Visual walkthrough — Parametric differentiation — dy - dx, d²y - dx²

4.1.23 · D2 · Maths › Calculus I — Limits & Derivatives › Parametric differentiation — dy - dx, d²y - dx²

Kuch bhi shuru hone se pehle, vocabulary ke do points jo hum baar baar use karenge:

Step 1 — Bug aur uski do speeds

KYA. Hum ek moving point (the "bug") ko ek curve par rakhte hain. Clock-time $t$ par uski position $(x(t),y(t))$ hai. Ek tick $dt$ mein woh thoda sa slide karta hai.

KYUN. Kyunki curve ka koi equation $y=f(x)$ nahi hai — yeh loop kar sakta hai, khud ko cross kar sakta hai, ya vertical khad ho sakta hai. Humare paas sirf clock $t$ ka handle hai. Toh pehle hum motion describe karte hain, aur baad mein usse geometry nikalte hain.

PICTURE. Chota sa step ek horizontal piece $dx=\dot{x}dt$ (the run) aur ek vertical piece $dy=\dot{y}dt$ (the rise) mein split hota hai. Figure mein, teal east arrow aur orange north arrow follow karo; saath milkar woh curve ke saath dashed plum step banate hain.

• $dx=\dot{x}dt$ — bug ek tick mein kitna east drift karta hai: east-speed $\dot{x}$ times tick $dt$. Agar $\dot{x}>0$ toh woh right jaata hai, agar $\dot{x}<0$ toh left back karta hai.
• $dy=\dot{y}dt$ — usi tick mein woh kitna north drift karta hai: north-speed times same $dt$.

Step 2 — Slope do speeds ka ratio hai

KYA. Path ki slope rise-over-run hai. Vertical step ko horizontal step se divide karo: $\frac{dy}{dx}=\frac{\dot{y}dt}{\dot{x}dt}=\frac{\dot{y}}{\dot{x}}$

KYUN. Slope depend nahi karni chahiye bug kitna fast chalta hai — sirf road ki shape par. Clock-tick $dt$ rise aur run dono mein appear karta hai, toh woh cancels ho jaata hai, aur ek pure ratio of speeds reh jaata hai.

• $\dot{y}$ (top) — north speed, "rise per tick".
• $\dot{x}$ (bottom) — east speed, "run per tick".
• $dt$ — shared tick jo divide out ho jaata hai; isi liye walking-speed irrelevant hai.

PICTURE. Do bugs, fast aur slow, same hill par. Alag arrow lengths, identical arrow angle — same slope.

Yeh pehla result poora first derivative hai. Ise $y^{\prime }$ kehte hain: $y^{\prime }\equiv \frac{dy}{dx}=\frac{\dot{y}}{\dot{x}}(\dot{x}\ne 0)$

Step 3 — Trap: "bas dobara kar do"

KYA. Second derivative feel hota hai jaise woh $\ddot{y}/\ddot{x}$ hona chahiye — Step 2 ka pattern copy karo lekin top aur bottom par accelerations ke saath.

KYUN yeh galat hai. Notation ko dhyan se dekho. Symbol $\frac{d^{2}y}{d{x}^2}$ shorthand hai "slope-taking operation $\frac{d}{dx}$ ko slope $\frac{dy}{dx}$ par apply karo": $\frac{d^{2}y}{d{x}^2}\equiv \frac{d}{dx}\left(\frac{dy}{dx}\right)$

• Outer $\frac{d}{dx}$ kehta hai "differentiate with respect to $\mathbf{x}$" — $t$ ke saath nahi. Yeh change per step east measure karta hai.
• $\ddot{y}/\ddot{x}$ time-accelerations ka ratio hoga. Woh ek alag sawaal ka jawab deta hai: "bug ka velocity vector per tick kitna turn karta hai?", yeh nahi ki "map-slope east mein move karte waqt kaise badalta hai."

PICTURE. Do labelled boxes. Teal box woh sawaal hai jo hum actually chahte hain (slope changing per unit of eastward distance). Plum box seductive impostor hai (accelerations per tick). Yeh simply ek hi measurement nahi hain.

Step 4 — Step-2 trick ko slope par hi reuse karo

KYA. Humein $\frac{d}{dx}(y^{\prime })$ chahiye, lekin hum sirf $t$ ke against differentiate karna jaante hain. Toh Step 2 ka exact same speed-ratio idea apply karo (chain-rule engine), ab $y$ ki jagah $y^{\prime }$ quantity par: $\frac{d}{dx}(y^{\prime })=\frac{d{y}^{\prime }/dt}{dx/dt}=\frac{\frac{d}{dt}\left(y^{\prime }\right)}{\dot{x}}$

KYUN. Jo bhi $t$ ke saath vary karta hai woh "rate-along-x = (rate-along-t) ÷ (x-speed)" follow karta hai. Slope $y^{\prime }$ sirf ek aur passenger hai jo clock par ride kar raha hai. Toh use wahi treatment milti hai jo $y$ ko mili thi.

• Numerator $\frac{d}{dt}(y^{\prime })$ — slope per tick kitna change karta hai.
• Denominator $\dot{x}$ — east speed jisse hum divide karte hain "per tick" ko "per unit east" mein convert karne ke liye. Yahi future mein extra power ka source hai.

PICTURE. Step-2 triangle ka ek relabelled version: vertical axis ab "slope value $y^{\prime }$" hai, horizontal axis still $x$ hai. Same divide-by-$\dot{x}$ move.

Step 5 — Quotient rule se slope differentiate karo

KYA. Ab $\frac{d}{dt}(y^{\prime })$ compute karo jahan $y^{\prime }=\frac{\dot{y}}{\dot{x}}$. Yeh ek fraction hai jiska top aur bottom dono $t$ par depend karte hain, toh Quotient rule sahi tool hai: $\frac{d}{dt}\left(\frac{\dot{y}}{\dot{x}}\right)=\frac{\stackrel{{\text{(top)}}^{\prime }\cdot \text{bottom}}{\overbrace{\ddot{y}\dot{x}}}-\stackrel{\text{top}\cdot (\text{bottom}{)}^{\prime }}{\overbrace{\dot{y}\ddot{x}}}}{{\dot{x}}^2}$

KYUN quotient rule aur kuch simpler nahi? Kyunki $y^{\prime }$ do changing cheezoon ka ratio hai. Agar hum sirf top differentiate karte toh hum pretend karte ki bottom ($\dot{x}$) frozen hai — lekin $\dot{x}$ khud $\ddot{x}$ rate se change ho raha hai. Quotient rule precisely woh bookkeeping hai jo dono changes ko credit deta hai.

Term by term:

• $\ddot{y}\dot{x}$ — "top differentiate karo, bottom rakho." North-acceleration east-speed se scaled.
• $-\dot{y}\ddot{x}$ — "top rakho, bottom differentiate karo." Yeh woh term hai jo tabhi vanish hoti hai jab $\ddot{x}=0$; ise drop karo aur cycloid example toot jaata hai.
• ${\dot{x}}^2$ — quotient rule hamesha denominator ko square karta hai. $\dot{x}$ ki power ka pehla appearance.

PICTURE — aise padho. Figure mein do ingredients corners par hain: $\dot{y}$ (orange, top) aur $\dot{x}$ (teal, bottom). Har diagonal arrow us corner ko differentiate karta hai jahan se woh leave karta hai:

• $\dot{y}$ se $\dot{x}$-corner tak jaane wala arrow product $\ddot{y}\dot{x}$ banata hai (top differentiated, times bottom);
• $\dot{x}$ se $\dot{y}$-corner tak jaane wala arrow $\dot{y}\ddot{x}$ banata hai (bottom differentiated, times top). Phir — crucially — subtraction ka order hai "top-diagonal minus bottom-diagonal": $\ddot{y}\dot{x}-\dot{y}\ddot{x}$. Inhe reverse karo aur har concavity sign flip ho jaata hai, toh minus ka dhyan rakho.

Step 6 — Ek baar aur divide karo: cube ka janam

KYA. Step 4 aur Step 5 ko saath rakkho. Step 4 ne kaha tha "$\frac{d}{dt}(y^{\prime })$ ko $\dot{x}$ se divide karo." Step 5 ne woh numerator diya. Toh: $\frac{d^{2}y}{d{x}^2}=\frac{1}{\dot{x}}\cdot \frac{\ddot{y}\dot{x}-\dot{y}\ddot{x}}{{\dot{x}}^2}=\boxed{\frac{\ddot{y}\dot{x}-\dot{y}\ddot{x}}{{\dot{x}}^3}}$

KYUN cube hai. Do divisions stack hote hain:

• ${\dot{x}}^2$ quotient rule se (Step 5),
• $\times \frac{1}{\dot{x}}$ "per tick" ko "per unit east" convert karne se (Step 4).

${\dot{x}}^2\cdot \dot{x}={\dot{x}}^3$. Yahi fingerprint cube ka poora origin hai. No cube ⇒ tum do mein se ek division bhool gaye.

PICTURE — bottom-up padho. Figure ek stack hai. Sabse neeche wali tile quotient-rule denominator ${\dot{x}}^2$ hai. Uske bilkul upar doosri tile hai, Step 4 se extra $÷\dot{x}$. Wo downward arrow follow karo jo do tiles ko merge karta hai: ${\dot{x}}^2\cdot \dot{x}$ collapse hokar basement mein single label ${\dot{x}}^3$ ban jaata hai, aur finished formula neeche print hoti hai. Picture ka point yeh hai ki division ke do alag floors hain, ek nahi.

Step 7 — Degenerate case: $\dot{x}=0$

KYA. Upar har step ne quietly $\dot{x}$ se divide kiya. Kya ho agar bug ki east-speed kisi instant par zero ho?

KYUN matters karta hai. $\dot{x}=0$ ka matlab hai bug us tick mein seedha upar ya neeche move kar raha hai — tangent vertical hai. Tab:

• $\frac{dy}{dx}=\frac{\dot{y}}{0}$ undefined hai (slope $\to \infty$),
• $\frac{d^{2}y}{d{x}^2}$ ke neeche ${\dot{x}}^3=0$ hai, woh bhi undefined.

Yeh geometry sach bol rahi hai, formula mein koi bug nahi: vertical tangent par genuinely koi finite "rise per run" nahi hoti. Aise points handle karne ke liye Tangents and normals dekho (roles swap karo: uski jagah $\frac{dx}{dy}$ use karo).

PICTURE. Ek unit circle $x=\cos t,y=\sin t$. $t=0$ par point $(1,0)$ hai, arrow seedha upar point karta hai, $\dot{x}=-\sin 0=0$ — vertical tangent, formula undefined. $t=\pi /2$ par point $(0,1)$ hai, $\dot{x}=-1\ne 0$, slope $=0$ (flat top). Dono cases dikhaye gaye hain.

The one-picture summary

Yeh single diagram saat steps compress karta hai: chaar raw numbers ($\dot{x},\dot{y},\ddot{x},\ddot{y}$) se shuru karo, ek division se slope banao, us slope ko quotient rule se differentiate karo (${\dot{x}}^2$ deta hai), phir ek baar aur divide karo (total ${\dot{x}}^3$ deta hai).

Active recall

Connections

• Parametric differentiation — dy - dx, d²y - dx² — parent result jise yeh page visually derive karta hai.
• Chain rule — "per tick" rates ko "per unit $x$" rates mein convert karta hai (Steps 2 & 4).
• Quotient rule — slope $\dot{y}/\dot{x}$ differentiate karta hai (Step 5).
• Concavity and second derivative — sign se concave up/down padhta hai (Step 7).
• Tangents and normals — vertical-tangent case handle karta hai jahan $\dot{x}=0$ ho.
• Cycloid, Parametric curves — curves jahan $\dot{y}\ddot{x}$ term truly matter karti hai.