Gradient perpendicular to level curves - surfaces: Visual walkthrough

4.4.11 · D2 · Maths › Multivariable Calculus › Gradient perpendicular to level curves - surfaces

Hum 2D mein kaam karte hain (ek level curve) kyunki yeh draw kiya ja sakta hai; 3D case (ek level surface) bilkul word-for-word same hai.

Step 0 — Woh picture jis pe sab kuch bana hai

Kisi bhi symbol se pehle, yeh agree karte hain ki hum dekh kya rahe hain.

Ek pahaadi imagine karo. Horizontal position $(x, y)$ pe zameen se uski unchai ek single number hai. Us rule ko jo woh number deta hai, $f$ bulao.

Ab flat map pe, har woh point draw karo jo same unchai pe hota hai. Points ka woh set ek curve hai.

Figure — Gradient perpendicular to level curves - surfaces

Figure mein peach shading pahaadi ko upar se dekha hua hai (gehра = zyada uooncha). Magenta loop ek level curve $f (x, y) = c$ hai. Us magenta loop pe har point exactly unchai $c$ pe hai. Hamari poori quest: us marked point $P$ pe us arrow ki direction dhundho jo pahaadi pe sabse tezi se chadhta hai, aur dikhao ki woh magenta loop ke perpendicular hai.

Step 1 — "Steepest increase" ka matlab kya hai: the gradient

Hume woh arrow chahiye jo sabse zyada uphill point kare. Pehle, kya batata hai ki jab hum $x$ ya $y$ ko thoda hilate hain toh height kaise badlti hai?

Un dono rates ko ek single arrow mein stack karo:

Kyun ek vector, sirf ek slope nahi? Kyunki ek pahaadi pe tum infinitely many directions mein chal sakte ho, sirf east ya west nahi. Dono partials "east rate" aur "north rate" hain; inhe ek arrow ke roop mein combine karne se hum baad mein kisi bhi direction ka rate ek dot product se pooch sakte hain. Yeh ek jaani-maani baat hai (from the Directional Derivative) ki yeh particular arrow steepest increase ki direction mein point karta hai — isliye hum ise build karte hain.

Orange arrow $P$ pe $\nabla f$ hai: yeh wahan point karta hai jahan shading sabse tezi se dark hoti hai — seedha pahaadi ke upar. Hamara claim hai ki yeh orange arrow magenta curve ke $9 0^{\circ}$ pe hai. Hume ise prove karna hai.

Step 2 — Ek trick: ek aisi curve pe savar ho jo level set pe rehti hai

Poore loop ke baare mein ek saath sochne ki jagah, hum ek tiny explorer ko magenta curve ke saath walk karte hue bhejte hain aur time ke through unki position track karte hain.

Yeh equation symbol by symbol kya kehti hai: moving position $r (t)$ ko height rule $f$ mein daalo; kyunki dot contour pe rehta hai, woh height hamesha same constant $c$ hai, chahe $t$ kuch bhi ho.

Yeh kyun karo? Kyunki is explorer ki velocity, by construction, ek aisi direction hai jo curve ke saath flat rehti hai. Agar hum dikhа saken ki gradient us velocity ke perpendicular hai — har explorer ke liye, travel ki har direction ke liye — toh gradient curve ke hi perpendicular hai.

Violet dot $r (t)$ walk ke beech mein hai; woh chota violet arrow dikhata hai woh aage kahan ja raha hai — uski velocity, Step 3 mein aane wali.

Step 3 — Explorer ki velocity tangent hai

Yeh curve ka tangent kyun hai? Velocity hamesha wahan point karti hai jahan tum jaane wale ho. Kyunki explorer kabhi magenta curve nahi chhodhta, "jaane wala" direction curve ko hug karta hai — woh us point pe tangent line ki direction hai.

Violet arrow $r^{'} (t)$ curve ke saath flat lie karta hai — yeh ise graze karta hai, bina cross kiye touch karta hua. Ab do arrows pakdo: orange $\nabla f$ (Step 1) aur violet $r^{'} (t)$ (yahan). Poora proof inke beech ke angle ke baare mein hai.

Step 4 — "Stays constant" equation ko differentiate karo

Yahan ek clever move hai. Hamare paas ek equation hai jo sab $t$ ke liye true hai: $f (r (t)) = c .$ Agar do cheezein sab $t$ ke liye equal hain, toh $t$ mein unke rates of change bhi equal hain. Toh dono sides differentiate karo.

Pehle right side — easy half: $\frac{d}{d t} (c) = 0.$ Kyun? $c$ ek fixed number hai; ek number jo kabhi nahi hilta, uska rate of change zero hota hai. Yeh zero poori story ka hero hai.

Left side — chain rule half. Height $f$ position pe depend karti hai, aur position time pe. "Ek moving point ka function" differentiate karne ke liye, hume Multivariable Chain Rule chahiye: $\frac{d}{d t} f (r (t)) = height position pe kaise react karta hai \nabla f (r (t)) \cdot position time pe kaise react karta hai r^{'} (t) .$

Chain rule kyun, koi aur tool kyun nahi? Hamare paas ek composition hai: time → position → height. Chain rule precisely woh machine hai jo "outer part ka rate" ko "inner part ke rate" se multiply karta hai. Many variables mein woh "multiply" ek dot product ban jata hai — gradient (height ki sensitivity har coordinate ke liye) dotted with velocity (time ke liye har coordinate ki sensitivity).

Figure mein dono rate-flows milते hue dikhte hain: orange gradient aur violet velocity ek dot product mein feed hote hue.

Step 5 — Right angle padho

Step 4 ke dono halves ko saath rakh do. Left rate = right rate, toh: $gradient \cdot tangent \nabla f (r (t)) \cdot r^{'} (t) = from \frac{d}{d t} c 0 .$

Ab Step 4 se dot product ka geometric meaning invoke karo: do nonzero arrows ka dot product zero hota hai tabhi jab unke beech ka angle $9 0^{\circ}$ ho. Isliye:

$P$ pe woh chota red square orange gradient aur violet tangent ke beech perfect right angle mark karta hai. Woh square poora theorem hai, ek corner se bana hua.

Step 6 — Degenerate case: agar $\nabla f = 0$ ho toh?

Har proof tumhe fine print bhi deta hai. Hamara conclusion " $\nabla f ⊥ r^{'}$ " quietly assume karta tha ki dono arrows nonzero hain — kyunki $cos θ = 0$ tabhi $9 0^{\circ}$ angle bata sakta hai jab actually do arrows ho angle karne ke liye.

Agar $\nabla f = (0, 0)$ ho toh? Toh $\nabla f \cdot r^{'} = 0$ ab bhi true hai, lekin "perpendicular" meaningless hai — zero vector ki koi direction nahi hoti. Yeh ek critical point hai: ek hilltop, ek valley bottom, ya ek saddle. Wahan, level "curve" ek smooth curve bilkul bhi nahi ho sakti (socho ek peak pe single point, ya ek saddle pe cross karti do lines).

Left: $f = x^{2} - y^{2}$ ke liye origin pe ek saddle, level set $f = 0$ do crossing lines hain — koi single tangent nahi, aur wahan $\nabla f = (0, 0)$ . Right: ek peak jahan level set ek single point hai. Dono mein, "perpendicular" ke liye kuch attach karne ko nahi hai.

Ek-picture summary

Ek canvas pe sab kuch: magenta level curve $f = c$ , violet explorer uske saath slide karta hua with velocity $r^{'} (t)$ (the tangent), orange gradient $\nabla f$ uphill point karta hua, aur red right-angle square jahan woh milte hain. Chain-rule chain time → position → height force karta hai $\frac{d}{d t} f (r (t)) = \nabla f \cdot r^{'} = 0$ , aur ek zero dot product hai hi ek right angle. Jab yeh picture tumhare paas ho, Tangent Plane to a Surface ( $\nabla f$ ko normal ki tarah use karo) aur Lagrange Multipliers (do gradients dono apne level sets ke normal hain toh parallel hone chahiye) turant follow karte hain.

Recall Feynman retelling simple shabdon mein

Ek pahaadi ko upar se contour lines ke map ki tarah draw karo — equal height ke loops. Ek tiny walker ko ek loop ke around bhejo. Kyunki woh loop pe rehta hai, unki height kabhi nahi badlti: jaise woh chalte hain, height ka rate of change exactly zero hai. Ab, "height ka rate of change jaise tum move karo" ka matlab hai tum kitne uphill point kar rahe ho — woh hai gradient dotted with tumhari walking direction. Agar woh zero ke equal hai aur tum truly move kar rahe ho, toh sirf ek possibility hai ki uphill arrow tumhari path ke perfect right angle pe point kare. Toh steepest-uphill arrow (the gradient) hamesha contour lines ko "T" ki tarah cross karta hai. Ek exception woh hai jab pahaadi ka top ya saddle ho, jahan zameen is sense mein flat ho ki uphill arrow kuch nahi reh jaata — perpendicular hone ke liye koi arrow hi nahi bachta.

Connections

Directional Derivative — $\nabla f \cdot u$ ; yeh curve ke saath ka zero hi angle ko pin karta hai.
Multivariable Chain Rule — Step 4 ka engine.
Tangent Plane to a Surface — $\nabla f$ ko normal vector ki tarah read karta hai.
Lagrange Multipliers — dono gradients apne level sets ke normal hain ⇒ woh line up karte hain.
Steepest Descent / Gradient Descent — $\nabla f$ ke against step karta hai, yaani level sets ke across.