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MidtermW05 - solutions

# MidtermW05 - solutions - University of Waterloo Math 119...

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Unformatted text preview: University of Waterloo Math 119 Midterm Examination Winter 2005 Duration: 75 minutw Date: February 7, 2005 Time: 19:00 - 20:15 Materials allowed: Hand-held Calculator Formula Sheet Family name (print): Given names: I.D. #: Signature: I—_ I-- 4— Please make your check mark m MATH 119 - Midterm Examination Winter 2005 [6] 1. The table below gives the valuw of the function y = sinh(:z) at the given nodes: In 0.5 0.6 0.7 0.8 yn 0.521 0.637 0.759 0.888 Find the Newton interpolating polynomial (using the forward difference formula) and use it to compute sinh(0.56) to 3 d.p. Fits?) gangland '50s; Al‘kuu Mo'- ké. = 05“ \ 04H: 2 N 1 o, 0.00; as 637x 0‘13 / S 0.00! 3;: 0.151 /’ ' \, 0M7 /' 33' 0:28}5 0"” Z :O.§3\, “”313. A3,: GHQ, 15:3,: 0.0%, I owl Aib.= 0.00l in'lb ‘qu, Sorwoqk A‘X;Q‘W S'UFMVXA 612,5 [wi‘l’L kEOJ / 1.: Oil XI: 0‘6, (.51) :3: 05% + O.\\6 (1-05) + 0.00; (1-0.s)(1—o.6) ’l‘. (on g a! (on; J. 0.001(x-05V1-QGXDL-OJ) ‘5‘, (0.03 9%.sz ‘Hh‘xs aft 1,: OSL 37m; \$14.. (0543:: OSCIOI +0 zduf Page 2 of 9 m.- u howbm.m“mnwﬁ .V MATH 119 — Midterm Examination Winter 2005 [8] 2. Show that the function f (x) = x3 — x — 1 has only one zero, and that it is located between at = 1 and x = 2. Then use Newton’s Method to identify that root, correct to 4 d.p. S'mce ¥|(1\=313'\; ‘H't crf’n‘cA vn‘vu c} S' are i=1 é. 61w; 4' is o. cv-hic.’ cud. HM. caulk“ 3' '1.) B Pusan, {(7%) nus‘V \n khan; msﬁmm; Lu" §r(_'7\-3)= —1.3\$’ 5. M CW"- 0‘" “\3 Max ——-7 New, Rn: —\ ) um: HAVE) ma She. R») n “At-Mm, a. Lkrukoke Value more» Sun-Mkes “d ‘ch Mugs" he. 0s Zero MW 3 M 5M0 Qo‘n‘rs. TX we. make ‘3... NRA Sun’s I,=\I owl “ﬂlj Neda}, Ma“) he EA 1E 1: 4:9 -. \- L—J = 2 W13 1 % =3_7/l 3 1; 3 741513 13-. L315; 1“ = L32H7 g ‘“\'\> 15 our rod] 3» “3+ 15 = \‘31H7 Page 3 of 9 n. a! umle»)mﬂ>—lnﬂﬁ_x-v .4 m MATH 119 - Midterm Examination Winter 2005 [6] 3. Find the value of 93(1) for the initial value problem da:__ 2: dt —2\/t+:t ' using Euler’s method with step size h = 0.1. SKI“; £11193 ’1; £(13‘A) ' ¥(X-) k I :1:(0.5) = 1. we, k”; Highs: Hx.)+\n¥2xo) (a. x (wk) 2: 1 (ms: Mm) RCA". up. so} 1““ = 1.3+ L‘ 1“ QJE-d'ln J AAA 1,: 1' 107.6)— Ca\cuu\3 I. = 104 L{ x“ QJ£.+1. = 1 + 01 [ l ] 2405+l '-‘ LOHoX 1017): 1; = 1.0407 + o.l[ \ 2W] - . = L071? x(o.z)¢e 13 = \.o7‘[1+o.l __L aim-1+ Lam] = !.\\73 1m)” 1H = 1.1173 + 0.x ..‘___ [ .Hms-r MU} = IASBH 1h) 2: 15 = 1,4591 4. o.»[ 1 QWH = U953 a Page 4 of 9 1.“, 1.1”;er , 1~mmA6AM~éew we: ~~ MATH 119 - Midterm Examination Winter 2005 [8} 4. 3) Find the midpoint and Simpson’s approximations for the integral 1 / sin(zz)d:c , using 10 intervals. 0 b) Find a number of intervals which would guarantee an error of less than 10—5 using the midpoint rule. a.) MAFIA: ISSiv-(LliA‘L 2: L {No.05}; “0.15M- + “0.1531 = OJ [51.(o.ools)+ 52409335) 4 5“- (°'°615)+ 51404115) + 5:“ (wash :1. (cisoash 5:- (mush 51.0.9625) + s}. (awash 51A (oaoasﬂ = 0-3018ké2't‘i . g-UK 125'. I _ a 9" iguana: z %E¥(o)i‘i£(o.\\+l¥(o.1\+ + mom M] = 9‘3" sum “15:- (opm as}. (0.0%) + 4 st. (om) + 2151mm) 4 Li \$1. (0.35) 4 2 51. (0,31,) + ‘i 5;, (aka) + 15L (0.0-0 +H 5:- (om A sum] O.3\02L09~3‘i , \\ what. \ FYI“ S k °" EQIL] . (“J ['15] = {010 hate.) TA aw, use, M1): Si" (1;), So {lily-'- 21055.1) WI): Janka) - H115?“ (1“). Twain \¥"(L)\ : lamifblix‘swz‘ﬂ 5 Hum + Hm (1‘) S G Oahu) 58% be, wt in“ iemis _L Lin: . To erxsun iEn\< [0-5, it“, we set —\— é i0“5 1 There‘s: Mimi) \S'i Mun» w“ 51w. 0. error oi' [as 54¢... \O'g, Page 5 of 9 MATH 119 - Midterm Examination Winter 2005 [8] 5. Calculate sin (g + 0.06) to within 10" by using Taylor’s Formula with remainder (or with Taylor’s Inequality), about the point 1:0 = :1: How many terms would have been necessary if you had used 10 = 0? WM. Hg: 51.1 Ml 1;.“th we, Lm -\:(>O= 5““ 3L lgﬁ: [—33 ‘FWIA: (.051 ¥l(.§{): % who: «u my: 11,; V‘llﬂ: —L051 tempt“): -l elk. Hou m3 lam: A. one, meal? Tank}: 1m\$uA33 sl-lm HA- \lzn(1)lé K h'la‘nu ulna; \¥(n+.)('t)\ e K 0A [19,13 . (mi-Ml, J 1" °"" 9*, l¥(ﬂ+u(£)\\$ 1, on 69.3 “ANNA, Xv 0'43 A, 5' 5M0“ \mmg bum” 93W. \ = Haw-004%“ (null = 50.06)“ («0! 1h,» < \o’Ll Kl n22. (‘lrid out war), 5. we M ﬁ “L Bl???) ‘ ﬁat-2‘ “My; Sues Sh (£+o.06) z JG 4. ébpg)- air} (0.0;); = 0.7H83 . I; u; \m) mm) 1.: D) we wow“ \rxutqe, med-A [mole (Ewpéjnﬁ mm, (Mm Tﬁm\ 0.} arm'- n=3- 3 OJ n13 '- 0,01 :1" 5:; W ma n=c u m . / CM. 7 lama) , n: L; 0.000% Page 6 of 9 MATH 119 — Midterm Examination Winter 2005 [7] 6. Solve the difference equation A211,. + Ayn — 2yn = 0, subject to the initial conditions yo = 1 and y1 = 1. 9mm. bat-5,5 Agmf'bgn) ‘Hﬁs c.“ h “wt-Aka a») (A‘jmn— 5‘5»)* [3‘3” " 1‘3" = O '\.Q. 0‘3““ “ 303”: O ‘ ﬁwﬁwm’ B‘jmx = ‘3M;‘ Kan“) 5» wt \nuz, ‘H‘E “lune/Io: rem». Saw" “3"“ d 35—: O. n _ M). n“ n __ Look}: SW (A Stimsa" “S J60, Xena kg“: 7\ Suit) 7\ — )\ " 3.1 "OJ ~LQ- 7e — X ' 1 = O (LMKAUT5\‘L 21»on => (x—tharﬂw 2" A51, 7w— -\. Hence, We azmraA So\uéﬁo— B an: CI {3}“ + 63(4)" . Page 7 of 9 MATH 119 - Midterm Examination Winter 2005 [7] 7. Consider the simple harmonic oscillator model (familiar from Math 117): i+z=0, z(0)=1, :i:(0)=0. Pretending that nobody knows how to solve this problem, suppose we still desire to ﬁnd an approximate solution, valid - at least — for short times Assume that the solution can be approximated by a Taylor Polynomial, i.e. z(t) z P5,o(t) = 00 + alt-1* (1th + a3t3 + a4t‘ + (155’, 100(0) kl ’ and determine the values of the coefﬁcients ak. where ak = SM= Subswwe Kyla m. k at». 1’," (£3: (1‘ + lull“ 3a,? + ‘lqtt’ + qut" l,,."l£\= new “3+, -l \qutz-ﬁ goq‘tl‘ I 5. lb. A2. \mm a 3 '1 5 ,_ (205+ “JV-“1%?“ loqst’l + (adage + q; + qﬁ + qﬂc + 01:75 )~0 1. . :- ' (”3+0“) l (GQJJ‘Q'H "L (12%“qu * (geaslqu + qq‘y‘l Q; is z 0 Mg“ \°°\':"‘3 X" 0* “efﬁ‘ﬂimsa’r it” slwl ‘l‘mcs, M‘s la: La << l, 5° ‘llw \Mx *éo 'lll’ns Co.- Le a~Io“¢) (“are nugln Smaller ‘l‘lw. *LQ o'l‘lutA. ’“w: we. M I (awalﬂ (gqsmfu + (laqi‘mﬂfﬁ (1cm5 “7‘le x O. \m Md His 3» lac Sodasl-‘nl TQQATLAN) (i.e. ivr all f) 5 a wan-ad have 108* Q.=D Carton“: lqu+qz=o 10q5+a3=o . “A5, Hi in?\-‘xm\ (”Mlxhhs “he“ us ‘l‘lxu’i Q°=l 0w) Q": 0, 50 “f '7‘1 “.4 Q3: 0 “ASA \- 5h“. Mu) q“ : 5‘3“ a.) q; 7- 0 TL», glam: \-§+ It“ = Milt” L” Page 8 of 9 W_ «F I unusual” ..--“»_rn.:.-.« ~. #4.». MATH 119 - Midterm Examination Winter 2005 This Page for Rough Work... Swami s-m P5,.m= now 1m + 3912 mm snowy 5mg 2'. 3‘. ‘1‘. Sf , we, a. Seem o. w 0. CW M3 am a. \c's, Fw’ckrm, xmuuﬁaﬁj H». \$3. in: i+i=0, So 5i.[°5= ‘51.“): 0 i; i=0) So i(°)= ‘i(°\= ‘ i», 2L= 0, so 91(0): -i(o\= 1(0): 0, Hm) P510“): l ‘ t} 4 "E“ Page 9 of 9 ...
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MidtermW05 - solutions - University of Waterloo Math 119...

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