Partial derivatives — notation, calculation, geometric meaning: Visual walkthrough

4.4.3 · D2 · Maths › Multivariable Calculus › Partial derivatives — notation, calculation, geometric meani

Poore derivation mein hum yeh running surface use karenge: $z = f (x, y) = x^{2} + y^{2}$ (ek round bowl) point $(a, b) = (1, 2)$ par. Yeh itna concrete hai ki draw ho sake aur itna simple ki har number haath se check ho sake.

Step 1 — "Floor ke upar height" ka matlab kya hai

KYA HAI. Do variables ka function $f (x, y)$ ek flat floor — $x y$ -plane — ke ek point ko leta hai aur ek single number return karta hai: ek height. Un saari heights ko stack karo aur ek landscape milta hai jo floor ke upar float karta hai: yeh hai surface $z = f (x, y)$ .

KYUN. Slopes ki baat karne se pehle picture fix karni hogi: do horizontal inputs $x, y$ (jahan tum khade ho) aur ek vertical output $z$ (zameen kitni oonchi hai). Baaki sab kuch is ek image ke upar bookkeeping hai.

PICTURE. Floor grid hai; point $(1, 2)$ se uthta arrow height $z = f (1, 2) = 1^{2} + 2^{2} = 5$ dikhata hai.

Figure — Partial derivatives — notation, calculation, geometric meaning

Equation $z = x^{2} + y^{2}$ ko term by term padho: $x^{2}$ kehta hai "centre se East–West jitna door, utna upar"; $y^{2}$ wohi baat North–South ke liye kehta hai; unka sum total height hai. $(1, 2)$ par woh height hai $1 + 4 = 5$ .

Step 2 — $y$ ko freeze karo: surface ko ek wall se kato

KYA HAI. $y$ ko $b = 2$ par hold karo. $y = 2$ wale har point ek vertical wall banate hain — plane $y = 2$ — jo seedha bowl ke through slice karta hai. Jahan wall surface se milti hai, ek single curve milta hai.

KYUN. Ek surface ke kisi point par slope ki infinitely many directions hoti hain, isliye "the slope" ka koi matlab nahi. Ek ordinary one-slope situation recover karne ke liye hum ek degree of freedom chhod dete hain: $y$ freeze karo. Ab sirf $x$ move kar sakta hai, aur surface ek plain curve ban jaata hai jise hum pehle se differentiate karna jaante hain — yeh poora trick hai single-variable thinking ko reuse karna.

PICTURE. Cyan wall $y = 2$ hai; amber curve woh hai jahan yeh bowl ko carve karta hai.

Note karo: $y$ freeze karne se $y$ delete nahi hua; $y^{2}$ number $4$ ban gaya. Woh constant $4$ curve ko raise karta hai par kisi bhi slope mein $0$ add karta hai.

Step 3 — $x$ -slope ek ordinary derivative hai

KYA HAI. Slice-curve $g (x) = x^{2} + 4$ par hum ek plain single-variable derivative lete hain aur $x = a = 1$ par evaluate karte hain. Yeh number define hota hai partial derivative $f_{x} (1, 2)$ ke roop mein.

KYUN. Humne Step 2 mein surface ko curve mein collapse kiya tha precisely isliye ki is step mein koi nayi machinery na chahiye — derivative ki limit definition jo tumhare paas pehle se hai woh saara kaam karti hai. Term by term limit yeh hai: $f_{x} (a, b) = lim_{h \to 0} \frac{f ( a + h , b ) - f ( a , b )}{h}$

$f (a + h, b)$ ::: ek chhota East step $h$ door ki height, wahi $y = b$ .
$f (a, b)$ ::: jahan tum khade ho wahan ki height.
divide by $h$ ::: rise over run — ek slope.
$lim_{h \to 0}$ ::: step ko ek point tak shrink karo taaki "run" ban jaye "us exact spot par tangent".

PICTURE. Amber tangent line slice-curve ko $x = 1$ par kiss karti hai; iska rise-over-run red slope triangle hai.

$g^{'} (x) = 2 x \Rightarrow f_{x} (1, 2) = g^{'} (1) = 2 \cdot 1 = 2 .$

Frozen $4$ ne $0$ contribute kiya, bilkul jaisa promise kiya tha. Bowl ka $x$ -slope $(1, 2)$ par $2$ hai.

Step 4 — Iske bajaye $x$ freeze karo: doosri wall

KYA HAI. Ab mirror move karo — $x$ ko $a = 1$ par hold karo, $y$ ko sweep karo. Plane $x = 1$ ek wall hai jo doosri taraf jaati hai, aur yeh ek alag curve carve karta hai.

KYUN. Bowl ki steepness East–West aur North–South mein zaruri nahi ki agree kare, isliye ek doosra, independent measurement chahiye. Setup ki symmetry ka matlab hai ki hum Steps 2–3 ko verbatim repeat karte hain jahan $x$ aur $y$ ke roles swap hote hain.

PICTURE. Doosri cyan wall $x = 1$ aur iska amber slice-curve, contrast ke liye pehle ke saath side by side.

$y$ -slope $4$ hai — $x$ -slope $2$ se steeper, kyunki $(1, 2)$ par hum $y$ direction mein center se zyada door hain, aur bowl centre se jitna door, utna teezi se chadhta hai. Yeh parent note ke Example 4 se match karta hai.

Step 5 — Do tangent lines, ek flat plane

KYA HAI. Har partial ne ek tangent line di jo apni wall ke andar rehti thi. Dono lines ek hi surface point $P = (1, 2, 5)$ se guzarti hain. Ek point par cross karti do lines exactly ek flat plane pin down karti hain — tangent plane.

KYUN. Ek plane woh flattest object hai jo ek curved surface ko touch kar sake aur uski direction share kare. Tangent hone ke liye use $x$ -direction aur $y$ -direction dono mein surface ka slope reproduce karna hoga. Woh do slopes precisely $f_{x}$ aur $f_{y}$ hain — toh plane forced hai; koi freedom nahi bachi.

PICTURE. $P$ se guzarti amber $x$ -tangent aur cyan $y$ -tangent, aur woh translucent plane jo unhe span karta hai.

Plane ka equation piece by piece banao: $z = height at P f (a, b) + x -slope f_{x} (a, b) (x - a) + y -slope f_{y} (a, b) (y - b)$

$f (a, b)$ ::: sahi height $5$ se shuru karo, taaki plane $P$ ko touch kare.
$f_{x} (a, b) (x - a)$ ::: har East step $(x - a)$ ke liye, East-slope $2$ ke hisaab se chadho.
$f_{y} (a, b) (y - b)$ ::: har North step $(y - b)$ ke liye, North-slope $4$ ke hisaab se chadho.

Apne numbers plug karo: $z = 5 + 2 (x - 1) + 4 (y - 2) .$

Check karo ki do slices agree karte hain. $y = 2$ set karo: $z = 5 + 2 (x - 1)$ , slope $2$ ✔ (Step 3 se match). $x = 1$ set karo: $z = 5 + 4 (y - 2)$ , slope $4$ ✔ (Step 4 se match). Plane dono partials reproduce karta hai — yahi "tangent" ka matlab hai.

Step 6 — Degenerate & edge cases (reader ko kabhi fall through mat karne do)

Real surfaces hamesha tame nahi hoti. Yeh hai ki har odd situation kaisi dikhti hai.

Case A — flat spot ( $f_{x} = f_{y} = 0$ ). Bowl ke bottom par, $(0, 0)$ : $f_{x} = 2 \cdot 0 = 0$ , $f_{y} = 2 \cdot 0 = 0$ . Dono slice-tangents horizontal hain, toh tangent plane flat lid $z = 0$ hai. Yeh ek minimum hai — surface ek level plane se upar curve karti hai.

Case B — ek slope zero, doosra nahi. $f = y^{2}$ jaisi valley ke trough par (koi $x$ nahi): $f_{x} = 0$ har jagah par lekin $f_{y} = 2 y$ . Plane sirf $y$ mein tilt karta hai; East–West level rehta hai. Ek valid tangent plane phir bhi exist karta hai.

Case C — corner/kink (koi bhi partial nahi). $f = x^{2} + y^{2}$ (ek ice-cream cone) ke liye tip $(0, 0)$ par, East step aur West step slopes $+ 1$ aur $- 1$ dete hain: limit left vs right se disagree karta hai, toh $f_{x} (0, 0)$ exist nahi karta. Koi tangent line nahi ⇒ koi tangent plane nahi. Partials ke liye zaruri hai ki slice-curve us point par smooth ho.

PICTURE. Bowl bottom (flat lid), valley (one-way tilt), cone tip (kink) — teen miniatures.

Ek picture mein summary

Sab kuch ek saath: bowl, do orthogonal cutting walls, do amber/cyan tangent lines jo $P = (1, 2, 5)$ par milte hain, aur flat tangent plane jo unhe span karta hai — slopes $2$ aur $4$ ke saath annotated.

Recall Feynman retelling — poora walk plain words mein

Ek hilly field par ek jagah khade ho; wahan tumhari height $5$ hai. Tum "the slope" nahi keh sakte kyunki tum kisi bhi direction mein face kar sakte ho. Toh tum honestly cheat karte ho: seedha East face karo aur ek tiny step lo — zameen $2$ steepness se aayi; yeh $f_{x}$ hai. Wapas aao, North face karo, tiny step — steepness $4$ ; yeh $f_{y}$ hai. Woh do chhote slope measurements do seedhi sticks hain jo hill par rakhi hain, ek East point karte hue, ek North, dono tumhare paon ko touch karte hue. Un do sticks par tikaa ek flat board tangent plane hai. Iska recipe hai "height $5$ se shuru karo, har East step ke liye $2$ add karo, har North step ke liye $4$ add karo." Agar zameen ke neeche tumhare paon par ek sharp point hota (ek cone tip), toh East step upar aur West step neeche agree nahi karte — koi single slope nahi, koi stick nahi, koi board nahi. Warna do partials hi plane hain.

Connections

Tangent plane and linear approximation — yeh page iska geometric derivation hai.
Gradient vector — do slopes $f_{x}, f_{y}$ ko ek arrow $\nabla f$ mein bundle karta hai.
Directional derivative — slope jab tum diagonal face karo, sirf E ya N nahi.
Total differential — plane $d z = f_{x} d x + f_{y} d y$ ke roop mein likha.
Chain rule (multivariable) — yeh slices kaise compose hote hain.
Single-variable derivative — har slice-curve ise wholesale reuse karta hai.
Clairaut's theorem — second-order slopes ki symmetry.