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211-2008&2009-1-F10-January2009

211-2008&2009-1-F10-January2009 - Kuwait University...

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Unformatted text preview: Kuwait University Math 211 22 January, 2009 Dept of Math & Comp Sci Final Exam Duration: 2 hours 10. . Find all the values of x for which the series 2 Calculators, pagers and mobile telephones are not allowed. °° e" (ct — 1)" -—-— converges. ”:1 n 2n . Let f(a:) = f; e"2dt. (a) Find a power series representation for f (x). (b) Find the 7th degree Taylor polynomial of f at O. . Let f (:12, 31,2) = in (my + yz + 2x) and A be the point (1,1,1). (a) Find the directional derivative of f at A in the direction of B (—1, -2, 3). (b) Find the direction in which f decreases most rapidly at A and the rate of decrease. . (a) Let z=f(a:,y), wherezr=et andy=t2+3t+2. Giventhat g:— =2myz—yand 62 2 dz 55—2my—m,findd—t whent—O. (b) If u (3,12) = sin (.9 + 2t) + ln (s — 2t), find a constant k such that u“ = k2uss. . Use the method of Lagrange multipliers to find the extreme values of the function f (m, y) = $2 + y3 + 2y2 on the ellipse 2:1:2 + y2 = 1. . Find the volume of the solid bounded by the surface 2 = 6 + (:1: -— 5)2 + 4y2, the planes x = 3 and y = 1, and the coordinate planes. . Evaluate the integral: 4 2 f / \ly3+1dydx. 0 fl . Use cylindrical coordinates to evaluate 2 V4‘1'5 2 f / f 2 sin (2:2 + 2/2) dzdydw. -2 —\/4——a:§ %(x2+y2) . Consider the vector field F (x, y) = (6:1: + 5y) i + (5:1: + 4y)j. (a) Find a function f (x, 3;) such that F =V f . (b) Evaluate f0 F-dr if C is given by r(t) = t2i -— t5", 0 S t S 2. Use Green’s theorem to evaluate the line integral f xe“23dx + ($4 + 2222312) dy c where O is the boundary of the region in the first quadrant between the circles given by:1:2+y2 =1 a.r1da:2+y2 =4. 00 CO QLZ' RQT /0Hfl/= l ’e—I“‘I/—'7§/X-II=L_ I L<I Ct) x6(,—'é'},+%‘> , m, MvI-I 2. j x— = I"a — : a/I’ermfiU Ila/Wonk, (C) j 1.: If“? :lLa/‘Mwmc (A) finsml‘; [—81 —) ”6’ La “(3" = (4),“ 2n __1 _.t:("£_ " "7' (—TJ“_"‘ 5."an x n L)?“ é t >52 = o *1?th oiéé's'én—r . I A)? #AJLT; 7 iii/al- q’j-O . Wx)- " - SL_ x3 :f- 7’;—z‘:—- .572 it}. _-.___ _ 7 EL; 4) VII. ex [(yn)t+Ie+x)J+(xq)kJ c) 71m): 5M) XJiJQi' A5="ZT—3J'*2:E =>M=le=ifi=>bi f{fl)= Vf' 'K’Fz/fi‘z/{B/ I”) u = :I/Vf/F'Vf IIVFII 5/55 => I7: ~j§(r+J+IT)I<bx I="’VI/Ij__ " ;‘ aji— —0 =2 x= I, it: , 2 — 31—3 75%; L"! “gel/II" 3 :It =zxfi+£Jfi=é+3=IE ' A) “s = 6’5(5+2I‘)+(5—2tf/ - “55 = —-m (s +24} [5.2” 2. i‘ at 32605 (Sf-21") *20—22'7”): Mu.“- ’451‘n(§+2f‘)— l/(S-lt‘) 2:1/[155 =>/<= Lg{g(x.y)= 2x2 +3 2-—/=0 2:) 212—+§:=I /2XCZ4U:‘~)I= = A V 2x — 2c "" x /-2 .9 VF <2 I 3a ”(I Ala Zaé3a++‘2'\)‘0 CueI. x=0 C;>J=:[email protected] )QO ~I)=I f{o,l) :jgxxa‘ 4 II: )0): Casez. 7c+0 ~=>A=a @>J(3(7+3)=o<‘7 CD Vi Vi ADV/E": [12X = 3: 7402/); MM" = é— =f(if 5,0) \Yc—I “’> =0 STOP. 3% V— I/fi+(z’57+43]5(x5{7 fI/ZgMflc/g =I§I _ . $.2’_- ff/zm ff} 3M/a/x /3W0931 —/J3+’)3//Z :22 2II 29 @II I: £51n(r2)/~a(24/MI9=I I (I fi”)“’“‘”70(”{" $34" : ”/fFZSMZLI—J-tz sIht20((-_ -: (I (i—S‘IVLH - 1.565514) &' 4) I1 5"“? ‘2 IMF 3* +87 +5141) -> If gum) =><7IQ0= AI £7: 5x+4g =>€IJJ= 242 +C. -> fag): 3x2 ffzJ-I-Zdz +c. A o 0 ~ F Coast/‘Vq'fi/e 7) r ___ M) 2%)): 4:13 I jFaI —,][(4’ —a) {mo}: /6. @IIIMHQQG— _ We ~P)5(,4 Ken; Hazy/,4} :LEIZ 907046)” ”Ir/6L: fiyc/gxé/‘Z FIJV=5 ”‘3’- 00 (MY .9 ...
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