Exercises — Parametric differentiation — dy - dx, d²y - dx²

Kya karte hain: har coordinate ko $t$ ke saath alag-alag differentiate karo.

• $\dot{x}=\frac{d}{dt}(3t+1)=3$ — $3t$ ka slope $3$ hai, constant $1$ kuch contribute nahi karta.
• $\dot{y}=\frac{d}{dt}(t^2-4)=2t$ — power rule; $-4$ gayab ho jaata hai.

Divide kyun karte hain: slope hota hai vertical-speed over horizontal-speed (shared $dt$ cancel ho jaata hai): $\frac{dy}{dx}=\frac{\dot{y}}{\dot{x}}=\frac{2t}{3}.$

$\frac{d^{2}y}{d{x}^2}$ ka matlab hai "$\frac{dy}{dx}$ ko $x$ ke saath differentiate karo, $t$ ke saath nahi," isliye tumhe $\frac{\dot{y}}{\dot{x}}$ par parametric trick dobara apply karni padti hai, jisse numerator mein quotient-rule aata hai aur denominator mein $=={\dot{x}}^3==$ hota hai — cube, na ki $\ddot{x}$.

• $\dot{x}=2t$, $\dot{y}=3t^2$.
• $\frac{dy}{dx}=\frac{3t^2}{2t}=\frac{3t}{2}$ ($t\ne 0$ ke liye valid, jahan $\dot{x}\ne 0$).
• $t=2$ par: $\frac{dy}{dx}=\frac{3\cdot 2}{2}=3.$

Tool: boxed second-derivative formula. Humare paas pehle se $\dot{x}=2t,\dot{y}=3t^2$ hai. Ab:

• $\ddot{x}=2$, $\ddot{y}=6t$.
• Plug in karo: $\frac{d^{2}y}{d{x}^2}=\frac{\ddot{y}\dot{x}-\dot{y}\ddot{x}}{{\dot{x}}^3}=\frac{(6t)(2t)-(3t^2)(2)}{(2t{)}^3}=\frac{12t^2-6t^2}{8t^3}=\frac{6t^2}{8t^3}=\frac{3}{4t}.$ "$dy/dx$ differentiate karo phir divide karo" route se cross-check: $\frac{dy}{dx}=\frac{3t}{2}$, toh $\frac{d}{dt}\left(\frac{3t}{2}\right)=\frac{3}{2}$; $\dot{x}=2t$ se divide karo: $\frac{3/2}{2t}=\frac{3}{4t}$. ✓ Same answer.

• $\dot{x}=-R\sin t$, $\dot{y}=R\cos t$.
• $\frac{dy}{dx}=\frac{R\cos t}{-R\sin t}=-\cot t.$ $R$ cancel ho jaata hai — bada circle bhi same angle par same slope rakhta hai.
• $t=\pi /6$ par: $-\cot (\pi /6)=-\sqrt{3}\approx -\mathrm{1.732.}$ ($R$ se independent hai, toh $R=2$ kuch change nahi karta.)

Horizontal tangent = slope $0$ = "vertical speed zero ho jabki horizontal speed na ho": $\dot{y}=0,\dot{x}\ne 0$.

• $\dot{y}=\cos t=0\Rightarrow t=\frac{\pi }{2},\frac{3\pi }{2}$. Dono par, $\dot{x}=-\sin t\ne 0$. Yeh hain top $(0,1)$ aur bottom $(0,-1)$ — figure mein red horizontal arrows dekho.

Vertical tangent = run zero hai = "horizontal speed zero ho jabki vertical speed na ho": $\dot{x}=0,\dot{y}\ne 0$.

• $\dot{x}=-\sin t=0\Rightarrow t=0,\pi$. Dono par, $\dot{y}=\cos t\ne 0$. Yeh hain right $(1,0)$ aur left $(-1,0)$ points, jahan $\frac{dy}{dx}$ undefined hai — yeh koi error nahi, bas geometry hai.

• $\dot{x}=1,\dot{y}=3t^2$; $\ddot{x}=0,\ddot{y}=6t$.
• $\frac{d^{2}y}{d{x}^2}=\frac{(6t)(1)-(3t^2)(0)}{1^3}=6t=6x.$

Case analysis ($6t$ ka sign decide karta hai — Concavity and second derivative dekho):

• $t>0$: positive → concave up (bowl upar ki taraf khulta hai).
• $t<0$: negative → concave down.
• $t=0$: zero → ek inflection point jahan concavity flip hoti hai, origin par.

• $\dot{x}=1-\cos t,\dot{y}=\sin t$, toh $\frac{dy}{dx}=\frac{\sin t}{1-\cos t}$.
• $t=\pi /2$ par: $\frac{\sin (\pi /2)}{1-\cos (\pi /2)}=\frac{1}{1-0}=1.$
• Limit $t\to 0^{+}$: upar aur neeche dono $\to 0$ ($\frac{0}{0}$). Small-angle pictures se, $\sin t\approx t$ aur $1-\cos t\approx \frac{t^2}{2}$, toh $\frac{dy}{dx}\approx \frac{t}{t^2/2}=\frac{2}{t}\to +\infty$. Cycloid har cusp par (jahan wheel zameen ko touch karta hai) vertical tangent ke saath rise karta hai — curve cusp mein seedha upar se aata hai.

Step 1 — point. $x=2\cos \frac{\pi }{4}=2\cdot \frac{\sqrt{2}}{2}=\sqrt{2}$; $y=3\sin \frac{\pi }{4}=\frac{3\sqrt{2}}{2}$. Step 2 — slope (Tangents and normals dekho).

• $\dot{x}=-2\sin t,\dot{y}=3\cos t$, toh $\frac{dy}{dx}=\frac{3\cos t}{-2\sin t}=-\frac{3}{2}\cot t$.
• $t=\pi /4$ par, $\cot (\pi /4)=1$, toh $m=-\frac{3}{2}$. Step 3 — line $(\sqrt{2},\frac{3\sqrt{2}}{2})$ se guzarti hai slope $-\frac{3}{2}$ ke saath: $y-\frac{3\sqrt{2}}{2}=-\frac{3}{2}\left(x-\sqrt{2}\right)\Rightarrow y=-\frac{3}{2}x+\frac{3}{2}\sqrt{2}+\frac{3\sqrt{2}}{2}=-\frac{3}{2}x+3\sqrt{2}.$ Toh $m=-\frac{3}{2},c=3\sqrt{2}\approx 4.243$.

• $\dot{x}=-2\sin t,\dot{y}=3\cos t$; $\ddot{x}=-2\cos t,\ddot{y}=-3\sin t$.
• Numerator $\ddot{y}\dot{x}-\dot{y}\ddot{x}=(-3\sin t)(-2\sin t)-(3\cos t)(-2\cos t)=6{\sin }^{2}t+6{\cos }^{2}t=6.$ (${\sin }^{2}t+{\cos }^{2}t=1$ use kiya.)
• Denominator ${\dot{x}}^3=(-2\sin t{)}^3=-8{\sin }^{3}t.$
• Isliye $\frac{d^{2}y}{d{x}^2}=\frac{6}{-8{\sin }^{3}t}=-\frac{3}{4{\sin }^{3}t}.$ $■$

• $\dot{x}=2t,\dot{y}=3t^2-3=3(t^2-1)$.
• (a) Horizontal ($\dot{y}=0,\dot{x}\ne 0$): $3(t^2-1)=0\Rightarrow t=\pm 1$; dono ke liye $\dot{x}=\pm 2\ne 0$. ✓
• (b) Vertical ($\dot{x}=0,\dot{y}\ne 0$): $2t=0\Rightarrow t=0$; wahan $\dot{y}=-3\ne 0$. ✓ Vertical tangent origin-side point $(0,0)$ par.
• (c) $t=1$ par Concavity. $\frac{d^{2}y}{d{x}^2}$ chahiye: $\ddot{x}=2,\ddot{y}=6t$. $\frac{d^{2}y}{d{x}^2}=\frac{(6t)(2t)-3(t^2-1)(2)}{(2t{)}^3}=\frac{12t^2-6t^2+6}{8t^3}=\frac{6t^2+6}{8t^3}=\frac{3(t^2+1)}{4t^3}.$ $t=1$ par: $\frac{3(2)}{4}=\frac{3}{2}>0$ → concave up.

Step 1 — target restate karo. $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)$ — slope ko $x$ ke saath differentiate karo. Step 2 — humare paas sirf $t$-derivatives hain, toh function $g=\frac{dy}{dx}=\frac{\dot{y}}{\dot{x}}$ par parametric trick apply karo: $\frac{d}{dx}(g)=\frac{dg/dt}{dx/dt}=\frac{1}{\dot{x}}\cdot \frac{d}{dt}\left(\frac{\dot{y}}{\dot{x}}\right).$ Step 3 — quotient rule $\frac{\dot{y}}{\dot{x}}$ par (top' · bottom − top · bottom', over bottom²): $\frac{d}{dt}\left(\frac{\dot{y}}{\dot{x}}\right)=\frac{\ddot{y}\dot{x}-\dot{y}\ddot{x}}{{\dot{x}}^2}.$ Step 4 — combine karo (Step 2 ke $\frac{1}{\dot{x}}$ se multiply karo): $\frac{d^{2}y}{d{x}^2}=\frac{1}{\dot{x}}\cdot \frac{\ddot{y}\dot{x}-\dot{y}\ddot{x}}{{\dot{x}}^2}=\frac{\ddot{y}\dot{x}-\dot{y}\ddot{x}}{{\dot{x}}^3}.■$ Quotient rule ka ${\dot{x}}^2$ aur Step 2 ka extra $\dot{x}$ — exactly yahan se cube janam leta hai.

Note karo ki curve hai $y=(e^t{)}^2=x^2$, toh hum jaante hain ki sahi answer hona chahiye $\frac{d^{2}y}{d{x}^2}=2$.

• $\dot{x}=e^t,\dot{y}=2e^{2t}$; $\ddot{x}=e^t,\ddot{y}=4e^{2t}$.
• (a) Sahi: $\frac{\ddot{y}\dot{x}-\dot{y}\ddot{x}}{{\dot{x}}^3}=\frac{(4e^{2t})(e^t)-(2e^{2t})(e^t)}{(e^t{)}^3}=\frac{4e^{3t}-2e^{3t}}{e^{3t}}=\frac{2e^{3t}}{e^{3t}}=2.$ ✓ ($y=x^2$ se match karta hai).
• (b) Naive galat: $\frac{\ddot{y}}{\ddot{x}}=\frac{4e^{2t}}{e^t}=4e^t=4x.$
• (c) Disagreement: sahi answer constant $2$ hai; naive answer $4x$ hai, jo $t$ ke saath change bhi karta hai. Woh sirf ek jagah coincide karte hain $4x=2\Rightarrow x=\frac{1}{2}$, ek single point par coincidence — proof ki naive formula genuinely galat hai.

Original ($x=t,y=t^2$): $\dot{x}=1,\dot{y}=2t$, toh $\frac{dy}{dx}=2t=2x.$ Reparametrized ($x=u^3,y=u^6$): yahan dots ka matlab $d/du$ hai.

• $\dot{x}=3u^2,\dot{y}=6u^5$.
• $\frac{dy}{dx}=\frac{6u^5}{3u^2}=2u^3.$
• Lekin $x=u^3$, toh $2u^3=2x.$ Identical. ✓ Parameter ek scaffold hai; curve aur uska slope real cheezein hain. Yahi deep reason hai ki shared $dt$ (ya $du$) cancel ho jaata hai — ek lesson jo saare Parametric curves mein jaata hai.