Gradient perpendicular to level curves - surfaces

4.4.11 · Maths › Multivariable Calculus

KYA claim kiya ja raha hai

KAISE derive karte hain (first principles se)

Sabse clean derivation ek level set ke andar rehne wali curve use karti hai.

Step 1 — Level set par ek curve ko parametrize karo. Maano $r (t)$ koi bhi smooth curve hai jo level set par rehti hai, toh $f (r (t)) = c for all t .$ Ye step kyun? Aisi curve ka tangent vector $r^{'} (t)$ , by definition, ek aisi direction hai jo level set ke saath-saath chalti hai. Agar $\nabla f ⊥ r^{'} (t)$ ho har aisi curve ke liye, tab $\nabla f$ poore level set ke perpendicular hai.

Step 2 — Dono sides ko $t$ ke respect mein differentiate karo. Left side ek composition hai, isliye multivariable chain rule use karo: $\frac{d}{d t} f (r (t)) = \nabla f (r (t)) \cdot r^{'} (t) .$ Right side: $\frac{d}{d t} (c) = 0$ . Ye step kyun? Chain rule exactly woh tool hai jo connect karta hai ki $f$ kaise change hota hai position ke change se. Ise set up karne se $f$ ki constancy kaam karti hai.

Step 3 — Conclusion padho. $\nabla f (r (t)) \cdot r^{'} (t) = 0.$ Zero ke barabar dot product ka matlab hai dono vectors perpendicular hain. Kyunki $r^{'} (t)$ level set ka tangent hai, $\nabla f$ level set ka normal hai. $■$

Figure — Gradient perpendicular to level curves - surfaces

Consequence: tangent line / plane ki equation

Kyunki $\nabla f$ normal vector hai, tum turant tangent objects nikal sakte ho:

Worked examples

Common mistakes

Flashcards

Why is

\nabla f

perpendicular to a level curve?

f (r (t)) = c

ko chain rule se differentiate karne par

\nabla f \cdot r^{'} (t) = 0

milta hai; tangent

\nabla f

ke saath zero dot product deta hai.

What does

\nabla f \cdot u = 0

tell you geometrically?

u

level set ke saath-saath direction hai (

f

mein koi change nahi), yaani uska tangent hai.

Direction of

\nabla f

relative to a level set?

Normal (perpendicular), increasing

f

ki taraf point karta hai (steepest ascent).

Tangent plane to

f (x, y, z) = c

x_{0}

\nabla f (x_{0}) \cdot (x - x_{0}) = 0

For

f = x^{2} + y^{2}

(3, 4)

, what is

\nabla f

and is it perpendicular to the circle?

(6, 8)

; haan, ye radial hai, tangent

(- 4, 3)

ke perpendicular.

Normal to the GRAPH

z = f (x, y)

(not the level curve)?

(f_{x}, f_{y}, - 1)

g = f (x, y) - z

se.

Implicit slope of

f (x, y) = c

in terms of partials?

d y / d x = - f_{x} / f_{y}

Recall Feynman: ek 12-saal ke bacche ko samjhao

Socho ek pahadi ka map jisme circles hain jo equal heights dikhati hain (jaise ek contour map). Agar tum ek circle ke saath-saath chalo, tum kabhi upar ya neeche nahi jaate — tum same height par rehte ho. Pahadi par seedha upar jaane ki direction hamesha un circles ko seedha cross karti hai, jaise ek "T". Ye kisi circle ke saath-saath nahi chal sakti, kyunki circle ke saath-saath height bilkul nahi badalti! Woh "seedha upar" wala arrow gradient hai, aur woh hamesha contour lines ke right angle par hota hai.

Connections

Directional Derivative — $\nabla f \cdot u$ ke roop mein define hota hai; level set ke saath zero hota hai.
Multivariable Chain Rule — derivation ka engine.
Tangent Plane to a Surface — gradient ko normal vector ki tarah use karta hai.
Lagrange Multipliers — $f$ aur constraint $g$ ke gradients parallel ho jaate hain (dono apne level sets ke normal hote hain).
Steepest Descent / Gradient Descent — $\nabla f$ ke against move karna, yaani level sets ke perpendicular.