Curl — definition, physical meaning (rotation): Worked examples

4.4.25 · D3 · Maths › Multivariable Calculus › Curl — definition, physical meaning (rotation)

Shuru karne se pehle, woh ek hi formula yaad karo jo humein chahiye, simple words mein:

Agar koi bhi symbol naya lagta hai, toh parent Curl — definition, physical meaning (rotation) har ek ko paddlewheel picture se build karta hai. Hum Line Integrals and Circulation (circulation ka matlab) aur Equality of Mixed Partial Derivatives (Clairaut) (kyun do identities vanish hoti hain) par bhi rely karte hain.

Scenario matrix

Har curl problem inhi cells mein se kisi ek mein aati hai. Right column us example ka naam batata hai jo usse best cover karta hai.

#	Case class	Tricky kya hai	Example
A	Positive curl (CCW spin)	sign $+$ hai	Ex 1
B	Negative curl (CW spin)	sign $-$ hai	Ex 2
C	Zero curl, seedhe arrows	expected zero, milta bhi zero	Ex 3
D	Zero curl, curved arrows	lagta hai spin hoga — hota nahi	Ex 4
E	Nonzero curl, straight arrows (shear)	lagta hai spin nahi hoga — hota hai	Ex 5
F	Position-dependent curl (point se point badlata hai)	answer ek number nahi, ek function hai	Ex 6
G	Full 3-D curl (tilted axis)	teeno components alive hain	Ex 7
H	Degenerate / constant field	zero kyunki kuch change hi nahi hota	Ex 8
I	Word problem (physical rotation)	reality ko field mein translate karo	Ex 9
J	Exam twist (ek identity prove karo)	symbolic, koi numbers nahi	Ex 10

Example 1 — Cell A: positive (counterclockwise) curl

Forecast: arrows ka picture socho: $(1, 0)$ par field $(0, 1)$ hai (upar point karta hai); $(0, 1)$ par yeh $(- 1, 0)$ hai (left point karta hai). Counterclockwise ghoom raha hai. Aage padhne se pehle $z$ -component ka sign guess karo.

Sirf $z$ -component nonzero ho sakta hai ( $z$ -dependence nahi, $F_{3} = 0$ ), toh $\partial_{x} F_{2} - \partial_{y} F_{1}$ compute karo. Yeh step kyun? $F_{1}, F_{2}$ mein koi $z$ nahi hai aur $F_{3} = 0$ , toh curl ke $x$ - aur $y$ -components saare $\partial (\cdot) 0$ ya $\partial_{z} (\cdot) = 0$ hain. Sirf flat-plane swirl bachta hai.
$\partial_{x} F_{2} = \partial_{x} (x) = 1$ aur $\partial_{y} F_{1} = \partial_{y} (- y) = - 1$ . Yeh step kyun? "Upar ki push $x$ kitni tezi se badhti hai jab main right chalta hun?" ( $= 1$ ) aur "right ki push $- y$ kitni tezi se badhti hai jab main upar chalta hun?" ( $= - 1$ ).
$(\nabla \times F)_{z} = 1 - (- 1) = 2$ . Toh $\nabla \times F = (0, 0, 2)$ .

Figure — Curl — definition, physical meaning (rotation)

Verify: dono effects reinforce karte hain (right chalna rise ko speed up karta hai, upar chalna leftward push ko speed up karta hai — same rotation), toh magnitude $> 1$ : $2$ ke saath consistent hai. Positive $z$ ⇒ CCW ⇒ forecast se match karta hai. ✓

Example 2 — Cell B: negative (clockwise) curl

Forecast: yeh Example 1 hai jisme har arrow reverse ho gaya hai, toh clockwise spin hona chahiye: ek negative $z$ -component expect karo.

$\partial_{x} F_{2} = \partial_{x} (- x) = - 1$ . Yeh step kyun? Neeche ki push $- x$ aur bhi negative hoti jaati hai jab hum right step lete hain.
$\partial_{y} F_{1} = \partial_{y} (y) = 1$ . Yeh step kyun? Right ki push $y$ badhti hai jab hum upar step lete hain.
$(\nabla \times F)_{z} = - 1 - 1 = - 2$ , toh $\nabla \times F = (0, 0, - 2)$ .

Verify: Ex 1 ke versus sign flip hua, magnitude same rahi — exactly wahi jo saare arrows reverse karne par hona chahiye. Negative $\hat{k}$ = clockwise = axle page ke andar point karta hai (right-hand rule). ✓

Example 3 — Cell C: zero curl, seedhe parallel arrows

Forecast: arrows saare $+ x$ ke along point karte hain, aur unki length $x$ ke along chalne par badlti hai, uske across nahi. Koi side-to-side speed difference nahi ⇒ zero guess karo.

$z$ -component: $\partial_{x} F_{2} - \partial_{y} F_{1} = \partial_{x} (0) - \partial_{y} (x) = 0 - 0 = 0$ . Yeh step kyun? $F_{1} = x$ mein koi $y$ nahi hai, toh $\partial_{y} (x) = 0$ : flow paddlewheel ke across (flow ke perpendicular) vary nahi karta.
$x$ - aur $y$ -components mein $F_{3} = 0$ aur $z$ -free fields ka $\partial_{z}$ aata hai — sab zero hain.

Verify: speed flow ke along change hoti hai (yeh divergence hai: $\partial_{x} (x) = 1 \neq = 0$ ), uske across nahi, toh curl vanish honi chahiye. Yeh prove karta hai ki curl aur divergence independent hain, parent ke Divergence — definition and physical meaning contrast se echo karta hai. ✓

Example 4 — Cell D: curved arrows lekin ZERO curl

Forecast: arrows literally origin ke around circle karte hain — yeh toh zaroor spin karega! Zyaadatar students kehte hain curl $= 2$ jaise Example 1 mein. Guess karo, phir dekho yeh kaise collapse hota hai.

$r^{2} = x^{2} + y^{2}$ likho. Tab $F_{2} = x / r^{2}$ , $F_{1} = - y / r^{2}$ . Yeh step kyun? $r^{2}$ naam rakhne se quotient rule readable rehta hai.
$\partial_{x} F_{2} = \partial_{x} (\frac{x}{x ^{2} + y ^{2}}) = \frac{( x ^{2} + y ^{2} ) - x ( 2 x )}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}}$ . Yeh step kyun? Quotient rule: (bottom·d(top) − top·d(bottom))/bottom². Yahan $\partial_{x} r^{2} = 2 x$ .
$\partial_{y} F_{1} = \partial_{y} (\frac{- y}{x ^{2} + y ^{2}}) = \frac{- ( x ^{2} + y ^{2} ) + y ( 2 y )}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}}$ . Yeh step kyun? Same rule; upar $\partial_{y} (- y) = - 1$ , $\partial_{y} r^{2} = 2 y$ .
$(\nabla \times F)_{z} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}} - \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}} = 0$ .

Verify: dono identical fractions exactly cancel ho jaate hain ⇒ curl $= 0$ har jagah except origin par, jahan field blow up karta hai. Whirling look is liye aata hai kyunki arrows $r$ badhne ke saath exactly sahi tarike se slow hote hain jo turning ko cancel kar de. (Yeh woh field hai jiski wajah se circulation nonzero ho sakti hai jabki curl zero ho — ek subtlety Stokes' Theorem aur Gradient and Conservative Fields ke liye.) ✓

Example 5 — Cell E: straight arrows, NONZERO curl (shear)

Forecast: har arrow horizontally point karta hai (bilkul bhi curve nahi). Intuition chilla raha hai "zero". Lekin paddlewheel feel karo: axle ke neeche flow ek taraf tez hai, upar doosri taraf tez hai…

$(\nabla \times F)_{z} = \partial_{x} (0) - \partial_{y} (- y) = 0 - (- 1) = 1$ . Yeh step kyun? Horizontal push $- y$ $x$ -axis ke neeche strongly rightward aur upar leftward hai — ek shear hai. $\partial_{y} (- y) = - 1$ woh top/bottom speed difference capture karta hai.

Verify: $x$ -axis par ek paddlewheel khada karo: neeche ka paani ek taraf dhakelta hai, upar ka doosri taraf — yeh ghoomta hai. Dead-straight arrows ke saath nonzero curl, confirm karta hai ki curl flow ke across shear padhta hai, visual curvature nahi. ✓

Example 6 — Cell F: curl jo position ke saath vary karta hai

Forecast: $x$ aur $y$ ke yeh products matlab swirl point to point change hota hai — answer ek single number nahi, ek formula hoga. Kahan yeh zero ho sakta hai?

$\partial_{x} F_{2} = \partial_{x} (y^{2} x) = y^{2}$ . Yeh step kyun? $x$ mein differentiate karte waqt $y$ ko constant mano; $y^{2} x \to y^{2}$ .
$\partial_{y} F_{1} = \partial_{y} (x^{2} y) = x^{2}$ . Yeh step kyun? $x$ ko constant mano; $x^{2} y \to x^{2}$ .
$(\nabla \times F)_{z} = y^{2} - x^{2}$ . Toh $\nabla \times F = (0, 0, y^{2} - x^{2})$ .
Yeh exactly lines $y = \pm x$ par zero hai. Yeh step kyun? $y^{2} - x^{2} = (y - x) (y + x) = 0$ jab $y = x$ ya $y = - x$ .

Verify: $(1, 0)$ pick karo: curl $= 0 - 1 = - 1$ (wahan CW). $(0, 1)$ pick karo: curl $= 1 - 0 = + 1$ (wahan CCW). Field alag-alag regions mein opposite taraf spin karta hai, diagonals $y = \pm x$ se separated — exactly wahan curl $= 0$ hai. ✓

Example 7 — Cell G: ek genuinely 3-D curl (tilted axis)

Forecast: teeno variables ek baar aate hain, toh shayad teeno curl components equal aayenge — ek diagonal axis.

$x$ -comp $= \partial_{y} F_{3} - \partial_{z} F_{2} = \partial_{y} (y) - \partial_{z} (x) = 1 - 0 = 1$ .
$y$ -comp $= \partial_{z} F_{1} - \partial_{x} F_{3} = \partial_{z} (z) - \partial_{x} (y) = 1 - 0 = 1$ .
$z$ -comp $= \partial_{x} F_{2} - \partial_{y} F_{1} = \partial_{x} (x) - \partial_{y} (z) = 1 - 0 = 1$ . Yeh steps kyun? Har row cyclic pattern $x \to y \to z \to x$ use karta hai: next field ki derivative minus previous field ki derivative.
Toh $\nabla \times F = (1, 1, 1)$ ; spin axis unit vector $\frac{1}{3} (1, 1, 1)$ hai.

Verify: magnitude $∣ (1, 1, 1) ∣ = 3$ . Har coordinate plane unit swirl carry karta hai, toh total rotation axis teeno axes ki taraf equally tilt hota hai — ek cube ka body diagonal. ✓

Example 8 — Cell H: degenerate constant field

Forecast: point to point kuch change nahi hota, toh kuch bhi paddlewheel ko twist nahi kar sakta. $0$ guess karo.

Kisi bhi constant ki har partial derivative $0$ hai, jaise $\partial_{x} (- 2) = 0$ , $\partial_{y} (7) = 0$ , etc. Yeh step kyun? Derivative change measure karta hai; constant kabhi change nahi hota.
Chhhon mein se har ek term $0$ hai, toh $\nabla \times F = (0, 0, 0)$ .

Verify: ek uniform flow (river jo ek rigid sheet ki tarah move kare) wheel ko bina ghumaye saath le jaata hai. Zero curl physics se match karta hai. Note karo ki yeh field ek gradient bhi hai ( $\nabla (5 x - 2 y + 7 z)$ ), toh parent ki identity se yeh irrotational hona hi chahiye — Gradient and Conservative Fields se ek cross-check. ✓

Example 9 — Cell I: word problem (ek spinning turntable)

Forecast: rigid rotation sabse purest spin hai, toh curl seedha upar ( $+ \hat{k}$ ) point karna chahiye aur $ω$ ke proportional hona chahiye. Constant guess karo.

$ω = 3$ substitute karo: $v = (- 3 y, 3 x, 0)$ . Yeh step kyun? Physics ko ("3 rad/s par spin karta hai") ek concrete field mein badlo.
$(\nabla \times v)_{z} = \partial_{x} (3 x) - \partial_{y} (- 3 y) = 3 - (- 3) = 6$ . Yeh step kyun? Same shear calculation jaise Example 1 mein, ab $ω$ se scale hua.
Toh $\nabla \times v = (0, 0, 6) = 2 ω \hat{k}$ . Yeh step kyun? $6 = 2 \times 3$ : curl angular speed se do guna hota hai.

Verify (units + meaning): $v$ ki units m/s hain, uski spatial derivative m/s per m $=$ 1/s, $ω$ ki rad/s se match karta hai. Famous rule "curl of a rigid rotation $= 2 ω$ " deta hai $2 (3) = 6$ . ✓

Example 10 — Cell J: exam twist (ek identity prove karo, koi numbers nahi)

Forecast: ek gradient "pure downhill flow" hai aur downhill flow swirl nahi kar sakta, toh exact zero expect karo — reason yeh hoga ki mixed partials commute karte hain.

Gradient likho: $\nabla ϕ = (\partial_{x} ϕ, \partial_{y} ϕ, \partial_{z} ϕ)$ , toh yahan $F_{1} = \partial_{x} ϕ$ , $F_{2} = \partial_{y} ϕ$ , $F_{3} = \partial_{z} ϕ$ . Yeh step kyun? Components ka naam rakho taki curl formula unpar act kar sake.
Curl ka $z$ -component: $\partial_{x} F_{2} - \partial_{y} F_{1} = \partial_{x} (\partial_{y} ϕ) - \partial_{y} (\partial_{x} ϕ)$ . Yeh step kyun? $F_{2}, F_{1}$ ko page ke top ke $z$ -formula mein plug karo.
Equality of Mixed Partial Derivatives (Clairaut) se, $\partial_{x} \partial_{y} ϕ = \partial_{y} \partial_{x} ϕ$ , toh yeh difference $0$ hai. Yeh step kyun? Clairaut kehta hai ki smooth $ϕ$ ke liye partial differentiation ka order matter nahi karta.
$x$ - aur $y$ -components same swap se identically vanish ho jaate hain ( $\partial_{y} \partial_{z} ϕ = \partial_{z} \partial_{y} ϕ$ , etc.). Isliye $\nabla \times (\nabla ϕ) = 0$ .

Verify: $ϕ = x^{2} y$ par test karo. Tab $\nabla ϕ = (2 x y, x^{2}, 0)$ aur curl ka $z$ -comp $= \partial_{x} (x^{2}) - \partial_{y} (2 x y) = 2 x - 2 x = 0$ . General proof is concrete case par hold karta hai. ✓

Recall Har cell ka ek-line recap

$z$ -curl ka sign CCW ( $+$ ) vs CW ( $-$ ) batata hai; curl flow ke across shear padhta hai, curved arrows nahi (Ex 4 vs Ex 5); yeh ek poora function ho sakta hai (Ex 6); rigid rotation $2 ω$ deta hai (Ex 9); ek gradient hamesha irrotational hota hai (Ex 10).

Active Recall

Connections

Parent: Curl — definition, physical meaning (rotation)
Line Integrals and Circulation · Stokes' Theorem · Green's Theorem
Divergence — definition and physical meaning · Gradient and Conservative Fields · Equality of Mixed Partial Derivatives (Clairaut)