Gradient vector ∇f — definition, properties: Visual walkthrough

4.4.9 · D2 · Maths › Multivariable Calculus › Gradient vector ∇f — definition, properties

Step 1 — Do variables ka function ek landscape hai

KYA. Ek rule $f$ lo jo do numbers $(x, y)$ khaata hai aur ek number return karta hai, uski "height". Height ko $z = f (x, y)$ kaho.

KYUN. Jab tak hum "slope" ya "steepest" ki baat nahi kar sakte, tab tak humein kuch chahiye jiska slope ho. Ek input ek curve deta hai; do inputs ek surface dete hain — ek pahaad. Pehle pahaad ko picture karna zaroori hai, kyunki baad ki har idea hai "jab main chalta hoon toh height kaise badlti hai?"

PICTURE. Neeche, input $(x, y)$ flat grey floor par rehta hai. Har floor point ke seedha upar, height $z$ ek smooth hill mein upar uthti hai. Yellow dot dekho: woh floor point $p = (p_{1}, p_{2})$ par baitha hai, aur vertical yellow stick uski height $f (p)$ measure karti hai.

Figure — Gradient vector ∇f — definition, properties

Step 2 — Do special walks: $x$ ke saath, $y$ ke saath

KYA. $y = p_{2}$ freeze karo aur sirf east chalo (sirf $x$ badlo). Height ek 1-D curve trace karti hai — pahaad ka ek slice. Uska slope partial derivative $f_{x}$ hai. Yahi north chalne par karo ( $x$ freeze karo): uska slope $f_{y}$ hai.

KYUN. Surface mein chalne ke liye infinitely many directions hain. Ek saath samajhna bahut mushkil hai. Toh hum do simplest walks se start karte hain — seedha har axis ke saath — kyunki wahan problem ordinary one-variable slope mein collapse ho jaati hai, jo hum pehle se samajhte hain.

PICTURE. Do coloured slices yellow point se guzarti hain. Blue slice east–west chalti hai (sirf $x$ badlta hai); $p$ par uski steepness $f_{x}$ hai. Green slice north–south chalti hai (sirf $y$ badlta hai); uski steepness $f_{y}$ hai. Har ek bas ek hill-cross-section hai — ek ordinary curve with an ordinary slope.

Step 3 — Lekin hum KISI BHI direction mein chalna chahte hain

KYA. Floor par koi bhi direction chuno aur use $\overset{u}{^} = (u_{1}, u_{2})$ kaho. Chhota hat matlab yeh ek unit vector hai: iska length exactly $1$ hai, yaani $u_{1}^{2} + u_{2}^{2} = 1$ . Hum poochte hain: agar main is taraf chalun toh height kitni tezi se badlegi?

Unit vector KYUN? "Direction" aur "kitna door" do alag sawaal hain. Hum chahte hain pure climb rate per metre walked. Agar humara arrow 2 metres lamba hota, toh hum secretly do metres ki walking measure kar rahe hote aur double-count kar rahe hote. Length $1$ fix karne se "kitna door" hat jaata hai aur sirf "kis taraf" bachta hai.

PICTURE. Yellow floor point se, length $1$ ka ek red arrow $\overset{u}{^}$ kisi chosen direction mein point karta hai. Blue ( $f_{x}$ ) aur green ( $f_{y}$ ) axis-arrows bhi length $1$ ke hain. Dhyaan do ki $\overset{u}{^}$ unke beech baitha hai — har direction "kuch east, kuch north" ka blend hai.

Step 4 — 2-D walk ko 1-D walk mein badlo

KYA. Ek single-variable function define karo $g (h) = f (p + h \overset{u}{^})$ — height sirf is baat ke function ke roop mein ki hum red arrow ke saath kitna, $h$ , chale hain. Tab $D_{\overset{u}{^}} f = g^{'} (0)$ : $g$ ka slope us waqt jab hum chalna shuru karte hain.

KYUN. Hum one-variable calculus achi tarah jaante hain. Trick yeh hai ki 2-D sawaal ko 1-D mein reduce karo seedhi red path ko ek number $h$ se parametrize karke. Ab " $\overset{u}{^}$ ke saath climb rate" literally " $g^{'} (0)$ " hai, ek ordinary derivative.

PICTURE. Floor par red path pahaad par lift hoti hai (surface par red curve). Apna view us path ke saath squash karo aur woh $g$ versus $h$ ka ek plain 2-D graph ban jaata hai: ek ordinary curve jiska slope $h = 0$ par chahiye.

Step 5 — Chain rule use karke crack karo

KYA. Jaise $h$ badhta hai, $x$ -coordinate $p_{1} + h u_{1}$ hai aur $y$ -coordinate $p_{2} + h u_{2}$ hai. Chain rule (multivariable) kehta hai total rate of height change hai: (rate along $x$ ) × (how fast $x$ moves) + (rate along $y$ ) × (how fast $y$ moves).

$g^{'} (0) = height’s sensitivity to x f_{x} \cdot = u_{1} (how fast x moves) \frac{d ( p _{1} + h u _{1} )}{d h} + height’s sensitivity to y f_{y} \cdot = u_{2} \frac{d ( p _{2} + h u _{2} )}{d h}$

Kyunki $\frac{d ( p _{1} + h u _{1} )}{d h} = u_{1}$ aur $\frac{d ( p _{2} + h u _{2} )}{d h} = u_{2}$ , yeh collapse ho jaata hai

$D_{\overset{u}{^}} f = f_{x} u_{1} + f_{y} u_{2}$

Chain rule KYUN aur kuch nahi? $\overset{u}{^}$ ke saath chalna $x$ aur $y$ dono ek saath move karta hai. Chain rule exactly woh tool hai jo kehta hai "jab kai inputs saath chalte hain, har input ka contribution jodo (uska apna slope times uski apni speed)." Simultaneous motion ka hisaab koi doosra rule nahi rakhta.

PICTURE. $h$ length ka red step floor par do shadows dalta hai: $h u_{1}$ length ka ek eastward shadow aur $h u_{2}$ length ka ek northward shadow. Height $f_{x}$ per eastward metre climb karti hai aur $f_{y}$ per northward metre. Dono climbs jodo — yahi equation hai.

Step 6 — Dot product pehchano; gradient paida hota hai

KYA. Expression $f_{x} u_{1} + f_{y} u_{2}$ hai "matching entries multiply karo, phir add karo". Yah pattern do vectors ka dot product hai: $D_{\overset{u}{^}} f = (f_{x}, f_{y}) \cdot (u_{1}, u_{2})$ Pehla vector, $(f_{x}, f_{y})$ , sirf pahaad par depend karta hai (is par nahi ki humne kis taraf chalna choose kiya). Isko ek naam dena chahiye: gradient $\nabla f$ .

Naam KYUN? Kyunki yeh cleanly factor hota hai: hill-info ( $\nabla f$ ) times my-choice ( $\overset{u}{^}$ ). Ek baar "pahaad kya offer karta hai" aur "main kya choose karta hoon" alag kar lo, tum pahaad ka sawaal pooch sakte ho — kaun sa choice best hai? — bina kuch recompute kiye.

PICTURE. $p$ se floor par do arrows: fixed gradient $\nabla f = (f_{x}, f_{y})$ (yellow) aur mera chosen $\overset{u}{^}$ (red). Dot product yellow arrow ka shadow hai jo red direction par pada hai, times $\overset{u}{^}$ ki length ( $= 1$ ).

Step 7 — Kaun sa direction sabse tezi se climb karta hai? (the payoff)

KYA. Ek dot product $∣\nabla f ∣ ∣ \overset{u}{^} ∣ cos θ$ bhi hota hai, jahan $∣\nabla f ∣$ gradient ki length hai, $∣ \overset{u}{^} ∣ = 1$ , aur $θ$ , $\nabla f$ aur $\overset{u}{^}$ ke beech ka angle hai. Toh $D_{\overset{u}{^}} f = ∣\nabla f ∣ cos θ .$ Sirf ek cheez hum control kar sakte hain woh hai $θ$ . Aur $cos θ$ tab sabse bada hota hai jab $θ = 0$ .

$cos$ KYUN? Kyunki "gradient ka kitna hissa meri walk ki taraf point karta hai" ek projection hai, aur projection hi unke beech ke angle ka $cos θ$ hota hai. Yeh exactly jawab deta hai: full uphill push ka kitna fraction meri chosen direction mein jaata hai?

PICTURE. Jaise red $\overset{u}{^}$ fixed yellow $\nabla f$ ke around swing karta hai, climb-rate $∣\nabla f ∣ cos θ$ dekho: maximal ( $+ ∣\nabla f ∣$ ) jab dono align hoon, zero jab perpendicular, most negative ( $- ∣\nabla f ∣$ ) jab opposite.

Step 8 — Degenerate case: $\nabla f = 0$

KYA. Agar $f_{x} = 0$ aur $f_{y} = 0$ ? Tab $\nabla f$ zero vector hai: length $0$ , koi direction nahi. Formula deta hai $D_{\overset{u}{^}} f = 0 \cdot \overset{u}{^} = 0$ har $\overset{u}{^}$ ke liye.

KYUN dikhayein? Contract: koi bhi scenario unpainted na chhodo. Ek flat spot par — hilltop, valley floor, ya saddle — koi uphill hota hi nahi. "Steepest ascent" ka koi jawab nahi kyunki $cos θ$ ko $∣\nabla f ∣ = 0$ se multiply kiya jaata hai. Reader ko pata hona chahiye ki formula abhi bhi kaam karta hai; bas "flat, har taraf" return karta hai.

PICTURE. Teen flat-gradient landscapes: ek rounded peak, ek bowl valley, aur ek saddle. Har ek par, gradient arrow ek dot mein shrink ho gaya hai — har walking direction ka zero initial slope hai. (Ye exactly woh points hain jo Lagrange multipliers aur optimisation dhundta hai.)

Ek picture mein summary

Sab kuch ek frame mein: pahaad, floor par pada yellow gradient arrow seedha uphill point karta hua, ek red chosen direction, uska shadow (dot product) climb-rate deta hua, aur level curve jise gradient right angle par kaatता hai.

Recall Feynman: poori walk plain words mein

Tum ek foggy pahaad par khade ho. Pehle humne notice kiya ki tum directly sirf do slopes feel kar sakte ho: agar tum due east step lo toh kitna steep hai ( $f_{x}$ ), aur due north kitna steep hai ( $f_{y}$ ). Lekin shayad tum diagonally chalna chahte ho. Toh humne poocha: agar tum kisi direction mein tiny step lo, toh kitna climb karoge? Humne us diagonal walk ko ek straight-line stroll mein badal diya jo ek number $h$ se measure hoti hai, aur chain rule ne bataya ki climb bas "east-slope times kitna east gaye, plus north-slope times kitna north gaye" hai. Woh sum secretly ek dot product hai — do lists align karo, multiply karo, add karo. Un lists mein se ek sirf pahaad par depend karti hai; humne use bottle kiya aur gradient kaha, ek arrow jo ground par pada hai. Kisi bhi direction mein climb gradient ka shadow hai us direction par. Shadows tab sabse lambe hote hain jab tum arrow ke seedhe saamne ho — toh sabse tezi se upar jaane ka raasta simply gradient ke saath hai, aur uski length steepness batati hai. Arrow ke across chalo aur uska shadow gayab ho jaata hai: tum level rehte ho. Aur agar arrow kabhi ek dot mein shrink ho jaaye, tum flat ground par ho — peak, valley, ya saddle — jahan koi direction bilkul bhi climb nahi karti.

Recall

$D_{\overset{u}{^}} f$ mein $\overset{u}{^}$ ki length 1 kyun honi chahiye? ::: Climb ko per unit distance measure karne ke liye; lamba vector jawab ko apni length se scale kar deta. Kaun sa ek object saari directional slopes pack karta hai? ::: Gradient $\nabla f = (f_{x}, f_{y})$ . Kaun sa rule diagonal walk ko $f_{x} u_{1} + f_{y} u_{2}$ mein convert karta hai? ::: Multivariable chain rule. $\nabla f$ steepest-ascent direction kyun hai? ::: $D_{\overset{u}{^}} f = ∣\nabla f ∣ cos θ$ , $θ = 0$ par max hota hai, yaani $\overset{u}{^}$ , $\nabla f$ ke saath aligned ho. $\nabla f = 0$ kya signal karta hai? ::: Ek first-order-flat point: peak, valley, ya saddle — koi steepest direction exist nahi karti.

Connections

Partial derivatives — $\nabla f$ ki do entries (Step 2).
Directional derivative — woh quantity jo humne derive ki (Step 3).
Chain rule (multivariable) — Step 5 ka engine.
Tangent plane and linearisation — linear picture jo $\nabla f$ build karta hai.
Level curves and surfaces — woh curve jise $\nabla f$ right angle par milta hai (Step 7).
Gradient descent — $- \nabla f$ ke saath neeche ride karta hai.
Lagrange multipliers — jahan $\nabla f = 0$ ya $\nabla f = λ \nabla g$ matter karta hai (Step 8).