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? 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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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