Parametric differentiation — dy - dx, d²y - dx²: Worked examples

4.1.23 · D3 · Maths › Calculus I — Limits & Derivatives › Parametric differentiation — dy - dx, d²y - dx²

Ye page parametric differentiation ka exercise gym hai. Parent note ne do formulas build kiye; yahan hum unhe har tarah ke situation se guzarte hain jo ek problem present kar sakta hai — acchi slopes, vertical tangents, degenerate points, negative signs, ek real-world word problem, aur ek exam twist.

Recall Do tools jo hum baar baar reuse karte hain (parent se)

Pehla derivative — speed ratio (vertical speed divided by horizontal speed): $\frac{d y}{d x} = \frac{y ˙}{x ˙} (\overset{x}{˙} \neq = 0)$ Doosra derivative — quotient rule plus ek extra $\overset{x}{˙}$ neeche: $\frac{d ^{2} y}{d x ^{2}} = \frac{y ¨ x ˙ - y ˙ x ¨}{x ˙ ^{3}}$ Yahan dot ka matlab hai "parameter $t$ ke saath rate of change": $\overset{x}{˙} = d x / d t$ , $\overset{x}{¨} = d^{2} x / d t^{2}$ , aur similarly $y$ ke liye. Agar kuch naya lage, parent dobara padho — ye page ise use karta hai, re-derive nahi karta. Andar ka engine Chain rule aur Quotient rule hai. Is poori page mein, hum in do displayed equations ko first-derivative formula aur second-derivative formula kehte hain.

Scenario matrix

Har parametric-differentiation problem in cells mein se kisi ek mein aata hai. Hamare worked examples neeche us cell ka label carry karte hain jo wo hit karte hain, isliye end tak koi cell empty nahi rahegi.

#	Cell (case class)	Tricky kyun hai	Example
A	Plain "nice" point, $\overset{x}{˙} \neq = 0$	kuch nahi — warm-up	Ex 1
B	$\overset{x}{˙} = 0$ → vertical tangent	slope undefined, error nahi	Ex 2
C	$\overset{y}{˙} = 0$ → horizontal tangent	slope exactly $0$	Ex 3
D	Sign of $d^{2} y / d x^{2}$ → concavity	up vs down, aur $\overset{x}{˙}$ ka sign matter karta hai	Ex 4
E	Degenerate point: $\overset{x}{˙} = \overset{y}{˙} = 0$ (cusp)	$0/0$ , limit chahiye	Ex 5
F	Real-world word problem (projectile)	units, sahi $t$ choose karo	Ex 6
G	Exam twist: wo $t$ dhundho jo required slope de	backwards solve karo	Ex 7
H	Limiting behaviour jab $t \to$ special value	asymptotic slope	Ex 8

Har cell dikhata hai ki same do formulas alag-alag tarike se behave karte hain. Khaas taur pe cells B, D, H mein $\overset{x}{˙}$ ka sign dekho — yahi quiet troublemaker hai.

Cell A — warm-up (nice point)

Worked example 1 —

$x = t^{2}, y = t^{3}$ ka slope aur concavity $t = 2$ par Forecast: Guess: $t = 2$ par slope positive hai ya negative? Curve wahan upar bend kar raha hai ya neeche? Padhne se pehle guess likh lo.

Har coordinate ko $t$ ke saath differentiate karo. Ye step kyun? formulas poori tarah $\overset{x}{˙}, \overset{y}{˙}, \overset{x}{¨}, \overset{y}{¨}$ se bane hain. $\overset{x}{˙} = 2 t, \overset{y}{˙} = 3 t^{2}, \overset{x}{¨} = 2, \overset{y}{¨} = 6 t$
First derivative $= \overset{y}{˙} / \overset{x}{˙}$ . Ye step kyun? slope vertical aur horizontal speed ka ratio hai. $\frac{d y}{d x} = \frac{3 t ^{2}}{2 t} = \frac{3 t}{2} \Rightarrow at t = 2 : \frac{3 ( 2 )}{2} = 3$
Second derivative is page ke top par recall ki gayi second-derivative formula ke zariye. Ye step kyun? hume $t$ ke through jaana hai, accelerations divide nahi karne — isliye $\overset{x}{˙}, \overset{y}{˙}, \overset{x}{¨}, \overset{y}{¨}$ ko $\frac{y ¨ x ˙ - y ˙ x ¨}{x ˙ ^{3}}$ mein feed karte hain. $\frac{d ^{2} y}{d x ^{2}} = \frac{( 6 t ) ( 2 t ) - ( 3 t ^{2} ) ( 2 )}{( 2 t ) ^{3}} = \frac{12 t ^{2} - 6 t ^{2}}{8 t ^{3}} = \frac{6 t ^{2}}{8 t ^{3}} = \frac{3}{4 t}$ $t = 2$ par: $\frac{3}{8}$ .

Verify: slope $3 > 0$ (rising) aur $d^{2} y / d x^{2} = 3/8 > 0$ (upward bending) — ye ek curve se match karta hai jo climb kar raha hai aur upar curve kar raha hai. Sanity check: $t$ eliminate karo ( $x = t^{2}, y = t^{3} \Rightarrow y = x^{3/2}$ ), toh $y^{'} = \frac{3}{2} x^{1/2}$ , aur $t = 2$ par, $x = 4$ : $y^{'} = \frac{3}{2} \cdot 2 = 3$ . ✓

Cell B — vertical tangent ( $\overset{x}{˙} = 0$ )

Neeche di figure mein lavender mein unit circle hai. Red dashed lines leftmost aur rightmost points par vertical tangents hain, red arrows seedha upar point karte hain ye stress karne ke liye ki wahaan motion purely vertical hai; mint dotted lines contrast ke liye horizontal tangents (top aur bottom) mark karti hain. Example 2 padhte waqt picture par do red points $(\pm 1, 0)$ dhundho.

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

Figure 1 — Unit circle $x = cos t, y = sin t$ . Red dashed = vertical tangents jahan $\overset{x}{˙} = 0$ (points $(\pm 1, 0)$ ); mint dotted = horizontal tangents jahan $\overset{y}{˙} = 0$ (points $(0, \pm 1)$ ).

Worked example 2 —

$x = cos t, y = sin t$ ke liye tangent vertical kahan hai? Geometric answer ke liye $t \in [0, 2 π)$ lo, lekin yaad rakho $t$ koi bhi real number ho sakta hai aur pattern har $2 π$ pe repeat hota hai. Forecast: circle. Ek circle par tangent line seedhi khadi kahan hoti hai? Compute karne se pehle points predict karo.

Derivatives. Ye step kyun? hume $\overset{x}{˙}$ chahiye taaki dekh saken kahan run vanish hoti hai. $\overset{x}{˙} = - sin t, \overset{y}{˙} = cos t$
$\overset{x}{˙} = 0$ set karo. Ye step kyun? vertical tangent ka matlab zero horizontal motion hai — pure up/down. $- sin t = 0 \Rightarrow sin t = 0 \Rightarrow t = nπ (n koi bhi integer)$ Primary lap $[0, 2 π)$ par ye $t = 0$ aur $t = π$ hai; baki saare solutions same do points par land karte hain kyunki circle repeat hoti hai.
Un points ko locate karo. $t = 0$ par: $(cos 0, sin 0) = (1, 0)$ . $t = π$ par: $(- 1, 0)$ . Ye step kyun? geometric spots naam karne ke liye.
Confirm karo ki slope blow up hoti hai — aur har side par uska sign check karo. Ye step kyun? undefined slope yahan vertical ki definition hai; lekin "kaun sa infinity" batata hai curve kaise lean karta hai. $t = 0$ ke paas, $\frac{d y}{d x} = \frac{c o s t}{- s i n t} = - cot t .$

Jab $t \to 0^{+}$ : $sin t > 0$ (small positive), $cos t \to 1 > 0$ , isliye $- cot t = - \frac{cos t}{sin t} \to - \infty$ .
Jab $t \to 0^{-}$ : $sin t < 0$ (small negative), $cos t \to 1 > 0$ , isliye $- cot t \to + \infty$ .

$sin t$ ka sign (denominator) hi infinity flip karta hai: tangent line dono taraf se vertical hai, lekin curve point ko ek taraf steeply neeche aur doosri taraf steeply upar chhod kar jaati hai. ( $t = π$ par bhi same one-sided flip hota hai, signs ki roles swap hoke.)

Verify: $(\pm 1, 0)$ unit circle ke leftmost/rightmost points hain; Figure 1 mein red vertical arrows dekho — wahan tangents exactly vertical hain. ✓ Formula ka suppose karna hai ki fail ho ( $\pm \infty$ de), aur wahi karta hai. Dekho Tangents and normals.

Cell C — horizontal tangent ( $\overset{y}{˙} = 0$ )

Worked example 3 — Usi circle ka top aur bottom

Same curve $x = cos t, y = sin t$ , phir se geometric answer ke liye $t \in [0, 2 π)$ lo (pattern har $2 π$ pe repeat hota hai). Forecast: tangent perfectly flat (slope $0$ ) kahan hai? Do points guess karo.

Is curve ke liye slope formula recall karo. Ye step kyun? hume chahiye kahan slope $0$ equals karta hai, isliye pehle ye establish karte hain ki slope kahan se aata hai. First-derivative formula use karte hue $\overset{x}{˙} = - sin t, \overset{y}{˙} = cos t$ ke saath: $\frac{d y}{d x} = \frac{y ˙}{x ˙} = \frac{c o s t}{- s i n t} = - cot t .$
$\overset{y}{˙} = 0$ set karo. Ye step kyun? flat tangent ka matlab zero vertical motion hai, jo slope ke numerator ko vanish karta hai. $cos t = 0 \Rightarrow t = \frac{π}{2} + nπ (n koi bhi integer)$ Primary lap $[0, 2 π)$ par ye $t = \frac{π}{2}$ aur $t = \frac{3 π}{2}$ hai.
Points: $t = \frac{π}{2} \to (0, 1)$ ; $t = \frac{3 π}{2} \to (0, - 1)$ . Ye step kyun? geometric spots naam karo.
Check karo ki slope $0$ hai, undefined nahi. $t = \frac{π}{2}$ par: $\frac{d y}{d x} = - cot (\frac{π}{2}) = 0$ . Ye step kyun? pakka karne ke liye ki hum "flat" hit kar rahe hain, "vertical" nahi — yahan $\overset{x}{˙} = - sin (\frac{π}{2}) = - 1 \neq = 0$ , isliye ratio ek clean $0$ hai (numerator zero, denominator non-zero).

Verify: $(0, \pm 1)$ circle ke top aur bottom hain — tangents wahan horizontally chalte hain, slope $0$ . ✓

Cell D — $d^{2} y / d x^{2}$ ke sign se concavity

Worked example 4 —

$x = cos t, y = sin t$ $t = \frac{π}{4}$ par concave up hai ya down? Forecast: tum circle ke upper-right par ho. Ye valley ki tarah curve karta hai ya hill ki tarah? Guess karo.

Circle ka second derivative scratch se derive karo. Ye step kyun? hume parabola ka answer borrow nahi karna chahiye — circle ko apni computation chahiye. $\overset{x}{˙} = - sin t, \overset{y}{˙} = cos t, \overset{x}{¨} = - cos t, \overset{y}{¨} = - sin t$ ko second-derivative formula mein feed karo: $\frac{d ^{2} y}{d x ^{2}} = \frac{y ¨ x ˙ - y ˙ x ¨}{x ˙ ^{3}} = \frac{( - s i n t ) ( - s i n t ) - ( c o s t ) ( - c o s t )}{( - s i n t ) ^{3}} = \frac{s i n ^{2} t + c o s ^{2} t}{- s i n ^{3} t} = \frac{- 1}{s i n ^{3} t},$ numerator mein $sin^{2} t + cos^{2} t = 1$ use karte hue.
$t = \frac{π}{4}$ plug karo. Ye step kyun? required point par sign evaluate karo. $sin \frac{π}{4} = \frac{1}{2} \Rightarrow sin^{3} \frac{π}{4} = \frac{1}{2 2} \Rightarrow \frac{d ^{2} y}{d x ^{2}} = \frac{- 1}{1/ ( 2 2 )} = - 22 \approx - 2.828$
Sign padho. Ye step kyun? negative second derivative ka matlab concave down hai (dekho Concavity and second derivative).

Verify: circle ke upper half par curve ek hill ki tarah arch karta hai — concave down — isliye negative value bilkul sahi hai. ✓ Contrast: lower half par $sin t < 0$ , isliye $- 1/ sin^{3} t > 0$ → concave up (valley-shaped). $sin t$ ka sign ( $sin^{3} t$ ke through) halves ke beech concavity flip karta hai — yahi cell D action mein hai.

Cell E — degenerate point (cusp, $\overset{x}{˙} = \overset{y}{˙} = 0$ )

Agli figure mein curve $x = t^{2}, y = t^{3}$ dikhti hai. Lavender branch $t \leq 0$ ke liye trace ki gayi hai, coral branch $t \geq 0$ ke liye; dono origin par slate dot par milti hain, ek sharp beak (cusp) banakar. Mint arrow common horizontal tangent direction dikhata hai. Example 5 padhte waqt, notice karo ki tangent horizontal hone ke bawajood, do branches origin se same line par nikal kar jaati hain — yahi ise smooth minimum ki bajaye cusp banata hai.

Figure 2 — $x = t^{2}, y = t^{3}$ ka cusp origin par. Dono branches ( $t \leq 0$ lavender, $t \geq 0$ coral) ek horizontal limiting tangent (mint) ke saath milti hain, lekin point smooth nahi hai.

Worked example 5 —

$x = t^{2}, y = t^{3}$ ka cusp $t = 0$ par Forecast: $t = 0$ par dono $\overset{x}{˙}$ aur $\overset{y}{˙}$ zero hain — formula $0/0$ deta hai. Curve wahan slope rakhta hai, ya sharp spike? Guess karo.

$t = 0$ par dono speeds check karo. Ye step kyun? blindly divide karne se pehle degeneracy detect karne ke liye. $\overset{x}{˙} = 2 t = 0, \overset{y}{˙} = 3 t^{2} = 0 at t = 0$ Dono zero — point $(0, 0)$ parametrisation ka stationary point hai; direct substitution forbidden $0/0$ hai.
Iske bajaaye limit lo. Ye step kyun? $0/0$ "koi answer nahi" nahi hai — iska matlab hai "nearby trend dekho." Ex 1 se, $\frac{d y}{d x} = \frac{3 t}{2}$ . $lim_{t \to 0} \frac{d y}{d x} = lim_{t \to 0} \frac{3 t}{2} = 0$
Interpret karo. Ye step kyun? limiting slope $0$ (horizontal) hai, lekin curve yahan actually direction reverse karta hai (ek cusp): $t < 0$ vs $t > 0$ se approach karte waqt bug ek hi tangent ke saath aata aur wapas jaata hai.

Verify: $x \geq 0$ ke liye $y = x^{3/2}$ deta hai $y^{'} = \frac{3}{2} x^{1/2} \to 0$ jab $x \to 0^{+}$ — limiting slope $0$ se match karta hai. Picture (Figure 2) origin par sharp beak dikhata hai: tangent horizontal hai lekin point smooth nahi hai. ✓ Isliye cell E ko hamesha limit chahiye, plug-in kabhi nahi.

Cell F — real-world word problem (projectile)

Worked example 6 — Ek pheka hua ball

Ek ball horizontal speed $u = 3 m/s$ aur initial upward speed $v = 4 m/s$ ke saath launch ki jaati hai; gravity $g = 10 m/s^{2}$ . Uska path ( $t$ seconds mein, positions metres mein) hai $x = 3 t, y = 4 t - 5 t^{2} .$ (a) $t = 0.2$ s par trajectory ka slope kya hai? (b) Kis time par ball momentarily flat hai (peak)? Forecast: launch par ball steeply rise karti hai; guess karo ki $t = 0.2$ par slope abhi bhi positive hai ya nahi.

Time ke saath differentiate karo. Ye step kyun? $t$ yahan literally time hai; $\overset{x}{˙}, \overset{y}{˙}$ velocity components hain (units m/s). $\overset{x}{˙} = 3, \overset{y}{˙} = 4 - 10 t$
Path ka slope $= \overset{y}{˙} / \overset{x}{˙}$ . Ye step kyun? map par slope do velocity components ka ratio hai — $d t$ (aur units s) cancel ho jaate hain, ek pure number bachta hai. $\frac{d y}{d x} = \frac{4 - 10 t}{3}$
(a) $t = 0.2$ par: $\frac{4 - 10 ( 0.2 )}{3} = \frac{4 - 2}{3} = \frac{2}{3} \approx 0.667$ . Ye step kyun? pooche gaye time par evaluate karo — abhi bhi positive, abhi bhi rising.
(b) Peak = horizontal tangent, $\overset{y}{˙} = 0$ . Ye step kyun? top par vertical velocity zero hoti hai (cell C disguise mein). $4 - 10 t = 0 \Rightarrow t = 0.4 s$

Verify: units check — $\overset{y}{˙} / \overset{x}{˙}$ hai $(m/s) / (m/s)$ = dimensionless slope. ✓ Peak height: $y (0.4) = 4 (0.4) - 5 (0.4)^{2} = 1.6 - 0.8 = 0.8$ m, aur $\overset{y}{˙} (0.4) = 0$ flatness confirm karta hai. ✓

Cell G — exam twist ( $t$ ke liye backwards solve karo)

Cell H — slope ka limiting behaviour

Final figure mein lavender mein cycloid ka ek arch dikhta hai, origin par cusp ke saath. Cusp ke paas teen short tangent segments drawn hain — $t = 0.6$ (mint), $t = 0.3$ (butter) aur $t = 0.12$ (coral). Dekho kaise wo steeper hote jaate hain jab $t$ $0$ ki taraf shrink karta hai: yahi hai Example 8 mein slope ke $+ \infty$ ki taraf jaane ka visual matlab.

Figure 3 — Cycloid $x = t - sin t, y = 1 - cos t$ . Tangent segments $t = 0.6, 0.3, 0.12$ par cusp ki taraf approach karte waqt vertical ki taraf steep hote hain origin par.

Worked example 8 — Cycloid

$x = t - sin t, y = 1 - cos t$ ka slope $t \to 0^{+}$ par Forecast: $t = 0$ par cycloid cusp ke saath zameen ko touch karta hai. Kya tangent vertical, horizontal, ya kuch beech mein ban jaata hai jab hum $t = 0$ se creep up karte hain? Guess karo.

Slope (parent Example 3 se). Ye step kyun? jaana hua kaam reuse karo. $\frac{d y}{d x} = \frac{s i n t}{1 - c o s t}$
$t = 0$ directly try karo: $\frac{sin 0}{1 - cos 0} = \frac{0}{0}$ . Ye step kyun? ye expose karne ke liye ki ye degenerate hai — limit chahiye (cell E ka cousin, lekin ab hum limiting value chahte hain).
Small- $t$ approximations. Ye step kyun? tiny $t$ ke liye, $sin t \approx t$ aur $1 - cos t \approx \frac{t ^{2}}{2}$ — ye $0/0$ ko kuch aisa banate hain jo hum padh saken. $\frac{d y}{d x} \approx \frac{t}{t ^{2} /2} = \frac{2}{t} t \to 0^{+} + \infty$
Interpret karo. Ye step kyun? infinite limiting slope ka matlab hai Cycloid ka cusp vertical tangent ke saath rise karta hai.

Verify: numerically $t = 0.01$ par: $\frac{sin 0.01}{1 - cos 0.01} \approx \frac{0.0099998}{0.00005} \approx 200$ , $2/ t = 2/0.01 = 200$ se match karta hai. ✓ Toh slope genuinely $+ \infty$ ki taraf jaata hai — cusp vertical hai, exactly jaisa Figure 3 ke steepening tangent lines dikhate hain.

Active recall

Kaun sa cell un points describe karta hai jahan

\overset{x}{˙} = 0

lekin

\overset{y}{˙} \neq = 0

Cell B — vertical tangent, slope undefined.

Kaun se cell ko plug-in ki jagah limit chahiye?

Cell E (cusp,

\overset{x}{˙} = \overset{y}{˙} = 0

0/0

deta hai) — aur cell H limiting slope ke liye.

x = t^{2}, y = t^{3}

ke liye,

d y / d x

aur

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

kya hai?

\frac{3 t}{2}

aur

\frac{3}{4 t}

Kis

t

par

x = cos t, y = sin t

mein vertical tangents hain?

t = nπ

(integer

n

), matlab points

(\pm 1, 0)

Kis

t

par

x = cos t, y = sin t

mein horizontal tangents hain?

t = \frac{π}{2} + nπ

(integer

n

), matlab points

(0, \pm 1)

Circle ke upper half par

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

ka sign?

Negative → concave down.

Projectile

x = 3 t, y = 4 t - 5 t^{2}

: path flat kab hai?

Jab

\overset{y}{˙} = 4 - 10 t = 0

, matlab

t = 0.4

s (peak par).

Connections

Parent: the two formulas — ye page unhe exercise karta hai.
Tangents and normals — cells B aur C poori tarah tangent lines ke baare mein hain.
Concavity and second derivative — cell D $d^{2} y / d x^{2}$ ka sign padhta hai.
Cycloid, Parametric curves — cells E aur H yahan rehte hain.
Chain rule, Quotient rule — har step ke andar ki machinery.

Scenario map

Neeche ka flowchart ek decision tree hai jise tum kisi bhi parametric point ke liye walk kar sakte ho. Apne $x (t), y (t)$ ke saath top se shuru karo. Pehle pucho kya $\overset{x}{˙}$ zero hai? Agar haan, tumhare paas vertical tangent hai (Cell B). Agar nahi, pucho kya $\overset{y}{˙}$ zero hai? — haan ka matlab slope $0$ ki horizontal tangent hai (Cell C). Agar dono zero nahi hain, pucho kya dono ek saath zero hain? — haan ka matlab limit lene wala cusp hai (Cell E, aur uska limiting-slope cousin Cell H); nahi ka matlab ordinary point hai jahan tum simply $\overset{y}{˙} / \overset{x}{˙}$ compute karte ho (Cell A). Us ordinary slope se phir tum concavity check karne (Cell D), real-world velocities interpret karne (Cell F), ya required slope ke liye backwards solve karne (Cell G) ki taraf branch kar sakte ho. Ek line mein: pehle zeros, phir ordinary slope, phir baaki sab.