Directional derivative — definition, formula: Visual walkthrough

4.4.8 · D2 · Maths › Multivariable Calculus › Directional derivative — definition, formula

Step 0 — Pehle yeh pictures padhna seekho

Shuru karne se pehle, teen plain-word ideas, har ek ek picture ke saath.

Figure — Directional derivative — definition, formula

Picture mein: floor pale grid hai, tumhare paon coral dot $a$ par hain, lavender arrow $u$ tumhara facing direction hai (length 1 draw kiya hai), aur upar floating surface pahaad $f$ hai. Neeche sab kuch ek number ke baare mein hai: jab tum us lavender arrow ki direction mein step lete ho toh pahaad kitni tezi se chadhta hai.

Step 1 — Poore pahaad ko ek seedhi walk mein freeze karo

$h$ distance chalne ke baad jo jagah hum pahunchte hain wo hai $start a + kitna door (metres) h kis taraf (length 1) u .$

Kyunki $u$ ki length 1 hai, $h$ number se move karna sach mein tumhe $h$ metres move karta hai — yahi poori wajah hai ki humne unit vector ki demand ki thi.

Step 2 — Walk ke saath height ko naam do: helper $g$

Term by term:

$h$ — input: abtak chali gayi distance (negative ho sakta hai — matlab ulti taraf chalna).
$a + h u$ — us waqt mere neeche ka floor spot.
$f (\dots)$ — wahan ki height.
$g (h)$ — ek single output number: meri current altitude.

Step 3 — Start par slope hi directional derivative hai

Har piece:

$g (h) - g (0)$ — rise: $h$ metres chalne ke baad altitude gain.
$h$ se divide karna — ise rise per metre banata hai, yaani ek slope.
$lim_{h \to 0}$ — step ko ek instant tak shrink karo, taaki hume exactly wahan instantaneous steepness mile jahan main khada hoon, na ki lambi walk ka average.
$g^{'} (0)$ — plain curve $g$ ka ordinary derivative, start $h = 0$ par evaluate kiya.

Toh mission ab concrete hai: $g^{'} (0)$ compute karo. Hum har baar limit grind nahi karna chahte — isliye hum ek aisa tool use karte hain jo already jaanta hai composed multivariable function ko differentiate karna.

Step 4 — Chain rule kyun, aur kuch nahi

Track karo ki har floor-coordinate $h$ par kaise depend karta hai. Pehle moving spot ko components mein likho, yaad karo $u = (u_{1}, u_{2})$ : $a + h u = (a_{1} + h u_{1}, a_{2} + h u_{2}),$ toh do moving coordinates hain $x_{1} (h) = a_{1} + h u_{1}, x_{2} (h) = a_{2} + h u_{2} .$ Har ek $h$ mein ek seedhi line hai, isliye iska speed constant hai: $\frac{d x _{1}}{d h} = u_{1}, \frac{d x _{2}}{d h} = u_{2} .$ Term by term: $u_{1}$ hai kitni tezi se mera east-coordinate har metre chalne par badhta hai; $u_{2}$ wahi north ke liye. Saath mein $(u_{1}, u_{2}) = u$ — facing arrow ke components literally walk ka "east-speed" aur "north-speed" hain.

Step 5 — Crank ghuma lo: chain rule sum

Yeh bas Step 4 ka general chain rule hai jisme $x^{'} (h) = u_{1}$ aur $y^{'} (h) = u_{2}$ plug kiye hain. Words mein recipe: (jitna $f$ change hota hai jab yeh coordinate move kare) $\times$ (jitni tezi se yeh coordinate move kare), har coordinate par summed. $\frac{\partial f}{\partial x _{i}}$ partial derivatives hain — pure east aur pure north steepness, jinka parent note mein pehle hi zikr hua tha.

Step 6 — Start par khade raho: $h = 0$ par evaluate karo

$D_{u} f (a) = g^{'} (0) = \frac{\partial f}{\partial x _{1}} (a) u_{1} + \frac{\partial f}{\partial x _{2}} (a) u_{2} .$

Ab do partials ko ek arrow mein bundle karo — gradient $\nabla f (a) = (\frac{\partial f}{\partial x _{1}}, \frac{\partial f}{\partial x _{2}})$ — aur notice karo ki right-hand side exactly "matching components multiply karke add karna" hai, jo dot product hai:

Yeh general hai: $f : R^{n} \to R$ ke liye wahi argument $n$ coordinates par ek sum deta hai, $D_{u} f = \sum_{i} \frac{\partial f}{\partial x _{i}} (a) u_{i} = \nabla f \cdot u$ .

Step 7 — Formula ko geometrically padho ( $cos θ$ wala face)

Hum $cos θ$ laate hain kyunki yeh exactly woh dial hai jo measure karta hai "mera facing kitna steepest direction ke saath aligned hai?" — value $1$ jab aligned ho, $0$ jab perpendicular ho, $- 1$ jab opposite ho.

Step 8 — Degenerate cases (koi gap mat chodho)

Intuition Har corner covered

(a) $\nabla f (a) = 0$ (flat spot / summit / saddle). Tab $D_{u} f = 0 \cdot u = 0$ har direction $u$ ke liye. Socho ek hilltop ya valley floor: kisi bhi taraf step lo, zameen momentarily level hai. Koi preferred direction exist nahi karta.

(b) Ulti taraf chalna, $- u$ . $D_{- u} f = \nabla f \cdot (- u) = - D_{u} f$ . Exact ulti taraf face karne par sign flip ho jaata hai: jo uphill tha wo ab utna hi downhill hai. Consistent, koi surprise nahi.

(c) $f$ , $a$ par differentiable nahi (ek sharp ridge ya crease). Step 4 chain rule ko $a$ par ek genuine tangent plane chahiye. Agar surface wahan kink karti hai, toh koi single flat tangent plane exist nahi karta, isliye neat sum galat ho sakta hai chahe raw limit $g^{'} (0)$ direction-by-direction exist bhi kare. Rule: boxed formula tabhi allowed hai jab $f$ differentiable ho (continuous partials guarantee karte hain).

Ek-picture summary

Ek panel poori story chain karta hai: pahaad $\to$ seedha slice $\to$ 1-D curve $g$ $\to$ tangent slope $g^{'} (0)$ $\to$ chain-rule sum $\to$ $\nabla f \cdot u$ $\to$ $∥\nabla f ∥ cos θ$ . Arrows follow karo aur tumne directional derivative ko ek hillside par ek step se re-derive kar liya.

Recall Feynman retelling — plain words mein poora walkthrough

Main ek pahaad par khada hoon aur kisi taraf dekh raha hoon (ek 1-metre arrow $u$ ). Main poore pahaad ko samajhne ki koshish nahi karta; main bas us taraf seedha chalta hoon aur apni height ko ek plain graph mein likhta hoon — neeche chali gayi distance, upar altitude. Woh graph $g$ hai. "Main jis taraf dekh raha hoon us direction mein zameen kitni steep hai?" bas $g$ ka slope start par hai, jo $g^{'} (0)$ hai. Woh slope limit grind kiye bina dhundhne ke liye, main notice karta hoon ki meri walk mujhe thoda east aur thoda north $u_{1}$ aur $u_{2}$ speeds par move karti hai. Chain rule kehta hai meri climb rate hai (east-steepness $\times$ east-speed) $+$ (north-steepness $\times$ north-speed). Do steepnesses ko ek arrow (gradient) mein aur do speeds ko $u$ mein bundle karne par, woh sum hi dot product $\nabla f \cdot u$ hai. Aakhir mein, kyunki $u$ 1 metre lamba hai, yeh $∥\nabla f ∥ cos θ$ ke barabar hai: main gradient ke seedha upar face karte hue fastest chadhta hoon ( $cos θ = 1$ ), sideways chalne par kuch nahi chadhta ( $cos θ = 0$ ), aur ulti taraf face karte hue fastest utarta hoon ( $cos θ = - 1$ ). Flat summit par gradient zero hai, isliye har direction level hai.

Recall Quick self-test

$g (h) = f (a + h u)$ mein reduce karna kyun help karta hai? ::: Yeh 2-D pahaad ko 1-D curve mein turn kar deta hai, isliye ordinary single-variable calculus (ek tangent slope) apply ho jaata hai. $g^{'} (0) = \sum_{i} f_{x_{i}} u_{i}$ mein, har $u_{i}$ kya represent karta hai? ::: Woh speed jis par coordinate $x_{i}$ $u$ ke along chalne ke har metre par change karta hai. $D_{u} f = ∥\nabla f ∥ cos θ$ kyun hai (koi $∥ u ∥$ nahi)? ::: Kyunki $u$ ek unit vector hai, $∥ u ∥ = 1$ . Agar $\nabla f (a) = 0$ ho, toh har $u$ ke liye $D_{u} f$ kya hai? ::: Zero — ek flat spot, har direction mein level. Dot-product formula kab fail ho sakta hai? ::: Jab $f$ , $a$ par differentiable nahi hai (ek kink/ridge), isliye koi tangent plane exist nahi karta.

Parent: Directional derivative — definition, formula (index 4.4.8) · Prereqs: Gradient vector, Partial derivatives, Multivariable chain rule, Dot product and cosine of angle, Level curves and level sets, Tangent planes and differentiability