ex225pract2-sols

ex225pract2-sols - Editing Assessments: 92 . " - f...

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Unformatted text preview: Editing Assessments: 92 . " - f Fk§f5~f..-’- iii?» ‘4' “Xx W- It; 1 "' Waive iififizrgfi E 8.] Name:? Class:??????? D3132? fr— ‘1" i .L u. 1;. ffdr,y) w. y“; {3‘ + 31,“). First approach {0,03 alongtho :rts-axis. Then ago} :: cyz" : a fora: 9e 0,30 {my} —, 0. New approach (0,0) aiong the y-axis. Thcnfory ?6 I}. f{0,y) :2 y‘ffiy‘ .=—~ 1,33, so firm) —1 133. Since f has two difi'ercm limits along ham different lines, the limit does am exist. {$22.21. h{:€,y} —. {flit-1:53)) rr {2: + 3y — 5}2 4- Va: + By — 6. Since I isapolynomiai, it is continuous an 13'.2 2:31:19 is E continuous on its domain {t 1‘. 2 0}. Thus it is continmms {an its domain I) wflm‘y) 12;: + 33: — 6 2 13} a: {(3.3) {y 2 —§:r. + 2}, whiohconsists ofaiipoims onwabovc the 11mg; m —-§z + 1 it). 27. (Kay) m in [2:3 + y“ - =1] :- y{f{z, y)} where f(z:,y} =12 + 1:2 - 4, oontinmis on flarandgfl) m Int. common its domain {15 at )v D}. Thus-G is continuum on its domain {{zgy} In +92 — 4 .3 0} w “3,3,0 I 12 + y2 y 4}. the exterior oftho circle I." + y“ a: 4. mayo _...._ ‘ a _ _ ’3 _ ii.2_31_ my, y) 1'. 22.2 + g? If (3‘31)? {a ) The first pioce of I is a {ali function definod evemvhorc except at tin: 1 if (my) m (0,9) origin, so I is continuous or; R“ anew-9053M}; at the originw Since :2 g 22:2 + y”, we have pays/{22:2 + 312)] 5 Eggs . 3" __, .u ' f ' | _ i .‘ a ' —— Roms that I9 I Gas {L21} —* {09m 55"!” lhcsql’emfimmm'{=,yi13imniflz' y) (mil-iiiemh“ +91 {1331. We 0.13m ' fiflfi) v 1-, so f is discontinuous $103113). Them-fore, f is oonfiouous cm the set {(r, y) | (3,9) 9&- (0:9)}. i4,— TYPE III This tests a combo of problems 11.4.3 and 11.5.2125. Here 2 will be implicitly defined by :c,y, you will have to find for example the tangent plane to the surface 2 = f (my). For credit, you need to Show implicit differentiation, not just solve and differentiate. 1. If sin xyz = 1/2 defines implicitly the function 2 = fix, 9], find an equation for the tangent plane to z = f(1‘, y) at the point where x = 3; =1 and 0 g 2 g rr/2. Solution. Use the formula for the tangent plane (2 on p. 770 of the text): Z - 30 = fz($oiyoJ($ — $0) + fy(-”~"o.yo)(y — 90) where z 2 fix, y) is given implicitly by the equation. Thus, formula 7 on p. 785 gives: 3F — = "g, etc, '6‘? where F{$,y, z) = sin xyz — 1/2 = 0 in our case. {37: _ yzcososyz _ 2 2—1:— _:cycosmyz I an E _ sczcoszcyz _ E @615 :cycosxyz y At the p0int,:t=y= 1, z=1r/6, the planehasequation: 1r 7r 7r ——=—~ —1—— —1. z 6 6(a: ) Sly ) Please note: The formulas must be remembered, they will not be printed on the exam... anlAs tn 1.) eLaJmmulfi‘ Mi'mgml-‘om is: 11 :55“: 10,019,219 C-x-xo) + {13 (40,15,154?) . “vane ref: (4'1) waiver» Inf-M33 «9,-2.- iaa FM,3,£): 42» “fire—I, 34'- W 2, f- tfl’ ; J A 5;, h M) 5;, _ mm + x7: “XMWNJWK. aTi—Lwfiokd: (95:) "" ‘ "‘""‘ . in if new: :1 as? at“ *3 3;? _ .O W haffo ole? Well} 'tkE-Méumen‘i- :JA. "Lei/tic Wig" never—4ft” 5’3“ 33- a; Q? ~ , . — - {-wa Wat. (am, a—«° T. ‘- - ii a“? $7" a a“ 37; §§ . a , 043-.) , Ole-«4)! tummy“ aflgbu) a ix-roJrozt, Rim-nmlfbo-tbe maan (1,0J°?(= , *-2~-—:33.' 4'335‘1i4. «:2 (“mm (x Erwhfl } 3 Eel-m ‘ 5[“‘%)* 52! )w‘dck hafimeMfiflzé’é‘”? ’1)? EM um ' B ' ° f L (gloat 47%.; normal [We ., )0) +t { .2! J giv- Bfa) ‘bakscn’c Plane 5 2—20 =.- (-11er +30.)(x-*o) + {n+93e56j ’51s) ‘NQM-‘i‘ 4- 340%” «£9395; //< “1),? { -1xo+3uf’:\ 50' on-bo =1 ‘1 z I)“ 6% '-' 0 ‘3' Iio" J go: *3 (q)le I) G») t— 365: 1 . :fixig) :2 Agfi max) V-:<i,1> 7 i a > Chi __ 4.. W -.;l'\i'\*:.,..»~—*—-—-— w wily»: "‘ ' ‘ . HQ w ”Gx m :C}+QW”>X\'>WJ€°~Z3* :0) "fl 1 m ' ad 2: (b 1;) w.» 2:: @292) i M IU‘ ' \ \ W 3 WM “{Efifi\ H vs, 2' 35mg)": fieKqL—Hxfifi “9m _ @fltwwfixx‘mx (2x ~62 3+“qu ) 2—, '1. 2% ewi‘wflmfi ). 2_ a. ‘1. 02' - (V 1') Um} H C291%&+9[\~03,2 6% .{V’irCV-Ols) EMM\ WWW": 3103.0 (/2 {Mung} at Era-Lad; ( :HHM Jmcfio'n <14 StEuPegf saw/1+ S Lg: _ ‘7— {im‘ésfik ©‘é_ Cf? my! \ m M Tm “’9 _ WELL-w TL 963m. 3‘31 Lv «Ex . fi—E nag/1 Goo/{51.7% 3v _ 10 [% K'E; %§ i x EK%FE_§XEQ ) u— )\(3i i 3, ® 8 If, 3 3 52¢ w) %¢‘€ T12.” 7:; e 2:) 173(‘6 +% 321g). €332? ‘\ 2.2% a) 3‘: . xx - My +SXZ“: m x06 +5%"):O @‘éfl wa - a x t? +59%) :0 7,) X10 fir”;— “Wst e” XT-lzé-m we Manam- ‘ew—r-fim Wm ng-fixg' W) ,3 g m Jean) ,3 (“fill EM) um 0485M MUM Ctr—74f. 2“ 1- b (31%} 7' jxx fag} wfixéj). $xe :— 4172+”) 2:) 9(FJB):QZXT’°J(@X+2) my; 1—, 2% 4331 '23 9% +2 ' . Te . -19“- 0mm 2° $6‘lLi): e 4: Since 43, is qn‘maeqsiirj of {1) 19 @563) wifl’fakf (famomax/mf‘fl/ a4 4L93qme fflmk a; 8053); “5179-53, f 5 89- 8“ 33 '0 =§ 3[«{32)*;j'“"0 :Qj—KJCt-3(3m>):o 51M x-Ié . -J. 2 1 0f Haj-.3 C039 “5; “0‘” MW”)? “awry: 5am; (Ore) ,1. J Case x25=*e,w Swamwfiflew (a) 5) :) «Elva—Vt? 5) «Vega 9‘9ka Solu'fing (“€ch9, {if Teret‘firé' I If,” 3’3" :5 ._l<0a‘{'!'(c‘>,0)| dud ‘3 a‘k‘lgfawfikaf’z (O 83’ awake gdmf Local m aximm ho local Wedbm QT saddle ram? {0 O) 1#2.? . I - ‘ Iced Mahmum af (_ gjinD) wig" ‘fiFI/‘bjéj; '8 , «em ~ M2101: W34; (1)03: FHCQ, (310 L1 (ti/WM we mmmwmdzmoe UMP -- )2 ‘2 2. I e "" 57 § @ a +(l1‘+a) {(2-2) = (peas 52$, V-9 in an :nalp 04: Mmé-Lm O‘DQILQQJ: (a4): JUrC—vva) 3 95) £053} IUQ+9+Q)Q+ {xx—9P 113” {mg L} {if 0 4x: 4&4) 13373 1' {213; {1.1" Q1- 3 3“} 3: z a O ‘ . ‘ n k 2 0 LT >0 J. «HM ¢5qjhamqup mrnemum SJ.“ $1 Ms Mpgfl; I+I+O = 9 4:23:05. (—1. me wmmwm am a (a y _ a -—16—- P FD“ gm! can “if? («M mu‘ftfao 4’50“ L ’Irxl‘ _ ‘ _ * , o are? L39 a .JMJ [4+ «,0; t-Mgmdmwafwuflecqade mo QM ' “(439 denwaewwm figm‘: 12;.“ £42: QMJ'umvl-QwMfi am 4 W0 '59.; :fl' d“ Help. 4H0” i 0 lg! I a tag; I ' 0a” who; @eme of 3%st chfl Men/(W50. 1’2"”? “MW: 49 +6") lam mmcqpamm a} _ Zaajjflujkg U O)!f a“ X3 erch :A’7LLLP ,anw W'OUEIS arm! 4%? ’3-qu , m.m w l. 3% I" “’1 “*9 air/egg): fiW-ngyai Ema) .2. 322:” 62} 1 ._ t: (“1) _. flESXa"4 bifim) yrséfaflejwi -:. w‘; QC)» @yéwa Editing Assessments: p2 fi1eWC:fDocuments‘ybzumm‘yawaeumgmum . Um H ’ggirf- 713g 1: fl , 35:25. Since 1' is. a polynomiai it is canth on D, so am absalute maximum and minimilm exist. Here It. — 4, J}, m »—5 so than: an: no critical points inside .0. Thus the absolute «new [111131 both occur or: the boundary. Along L1, 9: 7 {J and fwd!) = l — 5;; Fort} f: :1 S 3, a anmctioniny, so the maximum vahsc is f({}.,0) m 1 and the minimum W is “0.3) m - 14. Alang be, 1; m Band {LT-,0) m 1 +41: forfl r. S 2, an increasing fimction in 3, so the minimum value is no. 0} a 1 and'the valac is f(2,0) = 9. Along 1,-3.3: —§2 + 3 and {(23 —§:r. + 3) :z 1:22: “ 14 for [J g x g 2, an increasiag fimciion 1:13.50 the. minimum wine is KELS) = —14 and the maximvzlu: is f(2._0} x 9.13m thcabso'kse mimrnuffon D isfllfl} = Qandthzabwlute minimal is $10.3} :2 —L4. 4H2?- Mw) a 2x + 22y, Mm) a 2:; +2“. and swim, a I» a 0 gives {0, D) as the only eliticai point in D, with f6}: 0} = 4. 021L113; m -1, Hz. -—J.) m 5, acclian on L2: 9: m 1, “1.93:: y” +3; + 5, aquadmtic iny which aufiinsitsmimum a:('1,1), K1. 1} a: 7 and its minimum 5t (1, a). {(15 —§} m %. . 0:: L3: 112,1) m 2:3 + 5 which min: its maximmai (—1. 1') and [1, 1} with {($14) a 7 and its. minim-at (a? 1}, Ho. 1) n 5. On £4: fl-Ly} Try” +1; + 5 with mafimum'aé (-1. l}. fl—L 1) = 7 and minimuma! (— 1‘ —§), f{-l_.—%) r: «if. Thus the absohfls maxim is attained nt'both (itt 1) with fan; 1) m 7 and gamma minimum on .D is attained at (0:0) with No.0} a 4. V? 3C 10(4)) 1C3“2~);2%‘3 V3 =2:— [Z-DC-fl) I}? 32%) W ffiva - {’(w‘) "Er-3‘1 wait—.24 Eel-x) '2: 3% (1&9) “W Q’WHFPO Luamm) uXL?) a?) “fit-'33” . (1E: RUE/4' W) 2%Ov1)‘:‘<3r (“92+ E51352} NW- amzfl'”? géf“ Q “#03323 2:. mg.ng watf’g'fl 2% H» - :2. “L.” L 1 Q— iwx . «96' X “a L—{37 :- {588377217} ’fix’ji’l)‘ L ’ '$(H1\E‘Lifii my“; (“3Q”qu Kr (“iii—'0 'L M L L%L~%1) 11(5ifl'flfb\t '1: “LBi-lfi. 2 h..- i f i w ‘2.— 2 :FC W10} Via] “" Egg/fl) f {alwg :; zLCEQPLw—“q “‘3 Lgéf’twgilp—L} ’21 -34; ...
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This note was uploaded on 04/07/2008 for the course MA 225 taught by Professor Previato during the Fall '07 term at BU.

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ex225pract2-sols - Editing Assessments: 92 . " - f...

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