Visual walkthrough — Line integrals — scalar and vector, work done

4.4.27 · D2 · Maths › Multivariable Calculus › Line integrals — scalar and vector, work done

Hum poori cheez zero se banate hain. Agar "dot product" jaisa koi word aaye, hum pehle use draw karte hain.

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Step 1 — Curve kya hoti hai, aur ek number $t$ tumhe us par kyun locate karta hai

Hum us ek number ko $t$ kehte hain. Jaise $t$ ek start value $a$ se end value $b$ tak tick karta hai, ek hilta hua dot curve trace karta hai. Dot ki position likhi jaati hai

$\mathbf{r}(t)=(x(t),y(t)).$

• $\mathbf{r}$ (bold) — position vector: origin se us jagah tak ka ek arrow jahan dot abhi baitha hai.
• $x(t),y(t)$ — dot ke coordinates clock $t$ ke functions ke roop mein.
• $t\in [a,b]$ — clock $a$ (start) se $b$ (finish) tak chalta hai.

Parametrise kyun karo? Kyunki jab curve ke har point ka ek address $t$ hota hai, tab wiggly curve par koi bhi sum single variable $t$ mein ek ordinary integral ban jaata hai — jo hum pehle se karna jaante hain.

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Step 2 — Velocity ${\mathbf{r}}^{\prime }(t)$: woh arrow jo curve ke saath point karta hai

• $\Delta$ ("delta") — seedha shorthand "mein ek chhoti change" ke liye. $\Delta \mathbf{r}$ ek vector hai: length aur direction dono wala ek arrow.
• $\Delta t$ — is step ke liye clock-time ka tiny slice.

$\Delta t$ se divide kyun karo? Ek aisi rate paane ke liye jo step ke shrink hone par zero nahi ho jaaye. $\Delta t\to 0$ shrink karte hue:

${\mathbf{r}}^{\prime }(t)={\lim }_{\Delta t\to 0}\frac{\Delta \mathbf{r}}{\Delta t}=(x^{\prime }(t),y^{\prime }(t)).$

• ${\mathbf{r}}^{\prime }(t)$ — velocity vector. Iska direction curve par tangent hai (dikhata hai dot kahan ja raha hai); iska length $|{\mathbf{r}}^{\prime }(t)|$ speed hai (kitni tezi se arc-length khatam ho rahi hai).
• $x^{\prime }(t),y^{\prime }(t)$ — har coordinate ke derivatives: $x$ aur $y$ alag-alag kitni tezi se change hote hain.

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Step 3 — Ek tiny step ko measure karne ke do tarike: length $ds$ vs displacement $d\mathbf{r}$

Yeh woh fork hai jo do alag integrals produce karta hai.

Tiny step ki length. Displacement arrow ki length lo:

$\Delta s=|\Delta \mathbf{r}|\approx |{\mathbf{r}}^{\prime }(t)|\Delta t.$

Shrinking se milta hai arc-length element $ds=|{\mathbf{r}}^{\prime }(t)|dt,|{\mathbf{r}}^{\prime }(t)|=\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}.$

• $ds$ — ek scalar, hamesha $\ge 0$ (yeh ek length hai; Arc length dekho).
• $|{\mathbf{r}}^{\prime }(t)|$ — speed, squared component-rates ke square-root se. Squares signs khatam karte hain, isliye direction yahan phek di jaati hai — yeh yaad rakho.

Tiny step ka displacement. Poora arrow rakho: $d\mathbf{r}={\mathbf{r}}^{\prime }(t)dt.$

• $d\mathbf{r}$ — ek vector. Yeh direction yaad rakhta hai. Apna safar reverse karo aur ${\mathbf{r}}^{\prime }$ sign flip kar leta hai, isliye $d\mathbf{r}$ bhi sign flip karta hai.

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Step 4 — Scalar line integral: (value)$\times$(tiny length) add karo

$\Delta s_i$ factor kyun? Kyunki mass = density $\times$ length, aur $\Delta s_i$ piece ki length hai. Pieces ko shrink hone do; sum integral ban jaata hai. Step-3 ka fact $ds=|{\mathbf{r}}^{\prime }(t)|dt$ substitute karo taaki sab kuch $t$ mein ho:

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Step 5 — Dot product: push ka sirf "forward" wala part nikalna

Work integral se pehle, humein woh tool banana hoga jo woh use karta hai. Woh tool hai dot product.

Woh number jo "kitna $\mathbf{F}$, $\mathbf{d}$ ke along hai" extract karta hai woh hai dot product (Dot product mein poori details):

$\mathbf{F}\cdot \mathbf{d}=F_x{d}_x+F_y{d}_y=|\mathbf{F}||\mathbf{d}|\cos \theta .$

• $\theta$ — dono arrows ke beech ka angle.
• $\cos \theta$ — woh dial jo $+1$ (same direction, full help) se $0$ (perpendicular, koi help nahi) se $-1$ (opposite, full fight) tak jaata hai.

Yeh tool kyun aur koi nahi? Akela multiplication direction ignore kar deta; dot product woh unique simple operation hai jo sirf aligned component rakhta hai — exactly wahi jo "work" ko chahiye.

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Step 6 — Work: (forward-push)$\times$(tiny displacement) add karo

Step 3 se $d\mathbf{r}={\mathbf{r}}^{\prime }(t)dt$ substitute karo aur add karo:

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Step 7 — Bridge: vector integral aslaan ek tangential scalar integral kyun hai

Phir, Step 3 ke dono facts ko saath use karke: $d\mathbf{r}={\mathbf{r}}^{\prime }dt=\underset{\mathbf{T}}{\underbrace{\frac{{\mathbf{r}}^{\prime }}{|{\mathbf{r}}^{\prime }|}}}\underset{ds}{\underbrace{|{\mathbf{r}}^{\prime }|dt}}=\mathbf{T}ds.$

Isliye work integral literally ek scalar line integral hai jiska density forward-component $\mathbf{F}\cdot \mathbf{T}$ hai:

${\int }_C\mathbf{F}\cdot d\mathbf{r}={\int }_C(\mathbf{F}\cdot \mathbf{T})ds.$

• $\mathbf{F}\cdot \mathbf{T}$ — har point par ek plain number: "field yahan kitni strongly forward push karta hai."

Isliye work "wire ki mass jaisi lagti hai" lekin ek signed density ke saath.

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Step 8 — Har case: orientation reverse karna, sideways fields, zero speed

Humein corner cases mein kya hota hai woh dikhana hoga, warna tum unprepared miloge.

Case A — curve reverse karo (use ulta chalo). Travel direction flip hoti hai, isliye ${\mathbf{r}}^{\prime }\to -{\mathbf{r}}^{\prime }$ aur isliye $d\mathbf{r}\to -d\mathbf{r}$ aur $\mathbf{T}\to -\mathbf{T}$.

• Work: $W\to -W$. Field ke khilaaf kheenchna work ko undo karta hai. Sign flip hota hai.
• Scalar: $ds=|{\mathbf{r}}^{\prime }|dt$ mein woh square-root hai, isliye $|-{\mathbf{r}}^{\prime }|=|{\mathbf{r}}^{\prime }|$ — koi change nahi. Wire ki mass same rehti hai chahe tum kisi bhi end se shuru karo.

Case B — force exactly motion ke perpendicular hai. Tab $\theta =90^{\circ }$, $\cos \theta =0$, isliye $\mathbf{F}\cdot d\mathbf{r}=0$ har step par: zero work, chahe $\mathbf{F}$ kitna bhi strong kyun na ho. (Ek magnetic-type push jo sirf steer karta hai, kabhi speed up nahi karta.)

Case C — degenerate: dot ruk jaata hai, ${\mathbf{r}}^{\prime }(t)=0$. Us instant par jahan velocity zero hai, $ds=0$ aur $d\mathbf{r}=0$: woh instant dono mein se kisi integral mein kuch contribute nahi karta. Dono formulas valid rehte hain; ek momentary pause na koi mass add karta hai na koi work. (Agar dot poore interval ke liye stuck hai, toh use skip karne ke liye reparametrise karo.)

Recall

$\mathbf{F}=(y,x)$ path-independent kyun hai? Kyunki $\mathbf{F}=\nabla \varphi$ with $\varphi =xy$, isliye $W=\varphi (\text{end})-\varphi (\text{start})$ — sirf endpoints. Fundamental Theorem for Line Integrals aur Gradient and conservative fields ka preview. Check $0\to 1$: $\varphi (1,1)-\varphi (0,0)=1-0=1.$ ✓

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Ek-picture summary

Ek tiny step do children mein split hoti hai: uski length $ds$ (direction square-root se mita di gayi) scalar/mass integral ko feed karti hai; uska displacement $d\mathbf{r}$ (direction rakhi gayi) $\mathbf{F}$ se dot hota hai work integral ko feed karne ke liye. Woh single fork har difference explain karta hai — including yeh kyun curve reverse karne se mass alag nahi hoti lekin work ka sign flip ho jaata hai.

Recall Feynman retelling (cover karo aur aloud explain karo)

Ek chinti ek moode hue wire par chalti hai. Safar ko itne short steps mein kaato ki har ek basically seedha ho. Mass ke liye: har step ki thodi si length hai; multiply karo wahan wire kitni heavy hai se; sab pieces add karo. Length ka koi arrow nahi hai, isliye wire ko ulta chalna same total deta hai — mass ko direction ki parwah nahi. Work ke liye: har point par ek wind hai. Har short step par main sirf wind ka woh part count karta hun jo ant ko forward push kar raha hai — wahi dot product hai, woh tool jo aligned part rakhta hai aur sideways part chod deta hai. "Forward-push × step-length" ko poore saath mein add karo. Ulta chalo aur forward-push, backward-fight ban jaata hai, isliye total sign flip karta hai — work ko direction ki parwah hoti hai. Dono totals usi chhote step se aate hain; woh sirf is mein differ karte hain ki kya hum step ka arrow ($d\mathbf{r}$) rakhte hain ya sirf uski length ($ds$).

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Connections

• Line integrals — scalar and vector, work done — parent note jise yeh walkthrough unfold karta hai.
• Parametric curves — Step 1 ka address system $\mathbf{r}(t)$.
• Arc length — Step 3/4 with $f=1$.
• Dot product — Step 5 mein bana engine.
• Gradient and conservative fields · Fundamental Theorem for Line Integrals — Step 8 ka path-independence.
• Green's Theorem — work integral ka closed-curve version.