Gradient vector ∇f — definition, properties: Question bank

4.4.9 · D5 · Maths › Multivariable Calculus › Gradient vector ∇f — definition, properties

Reminders taaki yahan har symbol earned ho:

$\nabla f$ (bolo "grad f") partial derivatives ka vector hai, jaise ki 2D mein $(f_{x}, f_{y})$ .
$f_{x} = \partial f / \partial x$ $f$ ki slope hai jab tum sirf $x$ ko wiggle karo aur $y$ ko freeze karo.
$D_{\overset{u}{^}} f = \nabla f \cdot \overset{u}{^}$ change ki rate hai jab tum unit direction $\overset{u}{^}$ mein step karo (length exactly $1$ ).
$∣\nabla f ∣$ gradient arrow ki length hai; $θ$ $\overset{u}{^}$ aur $\nabla f$ ke beech ka angle hai.

True or false — justify

The gradient of a two-variable function is a vector, not a number.

True. Ek single number sirf ek slope encode kar sakta hai, lekin walk karne ke liye infinitely many directions hain; tumhe ek poora vector

(f_{x}, f_{y})

chahiye taaki

\nabla f \cdot \overset{u}{^}

ke zariye kisi bhi direction mein slope reconstruct kar sako.

\nabla f

us direction mein point karta hai jis direction mein level curve

f = c

chalti hai.

False. Yeh level curve ke perpendicular (normal) point karta hai. Curve ke saath chalte rehne par

f

constant rehta hai, toh

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

wahan — zero dot product ka matlab hai vectors right angles par hain.

f

ki maximum rate of increase ek point par

\nabla f

ke barabar hoti hai.

False (type error). Maximum rate number

∣\nabla f ∣

hai; ise achieve karne wali direction vector

\nabla f

hai. Rate ek scalar hai, direction ek vector hai — inhe equate mat karo.

Agar kisi point par

\nabla f = 0

ho, toh

f

wahan har direction mein flat hai.

True. Har directional derivative

\nabla f \cdot \overset{u}{^} = 0 \cdot \overset{u}{^} = 0

hai, toh pehle order par koi direction upar ya neeche nahi jaati — yeh exactly critical point (max, min, ya saddle) ki condition hai.

f

ki har value ko double karne se gradient bhi double ho jaata hai.

True, linearity se:

\nabla (2 f) = 2\nabla f

. Hill har jagah twice as steep ho jaati hai, toh har slope, aur isliye poora arrow,

2

se scale ho jaata hai.

Gradient hamesha graph surface

z = f (x, y)

ke tangent hota hai.

False.

\nabla f

input plane (yaani

x y

-domain) mein rehta hai, ground par flat leta hua uphill point karta hai — yeh bilkul bhi 3D surface par vector nahi hai. Domain arrow ko surface direction se confuse karna ek common slip hai.

Steepest descent ki direction mein,

D_{\overset{u}{^}} f = ∣\nabla f ∣

hota hai.

False — yeh

- ∣\nabla f ∣

hai. Steepest descent

\overset{u}{^} = - \nabla f /∣\nabla f ∣

hai, jisse

θ = π

milta hai aur

cos π = - 1

, toh rate negative hai: tum utni hi tezi se neeche jaate ho jितनी tezi se ascent upar jaati hai.

Agar do functions ka gradient har jagah same ho, toh woh same function hain.

False. Woh ek constant se differ kar sakte hain, jaise

f

aur

f + 7

ke saare partial derivatives same hain. Gradient slopes dekhta hai, aur constant add karne se height shift hoti hai bina kisi slope ko change kiye.

Spot the error

(1, 2)

(4, 6)

ki taraf

D_{\overset{u}{^}} f

: bas

\nabla f

(3, 4)

se dot karo."

Error:

(3, 4)

ki length

5

hai,

1

nahi. Non-unit vector se dot karne par true rate

∣ v ∣ = 5

se scale ho jaati hai. Fix: pehle

\overset{u}{^} = (3/5, 4/5)

par normalise karo.

"Steepest direction

- \nabla f

hai kyunki hum dot product ko minimise karte hain."

Error: steepest ascent

\nabla f \cdot \overset{u}{^}

ko maximise karta hai, jo

θ = 0

par achieve hota hai, yaani

+ \nabla f

ke saath.

- \nabla f

steepest descent hai. Sign flip ho gaya.

\nabla f \cdot \overset{u}{^} = ∣\nabla f ∣ ∣ \overset{u}{^} ∣ cos θ = ∣\nabla f ∣ cos θ

— toh

∣ \overset{u}{^} ∣

free mein disappear ho jaata hai."

Free mein nahi — yeh disappear hota hai kyunki assumption se

∣ \overset{u}{^} ∣ = 1

hai. Agar tum non-unit vector feed karo toh

∣ \overset{u}{^} ∣

factor wahan rehta hai aur tumhara answer multiply karta hai.

"Circle

x^{2} + y^{2} = 25

par

(3, 4)

par, gradient

(6, 8)

tangent hai, kyunki tangent

(- 4, 3)

hai aur

(6, 8)

'iske saath chalti hai'."

Error:

(6, 8) \cdot (- 4, 3) = - 24 + 24 = 0

, toh

(6, 8)

tangent ke perpendicular hai, yaani yeh normal (radial) direction hai — tangent nahi.

(1, 2)

par

f_{x} = 2 x y

2 \cdot 1 \cdot 2 = 4

hai, aur kyunki hum gradient chahte hain toh

f_{y}

dhundhne se pehle

∣\nabla f ∣ = 4

lete hain."

Error: magnitude saare components use karta hai. Tumhe

f_{y} = x^{2} = 1

bhi compute karna hoga, phir

∣\nabla f ∣ = 4^{2} + 1^{2} = 17

. Adhe-bane vector ka magnitude kabhi mat lo.

"Constant height par move karne ke liye,

\nabla f

direction mein chalo."

Error:

\nabla f

ke saath chalna sabse tezi se climb karta hai. Constant height ke liye

D_{\overset{u}{^}} f = 0

chahiye, yaani

\overset{u}{^} ⊥ \nabla f

— arrow ke across chalo, level curve ke saath.

"Kyunki

\nabla (f + g) = \nabla f + \nabla g

, toh

\nabla (f g) = \nabla f \nabla g

bhi."

Error: gradient linear hai (sum rule hold karta hai) lekin multiplicative nahi. Product rule deta hai

\nabla (f g) = f \nabla g + g \nabla f

, dono ka mix, har partial par product rule use karke.

Why questions

Chain rule kyun force karta hai ki directional derivative ek dot product ho?

Direction

\overset{u}{^} = (u_{1}, u_{2})

mein step karne par

x

rate

u_{1}

se move karta hai aur

y

rate

u_{2}

se; chain rule har partial slope ko uski rate se multiply karke sum karta hai:

f_{x} u_{1} + f_{y} u_{2}

, jo exactly

(f_{x}, f_{y}) \cdot (u_{1}, u_{2})

hai. Dot product choose nahi kiya gaya — yeh chain rule ki shape hai.

Steepest-ascent direction

\nabla f

hone ki guarantee kyun hai, kuch aur kyun nahi?

Kyunki

D_{\overset{u}{^}} f = ∣\nabla f ∣ cos θ

aur

cos θ

sabse bada (

= 1

) tab hota hai jab

θ = 0

, yaani jab

\overset{u}{^}

\nabla f

ke same direction mein point kare. Koi aur unit direction gradient ke seedha along point karne ko beat nahi kar sakta.

D_{\overset{u}{^}} f

mein

\overset{u}{^}

unit vector kyun hona chahiye?

Hum per unit length travelled rate chahte hain. Lamba vector har "step" mein zyada ground count karega aur rate ko inflate karega, isliye hum step length ko exactly

1

fix karte hain taaki number ek fair, comparable slope ho.

Gradient level sets ke perpendicular kyun hota hai, parallel kyun nahi?

Level set ke saath

f

constant hota hai, toh change zero hai:

\nabla f \cdot \overset{u}{^}_{tangent} = 0

. Zero dot product ka matlab orthogonal hai, toh

\nabla f

curve ke right angles par bahar niklata hai, iske saath nahi.

Smooth function ek point ke paas linear kyun dikhti hai, aur ek vector usse describe karne ke liye kaafi kyun hota hai?

Kisi bhi smooth hill ko zoom in karne par woh ek tilted plane mein flatten ho jaati hai, aur plane ka tilt poori tarah se uski do axis slopes

(f_{x}, f_{y})

se capture hota hai — ek single vector. Yahi local linearity hai jis ki wajah se gradient akele hi saare nearby behaviour ko determine karta hai.

Ek function ke many alag-alag points par same gradient value kyun ho sakti hai?

Gradient ek vector field hai — har point par ek alag arrow. Do points jinki local steepness aur climb direction same ho (jaise ek hill par do identical slopes) woh same gradient value share karte hain chahe woh alag locations hon.

Edge cases

f

ke local maximum par

\nabla f

kya hota hai aur kyun?

Yeh

0

hota hai. Bilkul top par koi direction climb nahi karti, toh har directional derivative zero hai, jo dono partials — aur isliye poora gradient — ko vanish karne par force karta hai.

Saddle point par, kya

\nabla f = 0

hai, aur kya iska matlab

f

max ya min hai?

Haan,

\nabla f = 0

(yeh critical point hai), lekin yeh na max hai na min:

f

kuch directions mein rise karta hai aur kuch mein fall. Zero gradient signal karta hai "first order par flat," na ki kaunsi shape.

Agar

\nabla f = 0

ho, toh "steepest ascent ki direction" kya hai?

Undefined. Steepest ascent

\nabla f /∣\nabla f ∣

hai, lekin

∣\nabla f ∣ = 0

se divide karna illegal hai — geometrically koi uphill nahi hai kyunki ground first order par saari directions mein level hai.

Constant function

f (x, y) = 5

ke liye, har jagah

\nabla f

kya hai?

Har point par zero vector

(0, 0)

. Flat sheet mein kisi bhi direction mein koi slope nahi hai, toh saare partials zero hain aur har level set poora plane hai.

Agar

\overset{u}{^}

aur

\nabla f

ke beech

θ = π /2

ho, toh

D_{\overset{u}{^}} f

kya hai aur physically tum kya kar rahe ho?

D_{\overset{u}{^}} f = ∣\nabla f ∣ cos (π /2) = 0

: tum level curve ke saath chal rahe ho, exactly same height par, na climb kar rahe na descend.

Kya

∣\nabla f ∣

negative ho sakta hai?

Nahi. Yeh ek length hai,

f_{x}^{2} + f_{y}^{2} \geq 0

ke roop mein compute hoti hai. Chosen direction mein rate negative ho sakti hai (downhill jaana), lekin gradient ki magnitude kabhi nahi.

Agar tum

f

- 1

se scale karo (hill ko valley mein flip karo), toh gradient ka kya hota hai?

Yeh reverse ho jaata hai:

\nabla (- f) = - \nabla f

. Har uphill downhill ban jaata hai, toh steepest-ascent arrow opposite direction mein point karne ke liye flip ho jaata hai — exactly yahi reason hai ki gradient descent

- \nabla f

follow karta hai.

Connections

Partial derivatives — woh components jo har trap upar manipulate karta hai.
Directional derivative — unit-vector rule yahan rehta hai.
Chain rule (multivariable) — dot-product form ka source.
Level curves and surfaces — perpendicularity traps.
Gradient descent — $\nabla f$ ka sign kyun matter karta hai.
Tangent plane and linearisation — local-linearity "why".
Lagrange multipliers — level sets ke normals, upar test kiye gaye.