Conservative Vector Fields — Question 1

Question 1

Determine whether $$\mathbf F(x,y)=\langle 2xy+3,\,x^2+4y\rangle$$ is conservative on $\mathbb R^2$. If it is, find a potential function $f$.

Tasks

1. Compare the relevant mixed partial derivatives.

2. Construct $f$ and determine the one-variable correction term.

3. Differentiate the result to verify the field.

Question 1 – Solution

Strategy. Test $P_y=Q_x$ on the simply connected domain, then integrate one component and recover the missing function.

Step 1: Test the field With $P=2xy+3$ and $Q=x^2+4y$, $$P_y=2x, \qquad Q_x=2x.$$ The components have continuous first partial derivatives on all of $\mathbb R^2$, which is simply connected. Thus the equality guarantees that $\mathbf F$ is conservative.

Step 2: Construct a potential Integrating $f_x=P$ with respect to $x$ gives $$f=x^2y+3x+g(y).$$ Then $f_y=x^2+g'(y)=Q=x^2+4y$, so $g'(y)=4y$ and $g(y)=2y^2+C$. $$\boxed{f(x,y)=x^2y+3x+2y^2+C}.$$

Verification Differentiation gives $\nabla f=\langle 2xy+3,x^2+4y\rangle=\mathbf F$. The arbitrary constant has zero gradient.