Parametric differentiation — dy - dx, d²y - dx²: Question bank

4.1.23 · D5 · Maths › Calculus I — Limits & Derivatives › Parametric differentiation — dy - dx, d²y - dx²

True or false — justify karo

True or false: Parametric curve ka slope

d y / d x

is baat par depend karta hai ki tum uske saath kitni tezi se chalte ho.

False. Dono

d y

aur

d x

mein ek common factor

d t

hota hai jo cancel ho jaata hai; slope ek pure ratio

\overset{y}{˙} / \overset{x}{˙}

hai, toh apni speed double karne se path ki steepness par koi fark nahi padta.

True or false: Agar har jagah

\overset{x}{˙} \neq = 0

ho, toh curve ka vertical tangent phir bhi ho sakta hai.

False. Vertical tangent ke liye run ka vanish hona zaroori hai, yaani

\overset{x}{˙} = 0

; agar

\overset{x}{˙}

kabhi zero nahi hota toh tangent kabhi vertical nahi hoga (halanki

\overset{y}{˙} = 0

phir bhi horizontal tangent de sakta hai).

True or false:

\frac{d ^{2} y}{d x ^{2}} = \frac{y ¨}{x ¨}

kyunki tum "usi tarah se do baar differentiate" karte ho.

False. Second derivative ka matlab hai

d y / d x

ko $x$ ke saath differentiate karna,

t

ke saath nahi;

t

ke through wapas jaane par

\overset{x}{˙}^{3}

ke upar ek quotient-rule expression banta hai, accelerations ka simple ratio nahi.

True or false: Formula

\frac{y ¨ x ˙ - y ˙ x ¨}{x ˙ ^{3}}

mein

\overset{x}{˙}^{3}

sirf Quotient rule ki wajah se hai.

False. Quotient rule

\overset{x}{˙}^{2}

deta hai; teesri power final step mein

\overset{x}{˙}

se ek extra division ki wajah se aati hai, yaani

\frac{d}{d x} (y^{'}) = \frac{d ( y ^{'} ) / d t}{d x / d t}

True or false: Agar kisi point par dono

\overset{x}{˙} = 0

aur

\overset{y}{˙} = 0

hon, toh curve wahan smooth hai.

False. Woh ek singular point hai (ek possible cusp ya corner, jaise Cycloid apne base par); tangent direction genuinely undefined hai aur

\overset{y}{˙} / \overset{x}{˙}

indeterminate

0/0

deta hai.

True or false:

x = t, y = f (t)

ke liye, parametric second derivative ordinary

f^{''} (x)

mein reduce ho jaati hai.

True. Tab

\overset{x}{˙} = 1, \overset{x}{¨} = 0

hoga, toh formula

\overset{y}{¨} /1 = \overset{y}{¨} = f^{''} (t) = f^{''} (x)

ban jaata hai — parametric machinery sahi se plain calculus reproduce karti hai.

True or false:

\frac{d ^{2} y}{d x ^{2}}

ka sign tumhe concave up ya down batata hai, chahe

t

kisi bhi direction mein increase ho.

True. Concavity

x y

-plane mein shape ki ek geometric property hai;

t

reverse karne se

\overset{x}{˙}

aur

\overset{y}{˙}

dono ka sign flip hota hai lekin concavity expression unchanged rehta hai (neeche "Why" section dekhein).

True or false: Ek parametric curve jahan kisi interval mein

\overset{x}{˙} < 0

ho, usse galat draw kiya gaya hai.

False.

\overset{x}{˙} < 0

ka matlab sirf ye hai ki point wahan leftward move kar raha hai; curve same

x

-values ke upar double back kar sakta hai, aur exactly isliye hum

y = f (x)

ki jagah parameter use karte hain.

Error dhundo

Student likhta hai

\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d t} (\frac{y ˙}{x ˙})

aur ruk jaata hai. Kya missing hai?

Unhone

t

ke saath differentiate kiya lekin

x

par wapas convert nahi kiya; unhe abhi bhi

\overset{x}{˙}

se divide karna hai, jisse woh extra factor aata hai jo

\overset{x}{˙}^{2}

\overset{x}{˙}^{3}

mein turn karta hai.

Student likhta hai

\frac{d y}{d x} = \frac{x ˙}{y ˙}

. Ye ulta kyun hai?

Jo variable tum ke liye solve karte ho (yahan

y

) woh upar jaata hai aur jo variable tum ke against differentiate karte ho (

x

) woh neeche;

\overset{y}{˙} / \overset{x}{˙}

rise-over-run hai, aur use flip karne par run-over-rise milta hai.

Student numerator ko

\overset{y}{˙} \overset{x}{¨} - \overset{y}{¨} \overset{x}{˙}

compute karta hai. Kya galat hua?

Dono terms ka sign swap ho gaya hai.

\overset{y}{˙} / \overset{x}{˙}

par Quotient rule hai (bottom · derivative-of-top − top · derivative-of-bottom), yaani

\overset{y}{¨} \overset{x}{˙} - \overset{y}{˙} \overset{x}{¨}

; use reverse karne se poora second derivative negate ho jaata hai.

Student

d t

ko "kyunki ye chhota hai" cancel karta hai, "kyunki ye common factor hai" ki wajah se nahi. Reasoning fix karo.

Chhotapan irrelevant hai —

d t

cancel hota hai kyunki ye identically dono

\overset{x}{˙}

aur

\overset{y}{˙}

ko multiply karta hai, toh ye algebraically divide out ho jaata hai; cancellation kisi bhi nonzero

d t

ke liye hold karti.

Student claim karta hai ki denominator mein

\overset{x}{˙}^{3}

ko "safe rehne ke liye"

∣ \overset{x}{˙} ∣^{3}

se replace kiya ja sakta hai. Kya ye sahi hai?

Nahi.

\overset{x}{˙}

ka sign genuinely matter karta hai — ye encode karta hai ki point right move kar raha hai ya left — aur derivation exactly

\overset{x}{˙}^{3}

deta hai; absolute value force karna concavity ke sign ko corrupt kar dega.

Student Chain rule ko

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

likhta hai. Error dhundo.

Chain rule multiply karta hai, add nahi:

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

. Usse

d y / d x

ke liye solve karna exactly wahi hai jo slope formula deta hai.

Student kehta hai horizontal tangent par "

\overset{x}{˙} = 0

." Unhe correct karo.

Horizontal tangent ke liye zero rise chahiye, yaani

\overset{y}{˙} = 0

(with

\overset{x}{˙} \neq = 0

). Zero run

\overset{x}{˙} = 0

vertical tangent ki condition hai — unhone dono ko swap kar diya hai.

Why questions

x

aur

y

ko is method ke kaam karne ke liye same parameter

t

share kyun karna chahiye?

Kyunki slope unke changes ka ratio hai usi instant par; agar ye alag clocks se driven hote toh cancel karne ke liye koi common

d t

nahi hota aur curve trace karne ke liye koi single moving point nahi hota.

Hume specifically second derivative ke liye Quotient rule kyun chahiye lekin first ke liye nahi?

First derivative ek single division

\overset{y}{˙} / \overset{x}{˙}

hai; second ke liye us poore fraction ko

t

ke saath differentiate karna padta hai, aur fraction ko differentiate karna exactly wahi hai jiske liye quotient rule hai.

Implicit differentiation ek valid alternative kyun hai jab curve parametrically given nahi hai?

Jab ek relation

F (x, y) = 0

ho lekin koi parameter nahi, implicit differentiation

y

x

ki unknown function maankar seedha

d y / d x

le aata hai — ye ek alag route se same "slope without

y = f (x)

" question ka jawab deta hai.

Condition

\overset{x}{˙} \neq = 0

first derivative mein kyun aata hai lekin

\overset{x}{˙}^{3} \neq = 0

(same cheez) second ko kyun haunt karta hai?

Dono exactly wahan fail karte hain jahan

\overset{x}{˙} = 0

; cube sirf second derivative ko faster blow up karta hai, lekin underlying geometry — vertical tangent ka koi finite slope nahi hai differentiate karne ke liye — identical hai.

Same curve ki do bilkul alag parametrisations ek shared point par same

d y / d x

kyun de sakti hain?

Kyunki slope curve ki shape ki ek intrinsic property hai, parametrisation ki nahi; reparametrising

\overset{x}{˙}

aur

\overset{y}{˙}

dono ko same chain-rule factor se rescale karta hai, jo ratio mein cancel ho jaata hai.

"Bottom par cube" ek correct second derivative ki reliable fingerprint kyun hai?

Koi bhi correct derivation

\overset{x}{˙}

se division ke saath khatam hoti hai quotient rule se already $\overset{x}{˙}^{2}$ produce hone ke baad;

\overset{x}{˙}^{3}

dekhna confirm karta hai ki dono steps hue, jabki

\overset{x}{˙}^{2}

bhuli hui final division ko betray karta hai.

Edge cases

Ek point par jahan

\overset{x}{˙} = 0

lekin

\overset{y}{˙} \neq = 0

ho, curve kaisa dikhta hai aur

d y / d x

kya hai?

Ek vertical tangent: run zero hai jabki rise nonzero, toh

d y / d x = \overset{y}{˙} /0

undefined hai (slope

\to \pm \infty

) — ye ek legitimate feature hai, koi mistake nahi.

Ek point par jahan

\overset{y}{˙} = 0

lekin

\overset{x}{˙} \neq = 0

ho, tangent kya hai?

Ek horizontal tangent:

d y / d x = 0/ \overset{x}{˙} = 0

, toh curve momentarily flat hai, jaise parent ke Example 1 mein

t = π /2

par circle ka top.

d y / d x

aur

d^{2} y / d x^{2}

ka kya hota hai jab

t

ek aisi value approach karta hai jahan

\overset{x}{˙} \to 0^{+}

versus

\overset{x}{˙} \to 0^{-}

Slope

+ \infty

- \infty

ki taraf diverge karta hai

\overset{y}{˙}

ke sign par depend karte hue, aur concavity (

\overset{x}{˙}^{3}

neeche ke saath) vertical tangent ke across sign flip kar sakti hai — ye ek genuine geometric transition hai, discontinuity error nahi.

Cycloid

x = t - sin t, y = 1 - cos t

ke liye, base point (

t = 2 π k

) ek cusp kyun hai aur smooth min kyun nahi?

Wahan dono

\overset{x}{˙} = 1 - cos t = 0

aur

\overset{y}{˙} = sin t = 0

hain, toh

\overset{y}{˙} / \overset{x}{˙} = 0/0

indeterminate hai; point momentarily rest par aata hai aur direction reverse karta hai, jisse ek rounded valley ki jagah sharp cusp banta hai.

Agar ek curve apne steps retrace kare (do alag

t

values par same

(x, y)

), kya wahan uske do alag slopes ho sakte hain?

Haan. Self-intersection par har pass par distinct tangent directions ho sakti hain kyunki

\overset{y}{˙} / \overset{x}{˙}

har

t

par alag se evaluate hota hai — ye exactly woh case hai jo

y = f (x)

kabhi describe nahi kar sakta.

x = a t, y = b t

(constants

a \neq = 0, b

) straight-line path ke liye

d^{2} y / d x^{2}

kya hai?

Zero. Yahan

\overset{x}{¨} = \overset{y}{¨} = 0

hain, toh numerator

\overset{y}{¨} \overset{x}{˙} - \overset{y}{˙} \overset{x}{¨} = 0

hai; ek line mein koi bending nahi hai, exactly jaisa concavity zero kehna chahiye.

Connections

Chain rule — ye reason hai ki $d y / d t = \frac{d y}{d x} \cdot \frac{d x}{d t}$ aur final $\div \overset{x}{˙}$ ka source.
Quotient rule — $\overset{y}{˙} / \overset{x}{˙}$ differentiate karta hai aur $\overset{x}{˙}^{2}$ supply karta hai.
Implicit differentiation — upar referenced parameter-free alternative.
Tangents and normals — jahan vertical/horizontal tangent edge cases matter karte hain.
Concavity and second derivative — $d^{2} y / d x^{2}$ ke sign ko interpret karta hai.
Cycloid, Parametric curves — cusp aur self-intersection edge cases yahan rehte hain.