Conservative Vector Fields — Question 7

Question 7 – Solution

Strategy. Subtract the two gradient equations and use paths inside the connected open region.

Step 1: Subtract gradients Since $\nabla f=\mathbf F=\nabla g$, $$\nabla(f-g)=\mathbf 0.$$ Let $A,B\in D$. An open connected region in $\mathbb R^n$ is path connected, so choose a piecewise smooth curve $C$ in $D$ from $A$ to $B$. Then $$(f-g)(B)-(f-g)(A) =\int_C\nabla(f-g)\cdot d\mathbf r=0.$$ Thus $(f-g)(B)=(f-g)(A)$ for every pair, and $$\boxed{f-g=K\text{ on }D}.$$

Step 2: Why connectedness matters On separate components, the same zero gradient permits a different constant on each component because no path in $D$ joins them.

Step 3: Uniqueness principle A potential on a connected region is unique up to one additive constant. Adding a constant changes function values but leaves every partial derivative unchanged.

Verification If $g=f+K$, then $\nabla g=\nabla f+\nabla K=\nabla f$, confirming that every additive constant really does produce another potential.