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 − 4x)i + 2xyj

Answer 1

Answer:

f = xy²-2x² satisfies that  F = ∇f.

Step-by-step explanation:

F(x,y) = (y² - 4x) i + 2xy j

We want f(x,y) such that

$f_x(x,y) = y^2 -4x\\f_y(x,y) = 2xy$

Lets find a primitive of y²-4x respect to the variable x. We need to think y (and y²) as constants here, so a primitive of y² would be xy² the same way that a primitive of k is xk (because we treat y² as constant). A primitive of x is x²/2, thus a primitive of 4x is 2x². Thus, a primitive of y²-4x is xy² - 2x². We can obtain any other primitive by summing a constant, however since we treated y as constant, then we have that

$f(x,y) = xy^2 - 2x^2 + c(y)$

where c(y) only depends on y (thus, it is constant repsect with x).

We will derivate the expression in terms of y to obtain information about c(y)

$2xy = f_y(x,y) = \frac{d}{dy} (xy^2 - 2x^2 + c(y) ) = 2xy -0 + \frac{d}{dy} c(y) = 2xy + \frac{d}{dy} c(y)$

Thus, $\frac{d}{dy}$ is constant. We can take f(x,y) = xy²-2x². This function f satisfies that  F = ∇f.

Answer 2

The equation f = xy²-2x² satisfies that  F = ∇f.

Calculations and Parameters:

F(x,y) = (y² - 4x) i + 2xy j

We want f(x,y) such that:

$f_x(x,y) = y^2 - 4x\\f_y(x,y) = 2xy$

Next, we would have to find a primitive of y²-4x with respect to the variable x.

We would have to take y (and y²) as constants so the primitives of y² would be xy² the same.

Hence, a primitive of x is x²/2, thus a primitive of 4x is 2x². Thus, a primitive of y²-4x is xy² - 2x²

$f(x,y) = xy^2 - 2x^2 + c(y)$

After differentiation,

$2xy + d/dy.c(y)$

Thus, $d/dy$  is a constant and we can take f(x,y) = xy²-2x².

This function f satisfies that F = ∇f.

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