Worked examples — Gradient perpendicular to level curves - surfaces

4.4.11 · D3 · Maths › Multivariable Calculus › Gradient perpendicular to level curves - surfaces

Kuch bhi shuru karne se pehle, teen symbols samajhne zaroori hain taaki pehli line par hi koi na fasse.

Neeche sab kuch yeh hai: $\nabla f$ compute karo, tangent ke liye $90^{\circ }$ trick lagao, aur confirm karo ki dot product $0$ hai.

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Scenario matrix

Har row ek class ki situation hai. "Quadrant" yahan hamesha uss jagah ko refer karta hai jahan gradient arrow $\nabla f$ point karta hai (uska sign pattern), na ki jahan woh point baitha hai. Aakhri column us worked example ka naam deta hai jo uspar land karta hai.

Cell	Ise alag kya banata hai	Worst-case danger	Example
A $\nabla f$ Quadrant I mein (dono entries $>0$)	dono partials $>0$	koi nahi — warm-up	Ex 1
B $\nabla f$ mein sign mix hai (ek entry $<0$)	koi partial $<0$	galat tangent sign	Ex 2
C Axis / ek partial $=0$	gradient purely ek axis ke saath point karta hai	tangent ek coordinate axis hai	Ex 3
D Degenerate: $\nabla f=0$	gradient zero ho jaata hai	"KISKE perpendicular?" toot jaata hai	Ex 4
E 3D level surface	tangent plane, line nahi	point-normal form use karni hogi	Ex 5
F Limiting behaviour	curve infinity tak jaati hai	kya normal "outward" point karna jaari rakhta hai?	Ex 6
G Real-world word problem	contour map / temperature	units padhna, ascent ki direction	Ex 7
H Exam twist: graph vs level set	$z=f(x,y)$ ka normal, level curve ka nahi	classic trap	Ex 8

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Example 1 — Cell A: gradient Quadrant I mein (warm-up)

Forecast: padhne se pehle guess karo — kya gradient ellipse ke andar point karega ya bahar, aur roughly kis direction mein (up-right, up-left…)?

1. Gradient compute karo. $\nabla f=(2x,6y)=(4,12)$. Yeh step kyun? Gradient hamaara normal vector hai; jab tak yeh nahi milta kuch nahi hoga. ${\partial }_x(x^2+3y^2)=2x$, ${\partial }_y(x^2+3y^2)=6y$. Dono entries positive hain, toh $\nabla f$ Quadrant I mein hai — yeh Cell A hai.
2. Tangent direction nikalo. $90^{\circ }$ trick $(a,b)\mapsto (-b,a)$ $(4,12)$ par lagao: yeh deta hai $(-12,4)$ (ya scaled, $(-3,1)$). Yeh step kyun? Yeh mechanical tarika hai "level set ke saath" produce karne ka "uske aakaar" se.
3. Dot-product check. $(4,12)\cdot (-12,4)=-48+48=0$. Yeh step kyun? Zero dot product hi yahan perpendicular ki definition hai — yeh poora theorem ek line mein hai.

Verify: Dono partials $(2,2)$ par positive hain, toh $\nabla f=(4,12)$ up-and-to-the-right point karta hai, ellipse se bahar zyada $f$ ki taraf (steepest ascent).

Figure 1 — Blue ellipse $x^2+3y^2=16$. $(2,2)$ par orange gradient arrow $(4,12)$ curve se bahar nikalta hai; green tangent arrow curve ke saath rehta hai. Dono arrows ek clean right angle par milte hain (Cell A).

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Example 2 — Cell B: gradient mein sign mix

Forecast: $\nabla f$ ka ek component negative hoga. Kaun sa, aur kya gradient phir bhi increasing $f$ ki taraf point karta hai?

1. Gradient. $\nabla f=(2x,-2y)=(2,-4)$. Yeh step kyun? Dhyaan do ${\partial }_y(-y^2)=-2y$: minus sign exactly wahan hai jahan students ise drop karte hain. Gradient $(2,-4)$ mein ek positive aur ek negative entry hai — sign mix, toh yeh Cell B hai.
2. Direction ki sign sanity. $(1,2)$ par, $f=1-4=-3$ ✅ (hum sahi curve par hain). $+x$ mein move karna $f$ badhata hai; $+y$ mein move karna $f$ ghataata hai. Toh uphill arrow right aur neeche ki taraf jhukta hai — $(2,-4)$ se match karta hai. Yeh step kyun? Cell B ka danger flipped sign hai; physics ko formula ke against check karo.
3. Tangent + dot check. $90^{\circ }$ trick $(a,b)\mapsto (-b,a)$ $(2,-4)$ par lagao: $(4,2)$. Dot: $(2,-4)\cdot (4,2)=8-8=0$. ✅ Yeh step kyun? Sign mix ke bawajood perpendicularity confirm hoti hai.

Verify: $x^2-y^2=-3$ ka implicit slope $\frac{dy}{dx}=-\frac{f_x}{f_y}=-\frac{2}{-4}=\frac{1}{2}$ hai. Tangent $(4,2)$ ka slope $2/4=\frac{1}{2}$ hai. ✅ Match.

Figure 2 — Saddle $x^2-y^2=-3$ ki blue level curve. $(1,2)$ par orange gradient $(2,-4)$ right-and-down point karta hai (iska $y$-entry negative hai); green tangent $(4,2)$ curve ke saath chalta hai, uske perpendicular (Cell B).

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Example 3 — Cell C: ek partial zero hai (gradient ek axis ke saath)

Forecast: circle ke bilkul upar, tangent kis direction mein honi chahiye? Horizontally ya vertically?

1. Gradient. $\nabla f=(2x,2y)=(0,10)$. Yeh step kyun? $x$-component exactly $0$ hai — yeh "ek partial zero ho jaata hai" wala case hai. Gradient seedha upar $y$-axis ki taraf point karta hai.
2. Tangent. $90^{\circ }$ trick $(0,10)$ par lagao: $(-10,0)$, yaani direction $(1,0)$ (ya $(-1,0)$): purely horizontal direction. Yeh step kyun? Jab normal vertical hota hai, tangent horizontal hona chahiye — Cell C tangent ko ek coordinate axis par force karta hai.
3. Dot check. $(0,10)\cdot (1,0)=0+0=0$. ✅ Yeh step kyun? Is special case mein bhi theorem bilkul hold karta hai.

Verify: Radius $5$ ke circle ka top $(0,5)$ hai; iska tangent line horizontal hai ($y=5$), direction $(1,0)$ se match karta hai. ✅

Figure 3 — Blue circle $x^2+y^2=25$. Top point $(0,5)$ par orange gradient $(0,10)$ seedha upar point karta hai; green tangent arrows horizontally left aur right jaate hain, uske perpendicular (Cell C).

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Example 4 — Cell D: degenerate case $\nabla f=0$

Forecast: kya theorem yahan apply bhi hota hai? Guess karo "yes / no / undefined."

1. Gradient. $\nabla f=(2x,2y)=(0,0)$ origin par. Yeh step kyun? Gradient zero vector hai. Aur $0\cdot \mathbf{u}=0$ har $\mathbf{u}$ ke liye.
2. Interpret karo. Perpendicularity ka matlab hai "sirf wahi direction zero dot karta hai jo normal se ho." Lekin har direction $0$ ke saath zero dot karta hai. Toh "perpendicular direction" well-defined nahi hai. Yeh step kyun? Yeh theorem ki honest limit hai — iske liye $\nabla f\ne 0$ chahiye. Jahan $\nabla f=0$ hote hain woh critical points hain; level set ek single point tak pinch ho sakta hai (yahan $f=0$ sirf origin hai, koi curve nahi).
3. Yaad rakhne ka rule. Perpendicularity result jahan bhi $\nabla f\ne 0$ wahan hold karta hai. $\nabla f=0$ par koi unique normal nahi hota. Yeh step kyun? Cell D wahi hai jise har exam umeed karta hai ki tum bhool jaoge.

Verify: $\nabla f(0,0)=(0,0)$, toh iska magnitude $0$ hai; normal undefined hai.

Figure 4 — Dashed gray circles level curves $f=1,4,9$ hain. Level set $f=0$ origin par single red dot hai. Faded orange arrows har direction mein fan out karte hain: kyunki $\nabla f=0$, koi bhi single ek "the" perpendicular nahi hai (Cell D).

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Example 5 — Cell E: ek 3D level surface (tangent plane)

Forecast: normal ab teen components ka hoga. Guess karo ki $z$-component positive hoga ya negative.

1. Gradient. $\nabla f=(2x,2y,-1)=(2,2,-1)$. Yeh step kyun? 3D mein gradient phir bhi normal hai; ${\partial }_z(-z)=-1$ negative third entry deta hai.
2. Point-normal plane. $\Delta \mathbf{x}=(x,y,z)-(1,1,2)$ likho apne point se plane par kisi general point tak ke displacement ke liye (toh $\Delta x=x-1$, $\Delta y=y-1$, $\Delta z=z-2$). Plane woh saare displacements hain jo normal ke perpendicular hain, $\nabla f\cdot \Delta \mathbf{x}=0$: $2(x-1)+2(y-1)-1(z-2)=0\Rightarrow 2x+2y-z=2.$ Yeh step kyun? Ek plane "point se woh saare displacements hain jo normal ke perpendicular hain" — exactly $\nabla f\cdot \Delta \mathbf{x}=0$ (dekho Tangent Plane to a Surface).
3. Tangent-vector spot check. Ek curve $\mathbf{r}(t)$ (ek moving point, jaise upar define kiya) lo jo surface par rehti hai aur $(1,1,2)$ se guzarti hai: $x=1+t$, $y=1$, phir $z=(1+t{)}^2+1$, toh $t=0$ par iska velocity ${\mathbf{r}}^{\prime }(0)=(1,0,2)$ hai. Normal ke saath dot: $(2,2,-1)\cdot (1,0,2)=2+0-2=0$. ✅ Yeh step kyun? Confirm karta hai ki gradient ek actual curve ke perpendicular hai jo surface mein hai — parent ka Step-1 argument, concrete banaya gaya.

Verify: Plane $2x+2y-z=2$, $(1,1,2)$ se guzarti hai: $2+2-2=2$ ✅.

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Example 6 — Cell F: limiting behaviour (curve infinity tak)

Forecast: jaise curve far out $x$-axis ki taraf flatten hoti hai, normal kis taraf tilt karta hai?

1. Gradient formula. $\nabla f=(y,x)$. Yeh step kyun? ${\partial }_x(xy)=y$, ${\partial }_y(xy)=x$ — normal literally swapped coordinates hi hai.
2. Teen points evaluate karo.
   • $(2,3)$ par: $\nabla f=(3,2)$ — up-right, moderate.
   • $(6,1)$ par: $\nabla f=(1,6)$ — ab steeply upar.
   • $(100,0.06)$ par: $\nabla f=(0.06,100)$ — almost seedha upar. Yeh step kyun? $x$-axis ke saath far out, curve almost horizontal hai, toh iska normal almost vertical hona chahiye — aur hai bhi.
3. Limit mein perpendicular. Tangent slope of $y=6/x$ hai $-6/x^2\to 0^{-}$ (curve flatten hoti hai). Gradient $(y,x)$ as an arrow ka slope $x/y=x/(6/x)=x^2/6\to \infty$ hai (vertical). Unka product $=(-6/x^2)(x^2/6)=-1$ har $x$ ke liye. ✅ Yeh step kyun? Do lines perpendicular hoti hain exactly tab jab slopes ka product $-1$ ho; yahan yeh sab $x$ ke liye hold karta hai, limit mein bhi.

Verify: $(2,3)$ par: tangent slope $-6/4=-1.5$, gradient-arrow slope $x/y=2/3\approx 0.667$; product $=(-1.5)(2/3)=-1$ ✅.

Figure 5 — Blue hyperbola $xy=6$. Orange gradient arrows $(y,x)$ $(2,3)$ aur $(6,1)$ par draw kiye hain. Jaise point $x$-axis ke saath bahar jaata hai curve flatten hoti hai, toh normal vertical ki taraf tilt karta hai (Cell F).

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Example 7 — Cell G: real-world word problem (temperature map)

Forecast: guess karo ki "warm up fastest" origin ki taraf point karta hai ya door?

1. Current value. $T(3,2)=100-9-8=83^{\circ }$C. Isotherm hai $100-x^2-2y^2=83$, yaani $x^2+2y^2=17$ (ek ellipse). Yeh step kyun? Woh level curve establish karta hai jis par bug khada hai.
2. Gradient = steepest-ascent direction. $\nabla T=(-2x,-4y)=(-6,-8)$ °C/cm. Yeh step kyun? Gradient hamesha fastest increase ki taraf point karta hai — yahan zyada temperature ki taraf (dekho Directional Derivative aur Steepest Descent / Gradient Descent). Kyunki dono entries negative hain, "warmer" origin ki taraf hai (down-left), jo sense banata hai: $T$ center par peak karta hai.
3. Fastest warming ka rate. $\Vert \nabla T\Vert =\sqrt{(-6{)}^2+(-8{)}^2}=\sqrt{36+64}=10$ °C/cm. Yeh step kyun? Gradient ka magnitude unit distance per maximum rate of change hai — units $={}^{\circ }$C per cm.
4. Constant-temperature direction. $(-6,-8)$ par $90^{\circ }$ trick lagao: jaise $(8,-6)$ ya $(-8,6)$. Yeh step kyun? Isotherm ke saath chalna matlab zero change, yaani level curve ka tangent = gradient ke perpendicular. Yeh part (c) ka jawab deta hai.

Verify: Part (a): $T(3,2)=100-9-8=83$ ✅, isotherm $9+2(4)=17$ ✅. Part (b): $\Vert \nabla T\Vert =\sqrt{36+64}=10$ °C/cm, direction $(-6,-8)$ origin ki taraf ✅. Part (c): dot check $\nabla T\cdot (8,-6)=(-6)(8)+(-8)(-6)=-48+48=0$ ✅ — constant-temperature direction genuinely gradient ke perpendicular hai.

Figure 6 — Plate par blue isotherm $x^2+2y^2=17$. Bug ki jagah $(3,2)$ par orange gradient $(-6,-8)$ hot center ki taraf down-left point karta hai (steepest warming); green tangent $(8,-6)$ isotherm ke saath chalta hai — constant temperature ka direction, gradient ke perpendicular (Cell G).

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Example 8 — Cell H: exam twist (graph vs level set)

Forecast: true ya false? Aur agar graph ka normal teen components ka hai, toh teesra kya hai?

1. Do objects ko alag karo. $\nabla f=(6,8)$ ek 2D vector hai input plane mein; yeh level curve $x^2+y^2=25$ ka normal hai, 3D surface $z=f(x,y)$ ka nahi. Yeh step kyun? Trap yeh hai ki level curve (map mein) aur graph (3D mein bowl) ko confuse kiya jaaye.
2. Graph ka normal properly banao. Graph ko ek nayi function $g(x,y,z)=f(x,y)-z=x^2+y^2-z$ ke level surface ki tarah likho. Iska level set $g=0$ hai graph. Yeh step kyun? Ab hum same theorem use kar sakte hain: $\nabla g$, $g=0$ ka normal hai.
3. Compute karo. $\nabla g=(f_x,f_y,-1)=(6,8,-1)$. Yeh step kyun? ${\partial }_z(-z)=-1$ woh crucial third component deta hai jo student bhool gaya.

Verify: Graph ka tangent $(3,4,25)$ par: moving point $\mathbf{r}(t)=(3+t,4,(3+t{)}^2+16)$ use karo, jiska $t=0$ par velocity ${\mathbf{r}}^{\prime }(0)=(1,0,6)$ hai. $(6,8,-1)$ ke saath dot: $6+0-6=0$ ✅ — 3D normal genuinely surface ke perpendicular hai.

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Poora matrix ek nazar mein

Flowchart ko upar se neeche padho as woh decision jo tum har point par lete ho: pehle gradient compute karo, phir pucho kya woh zero hai, phir pucho kitne variables hain. Har leaf un cells ka naam leta hai jo woh handle karta hai.

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Recall Self-test (guess karne ke baad reveal karo)

$x^2+y^2=25$ par $(0,5)$ par tangent direction kya hai? ::: Horizontal, $(1,0)$ — kyunki $\nabla f=(0,10)$ vertical hai. Perpendicularity theorem $x^2+y^2$ ke origin par kyun fail karta hai? ::: $\nabla f=0$; har direction zero dot karta hai, toh koi unique normal nahi. Real-world: kisi point par temperature change ka max rate equal hai ::: $\Vert \nabla T\Vert$ ke, degrees per unit distance mein. Normal to the graph $z=x^2+y^2$ at $(3,4,25)$? ::: $(6,8,-1)$, $g=f-z$ se. $f=xy$ ke liye, gradient $x$-axis ke saath far out kyun almost vertical hota hai? ::: Curve flatten hoti hai ($\text{slope}\to 0$), toh iska normal $(y,x)$ vertical ki taraf tilt karta hai.

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Connections

• Directional Derivative — Ex 7: steepest ascent ka rate $\Vert \nabla T\Vert$ hai.
• Multivariable Chain Rule — Ex 5 ka curve-through-surface check parent derivation ko concrete banata hai.
• Tangent Plane to a Surface — Ex 5, Ex 8 point-normal form use karte hain.
• Lagrange Multipliers — Ex 4 ka $\nabla f=0$ exception tab matter karta hai jab constraint gradients vanish hon.
• Steepest Descent / Gradient Descent — Ex 7: $\nabla T$ ke against walk karna sabse jaldi cool karta hai.