Is work path independent? Stewart - Calculus



It does not appear that work being change of kinetic energy depends on $C$ or that $F$ being conservative is assumed for the result in the red box. Hence, work is path independent? Or are there some assumptions made that are too advanced for the basic calculus reader?
 A: Your equation and its derivation are using the restriction: $c$ solves $F=m\ddot{c}$ (this is physically equivalent to saying that you are taking the specific case where $F$ is the resultant force acting on a particle, and you are computing the work of this force over its trajectory).
In this case, it doesn't even make sense to ask about path-independency. Your path is a particular one, the solution to a differential equation.
That specific case aside, the work of a vector field does depend on the path. Take $F=\frac{1}{x^2+y^2}(-y,x)$, and verify that the work done by this vector field over the top part of the circle is different from the lower part of the circle (both connecting $(1,0)$ to $(-1,0)$).
A: It depends against which force you are doing the work. If you have a curl free force field then and only then it is path independent. If the concept of curl is still not introduced then you can just remember that workis path independent in only certain cases like gravity, electrostatics. Simple counter example is imagine you have to move a body along two curves. One has more friction than other, then work done is not path independent.
