Question

Determine whether or not F is a conservative vector field. If it is, find a function f such that
\[F=\nabla f\]
\[F(x,y)=(2x-3y)i+(-3x+4y-8)j\]

Answer 1

If F is conservative, then by the equality of mixed partials the derivative of the x component of F wrt y should equal the derivative of the y component wrt x.
\[\frac{\partial}{\partial y}[2x-3y]=-3\]\[\frac{\partial}{\partial x}[-3x+4y-8]=-3\]So it is conservative.
\[f_x=2x-3y \Rightarrow f=x^2-3xy+p(y)\]\[f_y=-3x+4y-8 \Rightarrow f=-3xy+2y^2-8y+q(x)\]Hopefully you understand those steps. Now comparing the two results, we see that they share a -3xy term. So we let 2y^2-8y equal p(y), and x^2 for q(x), so we have our answer:
\[f(x,y)=x^2-3xy+2y^2-8y+C\]And you can easily check that ∇f=F.

Answer 2

Thank you so much @Herp_Derp

Answer 3

@MathSofiya It gets a little tricky when you try to solve for p and q. Often it works to just subtract the two equations, though I didn't show that above :)

Answer 4

twin lookin at twin :D

Answer 5

Awww

Answer 6

and will keep looking

Answer 7

@Herp_Derp I walked through the problem today paying attention to every little detail, and it makes perfect sense. Thanks again for the thorough explanation.

Answer 8

Sorry, but, herp_derp made a important mistake. The equality of mixed parcials is just a SUFFICIENT CONDITION, not a necessary condition. Take this example:
\[f(x,y)= \left( \frac{ -y }{x^2+y^2 }, \frac{ x }{x^2+y^2} \right)\]
You can see that the mixed parcials are in deed the same, BUT if you calculate the circulation of f(x,y) through a closed curve, g(t)=(cos t, sin t), the result is 2*pi.
As we know that the circulation of a conservative field through a closed curve must be 0. So f(x,y) is not a conservative field, although that the necessary condition is fulfilled.