Stationary points occur where the derivative is zero. The denominator is positive for all
, so we only need to worry about the numerator.
for all
, so we can divide through:
But
for all
, so this function has no stationary points...
I suspect there may be a typo in the question.
I believe the anwser would be -B
For to be conservative, we need to have
Integrate the first PDE with respect to :
Differentiate with respect to :
Now differentiate with respect to :
So we have
so is indeed conservative.