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exam 3 fall 2008 solutions

# exam 3 fall 2008 solutions - Engineering Mathematics(ESE 31...

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Unformatted text preview: Engineering Mathematics (ESE 31'?) Exam 3 November 5, 2008 This exam contains seven muitiple~choice problems worth two points each, seven true-false probiems worth one point each! and three freenresponse problems worth 19 points aitogether, for an exam iota} of 40 points. Part I. MultipleeChoiee Clearly circle the only correct response. Each is worth two points. 1. Let E; (3:, y: z) x EFF, y, a: + z}, and Eei: C be the line segment from the point (8, O, 0) to the point (1, 2., 3), Find the value of f '1?- d “if. C Hint. The line containing this segment can be parametrized by the vector function —§ 1 r (23) m [2%, 2t,3t§. (A) 8+2 \$475):- [eyzeeej one: :3? ’ "é: . ' (B) 8+3 e mm 2 {e } gage} (C) 6+5 ‘7 (D) 8+6 fgz \1'3552'233 (E) 8+7? :3? 5. i n g Cm; §wﬂ 1’ L, (”-53% “204?; E; (G) 6—5—85 M E ‘e ,5 ¢ W 1.5.. .2 {N (H) e+9.5 M 5a (/6 {wiiééléﬁ’ M {(53 +313 > o (I) e+15 w . (q M (3) e+16 , {a “f" 53 E 2. Let ﬂay, z) m say2 — 2:22, and note that (Vf)(x, y? z) x: [:92 — zg, 2:63;, “2222]. Find the value of the foliowing line integral. M (:3) [(92 “ 22% + 2:82: dy M 2m dz] 3‘ (A) 25; s V (3)44 révﬁgeﬁ ejCL/gjzjfj (e) 52 . __ (0)83 ‘35%&W4§3~532e3> (E) 88 : BS'W 5% n 2:; (F) 110 (G) 149 (H) 681 Let i? be the vector function (in the plane) given by 33E»? y) x {261cm 3;, efsin y], and let G be the curve which foiiows the straight fine segments from (O, 0) to (2, 2) to (1, 2) back to ((33 8). Which of the following expressions would give the correct value of the line integral § “E. d E”? C (”W (A) jg} f2 Bersrnydxdy } 2 _. M (B) I: [2: 361.8%} :1; d3; (it (C) fig" (3) f0 ff 36333111 y dy dz: g{Bexsin :1; dm dy (E) (if; execs 3: d1: (13; (F) f: 1:; cameos y dy (13: (G) [iffy excos 3; dz: dy (H) 1232!? execs y d3; (12: Find the correct parametrization for the folkowing surface. x2+y2\$4, 2m4 M Wi‘f:£\, ijW 2. (A) ?(u 202 [cosu.sinu,2v] 05115329? 031152 (B) ?(u ._v)m[cosu sinu 4v] 0321,3271“ 03352 (C) ?(u v): {mesa vsinu 2] Ogu§2w 0<v<2 ?(u, v)—— [uvcosu vsinu 4] 0 g HIS 212' 0 113953“; (E) ?(u, M) {ecosu vsmu 2v; OguSZW 03:42 (F) ?(u,v): [vcosu,esinu 4v] 0 S u g 27? 0 g 22 g 2 (G) ?(u,v)z {cosu sﬁnu 222} 0 g u g 271' O S 2; g 4: (H) ?(u,v): {cosu sinu 4e} 0 g u g 27%“ G g 1; g 4 (I) ?(n,v)m{vcosu,vsinu,2] {)guSZ’K Ogvgai (J) ?(u,v)::[veosu vsinu 4} 031L325? 051234 (K) ?(U,’U)E fvcosu. esinu 21)} 05155277 95654. (L) ?(ufv)m [ecosu vsimz: 41;] 0 g u g 272' D g 2; g 4 What information can be obtained from. the pail—off terms which resuit from the foliowing equation? (Yew answer shouid include exactly the information obtaineé from them ——~ no more and no less.) 00 co 00 Z m(m m 1)amxm”2 + 2 Z mamxmml ~§~ Z amxmﬂ m £3 72429 mgg mug “ <2, i 4» g (A) 018 I 0 \$.43 ““2 5K3 . 4?;1‘2 5‘? ’7‘?in (B) 62120 im[o“i§5€?&% «P 25.. {woa’iwﬁ + : iimﬁ :1 a} (C) a 9 W5 ‘ gm 2 m (D) a3=0anda1m0 2&2"? 252} 3 5:; (E) aﬁ\$oanda2=0 : m {3? (F) alﬂor‘lﬁdagmﬂ £1 ‘3‘; (G) (11 m “€39 (H) 532:““043 (C) agm—al E (J) The pull—off terms give no information. Consider the differential equation By” W 4my’ + (4 — 2:)y m 0. Find the form ofa basis of soiutions about 3:0 -—-_ 0. 00 34) '3: Z} :A wg/A’LA . 3 32'“? Z Amxm g m (A) m = x“? Z amxm 212 Lil mat) me N "X _ , 0° 00 if w “E; 2:! 4’ W ”*3 (B) y; m 117” Z amﬂfm' w W yllna? + ZEWZ Z Ammm 771:9 mml :j xii (DO 00 AM ““‘ (C) 211 m ﬁggamﬂim 3J2 == mggeAmxm "251%5‘ 4: 1X? 00 i * (D) ygmmgzamxm yzry1ln\$+\$22f1m\$m g} 3 lat" 6%“32‘3B 7“ 4 msf} 711:1 x: «39\$ (E) gm“: mm yrmx“‘4 ’Aggm ~,_ 1 11;!) m 2 7712336 m yiéﬁw?) ”493i 441! ”’” Q (F) 3}: = 271 EGGmJIIm ”g: = kyglnx +\$W4Z Ammm “A? ,. gk‘ .3, 20,1" .: g} ml" mmﬁ (G) :11 2 \$4 E: €1me 3J2. 2: a: Z Ammm f1, :3: E 3 4} mzi’} mza M W”‘”‘"““‘“” ”WW-m WWW W, M ‘ W me4z€lm\$m wky1En\$+m f}; W 2;! J21. _ ! mZO mwﬁ 2,! WWW-WWW . . Mk ., . . WM 7. According to the appiieabie convergence theorem, any series about 0 in the solution of the following differentiai equation must converge at Eeast on what interval? (13) (—3.3) "921\$ : 2M mxiingM/z. . (own , . g .. egoﬁégM (a) (0,1) “312.. Mm; WW Em ”M ”WT? «MM @ 5W M ,,3_ M 54:3. (F) (0m) (G) (13) (H) (Lee) (1) (3,00) Part II. True—Faise Write out the word “true” or “false” for each of the following. Each is worth one point. 8. Consider the motion of a ﬂuid in a region around a point P in threemspace, and let 3(m,y,z,t) : p(3:.y,.z,t)?(a:,y, 2,15), Where p is the density function and "ii is the veiocity function. If the ﬂuid is compressible and div "1i 2 0 at P, then P is neither a source nor a sink. , ‘ 2’ .3 J; Mk; M MMMMM 9 MM. {W MWJMMLMéimamﬁm. MW M M «A; w 9. The following differentiai form is exact. E . sin awn-3’ dx+xcos 63“in "ﬁm of“ ML 3; 5:“: Fl 3;”? ﬁﬁiﬁg withij W 3 Meg . & xwgg + wsﬁgﬁ ‘95} ”5F: ’xﬁfﬁ-g r .- 76.}; jg mix ﬁﬁﬂg L: .: yMs‘ :5: ‘5sz "'3” my 533% j {3’ For problems 1042, suppese the vector ﬁeld ? satisﬁes E; m Vf for some scalar functien f M t -. throughout a domain D. (You may assume that F ES a continuous vecter functmn With continuous partial derivatives, so that there are no tricky technicalities to consider.) ——} 10. div F :8“ 3%; jﬁfgﬁamj ﬂan“: C; "-9 u I l I 1 . F 18 irrotatlonai. W ,, I; «a? ,‘ {i}; W ﬂbé avg; F m WM% 13. The differential equation (1— \$2M” w Emy’ ~+ 303; m 0 has at ieast one poiynomial solutian. _ s m: we éfmwlam; WQZTM maﬁi 1g: 3’: gm y” 14. The inierval of convergence of each of the Legendre pelynomials is (woe9 so). jute,» 322: emiteﬁtj f C’gﬁﬁwdfjﬁ (:2. V1497 a? ﬁ ffiﬂ_ﬂma~j; £4, [”m- M) Part 111. Free Resgonse Follow directions careﬁliiy, and Show ail the steps needed to arrive at your soiution. The paint value for each problem is shown to its left (10) 15. Let F be the vector function given by F (\$.31, z)m My 6“ 21y, 2}, and let G be the curve which fOIIows the straight line segments from (0 0 8) to (O 4:, 2) to (2,4, 2) to (2, 0,0) back to (0, e, 0). Use Stokes’s Theorem to ﬁnd the vaiue of the line integral 3? "'13- d ‘r’. (I Hint: C is a rectangle in the plane y := 22. e e E 1;? J m m f: r :e e E ﬁx 5%? (33% New?” #3..) w i r? f M M f i e: a g 5:} a ’A. i Week/ELVK 5:5 kw; M W“ M, 3&2 § E)? r? 1" £313 25433 em 5:; _ , .; 3&2 Zaﬁaljf iv: f. (4) (5) 36. 17. Let T be the solid cyiinder given by 332 «e y; g 16, 0 g 2 g 2? and. let S be the compiete boundary ofT. Let “P? :2 [3:2, ya, :82]. Begin the process of using the Divergence Theorem to evaluate the surface integral} ff{ “13 - ‘3 MA. Continue the process up to the point where 9* the very next step would be integration, but do not integrate. (S is oriented as it should be with the positive direction outward, so you do not need to eonsider the issue of orientation.} 5§5§ﬁ\$>JA¢5§§grgsv :2 j§§{g +2§>33ng do one“; QMMLS "2 2e” ‘é‘i g ‘ _ £3 fa £3 (/% t” 2Jz55ﬁ§>ﬂwﬂg\$§k¥§ . a; e. “Mega; .2 n 1 ; 5;} § £3 {£31 e 2.}: €5VL€§ gel/”2 efﬁgy}?- Find a basis of soiutioos in series about 3:0 = 0 for the following differential equation. As you work through the process, ﬁnd the constants am up to and ineiuding a3. y”+(a:~2)y’+2y20 (:3 W W .x Major Hint. There are no pull—off terms, and the recurrence relation is as follows. Start the problem with this information; do not waste time by starting at the beginning! 2 1 For m 2 2, am 2 %Gm_1 -- m_1ame2 59:} £524; Q}: 59‘ £1»; iniﬁ ‘33 ire” ééﬁ I j’ééreb Mei: : ﬁe,“ 34; w e”: , :2 4 griem:eeee%eez»“‘ F ,ﬁté ‘ ‘ t ‘5- Z X 3 :3 £43 *6???” {\$5439th (/3 5% ”’Kéirk em ...
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