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

1 Answer
Jul 14, 2016

Answer:

NO

Explanation:

if the field #vec F# is conservative, then there exists a potential function, #f#, such that #vec F = - nabla f#

and as #nabla times nabla f = 0#, curl of gradient, it follows that if #f# exists then #nabla times vec F = 0# also

so we can test the curl of the vector field to see if it is indeed zero, as follows

#nabla times vec F = det[(hat x ,hat y ,hat z) , (del_x, del_y, del_z),(xy + z,xy-x,2y-zx)]#

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

# = [(2), (z+1), (y - x)] ne 0#

This is not conservative. It is necessary that the #nabla times vec F = 0 # for #vec F # to be conservative