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I know that a differential $dA=B(x,y)dx+C(x,y)dy$ is exact iff

$$\tag{1} \left( \frac{\partial B}{\partial y} \right)_x = \left( \frac{\partial C}{\partial x} \right)_y$$

I do not understand how to interpret this relation. The rate of change of $B$ with respect to $y$ equals to the rate of change of $C$ with respect to $x$?

I understand that if differential of a function $A(x,y)$ is exact, then the function $A$ is a state function, ie. independent of path. For example, for a closed loop

$$\oint dA=0$$

which implies

$$\tag{2} \oint B(x,y)dx=-\oint C(x,y)dy$$

It seems like another condition for exact differential. I am wondering how (1) is related to (2).

Ray Siplao
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  • In short: (2) implies (1). When the domain of $B$ and $C$ is simply connected then the reverse holds. Read a good source about closed and exact differential forms and the Poincare lemma. – Kurt G. Apr 17 '23 at 12:21
  • FYI - obviously you come from a physics background. In mathematics what you call a "state function" is just called a "function". A "function" that changed its value at $(x,y)$ depending on some path would not be a function of $(x,y)$, but a function of the paths, and thus we would call it $A(\gamma)$ (where $\gamma$ is the path), not $A(x,y)$. In fact, even in physics the concept of a "state function" is rather dodgy. A state variable is well-defined, but a "state function" implies a fixed dependency on other states, and paths in the state space are not states themselves. – Paul Sinclair Apr 18 '23 at 01:28

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