Directional derivative — definition, formula: Worked examples

4.4.8 · D3 · Maths › Multivariable Calculus › Directional derivative — definition, formula

Ye note parent note ki companion hai. Wahan humne formula banaya tha $D_{u} f (a) = \nabla f (a) \cdot u, ∥ u ∥ = 1.$ Yahan hum ise drill karenge. Plan yeh hai: pehle ek scenario matrix banao — ek checklist har tarah ke case ki jo is topic mein aa sakti hai — phir aisi examples work karo jo milke har cell ko tick kar den.

The scenario matrix

Har row ek "case class" hai — ek aisi situation jiska apna trap hai. Last column us example ka naam deta hai jo use cover karta hai.

#	Case class	Isme tricky kya hai	Covered by
C1	Direction do points ke roop mein di gayi hai	subtract karna padega, phir normalise	Ex 1
C2	Direction ek angle ke roop mein di gayi hai	pehle se unit hai — dobara normalise mat karo	Ex 2
C3	Negative directional derivative (downhill chalna)	answer ka sign meaningful hai	Ex 3
C4	Zero directional derivative ( $u ⊥ \nabla f$ )	tum level curve par ho	Ex 4
C5	Teen variables, $u$ mein mixed signs	minus sign ke saath normalising	Ex 5
C6	Maximum / minimum rate, aur special directions	$\pm ∥\nabla f ∥$	Ex 6
C7	Degenerate: point par $\nabla f = 0$	har direction se $0$ milta hai	Ex 7
C8	Word problem (temperature / hill), units ke saath	words ko vectors mein translate karo	Ex 8
C9	Exam twist: do directions mein $D_{u} f$ diya hai, $\nabla f$ nikalo	dot product ulta karo	Ex 9

Prerequisites jo aap khule rakhna chahein: Gradient vector, Partial derivatives, Dot product and cosine of angle, Level curves and level sets.

C1 — Do points se direction

Worked example Ex 1 — subtract karo, phir unit-ize karo

$a = (1, 2)$ par $f (x, y) = x^{2} + x y$ , $(1, 2)$ se $(4, 6)$ ki taraf chalna hai.

Forecast: target upar-aur-daayein hai, aur gradient (hum dekhenge) strongly rightward point karta hai — toh expect karo ek solidly positive number, kuch units ke size mein.

Gradient. $f_{x} = 2 x + y, f_{y} = x$ , isliye $\nabla f = (2 x + y, x)$ . $(1, 2)$ par: $\nabla f = (4, 1)$ . Yeh step kyun? Formula $\nabla f \cdot u$ hai; hume hamesha pehle specific point par gradient chahiye.
Direction vector. $(4, 6) - (1, 2) = (3, 4)$ . Yeh step kyun? "Ek point ki taraf" ka matlab hai arrow yahan se wahan tak — start ko end se subtract karo.
Unit-ize. $∥ (3, 4) ∥ = 9 + 16 = 5$ (nonzero, isliye legal hai), toh $u = (3/5, 4/5)$ . Yeh step kyun? Directional derivative change per unit distance measure karta hai, isliye heading ki length $1$ honi chahiye; warna slope $5$ se scale ho jaayega.
Dot. $D_{u} f = (4, 1) \cdot (3/5, 4/5) = \frac{12 + 4}{5} = \frac{16}{5} = 3.2$ . Yeh step kyun? $\nabla f \cdot u$ matching components ko multiply karke add karta hai — ye "f har axis par kitna change hota hai" ko hamare heading ke liye ek slope mein collapse karta hai.

Verify: $∥ u ∥ = (3/5)^{2} + (4/5)^{2} = 9/25 + 16/25 = 1$ ✓. Positive, moderate — forecast se match karta hai.

C2 — Angle se direction

Worked example Ex 2 — angle pehle se ek unit vector hai

$(1, 0)$ par $f (x, y) = sin (x y)$ , direction $+ x$ -axis se $3 0^{\circ}$ upar.

Forecast: $(1, 0)$ par " $x y$ " sine ke andar $0$ hai, toh cheezein apne fastest change ke paas hain — lekin kis taraf? Aage padhne se pehle guess karo.

Angle ko $u$ mein baadlo. $u = (cos 3 0^{\circ}, sin 3 0^{\circ}) = (\frac{3}{2}, \frac{1}{2})$ . Yeh step kyun? Unit circle par angle $θ$ par ek point $(cos θ, sin θ)$ hota hai — automatically length $1$ , isliye kuch normalise nahi karna. Figure 1 mein yeh arrow unit circle ki ek radius hai: uski tip circle par baithti hai, jo "length $1$ " ka visual meaning hai.
Gradient. $f_{x} = y cos (x y), f_{y} = x cos (x y)$ , isliye $\nabla f = (y cos (x y), x cos (x y))$ . $(1, 0)$ par: $cos (0) = 1$ , toh $\nabla f = (0, 1)$ . Yeh step kyun? Dot karne se pehle hume diye gaye point par gradient chahiye; chain-rule factor $cos (x y)$ dono partials mein aata hai kyunki ye outer $sin$ ko differentiate karne se aata hai.
Dot. $D_{u} f = (0, 1) \cdot (\frac{3}{2}, \frac{1}{2}) = \frac{1}{2}$ . Yeh step kyun? Dotting heading aur gradient ke beech overlap select karta hai. Yahan $x$ -rate $0$ hai, isliye sirf walk ka vertical lean ( $\frac{1}{2}$ ) contribute karta hai.

Verify: $∥ u ∥^{2} = \frac{3}{4} + \frac{1}{4} = 1$ ✓. Answer $\frac{1}{2} = 0.5$ , positive aur max $∥\nabla f ∥ = 1$ se kam — consistent hai.

Figure — Directional derivative — definition, formula

Figure 1 padhna. Plum circle unit circle hai. Orange arrow $3 0^{\circ}$ par $u$ hai — notice karo uski tip circle par exactly land karti hai, bina re-normalising ke length $1$ confirm karta hai. Teal arrow $\nabla f = (0, 1)$ hai, seedha upar point karta hai; directional derivative $\frac{1}{2}$ precisely hai kitna $u$ us teal arrow ki taraf lean karta hai.

C3 — Negative answer (downhill chalna)

Worked example Ex 3 — sign geometry carry karta hai

$(3, 4)$ par $f (x, y) = x^{2} + y^{2}$ , direction $v = (- 1, 0)$ (due west).

Forecast: yeh bowl origin se door rise karta hai. $(3, 4)$ par khade hokar west ki taraf jaana matlab centre ki taraf wapas jaana — toh downhill — expect karo ek negative slope.

Gradient. $f_{x} = 2 x, f_{y} = 2 y$ , isliye $\nabla f = (2 x, 2 y) = (6, 8)$ $(3, 4)$ par. Yeh step kyun? Hume local uphill direction aur uski steepness chahiye pehle, taaki hum koi bhi heading uske against measure kar sakein.
$v$ ko Unit-ize karo. $∥ (- 1, 0) ∥ = 1$ (nonzero, aur pehle se length $1$ hai), isliye $u = (- 1, 0)$ . Yeh step kyun? Hume length check karni chahiye assume karne ki jagah — yahan woh $1$ nikli, isliye koi division nahi chahiye.
Dot. $D_{u} f = (6, 8) \cdot (- 1, 0) = - 6$ . Yeh step kyun? Dot product yeh nikalta hai ki heading $\nabla f$ ke saath kitna align karti hai; heading gradient ke $+ x$ part ke opposite point karti hai, minus sign produce karta hai.

Verify: Ek axis ke along directional derivative $\pm$ partial ke barabar hoti hai: $- x$ direction mein jaana $- f_{x} = - 6$ deta hai ✓. Minus sign ka literal matlab hai " $f$ west step karne par decrease hoti hai" — tum bowl mein neeche ja rahe ho.

C4 — Zero answer (level curve par)

Worked example Ex 4 — gradient ke perpendicular

Wohi $f = x^{2} + y^{2}$ $(3, 4)$ par, lekin $v = (- 4, 3)$ direction mein chalo.

Forecast: $\nabla f = (6, 8)$ origin se seedha bahar point karta hai. Direction $(- 4, 3)$ — kya ye perpendicular hai? $(6) (- 4) + (8) (3) = - 24 + 24 = 0$ . Toh expect karo exactly zero: tum constant height ke circle ke around slide kar rahe ho.

Gradient. $\nabla f = (2 x, 2 y) = (6, 8)$ $(3, 4)$ par. Yeh step kyun? Zero slope exactly tab hoti hai jab heading $\nabla f$ ke perpendicular ho, isliye perpendicularity test karne ke liye hume $\nabla f$ chahiye.
Unit-ize. $∥ (- 4, 3) ∥ = 16 + 9 = 5$ (nonzero), isliye $u = (- 4/5, 3/5)$ . Yeh step kyun? Normalising perpendicularity nahi badalta, lekin rate ko "per unit distance" mean karne ke liye unit heading zaroori hai.
Dot. $D_{u} f = (6, 8) \cdot (- 4/5, 3/5) = \frac{- 24 + 24}{5} = 0$ . Yeh step kyun? Zero dot product right angle ka algebraic signature hai — heading aur gradient perpendicular hain.

Verify: $(3, 4)$ se guzarne waali $x^{2} + y^{2}$ ki level curve radius $5$ ka circle hai. Us circle par $(3, 4)$ par tangent radius $(3, 4)$ ke perpendicular hoti hai — aur $(- 4, 3)$ exactly woh tangent hai. Zero slope = constant height par chalna.

Figure 2 padhna. Plum circles $f$ ki level curves hain (constant height). $(3, 4)$ par orange arrow $\nabla f$ radially bahar point karta hai — uphill direction. Teal arrow hamaari heading $u = (- 4/5, 3/5)$ hai: ye plum circle ke along (tangent) lie karta hai, orange se right angle par. Kyunki ye kabhi ek circle nahi chodta, height nahi badlti — yahi geometric reason hai ki answer $0$ hai.

C5 — Teen variables, mixed signs

Worked example Ex 5 — unit vector ke andar ek minus sign

$(1, 1, 1)$ par $f (x, y, z) = x^{2} y + z^{3}$ , direction $v = (2, - 1, 2)$ .

Forecast: $z$ -slope bada hai ( $3 z^{2} = 3$ ) aur $x$ -slope bhi upar hai, lekin $u$ ka $- y$ component $+ y$ -slope se lad raha hai. Net positive lekin kam hona chahiye.

Gradient. $f_{x} = 2 x y, f_{y} = x^{2}, f_{z} = 3 z^{2}$ , isliye $\nabla f = (2 x y, x^{2}, 3 z^{2}) = (2, 1, 3)$ $(1, 1, 1)$ par. Yeh step kyun? Har partial yeh jawab deta hai "agar main sirf is ek variable ko nudge karun toh $f$ kitni tez change hoti hai?"; inhe stack karke poora gradient milta hai jise hum dot karte hain.
Unit-ize. $∥ (2, - 1, 2) ∥ = 4 + 1 + 4 = 3$ (nonzero), isliye $u = (\frac{2}{3}, - \frac{1}{3}, \frac{2}{3})$ . Yeh step kyun? Minus apne component ke saath rehta hai; normalising teeno ko same length $3$ se divide karta hai, heading preserve karta hai lekin length $1$ fix karta hai.
Dot. $D_{u} f = (2, 1, 3) \cdot (\frac{2}{3}, - \frac{1}{3}, \frac{2}{3}) = \frac{4}{3} - \frac{1}{3} + 2 = 1 + 2 = 3$ . Yeh step kyun? Same recipe jaise 2-D mein — matching components multiply karo aur add karo; $- \frac{1}{3}$ woh $y$ -slope hai jo hamare against kaam kar rahi hai.

Verify: $∥ u ∥^{2} = \frac{4}{9} + \frac{1}{9} + \frac{4}{9} = \frac{9}{9} = 1$ ✓. $- \frac{1}{3}$ term total ko neeche kheench raha hai; phir bhi solidly positive, jaise forecast mein tha.

C6 — Maximum aur minimum rate

Worked example Ex 6 — do extreme directions

$(1, 2)$ par $f = x^{2} + x y$ ke liye (toh $\nabla f = (4, 1)$ ), greatest aur least possible directional derivatives nikalo, aur woh directions jo unhe achieve karte hain.

Forecast: sabse bada slope gradient ki length hai; sabse negative minus woh hai.

Max value. $D_{u} f = ∥\nabla f ∥ cos θ$ , $θ = 0$ par maximize hota hai: value $= ∥\nabla f ∥ = 4^{2} + 1^{2} = 17$ . Yeh step kyun? $θ$ ke saath $\nabla f$ aur $u$ ke beech angle hai, $∥\nabla f ∥ cos θ$ ka sirf free part $cos θ$ hai, jo $1$ par peak karta hai jab $θ = 0$ (exactly gradient ke along chalo).
Max ki direction. $u = \frac{1}{17} (4, 1)$ . Yeh step kyun? $cos θ = 1$ matlab $u$ $\nabla f$ ke same direction mein point karta hai; $\nabla f$ ko unit-ize karne se woh heading milti hai.
Min value. $θ = π$ par, $cos θ = - 1$ : value $= - 17$ , direction $- \frac{1}{17} (4, 1)$ . Yeh step kyun? Sabse negative slope tab achieve hoti hai jab aap exactly gradient ke opposite face karo — steepest descent.

Verify: $17 \approx 4.123$ . Check karo Ex 1 ka answer $3.2$ is cap se neeche hai ( $3.2 < 4.123$ ) ✓ — koi direction gradient ki length se zyada beat nahi kar sakti.

C7 — Degenerate: gradient zero hai

Worked example Ex 7 — flat spot, har direction se

$0$ milta hai Origin $(0, 0)$ par $f (x, y) = x^{2} + y^{2}$ , koi bhi (nonzero) direction $u$ .

Forecast: origin bowl ka bottom hai — first order tak bilkul flat. Chahe tum kisi bhi taraf face karo, instantaneous slope $0$ hona chahiye.

Gradient. $f_{x} = 2 x, f_{y} = 2 y$ , isliye $\nabla f = (2 x, 2 y) = (0, 0)$ origin par. Yeh step kyun? Gradient ki length maximum possible slope hai; agar woh length $0$ hai, toh koi direction koi first-order rise produce nahi kar sakti.
Dot. $D_{u} f = (0, 0) \cdot u = 0$ har unit $u$ ke liye. Yeh step kyun? Zero vector ko kisi bhi cheez se dot karo toh $0$ aata hai — heading irrelevant ho jaati hai, jo exactly "flat spot" ka matlab hai.

Verify: Parent note ki limit definition se seedha, $f (h u) = h^{2} (u_{1}^{2} + u_{2}^{2}) = h^{2}$ , isliye $\frac{f ( h u ) - f ( 0 )}{h} = \frac{h ^{2}}{h} = h \to 0$ jab $h \to 0$ ✓. Surface har direction mein upar curve karti hai, lekin first-order rate $0$ hai — ek genuine flat spot (ek minimum).

C8 — Units ke saath word problem

Worked example Ex 8 — metal plate par temperature

Ek plate par point $(x, y)$ par temperature ( $^{\circ} C$ mein) $T (x, y) = 100 - x^{2} - 2 y^{2}$ hai, jahan $x, y$ metres mein hain. Ek cheenti $(2, 1)$ par baithti hai aur vector $(3, 4)$ ki direction mein march karti hai. Woh temperature change kitna tez feel karti hai, per metre chalne par?

Forecast: $(3, 4)$ bahar ki taraf point karta hai, jahan $- x^{2} - 2 y^{2}$ $T$ ko chota banata hai — expect karo ek negative rate (thanda ho raha hai), kuch $^{\circ} C$ per metre.

Gradient. $T_{x} = - 2 x, T_{y} = - 4 y$ , isliye $\nabla T = (- 2 x, - 4 y) = (- 4, - 4)$ $(2, 1)$ par. Yeh step kyun? Gradient har axis ke along temperature ki rate collect karta hai ( $^{\circ} C$ per metre mein) — kisi bhi heading ke liye raw material.
Unit-ize. $∥ (3, 4) ∥ = 9 + 16 = 5$ (nonzero), isliye $u = (3/5, 4/5)$ . Yeh step kyun? Question per metre pooch raha hai, isliye heading khud ek metre lambi honi chahiye — matlab length $1$ .
Dot. $D_{u} T = (- 4, - 4) \cdot (3/5, 4/5) = \frac{- 12 - 16}{5} = \frac{- 28}{5} = - 5.6$ . Yeh step kyun? Dotting dono cooling rates combine karta hai, is hisaab se weight deke ki cheenti ka path har axis ke along kitna lean karta hai.

Verify: Units: $\nabla T$ $^{\circ} C / m$ hai, $u$ dimensionless hai, isliye answer $^{\circ} C / m$ hai. Result $- 5.6^{\circ} C / m$ — negative (cooling), jaise forecast mein tha. Cheenti $(3, 4)$ ke along har metre chalne par $5. 6^{\circ}$ thandi hoti hai.

C9 — Exam twist: gradient recover karo

Worked example Ex 9 — dot product ulta karo

Ek differentiable $f (x, y)$ satisfy karta hai $D_{u} f = 1$ direction $u_{1} = (1, 0)$ mein aur $D_{u} f = 2$ direction $u_{2} = (0, 1)$ mein, dono same point $a$ par. $\nabla f (a)$ nikalo, phir $u_{3} = (3/5, 4/5)$ ke along directional derivative.

Forecast: axes ke along directional derivatives wohi partials hain, isliye $\nabla f$ simply $(1, 2)$ hona chahiye.

Partials read off karo. $D_{(1, 0)} f = \nabla f \cdot (1, 0) = f_{x} = 1$ aur $D_{(0, 1)} f = \nabla f \cdot (0, 1) = f_{y} = 2$ . Yeh step kyun? Gradient ko $(1, 0)$ se dot karne par uska pehla component select hota hai; $(0, 1)$ se dot karne par, doosra — isliye ye do special headings hume directly partials de deti hain.
Assemble karo. $\nabla f (a) = (1, 2)$ . Yeh step kyun? Gradient defined hai partials ke vector ke roop mein, isliye hum sirf step 1 ke numbers stack karte hain.
Use karo. $D_{u_{3}} f = (1, 2) \cdot (3/5, 4/5) = \frac{3 + 8}{5} = \frac{11}{5} = 2.2$ . Yeh step kyun? $\nabla f$ pata hone aur $f$ differentiable hone se, dot-product formula kisi bhi unit heading ke liye kaam karta hai.

Verify: $∥ u_{3} ∥ = 9/25 + 16/25 = 1$ ✓. Aur $2.2 < ∥\nabla f ∥ = 1^{2} + 2^{2} = 5 \approx 2.236$ ✓ — max se neeche, jaise hona chahiye. Yeh kaam karta hai kyunki $f$ differentiable hai, isliye linear (dot-product) formula har direction mein hold karta hai.

Wrapping up — har cell tick ho gayi

Recall

Recall Kaun sa cell kaun sa hai?

C3 mein bowl mein origin ki taraf chalne par $D_{u} f$ ka sign poochha jaata hai. Kaun sa sign, aur kyun? ::: Negative — origin (minimum) ki taraf jaate waqt height decrease hoti hai. C4 mein answer exactly $0$ kyun hai? ::: Direction $\nabla f$ ke perpendicular hai, yaani level curve ki tangent hai, isliye height momentarily constant hai. C7 mein gradient $0$ hai. General $u$ ke liye $D_{u} f$ kya hai? ::: $0$ har direction ke liye — zero vector ko kisi bhi cheez se dot karo toh $0$ aata hai. C9 mein do axis-direction derivatives se $\nabla f$ kaise recover karte hain? ::: Woh wohi partials hain: $D_{(1, 0)} f = f_{x}$ , $D_{(0, 1)} f = f_{y}$ , isliye $\nabla f = (f_{x}, f_{y})$ . $D_{u} f = ∥\nabla f ∥ cos θ$ mein $θ$ ka kya matlab hai? ::: Gradient $\nabla f$ aur heading $u$ ke beech ka angle. Hum "direction $0$ mein" directional derivative kyun nahi le sakte? ::: Zero vector ki koi heading nahi hoti aur $∥ 0 ∥ = 0$ unit-izing ko divide by zero bana deta hai — question undefined hai.