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

In summary, the solution shows that T=T(P,V) and then F( P,V,T(P,V) ) and the purpose is to derive a "cancellation law" similar to df/dy= df/dx / fy/dx .
  • #1
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Homework Statement


The question is attached along with its solution.


Homework Equations


Partial differentiation and the implicit function theorem.


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!
 

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  • #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.
 
  • #3
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?
 
  • #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.
 
  • #5
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?
 
  • #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:

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

1. What does it mean to "show that (product of these three partial derivatives) = -1"?

This means that the product of the three partial derivatives of a function is equal to -1.

2. Why is it important to show that the product of these three partial derivatives equals -1?

This is important because it helps to understand the behavior of the function and its derivatives. It also allows us to make predictions and solve problems in various fields such as physics, economics, and engineering.

3. How do you calculate the product of these three partial derivatives?

The product of three partial derivatives can be calculated by taking the partial derivatives of a function with respect to each variable, and then multiplying them together.

4. What are some real-life applications of showing that the product of these three partial derivatives equals -1?

Real-life applications include optimizing production processes in manufacturing, predicting the behavior of financial markets, and understanding the dynamics of chemical reactions.

5. Can you provide an example of showing that the product of these three partial derivatives equals -1?

Sure, let's say we have a function f(x,y,z) = x^2 + y^2 + z^2. The partial derivatives of this function are f_x = 2x, f_y = 2y, and f_z = 2z. When we multiply them together, we get 2x * 2y * 2z = 8xyz. So, for any x, y, and z values, the product of these three partial derivatives is equal to -1.

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