Curl — definition, physical meaning (rotation): Exercises

4.4.25 · D4 · Maths › Multivariable Calculus › Curl — definition, physical meaning (rotation)

Reminder of the only formula you need (from the parent): $\nabla \times F = (\partial_{y} F_{3} - \partial_{z} F_{2}, \partial_{z} F_{1} - \partial_{x} F_{3}, \partial_{x} F_{2} - \partial_{y} F_{1})$ Yahan $\partial_{x}$ ka matlab hai "cheez kitni tezi se change hoti hai jab tum $x$ ko thoda nudge karte ho" (ek partial derivative), aur $(F_{1}, F_{2}, F_{3})$ field ke teen arrow-lengths hain $x$ , $y$ , $z$ directions mein.

Level 1 — Recognition

Goal: definition padho aur plug in karo. Koi cleverness required nahi.

Problem 1.1

$F = (3 x, 5 y, 0)$ ke liye $z$ -component $(\nabla \times F)_{z} = \partial_{x} F_{2} - \partial_{y} F_{1}$ compute karo.

Recall Solution 1.1

Yahan $F_{1} = 3 x$ aur $F_{2} = 5 y$ hai.

$\partial_{x} F_{2} = \partial_{x} (5 y) = 0$ ( $5 y$ ke andar koi $x$ nahi hai, isliye $x$ ko nudge karne se kuch nahi badlega).
$\partial_{y} F_{1} = \partial_{y} (3 x) = 0$ ( $3 x$ ke andar koi $y$ nahi hai). $(\nabla \times F)_{z} = 0 - 0 = 0$ Ek field jo sirf apne khud ke axes ke along stretch karti hai, kabhi swirl nahi karti.

Problem 1.2

$F = (- 2 y, 2 x, 0)$ ke liye $(\nabla \times F)_{z}$ nikalo. Kya field clockwise ghum rahi hai ya counterclockwise?

Recall Solution 1.2

$F_{1} = - 2 y$ , $F_{2} = 2 x$ .

$\partial_{x} F_{2} = \partial_{x} (2 x) = 2$ .
$\partial_{y} F_{1} = \partial_{y} (- 2 y) = - 2$ . $(\nabla \times F)_{z} = 2 - (- 2) = 4$ Positive $\hat{k}$ ka matlab hai counterclockwise (right-hand rule: ungli us taraf curl karo jis tarah flow spin karta hai, thumb $+ \hat{k}$ page se bahar tumhari taraf nikalti hai).

Problem 1.3

Inme se kis field ka $(\nabla \times F)_{z} = 0$ hai? (a) $F = (y, x, 0)$ (b) $F = (- y, x, 0)$

Recall Solution 1.3

(a) $\partial_{x} (x) - \partial_{y} (y) = 1 - 1 = 0$ . Zero curl. (b) $\partial_{x} (x) - \partial_{y} (- y) = 1 - (- 1) = 2$ . Nonzero curl. Answer: (a) ka zero $z$ -curl hai. $y$ aur $- y$ ke beech mein sirf ek chhota sa sign flip hai — yahi "no spin" aur "spins" ke beech ka poora fark hai.

Level 2 — Application

Goal: poora 3-component curl vector compute karo.

Problem 2.1

$F = (y, z, x)$ ke liye $\nabla \times F$ compute karo.

Recall Solution 2.1

Har component ke liye cyclically $\partial_{next} F_{next} - \partial_{prev} F_{prev}$ use karo:

$x$ -comp: $\partial_{y} F_{3} - \partial_{z} F_{2} = \partial_{y} (x) - \partial_{z} (z) = 0 - 1 = - 1$ .
$y$ -comp: $\partial_{z} F_{1} - \partial_{x} F_{3} = \partial_{z} (y) - \partial_{x} (x) = 0 - 1 = - 1$ .
$z$ -comp: $\partial_{x} F_{2} - \partial_{y} F_{1} = \partial_{x} (z) - \partial_{y} (y) = 0 - 1 = - 1$ . $\nabla \times F = (- 1, - 1, - 1)$ Har swirl-plane mein $- 1$ hai: spin axis diagonal $(- 1, - 1, - 1)$ hai.

Problem 2.2

$F = (x^{2}, y^{2}, z^{2})$ ke liye $\nabla \times F$ compute karo.

Recall Solution 2.2

$F$ ka har component sirf apne khud ke variable par depend karta hai, isliye har cross-derivative zero ho jaati hai:

$x$ -comp: $\partial_{y} (z^{2}) - \partial_{z} (y^{2}) = 0 - 0 = 0$ .
$y$ -comp: $\partial_{z} (x^{2}) - \partial_{x} (z^{2}) = 0 - 0 = 0$ .
$z$ -comp: $\partial_{x} (y^{2}) - \partial_{y} (x^{2}) = 0 - 0 = 0$ . $\nabla \times F = (0, 0, 0)$ Is tarah ka "separable" field irrotational hota hai.

Problem 2.3

$F = (0, 0, x^{2} + y^{2})$ ke liye $\nabla \times F$ compute karo.

Recall Solution 2.3

Sirf $F_{3} = x^{2} + y^{2}$ nonzero hai.

$x$ -comp: $\partial_{y} (x^{2} + y^{2}) - \partial_{z} (0) = 2 y - 0 = 2 y$ .
$y$ -comp: $\partial_{z} (0) - \partial_{x} (x^{2} + y^{2}) = 0 - 2 x = - 2 x$ .
$z$ -comp: $\partial_{x} (0) - \partial_{y} (0) = 0$ . $\nabla \times F = (2 y, - 2 x, 0)$ Upar ki taraf push $F_{3}$ jaise-jaise tum $z$ -axis se door jaate ho badhta hai, isliye ek sideways tilka hua paddlewheel torqued ho jaata hai — curl horizontal plane mein rehti hai.

Level 3 — Analysis

Goal: number ko physical picture aur paddlewheel se connect karo.

Problem 3.1

Ek paddlewheel ko 2D field $F = (0, x, 0)$ mein daala jaata hai (arrows sab vertically point karte hain, $x$ badhne par lambe hote hain — figure dekho). Pehle spin direction predict karo, phir curl se confirm karo.

Figure — Curl — definition, physical meaning (rotation)

Recall Solution 3.1

Physical prediction: dayi taraf ( $x$ bada) oopar ke arrows lambe/tez hain; bayi taraf chhote hain. Isliye wheel ka right side left se zyada oopar push hota hai → ye counterclockwise rotate karta hai ( $+ \hat{k}$ ). Curl check: $(\nabla \times F)_{z} = \partial_{x} (x) - \partial_{y} (0) = 1 - 0 = 1$ . Positive, counterclockwise prediction se match karta hai. Note karo arrows bilkul seedhe hain — phir bhi wheel spin karti hai. Ye shear hai, curved flow nahi.

Problem 3.2

Kaunsi constant $c$ ke liye $F = (cy, x, 0)$ irrotational (curl $= 0$ ) hai?

Recall Solution 3.2

$(\nabla \times F)_{z} = \partial_{x} (x) - \partial_{y} (cy) = 1 - c$ . Zero set karo: $1 - c = 0 \Rightarrow c = 1$ . $c = 1$ ke saath humein $F = (y, x, 0)$ milta hai, jo $ϕ = x y$ ka gradient hai — aur gradients hamesha irrotational hote hain (parent identity $\nabla \times \nabla ϕ = 0$ ; dekho Gradient and Conservative Fields).

Problem 3.3

Field $F = (- y, x, 0)$ ka curl $(0, 0, 2)$ hai. "Magnitude of curl $= 2 \times$ angular speed" use karke origin pe ek tiny floating chip ki angular speed $ω$ nikalo.

Recall Solution 3.3

$∣\nabla \times F ∣ = 2$ , aur $∣ curl ∣ = 2 ω$ , isliye $ω = \frac{2}{2} = 1 radian per unit time.$ Field sach mein ek rigid turntable ki tarah $1$ rad/s pe rotate karti hai; curl angular speed ko exactly $2$ ke factor se over-count karta hai (do cross-terms mein se har ek ek $ω$ contribute karta hai).

Level 4 — Synthesis

Goal: curl ko divergence, gradients aur Stokes/Green ke saath combine karo.

Problem 4.1

$F = (x, y, 0)$ lo. Iska divergence $\nabla \cdot F = \partial_{x} F_{1} + \partial_{y} F_{2} + \partial_{z} F_{3}$ aur iska curl dono compute karo. Answers ki pair kya prove karti hai?

Recall Solution 4.1

Divergence: $\partial_{x} (x) + \partial_{y} (y) + \partial_{z} (0) = 1 + 1 + 0 = 2$ . Curl: har cross-derivative $0$ hai, isliye $\nabla \times F = (0, 0, 0)$ . Ek field mein strong divergence ho sakti hai lekin zero curl bhi. Ye prove karta hai ki divergence aur curl independent sawaalon ke jawab dete hain — dekho Divergence — definition and physical meaning. Ye field sirf baahir phailti hai (ek source), kabhi swirl nahi karti.

Problem 4.2

Maano $ϕ = x^{2} y + z$ . $F = \nabla ϕ = (\partial_{x} ϕ, \partial_{y} ϕ, \partial_{z} ϕ)$ compute karo aur phir $\nabla \times F$ . Mixed partials ke through result explain karo.

Recall Solution 4.2

$\nabla ϕ = (2 x y, x^{2}, 1)$ . Curl:

$x$ -comp: $\partial_{y} (1) - \partial_{z} (x^{2}) = 0 - 0 = 0$ .
$y$ -comp: $\partial_{z} (2 x y) - \partial_{x} (1) = 0 - 0 = 0$ .
$z$ -comp: $\partial_{x} (x^{2}) - \partial_{y} (2 x y) = 2 x - 2 x = 0$ . $\nabla \times \nabla ϕ = (0, 0, 0)$ $z$ -component exactly $\partial_{x} \partial_{y} ϕ - \partial_{y} \partial_{x} ϕ$ hai, jo Clairaut's theorem se $0$ hota hai. Har gradient field irrotational hoti hai.

Problem 4.3

Green's Theorem use karke unit circle $C$ (radius $1$ , counterclockwise) ke around $F = (- y, x, 0)$ ki circulation $\oint_{C} F \cdot d r$ nikalo. Green: $\oint_{C} F \cdot d r = \iint_{D} (\partial_{x} F_{2} - \partial_{y} F_{1}) d A$ .

Recall Solution 4.3

Integrand $z$ -curl hai: $\partial_{x} (x) - \partial_{y} (- y) = 1 - (- 1) = 2$ (constant). $\oint_{C} F \cdot d r = \iint_{D} 2 d A = 2 \cdot (area of unit disk) = 2 π .$ $2 π$ Kyunki curl constant hai, "circulation = curl $\times$ area" — tiny-loop definition scaled up. Ye Green's Theorem hai, Stokes' Theorem ka 2D face.

Level 5 — Mastery

Goal: general statements prove karo aur degenerate cases handle karo.

Problem 5.1

Prove karo ki kisi bhi smooth $F$ ke liye, $\nabla \cdot (\nabla \times F) = 0$ .

Recall Solution 5.1

$\nabla \times F = (\partial_{y} F_{3} - \partial_{z} F_{2}, \partial_{z} F_{1} - \partial_{x} F_{3}, \partial_{x} F_{2} - \partial_{y} F_{1})$ likho. Iska divergence lo (teen components ke $\partial_{x}$ , $\partial_{y}$ , $\partial_{z}$ ka sum): $\partial_{x} (\partial_{y} F_{3} - \partial_{z} F_{2}) + \partial_{y} (\partial_{z} F_{1} - \partial_{x} F_{3}) + \partial_{z} (\partial_{x} F_{2} - \partial_{y} F_{1}) .$ Chhe terms ko pairs mein group karo: $(\partial_{x} \partial_{y} F_{3} - \partial_{y} \partial_{x} F_{3}) + (\partial_{z} \partial_{x} F_{2} - \partial_{x} \partial_{z} F_{2}) + (\partial_{y} \partial_{z} F_{1} - \partial_{z} \partial_{y} F_{1}) .$ Har parenthesis Clairaut's theorem se zero hai (mixed partials ka order matter nahi karta). Isliye total $0$ hai. "Ek pure swirl mein koi source nahi hota."

Problem 5.2

Har constant $a$ dhundho jisse $F = (a y + z, x + 2 z, x + 3 y)$ irrotational ho, ya dikhao ki koi bhi/sab kaam karte hain.

Recall Solution 5.2

Teen curl components compute karo:

$x$ -comp: $\partial_{y} (x + 3 y) - \partial_{z} (x + 2 z) = 3 - 2 = 1$ .
$y$ -comp: $\partial_{z} (a y + z) - \partial_{x} (x + 3 y) = 1 - 1 = 0$ .
$z$ -comp: $\partial_{x} (x + 2 z) - \partial_{y} (a y + z) = 1 - a$ . $x$ -component $1 \neq = 0$ hai $a$ ki parwah kiye bina. Isliye curl kabhi zero nahi ho sakta. $No value of a makes it irrotational.$ (Agar hum $z$ -component ko kill karne ke liye $a = 1$ choose bhi karein, to stubborn $x$ -component bachha rehta hai.) Sirf ek nonzero component irrotationality ko break karne ke liye kaafi hai.

Problem 5.3

Degenerate cases. Har ek ke liye, curl batao aur interpret karo. (a) Zero field $F = (0, 0, 0)$ . (b) Ek uniform field $F = (c_{1}, c_{2}, c_{3})$ constants $c_{i}$ ke saath. (c) Sirf ek nonzero constant component wala field $F = (0, 0, K)$ .

Recall Solution 5.3

(a) $0$ ki har derivative $0$ hai: $\nabla \times F = (0, 0, 0)$ . Koi flow nahi, koi spin nahi — trivial baseline. (b) Constants ki derivatives $0$ hoti hain, isliye saare chhe terms zero ho jaate hain: $\nabla \times F = (0, 0, 0)$ . Ek uniform "hawa jo har jagah same direction mein chalti hai" paddlewheel ko bodily push karti hai lekin kabhi spin nahi karti. (c) Same reasoning — $K$ constant hai, saari derivatives $0$ : $\nabla \times F = (0, 0, 0)$ . Constant fields hamesha irrotational hote hain; curl sirf change dekhta hai, aur constants kabhi change nahi karte.

Recall Final self-check (answers cover karo)

Curl of $(- 2 y, 2 x, 0)$ ? ::: $(0, 0, 4)$ Curl of $(y, z, x)$ ? ::: $(- 1, - 1, - 1)$ Circulation of $(- y, x, 0)$ around the unit circle? ::: $2 π$ Kya koi $a$ hai jo $(a y + z, x + 2 z, x + 3 y)$ ko irrotational banata hai? ::: Nahi — curl ka $x$ -component har $a$ ke liye $1$ rehta hai. Kisi bhi constant field ka curl? ::: $(0, 0, 0)$

Connections

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