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

The force field is not conservative,

If *iff*

As stated above, the curl is given by the cross product of the gradient of

We have

The curl of the vector field is then given as:

We take the cross product as we usually would, except we'll be taking partial derivatives each time we multiply by a partial differential.

For the

We can tell immediately that the product will not be

For the

(Remember that if we take the partial of a function with respect to some variable which is not present, the partial derivative is

For the

This gives a final answer of

*not* conservative

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To determine if the force field is conservative, check if the curl of the vector field is zero. Calculate the curl as follows:

[ \nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) \mathbf{i} + \left( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \mathbf{j} + \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \mathbf{k} ]

If (\nabla \times \mathbf{F} = \mathbf{0}), then the force field is conservative.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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