Second derivative test — Hessian determinant: Worked examples

4.4.13 · D3 · Maths › Multivariable Calculus › Second derivative test — Hessian determinant

Scenario matrix

Kisi bhi example se pehle, un sab cheez ka poora space rakh lete hain jo ho sakti hain. Neeche har worked example us cell ke saath tagged hai jo woh cover karta hai.

Cell	Situation	Kya tricky hai	Example
A	$D > 0, f_{xx} > 0$	saaf minimum	Ex 1
B	$D > 0, f_{xx} < 0$	saaf maximum	Ex 2
C	$D < 0$	saaf saddle	Ex 3
D	Alag-alag quadrants mein kai critical points	har ek ko alag se classify karna padega	Ex 4
E	Point par $f_{xx} = 0$	$u$ mein square complete nahi ho sakta — $f_{y y}$ ki taraf pivot karo	Ex 5
F	$D = 0$ (degenerate)	test fail — haath se investigate karo	Ex 6
G	$D = 0$ lekin actually ek saddle hai	dikhata hai ki $D = 0$ ≠ "flat valley"	Ex 7
H	Real-world word problem (optimisation)	words ko $f$ mein translate karo, units ka dhyan rakho	Ex 8
I	Exam twist: ek parameter jo classification palat deta hai	ek constant ke function ke roop mein classify karo	Ex 9

Example 1 — Cell A: saaf minimum

Worked example

$f (x, y) = x^{2} + x y + y^{2} + 3 x$ Pehle forecast karo: $x^{2}$ aur $y^{2}$ dono terms upar curve karti hain, toh guess karo minimum. Lekin kahan? $+ 3 x$ cheez tilts karta hai — low point origin par nahi hoga.

Step 1 — Critical points dhundo. Dono first partials ko zero karo. $f_{x} = 2 x + y + 3 = 0, f_{y} = x + 2 y = 0$ Ye step kyun? Har extremum ya saddle wahan hota hai jahan gradient vanish kare; tilt term $3 x$ solution ko off-origin move karta hai. $f_{y} = 0$ se: $x = - 2 y$ . $f_{x} = 0$ mein sub karo: $2 (- 2 y) + y + 3 = 0 \Rightarrow - 3 y + 3 = 0 \Rightarrow y = 1$ , phir $x = - 2$ . Critical point: $(- 2, 1)$ .

Step 2 — Second partials. $f_{xx} = 2, f_{y y} = 2, f_{x y} = 1$ Ye step kyun? Ye quadratic form ke coefficients hain jo shape decide karte hain.

Step 3 — Discriminant. $D = (2) (2) - (1)^{2} = 3 > 0$ Ye step kyun? $D > 0$ kehta hai "extremum, saddle nahi"; iski sign pehla gate hai.

Step 4 — Min vs max chuno. $f_{xx} = 2 > 0$ , toh $(- 2, 1)$ par local minimum hai. Ye step kyun? $D > 0$ force karta hai ki $f_{xx}$ aur $f_{y y}$ ek hi sign share karein; positive matlab bowl upar ki taraf khulta hai.

Verify karo: wahan value hai $f (- 2, 1) = 4 - 2 + 1 - 6 = - 3$ . Nudge karo: $f (- 2, 1.1) \approx 4 - 2.2 + 1.21 - 6 = - 2.99 > - 3$ . Har taraf se upar hai ✓. Minimum ke consistent hai.

Example 2 — Cell B: saaf maximum

Worked example

$f (x, y) = - x^{2} - y^{2} + 4 x + 6 y$ Forecast: dono squares negative hain (neeche khulte hain) — ye ek dome hai, maximum expect karo.

Step 1 — Critical point. $f_{x} = - 2 x + 4 = 0 \Rightarrow x = 2, f_{y} = - 2 y + 6 = 0 \Rightarrow y = 3$ Critical point $(2, 3)$ . Ye step kyun? Gradient wahan hai jahan tangent plane flat ho jaata hai — woh akela jagah jahan max/min/saddle reh sakta hai.

Step 2 — Second partials. $f_{xx} = - 2, f_{y y} = - 2, f_{x y} = 0$ . Ye step kyun? Ye teen numbers woh poori input hain jo test ko chahiye; ye curvature measure karte hain har direction mein aur unke beech ki tilt bhi.

Step 3 — Discriminant. $D = (- 2) (- 2) - 0 = 4 > 0$ . Ye step kyun? $D > 0$ saddle ko rule out karta hai, sirf min ya max bachta hai.

Step 4 — Verdict. $f_{xx} = - 2 < 0$ → local maximum. $f_{xx}$ kyun? $D > 0$ se same-sign force hoti hai; negative matlab frowning (dome).

Verify: $f (2, 3) = - 4 - 9 + 8 + 18 = 13$ . Nudge $f (2.1, 3) = - 4.41 - 9 + 8.4 + 18 = 12.99 < 13$ ✓. Peak hai.

Example 3 — Cell C: saaf saddle

Worked example

$f (x, y) = x y$ Forecast: $y = x$ line ke saath hame $x^{2}$ milta hai (upar), $y = - x$ ke saath $- x^{2}$ (neeche). Ek taraf upar, doosri taraf neeche → saddle.

Step 1 — Critical point. $f_{x} = y = 0, f_{y} = x = 0 \Rightarrow (0, 0)$ . Ye step kyun? Sirf wahan surface flat hoti hai — test ke liye prerequisite hai.

Step 2 — Second partials. $f_{xx} = 0, f_{y y} = 0, f_{x y} = 1$ . Ye step kyun? Hume curvature data chahiye; dono pure curvatures yahan zero hain, toh saari shape information cross-term $f_{x y}$ mein hai.

Step 3 — Discriminant. $D = (0) (0) - (1)^{2} = - 1 < 0$ . Yahan kyun ruko? $D < 0$ akela saddle declare karta hai — $f_{xx}$ check karne ki zaroorat nahi.

Step 4 — Verdict. $(0, 0)$ par Saddle point. Ye step kyun? Ye woh conclusion state karta hai jo $D$ ki sign force karti hai, classification close karte hue.

Neeche ki figure is saddle ko ek contour map ke roop mein dikhati hai: blue curves woh jagah hain jahan $f = x y$ positive hai (point se door uthti hai), pink curves jahan negative hai (girti hai), aur yellow dashed axes $f = 0$ lines hain. Dekho ki dono signs us akele white critical point ko gherate hain — "upar" aur "neeche" directions ka yeh co-existence $D < 0$ ka visual fingerprint hai.

Verify karo: $f (0.1, 0.1) = 0.01 > 0$ lekin $f (0.1, - 0.1) = - 0.01 < 0$ — point ke paas dono signs hain ✓. Yahi two-sign behaviour hai jo $D < 0$ predict karta hai.

Example 4 — Cell D: kai critical points, charon quadrants

Worked example

$f (x, y) = x^{3} - 12 x + y^{3} - 3 y$ Forecast: cubics $x$ aur $y$ dono mein do turning points dete hain, toh chaar critical points expect karo jo quadrants mein bikhre hain — verdicts ka mix.

Step 1 — Critical points. Variables separate ho jaate hain: $f_{x} = 3 x^{2} - 12 = 0 \Rightarrow x = \pm 2, f_{y} = 3 y^{2} - 3 = 0 \Rightarrow y = \pm 1$ Ye step kyun? Kyunki $f$ ek $x$ -part aur $y$ -part mein split hota hai, gradient conditions decouple ho jaati hain; $x = \pm 2$ aur $y = \pm 1$ ka har combination ek critical point hai. Chaar points: $(2, 1), (2, - 1), (- 2, 1), (- 2, - 1)$ — har quadrant mein ek.

Step 2 — Second partials. $f_{xx} = 6 x, f_{y y} = 6 y, f_{x y} = 0$ , toh $D = 36 x y$ . Ye step kyun? $f_{x y} = 0$ ke saath, discriminant sirf dono diagonal curvatures ka product hai — $x y$ ki sign sab decide karta hai.

Step 3 — Har ek classify karo. (Yaad karo $D = 36 x y$ , aur agar $D > 0$ to $f_{xx} = 6 x$ padho.)

Point	$D = 36 x y$	$f_{xx} = 6 x$	Verdict
$(2, 1)$	$+ 72$	$+ 12$	min
$(2, - 1)$	$- 72$	—	saddle
$(- 2, 1)$	$- 72$	—	saddle
$(- 2, - 1)$	$+ 72$	$- 12$	max

Ye table kyun? Har quadrant alag verdict produce karta hai — yahi woh case class hai jahan aapko ek baar classify karke reuse bilkul nahi karna.

Verify: $(2, 1)$ par, $f = 8 - 24 + 1 - 3 = - 18$ ; $(- 2, - 1)$ par, $f = - 8 + 24 - 1 + 3 = 18$ . Min value $<$ max value, aur dono saddles $(2, - 1), (- 2, 1)$ par $f = 8 - 24 - 1 + 3 = - 14$ aur $- 8 + 24 + 1 - 3 = 14$ dete hain, beech mein baithte hain — geometrically sensible ✓.

Example 5 — Cell E: $f_{xx} = 0$ ka pivot

Worked example

$f (x, y) = y^{2} + 4 x y$ (dhyan se — derivation ne assume kiya tha $f_{xx} \neq = 0$ ) Forecast: koi $x^{2}$ term nahi, toh $f_{xx} = 0$ hai. Definition box se yaad karo ki parent derivation ne $u = x - a, v = y - b$ set kiya aur square $u$ mein complete kiya, ek move jisne quietly $f_{xx}$ se divide kiya. Kya test tab bhi verdict deta hai jab $f_{xx} = 0$ ho?

Step 1 — Critical point. $f_{x} = 4 y = 0 \Rightarrow y = 0, f_{y} = 2 y + 4 x = 0 \Rightarrow x = 0$ Critical point $(0, 0)$ . Ye step kyun? Test sirf wahan apply hota hai jahan gradient vanish kare; second derivatives touch karne se pehle woh point locate karna zaroori hai. (Yahan $(a, b) = (0, 0)$ hai, toh shift variables simply $u = x, v = y$ hain.)

Step 2 — Second partials. $f_{xx} = 0, f_{y y} = 2, f_{x y} = 4$ . Ye step kyun? Ye curvatures discriminant ko feed karte hain; note karo flagged feature — $f_{xx} = 0$ — jo bilkul wahi degeneracy hai jo ye example test karne ke liye banayi gayi hai.

Step 3 — Discriminant. $D = (0) (2) - (4)^{2} = - 16 < 0$ . Ye humein kyun bachata hai: $D < 0$ directly saddle deta hai aur kabhi $f_{xx}$ use nahi karta. "Complete the square in $u$ " wali chinta sirf $D > 0$ min/max split ke liye thi, jahan $f_{xx}$ se divide karna zaruri tha. Jab $f_{xx} = 0$ lekin $D < 0$ ho, toh hum simply $v$ mein square complete karte hain ( $f_{y y} = 2 \neq = 0$ se divide karte hue) — geometry, do opposite-sign curvatures, unchanged hai.

Step 4 — Verdict. $(0, 0)$ par Saddle. Ye step kyun? Ye woh conclusion record karta hai jo $D < 0$ dictate karta hai, dikhate hue ki degenerate pivot ne answer block nahi kiya.

Verify: $y = 0$ ke saath: $f = 0$ flat; $x = - y$ ke saath: $f = y^{2} - 4 y^{2} = - 3 y^{2} < 0$ ; $x = y$ ke saath: $f = y^{2} + 4 y^{2} = 5 y^{2} > 0$ . Dono signs appear hote hain ✓ — saddle confirmed.

Example 6 — Cell F: genuine $D = 0$ , inconclusive phir resolved

Worked example

$f (x, y) = x^{4} + y^{4}$ Forecast: fourth powers $0$ ke paas quadratics se zyada flat hote hain. Test andha ho sakta hai ( $D = 0$ ), bhale hi intuitively ye ek minimum ho.

Step 1 — Critical point. $f_{x} = 4 x^{3} = 0, f_{y} = 4 y^{3} = 0 \Rightarrow (0, 0)$ . Ye step kyun? Yahan gradient zero hai, ye confirm karna zaroori hai second derivative test ke baat karne se pehle.

Step 2 — Second partials. $f_{xx} = 12 x^{2}, f_{y y} = 12 y^{2}, f_{x y} = 0$ . $(0, 0)$ par sab $0$ hain. Ye step kyun? Hum curvature critical point par evaluate karte hain; har entry zero hona warning sign hai ki quadratic approximation flat hai.

Step 3 — Discriminant. $D = (0) (0) - 0 = 0$ → inconclusive. Test kyun fail karta hai: second-order Taylor information bilkul flat hai; asli shape fourth-order terms mein chhupi hai jo Hessian dekh nahi sakta.

Step 4 — Haath se investigate karo. Kisi bhi $(x, y) \neq = (0, 0)$ ke liye, $f = x^{4} + y^{4} > 0 = f (0, 0)$ . Toh ye ek strict local (aur global) minimum hai. Ye step kyun: test ke chup rehne par, hum quadratic form bypass karte hain aur $f - f (0, 0)$ ki sign directly check karte hain — minimum ki definition.

Verify: $f (0.1, 0) = 0.0001 > 0$ aur $f (0, 0.1) = 0.0001 > 0$ ; origin ke paas $f$ kahin negative nahi hai ✓. Minimum, $D = 0$ ke bawajood.

Example 7 — Cell G: $D = 0$ jo actually ek saddle hai

Worked example

$f (x, y) = x^{4} - y^{4}$ Forecast: Example 6 jaisa dikhta hai lekin sign flip ke saath. Woh flip "min" ko "saddle" mein badal dena chahiye — prove karta hai ki $D = 0$ koi guaranteed verdict nahi deta.

Step 1 — Critical point. $f_{x} = 4 x^{3}, f_{y} = - 4 y^{3} \Rightarrow (0, 0)$ . Ye step kyun? Wahi rule jaisa hamesha — gradient vanish hone ki jagah locate karo classify karne se pehle.

Step 2 — $(0, 0)$ par second partials. $f_{xx} = 12 x^{2} = 0, f_{y y} = - 12 y^{2} = 0, f_{x y} = 0$ . Ye step kyun? Hum point par curvature padhte hain; phir se sab zero matlab Hessian ek flat quadratic dekh raha hai aur uninformative hoga.

Step 3 — Discriminant. $D = 0$ → phir se inconclusive. Ye step kyun? $D$ compute karna phir bhi pehla move hai; yahan ye sirf confirm karta hai ki test decide nahi kar sakta, by-hand dekhna zaroori hai.

Step 4 — By hand. $x$ -axis ke saath ( $y = 0$ ): $f = x^{4} > 0$ (upar curve karta hai). $y$ -axis ke saath ( $x = 0$ ): $f = - y^{4} < 0$ (neeche curve karta hai). Do signs → saddle. Ye kyun matter karta hai: Examples 6 aur 7 mein ek hi $D = 0$ hai phir bhi opposite geometry — pakka proof ki $D = 0$ hands-on investigation maangta hai, kabhi lazy default nahi.

Verify: $f (0.1, 0) = 0.0001 > 0$ lekin $f (0, 0.1) = - 0.0001 < 0$ ✓ — genuine saddle.

Example 8 — Cell H: real-world optimisation (units ka dhyan rakho)

Worked example Ek box factory

Ek open-top rectangular box mein $V = 32 m^{3}$ hold karna zaroori hai. Base $x \times y$ (metres) aur height $h$ ke saath, hamein $h = 32/ (x y)$ milta hai. Use ki gayi material (in $m^{2}$ ) base plus chaar walls hai: $S (x, y) = x y + 2 x h + 2 y h = x y + \frac{64}{y} + \frac{64}{x}$ Forecast: surface minimize karne par almost square base milna chahiye — $x = y$ expect karo.

Step 1 — Critical point. $S_{x} = y - \frac{64}{x ^{2}} = 0, S_{y} = x - \frac{64}{y ^{2}} = 0$ Ye step kyun? Sabse kam material ek constrained extremum par hoti hai, yahan $h$ substitute karke unconstrained bana diya gaya hai. Pehle se: $y = 64/ x^{2}$ . Doosre mein sub karo: $x = 64/ y^{2} = 64 x^{4} /6 4^{2} = x^{4} /64 \Rightarrow x^{3} = 64 \Rightarrow x = 4$ . Phir $y = 64/16 = 4$ . Critical point $(4, 4)$ , height $h = 32/16 = 2 m$ .

Step 2 — Second partials. $S_{xx} = \frac{128}{x ^{3}}, S_{y y} = \frac{128}{y ^{3}}, S_{x y} = 1$ $(4, 4)$ par: $S_{xx} = 128/64 = 2, S_{y y} = 2, S_{x y} = 1$ . Ye step kyun? Ye curvatures batate hain ki critical point sach mein sabse sasta design hai ya disguise mein saddle — $64/ x$ jaisi rational functions ke saath genuine risk hai.

Step 3 — Discriminant. $D = (2) (2) - 1^{2} = 3 > 0$ , $S_{xx} = 2 > 0$ → local minimum ✓ (sabse kam material). $f_{xx} > 0$ kyun matter karta hai: confirm karta hai ki humne sabse sasta box dhundha, koi costly saddle nahi.

Verify (units + numbers): dimensions $4 \times 4 \times 2 m$ , volume $= 4 \cdot 4 \cdot 2 = 32 m^{3}$ ✓ (constraint meet hui). Surface $S = 16 + 64/4 + 64/4 = 16 + 16 + 16 = 48 m^{2}$ . Saare terms areas hain ( $m^{2}$ ) ✓ — dimensionally consistent hai.

Example 9 — Cell I: parameter ke saath exam twist

Worked example

$f (x, y) = x^{2} + k x y + y^{2}$ ke liye $(0, 0)$ ko classify karo jab $k$ vary kare Forecast: chhota $k$ bowl rakhna chahiye; bada $k$ saddle mein tip kar dena chahiye. Ek threshold value honi chahiye $k$ ki jahan wo palat jaaye.

Step 1 — Critical point. $f_{x} = 2 x + k y, f_{y} = k x + 2 y$ . $(0, 0)$ par dono har $k$ ke liye vanish hote hain, toh $(0, 0)$ hamesha critical hai. Ye step kyun? Koi linear terms nahi — origin $k$ ki parwah kiye bina critical point ke roop mein fixed hai, toh hum ek point study kar sakte hain jab $k$ move kare.

Step 2 — Second partials. $f_{xx} = 2, f_{y y} = 2, f_{x y} = k$ . Ye step kyun? Yahan tilt term $f_{x y} = k$ tunable knob hai; ise isolate karne se dikhta hai ki parameter test mein sirf cross-curvature ke through enter karta hai.

Step 3 — $k$ ke function ke roop mein discriminant. $D (k) = (2) (2) - k^{2} = 4 - k^{2}$ Ye step kyun? Poora verdict ab $4 - k^{2}$ ki sign par depend karta hai.

Step 4 — $k$ par case split.

$∣ k ∣ < 2$ → $D > 0$ aur $f_{xx} = 2 > 0$ → minimum. Kyun: chhoti tilt dono curvatures ko dominant aur same-signed rakhti hai, toh bowl survive karta hai.
$∣ k ∣ > 2$ → $D < 0$ → saddle. Kyun: bada cross-term diagonal curvatures ko overpower karta hai, opposite-sign directions force karta hai.
$∣ k ∣ = 2$ → $D = 0$ → inconclusive, haath se check karna padega. Do symmetric sub-cases:
- $k = + 2$ : $f = x^{2} + 2 x y + y^{2} = (x + y)^{2} \geq 0$ , ek degenerate minimum hai, line $x = - y$ ke saath flat (jahan $x + y = 0$ ). Wahan flat kyun: us line par $f = 0 = f (0, 0)$ , ek valley floor na ki ek akela low point.
- $k = - 2$ : $f = x^{2} - 2 x y + y^{2} = (x - y)^{2} \geq 0$ , bhi ek degenerate minimum hai, $x = y$ ke saath flat (jahan $x - y = 0$ ). Mirror kyun: $k$ ki sign flip karne se flat valley doosre diagonal par mirror ho jaati hai.

Figure $D (k) = 4 - k^{2}$ ko $k$ ke against plot karta hai. Blue region ( $D > 0$ ) woh hai jahan origin minimum hai; pink regions ( $D < 0$ ) saddles hain; yellow dots $k = \pm 2$ par wo $D = 0$ thresholds mark karte hain jahan verdict palat jaata hai aur valley flat ho jaati hai. Left se right padhne par tum dekhte ho ki akela parameter point ko saddle → min → saddle le jaata hai.

Verify: $k = 1$ par, $D = 3 > 0$ (min); $k = 3$ par, $D = 4 - 9 = - 5 < 0$ (saddle); $k = 2$ par, $D = 0$ aur $f = (x + y)^{2} \geq 0$ flat along $x + y = 0$ ; $k = - 2$ par, $D = 0$ aur $f = (x - y)^{2} \geq 0$ flat along $x - y = 0$ ✓. Threshold $∣ k ∣ = 2$ par confirmed.

Recall Kaunsa cell sabse mushkil tha, aur kyun?

Cells F aur G — kyunki $D = 0$ Hessian matrix ko andha kar deta hai. Question ::: Hessian sirf second-order curvature dekhta hai; jab dono eigenvalues $0$ houn, asli shape higher-order terms mein chhupi hoti hai aur aapko $f$ directly inspect karna padta hai.

Active recall

D < 0

kya verdict deta hai,

f_{xx}

ki parwah kiye bina?

Saddle point.

f = x^{2} + x y + y^{2} + 3 x

ke liye, critical point kahan hai aur uski type kya hai?

(- 2, 1)

, local minimum (

D = 3 > 0

f_{xx} = 2 > 0

f = x^{3} - 12 x + y^{3} - 3 y

ke liye,

(2, - 1)

classify karo.

Saddle (

D = 36 x y = - 72 < 0

Jab

f_{xx} = 0

lekin

D < 0

ho, kya test valid hai?

Haan —

D < 0

saddle deta hai aur kabhi

f_{xx}

use nahi karta.

D > 0

aur

f_{xx} = 0

saath kyun exist nahi kar sakte?

Tab

D = - f_{x y}^{2} \leq 0

hota hai, jo

D > 0

se contradict karta hai.

f = x^{4} + y^{4}

ke liye origin par:

D

aur asli type?

D = 0

; genuine minimum (

f > 0

inspect karke mila).

f = x^{4} - y^{4}

ke liye origin par:

D

aur asli type?

D = 0

; saddle (

x

ke saath upar,

y

ke saath neeche).

Volume 32 minimize karte hue surface ka box: dimensions?

4 \times 4 \times 2

m, surface

48 m^{2}

f = x^{2} + k x y + y^{2}

ke liye, threshold

∣ k ∣

jo min→saddle flip kare?

∣ k ∣ = 2

(

D = 4 - k^{2}

Connections

Second derivative test — Hessian determinant — parent rule jise ye examples exercise karte hain.
Critical points and gradient — har example $\nabla f = 0$ solve karke start hota hai.
Taylor series multivariable — kyun $D = 0$ higher-order terms mein answer chhupta hai.
Hessian matrix — woh object jiska determinant $D$ hai.
Quadratic forms and definiteness — $∣ k ∣ < 2$ / $∣ k ∣ > 2$ split action mein definiteness hai.
Lagrange multipliers — box problem ke peeche constrained cousin.

Worked examples — Second derivative test — Hessian determinant

Scenario matrix

Example 1 — Cell A: saaf minimum

Example 2 — Cell B: saaf maximum

Example 3 — Cell C: saaf saddle

Example 4 — Cell D: kai critical points, charon quadrants

Example 5 — Cell E: $f_{xx} = 0$ ka pivot

Example 6 — Cell F: genuine $D = 0$ , inconclusive phir resolved

Example 7 — Cell G: $D = 0$ jo actually ek saddle hai

Example 8 — Cell H: real-world optimisation (units ka dhyan rakho)

Example 9 — Cell I: parameter ke saath exam twist

Active recall

Connections

Concept Map