Tangent planes and linear approximations to surfaces: Worked examples

4.4.5 · D3 · Maths › Multivariable Calculus › Tangent planes and linear approximations to surfaces

Yeh ek formula kahan se aata hai

Is page ki har cheez ek single equation par tikhi hai. Pehle hum ise use karenge, pehle ise banayenge, taaki baad mein kuch bhi mystery na rahe.

Step 1 — 1-D idea se shuru karo. Ek variable mein tangent line hai $y \approx f (a) + f^{'} (a) (x - a)$ : height, plus slope times step. Hum 2-D twin chahte hain.

Step 2 — base point se guzarne wala sabse general non-vertical plane likho. $z = f (a, b) + A (x - a) + B (y - b)$ Yeh form kyun? $x = a, y = b$ set karne par $z = f (a, b)$ force hota hai, toh yeh automatically base point se guzarta hai. Unknowns $A, B$ plane ke $x$ - aur $y$ -directions mein slopes hain — ab hum inhe surface se match karne ke liye choose karte hain.

Step 3 — x-slice match karo. $y = b$ freeze karo. Plane collapse hoke $z = f (a, b) + A (x - a)$ ban jaata hai, slope $A$ ki ek line. Surface ke x-slice ki slope $f_{x} (a, b)$ hai. Plane ko surface se hug karwane ke liye, $A = f_{x} (a, b)$ .

Step 4 — y-slice match karo. Usi tarah $x = a$ freeze karo: $B = f_{y} (a, b)$ .

Figure — Tangent planes and linear approximations to surfaces

Ab kyunki hum jaante hain formula kyun kaam karta hai, baaki page cases exhaust karne mein spend karte hain.

Scenario matrix

Har tangent-plane problem inhi cells mein se ek mein aata hai. Niche har example ek cell fill karta hai.

#	Cell (kya alag hai)	Example
A	Dono slopes positive — plain paraboloid	Ex 1
B	Mixed sign slopes (ek $+$ , ek $-$ )	Ex 2
B $^{'}$	Dono slopes negative — four-quadrant sign story complete karna	Ex 2b
C	Ek zero slope — ek direction mein flat, doosre mein tilted	Ex 3
D	Degenerate surface: $f$ pehle se hi ek plane hai (approximation exact hai)	Ex 4
E	Breakdown case: partials exist karte hain lekin surface differentiable NAHI hai (koi tangent plane nahi)	Ex 5
F	Real-world word problem — measurement error ke liye differentials	Ex 6
G	Exam twist: implicit level surface $F (x, y, z) = 0$ diya hai, normal vector use karo	Ex 7
H	Limiting behaviour: error distance $^{2}$ ki tarah badhta hai — trust quantify karna	Ex 8

Example 1 — Cell A: dono slopes positive

Worked example Paraboloid

$f (x, y) = x^{2} + y^{2}$ at $(1, 2)$ , $f (1.05, 2.1)$ estimate karo Forecast: Dono partials $2 x, 2 y$ at $(1, 2)$ positive hain, toh $+ x$ aur $+ y$ mein step karne se dono estimate ko $f (1, 2) = 5$ se upar uthana chahiye. Guess: $5$ se thoda upar.

Yahan base point $(a, b) = (1, 2)$ hai.

Step 1 — height. $f (1, 2) = 1 + 4 = 5$ . Yeh step kyun? Plane ko surface se base point par touch karna chahiye; woh height constant term hai.

Step 2 — slopes. $f_{x} = 2 x \Rightarrow f_{x} (1, 2) = 2$ ; $f_{y} = 2 y \Rightarrow f_{y} (1, 2) = 4$ . Yeh step kyun? Yeh do tilts hain jo plane ko match karni hain.

Step 3 — steps. $x - a = 1.05 - 1 = 0.05$ ; $y - b = 2.1 - 2 = 0.1$ . Yeh step kyun? Formula har slope ko base point se offset se multiply karta hai, $x$ ya $y$ se nahi.

Step 4 — assemble karo. $L = 5 + 2 (0.05) + 4 (0.1) = 5 + 0.1 + 0.4 = 5.5$ Yeh step kyun? Hum simply flat plane par chal rahe hain: height se shuru karo, phir har slope ko uska matching step se multiply karke add karo. Woh sum hi $(1.05, 2.1)$ ke upar plane ki height hai — surface ka hamara estimate.

Verify: True value $f (1.05, 2.1) = 1.1025 + 4.41 = 5.5125$ . Error $= 0.0125$ , tiny — aur $5$ se upar jaise forecast tha. ✓

Figure s02 flat red patch ko $(1, 2, 5)$ par bowl se touch karte, dono taraf upar tilting karte dikhata hai.

Example 2 — Cell B: mixed-sign slopes

Worked example Saddle

$f (x, y) = x^{2} - y^{2}$ at $(2, 1)$ , $f (2.1, 0.9)$ estimate karo Forecast: $f_{x} = 2 x > 0$ lekin $f_{y} = - 2 y < 0$ . $+ x$ step karne se utha, $- y$ step karne se bhi utha (negative slope times negative step). Guess: clearly $f (2, 1) = 3$ se upar.

Base point $(a, b) = (2, 1)$ .

Step 1 — height. $f (2, 1) = 4 - 1 = 3$ . Yeh step kyun? Constant term = base point par surface height.

Step 2 — slopes. $f_{x} = 2 x \Rightarrow 4$ ; $f_{y} = - 2 y \Rightarrow - 2$ . Yeh step kyun? Yeh ek saddle hai: $x$ mein upar curve karta hai, $y$ mein neeche, toh dono slopes ke signs opposite hain. Yahan sign galat karna classic error hai.

Step 3 — steps. $x - 2 = 0.1$ ; $y - 1 = - 0.1$ . Yeh step kyun? Pehle jaisi hi offset rule — lekin dhyan do $y$ -step negative hai kyunki hum $y = 1$ se $0.9$ pe move kiye. Step ka sign utna hi important hai jitna slope ka sign.

Step 4 — assemble karo. $L = 3 + 4 (0.1) + (- 2) (- 0.1) = 3 + 0.4 + 0.2 = 3.6$ Yeh step kyun? Height plus slope-times-step, do baar. Negative slope times negative step ek positive contribution deta hai — exactly isiliey estimate $3$ se upar land karta hai, forecast se match karta hai.

Verify: True $f (2.1, 0.9) = 4.41 - 0.81 = 3.6$ . Error $= 0$ do decimals tak — $3$ se upar jaise forecast tha. ✓

Figure s03 saddle aur uska tilted tangent plane dikhata hai — red plane surface ko cut karta hai, sirf upar nahi tika.

Example 2b — Cell B $^{'}$ : dono slopes negative

Worked example

$f (x, y) = 10 - x^{2} - y^{2}$ (downward bowl) at $(1, 1)$ , $f (1.1, 1.2)$ estimate karo Forecast: $f_{x} = - 2 x < 0$ aur $f_{y} = - 2 y < 0$ — dono slopes negative. $+ x$ aur $+ y$ step karne se (dono positive steps) estimate $f (1, 1) = 8$ se neeche aana chahiye. Yeh four-quadrant sign story complete karta hai: ab hum $(+, +)$ , $(+, -)$ , aur $(-, -)$ dekh chuke hain.

Base point $(a, b) = (1, 1)$ .

Step 1 — height. $f (1, 1) = 10 - 1 - 1 = 8$ . Yeh step kyun? Plane dome ko base point par touch karta hai; yeh height constant term hai.

Step 2 — slopes. $f_{x} = - 2 x \Rightarrow - 2$ ; $f_{y} = - 2 y \Rightarrow - 2$ . Yeh step kyun? Yeh ek ulta paraboloid hai, toh dono directions mein neeche slope karta hai — dono partials negative hain.

Step 3 — steps. $x - 1 = 0.1$ ; $y - 1 = 0.2$ . Yeh step kyun? Dono steps positive hain (hum bade $x$ aur $y$ par gaye); negative slopes estimate ko neeche kheenchenge.

Step 4 — assemble karo. $L = 8 + (- 2) (0.1) + (- 2) (0.2) = 8 - 0.2 - 0.4 = 7.4$ Yeh step kyun? Height plus slope-times-step, do baar. Do negative contributions estimate ko $8$ se neeche kheenchti hain, exactly jaisi downward dome demand karti hai.

Verify: True $f (1.1, 1.2) = 10 - 1.21 - 1.44 = 7.35$ . Error $= 0.05$ , small — aur $8$ se neeche jaise forecast tha. ✓

Example 3 — Cell C: ek zero slope

Worked example

$f (x, y) = sin x + y^{2}$ at $(\frac{π}{2}, 1)$ , $f (\frac{π}{2} + 0.1, 1.1)$ estimate karo Forecast: $f_{x} = cos x$ , aur $cos \frac{π}{2} = 0$ . Sine wave ke top par surface $x$ -direction mein momentarily flat hai, toh $x$ -step (almost) kuch contribute nahi karta. Guess: change mostly $y$ se aayegi.

Base point $(a, b) = (\frac{π}{2}, 1)$ .

Step 1 — height. $f (\frac{π}{2}, 1) = sin \frac{π}{2} + 1 = 1 + 1 = 2$ . Yeh step kyun? Constant term surface ki height hai base point par — woh level jahan se dono slope contributions measure hote hain.

Step 2 — slopes. $f_{x} = cos x \Rightarrow cos \frac{π}{2} = 0$ ; $f_{y} = 2 y \Rightarrow 2$ . Yeh step kyun? Ek zero partial legal aur important hai: iska matlab hai tangent plane us axis ke along level hai. Zero ek slope hai, missing nahi.

Step 3 — steps. $x - a = 0.1$ ; $y - b = 0.1$ . Yeh step kyun? Base point se offsets, har input ke liye ek. Bhale hi $x$ -slope $0$ ho, hum phir bhi honestly uska step record karte hain — formula khud usse zero se multiply kar dega.

Step 4 — assemble karo. $L = 2 + 0 (0.1) + 2 (0.1) = 2 + 0 + 0.2 = 2.2$ Yeh step kyun? Height plus slope-times-step, do baar. Zero $x$ -slope $x$ -contribution ko kill kar deta hai, toh poora change $y$ -term se aata hai — flat-in- $x$ picture numerical ho gayi.

Verify: True $f = sin (1.6708) + 1.21 = 0.99500 + 1.21 = 2.20500$ . Tiny $0.005$ gap exactly woh curvature hai jo flat $x$ -slope ne ignore ki. Forecast confirm: almost sab change $y$ se aaya. ✓

Example 4 — Cell D: degenerate surface (pehle se hi plane)

Worked example

$f (x, y) = 3 x - 2 y + 7$ at $(4, 5)$ , $f (4.3, 4.6)$ estimate karo Forecast: Surface hai hi flat plane. Ek plane ka tangent plane woh plane khud hi hota hai — toh "approximation" exact honi chahiye, error zero, chahe kitna bhi step karo.

Base point $(a, b) = (4, 5)$ .

Step 1 — height. $f (4, 5) = 12 - 10 + 7 = 9$ . Yeh step kyun? Hamesha ki tarah — constant term base point par surface value hai; flat surface iss rule ko nahi badlti.

Step 2 — slopes. $f_{x} = 3$ , $f_{y} = - 2$ — har jagah constant. Yeh step kyun? Linear function ke liye partials point par depend nahi karte; plane kabhi curve nahi karta.

Step 3 — steps. $x - 4 = 0.3$ ; $y - 5 = - 0.4$ . Yeh step kyun? Hamesha ki tarah base point se offsets; yahan $y$ -step negative hai kyunki hum $y = 5$ se $4.6$ par gire.

Step 4 — assemble karo. $L = 9 + 3 (0.3) + (- 2) (- 0.4) = 9 + 0.9 + 0.8 = 10.7$ Yeh step kyun? Height plus slope-times-step, do baar. Kyunki true surface mein koi curvature nahi, yeh "plane par chalna" surface ko exactly reproduce karta hai, approximately nahi.

Verify: True $f (4.3, 4.6) = 12.9 - 9.2 + 7 = 10.7$ . Error bilkul $0$ — woh degenerate case jahan linear approximation approximation hai hi nahi. ✓

Example 5 — Cell E: breakdown (koi tangent plane nahi)

Worked example Kya

$f (x, y) = \frac{x y}{x ^{2} + y ^{2}}$ (jahan $f (0, 0) = 0$ ) ka origin par koi tangent plane hai? Forecast: Parent ki warning kehti hai "partials exist karna kaafi nahi." Dekhte hain yeh fail kyun hota hai.

Base point $(a, b) = (0, 0)$ .

Step 1 — origin par partials compute karo. $x$ -axis ke along ( $y = 0$ ): $f (x, 0) = 0$ sabhi $x$ ke liye, toh $f_{x} (0, 0) = 0$ . $y$ -axis ke along ( $x = 0$ ): $f (0, y) = 0$ , toh $f_{y} (0, 0) = 0$ . Yeh step kyun? Dono partials exist karte hain aur $0$ ke barabar hain.

Step 2 — candidate plane $L$ likho. $f (0, 0) = 0$ , $f_{x} (0, 0) = 0$ , $f_{y} (0, 0) = 0$ apne derived formula mein daalne par: $L (x, y) = 0 + 0 \cdot (x - 0) + 0 \cdot (y - 0) = 0.$ Yeh step kyun? Yeh wahi plane hai jo formula produce kar sakta hai — flat sheet $z = 0$ . Differentiability test ab poochta hai: kya yeh candidate actually surface ko hug karta hai?

Step 3 — diagonal $y = x$ test karo. $f (x, x) = \frac{x \cdot x}{x ^{2} + x ^{2}} = \frac{x ^{2}}{2 x ^{2}} = \frac{1}{2}$ Yeh step kyun? $y = x$ ke along function $\frac{1}{2}$ par atka hua hai, origin ke kitna bhi paas — lekin candidate plane $L$ wahan $z = 0$ kehta hai. Woh base point ke arbitrarily close $\frac{1}{2}$ se disagree karte hain.

Step 4 — differentiability limit check karo. Error over distance hai $\frac{f - L}{x ^{2} + y ^{2}} = \frac{1/2 - 0}{2 ∣ x ∣} \to \infty$ jab $x \to 0$ $y = x$ ke along. Yeh step kyun? Differentiability demand karta hai ki yeh ratio $\to 0$ ho. Yahan yeh blow up karta hai. Toh $f$ at $(0, 0)$ differentiable nahi hai.

Verify: $f$ origin par continuous bhi nahi hai (axes par value $0$ , diagonal par $\frac{1}{2}$ ), toh differentiable nahi ho sakta, toh koi tangent plane exist nahi karta — bhale hi dono partials $0$ hoon. ✓

Figure s04 do paths ko origin par alag heights par milte dikhata hai — surface ko yahan flatten nahi kiya ja sakta.

Example 6 — Cell F: real-world word problem

Worked example Ek closed metal box ki length

$x = 50$ cm, width $y = 30$ cm, height $h = 20$ cm hai, har ek $0.2$ cm tak measurable. Computed volume $V = x y h$ mein maximum error estimate karo. Forecast: Error ke teen sources, sab add up honge. Kyunki har side large hai, $0.2$ cm ki bhi slip ek chunky volume error mein multiply hoti hai. Guess: kuch sau cm³.

Step 1 — partials (sensitivities). $V_{x} = y h = 30 \cdot 20 = 600$ , $V_{y} = x h = 50 \cdot 20 = 1000$ , $V_{h} = x y = 50 \cdot 30 = 1500$ . Yeh step kyun? Har partial kehta hai "us dimension mein error ke har cm par $V$ kitne cm³ change karta hai." Yahan gradient use karna exactly 3 inputs mein tangent-plane idea hai.

Step 2 — differential. $d V = V_{x} d x + V_{y} d y + V_{h} d h$ Yeh step kyun? $d V$ change $Δ V$ ka flat-plane prediction hai; valid kyunki $0.2$ cm har side ke compared mein small hai.

Step 3 — worst case (sab errors same sign). $d x = d y = d h = 0.2$ lo: $d V = 600 (0.2) + 1000 (0.2) + 1500 (0.2) = 120 + 200 + 300 = 620 cm^{3}$ Yeh step kyun? Differential slope-times-step terms ka ek sum hai — wahi tangent-plane recipe. Yeh sabse bada tab hota hai jab har step ek hi direction mein push kare (sab $+ 0.2$ ), toh sab errors positive lena worst-case (maximum) volume error deta hai jo humse poocha gaya tha. Mixed signs partly cancel hoke ise shrink kar denge.

Verify (units + sanity): $V = 50 \cdot 30 \cdot 20 = 30000$ cm³, aur $620/30000 \approx 2.07%$ . Har partial ki units cm² (ek area) hain, cm se multiply hone par cm³ milte hain — sahi. Percentage rough rule $\frac{d V}{V} = \frac{d x}{x} + \frac{d y}{y} + \frac{d h}{h} = 0.004 + 0.0067 + 0.01 = 0.0207$ se match karta hai. ✓

Example 7 — Cell G: exam twist, implicit level surface

Kabhi kabhi $z$ ko cleanly isolate nahi kiya ja sakta. Tab hum gradient aur normal vector use karte hain. Pehle general tool, phir example.

Worked example Ellipsoid

$x^{2} + 2 y^{2} + 3 z^{2} = 20$ ka tangent plane at $(3, - 1, 3)$ Forecast: $z = f (x, y)$ ke liye cleanly solve karna mushkil hai. Iske bajaye ise $F = 0$ treat karo aur normal $n = \nabla F$ use karo, phir $n \cdot (X - X_{0}) = 0$ apply karo.

Step 1 — $F = 0$ ki tarah likho. $F (x, y, z) = x^{2} + 2 y^{2} + 3 z^{2} - 20$ . Yeh step kyun? Sab kuch ek side par move karna ise level surface banata hai, toh $\nabla F$ $z$ isolate kiye bina normal deta hai.

Step 2 — gradient (normal). $\nabla F = ⟨ 2 x, 4 y, 6 z ⟩ (3, - 1, 3) n = ⟨ 6, - 4, 63 ⟩ .$ Yeh step kyun? Yeh vector us point par surface ke perpendicular hai — woh $n$ jo hamari construction ko chahiye.

Step 3 — check karo ki point actually surface par hai. $9 + 2 (1) + 3 (3) = 9 + 2 + 9 = 20$ ✓. Yeh step kyun? Ek tangent plane meaningless hai agar point surface par nahi — yeh ek favourite exam trap hai.

Step 4 — $n \cdot (X - X_{0}) = 0$ apply karo aur simplify karo. $6 (x - 3) - 4 (y - (- 1)) + 63 (z - 3) = 0$ matlab $6 (x - 3) - 4 (y + 1) + 63 (z - 3) = 0$ . Har piece expand karo: $6 x - 18 - 4 y - 4 + 63 z - 18 = 0$ , toh $6 x - 4 y + 63 z = 40 .$ Yeh step kyun? Yeh dot-product-equals-zero plane likha gaya hai — "normal ke perpendicular, base point se guzarta hua." Kuch assume nahi kiya; yeh upar ke callout ki general construction hai.

Verify: Point $6 x - 4 y + 63 z$ mein daalo: $18 + 4 + 63 \cdot 3 = 18 + 4 + 18 = 40$ . ✓ Aur coefficients $⟨ 6, - 4, 63 ⟩$ $\nabla F$ se match karte hain, toh plane sahi oriented hai. ✓

Example 8 — Cell H: kitna trust kar sakte ho? (error $\sim$ distance²)

Worked example Ex 1 se

$f (x, y) = x^{2} + y^{2}$ at $(1, 2)$ ke liye, error $y$ ke along $0.1, 0.2, 0.4$ amounts se step karne par kaisa behave karta hai? Forecast: Parent ne kaha "error distance ke square ki tarah badhta hai." Step double karne se error roughly chaar guna hona chahiye. Dekho.

Step 1 — exact error formula. $L = 2 x + 4 y - 5$ (Ex 1 se) aur $f = x^{2} + y^{2}$ ke saath, $x = 1$ freeze karo: $E = f - L = (1 + y^{2}) - (2 + 4 y - 5) = y^{2} - 4 y + 4 = (y - 2)^{2}$ Yeh step kyun? Error ko closed form mein likhna hume distance $^{2}$ law ko directly dekhne deta hai iske bajaye sirf trust karein: error literally $y$ -step $s = y - 2$ ka square hai.

Step 2 — tabulate karo. $s = y - 2$ step ho:

$s = 0.1$ : $E = (0.1)^{2} = 0.01$
$s = 0.2$ : $E = (0.2)^{2} = 0.04$
$s = 0.4$ : $E = (0.4)^{2} = 0.16$ Yeh step kyun? $s$ ka har doubling $E$ ko $4$ se multiply karta hai — promised quadratic growth, exactly.

Verify: $0.04/0.01 = 4$ aur $0.16/0.04 = 4$ . Error literally $s^{2}$ hai; linear approximation sirf small $s$ ke liye trustworthy hai. ✓

Figure s05 true parabola ko uski tangent line ke saath plot karta hai aur barhta hua gap shade karta hai.

Recall Matrix ke across quick self-test

Kisi bhi distance par zero error kaun si cell mein hai? ::: Cell D — surface pehle se hi ek plane hai (zero curvature). Dono partials $0$ hain lekin koi tangent plane nahi — kaun si cell aur kyun? ::: Cell E — partials sirf axes probe karte hain; diagonal ke along $f$ jump karta hai, toh differentiability limit blow up karta hai. Zero partial ka matlab hai surface... ::: us ek axis ke along us point par level (flat) hai — Cell C. Dono slopes negative aur positive steps estimate ko... ::: base height se neeche bhejte hain (downward dome) — Cell B $^{'}$ . Level surface $F = 0$ ke liye, tangent plane perpendicular hai... ::: gradient $\nabla F$ ke — Cell G. Base point se apni distance double karne par error roughly... ::: chaar guna hoti hai (error $\sim$ distance $^{2}$ ) — Cell H.

Connections

Parent topic — woh derivation jinhe yeh examples exercise karte hain.
Partial derivatives — har example $f_{x}, f_{y}$ compute karne se shuru hota hai.
The gradient vector — Ex 7 $\nabla F$ ko normal ki tarah use karta hai.
Differentiability of multivariable functions — Ex 5 breakdown case hai.
Tangent line and linear approximation (single variable) — woh 1-D idea jise Ex 8 ka error law mirror karta hai.
The chain rule (multivariable) — surfaces mein $\cdot$ aur $sin$ differentiate karne ke liye use hota hai.
Directional derivatives — Ex 7 ka "normal ke perpendicular" ek dot-product condition hai.

#flashcards/maths

Where does

L (x, y) = f (a, b) + f_{x} (x - a) + f_{y} (y - b)

come from?

General plane

z = f (a, b) + A (x - a) + B (y - b)

ko surface ke do axis slices se match karo,

A = f_{x} (a, b)

B = f_{y} (a, b)

force karke.

Tangent plane estimate of

x^{2} + y^{2}

(1, 2)

for

(1.05, 2.1)

L = 5.5

(true

5.5125

Saddle

x^{2} - y^{2}

(2, 1)

ke liye, do partial slopes kya hain?

f_{x} = 4

f_{y} = - 2

10 - x^{2} - y^{2}

(1, 1)

ke liye

(1.1, 1.2)

ka estimate?

L = 7.4

(true

7.35

) — dono slopes negative.

f = \frac{x y}{x ^{2} + y ^{2}}

ka origin par koi tangent plane kyun nahi hai?

Dono partials

0

hain lekin

y = x

ke along value

\frac{1}{2} \neq = 0

hai; error/distance ratio

\to \infty

, toh

f

differentiable nahi hai.

V = x y h

(

50, 30, 20

) ka volume error

\pm 0.2

each ke saath?

d V = 620

cm³ (lagbhag

2.07%

General implicit-surface tangent plane at

X_{0}

n \cdot (X - X_{0}) = 0

jahan

n = \nabla F

; yahan

6 x - 4 y + 63 z = 40

x^{2} + y^{2}

ka error

x = 1

ke along step

s = y - 2

ke function ke roop mein kaisa badhta hai?

E = s^{2}

— step mein exactly quadratic.