Directional derivative — definition, formula: Question bank

4.4.8 · D5 · Maths › Multivariable Calculus › Directional derivative — definition, formula

Prerequisite ideas jinpar tum rely karoge: Gradient vector, Partial derivatives, Dot product and cosine of angle, Level curves and level sets, Tangent planes and differentiability, aur Multivariable chain rule.

True or false — justify

A directional derivative ek vector hota hai jo us direction mein point karta hai jismein tum chale.

False — ye ek scalar hai, ek single slope number. Woh vector jo "most uphill" point karta hai woh

\nabla f

hai, na ki

D_{u} f

Agar

∥ u ∥ \neq = 1

ho, to formula

\nabla f \cdot u

phir bhi directional derivative deta hai.

False — ye rate ko

∥ u ∥

se scale karke deta hai. Kyunki limit mein

h

true distance measure karta hai, sirf ek unit vector "change per unit distance" report karta hai.

D_{u} f

aur

D_{- u} f

hamesha equal hote hain.

False — ye ek doosre ke negatives hote hain:

\nabla f \cdot (- u) = - (\nabla f \cdot u)

. Ek direction mein chalna utni hi tezi se chadhna hai jitni tezi se opposite direction mein utarna.

Ek aisi point par jahan

\nabla f = 0

ho, har directional derivative zero hota hai.

True —

0 \cdot u = 0

sabhi

u

ke liye. Aise critical points (peaks, valleys, saddles) har direction mein first order par flat dikhte hain.

D_{u} f

ki maximum possible value ek point par

∥\nabla f ∥

ke barabar hoti hai.

True — kyunki

D_{u} f = ∥\nabla f ∥ cos θ

aur

cos θ \leq 1

, isliye sabse bada slope

∥\nabla f ∥

hai, jo tab milta hai jab

u

ka direction

\nabla f

ke saath align ho.

(0, 1)

ke along directional derivative partial derivative

f_{y}

ke barabar hota hai.

True —

\nabla f \cdot (0, 1) = f_{y}

. Ek coordinate axis ke along directional derivative exactly wahi partial derivative hota hai.

Agar

f

ke sabhi directional derivatives

a

par exist karte hain, to

f

wahan differentiable hai.

False — har directional derivative ka exist karna differentiability imply nahi karta (classic counterexamples mein origin par ek sharp ridge hoti hai). Differentiability ek stronger condition hai.

Do functions jinka

a

par same gradient ho, unka har direction mein

a

par same directional derivative hoga.

True —

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

sirf us point par gradient par depend karta hai, isliye equal gradients equal directional derivatives force karte hain.

Spot the error

"Direction hai

(3, 4)

, gradient hai

(4, 1)

, to

D_{u} f = (4, 1) \cdot (3, 4) = 16

Vector

(3, 4)

ko unit-ize nahi kiya gaya tha (

∥ (3, 4) ∥ = 5

). Sahi answer

16/5

hai; normalisation skip karne se ye paanch guna badh gaya.

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

, to main

∥ u ∥

ko kisi bhi case mein drop kar sakta hoon."

Tum

∥ u ∥

tabhi drop kar sakte ho kyunki

∥ u ∥ = 1

assume kiya gaya tha. Agar

u

unit nahi hai, to woh step secretly

∥ u ∥

factor kho deta hai.

"Gradient woh direction hai jismein main chal raha hoon, isliye

D_{u} f

ek vector hai."

Do objects ko confuse kar rahe hain:

\nabla f

ek vector hai, lekin

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

unka dot product hai, isliye ye ek scalar slope hai.

f

a

par ek corner hai lekin sabhi directional derivatives exist karte hain, isliye formula

\nabla f \cdot u

apply hota hai."

Formula ko differentiability chahiye, na sirf directional derivatives ka exist karna. Ek corner par

f

differentiable nahi ho sakta aur shortcut galat values de sakta hai.

"Kyunki

u

ke along

D_{u} f = 0

hai, to function us poori line par constant hai."

Ek zero directional derivative sirf point

a

par ek instantaneous fact hai; ye kehta hai ki tum wahan ek level curve ke tangent ho, na ki

f

poori straight line par constant rehta hai.

"Steepest descent direction dhundne ke liye main

\nabla f \cdot u

ko maximise karta hoon."

Steepest descent ka matlab dot product ko minimise karna hai, jisse

u = - \nabla f /∥\nabla f ∥

aur value

- ∥\nabla f ∥

milti hai; maximise karne se ascent milta hai.

Why questions

Definition mein

u

unit vector kyun hona chahiye?

Kyunki parameter

h

actual travel ki distance measure karta hai; agar

∥ u ∥ \neq = 1

ho, to size

h

ka ek step

h ∥ u ∥

real distance cover karta hai aur reported slope

∥ u ∥

se stretch ho jaata hai.

Gradient steepest ascent ki direction mein kyun point karta hai?

D_{u} f = ∥\nabla f ∥ cos θ

likhne par, slope tab sabse bada hota hai jab

cos θ = 1

ho, yaani jab

u

ka direction

\nabla f

ke along ho; har doosra heading sirf partially uski taraf jhukta hai aur kam chadhta hai.

Jab

u ⊥ \nabla f

ho to directional derivative zero kyun hota hai?

Perpendicular ka matlab

cos θ = 0

hai, isliye

\nabla f \cdot u = 0

; geometrically tum ek level curve ke along chal rahe ho jahan height first order par nahi badhlati.

Formula derive karte time chain rule kyun appear hoti hai?

Line

g (h) = f (a + h u)

par restrict karne se har coordinate

h

ka function ban jaata hai; Multivariable chain rule har partial ko uske coordinate ki rate

u_{i}

se multiply karke sum karta hai, jisse

\sum f_{x_{i}} u_{i} = \nabla f \cdot u

milta hai.

Hum hamesha limit definition use karne ki jagah formula ki taklif kyun uthate hain?

Limit ko har nayi direction ke liye recompute karna padta hai, jabki

\nabla f \cdot u

ek gradient ko sabhi directions ke liye reuse karta hai — partials ek baar compute karo, kisi bhi

u

ke saath dot karo.

Maximum slope

∥\nabla f ∥

kyun hai aur

u

se related kuch kyun nahi?

Kyunki

∥ u ∥ = 1

fixed hai, isliye

D_{u} f = ∥\nabla f ∥ cos θ

sirf angle par depend karta hai; iska peak

∥\nabla f ∥

gradient ki length ki property hai, independent of which unit direction tum choose karo.

Ek axis ke along directional derivative partial derivative mein kyun reduce ho jaata hai?

Unit vector

(0, \dots, 1, \dots, 0)

dot product mein

\nabla f

ka exactly ek component pick karta hai, jo precisely us partial derivative ki definition hai.

Edge cases

Ek aisi point par

D_{u} f

kya hai jahan

\nabla f = 0

ho?

Har direction mein zero, kyunki

0 \cdot u = 0

; first order par surface wahan flat hai, isliye sirf slope information peak, pit, ya saddle mein distinguish nahi kar sakti.

Agar

f

har jagah constant ho, to uske directional derivatives kya hain?

Sabhi zero —

\nabla f = 0

har point par hai, isliye chalne ki koi bhi direction height nahi badhlati.

Ek level curve ke along, us curve ke tangent

u

ke liye

D_{u} f

kya hoga?

Zero, kyunki level curve ka tangent

\nabla f

ke perpendicular hota hai; isliye gradients hamesha level curves ke normal hote hain.

Ek linear function

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

ke liye,

D_{u} f

point se point tak kaise vary karta hai?

Ye position ke saath vary nahi karta —

\nabla f = (a, b)

constant hai, isliye ek fixed

u

ke along directional derivative har jagah same hota hai; sirf direction number ko change karti hai.

Kya do alag directions same directional derivative value de sakti hain?

Haan — koi bhi do unit directions jo

\nabla f

ke saath same angle

θ

banati hain woh value

∥\nabla f ∥ cos θ

share karti hain (jaise gradient ke dono taraf symmetrically straddling karne wali do directions).

Jab

u

ek point ke around 2D mein poora circle rotate karta hai to

D_{u} f

ka kya hota hai?

∥\nabla f ∥ cos θ

trace karta hai, smoothly oscillate karta hai maximum

+ ∥\nabla f ∥

(

\nabla f

ke along),

0

se hokar (perpendicular),

- ∥\nabla f ∥

tak (opposite), aur waapas.

Recall Traps ki one-line summary

Direction ko normalize karo, yaad rakho answer ek scalar hai, aur $\nabla f \cdot u$ ko differentiability ke bina invoke mat karo. Reason ::: Ye teen habits zyaadatar directional-derivative mistakes ko defuse kar deti hain.