Visual walkthrough — Related rates — setting up and solving

4.1.25 · D2 · Maths › Calculus I — Limits & Derivatives › Related rates — setting up and solving

Hum sirf yeh ek sawaal ka jawab denge:

Ek seedhi ladder wall ke saath tikee hui hai. Koi uska foot wall se steady speed pe dur kheench raha hai. Har instant par ladder ka top wall ke saath kitni fast neeche slide ho raha hai?

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Step 1 — Scene draw karo aur jo cheezein move karti hain unhe naam do

KYA: hum wall (vertical), zameen (horizontal), aur ladder ko un dono ke beech slanted line ke roop mein sketch karte hain, phir har length ko ek letter lagate hain.

KYUN: hum letters se, numbers se nahi, label karte hain, kyunki poori baat yeh hai ki $x$ aur $y$ move karte hain. Agar hum abhi "$x=3$" likhte, toh picture freeze ho jaati aur motion kho jaata — exactly wahi deadly early-substitution error jiske baare mein parent note warn karta hai.

PICTURE: ladder ek right triangle banata hai. Corner mein chhota sa square dekho — yeh guarantee hai ki wall aur zameen perfect right angle par milte hain. Dhyan rakho ki hum horizontal length ko $x$ aur vertical length ko $y$ kehte hain.

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Step 2 — Dono motions ko arrows ke roop mein dekho

KYA: hum ladder ko dobara draw karte hain jisme foot par ek arrow wall se door ki taraf point karta hai, aur top par ek arrow wall ke saath neeche ki taraf point karta hai.

KYUN $\frac{d}{dt}$ aata hai aur kuch nahi? Kyunki sawaal literally hai "kitna fast" — ek speed. Speed hai change ÷ time, aur woh tool jo instantaneous change per unit time measure karta hai woh hai time ke saath respect mein derivative. Koi aur tool "abhi kitna fast" ka jawab nahi deta.

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Step 3 — Woh equation dhundho jo $x$ aur $y$ ko chain kare

KYA: hum woh ek equation likhte hain jo dono moving lengths ko ek doosre se tie karti hai.

KYUN Pythagorean theorem aur koi angle formula nahi? Kyunki yeh sirf lengths use karta hai — aur lengths exactly woh quantities hain jinki speeds chahiye hain. Iska right-hand side bhi fixed hai ($L^2$ kabhi nahi badalta), jo Step 5 mein bahut convenient hoga.

PICTURE: har side par bane hue shaded squares. $x$ par blue square plus $y$ par violet square ka total area ladder par magenta square ke barabar hai — har instant, chahe triangle deform ho raha ho.

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Step 4 — Realize karo ki sab kuch secretly time ka function hai

KYA: hum $x$ aur $y$ ko $t$ ke functions ki tarah reinterpret karte hain, taaki $x^2+y^2=L^2$ sabhi times ke liye hold kare, sirf ek photo ke liye nahi.

KYUN: kyunki rate ($\frac{dx}{dt}$) sirf tab meaningful hai jab kuch $t$ ke saath change hota hai. Time mein differentiate karne ke liye, woh cheez pehle $t$ par depend karni chahiye. Yeh wahi viewpoint hai jaise Implicit differentiation: hum kabhi $y$ ko $t$ ke terms mein solve nahi karte — hum bas accept karte hain ki yeh $t$ ka function hai aur poori relation ko differentiate karte hain.

PICTURE: teen stacked freeze-frames — ladder tip over ho raha hai jaise $t$ badhta hai — $x$ grow kar raha hai aur $y$ shrink kar raha hai.

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Step 5 — Link ko time ke saath differentiate karo

Ab $x^2+y^2=L^2$ ki dono sides par $\frac{d}{dt}$ apply karo.

Chain rule yahan star hai. $x^2$ differentiate karne ke liye jahan $x$ khud $t$ par depend karta hai, hum $x$ ke through jaate hain: $\frac{d}{dt}\left(x^2\right)=\underset{\text{outer: }\frac{d}{dx}(x^2)}{\underbrace{2x}}\cdot \underset{\text{inner: rate of }x}{\underbrace{\frac{dx}{dt}}}.$

Har term ke saath yeh karte hue:

KYA: humne lengths ke beech ki equation ko rates ke beech ki equation mein badal diya.

Right side par $0$ kyun? $L$ kabhi nahi badalta, toh $L^2$ ek fixed number hai, aur kisi bhi fixed number ki rate of change $0$ hoti hai. Ladder ki constancy hi woh cheez hai jo dono rates ko zero mein add karati hai — unhe exactly cancel karna hi padega.

PICTURE: equation ko ek balance beam ki tarah draw kiya gaya. $x$-term ek taraf push karta hai, $y$-term doosri taraf; unka sum kuch nahi hona chahiye, toh $x$ mein growth $y$ mein shrinkage ko force karti hai.

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Step 6 — Unknown rate ke liye solve karo

Hamare paas $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$ hai aur hum $\frac{dy}{dt}$ ko ek side par akela chahte hain.

Teen moves carry out karte hue: $y\frac{dy}{dt}=-x\frac{dx}{dt}\stackrel{÷y,y\ne 0}{⟹}\boxed{\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}}$

Ek instant par plug in karo. Maano $L=5$ m, foot $\frac{dx}{dt}=1$ m/s se move kar raha hai, aur abhi $x=3$ m hai. Link se, $y=\sqrt{L^2-x^2}=\sqrt{25-9}=4$ m (safely nonzero, toh $y$ se divide karna theek tha). Toh $\frac{dy}{dt}=-\frac{3}{4}(1)=-0.75\text{m/s}.$

Numbers abhi kyun substitute kiye? Agar humne Step 3 mein $x=3$ set kar diya hota, toh $x$ ek constant hota, $\frac{dx}{dt}$ zero hota, aur haara formula collapse ho jaata. Numbers sabse last mein jaate hain — hamesha.

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Step 7 — Degenerate cases (reader ko kabhi stranded mat chhodna)

Formula $\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$ mein ek landmine hai: agar $y=0$ ho toh? Exactly yahi woh case hai jisme Step 6 ne divide karne se mana kiya tha. Har extreme chalte hain.

Case B ko numerically verify karte hain. $L=5$, $\frac{dx}{dt}=1$ ke saath (first quadrant mein rehte hue, $y>0$):

$x$ (m)	$y=\sqrt{25-x^2}$	$\frac{dy}{dt}=-x/y$ (m/s)
$3$	$4.00$	$-0.75$
$4$	$3.00$	$-1.33$
$4.9$	$0.995$	$-4.93$
$4.99$	$0.316$	$-15.8$

Speed blow up karti hai — forecast confirm hua.

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Ek-picture summary

Ek diagram saate saaton steps compress karta hai: labelled triangle, dono velocity arrows, Pythagorean squares, aur final formula apne sign ke saath, sab ek frame mein.

Recall Feynman retelling — poora walkthrough simple words mein

Wall par ek ladder tikao. Corner ko origin par rakho, rightward ko positive $x$ aur upward ko positive $y$ mano; dono first quadrant mein rehte hain. Floor-gap ko $x$ aur wall-height ko $y$ kaho; ladder ki length kabhi nahi badlati. Arrows draw karo: foot right taraf creep karta hai (yeh ek positive speed hai, $\dot{x}>0$), toh top neeche creep karna chahiye (ek negative speed, $\dot{y}<0$). Ab magic sentence: wall, floor aur ladder hamesha ek right triangle banate hain, toh $x^2+y^2$ hamesha ladder squared ke barabar hota hai — ek number jo kabhi move nahi karta. Kyunki $x$ aur $y$ actually time ki movies hain, poochho "us equation ki har side kitni fast change ho rahi hai?" Left par, har squared length (twice-the-length) × (uski apni speed) rate par change hoti hai — yahi chain rule hai jo "per-second" toll pay kar raha hai. Right par, ek constant zero par change hoti hai. Toh $2x\dot{x}+2y\dot{y}=0$: dono rates cancel karni chahiye. Do se divide karo, $x$-term ko doosri side le jao, aur $y$ se divide karo (tabhi allowed jab $y$ zero na ho) aur milega: top ki speed $-\frac{x}{y}$ times foot ki speed hai. Minus kehta hai "neeche"; fraction $\frac{x}{y}$ kehta hai "lamba base, chhoti height ⇒ terrifyingly fast." Aur jab ladder almost flat ho, $y$ tiny hai, fraction explode karta hai, aur top neeche rocket karta hai — equation tum ko pehle warn kar deti hai jab asli ladder kabhi nahi kar sakti. Yaad rakhne wala ek rule: letters ko differentiate karte waqt alive rakho, aur numbers sirf bilkul end mein dalo.

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Connections

• Related rates — setting up and solving — parent recipe jise yeh page pictures mein derive karta hai.
• Chain rule — woh engine jo Step 5 mein "per-second" toll pay karta hai.
• Implicit differentiation — woh viewpoint jo humein $y$ ke liye solve kiye bina $x^2+y^2=L^2$ differentiate karne deta hai.
• Derivatives as rates of change — kyun $\frac{dx}{dt}$ ek velocity arrow hai.
• Pythagorean theorem — Step 3 ki linking equation.
• Similar triangles — cone/shadow problems ke liye analogous elimination trick.
• Optimization — related rates ke baad next derivative application.