Gradient vector ∇f — definition, properties: Worked examples

4.4.9 · D3 · Maths › Multivariable Calculus › Gradient vector ∇f — definition, properties

Shuru karne se pehle, teen chhoti reminders taaki har symbol earned lage:

Scenario matrix

Har gradient problem in cells mein se ek mein hoti hai. Neeche har example us cell ke saath tagged hai jo woh cover karta hai.

Cell	Scenario class	Isme kya tricky hai	Example
A	Sab positive components, plain compute	kuch nahi — yeh warm-up hai	Ex 1
B	Mixed signs $(+, -)$ , direction downhill point karta hai	negative $D_{\overset{u}{^}} f$ ; $cos θ$ ka sign	Ex 2
C	Degenerate: $\nabla f = 0$ (ek flat/critical point)	har direction rate $0$ deta hai; steepest direction undefined	Ex 3
D	Non-circular level curve ke saath perpendicularity	tangent obvious nahi; $\nabla f \cdot (tangent) = 0$ use karna padega	Ex 4
E	Non-unit vector diya gaya hai (yeh trap hai)	pehle normalise karna zaroori	Ex 5
F	Real-world word problem (temperature/hill)	words ko $\nabla f$ mein translate karo, units	Ex 6
G	Limiting/extreme direction: fastest descent aur zero-change direction	teeno key angles ke signs $0, π, π /2$	Ex 7
H	Exam twist: backwards kaam karo — ek directional derivative diya, unknown dhundho	ek component ke liye solve karo	Ex 8

Har $f_{x}, f_{y}$ compute karne ke liye Partial derivatives dekho, aur dot-product rule ke liye Directional derivative dekho.

Worked examples

Worked example Ex 1 — Cell A: all-positive warm-up

$f (x, y) = 3 x^{2} + y^{3}$ . $(1, 2)$ par $\nabla f$ aur steepest-ascent rate nikalo.

Forecast: andaaza lagao — kya steepest rate $10$ se badi hogi ya chhoti?

$f_{x} = 6 x$ , $f_{y} = 3 y^{2}$ . Yeh step kyun? Partial derivative doosre variable ko freeze karti hai, isliye $3 x^{2}$ ek single-variable function ki tarah differentiate hoti hai aur $y^{3}$ constant treat hoti hai (uski $x$ -derivative $0$ hai).
$(1, 2)$ par: $\nabla f = (6 \cdot 1, 3 \cdot 2^{2}) = (6, 12)$ . Kyun? Bas point ko har slope mein plug karo.
Steepest rate $= ∣\nabla f ∣ = 6^{2} + 1 2^{2} = 180 = 65 \approx 13.42$ . Kyun? Property 2: max rate gradient ki length ke barabar hoti hai, aur length $= sum of squares$ .

Verify: dono components positive hain, isliye arrow up-and-right first quadrant mein point karta hai, aur $180 > 10$ jaisa expect tha. Climb ki direction $= (6, 12) / 180$ , ek unit vector: $(6^{2} + 1 2^{2}) /180 = 1$ . ✓

Worked example Ex 2 — Cell B: mixed signs, downhill step karna

$f (x, y) = x^{2} - y^{2}$ at $(2, 1)$ . Direction $\overset{u}{^} = (0, 1)$ mein step karo ( $y$ mein seedha upar). $D_{\overset{u}{^}} f$ nikalo aur batao uphill ho ya downhill.

Forecast: $y^{2}$ mein minus sign hai. $+ y$ mein move karne par height milegi ya khoegi?

$\nabla f = (2 x, - 2 y)$ , toh $(2, 1)$ par: $\nabla f = (4, - 2)$ . Minus kyun? $- y^{2}$ differentiate hoke $- 2 y$ deta hai; sign carry through hota hai.
$\overset{u}{^} = (0, 1)$ ki length $0^{2} + 1^{2} = 1$ hai — pehle se unit hai. Check kyun karo? $D_{\overset{u}{^}} f$ tabhi sahi "per-step" rate hai jab unit vector ho.
$D_{\overset{u}{^}} f = (4, - 2) \cdot (0, 1) = 4 \cdot 0 + (- 2) \cdot 1 = - 2$ . Kyun? Dot product matching pieces multiply karke add karta hai; yahan sirf $y$ -piece bachta hai.

Verify: result negative hai, matlab $+ y$ mein move karna tumhe downhill le jaata hai — yeh samajh aata hai kyunki $- y^{2}$ decrease hoti hai jab $∣ y ∣$ badhta hai. Aur $∣ D_{\overset{u}{^}} f ∣ = 2 \leq ∣\nabla f ∣ = 20 \approx 4.47$ , yeh "koi direction gradient ki length se nahi jeet sakti" rule maanti hai. ✓

Worked example Ex 3 — Cell C: degenerate, gradient zero hai

$f (x, y) = x^{2} + y^{2}$ at origin $(0, 0)$ . $\nabla f$ aur steepest-ascent direction nikalo.

Forecast: bowl ka bottom origin par hai. Ekdum bottom se "uphill" kaun si taraf hogi?

$\nabla f = (2 x, 2 y)$ , toh $(0, 0)$ par: $\nabla f = (0, 0) = 0$ . Kyun? Dono slopes zero ho jaate hain — surface low point par momentarily flat hai.
Steepest rate $= ∣ 0 ∣ = 0$ . Kyun? Zero vector ki length $0$ hoti hai.
Steepest direction undefined hai. Yeh step kyun matter karta hai? Har direction $D_{\overset{u}{^}} f = 0 \cdot \overset{u}{^} = 0$ deti hai; koi "most uphill" raasta nahi hai kyunki yahan har taraf equally flat hai. Yeh exactly ek critical point hai — Gradient descent aur Lagrange multipliers dono isi ko dhundhte hain.

Verify: koi bhi $\overset{u}{^} = (u_{1}, u_{2})$ try karo: $D_{\overset{u}{^}} f = 0 \cdot u_{1} + 0 \cdot u_{2} = 0$ sabke liye. Consistent — koi preferred uphill nahi. ✓

Worked example Ex 4 — Cell D: non-circular level curve ke perpendicular

$f (x, y) = x^{2} + 4 y^{2}$ . $(2, 1)$ par (jo $f = 8$ , ek ellipse par lie karta hai), dikhao ki $\nabla f$ level curve ke perpendicular hai.

Forecast: level curve ek squashed circle hai. Kya $\nabla f$ phir bhi exactly iske across point karega?

$\nabla f = (2 x, 8 y)$ , toh $(2, 1)$ par: $\nabla f = (4, 8)$ . $8$ kyun? $4 y^{2}$ differentiate hoke $8 y$ deta hai.
Ellipse par $(2, 1)$ par ek tangent dhundho. $x^{2} + 4 y^{2} = 8$ ko implicitly differentiate karo: $2 x + 8 y y^{'} = 0 \Rightarrow y^{'} = - \frac{2 x}{8 y} = - \frac{2 \cdot 2}{8 \cdot 1} = - \frac{1}{2}$ . Toh ek tangent direction hai $(1, - \frac{1}{2})$ , ya fraction clear karke $t = (2, - 1)$ . Implicit differentiation kyun? Curve koi function $y (x)$ nahi hai jise hum solve kar sakein, isliye equation ko as-is differentiate karte hain slope paane ke liye.
Gradient ko tangent se dot karo: $\nabla f \cdot t = (4, 8) \cdot (2, - 1) = 8 - 8 = 0$ . Kyun? Property 3 kehti hai $\nabla f ⊥$ level set; zero dot product exactly perpendicularity hai.

Verify: dot $0$ hai, isliye red arrow (gradient) ellipse se right angle par milta hai chahe curve stretched ho. Level curves and surfaces dekho. ✓

Worked example Ex 5 — Cell E: non-unit trap

$f (x, y) = x y$ at $(3, 2)$ . Point $(6, 6)$ ki taraf rate of change nikalo.

Forecast: $(6, 6)$ tak step vector $(3, 4)$ hai, length $5$ . Agar normalise skip karo, toh tumhara answer $5$ se scale ho jaayega — galti pakdo.

$\nabla f = (y, x)$ , toh $(3, 2)$ par: $\nabla f = (2, 3)$ . $(y, x)$ kyun? $\partial (x y) / \partial x = y$ ( $y$ constant treat karo), $\partial (x y) / \partial y = x$ .
Step vector $v = (6, 6) - (3, 2) = (3, 4)$ , aur $∣ v ∣ = 3^{2} + 4^{2} = 5$ . Kyun? Direction = destination minus start.
Normalise karo: $\overset{u}{^} = (3, 4) /5 = (0.6, 0.8)$ . Yeh step kyun? $D_{\overset{u}{^}} f$ tabhi genuine per-unit-length rate hai jab length- $1$ arrow ho (parent note ki mistake (a)).
$D_{\overset{u}{^}} f = (2, 3) \cdot (0.6, 0.8) = 1.2 + 2.4 = 3.6$ . Kyun? Matching pieces multiply karo, add karo.

Verify: galat (unnormalised) answer hoga $(2, 3) \cdot (3, 4) = 6 + 12 = 18 = 5 \times 3.6$ — exactly $5 \times$ zyada, jo confirm karta hai ki scale factor $∣ v ∣ = 5$ hai. ✓

Worked example Ex 6 — Cell F: real-world word problem

Ek metal plate ka temperature $T (x, y) = 100 - x^{2} - 3 y^{2}$ degrees hai, jahan $x, y$ metres mein hain. Ek cheenti $(1, 2)$ par baithi hai. (a) Ushe sabse tezi se garam hone ke liye kaun si taraf chalna chahiye, aur kitni tezi se (°C per metre)? (b) Same temperature rakhne ke liye kaun si taraf?

Forecast: plate sabse garam centre $(0, 0)$ par hai; cheenti off-centre hai. Kya "warmest way" centre ki taraf point karega ya door?

$\nabla T = (- 2 x, - 6 y)$ , $(1, 2)$ par: $\nabla T = (- 2, - 12)$ . Negatives kyun? Temperature badhne par girti hai, isliye slopes negative hain.
Fastest warming = $\nabla T = (- 2, - 12)$ ki direction; rate $= ∣\nabla T ∣ = (- 2)^{2} + (- 12)^{2} = 148 = 237 \approx 12.17$ °C/m. Kyun? Property 2 phir — steepest ascent gradient ke saath hai, rate uski length hai. Units: (°C)/(m) kyunki $T$ °C mein hai aur steps metres mein hain.
Constant-temperature direction: $\nabla T$ ke perpendicular. Components swap karo aur ek negate karo: $(- 12, 2)$ (ya $(12, - 2)$ ). Kyun? $θ = π /2$ wali property $D_{\overset{u}{^}} T = 0$ deti hai; $(a, b)$ ke perpendicular vector $(- b, a)$ hota hai.

Verify: warming direction $(- 2, - 12)$ hot centre ki taraf point karta hai — sensible hai. Perpendicular check: $(- 2, - 12) \cdot (- 12, 2) = 24 - 24 = 0$ . ✓

Worked example Ex 7 — Cell G: teeno limiting angles ek saath

$f (x, y) = 2 x + y$ (ek tilted plane) kisi bhi point par. Fastest-ascent, fastest-descent, aur zero-change directions rates ke saath do.

Forecast: ek plane ke liye gradient har jagah same hota hai. Teeno special rates kya hain?

$\nabla f = (2, 1)$ , constant hai. Constant kyun? $2 x + y$ ki partial derivatives constants $2$ aur $1$ hain.
$∣\nabla f ∣ = 2^{2} + 1^{2} = 5 \approx 2.236$ . Fastest ascent: direction $(2, 1) / 5$ , rate $+ 5$ ( $θ = 0$ , $cos θ = 1$ ). Kyun? $D_{\overset{u}{^}} f = ∣\nabla f ∣ cos θ$ tab max hota hai jab step $\nabla f$ ke saath align ho.
Fastest descent: direction $- (2, 1) / 5$ , rate $- 5$ ( $θ = π$ , $cos θ = - 1$ ). Kyun? Opposite direction sign flip karta hai; yahi arrow hai jise Gradient descent follow karta hai.
Zero change: koi bhi $\overset{u}{^} ⊥ (2, 1)$ , jaise $(1, - 2) / 5$ , rate $0$ ( $θ = π /2$ , $cos θ = 0$ ). Kyun? Gradient ke perpendicular rehna tumhe level line par rakhta hai.

Verify: zero direction check karo: $(2, 1) \cdot (1, - 2) / 5 = (2 - 2) / 5 = 0$ . Aur ascent rate: $(2, 1) \cdot (2, 1) / 5 = 5/ 5 = 5$ . ✓

Worked example Ex 8 — Cell H: exam twist, backwards kaam karo

Ek point $P$ par, $f_{x} = 3$ aur $f_{y} = a$ unknown hai. Tumhe bataya gaya hai ki direction $\overset{u}{^} = (\frac{3}{5}, \frac{4}{5})$ mein directional derivative $\frac{29}{5}$ hai. $a$ nikalo, phir $P$ par steepest-ascent rate nikalo.

Forecast: tum dot product ka output jaante ho aur missing input recover karna hai. Ek linear equation, ek unknown — solvable hai.

Rule likho: $D_{\overset{u}{^}} f = \nabla f \cdot \overset{u}{^} = 3 \cdot \frac{3}{5} + a \cdot \frac{4}{5} = \frac{9}{5} + \frac{4 a}{5}$ . Kyun? Same dot-product formula, bas $a$ symbolic hai.
Diye gaye value ke barabar karo: $\frac{9}{5} + \frac{4 a}{5} = \frac{29}{5} \Rightarrow \frac{4 a}{5} = \frac{20}{5} = 4 \Rightarrow a = 5$ . Kyun? $a$ ke liye single linear equation solve karo.
Ab $\nabla f = (3, 5)$ , toh steepest rate $= ∣\nabla f ∣ = 3^{2} + 5^{2} = 34 \approx 5.83$ . Kyun? Ek baar gradient pata chal jaaye, uski length max rate hai.

Verify: $a = 5$ wapas rakh kar check karo: $(3, 5) \cdot (\frac{3}{5}, \frac{4}{5}) = \frac{9}{5} + \frac{20}{5} = \frac{9}{5} + \frac{20}{5} = \frac{29}{5}$ . ✓ Diye gaye data se match karta hai.

Recall Quick self-test (andaaza lagaane ke baad reveal karo)

"Temperature map par coldest-approach direction dhundho" kaun se cell mein aayega? ::: Cell F/G — fastest descent, direction $- \nabla T$ . Agar $P$ par $\nabla f = (0, 0)$ ho, toh har $\overset{u}{^}$ ke liye $D_{\overset{u}{^}} f$ kya hoga? ::: Hamesha $0$ ; point critical hai. Kisi target point ki taraf ek direction diya ho, toh sabse PEHLE kya karna chahiye? ::: Step vector ko length $1$ tak normalise karo.

Connections

4.4.09 Gradient vector ∇f — definition, properties (Hinglish) — parent topic.
Partial derivatives — har example inhi ko compute karke shuru hota hai.
Directional derivative — dot-product rule jo throughout use hua hai (Ex 2, 5, 8).
Chain rule (multivariable) — us dot-product rule ka origin.
Level curves and surfaces — Ex 4 ki ellipse aur Ex 7 ki level lines.
Tangent plane and linearisation — Ex 7 ka plane.
Gradient descent — Ex 3 ka critical point aur Ex 7 ka descent direction.
Lagrange multipliers — Ex 4 jaisi perpendicularity use karta hai.