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Math_138__Assignment_1_Solutions

Math_138__Assignment_1_Solutions - ‘ t MATH 138 SOLUTIONS...

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Unformatted text preview: ' ‘ t MATH 138 SOLUTIONS T0 ASSIGNMENT 1 PAGE fl. 1- C) To Lomcuh S «‘SMKM, gum um. In: x 4,3»me $44.: 2x6»; Md; ru-=-c.osx. Wu Iblfip 05m 89‘3““7‘54‘“ ‘1 (-55“? SQC‘PS") 27"” = -1msx+ 251%st Non) ‘ MgpeMj 1:53P ujrfln M: X dw' -- CDS'X ELM , $0 (in. =d¢€ \f: smx O‘Luinfi qu’sm-xdm- —-)c.c,oslx+ 2(7c5u'mu — Ssm'x and = -3L coslx + 2x3wvx+ Zcosrx + Q _____________#_________._———-———~— e) Sfiecgig “Chm a. new cbmm Mme. MGMtpMm—vs \rlfidie {swag = 2520.9 (mace +TM93= bee B + Mag tome ( chmme ) m9 J.- .5ch m S McGoLG = S “-6.9 +¢£¢BWG o‘tfi of“ M=Tam9 +£2.09, fame «+ 501.9 =3, °LLL=Cb€CzG+AECm—am9)dfi 90+“— 5”= 5% =enm+c, ".9- SAM-9:13 = leam®+bacl9l -\- C. ___.—._.-—-———_—-———-"—‘““—-—"-———‘—_—-—-‘——I_I——-fi 2x13 S 756de uue ube. 1‘3? “gm Q 1. h)" To “mutt O . u= a?“ iv --= e449: so cLu= 2mm, «x: .e“x caiumca 69(51an 1 Six excl.“ -—['X 1 delj“: gage’xwx : —-(em33€_en . aq-Zj X2 xdx_ ‘ 0 0 Re— —L.L$\nC8 153’?) UJIT‘fl ru—"x. , :1»: (MC, and. W=G‘XAJXJU=—éj(_ cue $0.1M. 9»? Bus {ms 5 S x'l'e cm =—[3“:___) +ZE XE 1° +13 each: = [email protected]~:$3___l __2.QM33 4‘;— «1m __ 3(- (em? -28“ 3 +4) L) To 9.0th 4393;: Multan 4'4 du-a‘fidaca ow:- 2(LH-Odu. 8mm32 , 35m CilgVTfiJn-M 2 MATH 138 SOLUTIONS TO ASSIGNNIENT 1 PAGE2 4 k3FO’L SQh’iM 3 @P’L‘a Th3“) U-HHA M"6\n¢(}'dJU'= id}: 8 DC. $dfl=ggd¢<, my: 1:3. - —‘.L Spiéd“: 3555'; 3 acid“ “3% fwd/U": uxf—SvoLu) x '= -J£€’.ny. —- "i + C . 3 4 my Eumah. 5M5%(:)dzx Clpp’bfl I Ind P . LU'TLV i ’2. h) Fm ECOS‘GM , 8m? 4.1.5" run—.53.) o-nnc’» m- éLJ'T-M l x J Le. ”:95” 3.2!;th Trwa, baa othFwTL'u-n 3 503554 M = [211. Cosudu _ New 14.1 low =Casudu 3b V: SQHLA.’ 2 Smacsu tin w wet I; P “- aw “PM if . 2(us2mu - jfibnudu) = 2(usvhu4r cosmfl + C Fufidbfil ,5thqu ruzd'f ta abfalm .4 .— jcesd‘fidm = 2d§<sénv§ + Ecosd'; +C. MATH 138 SOLUTIONS 'To ASSIGNMENT 1 PAGE '5 o} 7' TGLUMtL ijQxZW no‘l‘e Tina? jxexdg 2 ex. 5° 7 .L 2 we. m'ltl. s 12’ z x1 7. 1 xe = j e 01x _ 1 K x j aux éLW.—%xe “fa“é‘éd’s‘Ofier) = _1_, 2 it"; I at“ _______________.__.———-—- :93 l sink-H ) ‘ ‘ note Fha’r _.L——- .-.- ———--—. —-—-—. P .Ffl S Sting“; ch): ) sfn'x-l 5"2'3‘" SW’XH “/3 “I = s_mx+‘: = Sun-n1 ' \ ab): :5 BSGn’X-i—l cix 5‘““’“ —-'-=05’x‘ Sewn-«I “20524: D D “f . = Sci Sunx‘__l_____ __ k M cosx cos'x Cos‘x D f! = ~— 5 3(th Toma + AECD’SDGLA‘ o a ‘6 = [be UK. + TM'X‘X w 931-“253 Usepafihafl S—rac‘um {'b aumata. 2".‘cM d»; 1 3C5+x ' NOEL SW5“? 2' 1341=ICIZ+D - Tenn: uJ-L WW 1(1) .=' g: + £2? E ACXin-I- Edda-t- C1 -: “1(A+‘53+CX+A' 'x'x+\ 'xCxfilB ‘ xQ'x’H-O Efiwrvkg-nwafims : 7m —:—_ «"(meh c1+A ' a ‘M ® Ba- —A= ~l, A = \ G) G ‘9 ’rfm.» M cmm in“ =_ +_ + —x+\ H S2 ‘ 2 “KM-X 3L x3+\ mu 3-1— — S J— —‘.‘L+\\ a 13+“ 2 t<3¢ + “1+! 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Noting “’10:" “flu chasm; eg'fiu (“In rug—m ‘5 4 X: +3smr’2. ‘ we} Wu Mménai'm' m 2 u». gaf- .0.ch— :1 id: _ X + 3* +7. 5-:_:+3x'3~'1x‘r\ =4, 9‘. r5£+25<+\ 1. l x4+3x5wazx x‘wamz I: :4, x‘+3x+z" O i-| Tim M” j z x A“ = x = 39.1.4, fi+fiyw =$Ax+A+Bvx+28= 1. #A+B=O :93” = 35-3- - €“\X%2\ + %\1+‘\+C PHZBS‘ Bh‘ MATH 138 SOLUTIONS T0 ASSIGNMENT 1 3 PAGE 5 ’60.) FJML cm‘iwni's A wok 3 “ch 11mm em gum (Anna + embody = ewsimb'x gm 5.11 :. LHS - aAemsén‘orx + bAe‘Mcosbx a ABEMCOSBK —bBcu>§<;n'M< . = (aA —-‘I=B3 emsdabx 4— (ha \- 0.83 ewcosbax . 1'?wa n+9; RHs xgg aA-bB=\ en? Ema-«3:0 Juan-5A,; WU: (Btu-U: «Ada-(4%) =\ =i> A(o.a+b1')-.—e.. aA=£zFE)B_;E§ b) .Q’MTQ%\O.‘V|T\OG bo‘flx aim *5 (BF) wlT’h {120ka To a: c3633 mu: Sea’édnbx M = A6” sfinbrx + Bewaos b—x = w . a??? iasmbm Hbcusbx} _ 11‘ N - . -1 . :— } c) “that 83 “sun“ M = [__e._1___£ -5uwx—coswfl {in ll: 0 (:IMUJ w = ‘ = ‘§_2_.F{-sfiwrr—cos1r} + Sé:{sfino +cosoB . .1: o -1; = g.__+ E9: = %(‘*e \f 5b) uh. wish To £"nd. ’dru. Coma. e344“; M£\En ECU/hm [3% 5=$- -, 5=;‘;—I, ma» he _ Panama“ fuel («with achual safidiugstém A1, gmafi'afifi ma. Mr“; ‘ {:Alx (Bu—5(341 = wax-wow SW «35'st legseJamoKTeh QCWCd an 51-90 1'0 SEQ-t [BI-dwa- A: gin-,0 2.x (WY-Ji-I-DAX M159. Sim: 6W 3:ngme - 3.3-de I ) I , MATH 138 SOLUTIONS TO ASSIGNMENT 1 PAGE (9 (a. A noohd Lfil‘s e—fi. 03F b=o amok —%o..o Vehckfii. ‘ mu}: _ 9‘“; .. aooo Qm (l __ 5?,51‘3 m /s- LO-n. pun-six Tc gLJ—xd flu. “Raisin? 91 TL: nochsd' {MW‘Uerm Suhu I'D-CD = girfiéét) 8m -9'L1t.3]1f Ad , um“. ‘91au-e. 1.} & 0‘45 ‘ .. _ 2 a; .. 9.8+. 30909n(1_ mt) ":3 use Etna (3 c—n‘riu mac-not tnnm , Lu'dla ILL=6M(I- 52-75% J +dv=d1 = #— ‘ - - . _. 3 3“ m 3%.) (it and \f—t’ 3“}an ”at: ~1— j’ana— Lew = tan-gag)- [iii—dizrgdt EL 3'45 i stew-LO— gifqfljoq—l—LthGL-t l—T—‘Zt = Ham— git) -Jc _ agiwtasifiehc =%(1-%§.Q {-t-- @53- - i: + C‘ ll Thu: n5: -4.9t‘—aooo WG—- §$Stj (t- 3%“: +2.ch +- 0 Ethan 3:0 wfixemt=o, Tim «Wei 0. is afiwwtud O: o—aooo€m(13(o-3~:__§)a-O+C‘ t'fi. Q=o. Fianall‘fi, wean 12:! mim: BO 5. cut-QM 3 1 =..4.9. o -3oooem l— £3.53. -316) “if I: C 3% (an __2_ +3cmmbo = "R1940 - l4?,5‘|§,‘i 4- 180_oo:: = 14344.1 and? res Q \‘3 Jmm. . {,0 LG Mi‘ernon-duj : {56503.15 ‘3 as. = [{gjfu 301m 1%(1— fif)}dfl O 1': .... : 14,944-1 ‘ O . ‘ ' i MATH 138 SOLUTIONS TO ASSIGNMENT 1 PAGE 4‘ km w To £n'fid 1m mam 3%“ gum»; 933+ $213 :1. We annuals swnmefru'cabou‘l'bork x'amcx :3 was , Smo. Qflalz x3}? Tenn: Luz wicnl-aé\hd flu. Qyna-fin 5-1 warm. fimdnwf; “ML The“ MuJJLCPQLfi 19% 4. fie Quncgum «:5; «ignited dam eff is ALE-v Axi‘. £3.31 Sofia. 8&5" chuadnmf ) X 5s“- "AX1+ Silaazaxz kg New??? 5—: 800:11— xy3)3/z_=> 53(1): 1357:”; |_X:L|-5)|/z - TEN-MD AL’R‘. Vi+tifl3 [50:5 =.- £2113 Ax. Smm‘hgm *8‘ 1.0». m Chin-w» Thai flu. MWAMNOL‘H F60 erg as. naSw'mfl Smalzfm g” is 93“” hi 1%C2rx-9-5) Fm = % %\——-—\L3-x—n)1- . + c. L041 usfsk To gt‘nd m pm‘hd gracing-Ins Ameos'nTkaa-fi 5306. Film" SIMML-i FCDQ: FC'X3: lg 9%. be“ 4— JE" en\19<+5\ — é. en. [BK-W + C- = é€m\1\+;€ em \2m5\ - Egmtsxafl *c -_— J€€\A‘-D‘-\ 4- t PM \Z’X-a-S‘t - %—€m.\‘3’x-L{ + C . SW32} $00 = Fl(7(\ Lb‘é Ibo. MLIK‘Vfc-vx b—Eom‘ficlMU-JGKLAH'): _ J— _1_\ .. _L—— . 1 £00" 2'! + 3 109:5 3x4 1: ”My _ e'XCZ'X-fi-EA i _ 2 <3 . TM- gm: W a. rat {er4- 53 {394—0 —W’“m“”" \2 at“ + '4 ac — \‘5 G m mad“; Swag g _ bzéxB-ifi'xfiflo i A. b ‘x MATH 138 SOLUTIONS TO ASSIGNMENT 1 PAGE 8 GHALLEM GE - x ‘ (”Th ' m‘n‘k O—AJ-O- o.- (an }s m nid'omcafit. (QaaAAsz a3 mud“ (CL: and bicarlni {(4). SI'mUmJLf, 1-1”. mcsn‘ma. w'lflm OJULO. high) {'5 m Managua (o, b, 8, gm) , ad. “3“er E; (11nd high“ “by b 0.. a A! b er) b 3““ 3 lb SQQ'OKQ-Dmmtmfl-ljllai‘eud E, 5:00] ‘¢§(1)=_ch(x)du +JJc-‘(Lpd‘dg Wot. Thai.“ b 541:) "' Q HQ) Lb A‘uqfl' 'fi’u. ma. (Lb-g IR“ ‘W‘Le. &on%m nJ-LTQmaDJ. minus, Tim. .1me M‘t CW MQIaMfiJQL 0% ma agfal. Thug 1+ is 5L0" firm. oumx gag 't‘buauwgancrg Fifi-Max“ usfng o. Pongk'san afifafl‘ AraaCRb = ETD 2783:5111. =: J £00 an . ’ ' ' an S‘mm'l ,G CmTLrLO-‘H EQQLECQ] Lu-ms [[email protected] =_ Qwa. _\ . ~— . A “M P (A‘j\ a? “fin“; ants-gems- \ AWOZS 61055 __JC.¢J:[email protected]]A3' 9 CA (1] %Mmfl I Eta P 'h: 3-90!) d,“ ’ ufit’h fl = JECK) + dV=dM %U-’U5 GLU- = ;ICK)M CLm-d. V23- ; Tm S§C¥3OLX= 153m _ Exg’mobx . 01> thL%naI-L‘ncfi rub JAbLLaJth-um gram IX: a. is fbflmw = MM: - j “xywa 5 = bum—day — fw'amx "‘o Co?umk flu m+ mammal) 1-0 “6: x=b (rum -:. r. 3 . M = ‘I/() a ' 3'6“. —-) ggfik :m =' (G) x 3" ”1‘ 73de f “5:? 6%, :maLfi‘b “’1 filmiica' 9” J °‘ ie x =5 #40 Mfrfifl-V’qmj (a) 4m :1 53m mwwa. _ —' $4,435ch Tkmm - fr 3 g 07130147] , _ commune; flu. meg. SOLUTIONS FOR MATH 138: Assignment 1 Maple Addendum Questions Winter 2006 Author: Barb Forrest > :as tart : QUESTION 1: INTEGRATION REVIEW Question 1(a): >int(1 + 3*}: - x"2, x=0. .4); 20 3 Question 1(b): > int(2*exp (x) + 4*cos (x) , x=0. .5) ; 2e5+4sin(5)—2 Question 1(a): > int(csc{x) *cot(x) , x=Pi/3. .Pi/2); 243 —‘1 3 >restart: QUESTION 2: PARTIAL FRACTIONS >rastart: Question 2(a): >integrand := {4*x‘3 — 27*x“2 + 5*): - 32)/(30*x"5 - 13*x"4 + 50*x"3 - 286*x"2 - 299*}: - 70); 4x3-27x1+5x—32 ‘t d:=—~——- m egm 30x5*13x4+50x3—286x2—299x—70 > convert (integrand , parfrac , x) ; 668 9438 22098 x + 48935 24110 _323(2x+1) ' 80155(3x-—7)+ 260015(x2+x+5)+ 4879(5x+2) > int (1/260015* (22098*x+43935) / (x“2+x+5) +24110/4879/ (5*x+2) - 9438/80155/(3*x-7)-668/323/ (2*x+1) , x); 334 3146 11049 1 Elnaxi-1)—-80155ln(3x-7)+§600l51n(x +x+5) 3988 (bum/W 4822 +260015 marctan[ 19 +——48791n(5x+2) > > >restart: Question 2(b): >integrand := (4.09*x + 1.24)/(2.97*x"2 + 0.37”}: - 4.15); _ 4.09x+1.24 mlegmnd2= 2.97 x2 + 0.37 x - 4.15 > convert (integrand, parfrao , x) ; 08286739976 05484303795 35 — 1.!21427628 + x 4- 1246006753 > int( . 8286739976] (x- 1 . 121427628) + . 5434303795/ (x+1 .246006753) ,x); 08286739976 ln(x — 1121427628) + 05484303795 ]n(x + 1246006753) QUESTION 3: ARC LENGTH > restart: Question 3(a): Since y = exp(x), we have dy/dx = exp(x) > int(sqrt(1 + exp (x) "2} , x=0. .1) ,- — 2 + arctanh[%]+411+e2 waretanh[ 1 2] 1+: > evalf (is) ; 2.003497112 Question 3(b): Use y = sqrt(3 - 3x“2), x ranges from 0.1 > restart: > f := x -> sqrt (3*3*x"2); fin x —> 3 — 3 x2 > diff (f (x) , It); 3x 3~--3J:2 >int(aqxt(1 + (9*x‘2/ (3-3*x"2) ) ) , x=0. .1) ,- E1lipticE(J21) > evalf (515) : 2.184438142 - 0.1 This is the perimeter a of one quaner of the ellipse. The perimeter of the entire ellipse is 4 times this perimeter, > (2 . 184438142-0 . *I) *4; > 8737752568 — 0. I The perimeter of the ellipse is 3737752563 ...
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