Visual walkthrough — Fundamental theorem for line integrals

4.4.28 · D2 · Maths › Multivariable Calculus › Fundamental theorem for line integrals

Hum assume karte hain ki tum sirf itna jaante ho: ek arrow (vector) kya hota hai, aur tiny pieces ko add karne ka kya matlab hai. Baki sab — dot product, gradient, "line integral" — sab kuch neeche ek picture se rebuild kiya jayega.

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Step 1 — "Field" aur "path" kya hota hai?

KYA. Ek vector field $\mathbf{F}$ ek aisi rule hai jo plane ke har point par ek chota arrow lagati hai. Ek path (curve) $C$ ek walking route hai start point $A$ se end point $B$ tak.

KYUN. Pehle hum field ko path ke saath integrate kar sakein, hume dono cheezein clearly samajhni hain. Baad mein sab kuch sirf yahi hai: "field kitna mere chalne ki direction mein push karta hai?"

PICTURE. Neeche, pale arrows field $\mathbf{F}$ hain (har grid point par ek). Dark curve hamara route $C$ hai, $A$ (bottom-left) se $B$ (top-right) tak jaata hua.

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Step 2 — Walk ko ek clock variable $t$ se describe karo

KYA. Hum curve ke saath apni position ko ek single number $t$ se label karte hain ("clock par time"), jo start par $t=a$ se chalke end par $t=b$ tak jaata hai. Clock-reading $t$ par position $\mathbf{r}(t)$ likhi jaati hai — ek aisa point jo $t$ par depend karta hai.

$\mathbf{r}(t)=(\underset{\text{across}}{\underbrace{x(t)}},\underset{\text{up}}{\underbrace{y(t)}}),\underset{\text{start}}{\underbrace{\mathbf{r}(a)=A}},\underset{\text{end}}{\underbrace{\mathbf{r}(b)=B}}.$

KYUN. Ek curve do dimensions mein wandering karti hai, jise directly integrate karna mushkil hai. Lekin agar ek single clock variable $t$ use drive kare, toh sab kuch ek variable ka function ban jaata hai — aur one-variable calculus ek solved problem hai. Yahi trick hai jo ordinary Fundamental Theorem of Calculus ko eventually kaam karne deti hai.

PICTURE. Wahi curve, ab clock-readings $t=a,t_1,t_2,\dots ,t=b$ uske saath ticked. Har tick ek point $\mathbf{r}(t)$ hai.

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Step 3 — Tiny step $d\mathbf{r}$ aur velocity arrow ${\mathbf{r}}^{\prime }(t)$

KYA. Ek tiny clock-interval $dt$ mein, teri position ek tiny displacement arrow $d\mathbf{r}$ se move karti hai. Woh arrow curve ke saath point karta hai, aur uski length teri speed times $dt$ hai:

$d\mathbf{r}=\underset{\text{velocity arrow}}{\underbrace{{\mathbf{r}}^{\prime }(t)}}\underset{\text{tiny time}}{\underbrace{dt}}.$

Yahan ${\mathbf{r}}^{\prime }(t)$ velocity hai: woh direction jisme tum ja rahe ho aur kitni tezi se, bilkul curve ke tangent par.

KYUN. Ek line integral poochta hai "field mera chalne ki direction mein kitna help karta hai?" Iska jawab dene ke liye hume chalne ki direction ek arrow ke roop mein chahiye — woh arrow ${\mathbf{r}}^{\prime }(t)$ hai. Use $dt$ se multiply karne par woh actual tiny hop mil jaata hai jo hum lete hain.

PICTURE. Ek tick par zoom-in: green tangent arrow ${\mathbf{r}}^{\prime }(t)$, aur tiny hop $d\mathbf{r}$ curve ke saath leta hua.

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Step 4 — Line integral assemble karo

KYA. Har point par hum field arrow $\mathbf{F}(\mathbf{r}(t))$ ko apne tiny step ke saath dot karte hain, aur saare pieces add karte hain:

${\int }_C\mathbf{F}\cdot d\mathbf{r}={\int }_a^b\underset{\text{field jahan main hun}}{\underbrace{\mathbf{F}(\mathbf{r}(t))}}\cdot \underset{\text{meri direction}}{\underbrace{{\mathbf{r}}^{\prime }(t)}}\underset{\text{tiny hop}}{\underbrace{dt}}.$

Wiggly ${\int }_C$ ek plain ${\int }_a^b$ clock ke upar ban gaya — ek sachcha ordinary integral.

KYUN. Har term $\mathbf{F}\cdot {\mathbf{r}}^{\prime }dt$ ek hop mein mila "path ke saath push" ka chhota sa hissa hai. Unhe sum karna (yahi $\int$ karta hai) poore trip mein field ki help ka total deta hai. Yeh general line integral hai — kisi bhi field ke liye sach, chahe conservative ho ya nahi.

PICTURE. Curve par do field arrows: ek step ke saath almost aligned (bada positive contribution, coral), ek almost perpendicular (tiny contribution, faded). Dot product har hop ko alignment se weight karta hai.

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Step 5 — Special case: field ek gradient hai

KYA. Ab maano field arbitrary nahi balki kisi scalar "height" function $f$ ka gradient hai:

$\mathbf{F}=\nabla f=(\underset{\text{slope in }x}{\underbrace{f_x}},\underset{\text{slope in }y}{\underbrace{f_y}}).$

$f(x,y)$ ko har point ke upar ek pahari ki height socho. Gradient arrow $\nabla f$ kisi bhi jagah seedha upar ki taraf point karta hai, aur jahan pahari steep hai wahan lambaa hota hai.

KYUN. Yahi woh ek extra assumption hai jo poori cheez collapse kar deti hai. Is tarah se bana field conservative kehlata hai, aur $f$ uska potential hai. Hume $f$ ka exist karna zaroori hai kyunki agla step secretly $f$ ka rate of change rebuild karta hai.

PICTURE. Ek pahari (shaded contours = height $f$). Teen points par, lavender $\nabla f$ arrows upar ki taraf point karte hain, contour lines ke perpendicular, steep zameen par lamba.

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Step 6 — Dil ki baat: integrand ek perfect derivative hai

KYA. Define karo $g(t)=f(\mathbf{r}(t))$: chalte waqt tumhari height, clock ka ek plain one-variable function. Multivariable chain rule kehti hai uska rate of change exactly hamara integrand hai:

$\frac{dg}{dt}=\underset{\text{x mein move karne se height change}}{\underbrace{f_x{x}^{\prime }(t)}}+\underset{\text{y mein move karne se height change}}{\underbrace{f_y{y}^{\prime }(t)}}=\underset{\text{Step 4 ka wahi integrand}}{\underbrace{\nabla f(\mathbf{r}(t))\cdot {\mathbf{r}}^{\prime }(t)}}.$

KYUN. Jab clock ticks karta hai, $f$ dono coordinates ke through change hota hai. Har coordinate contribute karta hai (uski slope)(woh coordinate kitni tezi se move karta hai) — aur un dono contributions ko add karna precisely ek dot product $\nabla f\cdot {\mathbf{r}}^{\prime }$ hai. Toh jo hum integrate kar rahe hain woh koi bhi function nahi hai: yeh $g^{\prime }(t)$ hai, tumhari height ka derivative. Yahi pura secret hai.

PICTURE. Left: pahari jisme path drawn hai. Right: height profile $g(t)=f(\mathbf{r}(t))$ clock $t$ ke against plotted — ek simple 1-D curve jiska har $t$ par slope hamara integrand hai.

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Step 7 — Ordinary FTC se finish karo

KYA. Hum ab ek plain derivative ko ek plain interval par integrate kar rahe hain, toh single-variable Fundamental Theorem of Calculus apply hota hai:

${\int }_a^b\underset{g^{\prime }(t)}{\underbrace{\frac{dg}{dt}}}dt=\underset{\text{end par height}}{\underbrace{g(b)}}-\underset{\text{start par height}}{\underbrace{g(a)}}.$

$g(t)=f(\mathbf{r}(t))$ aur $\mathbf{r}(a)=A,\mathbf{r}(b)=B$ substitute karne par:

$\boxed{{\int }_C\nabla f\cdot d\mathbf{r}=f(B)-f(A)}■$

KYUN. FTC kehta hai "ek derivative ka total accumulated change original ka net change hota hai." Kyunki hamara integrand hi $g^{\prime }$ hai, path ke saath total pushed-along equals sirf (end height) − (start height). Saare beech ke wiggles cancel ho gaye.

PICTURE. Height profile $g(t)$: $g^{\prime }(t)$ ke neeche shaded area $g(a)$ se $g(b)$ tak ke vertical drop ke barabar hai — sirf do endpoints ki heights bachti hain.

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Step 8 — Har case: do paths, aur closed loop

KYA. Kyunki answer $f(B)-f(A)$ hai aur usme route ka koi nishan nahi:

1. Do alag paths $C_1,C_2$ from $A$ to $B$ same value dete hain — path independence.
2. Ek closed loop ($A=B$) deta hai $f(A)-f(A)=0$.
3. Degenerate path (jagah par raho, $A=B$, zero-length curve) bhi $0$ deta hai — consistent, koi special case nahi chahiye.

KYUN. Ye nayi theorems nahi hain; yeh wahi ek formula hai alag alag situations mein padha gaya. Sabko cover karo taaki koi bhi scenario surprise na kare. (Parent se liya gaya warning: iske liye $f$ ka region mein har jagah actually exist karna zaroori hai — wahan vortex counterexample dekho, jo Green's theorem aur curl test se juda hua hai.)

PICTURE. Ek hi pahari par $A$ se $B$ tak do coloured trails (lavender, coral) — same height difference. Neeche, ek closed loop $A$ par wapas jaata hua: net climb $=0$.

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Ek-picture summary

Ek hi canvas par sab kuch: field arrows, wiggly path, hidden height-hill, aur do endpoint heights mein collapse.

Recall Feynman: poori walk ko simple words mein batao

Ek pahari socho; tumhari height $f$ hai. Har jagah ek arrow seedha upar ki taraf point karta hai — woh gradient field $\nabla f$ hai. Ab koi bhi trail chalo ek neeche wale cabin $A$ se ek upar wale lodge $B$ tak. Har tiny step par tum poochte ho, "main kitna chadha?" — woh hai field arrow dotted with your step. Un saare tiny climbs ko add karo. Lekin yahan magic hai: har tiny climb exactly us step par tumhari height mein change hai. Toh unhe sab add karna sirf height changes ka total hai — jo telescope karke (final height) − (starting height) ban jaata hai. Beech ka messy hissa completely cancel ho jaata hai. Seedha trail lo, loop-the-loop trail lo: same answer, kyunki pahari ko parwah nahi ki tum kaise pahunche. Aur ghar wapas ek full loop chalo? Tum usi height par end karte ho jahan se chale the — net climb zero.

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Connections

• Fundamental theorem for line integrals — woh parent result jin pictures se yeh derive hota hai.
• Line integrals of vector fields — Step 4 ka general integral gradient trick se pehle.
• Gradient — Step 5 ka uphill-arrow operator.
• Potential functions — height function $f$.
• Conservative vector fields — woh fields jahan yeh collapse hota hai.
• Fundamental Theorem of Calculus — Step 7 mein 1-D engine.
• Green's theorem · Curl — govern karte hain ki ek valid $f$ exist kab karta hai.