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If there is a differential form $ A(x,y,z) dx + B(x,y,z) dy + C(x,y,z) dz$ where there exists some function $\psi(x,y,z)$

Let $ \psi = \psi (x,y,z)$

Then the total differential is $ d \psi = \left(\frac{\partial \psi}{\partial x} \right) dx + \left(\frac{\partial \psi}{\partial y} \right) dy + \left(\frac{\partial \psi}{\partial z} \right) dz$

If $ A = \left(\frac{\partial \psi}{\partial x} \right) $ , $ B = \left(\frac{\partial \psi}{\partial y} \right)$ , $ C = \left(\frac{\partial \psi}{\partial z} \right)$

Then the form $ A(x,y,z) dx + B(x,y,z) dy + C(x,y,z) dz$ is said to be 'exact'.

First, what does being 'exact' mean?

Also what does this have to do with independence of path if we integrate $ \int_i^f d\psi = \psi(f) - \psi(i)$

Why does an exact differential imply a conservative field?

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    $\begingroup$ $A(x,y,z)\, dx + B(x,y,z)\, dy + C(x,y,z) \,dz$ is not an ‘equation’, as you call it (there is no unknown to be determined) but a differential form (more precisely a differential $1$-form). A differential $1$-form which is the differential of a function is said to be exact. $\endgroup$ – Bernard Feb 6 at 18:35

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