# Very Basic Question: Relationship between Potential, Conservative Forces and Path Independency

I am studying for my exam and wanted to clarify if the following I got from Taylor are true, because we have written something different in my lectures:

$$\nabla\times\vec F=0\ \ \Leftrightarrow\ \ \oint\limits_C\vec F\cdot d\vec r =0$$

If above holds for a force, then we are mathematically allowed to find a scalar potential field, such that: $$\vec F=-\nabla V$$

And if, in addition to above, the force is also non-dependent on velocity $$\dot r$$ and time $$t$$, then the mechanical energy of the system is conserved, right?

Is there any relationship between the force being dependent on $$\dot r$$ or $$t$$, and the force having a corresponding potential field? Can a force dependent on $$\dot r$$ also have a corresponding scalar potential field?

If the force is a function of $$t$$ or $$v$$, it is not possible to write: $$F = -\nabla V = - \left (\frac{\partial V}{\partial x} , \frac{\partial V}{\partial y} , \frac{\partial V}{\partial z} \right )$$ because for the same point $$x,y,z$$ it would be possible different values for $$F$$. The above expression states that $$F = F(x,y,z)$$