Parametric surfaces — tangent planes, surface area: Worked examples

4.4.30 · D3 · Maths › Multivariable Calculus › Parametric surfaces — tangent planes, surface area

Yeh page Parametric surfaces — tangent planes, surface area ka "sab kuch daal do" companion hai. Pehle hum ek matrix banate hain — har tarah ki situation jo in problems mein aa sakti hai — phir examples karte hain jab tak us matrix ka har cell touch na ho jaye. Yahan kuch bhi assume nahi kiya gaya sivaaye is ke: tum differentiate kar sakte ho, tum jaante ho ki Cross Product do vectors ke perpendicular ek vector deta hai, aur tum double integral kar sakte ho.

Shuru karne se pehle, ek-line ka reminder jis par sab kuch tika hai:

The scenario matrix

Is topic ka har exam problem in hi cells mein se ek hai. Column batata hai kya poocha gaya hai; row batata hai kya mushkil hai.

Complication ↓ \ Task →	Tangent plane	Surface area
Clean / textbook	(a) paraboloid	(b) cone piece
Sign / orientation matters	(c) $n$ kis taraf point karta hai?	—
Degenerate point ( $r_{u} \times r_{v} = 0$ )	(d) cone tip	(d) tip contributes zero
Graph shortcut $z = f (x, y)$	(e) $⟨ - f_{x}, - f_{y}, 1 ⟩$ reuse karo	(e) $f_{x}^{2} + f_{y}^{2} + 1$
Non-orthogonal grid ( $θ \neq = 9 0^{\circ}$ )	—	(f) sheared plane
Real-world word problem	—	(g) dome par paint
Exam twist (limiting value)	—	(h) area $\to$ known formula

Ab hum har cell hit karte hain. Labels dekho — har example batata hai woh kaunsa cell fill kar raha hai.

(a) Clean tangent plane — cell "textbook / plane"

Step 1. Tangent vectors compute karo. $r_{u} = ⟨ 1, 0, 2 u ⟩ = ⟨ 1, 0, 4 ⟩, r_{v} = ⟨ 0, 1, - 2 v ⟩ = ⟨ 0, 1, - 2 ⟩ .$ Yeh step kyun? $r_{u}$ woh velocity hai jab tum $u$ -direction mein slide karte ho; yeh surface ke andar lie karta hai, isliye yeh tangent plane mein bhi lie karna chahiye. $r_{v}$ ke liye bhi yahi. Ek plane ke liye do in-plane directions kaafi hain.

Step 2. Normal ke liye cross karo. $n = r_{u} \times r_{v} = ⟨ (0) (- 2) - (4) (1), (4) (0) - (1) (- 2), (1) (1) - (0) (0) ⟩ = ⟨ - 4, 2, 1 ⟩ .$ Yeh step kyun? Ek plane ek point aur ek perpendicular direction se define hota hai. Cross Product exactly woh tool hai jo do diye gaye vectors ke perpendicular vector banata hai.

Step 3. $(2, 1, 3)$ se guzarne wala plane assemble karo. $- 4 (x - 2) + 2 (y - 1) + 1 (z - 3) = 0 ⟹ - 4 x + 2 y + z + 3 = 0 ⟹ z = 4 x - 2 y - 3.$ Yeh step kyun? $(2, 1, 3)$ se plane ke kisi bhi doosre point $(x, y, z)$ tak ka vector $n$ ke $⊥$ hai, isliye uska $n$ ke saath dot product zero hai — wahi dot product equation hai.

Verify: point daalo: $z = 4 (2) - 2 (1) - 3 = 8 - 2 - 3 = 3$ ✓. Aur $n \cdot r_{u} = - 4 + 0 + 4 = 0$ , $n \cdot r_{v} = 0 + 2 - 2 = 0$ — normal sach mein dono tangents ke perpendicular hai ✓.

Figure — Parametric surfaces — tangent planes, surface area

(b) Clean surface area — cell "textbook / area"

Step 1. Tangent vectors. $r_{u} = ⟨ cos θ, sin θ, 1 ⟩, r_{θ} = ⟨ - u sin θ, u cos θ, 0 ⟩ .$ Kyun? $r_{u}$ cone ki straight ruling ke along outward-and-up point karta hai; $r_{θ}$ height $u$ par circle ke around point karta hai.

Step 2. Cross product. $r_{u} \times r_{θ} = ⟨ - u cos θ, - u sin θ, u ⟩ .$ Kyun? Hum area stretch factor $∣ r_{u} \times r_{θ} ∣$ chahte hain; Double Integrals machinery ko yeh scalar chahiye taaki har tiny $(u, θ)$ rectangle apna actual curved area count kare, flat area nahi.

Step 3. Magnitude. $∣ r_{u} \times r_{θ} ∣ = u^{2} cos^{2} θ + u^{2} sin^{2} θ + u^{2} = u^{2} + u^{2} = u 2 .$ Kyun? $cos^{2} θ + sin^{2} θ = 1$ se $θ$ -dependence collapse ho jaati hai, ek clean stretch factor milta hai.

Step 4. Integrate karo. $A = \int_{0}^{2 π} \int_{0}^{2} u 2 d u d θ = 2 (2 π) [\frac{u ^{2}}{2}]_{0}^{2} = 2 (2 π) (2) = 42 π .$

Verify: $42 π \approx 17.77 > 4 π \approx 12.57$ — flat disk se bada, jaise forecast kiya tha ✓. Sanity: slant $ℓ$ aur base radius $R$ wale cone ka lateral area $π R ℓ$ hota hai; yahan $R = 2$ , $ℓ = 22$ , to $π (2) (22) = 42 π$ ✓.

(c) Sign / orientation — cell "normal kis taraf point karta hai?"

Step 1. $(\frac{π}{2}, 0)$ par tangent vectors. $r_{ϕ} = ⟨ cos ϕ cos θ, cos ϕ sin θ, - sin ϕ ⟩ = ⟨ 0, 0, - 1 ⟩,$ $r_{θ} = ⟨ - sin ϕ sin θ, sin ϕ cos θ, 0 ⟩ = ⟨ 0, 1, 0 ⟩ .$ Kyun? Hum specific point par evaluate karte hain kyunki orientation ek local question hai.

Step 2. Diye gaye order $r_{ϕ} \times r_{θ}$ mein cross karo. $⟨ 0, 0, - 1 ⟩ \times ⟨ 0, 1, 0 ⟩ = ⟨ (0) (0) - (- 1) (1), (- 1) (0) - (0) (0), (0) (1) - (0) (0) ⟩ = ⟨ 1, 0, 0 ⟩ .$ Kyun? Cross Product mein factors ka order sign flip karta hai; flux aur orientation ke liye sign hi sab kuch hai, isliye diye gaye order ko respect karna zaroori hai.

Step 3. Outward direction $⟨ 1, 0, 0 ⟩$ se compare karo: same hai → yeh ordering outward normal deta hai.

Verify: $r_{ϕ} \times r_{θ} = ⟨ 1, 0, 0 ⟩$ $(1, 0, 0)$ par origin se radially outward point karta hai ✓. $r_{θ} \times r_{ϕ}$ reverse karne par $⟨ - 1, 0, 0 ⟩$ milega — inward normal, jo confirm karta hai ki order sign control karta hai.

(d) Degenerate point — cell " $r_{u} \times r_{v} = 0$ "

Step 1. $u = 0$ par cross product evaluate karo. (b) se, $r_{u} \times r_{θ} = ⟨ - u cos θ, - u sin θ, u ⟩$ . $u = 0$ par yeh $⟨ 0, 0, 0 ⟩$ hai. Kyun? Surface ka smoothness test exactly $r_{u} \times r_{v} \neq = 0$ hai. Yahan fail ho jaata hai.

Step 2. Interpret karo. Zero normal matlab do tangent vectors independent nahi hain (tip par $r_{θ} = ⟨ 0, 0, 0 ⟩$ — poora circle ek point mein collapse ho gaya hai). Koi 2D plane span nahi hoti, isliye apex par koi tangent plane nahi hai. Kyun? Ek plane ke liye do independent in-surface directions chahiye; cusp par woh hote nahi.

Step 3. Kya yeh (b) ke area integral ko kharab karta hai? Nahi — bad point sirf ek value $u = 0$ hai, ek zero-width set. Integrand $u 2 \to 0$ wahan bhi jaata hai, isliye kuch contribute nahi karta.

Verify: $u \to 0 lim ∣ r_{u} \times r_{θ} ∣ = u \to 0 lim u 2 = 0$ ✓ — degenerate point integrable hai aur (b) ka area $42 π$ theek hai.

(e) Graph shortcut — cell " $z = f (x, y)$ , dono kaam ek saath"

Part (i).

Step 1. $f_{x} = 2 x, f_{y} = 2 y$ . $(1, 2)$ par: $f_{x} = 2, f_{y} = 4$ , point $(1, 2, 5)$ . Kyun? Graph parametrization pre-compute karta hai $r_{x} = ⟨ 1, 0, f_{x} ⟩$ , $r_{y} = ⟨ 0, 1, f_{y} ⟩$ , isliye normal ka fixed pattern hai $⟨ - f_{x}, - f_{y}, 1 ⟩$ .

Step 2. Normal $= ⟨ - 2, - 4, 1 ⟩$ . Plane: $- 2 (x - 1) - 4 (y - 2) + 1 (z - 5) = 0 ⟹ z = 2 x + 4 y - 5.$ Verify (i): $(1, 2)$ par, $z = 2 + 8 - 5 = 5$ ✓.

Part (ii).

Step 3. Stretch factor $f_{x}^{2} + f_{y}^{2} + 1 = 4 x^{2} + 4 y^{2} + 1$ . Polar mein ( $x^{2} + y^{2} = ρ^{2}$ , $d A = ρ d ρ d θ$ , dekho Change of Variables and the Jacobian): $A = \int_{0}^{2 π} \int_{0}^{1} 4 ρ^{2} + 1 ρ d ρ d θ .$ Polar kyun? Stretch factor sirf $ρ$ par depend karta hai, isliye polar se inner integral ek clean substitution ban jaata hai.

Step 4. Maano $w = 4 ρ^{2} + 1$ , $d w = 8 ρ d ρ$ : $\int_{0}^{1} 4 ρ^{2} + 1 ρ d ρ = \frac{1}{8} \int_{1}^{5} w d w = \frac{1}{8} \cdot \frac{2}{3} [w^{3/2}]_{1}^{5} = \frac{1}{12} (5^{3/2} - 1) .$ $A = 2 π \cdot \frac{5 ^{3/2} - 1}{12} = \frac{π}{6} (55 - 1) .$

Verify (ii): $\frac{π}{6} (55 - 1) \approx \frac{π}{6} (11.18 - 1) \approx 5.33 > π \approx 3.14$ — curved surface flat disk se bada, jaise forecast kiya tha ✓.

(f) Non-orthogonal grid — cell " $θ \neq = 9 0^{\circ}$ , dot vs cross matters"

Step 1. Tangents: $r_{u} = ⟨ 1, 0, 0 ⟩$ , $r_{v} = ⟨ 1, 1, 0 ⟩$ . Kyun? Simple differentiation; note karo ki yeh alag, non-perpendicular directions mein point karte hain.

Step 2. Galat naive factor $∣ r_{u} ∣∣ r_{v} ∣ = 1 \cdot 2 = 2$ . Galat wala kyun dikhayein? Students assume karte hain ki stretch factor length-times-length hai. Yeh tabhi sahi hai jab grid orthogonal ho.

Step 3. Sahi cross product: $r_{u} \times r_{v} = ⟨ 0, 0, 1 ⟩, ∣ r_{u} \times r_{v} ∣ = 1.$ Cross kyun, length product kyun nahi? $∣ r_{u} \times r_{v} ∣ = ∣ r_{u} ∣∣ r_{v} ∣ sin θ$ . Yahan tangents ke beech angle $θ$ satisfy karta hai $cos θ = \frac{r _{u} \cdot r _{v}}{∣ r _{u} ∣∣ r _{v} ∣} = \frac{1}{2}$ , to $θ = 4 5^{\circ}$ aur $sin θ = \frac{1}{2}$ . Tab $2 \cdot \frac{1}{2} = 1$ .

Step 4. Area $= \int_{0}^{1} \int_{0}^{1} 1 d u d v = 1$ .

Verify: cross-product area $= 1$ ; naive product ne $2 \approx 1.414$ diya — $1/ sin 4 5^{\circ}$ factor se over-count. Forecast of $1$ confirmed ✓.

(g) Real-world word problem — cell "dome par paint"

Step 1. Upar wala aadha parametrize karo: $ϕ \in [0, \frac{π}{2}]$ , $θ \in [0, 2 π]$ , $R = 3$ . Kyun? $ϕ = 0$ top hai; $ϕ = \frac{π}{2}$ equator hai — exactly upper half.

Step 2. Sphere stretch factor $∣ r_{ϕ} \times r_{θ} ∣ = R^{2} sin ϕ = 9 sin ϕ$ reuse karo (parent note mein derive kiya gaya hai $sin^{2} + cos^{2} = 1$ se).

Step 3. Area: $A = \int_{0}^{2 π} \int_{0}^{π /2} 9 sin ϕ d ϕ d θ = 9 (2 π) [- cos ϕ]_{0}^{π /2} = 9 (2 π) (1) = 18 π m^{2} .$ Limit $π /2$ kyun? Sirf upar wala aadha paint hota hai, isliye $ϕ$ equator par ruk jaata hai.

Step 4. Litres $= \frac{18 π}{8} = \frac{9 π}{4} \approx 7.07$ litres.

Verify: $18 π \approx 56.5$ m² $2 π R^{2}$ se match karta hai ✓; $9 π /4 \approx 7.07$ litres, forecast se match karta hai. Units: $m^{2} \div (m^{2} / litre) = litres$ ✓.

(h) Exam twist — limiting value — cell "area $\to$ known formula"

Step 1. $z = c$ ke liye: $f_{x} = 0, f_{y} = 0$ , to $f_{x}^{2} + f_{y}^{2} + 1 = 1 = 1$ . Kyun? Yeh sanity check hai jo har formula ko pass karna chahiye — flat surface stretch nahi hona chahiye.

Step 2. $A = \iint_{D} 1 d A = area (D)$ . ✓ Formula theek se degenerate hota hai.

Step 3. Unit square $[0, 1]^{2}$ ke upar $z = k x$ ke liye: $f_{x} = k, f_{y} = 0$ , factor $= 1 + k^{2}$ , constant hai, to $A = 1 + k^{2} \cdot area = 1 + k^{2} .$ Kyun? Ek unit square ko slope $k$ se tilt karne par uski length $x$ ke along exactly $1 + k^{2}$ se stretch hoti hai (rise- $k$ , run- $1$ triangle ka hypotenuse) — pure Arc Length logic jo surface par promote ho gaya.

Step 4. Limits: $k \to 0 \Rightarrow A \to 1$ (flat); $k \to \infty \Rightarrow A \to \infty$ (vertical wall, $D$ par project karne par infinite).

Verify: $k = 0$ par, $A = 1$ ✓; $k = 1$ par, $A = 2 \approx 1.414$ ✓; $∣ k ∣$ mein monotone increasing ✓.

Recall check

Recall Kaun sa product kaun se kaam ke liye?

Area ke liye $r_{u}, r_{v}$ ka kaun sa product chahiye, aur doosra kyun nahi? ::: Cross product ka magnitude $∣ r_{u} ∣∣ r_{v} ∣ sin θ$ — yeh spread measure karta hai aur parallel edges ke liye zero hai; dot product alignment measure karta hai aur perpendicular edges ke liye zero hai (area ko jo chahiye uska opposite).

Recall Degenerate points

$r_{u} \times r_{v} = 0$ kya batata hai? ::: Tangents dependent hain (ek cusp/tip); wahan koi tangent plane exist nahi karta, lekin ek aisa point area integral mein zero contribute karta hai.

Recall Non-orthogonal grid trap

Kab $∣ r_{u} ∣∣ r_{v} ∣$ true area factor ke barabar hota hai? ::: Sirf jab grid orthogonal ho ( $θ = 9 0^{\circ}$ , $sin θ = 1$ ); warna yeh $1/ sin θ$ factor se overestimate karta hai.