Partial Derivatives for P, S, T in Differential Calculus

In summary, the conversation discusses finding the partial derivative of P with respect to V at fixed T and fixed S, using the given equation and the chain rule. The final result is \frac{\partial{P}}{\partial{V}}=-\frac{TS}{4V^2} at constant T and constant S. The conversation also briefly touches on the possibility of finding the derivative at constant T and varying S, or vice versa.
  • #1
Bill Foster
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Homework Statement




I'm looking for [tex]\frac{\partial{P}}{\partial{V}}[/tex] at fixed T and fixed S.


Homework Equations



[tex]P=\frac{TS}{4V}[/tex]

The Attempt at a Solution



[tex]\frac{dP}{dV}=\frac{\partial{P}}{\partial{V}}+\frac{\partial{P}}{\partial{T}}\frac{dT}{dV}+\frac{\partial{P}}{\partial{S}}\frac{dS}{dV}[/tex]

[tex]\frac{\partial{P}}{\partial{V}}=-\frac{TS}{4V^2}[/tex]

[tex]\frac{\partial{P}}{\partial{T}}=\frac{S}{4V}[/tex]

[tex]\frac{\partial{P}}{\partial{S}}=\frac{T}{4V}[/tex]

[tex]\frac{dP}{dV}=\frac{\partial{P}}{\partial{V}}+\frac{\partial{P}}{\partial{T}}\frac{dT}{dV}+\frac{\partial{P}}{\partial{S}}\frac{dS}{dV}=-\frac{TS}{4V^2}+\frac{S}{4V}\frac{dT}{dV}+\frac{T}{4V}\frac{dS}{dV}[/tex]

At constant T, I get this: [tex]\frac{dP}{dV}=-\frac{TS}{4V^2}+\frac{T}{4V}\frac{dS}{dV}[/tex]

At constant S, I get this: [tex]\frac{dP}{dV}=-\frac{TS}{4V^2}+\frac{S}{4V}\frac{dT}{dV}[/tex]

What do I do about the other differentials: [tex]\frac{dS}{dV}[/tex] and [tex]\frac{dT}{dV}[/tex]?

Wouldn't this also be true?

[tex]\frac{dS}{dV}=\frac{\partial{S}}{\partial{V}}+\frac{\partial{S}}{\partial{T}}\frac{dT}{dV}+\frac{\partial{S}}{\partial{P}}\frac{dP}{dV}[/tex]

[tex]\frac{dT}{dV}=\frac{\partial{T}}{\partial{V}}+\frac{\partial{T}}{\partial{S}}\frac{dS}{dV}+\frac{\partial{T}}{\partial{P}}\frac{dP}{dV}[/tex]
 
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  • #2


You want to find [tex]
\frac{\partial{P}}{\partial{V}}
[/tex]

and your second line is

[tex]
\frac{\partial{P}}{\partial{V}}=-\frac{TS}{4V^2}
[/tex]

So... you're done?
 
  • #3


Office_Shredder said:
You want to find [tex]
\frac{\partial{P}}{\partial{V}}
[/tex]

and your second line is

[tex]
\frac{\partial{P}}{\partial{V}}=-\frac{TS}{4V^2}
[/tex]

So... you're done?


Yes. That's at constant T and constant S.

Is it the same if T is constant and S is not constant? Or vice-versa?
 

FAQ: Partial Derivatives for P, S, T in Differential Calculus

What is a differential?

A differential is a mathematical concept that is used to measure the rate of change of a function at a specific point. It can also be thought of as the slope of a curve at a given point.

How is a differential calculated?

The differential is calculated by taking the derivative of the function at the given point. This involves finding the slope of the tangent line to the curve at that point.

What is the difference between a differential and a derivative?

A differential is a single value that represents the rate of change at a specific point, while a derivative is a function that represents the rate of change at every point on the curve.

How are differentials used in real life?

Differentials are used in many areas of science and engineering, such as physics, economics, and engineering. They can be used to model and solve problems involving rates of change, such as motion, growth, and decay.

What is the relationship between differentials and integrals?

Differentials and integrals are closely related concepts. The integral of a function is the reverse process of finding its differential. In other words, the integral is the sum of all the differentials along a given interval.

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