Critical points — finding, classifying: Visual walkthrough

4.4.12 · D2 · Maths › Multivariable Calculus › Critical points — finding, classifying

Step 1 — "Har direction mein flat" ka asli matlab

KYA. Hum ek special point $(a, b)$ se shuru karte hain jahan zameen bilkul seedhi hai. "Seedhi" ka matlab hai: agar main kisi bhi horizontal direction mein ek choti si step loon, to height upar ya neeche nahi jaati. Do sabse basic directions hain — $x$ ke saath aur $y$ ke saath.

KYUN. $x$ ke along slope ko $f_{x}$ likhte hain — iska matlab hai "height kitni tezi se badalti hai agar main $x$ ko nudge karun aur $y$ ko freeze rakhun." $y$ ke along slope $f_{y}$ hai. Kisi hilltop, valley, ya pass par, ye dono zero hote hain. Isi ko parent note $\nabla f = 0$ kehta hai; symbol $\nabla f = (f_{x}, f_{y})$ inhi do slopes ka ek bundle hai.

PICTURE. Do coloured tangent lines dekho. Blue line $x$ -direction mein jaati hai, orange $y$ -direction mein. Dono bilkul horizontal hain — yahi hamari starting condition hai.

Figure — Critical points — finding, classifying

Step 2 — Zoom karke dekho: height difference ek pure quadratic hai

KYA. Flat point se ek choti si step $(h, k)$ lo: $h$ east, $k$ north. Hum poochte hain ki height kitni badalti hai, $Δ = f (a + h, b + k) - f (a, b)$ . Ise hum ek Taylor expansion se approximate karte hain.

KYUN yeh tool. Ek Taylor expansion kisi bhi smooth surface ko ek point ke paas likhti hai as: (uski height) + (linear slope terms) + (curvature terms) + (aur choti cheezein). Hum ise isliye use karte hain kyunki yeh ek darauney curved surface ko ek polynomial mein badal deta hai jise hum haath se analyse kar sakte hain. Yahan linear slope terms $f_{x} h + f_{y} k$ hain — aur dono slopes zero hain (Step 1). To pehla surviving piece curvature piece hai.

$Δ \approx = 0 at a flat point f_{x} h + f_{y} k + \frac{1}{2} (f_{xx} h^{2} + 2 f_{x y} hk + f_{y y} k^{2}) .$

Surviving bracket ke term by term:

$f_{xx}$ — $x$ ke along curvature (kya east-west slice muskurata hai ya frownta hai?), $h^{2}$ se multiply.
$f_{y y}$ — $y$ ke along curvature (north-south slice), $k^{2}$ se multiply.
$f_{x y}$ — twist: east-west slope $f_{x}$ kitna badalta hai jab tum north ( $y$ mein) chalte ho. $2 hk$ se multiply.
$\frac{1}{2}$ — Taylor formula se ek bookkeeping constant; yeh poore bracket ko scale karta hai, lekin ek fixed positive number hone ki wajah se sign kabhi nahi badalta.

PICTURE. Flat landscape ko, point ke paas, ek smooth "cup ya cap ya saddle" bowl ne replace kar liya hai jiske sirf ingredients hain ye teen second-derivative numbers.

Step 3 — Kyun hum sirf $f_{xx}$ aur $f_{y y}$ nahi dekh sakte

KYA. Bholi umeed yeh hai: "agar $x$ -slice upar curve kare ( $f_{xx} > 0$ ) aur $y$ -slice upar curve kare ( $f_{y y} > 0$ ), to zaroor bowl hai." Hum dikhate hain ki yeh umeed fail ho sakti hai.

KYUN fail hoti hai. Twist term $2 f_{x y} hk$ diagonal direction mein (jahan $h$ aur $k$ opposite signs mein ho) itna strongly negative ho sakta hai ki $Q$ ko zero se neeche le jaaye — tab bhi jab dono pure curvatures positive hon.

PICTURE. Counterexample $f = x^{2} + y^{2} - 3 x y$ dekho. Axes ke along (blue aur orange slices) zameen upar curve karti hai. Lekin diagonal $y = x$ ke along (red slice) twist jeet jaata hai aur zameen neeche curve karti hai. Dono axis-slices jhooth bolti hain: yeh ek saddle hai.

Step 4 — Trick: complete the square

KYA. Hum $Q$ ko rewrite karte hain taaki twist term $2 f_{x y} hk$ ek perfect square mein absorb ho jaaye. Abhi ke liye assume karo ki $f_{xx} \neq = 0$ .

KYUN complete the square. Ek perfect square $(\dots)^{2}$ kabhi negative nahi ho sakta. Agar hum $Q$ ko (positive-ya-negative number) $\times$ (square) + (ek aur number) $\times$ (square) ki form mein force kar sakein, to do multipliers ke signs humein sab kuch bata denge — kisi direction-hunting ke bina. Completing the square ek standard algebraic move hai jo $a h^{2} + bhk + c k^{2}$ ko squares mein badalta hai.

$f_{xx} h^{2} + 2 f_{x y} hk + f_{y y} k^{2} = f_{xx} (new variable u h + \frac{f _{x y}}{f _{xx}} k)^{2} + (f_{y y} - \frac{f _{x y}^{2}}{f _{xx}}) k^{2} .$

Isko padho:

$f_{xx}$ — pehle square $u^{2}$ ke aage coefficient.
$u = h + \frac{f _{x y}}{f _{xx}} k$ — ek shifted, tilted direction; square $u^{2} \geq 0$ hamesha.
$(f_{y y} - \frac{f _{x y}^{2}}{f _{xx}})$ — doosre square $k^{2}$ ke aage coefficient; ise "leftover" kaho.

PICTURE. Socho jaise apne coordinate axes ko rotate/tilt kar rahe ho taaki twist gayab ho jaaye. Naye tilted axes mein, bowl mein koi cross-term nahi hai — yeh do independent curvatures ka ek clean sum hai, jise hum seedha padh sakte hain.

Step 5 — Leftover coefficient hi discriminant $D$ hai

KYA. Us leftover coefficient ko common denominator par simplify karo: $f_{y y} - \frac{f _{x y}^{2}}{f _{xx}} = \frac{f _{xx} f _{y y} - f _{x y}^{2}}{f _{xx}} = \frac{D}{f _{xx}}, where D = f_{xx} f_{y y} - f_{x y}^{2} .$

KYUN yeh important hai. To poora quadratic form ab yeh hai: $Q = f_{xx} u^{2} + \frac{D}{f _{xx}} k^{2} .$ $u^{2}$ aur $k^{2}$ dono $\geq 0$ hain. Sirf do cheezein signs flip kar sakti hain: do coefficients — $f_{xx}$ aur $D / f_{xx}$ . Number $D$ aa gaya — magic se nahi, balki completing the square ke baad surviving coefficient ki tarah.

$D$ ko theek se naam do. Charon second derivatives ko ek $2 \times 2$ grid mein rakho jise Hessian matrix kehte hain, $H$ likhte hain: $H = (f_{xx} f_{y x} f_{x y} f_{y y}) = Clairaut (f_{xx} f_{x y} f_{x y} f_{y y}) .$

Top-left $f_{xx}$ , bottom-right $f_{y y}$ — do pure curvatures.
Off-diagonal $f_{x y} = f_{y x}$ — twist, Clairaut's theorem se equal, isliye $H$ symmetric hai.

Is $2 \times 2$ matrix ka determinant (top-left times bottom-right, minus do off-diagonals ka product) exactly humara $D$ hai: $det H = f_{xx} f_{y y} - f_{x y} f_{y x} = f_{xx} f_{y y} - (f_{x y})^{2} = D .$

$Q$ mein term by term:

$f_{xx} u^{2}$ — pehle tilted axis ke along curvature. Sign $=$ $f_{xx}$ ka sign.
$\frac{D}{f _{xx}} k^{2}$ — doosre tilted axis ke along curvature. Sign $=$ $D / f_{xx}$ ka sign.

PICTURE. Do independent parabolic slices, ek per tilted axis, har ek ka apna upward ya downward opening jo ek single coefficient se control hota hai.

Step 6 — Har sign padho: saare cases

KYA. Ab hum do coefficients $f_{xx}$ aur $D / f_{xx}$ ke har possible sign combination se guzarte hain.

KYUN. Ek judge-proof test reader ko kabhi confused nahi chhod sakta. Yeh hain saare cases.

Case A — $D > 0$ aur $f_{xx} > 0$ (bowl / minimum). Tab $D / f_{xx} > 0$ bhi. Dono coefficients positive $\Rightarrow Q = (+) u^{2} + (+) k^{2} > 0$ har nonzero step ke liye. Har direction upar jaati hai ⇒ local minimum.

Case B — $D > 0$ aur $f_{xx} < 0$ (dome / maximum). Agar $D > 0$ lekin $f_{xx} < 0$ , to $D / f_{xx} < 0$ bhi. Dono coefficients negative $\Rightarrow Q < 0$ har jagah ⇒ local maximum. (Note: $D > 0$ force karta hai ki $f_{y y}$ $f_{xx}$ ka sign share kare, kyunki $f_{xx} f_{y y} = D + f_{x y}^{2} > 0$ . To $f_{y y}$ ko alag se check karne ki zaroorat nahi.)

Case C — $D < 0$ (saddle). Tab $f_{xx}$ aur $D / f_{xx}$ ke opposite signs hain. Ek tilted axis upar curve karti hai, doosri neeche $\Rightarrow Q > 0$ kuch directions mein, $Q < 0$ doosron mein ⇒ saddle, chahe $f_{xx}$ kuch bhi ho.

Case D (degenerate) — $D = 0$ . Leftover coefficient $D / f_{xx} = 0$ . Tab $Q = f_{xx} u^{2}$ — yeh poori line $u = 0$ ke along flat (exactly zero) hai. Quadratic ab decide nahi kar sakta; higher-order terms le lete hain. Inconclusive. (Example $f = x^{4} + y^{4}$ : yahan $Q \equiv 0$ , phir bhi quartic ise genuine minimum banata hai — sirf direct inspection se pata chalta hai.)

Case E (degenerate) — $f_{xx} = 0$ . Hamara completing-the-square assume karta tha $f_{xx} \neq = 0$ , to ise alag handle karo. Agar $f_{xx} = 0$ to $D = - f_{x y}^{2} \leq 0$ .

Agar $f_{x y} \neq = 0$ : $D < 0$ ⇒ saddle (twist akela opposite curvatures banata hai).
Agar $f_{x y} = 0$ bhi: $D = 0$ ⇒ inconclusive; $f_{y y} k^{2}$ aur higher terms par wapas jao.

To $D$ -first rule tab bhi kaam karta hai jab $f_{xx} = 0$ ho: koi case uncovered nahi rehta.

PICTURE. Char miniature landscapes side by side — bowl, dome, saddle, aur $D = 0$ ke liye ek flat trough — har ek apne sign pattern ke saath tagged.

Step 7 — Ek real example par poori run

KYA. Banaye hue machine ko $f = x^{3} - 3 x + y^{2}$ par apply karo, exactly jaisa parent ke Example 2 mein hai.

KYUN. Dekho ki $D$ aur $f_{xx}$ ek function par point se point par kaise badalte hain.

Flat points dhundo: $f_{x} = 3 x^{2} - 3 = 0 \Rightarrow x = \pm 1, f_{y} = 2 y = 0 \Rightarrow y = 0.$ Do flat points: $(1, 0)$ aur $(- 1, 0)$ .

Second derivatives: $f_{xx} = 6 x$ , $f_{y y} = 2$ , $f_{x y} = 0$ , to $D = f_{xx} f_{y y} - f_{x y}^{2} = (6 x) (2) - 0 = 12 x .$

$(1, 0)$ par: $D = 12 > 0$ aur $f_{xx} = 6 > 0$ → minimum (Case A).
$(- 1, 0)$ par: $D = - 12 < 0$ → saddle (Case C), $f_{xx}$ chahe kuch bhi ho.

PICTURE. Ek landscape dono flat points ko hold kiye hue: $(1, 0)$ par ek valley aur $(- 1, 0)$ par ek pass.

Ek-picture summary

Sab kuch ek flow mein compress hota hai: complete the square → leftover coefficient $D$ hai → $D$ decide karta hai, phir $f_{xx}$ describe karta hai.

Recall Feynman retelling — poori walkthrough plain words mein

Ek hilly field ke flat spot par khade ho. Flat ka matlab hai har direction seedhi shuru hoti hai, to slope akela nahi bata sakta ki tum bowl mein ho ya saddle par. To tum curving feel karte ho. Teen curving-numbers hain: east-west kaise bend karta hai, north-south kaise bend karta hai, aur ek sneaky "twist" number jo dono ke mix ke liye hai (sirf ek twist number hai kyunki Clairaut's theorem kehta hai dono mixed rates equal hain). Seedhe inhe compare nahi kar sakte, kyunki twist confuse karta hai. Clever move: apna sir tilao (complete the square) jab tak twist gayab na ho — ab zameen sirf do clean parabolas hain right angles par. Do numbers bachte hain: ek $f_{xx}$ hai, doosra $D / f_{xx}$ hai jahan $D = f_{xx} f_{y y} - f_{x y}^{2}$ . Agar dono parabolas upar khulte hain, yeh bowl hai (minimum). Dono neeche, dome (maximum). Ek upar ek neeche, saddle. Aur yeh exactly tab hota hai jab $D < 0$ . Agar $D = 0$ , ek parabola flat ho gaya aur hum sirf curving se nahi bata sakte — hume aur gehrai se dekhna hoga. To: $D$ decide karta hai ki kya, $f_{xx}$ batata hai kaunsa.

Connections

Critical points — finding, classifying — parent note; yeh page uska central result derive karta hai.
Taylor Series in Several Variables — Step 2 ka quadratic approximation.
Quadratic Forms and Definiteness — $Q$ definite/indefinite ↔ min/max/saddle.
Hessian Matrix — $D = det H$ ; eigenvalues tilted-axis curvatures hain.
Gradient and Directional Derivatives — $\nabla f = 0$ Step 1 ke flat point ko mark karta hai.
Single-variable Second Derivative Test — "curvature check karo" ka one-variable ancestor.