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# test#1 - Page 1 of 5 CEE 305 Civil 8 Environmental...

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Unformatted text preview: Page 1 of 5 CEE 305: Civil 8:. Environmental Engineering Computation (Fall 2011) Professor Duc T. Nguyen; 135 Kaufman (Ofﬁce #683-3761) Wed.‘ October 12: 2011; 3:009:11 - 4:159:11; '1012 BAL Closed Books/Notes; calculator is allowed Computer 8:. internet are "NOT” allowed 1-page {both sides 8.5" x 11") formulas/examples etc...is allowed TOPIC 1: IINITRODUCI'IONﬂ APPROXIMATION & ERRORS Problem 01 Given the following function: y = 3xe2x —4x2 + 2 Find the ”exact" ﬁrst derivative of y (=v'} at location x=1? a". (“+9 )J’Zax gm Melts =— Problem 02 Given the following function: y = 3x3” —4)«:2 + 2; and the step-size = dx-= 0.01. Using FORWARD DIVIDED DIFFERENCE formula, ﬁnd the ”approximated” ﬁrst derivative ofy(= y’) at the location x-l? 7KIW AY) —f(X) :20. 7599?— 29.1397?- _ 156.35DDI AK . p'pri‘ Problem 03 Based on the results obtained from Problems #01 & #02, find the "absolute relative TRUE error" in calculating the ﬁrst derivative y', at location x=1? E5 ___ 7g“ L VALHt ‘_ #9me {GM—WE — ' lée l= E5 / 77wa mama] x100 E6 - 23.9925 459.3900 I QEDIWS’S 5* =l[ﬁ0.7V35/§%.soiﬁ [XIOD= wry; Problem04 Express the value of 41 {using "’base 10”) by using "base 2" (or BINARY method)?? LII.” —. (10100 02 Page 2 of5 Problem 05 i (03) = 11.37. '73 GiVen the foilowing function: I‘ ' " I ‘ (7%033 - 32:25:??? y=3xe2"—4x2+2 ,7! =(pr f‘ 3052" ~ g % ”-{Qx * EkZXLgX %"(0-3} = ¢3ﬁ35\$ Using “2"" order TAYLOR series approximation”, compute the value of y (at x=1), based on the knowledge of the function y, and all its derivatives at the location x=0.8? got) = n. 9205+ \$21234 (x493; 1, 43.9355 bit—0.35); ._'- / 2. TOPIC 2: DIFFERENTIATION ' —_ y“) = fi-329%+ 32,2453“? (no.3) +6ges’55’U—033" Problem 06- a i Z The velocity of a roCket (as a function of time) can be given as: 3 U ) 7'; ’6 ' ’7ng ‘ v(t) = :3 +4:2 —'2r For the given step-size cit=15, and using the FORWARD DIVIDED DIFFERENCE formula, ﬁnd the approximated , acceleration ah}, at t=65?? a Lb) —_ v(')).-v( b) “_ [-7 3+ ‘1’ [7)l—2L7)] .. [02‘ Lil], )1. 2:15)] / _ 1 Problem07 4“?) ééit/‘Ziﬂ 3: l79-1n/sz ‘ The velocity of a rocket (as a function of time) can be given as: v(a:)=z3 +4r2—2: ' ' ' . For the given step-size clt=15, and using the CENTRAL DIFFERENCE formula, find the approximated acceleration ' a(t),att=6s?? no): Vi’i)-‘VL5) :52; #[63+llcé);— 266)]_ 2H) 2 Problem 08 @— Lb} ?625 n 215’ F; 2, The values of velocity {=v) versus time {=t), at some given discrete time is given in the following table: Tin seconds v=ve|ocit in inches second i 0 0 - W) " ﬂitting + Lb—ZMir‘OMZ + Cw“? 1 4 L2H?) (2—H) (W927 (3A1) 6-1-7.) (tr 3) 2 9 _ i i: We) = ’7—t2+2-2e r7710 Using ”LAGRANGE 2'1d order interpolation polynomial function”, ﬁnd the approximated value of veiocity at the time t=2.6 5?? NOTE: 2"d order interpolation polynomial function = quadratic function viz. a) a —2 inf-v 224m) ’35 Page 3 of 5 Problem 09 The values of velocity (=v) versus time (=t), at some given discrete time is given in the following table: Tin seconds v=ve|oci in inches second 0 0 1 4 .9 2 «to 9 V35» --> 3 e, 12 v}: 4 18 Using “LAGRANGE 1St order interpolation polynomial function”, ﬁnd the approximated value of ACCELERATION at the time t=2.6 5?? - NOTE: it“;t order interpolation polynomial function = linear function VH1) *Qﬁi) x0, _._L:_Zé_) x17, a 3K+77 (1'33 (34.) V'Le) e aux) ‘3 ’7 “3:151, I VLtD¢37K+3 l l Problem 10 The values of velocity (=v) versus time (=t), at some given discrete time is given in the following table: Tin v=veiocit in inches second ““5" . . y¢KJ=f£x>:f¢’:;€:‘J 0 0 mc 1 4 ' xH .32 gaflxr‘) “(9); vgt)- ”twat? xi ——>3 - mafia) 211‘: l 4 18 Using "BACKWARD DIFFERENCE formula", ﬁnd the approximated value of ACCELERATEON at the time t=3 5?? NOTE: 2"d order interpolation polynomial function = quadratic function are) -... {24 , 1 I [£27 ém — Page 4 of 5 TOPIC 3: [ROOTS OF] NON LINEAR EQUATIONS Problem 11 ' ﬂ ,. @ Xw ' ’9: ”w = 3—14 eloo’l Given the following non-linear function: ' 2 ,2 "F...- y=x3+x2—2x ' I J5K13=-31+(-.3)1--2L-3):42 4’5“») = —o.o25 And the initial guesses: x_lower bound = -3 and x__upper bound = +4 4,2 (—0.49% ) “HTS—>0 Using the BISECTION method, please complete the following table: i Iteration #i x_lower bound x_upper_bOtind predicted root absolute relative approx. error ' o -3 +4 ?? or iii/A EA n learn «saw-"=1— o. s or! l ' . - /. an no 5 1 '3? 0f ?? if .7? 2,21“ , . [QHDeZei/xloo . 2 ?? of ?? 1.2,? ' ”1375 ?? (”aloe/D =- @ m=0§ev.,.2.¢e @ xw w 2-251 f in 3 wig-tarl— 7.60.5“) = -0- (975 7 {ate ) 1 ‘01::5 : ﬁlm) 7 2:2I5—t 2.159 zrmr) - ”-35%! ' ﬂu.) mm; 3+ {.30521—I3‘7J‘K2) ,o.oe€(ll.453l)#474740 = "75"”? i Problemlz ' - o.ozs‘(l.o¢0L) whosvééo Given thefollowing non—linear function: '1 " Fﬂxmx (mu/7. And the initial guesses: x_lower bound = -3 and x_upper bound = +4 7:: ‘1’ ) t 72/ Using the FALSE POSITION method, please complete the following table: Iteration #i x_lower bound xmupper bound predicted root . o —3 +4 ?? —fL hm e EH41) '(',Z)C’72) :- (- l 2 ~ 02 ) — Page 5 of 5 Problem 13 Given the following non-linear function: . rm s'x-ﬂi? _ —,£-(—/2.)—.—2.3e3 And the initial guesses: x = -3 Using the NEWTON RAPHSON method, please complete the following table: ' Iteration #i Xi=initia| (and/or estimated root) 0 -3 1 :2? -— 2: 3 b 3 '4 Problem 14 Based on the data, the results you have already obtained in Problem 11 (using the BISECT ION method), how many SIGNIFICANT digits can trust in your answer at the end of iteration 2?? /55\l'\$’0.3’¥/°2-m7“ WW" = ”2”” 0-{70 MEvD. 10"!!— 5ﬁ [.1 No SKNMMAJT' :DiélTJ' Meawcri The SECANT method (which requires 2 initial guesses for finding a root of a given non-linear equation) can be considered as the ”BRACKETING” methodil TRUE?? or Problem 15 ...
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