Conservative Vector Fields — Question 10

Question 10 – Solution

Strategy. Compare the constant cross partials, then integrate symbolically without assigning numerical coefficients.

Step 1: Classify Let $P=ax+by$ and $Q=cx+dy$. Then $$P_y=b, \qquad Q_x=c.$$ Because $\mathbb R^2$ is simply connected, equality is both necessary and sufficient. Hence $$\boxed{\mathbf F\text{ is conservative exactly when }b=c}.$$

Step 2: Construct the potential Assume $b=c$. Integrating $f_x=ax+by$ gives $$f=\frac a2x^2+bxy+g(y).$$ Since $f_y=bx+g'(y)=bx+dy$, we have $g'(y)=dy$, so $$\boxed{f(x,y)=\frac a2x^2+bxy+\frac d2y^2+C}.$$

Step 3: Compute the work The Fundamental Theorem for Line Integrals gives $$\int_C\mathbf F\cdot d\mathbf r =f(p,q)-f(0,0) =\boxed{\frac a2p^2+bpq+\frac d2q^2}.$$

Verification Differentiating the potential gives $\langle ax+by,bx+dy\rangle$, which equals the original field precisely because $c=b$. The additive constant cancels from the work.