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final persson fall 09 solutions

final persson fall 09 solutions - Name SID Section Sec Time...

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Unformatted text preview: Name: SID Section: Sec Time 01 MW 8am - 9am 02 MW 83m — Sam 03 MW 10am - 113.111 04 MW 10am - llam 05 MW 11am - 12pm 05 MW 12pm - 1pm 07 MW 1pm — 2pm 09 MW 2pm — 3pm 10 MW 3pm - 4pm 11 MW 4pm - 5pm 12 TT 11:30am — 2pm UCB Math 1B, Fall 2009: Final Exam Prof. Persson, December 18, 2009 Other / none, explain: Instructions: 0 One double-sided sheet of notes, no books, no calculators. Sebfibn: Room 75 Evans 5 Evans 75 Evans 3113 Etcheverry 81 Evans 5 Evans 2 Evans 247 Dwinelle 4 Evans 3113 Etcheverry 2300 Stephens Circle your discussion section below: GSI G. Melvin T. Wilson . Cristofaro—Gardiner Kim . Melvin . Wilson . Tilley . Cristofaro—Gardiner Kim . Tilley L. Martirosyan >EUD>I€CD§EU 0 Exam time 180 minutes, do all of the problems. 0 You must justify your answers for full credit. 0 Write your answers in the space below each problem. 0 If you need more space, use reverse side or scratch pages. Indicate clearly where to find your answers. OONODWI-llwm Grading / 10 / 10 / 10 / 15 /15 / 15 / 15 /100 1. (10 points) Solve the initial value problem y'+y cos ac— — y2 (:03 '33, y(0)— — 2 Sef‘l’dble'. y’: 505%.Cy2-y) 1‘1 l Sienx C :63 I 6053(le ycv-n l‘ACS‘M‘ l __ \ :~_ sunk-t4 2 l ___ ____ (I: 7017 7(0) —-—-._. .. )A-J/A 2. (10 points) Find a particular solution of the differential equation yfl+y=CSC$, 0<33<7r/2. Va 2 c cosyu- casinx 2) \/|:- Cosy ylzjinfi vafld'l'cm‘ a‘FFflfdm‘C'l‘Ct‘ Q7 “4+; 505’, u ladsfi-fHJ max: 0 ”flow, cosxsmxwfiswao l “Ml! Stnx +M9I CO OS )( :C9C'7k @“UI (05151.5)L+\43'Cosj‘(:(4fq r\'—_\' dialCDmN‘G) +0 32+ \A “(Sm x+ affirm 5 art—ca z) \n \-_-_ 60+Ciq "A \A ijscmq I U =“MJSIAX _: l Sin/CA“- lnl 5mle J at} :3) Milli-fl 7‘) MI "‘“hX 3. (10 points) Find the interval of convergence, including determination of the convergence at the end points, for the power series 2”(a: 1)” “:2:(—— 73(—(1nn_;2) ' =1qu ‘ . am (—93” burn” . n QM n31 \ 7 . LM \ \m = (“q—n (1M (wt-«“1 k— ZTUUHY‘ zfinau—\\2..n . lino '1') A a: (F Auk“ \ V\ LAVA . M ”" 7- \ m JAM —-——-—- AGO n+1 V‘ “4"” Rube“ m" .- \ ._._ /v\ w“ T '- J 7' w- cm» “a” 4. (10 points) Consider the function f (a?) = / cos(t2) dt. ' 0 (a) Find the Maclaurin series of f (x) WE: We, dye/ax: 2i 6')»in #130 (2“), 30 me2e 20.: (4)4th I720 -l X x 00 w Q" ~ 00 ." 4:an - ? . F (Pat £ :1 (:9- '__. o 0 ‘ (b) Estimate the accuracy of the approximation flat) a: T5 (£8) (the Maclau— rin polynomial of degree 5) for [cal g 1/2. F - e - W“ P“ W T500: 21:. 40W (aw/WMmf éeu'ea flat 1 . . . X9 ”‘9 ‘1 - ra— -—/"’ 9 mp!” S 4‘7””; ‘ «(t-9 ‘ 4W2 5. (15 points) Evaluate the integral or Show that it is divergent (continued on next page). mlnxd wklx‘ bk Infix at"; “Gigi (a) —2 117 C1 1 5“ {Low J- ,V, ”L X ’ X J ’4' 00 4; {ram 4,- . _L m 1 1/9: 7 ___ _' L a-flcl 6 X56 (1)) ewgdas WM» Mdgféu' x1; ’ e A :3 le “£— ___ ll‘M — M6 elm-I: “u #3:;— $ Lady X3 {750;} (iv/:4“ w: 6 Q 't l: l l l I [A m [u u + [QM U- _.- l v— C :- W ’[W’] J” e J“ " +401 4; 1‘; (continued from previous page) () 17/2 COSt dt WILL“ U": 51'1”": J Q1“; C95 ”b) 7 C /0 (1+sin2t)5/2 I; 1 Jo @ Z C0547 CH; ALI _ .— a; a” W : W ._ 26,49 0 _ Q (LIV: 556 1T: , 1? if tr 1‘? “P z a — 56026? » fl: @5636” map-” (#792 56559 (MM?) 0 0 ‘9 a ‘1}; E’UZ 6. (15 points) Determine if the series below are absolutely convergent (AC), conditionally convergent (CC), or divergent (D) (continued on next page). 51 ( (8)2:— ' :- 60 “(S 14*)” 11V”3+3” me my. d; IS QVQA‘l‘V’bI‘?’ lge—LJ-6filzfifi:j (Ca-pm CLL‘éCIL “('LLES iteerw'l—mew so 7 +1“. AsT th‘e, . re?" 6/", I III/Mm “V514 [ ""_>[ ’i 2?? _________,_,— l/vfi‘ - é_mi #4:! film/any?) so “’H‘a SU‘W/lfi ‘5 (E by hmt‘l’ Cam/)‘b’K-iéah b) insin(n_3/2) 'nzl M 4““(“3/1‘» Ly] VFW ] ”a @l M m -—‘7¢><3 “fit/”L .... ng/I (beau/USE (KL/glint) l we: K '30); W .-— {I “"1 jive/3%, 6a ”lie 90““ (continued from previous page) <0) >54)” (E - %) n=1 7. (15 points) Consider an undamped mass—spring system with mass m and spring constant k, subject to an external force F (t) 2 F0 coswt Where w = \/ k/m. Find the position $(t) of the mass at time 75, relative to the equilibrium position, given that it starts from the position 513(0) 2 930 With velocity ct’(0) = '00. Mr"+— kzu :9 Y": iLE: iiw Xctt) = C. com-u- GLSTHWt YP (t) -.:-.. Aha» wt 1— Bts'inwt fink-t) z: Acosw‘t-i-[Sfinwt -— Awe-.5 i: sinwt +Bwtccsw-i: XP"(-L—) -= —Awsmwt *wabwt -—Aw s’inwt +ch~swt .. 1 AW "Ecoswt wa‘t smw-t QOW‘GBN): F9 © -1AWIM1 :o [3: J31 ’ m” :Ar-O. Nit)“— E‘: 'a'izs‘inwt MW Wt): Mt) Home) = Claoswt-iolsinwti'éwtflhwt 10.): Cl: X D X/( 9): CLW :- v. C1: VD W Ht): Yo Cosw-t + £3 sinwt + 5. 4:51nut. W mu» 8. (15 points) (a) Use power series methods to solve the initial value problem 9” — wy’ — 2y = 41:32, 9(0) = 1, y’(0) z 1- a”: 2(IH 001?”sz X" h=O (b) Write the solution in terms of elementary functions. 1. 15/1 6“: “53 .11- =) (a: {+11‘1' Va 1‘ ken C6: C1,“ l J. l ...
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