Gradient perpendicular to level curves - surfaces: Question bank

4.4.11 · D5 · Maths › Multivariable Calculus › Gradient perpendicular to level curves - surfaces

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$f (x, y) = c$ ek level curve hai: input points ka woh set jahan $f$ same value $c$ leta hai. Iske saath chalo aur $f$ kabhi nahi badalta.
$\nabla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$ gradient hai: input plane mein ek vector jo direction batata hai jis taraf $f$ sabse tez badhta hai.
$\nabla f \cdot u$ (dot product) $= 0$ ka matlab hai dono vectors perpendicular hain. Yahi poora engine hai, Directional Derivative aur Multivariable Chain Rule ke zariye.

True or false — justify karo

Level curve par

f

ki value constant hoti hai, isliye tangent direction mein Directional Derivative zero hota hai.

True — woh zero directional derivative

\nabla f \cdot t = 0

exactly yahi kehta hai ki

\nabla f

, tangent ke perpendicular hai, yaani level curve ke perpendicular hai.

\nabla f

hamesha us direction mein point karta hai jis taraf level curve jaati hai.

False — yeh level curve ke across (perpendicular) point karta hai, increasing

f

ki taraf; curve ke saath

f

nahi badalta, isliye "most-change" arrow uske saath nahi chal sakta.

f (x, y) = c

ki tangent line ki direction

\nabla f

hoti hai.

False —

\nabla f

normal hai; tangent direction koi bhi nonzero vector hota hai jo

\nabla f

ke perpendicular ho, jaise

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

Agar level set par kisi point par

\nabla f = 0

ho, toh perpendicularity ka claim ab bhi tangent direction batata hai.

False — zero vector sabse perpendicular hota hai, isliye woh koi tangent direction select nahi karta; statement vacuous ho jaata hai aur woh point level set ka singular/critical point ho sakta hai.

\nabla f

, graph surface

z = f (x, y)

ke perpendicular hota hai.

False —

\nabla f

2D input plane mein rehta hai aur wahan level curves ka normal hai; graph ka normal

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

hota hai, jo

g = f (x, y) - z

se milta hai.

f

2 f

se scale karne par gradient rotate hota hai, isliye level curve ke saath perpendicularity toot sakti hai.

False —

\nabla (2 f) = 2\nabla f

same direction mein point karta hai (bas thoda lamba), aur level curves as sets unchanged rehti hain, isliye yeh perpendicular rehta hai.

f

f^{3}

se replace karne par gradient us point par same line mein point karta hai jahan

f \neq = 0

True — chain rule se

\nabla (f^{3}) = 3 f^{2} \nabla f

milta hai, jo

\nabla f

ka ek positive-ya-zero scalar multiple hai, isliye jahan

f \neq = 0

wahan direction unchanged rehti hai (aur perpendicularity bani rehti hai).

Do alag functions ek point se same level curve share kar sakte hain lekin wahan unke gradient directions alag ho sakte hain.

False direction ke liye — dono gradients shared curve ke normal hain, isliye woh same line par hain; sirf unki lengths aur shayad signs (inward/outward) alag ho sakte hain.

Error dhundho

f (x, y) = x y = 6

(2, 3)

: kyunki

\nabla f = (3, 2)

hai, tangent slope

3/2

hai."

Error —

\nabla f

normal hai, tangent nahi. Tangent direction

(3, 2)

ke perpendicular hai, jaise

(- 2, 3)

, jisse slope

3/ (- 2) = - 1.5

milti hai, jo

- f_{x} / f_{y} = - 3/2

se match karta hai.

z = f (x, y)

ka tangent plane paane ke liye

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

use karo."

Error — woh formula level set

f = c

ka tangent deta hai. Graph

z = f

ke liye,

g = f - z

likhkar normal

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

use karo; dekho Tangent Plane to a Surface.

D_{u} f = \nabla f \cdot u = 0

ka matlab hai

\nabla f = 0

Error — dot product zero hai kyunki

u

\nabla f

ke perpendicular hai (level set ka tangent), na ki isliye ki

\nabla f

zero ho gaya.

"Kyunki gradient level curve ke perpendicular hai, toh yeh

x

-axis ke bhi perpendicular hona chahiye."

Error — perpendicularity level curve ke saath hai us point par, jiska direction generally

x

-axis nahi hota; sirf tabhi yeh follow hota agar curve wahan

x

-axis ke saath chal raha hota.

"Ek saddle-shaped contour crossing par (jahan level set khud se cut karta hai),

\nabla f

level set ke perpendicular hai."

Error — aisi jagah

\nabla f = 0

hota hai, isliye koi well-defined single tangent direction nahi hota; perpendicularity statement ke liye

\nabla f \neq = 0

chahiye.

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

ke liye

(3, 4)

par,

\nabla f = (6, 8)

origin ki taraf inward point karta hai."

Error —

(6, 8) = 2 (3, 4)

radially outward point karta hai, increasing

f

ki direction mein;

f

badhta hai jab tum origin se door jaate ho.

Poore level set ke saath perpendicularity ka matlab hai har tangent direction ke saath perpendicularity; ek akela curve sirf ek direction rule out karta hai, isliye hume yeh sab curves ke liye chahiye.

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

" perpendicularity kyun prove karta hai, sirf "no change" nahi?

Nonzero vectors ke beech zero dot product perpendicular ki definition hai; kyunki

r^{'} (t)

tangent hai,

\nabla f

ka usse orthogonal hona exactly "level set ka normal hona" hai.

Yahi argument level curves se level surfaces tak 3D mein kyun extend ho jaata hai?

Proof ne kabhi dimension use nahi ki:

f (x) = c

ke andar kisi bhi curve

r (t)

ke liye hume ab bhi

\nabla f \cdot r^{'} = 0

milta hai, aur itni curves tangent plane span kar leti hain, isliye

\nabla f

surface ka normal hai.

Why do Lagrange Multipliers set

\nabla f ∥ \nabla g

Har gradient apni level set ka normal hai; constrained optimum par dono level sets tangent hoti hain, isliye unke normals same line par point karte hain, yaani

\nabla f = λ \nabla g

Steepest Descent / Gradient Descent guaranteed kyun hai ki contours ke across move kare, unke saath nahi?

Yeh

- \nabla f

direction mein step leta hai, jo level set ke perpendicular hai; contour ke saath ek step ka

\nabla f \cdot step = 0

hoga aur

f

bilkul reduce nahi hoga, isliye descent usse avoid karta hai.

Edge cases

f

ke local maximum par, jahan

\nabla f = 0

hai, kya gradient level set ke perpendicular hai?

Statement vacuous hai — wahan "level set" ek single point ya degenerate ho sakti hai, aur zero vector ki koi direction nahi, isliye perpendicularity se koi information nahi milti.

Linear function

f (x, y) = a x + b y

ke liye, level curves parallel straight lines hain; kya

\nabla f

ab bhi unke normal hai?

Haan —

\nabla f = (a, b)

har jagah constant hai aur parallel lines

a x + b y = c

mein se har ek ke perpendicular hai; straight level "curve" sabse simple case hai, koi exception nahi.

Agar kisi level curve mein sharp corner ho (jaise implicit equation

∣ x ∣ + ∣ y ∣ = 1

se), toh kya perpendicularity corner par hold karti hai?

Nahi — corner par tangent direction undefined hota hai aur

f

wahan differentiable nahi hai, isliye

\nabla f

(aur hence claim) us point par exist nahi karta, sirf smooth pieces par karta hai.

Socho

f (x, y) = x^{2}

, jiska level set

x^{2} = 1

do vertical lines

x = \pm 1

ki pair hai; kya

\nabla f

unke perpendicular hai?

Haan —

\nabla f = (2 x, 0) = (\pm 2, 0)

horizontal hai, vertical lines ke perpendicular; level set ka disconnected hona local perpendicularity nahi todta.

f (x, y) = y - x^{2}

ke level set

f = 0

par (ek parabola), kya perpendicularity value

c = 0

par depend karti hai?

Nahi — wahi chain-rule argument har

c

ke liye kaam karta hai; har point par

\nabla f = (- 2 x, 1)

, parabola

y = x^{2} + c

ka normal hai chahe level kaun sa bhi chuna ho.

Kya hoga perpendicularity claim ke saath ek aisi level set par jahan

f

poore 2D region par constant ho (ek "flat" plateau)?

Wahan

\nabla f = 0

poore jagah hai, isliye koi distinguished normal direction nahi; level "set" ek area hai, curve nahi, aur statement ab applicable nahi rehta.

Recall Yahan har trap ki one-line summary

$\nabla f$ input plane mein ek normal vector hai, defined aur level set ke perpendicular sirf wahan jahan $f$ differentiable ho aur $\nabla f \neq = 0$ ho; yeh kabhi tangent slope nahi, kabhi graph ka normal nahi, aur iske length ko rescale karna right angle nahi todta.

Connections

Directional Derivative — level set ke saath zero hona perpendicularity ka seed hai.
Multivariable Chain Rule — $\nabla f \cdot r^{'} = 0$ ke peeche ka engine.
Tangent Plane to a Surface — $\nabla f$ ko normal ki tarah sahi use karna.
Lagrange Multipliers — gradients align karte hain kyunki level sets touch karti hain.
Steepest Descent / Gradient Descent — contours ke perpendicular steps.

Question bank — Gradient perpendicular to level curves - surfaces

True or false — justify karo

Error dhundho

Why questions

Edge cases

Connections