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

1 Answer
Aug 10, 2016

Answer:

Non-conservative

Explanation:

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

And because #vec F = \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((hat x, hat y, hat z),(del_x, del_y, del_z),(xy, 2z-y+x, 2y-z))#

#= hat x (2-2) - hat y (0-0) + hat z (1-x) ne 0#

Non-conservative