Gradient as direction of steepest ascent: Worked examples

4.4.10 · D3 · Maths › Multivariable Calculus › Gradient as direction of steepest ascent

Ye page assume karta hai ki tumne parent note padh liya hai. Hum usme se sirf teen cheezein reuse karte hain, aur kuch nahi:

$\nabla f = (f_{x}, f_{y})$ , vector of partial derivatives (slopes in $x$ aur $y$ directions mein).
$D_{u} f = \nabla f \cdot u$ ek unit direction $u$ ke liye — dekho Directional Derivative aur Dot Product and Angle.
Steepest ascent $u^{⋆} = \nabla f /∥\nabla f ∥$ ke along, rate $∥\nabla f ∥$ pe.

Scenario matrix

Is topic ke har problem inka koi ek cell mein aata hai. Neeche har example ko uss cell ke saath tag kiya gaya hai jo wo kill karta hai.

#	Case class	Tricky kyu hai	Example
A	Dono partials positive	"clean" textbook case	Ex 1
B	$\nabla f$ mein mixed signs	direction quadrant II/IV mein jaati hai	Ex 2
C	Direction downhill di gayi	dot product negative aata hai	Ex 3
D	Zero gradient (flat point / peak)	steepest direction undefined hai	Ex 4
E	Level curve ke perpendicular	dot product exactly $0$ hona chahiye	Ex 5
F	Real-world word problem (units!)	temperature/height ko $f$ mein translate karo	Ex 6
G	Exam twist — "kaunsa raasta $f$ ko constant rakhta hai?"	$9 0^{\circ}$ direction dhundhni padegi	Ex 7
H	Limiting / degenerate ridge (partial = 0)	gradient ek axis ke along point karta hai	Ex 8

Case A — dono partials positive

Worked example Ex 1 (cell A) — steepest ascent, clean case

$f (x, y) = x^{2} + 3 y$ point $P = (2, 1)$ pe. Steepest ascent ki direction aur maximum rate nikalo.

Forecast: compute karne se pehle guess karo — arrow $x$ ki taraf zyada jhukega ya $y$ ki taraf? (At $x = 2$ the $x$ -slope $2 x = 4$ beats the constant $y$ -slope $3$ , so $x$ ki taraf jhukna chahiye.)

$f_{x} = 2 x, f_{y} = 3$ . Yeh step kyun? gradient partial derivatives ka vector hai; $f$ ko $x$ mein differentiate karo $y$ ko fixed rakhke, phir $y$ mein $x$ ko fixed rakhke.
$\nabla f (2, 1) = (4, 3)$ . Kyun? point ko har partial mein plug karo.
$∥\nabla f ∥ = 4^{2} + 3^{2} = 25 = 5$ . Kyun? increase ki max rate gradient ki length ke barabar hoti hai (parent ke "three big facts").
$u^{⋆} = \frac{( 4 , 3 )}{5} = (0.8, 0.6)$ . Kyun? steepest ascent gradient ko unit length pe normalise karke milta hai.

Verify: $∥ u^{⋆} ∥ = 0. 8^{2} + 0. 6^{2} = 0.64 + 0.36 = 1$ ✓, aur $\nabla f \cdot u^{⋆} = 4 (0.8) + 3 (0.6) = 3.2 + 1.8 = 5 = ∥\nabla f ∥$ ✓ — gradient ko apne unit direction se dot karo toh uski length wapas aati hai, exactly jaisa steepest ascent demand karta hai. Arrow $x$ ki taraf jhukta hai ( $0.8 > 0.6$ ), forecast se match karta hai.

Figure — Gradient as direction of steepest ascent

Figure mein amber gradient arrow $(4, 3)$ $P$ pe dikhaya gaya hai, uska cyan unit version, aur do white component slopes jinse woh bana hai.

Case B — gradient mein mixed signs

Worked example Ex 2 (cell B) — ek gradient jo quadrant IV mein point karta hai

$f (x, y) = x^{2} - y^{2}$ at $P = (3, 2)$ . Steepest ascent ki direction?

Forecast: $f$ badhta hai jab $x$ badhta hai lekin badhta hai jab $y$ ghatta hai ( $- y^{2}$ ki wajah se). Toh uphill arrow right aur neeche point karna chahiye.

$f_{x} = 2 x, f_{y} = - 2 y$ . Kyun? differentiate karo; $y^{2}$ pe minus sign carry through hota hai.
$\nabla f (3, 2) = (6, - 4)$ . Kyun? $P$ substitute karo. Negative $y$ -component matlab "chadhne ke liye $y$ mein neeche jao."
$∥\nabla f ∥ = 6^{2} + (- 4)^{2} = 36 + 16 = 52 = 213$ . Kyun? magnitude mein squares use hote hain, toh signs gayab ho jaate hain — length phir bhi well-defined hai.
$u^{⋆} = \frac{( 6 , - 4 )}{2 13} = \frac{( 3 , - 2 )}{13}$ . Kyun? normalise karo; 2 ka factor cancel ho jaata hai.

Verify: $∥ u^{⋆} ∥^{2} = \frac{9 + 4}{13} = 1$ ✓. Arrow quadrant IV mein point karta hai (right, neeche), forecast se match karta hai — ek common trap sign ignore karke up-right point karna hai.

Figure mein $P = (3, 2)$ dikhaya gaya hai amber gradient $(6, - 4)$ ke saath jo quadrant IV mein down-right jhuk raha hai, aur uska cyan unit version.

Case C — ek direction di gayi, aur wo downhill hai

Worked example Ex 3 (cell C) — negative directional derivative

$f (x, y) = x^{2} + 3 y$ at $P = (2, 1)$ ke saath (Ex 1 jaisa, toh $\nabla f = (4, 3)$ ), agar tum towards $Q = (- 2, - 2)$ chalte ho toh $f$ kitni tezi se change hoga?

Forecast: $Q$ , $P$ ke down-and-left mein hai, roughly uphill arrow $(4, 3)$ ke opposite. Ek negative rate expect karo (tum neeche ja rahe ho).

Raw direction $v = Q - P = (- 2 - 2, - 2 - 1) = (- 4, - 3)$ . Kyun? tip minus tail travel ki direction deta hai.
$∥ v ∥ = 16 + 9 = 5$ , toh $u = \frac{( - 4 , - 3 )}{5} = (- 0.8, - 0.6)$ . Kyun? tumhe normalise karna hi padega, warna rate $∥ v ∥$ se scale ho jaata hai aur "per unit distance" nahi rehta (parent ki headline mistake).
$D_{u} f = \nabla f \cdot u = 4 (- 0.8) + 3 (- 0.6) = - 3.2 - 1.8 = - 5$ . Kyun? $D_{u} f = \nabla f \cdot u$ apply karo.

Verify: $u = - u^{⋆}$ Ex 1 se, yaani exactly steepest-descent direction, toh hume $- ∥\nabla f ∥ = - 5$ milna chahiye ✓. Ek negative directional derivative ka matlab sirf yeh hai ki " $f$ step karne pe ghatta hai" — sign information hai, error nahi.

Figure mein $P$ pe amber uphill gradient dikhaya gaya hai aur cyan walk-direction $Q$ ki taraf opposite mein point karta hai — unke beech ka $18 0^{\circ}$ hi wajah hai ki rate $- 5$ hai.

Case D — zero gradient (degenerate!)

Worked example Ex 4 (cell D) — har direction equally steep hai

$f (x, y) = x^{2} + y^{2}$ origin $P = (0, 0)$ pe. Steepest ascent kaunsi taraf hai?

Forecast: origin ek bowl ka bottom hai. Har direction equally uphill jaati hai — toh kya koi unique steepest way hai? (Nahi.)

$f_{x} = 2 x, f_{y} = 2 y \Rightarrow \nabla f (0, 0) = (0, 0)$ . Kyun? dono partials bowl ke centre pe vanish ho jaate hain.
$∥\nabla f ∥ = 0$ . Kyun? increase ki max rate first order pe zero hai — surface bilkul bottom pe flat hai.
$u^{⋆} = \frac{( 0 , 0 )}{0}$ form nahi ho sakta. Kyun? zero vector ko normalise nahi kar sakte, toh formula koi single preferred direction nahi deta.

Verify: koi bhi unit direction check karo, jaise $u = (1, 0)$ : $D_{u} f = \nabla f \cdot u = 0 (1) + 0 (0) = 0$ . $(0, 1)$ , $(\frac{1}{2}, \frac{1}{2})$ , kuch bhi repeat karo — hamesha $0$ milega ✓.

The nuance: "undefined" ek statement hai formula $\nabla f /∥\nabla f ∥$ ke baare mein (division by zero). Geometrically honest reading yeh hai ki "har direction equally steep hai" — first-order rate sab mein $0$ hai, toh koi direction jeet nahi sakti aur koi haar nahi sakti. Yeh ek critical point hai; ise report karo: " $\nabla f = 0$ , all directions tie at rate $0$ ."

Case E — level curve ke perpendicular

Worked example Ex 5 (cell E) —

$9 0^{\circ}$ fact, figure se confirm kiya $f (x, y) = x y$ at $P = (2, 3)$ ; usme se guzarne wali level curve hai $x y = 6$ . Dikhao ki $\nabla f$ us curve ke perpendicular hai.

Forecast: parent note claim karta hai ki gradient level curves ke $⊥$ hota hai, toh $\nabla f$ aur tangent ka dot product exactly $0$ hona chahiye.

$\nabla f = (y, x) = (3, 2)$ at $P$ . Kyun? $\partial_{x} (x y) = y$ , $\partial_{y} (x y) = x$ .
$x y = 6$ ka tangent: implicitly differentiate karo, $y + x y^{'} = 0 \Rightarrow y^{'} = - \frac{3}{2}$ . Kyun? ek level curve ke along $f$ constant hota hai, toh uska total change zero hai; $y^{'}$ curve ka slope hai. Ek tangent vector apne components ko $(run, rise) = (1, y^{'}) = (1, - \frac{3}{2})$ read karta hai. Fraction clear karne ke liye hum 2 se scale karte hain (ek direction ko positive number se scale karne se direction nahi badalti), jo tidy integer form $(2, - 3)$ deta hai.
$\nabla f \cdot (2, - 3) = 3 (2) + 2 (- 3) = 6 - 6 = 0$ . Kyun? agar dot product $0$ hai, toh unke beech ka angle $9 0^{\circ}$ hai.

Verify: $cos θ = \frac{\nabla f \cdot tangent}{∥\nabla f ∥ ∥ tangent ∥} = \frac{0}{13 \cdot 13} = 0 \Rightarrow θ = 9 0^{\circ}$ ✓. Tangent ke along directional derivative $0$ hai (tum $x y = 6$ pe rehte ho), toh $f$ change nahi hota — exactly isliye uphill arrow curve ko right angle pe chodna chahiye.

Figure mein hyperbola $x y = 6$ dikhaya gaya hai, $P$ pe uska cyan tangent, aur amber gradient exactly $9 0^{\circ}$ pe us se nikalta hua.

Case F — real-world word problem (units ka dhyan rakho!)

Worked example Ex 6 (cell F) — ek hiker temperature field pe

Ek metal plate pe, temperature hai $T (x, y) = 20 + 4 x - 3 y$ degrees Celsius, jahan $x, y$ metres mein hain. Ek bug $(5, 5)$ pe baitha hai. (a) Use garam hone ke liye kaunsi taraf crawl karna chahiye? (b) Kis rate pe (°C per metre)? (c) Wo instead east crawl karta hai, direction $(1, 0)$ — uska temperature kitni tezi se change hota hai?

Forecast: $T$ $x$ ke saath badhta hai aur $y$ ke saath ghatta hai, toh "sabse garam raasta" east-and-south point karega. Rate per metre $4^{2} + 3^{2} = 5$ ke aas paas honi chahiye.

$T_{x} = 4, T_{y} = - 3$ , toh $\nabla T = (4, - 3)$ har jagah (linear field). Kyun? differentiate karo; field planar hai, toh gradient constant hai. Units: $^{\circ} C / m$ .
Sabse garam direction $u^{⋆} = \frac{( 4 , - 3 )}{16 + 9} = \frac{( 4 , - 3 )}{5} = (0.8, - 0.6)$ . Kyun? gradient normalise karo — $T$ ka steepest ascent.
Max rate $= ∥\nabla T ∥ = 5^{\circ} C/m$ . Kyun? gradient ka magnitude increase ki fastest possible rate hai.
East crawl karna: $D_{(1, 0)} T = \nabla T \cdot (1, 0) = 4 (1) + (- 3) (0) = 4^{\circ} C/m$ . Kyun? gradient ko di gayi (already unit) direction se dot karo.

Verify: $u^{⋆}$ ek unit vector hai ( $0. 8^{2} + 0. 6^{2} = 1$ ) ✓. East $4 < 5$ deta hai ✓ — gradient direction ko koi beat nahi kar sakta, aur units throughout $^{\circ} C per metre$ rehti hain, confirm karta hai ki humne sahi normalise kiya (ek non-unit direction nonsense units deta).

Figure mein bug $(5, 5)$ pe dikhaya gaya hai, amber warmest-way arrow $(4, - 3)$ east-south point karta hua, aur cyan "east" arrow jiska slower rate $4 < 5$ uski shorter projection se padha ja sakta hai.

Case G — exam twist: "kaunsa raasta level rehta hai?"

Worked example Ex 7 (cell G) — zero change ki direction

$f (x, y) = 3 x + 4 y$ at $P = (1, 1)$ . Ek unit direction dhundo jiske along $f$ nahi badalti, aur confirm karo ki yeh valid answer hai.

Forecast: "koi change nahi" matlab $\nabla f$ ke perpendicular. Do aisi directions honi chahiye (level line ke along ek doosre ke opposite).

$\nabla f = (3, 4)$ . Kyun? linear field ke partials uske coefficients hain.
$(3, 4)$ ke perpendicular vector hai $(4, - 3)$ (swap karo aur ek negate karo). Kyun? $(3) (4) + (4) (- 3) = 0$ , toh yeh $9 0^{\circ}$ angle banata hai — flat direction.
Normalise karo: $u = \frac{( 4 , - 3 )}{16 + 9} = \frac{( 4 , - 3 )}{5} = (0.8, - 0.6)$ . Uska opposite $(- 0.8, 0.6)$ bhi equally kaam karta hai. Kyun? direction unit length honi chahiye; dono signs level curve ke along hain.

Verify: $D_{u} f = \nabla f \cdot u = 3 (0.8) + 4 (- 0.6) = 2.4 - 2.4 = 0$ ✓. Zero rate matlab tum plane ki ek level line ke along chalte ho, toh $f$ genuinely constant hai — twist sirf $9 0^{\circ}$ fact ulta padha gaya hai.

Figure mein amber gradient $(3, 4)$ aur do cyan level directions $\pm (0.8, - 0.6)$ dikhaye gaye hain jo ise $9 0^{\circ}$ pe cross karte hain — kisi bhi taraf chalo aur $f$ constant rehta hai.

Case H — degenerate ridge (ek partial zero hai)

Worked example Ex 8 (cell H) — gradient ek axis ke along

$f (x, y) = x^{2} + cos y$ at $P = (1, 0)$ . Steepest ascent direction aur rate?

Forecast: $cos y$ $y = 0$ pe apne peak pe hai, toh wahan uska slope $0$ hai. Uphill arrow purely $x$ ke along point karna chahiye.

$f_{x} = 2 x, f_{y} = - sin y$ . Kyun? $\frac{d}{d y} cos y = - sin y$ .
At $P = (1, 0)$ : $\nabla f = (2, - sin 0) = (2, 0)$ . Kyun? $sin 0 = 0$ , toh $y$ -slope vanish ho jaata hai — $y$ -direction mein chalti ek ridge.
$∥\nabla f ∥ = 2^{2} + 0^{2} = 2$ . Kyun? zero component ke saath magnitude sirf surviving component hai.
$u^{⋆} = \frac{( 2 , 0 )}{2} = (1, 0)$ . Kyun? normalise karo; arrow purely $+ x$ ke along point karta hai, forecast confirm karta hai.

Verify: $D_{(1, 0)} f = \nabla f \cdot (1, 0) = 2$ = $∥\nabla f ∥$ ✓, aur perpendicular direction $(0, 1)$ deta hai $D_{(0, 1)} f = 2 (0) + 0 (1) = 0$ ✓ — ridge ke along chalne se $f$ momentarily flat rehta hai, exactly jaisa ek zero partial predict karta hai.

Figure mein $P = (1, 0)$ dikhaya gaya hai amber gradient $(2, 0)$ ke saath jo purely $+ x$ ke along point karta hai aur cyan ridge direction $(0, 1)$ jiske along $f$ flat rehta hai.

Recall Kaun sa cell kaun sa tha?

A positive-clean ::: Ex 1 B mixed signs ::: Ex 2 C downhill (negative $D_{u} f$ ) ::: Ex 3 D zero gradient (all directions tie) ::: Ex 4 E perpendicular to level curve ::: Ex 5 F word problem with units ::: Ex 6 G "stay level" twist ::: Ex 7 H degenerate ridge (a partial $= 0$ ) ::: Ex 8

Connections

Gradient as direction of steepest ascent (parent)
Directional Derivative
Partial Derivatives
Dot Product and Angle
Level Curves and Contour Maps
Gradient Descent (Machine Learning)
Tangent Plane and Linearization