# Question #36742

Mar 6, 2017

See below

#### Explanation:

The work done in a conservative force field is guaranteed to be zero only over a closed path.

By definition, a vector field is conservative if and only if there exists a potential function $V \left(m a t h b f x\right)$ such that $m a t h b f F = - \nabla V \left(m a t h b f r\right)$.

For a force field, we have the definition of work as:

$\delta W = m a t h b f F \cdot \delta m a t h b f r$

$\implies W = {\int}_{\Gamma \left(m a t h b f p , m a t h b f q\right)} m a t h b f F \cdot d m a t h b f r$

....where $\Gamma$ is a curve from $m a t h b f p$ to $m a t h b f q$ in the field.

The Gradient Theorem , aka the Fudamental Theorem for Line Integrals, states that:

${\int}_{\Gamma \left(m a t h b f p , m a t h b f q\right)} \nabla V \left(m a t h b f r\right) \cdot d m a t h b f r = V \left(m a t h b f q\right) - V \left(m a t h b f p\right)$

This is where the idea of path independence comes from.

Now, if $m a t h b f p = m a t h b f q$, then this becomes:

$\oint \nabla V \left(m a t h b f r\right) \cdot d m a t h b f r = 0$