Fundamental theorem for line integrals: Question bank

4.4.28 · D5 · Maths › Multivariable Calculus › Fundamental theorem for line integrals

Intuition Is page ko kaise use karein

Neeche diye gaye har item mein ek trap hai jo Fundamental Theorem for Line Integrals (FTLI) tumhare liye set karta hai. Maqsad ye nahi hai ki numbers grind karo — maqsad ye hai ki reasoning ki galti ko pakdo pehle, warna poora problem kharab ho jaata hai. Har answer tab hi reveal karo jab tum apni justification zor se bol chuke ho. Agar justify nahi kar sakte, toh samjho abhi pata nahi hai.

Ek line ka reminder jis par sab kuch tika hai: agar $F = \nabla f$ (ek conservative field jiska potential $f$ ho), toh $\int_{C} \nabla f \cdot d r = f (B) - f (A),$ jahan $C$ start point $A$ se end point $B$ tak jaata hai. Yahan $\nabla f$ (gradient) partial derivatives ka vector hai $(f_{x}, f_{y})$ , aur $\int_{C} F \cdot d r$ line integral hai — field ko tiny displacement ke saath dot karke curve ke along sum kiya jaata hai.

True or false — justify

Recall T/F: Kisi bhi vector field ka line integral ek curve par sirf endpoints par depend karta hai.

False ::: Sirf conservative fields ( $F = \nabla f$ ) path independent hote hain; ek general field jaise $F = (- y, x)$ same do points ke beech alag-alag paths par alag-alag answers deta hai.

Recall T/F: Agar

$\oint_{C} F \cdot d r = 0$ kisi ek particular closed loop $C$ ke liye, toh $F$ conservative hai. False ::: Conservativeness ke liye zaroori hai ki har closed loop zero de; ek akela vanishing loop accident se ho sakta hai (jaise symmetry ki wajah se) jabki doosre loops nonzero ho sakte hain.

Recall T/F: Agar

$F = \nabla f$ , toh domain mein har closed curve $C$ ke liye $\oint_{C} F \cdot d r = 0$ hoga. True ::: Ek closed curve ka start point $=$ end point hota hai, isliye $A = B$ aur $f (B) - f (A) = 0$ ; ye sab loops ke liye tab hi hold karta hai jab domain par har jagah potential exist karta ho.

Recall T/F: Condition

$P_{y} = Q_{x}$ guarantee karta hai ki $F = (P, Q)$ conservative hai. False ::: Ye sirf simply connected region par conservativeness guarantee karta hai; ek aisi domain par jo hole ke saath ho (jaise plane minus origin), $P_{y} = Q_{x}$ hold kar sakta hai phir bhi hole ke around ek loop nonzero integral de sakta hai.

Recall T/F:

$A$ se $B$ tak ki do curves jo ek doosre ko cross karti hain, unka line integral ek conservative field ke liye same hoga. True ::: Path independence ka matlab hai ki $A$ se $B$ tak sab curves $f (B) - f (A)$ deti hain; ye cross karein, wiggle karein, ya length share karein — koi farq nahi padta.

Recall T/F: Agar

$f$ aur $f + 7$ dono $F$ ke potentials hain, toh ye $\int_{C} F \cdot d r$ ki alag-alag values denge. False ::: Constant cancel ho jaata hai difference mein: $(f (B) + 7) - (f (A) + 7) = f (B) - f (A)$ , isliye integration ka constant kabhi bhi integral ko affect nahi karta.

Recall T/F: Ek conservative field ka open path par nonzero line integral ho sakta hai.

True ::: Path independence ka matlab zero nahi hota — matlab hota hai $f (B) - f (A)$ , jo nonzero hoga jab bhi endpoint potentials alag hon.

Spot the error

Recall "

$F = (- y, x)$ ; kyunki ye smooth aur rotational lagta hai, main FTLI use karunga aur bas endpoints subtract kar dunga." Error ::: Ye field conservative nahi hai: $P_{y} = - 1 \neq = 1 = Q_{x}$ , isliye koi potential exist nahi karta aur FTLI apply nahi hota — integral sach mein path par depend karta hai.

Recall "Maine

$f_{x} = P$ find kar liya isliye $f = \int P d x$ ; ho gaya, yahi potential hai." Error ::: Integration ka constant doosre variable par depend kar sakta hai, isliye $f = \int P d x + h (y)$ ; tumhe abhi bhi $y$ mein differentiate karna hoga aur $h (y)$ pin down karne ke liye $f_{y} = Q$ se match karna hoga.

Recall "Vortex field

$(\frac{- y}{x ^{2} + y ^{2}}, \frac{x}{x ^{2} + y ^{2}})$ ke liye $P_{y} = Q_{x}$ hai, isliye unit circle ke around iski integral $0$ hai." Error ::: Domain (plane minus origin) simply connected nahi hai; loop hole ko enclose karta hai aur integral $2 π$ hai, $0$ nahi — equal-partials test yahan conservativeness guarantee karne mein fail karta hai.

Recall "Curve

$C$ complicated hai, isliye mujhe $r (t)$ parametrise karna hoga aur integral grind karna hoga, bhale hi $F = \nabla (x^{2} y)$ diya ho." Error ::: Kyunki $F$ ek gradient ke roop mein diya gaya hai, FTLI tumhe parametrisation bilkul skip karne deta hai aur bas $f (B) - f (A)$ compute karo — messy path irrelevant hai.

Recall "FTLI hai

$\int_{C} \nabla f \cdot d r = f (A) - f (B)$ , end minus... wait, start minus end." Error ::: Sahi orientation hai $f (B) - f (A)$ = end minus start (FTC ke $g (b) - g (a)$ se match karta hai); ise ulta karna sign flip kar deta hai, jo bilkul wahi effect hai jo $C$ ko backwards traverse karne se hota hai.

Recall "

$F$ conservative hai, isliye $(0, 0)$ se $(1, 1)$ tak is open path par $\int_{C} F \cdot d r = 0$ hoga." Error ::: Conservative ka matlab open paths par zero nahi hota; ye $f (1, 1) - f (0, 0)$ ke barabar hai, jo zero tab hi hoga jab un dono potential values mein ittefaq se coincide ho.

Why questions

Recall FTLI ka proof ordinary

FTC tak kyun reduce ho jaata hai? Kyunki integrand $\nabla f (r (t)) \cdot r^{'} (t)$ bilkul $\frac{d}{d t} f (r (t))$ hai chain rule ki wajah se ::: isliye ye ek perfect derivative $g^{'} (t)$ hai, aur FTC $\int_{a}^{b} g^{'} (t) d t$ ko $g (b) - g (a) = f (B) - f (A)$ mein badal deta hai.

Recall FTLI hold karne ke liye

$C$ par $\nabla f$ continuous kyun hona chahiye? Continuity ensure karti hai ki $g (t) = f (r (t))$ differentiable hai aur uska derivative integrable hai ::: iske bina FTC step fail ho sakta hai — $C$ ke along koi jump ya singularity "integral of a derivative" argument ko tod deta hai.

Recall Ek conservative field ka closed-loop integral kyun zero hota hai, intuitively?

Kyunki tum apni starting height par wapas aa jaate ho ::: potential elevation ki tarah hai, aur start par wapas jaane wale kisi bhi loop par net elevation change zero hota hai, isliye $f (B) - f (A) = 0$ .

Recall Vortex counterexample mein asli culprit domain mein "hole" hai, formula nahi — kyun?

Kyunki ek potential ko ek single continuous "angle" function hona hoga, lekin angle har baar hole ke around circle karne par $2 π$ jump karta hai ::: punctured plane par koi single-valued $f$ exist nahi karta, isliye FTLI ki hypothesis quietly fail ho jaati hai.

Recall Mixed-partials test

$P_{y} = Q_{x}$ potentials se aata hi kyun hai? Agar $F = \nabla f$ toh $P = f_{x}$ aur $Q = f_{y}$ , aur continuous mixed partials commute karte hain ::: isliye $P_{y} = f_{x y} = f_{y x} = Q_{x}$ — ye test sirf Clairaut's equality hai, jo potential ke exist hone ke liye necessary hai.

Recall Hum specific path ignore kar sakte hain lekin endpoints kabhi nahi — kyun?

Path-dependent contributions exactly cancel ho jaate hain (yahi toh theorem ka content hai) ::: lekin $f (B) - f (A)$ ko phir bhi dono endpoint values chahiye, isliye endpoints mein hi saari surviving information hoti hai.

Recall FTLI ko ordinary FTC ka "multivariable twin" kyun kaha jaata hai?

| ::: Dono kehte hain "ek derivative ko integrate karo, boundary values lo": FTC deta hai $\int_{a}^{b} g^{'} d x = g (b) - g (a)$ ek interval ke endpoints par, aur FTLI deta hai $\int_{C} \nabla f \cdot d r = f (B) - f (A)$ ek curve ke endpoints par.

Edge cases

Recall Edge case:

$\int_{C} \nabla f \cdot d r$ kya hoga jab $C$ ek single point ho (start = end, zero length)? Zero ::: $A = B$ ke saath hume milta hai $f (B) - f (A) = 0$ , jo degenerate "loop" of no length ke saath consistent hai; integrate karne ke liye kuch hai hi nahi.

Recall Edge case:

$F = 0$ (zero field). Kya ye conservative hai, aur iski koi bhi line integral kya hai? Haan, constant potential ke saath conservative hai ::: $0 = \nabla (c)$ kisi bhi constant $c$ ke liye, aur har line integral $c - c = 0$ hai chahe path ya endpoints kuch bhi ho.

Recall Edge case: Curve

$C$ beech mein $f$ ki domain se bahar chali jaati hai (jaise ek aisi point se guzarti hai jahan $\nabla f$ undefined ho). Kya hum phir bhi FTLI apply kar sakte hain? Nahi ::: FTLI ko chahiye ki $\nabla f$ $C$ ke puri par continuous ho; agar $C$ kisi singularity ko cross kare toh theorem apply nahi hota aur integral converge bhi nahi ho sakta.

Recall Edge case: Kya hoga agar hum same conservative field ko

$C$ along aur phir $C$ reversed along traverse karein? Dono integrals negative hain aur sum zero hota hai ::: reverse karne se $A$ aur $B$ swap ho jaate hain, milta hai $f (A) - f (B) = - (f (B) - f (A))$ ; saath mein ye ek closed loop banate hain jiska total $0$ hai.

Recall Edge case: Ek field right half-plane

$x > 0$ par conservative hai lekin tum ek aisi path par integrate karte ho jo $x > 0$ mein hi rehti hai. Kya kahin aur ka global "hole" matter karta hai? Nahi ::: jo matter karta hai wo ye hai ki $C$ aur ek potential ek simply connected sub-region mein hon; kyunki right half-plane simply connected hai aur $C$ ko contain karta hai, FTLI wahan apply hota hai chahe field door kahin bhi bura bartaav kare.

Recall Edge case: Kya ek constant nonzero field jaise

$F = (3, 4)$ conservative hai, aur iska potential kya hai? Haan ::: $F = \nabla (3 x + 4 y)$ , isliye $f = 3 x + 4 y$ ek potential hai aur koi bhi line integral $f (B) - f (A)$ ke barabar hoga; constant fields $R^{2}$ par hamesha conservative hote hain.

Connections

Fundamental theorem for line integrals — wo parent theorem jise yahan har trap test karta hai.
Conservative vector fields — wo exact class jahan path independence rehti hai.
Potential functions — wo $f$ jiske endpoint values answer hain.
Green's theorem — explain karta hai kyun "hole" $P_{y} = Q_{x}$ test ko tod deta hai.
Fundamental Theorem of Calculus — wo 1D ancestor jis tak proof reduce hota hai.
Gradient — wo operator $\nabla$ jo conservative fields produce karta hai.
Curl — $P_{y} = Q_{x}$ test ka 3D generalization.