Directional derivative — definition, formula

4.4.8 · Maths › Multivariable Calculus

YEH HAI KYA?

$u$ unit vector kyun hona chahiye? Kyunki limit mein $h$ asal mein taye kiya gaya distance measure karta hai. Agar $∥ u ∥ = 2$ ho, toh " $u$ -steps" mein $h$ move karne par $2 h$ real distance cover hoti hai, aur slope artificially double ho jaata. $∥ u ∥ = 1$ force karne se "rate per unit distance" meaningful ban jaata hai.

WORKING FORMULA KAISE MILTI HAI? (Scratch se derivation)

Hum har baar woh limit calculate nahi karna chahte. Isko kisi aisi cheez mein reduce karte hain jo hum pehle se jaante hain — partial derivatives.

Step 1 — $f$ ko $a$ se direction $u$ mein line par restrict karo. Ek single-variable helper define karo: $g (h) = f (a + h u) .$ Yeh step kyun? Seedhi line par chalna multivariable problem ko 1-D problem mein badal deta hai. Directional derivative exactly $g^{'} (0)$ hai: $D_{u} f (a) = lim_{h \to 0} \frac{g ( h ) - g ( 0 )}{h} = g^{'} (0) .$

Step 2 — Chain rule se $g$ ko differentiate karo. $a + h u = (a_{1} + h u_{1}, \dots, a_{n} + h u_{n})$ likho. Har coordinate $x_{i} (h) = a_{i} + h u_{i}$ hai, isliye $\frac{d x _{i}}{d h} = u_{i}$ . Multivariable chain rule se: $g^{'} (h) = \sum_{i = 1}^{n} \frac{\partial f}{\partial x _{i}} (a + h u) \cdot \frac{d x _{i}}{d h} = \sum_{i = 1}^{n} \frac{\partial f}{\partial x _{i}} \cdot u_{i} .$ Yeh step kyun? Chain rule "har coordinate ke through $f$ kaise change hoti hai" $\times$ "woh coordinate kitni tezi se move karta hai" ko sum karta hai.

Step 3 — $h = 0$ par evaluate karo (taaki point $a$ ho). $D_{u} f (a) = i = 1 \sum n \frac{\partial f}{\partial x _{i}} (a) u_{i} = \nabla f (a) \cdot u$

Figure — Directional derivative — definition, formula

Dot product ke through geometric meaning

Kyunki $D_{u} f = \nabla f \cdot u = ∥\nabla f ∥ ∥ u ∥ cos θ = ∥\nabla f ∥ cos θ$ (kyunki $∥ u ∥ = 1$ ):

Worked examples

Forecast-then-Verify

Recall Computing se pehle forecast karo

$f (x, y) = x^{2} + y^{2}$ ke liye $(3, 4)$ par, $u = (0, 1)$ ke saath $D_{u} f$ predict karo, phir check karo. Forecast: purely $+ y$ direction mein move karte hue, slope $f_{y} = 2 y = 8$ ke barabar hona chahiye. Verify: $\nabla f = (2 x, 2 y) = (6, 8)$ ; $D_{u} f = (6, 8) \cdot (0, 1) = 8.$ ✓ (Kisi axis ke saath directional derivative = woh partial derivative.)

Common mistakes (Steel-manned)

Recall Feynman: 12-saal ke bachche ko explain karo

Socho ek pahaadi maidan hai aur tum kahin khade ho. Agar tum bilkul East ki taraf face karo aur ek tiny step lo, tum upar ja sakte ho. Agar tum North ki taraf face karo, shayad tum aur tezi se upar jao, ya neeche bhi ja sako. "Directional derivative" sirf ek number hai jo kehta hai: jis taraf main abhi face kar raha hoon, jab main aage kadam barhata hoon toh zameen kitni steep hoti hai? Ek special direction hai — seedha pahadi ke upar — jahan tum sabse tezi se chadhte ho. Woh arrow gradient kehlata hai. Kisi aur direction ke liye, tum kam chadhte ho, ek aise amount se jo dikhata hai ki woh direction gradient ki taraf kitna jhukta hai.

Flashcards

Define the directional derivative of

f

a

in direction

u

(limit form).

D_{u} f (a) = lim_{h \to 0} \frac{f ( a + h u ) - f ( a )}{h}

, with

∥ u ∥ = 1

Why must

u

be a unit vector?

Taaki

h

actual distance travelled measure kare; warna slope

∥ u ∥

se scale ho jaata hai.

Working formula for

D_{u} f

(differentiable

f

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

How is the formula derived?

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

set karo; phir

D_{u} f = g^{'} (0)

aur chain rule deta hai

g^{'} (0) = \sum f_{x_{i}} u_{i} = \nabla f \cdot u

In which direction is

D_{u} f

largest, and what is its value?

u = \nabla f /∥\nabla f ∥

ke saath; max value

= ∥\nabla f ∥

What is

D_{u} f

when

u ⊥ \nabla f

Zero — tum ek level curve ke saath move kar rahe ho.

Is the directional derivative a scalar or a vector?

Ek scalar hai (ek single slope number).

When can the dot-product formula fail despite the limit existing?

Jab

f

a

par differentiable nahi hai (jaise ek sharp point par), chahe saare directional derivatives exist karein.

Relation between

D_{u} f

and partial derivatives along axes?

D_{e_{i}} f = \partial f / \partial x_{i}

(axis direction partial derivative wapas deta hai).

Connections

Gradient vector — steepest ascent ki direction; $D_{u} f = \nabla f \cdot u$ .
Partial derivatives — special case jab $u$ ek coordinate axis ho.
Multivariable chain rule — derivation ke peeche ka engine.
Level curves and level sets — jahan $D_{u} f = 0$ ho.
Tangent planes and differentiability — woh condition jo formula ko validate karti hai.
Dot product and cosine of angle — $∥\nabla f ∥ cos θ$ interpretation deta hai.