Critical points — finding, classifying: Exercises

4.4.12 · D4 · Maths › Multivariable Calculus › Critical points — finding, classifying

Shuru karne se pehle, wo do tools jo poori tarah hum use karte hain:

Yahaan $f_{x}$ ka matlab hai "agar aap sirf $x$ -direction mein chalte hain aur $y$ freeze karte hain to $f$ ka slope", aur $f_{xx}$ ka matlab hai "woh $x$ -slope khud kitna badal raha hai" — yani $x$ ke along curvature. $f_{x y}$ cross term hai: $x$ -slope kitna badalta hai jab aap $y$ shift karte hain. Agar ye words naye lagte hain, pehle parent Critical points — finding, classifying dobara padho; derivation Taylor Series in Several Variables aur Quadratic Forms and Definiteness mein hai.

Figure — Critical points — finding, classifying

Upar ki picture recipe ke step 3 ki answer key hai. Isko left se right padho:

Left (bowl): dono curvatures upar point karti hain, woh agree karti hain → $D = f_{xx} f_{y y} - (f_{x y})^{2} > 0$ with $f_{xx} > 0$ . Yeh Exercise 3 ka $(4, - 2)$ aur Exercise 8 ka $(0, 0)$ hai.
Middle (dome): dono curvatures neeche point karti hain, phir bhi agree kar rahi hain → $D > 0$ lekin $f_{xx} < 0$ . Yeh Exercise 7(b) ka $g = - x^{4} - y^{4}$ origin par hai — ek local maximum.
Right (saddle): dono curvatures fight karti hain — ek axis ke along upar, doosri ke along neeche → $D < 0$ . Yeh Exercise 5 ka $(0, 0)$ aur har saddle jo aap neeche miloge.

To jab aap $D$ compute karte hain, aap literally picture se pooch rahe hain: "kya curvatures agree karti hain (extremum) ya fight karti hain (saddle)?" Amber dot har panel mein critical point mark karta hai — woh ek flat spot jahaan recipe apply hoti hai.

L1 — Recognition

Recall Solution 1

Hum kya karte hain: bas $D = f_{xx} f_{y y} - (f_{x y})^{2}$ mein plug karo. Koi gradient nahi, koi Taylor nahi — second derivatives humein diye gaye hain. $D = (4) (9) - (2)^{2} = 36 - 4 = 32 > 0.$ Kyunki $D > 0$ hai, yeh ek extremum hai, saddle nahi. Ab $f_{xx} = 4 > 0$ hai ("cup, smiley"), to yeh upar ki taraf curve karta hai. Ise upar ki figure ke left panel se match karo. Conclusion: local minimum.

Recall Solution 2

$D = - 7 < 0$ . Jis pal $D < 0$ hota hai, dono curvatures ke opposite signs hote hain — ek taraf upar, doosri taraf neeche — to yeh ek saddle point hai, bas (figure ka right panel). $f_{xx}$ ki value yahaan irrelevant hai; ise check karna waste of effort hai. Student ki galti: unhone max vs min pick karne ke liye $f_{xx}$ use karne ki koshish ki, lekin woh branch tabhi exist karti hai jab $D > 0$ ho. $D < 0$ ke saath describe karne ke liye koi max/min hai hi nahi. Conclusion: saddle.

L2 — Application

Recall Solution 3

Step 1 — gradient (kyun: extrema ko har direction mein zero slope chahiye). $f_{x} = 2 x + y - 6, f_{y} = x + 2 y .$ Step 2 — system solve karo. $f_{y} = 0$ se: $x = - 2 y$ . $f_{x} = 0$ mein substitute karo: $2 (- 2 y) + y - 6 = 0 \Rightarrow - 3 y - 6 = 0 \Rightarrow y = - 2, x = - 2 (- 2) = 4.$ Ek critical point: $(4, - 2)$ . Step 3 — discriminant. $f_{xx} = 2, f_{y y} = 2, f_{x y} = 1$ (constants — yeh $f$ ek pure quadratic hai, to curvature har jagah same hai). $D = (2) (2) - (1)^{2} = 3 > 0, f_{xx} = 2 > 0.$ Conclusion: local minimum at $(4, - 2)$ — bowl (figure ka left panel).

Recall Solution 4

Step 1. $f_{x} = 3 x^{2} - 12, f_{y} = 2 y - 4$ . Step 2. $3 x^{2} - 12 = 0 \Rightarrow x = \pm 2$ ; $2 y - 4 = 0 \Rightarrow y = 2$ . Do candidates: $(2, 2)$ aur $(- 2, 2)$ . Step 3. $f_{xx} = 6 x, f_{y y} = 2, f_{x y} = 0$ , to $D = 12 x$ .

$(2, 2)$ par: $D = 24 > 0$ , $f_{xx} = 12 > 0$ → local minimum.
$(- 2, 2)$ par: $D = - 24 < 0$ → saddle. Kyun woh different hain: $x$ -curvature $f_{xx} = 6 x$ ka sign $x$ ke saath flip hota hai. Jahaan yeh upar curve karta hai (positive $x$ ) woh $y$ -bowl se agree karta hai → min; jahaan yeh neeche curve karta hai (negative $x$ ) woh fight karta hai → saddle.

L3 — Analysis

Recall Solution 5

Step 1. $f_{x} = 2 x - 3 y, f_{y} = 2 y - 3 x$ . Step 2. $2 x = 3 y$ aur $2 y = 3 x$ . $x = \frac{3}{2} y$ substitute karo $2 y = 3 x$ mein: $2 y = \frac{9}{2} y \Rightarrow y (2 - \frac{9}{2}) = 0 \Rightarrow y = 0 \Rightarrow x = 0$ . Sirf $(0, 0)$ . Step 3. $f_{xx} = 2, f_{y y} = 2, f_{x y} = - 3$ . $D = (2) (2) - (- 3)^{2} = 4 - 9 = - 5 < 0.$ Conclusion: saddle at $(0, 0)$ — even though $f_{xx} = 2 > 0$ aur $f_{y y} = 2 > 0$ . Bada cross term $f_{x y} = - 3$ dono ko overpower karta hai, to $x = y$ line ke along surface actually dip karta hai (figure ka right panel, chahe dono diagonal curvatures positive read ho rahi hon).

Recall Solution 6

Step 1. $f_{x} = 3 x^{2} - 3 y, f_{y} = 3 y^{2} - 3 x$ . Step 2. $f_{x} = 0$ se: $y = x^{2}$ . $f_{y} = 0$ mein sub karo: $3 (x^{2})^{2} - 3 x = 0 \Rightarrow 3 x^{4} - 3 x = 0 \Rightarrow 3 x (x^{3} - 1) = 0$ . To $x = 0$ (→ $y = 0$ ) ya $x = 1$ (→ $y = 1$ ). Critical points $(0, 0)$ aur $(1, 1)$ . Step 3. $f_{xx} = 6 x, f_{y y} = 6 y, f_{x y} = - 3$ , to $D = 36 x y - 9$ .

$(0, 0)$ par: $D = 0 - 9 = - 9 < 0$ → saddle.
$(1, 1)$ par: $D = 36 - 9 = 27 > 0$ , $f_{xx} = 6 > 0$ → local minimum.

L4 — Synthesis

Recall Solution 7

Teeno ke liye, $(0, 0)$ par $f_{x}, f_{y}, f_{xx}, f_{y y}, f_{x y}$ sab zero ho jaate hain (har term degree 4 ka hai, to wahaan first aur second derivatives zero hain). Isliye teeno ke liye $D = 0$ hai — test inconclusive hai. $D = 0$ ka matlab "inconclusive" kyun hai (key idea, sirf rule nahi): $D$ Taylor expansion ke second-order (quadratic) piece se bana hai — woh bowl/dome/saddle shape jo surface ko point ke paas approximate karta hai. Jab har second derivative zero ho, woh quadratic approximation completely flat hoti hai: second order tak ye teeno surfaces identical dikhti hain (ek flat table). Shape apne aap ko sirf quartic ( $x^{4}, y^{4}$ ) terms mein reveal karta hai, jo Hessian kabhi nahi dekhta. To humein woh higher-order terms haath se inspect karne padte hain. Hum seedha karte hain:

(a) $x^{4} + y^{4} \geq 0$ , zero sirf origin par → local (global) minimum (bowl, lekin $4$ th powers se bana).
(b) $- x^{4} - y^{4} \leq 0$ , zero sirf origin par → local (global) maximum — yeh concrete dome case hai, upar ki figure ke middle panel se match karta hai.
(c) $x^{4} - y^{4}$ : $x$ -axis ke along yeh $+ x^{4} > 0$ hai; $y$ -axis ke along yeh $- y^{4} < 0$ hai → ek taraf upar, doosri taraf neeche → saddle. Lesson: jab $D = 0$ ho to quadratic (2nd-order) picture flat hoti hai; quartic terms decide karte hain, aur aapko unhe haath se dekhna padhta hai.

Recall Solution 8

Step 1. Product rule use karo. $u = x^{2} + y^{2}$ , $v = e^{- x}$ ke saath: $f_{x} = (2 x) e^{- x} + (x^{2} + y^{2}) (- e^{- x}) = e^{- x} (2 x - x^{2} - y^{2}),$ $f_{y} = (2 y) e^{- x} .$ Step 2. $e^{- x}$ kabhi zero nahi hota, to $f_{y} = 0 \Rightarrow y = 0$ ; phir $f_{x} = 0 \Rightarrow 2 x - x^{2} = 0 \Rightarrow x (2 - x) = 0 \Rightarrow x = 0$ ya $x = 2$ . Critical points $(0, 0)$ aur $(2, 0)$ . Step 3 — second derivatives. Dobara differentiate karo (carefully, kyunki dono factors $x$ par depend karte hain): $f_{xx} = e^{- x} (x^{2} - 4 x + 2 + y^{2}), f_{y y} = 2 e^{- x}, f_{x y} = - 2 y e^{- x} .$

$(0, 0)$ par: $f_{xx} = 1 \cdot (0 - 0 + 2 + 0) = 2$ , $f_{y y} = 2$ , $f_{x y} = 0$ . $D = (2) (2) - 0 = 4 > 0$ , $f_{xx} = 2 > 0$ → local minimum (value $0$ ).
$(2, 0)$ par: $f_{xx} = e^{- 2} (4 - 8 + 2 + 0) = - 2 e^{- 2}$ , $f_{y y} = 2 e^{- 2}$ , $f_{x y} = 0$ . $D = (- 2 e^{- 2}) (2 e^{- 2}) = - 4 e^{- 4} < 0$ → saddle.

L5 — Mastery

Recall Solution 9

Method — Lagrange multipliers (constraint present hai; dekho Lagrange Multipliers). $F = x^{2} + y^{2} + z^{2}$ minimize karo $g = x + 2 y + 3 z - 6 = 0$ ke subject to. $\nabla F = λ \nabla g$ set karo: $2 x = λ (1), 2 y = λ (2), 2 z = λ (3) .$ To $x = \frac{λ}{2}, y = λ, z = \frac{3 λ}{2}$ . Constraint mein plug karo: $\frac{λ}{2} + 2 λ + 3 \cdot \frac{3 λ}{2} = 6 \Rightarrow \frac{λ}{2} + 2 λ + \frac{9 λ}{2} = 6 \Rightarrow 7 λ = 6 \Rightarrow λ = \frac{6}{7} .$ Phir $x = \frac{3}{7}, y = \frac{6}{7}, z = \frac{9}{7}$ . Minimum distance: $x^{2} + y^{2} + z^{2} = \frac{9 + 36 + 81}{49} = \frac{126}{49} = \frac{126}{7} = \frac{3 14}{7}$ . Yeh minimum kyun hai: squared distance plane par door jaane par unbounded grow karti hai, to Lagrange jo single stationary point return karta hai woh global minimum hona chahiye. (Sanity check point–plane formula se $\frac{∣ - 6 ∣}{1 + 4 + 9} = \frac{6}{14} = \frac{3 14}{7}$ ✓.)

Recall Solution 10

Step 1 — gradient system honestly solve karo. $f_{x} = 2 x + cy, f_{y} = c x + 2 y$ . Dono ko zero set karne par linear system milta hai: ${2 x + cy = 0 c x + 2 y = 0 ⟺ (2 c c 2) (x y) = (00) .$ Ek homogeneous linear system ka sirf trivial solution $(0, 0)$ hota hai jab uska matrix invertible ho, yaani jab $det (2 c c 2) = 4 - c^{2} \neq = 0 ⟺ c \neq = \pm 2.$ Jab $det = 0$ ho (yaani $c = \pm 2$ ) to dono equations ek doosre ke multiples ban jaati hain, to woh ek single line of solutions mein collapse ho jaati hain — infinitely many critical points. (Notice karo ki yeh determinant exactly $D$ hai; ek pure quadratic ke liye jiska Hessian yahi matrix hai, yeh coincidence nahi hai.) Step 2 — discriminant. $f_{xx} = 2, f_{y y} = 2, f_{x y} = c$ , to $D = (2) (2) - c^{2} = 4 - c^{2} .$ Step 3 — $c$ ke har case ko cover karo:

$D > 0 ⟺ - 2 < c < 2$ : unique critical point $(0, 0)$ , aur $f_{xx} = 2 > 0$ → local minimum.
$D < 0 ⟺ c < - 2$ ya $c > 2$ : unique critical point $(0, 0)$ → saddle.
$c = + 2$ (to $D = 0$ ): gradient poori line $x + y = 0$ ke along vanish karta hai. Directly, $f = x^{2} + 2 x y + y^{2} = (x + y)^{2} \geq 0$ , us line par exactly $0$ ke barabar → ek non-strict (degenerate) minimum valley; $x + y = 0$ ka har point ek (non-isolated) critical point hai.
$c = - 2$ (to $D = 0$ ): gradient $x - y = 0$ ke along vanish karta hai. Directly, $f = (x - y)^{2} \geq 0$ → $x = y$ ke along ek valley. Answer: $∣ c ∣ < 2$ ke liye minimum (unique point); $∣ c ∣ > 2$ ke liye saddle (unique point); $c = \pm 2$ par system degenerate hokar critical points ki poori ek line ban jaati hai jo ek flat, non-strict valley form karti hai (ek positive-semidefinite form).

Common mistake L5 trap: yeh bhool jaana ki constrained point minimum nahi bhi ho sakta, aur degenerate boundary skip karna.

Kyun sahi lagta hai: Lagrange ek clean point deta hai, to woh "answer hona chahiye"; aur $c$ -family ke liye, $∣ c ∣ < 2$ vs $∣ c ∣ > 2$ sab kuch cover karta lagta hai. Kyun galat hai: Lagrange stationary points dhundhta hai — aapko phir bhi argue karna hota hai ki yeh woh minimum hai jo aap chahte hain (Ex 9: squared distance $\to \infty$ door jaane par, to akela stationary point min hai). Aur strict inequalities $c = \pm 2$ skip kar deti hain, jahaan Hessian ka determinant zero hai, linear system degenerate ho jaata hai, aur aapko critical points ki poori ek line milti hai — ek real, testable case. Fix: hamesha bataao ki kyun ek Lagrange stationary point woh extremum hai jo aap chahte hain (boundedness/geometry), aur jab $D = det H = 0$ ho to gradient system explicitly solve karo — uske infinitely many solutions ho sakte hain.

Active Recall

Recall Second-derivative test mein operations ka order kya hai?

Pehle $D$ compute karo; sirf tab jab $D > 0$ ho tab $f_{xx}$ padho (positive → min, negative → max). $D < 0$ → seedha saddle; $D = 0$ → seedha inspect karo.

Recall Bada

$∣ f_{x y} ∣$ saddles kyun create karta hai even jab $f_{xx}, f_{y y} > 0$ hon? Mixed term cross-curvature measure karta hai; $D = f_{xx} f_{y y} - f_{x y}^{2}$ mein bada $f_{x y}^{2}$ $D$ ko negative push kar sakta hai, matlab koi diagonal direction neeche curve karti hai — ek indefinite quadratic form.

Recall Jab

$D = 0$ ho to aap kya karte hain? Hessian silent hai. $f$ ko seedha bound karo, axes aur diagonals ke along test karo, ya higher order tak expand karo (Taylor). $D = 0$ ek min, max, saddle, ya (pure quadratic ke liye) critical points ki poori line chhupa sakta hai.

Connections

Critical points — finding, classifying — parent recipe jo ye exercises drill karti hain.
Gradient and Directional Derivatives — har problem ka Step 1: $\nabla f = 0$ .
Hessian Matrix — jahaan $D = det H$ rehta hai; zero determinant degeneracy signal karta hai (Ex 10).
Quadratic Forms and Definiteness — kyun positive diagonal necessarily positive definite nahi hota (Ex 5, 10).
Taylor Series in Several Variables — $D = 0$ escape hatch (Ex 7).
Lagrange Multipliers — constrained extrema (Ex 9).
Single-variable Second Derivative Test — 1-D reflex jo L1 par mislead karta hai.