Exercises — Functions of several variables — graphs, level curves, level surfaces

Humein kya chahiye: ek real square root ke andar ka expression $\ge 0$ hona chahiye — negative number ka real root nahi le sakte.

• Require $16-x^2-y^2\ge 0$, i.e. $x^2+y^2\le 16$.
• Yeh kaisa dikhta hai: $x^2+y^2\le 16$ origin pe centred radius $4$ ki ek filled disk hai (Figure s01 dekho). To domain $=\{(x,y):x^2+y^2\le 16\}$.
• Range: $16-x^2-y^2$ rim pe $0$ se lekar centre pe $16$ tak jaata hai. Iska square root $0$ se $4$ tak jaata hai. To range $=[0,4]$.

Har ek ko constant $c$ ke barabar set karo aur standard shape padho:

• (a) $3x+2y=c\Rightarrow y=\frac{c-3x}{2}$ — straight lines (sab parallel, slope $-\frac{3}{2}$).
• (b) $x^2+y^2=c$ — radius $\sqrt{c}$ ke circles (for $c>0$).
• (c) $x^2-y^2=c$ — hyperbolas.
• (d) $y-x^2=c\Rightarrow y=x^2+c$ — vertically shifted parabolas.

$f=c$ kyun set karte hain? Definition ke anusaar level curve barabar output wale inputs ko collect karti hai.

• $\sqrt{x^2+y^2}=c$. Kyunki square root kabhi negative nahi hoti, humein $c\ge 0$ chahiye.
• Dono sides ko square karo (valid hai kyunki dono sides $\ge 0$ hain): $x^2+y^2=c^2$.
• Yeh kaisa dikhta hai: radius $c$ ka circle (na ki $\sqrt{c}$!). To $c=1,2,3$ radius $1,2,3$ ke circles dete hain; $c=0$ origin pe single point deta hai.
• Spacing: yahan radius $=c$ hai, isliye equal $c$-steps se equally spaced circles milte hain — yeh cone constant steepness se chadht hai. Isko $x^2+y^2$ se compare karo jiska radius $=\sqrt{c}$ hai jo outward crowd hota hai. (Figure s02 dono ko side by side dikhata hai.)

Humein kya chahiye: logarithm ka input strictly positive hona chahiye — zero ya negative number ka $\ln$ undefined hai.

• Require $x-y^2>0$, i.e. $x>y^2$.
• Yeh kaisa dikhta hai: curve $x=y^2$ ek sideways parabola hai jo right ki taraf open hoti hai. Condition $x>y^2$ us parabola ke right side ka (andar ka) region hai, boundary shaamil nahi (strict inequality).
• Domain $=\{(x,y):x>y^2\}$.

• $2x+3y+6z=c$ set karo. Teen unknowns mein ek linear equation se ek degree of freedom remove hoti hai, ek 2D object bachti hai — ek plane.
• Yeh sab parallel planes hain (same left-hand side, alag $c$). Inki common normal direction $(2,3,6)$ hai.
• Jaise $c$ equal steps mein badhta hai, planes equal distances shift hoti hain, kyunki equation linear hai (koi crowding nahi). $c$ aur $c+\Delta c$ planes ke beech distance $=\frac{\Delta c}{\sqrt{2^2+3^2+6^2}}=\frac{\Delta c}{7}$.

• Radius $=\sqrt{c}$: $c=1,4,9,16$ ke liye radii $1,2,3,4$ milti hain.
• Gaps (outer minus inner radius): $2-1=1$, $3-2=1$, $4-3=1$. Radius gaps yahan equal hain kyunki humne perfect squares choose kiye — lekin note karo ki $c$-jumps $3,5,7$ the, yaani unequal. Equal radius-steps ever-larger height jumps se aayi.
• Reading ko ulta karo: equal height jumps ke liye (maanlo $c=1,2,3,4$) radii $1,\sqrt{2},\sqrt{3},2\approx 1,1.41,1.73,2$ hongi, jiske gaps $0.41,0.32,0.27$ shrink karte hain. Circles outward crowd hoti hain → outer bowl steepest hai. Crowded level curves $\iff$ steep terrain.

• $c>0$: $x^2-y^2=c\Rightarrow \frac{x^2}{c}-\frac{y^2}{c}=1$ — hyperbolas left/right open hoti hain (yeh $x$-axis cross karti hain).
• $c<0$: $c=-k$ likho, $k>0$: $x^2-y^2=-k\Rightarrow \frac{y^2}{k}-\frac{x^2}{k}=1$ — hyperbolas up/down open hoti hain (yeh $y$-axis cross karti hain).
• $c=0$: $x^2=y^2\Rightarrow y=\pm x$ — do crossing diagonal lines. Yeh degenerate hyperbola hai: yeh apne asymptotes par collapse ho gayi hai. (Figure s03.)
• Crossing lines se separated nested hyperbolas saddle ki fingerprint hain.

• $100-x^2-4y^2=64$ set karo $\Rightarrow x^2+4y^2=36$.
• $36$ se divide karo: $\frac{x^2}{36}+\frac{y^2}{9}=1$. Yeh kaisa dikhta hai: ek ellipse jiska $x$ direction mein semi-axis $6$ aur $y$ direction mein $3$ hai.
• Interpretation: is ellipse par sabhi points exactly $64^{\circ }$ par hain. Sabse hot point centre hai ($T=100$); isotherms nested ellipses hain jo iske taraf shrink hoti hain.

• Parallel lines $y=2x+c$: rearrange karo $c=y-2x$. To $f(x,y)=y-2x$ exactly yahi level curves deta hai — $f=c$ set karo aur $y=2x+c$ wapas mil jaata hai. ✅
• $(1,-2)$ pe centred circles: us centre ka circle $(x-1{)}^2+(y+2{)}^2=r^2$ hai. Isko level set banane ke liye, $f(x,y)=(x-1{)}^2+(y+2{)}^2$ lo; tab $f=c$ (for $c>0$) $(1,-2)$ pe centred radius $\sqrt{c}$ ka circle hai. ✅

• Cone: level curve $\sqrt{x^2+y^2}=c\Rightarrow$ radius $=c$. Equal height-steps → equally spaced circles → constant steepness har jagah.
• Paraboloid: radius $=\sqrt{c}$. Equal height-steps → circles outward crowd hoti hain → steepness increases without bound jaise aap bahar jaate ho.
• Conclusion: axis se door paraboloid steeper hai (iske circles zyada crowded hain). Axis ke paas paraboloid actually gentler hai (iske circles shuru mein widely spaced hain). Crossover wahan hai jahan $\sqrt{c}=c$ hai, i.e. $c=1$ (radius $1$).

• $x^2+y^2+z^2-2z=c$ set karo. $z$ mein complete the square karo ($1$ add aur subtract karo): $x^2+y^2+(z-1{)}^2-1=c$, to $x^2+y^2+(z-1{)}^2=c+1$.
• Yeh kaisa dikhta hai: $(0,0,1)$ pe centred radius $\sqrt{c+1}$ ke spheres, valid jab $c+1>0$ yaani $c>-1$. $c=-1$ par surface single point $(0,0,1)$ hai; $c<-1$ ke liye yeh empty hai.
• To $g$ ke level surfaces $f$ ke jaisi hi spheres hain, sirf $1$ se upar shift huin aur re-label hui hain.

$A{x}^2+B{y}^2=c$ likho aur signs se reason karo.

• $A>0,B>0$: left side $\ge 0$.
  • $c>0$: ellipse $\frac{x^2}{c/A}+\frac{y^2}{c/B}=1$ (circle iff $A=B$).
  • $c=0$: sirf $x=y=0$ kaam karta hai → single point origin.
  • $c<0$: non-negatives ka sum negative nahi ho sakta → empty.
• $A<0,B<0$: mirror image. $c<0$: ellipse; $c=0$: point; $c>0$: empty.
• $A>0,B<0$ (opposite signs), yaani saddle-type:
  • $c\ne 0$: hyperbola ($c>0$ ho to $x$ along open, $c<0$ ho to $y$ along open).
  • $c=0$: $A{x}^2=-B{y}^2$, yaani $y=\pm \sqrt{A/|B|}x$ → do crossing lines (degenerate hyperbola).
• Ek coefficient zero, maanlo $B=0,A>0$: $A{x}^2=c$.
  • $c>0$: $x=\pm \sqrt{c/A}$ → do parallel vertical lines.
  • $c=0$: $x=0$ → single line ($y$-axis).
  • $c<0$: empty. Ab har case cover ho gaya: ellipse, point, empty, hyperbola, crossing lines, parallel lines, single line.

• Idea: equation $f(x_1,\dots ,x_n)=c$ $n$ free numbers par ek constraint hai. Generically ek equation exactly ek degree of freedom remove karti hai, isliye bacha hua object dimension $n-1$ ka hota hai.
• $n=1$: $f(x)=c$ isolated points select karta hai (dimension $0$) — e.g. $x^2=4\Rightarrow x=\pm 2$.
• $n=2$: $f(x,y)=c$ ek curve hai (dimension $1$) — level curve.
• $n=3$: $f(x,y,z)=c$ ek surface hai (dimension $2$) — level surface.
• "Generically" un degenerate exceptions ko flag karta hai jo humne dekhe (level set ek point tak shrink ho sakti hai ya vanish ho sakti hai), lekin typical dimension hamesha $n-1$ hoti hai.

L1 formula se shape ka naam lo · L2 formula ko circle/plane/parabola ki actual equation mein turn karo · L3 curves ki crowding se steepness padho · L4 isko ulta chalao — wo $f$ banao jiska level set tum chahte ho · L5 sign se har case classify karo, aur jaano ki dimension exactly ek se kyun girti hai.