Gradient vector ∇f — definition, properties: Exercises

4.4.9 · D4 · Maths › Multivariable Calculus › Gradient vector ∇f — definition, properties

Reminders jinpe tum rely karoge (sab parent mein banaye gaye hain):

$\nabla f = (f_{x}, f_{y})$ — partial derivatives ka vector.
$D_{\overset{u}{^}} f = \nabla f \cdot \overset{u}{^} = ∣\nabla f ∣ cos θ$ — directional derivative, jahan $θ$ step direction $\overset{u}{^}$ aur $\nabla f$ ke beech ka angle hai.
$\overset{u}{^}$ ek unit vector hona chahiye (length $1$ ), warna answer scale ho jaata hai.
$\nabla f$ sabse steep uphill point karta hai; $- \nabla f$ sabse steep downhill; $\nabla f ⊥$ level curves.

Ek picture jo poore page ke liye dimaag mein rakhni hai:

Figure — Gradient vector ∇f — definition, properties

Coral arrow $\nabla f$ hai (sabse steep uphill). Ise $9 0^{\circ}$ ghumao aur tumhe mint arrow milta hai, level curve ka tangent jahan $f$ kabhi nahi badalta. Coral arrow ko flip karo aur tumhe sabse steep downhill milta hai. Neeche ke almost har exercise mein yahi teen moves hain.

Level 1 — Recognition

L1.1

$f (x, y) = 3 x - 5 y + 7$ ke liye, $\nabla f$ likho.

Recall Solution

Kya. Gradient partial slopes ki list hai $(f_{x}, f_{y})$ . Kyun. $f_{x}$ = slope jab hum $x$ ko wiggle karte hain aur $y$ ko freeze karte hain; $y$ ko constant treat karo. Constant $7$ aur $- 5 y$ term $\partial / \partial x$ ke under vanish ho jaate hain.

$f_{x} = 3$ , $f_{y} = - 5$ . $\nabla f = (3, - 5) .$ Note karo ki yeh har jagah same hai — ek linear function ka gradient constant hota hai (uski picture ek flat tilted plane hai).

L1.2

Inme se kaun sa $f (x, y) = x^{2} + y^{2}$ ka gradient hai: (a) scalar $2 x + 2 y$ , (b) vector $(2 x, 2 y)$ , (c) number $∣\nabla f ∣$ ?

Recall Solution

(b). Gradient ek vector hai — har input variable ke liye ek number. Choice (a) do alag slopes ko illegally ek scalar mein add karta hai; (c) gradient ki magnitude hai, ek length, gradient khud nahi.

L1.3

Sach ya jhooth: $\nabla f$ , level curve $f = c$ ka tangent hai.

Recall Solution

Jhooth. $\nabla f$ level curve ke perpendicular (normal) hai. Level curve ke saath saath chalna $f$ ko constant rakhta hai, toh wahan directional derivative $0$ hai; aur $0 = \nabla f \cdot \overset{u}{^}$ force karta hai ki tangent direction $\overset{u}{^}$ ke liye $\nabla f ⊥ \overset{u}{^}$ ho. (Yeh exactly upar ki picture mein coral-vs-mint relationship hai.)

Level 2 — Application

L2.1

$f (x, y) = x^{2} y$ . $(2, 3)$ par $\nabla f$ compute karo.

Recall Solution

$f_{x} = 2 x y$ ( $y$ freeze karo; $\frac{d}{d x} x^{2} = 2 x$ times the constant $y$ ).
$f_{y} = x^{2}$ ( $x$ freeze karo; $\frac{d}{d y} y = 1$ times the constant $x^{2}$ ). $(2, 3)$ par: $f_{x} = 2 (2) (3) = 12$ , $f_{y} = 2^{2} = 4$ . $\nabla f (2, 3) = (12, 4) .$

L2.2

Usi $f$ ke liye $(2, 3)$ par, maximum rate of change aur usse achieve karne wali direction nikalo.

Recall Solution

Max rate $= ∣\nabla f ∣ = 1 2^{2} + 4^{2} = 144 + 16 = 160 = 410 \approx 12.649$ . Direction = gradient ka apna unit vector: $\hat{d} = \frac{( 12 , 4 )}{4 10} = (\frac{3}{10}, \frac{1}{10}) .$ $∣\nabla f ∣$ kyun? Kyunki $D_{\overset{u}{^}} f = ∣\nabla f ∣ cos θ$ tab sabse bada hota hai jab $cos θ = 1$ , yaani $θ = 0$ ( $\nabla f$ ke saath seedha chalo); tab rate $∣\nabla f ∣$ ke barabar hoti hai. Neeche ki figure mein dikhaya gaya hai ki yeh "shadow" $\nabla f cos θ$ kaise shrink hoti hai jab $\overset{u}{^}$ gradient se door swing karta hai.

L2.3

$f (x, y) = x^{2} y$ , point $(2, 3)$ , direction $(2 + 3, 3 + 4) = (5, 7)$ ki taraf. $D_{\overset{u}{^}} f$ nikalo.

Recall Solution

Step 1 — direction banao. Displacement $= (5 - 2, 7 - 3) = (3, 4)$ , length $3^{2} + 4^{2} = 5$ . Step 2 — normalise karo (unit length honi chahiye): $\overset{u}{^} = (3/5, 4/5)$ . Step 3 — gradient ke saath dot karo (L2.1 se, $\nabla f = (12, 4)$ ): $D_{\overset{u}{^}} f = (12, 4) \cdot (3/5, 4/5) = \frac{36}{5} + \frac{16}{5} = \frac{52}{5} = 10.4.$

L2.4

$f (x, y, z) = x y z$ . $(1, 2, 3)$ par $\nabla f$ nikalo.

Recall Solution

Teen variables mein bhi same rule — gradient ko bas ek teesra slot milta hai.

$f_{x} = y z$ , $f_{y} = x z$ , $f_{z} = x y$ . $(1, 2, 3)$ par: $f_{x} = 6$ , $f_{y} = 3$ , $f_{z} = 2$ . $\nabla f (1, 2, 3) = (6, 3, 2) .$

Level 3 — Analysis

L3.1

$f (x, y) = x^{2} + y^{2}$ . $(3, 4)$ par dikhao ki $\nabla f$ level circle ke perpendicular hai, aur wahan circle ka ek unit tangent do.

Recall Solution

$\nabla f = (2 x, 2 y) = (6, 8)$ at $(3, 4)$ — yeh origin se radially outward point karta hai. Yeh page ki opening figure mein coral arrow hai: ek radius apne circle ke perpendicular hota hai, toh gradient normal hai. Ek tangent ko satisfy karna hoga $\nabla f \cdot \overset{u}{^} = 0$ . $(6, 8)$ ko $9 0^{\circ}$ rotate karo: $(- 8, 6)$ . Check: $(6) (- 8) + (8) (6) = - 48 + 48 = 0$ ✓ — yeh mint arrow hai. Unit tangent: $\frac{( - 8 , 6 )}{64 + 36} = \frac{( - 8 , 6 )}{10} = (- 4/5, 3/5)$ .

L3.2

$f = x^{2} + y^{2}$ ke liye $(3, 4)$ par, $f$ sabse fast decrease kis direction mein karta hai, aur kitni rate se?

Recall Solution

Sabse steep descent hai $- \nabla f = (- 6, - 8)$ — seedha origin ki taraf point karta hai (bowl mein downhill), coral arrow ka flip. Rate $= - ∣\nabla f ∣ = - 36 + 64 = - 100 = - 10$ . Negative kyun? $D_{\overset{u}{^}} f = ∣\nabla f ∣ cos θ$ sabse chhota $θ = π$ par hota hai ( $cos π = - 1$ ), jo $- ∣\nabla f ∣$ deta hai.

L3.3

$f (x, y) = x^{2} y$ , point $(2, 3)$ , $\nabla f = (12, 4)$ . Ek unit direction nikalo jis par $f$ momentarily unchanged rahe ( $D_{\overset{u}{^}} f = 0$ ).

Recall Solution

Humein chahiye $\overset{u}{^} ⊥ \nabla f$ , yaani $12 u_{1} + 4 u_{2} = 0 \Rightarrow u_{2} = - 3 u_{1}$ . $(1, - 3)$ lo, length $1 + 9 = 10$ , toh $\overset{u}{^} = (\frac{1}{10}, \frac{- 3}{10}) (ya uska negative) .$ Kaisa dikhta hai: yeh $\overset{u}{^}$ $(2, 3)$ se guzarne wale $f$ ke level curve ka tangent hai — us taraf chalo aur tumhari height nahi badlegi (phir se mint-arrow wala role).

L3.4

$f = x^{2} + y^{2}$ ke liye $(3, 4)$ par, $\nabla f$ se $6 0^{\circ}$ angle banaane wale $\overset{u}{^}$ ke liye $D_{\overset{u}{^}} f$ compute karo, aur check karo ki yeh dot-product form se match karta hai.

Recall Solution

Cosine form: $D_{\overset{u}{^}} f = ∣\nabla f ∣ cos 6 0^{\circ} = 10 \cdot \frac{1}{2} = 5.$ Yeh fast route hai: hume sirf magnitude aur angle chahiye, exact $\overset{u}{^}$ nahi. Koi bhi $\overset{u}{^}$ jo $(6, 8)$ ke saath $6 0^{\circ}$ banaye wahi value deta hai — geometry (s02 figure ki tarah $\overset{u}{^}$ par $\nabla f$ ka shadow) hi sab kuch hai.

Level 4 — Synthesis

L4.1

Point $p = (2, 3)$ par $z = f (x, y) = x^{2} y$ ka tangent plane nikalo.

Recall Solution

Pehle notation. Base point ke liye $p = (2, 3)$ aur nearby point ke liye $x = (x, y)$ likho, toh $x - p = (x - 2, y - 3)$ $p$ se door chhota step hai. Yeh formula kyun. $p$ ke paas ek smooth surface flat (ek plane) dikhti hai. Ek plane ki height step ke saath linearly badlti hai: $x$ mein $Δ x$ move karo aur height (slope in $x$ ) $\times Δ x$ badh jaati hai, yaani $f_{x} Δ x$ ; similarly $f_{y} Δ y$ . Dono contributions add karna precisely dot product $\nabla f (p) \cdot (x - p)$ hai. Known height $f (p)$ se shuru karo aur yeh linear rise add karo (dekho Tangent plane and linearisation): $z = f (p) + \nabla f (p) \cdot (x - p) .$ Plug in. $f (2, 3) = 2^{2} \cdot 3 = 12$ aur $\nabla f (2, 3) = (12, 4)$ (L2.1 se). $z = 12 + 12 (x - 2) + 4 (y - 3) = 12 + 12 x - 24 + 4 y - 12,$ $z = 12 x + 4 y - 24 .$ Sanity check at $p = (2, 3)$ : $12 (2) + 4 (3) - 24 = 24 + 12 - 24 = 12 = f (2, 3)$ ✓ — plane surface ko exactly base point par touch karta hai.

L4.2

Ek hiker hill $h (x, y) = 4 - x^{2} - y^{2}$ par $(1, 1)$ par khada hai. Kaun sa compass direction (unit vector) sabse steep uphill hai, aur kitna steep? Kis direction mein hiker descend karta?

Recall Solution

$\nabla h = (- 2 x, - 2 y) = (- 2, - 2)$ at $(1, 1)$ . Ascent ki direction = $\nabla h$ khud: unit form $\frac{( - 2 , - 2 )}{8} = (- \frac{1}{2}, - \frac{1}{2})$ — origin ki taraf ( $(0, 0)$ par summit). Steepness $= ∣\nabla h ∣ = 4 + 4 = 8 = 22 \approx 2.828$ . Descent = $- \nabla h = (2, 2)$ , unit $(\frac{1}{2}, \frac{1}{2})$ — summit se door. Yeh $- \nabla h$ step exactly wahi hai jo Gradient descent downhill jaane ke liye follow karta hai.

L4.3

$f (x, y) = e^{x} sin y$ . $(0, π /2)$ par $\nabla f$ nikalo aur wahan max rate of increase nikalo.

Recall Solution

$f_{x} = e^{x} sin y$ ( $sin y$ , $x$ ke saath constant hai; $\frac{d}{d x} e^{x} = e^{x}$ ).
$f_{y} = e^{x} cos y$ ( $e^{x}$ , $y$ ke saath constant hai; $\frac{d}{d y} sin y = cos y$ ). $(0, π /2)$ par: $e^{0} = 1$ , $sin (π /2) = 1$ , $cos (π /2) = 0$ . $\nabla f = (1 \cdot 1, 1 \cdot 0) = (1, 0) .$ Max rate $= ∣\nabla f ∣ = 1$ , direction $(1, 0)$ mein (the $+ x$ direction).

L4.4

Do functions: $f = x^{2}$ , $g = y^{3}$ . Verify karo ki $(1, 1)$ par $\nabla (2 f + 3 g) = 2\nabla f + 3\nabla g$ .

Recall Solution

Left side. $2 f + 3 g = 2 x^{2} + 3 y^{3}$ , toh $\nabla (2 f + 3 g) = (4 x, 9 y^{2}) = (4, 9)$ at $(1, 1)$ . Right side. $\nabla f = (2 x, 0) = (2, 0)$ ; $\nabla g = (0, 3 y^{2}) = (0, 3)$ . $2 (2, 0) + 3 (0, 3) = (4, 0) + (0, 9) = (4, 9)$ ✓. Kyun kaam karta hai: har partial derivative linear hai, aur linear operations ko ek vector mein stack karna unhe linear rakhta hai — yeh wohi linearity hai jo tangent plane ko linear parts ke liye exact banati hai.

Level 5 — Mastery

L5.1

Ellipse-like surface par, gradients use karke circle $x^{2} + y^{2} = 25$ par woh point(s) nikalo jahan $f (x, y) = x + 2 y$ sabse bada ho. (Lagrange multipliers ka ek glimpse.)

Recall Solution

Idea. Constraint circle ke saath maximum par, circle ke saath move karne se $f$ increase nahi ho sakta, toh $\nabla f$ circle ke normal $\nabla g$ ke parallel hona chahiye, jahan $g = x^{2} + y^{2}$ .

$\nabla f = (1, 2)$ (constant).
$\nabla g = (2 x, 2 y)$ . Set karo $\nabla f = λ \nabla g$ : $1 = 2 λ x$ , $2 = 2 λ y \Rightarrow y = 2 x$ . $x^{2} + y^{2} = 25$ mein substitute karo: $x^{2} + 4 x^{2} = 25 \Rightarrow x^{2} = 5 \Rightarrow x = \pm 5$ , $y = \pm 25$ . Maximum $+$ branch pick karta hai ( $\nabla f$ ke saath aligned): $(5, 25)$ , deta hai $f = 5 + 2 (25) = 55 \approx 11.180.$ Minimum $(- 5, - 25)$ par hai jahan $f = - 55$ .

L5.2

$f (x, y) = x^{2} y$ . Prove karo ki $(2, 3)$ par directional derivative kabhi bhi $410$ se zyada nahi ho sakta, chahe tum koi bhi unit direction lo.

Recall Solution

$D_{\overset{u}{^}} f = \nabla f \cdot \overset{u}{^} = ∣\nabla f ∣ ∣ \overset{u}{^} ∣ cos θ = ∣\nabla f ∣ cos θ$ (kyunki $∣ \overset{u}{^} ∣ = 1$ ). Kyunki $cos θ \leq 1$ har angle ke liye, $D_{\overset{u}{^}} f \leq ∣\nabla f ∣ = ∣ (12, 4) ∣ = 160 = 410$ . Equality sirf tab hoti hai jab $θ = 0$ , yaani $\overset{u}{^}$ $\nabla f$ ke saath point kare. Toh $410$ ek hard ceiling hai — max rate. Yeh disguise mein Cauchy–Schwarz bound hai.

L5.3

Temperature $T (x, y) = 100 - x^{2} - 2 y^{2}$ hai. $(1, 1)$ par ek bug sabse fast cool down karna chahta hai. Uski motion ki unit direction aur cooling rate do. Phir ek aisi direction nikalo jis par uska temperature constant rahe.

Recall Solution

$\nabla T = (- 2 x, - 4 y) = (- 2, - 4)$ at $(1, 1)$ . Sabse fast cool = steepest descent = $- \nabla T = (2, 4)$ . Unit: $\frac{( 2 , 4 )}{20} = (\frac{1}{5}, \frac{2}{5})$ . Cooling rate $= - ∣\nabla T ∣ = - 4 + 16 = - 20 = - 25 \approx - 4.472$ (temperature $25$ per unit length se drop karta hai). Constant-temperature direction: $\nabla T = (- 2, - 4)$ ke perpendicular. Solve karo $- 2 u_{1} - 4 u_{2} = 0 \Rightarrow u_{1} = - 2 u_{2}$ ; lo $(- 2, 1)$ , unit $(- \frac{2}{5}, \frac{1}{5})$ . Yeh bug ke through isotherm (level curve) ka tangent hai.

L5.4

$f (x, y) = ln (x^{2} + y^{2})$ . Dikhao ki $∣\nabla f ∣$ sirf origin se distance $r = x^{2} + y^{2}$ par depend karta hai, aur woh dependence state karo.

Recall Solution

$r^{2} = x^{2} + y^{2}$ lo. $\frac{d}{d x} ln (u) = \frac{u _{x}}{u}$ use karke:

$f_{x} = \frac{2 x}{x ^{2} + y ^{2}}$ , $f_{y} = \frac{2 y}{x ^{2} + y ^{2}}$ . $∣\nabla f ∣ = \frac{( 2 x ) ^{2} + ( 2 y ) ^{2}}{x ^{2} + y ^{2}} = \frac{2 x ^{2} + y ^{2}}{x ^{2} + y ^{2}} = \frac{2}{x ^{2} + y ^{2}} = \frac{2}{r} .$ Toh $∣\nabla f ∣ = 2/ r$ : steepness badh jaati hai jab tum origin ke paas aate ho aur door jaate waqt $0$ ho jaati hai — yeh level curves (circles) ke center ke paas crowded hone se match karta hai. ( $r = 0$ par undefined, jahan $ln$ blow up karta hai.)

Recall Self-test checklist

Kya 2D aur 3D mein $\nabla f$ compute kar sakte ho? ::: L1–L2, L4.3 Kya target point ko unit $\overset{u}{^}$ mein convert karke $D_{\overset{u}{^}} f$ nikal sakte ho? ::: L2.3 Kya steepest ascent/descent directions aur rates nikal sakte ho? ::: L3.2, L4.2, L5.3 Kya zero change ki directions (level set ka tangent) nikal sakte ho? ::: L3.3, L5.3 Kya $\nabla f$ se tangent plane bana sakte ho? ::: L4.1 Kya $D_{\overset{u}{^}} f$ bound kar sakte ho aur $\nabla f ∥ \nabla g$ use kar sakte ho? ::: L5.2, L5.1

Connections

Partial derivatives — har gradient component ek hai.
Directional derivative — poore mein use hone wala $\nabla f \cdot \overset{u}{^}$ engine.
Chain rule (multivariable) — linearity checks ke peeche.
Tangent plane and linearisation — L4.1.
Level curves and surfaces — L3.1, L3.3, L5.3.
Gradient descent — L4.2, L5.3 ( $- \nabla f$ ke saath move karna).
Lagrange multipliers — L5.1.