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

##### 1 Answer

#### Answer:

The force field is conservative;

#### Explanation:

If *iff*

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

We have

#P=x#

#Q=z#

#R=y#

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

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

This tells that our force field has the potential to be conservative (pun intended). We now must attempt to produce the potential function of the force field. To find the potential function of the vector field, we find

We are given

#f_x(x,y,z)=x#

#f_y(x,y,z)=z#

#f_z(x,y,z)=y#

We now find the antiderivative of one of these components. Which does not matter, so I'll start with

#f_x(x,y,z)=x#

#int(x)dx=1/2x^2+g(y,z)#

(Note we include

#=>f(x,y,z)=1/2x^2+g(y,z)#

We now take the

#del/(dely)(1/2x^2+g(y,z))=0+g_y(y,z)#

We set this equal to our original

#z=g_y(y,z)#

#=>g(y,z)=zy+h(z)#

Now we take the partial of the above function

#del/(delz)(zy+h(z))=y+h'(z)#

#=>y=y+h'(z)#

#=>h'(z)=0#

#=>h(z)=0#

Revisiting our function

We then have satisfied both conditions:

*is* conservative