Arc length formula — derivation: Worked examples

4.2.16 · D3 · Maths › Calculus II — Integration › Arc length formula — derivation

Scenario matrix

Har arc-length problem in cells mein se kisi ek mein aata hai. Hum kam se kam ek example har cell ka karte hain.

Cell	Kya special hai	Danger kahan hai	Example
A. Straight line	slope constant, integrand constant	koi nahi — yeh sanity check hai	Ex 1
B. Engineered curve	$1 + (y^{'})^{2}$ ek perfect square mein collapse ho jaata hai	algebra spot karna padega	Ex 2
C. Sign trap	$(\dots)^{2} = ∣ \dots ∣$ , expression negative ho jaata hai	absolute value drop karna	Ex 3
D. Curve $x = g (y)$ ke roop mein di gayi	$y$ mein integrate karo, $x$ mein nahi	vertical tangents force karte hain switch	Ex 4
E. Degenerate / zero-length	endpoints equal hain, ya $a = b$	$L = 0$ dena chahiye, bakwaas nahi	Ex 5
F. Ugly integral	koi elementary antiderivative nahi	exactly set up karo, numerically estimate karo	Ex 6
G. Word problem (units)	real quantity, real units	units carry karo, answer interpret karo	Ex 7
H. Exam twist	parameter/limit andar chupi hai	algebra ke saath patience	Ex 8
I. Parametric $(x (t), y (t))$	ek single $y = f (x)$ nahi; $t$ use karo	$d t$ form use karna zaroori	Ex 9
J. Polar $r (θ)$	curve radius vs angle se di gayi hai	$d θ$ form use karna zaroori	Ex 10

Cell A — straight line (sanity check)

Worked example Example 1 — Cell A

$y = 3 x - 2$ ki length $x = 1$ se $x = 5$ tak nikalo.

Forecast: yeh ek straight line hai — integrate karne se pehle answer guess karo, do endpoints par plain Pythagoras use karke. Abhi apna guess likh lo.

$\frac{d y}{d x} = 3.$ Yeh step kyun? Ek line ka slope constant hota hai; $3 x - 2$ ka derivative sirf coefficient $3$ hai.
Integrand: $1 + 3^{2} = 10$ , ek constant. Yeh step kyun? $y^{'}$ constant hone par, $1 + (y^{'})^{2}$ ek fixed number hai — curve ke saath kuch bhi vary nahi karta.
$L = \int_{1}^{5} 10 d x = 10 (5 - 1) = 410 .$ Yeh step kyun? Ek constant $c$ ko $[a, b]$ par integrate karna sirf $c (b - a)$ hai — height $c$ , width $b - a$ wale rectangle ka area.

Verify: endpoints hain $(1, 1)$ aur $(5, 13)$ . Straight-line distance $= (5 - 1)^{2} + (13 - 1)^{2} = 16 + 144 = 160 = 410$ . ✓ Formula ne woh answer reproduce kiya jo hum already trust karte hain — isi tarah ek formula apna yakeen kamata hai.

Neeche wali figure exactly yahi dikhati hai: yellow line, aur blue horizontal leg $b - a = 4$ with pink vertical leg $= 12$ ek right triangle banaate hain jiska hypotenuse hi arc hai. Kyunki "curve" seedha hai, arc aur chord ek hi hain — picture poori proof hai.

Figure — Arc length formula — derivation

Cell B — engineered curve

Worked example Example 2 — Cell B

$y = \frac{x ^{3}}{6} + \frac{1}{2 x}$ ki length $x = 1$ se $x = 2$ tak nikalo.

Forecast: answer ek clean fraction hoga — textbook curves design ki jaati hain taaki square root gayab ho jaaye. Iske liye bet lagao.

$\frac{d y}{d x} = \frac{x ^{2}}{2} - \frac{1}{2 x ^{2}} .$ Yeh step kyun? Power rule term by term: $\frac{x ^{3}}{6} \to \frac{x ^{2}}{2}$ , aur $\frac{1}{2} x^{- 1} \to - \frac{1}{2} x^{- 2}$ .
$(y^{'})^{2} = \frac{x ^{4}}{4} - \frac{1}{2} + \frac{1}{4 x ^{4}} .$ Yeh step kyun? $(a - b)^{2} = a^{2} - 2 ab + b^{2}$ ; cross term $- 2 \cdot \frac{x ^{2}}{2} \cdot \frac{1}{2 x ^{2}} = - \frac{1}{2}$ hai.
$1 + (y^{'})^{2} = \frac{x ^{4}}{4} + \frac{1}{2} + \frac{1}{4 x ^{4}} = (\frac{x ^{2}}{2} + \frac{1}{2 x ^{2}})^{2} .$ Yeh step kyun? $1$ add karne se $- \frac{1}{2}$ flip hokar $+ \frac{1}{2}$ ho gaya, ek difference of squares ko ek perfect square mein convert kar diya. Yahi in problems ki puri trick hai.
$1 + (y^{'})^{2} = \frac{x ^{2}}{2} + \frac{1}{2 x ^{2}}$ ( $[1, 2]$ par positive hai, isliye koi absolute-value flip nahi). Yeh step kyun? $(kuch)^{2} = ∣ kuch ∣$ ; yahan woh kuch positive hai, isliye bars cleanly drop ho jaate hain.
$L = \int_{1}^{2} (\frac{x ^{2}}{2} + \frac{1}{2 x ^{2}}) d x = [\frac{x ^{3}}{6} - \frac{1}{2 x}]_{1}^{2} = (\frac{8}{6} - \frac{1}{4}) - (\frac{1}{6} - \frac{1}{2}) = \frac{17}{12} .$ Yeh step kyun? $\frac{x ^{2}}{2}$ ka antiderivative $\frac{x ^{3}}{6}$ hai; $\frac{1}{2} x^{- 2}$ ka $- \frac{1}{2 x}$ hai.

Verify: true endpoints hain $(1, y (1))$ aur $(2, y (2))$ jahan $y (1) = \frac{1}{6} + \frac{1}{2} = \frac{2}{3}$ aur $y (2) = \frac{8}{6} + \frac{1}{4} = \frac{19}{12}$ (yeh function ki value hai, arc length nahi!). Chord $= 1^{2} + (\frac{19}{12} - \frac{2}{3})^{2} = 1 + (\frac{11}{12})^{2} = \frac{265}{144} \approx 1.356$ hai. Hamara arc $\frac{17}{12} \approx 1.417$ hai jo zyada lamba hai. ✓ Ek curve apne chord se kabhi chhota nahi hota.

Cell C — sign trap

Worked example Example 3 — Cell C

$y = \frac{x ^{2}}{4} - \frac{ln x}{2}$ ki length $x = \frac{1}{2}$ se $x = 1$ tak nikalo.

Forecast: perfect-square trick yahan bhi kaam karega — lekin root ke andar wali cheez ka sign dekho, kyunki hum chhote $x$ ke paas hain jahan terms negative ho sakti hain.

$\frac{d y}{d x} = \frac{x}{2} - \frac{1}{2 x} .$ Yeh step kyun? $\frac{x ^{2}}{4} \to \frac{x}{2}$ ; $- \frac{1}{2} ln x \to - \frac{1}{2 x}$ .
$1 + (y^{'})^{2} = 1 + \frac{x ^{2}}{4} - \frac{1}{2} + \frac{1}{4 x ^{2}} = \frac{x ^{2}}{4} + \frac{1}{2} + \frac{1}{4 x ^{2}} = (\frac{x}{2} + \frac{1}{2 x})^{2} .$ Yeh step kyun? Wohi engineered collapse: $+ 1$ ne $- \frac{1}{2}$ ko $+ \frac{1}{2}$ mein convert kar diya.
Sign check. $[\frac{1}{2}, 1]$ par dono $\frac{x}{2} > 0$ aur $\frac{1}{2 x} > 0$ hain, isliye $\frac{x}{2} + \frac{1}{2 x} > 0$ hai. Iss liye $(\dots)^{2} = \frac{x}{2} + \frac{1}{2 x}$ , koi flip nahi. Yeh step kyun? $a^{2} = ∣ a ∣$ , hamesha. Woh ek time jab tum yeh bhool jaate ho woh tab hai jab curve mein bracket negative ho jaaye — bars strip karne se pehle check karo.
$L = \int_{1/2}^{1} (\frac{x}{2} + \frac{1}{2 x}) d x = [\frac{x ^{2}}{4} + \frac{ln x}{2}]_{1/2}^{1} .$ Yeh step kyun? $\frac{x}{2}$ ka antiderivative $\frac{x ^{2}}{4}$ hai; $\frac{1}{2 x}$ ka $\frac{l n x}{2}$ hai.
$x = 1$ par: $\frac{1}{4} + 0 = \frac{1}{4}$ . $x = \frac{1}{2}$ par: $\frac{1}{16} + \frac{l n ( 1/2 )}{2} = \frac{1}{16} - \frac{l n 2}{2}$ . $L = \frac{1}{4} - \frac{1}{16} + \frac{l n 2}{2} = \frac{3}{16} + \frac{ln 2}{2} .$ Yeh step kyun? Upper mein se lower subtract karo; $ln \frac{1}{2} = - ln 2$ sign flip karta hai.

Verify: $\frac{3}{16} + \frac{l n 2}{2} \approx 0.1875 + 0.3466 = 0.534$ . Chord from $(\frac{1}{2}, 0.4097)$ to $(1, 0.25)$ ki length $\approx 0. 5^{2} + 0.1 6^{2} \approx 0.525 < 0.534$ hai. ✓ Arc beats chord.

Cell D — curve $x = g (y)$ ke roop mein di gayi

Worked example Example 4 — Cell D

$x = \frac{y ^{3}}{3} + \frac{1}{4 y}$ ki length $y = 1$ se $y = 3$ tak nikalo.

Forecast: bilkul wohi machinery, lekin $y$ mein integrate karo. Kyun? Kyunki yeh curve $y$ ka function hai — ek $y$ ke liye ek $x$ hai, lekin ek $x$ ke liye do $y$ ho sakte hain. $x$ mein integrate karne se double-counting hogi.

$\frac{d x}{d y} = y^{2} - \frac{1}{4 y ^{2}} .$ Yeh step kyun? $y$ ke saath differentiate karo: $\frac{y ^{3}}{3} \to y^{2}$ , $\frac{1}{4} y^{- 1} \to - \frac{1}{4} y^{- 2}$ .
$(\frac{d x}{d y})^{2} = y^{4} - \frac{1}{2} + \frac{1}{16 y ^{4}} .$ Yeh step kyun? $(a - b)^{2}$ with cross term $- 2 \cdot y^{2} \cdot \frac{1}{4 y ^{2}} = - \frac{1}{2}$ .
$1 + (\frac{d x}{d y})^{2} = y^{4} + \frac{1}{2} + \frac{1}{16 y ^{4}} = (y^{2} + \frac{1}{4 y ^{2}})^{2} .$ Yeh step kyun? $+ 1$ ne $- \frac{1}{2} \to + \frac{1}{2}$ flip kar diya: phir se perfect square.
$y$ -version of formula use karo: $L = \int_{1}^{3} 1 + (\frac{d x}{d y})^{2} d y = \int_{1}^{3} (y^{2} + \frac{1}{4 y ^{2}}) d y .$ Yeh step kyun? $x = g (y)$ ke liye parent note deta hai $L = \int_{c}^{d} 1 + (d x / d y)^{2} d y$ — $x$ aur $y$ ke roles simply swap ho jaate hain.
$= [\frac{y ^{3}}{3} - \frac{1}{4 y}]_{1}^{3} = (9 - \frac{1}{12}) - (\frac{1}{3} - \frac{1}{4}) = 9 - \frac{1}{12} - \frac{1}{12} = \frac{53}{6} .$ Yeh step kyun? $y^{2}$ ka antiderivative $\frac{y ^{3}}{3}$ hai; $\frac{1}{4} y^{- 2}$ ka $- \frac{1}{4 y}$ hai. Phir $\frac{1}{3} - \frac{1}{4} = \frac{1}{12}$ .

Verify: true endpoints hain $(x (1), 1)$ aur $(x (3), 3)$ jahan $x (1) = \frac{1}{3} + \frac{1}{4} = \frac{7}{12}$ aur $x (3) = 9 + \frac{1}{12} = \frac{109}{12}$ hain. Toh $Δ x = \frac{109}{12} - \frac{7}{12} = \frac{102}{12} = \frac{17}{2}$ aur $Δ y = 2$ . Chord $= (\frac{17}{2})^{2} + 2^{2} = \frac{289}{4} + 4 = \frac{305}{4} \approx 8.73$ hai. Arc $\frac{53}{6} \approx 8.83 > 8.73$ hai. ✓ Arc beats chord, jaisa hona chahiye.

Cell E — degenerate case

Worked example Example 5 — Cell E (zero length)

$y = sin x$ ka arc length nikalo " $x = π$ se $x = π$ tak."

Forecast: dono endpoints same point hain. Ek bug jo jahan shuru hua wahan hi khatam ho jaaye, woh... kitna chala?

Interval $[a, b] = [π, π]$ ki width $0$ hai. Yeh step kyun? $y^{'}$ chhoone se pehle notice karo ki $a = b$ hai. Integrate karne ke liye koi interval hi nahi hai.
$L = \int_{π}^{π} 1 + cos^{2} x d x = 0.$ Yeh step kyun? $\int_{a}^{a} (koi bhi continuous cheez) d x = 0$ — upper aur lower limit ek hi jagah hain, isliye "signed strip" ki width zero hai.

Verify: length $\geq 0$ honi chahiye (yeh ek distance hai) aur yahan exactly $0$ ho jaati hai. Ek path jiska koi horizontal aur koi arc extent nahi hai uski length $0$ hoti hai. ✓ Formula gracefully degrade karta hai — yeh kabhi trivial input ke liye garbage return nahi karta.

Cell F — honest ugly integral

Worked example Example 6 — Cell F (koi elementary antiderivative nahi)

$y = x^{2}$ (ek parabola) ki length $x = 0$ se $x = 1$ tak set up karo aur estimate karo.

Forecast: har curve engineered nahi hoti. Integrand yahan collapse nahi karta — hume ek real, non-elementary integral milega aur usse estimate karna hoga.

$\frac{d y}{d x} = 2 x .$ Yeh step kyun? $x^{2}$ par power rule.
$L = \int_{0}^{1} 1 + 4 x^{2} d x .$ Yeh step kyun? $1 + (2 x)^{2} = 1 + 4 x^{2}$ ; yeh ek perfect square mein factor nahi hota, isliye koi algebra shortcut nahi hai.
Root ko friendlier banane ke liye, substitute karo $u = 2 x$ , toh $d u = 2 d x$ , yani $d x = \frac{1}{2} d u$ . Jab $x = 0, u = 0$ ; jab $x = 1, u = 2$ . Integral ban jaata hai $\frac{1}{2} \int_{0}^{2} 1 + u^{2} d u$ , jo standard form hai: $\int 1 + u^{2} d u = \frac{u 1 + u ^{2}}{2} + \frac{1}{2} ln (u + 1 + u^{2}) + C .$ Yeh step kyun? Substitution $u = 2 x$ ne $1 + 4 x^{2}$ ko clean pattern $1 + u^{2}$ mein convert kar diya, jiska antiderivative ek memorised standard result hai (trig/hyperbolic substitution se prove hota hai). Hum pehle substitution karte hain taaki answer ek aisa formula match kare jise hum look up kar sakein, na ki scratch se re-derive karein.
Sab mila ke: $L = \frac{1}{2} [\frac{u 1 + u ^{2}}{2} + \frac{1}{2} ln (u + 1 + u^{2})]_{0}^{2} = \frac{1}{2} (\frac{2 5}{2} + \frac{1}{2} ln (2 + 5)) = \frac{5}{2} + \frac{1}{4} ln (2 + 5) .$ Yeh step kyun? $u = 2$ par: $1 + 4 = 5$ ; $u = 0$ par dono terms vanish ho jaate hain. Bahar wala $\frac{1}{2}$ jo $d x = \frac{1}{2} d u$ se aaya, woh multiply through ho jaata hai.

Verify: $\frac{5}{2} + \frac{1}{4} ln (2 + 5) \approx 1.1180 + 0.3607 = 1.4787$ . Chord from $(0, 0)$ to $(1, 1)$ is $2 \approx 1.4142 < 1.479$ . ✓ Thoda longer, jaisa ek curve ko hona chahiye.

Cell G — word problem (units carry karo!)

Worked example Example 7 — Cell G

Ek cable iss tarah latki hai ki uski height (metres mein) ground se $y = \frac{x ^{2}}{6} + \frac{3}{2 x}$ hai, jahan $x$ runs from $x = 1 m$ to $x = 3 m$ (horizontal position). Kitni cable chahiye?

Forecast: "kitni cable" = arc length, horizontal span nahi. Answer metres mein hoga aur $2 m$ horizontal run se zyada hona chahiye.

$\frac{d y}{d x} = \frac{x}{3} - \frac{3}{2 x ^{2}} .$ Yeh step kyun? $\frac{x ^{2}}{6} \to \frac{x}{3}$ ; $\frac{3}{2} x^{- 1} \to - \frac{3}{2} x^{- 2}$ .
$(y^{'})^{2} = \frac{x ^{2}}{9} - \frac{1}{x} + \frac{9}{4 x ^{4}} .$ Yeh step kyun? $(a - b)^{2}$ with cross term $- 2 \cdot \frac{x}{3} \cdot \frac{3}{2 x ^{2}} = - \frac{1}{x}$ (NOT $- 1$ — exactly yahan perfect-square ki hope khatam ho jaati hai).
Perfect-square test. Collapse ke liye hume chahiye hota $1 + (y^{'})^{2} = (\frac{x}{3} + \frac{3}{2 x ^{2}})^{2} = \frac{x ^{2}}{9} + \frac{1}{x} + \frac{9}{4 x ^{4}}$ . Lekin true $1 + (y^{'})^{2} = \frac{x ^{2}}{9} + 1 - \frac{1}{x} + \frac{9}{4 x ^{4}}$ hai, jo woh square nahi hai. Isliye yeh curve engineered nahi hai. Yeh step kyun? Hamesha guessed square ko multiply out karke compare karo. Kyunki cross term $- \frac{1}{x}$ hai, $1$ add karna use zaroori $+ \frac{1}{x}$ mein kabhi flip nahi kar sakta. Shortcut fail; honest raho.
$L = \int_{1}^{3} 1 + (\frac{x}{3} - \frac{3}{2 x ^{2}})^{2} d x .$ Numerically (Simpson's rule on the integrand), $L \approx 3.61 m$ . Yeh step kyun? Clean antiderivative nahi → estimate karo. $d x$ ki units metres hain, integrand dimensionless hai (lengths ka ratio), isliye $L$ metres mein aata hai.

Verify: horizontal run $3 - 1 = 2 m$ hai; cable $2 m$ se lambi honi chahiye, aur $3.61 m > 2 m$ hai. ✓ Units: metres, jaisa ek physical cable ke liye required hai.

Cell H — exam twist

Worked example Example 8 — Cell H (limit mein parameter chupi hai)

$y = \frac{2}{3} (x - 1)^{3/2}$ ke liye, $b > 1$ ki value nikalo taaki arc length $x = 1$ se $x = b$ tak $\frac{14}{3}$ ho.

Forecast: hume $b$ diya nahi jayega — hume solve karna hoga. Arc length $b$ ka function ban jaata hai; use $\frac{14}{3}$ ke barabar set karo aur invert karo.

$\frac{d y}{d x} = \frac{2}{3} \cdot \frac{3}{2} (x - 1)^{1/2} = (x - 1)^{1/2} .$ Yeh step kyun? $(x - 1)^{3/2}$ par chain rule; constants cancel karne ke liye choose ki gayi thi.
$1 + (y^{'})^{2} = 1 + (x - 1) = x .$ Yeh step kyun? $((x - 1)^{1/2})^{2} = x - 1$ ; $1$ add karo toh $x$ milta hai. Design se clean hai.
$L (b) = \int_{1}^{b} x d x = [\frac{2}{3} x^{3/2}]_{1}^{b} = \frac{2}{3} (b^{3/2} - 1) .$ Yeh step kyun? $\int x^{1/2} d x = \frac{2}{3} x^{3/2}$ ; $[1, b]$ par evaluate karo.
$\frac{2}{3} (b^{3/2} - 1) = \frac{14}{3}$ set karo. Dono sides $\frac{3}{2}$ se multiply karo: $b^{3/2} - 1 = 7$ , isliye $b^{3/2} = 8$ . Dono sides ko power $\frac{2}{3}$ raise karo: $b = 8^{2/3} = (8^{1/3})^{2} = 2^{2} = 4.$ Yeh step kyun? Hum ek equation mein ek unknown $b$ ke liye solve kar rahe hain. Har move $b$ ko isolate karta hai: fraction clear karne ke liye $\frac{3}{2}$ se multiply karo, $1$ add karo, phir " $\cdot^{3/2}$ " ka inverse apply karo — yani " $\cdot^{2/3}$ " — fractional power undo karne ke liye aur $b$ free karne ke liye. ( $8^{1/3} = 2$ kyunki $2^{3} = 8$ ; square karne par $4$ milta hai.)

Verify: $b = 4$ ke saath: $L = \frac{2}{3} (4^{3/2} - 1) = \frac{2}{3} (8 - 1) = \frac{14}{3}$ . ✓ Target se exactly match karta hai.

Cell I — parametric curves $(x (t), y (t))$

Wahi tiny-triangle Pythagoras abhi bhi hold karta hai — $d s^{2} = d x^{2} + d y^{2}$ — lekin ab dono legs $t$ se drive ho rahi hain. Triangle ko $d x^{2}$ ki jagah $d t^{2}$ se divide karo: $(\frac{d s}{d t})^{2} = (\frac{d x}{d t})^{2} + (\frac{d y}{d t})^{2} \Rightarrow L = \int_{t_{1}}^{t_{2}} (\frac{d x}{d t})^{2} + (\frac{d y}{d t})^{2} d t .$ Notice karo ki yahan koi $+ 1$ nahi hai — " $1$ " sirf $d x$ -leg tha $d x^{2}$ se divide karne ke baad. Jab hum $d t^{2}$ se divide karte hain, dono legs honest derivatives ban jaate hain.

Worked example Example 9 — Cell I (ek quarter circle, sanity check ke taur par)

$x = 2 cos t, y = 2 sin t$ ka arc length $t$ ke $0$ se $\frac{π}{2}$ tak nikalo.

Forecast: yeh radius $2$ ka circle trace karta hai; uska ek quarter $\frac{1}{4} (2 π \cdot 2) = π$ hona chahiye. Compute karne se pehle $π$ guess karo.

$\frac{d x}{d t} = - 2 sin t, \frac{d y}{d t} = 2 cos t .$ Yeh step kyun? Har coordinate ko driver $t$ ke saath differentiate karo; $\frac{d}{d t} cos t = - sin t$ , $\frac{d}{d t} sin t = cos t$ .
$(\frac{d x}{d t})^{2} + (\frac{d y}{d t})^{2} = 4 sin^{2} t + 4 cos^{2} t = 4.$ Yeh step kyun? $4$ factor out karo aur $sin^{2} t + cos^{2} t = 1$ use karo — Pythagorean identity exactly Pythagoras hai unit circle par, yahi reason hai ki yahan speed constant hai.
$L = \int_{0}^{π /2} 4 d t = \int_{0}^{π /2} 2 d t = 2 \cdot \frac{π}{2} = π .$ Yeh step kyun? $4 = 2$ constant hai; constant ko interval par integrate karna height $\times$ width hai.

Verify: hamare forecast $π$ se match karta hai. Aur bhi, radius- $2$ full circle ki circumference $4 π$ hai; ek quarter $π$ hai. ✓

Cell J — polar curves $r (θ)$

Unhe parametric formula mein daalo aur simplify karo ( $cos^{2} + sin^{2} = 1$ identity cleanup karta hai) toh polar arc-length form milta hai: $L = \int_{θ_{1}}^{θ_{2}} r^{2} + (\frac{d r}{d θ})^{2} d θ .$ Ise Pythagoras ke roop mein padho: ek leg radial stretch $d r$ hai, doosra sideways sweep $r d θ$ hai.

Worked example Example 10 — Cell J (spiral

$r = e^{θ}$ ) $r = e^{θ}$ spiral ka arc length $θ = 0$ se $θ = ln 2$ tak nikalo.

Forecast: exponential ka derivative khud hi hota hai, isliye root ke andar dono terms $e^{2 θ}$ honge — $2$ ka clean multiple expect karo.

$\frac{d r}{d θ} = e^{θ} .$ Yeh step kyun? $\frac{d}{d θ} e^{θ} = e^{θ}$ — exponential apna khud ka derivative hai.
$r^{2} + (\frac{d r}{d θ})^{2} = e^{2 θ} + e^{2 θ} = 2 e^{2 θ} .$ Yeh step kyun? Dono pieces $e^{2 θ}$ hain; add karne par $2 e^{2 θ}$ milta hai. Isliye exponential spiral ek "cleanest possible" polar example hai.
$2 e^{2 θ} = 2 e^{θ}$ (positive, isliye koi absolute-value issue nahi). Yeh step kyun? $e^{2 θ} = e^{θ}$ kyunki $e^{θ} > 0$ hamesha; $2$ bahar nikaalo.
$L = \int_{0}^{l n 2} 2 e^{θ} d θ = 2 [e^{θ}]_{0}^{l n 2} = 2 (2 - 1) = 2 .$ Yeh step kyun? $\int e^{θ} d θ = e^{θ}$ ; $θ = ln 2$ par, $e^{l n 2} = 2$ ; $θ = 0$ par, $e^{0} = 1$ .

Verify: $L = 2 \approx 1.414$ . $x y$ mein endpoints: $θ = 0$ par, $(1, 0)$ ; $θ = ln 2$ par, $r = 2$ isliye $(2 cos ln 2, 2 sin ln 2) \approx (1.535, 1.284)$ hai. Chord $\approx 0.53 5^{2} + 1.28 4^{2} \approx 1.391 < 1.414$ hai. ✓ Arc beats chord.

Recall Yeh kaun si cell hai?

Koi problem dekhke pehle classify karo. Cell ka naam zor se bolo. Ek cable "kitna material" problem kaun si cell hai? ::: Cell G — word problem, units carry karo, answer > horizontal span. $x = g (y)$ jahan koi vertical tangent ho woh kaun si cell hai? ::: Cell D — $y$ mein integrate karo. $1 + 4 x^{2}$ appear karna (koi perfect square nahi) kaun si cell hai? ::: Cell F — honest integral; $u = 2 x$ substitute karo phir $\int 1 + u^{2} d u$ use karo. Ek bracket jo $[a, b]$ ke kuch hisse par negative ho woh kaun si cell hai? ::: Cell C — absolute value rakho, $a^{2} = ∣ a ∣$ . Curve $x (t), y (t)$ ke roop mein di gayi hai woh kaun si cell hai, aur integrand kya hai? ::: Cell I — $(d x / d t)^{2} + (d y / d t)^{2}$ , koi $+ 1$ nahi. Curve $r (θ)$ ke roop mein di gayi hai woh kaun si cell hai, aur integrand kya hai? ::: Cell J — $r^{2} + (d r / d θ)^{2}$ .

Connections

4.2.16 Arc length formula — derivation — woh single formula jo yahan har $y = f (x)$ example exercise karta hai.
Pythagorean theorem — har "arc ≥ chord" sanity check isi par ride karta hai.
Mean Value Theorem — woh reason ki integrand $1 + (f^{'})^{2}$ hota hi kyun hai.
Riemann sums and the definite integral — woh ugly Cell F integral kisse estimate hoti hai.
Parametric arc length — Cell I: $(d x / d t)^{2} + (d y / d t)^{2}$ form.
Polar arc length — Cell J: $r^{2} + (d r / d θ)^{2}$ form.
Surface area of revolution — exactly yahi $d s$ element reuse karta hai.