06aut-fsols - Mk 5! Fa” 2006 Fm 53mm 307. OOH CON...

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Unformatted text preview: Mk 5!» Fa” 2006 , Fm/ 53mm 307%.»; OOH CON Ob—lO Ol—‘H CNN rhODNJr—I H l LONH [—1 1. Suppose rref (A) = [ :l and suppose you know that A —1 (a) [5] Write in parametric form all solutions of the system of equations Ax = l: 7 :| . msolwl’ions will [Se A‘fr‘ans/al’im 5P 1%; ”HI/Space a? A by a Pflrficu/m- 5010470“ 224?; WWW Mew/wk 25m W )(3 +Xq+2><550 t *0 A ' , :: - X; Xl" ‘RXF’QFRX; ‘1 “I “Z *3 _ : ‘ o o :1 {3— 39-sz X g + y _, +X _2 5 ' o O O I l X. l ‘1 "l ~2 we a1 + ,2 i 2 . ' o o l“ 3 Ill/28b I: - ;; ff 3 +64 7‘ TV “3 ’6)M)V6E 4;} #48 SOldihn se’lL -5 x3 —5 O o I 2 1 (b) [5] Denote the i—th column of A by ai. Suppose 32 = {4] and a4 : [—1] . Find A. 6 —1 [Hint: Make use of some linear dependence relations between the columns of A] We,&epemdlenc<tes amaaj ‘HI, Columns 5? NYfYA) W1 ‘Hle some 018 +5086 Gin/10143 ‘HLa calm";- cl“ A 50 ”grins’hnaa We, know ‘HML 3235a; mole 52M} =34 We! 523,4‘52542—13 ) J 5 l \ ' mg, an“ i an agar: = J. i ,' = f: . / 3 , 3 ‘7 'I ‘- >5 :' 3i 41 > .4 am)? a =33+Q§= Z O 2 l 5 l 3 (l ‘l‘ ‘G 5 "" _ (ll/flake ”495—26 —~ 2 -' ) 6 ~53 “2. -I | Z O l Z / ,- ng A“ 2 ‘f *3 “I ’2 , 3) G -L/ —i —2 2. For each of the following subsets S of R3 determine if S is a subspace of R3. If not, give a reason. If S is a subspace you don’t need to prove that, but give a basis of S. (a) [2] S = { H Ntfll'a Subspace; O 4 S bfl‘twse, 0‘3 ~O+3‘O¢SZ\ x—2y+3z=2} (b) [4] S: { A11 H orthogonal to both H and [4]} z 0 3 17425 is a subs/care, {Hé or‘flwaom" comP/pzmrfli' 5p SFMflfl'FB’ or afuiva/em‘é. «th ”Of/SPMCQ 0‘? [i Z 0]) a ) I ' 3 0 ~{ 5 a basis a? S is EYE]? ' i A55 55 a 505%)an 190+ +lmxe 55 A Jep—emémy Him: SFflnfl}flj 55f; E}: 3W 3W ' [1W 3 A ‘ t O ‘ I 3.: has [’75 (A E SIS ’HLé 56+, {of} [! I§ . WQE’ILWWCEIS aft *1 f! (War ind 5/ vii/fag (#9 "0f Maj swim muHifl’esdp PRC/4 o+her.) 0 1 1 3. (a) [4] For which choice(s) of constant k is the matrix [1 2 k:| not invertible? 1 4 k2 1776, JEfQI‘Mh/I 214+ 0 I I 12 k :2 l k (2 “ ., 2 [aka] I if! {I at ~ (hmj‘aflz) = 4:2“!va Which is zero rpnmj up] rt 1% Wm+rix ’S Mfmver’fib/e Solver O=+§k+z=~(k?*k*2)=‘(k~2)(kfl)) fie. My! / (b) [3] Let A = [0989 — sine . —1 31110 c050 ] Find det(A) and A . O 62 " M 'a 2 AMA = :59 (:3: = C0579 ~{~sm29)= COSZIQKMQ ‘EL ’mmflmzys ’fo 19.40! A.’; APPIy+l¢ "pow/10,4 [A 5‘1"” 01 -13 ) [c0363 snag—I c 02 “WA ‘5 0i [17:16? “Mr-0:6! 0r FOW~f€£00C€+llQ (1U Wn‘kgol Ma‘l'v‘lx C630 ”fing ’ O to! *1, “J 3 we («>36 0 i Wei film 3 XL! (‘0!S( 3mm NEH) .OI) {Me Mad, A '5 ‘H’Q MNX 0‘? 4!” 4M%£W""°”‘ 0” [P2 #75} Mi?" €S coum'ferclmkwise by 6) SO fluff A" S/MUM Ba #1 ”45(er ‘Qr ro+é+;am AV'Q 1 A"_ (”3519) ‘Sin(-§)]a (c) [3] If B is an n X n matrix, find a formula for det(3B) in terms of det(B). N0+€ 3B ‘ (30103 J 30 4&(38) = 07ei{(3.r,.)5) : M81”) 04%.]? j 33 O ” -. : 3"J 50 "Mani g/efb’B) 1' 3 Jé‘fg , amp J€+{31n): K. O ‘3 4. Let R9 be the linear transformation that rotates R3 about the y—axis by 6 radians in the direction taking the positive x—axis toward the positive z—axis. (a) [4] Find the matrix for R9 with respect to the standard basis of R3. W yvcwpmn+ 6p m vec’ior We.” I36 Unchanjeal?‘ we can o’mw W Sift/{hm cm 442 )E-fz/ane. it My @ng m be at; +0 (59¢) [”39] ) We Sins? .5 , 0 ' .5 —-Sm6’ 63 ‘[?1 WI” wick +0 @(83)‘[C00$§]. AS wifeaij we have R9 (3;) :é;3[§]‘ Thus ‘Hm Mairix P R w’rfls ms ’P‘fo g” A ‘5 8’59 0 ”5M9 J o 19 fev— 6'13742323’3 Ede) R6?) 85%;): f 0 \/§ 0 —1 sing) O Cass; (b) [6] Compute A99 Where A = 0 2 0 t mi [Hintz Think geometrically Note sin(%) 2 — What is 1A7] (20 “y? ms” 0 ”SM "6 fi 444; hm" SUjIjflflls foolimj 0+ LA: [32)! f o ];[ O C) l {/)J A o C/ ~ 0 J 2’ 5’" {Tr/E) O (303(77/6» which }S ’54? m‘irix a? Ewe WTM NSfeal 46 ‘W 3+0qu basis 09 IRE. Since Compositwém 010 [Wow trams-(iirwdh‘amg camslpaawps ”to Witt/MAW 04‘ ”MW-85) M know ‘H/WL (ii—A >M is 11]” MATH 0‘? ( 3,31%)”.1'1 'w/resp.+o flu 5H 1001535 6? i133. Wail/fired ”to whim 1’) 1: (snipe 3.21%ng 0M? lgfikqmflf’o’i if 21,.) . fl? _ 09a“ m “I? 002’ r to?” flog éqio}.Aqq :15 _fim [44(5er 0‘? Rw/z ) w’ficé‘ ['5 [a l 0'], fl‘pai/yj A !:a’z ”[0 i O]:[O (2% O o ‘00 5. As a reward for this problem, you will find an explicit formula for the Fibonacci sequence (10, a1, a2, a3, .. .defined recursively by (10 = 0, a1 = 1, an = an_1 + an_2 (so the terms go 0, 1, 1, 2, 3, 5, 8, 13, ....) (a) [3] Let xn = [ an ] . Circle a matrix A so that Axn_1 = xn. an—l 0 1 0 —1 Al ll A—L ll 1 —1 1 1 1 —1 A—l—l ol A-lo ll A-lo ll n~l “WV-1 ) M A o‘n—J ~ 0 p ) SO fie Mid—PIX A in- mud Sabra, AK: ] 5 {02: W14]. > (b) [2] Find the eigenvalues of A. BE ME Clmrfcilyfi‘p A = (Idfipl 1] ‘ A0")‘(“/X'l) = xlz‘ll‘j. AZ—/\~I=o <=> /\= ’f “WW-D: ME 2 Z . [Problem 5 continued] To correctly answer the remaining questions it is not necessary that 1 you have correctly found A and ,u. You may assume that l’M is an eigenvector of A with l 1 ——>\ 1 lol- (11 GO eigenvalue A, and [ J is an eigenvector for A with eigenvalue ,u. Note that x1 = [ (c) [3] Use the diagonalization idea to solve for xn and circle the correct answer. i. xn =An-1 [(1) = 0-1Dn-10 [[1]] 0: [3M 3A , D — [’6 2] ii. xn = An-1 [5 = o-lpn-lc [(1)] 0: [3M 31 , D = [3 2] _ 1‘ _ _ 1 1 1 ‘ M 0 _ n 1 _ n 1 1 _ _ 111. Xn—A [0 —CD C [0], C—[ _)\, D—lO A] (Cawghecl: 5y muHi’o/yphj by C) a In M I 0 M j Sum yea get (I J The punch line of this problem, obtained by combining parts (a)—(d), is the formula: an=:\-;1-_—;(/\”—M‘ _F{(I+E’1) ‘6: 5)") (Coo/i) (m5 awakens hemuse >95 [41% A""[o'=l CD""C"lo l " CD""-,\:'/;[ZL'J[Ll “ Con-'I‘V/(A-fll] : C-[lO‘M JF/Owfl : [At/(A?) - l W l I __[ /‘ 7907“) 0% ”WA/4 C fl/KJ/«Ll [7n -ll][-:"/0:lc—l '17“ ."1/4 741"] "l \ 6. Let T1 and T2 be the linear transformations that are reflections in R3 across the planes V1 and V2 respectively, Where V1 is given by the equation :3 + y — z = 0 and V2 is given by the equation 2:1: — y + z = 0. (a) [1] Find normal vectors ml to V1 and n2 to V2. (They do not need to be unit vectors.) n, = {554) flz = {2)‘l)l) (b) [1] Verify that n1 6 V2 and n2 6 V1. Check Rid/Z : Rx~y+€= a-l‘i+(~l)=0 \/ Clam/t 329W 3 X'ry-z? = SHE/)4 = o w/ (c) [1} Two planes are said to be orthogonal if their normal vectors are orthogonal. Verify that V1 and V2 are orthogonal. VIZ-74:: (Lg-0454);): a~I~I=Q / (80f ‘H’OlS Was clear from pafifséQZQ/L) : We knew ‘H’h‘l V. 1' spaaflfil amp! Since ‘VZQV’ ) 37“ may} 1% fits-f 321. )7): 80+ #3: also Wassuri‘mj )03+ 423 file #16 fi’o‘rL lowed/cf) [Problem 6 continued] (d) [3] Find a nonzero vector n3 6 V1 0 V2. MMI: wax/Z: §(x3y]%)l;_>;:: :§_ NOW I —_ID WT :3.ng =~II4 22m ”Imus, In so ””240! I» J) Mala“ .: SIIIce we. wIIIII fise V, ) we, neeal 713$?” and! Since, we ) *3 .2. —> , ~ We H3GVL) we need? Islnz. Smae we II. 933) 4mg III my Way +0 «(II/III $0ch an 71; : Ie‘I 7'12: EXT/I: I (We ‘0on @szi-Sj) (e) [2] Find one basis 15’ of R3 consisting of three vectors that are simultaneously eigenvec- tors of both T1 and T2. (Remember T1 and T2 are the reflections across V1 and V2 respectively.) smear Is a WIMMWS v, we know T I v) “II II veV, M l,/w)=*w I’m Iva/,1. [eeml'lwr Vi): '&(?*Pr03\I/>‘<’))= —§+;u>mjv.{7) 1% any 3?) ms, firth W. J while 17m): “It MI wept; MeaanI Ia III mtjm pmper‘I’ (II 1‘ Imp/.5: IIIq‘I’I‘ 13:2) ’rzjnlIrl’l)’ ”M EOE-)7? : 3, 7ZUS' gr, SivaIaneoust €Ij€nWCIor3 {ISI— T, 872 . ’I7Iey «I; (f) [2] Show that T1 0T2: T2 0 T1. ‘ “I'IZ Will/e, ”'42) ”3g IS a SCI 0‘? \Iec‘I‘ers 1340+ mtg rm a basis smce Ikey mt Imearly mae/cynaIenI (aII mt mv’I‘uaII/ oriiafiomI) MOI 'I’Inere me 3 SP flew? ’IIIL Morl’rs‘x 1(ng Tw/Iesp +0 ‘Ibe Ioms‘Is Hufijfigg i3 [73’ (I) g] I, PM+ (e) O O i I! ) moi ‘I’Iu IWI‘I'Nx {Em/It WW/Ifsp “I“o éfi‘uh’ n1) ’E%I5 [I 03 ‘ I (Was was £42.95) 4M M MAW“ 10” TOT a“! TOT Am MI, [E ' loo] , I2 MyI‘I‘EfIfIQj ‘Il’Ii O I ‘I’Wa oIlajowaI mafI'rIces M ‘IIV, ‘I'wo craters. Smac’r IO 7;an $907; IMVL‘I‘Ize Saw-a WQ’ILr‘iXJ ‘H‘éy ”We t7UQL 8 7. Let B be the orthonormal basis of R3 given in standard coordinates by 1/3 2/3 2/3 = [2/3] V2 = [1/3] v1 = [—2/3] . 2/3 —2/3 1/3 Let V = span{v1, v2} and let P : R3 —> R3 be the orthogonal projection onto the plane V. (a) [3] Write down the matrix B for P with respect to the orthonormal basis 15’ 2 {V1, V2, V3}. Since PO? 1? 19w 66V) we know PNJL‘V am? [DO/2):?) . J “‘3 4L A , u.) «A ‘3 Sit/1C9, P{w)=o 4hr 736V ) (moi S‘mce 1736 V: 630:”th "\73‘LV, Mo! v3 _LvL mam, 1367:1023, [ .3 5 flUS PIVflg P}: A : ) 6‘ OOO [ O (b) [3] Write down the change of basis matrix C with Cej = Vj where {witwtyesit} is the standard basis of R3. Write down 0—1. [Hintz no computation needed] t l J. ‘ “/3 2/3 2/3_ -1 7/ 2/ Z/ ‘ C: [Q V: [/31]: Z/3 73 “‘73 ) a”? C :07: 2/3 1/3 4:] L g l ' 1/3 _%/3 V3 L?) 2/3 — 3 iSJ Ci: "-3 2/3 /3 a I dingy/ml (c) [4] Find the matrix A for the projection P with respect to the standard basis of R3. , av, I/gz/sz/3 loo fez/3% széO‘ng/si’s Va ”/c “74 A- CBC , Z , i , 7 43V3‘9g0102/36fié: Z/ngOZ/373‘z2H/q5/ciZ/q _ i Z .. ; Ma’s/3°00 ’32/3/3 V5330 3573/3] mags/3 , ’ a _ ,_ / 2/ (Nair! weaulo] also Calico/4+9 A L7 4413-113 4%: mtrifivx X~ [2,: 7:] of £013); vac/’01: [or V (Mag Us»? ‘Hu ”Fad; Hm] A: X'XT ) 8. (a) [3] Let V C R" be a subspace and let P : R” —> V be the orthogonal projection. Regard P as a linear transformation R" ——> R”. What real numbers are possible eigenvalues of P? Only 0am? / are {>055er Eigenvalues op 10. 43 See‘l’lalsJ comm/4r a 13mm g «9r IR” 1% was dpvedors gar-fir}; Mr 42m 0 use: ‘9» v) M Mm 2%,}...ng m «cm rm a vt (we was itvmtvb n) W: m WW5 each a? v) VL [0645 (1mm 2 o}. aerial 1m 0414” Cases (DU/or be WM ems/y) . '-._ o , , 1710‘ ’Hw MWHI’X [or PWi‘HwY‘CSfecf‘fe @ ‘3 [Ol0.._0]}wlyic[q has [i [S mm! "+05 014 ’HH; diagonal 0W9? Wl‘nlmj efl+'*lt$ O SlflCe {MS )5 alfajphqh we can WM ’fl‘i eljrwa/mg 0‘? p as jusf 0 «no! I, (b) [3] If T : R” —> R” is a linear transformation that satisfies T3 = T, what real numbers are possible eigenvalues of T? swim is damn A) WA away v: m «we ) 190+ also A6 = T7 = 73-? = T{T{T$l)) = ”mm a» : 17H ) : Ma: 4%; 50 (A3403 =5. Since @758 L}, 4$1#2é-w 019 an arm‘s/‘05 we we} Wm [\3_,\=o) So A{A~;)(,\+/)=Q So Ago) If/‘m +44 ml, Weakest/L (50% a“? ”dust (an be mile/val ”[3? 50m T: Judi Consider 0'1”) fm}nm” “[1; a) (c) [4] Show that any orthonormal set of vectors {V1,v2, . . . 7vk} in R” must be linearly independent. [Recall that orthonormal means v1- -vi = 1 and vi -vj = 0 if 2’ # j] (Noieiéal' we 1%,)“ch used 4117}: ”get in Rog/em Ge“, ) A A A Q SUPPOSQ C'V’ +9vL+H ~+ Ckvk= 0 gr Sm’ars CU”) Ck . gm ’[hlq ‘l’hc ”’0"; Pmopucl O‘p cacti Slap! wh‘ll “V; (igrSOVM “fixeoyi Hm» I m1k>) W9 OHM (5% : 2 (cwaZT’M -+ck“\7k).T/: :Gll-‘Zicz‘Z'Vr+ *CkZfi : Ci 9:9: :: C; ) Sirlcc Rafi/2:0 w/tzueverjfiz 5m C140 42,, at a aim {W 1:) we’ve saw «are our amt/Mam of an ‘Hm’l qua/S 5 6; -H.L +r;via/ Combnor/“iomJ So +57% :3, )UJVL, an linear/7 inc/apemleuf 10 9. Let f : R2 —-> R be a differentiable function satisfying: f5( 6)= f(5,6~2) = 6 f(5.1,6)= 6 05 f(5,6.1) = 5.5 f.(5 01, 6)— _ 5.1005 f(5, 5.99) = 4.95 )— f.(5 001, 6 — 5. 010005 (a) [2] Use all of the above data to give the best value of the partial derivative fm(a:, y) at the point (any) = (5, 6). Since £(S,€)= "M “5“) +7553 Wamsioler +11 611601095. [Mo h J J'rp'tévmoe cit/609%] [2w 5m” [2: 50 +51; WlUQS appear/lo lee, +QHJMj ’l‘awaro] nPPmXImee/y: [0 (b) [2] Use all of the above data to give the best value of the partial derivative fy(x,y) at the point (:13, y): (5, 6). 5mm [7516]): ['M W we @510sz— ‘ML oh‘mfimhce fuatiw+s [1‘90 A ‘FOV Sin/15]] - 50+va Var/912mm My; [/ng #10413 +051an 5 (c ) [6] Give a linear approximation of the function f. Use your approximation to estimate f ( 4-) 1%.») :6 +7539 T105960 + 16/59 M) ‘2 5 + /0/x~5) +5[7’€) ' 30 W99)?- 5+IO{€-5)+ S[5‘~€) = 514040 25:] 11 10. (a) [5] Let f(a:,y,z) 2 am2 + by2 + cz2 Where a, b, and c are constants. Suppose at the point (—3, 1, 13) f (110,31, 7:) decreases most rapidly in the direction (6, —7, 5). What are the possible values of a, b, and c? _ 2 ,4 4333mm M9; +415 (5313+ vie/3;) 5.3) 3 (353.5) M V10 We; in 4% diretkon op mosf mph)! I'm-Mouse 0‘“ “F (and Ffi-st Fofmfs in 1%: fill-“65741;” 0‘? wmd‘ m/m‘l damage) New) SM‘Q ‘P sax +I9YZ+631 we alSo [7th E? 1 (QQX) REV) QC?) ) ml “103 WM 13): (‘63 2A ace) aw 333,266) 432-5), we W Li; "2% ° (b) [5] Suppose f : R2 —> R and g : R2 —> R are differentiable with f(2,3) = 6 and Vf(2, 3) = (—1,4), and with g(2,3) = 10 and Vg(2,3) = (3,9). In What direction at (2, 3) does the product f 9 increase most rapidly? We meal VP?) m‘ ‘HLgPO/t‘all (2/3), 039 version m0 ‘flz 93047qu rule in MUH‘I'fJ/g WIN/{u “ 54"} A e A Q I ‘ a 3 es W33) , 3V? + 7“. :75 J. 35%)(313): 3&3} Wash «Ck»??? = to-(‘mfi (3&3) 2" (—mfio) + (/2357) 2 [(53 99;. (AH‘trml‘Sve/y ) VW 399k £93) ”"0? Six—(1a)) ml (2/3), 50 Ly +14: produclrule i(4‘){23) 2 1"»;(3 3) 3(23)+4\{23) 3623) j - /0+6‘ 3: X will/e ffllflgyga) 49/23) 0(12 3)*472,3)j [2 3) = ’i [0+ € 7: 97) 12 11. If F : R” —> R” is a vector function and you Wish to find solutions of F(v) = 0, Newton’s method begins with a first approximation v0 6 R”, then produces a (hopefully) more accurate approximation v1 6 R“ given by V1 = V0 — (DFVO)_1F(V0) where DFVO is the n x n derivative matrix of F at V0. Suppose F ([56]) = [ x2+2y—-2] 3/ $33; — 1 and suppose v — [x0] - [—1 0 yo —1 equations 3:2 + 2y —— 2 = 0 and fig — 1 = O. is a first approximation to a solution of the simultaneous (a) [3] Find DFVO. CW. 33/ 2x 1 “(EL 51 DF={ AX J7 : So DFV: “ 911/31 Mi/éy 3X17 X3 ) o _3 “I (b) [3] Find (DFVOVI. (M95! 1 [j 13 12. (a) [5] Recall that in R2, the relation between polar coordinates (7", 0) and rectangular coordinates (33,31) is given by ac = 7“ cos 0 and y = rsin 0. If f (3:, y)— —— :c 3,;L/+y2332 express (fl 8T , 80) as a product of two matrices With entries expressed in terms of 7" and 0. (A “matrix” is allowed to have only one row or column.) :5? SGRZ—a E1 is Wfiwéfiom defined 57 (33(59): [raga] rs'mé) {= (X)y))) ‘H/reh we. Seek “HM Formal olean’l‘rveS {w/vesp ’l‘G FZQ) o‘p ‘Hm {LAW +23? Zero/)3 . By M Cbmm rule, Bi; To] D/{tgyrg 9): DP(3(«;¢))- DjIro) = D9692) D3656?) 0 A 3m; coo mo WE have 3: [Ail/Ar 331/ggnging ran-£9] m0! M: [6x 6y]: [WA/2x *3+3y><]) So 395; a] [WWW Wadfmé’fimé’] 3mg) I‘CDSQ : [3 FCOSZQSMQ + 3‘35ngch rgaasgg JrQrKSInQCele] [as 6) ”TS/M 6’] f7“? @0319» v (b) [5] Suppose f : R2 —> R2 and g : R2 —> R2 are differentiable and let h = g o f be the composition function. Suppose f(57 8) = (6a 7) g(67 7) = (57 8) f(67 7) : (5a 8) g(5’ 8) Z (67 7) Dg(o, 7) = [g i] Dg(5, 8) = E i] Dh(5,8)= [; :] Dh(6,7)= B 2] Find Df(5, 8). (WARNING: h is the composition, not f.) Smog #36?) fire doom w/e, lMPiIES ital" was) s M3010) (5)3) = 03mm» e NYE/3) = woo; W532) 3:. W52) = WM“ Wig) = [3 ill—1 l; 3l 14 13. Let f : R2 —> R be a differentiable function, and assume that the picture below shows for c = —2, —1, 0, 1, 2 the entire level curves f (m,y) = c in the region depicted. Each axis is drawn to the same scale and the positive a: and y directions are as usual. (a) [2] Circle all possible values of the directional derivative DV f at the point A = (a, —a) in the direction v = («LT —%) ® 1 —1 (1,—1) (—1,1) (b) [2] Circle all possible values of the gradient V f of f at the point (0, 0). (1,1) (1,—1) (—1,1) (~1, —1) (d) [3] Circle all possible values of the partial derivative fx at the point D = ((1,0) pictured. Paid (a): Dimci'ioml o’era’v. M dive drmtl‘iom 0? A level com Must [33 Zero FF is 9537‘. 2/92/25 4mg] 5d) mm: W» m“ be ”Mom! +0 ievei 56+ “HA/s pm”! 20+ 1729 mm cams are a: Pm‘l'k)‘ Wk)” is offAc‘joml +0 ‘Hte level 36" M pm'utr in Jivecfma o‘F imcyeagiaj ‘10} 50 on? om choicg‘ FM‘HJ): 4); = 006;? is posi’lI‘IN (2+ (060)) since innj in ‘l/v p19. X-cr/Iiecvliovi low/s 1‘0 incl/vase )h-P‘ 15 14. Let f(zr;,y) 2 6222—24 (a) [5] Evaluate the Hessian of f at (1,1). ’l; = Qxexi’y ”P7 -_~ —ex1‘ 7 £x : 2 8X17 + lezexa/ fly: ‘Qx exz‘)’ = a?” “077 2 8x2“)! (b) [5] Find the second—order Taylor polynomial of f at (1,1). mm =~ w + Mtwwh mum-o J“ ELl;x(’)’l’(><")Z+ aflb'lb/r’X/W) + ilrflblyé‘i‘ Ill 1 H a("")*(")/l"’)+ 3(X»t)L'Bl{X~IX7’I)+§i/y~l)z . 16 15. (a) [4] Find all critical points of the function f (any) 2 x2 — y3 — 3323/ + 123;, that is find all points Where V f = (0,0). Sel' 4:50 5?) gx[[»y):0 a) x50 an: y'zl 'Case X507 Them 4;, Casef! 3 ’flwfifo "39 ‘XZ+ 9:0) ‘4) $4720 =>x=i3. mg) an arm! Pam‘s m (03)) (0)11) ) (3)0 ) (a) I)_ (b) The origin is a critical point of each of the following functions. Classify it as a local max, a local min, or neither. i. [3] g(:c,y) = $2 +433y+3y2 Have jx=62x+47 3 fiY=LIx+Cy) so axx=2) 3,9,sz :3)“ ) $72G 17w: lESSim/x ol' (0)0) {5 [2 H15 Ll Q =0 4‘; ~3yz+laco => yaw/=0 ='> Fiat “we m 44:2] (244: ~‘/<O\ [idles Slam CNACNOx/x implies ‘MWL 3 [40:5 Milleramqf YM 01 mm all /0 0), {Le ASA/fie P0 4+) ii. [3] Mac, y, z) = 39:2 + 2963; + 332 + z2 ”51% kx: 6X+62y+iz) Aydax) IQ: x—p-QZJ SO tux: elk)? :2: [a l" 2'2!” p.) xa 2x; _ ’Wco)“wfldn)bflza‘ flu; Hessian arl (0920);: g: g (l .J O <1 9 Since 1"» x099 €>0 wlile, d'd’ Bow”)? L W(0”9/(: Old): 02?: FLI<OI wean; 671W} Laho»www ' ' 7 389. +14% ’ll’fl 355% crrlerlgm lnAPlI€s ‘l’lmf‘l k [ms nen‘wer- a MW Mor 6? MM J (00 O) 17 16. (a) [5] Maximize the function 339/223 on the part of the plane a: + y + z = 6 with x 2 0, y 2 O, and 2 2 0. Explain your reasoning. Wigwam“ £65 )E)=x72-33 won/I ”I‘mIfe on or (MAXIMUM WIW’! X20) 0» wI'eu yco) or when 2: 0) since I" cm Lg Posh‘n'm as Ioaaj or; x>0, y>o) %>0‘ 3; IeI§ msIm‘c’I’ wr‘ a‘IIa/rfiém ‘I’o ‘I’Iiis “9ij OMI 33mm: ‘IIM LnUWIé‘Yy‘ 77¢ Mod a? lAjmn?e MUI‘I‘SPIIérS cm Is»: USeo/I ’Ib ic/eanS/ EXIVEIWL 91‘” ‘ID SOAJ‘QC‘IL ”Io ’II‘e comsIm‘MI 30 yfl): Mfr? C= O we mud «IVeIIzsys-Im gw: ”“73 E Wt Imam VI“: (I; fig): (Vii; “sz 302121) amt! $3: (I) I)’)) 50 we have I“? = :9]?ng =2) 'eI‘I'Aér A10 01- A: E 'EE’J 1': —D EI'IlAErJ: 0 Or [:14- LE A.“ ’"IZJ’A /\ 2X73} 32f {.2 3x732 2" 3X J x+ +2 4Q ' , _ A y 50 eIIIu» A-O or 6«- >(’+y+-Z = x‘f Zx’fgx 1‘ QXI. [4: IeoI: (2+ case Waco Infi' 'IIqs (MIL occv 801" A: OIMPINS ’Iha‘I enema/we OI: gyf is Zero wIzICI‘ W: as?" ,‘jvwe, flaw momma 30 M Imvi 63(36) WM“ X=I “MI y: :ch 02) and 3:3)(: 3 jgwIwap‘F wIncL mus‘I‘ Lean/m7?) Is I >712 -23= (b) [5] Find the maximum and minimum values of the function exz‘y on the unit circle 2 2 a? +3; 2 1. LEI” pay): (2)3 70.on 30 y): x2+yz~l USiia3’I’I-A WIIMJ Sp Lnj‘ranjfi My” IflI neg) WIS SWIVEL ‘I’Ifle Sflem W‘s/‘75.; We I’MVQ V? = {IL}, )= (52x37 DIME-2‘7]: RAX x7; axe V a» z, x , x2 , A z = " —-'-=-~ '33 z :3 z r, 1;;— / ne‘hguy ‘7 «cw 2); =59 —: y / 3W" O , X 0:7 I'm X24312“ =0 {Alfie/HO) :EI‘ x=0 we Imvwe muoIlpIaIe mhIS (0 I) nmI /0*I) ”III Ciro/g} rp ys’IZ/ We IIMW (lamb/Me Pom-é; (\E ‘1) m,j{-=;r;) 3 ~l 0m Illi drag“ fifI'mj 909+ @4011 flp‘fste Pog'nj; W? —#;,a( 4‘f0 I) g ,— )__{\(0 AI): 8 )‘I-\(E)’i);~€ 5/, 3;); i; 1-2,):6 59 /Ist In minimum, WIN 15/r ) A: (2on #2 ”MW/470m I5 I MMI‘ £4???ij aIsc f‘t’fi’IrtS US +0 c/gecIf PI: on aim/e qum/ij'o Q bf gem mt no POM—ts MIN #13 IS Ffm‘) ICCISMmI £0nt i Lagrange “ISO Miblvfl US 1 f ...
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