Worked examples — Related rates — setting up and solving

4.1.25 · D3 · Maths › Calculus I — Limits & Derivatives › Related rates — setting up and solving

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Scenario matrix

Har related-rates problem inn case classes mein se ek mein aata hai. Table mein listed hain; neeche har example ko us cell se tag kiya gaya hai jo woh fill karta hai.

Cell	Case class	Kya special hai	Example
A	Growing quantity (positive rate)	$\frac{dQ}{dt}>0$, area/volume expand karta hai	Ex 1 — balloon
B	Shrinking quantity (negative rate)	answer $<0$ aata hai, sign padhna zaroori hai	Ex 2 — melting snowball
C	Rate = 0 ek special instant pe	derivative vanish ho jaata hai chahe cheezein move kar rahi hon	Ex 3 — ladder at the symmetric point
D	Rate → ∞ (degenerate / limiting input)	denominator → 0, speed blow up karti hai	Ex 4 — ladder near flat
E	Do variables dono change hote hain → ek eliminate karo	pehle Similar triangles chahiye	Ex 5 — walking shadow
F	Angle changes (trig linking equation)	$\tan$, aur kyun hum isse choose karte hain	Ex 6 — rotating searchlight
G	Real-world word problem (distance closing)	Pythagoras between two moving points	Ex 7 — two cars
H	Exam twist: value poochhi jaati hai, rate nahi	ek given rate se length/time nikalo	Ex 8 — when is depth rising fastest

Ab hum har cell ko walk karte hain. Forecast padho aur sign/behaviour guess karo steps padhne se pehle — yahi habit poori skill hai.

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Cell A — ek growing quantity

Forecast: Volume badh raha hai, toh radius bhi badhe ga → answer positive. Aur kyunki ek bade balloon ko radius mein 1 cm gain karne ke liye bahut air chahiye (surface area bahut badi hai), hum predict karte hain ki radius rate chhoti hai aur $r$ badhne ke saath chhoti hoti jaati hai.

Neeche ki figure do balloon outlines dikhati hai — ek chhota (blue) aur $r=5$ wala (orange). Dekho orange circle kitna lamba hai blue se: woh circumference incoming air ko spread hone wale surface area ko represent karti hai. Lal arrow woh ek changing length $r$ hai, aur green label humein yaad dilata hai ki air fixed $100{\text{cm}}^3/\text{s}$ pe enter karti hai. Yeh takeaway steps mein le jaao: usi air ko badi skin mein pump karne se radius kam uthti hai.

1. Link. Sphere ke liye, $V=\frac{4}{3}\pi r^3$. Yeh step kyun? Volume sirf ek changing length $r$ pe depend karta hai — ek single-variable link, sabse simple tarah ka.
2. $t$ ke w.r.t. differentiate karo (chain rule, kyunki $r=r(t)$): $\frac{dV}{dt}=\frac{4}{3}\pi \cdot 3r^2\cdot \frac{dr}{dt}=4\pi r^2\frac{dr}{dt}.$ Yeh step kyun? Hum $\frac{dr}{dt}$ chahte hain, isliye hum poori equation ko time mein differentiate karte hain; $\frac{d}{dt}(r^3)=3r^2\frac{dr}{dt}$ chain rule hai jo kaam aa raha hai.
3. Unknown rate ke liye solve karo pehle, symbolically: $\frac{dr}{dt}=\frac{1}{4\pi r^2}\frac{dV}{dt}.$ Yeh step kyun? Notice karo ki $4\pi r^2$ exactly surface area hai — toh radius-speed = pump-speed ÷ surface area. Yeh forecast confirm karta hai aur figure ki big-skin-vs-small-skin picture se match karta hai: badi surface ⇒ slower growth.
4. Instant substitute karo $r=5,\frac{dV}{dt}=100$: $\frac{dr}{dt}=\frac{100}{4\pi (25)}=\frac{100}{100\pi }=\frac{1}{\pi }\approx 0.318\text{cm/s}.$ Ab kyun? Numbers last mein jaate hain, differentiate karne ke baad — warna $r$ constant ban jaata hai aur $\frac{dr}{dt}$ gayab ho jaata hai.

Verify: Units: $\frac{{\text{cm}}^3/\text{s}}{{\text{cm}}^2}=\text{cm/s}$ ✓. Sign positive ✓ (growing). Value chhoti ✓ jaise forecast tha.

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Cell B — ek shrinking quantity

Forecast: Volume shrink ho raha hai, toh radius bhi shrink hoga → answer negative. Magnitude bhi dekho: jaise ball pighalta hai chhota hota hai, uski skin (surface area) chhoti hoti hai, aur wohi cm³ hatane se radius tezi se andar aata hai — toh shrink-speed badhni chahiye jab $r$ drop kare.

Figure snowball ko do moments pe dikhata hai: bada $r=4$ ball (blue) aur baad mein, chhota ball (orange). Lal arrow woh ek changing length $r$ hai, aur green label melting rate $\frac{dV}{dt}=-8$ mark karta hai. Dekho orange circle kitni chhoti hai — chhoti skin matlab har cm³ hatane pe radius zyada andar khichta hai, jo exactly woh magnitude behaviour hai jo humne forecast kiya tha.

1. Link & differentiate — Ex 1 jaisa hi: $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\Rightarrow \frac{dr}{dt}=\frac{1}{4\pi r^2}\frac{dV}{dt}.$ Yeh step kyun? Geometry nahi badli; sirf given rate ka sign badla.
2. Substitute $r=4,\frac{dV}{dt}=-8$: $\frac{dr}{dt}=\frac{-8}{4\pi (16)}=\frac{-8}{64\pi }=-\frac{1}{8\pi }\approx -0.0398\text{cm/min}.$ Yeh step kyun? Negative input se negative output milta hai — formula honestly report karta hai "shrinking."

Verify: Sign negative ✓ "melting" se match karta hai. Ab clean magnitude rule: kyunki $\frac{dr}{dt}=\frac{1}{4\pi r^2}\frac{dV}{dt}$, factor $\frac{1}{r^2}$ se $|\frac{dr}{dt}|$ chhote $r$ ke liye bada hota hai. $r=8$ pe check karo (ek bada ball): $\frac{dr}{dt}=\frac{-8}{4\pi (64)}=\frac{-8}{256\pi }\approx -0.00995$ — $r=4$ pe se chhota magnitude ✓. Toh shrinking ball apni radius-loss $r$ badhne ke saath speed up karti hai, forecast se match karta hai.

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Cell C — ek rate jo exactly zero ho

Forecast: Jab ladder seedha upar khada hota hai ($x=0$), top bilkul peak pe hota hai. Jaise ek ball seedha upar phenko toh top pe momentarily ruk jaata hai, hum predict karte hain ki top ki downward speed us instant pe exactly 0 hogi.

Figure do variables ko aankhon mein fix karta hai: vertical orange ladder, red dot jo top ko height $y=10$ pe mark karta hai, aur blue arrow jo base ko bahar slide hote dikhata hai ($\frac{dx}{dt}=2$). Red dot dekho — is vertical instant pe uske paas na aur upar jaane ki jagah hai aur neeche girna shuru nahi kiya hai, toh uski up/down speed momentarily zero hai chahe base already move kar raha ho.

1. Variables name karo. $x=$ wall se ladder ke base tak ki distance; $y=$ wall pe ladder ke top ki height. Dono time ke saath change karte hain. Yeh step kyun? Linking equation mein har symbol ko pehle kuch concrete mean karna chahiye — $x$ aur $y$ right triangle ke do legs hain.
2. Link (Pythagorean theorem): $x^2+y^2=10^2$. Yeh step kyun? Wall, ground, ladder fixed hypotenuse (10 m ladder) ka right triangle banate hain, toh do legs $x$ aur $y$ ek doosre se tied hain.
3. $t$ ke w.r.t. differentiate karo: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\Rightarrow \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}.$ Yeh step kyun? Right side ($100$) constant hai, toh uski rate $0$ hai — yeh disguise mein Implicit differentiation hai.
4. Instant substitute karo $x=0$ (toh $y=10$): $\frac{dy}{dt}=-\frac{0}{10}(2)=0\text{m/s}.$ Yeh step kyun? Numerator mein $x$ ka factor rate ko maar deta hai. Geometrically top momentarily vertically move nahi kar raha, exactly figure ke red dot ki tarah.

Verify: Formula $\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$ $0$ hai ⇔ $x=0$ (kyunki $\frac{dx}{dt}\ne 0,y\ne 0$) ✓. Physically top girne se pehle apni highest point pe hover karta hai — ek genuine zero-rate instant, humara Cell C.

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Cell D — ek rate jo infinity tak blow up kare

Forecast: Jaise ladder flat hota hai, top lagbhag ground pe hota hai ($y\to 0$). Chhote $y$ se divide karne se $\frac{dy}{dt}$ bahut bada aur negative ho jaata hai → top bahut tezi se crash karta hai speed → ∞.

Yeh Ex 3 se motion ka opposite end hai. Figure orange ladder ko almost ground pe dikhata hai: lal top-dot floor ke thoda upar baith raha hai toh $y$ almost zero hai, jabki base $x=9.99$ pe bahut door hai (blue arrow still sliding). Woh near-flat picture dhyaan mein rakho — woh chhota denominator $y$ hi sab kuch explode karta hai.

1. Rate law reuse karo $\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$. Yeh step kyun? Same geometry — sirf instant ab degenerate hai.
2. $y$ nikalo $x=9.99$ pe: $y=\sqrt{100-9.99^2}=\sqrt{100-99.8001}=\sqrt{0.1999}\approx 0.4471$ m. Yeh step kyun? Hume plug in karne ke liye current $y$ chahiye; yeh almost zero hai.
3. Substitute karo: $\frac{dy}{dt}=-\frac{9.99}{0.4471}(2)\approx -44.69\text{m/s}.$ Yeh step kyun? Chhota denominator → enormous speed.
4. Limit jab $x\to 10^{-}$: $y\to 0^{+}$, toh $\frac{dy}{dt}\to -\infty$. Yeh step kyun? Yeh woh degenerate limiting behaviour hai jise expose karne ke liye Cell D exist karta hai.

Verify: $x=9.99$ pe, computed $\approx -44.7$ m/s ✓ (huge, negative). Denominator $y\to 0$ magnitude ko $\to \infty$ force karta hai ✓. Physically model toot jaata hai (ladders infinite speed nahi reach karti) — ek lesson ki maths theek hai lekin idealisation boundary pe fail ho jaata hai.

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Cell E — do changing variables (pehle ek eliminate karo)

Forecast: Shadow tip insaan se tezi se move karti hai (shadow aage stretch hoti hai), toh uski speed 1.5 m/s se zyada hogi.

Figure mein tall blue post aur chhota orange insaan ek light ray (gray) cast karte hain jo ground pe red shadow tip pe lagti hai. Do right triangles — bada (post) aur chhota (insaan) — same ray angle share karte hain, isliye unke sides proportion mein hain. Do ground segments labelled $x$ (insaan se post) aur $s$ (insaan se shadow tip) notice karo: yeh do changing lengths hain jinhe hum differentiate karne se pehle relate karna chahte hain.

1. Variables set karo. $x$ = insaan-se-post distance, $s$ = shadow ki length (insaan se shadow tip). Post se tip position $=x+s$. Yeh step kyun? Do lengths change hoti hain ($x$ aur $s$); hume inhe relate karna hai.
2. Similar triangles se link karo. Bada triangle (post, ground, ray) aur chhota triangle (insaan, ground, ray) ray ka angle share karte hain: $\frac{6}{x+s}=\frac{1.8}{s}.$ Yeh step kyun? Equal-angled right triangles ke proportional sides hote hain — yeh fixed geometry hai jo $s$ ko $x$ se tie karti hai.
3. Differentiate karne se pehle $s$ eliminate karo: cross-multiply $6s=1.8(x+s)\Rightarrow 6s-1.8s=1.8x\Rightarrow 4.2s=1.8x\Rightarrow s=\frac{1.8}{4.2}x=\frac{3}{7}x.$ Yeh step kyun? Agar hum dono $x$ aur $s$ rakhen aur differentiate karein, toh humein unknown rate $\frac{ds}{dt}$ chahiye hogi — woh rate jise kisi ne nahi diya. Differentiate karne se pehle $s$ ko $x$ mein likhne se woh extra unknown completely remove ho jaata hai; yeh parent note ke cone example se eliminate-first lesson hai.
4. Tip distance $D=x+s=x+\frac{3}{7}x=\frac{10}{7}x$. Differentiate karo: $\frac{dD}{dt}=\frac{10}{7}\frac{dx}{dt}.$ Yeh step kyun? $D$ ab $x$ ka plain multiple hai; chain rule trivial hai.
5. Substitute $\frac{dx}{dt}=1.5$: $\frac{dD}{dt}=\frac{10}{7}(1.5)=\frac{15}{7}\approx 2.14\text{m/s}.$ Yeh step kyun? Numbers last, jaise hamesha.

Verify: $2.14>1.5$ ✓ (tip insaan se aage nikal jaati hai, forecast match karta hai). Note karo $\frac{dD}{dt}$ $x$ se independent hai — tip constant speed se move karti hai chahe kitna bhi door ho — ek clean, checkable surprise ✓.

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Cell F — ek angle changes (trig link)

Forecast: Jaise $\theta$ 90° ki taraf badhta hai, beam almost wall ke parallel ho jaati hai aur spot race karta chala jaata hai — toh spot-speed $\theta$ ke saath increase karti hai; 45° pe yeh already ek healthy number hona chahiye.

$p$ ko perpendicular ke foot se wall ke saath measure ki gayi bright spot ki position maano. Sign convention: hum $p>0$ us direction mein lete hain jis direction mein beam sweep karta hai (figure mein wall ke upar), toh $\frac{dp}{dt}>0$ matlab spot woh taraf climb karta hai. Domain: yeh setup sirf $-\frac{\pi }{2}<\theta <\frac{\pi }{2}$ ke liye sense make karta hai; exactly $\theta =\pm \frac{\pi }{2}$ pe beam wall ke parallel hoti hai aur kabhi nahi milti, isliye speed blow up hoti hai jab $\theta \to \frac{\pi }{2}$.

Figure mein searchlight (blue dot) vertical gray wall se fixed $L=200$ m door baith hai; orange beam wall pe red spot pe strike karta hai jo wall ke upar $p$ height pe hai. Dashed gray line perpendicular hai (length $L$ wali adjacent side), aur $p$ opposite side hai. Dekho kya hota hai jab green angle $\theta$ 90° ki taraf khulta hai: spot wall pe upar shoot karta hai.

1. Link: $p$ ko wall ke saath spot ki position maano, $\tan \theta =\frac{p}{L}\Rightarrow p=L\tan \theta =200\tan \theta .$ Yeh step kyun? Yeh moving length $p$ ko moving angle $\theta$ ke terms mein isolate karta hai.
2. $t$ ke w.r.t. differentiate karo. Yaad karo $\frac{d}{d\theta }\tan \theta ={\sec }^2\theta$, toh chain rule se ($\theta =\theta (t)$): $\frac{dp}{dt}=200{\sec }^2\theta \frac{d\theta }{dt}.$ Yeh step kyun? $\theta$ $t$ pe depend karta hai, isliye $\tan \theta$ differentiate karne se extra $\frac{d\theta }{dt}$ factor aata hai — chain rule phir se.
3. Substitute $\theta =\frac{\pi }{4}$. Yahaan $\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$, toh $\sec \frac{\pi }{4}=\frac{1}{\cos (\pi /4)}=\sqrt{2}$ aur ${\sec }^2\frac{\pi }{4}=2$; $\frac{d\theta }{dt}=0.1$ ke saath: $\frac{dp}{dt}=200(2)(0.1)=40\text{m/s}.$ Yeh step kyun? Numbers last; growing factor ${\sec }^2\theta$ exactly woh reason hai jo spot race karta hai, forecast confirm karta hai.

Verify: Units: $\text{m}\cdot (\text{unitless})\cdot (\text{rad/s})=\text{m/s}$ ✓. Positive ✓ (spot beam-sweep direction mein climb karta hai). Value $40$ m/s ✓. Jaise $\theta \to {\frac{\pi }{2}}^{-}$, ${\sec }^2\theta \to \infty$, toh $\frac{dp}{dt}\to \infty$ ✓ — domain restriction $\theta <\frac{\pi }{2}$ se match karta hai.

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Cell G — real-world closing distance (do moving points)

Forecast: Dono cars crossing ki taraf ja rahi hain, toh unke beech ka gap shrink ho raha hai → answer negative.

Figure intersection ko origin pe rakhta hai, Car A (blue) directly north pe $a=0.3$ pe aur Car B (orange) directly east pe $b=0.4$ pe. North aur east roads right angle pe milti hain, toh do cars aur intersection ek right triangle banate hain jiska hypotenuse (green) cars ke beech ki distance $D$ hai. Arrows dono cars ko corner ki taraf badhte dikhate hain, isliye dono leg-lengths decrease ho rahe hain.

1. Variables. $a$ = A ki north-of-intersection distance, $b$ = B ki east distance, $D$ = cars ke beech ki distance. Dono decrease ho rahe hain: $\frac{da}{dt}=-60$, $\frac{db}{dt}=-80$ (approach ho rahe hain ⇒ distances shrink). Yeh step kyun? Signs ko "paas aana" encode karna chahiye — $0$ ki taraf jaana matlab negative rate.
2. Link (Pythagorean theorem, right angle intersection pe): $D^2=a^2+b^2$. Yeh step kyun? North aur east directions perpendicular hain, toh cars aur intersection ek right triangle banate hain.
3. $t$ ke w.r.t. differentiate karo: $2D\frac{dD}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}\Rightarrow \frac{dD}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{D}.$ Yeh step kyun? Poora relation differentiate karo; jo ek rate chahiye uske liye solve karo.
4. $D$ nikalo: $D=\sqrt{0.3^2+0.4^2}=\sqrt{0.09+0.16}=\sqrt{0.25}=0.5$ km. Yeh step kyun? Denominator mein plug karne ke liye current $D$ chahiye (0.1 se scale kiya 3-4-5 triangle).
5. Substitute: $\frac{dD}{dt}=\frac{(0.3)(-60)+(0.4)(-80)}{0.5}=\frac{-18-32}{0.5}=\frac{-50}{0.5}=-100\text{km/h}.$ Yeh step kyun? Numbers last; negative confirm karta hai gap close ho raha hai.

Verify: $-100$ km/h, negative ✓ (closing). Magnitude sanity: yeh har car ki speed se zyada hai kyunki dono contribute karti hain — reasonable ✓. Units km/h ✓.

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Cell H — exam twist: rate nahi, value ke liye solve karo

Forecast: Parent note se, $h=4$ pe rise $\frac{1}{2\pi }\approx 0.159$ m/min tha. Hum ek tezi rise chahte hain ($0.5$), aur level tezi se badhta hai jab surface chhoti hoti hai — toh hum ek chhali depth expect karte hain, $h<4$.

1. Rate law (parent se): $V=\frac{\pi }{12}{h}^3$ differentiate karo: $\frac{dV}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt}.$ Yeh step kyun? Yeh do jaani/target quantities ke beech general link hai.
2. Twist — rate nahi, $h$ ke liye solve karo. Desired rate plug in karo aur rearrange karo: $h^2=\frac{4}{\pi \frac{dh}{dt}}\frac{dV}{dt}\Rightarrow h=\sqrt{\frac{4\frac{dV}{dt}}{\pi \frac{dh}{dt}}}.$ Yeh step kyun? Unknown ab ek length hai; rate law likhne ke baad ordinary algebra $h$ isolate kar deta hai.
3. Substitute $\frac{dV}{dt}=2,\frac{dh}{dt}=0.5$: $h=\sqrt{\frac{4(2)}{\pi (0.5)}}=\sqrt{\frac{8}{0.5\pi }}=\sqrt{\frac{16}{\pi }}\approx 2.257\text{m}.$ Yeh step kyun? Numbers last; value real aur positive aata hai, jaisa ek depth hona chahiye.

Verify: $h\approx 2.257<4$ ✓ (shallower, forecast match karta hai). Plug back karo: $\frac{dh}{dt}=\frac{4(2)}{\pi (2.257{)}^2}=\frac{8}{\pi (5.094)}\approx 0.5$ ✓ round-trip confirmed. Units m ✓.

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Recall Har cell ka ek-line summary

Answer ka sign growing vs shrinking batata hai; numerator mein $0$ ki taraf jaata factor zero rate deta hai; denominator mein $0$ ki taraf jaata factor blow-up deta hai; do moving lengths ko ek eliminate karne ke liye Similar triangles chahiye; angles $\tan$/${\sec }^2$ laate hain; closing-distance problems Pythagoras use karti hain; aur exam twists sirf usi rate law ko ek alag unknown ke liye re-solve karti hain.

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Connections

• Related rates — setting up and solving — parent recipe jinhe yeh examples exercise karte hain.
• Chain rule — har differentiation mein $\frac{dr}{dt},\frac{d\theta }{dt}$ factors deta hai.
• Implicit differentiation — $x^2+y^2=L^2$ aur $D^2=a^2+b^2$ kaise differentiate karte hain.
• Derivatives as rates of change — signs aur magnitudes ko physically padhna.
• Similar triangles — shadow aur cone problems mein extra variable eliminate karta hai.
• Pythagorean theorem — ladder aur closing-distance cases mein link.
• Optimization — agla jagah jahaan yeh setups phir se aate hain.