Gradient as direction of steepest ascent: Question bank

4.4.10 · D5 · Maths › Multivariable Calculus › Gradient as direction of steepest ascent

Ye items parent note ke ideas ko probe karte hain: $\nabla f$ ka matlab (the gradient, partial derivatives ka vector), directional derivative $D_{u} f = \nabla f \cdot u$ (valid sirf unit vector $u$ ke liye, matlab $∥ u ∥ = 1$ ), iska dot product se connection, aur $\nabla f$ level curves ke saath kaise relate karta hai.

True or false — justify

The gradient hamesha function ke maximum (sabse unche peak) ki taraf point karta hai.

False. Yeh usi ek point par locally fastest increase ki direction mein point karta hai; door ke peaks ke baare mein isse kuch pata nahin. Yeh ek local slope arrow hai, summit ka GPS nahin.

Agar kisi point par

\nabla f = 0

hai, toh

f

ka wahan maximum hai.

False. Zero gradient ka matlab hai ek critical point — max, min, ya saddle kuch bhi ho sakta hai. Bas itna pata hai ki har directional derivative wahan

0

hai, toh surface momentarily flat hai.

Ek unit

u

ke liye,

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

ki maximum value sabhi aisi

u

par

∥\nabla f ∥

ke barabar hoti hai.

True. Kyunki

D_{u} f = ∥\nabla f ∥ cos θ

(using

∥ u ∥ = 1

) aur

cos θ \leq 1

, isliye sabse badi value

∥\nabla f ∥

hai, jo tab milti hai jab

u

aur

\nabla f

align ho jaayein (

θ = 0

Gradient level curve ka tangent hota hai us point par.

False. Yeh level curve ke perpendicular hota hai. Curve ke along

f

nahi badalta, isliye

D_{u} f = 0 \Rightarrow cos θ = 0 \Rightarrow θ = 9 0^{\circ}

Agar do directions ka directional derivative same hai, toh woh same direction honi chahiye.

False.

cos θ

symmetric hota hai, isliye do unit vectors jo

\nabla f

ke dono taraf equal angles banate hain unka

D_{u} f

same hoga. Yeh gradient line ke across mirror images hain.

Steepest descent ek alag, independent direction hai jise alag se compute karna padta hai.

False. Yeh simply

- \nabla f

hai, ekdum ulta arrow (

θ = π

cos θ = - 1

). Ek computation se dono mil jaate hain.

f

2 f

se replace karne par steepest ascent ki direction badal jaati hai.

False.

\nabla (2 f) = 2\nabla f

— same direction, double length. Steepest ascent ki direction nahi badlti; sirf rate

∥\nabla f ∥

double ho jaata hai.

D_{u} f

(unit

u

ke liye) negative ho sakta hai.

True. Agar tum "downhill" step karo, toh

cos θ < 0

isliye

D_{u} f < 0

; sabse zyada negative

- ∥\nabla f ∥

hota hai (steepest descent).

Spot the error

D_{v} f = \nabla f \cdot v

jahan

v = (3, 4)

woh direction hai jo main chahta hoon."

Error:

v

ek unit vector nahin hai (

∥ v ∥ = 5

). Formula

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

∥ u ∥ = 1

chahiye, isliye pehle normalize karo:

u = (3, 4) /5

. Warna rate

5

ke factor se inflate ho jaata hai.

"Kyunki

\nabla f

uphill point karta hai,

\nabla f

ke along chalna mujhe same contour line par rakhta hai."

Error: contour lines level sets hain jahan

f

constant hota hai;

\nabla f

ke along chalna

f

ko maximally change karta hai. The gradient contours ke perpendicular hota hai, unke along nahin.

"The directional derivative sirf direction par depend karta hai, point par nahin."

Error:

\nabla f

ko at the point

x = (x, y)

evaluate kiya jaata hai, isliye

D_{u} f = \nabla f (x) \cdot u

jab

x

move karta hai toh change hota hai. Point aur direction dono matter karte hain.

∥\nabla f ∥

x

-direction mein slope hai."

Error:

x

-direction mein slope partial

f_{x}

hai. Magnitude

∥\nabla f ∥ = f_{x}^{2} + f_{y}^{2}

sabhi directions mein maximum slope hai, jo generally

∣ f_{x} ∣

se bada hota hai.

"Maine

D_{u} f = 5

find kiya (unit

u

) lekin

∥\nabla f ∥ = 4

— great."

Error: impossible. Koi bhi directional derivative unit direction se

∥\nabla f ∥

se zyada nahi ho sakta, kyunki

D_{u} f = ∥\nabla f ∥ cos θ \leq ∥\nabla f ∥

. Recomputation zaroori hai.

f (x, y, z)

ka gradient ek scalar hai jo overall steepness deta hai."

Error: gradient ek vector hai

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

. Iska magnitude scalar steepness hai; vector direction bhi carry karta hai.

"Steepest descent ke liye main

1/\nabla f

leta hoon."

Error: vector se divide nahi kar sakte. Steepest descent

- \nabla f

hai (isse negate karo), aur unit descent direction

- \nabla f /∥\nabla f ∥

hai.

Why questions

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

mein

u

unit vector kyun hona chahiye?

Taaki rate travel ki har unit distance par measure ho. Agar

∥ u ∥ \neq = 1

, toh dot product bhi

∥ u ∥

se scale ho jaata hai, "kitna door" aur "kitna fast" mix ho jaate hain.

Steepest ascent direction

\nabla f /∥\nabla f ∥

kyun hai,

\nabla f

khud kyun nahin?

Direction unit-length version hoti hai;

∥\nabla f ∥

se divide karne par magnitude hatt jaata hai aur sirf "which way" bachta hai.

\nabla f

same taraf point karta hai lekin unit length nahin hai.

Level curves ke saath perpendicularity dot-product formula se kyun follow karti hai?

Level curve ke along

f

constant hota hai, isliye change ki rate

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

. Zero dot product (nonzero vectors ke saath) matlab

9 0^{\circ}

angle.

Har baar limit ke bajaay

D_{u} f

derive karne ke liye chain rule kyun use karte hain?

Limit ko har direction ke liye alag se compute karna padta. Chain rule

D_{u} f

ko ek baar express karta hai, partials ke terms mein jo pehle se hain, sabhi directions ke liye valid.

Gradient descent algorithm

- \nabla f

mein step kyun karta hai?

Kyunki

- \nabla f

fastest decrease ki direction hai, jo algorithm ko har step mein loss jitna ho sake utna quickly reduce karne deta hai — dekho Gradient Descent (Machine Learning).

Maximum rate gradient ka magnitude kyun hai, yeh ek purely geometric fact?

Kyunki

\nabla f \cdot u = ∥\nabla f ∥ cos θ

aur

cos θ

sabse zyada

1

ho sakta hai; direction ka detail gayab ho jaata hai, sirf length

∥\nabla f ∥

bachti hai.

Edge cases

Ek critical point par jahan

\nabla f = 0

hai, steepest ascent ki direction kya hai?

Undefined. Har direction mein

D_{u} f = 0

deta hai, aur

\nabla f /∥\nabla f ∥ = 0 /0

meaningless hai — surface first order mein sabhi directions mein flat hai.

Ek linear field

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

ke liye,

\nabla f

point se point kaise change karta hai?

Nahi karta:

\nabla f = (a, b)

har jagah hai. Steepness aur steepest direction poore plane mein constant hai (ek tilted flat sheet).

Agar

∥\nabla f ∥

bahut chota lekin nonzero hai, toh terrain ke baare mein yeh kya batata hai?

Slope almost flat hai — steepest uphill direction bhi bahut slowly rise karta hai. Gradient direction phir bhi well-defined hai, bas short hai.

Kya do perpendicular directions dono positive directional derivatives de sakti hain?

Haan, jab tak koi bhi level curve ke along na ho. Agar

\nabla f

unke beech strictly lie kare, toh dono

\nabla f

ke saath

9 0^{\circ}

se kam angle banate hain, isliye dono mein

cos θ > 0

Ek level curve par kisi point par, kya

D_{u} f

sirf tangent direction ke liye zero hai ya ek poore range ke liye?

Sirf do tangent directions ke liye (

\pm

curve ke along). Har doosri direction mein

cos θ \neq = 0

hota hai, jo nonzero rate deta hai — uphill side par positive, downhill side par negative.

Ek perfectly flat region (

f = const

har jagah) par steepest ascent ka kya hota hai?

\nabla f = 0

throughout, isliye increase ki koi direction nahin — har step

f

same rakhta hai. Wahan "steepest ascent" meaningless hai.

Agar

u

aur

\nabla f

ke beech

6 0^{\circ}

ka angle hai, toh maximum rate ka kitna fraction milta hai?

cos 6 0^{\circ} = \frac{1}{2}

, isliye

∥\nabla f ∥

ka exactly half. Sirf angle hi fraction determine karta hai, gradient kitna bada hai yeh matter nahin karta.

Recall Ek-line self-test

Gradient ki direction jawab deti hai "kaun si taraf upar?"; iska length jawab deta hai "kitna steep?"; iska dot with a unit direction jawab deta hai "is way mein kitna fast?".

Connections

Gradient as direction of steepest ascent
Directional Derivative
Partial Derivatives
Multivariable Chain Rule
Dot Product and Angle
Level Curves and Contour Maps
Gradient Descent (Machine Learning)
Tangent Plane and Linearization