SolFyll_1_06analH_1 - 56431')(& ¦¤¢$%$ '2 0 '...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 56431')(& ¦¤¢$%$ '2 0 ' ! £ ¡ # !" ©¦¦ ©¨¥¦¤¡¢ ¡ §   ¡  £§ £ q EQ g U d £ 2 E X 6apiDhfe6ccbaQ`Y VW2 RSTRPHFDB@A89§ 7 U Q I G EC 8 ς ˆ‰‡…ƒRx6wBtr †„‚€y vu s wu9l † p – “ r94…‚ 4i n ”’l kj„ i`h(ed˜ ™— q u y u p ‘ “o‘ m9 † y g f ˜ n ˆ©4u –• y 2 − 4x + 8 > 0 ⇔ y 2 > 4(x − 2). D1 = {(x, y) ∈ R2 : y 2 > 4(x − 2)} D2 = ∪ {(x, y) ∈ R2 : 2kπ ≤ x2 + y 2 ≤ (2k + 1)π} vs v ©q k∈N D3 = {(x, y) ∈ R2 : (x + 1)2 + y 2 ≥ 1 y 2 = 4(x − 2). (x − 1)2 + y 2 > 1}. P€49t~|`4zw4vou6t ‚ }€ } {y xw ‰wˆ‡ … „ w4…ou † )pƒ = g(x, c) ⇒ y = 27x2 (x−1)+4 27(x−1) x± 2 = x± 1 27(x + 3 )(x − 2 )2 27x2 (x − 1) + 4 1 3 = ≥0⇔x≤− 27(x − 1) 27(x − 1) 3 x= 2 3 Ž  ‰ ”‹…Œ ‡ Ž ‰ ”‹Ž Šx † „ w ” )p˜ ‡ ‡‰ww4€}—‡j–w(h4wwjww ”  w “‡ ‰ 4x ”‹ ~| …  Ž x 1 −3 < x < 1 2 1 ( 3 , 3 ). y= 1 3 • k(4’4€ Ž hwx9 x ” “‘  }Š ϕ(x) = c=1/27 v 2 4c x−1 ϕ(x) 2 x>1 L , 2 x= . 3 œ v‘‡šx ƒ›ww”9h4x Œ ™ ‡ Ž  ‹ wh‡‰9”4x ‡ Š v v x2 + =0⇒ …‡ w‹ ˆ Œ Šx w9 y= x± c x−1 v x=1 y(x − 1)(y − x) = c ⇒ y 2 − xy − y≥x y≤x y≥x v i) x, y ∈ (−∞, 0] κ ii) x, y ∈ [0, 1] κ f (x, y) ≥ 0 ⇔ iii) x, y ≥ 1 κ g(x) = 1+x+x2 = (x+1/2)2 +(3/4) ⇒ y −(3/4) = (x+1/2)2 ˆ©ª…rw©¨ †ky…r£r kj„Rx)v s §• vs v ©q vu s wB" y √ x −1 = c, x > o ⇒ y 2 = c2 x, x > 0, c ∈ R sin xy = c1 ⇒ xy = sin c1 = c, |c| ≤ 1 23x+5y−2z = c1 = 2c ⇒ 3x+5y −2z = c, c ∈ R D = {(x, y, z) ∈ R3 : x2 + y 2 + z 2 < 1} ln c > 0 ⇒ x2 + y 2 + z 2 = ( c−1 )2 , c > 1 c+1 Å m …‚ ¶ Äà ¿ pwÁW‰À6¾ ž ‡„o4…”j³r k…—r iD(g jw…r…ƒ„w9l † ºy¦‚¢ ¡„• †ym p ¡€© ‚ †u Å „ i …wr ¥ m‚ Ä p ¿ ž ‡o4…w”Á‚æ W‰À¡j6•¾ †„ºy  ¢ „ Å ¥ r k…Rr i y ¶ Fž †ym u µ ÄÃÁ ¿ pw‰À6¾ s ½ r k…m—r i ‡oº4¦…”j„• ¥ †y †„ y ‚¢ ¡ ž §„ ¹¢ † ‘ ¥ oº 9r i §¨ i 4»w4…oºj€ – † • ‡„…‚ d¼kp „ ‡hkj„ …© B¥ u † ¡u „©¢r †u s u †9w© Ÿ 3ž u ‡hkj„ …© ‘9¥ r µ §„ ¹¢ † ‰ˆwu¡ ‡„ † § †u ´ p¸ s ˆ tr i †u §„ r¡„© ‚u ¬„ § †•u¡r© w¡ ‡—jw…m i ”­jzr wu k44w³r ·)ž w©44¦ k—jw©…m i ”­j„ ¥ u ¨¢¡ †• r¡„ ‚u ¬ † wu9l ˆ‰w¡ ‡Puy ¶ fž †u §„ µ ´ r ‡w© ‘ zr i w4…hy w…‚ §„ ¨ „©¢r¡ §u †u Dž ¡ ‘i s i m ‚ ¥ ²†u „ …wr l u † u ‰ˆw¡ ‡³…r i –4u i 44…³9l ky i w4i ”9³ †u 4i 9±’p u §¡ i ”Fw9l † u § „ ¢ ‘ ¨ ¨ ¦ y u ‚ – † „ © u y u ‚ p ‘ž “ – 9”…°¯ p® rqry‚ s ¢r ¡ …y i j€9l p r “ ”ª9l l †u § w¡ ‡„ j…‚ §©ƒ™r ¨ r•„ ‚„ ž ¡ ¤ y w­¬£«u‡j€ s r9l 9r ¥ †u ¨ †„¡ † „ i …wr ¥ m‚ ž ¡ g j…‚ †9ƒ1¡ 4”wž 4“ 4…‚Du ‡j€ s 9l 9r ¥ Ÿ ž • € ‚ „ g “ ‘ “ ¦ 4y “ u † „ ¡ r † ¡ ¤ 9‡j…”£u †‡j€ s r9 †l rqy„‚¢ „¡ x2 +y 2 = 4, z = 0 (0, 0) x = 2 cos t, y = 2 sin t, t ∈ (c) : z = 1 − 2 cos t − 2 sin t t=π (c) x0y [0, 2π] P (−2, 0, 3) r(t) = (2 cos t, 2 sin t, 1 − 2 cos t − 2 sin t) 2 cos t) ⇒ r (π) = (0, −2, 2) 1 1 n = (0, − √2 , √2 ). c= 1 sin 2θ = 2 = sin π ⇒ θ = 3 π 2π 3π θ = 4, 3 , 4 1 2 lim π f (P ) P →0,θ= 4 f2 r (t) = (−2 sin t, 2 cos t, 2 sin t − π 6 c = 1, − 1 , −1 2 lim π f (P ) = 1 = 1 = 2 P →0,θ= 6 lim f (P ). f1 f1 (0, 1 ) n lim (x,y)→(0,0) =0→0 f2 (x, y) = P →0 1 1 f1 ( n , n ) lim (x,y)→(0,0) = n 2 → +∞ x2 + y 2 ≤ (|x| + |y|)2 x2 + y 2 f3 (x, y) = 0. x2 + y 2 |x| + |y| ≤ δ = ε/2 ≤ √ (|x| + |y|)2 x2 + y 2 2 ˆ„©¢rº w4…oj€ p u i 4¡ ¡r ª…­4u Ç r¬ ε 2 =ε 2 (1+x2 ) sin y y (x,y)→(0,0) lim = lim (x,y)→(0,0) (1 + (|x| + |y|) √ 2(|x| + |y|) ≤ 2(|x| + |y|), =√ 2 x2 + y 2 wy…P¡ ’p u i ¡ ¤ y g h¹ ¥ v u© m¡ ‘i „ Å (x + y)2 =1 …”wu9l m‚u† ÄpÃwƉÀ6¾ Á ¿ „©¢r w4…oºj€ (x,y)→(0,0) sin y 2 x ) lim =1·1 (x,y)→(0,0) y f4 (x, y) = (1, 0) „¹ l hk¦9³ †u s ¥ lim (x,y)→(0,0) È §u ‡y w…‚ f4 f1 (x, y) (|x| + |y|) √ 2(|x| + |y|) ≤ 2(|x| + |y|) → 0, =√ 2 x2 + y 2 ª…­4u ¥ r¬ lim lim (x,y)→(0,0) 1 f2 (e−n , n ) = e−1 → e−1 ey ln x 1 f2 (e−2n , n ) = e−2 → e−2 f3 (x + y)2 f4 (x, y, z) = c1 = x2 + y 2 ε>0 2 (x+y)2 ∀ (x, y) : |x| + |y| < δ = ε/2 ⇒ | √ 2 2 | ≤ 2(|x| + |y|) < x +y ‘ † i mF‡„ i ”ª9w…‚ p u ‡„oj€4¢¨ p4i mw©¨ ky…rP4r i †w9l ‘ ¨ ‘ t¨4¦ ‘ j„w4…r s ‘ y…’¡ i …‚ ¶ pÊ ‘ † ‚ lry † º ¡ † ¡ u l ¡ ©ª r ‚ ‘ m s †ž „ y ‚ ¡ ‡oº4¦…”¢j„• Å µ ¥ r ky…mÉr i uy ¶ ³ž † †wu §¡ d ž pwY‰À6¾ ÄÃÁ ¿ Å ÄÃÁ ¿ „ i …wr ¥ m‚ †u 4 §¡ d pw‰À6¾ p ¡€©r d (g jw……‚ (ž f5 0 = f (0, 0) f6 lim f6 (x, y) √ x2 +y 2 =0 ž lim (x,y)→(0,0) (x+y)2 2xy 1 1 1 |f5 (x, y)| = | 2 xy x2 +y2 ≤ 2 |x| |y| 2|x| |y| ≤ 2 |x| |y| x2 +y 2 f6 (1/n, 1/n) = 0 → 0 f6 (2/n, 1/n) = 15 17 → lim (x,y)→(0,0) f5 (x, y) = 15 17 (x,y)→(0,0) f5 ž ¢ §yr ¢ i ‘ yu …r k…’…r f 4…‚”ª ‘ ¡ i †u s 49l †hw¡ ‡„j’¡ g §k…r’¡ 4”w4“ ¦…ƒ„¡ ‡i u ”i  †¨ ‘ w9l ‘f 4…”%)ˆ©w4uj…44¢ Í –¡u †u § ¡„• y g“‘“ ‚ g i‘ m †u yu‚ª ¯ u© ¨ y€‚© žv † wu9l p „ §pk…r “ r9”u p † i „ i ¨ ‡y g „h¹ Ç ‡„o4¦…”¢Éj• ¹¢ l §„ † ºy ‚ ¡„ r k…mcr »p(g © Ÿ µ †y i ž u¨ §hw¡ §‡£u ky…m’u ”w4“ …‚ƒPu i 4u ”‘ • ¥ “‘“ ¦ „ –• “ †u „ † „ i …wr ¥ m‚ †u w9l „ i m…wr ¥ ‚ „w4¢…roºj€ ˆ © ¡ ‘i †u ´ pÌ s ž wu¡ ‡j„• ˤ j„ kp oj„4w¡ ‡„ ‘ ¡ ¥ –º ¡¢¨†u § † §„ ¡ (1/n, 1/n) (x,y)→(0,0) lim f (x, y) (xn , yn ) = (2/n, 1/n) (x,y)→(0,0) lim x→0 f (x, y) (xn , yn ) = y→0 x→0 lim f (x, y) = 0 f (x, y) = lim lim f (x, y) = 0 x2 y 2 x2 y 2 +(x−y)2 y→0 lim f (x, y) = 0 x→0 y→0 lim lim f (x, y) = 0 f6 ...
View Full Document

This note was uploaded on 10/02/2009 for the course G 001 taught by Professor Shmmygr during the Spring '07 term at National Technical University of Athens, Athens.

Ask a homework question - tutors are online