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Apparently a gradient field has special properties that not all vector fields have one being that it is conservative. However the gradient field only exists if there is an exact differential and from what I understand this is a sufficient and necessary condition.

Is there property of a differential that is the same as conservatism? ( i.e. A differential is a place holder for the addition of partial derivatives and the gradient vector field is that addition resulting in a vector that points in the direction of maximum increase of the slope. It is not surprising that it is the direction of maximum increase, since if you skew it as can easily be done by a directional derivative away from the original basis vectors the Pythagorean theorem shows the rise over run will not be as steep.

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You are right to talk about steepest increase etc.

One kind of physical/geometric way to think about it is that if you are walking on the graph of $f$ in the direction $\vec{v}$, then $\vec{v}\cdot \nabla f$ is giving you the rate of change in 'altitude'. So if you integrate around a closed loop (and hence come back to where you started), it's obvious that overall your altitude should not have changed.

Maybe seeing the 'proof' will help a bit, i.e. one way to 'motivate' integrating around loops and looking for zero is to see that it is the same as saying the integral along a path $\gamma$ that goes from $a$ to $b$ depends only on the end points $a$, $b$ and not on the particular route $\gamma$ that you take. Given this, and assuming you are on a nice (simply connected) domain, you can 'integrate' a conservative vector field $\mathbf{v}$ by just fixing a single point $a$ in the domain and defining $$ \phi(x) = \phi(a) + \int_{\gamma} \mathbf{v} \cdot d\mathbf{s}. $$ where $\gamma$ is any path from $a$ to $x$. This is well-defined precisely because $\mathbf{v}$ is conservative. So

Conservative $\Rightarrow\ $Path integrals only depend on endpoints $\Rightarrow $ You can integrate and find a function whose gradient is your field.

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  • $\begingroup$ Ok T_M I follow you but What is the motivation for integrating around a closed loop? ( I do understand the area can be allowed to represent work.) $\endgroup$ – Sedumjoy Jun 14 '18 at 20:40

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