Arc length formula — derivation: Visual walkthrough

4.2.16 · D2 · Maths › Calculus II — Integration › Arc length formula — derivation

Step 1 — Hum actually maap kya rahe hain?

Intuition WHAT / WHY / PICTURE

WHAT. Hamare paas ek curve hai — ek tedhi-medhi line jo kisi rule $y = f (x)$ se bani hai. Yeh rule ek machine hai: horizontal position $x$ daalo, height $y$ milti hai. Hum chahte hain curve ki length khud, woh distance jo ek choti si chinti crawl karegi curve ke saath chalne mein, left end $x = a$ se right end $x = b$ tak. WHY yeh mushkil hai. Ek straight line ke liye hum seedha end-to-end maap sakte hain. Curve ke liye aisa koi shortcut nahi — uski length uske ends ke beech ki straight distance se zyada hoti hai, kyunki chinti har bump ke upar aur neeche jaati hai. Koi ek memorised formula nahi hai, isliye humein ek banana padega. PICTURE. Neeche, cyan curve woh path hai. Dashed amber line ends ke beech ki straight "as-the-crow-flies" distance hai. Notice karo cyan path clearly longer hai — woh gap exactly wahi problem hai jo hum solve kar rahe hain.

Figure — Arc length formula — derivation

Yahan $x$ = kitna right ho tum, $y = f (x)$ = rule jo height deta hai, $a$ = start, $b$ = end. Letter $L$ ka matlab hoga cyan curve ki sach mein length.

Step 2 — Curve ko tiny straight pieces mein kaato

Har dot ek point $(x_{i}, y_{i})$ hai jahan $y_{i} = f (x_{i})$ . $i$ sirf ek counter hai: 1st dot, 2nd dot, aur aise $n$ -th tak.

Step 3 — Ek tiny triangle, aur uski tilted side kitni lambi hai

Intuition WHAT / WHY / PICTURE

WHAT. Dot $i - 1$ aur dot $i$ ke beech wale chord par focus karo. Ek se doosre tak jaane mein tum right move karte ho ek amount se jise hum $Δ x_{i}$ kehte hain aur upar ya neeche ek amount $Δ y_{i}$ se. Chhota symbol $Δ$ (Greek "delta") ka matlab sirf "mein change" hai — cheez kitni badi (ya choti) hui. Is chord ki straight-line length ko hum symbol $L_{i}$ ka naam dete hain — yeh uski definition hai, woh number jo hum abhi compute karne wale hain. WHY right triangle. Sideways aur vertical moves ek doosre se $9 0^{\circ}$ par hain, isliye chord, sideways move, aur vertical move ek right-angled triangle banate hain. Yahi woh shape hai jiske liye Pythagoras bana tha. PICTURE. Neeche, upar: ek rising chord, horizontal leg $Δ x_{i}$ labeled, vertical leg $Δ y_{i}$ , slanted side (chord length) $L_{i}$ labeled. Neeche: ek falling chord jahan curve neeche jaata hai.

Pythagoras ko scratch se dobara banana (kuch memorise karne ki zaroorat nahi). Apne triangle ki char copies ko ek tilted square ke around rakho jiska side chord $L_{i}$ hai. Bada bahari square ka side $Δ x_{i} + Δ y_{i}$ hai. Uska area tilted inner square ( $L_{i}^{2}$ ) aur char triangles ( $4 \times \frac{1}{2} Δ x_{i} Δ y_{i}$ ) ke barabar hai:

big square (Δ x_{i} + Δ y_{i})^{2} = tilted square L_{i}^{2} + four triangles 4 \cdot \frac{1}{2} Δ x_{i} Δ y_{i}

Left side expand karo: $(Δ x_{i})^{2} + 2Δ x_{i} Δ y_{i} + (Δ y_{i})^{2}$ . $2Δ x_{i} Δ y_{i}$ triangles ko cancel kar deta hai, aur milta hai

L_{i} = (Δ x_{i})^{2} + (Δ y_{i})^{2} .

$(Δ x_{i})^{2}$ — sideways travel, squared.
$(Δ y_{i})^{2}$ — vertical travel, squared.
— squaring ko undo karta hai taaki actual length mile. Dekho Pythagorean theorem.

Step 4 — Sideways step $Δ x_{i}$ ko root se bahar nikalo

L_{i} = (Δ x_{i})^{2} + (Δ y_{i})^{2} = (Δ x_{i})^{2} (1 + \frac{( Δ y _{i} ) ^{2}}{( Δ x _{i} ) ^{2}}) = 1 + (chord ki tilt \frac{Δ y _{i}}{Δ x _{i}})^{2} Δ x_{i}

Hum $Δ x_{i}$ ko root se bahar nikal sake kyunki $Δ x_{i} > 0$ (hum hamesha left-to-right chalte hain — yeh exactly Step 1 ka "no doubling back, advance rightward" assumption hai — toh $(Δ x_{i})^{2} = Δ x_{i}$ , koi minus sign nahi).
$1$ sideways leg se bacha hua hai — yeh wahi hai jo humein horizontal travel bhoolne se rokta hai.
$\frac{Δ y _{i}}{Δ x _{i}}$ = vertical-move ÷ right-move = chord ka slope (ek "secant slope"). Agar chord gire, yeh ratio negative hai — lekin aage square ho jaata hai, toh phir sign wash out ho jaata hai.

Step 5 — Chord ke slope ko ek sach mein derivative banao (Mean Value Theorem)

Un conditions ke under, Mean Value Theorem ek point $x_{i}^{*}$ promise karta hai $x_{i - 1}$ aur $x_{i}$ ke beech strictly, jahan

tangent slope at x_{i}^{*} f^{'} (x_{i}^{*}) = chord ka average slope \frac{f ( x _{i} ) - f ( x _{i - 1} )}{x _{i} - x _{i - 1}} = \frac{Δ y _{i}}{Δ x _{i}} .

$f^{'}$ matlab "slope function" — curve kitni steep hai ek point par.
$x_{i}^{*}$ (star) chhote interval ke andar koi jagah hai; MVT kehta hai yeh exist karta hai, hume ise dhundhna nahi.

Step 4 ke result mein substitute karo:

L_{i} = 1 + (f^{'} (x_{i}^{*}))^{2} Δ x_{i} .

Step 6 — Sab add karo, shrink karo, aur integral padho

L = n \to \infty lim i = 1 \sum n height g (x_{i}^{*}) 1 + (f^{'} (x_{i}^{*}))^{2} strip width Δ x_{i} = \int_{a}^{b} 1 + (f^{'} (x))^{2} d x .

Sample height $g (x_{i}^{*}) = 1 + (f^{'} (x_{i}^{*}))^{2}$ integrand $g (x)$ ban jaata hai.
$Δ x_{i}$ ban jaata hai $d x$ , ek infinitely thin width.
$\sum$ ban jaata hai $\int$ — wahi idea "sab add karo," ab continuous.

Step 7 — Edge cases: kya picture tab bhi hold karti hai?

Flat curve ( $f^{'} = 0$ ). Vertical leg $Δ y_{i} = 0$ ; triangle ek horizontal segment mein collapse ho jaata hai. Integrand $= 1 + 0 = 1$ , toh $L = \int_{a}^{b} 1 d x = b - a$ — exactly width. ✓ Ek flat line ki length hai hi uska horizontal span.
Vertical wall ( $f^{'} \to \infty$ ). Slope blow up karta hai; $y = f (x)$ formula choke karta hai kyunki vertical piece mein $Δ x_{i} = 0$ hai (bahar nikalne ke liye kuch nahi). Fix: $x = g (y)$ form par flip karo aur $y$ mein integrate karo — vertical leg "safe" direction ban jaata hai.
Corner (slope jumps). Ek sharp kink par $f^{'}$ exist nahi karta — yeh us ek point par MVT ki differentiability condition violate karta hai. Kyunki ek bura point zero width ka hota hai, integral unaffected rehta hai — bas corner par interval split karo aur do pieces add karo.

Ek picture mein summary

Film strip ko left-to-right padho: curve → chords → ek triangle → $d x$ bahar nikalo → tangent chord se match karta hai → strips sum hokar integral bante hain. Har arrow upar ek step hai.

Recall Feynman: plain words mein poora walkthrough

Tum ek bumpy path ki length chahte ho. Curve ko directly maap nahi kar sakte, toh tum iske saath chhoti straight sticks end-to-end rakhte ho (Step 2). Har stick ek chote right triangle ki lambi side hai jiske legs hain "thoda sa rightward" aur "thoda sa upar ya neeche"; corner-square trick (Pythagoras) kehti hai stick ki length $right^{2} + vertical^{2}$ hai — aur kyunki vertical bit square hoti hai, koi farq nahi ki curve climb kare ya gire (Step 3). Sab kuch rightward steps par ek sum mein collect karne ke liye, rightward bit ko factor out karo aur milta hai $1 + slope^{2}$ times the rightward step (Step 4). "Stick ka slope" curve ka real slope nahi hai bilkul, lekin Mean Value Theorem — jab tak curve smooth ho (continuous aur differentiable) — guarantee karta hai ki andar kahin real tangent exactly utni hi tilted hai, toh tum ise $f^{'}$ se replace karo (Step 5). Ab har stick padhti hai $1 + (f^{'})^{2} Δ x$ ; infinitely many infinitely short ones add karna exactly wahi hai jo integral sign matlab hai, deta hai $L = \int_{a}^{b} 1 + (f^{'})^{2} d x$ (Step 6). Ise flat road par test karo (length = width), vertical cliff par (instead $y$ mein integrate karo), aur sharp corner par (wahan split karo) — teeno se survive karta hai (Step 7). Wahi $d s = d x^{2} + d y^{2}$ baad mein surfaces of revolution, parametric aur polar lengths banata hai.

Flashcards

Har chord ko uski length kaun sa chhota shape deta hai?

Ek right triangle jiske legs

Δ x_{i}, Δ y_{i}

hain; chord

L_{i}

hypotenuse hai

(Δ x_{i})^{2} + (Δ y_{i})^{2}

Arc-length formula descending curves ke liye kyun kaam karta hai jahan

Δ y_{i} < 0

ho?

Kyunki

Δ y_{i}

squared enter karta hai, aur

(Δ y_{i})^{2} = ∣Δ y_{i} ∣

; sign wash out ho jaata hai, toh length same hai chahe climb karo ya fall.

Δ x_{i}

ko square root se

\pm

ke bina bahar kyun nikal sakte hain?

Kyunki

Δ x_{i} > 0

(hum left to right chalte hain, koi doubling back nahi), toh

(Δ x_{i})^{2} = Δ x_{i}

Mean Value Theorem ko har subinterval par kaun si do hypotheses chahiye?

f

closed

[x_{i - 1}, x_{i}]

par continuous aur open

(x_{i - 1}, x_{i})

par differentiable.

MVT kya replace karta hai, aur kisse?

Chord ka average slope

Δ y_{i} /Δ x_{i}

ko ek exact derivative

f^{'} (x_{i}^{*})

se ek interior point par.

Final sum ko integral mein kya convert karta hai?

Yeh ek Riemann sum

\sum g (x_{i}^{*}) Δ x_{i}

hai;

Δ x_{i} \to 0

hone par uska limit

\int_{a}^{b} g d x

hai.

Arc-length element

d s

kya hai?

d s = d x^{2} + d y^{2} = 1 + (d y / d x)^{2} d x

; phir

L = \int d s

Ek vertical piece jahan

f^{'} \to \infty

ho, wahan kya karte hain?

y

mein integrate karo

L = \int_{c}^{d} 1 + (d x / d y)^{2} d y

use karke.

Corner wale curve ke liye, non-differentiable point ko kaise handle karte hain?

Interval ko corner par split karo aur do arc lengths add karo.

Connections

Arc length formula — derivation — woh parent jise yeh page visualise karta hai.
Pythagorean theorem — har chord ki length ka engine (Step 3 mein scratch se dobara banaya).
Mean Value Theorem — secant slope ko ek true derivative se swap karta hai (Step 5).
Riemann sums and the definite integral — sum ko integral mein convert karta hai (Step 6).
Surface area of revolution, Parametric arc length, Polar arc length — jahan $d s$ aage jaata hai.