Exercises — Arc length formula — derivation

Move 1 — differentiate karo. $\frac{dy}{dx}=3x^2$ (power rule). Move 2 — integrand banao. ${\left(\frac{dy}{dx}\right)}^2=9x^4$, toh $1+(y^{\prime }{)}^2=1+9x^4$. Move 3 — integral assemble karo diye gaye limits ke beech: $L={\int }_1^4\sqrt{1+9x^4}dx.$ Yeh ek correct setup hai. (Is particular case ka koi nice closed form nahi hai — recognition problems bas yahi test karti hain ki kya aap integral bana sakte ho, evaluate nahi.)

${\left(\frac{1}{2x}\right)}^2=\frac{1}{4x^2}$, toh $1+(y^{\prime }{)}^2=1+\frac{1}{4x^2}=\frac{4x^2+1}{4x^2}.$ Common denominator par kyon likho? Kyunki agla step hamesha $\sqrt{\cdot }$ hota hai, aur fractions ki roots $\frac{\sqrt{4x^2+1}}{2x}$ ki tarah split hoti hain — yeh spot karna asaan hai ki upar perfect square hai ya nahi.

Move 1. $\frac{dy}{dx}=\frac{2}{3}\cdot \frac{3}{2}(x-1{)}^{1/2}=(x-1{)}^{1/2}$. Itna clean kyon? Constants $\frac{2}{3}$ ko precisely isliye choose kiya gaya tha taaki power rule se $\frac{3}{2}$ cancel ho jaaye. Move 2. $(y^{\prime }{)}^2=x-1$, toh $1+(y^{\prime }{)}^2=1+(x-1)=x$. Ek perfect... well, bas $x$. Aur bhi nicer. Move 3. $L={\int }_1^5\sqrt{x}dx={\left[\frac{2}{3}{x}^{3/2}\right]}_1^5=\frac{2}{3}\left(5^{3/2}-1\right)=\frac{2}{3}\left(5\sqrt{5}-1\right).$ Numerically $\approx 6.79$. Sanity check: endpoints hain $(1,0)$ aur $(5,\frac{2}{3}\cdot 8)=(5,\frac{16}{3})\approx (5,5.33)$; chord $=\sqrt{16+28.4}\approx 6.66<6.79$. Curve apne chord se longer hai ✓. Neeche figure dekho.

$x=g(y)$ formula kyon? Kyunki curve humein $x$ as depending on $y$ di gayi hai, toh root se $dy$ nikalna natural hai. Move 1. $\frac{dx}{dy}=\frac{1}{3}\cdot \frac{3}{2}{y}^{1/2}=\frac{1}{2}{y}^{1/2}$. Move 2. ${\left(\frac{dx}{dy}\right)}^2=\frac{y}{4}$, toh $1+\frac{y}{4}=\frac{4+y}{4}$, aur $\sqrt{\cdot }=\frac{\sqrt{4+y}}{2}$. Move 3. $L={\int }_0^3\frac{\sqrt{4+y}}{2}dy=\frac{1}{2}\cdot \frac{2}{3}(4+y{)}^{3/2}{|}_0^3=\frac{1}{3}\left(7^{3/2}-4^{3/2}\right)=\frac{1}{3}\left(7\sqrt{7}-8\right).$ Numerically $\approx 3.51$.

Move 1. $\frac{dy}{dx}=\frac{x^2}{2}-\frac{1}{2x^2}$. Term-by-term kyon? Sum rule; $\frac{1}{2x}=\frac{1}{2}{x}^{-1}$ differentiate hokar $-\frac{1}{2}{x}^{-2}$ deta hai. Move 2 — analysis step. Slope ko $P-Q$ likho jahan $P=\frac{x^2}{2}$ aur $Q=\frac{1}{2x^2}$ hai. Square karo, cross term dhyan se dekhte hue: $(P-Q{)}^2=P^2-2PQ+Q^2=\frac{x^4}{4}-2\cdot \frac{x^2}{2}\cdot \frac{1}{2x^2}+\frac{1}{4x^4}=\frac{x^4}{4}-\frac{1}{2}+\frac{1}{4x^4}.$ Cross term $-2PQ=-\frac{1}{2}$ hai. Ab formula se mandatory $1$ add karo: $1+(y^{\prime }{)}^2=\frac{x^4}{4}+\left(-\frac{1}{2}+1\right)+\frac{1}{4x^4}=\frac{x^4}{4}+\frac{1}{2}+\frac{1}{4x^4}.$ Yeh ab perfect square kyon hai? Perfect square $(P+Q{)}^2=P^2+2PQ+Q^2$ ko middle term $+2PQ=+\frac{1}{2}$ chahiye. Humne $-\frac{1}{2}$ se shuru kiya, aur $1$ add karne par exactly $+\frac{1}{2}$ milta hai — kyunki $-\frac{1}{2}+1=+\frac{1}{2}$. Toh "$+1$" exactly woh amount hai jo middle term ko $-2PQ$ se $+2PQ$ mein flip karta hai. Isliye $1+(y^{\prime }{)}^2={\left(\frac{x^2}{2}+\frac{1}{2x^2}\right)}^2=(P+Q{)}^2.$ Yeh tabhi kaam karta hai jab in curves ko engineer kiya gaya ho taaki $2PQ=\frac{1}{2}$ exactly ho; phir formula ka $+1$ sign flip free mein kar deta hai. Move 3 — root lo ($[1,2]$ par positive): $L={\int }_1^2\left(\frac{x^2}{2}+\frac{1}{2x^2}\right)dx={\left[\frac{x^3}{6}-\frac{1}{2x}\right]}_1^2=\left(\frac{8}{6}-\frac{1}{4}\right)-\left(\frac{1}{6}-\frac{1}{2}\right)=\frac{13}{12}+\frac{1}{3}=\frac{17}{12}.$ Numerically $\approx 1.417$.

Move 1. $\frac{dy}{dx}=\frac{-\sin x}{\cos x}=-\tan x$. Kyon? Chain rule: $\ln u$ ka derivative $u^{\prime }/u$ hota hai jahan $u=\cos x$. Move 2. $(y^{\prime }{)}^2={\tan }^{2}x$, aur Pythagorean identity $1+{\tan }^{2}x={\sec }^{2}x$ deti hai $1+(y^{\prime }{)}^2={\sec }^{2}x$. Move 3 — absolute value dhyan rakhna. $\sqrt{{\sec }^{2}x}=|\sec x|$. $[0,\frac{\pi }{4}]$ par, $\cos x>0$ toh $\sec x>0$ aur $|\sec x|=\sec x$. Isliye $L={\int }_0^{\pi /4}\sec xdx=[\ln |\sec x+\tan x|{]}_0^{\pi /4}=\ln (\sqrt{2}+1)-\ln (1)=\ln (1+\sqrt{2}).$ Numerically $\approx 0.881$.

$ds$ yahan kyon aata hai. Ek surface of revolution tab banta hai jab length $ds$ ka ek tiny arc piece radius $y$ ke circle mein travel karta hai; har band ki area $2\pi yds$ hoti hai. Toh hum wahi $ds=\sqrt{1+(y^{\prime }{)}^2}dx$ reuse karte hain jo parent derivation mein banaya tha. Move 1. $y=x^{1/2}\Rightarrow \frac{dy}{dx}=\frac{1}{2}{x}^{-1/2}=\frac{1}{2\sqrt{x}}$. Move 2. $1+(y^{\prime }{)}^2=1+\frac{1}{4x}=\frac{4x+1}{4x}$, toh $ds=\frac{\sqrt{4x+1}}{2\sqrt{x}}dx$. Move 3 — surface integral assemble karo radius $y=\sqrt{x}$ ke saath: $S={\int }_0^{1}2\pi yds={\int }_0^{1}2\pi \sqrt{x}\cdot \frac{\sqrt{4x+1}}{2\sqrt{x}}dx=\pi {\int }_0^1\sqrt{4x+1}dx.$ Radius ka $\sqrt{x}$ aur denominator mein $\sqrt{x}$ cancel ho jaate hain — ek single clean square root bachta hai. (Evaluate karne par: $\pi \cdot \frac{1}{4}\cdot \frac{2}{3}(4x+1{)}^{3/2}{|}_0^1=\frac{\pi }{6}(5\sqrt{5}-1)\approx 5.33$.)

Kyon naya-dikhta formula, same idea. Tiny $dt$ par, point $dx=\frac{dx}{dt}dt$ across aur $dy=\frac{dy}{dt}dt$ upar move karta hai. Us infinitesimal triangle par Pythagoras: $ds=\sqrt{d{x}^2+d{y}^2}=\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt$. Parent se same, bas $t$ se parameterised. Move 1. $\frac{dx}{dt}=2t,\frac{dy}{dt}=3t^2$. Move 2. ${\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2=4t^2+9t^4=t^2(4+9t^2)$, toh $ds=t\sqrt{4+9t^2}dt$ ($t\ge 0$ on $[0,1]$ ke saath, toh $\sqrt{t^2}=t$, koi $|t|$ trouble nahi). Move 3. $u=4+9t^2$, $du=18tdt$ substitute karo: $L={\int }_0^{1}t\sqrt{4+9t^2}dt=\frac{1}{18}{\int }_4^{13}\sqrt{u}du=\frac{1}{18}\cdot \frac{2}{3}{u}^{3/2}{|}_4^{13}=\frac{1}{27}\left(13\sqrt{13}-8\right).$ Numerically $\approx 1.44$.

Woh polar $ds$ kyon. Polar coordinates mein ek tiny step mein ek radial part $dr$ aur ek sideways arc part $rd\theta$ (arc = radius × angle) hota hai. Ye dono perpendicular hain, toh Pythagoras deta hai $ds=\sqrt{(dr{)}^2+(rd\theta {)}^2}=\sqrt{{\left(\frac{dr}{d\theta }\right)}^2+r^2}d\theta$ — same Pythagorean skeleton. Neeche figure dekho. Move 1. $\frac{dr}{d\theta }=-\sin \theta$. Move 2. $r^2+{\left(\frac{dr}{d\theta }\right)}^2=(1+\cos \theta {)}^2+{\sin }^2\theta =1+2\cos \theta +{\cos }^2\theta +{\sin }^2\theta =2+2\cos \theta .$ Half-angle identity $1+\cos \theta =2{\cos }^2\frac{\theta }{2}$ use karo, toh $2+2\cos \theta =4{\cos }^2\frac{\theta }{2}$ aur $\sqrt{2+2\cos \theta }=2\left|\cos \frac{\theta }{2}\right|.$ Move 3 — sign check karo. $\theta \in [0,\pi ]$ par, $\frac{\theta }{2}\in [0,\frac{\pi }{2}]$, jahan $\cos \frac{\theta }{2}\ge 0$, toh bars safely drop ho jaate hain: $L={\int }_0^{\pi }2\cos \frac{\theta }{2}d\theta =2\cdot \frac{\sin (\theta /2)}{1/2}{|}_0^{\pi }=4\sin \frac{\theta }{2}{|}_0^{\pi }=4(\sin \frac{\pi }{2}-0)=4.$ Poora cardioid ($0$ se $2\pi$ tak) symmetry se is ka double, $8$, hoga.

Move 1 — differentiate karo. $y=\frac{1}{2(a+1)}{x}^{a+1}+\frac{1}{2(a-1)}{x}^{-(a-1)}$ ke saath: $\frac{dy}{dx}=\frac{a+1}{2(a+1)}{x}^a+\frac{-(a-1)}{2(a-1)}{x}^{-a}=\frac{1}{2}{x}^a-\frac{1}{2}{x}^{-a}.$ Constants kyon matter karte hain: denominators $2(a\pm 1)$ isliye choose kiye gaye hain taaki exponent factors $(a\pm 1)$ cancel ho jaayein, clean $\frac{1}{2}{x}^{\pm a}$ bachta hai. Move 2 — square karo, cross term dekho. $P=\frac{1}{2}{x}^a$, $Q=\frac{1}{2}{x}^{-a}$ ke saath: $(y^{\prime }{)}^2=(P-Q{)}^2=P^2-2PQ+Q^2,2PQ=2\cdot \frac{1}{2}{x}^a\cdot \frac{1}{2}{x}^{-a}=\frac{1}{2}.$ Toh $(y^{\prime }{)}^2=\frac{1}{4}{x}^{2a}-\frac{1}{2}+\frac{1}{4}{x}^{-2a}$. 1 add karo: middle term $-\frac{1}{2}$ ban jaata hai $-\frac{1}{2}+1=+\frac{1}{2}=+2PQ$: $1+(y^{\prime }{)}^2=\frac{1}{4}{x}^{2a}+\frac{1}{2}+\frac{1}{4}{x}^{-2a}=(P+Q{)}^2={\left(\frac{1}{2}{x}^a+\frac{1}{2}{x}^{-a}\right)}^2.$ Conclusion. $\sqrt{1+(y^{\prime }{)}^2}=\frac{1}{2}{x}^a+\frac{1}{2}{x}^{-a}$ ($x>0$ ke liye positive). Isliye textbook arc-length curves hamesha "power plus reciprocal power" mein split hoti hain: mandatory $+1$ difference-of-squares cross term $-\frac{1}{2}$ ko sum-of-squares $+\frac{1}{2}$ mein turn kar deta hai. Yeh woh fingerprint hai jo tum ab instantly pehchaan sakte ho.

1. Kya maine pehle differentiate kiya, phir square kiya? (L1 trap)
2. Curve $y(x)$, $x(y)$, $(x(t),y(t))$, ya $r(\theta )$ ke roop mein di gayi hai? Matching $ds$ pick karo. (L2/L4 traps)
3. $\sqrt{(\text{stuff}{)}^2}$ ke baad, kya stuff poore interval mein $\ge 0$ hai? Agar nahi, toh integral split karo jahan sign change ho, ya $|\text{stuff}|$ carry karo. (L3/L5 traps)
4. Sanity: kya mera answer endpoints ke beech straight chord se $\ge$ hai? Length kabhi straight line se choti nahi ho sakti.