Exercises — Parametric surfaces — tangent planes, surface area

${\mathbf{r}}_u$ ka matlab hai "har component ko $u$ ke respect mein differentiate karo, $v$ ko fixed rakho."

• $x=u\Rightarrow x_u=1$; $y=v\Rightarrow y_u=0$; $z=3u-2v\Rightarrow z_u=3$. ${\mathbf{r}}_u=⟨1,0,3⟩$ $v$ ke liye bhi same (hold $u$ fixed):
• $x_v=0,y_v=1,z_v=-2$. ${\mathbf{r}}_v=⟨0,1,-2⟩$

Component rule se cross product compute karo $\mathbf{a}\times \mathbf{b}=⟨a_2{b}_3-a_3{b}_2,a_3{b}_1-a_1{b}_3,a_1{b}_2-a_2{b}_1⟩.$ ${\mathbf{r}}_u=⟨1,0,3⟩$, ${\mathbf{r}}_v=⟨0,1,-2⟩$ ke saath: ${\mathbf{r}}_u\times {\mathbf{r}}_v=⟨(0)(-2)-(3)(1),(3)(0)-(1)(-2),(1)(1)-(0)(0)⟩=⟨-3,2,1⟩.$ Yeh zero vector nahi hai, isliye haan, surface har jagah smooth hai. (Yeh actually ek flat plane hai, jo ki jitna smooth ho sakta hai utna hai.)

Point. $\mathbf{r}(2,1)=⟨2,1,2^2-1^2⟩=⟨2,1,3⟩$, isliye $P_0=(2,1,3)$.

Tangent arrows. ${\mathbf{r}}_u=⟨1,0,2u⟩=⟨1,0,4⟩,{\mathbf{r}}_v=⟨0,1,-2v⟩=⟨0,1,-2⟩.$

Normal = cross product: $\mathbf{n}={\mathbf{r}}_u\times {\mathbf{r}}_v=⟨(0)(-2)-(4)(1),(4)(0)-(1)(-2),(1)(1)-0⟩=⟨-4,2,1⟩.$

Plane. Point $P_0$ aur normal $⟨-4,2,1⟩$ milake $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$: $-4(x-2)+2(y-1)+1(z-3)=0⟹-4x+2y+z+3=0,$ ya $z$ ke liye solve karke: $z=4x-2y-3$.

${\mathbf{r}}_u=⟨\cos v,\sin v,1⟩,{\mathbf{r}}_v=⟨-u\sin v,u\cos v,0⟩.$ Cross product: ${\mathbf{r}}_u\times {\mathbf{r}}_v=⟨(\sin v)(0)-(1)(u\cos v),(1)(-u\sin v)-(\cos v)(0),(\cos v)(u\cos v)-(\sin v)(-u\sin v)⟩$ $=⟨-u\cos v,-u\sin v,u{\cos }^{2}v+u{\sin }^{2}v⟩=⟨-u\cos v,-u\sin v,u⟩.$ Length: $|{\mathbf{r}}_u\times {\mathbf{r}}_v|=\sqrt{u^2{\cos }^{2}v+u^2{\sin }^{2}v+u^2}=\sqrt{u^2+u^2}=\sqrt{2}u.$ (Humne ${\cos }^{2}v+{\sin }^{2}v=1$ do baar use kiya — har trig-parametrized surface ke liye workhorse identity yahi hai.)

${\mathbf{r}}_u=⟨2u,v,0⟩,{\mathbf{r}}_v=⟨0,u,2v⟩.$ ${\mathbf{r}}_u\times {\mathbf{r}}_v=⟨(v)(2v)-(0)(u),(0)(0)-(2u)(2v),(2u)(u)-(v)(0)⟩=⟨2v^2,-4uv,2u^2⟩.$ Yeh $0$ tab hoga jab teeno components vanish karein: $2v^2=0,-4uv=0,2u^2=0⟹u=0\text{and}v=0.$ Sirf ek parameter point $(u,v)=(0,0)$ fail karta hai, jo surface point $(0,0,0)$ par map hota hai.

Geometry: $(0,0)$ par dono tangent arrows zero vector mein collapse ho jaate hain (${\mathbf{r}}_u={\mathbf{r}}_v=0$). Koi parallelogram nahi hai, toh koi plane span nahi hoti — parametrization ek point par pinch kar jaati hai (ek degenerate patch). Surface wahan bhi dikhne mein theek lag sakti hai; yeh map hai jo singular hai.

${\mathbf{s}}_u=⟨1,1,0⟩,{\mathbf{s}}_v=⟨1,-1,0⟩.$ ${\mathbf{s}}_u\times {\mathbf{s}}_v=⟨(1)(0)-(0)(-1),(0)(1)-(1)(0),(1)(-1)-(1)(1)⟩=⟨0,0,-2⟩,$ toh $|{\mathbf{s}}_u\times {\mathbf{s}}_v|=2$.

Area phir bhi kyun match karta hai: yeh factor $2$ B ke map ka local area-stretch hai — yeh exactly B ke linear substitution $(x,y)=(u+v,u-v)$ ka Jacobian hai. B ka parameter region $D_B$ A ke unit square se $2$ ke factor se chhota hai (kyunki B ka map $(u,v)$ grid ko square par squash/rotate karta hai). Us half-sized region par stretch factor $2$ integrate karne se $2\times \frac{1}{2}=1$ milta hai: wahi physical area jo A ko ${\iint }_{[0,1{]}^2}1dA=1$ se milta hai. Stretch factor aur region size hamesha compensate karte hain.

Graph parametrization $\mathbf{r}(x,y)=⟨x,y,x^2+y^2⟩$ use karo. $f=x^2+y^2$ ke saath: $f_x=2x,f_y=2y,|{\mathbf{r}}_x\times {\mathbf{r}}_y|=\sqrt{f_x^2+f_y^2+1}=\sqrt{4x^2+4y^2+1}.$ Toh $A={\iint }_{x^2+y^2\le 4}\sqrt{4x^2+4y^2+1}dA.$ Integrand sirf $r^2=x^2+y^2$ par depend karta hai, toh polar coordinates mein switch karo ($x=r\cos \theta ,y=r\sin \theta$, area element $dA=rdrd\theta$ — yeh factor $r$ khud ek Jacobian hai, dekho Double Integrals): $A={\int }_0^{2\pi }{\int }_0^2\sqrt{4r^2+1}rdrd\theta .$ Inner integral: let $w=4r^2+1$, toh $dw=8rdr\Rightarrow rdr=\frac{dw}{8}$. Jab $r:0\to 2$, $w:1\to 17$. ${\int }_0^2\sqrt{4r^2+1}rdr=\frac{1}{8}{\int }_1^{17}\sqrt{w}dw=\frac{1}{8}\cdot \frac{2}{3}{w}^{3/2}{|}_1^{17}=\frac{1}{12}(17^{3/2}-1).$ ${\int }_0^{2\pi }d\theta =2\pi$ se multiply karo: $\boxed{A=\frac{2\pi }{12}(17^{3/2}-1)=\frac{\pi }{6}(17\sqrt{17}-1)\approx 36.18}$

${\mathbf{r}}_u=⟨\cos v,\sin v,0⟩,{\mathbf{r}}_v=⟨-u\sin v,u\cos v,1⟩.$ ${\mathbf{r}}_u\times {\mathbf{r}}_v=⟨(\sin v)(1)-(0)(u\cos v),(0)(-u\sin v)-(\cos v)(1),u{\cos }^{2}v+u{\sin }^{2}v⟩=⟨\sin v,-\cos v,u⟩.$ $|{\mathbf{r}}_u\times {\mathbf{r}}_v|=\sqrt{{\sin }^{2}v+{\cos }^{2}v+u^2}=\sqrt{1+u^2}.$ $A={\int }_0^{\pi }{\int }_0^1\sqrt{1+u^2}dudv.$ Standard antiderivative $\int \sqrt{1+u^2}du=\frac{1}{2}(u\sqrt{1+u^2}+\mathrm{arcsinh}u)$ (yeh Arc Length integrand ka cousin hai). $0\to 1$ evaluate karo: ${\int }_0^1\sqrt{1+u^2}du=\frac{1}{2}(\sqrt{2}+\ln (1+\sqrt{2})).$ Phir ${\int }_0^{\pi }dv=\pi$ se multiply karo: $\boxed{A=\frac{\pi }{2}(\sqrt{2}+\ln (1+\sqrt{2}))\approx 3.606}$

Cross product compute karo. ${\mathbf{r}}_x=⟨1,0,f_x⟩,{\mathbf{r}}_y=⟨0,1,f_y⟩.$ ${\mathbf{r}}_x\times {\mathbf{r}}_y=⟨(0)(f_y)-(f_x)(1),(f_x)(0)-(1)(f_y),(1)(1)-(0)(0)⟩=⟨-f_x,-f_y,1⟩.$ Length. $|{\mathbf{r}}_x\times {\mathbf{r}}_y|=\sqrt{(-f_x{)}^2+(-f_y{)}^2+1^2}=\sqrt{f_x^2+f_y^2+1}.$ The inequality. Kyunki $f_x^2+f_y^2\ge 0$ hamesha, integrand satisfy karta hai $\sqrt{f_x^2+f_y^2+1}\ge \sqrt{0+0+1}=1,$ equality sirf wahan hogi jahan $f_x=f_y=0$ (ek locally flat, horizontal jagah). $D$ par integrate karo: $A(S)={\iint }_D\sqrt{f_x^2+f_y^2+1}dA\ge {\iint }_{D}1dA=\text{area}(D).$ Agar $f$ non-flat hai, toh positive area ke ek region par uska slope hai, toh inequality wahan strict hai aur $A(S)>\text{area}(D)$: surface ko tilt karna sirf area add kar sakta hai — flat shadow hamesha sabse chhota hota hai. (${\mathbf{r}}_x\times {\mathbf{r}}_y$ ka $z$-component $1$ hona exactly yahi batata hai: iska matlab hai normal hamesha vertical ki taraf "lean back" karta hai, aur length $\ge$ koi bhi single component.)

Magnitudes agree kyunki $|d\mathbf{S}|=|({\mathbf{r}}_u\times {\mathbf{r}}_v)dA|=|{\mathbf{r}}_u\times {\mathbf{r}}_v|dA=dS$ (scalar $dA>0$ magnitude se seedha bahar aa jaata hai).

• Area measure karta hai ki kitni surface hai — ek size, koi direction nahi — isliye yeh length $|{\mathbf{r}}_u\times {\mathbf{r}}_v|$ use karta hai, har chhote parallelogram ka size.
• Flux measure karta hai ki ek vector field ka kitna surface se pass through hota hai, jiske liye ek direction chahiye yeh batane ke liye ki "kis taraf se through." Isliye yeh poora vector ${\mathbf{r}}_u\times {\mathbf{r}}_v$ rakhta hai, jiska direction normal hai aur length phir bhi patch area hai. Ek quantity, do uses: length how-much ke liye, full vector through-which-way ke liye.