Question

Determine whether or not f is a conservative vector field. If it is, find a function f such that f = ∇f. (if the vector field is not conservative, enter dne. ) f(x, y) = (y2 − 6x)i 2xyj

Answer 1

It looks like the vector field is

`\vec f(x,y) = (y^2-6x) \, \vec\imath + 2xy \, \vec\jmath`

`\vec f` is conservative if we can find a scalar function
`f(x,y)` whose gradient is
`\vec f`. This entails solving the partial differential equations

`(\partial f)/(\partial x) = y^2 - 6x`

`(\partial f)/(\partial y) = 2xy`

Integrate both sides of the second PDE with respect to y :

`\displaystyle \int (\partial f)/(\partial y) \, dy = \int 2xy \, dy \implies f(x,y) = xy^2 + g(x)`

Differentiate with respect to x and solve for
`g(x)` :

`(\partial f)/(\partial x) = y^2 - 6x = y^2 + (dg)/(dx) \implies (dg)/(dx) = -6x \implies g(x) = -3x^2+C`

It follows that
`\vec f` is indeed conservative with potential function

`f(x,y) = xy^2 - 3x^2 + C`