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

1 Answer
Feb 25, 2017

Answer:

The force field is not conservative; #curl(vecF)!=0#.

Explanation:

If #vecF# is a vector (force) field in #RR^3#, then the curl of #vecF# is the vector #curl(vecF)# and is written as #gradxxvecF#. The vector field #vecF# is conservative if and only if #vecF=gradf# for some potential function. If #vecF# is conservative, #curl(vecF)=vec0#. However, this does not mean that a vector field with #curl=vec0# is absolutely conservative; the vector field must also have a potential function to be conservative, but the first prerequisite is that #curl(vecF)=vec0#.

As stated above, the curl is given by the cross product of the gradient of #vecF# (in the form #< P, Q, R >#) and itself.

lamar.edu

We have #vecF=< xy, 2z-y^2+x, 2y-zx ># where

#P=xy#

#Q=2z-y^2+x#

#R=2y-zx#

The curl of the vector field is then given as:

#abs((veci,vecj,veck),(del/(delx),del/(dely),del/(delz)),(xy,2z-y^2+x,2y-zx))#

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

For the #veci# component, we have #del/(dely)(2y-zx)-del/(delz)(2z-y^2+x)#

#=>(2-2)=0#

So far, so good.

For the #vecj# component, we have #del/(delx)(2y-zx)-del/(delz)(xy)#

#=>-(-z-0)=z#

Remember that if we take the partial of a function with respect to some variable which is not present, the partial derivative is #0# as we treat all other variables as constants.

For the #veck#component, we have #del/(delx)(2z-y^2+x)-del/(dely)(xy)#

#=>(1-x)=1-x#

This gives a final answer of #curl(vecF)=<0,z,1-x> !=vec0#

#:.# Force field is not conservative