What exactly is conservative vector field? I'm studying calculus, but since the example involved a physical concept. I will ask here:
This is how it goes:

This means that in a conservative force field, the amount of work
  required to move an object from point a to point b depends only on
  those points, not on the path taken between them.

But what does it mean? How does it not depend on the path? 
 A: If $\vec{F}$ is a conservative force field, then it satisfies the property
$$
\tag{1}
\vec{\nabla} \times \vec{F} = 0,
$$
and it can be written as
$$
\tag{2}
\vec{F} = \vec{\nabla}V,
$$
for a scalar function $V$ (which corresponds to potential function in physics).
Note that, when you put $(2)$ into $(1)$ it becomes a "curl of a gradient" and is automatically vanishes. You can derive this result by using simple mathematical knowledge.
In physics, work done on a particle by applying a force $\vec{F}$ along a path is defined as
$$
\tag{3}
W = \int_C \vec{F} \cdot d\vec{s},
$$
where $C$ is any path connecting two points in the space, call $A$ (initial point) and $B$ (end point). These are the points we start/finish applying the force. Regarding this, we can rewrite $(3)$ as
$$
W = \int_A^B \vec{F} \cdot d\vec{s}.
$$
Now, if $\vec{F}$ is conservative then we can also use property $(2)$, which gives
\begin{align}
W & = \int_A^B \vec{\nabla}V \cdot d\vec{s}
\\
& = \int_A^B \frac{\partial}{\partial \vec{s}}V \cdot d\vec{s}
\\
& = V(B) - V(A).
\end{align}
Thus, with the given property that force field is conservative we find work done on a particle by exerting this force field only depends on the end points but not on the path we choose.

NOTE: Conventionally, in physics we write $(2)$ with a minus sign in front,
$$
\vec{F} = -\vec{\nabla}V.
$$
However, in the above text, I used it in the formal mathematical description without regarding any physical concerns.
