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Su95Ex1sol - Su 95 EGN 342a EXAM 1 Name 14 e:3 sﬂew ALL...

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Unformatted text preview: Su 95 EGN 342a EXAM 1 Name 14 e :3 sﬂew ALL WGRK! Problem 1 {35 pts) A} Find fﬁx), the second order unnamed Taylor Series Expansion of the function f(x}=xsin1 about the point xﬂ. Ieave yenr answer in terms of x ane x0. 13)me = m? find thetrne relative error atx=1.1x0. g I . g g I g g I q g g I I g I q I I . g . I g . I . g I q g I g g . . g . . . . . . . . a n a a I a o I a n I a n 1 a n 1 I I 1 n I 1 n I 1 I I 1 I I 1 v n 1 v n I v v I I v I e v I r v e I v v I e v I e In ESP 3‘11“: E341!“ 1 "Su'ﬁu'kynns .5 as, (D gcgy: \$31.11}: a”) sun} -_ gun) + ﬂ‘Lmu-x.) + 5"”; (vhf aura—I— 1.‘ 8 CH = KLuSKiFan‘L 5: .110 = “Vi-MK 1:351. art-53¢. = zcusx -‘1L'Sm7( _ L *3 \$1.00“ Kuhn * {iacusmsmmu—x.) + iﬁcosxﬂémmﬁrm B3 15 «[1 «E, x= mi? 1": \$4,: o-tﬁa= and “(L5 1. g (.1 Hi 3 - 1T “- n- L - I“ a. _. Sin ('35 '5‘ IT" "i 11' IT IT I1- 1... ‘1 “L 1 “‘1 C " [11+LC‘LC35vhpsln-F)fﬂ«lu) : E(:]+(D+l +£(o—g)(o.0!21) hr 1, Z Tf< - ) Lt " Lit-E __ E (O'DtT-L L Ht 1', = mas-“irr- n3 time t I‘ I d _ '1' ET * mmmw 5n... U‘ “A =0Jﬁ\$n£> - nwas' L ....___.—_._________ ‘rCL‘TIL‘l 1.1 15 Mixno'lﬂ 3 L L- : 1.7666. — 1.?935: - ,{lgq I‘l L'Juéﬂ. So 95 EGN 342C! EXAM 1 SHOW ALL WORK! Problem 2 [35 pts} A cantilevered beam extending from its clamped end (x={l} to its free end (x=LJ has a maximum deﬂection Em“ at the end of the beam, Le. at x=L. The deﬂection ti at location x=aL [ﬂ{a< 1} is related to 5m: by the equation i'ttt)=tt‘1 ~—4::::3‘-t~llﬁl:t2 -35!5M :13 Use the Eiaeetion Method to solve for the value of 0'. at which ﬁfﬁm“ is equal to DES. Start with an initial bracket of otL = {1.5 and otu = LU. Complete 3 iterations and ﬁll in the table below. Round all answers to 5 placeg. after the deeimal point. ____ :12) gnaw!» AL=D.S‘ Idsuzi d 1 GIL {-414 .- R I. —Q‘§+1 10.75 7‘ 7. “‘0 ’ \$02.53 : u‘k‘lE‘Jﬂs 23\$L\$L}?L°kgd ‘- Q'}(—) 3' a swig =. ﬁrghjg) : ub‘L‘lhl "'7‘? OIIL=D1?SF = ﬁlﬂﬂf 1" 1.. 1E- k 1 9L“: ‘J‘I-D "" a; a! ...d. d '1 -E).1h 1L 1:. \ 1-. \D B S 1 0.14135 "L QKJ-L‘JW 1..“ : L'3L*j L G Sicily t. gﬁﬂ‘j‘S‘) ‘= nil. L‘lﬁul‘ 13 1") 91.“: D.%TS' CU“: 191(03‘13') '4 515:3?- J‘ﬁl 6“ HI“ = DJS+°r31§ :D'EILS: 'L 2.- 16:“: onatLS —c: "r '8 SM \ "1' 0.07691.— 0.2.ng {A ._.,_ a: .. {r .3, thmsﬁ 93““ kggdbiﬂdpl = (-‘n H (- a Hot x=\$££>nb as = 0‘0 '1 R " 1 ° 2“: du=o.%11'~3 AGAT- nip-+4.; 1 0,127+ 0. Evil? : enema? E. 2.. OJBILS— claim? 1 t. 0.0% engulf SL1 95 HEN 342i] EXAM 1 SHOW ALL WORK! Name V:- E % Problem 3 {3D pta) Use the Newton—Raphaon Method to ﬁnd the value of x where the linear function y =x+% and the trigonometric mnetion y=eoa(x) are equal. Start with an initial guess of x9 = 0. Fill in the table below and round the answers for x1 and xi“ to 4 plaeea after the decimal point. Use the rounded answers for xi to calculate xi+1. Stop when there is no change in it; (rounded to 4 places after the decimal point). I . c‘ {'3‘} =1+Stmﬁ '*:4l= 1; ~ ngf3 I 9 iii) )Ea:ﬂ man? #1.":— 115‘ \$1553.} = ﬂ __ 9—14,ch 1 0.5 . 9km "la-<21 21:! 1 __ 3 it '5. 34%.} = 0.3 __ G‘SFL%LQ.S) fﬂ-g ‘ _.._._.__..——-—-—— =Q¥fl73 SUN) tarauntoﬁﬁ f= 2*, \$ '3. 1" 7%- 9WL} : cam-h -— n.~n-r5- L=S{D.H|11} 1-0-5 __ 13 \ 0.01mi 1 ir\$tn(D-Hu15) 9Hka 11-3) '11“ —_)(_b___ gal) 1-OHISI = 031% l - 03.5% *La5[ﬂ.Ha§I)1-O..Sr 91113} ‘f3inio.~nii) ...
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Su95Ex1sol - Su 95 EGN 342a EXAM 1 Name 14 e:3 sﬂew ALL...

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