hw8_sol - 17.2 (a) The analytical aolntion can be evaluated...

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Unformatted text preview: 17.2 (a) The analytical aolntion can be evaluated a9 '4 —*-x 4x 499 Jan} Icl (1—9 ‘ )dr=[x+tl.5a ' E=4+Ufia ' —tl— 6.56? ' =3.5{]{]16T?31 (h) aingle application of the trapezoidal rule . 5 . (4 — mm =199329 (a: = 42.33%) 1- {c} conipoaite trapezoidal rule 14:2: {l + 2131981634) + {1.999665 _ ,3 [4—9) 4 .96363 [at =15.35%} 14:4: {l + 2(6.36466 + 6.981634 + 6.99752) + {1.999665 {4—H} 8 = 3.343? (a. = 4.47%} (d) aingle application of Sinlp-zwon‘a~ 1.53 rule CI + 4(6981684} + {1.999665 (4-0} 6 = 3.2342? (gr = 5.1.799) {e} colnpoaite Silnpaon‘s 1.33 rule {a = 4} {l + 413336466 + 6.99752) + £03.98 1634} + 0.999665 (4—0) 1,) = 3.49959 (gr = 0.34%} {i} Simpson‘s 33 rule. 0 + 31:6.930517 + (3.99517?) + [1.999665 (4-0} 3 =3.33:~3355 (sf =3.19%) 17.5 (a) mmlfiical mlutiml: :- I116 In I- ; 0.5 23—1'5xdx= [—1.555559'1-5‘h =—n.542{)9 — {—1.33335} = 9.59124 {11) Trapezoidal rule 2 1.3555 1.3555 1.595 1.55? 1.3745 5:03.05—Q)++(fl.15—D.05)++(0.25—0.15)+ .3? . . . 1 46:11331+{DI4T5_GI35)11331:09303 1- 1- 0.9303 0.3151 + (0.5 — 0.455}+ = 0.59234 +(fl.35—D.25j D.79124—0.79234 5,. = — :-< lfi{]% = {12022% {1.79124 {:2} Trapezoida].-"Sinlpsml‘5 rulea 2+1.:~5555 1.3555+31.555+1.5545 +1.1351 I = (0.05 — [Bf + ((5.35 — 0.03% 1.1331+ 4(0.98D8)+ {1.8131 +(D.6—fl.35) 5 = UTQIESE CL??? 124 — 0.791282 :-< lflfl‘i’o = UOOSZEH: 5].: [139124 17.6 {11) Analytical solution: '1 '3 '~ '2: 1 .TE '3 x2 _ 4—7 :' :3 := ——7‘ I' :3— : j_1jfit.r -} +36 )dxd} L [ 3 -Jr x+J 2 L d} 1 -1 3 a 3 3 4 3 1 4] = ——4:-+2- 4v: — -——- +—v =2.666657 J—l 3 '1 Jr I, 2“ _1 (h) Sweeping aeroag the I dimension: y=—L _'} '3‘_”J I=(2_gjfl=_3 4 126: I1] 21 4 I=(-—6)—+ 0+ =3 4 1:1: —2 It) 4 ;=(__g}#=1 Theae value-3 can then be integrated along the y dinlen-siml: —3 23 1 ;=(1_[_1}};)+=2 4 ’J 7_ g = —"66555' 2 mam/5:25:34 I 2.66666? {e} Sweeping across the I dfinemien: I = (2 — my = 5.33333 I = (_ — miw=255555t I = (2 — 51)%(m+4 = 5.55555? The-5e value-s can then be integrated along the y dimensiml: — 3.33333 + 4(1666667) + {3.156666?’ I5 I = (1 — (—1)) = 2.555557 Which is perfect. 18.3 Although it‘s not required. the analytical solution can be evaluated simply as I=Ji:.m* dx=[e‘(x—1)E=41.1no?535 (a) The tableau depicting the implementation of Rotnherg integration to 5:; = 0.5% is floranon-—> ‘I 2 3 est—z» 119535994 5.934992: 9.192991: 59—) 29.953993 9.35?9% 1 99394911315 43.5?337’269 4121395531 2 55.2?625849 41.313957’5’14 4 4433949599 which represents a percent relative error of0.102 %. (b) The transformations can be computed as 5+0 + .s—o 3-0 =M=15+15xd div: 1 dxd=1.5dxd x 7 These can be substituted to yield 1.5—1.5xd ] I =J_1[(1.5 +1.5xd}c 1.5dxd The transformed function can be evaluated using the values from Table 13.1 I: f(—9.577359269}+ f(9.57’7359269)2396975958 which represents a percent relative error of 3.3 “fa. (c) Using LEIATLAB >3- f-nrrrat long }} I = quadtinlinet'x.*exptx}'],D,J} I: 41.1?197355999233 which represent-s a percent relative error «of l.1><19_S 9rd. }} I = quadltinlineth.*exp{x]'],Drfl I: 41.1?19T466399175 . . _.5 which represent-s a percent rela’m‘e error if 2x19 ‘39. 19.1 X jay-.2 9.251T99395 9955925925 XH 9.523593T?5 9955925494 X,- 9.?35393153 9.?9?‘|95?8’l KM 1.94?‘|9?55’1 9.5 X- 3 “1398995939 9258319945 true = —sh1(;'r—1) = —9.?97195T3 19.4 The true value 1'5, —ai11(rr 4} = —9.79T196T3. — .25 — . 5 2 Dfir 3}=—£l 332 996 9 5:453:17? 2(1.r:14?193) ’3 _ ‘1 DOT 6)=9._58319 9.96_.926=_fl-6Tfl4 ageszasgg} D = 31995324} — g {—9. 5347’?) = —9.T95 3 9 ...
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hw8_sol - 17.2 (a) The analytical aolntion can be evaluated...

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