Tangent planes and linear approximations to surfaces: Exercises

4.4.5 · D4 · Maths › Multivariable Calculus › Tangent planes and linear approximations to surfaces

Level 1 — Recognition

Kya tum formula padh ke usme se sahi numbers nikal sakte ho?

L1.1

Given $f (x, y) = 3 x - 2 y + 7$ . $f (2, 1)$ , $f_{x} (2, 1)$ , $f_{y} (2, 1)$ , aur normal vector $n$ likho.

Recall Solution L1.1

$f (2, 1) = 3 (2) - 2 (1) + 7 = 6 - 2 + 7 = 11$ .
$f_{x} = 3$ ( $x$ ka coefficient), to $f_{x} (2, 1) = 3$ .
$f_{y} = - 2$ ( $y$ ka coefficient), to $f_{y} (2, 1) = - 2$ .
Normal $n = ⟨ f_{x}, f_{y}, - 1 ⟩ = ⟨ 3, - 2, - 1 ⟩$ .

Humne kya kiya: kyunki $f$ pehle se hi linear hai, iske partials bas constant coefficients hain. Ye kaisa dikhta hai: surface apna khud ka tangent plane hai — ek flat sheet, $x$ mein $3$ aur $y$ mein $- 2$ tilt hai.

L1.2

Ek surface ke liye $(a, b) = (0, 3)$ par $f (a, b) = 4$ , $f_{x} (a, b) = - 1$ , $f_{y} (a, b) = 5$ hai. Tangent plane equation likho.

Recall Solution L1.2

Seedha $L$ mein plug karo: $z = 4 + (- 1) (x - 0) + 5 (y - 3) = 4 - x + 5 y - 15 = - x + 5 y - 11.$ $- 3$ kyun aaya: y-step hai $(y - b) = (y - 3)$ , sirf $y$ nahi. Base point $b = 3$ hai.

Level 2 — Application

Plane compute karo aur usse ek value approximate karne ke liye use karo.

L2.1

$f (x, y) = x^{2} y$ ka $(2, 1)$ par tangent plane nikalo.

Recall Solution L2.1

Step 1 (height): $f (2, 1) = 2^{2} \cdot 1 = 4$ . Step 2 (slopes): $f_{x} = 2 x y \Rightarrow f_{x} (2, 1) = 2 \cdot 2 \cdot 1 = 4$ ; $f_{y} = x^{2} \Rightarrow f_{y} (2, 1) = 4$ . Step 3 (assemble): $z = 4 + 4 (x - 2) + 4 (y - 1) = 4 x + 4 y - 8.$

L2.2

L2.1 ke plane ka use karke $f (1.9, 1.05) = (1.9)^{2} (1.05)$ estimate karo.

Recall Solution L2.2

Base se steps: $d x = 1.9 - 2 = - 0.1$ , $d y = 1.05 - 1 = 0.05$ . $L = 4 + 4 (- 0.1) + 4 (0.05) = 4 - 0.4 + 0.2 = 3.8.$ True value $= 3.61 \times 1.05 = 3.7905$ . Error $\approx 0.0095$ — chhota, kyunki dono steps chhote hain. ✓

L2.3

$f (x, y) = e^{x} cos y$ ka $(0, 0)$ par tangent plane nikalo aur usse $e^{0.1} cos (0.05)$ approximate karne ke liye use karo.

Recall Solution L2.3

Step 1: $f (0, 0) = e^{0} cos 0 = 1$ . Step 2: $f_{x} = e^{x} cos y \Rightarrow f_{x} (0, 0) = 1$ . $f_{y} = - e^{x} sin y \Rightarrow f_{y} (0, 0) = - 1 \cdot 0 = 0$ . Step 3: $z = 1 + 1 (x - 0) + 0 (y - 0) = 1 + x$ . Approximate: $d x = 0.1$ , $d y = 0.05$ : $L = 1 + 0.1 = 1.1$ . True value $= e^{0.1} cos (0.05) \approx 1.10517 \times 0.99875 \approx 1.1038$ . Error $\approx 0.0038$ . ✓ $f_{y}$ kyun gayab hua: $cos y$ apne peak $y = 0$ par flat hai — $y$ mein move karne se height first order mein nahi badlti, to plane wahan aage-peeche level hai.

Level 3 — Analysis

Geometry interpret karo, differentials use karo, chained functions handle karo.

L3.1

$f (x, y) = x^{2} + y^{2}$ ke liye $(3, 4)$ par tangent plane nikalo, phir verify karo ki normal vector $n = ⟨ f_{x}, f_{y}, - 1 ⟩$ dono cross-section tangent directions $⟨ 1, 0, f_{x} ⟩$ aur $⟨ 0, 1, f_{y} ⟩$ ke genuinely perpendicular hai.

Recall Solution L3.1

Slopes: $f_{x} = \frac{x}{x ^{2} + y ^{2}} = \frac{3}{5}$ , $f_{y} = \frac{4}{5}$ (parent Example 2 dekho). Plane: $z = 5 + \frac{3}{5} (x - 3) + \frac{4}{5} (y - 4) .$ Normal $n = ⟨ \frac{3}{5}, \frac{4}{5}, - 1 ⟩$ . Perpendicularity check (dot product = 0 matlab right angle): $n \cdot ⟨ 1, 0, \frac{3}{5} ⟩ = \frac{3}{5} + 0 + (- 1) \frac{3}{5} = 0. ✓$ $n \cdot ⟨ 0, 1, \frac{4}{5} ⟩ = 0 + \frac{4}{5} + (- 1) \frac{4}{5} = 0. ✓$ Matlab kya hai: normal dono tangent lines ke right angle par hai, to ye poore plane ke perpendicular hai jise wo span karte hain. Yahi reason hai ki The gradient vector wala view kaam karta hai.

L3.2

Ek cylinder ka radius $r = 5$ cm aur height $h = 12$ cm hai, dono $0.05$ cm tak measure ho sakte hain. Volume $V = π r^{2} h$ . Maximum error $d V$ estimate karo.

Recall Solution L3.2

Partials: $V_{r} = 2 π r h$ , $V_{h} = π r^{2}$ . $(5, 12)$ par: $V_{r} = 2 π (5) (12) = 120 π$ , $V_{h} = π (25) = 25 π$ . Differential (worst case, dono errors positive): $d V = ∣ V_{r} ∣ d r + ∣ V_{h} ∣ d h = 120 π (0.05) + 25 π (0.05) = 6 π + 1.25 π = 7.25 π \approx 22.78 cm^{3} .$ Absolute values kyun: error bound karne ke liye hum assume karte hain ki dono contributions ek saath add hoti hain, cancel nahi hoti.

L3.3

Plane $z = 2 x + 4 y - 5$ , $f = x^{2} + y^{2}$ ka $(1, 2)$ par tangent hai (parent Example 1). Partials recompute kiye bina explain karo ki $f (1.5, 2.5)$ ka linear estimate $f (1.1, 1.9)$ se worse kyun hai, aur roughly quantify karo.

Recall Solution L3.3

Is paraboloid ke liye error $E = f - L$ exactly $(x - 1)^{2} + (y - 2)^{2}$ hai (algebra neeche), yaani base point se squared distance.

$(1.1, 1.9)$ par: distance $^{2} = 0. 1^{2} + 0. 1^{2} = 0.02$ .
$(1.5, 2.5)$ par: distance $^{2} = 0. 5^{2} + 0. 5^{2} = 0.50$ . Ratio $= 0.50/0.02 = 25$ : door wale point ka error 25× zyada hai. Algebra: $f - L = (x^{2} + y^{2}) - (5 + 2 (x - 1) + 4 (y - 2)) = x^{2} - 2 x + 1 + y^{2} - 4 y + 4 = (x - 1)^{2} + (y - 2)^{2}$ . ✓ Ye "error distance ke square ki tarah badhta hai" ka concrete roop hai.

Level 4 — Synthesis

Ideas combine karo: unknowns solve karo, backwards kaam karo, related surfaces cross karo.

L4.1

Kisi $f (x, y)$ ka $(1, - 1)$ par tangent plane $z = 2 + 3 (x - 1) - 4 (y + 1)$ hai. $f (1.02, - 0.97)$ estimate karo aur $f (1, - 1)$ , $f_{x} (1, - 1)$ , $f_{y} (1, - 1)$ batao.

Recall Solution L4.1

Plane ko formula $L = f (a, b) + f_{x} (x - a) + f_{y} (y - b)$ ki tarah $(a, b) = (1, - 1)$ ke saath padhein: $f (1, - 1) = 2$ , $f_{x} (1, - 1) = 3$ , $f_{y} (1, - 1) = - 4$ (note $y - b = y - (- 1) = y + 1$ ). Steps: $d x = 0.02$ , $d y = - 0.97 - (- 1) = 0.03$ . $f (1.02, - 0.97) \approx 2 + 3 (0.02) - 4 (0.03) = 2 + 0.06 - 0.12 = 1.94.$

L4.2

Paraboloid $z = x^{2} + y^{2}$ par wo point nikalo jahan tangent plane, plane $z = 6 x - 4 y + 1$ ke parallel ho.

Recall Solution L4.2

Idea: do non-vertical planes tab parallel hote hain jab unke $x$ - aur $y$ -slopes match karein. Target plane ka $x$ -slope $6$ , $y$ -slope $- 4$ hai. Surface slopes equal set karo: $f_{x} = 2 x = 6 \Rightarrow x = 3, f_{y} = 2 y = - 4 \Rightarrow y = - 2.$ Height: $z = 3^{2} + (- 2)^{2} = 9 + 4 = 13$ . Point: $(3, - 2, 13)$ . Kaisa dikhta hai: paraboloid axis se jitna door jaao utna steep tilta hai; demanded tilt wala unique spot $(3, - 2)$ hai.

L4.3

Dikhao ki sphere piece $f (x, y) = 9 - x^{2} - y^{2}$ (upper hemisphere, radius 3) ke har point par, surface normal $n = ⟨ f_{x}, f_{y}, - 1 ⟩$ radius direction mein point karta hai. $(1, 2)$ par verify karo.

Recall Solution L4.3

Partials (square root par chain rule): $f_{x} = \frac{- x}{9 - x ^{2} - y ^{2}}, f_{y} = \frac{- y}{9 - x ^{2} - y ^{2}} .$ $(1, 2)$ par: $9 - 1 - 4 = 4 = 2$ , to $z = 2$ aur $f_{x} = - \frac{1}{2}, f_{y} = - 1$ . $n = ⟨ - \frac{1}{2}, - 1, - 1 ⟩$ . $- 2$ se multiply karo: $⟨ 1, 2, 2 ⟩$ — exactly position vector $(x, y, z) = (1, 2, 2)$ centre se. Kyun: sphere par outward radius hi normal hota hai, to ye confirm karta hai ki tangent-plane machinery classical geometry reproduce karti hai. Dekho The gradient vector.

Level 5 — Mastery

Prove karo, generalise karo, aur degenerate cases face karo.

L5.1

Consider $f (x, y) = ⎩ ⎨ ⎧ \frac{x y}{x ^{2} + y ^{2}}, 0, (x, y) \neq = (0, 0) (x, y) = (0, 0) .$ (a) Dikhao ki dono partials $f_{x} (0, 0)$ aur $f_{y} (0, 0)$ exist karte hain aur $0$ ke barabar hain, to candidate tangent plane $z = 0$ hai. (b) Dikhao ki $f$ $(0, 0)$ par differentiable nahi hai, to koi true tangent plane exist nahi karta — diagonal $y = x$ test karke.

Recall Solution L5.1

(a) Origin par Partials — har axis ke saath limit definition use karo. $y = 0$ ke saath: $f (x, 0) = \frac{x \cdot 0}{x ^{2} + 0} = 0$ sabhi $x$ ke liye. To $f_{x} (0, 0) = lim_{h \to 0} \frac{f ( h , 0 ) - f ( 0 , 0 )}{h} = lim_{h \to 0} \frac{0 - 0}{h} = 0.$ Symmetry se $f_{y} (0, 0) = 0$ . Candidate plane: $L = 0 + 0 \cdot x + 0 \cdot y = 0$ .

(b) Differentiability fail hoti hai. Precise test (parent note): kya $\frac{f ( x , y ) - L ( x , y )}{x ^{2} + y ^{2}} \to 0 ?$ $y = x$ ke saath approach karo: $f (x, x) = \frac{x \cdot x}{x ^{2} + x ^{2}} = \frac{x ^{2}}{2 x ^{2}} = \frac{1}{2}$ (constant!). Aur $L = 0$ . Distance $= x^{2} + x^{2} = ∣ x ∣ 2$ . To ratio hai $\frac{\frac{1}{2} - 0}{∣ x ∣ 2} = \frac{1}{2 2 ∣ x ∣} \to \infty \neq = 0.$ Error distance se tez nahi shrink hoti — balki blow up hoti hai. To koi tangent plane exist nahi karta chahe dono partials $0$ hoon. Kaisa dikhta hai (Figure): surface dono axes ke saath level hai lekin diagonal ke saath height $\frac{1}{2}$ tak jump karti hai — ek twisted saddle jise flat plane kabhi hug nahi kar sakti.

Figure — Tangent planes and linear approximations to surfaces

L5.2

Prove karo ki kisi bhi function ke liye jo $f (x, y) = g (x) + h (y)$ form mein ho (separable), $(a, b)$ par tangent plane do single-variable tangent lines ka sum hai. $f = sin x + y^{3}$ ko $(0, 1)$ par verify karo.

Recall Solution L5.2

Proof. $f_{x} = g^{'} (x)$ ( $h (y)$ term $x$ mein constant hai), similarly $f_{y} = h^{'} (y)$ . To $L = g (a) + h (b) + g^{'} (a) (x - a) + h^{'} (b) (y - b) .$ Variable ke hisaab se group karo: $L = tangent line to g [g (a) + g^{'} (a) (x - a)] + tangent line to h [h (b) + h^{'} (b) (y - b)] .$ Ye seedha Tangent line and linear approximation (single variable) se link karta hai: 2D plane do independent 1D approximations mein decouple ho jaata hai. Verify $f = sin x + y^{3}$ $(0, 1)$ par: $f (0, 1) = 0 + 1 = 1$ . $f_{x} = cos x \Rightarrow 1$ ; $f_{y} = 3 y^{2} \Rightarrow 3$ . $L = 1 + 1 (x - 0) + 3 (y - 1) = 1 + x + 3 y - 3 = x + 3 y - 2.$ $g$ -part $0 + 1 \cdot x$ , $sin x$ ki $0$ par tangent line hai; $h$ -part $1 + 3 (y - 1)$ , $y^{3}$ ki $1$ par tangent line hai. ✓

L5.3

Surface $z = f (x, y)$ ka $(1, 2)$ par tangent plane $z = 5 + 2 (x - 1) - 3 (y - 2)$ hai. Ek bug $(1, 2)$ se unit vector $u = ⟨ \frac{3}{5}, \frac{4}{5} ⟩$ ki direction mein chalta hai. Sirf tangent plane use karke bug ka rate of height gain (directional derivative) nikalo — aur dikhao ki ye $\nabla f \cdot u$ ke barabar hai.

Recall Solution L5.3

Plane se: $f_{x} (1, 2) = 2$ , $f_{y} (1, 2) = - 3$ , to $\nabla f = ⟨ 2, - 3 ⟩$ . Tangent plane par ek chhote step $(d x, d y)$ mein height change $d z = f_{x} d x + f_{y} d y$ hoti hai. $u$ ke saath distance $t$ move karne par $d x = \frac{3}{5} t$ , $d y = \frac{4}{5} t$ , to $d z = 2 \cdot \frac{3}{5} t + (- 3) \cdot \frac{4}{5} t = (\frac{6}{5} - \frac{12}{5}) t = - \frac{6}{5} t .$ Rate of height gain $= d z / t = - \frac{6}{5} = - 1.2$ (bug neeche ja raha hai). Dot product se check karo (ye definition hai, dekho Directional derivatives): $\nabla f \cdot u = ⟨ 2, - 3 ⟩ \cdot ⟨ \frac{3}{5}, \frac{4}{5} ⟩ = \frac{6}{5} - \frac{12}{5} = - \frac{6}{5} . ✓$ Kyun agree karte hain: directional derivative flat tangent plane ka slope $u$ ke saath padhna hi hai — plane mein pehle se har directional slope contained hai.

Recall One-line self-check

Tangent plane equation ::: $z = f (a, b) + f_{x} (a, b) (x - a) + f_{y} (a, b) (y - b)$ "Plane P ke parallel" matlab match karo ::: sirf $x$ - aur $y$ -slopes (constant nahi) Linear approximation ka error badhta hai jaise ::: $(a, b)$ se distance ka square Plane se directional derivative ::: $\nabla f \cdot u = f_{x} u_{1} + f_{y} u_{2}$

Connections

Partial derivatives — upar har jagah $f_{x}, f_{y}$ .
The gradient vector — L3.1, L4.3, L5.3 mein normals.
Differentiability of multivariable functions — L5.1 failure.
Tangent line and linear approximation (single variable) — L5.2 decoupling.
Directional derivatives — L5.3.
The chain rule (multivariable) — $\cdot$ partials mein use hua.