# A force field is described by <F_x,F_y,F_z> = < xy , 2z-y +x, 2y -z > . Is this force field conservative?

Aug 10, 2016

Non-conservative

#### Explanation:

The definitive evidence of a conservative field $\vec{F} \left(x , y , z\right)$ is the existence of potential function $f \left(x , y , z\right)$ which defines the potential at every point and makes the transitions between points in the field path-independent.

And because $\vec{F} = \setminus \nabla f$ and thus $\nabla \times \vec{F} = \nabla \times \nabla f = 0$ by definition as curl grad is zero....

.....then a necessary but insufficient condition is that the curl of the field be zero. (IOW the curl could be zero but the field still be non -conservative, but if the field is conservative the curl has to be zero)

We test that here:

$\nabla \times \vec{F} = \det \left(\begin{matrix}\hat{x} & \hat{y} & \hat{z} \\ {\partial}_{x} & {\partial}_{y} & {\partial}_{z} \\ x y & 2 z - y + x & 2 y - z\end{matrix}\right)$

$= \hat{x} \left(2 - 2\right) - \hat{y} \left(0 - 0\right) + \hat{z} \left(1 - x\right) \ne 0$

Non-conservative