Related rates — setting up and solving — Hinglish notes

4.1.25 · Maths › Calculus I — Limits & Derivatives

Iske andar ka engine chain rule hai: agar $V = f (r)$ aur $r = r (t)$ , to $\frac{d V}{d t} = \frac{d V}{d r} \cdot \frac{d r}{d t} .$

Ise HOW setup karte hain (6-step recipe)

Figure — Related rates — setting up and solving

Worked Example 1 — Expanding circle (oil spill)

Ek oil spill ek circle hai jiska radius $\frac{d r}{d t} = 2$ m/s ki speed se badhta hai. Jab $r = 10$ m ho, to area kitni fast badh rahi hai?

Step 3 — link: $A = π r^{2}$ . Kyun? Area ek circle ke liye sirf radius par depend karta hai.

Step 4 — $t$ mein differentiate karo: $\frac{d A}{d t} = π \cdot 2 r \cdot \frac{d r}{d t} = 2 π r \frac{d r}{d t} .$ Yeh step kyun? $r$ , $t$ ka function hai, isliye $\frac{d}{d t} (r^{2}) = 2 r \frac{d r}{d t}$ (chain rule).

Step 5 — instant substitute karo: $r = 10, \frac{d r}{d t} = 2$ : $\frac{d A}{d t} = 2 π (10) (2) = 40 π \approx 125.7 m^{2} / s .$ Ab kyun? Differentiate karne ke baad hi $r = 10$ fix karna safe hai.

Worked Example 2 — Sliding ladder

Ek 5 m ki ladder wall se lagi hai. Iska base $\frac{d x}{d t} = 1$ m/s ki speed se door slide kar raha hai. Jab base wall se $x = 3$ m door ho, to top kitni fast neeche slide kar rahi hai?

Step 3 — link (Pythagoras): $x^{2} + y^{2} = 5^{2}$ . Kyun? Wall, ground, ladder ek right triangle banate hain; ladder ki length (5) constant hai.

Step 4 — differentiate karo: $2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0 \Rightarrow \frac{d y}{d t} = - \frac{x}{y} \frac{d x}{d t} .$ Yeh step kyun? RHS constant hai $\Rightarrow$ uska derivative $0$ hai. Minus sign batayega ki $y$ decrease ho rahi hai.

Us instant par $y$ nikalo: $y = 25 - 9 = 4$ m.

Step 5 — substitute karo: $\frac{d y}{d t} = - \frac{3}{4} (1) = - 0.75 m/s .$ Negative sign matlab top neeche ja rahi hai — physically sahi hai. Kyun? Jab base bahar jaata hai, top ko girna hi padta hai.

Worked Example 3 — Conical tank (jahaan dhyaan rakhna padta hai)

Paani ek cone (apex neeche) mein bhara ja raha hai, top par radius $R = 4$ m, height $H = 8$ m, $\frac{d V}{d t} = 2$ m³/min ki speed se. Jab $h = 4$ m ho, depth $h$ kitni fast badh rahi hai?

Step 3 — link: $V = \frac{1}{3} π r^{2} h$ . Lekin $r$ aur $h$ dono change ho rahe hain — similar triangles use karke $r$ ko eliminate karo: $\frac{r}{h} = \frac{R}{H} = \frac{4}{8} = \frac{1}{2}$ , isliye $r = \frac{h}{2}$ . Eliminate kyun? Hume sirf $h$ chahiye; kam variables matlab kam unknown rates.

$V = \frac{1}{3} π (\frac{h}{2})^{2} h = \frac{π}{12} h^{3} .$

Step 4 — differentiate karo: $\frac{d V}{d t} = \frac{π}{12} \cdot 3 h^{2} \frac{d h}{d t} = \frac{π h ^{2}}{4} \frac{d h}{d t} .$

Step 5 — substitute karo $h = 4$ , $\frac{d V}{d t} = 2$ : $2 = \frac{π ( 16 )}{4} \frac{d h}{d t} = 4 π \frac{d h}{d t} \Rightarrow \frac{d h}{d t} = \frac{1}{2 π} \approx 0.159 m/min .$

Recall Feynman: 12-saal ke bachche ko explain karo

Socho tum ek balloon phulaa rahe ho. Balloon ki skin (area) aur andar ki hawa (volume) dono badhte hain, lekin woh balloon ki shape se ek doosre se tied hain. Agar main tumhe bataaun ki tum hawa kitni fast pump kar rahe ho, to tum pata kar sakte ho ki skin kitni fast stretch ho rahi hai — kyunki ek equation unhe link karti hai. Trick yeh hai: maan lo sab kuch time par depend karta hai, poori equation ki rate lo, aur woh rate solve karo jo tumhe nahi pata. Bas size ko zyada jaldi lock mat karo, warna bhool jaaoge ki woh change ho raha tha!

Flashcards

Related rates mein engine ke roop mein kaunsa calculus rule use hota hai?

Chain rule, kyunki har variable time

t

ka function hai.

Numbers differentiate karne ke BAAD substitute kyon karne chahiye, pehle kyun nahi?

Jaldi substitute karne se changing variable constant ban jaata hai, to uski derivative (woh rate jo chahiye) galat se 0 ho jaati hai.

Circle area

A = π r^{2}

ke liye,

\frac{d A}{d t}

kya hai?

\frac{d A}{d t} = 2 π r \frac{d r}{d t}

Length

L

ki ladder:

x^{2} + y^{2} = L^{2}

t

mein differentiate karo.

2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0

, isliye

\frac{d y}{d t} = - \frac{x}{y} \frac{d x}{d t}

Cone problem mein differentiate karne se pehle

r

ko eliminate kyun karte hain?

Kyunki

r

aur

h

dono change hote hain; similar triangles se

r

eliminate karne par unknown rate

\frac{d r}{d t}

se bachte hain.

\frac{d y}{d t}

ka negative answer physically kya matlab hai?

Quantity

y

us instant par decrease ho rahi hai.

Kisi bhi related-rates problem ka pehla step kya hai?

Variables se label ki gayi ek picture draw karo (changing quantities ke liye fixed numbers nahi).

Connections

Chain rule — related rates ko enable karne wala core mechanism.
Implicit differentiation — same technique, ek relation ko $t$ ke respect mein differentiate karna.
Derivatives as rates of change — $\frac{d x}{d t}$ ko velocity/speed ki tarah interpret karna.
Similar triangles — cone/shadow problems mein variables eliminate karne ke liye use hota hai.
Pythagorean theorem — ladder/distance problems mein links.
Optimization — modelling mein derivatives ka agla application.