Conservative vector fields — potential functions: Exercises

4.4.26 · D4 · Maths › Multivariable Calculus › Conservative vector fields — potential functions

Intuition Is page ko kaise use karein

Yeh ek self-test hai. Ek problem padho, solution chhupao, aur pehle khud paper par try karo. Tabhi collapsible [!recall]- kholo check karne ke liye. Problems paanch rungs chadhti hain:

L1 Recognition — kya tum ek conservative field pehchaan sakte ho?
L2 Application — kya tum $f$ dhundhne ki recipe run kar sakte ho?
L3 Analysis — kya tum theory ko use kar sakte ho (path-independence, loops)?
L4 Synthesis — kya tum ideas combine kar sakte ho (holes, curl, 3D)?
L5 Mastery — kya tum edge par reason kar sakte ho jahan naive rules toot jaate hain?

Yahan use kiya gaya har symbol parent note Conservative Vector Fields — Potential Functions mein zero se build kiya gaya hai. Agar koi notation unfamiliar lage, woh note use zero se construct karta hai.

Recall Quick reference (woh teen tools jo tum baar baar reuse karoge)

1. 2D test. $F = (P, Q)$ ke liye: conservative $⟺ P_{y} = Q_{x}$ (ek aisi domain par jo holes-free ho). 2. Recipe. $P$ ko $x$ mein integrate karo taaki $f = \int P d x + g (y)$ mile; phir $f_{y} = Q$ force karo $g$ solve karne ke liye. Dekho Gradient and directional derivative. 3. Shortcut. Conservative field ke liye, $\int_{C} F \cdot d r = f (B) - f (A)$ — yeh Fundamental Theorem of Line Integrals hai, aur yeh Line integrals of vector fields ko trivial bana deta hai jab tumhare paas $f$ ho.

Intuition "3D curl test"

$P_{y} = Q_{x}, P_{z} = R_{x}, Q_{z} = R_{y}$ asal mein kya hai $F = (P, Q, R)$ ka curl ek vector hai jo cross-slopes ke differences se bana hai: $curl F = (R_{y} - Q_{z}, P_{z} - R_{x}, Q_{x} - P_{y}) .$ " $curl F = 0$ " kehna isliye simply yeh kehna hai ki woh saare teen differences zero hain, yaani $Q_{z} = R_{y}$ , $P_{z} = R_{x}$ , $Q_{x} = P_{y}$ — bilkul wahi teen equalities jo hum check karte hain. Yeh particular pairings kyun? Agar $F = \nabla f$ , toh $P = f_{x}, Q = f_{y}, R = f_{z}$ , isliye $P_{y} = f_{x y}$ aur $Q_{x} = f_{y x}$ — aur Clairaut's theorem (equality of mixed partials) se dono mixed partials equal hain. Teen equalities mein se har ek Clairaut ka ek instance hai, ek har pair of variables ke liye $(x, y), (x, z), (y, z)$ . Poori kahani ke liye Curl of a vector field dekho; yahan tumhe sirf teen equalities ki zaroorat hai.

Level 1 — Recognition

Yahan $P$ pehla component hai $F = (P, Q)$ ka aur $Q$ doosra. Symbol $P_{y}$ ka matlab hai " $P$ ko $y$ ke saath differentiate karo, $x$ ko fixed rakhte hue." Neeche figure dekho: yeh dono behaviours side by side dikhata hai — left par, ek conservative field jiske arrows seedha ek valley se bahar point karte hain (ek sachcha gradient); right par, ek swirling field nonzero curl ke saath jo gradient nahin hai. Tumhari aankh dono shapes seekhti hai, lekin test $P_{y}$ vs $Q_{x}$ final word hai.

Figure — Conservative vector fields — potential functions

Recall Solution 1.1

Ek hi tool: $P_{y}$ aur $Q_{x}$ compute karo aur compare karo. (a) $P = 2 x \Rightarrow P_{y} = 0$ . $Q = 2 y \Rightarrow Q_{x} = 0$ . Equal $\Rightarrow$ conservative. (Yeh $\nabla (x^{2} + y^{2})$ hai — figure ke left wala perfect bowl.) (b) $P = - y \Rightarrow P_{y} = - 1$ . $Q = x \Rightarrow Q_{x} = 1$ . Equal nahin $\Rightarrow$ not conservative. Yeh figure ke right wala swirl hai. (c) $P = 3 x^{2} y \Rightarrow P_{y} = 3 x^{2}$ . $Q = x^{3} \Rightarrow Q_{x} = 3 x^{2}$ . Equal $\Rightarrow$ conservative. (Yeh $\nabla (x^{3} y)$ hai.)

Recall Solution 1.2

False. Equivalence require karti hai ki loop integral har closed loop ke liye vanish ho, sirf ek lucky loop ke liye nahin. Ek swirling field phir bhi zero de sakta hai ek aise loop ke around jo symmetric ho. Ek zero coincidence hai; sab zeros theorem hai.

Level 2 — Application

Ab hum potential $f$ banate hain. Recipe yaad karo: $P$ ko $x$ mein integrate karo, ek unknown $g (y)$ add karo ( $x$ mein constant lekin possibly $y$ par depend karta hai), phir $y$ mein differentiate karo aur $Q$ se match karo.

Recall Solution 2.1

Test: $P_{y} = 2 x$ , $Q_{x} = 2 x$ . Equal ✓. $P$ ko $x$ mein integrate karo: kyun? $P$ supposed to be $f_{x}$ hai, toh $x$ -derivative ko undo karna $f$ rebuild karta hai — kuch bhi flat-in- $x$ tak. $f = \int 2 x y d x = x^{2} y + g (y) .$ Differentiate karo, $Q$ se match karo: kyun? Rebuilt $f$ ko bhi $f_{y} = Q$ hona chahiye; woh doosri slope equation leftover hill $g (y)$ pin down karti hai. $f_{y} = x^{2} + g^{'} (y) =! x^{2} + 3 y^{2} \Rightarrow g^{'} (y) = 3 y^{2} \Rightarrow g (y) = y^{3} + C .$ 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 ✓. $P$ ko $x$ mein integrate karo: kyun? us height ko recover karo jiska $x$ -slope $P$ hai. $f = \int e^{x} cos y d x = e^{x} cos y + g (y) .$ $Q$ se match karo: kyun? us height ke $y$ -slope ko given $Q$ se agree karao. $f_{y} = - e^{x} sin y + g^{'} (y) =! - e^{x} sin y \Rightarrow g^{'} (y) = 0 \Rightarrow g = C .$ Answer: $f = e^{x} cos y + C$ .

Recall Solution 2.3

Test (teen curl equalities — upar definition box dekho): $P_{y} = 2 x = Q_{x}$ ✓; $P_{z} = 2 z = R_{x}$ ✓; $Q_{z} = 0 = R_{y}$ ✓. Teeno differences vanish karte hain, toh $curl F = 0$ . $P$ ko $x$ mein integrate karo: kyun? $f$ ko uske $x$ -slope se rebuild karo; 3D mein flat-in- $x$ leftover dono $y$ aur $z$ par depend kar sakta hai. $f = \int (2 x y + z^{2}) d x = x^{2} y + x z^{2} + g (y, z) .$ $Q$ se match karo: kyun? $f$ ka $y$ -slope $Q$ equal hona chahiye, jo fix karta hai ki $g$ $y$ par kaise depend karta hai. $f_{y} = x^{2} + g_{y} =! x^{2} \Rightarrow g_{y} = 0 \Rightarrow g = h (z) .$ $R$ se match karo: kyun? last slope, $f_{z} = R$ , last $z$ -only piece ko kill karta hai. $f_{z} = 2 x z + h^{'} (z) =! 2 x z \Rightarrow h^{'} (z) = 0 \Rightarrow h = C .$ Answer: $f = x^{2} y + x z^{2} + C$ .

Level 3 — Analysis

Yahan hum conservativeness ko use karte hain line integral shortcut karne ke liye. Poora point: agar $F = \nabla f$ , toh path bhool jao aur sirf endpoint values subtract karo.

Recall Solution 3.1

Humare paas pehle se $f = x^{2} y + y^{3}$ hai. Yeh shortcut kyun kaam karta hai: field ek gradient hai, toh work sirf dono ends ke beech climbed height hai. $f (B) = f (2, 1) = 4 \cdot 1 + 1 = 5$ . $f (A) = f (1, 0) = 0 + 0 = 0$ . $\int_{C} F \cdot d r = f (B) - f (A) = 5 - 0 = 5$ . Koi parametrization nahin chahiye — yeh path-independence ka payoff hai.

Recall Solution 3.2

Test: $P_{y} = cos x + cos y$ ; $Q_{x} = cos x + cos y$ . Equal ✓. $P$ ko $x$ mein integrate karo: kyun? $x$ -slope ko undo karo height $f$ rebuild karne ke liye. $f = \int (y cos x + sin y) d x = y sin x + x sin y + g (y) .$ $Q$ se match karo: kyun? $f_{y} = Q$ force karo $y$ -only leftover fix karne ke liye. $f_{y} = sin x + x cos y + g^{'} (y) =! sin x + x cos y \Rightarrow g^{'} (y) = 0.$ Toh $f = y sin x + x sin y$ . 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

Field conservative hai, toh kisi bhi closed loop ke around integral $0$ hai (start point $=$ end point $\Rightarrow f (B) - f (A) = 0$ ). Kyun: ek closed loop same height par wapas aata hai, toh net climb zero hai. $0$ .

Level 4 — Synthesis

Ab hum test, recipe, aur domain caveat combine karte hain. Domain ko dhyan se dekho.

Recall Solution 4.1

(a) Maano $r^{2} = x^{2} + y^{2}$ . $P_{y} = \frac{\partial}{\partial y} \frac{- y}{x ^{2} + y ^{2}} = \frac{- ( x ^{2} + y ^{2} ) - ( - y ) ( 2 y )}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}}$ . $Q_{x} = \frac{\partial}{\partial x} \frac{x}{x ^{2} + y ^{2}} = \frac{( x ^{2} + y ^{2} ) - x ( 2 x )}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}}$ . Equal ✓ har jagah origin ko chhod kar (jahan $F$ undefined hai). (b) Parametrize karo $x = cos t, y = sin t$ , $t \in [0, 2 π]$ . Tab $r^{2} = 1$ , toh $F = (- sin t, cos t)$ aur $d r = (- sin t, cos t) d t$ . $F \cdot d r = (sin^{2} t + cos^{2} t) d t = d t$ . Integral $= \int_{0}^{2 π} d t = 2 π$ . (c) Koi contradiction nahin. Test sirf simply connected (hole-free) domain par conservativeness guarantee karta hai. Yahan origin ek puncture hai, toh domain mein ek hole hai, loop use encircle karta hai, aur integral $2 π \neq = 0$ hai. Field $P_{y} = Q_{x}$ ke bawajood globally conservative nahin hai.

Recall Solution 4.2

Test force karo: $P_{y} = 2 a y$ , $Q_{x} = 6 y$ . Conservative chahiye $2 a y = 6 y$ har $y$ ke liye $\Rightarrow a = 3$ . Toh $F = (3 x^{2} + 3 y^{2}, 6 x y)$ . $P$ integrate karo: kyun? height recover karne ke liye $x$ -slope undo karo. $f = \int (3 x^{2} + 3 y^{2}) d x = x^{3} + 3 x y^{2} + g (y) .$ $Q$ se match karo: kyun? leftover $g$ fix karne ke liye $f_{y} = Q$ force karo. $f_{y} = 6 x y + g^{'} (y) =! 6 x y \Rightarrow g^{'} (y) = 0.$ Answer: $a = 3$ , $f = x^{3} + 3 x y^{2} + C$ .

Recall Solution 4.3

Test (teen curl equalities): $P_{y} = 2 x z = Q_{x}$ ✓; $P_{z} = 2 x y = R_{x}$ ✓; $Q_{z} = x^{2} = R_{y}$ ✓. Teeno differences vanish karte hain, toh $curl F = 0$ . $P$ ko $x$ mein integrate karo: kyun? $f$ ko uske $x$ -slope se rebuild karo; flat-in- $x$ leftover $y, z$ par depend karta hai. $f = \int 2 x y z d x = x^{2} y z + g (y, z) .$ $Q$ se match karo: kyun? $y$ -slope $Q$ equal hona chahiye, $g$ ki $y$ -dependence fix karta hai. $f_{y} = x^{2} z + g_{y} =! x^{2} z + 2 y \Rightarrow g_{y} = 2 y \Rightarrow g = y^{2} + h (z) .$ $R$ se match karo: kyun? $z$ -slope $R$ equal hona chahiye, last $z$ -only piece kill karta hai. $f_{z} = x^{2} y + h^{'} (z) =! x^{2} y \Rightarrow h^{'} (z) = 0.$ Answer: $f = x^{2} y z + y^{2} + C$ .

Level 5 — Mastery

Edge cases, reasoning, aur physics se stitching.

Recall Solution 5.1

Round trip: start $=$ end, toh $\oint F \cdot d r = f (3, 1) - f (3, 1) = 0$ . Path irrelevance guarantee karta hai ki dono legs exactly cancel karenge. Answer $0$ . (Yeh Conservation of energy in physics hai — ek loop se koi free energy nahin.) 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: $P_{y} = 0 = Q_{x}$ ✓, toh conservative. Koi bhi constant $f = C$ kaam karta hai, kyunki $\nabla C = (0, 0)$ . Potential unique hai sirf us additive constant tak — jaise hamesha hota hai. Constant field $F = (a, b)$ : $P_{y} = 0 = Q_{x}$ ✓, conservative. Integrate karo: $f = \int a d x = a x + g (y)$ ; match karo $f_{y} = g^{'} (y) = b \Rightarrow g = b y$ . Toh $f = a x + b y + C$ . Geometrically yeh ek tilted flat plane hai — iska gradient constant "steepest ascent" direction $(a, b)$ hai.

Recall Solution 5.3

(a) Conservative field. $F = (2 x y, x^{2})$ : yahan $P = 2 x y, Q = x^{2}$ , toh $P_{y} = 2 x$ aur $Q_{x} = 2 x$ . Equal ✓ $\Rightarrow$ conservative. Potential recover karo: $P$ ko $x$ mein integrate karo, $f = \int 2 x y d x = x^{2} y + g (y)$ ; phir $f_{y} = x^{2} + g^{'} (y) =! x^{2} \Rightarrow g^{'} (y) = 0$ , toh $f = x^{2} y$ . Kyunki $F = \nabla f$ , Fundamental Theorem force karta hai $\oint_{C} F \cdot d r = 0$ har closed $C$ ke liye — unit circle bhi sameta, bina kisi computation ke. (b) Impostor. $G = (0, x^{2})$ : yahan $P = 0, Q = x^{2}$ , toh $P_{y} = 0$ lekin $Q_{x} = 2 x$ . Kyunki $0 \neq = 2 x$ (sirf line $x = 0$ par chhod kar), test fail hota hai $\Rightarrow$ $G$ not conservative hai. Ab unit circle $x = cos t, y = sin t$ , $t \in [0, 2 π]$ ke around $G$ integrate karo, jahan $d r = (- sin t, cos t) d t$ : $G \cdot d r = 0 \cdot (- sin t) + cos^{2} t \cdot cos t = cos^{3} t .$ $\oint_{x^{2} + y^{2} = 1} G \cdot d r = \int_{0}^{2 π} cos^{3} t d t = 0.$ Toh $G$ is ek loop par zero deta hai lekin genuinely non-conservative hai. (c) Moral. Ek single vanishing loop kuch prove nahin karta: $G$ is loop mein symmetry se pass ho gaya ( $cos^{3} t$ ek full turn mein zero integrate hota hai) jabki real criterion $P_{y} = Q_{x}$ fail kiya. Sirf " $\oint = 0$ har loop ke liye" — equivalently hole-free domain par $P_{y} = Q_{x}$ — conservativeness certify karta hai. Exactly isliye parent note ki equivalence list har closed loop par insist karti hai, aur yeh Exercise 1.2 ko mirror karta hai.

Recall Solution 5.4

Physics convention: force potential energy ka minus gradient hai, $F = - \nabla U$ , toh $\nabla U = (k x, k y)$ . $(k x, k y)$ par test: $P_{y} = 0 = Q_{x}$ ✓, conservative. Integrate karo: kyun? $U$ ka $x$ -slope undo karo. $U = \int k x d x = \frac{1}{2} k x^{2} + g (y)$ ; match karo $U_{y} = g^{'} (y) = k y \Rightarrow g = \frac{1}{2} k y^{2}$ . Toh $U = \frac{1}{2} k (x^{2} + y^{2})$ — familiar spring energy $\frac{1}{2} k r^{2}$ . $F$ dwara kiya gaya work $O = (0, 0)$ se $B = (3, 4)$ tak. Minus kyun: kyunki $F = - \nabla U$ , iska Fundamental-Theorem value $- (U (B) - U (O))$ hai. $W = \int_{C} F \cdot d r = - (U (B) - U (O)) = - \frac{1}{2} k (9 + 16 - 0) = - \frac{25}{2} k .$ Negative: spring stretching ka oppose karta hai, jaise hona chahiye. Yeh Conservation of energy in physics action mein hai.

Recall Master checklist (har problem se pehle yeh kaho)

Test first ::: pehle $P_{y}$ vs $Q_{x}$ compute karo (ya 3D mein teen curl equalities) kuch bhi karne se pehle. Domain check ::: koi punctures hain? agar loop hole wrap karta hai, toh test jhooth bol sakta hai. Recipe ::: $P$ ko $x$ mein integrate karo, $+ g (y)$ carry karo, $f_{y} = Q$ match karo. Shortcut ::: conservative $\Rightarrow$ line integral $= f (B) - f (A)$ ; loop $= 0$ . Physics sign ::: $F = - \nabla U$ , nahin $+ \nabla U$ .