MATH 201 Exam 2 with Solution, Fall 2011

American University of Beirut · Calculus and Analytical Geometry III · 8 pages · Prof. Kamal Makdisi · with solution

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Math 201 - Fall 2011-12 Calculus and Analytic Geometry III, all sections Quiz 2, November 19 - Duration : 80 minutes GRADES· 1 (121) 2 (/12) 3 (/16) 4 (/12) 5 (/1 5) 6 (/12) 7 (/12) TOTAL GRADE YOUR NAME: YOUR AVE ID#: PLEASE CIRCLE YOUR SECTION: Section 1 MWF 3, Kobeissi Recitation F 11 Section :5 MWF 10, Abi-Khuzam Recitation T 11 Section 9 MWF 11 , Brock Recitation T 12:30 Section 13 MWF 2, Nahlus Recitation Tb 11 Section 17 MWF 8, Makdisi Recitation F 2 Section 21 MWF 1, Raji Recitation M 8 Section 24 MWF 10, Egei1eh Recitation F 11 Section 2 MWF 3, Kobeissi Recitation F 5 SectioD 6 MWF 10, Abi-Khuzam Recita.tion T 3:30 Section 10 MWF 11 , Brock Recitation T 2 Section 14 MWF 2, Nablu, Recitation Th 3:30 Section 18 MWF 8, Makdisi Recitation Th 8 Section 22 MWF 1, Raji Recitation M 9 Section 25 MWF 10, Egeileh Recitation F 2 Section 3 MWF 3, Kobeissi Red tation F 4 Section 7 MWF 10, Abi-Khuzam Recitation T 5 Section 11 MWF 11 , Brock Recitation T 11 Section 15 MWF 2, Nablu, Recitation Th 8 Section 19 MWF 8, Makdisi Reci tatioo Th 2 Section 23 MWF 1, Raji Recitation M 4 Section 26 MWF 10, Egeileh Recitatjon F 3 Section 4 MWF 3, Kobeissi Recitation F 10 Section 8 MWF 10, Abi·Khuzam Reeita.tion T 2 Section 12 MWF 11 , Brock Recitation T 3:30 Section 16 MWF 2, Nahlu, Recitation Th 5 Section 20 MWF 8, Makdisi Recitation Th 3:30 INSTRUCTIONS: 1. Write your NAME and AUB ID number, and circle your SECTION above. 2. Solve the problems inside this white booklet. Explain your steps precisely and clearly to ensure full credit. Partial solutions will receive partial credit. 3. You may also use the back of any page for solutions. If you need to continue a solu­ tion on another page, INDICATE CLEARLY WHERE THE GRADER SHOULD CONTINUE READING. 4. Closed book and notes. NO CALCULATORS ALLOWED. Turn OFF and put away any cell phones. GOOD LUCK! 1. (7 pts each part, 21 pts total) Given the level surface S of equation , , , x - y - z = -4. (a) Find an equation for the tangent plane to the surface S at the point P(3 , 2, -3), II./J ' ~ < -< S Go cL.K1I.N1)(.. ~U- ~ uCtl1.' fwn-clWw i (;t-//t:?} = ;<; -;y - R ' ~f(:Y, Ji)~ J.(t: - -;.~i-~i! r 'i7f(~~:-.3) -= 6 1 -'Il-t {l ~v/~COh~~d(H ,J) _', ~ 0; ~~ d P[3/~-3) "'. b (X-3) - LtIJj-p) f b (~1'3)= (> 6 ?"- 't'J -r (, r j- 5' = 0 , (b) Find parametric equations of the norma.l line to the surface S at P(3 , 2, -3), ..... ~ -1 h-u-m-~ (~) , V-f(3/~-3) = 6i -'fJ -t 6k/ t01AIIu" ,,; c. Vi ~ rM rdid,to -ik- n-~Wl J kM., d ( 3,. ~ - 3 ) . 'fI,.... PM-MV'-t£,..,'<- Uf'~ ') Ak -n.--.'wwJ kL ~r p""" 'X-3 -= c;,l -Y--:l=--It t ~ r 3~ ct / _ '"'r'"''''''' ad (c) Does the normal line in part (b) intersect the surface S in a point other than P? If yes, find the coordinates of the other point of intersectioD. 'rh& ~w.tU 4",- ~M ,S /{ ~ ~ 4 ,g::. f:W'P;", (3rtt/ .:l_ '1t/ _3,6/) )W.rn. ~~ S h .(J<7mevatw.-sj t . , .e (3f t /) Q _ {;l-Ift/- ( _3-{-6tJ < =- _4/ L (91- 3etl- 3 (, t )_ ( If - itt- -(- let ')- (1- 3tt: db! ') ='" - t /(,{2+8!t.=o " {- (_ICf-I-U) =0 , i = 0 ~J t'~" ,...fk ~ P(~ ':; _3/ > L- _ :li- ' ,< .,. Q(31'"3.J/ .:<- J<..f< - 1/). 3,.. ' c..-"z / 5.. A~ 'YUJW/1~ ke r:ivu ~H.MecJ- u;;.. /)~~ S td a?j.rl/,u ~>rd & ( 3&/ -..2<) 3 0 ) , 2. (6 pts each part, 12 pts total) Given the two level surraces S, ~ {(x, y, z) : x' + y' + z, ~ 4}, and S, = {(x, y , z) : x + y = OJ (a) Find a vector normal to the surrace S, at the point P( I , - I, 12). And find a second vector normal to the surrace S, at tbe same point P( I, - I , 12). S{ C1MJ 52 cw.. hef ~ ~ ~ At;... -fw' ~ -/(7;:t t/ -:; 7 ~.'J ~ i '- tl?Iti g (~). V ~ 7 r :J / ~(.kvdg.. ~ ~ P I _ <J-I(~1'1/-:;: ,:2i"<"'.:2Ji"'~'''- / c7J(/>/.!J t./.=. :tf- 8 ~ W('1- ~ Ii.)=.:<;;-.$/".2.tii t ./ 08((1- I, Ii) -= /'.. { 7J..<u ~.f(I, _ ~ 17./ ~ .;2;;: _:I~j:'... .2.tl-/... '" '1MYWvI.af wSj aJ- ~ rA?tA '3~ (~ _ ~ fL/ ". -:;+- i ~ <, .52 cd P. (b) Find parametric equations for the line that is tangent to the curve of intersection or the two surfaces S" and S, at tbe same point P(1 , - 1, 12). A ~ V p.tt;t4dd 4z, 4"1"- ~(.J k(/~d-~ '"' - "- ­ -'> --" 1 . J -h -'l V= -of(p) '/. VJ (f) :;:;,2 _ .:l .:;.J2 I i o "- --" ~ _ :){?>. t 7' ti.,J;; a" + I; {!. i) Al...- -Uw.~ -tw,.',- j;1 ~ Uf4-!N'-­ ~kcUJ~ ti'JI,z..' 11 ~iJ.ft..>,- ~90~~ p(I,_ ~.r,: , _ ,2./£ I: ) 'X _ I "" ;;lit J rl "' 1 ;i'-I:J = - . - ~ j::"-. .-~-:: ~ "-~- - .- ----7 . ":- .--: ~ 3. (8 pts each part, 16 pts total) Given a function f(x, y) which satisfies 1(2,1) = 10, "/1(2,1) = 4i+3j, "/1(.,3) = 6i+5j, "fi(6,5) = 2i+j . (8.) Find the directional derivative of f at the pOint P(6, 5) in the direction of the vector v = 4i + 3j. /-I"orn- :;)..f(6,S)=fii+/" p [,.({,/S)= ~/ 1~ !6/"-)-::1. Wl. It U/n1i- Vt~ ";'Alfv...- ~e~ S) .4h- "uWt- ~ ",, ' :rvw,--?" .' _) -7" -.J) rt--""--- 4t./- 3{ b. ~ 3 ~ - ~ - -= L..,-i---:=3­ / V I .f/i';:1 5 '" T7u. dM.~ dtw;-JCv-- ~f cd PI(- 5-) WI I£, (b)Find an approximate value of 1(1.99, 1.02). rY'fnrl- -ef!~/) = 4-::1" 3/ M P 1/1 (~ J! :,4/ I,,!,;; 1/ =- 3. IVn. , ~~ ....,.,.~ _v~ ~ / ttL w-j -t..:.""", "f'JI4"l"';"..t.,;n; / -f(iQQ/d.2) _j(.:J,I) Z (1.9r-~)~(:l,/)f-(IO~-')-&(-}I) (_o .OJ)'Ir (0 .0>-)3= _°.04-".0 . 0 (",,0. 0 ::(. /0 _02 . .f (I. q"I) 1 . • ,2} 'Z -f(~ J) I- 0 , <>:; -= J0 r 0 . 0 2 -= . . 4. (12 ptS) Use the method of Lagrange multipliers to find the ma.'(jmum value of J(x ,y) ~ x' + y' - 3x - 2y on the curve x' + y' ~ 13. Wt.- ~ ~ 41~?p)p~ Q.ffrt,Y! : ~ ~J'tx/~) / ~ d (;> (7)} =- ;:t "Z-f-l L - 13 ' We rrU1a,"~ ! !?< - "3) i f (.?:;--o/; ~ )I (.27;'" 2lj-":) ) ~ w;U; j ( ;>/;;1 '" D ThM ;h-t> v ~ ~vv ~;x _3 =- 2). 'X :l.--;_ (l=. :v.;:; 1. 1. 3 :X+(J-=- I "'- ) ~?< (1-;>') =' 3 I .9if (I -).) ~ :I ';ift '-."./3 (f. ".I. ,/ ~ J\ 1= ' I ~ 4V ;:<: =- 3 - I ;;; : .....:/_ ­ ~~-iJ ,:((1 -;>' ) (--'\ fM.v ';; ~ ",,' ~ ~,1 ~ P hI< n .., 2­ )L ~ ') -f- ( . 1 _ _ 13 9 f- '- - 13.) ( ,P,(I-A) 1-;:>' - ) 'I (I-A) 2. (I- A) / -), - + .1. 13 = 13, - - .x. 'i(p.) L I x = t i ~1"1l.G ')? -~ / ; /..a,v, #-u- ~ M-r/ fL- /Wl?> 4--& ¥~ fW»,{b '<A.'-. (3/ .2) M- I-~ _.2) - TJwv. -I (~.2) -=- /3 - '} - 4' ::: " 1(-}-2)~ 131'91-'( =-2' . 71u. ~/fmvfrt ~ 11 ",.. b U'W~ ., .pC. 5. (15 pts) Find the maximum and minimum values of the function f(x,y)=x2+y2 -x-yonD={(x,y):x'+y':S2, y~O}, 1'. ~ . . '-I~A tJ{, A4-r"'';';;;~ ~".I_. ;><7 ~ V'iif1 C-.t.l ~. Mt- o1/.tv..,.e4 T'v,"v "- " ;--- - - 'J"~" :;~!:: . ~ ~.;, Oh.<-~;..J ~11/ 'llA?nd-;' /fA! w/?d. 'd.... v. Th<.. .i~'l J) ~ ') ~ ~/ ~M3 C(o, li, ~ oW. ACiO. '/~ / (3 ow"'7 kf. . , '/ '/ J) • ''''~ ~::;o ; -6sx -<:n/ I(x,o}~ ':I: - ')C, A(.f),o) 0 liJ<,," ~ ~ ".1, ~ti<lW<. V'~ h<->.J x= ~ do" • so" ~' k-I.... vy;~~ ...... A:8 ML (·0 b ), fLo); Irz; o}. /" 861W1~ AW.. :J.:::: Vi> .?< ;:-' - /2: '" )< '<i J2. . -f(~ fe2-,,- )=, JI-?<-j.'2-~. ~d-1'v< ~ - I + 1( • .1.2 ;:, .' """?c" L 2. $v ~ V) J"-«kh v<- tl.".,I -,,'7 .:1-')<-, /< '" ~ ~ -:: !: I . /#-~~ ".. = 1/ tv>u( flu. ~;1..~ v,rf1#y,/VM. ~." ;IA<.. ,4 cO M.I... (- J"i,o) (I, I), (n>,,) . f'I~ W< ~ 4tV- ~ '1 I (~jl tt4 44... fw.'h,i0.­ (//~""" ~.' =f(J... 1/ - fr I q I _ L , 2- } 2-(;:: - 2- z. -{(.,(i,o) '=' ,:lfV2· fr -&1"I = ~ - ; := - f I(~ 0/ . .:<. IZ . f f ~II - - I. I = <>. 1f'1 He.n<e.-, 14. ~I ~YVL vd,M. ~ /JI'VJ""~ • , , 6. (6 p ts each p art , total 12 p ts) (a) If ~ = !({j,];)' and x = TCOSs , y = rsiDs, use the chain rule to compute - fh and - 80' ?w ­ - 8'w (b) Compute fh' and simplify your answer. F¥"h- 1>Mi !t?}, ?JW _ ', ''''' b -"-.I .A>0'!'J-. ') 'I. - 'rX <N~I ~ , . WtT ( l;KfM>-'&) f';! - l;tx c.M 8' + + ffih" b ( '1J' ~ lJ f- I~7- "",,'~/ (~IT ) ix?' ~ &418C<Y<lO (11''8- l' f1'X-/ +{#" {tJ} )~)'­ 1 !~ t~ . ~rJ,;""" ~ bt.f>1,hi#(.(JJt/>,t ~~­ fIIJn.1.)(4I. ~ Mt P/fl4.tJ- MvI J1/L ~ /U'Yv'j-hf'l -1.v (tp;;t it1f7?J ,R(aM-Ut¢ t7 ) 1;.,:r r- ~ ',)') f~!! ­ 7, (6 pts each part, 12 pts total) Consider the function x'y I(x,y) = 2x' +3 sin' y' if(x, y) # (0, 0) { 0, if (x, y) = (0, 0) . (a) Use the defin.ition of differentiability to prove that I is NOT d'~ r­ entlable at (0,0). I e For this part, note that 1(0,0) = 0, You may use without proof that 1,(0,0) = 0 and 1,(0, 0) = 0, WL 1'ItId $F ~~ ~ -!1f)f~!)f'J.) _ :{(D/ o) - M'/0°) - ~ f-r/ vr : (f£)-?/~ .I .; nfl:. L p .~ - fot.n.- _11~_-:L"_. I (9)"119") /) t, ',t3~LA (-;:~==f==t..=-- IJj A. =- () I Mid ~ -';> 0 , A;f."" ~J ~f I/> 0 , '1 -k ~ -Ie, ~ -It -"" 0 I PI .M1' 'f to '? oJ) ",,'l. .1~ -fJ3 ,..L.~~ I . -t. -7 0 21. 't- J,()V~'I. Ii.' I~ I .;.f-or <7 P~3 (~!:/J(j 11'70 ? V rJ{-n.<e,. vCk- J~ tWm<-~ t~ '7'i'tP o/-j anA f ,,,?VP+ fi-, ift1.hJv~ 1-1- IOf 0/ ­ (b) Prove that J is continuous at (0,0), ~. u ......,,; 'if- ~ ll/ iN< atw7 ~<- t... 2.. '1.­ 6~< 1L ~ :2:x f- 3,w.-. if-' &-v/ ,'/ 17/:;/:(:1%)/ ~ .2?'f-3MM~:FO ) 2/ !f f !4f7rJ) I:::: ~, !2?'~7 1hn VW''r-? '01 4/Vvf Jrw, ' lIl'/!1! ~ 0 (~JI4 ( ~(}) 130f1 1-1°/ vi :::: '"/ /.l-O b m' jI?«';j/ ~o/ 0/ ! 11/ .;1';'1 ~w: f; uf)lh~ tV