Conservative vector fields — potential functions: Exercises

Recall Quick reference (the three tools you will reuse)

1. The 2D test. For $F = (P, Q)$ : conservative $⟺ P_{y} = Q_{x}$ (on a domain with no holes). 2. The recipe. Integrate $P$ in $x$ to get $f = \int P d x + g (y)$ ; then force $f_{y} = Q$ to solve for $g$ . See Gradient and directional derivative. 3. The shortcut. For a conservative field, $\int_{C} F \cdot d r = f (B) - f (A)$ — this is the Fundamental Theorem of Line Integrals, and it makes Line integrals of vector fields trivial once you have $f$ .

Level 1 — Recognition

Here $P$ is the first component of $F = (P, Q)$ and $Q$ the second. The symbol $P_{y}$ means "differentiate $P$ with respect to $y$ , holding $x$ fixed." Look at the figure below: it shows both behaviours side by side — on the left, a conservative field whose arrows point straight out of a valley (a true gradient); on the right, a swirling field with nonzero curl that is not a gradient. Your eye learns the two shapes, but the test $P_{y}$ vs $Q_{x}$ is the final word.

Figure — Conservative vector fields — potential functions

Recall Solution 1.1

The only tool: compute $P_{y}$ and $Q_{x}$ and compare. (a) $P = 2 x \Rightarrow P_{y} = 0$ . $Q = 2 y \Rightarrow Q_{x} = 0$ . Equal $\Rightarrow$ conservative. (It is $\nabla (x^{2} + y^{2})$ — the perfect bowl on the left of the figure.) (b) $P = - y \Rightarrow P_{y} = - 1$ . $Q = x \Rightarrow Q_{x} = 1$ . Not equal $\Rightarrow$ not conservative. This is the swirl on the right of the figure. (c) $P = 3 x^{2} y \Rightarrow P_{y} = 3 x^{2}$ . $Q = x^{3} \Rightarrow Q_{x} = 3 x^{2}$ . Equal $\Rightarrow$ conservative. (It is $\nabla (x^{3} y)$ .)

Recall Solution 1.2

False. The equivalence requires the loop integral to vanish for every closed loop, not just one lucky one. A swirling field can still give zero around a loop that happens to be symmetric. One zero is a coincidence; all zeros is the theorem.

Level 2 — Application

Now we build the potential $f$ . Recall the recipe: integrate $P$ in $x$ , add an unknown $g (y)$ (constant in $x$ but possibly depending on $y$ ), then differentiate in $y$ and match $Q$ .

Recall Solution 2.1

Test: $P_{y} = 2 x$ , $Q_{x} = 2 x$ . Equal ✓. Integrate $P$ in $x$ : why? $P$ is supposed to be $f_{x}$ , so undoing the $x$ -derivative rebuilds $f$ — up to anything that was flat in $x$ . $f = \int 2 x y d x = x^{2} y + g (y) .$ Differentiate, match $Q$ : why? The rebuilt $f$ must also have $f_{y} = Q$ ; that second slope equation pins down the leftover hill $g (y)$ . $f_{y} = x^{2} + g^{'} (y) =! x^{2} + 3 y^{2} \Rightarrow g^{'} (y) = 3 y^{2} \Rightarrow g (y) = y^{3} + C .$ Answer: $f (x, y) = x^{2} y + y^{3} + C$ . Check: $\nabla f = (2 x y, x^{2} + 3 y^{2}) = F$ ✓.

Recall Solution 2.2

Test: $P_{y} = - e^{x} sin y$ , $Q_{x} = - e^{x} sin y$ . Equal ✓. Integrate $P$ in $x$ : why? recover the height whose $x$ -slope is $P$ . $f = \int e^{x} cos y d x = e^{x} cos y + g (y) .$ Match $Q$ : why? force the $y$ -slope of that height to agree with the given $Q$ . $f_{y} = - e^{x} sin y + g^{'} (y) =! - e^{x} sin y \Rightarrow g^{'} (y) = 0 \Rightarrow g = C .$ Answer: $f = e^{x} cos y + C$ .

Recall Solution 2.3

Test (the three curl equalities — see the definition box above): $P_{y} = 2 x = Q_{x}$ ✓; $P_{z} = 2 z = R_{x}$ ✓; $Q_{z} = 0 = R_{y}$ ✓. All three differences vanish, so $curl F = 0$ . Integrate $P$ in $x$ : why? rebuild $f$ from its $x$ -slope; in 3D the flat-in- $x$ leftover can depend on both $y$ and $z$ . $f = \int (2 x y + z^{2}) d x = x^{2} y + x z^{2} + g (y, z) .$ Match $Q$ : why? the $y$ -slope of $f$ must be $Q$ , which fixes how $g$ depends on $y$ . $f_{y} = x^{2} + g_{y} =! x^{2} \Rightarrow g_{y} = 0 \Rightarrow g = h (z) .$ Match $R$ : why? the last slope, $f_{z} = R$ , kills the final $z$ -only piece. $f_{z} = 2 x z + h^{'} (z) =! 2 x z \Rightarrow h^{'} (z) = 0 \Rightarrow h = C .$ Answer: $f = x^{2} y + x z^{2} + C$ .

Level 3 — Analysis

Here we use conservativeness to shortcut a line integral. The whole point: if $F = \nabla f$ , forget the path and just subtract endpoint values.

Recall Solution 3.1

We already have $f = x^{2} y + y^{3}$ . Why this shortcut works: the field is a gradient, so the work is just the height climbed between the two ends. $f (B) = f (2, 1) = 4 \cdot 1 + 1 = 5$ . $f (A) = f (1, 0) = 0 + 0 = 0$ . $\int_{C} F \cdot d r = f (B) - f (A) = 5 - 0 = 5$ . No parametrization needed — that is the payoff of path-independence.

Recall Solution 3.2

Test: $P_{y} = cos x + cos y$ ; $Q_{x} = cos x + cos y$ . Equal ✓. Integrate $P$ in $x$ : why? undo the $x$ -slope to rebuild the height $f$ . $f = \int (y cos x + sin y) d x = y sin x + x sin y + g (y) .$ Match $Q$ : why? force $f_{y} = Q$ to fix the $y$ -only leftover. $f_{y} = sin x + x cos y + g^{'} (y) =! sin x + x cos y \Rightarrow g^{'} (y) = 0.$ So $f = y sin x + x sin y$ . Endpoints: $f (π /2, π /2) = \frac{π}{2} sin \frac{π}{2} + \frac{π}{2} sin \frac{π}{2} = \frac{π}{2} + \frac{π}{2} = π$ . $f (0, 0) = 0$ . $\int_{C} F \cdot d r = π - 0 = π$ .

Recall Solution 3.3

The field is conservative, so around any closed loop the integral is $0$ (start point $=$ end point $\Rightarrow f (B) - f (A) = 0$ ). Why: a closed loop returns to the same height, so the net climb is zero. $0$ .

Level 4 — Synthesis

Now we combine the test, the recipe, and the domain caveat. Watch the domain carefully.

Recall Solution 4.1

(a) Let $r^{2} = x^{2} + y^{2}$ . $P_{y} = \frac{\partial}{\partial y} \frac{- y}{x ^{2} + y ^{2}} = \frac{- ( x ^{2} + y ^{2} ) - ( - y ) ( 2 y )}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}}$ . $Q_{x} = \frac{\partial}{\partial x} \frac{x}{x ^{2} + y ^{2}} = \frac{( x ^{2} + y ^{2} ) - x ( 2 x )}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}}$ . Equal ✓ everywhere except the origin (where $F$ is undefined). (b) Parametrize $x = cos t, y = sin t$ , $t \in [0, 2 π]$ . Then $r^{2} = 1$ , so $F = (- sin t, cos t)$ and $d r = (- sin t, cos t) d t$ . $F \cdot d r = (sin^{2} t + cos^{2} t) d t = d t$ . Integral $= \int_{0}^{2 π} d t = 2 π$ . (c) No contradiction. The test only guarantees conservativeness on a simply connected (hole-free) domain. Here the origin is a puncture, so the domain has a hole, the loop encircles it, and the integral is $2 π \neq = 0$ . The field is not globally conservative despite $P_{y} = Q_{x}$ .

Recall Solution 4.2

Force the test: $P_{y} = 2 a y$ , $Q_{x} = 6 y$ . Conservative needs $2 a y = 6 y$ for all $y \Rightarrow a = 3$ . So $F = (3 x^{2} + 3 y^{2}, 6 x y)$ . Integrate $P$ : why? undo the $x$ -slope to recover the height. $f = \int (3 x^{2} + 3 y^{2}) d x = x^{3} + 3 x y^{2} + g (y) .$ Match $Q$ : why? force $f_{y} = Q$ to fix the leftover $g$ . $f_{y} = 6 x y + g^{'} (y) =! 6 x y \Rightarrow g^{'} (y) = 0.$ Answer: $a = 3$ , $f = x^{3} + 3 x y^{2} + C$ .

Recall Solution 4.3

Test (the three curl equalities): $P_{y} = 2 x z = Q_{x}$ ✓; $P_{z} = 2 x y = R_{x}$ ✓; $Q_{z} = x^{2} = R_{y}$ ✓. All three differences vanish, so $curl F = 0$ . Integrate $P$ in $x$ : why? rebuild $f$ from its $x$ -slope; the flat-in- $x$ leftover depends on $y, z$ . $f = \int 2 x y z d x = x^{2} y z + g (y, z) .$ Match $Q$ : why? the $y$ -slope must equal $Q$ , fixing $g$ 's $y$ -dependence. $f_{y} = x^{2} z + g_{y} =! x^{2} z + 2 y \Rightarrow g_{y} = 2 y \Rightarrow g = y^{2} + h (z) .$ Match $R$ : why? the $z$ -slope must equal $R$ , killing the last $z$ -only piece. $f_{z} = x^{2} y + h^{'} (z) =! x^{2} y \Rightarrow h^{'} (z) = 0.$ Answer: $f = x^{2} y z + y^{2} + C$ .

Level 5 — Mastery

Edge cases, reasoning, and stitching to physics.

Recall Solution 5.1

Round trip: start $=$ end, so $\oint F \cdot d r = f (3, 1) - f (3, 1) = 0$ . Path irrelevance guarantees the two legs cancel exactly. Answer $0$ . (This is Conservation of energy in physics — no free energy from a loop.) Outbound leg $(3, 1) \to (5, 4)$ : $f (5, 4) = 25 - 16 = 9$ , $f (3, 1) = 9 - 1 = 8$ , work $= 9 - 8 = 1$ .

Recall Solution 5.2

Zero field: $P_{y} = 0 = Q_{x}$ ✓, so conservative. Any constant $f = C$ works, since $\nabla C = (0, 0)$ . The potential is unique only up to that additive constant — as always. Constant field $F = (a, b)$ : $P_{y} = 0 = Q_{x}$ ✓, conservative. Integrate: $f = \int a d x = a x + g (y)$ ; match $f_{y} = g^{'} (y) = b \Rightarrow g = b y$ . So $f = a x + b y + C$ . Geometrically this is a tilted flat plane — its gradient is the constant "steepest ascent" direction $(a, b)$ .

Recall Solution 5.3

(a) The conservative field. $F = (2 x y, x^{2})$ : here $P = 2 x y, Q = x^{2}$ , so $P_{y} = 2 x$ and $Q_{x} = 2 x$ . Equal ✓ $\Rightarrow$ conservative. Recover the potential: integrate $P$ in $x$ , $f = \int 2 x y d x = x^{2} y + g (y)$ ; then $f_{y} = x^{2} + g^{'} (y) =! x^{2} \Rightarrow g^{'} (y) = 0$ , so $f = x^{2} y$ . Because $F = \nabla f$ , the Fundamental Theorem forces $\oint_{C} F \cdot d r = 0$ for every closed $C$ — the unit circle included, with no computation. (b) The impostor. $G = (0, x^{2})$ : here $P = 0, Q = x^{2}$ , so $P_{y} = 0$ but $Q_{x} = 2 x$ . Since $0 \neq = 2 x$ (except on the line $x = 0$ ), the test fails $\Rightarrow$ $G$ is not conservative. Now integrate $G$ around the unit circle $x = cos t, y = sin t$ , $t \in [0, 2 π]$ , with $d r = (- sin t, cos t) d t$ : $G \cdot d r = 0 \cdot (- sin t) + cos^{2} t \cdot cos t = cos^{3} t .$ $\oint_{x^{2} + y^{2} = 1} G \cdot d r = \int_{0}^{2 π} cos^{3} t d t = 0.$ So $G$ gives zero on this one loop yet is genuinely non-conservative. (c) The moral. A single vanishing loop proves nothing: $G$ passed this loop by symmetry ( $cos^{3} t$ integrates to zero over a full turn) while failing the real criterion $P_{y} = Q_{x}$ . Only " $\oint = 0$ for every loop" — equivalently $P_{y} = Q_{x}$ on a hole-free domain — certifies conservativeness. This is exactly why the equivalence list in the parent note insists on every closed loop, and it mirrors Exercise 1.2.

Recall Solution 5.4

Physics convention: force is minus the gradient of potential energy, $F = - \nabla U$ , so $\nabla U = (k x, k y)$ . Test on $(k x, k y)$ : $P_{y} = 0 = Q_{x}$ ✓, conservative. Integrate: why? undo the $x$ -slope of $U$ . $U = \int k x d x = \frac{1}{2} k x^{2} + g (y)$ ; match $U_{y} = g^{'} (y) = k y \Rightarrow g = \frac{1}{2} k y^{2}$ . So $U = \frac{1}{2} k (x^{2} + y^{2})$ — the familiar spring energy $\frac{1}{2} k r^{2}$ . Work done by $F$ from $O = (0, 0)$ to $B = (3, 4)$ . Why the minus: since $F = - \nabla U$ , its Fundamental-Theorem value is $- (U (B) - U (O))$ . $W = \int_{C} F \cdot d r = - (U (B) - U (O)) = - \frac{1}{2} k (9 + 16 - 0) = - \frac{25}{2} k .$ Negative: the spring opposes stretching, as it must. This is Conservation of energy in physics in action.

Recall Master checklist (say it before every problem)

Test first ::: compute $P_{y}$ vs $Q_{x}$ (or the three curl equalities in 3D) before anything else. Domain check ::: any punctures? if the loop wraps a hole, the test can lie. Recipe ::: integrate $P$ in $x$ , carry $+ g (y)$ , match $f_{y} = Q$ . Shortcut ::: conservative $\Rightarrow$ line integral $= f (B) - f (A)$ ; loop $= 0$ . Physics sign ::: $F = - \nabla U$ , not $+ \nabla U$ .