Conservative vector fields — potential functions: Worked examples

Before we start, one reminder of the three tools we lean on, and why each:

The gradient $\nabla f = (f_{x}, f_{y}, f_{z})$ — the arrow field built from a hill $f$ (see Gradient and directional derivative). We reverse it to find the hill.
The cross-partial / curl test — a fast yes/no filter so we don't hunt for a hill that doesn't exist (see Curl of a vector field, Clairaut's theorem (equality of mixed partials)).
The Fundamental Theorem of Line Integrals (FTLI) — turns work integrals into subtraction $f (B) - f (A)$ (see Line integrals of vector fields).

The scenario matrix

Every conservative-field problem falls into one of these case classes. Each row is a "cell" the exam can hit; the last column says which example below covers it.

Cell	Case class	What makes it tricky	Example
C1	Plain 2D, both signs positive	baseline recipe	Ex 1
C2	2D with negative / mixed-sign components	sign-tracking in $g (y)$	Ex 2
C3	Not conservative (curl $\neq = 0$ )	recipe must fail loudly	Ex 3
C4	Degenerate / punctured domain (hole)	test passes but field is NOT conservative	Ex 4
C5	FTLI to compute work over a path	endpoints only, ignore the curve	Ex 5
C6	3D field, integrate three times	leftover depends on 2 vars	Ex 6
C7	Zero field / constant field (limiting/degenerate input)	boundary of the theory	Ex 7
C8	Real-world word problem (physics energy)	translate force $\to$ potential	Ex 8
C9	Exam twist: closed loop + "which path?" trap	loop $= 0$ vs. hole	Ex 9

We now hit each cell.

Ex 1 — Cell C1: the clean baseline

Worked example Statement

$F = (2 x + y, x + 6 y)$ . Find a potential $f$ , or show none exists.

Forecast: Both parts look "mixed" — the $y$ appears in $P$ and the $x$ appears in $Q$ symmetrically. Guess: conservative. Guess the hill before reading on.

Test the curl. $P = 2 x + y \Rightarrow P_{y} = 1$ . $Q = x + 6 y \Rightarrow Q_{x} = 1$ . Equal ✓. Why this step? $P_{y} = Q_{x}$ is the 2D curl test; if it fails we stop immediately — no hill exists.
Integrate $P$ in $x$ . $f = \int (2 x + y) d x = x^{2} + x y + g (y)$ . Why this step? $f_{x}$ must equal $P$ ; integrating undoes the $x$ -derivative, but anything constant in $x$ (a function $g (y)$ ) is invisible to $\partial / \partial x$ , so we carry it.
Differentiate in $y$ , match $Q$ . $f_{y} = x + g^{'} (y) =! x + 6 y \Rightarrow g^{'} (y) = 6 y \Rightarrow g (y) = 3 y^{2} + C$ . Why this step? The still-unknown $g$ is pinned by demanding $f_{y} = Q$ .

Answer: $f (x, y) = x^{2} + x y + 3 y^{2} + C .$

Verify: $\nabla f = (2 x + y, x + 6 y) = F$ ✓. Forecast confirmed.

Ex 2 — Cell C2: mixed signs everywhere

Ex 3 — Cell C3: the recipe must fail

Ex 4 — Cell C4: the punctured-domain trap (degenerate domain)

Figure — Conservative vector fields — potential functions

Compute $P_{y}$ . With $P = \frac{- y}{x ^{2} + y ^{2}}$ , quotient rule gives $P_{y} = \frac{- ( x ^{2} + y ^{2} ) - ( - y ) ( 2 y )}{( x ^{2} + y ^{2} ) ^{2}} = \frac{- x ^{2} - y ^{2} + 2 y ^{2}}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}} .$

Compute $Q_{x}$ . With $Q = \frac{x}{x ^{2} + y ^{2}}$ , $Q_{x} = \frac{( x ^{2} + y ^{2} ) - x ( 2 x )}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}} .$ Why these steps? We must see with our own hands that $P_{y} = Q_{x}$ everywhere the field is defined — so the naive test says "conservative."

Take the loop integral around the unit circle $r (t) = (cos t, sin t)$ , $t \in [0, 2 π]$ . Here $x^{2} + y^{2} = 1$ , so $F = (- sin t, cos t)$ and $r^{'} = (- sin t, cos t)$ . $\oint F \cdot d r = \int_{0}^{2 π} (sin^{2} t + cos^{2} t) d t = \int_{0}^{2 π} 1 d t = 2 π \neq = 0.$ Why this step? A conservative field has zero loop integral (FTLI, $A = B$ ). A nonzero loop is a hard disproof.

Answer: Not conservative on the punctured plane — the hole at the origin breaks the "simply connected" requirement (look at the red loop enclosing the missing point in the figure). See Green's theorem for why the enclosed hole matters.

Verify: $P_{y} = Q_{x} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}}$ symbolically, yet $\oint = 2 π$ .

Ex 5 — Cell C5: FTLI to compute work along a path

Worked example Statement

Using $F = (2 x + y, x + 6 y)$ from Ex 1, compute $\int_{C} F \cdot d r$ from $A = (- 1, 0)$ to $B = (2, - 1)$ along any curve.

Forecast: Since Ex 1 proved it conservative, the wiggly path is irrelevant — only two numbers matter. Guess whether the work is positive or negative first.

Recall the hill. $f (x, y) = x^{2} + x y + 3 y^{2}$ (drop $C$ ; it cancels in subtraction). Why this step? FTLI says work $= f (B) - f (A)$ , so we need $f$ at the two endpoints only.
Evaluate at $B = (2, - 1)$ . $f (2, - 1) = 4 + (2) (- 1) + 3 (1) = 4 - 2 + 3 = 5.$
Evaluate at $A = (- 1, 0)$ . $f (- 1, 0) = 1 + 0 + 0 = 1.$
Subtract. $\int_{C} F \cdot d r = f (B) - f (A) = 5 - 1 = 4.$ Why this step? $f (B) - f (A)$ is the total "downhill/uphill" change; the path in between is invisible.

Answer: $4$ .

Verify: the $+ C$ in $f$ cancels ( $(5 + C) - (1 + C) = 4$ ), so the answer is path- and constant-independent ✓.

Ex 6 — Cell C6: full 3D reconstruction

Worked example Statement

$F = (2 x y + z^{3}, x^{2}, 3 x z^{2})$ . Find $f$ in 3D.

Forecast: Three components mean three matching conditions. Guess: conservative, and $f$ has an $x^{2} y$ piece plus an $x z^{3}$ piece.

Curl test (all three pairs).
- $P_{y} = 2 x$ , $Q_{x} = 2 x$ ✓
- $P_{z} = 3 z^{2}$ , $R_{x} = 3 z^{2}$ ✓
- $Q_{z} = 0$ , $R_{y} = 0$ ✓ Why this step? In 3D, $curl F = 0$ needs all three cross-partial equalities (see Curl of a vector field).
Integrate $P$ in $x$ . $f = \int (2 x y + z^{3}) d x = x^{2} y + x z^{3} + g (y, z)$ . Why this step? Now the invisible leftover depends on both remaining variables $y$ and $z$ .
Match $Q$ . $f_{y} = x^{2} + g_{y} =! x^{2} \Rightarrow g_{y} = 0$ , so $g = h (z)$ (no $y$ ).
Match $R$ . $f_{z} = 3 x z^{2} + h^{'} (z) =! 3 x z^{2} \Rightarrow h^{'} (z) = 0$ , so $h = C$ . Why these steps? Each further derivative peels away one variable's dependence from the leftover until only a constant remains.

Answer: $f = x^{2} y + x z^{3} + C .$

Verify: $\nabla f = (2 x y + z^{3}, x^{2}, 3 x z^{2}) = F$ ✓.

Ex 7 — Cell C7: the degenerate limiting inputs

Ex 8 — Cell C8: real-world energy problem

Worked example Statement

A particle feels the force $F = (- k x, - k y)$ (a 2D spring pulling toward the origin, $k > 0$ , units N/m·m). It moves from $A = (3, 0)$ to $B = (0, 4)$ . How much work does $F$ do? Take $k = 2$ .

Forecast: The particle starts and ends at the same distance $5$ … wait, $∣ A ∣ = 3$ , $∣ B ∣ = 4$ . It moves to a point closer/farther? Guess the sign of the work first (does the spring help or resist?).

Test / recognise the potential. $P = - k x, Q = - k y$ : $P_{y} = 0 = Q_{x}$ ✓ conservative. Integrate: $f = \int - k x d x = - \frac{k}{2} x^{2} + g (y)$ ; $f_{y} = g^{'} (y) =! - k y \Rightarrow g = - \frac{k}{2} y^{2}$ . So $f = - \frac{k}{2} (x^{2} + y^{2})$ . Why this step? Spring force is minus the gradient of energy; $f$ here is the negative of the usual potential energy $U = \frac{k}{2} r^{2}$ . This links to Conservation of energy in physics.
Apply FTLI. Work $= f (B) - f (A) = - \frac{k}{2} (0^{2} + 4^{2}) - (- \frac{k}{2} (3^{2} + 0^{2})) = - \frac{k}{2} (16) + \frac{k}{2} (9) = - \frac{k}{2} (7)$ . Why this step? Only the endpoints' distances from origin matter — the path is irrelevant for this conservative spring.
Plug $k = 2$ . Work $= - \frac{2}{2} (7) = - 7$ joules. Why this step? Negative work: the particle ended farther from the origin ( $4 > 3$ ), so the inward spring resisted the motion. Sign matches intuition.

Answer: $- 7$ J.

Verify: units: $[k]$ ·length $^{2}$ = (N/m)·m $^{2}$ = N·m = J ✓. Sign negative because $∣ B ∣ > ∣ A ∣$ (moved against the pull) ✓.

Ex 9 — Cell C9: the exam-style closed-loop twist

Worked example Statement

Let $F = (2 x y + 3, x^{2} - 4 y)$ (conservative, from the parent note). A student computes $\oint_{C} F \cdot d r$ around the boundary of the triangle $(0, 0) \to (1, 0) \to (1, 2) \to (0, 0)$ . They spend 20 minutes parametrising three edges. What's the shortcut answer, and where's the trap?

Forecast: Closed loop + conservative. Guess the answer instantly before any integral.

Check conservativeness. $P_{y} = 2 x = Q_{x}$ ✓, domain all of $R^{2}$ (simply connected, no hole). Why this step? The C4 trap only bites when a hole is enclosed; here there is none, so the shortcut is legal.
Use loop $= 0$ . For a conservative field on a hole-free domain, $\oint_{C} F \cdot d r = f (B) - f (A)$ with $A = B$ , giving $0$ . Why this step? FTLI with equal endpoints — no parametrisation needed.

Answer: $\oint_{C} F \cdot d r = 0.$

Verify: direct sum of the three edge integrals of $F = (2 x y + 3, x^{2} - 4 y)$ around the triangle totals $0$ (checked numerically).

Recall

Recall Which cell am I in?

Given a new field, what is your very first move? ::: Run the cross-partial / curl test — it filters out non-conservative fields before you waste time hunting for $f$ . The test passes but the domain has a hole around which the loop winds — conclusion? ::: You may NOT declare it conservative; check a loop integral (Cell C4). Closed loop, conservative field, no enclosed hole — value? ::: Exactly $0$ (Cell C9). Constant nonzero force field — is it conservative? ::: Yes; its potential is a tilted plane $a x + b y$ (Cell C7).