Conditions for a force to be conservative Taylor's classical mechanics ,chapter 4, states:
A force is conservative,if and only if it satisfies two conditions:

*

*$\vec{F}$ is a function of only the position. i.e $\vec{F}=\vec{F}(\vec{r})$.


*The work done by the force is independent of the path between two points.
Questions:

*

*Doesn't $1$ automatically imply $2$? : Since from 1, we can conclude that $\vec{F}=f(r)\hat{r}$, for some function $f$. Then, if $A$ is the antiderivative of $f$, we can say that $\vec{F}=\nabla{A}$, and therefore the work (line integral) will depend on the  final and initial positions only.  Or even simply put, $\vec{F}.d\vec{r}$ is a simple function of $r$ alone, so the integral will only depend on initial and final $r$.

*I have seen in many places, only "2" is the definition of a conservative force. In light of this, I cant think of why 1 has to be true: i.e how is it necessary that path independence implies $\vec{F}=f(r)\hat{r}$.

It could be that my interpretation of 1 as $\vec{F}=f(r)\hat{r}$ is wrong, on which my entire question hinges. Taylor writes $\vec{F}=\vec{F(\vec{r})}$ , which I interpreted as : "since F is a function of position vector, F is a function of both the magnitude and direction, and hence $\vec{F}=f(r)\hat{r}$".
 A: Your conclusions are not correct.
Here is a simple counter-example.
Consider this force
$$\vec{F}=k(x\hat{y}-y\hat{x})$$
where $\hat{x}$ and $\hat{y}$ are the unit-vectors
in $x$ and $y$-direction, and $k$ is some constant.
From this definition we see, the magnitude
of the force is $F=k\sqrt{x^2+y^2}=kr$,
and its direction is at right angle to
$\vec{r}=x\hat{x}+y\hat{y}$.
So we can visualize this force field like this:

The force circulates the origin in a counter-clockwise sense.
This force clearly satifies your first condition


*

*$\vec{F}$ is a function of only the position,
i.e. $\vec{F}=\vec{F}(\vec{r})$

But it is not of the form $\vec{F}=f(r)\hat{r}$.
And this force violates your second condition



*The work done by the force is independent of the
path between the two point.


To prove this consider the following two paths:

*

*Path A (in green): beginning on the right
at $(x=R,y=0)$, doing a half circle counterclockwise,
to the point on the left $(x=-R,y=0)$.

*Path B (in red): beginning on the right
at $(x=R,y=0)$, doing a half circle clockwise,
to the point on the left $(x=-R,y=0)$.


Then the work for path A is (because here $\vec{F}$ is always parallel to $d\vec{r}$)
$$W_A=\int \vec{F}(\vec{r}) d\vec{r}=kR\cdot\pi R=\pi k R^2.$$
Then the work for path B is (because here $\vec{F}$ is always antiparallel to $d\vec{r}$)
$$W_B=\int \vec{F}(\vec{r}) d\vec{r}=-kR\cdot\pi R=-\pi k R^2.$$
You see, the work is different for the two paths, although
the start and end point of the paths are the same.
This is a simple example of a non-conservative force.
The non-conservativeness can easily be checked
by calculating its curl and finding it is non-zero.
$$\vec{\nabla}\times\vec{F}
 =\vec{\nabla}\times k(x\hat{y}-y\hat{x})
 =2k\hat{z}
 \ne \vec{0}$$
A: If the function is (as you assumed) one of distance, then you are right. But there are many functions of position coordinates whose curl is not zero, hence non-conservative.
Edit: Try for example $$F = (xy,-xy,0)$$
A: The central forces with spherical symmetry are also conservative forces.
You can show this proving that the work does not depend on the path.
HP: $$\vec{F}=F(r)\vec{r}$$ where $\vec{r}$ is the unitary vector to the position vector.
So you have: $$W_{AB}=\int_A^BF(r)\vec{r}\cdot ds$$ Since in polar coordinates elementar displacement in $d\vec{s}=d\vec{r}=dr\vec{r}+rd\theta\vec{\theta}$, have $$W_{AB}=\int_{r_A}^{r_B}F(r)dr$$ So the work done by the force depends only by the initial and final distance.
You can also show that a central force with spherical symmetry is irrotational calculating the curl of the force but in polar coordinates.
Remember that the fact that the curl of the force is $0$ isn't enough for the conservation but the function must to be defined on a simply connetted domain, like gravitational and Coulomb force.
