Exercises — Second derivative test — Hessian determinant

HUM KYA karte hain: $D=f_{xx}{f}_{yy}-(f_{xy}{)}^2$ mein plug in karo. $D=(4)(9)-0^2=36>0.$ $f_{xx}$ kyun check karein: $D>0$ kehta hai "extremum hai", lekin kaun sa nahi. $f_{xx}=4>0$ ka sign kehta hai surface upar ki taraf muskurati hai. Verdict: local minimum.

$D=(-2)(-5)-(1{)}^2=10-1=9>0.$ Dono diagonal entries negative hain, aur waqai $D>0$ unhe same sign share karne par majboor karta hai. $f_{xx}=-2<0$ → surface jhukti hai. Verdict: local maximum.

$D=(3)(1)-(4{)}^2=3-16=-13<0.$ Kyun ruk jaate hain: $D<0$ akele hi settle kar deta hai — $f_{xx}$ dekhne ki zarurat nahi. Geometrically cross-term $f_{xy}=4$ itna bada hai ki bowl ko Pringle mein tilt kar deta hai. Verdict: saddle point.

Step 1 — dhundho jahan zameen flat hai. KYU: test sirf $\nabla f=0$ par apply hota hai. $f_x=2x-2=0\Rightarrow x=1,f_y=8y+8=0\Rightarrow y=-1.$ Critical point: $(1,-1)$. Step 2 — second derivatives (yahan sab constants hain): $f_{xx}=2,f_{yy}=8,f_{xy}=0.$ Step 3 — discriminant: $D=(2)(8)-0^2=16>0,f_{xx}=2>0.$ Verdict: $(1,-1)$ par local minimum.

Step 1: $f_x=2x-y=0,f_y=-x+2y-3=0.$ Pehli equation se, $y=2x$. Doosri mein substitute karo: $-x+2(2x)-3=0\Rightarrow 3x=3\Rightarrow x=1$, toh $y=2$. Critical point: $(1,2)$. Step 2: $f_{xx}=2,f_{yy}=2,f_{xy}=-1$. Step 3: $D=(2)(2)-(-1{)}^2=4-1=3>0,f_{xx}=2>0.$ Verdict: $(1,2)$ par local minimum.

Step 1: $f_x=y=0$ aur $f_y=x=0$ → sirf ek critical point $(0,0)$. Step 2: $f_{xx}=0,f_{yy}=0,f_{xy}=1$. Step 3: $D=(0)(0)-(1{)}^2=-1<0.$ Verdict: $(0,0)$ par saddle point. Figure dekho: $y=x$ ke saath $xy$ product chadhta hai, $y=-x$ ke saath girta hai — saddle ki pehchaan.

Step 1: $f_x=3x^2-12=0\Rightarrow x^2=4\Rightarrow x=\pm 2$. Aur $f_y=2y=0\Rightarrow y=0$. Do critical points: $(2,0)$ aur $(-2,0)$. KYU do: $x^3$ $f_x$ ko quadratic banata hai, toh do roots. Step 2: $f_{xx}=6x,f_{yy}=2,f_{xy}=0$. Dhyaan do $f_{xx}$ point par depend karta hai. Step 3, $(2,0)$ par: $f_{xx}=12$, toh $D=(12)(2)-0=24>0$, $f_{xx}>0$ → local minimum. Step 3, $(-2,0)$ par: $f_{xx}=-12$, toh $D=(-12)(2)-0=-24<0$ → saddle point. Lesson: wahi formula alag alag points par alag verdicts deta hai kyunki $f_{xx}=6x$ sign badalta hai.

Step 1: $f_x=3x^2-3y=0\Rightarrow y=x^2$; $f_y=3y^2-3x=0\Rightarrow x=y^2$. $y=x^2$ substitute karo: $x=(x^2{)}^2=x^4\Rightarrow x^4-x=0\Rightarrow x(x^3-1)=0\Rightarrow x=0\text{ or }x=1$. Critical points: $(0,0)$ aur $(1,1)$. Step 2: $f_{xx}=6x,f_{yy}=6y,f_{xy}=-3$. $(0,0)$ par: $D=(0)(0)-(-3{)}^2=-9<0$ → saddle point. $(1,1)$ par: $D=(6)(6)-(-3{)}^2=36-9=27>0$, $f_{xx}=6>0$ → local minimum.

Step 1: $f_x=2x+ky$, $f_y=kx+2y$; dono $(0,0)$ par zero hote hain, toh yeh har $k$ ke liye critical hai. Step 2: $f_{xx}=2,f_{yy}=2,f_{xy}=k$. Step 3: $D=(2)(2)-k^2=4-k^2$. Saddle ke liye $D<0$ chahiye: $4-k^2<0\Rightarrow k^2>4\Rightarrow |k|>2$. Verdict: $k>2$ ya $k<-2$ par saddle. ($k=\pm 2$ par, $D=0$: inconclusive. $|k|<2$ par: minimum.)

Taylor ka picture. Ek point $(a,b)$ ke paas, $u=x-a,v=y-b$ ke saath second order tak expand karo: $f(a+u,b+v)\approx f(a,b)+\underset{\text{linear}}{\underbrace{f_{x}u+f_{y}v}}+\frac{1}{2}(f_{xx}{u}^2+2f_{xy}uv+f_{yy}{v}^2).$ Test sirf quadratic part se shape padhta hai. Yeh tabhi valid hai jab linear part zero ho — yaani $f_x=f_y=0$. Warna linear term $f_{x}u+f_{y}v$ chhote $u,v$ ke liye dominant hota hai (woh quadratic ko point ke paas beat karta hai), toh surface tilted hai, aur "min/max/saddle" sahi sawaal bhi nahi hai — sirf ek slope hai. $(1,0)$ par $f=x^2+y^2$ ke liye: $f_x=2x=2\ne 0$. Gradient nonzero hai; surface bas $x$ mein upar ki taraf slope karti hai. Andhaa dhundh $D=f_{xx}{f}_{yy}-f_{xy}^2=4>0$ compute karna aur "minimum" declare karna meaningless hai — $(1,0)$ bowl ke neeche nahi, side par hai. Pehle hamesha $\nabla f=0$ solve karo.

(a) Direct: $D=f_{xx}{f}_{yy}-f_{xy}^2=(3)(3)-(1{)}^2=9-1=8$. (b) Eigenvalues $\det (H-\lambda I)=0$ solve karte hain: $\det \left(\begin{array}{cc}3-\lambda & 1\\ 1& 3-\lambda \end{array}\right)=(3-\lambda {)}^2-1=0\Rightarrow 3-\lambda =\pm 1\Rightarrow \lambda =2\text{ or }4.$ Product: ${\lambda }_1{\lambda }_2=2\times 4=8=D$. ✓ Geometry: dono eigenvalues positive hain → surface dono principal axes ke saath upar curve karti hai → positive definite → local minimum. Eigenvalues $2$ aur $4$ bowl ke do principal directions mein curvatures hain.

HUM KYA chahiye: $(2,-1)$ par critical point, neeche ki taraf curvature, koi cross-term nahi. Ek shifted downward paraboloid kaam karega: $f(x,y)=-3(x-2{)}^2-2(y+1{)}^2.$ Critical point check karo: $f_x=-6(x-2)=0\Rightarrow x=2$; $f_y=-4(y+1)=0\Rightarrow y=-1$. ✓ Derivatives check karo: $f_{xx}=-6$ ✓, $f_{yy}=-4$, $f_{xy}=0$ ✓. Test: $D=(-6)(-4)-0^2=24>0$ aur $f_{xx}=-6<0$ → local maximum. ✓ Construction successful.

Step 1: $f_x=4x^3=0$, $f_y=4y^3=0$ → $(0,0)$ par critical hai. Step 2: $f_{xx}=12x^2$, $f_{yy}=12y^2$, $f_{xy}=0$. $(0,0)$ par: $f_{xx}=0$, $f_{yy}=0$, $f_{xy}=0$. Step 3: $D=(0)(0)-0^2=0$ → inconclusive. Quadratic Taylor part bilkul flat hai, toh second-order info kuch nahi batati. Haath se: $f(x,y)=x^4+y^4\ge 0$ sab $(x,y)$ ke liye, equality sirf $(0,0)$ par (dono fourth powers $\ge 0$ hain). Toh $f(0,0)=0$ woh sabse chhoti value hai jo $f$ kahi bhi leta hai → global (hence local) minimum. Curvature quartic hai, quadratic nahi, yehi exact reason hai ki second-derivative test ise nahi dekh sakta.

Step 1: $f_x=3x^2=0\Rightarrow x=0$; $f_y=0$ sab $y$ ke liye. Toh poori line $x=0$ critical hai! $(0,0)$ par focus karo. Step 2: $f_{xx}=6x$, $f_{yy}=0$, $f_{xy}=0$. $(0,0)$ par: sab zero. Step 3: $D=0$ → inconclusive. Haath se: $+x$ mein move karo, $f=x^3>0$; $-x$ mein move karo, $f=x^3<0$. Toh $f$ ek taraf badhta hai aur doosri taraf ghatta hai usi axis ke saath — jaise $y=x^3$ ka inflection point ek surface mein lift ho gaya. Yeh strict min nahi hai (neeche values exist karti hain), strict max bhi nahi (upar values exist karti hain). $y$ mein move karte waqt, $f$ flat rehta hai. Yeh ek degenerate inflection ridge hai, genuine saddle nahi (saddle ko opposite signs ki curvature chahiye; yahan $x$-behaviour cubic hai, quadratic nahi). Yahi exact "$D=0$ = haath se investigate karo" case hai.

Step 1 — critical points: $f_x=4x^3-4x=4x(x^2-1)=0\Rightarrow x=0,1,-1$. Aur $f_y=2y=0\Rightarrow y=0$. Teen critical points: $(-1,0),(0,0),(1,0)$. Step 2 — derivatives: $f_{xx}=12x^2-4$, $f_{yy}=2$, $f_{xy}=0$. Toh $D=(12x^2-4)(2)$. $(0,0)$ par: $f_{xx}=-4$, $D=(-4)(2)=-8<0$ → saddle point. $(1,0)$ par: $f_{xx}=12-4=8$, $D=(8)(2)=16>0$, $f_{xx}>0$ → local minimum. $(-1,0)$ par: $f_{xx}=8$, $D=16>0$, $f_{xx}>0$ → local minimum. Shape: ek "double well" — do valleys $(\pm 1,0)$ par jo origin ke upar ek saddle ridge se separated hain (classic $x^4-2x^2$ double-well profile, $y^2$ bowl se lift kiya gaya).