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exam 3 spring 2009 solutions - Engineering Mathematics(ESE...

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Unformatted text preview: Engineering Mathematics (ESE 317) Exam 3 March 25, 2009 This exam contains seven multiplewchoice problems worth two points each, 9 true—false problems worth one point each, three short-answer problems worth one point each, and four free~response problems worth 14 points altogether, for an exam total of 40 points. Fart I. Multiple—Choice Clearly fill in the oval on your answer card which corresponds to the only correct response. Each is worth two points. i «“5 t i 3e: 1. Let like, y, z) x [y, six], and let G be the curve from (1 1 0) to (e 1 l) which is parametrized by the vector function r(t)— [13,1 hit] Findthe value of the hue Integral f F- d r (A) 1 {3Z6} [1: id— e] (13) 2 ((3)5 F§FZ%'>‘ [if ‘25] e (D)2e fifaazfijei/feflfigigfi (E) eel C 5 (F) e+1 : £75 jji :: 2e-~2 ((o) 2e~2§ (H) 26 + 2 2. Let it 2 [2, 0, £8], and let 8 be the surface which is parametrized as follows: ?(u,o):[3cosu,3sinu,v] ogugg 032234 Find the value of the surface integral ff( l; ' "3) dA. s (A) 0 WWW» e“ [ego 9 3mm} (B) 4 ,2; ,3; «e (C) 4% e? f s j 11: E ? i a] (D) 8 17% m 35:52,: Beebe o 3“ ngeu Jeweler) (E) 8??" {3 a s (F) 12 (a) m 5555?» flats «riff Egmflgvgggf (H) E 5 we; ) ”‘lF 7w; f] f [gaz deflydcgwj gig? it?" (1) 24h" For probiems 3 and 4, consider the following theorem and proof from section 30.2. A A . . l . Theorem. Let F (1:, y, z) be a continuous vector fimction mth continuous partzat derivatives, amt let D ‘ —* v , 9 u M I 1 be a domain in R3. If F ES conservative, then the fine integral f F - d ? 15 path Independent. 0 Proof. Let C be a path in D from a point A to a point B, parametrized by ?(t) m [:c(t), 3,105), 205)] fromt :2 atot: b, (so ?(a) = A and?(b) : B). Then... ffid? C 1; (A) W}? ‘ were " .» , W...” = f<ngx+F2dy+F3dz) $7 wemeeaiom W fl»? f” firm]; C MB) 2 MWW'QfiM$3M%d 8f af 8f 157’ng Ffiggwgif'3fg’i m£(5;dx+3§dy+5;dz) i 53% } 2 £572 3 3 <55“ U (C) uni—LL) 7% m m M w W m t (9) “55¢ efwfit gmmfim PM“ W“ 2 béfidt a)? éwa wfiaé {3! file egg/3'14 We?!) ii (E) m meme), 2(2)) U (F) m f($(b),y(b), 3(5)) .... f($(a),y(a), 3(a)) U (G) mf(B)mf(/1) ti CH) Thus the value of the fine integral is the same regardless of the path. - , -+ u u e 3. Whmh step of the proof uses the assumption that F 15 conservatlve? (£35 4. Which step of the proof uses the chain ruie‘? (1 2}} 5. What series approach, if any, should be used to solve the following differential equation about 1130 It 0? xy” +3$2y’+:cy m 0 AQMW wgfifi'ie’é’i-i j”? (3%}.st 33$ @Wawfeaflw‘i (A) There 13 no need to use series to solve this differential equation. {WW (B) The power series method should be used. (C) The Frobenius method should be used, and there will not be a natural logarithm term. (E3) The Frobenius method should be used, and there might be or will be a natural logarithm term. 6. According to the applicable convergence theorem, any series about 0 in the solution of the following differential equation must converge at least on What interval? y” + figy’ + fizigy 2 0 (m Hmmm ‘ rsfielfweaé: $2?) $43 (B) (“”93 9) W ©(em $12 @Hfifi. ‘ (fiwe (F) (032) . (G) (0:3) (H) (034) (D (0» 9) (J) (0: 00) ’7. Consider the fotiowing difi‘erential equation. 3:23!” + 5133/ + (4:32 + 3);; m 0 The foiiowing information is known about the solutions of this differential equation about the singular point 2:0 2 0. - The DE has a Froben-ius series solution, but it does n_ot have two iineariy independent Frobenius series soiutions. (The second “basis solution” has a naturat logarithm term.) - For the Frobenins series soiutiom the following is known about the constants. 1 1 i alfiagzafiz...:0 ag=w§ao $421504} aeflmmafi Your task is to find the roots of the indicial equation and to use them together with the above information to select the Frobeoius series solution from the following choices. is ii‘ "P , (A) ym—éxw3+f§x“1_%x+-H pfizfiwg’;§ jflgfigflxisjeg XWXE 1 i 1 (C) ym—§+Ex2—mx4+m i l i 1 m) y=w§x+fif€gwm$5+‘” Wis _1_s___1_7 :Mi J?» 7"“ __ -i i — so mt“ a; “to g ,4 (F) yfaréwlw1+12xml44xq+l" $3 :3 Y 2 Qflx “‘3 ,. {gist/C : - Wt” IV?“ = w a Z 2:: sax sees +4322: 4’"ng 4*” 3 V .3 i 5”” PartII.True-Faise gjir g5*fwéxqs;%o( wgetigx sh“ Mark “A” on your answer card if the statement is true; mark “B” if it is faise. Each is worth one point. 8. The curve which traverses the boundary of a square is piecewise smooth but not smooth. W 9. Let 133,31, 2) he a continuous vector function with continuous partial derivatives. Let A and B he points in a oomain D, iet C, be a. path in D from A to B, anti iet 02 be a path in i0 ii‘om B to A Then j'?-d?=wa-d?. e C; (“’1 ( x 2., 4 i a .- )LgAs, fimeefit) 4 “seize same, g%Z;?/mf/ 4 ifs Ms); wiefamwéfitij Vim / M , we i we V 5 PW? swieea‘sf 5’2 Ct 10. The following éifi'erentiai form is exact. f s? W (3: sin(xy)) dx + (y sin(xy)) dy + (2 sin 2) dz ,;%»:2; I: if} -: ‘gg gig 9/; 53‘ 32 $5 3 F; m 5} .2; 53> 5:3 f 55 g. .9, 4‘3 X W e11”; ; 9M “6.12:. My 2 -4 _ “5% a gutéc’;) a: ’2’: gm (/2202) fie, WfQ Wf? 11. 0 f; cosdxsin4ydydx = [ 0 cos4scdw] [jg sin4ydy] we awe/gage MW WtééeéM/féwf E W I}; w” Eff/d, iii/@Me 32%. yim’wdléhflm 12. Let C be the curve in threewspace which goes in stratght segments from (0,0,0) to (1,0,0) to (O, 0, 1) and back to (0, 0, 0), and let S be the triangular surface that is bounded by C. (Note that S ~+ —-+ lies in the mz~plane, so that both j and — j are normal to it.) For the purposes of Stokes‘s _.) Theorem, the orientation given by j is correct. T g M fig, wwwfiw 54““ .5,» x} ‘ 4:; M? 5,5; maefila v 13. The function flat) m (at: _ 2)3 is analytic at .120 m 2‘ r }. J E .g F “ :juy’z cyfpij; WWM Ma M M J a; jet 15. Every Legendre polynomial passes through the points (*1, —~ It) and (1, 1). , f g I - . ; - , : r? 3 #fi“&%”fié/ aémjv 3:5 iiflw‘fi-éafiiw @xfliég/fw {:7 3:3 5 E r ”m if I r 1‘ 5 » U 9 , Wm fizgfli. gynaed. fifiugfléw “fig/MW» {“55“} 3 mg viii“ fiéekgfiw W'W— feat/eats. My”? fi${~% L . J 5‘ :E a 16. Suppose that O is a singuiar point of the differential equation M + p(:z:)y’ ~+~ 91(21):; = O, and that the fiinctions asp(a:) and x2q(:c) are both analytic at 0‘ Then the differential equation has at least one F robenius series soiution about 0‘ Part III, Short Answer Each of the following is worth one point. Each answer is either right or wrong: no work is required, and no partiai credit will be given. 17. Consider the double integral ff f: f (any) dz: (13:. Shade in the region in the my~plane over which this integration is being done. (Make sure your shading is clear. Indistiact answers wili not receive credit.) ; f g m afi'mfifi/Lgajfi. gnaw a: if”? «a: "awe “M {J 18. What type of surface is represented by the parametrization shown in problem 2? (Your answer must be precise in order to receive credit.) 8 one w enewtfga if e. eeiteauéiae a» “fiat: - e y; 19. The function fix) = 13%; is correctiy represented by the power series 1 + :5 + 2:2 + 9:3 + However, f(2) m $13 .1 —1, but the series E + 2 + 22 + 23 + cieariy does not add up to m1. How can this apparent contradiction be expiained? (The reason can be stated simply and briefly. If you don’t know what it is, please don’t waste your time making something up, and don't give multiple answers, hoping that one of them will “hit.”) i ,. m ,, r ‘3 i" " e w {1? X 4“” i3» tar-r “rent?” item “ii/fee. figigggwifagaeaa, Mtg/ii 9 g; Part IV. Free Response Follow directions oarefially, and Show all the steps needed to arrive at your solution. The point value for each problem is shown to its left. (3) 20. Let Etc, 3;) -— My, my}, and let G be the curve in the cry-plane which goes straight from (0, 0) to (2, 0), then follows the circle $2 +112 m “"4 counter- clockwise from (2 O) to (V__ Vi), then follows the line y z: a: home (Vi V§) )straight back to (0, 0) Set up Whatever integral or integrals wouid be needed to find the value of the line integral f F- d“. I. Do the compiete set-up so that the very next step would be integration, but do not do the integration or finish the problem, J'P’L, We 4;?) weweej/wwwwmfim :9 “ii/Waste wewee am 32ng e ,3 fimfiw I §?« at? 3 jfigexfjeé’l Ki 3.: j if} {fifiggig WWI’Lfivflfi‘é’gwEAJ-Jggg £3 - e (3) 21. Let i? z [cos4y, sin4y,$ln(4z + 1)]. LetSbethesurfaoeofthe“box” 0 S :1: < 1, 0 < y < E, 0 < z < 5. Set 119 Whatever integral or integrals would be needed to find the value of the surface integral ffl F - n ) dA Do the complete set up so that the very next step would be integration, but do not do the integration or finish the problem. (Do not wony about the issue of orientation ) j’fi flag—fie?“ $3 etewé feWiL-w M? fire we: gng& j “4%; gm W {Y’Lée gee ewe mem w, 3 sf ‘, “Kiwi“ S3552! we; 3:? "Tia?“jefi’g {a}? ”7377;”? : {25311 :3: j‘j g W/grflwwfifig % )Cy‘XCiE}: CZQ’ «32-»! (2) 22. (6) 23. Let C’ be the intersection of the cylinder :52 + 312 = l and the plane 3 = y + 1. This Es an ellipse, and it bounds an elliptical disk. What is an appropriate parametrization for this (lisk? (Hint: This is the parametrization that would be needed if Stokes’s Theorem were to be appiied, as in homework 6.) Be sure to supply not only the formula but also the intervals for u and v. ?(u,o): [vaofiz‘i ) 95%:14’) Vfisi’tééeiwij ages Ze’ d§S63[ Consider the following difi‘erentiat equation. y”+2y’~6my=0 Since :30 m 0 is a regular point, there are power series solutions. The first three steps in the solving process are differentiating, plugging the defivatives into the DE, and multiplying in the variable coefficients. After these steps have been completed, the following equation results. 00 00 oo 2 ml??? “ Marni/“m“? + 2 Z mamxm‘1 w 6 Z amx7"'"{”'l m 0 71230 "133 771:0 Your task is to continue the solving process until you have written the coefficient (13 in terms of on and/or (11. You do not need to work out more constants or finish the solution. ‘Lé- 3L3 M ”Wt“? ‘ g Q ”xetvz C3 ‘53"? - wz ..... “ , w .. 2: . Z ”hiffiw’i) QM XW we iwgiéfm MtEé’Mfifi £5ng Wt 3 W3. 734:2 {it 2&}% tZeifi§ ...
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