Line integrals — scalar and vector, work done: Exercises

4.4.27 · D4 · Maths › Multivariable Calculus › Line integrals — scalar and vector, work done

Shuru karne se pehle, un do tools ki ek-line reminder jo baar baar kaam aayenge.

Figure — Line integrals — scalar and vector, work done

Level 1 — Recognition

Goal: sahi tool choose karo aur integral set up karo. Abhi koi cleverness nahi chahiye.

Problem 1.1

Ek wire $r (t) = (3 t, 4 t)$ ko follow karta hai $t \in [0, 1]$ ke liye, aur uska constant density $f = 1$ hai. Iska total mass sirf uski length hai. Woh length nikaalo.

Recall Solution 1.1

Kaunsa tool? Wire ke saath spread density → scalar line integral. $f = 1$ ke saath yeh Arc length hai.

Velocity: har component ko differentiate karo. $r^{'} (t) = (3, 4)$ .
Speed: $∣ r^{'} (t) ∣ = 3^{2} + 4^{2} = 25 = 5$ . Yeh step kyun? $d s = ∣ r^{'} ∣ d t$ ek step in $t$ ko length mein ek step mein convert karta hai.
$L = \int_{0}^{1} 1 \cdot 5 d t = 5 \cdot (1 - 0) = 5 .$ Sanity check: curve $(0, 0)$ se $(3, 4)$ tak ek straight segment hai, jiski ordinary distance $3^{2} + 4^{2} = 5$ hai. ✓

Problem 1.2

$F = (2, 3)$ (ek constant force) aur straight path $r (t) = (t, t)$ , $t \in [0, 1]$ ke liye, $\int_{C} F \cdot d r$ set up aur evaluate karo.

Recall Solution 1.2

Kaunsa tool? Ek force jo path ke saath kheench raha hai → vector line integral (work).

$r^{'} (t) = (1, 1)$ .
$F (r (t)) = (2, 3)$ — constant hai, toh koi substitution ki zaroorat nahi.
Dot product: $F \cdot r^{'} = 2 (1) + 3 (1) = 5$ . Dot kyun? Sirf $F$ ka woh part jo motion ke along hai, wahi work karta hai.
$W = \int_{0}^{1} 5 d t = 5 .$

Level 2 — Application

Goal: ek genuine parametrisation substitute karo aur integrate karo.

Problem 2.1

Density $f (x, y) = x + y$ quarter unit circle $r (t) = (cos t, sin t)$ , $t \in [0, π /2]$ ke saath. Mass nikaalo.

Recall Solution 2.1

$r^{'} (t) = (- sin t, cos t)$ , toh $∣ r^{'} (t) ∣ = sin^{2} t + cos^{2} t = 1$ . Unit circle → unit speed.
$f (r (t)) = x + y = cos t + sin t$ . Density mein parametrisation substitute karo.
Mass $= \int_{0}^{π /2} (cos t + sin t) \cdot 1 d t = [sin t - cos t]_{0}^{π /2} = (1 - 0) - (0 - 1) = 2 .$

Problem 2.2

$F = (x, y)$ along $r (t) = (t, t^{2})$ , $t \in [0, 1]$ . Work $W$ nikaalo.

Recall Solution 2.2

$r^{'} (t) = (1, 2 t)$ .
$F (r (t)) = (x, y) = (t, t^{2})$ .
$F \cdot r^{'} = t (1) + t^{2} (2 t) = t + 2 t^{3}$ .
$W = \int_{0}^{1} (t + 2 t^{3}) d t = [\frac{t ^{2}}{2} + \frac{t ^{4}}{2}]_{0}^{1} = \frac{1}{2} + \frac{1}{2} = 1 .$

Level 3 — Analysis

Goal: yeh sochna ki setup badalne par kya badalta hai — orientation, speed, parametrisation.

Problem 3.1 (orientation)

$F = (x, y)$ ke liye, reversed parabola ke saath work compute karo: $(1, 1)$ se $(0, 0)$ tak, yani $\tilde{r} (t) = (1 - t, (1 - t)^{2})$ , $t \in [0, 1]$ . Problem 2.2 se compare karo.

Recall Solution 3.1

$\tilde{r}^{'} (t) = (- 1, - 2 (1 - t))$ . Har component par chain rule.
$F (\tilde{r} (t)) = (1 - t, (1 - t)^{2})$ .
Dot: $(1 - t) (- 1) + (1 - t)^{2} \cdot (- 2 (1 - t)) = - (1 - t) - 2 (1 - t)^{3}$ .
$u = 1 - t$ let karo, $d u = - d t$ ; jab $t : 0 \to 1$ , $u : 1 \to 0$ . $W = \int_{0}^{1} [- (1 - t) - 2 (1 - t)^{3}] d t = \int_{1}^{0} [- u - 2 u^{3}] (- d u) = \int_{1}^{0} (u + 2 u^{3}) d u = - \int_{0}^{1} (u + 2 u^{3}) d u = - 1.$ Toh $W = - 1$ , exactly Problem 2.2 ka negative. Kyun? Orientation reverse karne se $d r$ flip hota hai, isliye poora dot-product integral flip ho jaata hai.

Problem 3.2 (scalar integrals ki reparametrisation invariance)

Problem 2.1 ka mass redo karo lekin quarter circle ko do guna tez chalte huye: $r (t) = (cos 2 t, sin 2 t)$ , $t \in [0, π /4]$ . Dikhao ki mass nahi badla.

Recall Solution 3.2

$r^{'} (t) = (- 2 sin 2 t, 2 cos 2 t)$ , toh $∣ r^{'} (t) ∣ = 4 sin^{2} 2 t + 4 cos^{2} 2 t = 2$ . Do guna speed.
$f (r (t)) = cos 2 t + sin 2 t$ .
Mass $= \int_{0}^{π /4} (cos 2 t + sin 2 t) \cdot 2 d t$ . $u = 2 t$ , $d u = 2 d t$ substitute karo: $= \int_{0}^{π /2} (cos u + sin u) d u = [sin u - cos u]_{0}^{π /2} = 2.$ Same answer, $2$ . Kyun? Extra speed factor $2$ exactly cancel ho jaata hai kyunki arc aadhe $t$ -interval mein cover ho raha hai — $d s$ physically same length hai chahe kuch bhi ho.

Level 4 — Synthesis

Goal: tool ko ek bade idea ke saath combine karo — conservative fields aur Fundamental Theorem.

Dikhao ki $F = (y, x)$ conservative hai ek potential $ϕ$ dhundh kar jiske liye $F = \nabla ϕ$ , phir Fundamental Theorem for Line Integrals use karo $(0, 0)$ se $(2, 3)$ tak kisi bhi path par work compute karne ke liye.

Recall Solution 4.1

$ϕ$ dhundho. Humein chahiye $\partial_{x} ϕ = y$ aur $\partial_{y} ϕ = x$ .

Pehle wale ko $x$ mein integrate karo: $ϕ = x y + g (y)$ .
$y$ mein differentiate karo: $\partial_{y} ϕ = x + g^{'} (y)$ ; $x$ se match karo ⇒ $g^{'} (y) = 0$ ⇒ $g$ constant hai. $ϕ = x y$ lo.
Gradient and conservative fields se check karo: $\nabla ϕ = (\partial_{x} ϕ, \partial_{y} ϕ) = (y, x) = F$ . ✓ Theorem apply karo. Conservative field ke liye, $\int_{C} F \cdot d r = ϕ (end) - ϕ (start)$ . $W = ϕ (2, 3) - ϕ (0, 0) = (2) (3) - 0 = 6 .$ Yeh kisi bhi path ke liye kyun valid hai? Kyunki work sirf endpoints par depend karta hai — yeh $F = \nabla ϕ$ ka direct consequence hai.

Problem 4.2

Problem 4.1 ka answer directly straight segment $r (t) = (2 t, 3 t)$ , $t \in [0, 1]$ ke saath verify karo.

Recall Solution 4.2

$r^{'} (t) = (2, 3)$ .
$F (r (t)) = (y, x) = (3 t, 2 t)$ .
Dot: $3 t (2) + 2 t (3) = 12 t$ .
$W = \int_{0}^{1} 12 t d t = [6 t^{2}]_{0}^{1} = 6$ . ✓ Endpoint computation se match karta hai.

Level 5 — Mastery

Goal: ek closed-curve computation, do tareekon se, yeh reveal karte huye ki shortcut kab fail hota hai.

Problem 5.1

$F = (- y, x)$ lo. Full unit circle counterclockwise ke around work compute karo, $r (t) = (cos t, sin t)$ , $t \in [0, 2 π]$ . Phir Green's Theorem use karke explain karo kyun ek conservative field $0$ deta.

Recall Solution 5.1

Direct computation.

$r^{'} (t) = (- sin t, cos t)$ .
$F (r (t)) = (- sin t, cos t)$ .
Dot: $(- sin t) (- sin t) + (cos t) (cos t) = sin^{2} t + cos^{2} t = 1$ .
$W = \int_{0}^{2 π} 1 d t = 2 π .$

Green's Theorem check. Ek positively oriented closed curve $C$ ke liye jo region $D$ bound karti hai, $\oint_{C} P d x + Q d y = \iint_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d A .$ Yahan $P = - y, Q = x$ , toh $\partial_{x} Q - \partial_{y} P = 1 - (- 1) = 2$ . Isliye $W = \iint_{D} 2 d A = 2 \cdot (area of unit disc) = 2 π .$ ✓ Same $2 π$ .

Nonzero kyun? Ek conservative field ek loop ke baad same potential value par wapas aata hai, jo $\oint = 0$ deta hai. Yahan $\partial_{x} Q - \partial_{y} P = 2 \neq = 0$ , toh $F$ conservative nahi hai — loop genuinely $2 π$ net work karta hai. Yeh swirling "rotation" field hai.

Problem 5.2 (two-readings identity ki mastery)

Same circle par $F = (- y, x)$ ke liye, verify karo ki $\int_{C} F \cdot d r = \int_{C} (F \cdot T) d s$ right-hand side compute karke.

Recall Solution 5.2

Unit tangent: $T = \frac{r ^{'} ( t )}{∣ r ^{'} ( t ) ∣} = \frac{( - sin t , cos t )}{1} = (- sin t, cos t)$ .
$F \cdot T = (- sin t) (- sin t) + (cos t) (cos t) = 1$ . Field yahan purely tangential hai — yeh exactly motion ke along point karta hai, magnitude $1$ .
$d s = ∣ r^{'} ∣ d t = 1 d t$ , toh $\int_{C} (F \cdot T) d s = \int_{0}^{2 π} 1 d t = 2 π .$ ✓ Bilkul same, jaisa $d r = T d s$ ne promise kiya tha.

Recall

Recall Rapid self-test

"Wire ka mass" ke liye tumne kaunsa integral use kiya? → scalar $\int_{C} f d s$ .
Curve reverse karne se tumhare kaunse answers badle? → sirf vector/work integral (P3.1, sign flip).
Parametrisation tez karne se mass badla? → nahi (P3.2), $d s$ invariant hai.
Shortcut $W = ϕ (end) - ϕ (start)$ ke liye kya condition chahiye? → $F = \nabla ϕ$ conservative ho.
$(- y, x)$ ka loop zero kyun nahi tha? → curl $= 2 \neq = 0$ ; conservative nahi hai.

Connections

Arc length — Problem 1.1 exactly $\int_{C} 1 d s$ hai.
Dot product — upar har vector integral mein.
Parametric curves — har $r (t)$ mein.
Gradient and conservative fields — Problems 4.1–4.2.
Fundamental Theorem for Line Integrals — endpoint shortcut.
Green's Theorem — Problem 5.1.