiki/Implicit_and_explicit_functions#Formula_for_two_variables>wikipedia</a>

    [edit] sorry i'm not sure that the sign change works.
     
    Last edited: Mar 24, 2012
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