Conservative vector fields — potential functions: Worked examples

4.4.26 · D3 · Maths › Multivariable Calculus › Conservative vector fields — potential functions

Shuru karne se pehle, teen tools ka ek reminder jinpar hum rely karte hain, aur har ek kyun:

Gradient $\nabla f = (f_{x}, f_{y}, f_{z})$ — ek hill $f$ se bana arrow field (dekho Gradient and directional derivative). Hum ise reverse karte hain hill dhundhne ke liye.
Cross-partial / curl test — ek fast yes/no filter taaki hum aisi hill ke peeche na bhaagein jo exist hi nahi karti (dekho Curl of a vector field, Clairaut's theorem (equality of mixed partials)).
Fundamental Theorem of Line Integrals (FTLI) — work integrals ko subtraction $f (B) - f (A)$ mein convert kar deta hai (dekho Line integrals of vector fields).

Scenario matrix

Har conservative-field problem inn case classes mein se ek mein aata hai. Har row ek "cell" hai jo exam hit kar sakta hai; last column batata hai ki neeche kaun sa example cover karta hai.

Cell	Case class	Kya tricky hai	Example
C1	Plain 2D, dono signs positive	baseline recipe	Ex 1
C2	2D mein negative / mixed-sign components	$g (y)$ mein sign-tracking	Ex 2
C3	Not conservative (curl $\neq = 0$ )	recipe loudly fail karni chahiye	Ex 3
C4	Degenerate / punctured domain (hole)	test pass hota hai but field conservative NAHI hai	Ex 4
C5	FTLI se path par work compute karna	sirf endpoints, curve ignore karo	Ex 5
C6	3D field, teen baar integrate karna	leftover 2 vars par depend karta hai	Ex 6
C7	Zero field / constant field (limiting/degenerate input)	theory ki boundary	Ex 7
C8	Real-world word problem (physics energy)	force $\to$ potential translate karo	Ex 8
C9	Exam twist: closed loop + "which path?" trap	loop $= 0$ vs. hole	Ex 9

Ab hum har cell ko tackle karte hain.

Ex 1 — Cell C1: clean baseline

Worked example Statement

$F = (2 x + y, x + 6 y)$ . Ek potential $f$ dhundho, ya dikhao ki koi exist nahi karta.

Forecast: Dono parts "mixed" lagte hain — $y$ appear karta hai $P$ mein aur $x$ appear karta hai $Q$ mein symmetrically. Guess: conservative. Aage padhne se pehle hill guess karo.

Curl test karo. $P = 2 x + y \Rightarrow P_{y} = 1$ . $Q = x + 6 y \Rightarrow Q_{x} = 1$ . Equal ✓. Yeh step kyun? $P_{y} = Q_{x}$ 2D curl test hai; agar yeh fail ho toh hum immediately ruk jaate hain — koi hill exist nahi karti.
$P$ ko $x$ mein integrate karo. $f = \int (2 x + y) d x = x^{2} + x y + g (y)$ . Yeh step kyun? $f_{x}$ ko $P$ ke equal hona chahiye; integrate karna $x$ -derivative ko undo karta hai, lekin jo cheez $x$ mein constant hai (ek function $g (y)$ ) woh $\partial / \partial x$ ko invisible hai, isliye hum usse carry karte hain.
$y$ mein differentiate karo, $Q$ se match karo. $f_{y} = x + g^{'} (y) =! x + 6 y \Rightarrow g^{'} (y) = 6 y \Rightarrow g (y) = 3 y^{2} + C$ . Yeh step kyun? Ab-bhi-unknown $g$ ko $f_{y} = Q$ demand karke pin kiya jaata hai.

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

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

Ex 2 — Cell C2: har jagah mixed signs

Worked example Statement

$F = (3 x^{2} - 2 y, - 2 x + 4 y^{3})$ . $f$ dhundho.

Forecast: $- 2 y$ aur $- 2 x$ minus sign ke saath ek doosre ko mirror karte hain. Kya negatives test mein cancel ho jaate hain? Guess haan → conservative.

Test. $P_{y} = \frac{\partial}{\partial y} (3 x^{2} - 2 y) = - 2$ . $Q_{x} = \frac{\partial}{\partial x} (- 2 x + 4 y^{3}) = - 2$ . Equal ✓. Yeh step kyun? Signs bhi match hone chahiye — ek common slip hai magnitudes compare karna aur minus bhool jaana.
$P$ ko $x$ mein integrate karo. $f = \int (3 x^{2} - 2 y) d x = x^{3} - 2 x y + g (y)$ . Yeh step kyun? $y$ ko constant maano; $\int 3 x^{2} d x = x^{3}$ , $\int - 2 y d x = - 2 x y$ .
$Q$ se match karo. $f_{y} = - 2 x + g^{'} (y) =! - 2 x + 4 y^{3} \Rightarrow g^{'} (y) = 4 y^{3} \Rightarrow g (y) = y^{4} + C$ . Yeh step kyun? $- 2 x$ already Step 2 se free mein aa jaata hai; sirf purely- $y$ leftover $4 y^{3}$ hi $g$ ko feed karta hai.

Answer: $f (x, y) = x^{3} - 2 x y + y^{4} + C .$

Verify: $f_{x} = 3 x^{2} - 2 y$ ✓, $f_{y} = - 2 x + 4 y^{3}$ ✓.

Ex 3 — Cell C3: recipe fail karni chahiye

Worked example Statement

$F = (x^{2} y, x y^{2})$ . Conservative hai?

Forecast: Harmless aur polynomial lagta hai. Lekin harmless-looking koi proof nahi hai. Pehle predict karo, phir test karo.

Test. $P = x^{2} y \Rightarrow P_{y} = x^{2}$ . $Q = x y^{2} \Rightarrow Q_{x} = y^{2}$ . Kya $x^{2} = y^{2}$ sabhi $(x, y)$ ke liye hai? Nahi — sirf lines $y = \pm x$ par. Yeh step kyun? Test ek identity ki tarah hold karna chahiye, domain mein har jagah, isolated points par nahi.

Answer: Not conservative. Koi potential exist nahi karta.

Verify: $(1, 2)$ par: $P_{y} = 1$ , $Q_{x} = 4$ , unequal — ek witness point hi ise kill kar deta hai.

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

Figure — Conservative vector fields — potential functions

$P_{y}$ compute karo. $P = \frac{- y}{x ^{2} + y ^{2}}$ ke saath, quotient rule deta hai $P_{y} = \frac{- ( x ^{2} + y ^{2} ) - ( - y ) ( 2 y )}{( x ^{2} + y ^{2} ) ^{2}} = \frac{- x ^{2} - y ^{2} + 2 y ^{2}}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}} .$

$Q_{x}$ compute karo. $Q = \frac{x}{x ^{2} + y ^{2}}$ ke saath, $Q_{x} = \frac{( x ^{2} + y ^{2} ) - x ( 2 x )}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}} .$ Yeh steps kyun? Hume apne haathon se dekhna hai ki $P_{y} = Q_{x}$ har jagah jahan field defined hai — taaki naive test kahe "conservative."

Unit circle ke around loop integral lo $r (t) = (cos t, sin t)$ , $t \in [0, 2 π]$ . Yahan $x^{2} + y^{2} = 1$ , isliye $F = (- sin t, cos t)$ aur $r^{'} = (- sin t, cos t)$ . $\oint F \cdot d r = \int_{0}^{2 π} (sin^{2} t + cos^{2} t) d t = \int_{0}^{2 π} 1 d t = 2 π \neq = 0.$ Yeh step kyun? Ek conservative field ka zero loop integral hota hai (FTLI, $A = B$ ). Ek nonzero loop ek hard disproof hai.

Answer: Punctured plane par conservative nahi hai — origin par hole "simply connected" requirement tod deta hai (figure mein red loop dekho jo missing point ko enclose karta hai). Dekho Green's theorem ki why enclosed hole matter karta hai.

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

Ex 5 — Cell C5: FTLI se path ke along work compute karna

Worked example Statement

Ex 1 se $F = (2 x + y, x + 6 y)$ use karke, $\int_{C} F \cdot d r$ compute karo $A = (- 1, 0)$ se $B = (2, - 1)$ tak kisi bhi curve ke along.

Forecast: Kyunki Ex 1 ne prove kiya tha ki yeh conservative hai, wiggly path irrelevant hai — sirf do numbers matter karte hain. Pehle guess karo ki work positive hogi ya negative.

Hill recall karo. $f (x, y) = x^{2} + x y + 3 y^{2}$ ( $C$ drop karo; subtraction mein cancel ho jaata hai). Yeh step kyun? FTLI kehta hai work $= f (B) - f (A)$ , isliye hume sirf do endpoints par $f$ chahiye.
$B = (2, - 1)$ par evaluate karo. $f (2, - 1) = 4 + (2) (- 1) + 3 (1) = 4 - 2 + 3 = 5.$
$A = (- 1, 0)$ par evaluate karo. $f (- 1, 0) = 1 + 0 + 0 = 1.$
Subtract karo. $\int_{C} F \cdot d r = f (B) - f (A) = 5 - 1 = 4.$ Yeh step kyun? $f (B) - f (A)$ total "downhill/uphill" change hai; beech ka path invisible hai.

Answer: $4$ .

Verify: $f$ mein $+ C$ cancel ho jaata hai ( $(5 + C) - (1 + C) = 4$ ), isliye answer path- aur constant-independent hai ✓.

Ex 6 — Cell C6: full 3D reconstruction

Worked example Statement

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

Forecast: Teen components matlab teen matching conditions. Guess: conservative, aur $f$ mein ek $x^{2} y$ piece plus ek $x z^{3}$ piece hai.

Curl test (teeno pairs).
- $P_{y} = 2 x$ , $Q_{x} = 2 x$ ✓
- $P_{z} = 3 z^{2}$ , $R_{x} = 3 z^{2}$ ✓
- $Q_{z} = 0$ , $R_{y} = 0$ ✓ Yeh step kyun? 3D mein, $curl F = 0$ ke liye teeno cross-partial equalities chahiye (dekho Curl of a vector field).
$P$ ko $x$ mein integrate karo. $f = \int (2 x y + z^{3}) d x = x^{2} y + x z^{3} + g (y, z)$ . Yeh step kyun? Ab invisible leftover dono remaining variables $y$ aur $z$ par depend karta hai.
$Q$ se match karo. $f_{y} = x^{2} + g_{y} =! x^{2} \Rightarrow g_{y} = 0$ , isliye $g = h (z)$ ( $y$ nahi).
$R$ se match karo. $f_{z} = 3 x z^{2} + h^{'} (z) =! 3 x z^{2} \Rightarrow h^{'} (z) = 0$ , isliye $h = C$ . Yeh steps kyun? Har aage ka derivative leftover se ek variable ki dependence peel away karta hai jab tak sirf ek constant nahi reh jaata.

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: degenerate limiting inputs

Ex 8 — Cell C8: real-world energy problem

Worked example Statement

Ek particle force feel karta hai $F = (- k x, - k y)$ (ek 2D spring origin ki taraf khichta hai, $k > 0$ , units N/m·m). Yeh $A = (3, 0)$ se $B = (0, 4)$ tak move karta hai. $F$ kitna work karta hai? $k = 2$ lo.

Forecast: Particle shuru aur khatam hota hai same distance $5$ par… ruko, $∣ A ∣ = 3$ , $∣ B ∣ = 4$ . Yeh ek aisi jagah move karta hai jo closer/farther hai? Pehle work ka sign guess karo (kya spring help karta hai ya resist?).

Test / potential pehchano. $P = - k x, Q = - k y$ : $P_{y} = 0 = Q_{x}$ ✓ conservative. Integrate: $f = \int - k x d x = - \frac{k}{2} x^{2} + g (y)$ ; $f_{y} = g^{'} (y) =! - k y \Rightarrow g = - \frac{k}{2} y^{2}$ . Isliye $f = - \frac{k}{2} (x^{2} + y^{2})$ . Yeh step kyun? Spring force usual potential energy $U = \frac{k}{2} r^{2}$ ke gradient ka minus hoti hai; yahan $f$ usual potential energy ka negative hai. Yeh Conservation of energy in physics se link karta hai.
FTLI apply karo. Work $= f (B) - f (A) = - \frac{k}{2} (0^{2} + 4^{2}) - (- \frac{k}{2} (3^{2} + 0^{2})) = - \frac{k}{2} (16) + \frac{k}{2} (9) = - \frac{k}{2} (7)$ . Yeh step kyun? Sirf endpoints ki origin se distances matter karti hain — yeh conservative spring ke liye path irrelevant hai.
$k = 2$ plug karo. Work $= - \frac{2}{2} (7) = - 7$ joules. Yeh step kyun? Negative work: particle farther end hua origin se ( $4 > 3$ ), isliye inward spring ne motion ko resist kiya. Sign intuition se match karta hai.

Answer: $- 7$ J.

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

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

Worked example Statement

Maano $F = (2 x y + 3, x^{2} - 4 y)$ (conservative, parent note se). Ek student $\oint_{C} F \cdot d r$ compute karta hai triangle $(0, 0) \to (1, 0) \to (1, 2) \to (0, 0)$ ki boundary ke around. Woh teen edges parametrize karne mein 20 minute lagaata hai. Shortcut answer kya hai, aur trap kahan hai?

Forecast: Closed loop + conservative. Koi bhi integral se pehle answer instantly guess karo.

Conservativeness check karo. $P_{y} = 2 x = Q_{x}$ ✓, domain pura $R^{2}$ hai (simply connected, koi hole nahi). Yeh step kyun? C4 trap tabhi kaat ta hai jab ek hole enclosed ho; yahan koi nahi hai, isliye shortcut legal hai.
Loop $= 0$ use karo. Ek hole-free domain par conservative field ke liye, $\oint_{C} F \cdot d r = f (B) - f (A)$ jahan $A = B$ , jo $0$ deta hai. Yeh step kyun? FTLI equal endpoints ke saath — koi parametrisation nahi chahiye.

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

Verify: triangle ke around $F = (2 x y + 3, x^{2} - 4 y)$ ke teen edge integrals ka direct sum $0$ hota hai (numerically checked).

Recall

Recall Main kis cell mein hoon?

Ek naaye field dene par, tumhara sabse pehla move kya hai? ::: Cross-partial / curl test run karo — yeh non-conservative fields ko filter karta hai tumhare $f$ dhundhne se pehle hi. Test pass hota hai lekin domain mein ek hole hai jiske around loop wind karta hai — conclusion? ::: Tum use conservative NAHI declare kar sakte; ek loop integral check karo (Cell C4). Closed loop, conservative field, koi enclosed hole nahi — value? ::: Exactly $0$ (Cell C9). Constant nonzero force field — kya yeh conservative hai? ::: Haan; uska potential ek tilted plane $a x + b y$ hai (Cell C7).