4.4.1 · D3Functions of several variables — graphs, level curves, level
Intuition Yeh page kis kaam ki haiParent note ne tumhe bataya ki level curves aur level surfaces hote kya hain . Yeh page unhe har us scenario mein drill karta hai jo tumhe mil sakta hai: positive/zero/negative level values, degenerate cases, unbounded shapes, ek real-world map problem, aur ek exam-style twist. Har example se pehle tum khud shape forecast karoge, phir hum compute karenge, phir hum verify karenge back-plugging se.
Har level-set problem inhi cells mein se ek mein fit hota hai. Neeche ke worked examples ko unse related cell(s) ke saath tag kiya gaya hai — milke yeh poora table cover karte hain.
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Cell (kya cheez ise special banati hai)
Example jo ise hit karta hai
A
c > 0 Ex 1, Ex 6
B
c = 0 degenerate (point / crossing lines)Ex 1, Ex 2
C
c < 0 empty set (impossible value)Ex 1, Ex 6
D
c shape family ko flip karta hai (hyperbola axis palat jaati hai)Ex 2
E
Unbounded level curve (line, parabola, whole strip)Ex 3
F
Domain restriction kaati hai (log, root, division)
Ex 4, Ex 7
G
Level surface 3D mein (n = 3 Ex 5
H
Real-world word problem (ek contour map)
Ex 6
I
Exam twist — shape ko bina clean formula ke pehchano
Ex 7
Definitions jinpar hum rely karte hain (parent se):
Definition Level curve / level surface (recap)z = f ( x , y ) value c hai {( x , y ) : f ( x , y ) = c } c x y w = f ( x , y , z ) value c hai {( x , y , z ) : f ( x , y , z ) = c } c label hai, kabhi coordinate nahi.
Common mistake Do facts jinpar hum rely karenge (yahan state kiye taaki yeh page self-contained rahe)Crowding = steepness. Jab c Level-set dimension = (#variables) − 1. Ek equation ek degree of freedom hatata hai, isliye f ( x , y , z ) surface hota hai, curve nahi. Hum iska use Ex 5 mein karte hain. (Dono facts woh "common mistakes" hain jinke baare mein parent note mein warning di gayi hai.)
f ( x , y ) = x 2 + y 2 c = − 4 , 0 , 1 , 4 , 9 Forecast: tumne yeh bowl dekha hai. Guess karo: in paanch values mein se kaun si real curve degi, kaun si single point, kaun si kuch nahi? Aage padhne se pehle apna guess likho.
Step 1. Set karo f = c x 2 + y 2 = c Yeh step kyun? Definition ke hisaab se level curve woh saare inputs collect karti hai jo output c c
Step 2. Circle template x 2 + y 2 = r 2 Yeh step kyun? Hum shape ka naam aur size chahte hain. Ek known template se match karne par radius r = c
Step 3. c
c = 9 9 = 3 c = 4 2 c = 1 1 c = 0 x 2 + y 2 = 0 x = y = 0 single point , origin. (cell B, degenerate)c = − 4 no points — empty set. (cell C)
Sign se split kyun karo? Kyunki c c ≥ 0 c = 0 c = 0
Step 4. Spacing note karo: radii 1 , 2 , 3 c = 1 , 4 , 9 c c Yeh note kyun karo? Crowded rings =
Figure 1 — f ( x , y ) = x 2 + y 2 c = 1 c = 4 c = 9 c = 0 c = − 4
Figure dekho: teen coloured rings (green c = 1 c = 4 c = 9 c = 0 c = − 4
Verify: ( 3 , 0 ) f 3 2 + 0 2 = 9 c = 9 ( 0 , 0 ) 0 c = 0 4 = 2 Verify kyun? Ek point ko f
f ( x , y ) = x 2 − y 2 c = 4 c = 0 c = − 4 Forecast: in teeno mein se ek smooth curve nahi hai. Kaun si, aur woh kya ban jaati hai?
Step 1. c = 4 x 2 − y 2 = 4 4 x 2 − 4 y 2 = 1 4 se divide kyun karo? Standard hyperbola form a 2 x 2 − b 2 y 2 = 1 + x 2 left–right khulta hai (x x = ± 2
Step 2. c = − 4 x 2 − y 2 = − 4 ⇒ 4 y 2 − 4 x 2 = 1 Signs flip kyun karo? − 1 + y 2 up–down khulta hai (y y = ± 2 c
Step 3. c = 0 x 2 − y 2 = 0 ⇒ ( x − y ) ( x + y ) = 0 ⇒ y = x or y = − x Factor kyun karo? Difference of squares factor hota hai; har factor = 0 degenerate hyperbola (apni hi asymptotes). (cell B)
Figure 2 — Saddle f ( x , y ) = x 2 − y 2 c = 4 x = ± 2 c = − 4 y = ± 2 y = x y = − x c = 0
Figure dekho: blue hyperbola left–right khulti hai (c = 4 c = − 4 c = 0 c = 0 wahi lines hain.
Verify: c = 4 ( 2 , 0 ) 4 − 0 = 4 c = − 4 ( 0 , 2 ) 0 − 4 = − 4 c = 0 ( 3 , 3 ) y = x 9 − 9 = 0 Verify kyun? Har level value par ek point test karna confirm karta hai ki har point sahi branch par pada hai (aur ki c = 0
f ( x , y ) = y − x 2 c = − 1 , 0 , 2 Forecast: y
Step 1. Set karo y − x 2 = c y y = x 2 + c y
Step 2. Padho: har level curve ek upward parabola y = x 2 c unbounded hain (koi largest x
c = 2 ( 0 , 2 ) c = 0 c = − 1 ( 0 , − 1 )
Step 3. Kya c Nahi. Har real c x = 0 , y = c Check kyun karo? Ek subtraction (sum of squares nahi) koi bhi real value hit kar sakti hai, isliye koi c
Figure 3 — Teen identical upward parabolas y = x 2 + c c = − 1 ( 0 , − 1 ) c = 0 c = 2 ( 0 , 2 ) ( 1 , 3 )
Figure dekho: identical shape ki teen parabolas vertically stacked — green (c = − 1 c = 0 c = 2 ( 1 , 3 )
Verify: point ( 1 , 3 ) c = 2 y = x 2 + 2 = 1 + 2 = 3 f ( 1 , 3 ) = 3 − 1 = 2 Verify kyun? Same point ko do tareekon se check karna (solved form se aur original f y
f ( x , y ) = ln ( x 2 + y 2 ) f f = 0 Forecast: domain mein ek hole hoga. Aur f = 0
Step 1. Domain: ln positive input chahiye, isliye require karo x 2 + y 2 > 0 Strict > ln 0 x 2 + y 2 = 0 origin . Domain = R 2 ∖ {( 0 , 0 )}
Step 2. Level curve f = 0 ln ( x 2 + y 2 ) = 0 Yahan log kyun aasaan hai? ln e ( ⋅ ) x 2 + y 2 = e 0 = 1 e 0 ln e x e 0 = 1
Step 3. Toh f = 0 radius 1 hai. Aur f = c x 2 + y 2 = e c e c /2 c e c > 0
Verify: ( 1 , 0 ) ln ( 1 2 + 0 ) = ln 1 = 0 c = 2 = e 1 ≈ 2.718 ( e , 0 ) ln ( e 2 ) = 2 Verify kyun? Kyunki log/exp step ko galat invert karna aasaan hai; point substitute karna confirm karta hai ki circle radius actually intended level value produce karta hai.
f ( x , y , z ) = x 2 + y 2 − z f = 0 f = 1 Forecast: teen inputs ke saath, ek equation surface deni chahiye (dimension 3 − 1 = 2
Step 1. f = 0 x 2 + y 2 − z = 0 ⇒ z = x 2 + y 2 z ( x , y ) z paraboloid hai (upar ki taraf khulaata bowl). (cell G)
Step 2. f = 1 x 2 + y 2 − z = 1 ⇒ z = x 2 + y 2 − 1 Kyun? Same bowl, 1 neeche shifted (vertex z = − 1 c
Step 3. Dimension rule cross-check karo: 3 variables, 1 equation ⇒ level set 3 − 1 = 2
Figure 4 — f ( x , y , z ) = x 2 + y 2 − z f = 0 z = x 2 + y 2 f = 1 z = x 2 + y 2 − 1 z = − 1
Figure dekho: 3D mein do stacked bowls — blue wala f = 0 z = x 2 + y 2 f = 1 z = − 1
Yeh paraboloids Quadric surfaces ke examples hain.
Verify: f = 0 ( 1 , 1 , 2 ) 1 + 1 − 2 = 0 f = 1 ( 1 , 1 , 1 ) 1 + 1 − 1 = 1 Verify kyun? 3D point ko misplace karna aasaan hai; substitute karna confirm karta hai ki woh sahi shell par pada hai na ki do bowls ke beech mein.
Worked example Ek hill ki heightH ( x , y ) = 100 − x 2 − 4 y 2 H L = 1 m x , y L H = ( 100 − x 2 − 4 y 2 ) m H = H = 100 , 64 , 0 H = 120 Forecast: x 2 4 y 2
Step 0 (units check). Bracket ke andar, 100 x 2 4 y 2 pure numbers hain (kyunki x , y L = 1 m 1 m 100 − x 2 − 4 y 2 Yeh step kyun? Tum sirf matching units wali quantities add ya subtract kar sakte ho; x , y
Step 1. Set karo H = c 100 − x 2 − 4 y 2 = c ⇒ x 2 + 4 y 2 = 100 − c Rearrange kyun karo? Saare square terms ko ek side move karne ke liye, left par positive squares ka sum deta hai — ek recognisable ellipse form jiska size single number 100 − c
Step 2. H = 100 x 2 + 4 y 2 = 100 − 100 = 0 ⇒ x = y = 0 single point , hill ka top. (cell B/A boundary)
Yeh point par kyun collapse hota hai? Positive squares ka sum 0 0 x = 0 y = 0
Step 3. H = 64 x 2 + 4 y 2 = 100 − 64 = 36 36 x 2 + 9 y 2 = 1 36 se divide kyun karo? Right-hand number se divide karne par standard ellipse form a 2 x 2 + b 2 y 2 = 1 a = 6 b = 3 x 6 y 3 y 4 y 2 x 2
Step 4. H = 0 x 2 + 4 y 2 = 100 − 0 = 100 100 x 2 + 25 y 2 = 1 a = 10 , b = 5 Yeh sabse wide kyun hai? Chota c 100 − c
Step 5. H = 120 x 2 + 4 y 2 = 100 − 120 = − 20 < 0 impossible , empty set. Hill kabhi 120 100 Negativity ka matlab empty kyun hai? Left side x 2 + 4 y 2
Figure 5 — Hill H = 100 − x 2 − 4 y 2 H = 64 6 3 H = 0 10 5 x y H = 100 H = 120 100
Figure dekho: blue ellipse (H = 64 H = 0 x y 4 y 2 H = 100 H = 120
Verify: ( 6 , 0 ) H = 100 − 36 − 0 = 64 H = 64 ( 10 , 0 ) H = 100 − 100 = 0 ( 0 , 0 ) H = 100 ( x , y ) H = 120 Verify kyun? Har ellipse ke corner points substitute karna dono height labels aur semi-axes (x 6 10
f ( x , y ) = y x y = 0 f = 2 f = 0 f = − 1 f Forecast: ek ratio constant hona — iska matlab straight-line relationship hona chahiye. Guess karo ki har level curve kis point se guzarti hai.
Step 1. Domain: y = 0 poora x — yeh ek domain-restriction case hai. (cell F)
Poora x x y = 0
Step 2. f = 2 y x = 2 ⇒ x = 2 y Multiply up kyun karo? Fraction clear karna (y y = 0 punctured out hai — yeh line x = 2 y minus point ( 0 , 0 ) Line origin se kyun guzrti hai? Kyunki constant ratio ka matlab hai x y
Step 3. f = 0 y x = 0 ⇒ x = 0 y = 0 y x = 0 y = 0 numerator zero ho aur denominator non-zero rahe; toh x = 0 , y = 0 y
Step 4. f = − 1 y x = − 1 ⇒ x = − y y = − x Kyun? Fraction clear karne par x = − y y = 0
Figure 6 — f ( x , y ) = x / y x = 2 y f = 2 y f = 0 y = − x f = − 1
Figure dekho: origin se teen straight lines — blue (x = 2 y y f = 0 y = − x removed hai. Saari lines us ek hole se fan out karti hain, jo constant ratio ka visual signature hai.
Toh x / y origin se guzarne wali ek line hai (origin removed), aur us line ki slope c Gradient vector ) explain karta hai ki yeh saari rays ek point se fan out kyun karti hain.
Verify: ( 2 , 1 ) f = 2 2/1 = 2 ( 0 , 5 ) f = 0 0/5 = 0 ( 3 , − 3 ) f = − 1 3/ ( − 3 ) = − 1 Verify kyun? Substituting confirm karta hai ki har test point y = 0
Recall Matrix par quick self-test"Empty set" kis cell mein aata hai, aur ek function+value do jo ise hit kare? ::: Cell C — jaise x 2 + y 2 = − 4 c f ( x , y , z ) = c 2 3 1 H = 120 x 2 + 4 y 2 = − 20 < 0 100
"Sign, Zero, Squeeze." Kisi bhi level-set problem ke liye c Sign check karo (kaun si family / empty?), Zero case (degenerate point ya crossing lines?), aur domain ka Squeeze (roots, logs, division holes kaatni hai).