Exercises — Related rates — setting up and solving

Draw: ek sphere jisme bahar ki taraf radius arrow $r$ point kar raha ho, aur ek chhota "air-in" arrow $\frac{dV}{dt}$ label karta ho. Link: $V=\frac{4}{3}\pi r^3$. $t$ mein Differentiate karo (chain rule on $r^3$, kyunki $r=r(t)$): $\frac{dV}{dt}=\frac{4}{3}\pi \cdot 3r^2\cdot \frac{dr}{dt}=4\pi r^2\frac{dr}{dt}.$ Yahi relation hai. Kuch bhi substitute nahi kiya gaya — Level‑1 "set up" task ke liye bilkul sahi.

Draw: ek right triangle jisme legs $x$, $y$ (letters, dono changing) aur hypotenuse fixed $13$ label ho. Right side ek constant hai, isliye uski rate $0$ hai: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0.$ $2$ se divide karo: $x\frac{dx}{dt}+y\frac{dy}{dt}=0.$ Yeh Implicit differentiation hai jo $t$ ko hidden variable maankar apply ki gayi hai.

Draw: ek circle jisme andar ki taraf radius arrow ho (shrinking), $\frac{dr}{dt}<0$ label karo. $r>0$ aur $2\pi >0$, isliye $2\pi r>0$. Ek positive number ko negative $\frac{dr}{dt}$ se multiply karne par $\frac{dA}{dt}<0$ milta hai: area shrink hota hai. Sign in, sign out — rate equation physical direction carry karta hai.

Draw: ek circle jisme bahar ki taraf radius arrow $\frac{dr}{dt}=0.5$ label ho; area $A$ shaded interior hai. Link: $A=\pi r^2$. Differentiate: $\frac{dA}{dt}=2\pi r\frac{dr}{dt}$. Substitute (ab safe hai): $r=4,\frac{dr}{dt}=0.5$: $\frac{dA}{dt}=2\pi (4)(0.5)=4\pi \approx 12.57{\text{m}}^2/\text{s}.$

Draw: ek cube jisme ek edge $s$ (ek letter) label ho aur $\frac{ds}{dt}$ ke liye ek tiny outward arrow ho. Link: $V=s^3$. Differentiate: $\frac{dV}{dt}=3s^2\frac{ds}{dt}$. Substitute: $3(5{)}^2(0.02)=3(25)(0.02)=1.5{\text{cm}}^3/\text{s}.$

Draw: upar wala right triangle — base $x$, height $y$ (dono letters), hypotenuse fixed $10$, base-arrow $\frac{dx}{dt}=2$ wall se door point karta hai. Link (Pythagorean theorem): $x^2+y^2=100$. Ab $y$ nikalo: $y=\sqrt{100-36}=8$ m. Differentiate: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\Rightarrow \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$. Substitute: $\frac{dy}{dt}=-\frac{6}{8}(2)=-1.5\text{m/s}.$ Negative $\Rightarrow$ top neeche jaata hai, jaise expected tha.

Draw: ek shrinking sphere jisme andar ki taraf radius arrow $\frac{dr}{dt}$ label ho aur $\frac{dV}{dt}<0$ ke liye "air/volume out" note ho. Link: $V=\frac{4}{3}\pi r^3$. Differentiate: $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$. $\frac{dr}{dt}$ ke liye solve karo: $\frac{dr}{dt}=\frac{1}{4\pi r^2}\frac{dV}{dt}$. Substitute: $\frac{dr}{dt}=\frac{-8}{4\pi (9)}=\frac{-8}{36\pi }=-\frac{2}{9\pi }\approx -0.0707\text{cm/min}.$

Draw: upar wala cone cross-section — upar fixed $R=3$, $H=6$; water triangle jisme depth $h$ aur surface radius $r$ (dono letters) hon. Link: $V=\frac{1}{3}\pi r^{2}h$, lekin $r$ bhi change hota hai. Similar triangles use karo: $\frac{r}{h}=\frac{R}{H}=\frac{3}{6}=\frac{1}{2}\Rightarrow r=\frac{h}{2}$. Pehle $r$ substitute karo: $V=\frac{1}{3}\pi {\left(\frac{h}{2}\right)}^{2}h=\frac{\pi }{12}{h}^3.$ Differentiate: $\frac{dV}{dt}=\frac{\pi }{12}\cdot 3h^2\frac{dh}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt}.$ $h=2$ substitute karo: $4=\frac{\pi (4)}{4}\frac{dh}{dt}=\pi \frac{dh}{dt}\Rightarrow \frac{dh}{dt}=\frac{4}{\pi }\approx 1.273\text{m/min}.$

Draw: origin par intersection, A ka path north (distance $a$), B ka path east (distance $b$), aur separation $D$ as slanted hypotenuse jo do cars ko join karta hai. Maano $a$ = A ki north distance, $b$ = B ki east distance, $D$ = separation. Link: $D^2=a^2+b^2$. Differentiate: $2D\frac{dD}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}\Rightarrow D\frac{dD}{dt}=a\frac{da}{dt}+b\frac{db}{dt}.$ $t=2$ h par: $a=120,b=160,D=\sqrt{120^2+160^2}=\sqrt{14400+25600}=\sqrt{40000}=200$ km. Substitute: $200\frac{dD}{dt}=120(60)+160(80)=7200+12800=20000\Rightarrow \frac{dD}{dt}=\frac{20000}{200}=100\text{km/h}.$

Draw: ek $6$ m lamppost, $2$ m ka insaan usse $x$ distance par, aur shadow length $s$ insaan ke paas se aage tip tak stretching; do nested right triangles uss tip ko share karte hain. Similar triangles: bada triangle (lamp, ground, tip) ki height $6$ aur base $x+s$; chhota triangle (head, ground, tip) ki height $2$ aur base $s$: $\frac{6}{x+s}=\frac{2}{s}\Rightarrow 6s=2(x+s)\Rightarrow 6s=2x+2s\Rightarrow 4s=2x\Rightarrow s=\frac{x}{2}.$ Tip position $T=x+s=x+\frac{x}{2}=\frac{3x}{2}$. Differentiate: $\frac{dT}{dt}=\frac{3}{2}\frac{dx}{dt}=\frac{3}{2}(1.5)=2.25\text{m/s}.$ Note: tip ki speed $x$ par depend nahi karti — yeh constant $2.25$ m/s par move karti hai.

Draw: Ex 2.3 jaisa hi ladder triangle, ab uska interior shaded hai — woh shaded region woh area $A$ hai jo hum track kar rahe hain. Humein dono $\frac{dx}{dt}$ aur $\frac{dy}{dt}$ chahiye. Ex 2.3 se: $x=6$ par, $y=8$, $\frac{dx}{dt}=2$, $\frac{dy}{dt}=-1.5$. Link: $A=\frac{1}{2}xy$. Differentiate (product rule): $\frac{dA}{dt}=\frac{1}{2}\left(\frac{dx}{dt}y+x\frac{dy}{dt}\right).$ Substitute: $\frac{1}{2}((2)(8)+(6)(-1.5))=\frac{1}{2}(16-9)=\frac{1}{2}(7)=3.5{\text{m}}^2/\text{s}.$ Positive: triangle us instant par abhi bhi badh raha hai chahe top gir raha ho.

Draw: observer ground level par $500$ m door pad se (ek fixed horizontal leg), rocket height $y$ par (vertical leg, ek letter), aur line of sight observer par angle $\theta$ banaati hai. Link: $\tan \theta =\frac{y}{500}$ (observer ke right triangle ka opposite over adjacent). Differentiate: ${\sec }^2\theta \frac{d\theta }{dt}=\frac{1}{500}\frac{dy}{dt}.$ (Left side $\frac{d}{d\theta }\tan \theta ={\sec }^2\theta$ use karta hai phir chain rule $\times \frac{d\theta }{dt}$). $y=500$ par: $\tan \theta =\frac{500}{500}=1\Rightarrow \theta =45^{\circ }$, aur identity se ${\sec }^2\theta =1+{\tan }^2\theta =1+1=2$. Solve: $\frac{d\theta }{dt}=\frac{1}{500{\sec }^2\theta }\frac{dy}{dt}=\frac{1}{500(2)}(100)=\frac{100}{1000}=0.1\text{rad/s}.$

Draw: cone cross-section jisme fixed $R=2$, $H=4$; water depth $h$ gir raha hai (level line par downward arrow), surface radius $r$ uske saath shrink ho raha hai. Similar triangles: $r=\frac{R}{H}h=\frac{2}{4}h=\frac{h}{2}$, isliye $V=\frac{1}{3}\pi {\left(\frac{h}{2}\right)}^{2}h=\frac{\pi }{12}{h}^3.$ Differentiate: $\frac{dV}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt}$. $h=3$ substitute karo: $\frac{dV}{dt}=\frac{\pi (9)}{4}(-0.1)=-\frac{0.9\pi }{4}=-0.225\pi \approx -0.7069{\text{m}}^3/\text{min}.$

Draw: $4$ m upar pulley (fixed vertical leg), boat horizontal distance $x$ par (ek letter), aur rope $L$ as slanted hypotenuse boat se pulley tak, ek reel-in arrow $L$ shrink karta hai. Link (Pythagorean theorem): $L^2=x^2+4^2=x^2+16.$ Reeling in matlab $\frac{dL}{dt}=-2$. Differentiate: $2L\frac{dL}{dt}=2x\frac{dx}{dt}\Rightarrow \frac{dx}{dt}=\frac{L}{x}\frac{dL}{dt}.$ $x=3$ par: $L=\sqrt{9+16}=5$. Isliye $\frac{dx}{dt}=\frac{5}{3}(-2)=-\frac{10}{3}\approx -3.33\text{m/s}.$ Limit $x\to 0$: $L\to 4$, aur $\frac{dx}{dt}=\frac{L}{x}\frac{dL}{dt}\to \frac{4}{x}(-2)\to -\infty$. Boat ki horizontal speed blow up ho jaati hai jab woh dock ke paas pahunchti hai — chahe rope steady $2$ m/s par aati ho. Physically, dock ke paas almost saari rope-shortening horizontal motion mein convert ho jaati hai.

Draw: ek sphere jisme steady outward radius arrow $\frac{dr}{dt}=1$ ho; equal-time snapshots imagine karo — add hone wale shells thicker lagte hain kyunki area $\propto r^2$ badhti hai. Link: $V=\frac{4}{3}\pi r^3\Rightarrow \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}=4\pi r^2$ (kyunki $\frac{dr}{dt}=1$). Isliye $\frac{dV}{dt}$ $r^2$ ki tarah badhta hai — yeh increase hota hai. Uski rate nikalne ke liye, phir se $t$ mein differentiate karo (yeh second derivative $\frac{d^{2}V}{d{t}^2}$ deta hai): $\frac{d^{2}V}{d{t}^2}=\frac{d}{dt}\left(\frac{dV}{dt}\right)=4\pi \cdot 2r\frac{dr}{dt}=8\pi r\frac{dr}{dt}.$ $r=5$ par: $8\pi (5)(1)=40\pi \approx 125.66{\text{cm}}^3/{\text{s}}^2.$ (Units: cm³ per second, per second.)

Draw: ek cone pile jisme height $h$ base radius $r$ ke barabar ho ($h=r$ mark karo), aur ek "sand in" arrow $\frac{dV}{dt}=6$ label karta ho. Link: $r=h$ ke saath, $V=\frac{1}{3}\pi h^{2}h=\frac{\pi }{3}{h}^3.$ Differentiate: $\frac{dV}{dt}=\pi h^2\frac{dh}{dt}\Rightarrow \frac{dh}{dt}=\frac{1}{\pi h^2}\frac{dV}{dt}=\frac{6}{\pi h^2}.$ (a) $h=2$: $\frac{dh}{dt}=\frac{6}{\pi (4)}=\frac{6}{4\pi }=\frac{3}{2\pi }\approx 0.477\text{m/min}.$ (b) $\frac{6}{\pi h^2}=0.5$ set karo: $h^2=\frac{6}{0.5\pi }=\frac{12}{\pi }\Rightarrow h=\sqrt{\frac{12}{\pi }}\approx 1.954\text{m}.$

Upar har problem usi skeleton ko follow karta tha — kya tum ise recite kar sakte ho? Chhe steps order mein ::: Draw, Link, Differentiate, Substitute, Solve, aur State units. "Draw" mein kya hona chahiye ::: ek labelled picture jo moving quantities ke liye variables (letters) use kare aur fixed numbers sirf constants ke liye, plus rate-arrows. Jahan $A=\frac{1}{2}xy$ jaisa product care mangta hai ::: product rule use karo jab do time-functions multiply hon. Kaise ek limiting case ($x\to 0$) ek blow-up reveal karta hai ::: denominator mein ek variable rate ko $\pm \infty$ par le jaata hai. Variable kab eliminate karein ::: differentiate karne se pehle, fixed geometry use karke (similar triangles / Pythagoras).