How to Determine if a Differential Form is Exact

In summary, the conversation discusses the concept of an exact differential and how to determine if a given differential form is exact. The participants also solve for the derivative of ln(x) and discuss how to find z given a differential form. Ultimately, it is determined that dz is an exact differential for z= xy-y+lnx+2 and any z(x,y)= xy- y+ ln(x)+ C satisfies the given differential form.
  • #1
zeshkani
29
0
how would i show that this is the exact differential

z=xy-y+lnx+2
dz= (y+ 0)dx + (x-1)dy (i hope the total differential is right)

so how would i show that dz is the exact differential ?
 
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  • #2
The definition of the total differential of a function f(x,y) is the expression

[tex]df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy[/tex]

Now with the z you're given, is the expression for dz correct? That is to say, is

[tex]\frac{\partial z}{\partial x}=(y+0)[/tex]

and

[tex]\frac{\partial z}{\partial y}=(x-1)[/tex]

??
 
  • #3
i believe dz is correct because i used partial differentials to solve it, and i hope its correct
 
  • #4
what is the derivative of ln(x)?
 
  • #5
Once you fixed the derivative of ln x thing, the fact that dz is the total derivative of z pretty much means it is an "exact" differential. That's the definition of "exact" differential!
 
  • #6
thx and the derivative of lnx is just 1/x
 
  • #7
Yes.
z=xy-y+lnx+2 so
dz= (y+ 1/x)dx + (x-1)dy

And that is an "exact" differential precisely because it is the differential of z.

Now suppose you were given the differential form (y+ 1/x)dx+ (x- 1)dy without having been given z. How would you determine whether it was an exact differential?
If
[tex](y+ 1/x)= \frac{\partial z}{\partial x}[/tex]
for some function z and
[tex]x-1= \frac{\partial z}{\partial y}[/tex]
then
[tex]\frac{\partial(y+ 1/x)}{\partial y}= \frac{\partial^2 z}{\partial x\partial y}[/tex]
and
[tex]\frac{\partial(x-1)}{\partial x}= \frac{\partial^2 z}{\partial y\partial x}[/tex]
and so must be equal. Fortunately, it is easy to see that each second derivative is 1 and so they are in fact equal.
Okay, now, how would we find z? Knowing that
[tex]x-1= \frac{\partial z}{\partial y}[/tex]
integrating with respect to y (treating x as a constant) we get z= xy- y+ C, except that, since we are treating x as a constant, that "constant of integration", C, may depend on x: z= xy- y+ C(x). Differentiating that with respect to x,
[tex]\frac{\partial z}{\partial x}= y+ C'(x)= y+ 1/x[/tex]
Notice that the "y" terms cancel (it was the "cross condition" above that guarenteed that) and so we have C'(x)= 1/x. Then C(x)= ln(x)+ C where "C" now really is a constant.
Any z(x,y)= xy- y+ ln(x)+ C satisfies dz= (y+ 1/x)dx+ (x-1)dy.
 

1. What is the definition of an exact differential?

An exact differential is a mathematical concept used in calculus to describe a function that has a unique derivative at every point. In other words, the derivative of an exact differential function is independent of the path taken between two points on the function.

2. How is an exact differential different from an inexact differential?

An inexact differential is a function that does not have a unique derivative at every point, meaning that the derivative can vary depending on the path taken. An exact differential, on the other hand, has a unique derivative at every point and is independent of the path taken.

3. What is the significance of exact differentials in thermodynamics?

In thermodynamics, exact differentials are used to describe the state of a system at a particular point in time. They are important because they allow us to calculate changes in a system without having to consider the path taken.

4. Can you give an example of an exact differential?

One example of an exact differential is the total differential of a function, such as dU = dQ - dW in thermodynamics. This equation is exact because the derivative of internal energy (U) is independent of the path taken between two points.

5. How are exact differentials used in solving problems?

Exact differentials are used in solving problems by simplifying the calculation of changes in a system. By using exact differentials, we can focus on the initial and final states of a system without having to consider the path taken, making problem-solving more efficient and accurate.

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