Parametric surfaces — tangent planes, surface area: Visual walkthrough

4.4.30 · D2 · Maths › Multivariable Calculus › Parametric surfaces — tangent planes, surface area

Step 1 — Do knobs ek surface paint karte hain

KYA. "Control paper" ki ek flat sheet par do dials hain. Horizontal dial ko hum $u$ kehte hain aur vertical dial ko $v$ . Unhe ghoomane se ek point $(u, v)$ ek flat region ke andar pick hota hai jise hum $D$ kehte hain (bas "wo dial settings ka set jo hum allow karte hain"). Ek machine jise $r$ kehte hain $(u, v)$ padhti hai aur 3D space mein ek point bahar nikalti hai: $r (u, v) = ⟨ x (u, v), y (u, v), z (u, v) ⟩$

Term by term: $x, y, z$ teen ordinary height/width/depth numbers hain, aur har ek apni recipe hai jo dono dials par depend karti hai. Angle brackets $⟨ ⟩$ teeno numbers ko origin se ek arrow mein bundle karte hain — uski tip surface point hai.

DO dials kyun, ek nahi? Ek curve ek wire hai — tum sirf aage ya peeche jaate ho, isliye ek dial ( $t$ ) kaafi hai. Ek surface ek bedsheet hai — tum do independent tarafo mein slide kar sakte ho. Do freedoms → do dials. Yahi poori wajah hai ki hum baad mein do baar differentiate karte hain, ek baar har dial ke liye.

PICTURE. Left: flat control paper $D$ jisme dial settings ka ek grid hai. Right: machine $r$ us grid ko space mein ek curvy sheet par warp karti hai.

Figure — Parametric surfaces — tangent planes, surface area

Step 2 — Ek dial freeze karo, ek aisi curve pao jis par ride kar sako

KYA. $v$ ko kisi value $v_{0}$ par rokke rakho aur sirf $u$ dial ghoomao. Ab sirf ek cheez change hoti hai, isliye $r$ ki tip surface par ek wire kheenchti hai. Hum ise $u$ -curve kehte hain. Symmetrically, $u = u_{0}$ freeze karke $v$ ghoomane se $v$ -curve milti hai.

YEH KYUN KARTE HAIN? Area ek 2D cheez hai, lekin hum motion sirf ek direction mein ek waqt mein measure karna jaante hain (ek velocity). Isliye hum 2D freedom ko do 1D rides mein todh dete hain. Har ride ek curve hai, aur curve ka ek velocity vector hota hai — wahi object jo Tangent Lines and Velocity Vectors mein hai.

$u$ -ride ki velocity $r_{u}$ likhi jaati hai aur iska matlab hai "surface point kitni tezi se aur kis direction mein move karta hai jab main sirf $u$ ko thoda sa nudge karun": $r_{u} = \frac{\partial r}{\partial u} = ⟨ \frac{\partial x}{\partial u}, \frac{\partial y}{\partial u}, \frac{\partial z}{\partial u} ⟩$

Curly $\partial$ (kaho "partial") ek ordinary derivative hai jo $v$ ko ek frozen constant maanta hai. Har slot ka jawab hai: "jab $u$ ek baal ke barabar badalta hai, to yeh coordinate kitna badalta hai?" Isi tarah $r_{v} = \partial r / \partial v$ $u$ ko freeze karta hai.

PICTURE. Ek point se guzarne wali do grid curves, unke tangent arrows $r_{u}$ (yellow) aur $r_{v}$ (blue) surface se chipke hue.

Step 3 — Ek tiny dial-rectangle ek curved tile ban jaata hai

KYA. Flat control paper par, ek tiny rectangle lo: width $Δ u$ ( $u$ dial ka ek chota step) aur height $Δ v$ ( $v$ dial ka ek chota step). Yahan $Δ$ (kaho "delta") ka matlab sirf "thoda sa" hai. Us rectangle ki flat area $Δ u Δ v$ hai. Iske charo corners ko $r$ mein daalo aur surface par ek chota curved tile milega.

ITNA CHOTA KYUN SHUROO KARTE HAIN? Ek poori curved surface ka koi simple area formula nahi hota — lekin ek chhota sa kaafi piece lagbhag flat hota hai, aur flat pieces hum measure kar sakte hain. Yahi Double Integrals ka master trick hai: kaato, crumb ko measure karo, jodo.

PICTURE. $D$ par ek flat $Δ u \times Δ v$ rectangle surface par ek warped tile mein map hota hua.

Step 4 — Tile ko ek parallelogram mein seedha karo

KYA. Curved tile ke do edges corner point $P_{0}$ se shuru hote hain. Ek edge $u$ -curve ke saath $Δ u$ step ke liye jaati hai; doosri $v$ -curve ke saath $Δ v$ step ke liye. Kyunki steps tiny hain, "curve ke saath move karna" lagbhag wahi hai jaise "tangent velocity ke saath seedha move karna." Isliye edges approximately do seedhe arrows hain: $edge 1 r_{u} Δ u and edge 2 r_{v} Δ v$

Har ek padhne par: $r_{u}$ $u$ -ride ki direction aur speed hai, aur chote number $Δ u$ se multiply karne par speed ek actual chota displacement ban jaati hai. $r_{v} Δ v$ ke liye bhi wahi.

YEH LEGAL KYUN HAI? Yeh linear (first-order) approximation hai: kisi point ke paas, ek smooth curve apni tangent line se alag nahi lagti — bilkul wahi idea jo Tangent Lines and Velocity Vectors aur Arc Length ke chote seedhe segments ke peeche hai. Do seedhe edges ek parallelogram banate hain jo curved tile ke saath chipka rehta hai.

PICTURE. Curved tile jisme do tangent edges seedhe kheeche gaye hain, ek flat parallelogram mein close ho rahe hain (chalk-pink).

Step 5 — Us parallelogram ki area ek cross product hai

KYA. Hume do arrows $a = r_{u} Δ u$ aur $b = r_{v} Δ v$ se bane parallelogram ki area chahiye. Rule (Cross Product se) yeh hai: $area = a \times b = ∣ a ∣ ∣ b ∣ sin θ$ jahan $θ$ do edges ke beech ka angle hai, aur $∣ \cdot ∣$ ka matlab hai "arrow ki length."

$sin θ$ kyun aata hai, $cos θ$ nahi? $b$ ki tail ko $a$ par slide karo. Parallelogram ki height $b$ ka woh hissa hai jo $a$ se sideways nikalta hai — woh hai $∣ b ∣ sin θ$ . Area = base $\times$ height = $∣ a ∣ \cdot ∣ b ∣ sin θ$ . Sideways wala hissa $sin$ hai, na ki along-wala hissa $cos$ .

CROSS KYUN, DOT NAHI? Dot product $cos θ$ use karta hai: yeh sabse bada hota hai jab arrows ek line mein hote hain aur zero hota hai jab perpendicular hote hain. Area ka mood ulta hai — perpendicular edges sabse zyada area fence off karti hain, parallel edges kuch bhi nahi. Sirf $sin θ$ (cross product) aise behave karta hai. Yeh "dot vs cross" wali galti hai jiske baare mein parent note warn karta hai, geometrically dekha gaya.

Chote numbers bahar nikalo. Kyunki $Δ u, Δ v$ sirf positive numbers hain: $r_{u} Δ u \times r_{v} Δ v = r_{u} \times r_{v} Δ u Δ v$ Isliye tiny tile ki area hai: $Δ S \approx stretch factor r_{u} \times r_{v} flat crumb area Δ u Δ v$

Ise padho: flat control-paper crumb ki area $Δ u Δ v$ thi; machine $r$ ne ise $∣ r_{u} \times r_{v} ∣$ number se stretch karke surface par real tile banaya.

PICTURE. Parallelogram ka base–height decomposition, jisme $∣ b ∣ sin θ$ true height ke roop mein dikha hai.

Step 6 — Degenerate cases: jab tile collapse ho jaata hai

KYA. Stretch factor $∣ r_{u} ∣ ∣ r_{v} ∣ sin θ$ ko broken situations mein dekho. Iske zero hone ke exactly teen tarike hain, aur har ek ek real geometric event hai:

Kya zero hota hai	Matlab	Tile ki picture
$sin θ = 0$ ( $θ = 0$ ya $18 0^{\circ}$ )	do edges parallel hain	tile ek line mein squash ho jaata hai — koi area nahi
$∣ r_{u} ∣ = 0$	$u$ -ride ruk jaati hai (surface pinched)	tile ki koi width nahi
$∣ r_{v} ∣ = 0$	$v$ -ride ruk jaati hai	tile ki koi height nahi

YEH KYUN MATTER KARTA HAI. Kisi bhi point par jahan $r_{u} \times r_{v} = 0$ hoga, surface smooth nahi hogi — wahan koi honest tangent plane nahi hai aur parallelogram picture toot jaati hai. Ek famous example: sphere ka north pole, jahan saare $θ$ -circles ek point par shrink ho jaate hain ( $sin ϕ = 0$ ), isliye $∣ r_{ϕ} \times r_{θ} ∣ = R^{2} sin ϕ = 0$ . Integral phir bhi kaam karta hai kyunki stretch factor exactly wahan zero hai — ek pinched line koi area contribute nahi karti — isliye sum mein kuch galat nahi hota. Degenerate points integral ke andar allowed hain jab tak woh ek thin set banate hain.

PICTURE. Teen collapsing tiles side by side: parallel edges, stalled $u$ , stalled $v$ .

Step 7 — Har tile ko jodo: sum ek integral ban jaata hai

KYA. $D$ ko in tiny rectangles se dhaanko, har ek ka ek stretched tile. Total surface area sabhi crumb-areas ka sum hai: $A (S) \approx \sum_{tiles} r_{u} \times r_{v} Δ u Δ v$ Ab har rectangle ko zero size ki taraf shrink karo. Do cheezein ek saath hoti hain: parallelogram approximation exact ho jaata hai, aur sum ek double integral ban jaata hai: $A (S) = \iint_{D} r_{u} \times r_{v} d A$

Term by term ek aakhri baar:

$\iint_{D}$ — "region $D$ mein har dial setting par sweep karo."
$∣ r_{u} \times r_{v} ∣$ — us setting par tile ka stretch number.
$d A = d u d v$ — control-paper area ka flat crumb jo stretch ho raha hai.

STRETCH FACTOR OPTIONAL KYUN NAHI HAI. Agar tum $\iint_{D} d A$ likhoge to sirf dial paper ki flat area milegi, surface ki nahi. Factor $∣ r_{u} \times r_{v} ∣$ precisely map $r$ ka Jacobian hai — yeh flat-paper area ko real curved-surface area mein convert karta hai. Ise drop karo aur tum galat sheet measure kar rahe ho.

PICTURE. Bahut saare small tiles surface ko tile karte hue, har ek kitna stretch hua iske hisaab se shaded, sum-arrow ke saath integral sign bante hue.

Reality check — sphere ek hi saansi mein

Ek-picture summary

Dial paper par ek flat crumb $d u d v$ → machine $r$ → $r_{u} d u$ aur $r_{v} d v$ se spanned ek tilted parallelogram → uski area hai $∣ r_{u} \times r_{v} ∣ d u d v$ → sabko $\iint_{D}$ se jodo.

Recall Feynman retelling — poori walk plain words mein

Ek flat graph paper ki sheet imagine karo jisme do dials hain, $u$ aur $v$ . Ek machine paper par har dot padhti hai aur use hawa mein kahin chipka deti hai, isliye flat sheet ek bumpy bedsheet ban jaati hai. Yeh jaanne ke liye ki kitni bedsheet hai, graph paper ko tiny squares mein kaato. Har square bedsheet par thoda tilted chhota diamond ban ke land karta hai. Ek diamond ki area paane ke liye, uske sides ke saath do arrows kheencho — ek woh taraf dikhata hai jab tum $u$ dial ghoomate ho ( $r_{u}$ ), ek $v$ dial ke liye ( $r_{v}$ ) — aur pucho ki woh kitna space fence off karte hain. Woh "fenced area" cross product ki length hai, $∣ r_{u} \times r_{v} ∣$ , times chote step sizes. Yeh bada hota hai jab arrows alag khade hote hain aur kuch nahi hota jab woh line up kar lete hain (ya jab ek arrow mar jaata hai, jaise globe ke poles par). Har diamond jodo aur, squares ko infinitely chote hone dete hue, sums ka pile double integral $\iint_{D} ∣ r_{u} \times r_{v} ∣ d A$ ban jaata hai. Yahi poori kahani hai: kaato, ek tilted crumb cross product se measure karo, sabko jodo.

Recall

Hum

r

ko do baar differentiate kyun karte hain (ek baar

u

mein, ek baar

v

mein)?

Ek surface mein move karne ke do independent directions hote hain; har dial ek velocity deta hai,

r_{u}

aur

r_{v}

Tiny tile ki area

sin θ

(cross) kyun use karti hai,

cos θ

(dot) nahi?

Area edges ka sideways spread hai — height

= ∣ r_{v} ∣ sin θ

; yeh maximum hota hai jab edges perpendicular hoti hain aur zero hota hai jab parallel hoti hain.

∣ r_{u} \times r_{v} ∣

integral mein physically kya represent karta hai?

Local area-stretch (Jacobian) jo flat crumb area

d u d v

ko real surface area mein convert karta hai.

Sphere ke pole par stretch factor

R^{2} sin ϕ = 0

hai — kya yeh problem hai?

Nahi; tile genuinely wahan ek point par collapse ho jaata hai, koi area contribute nahi karta, isliye integral unaffected rehta hai.