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211-2007&2008-3-M20-July2008

211-2007&2008-3-M20-July2008 - Kuwait University...

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Unformatted text preview: Kuwait University Math. 211 July 19, 2008 Dept. of Math. 85 Comp. Sci. Second Exam. Time: 90 min. AnSWer all questions. Calculators and Mobile Phones are not allowed. 1. [2 pts.] Describe and sketch the domain of ln(4 — 4:172 - if) f(:l;,y)z may 2. [2 pts.] Find the following limit, if it exists lirn xy+2$—3y—6 (mm—43:2) $2 + y2 — 6:13 + 43; + 13 3. [3 pts.] Use the definition of partial derivative to find %(0, 2) of the function f(:1:,y) = VfiTyg ezymmy + 1/3: + ytan cry (2+1) ).pts] The surface S 15 given by :1:2 — my— _ 22 tan my — a -[ (a) Find the parameteric equations ofthe normal line to S at the point (1 1 ,—2). (b) Find the intersection of this normal line with 312— plane 5. [3 pts.] Use differentials to find an approximate value of 5 (3s)2 + 2(2.1)3. 6. {9+2} pts.] Let 2 : f(:c,y) is determined implicitly by $2+zzcos(E)—1I$2+y2. Z (a) Findg— :and 3" at the point PH) 1 ,.1) (b) Find the rate Bof change of z at the point P in the direction of a = 43’ — 33' 7. [(4+4) pts.] (a) Find the minimum of the function f(:c, y) = 2:122 + y2 - 6 subject to the constraint :32 + y2 = 4. (b) Find local extrema and saddle points (if any) of fuzzy) = 583 + 2113 - 3:82 1. 9212 +121; 99 Key Solution for 2"“ Exam. Math. 211, july 19, 2008 I. Df={(3:,y) 352+? < l&:c>y2} . . . . (m—3)(y+2) 2. The given ilm1t equal (m,y)lrl+%,—2) (W . Set. {L‘- 3 = m(y+ 2), 2 we obtain lim m(y + 2) = m .We have different values for are—2 m2(y + 2)2 + (y + 2)3 m2 + 1 the limit with different paths. Thus, the limit DNE W em _ e4 3' 335w: 2) : $13—13) k = £1131) —k— = 264 L’Hopital’s rule. 4a. F{J:,y,z) = 3:2 W my * 22tan’1xy +7r = 0. F$|p : 23; _ y m Iii—jg? : _1 2 Fill? = _'73 ._ 1:33:23; I —3 Fz|p = —22 tan—13:3; = T. The parametric equations of the normal line are :c:1—t,y=1—3t,z=—2+irt. 4b. The intersection point is (O, —2, —2 + ’fl'). 5. f{:r:,y) = ffxi + 2195,27 = 4,1; = 2,Aa: : w0.2,A'y = 0.1. 5 (a: + A502 + 2(y —1— A9) = flat, y) + fmAa: + fyA'y ‘5/(315)2 + 2(2.1)3 = 2+§$($2+2y3)'§A$+§y2($2+2y3)‘%fliy : 2—0.02+0.03 : 2.01. 63. F(a3,y,z) = $2 + z — (3053f) — #212 +9!2 3' EH 1' §E__£__ 2$+Z51n(z)+ f—‘—'x2+y2 MP 2 _0 82:”— Fz— 1—%sin(£zfl) 33— 2' £1; _L @__fl__ ZSln(z)+V”2+y2 Ath——1 ay_ FZ— 1—igsin(£§i) ' ’y 6b. Duz = 293m + zyug, u = (4?; — 330. Thus, Duelip = g. l 5 own m 7a. 7b. F($,y,/\) = 23:2 +392 — 6 — M562 + y2 — 4). 4a: — 2A2: VF: ( 2yu2Ay ) =0. 4 Thus,:c=00r)1:2. For :2: = 0 implies y = i2. For A = 2,31,: = 0. Thus, a: : :I:2 The points are (0, 2), (0;. —2), (2, 0], (— ,0). —2 The values of f are f(2,U) : f( The min. value for f is “2. ,0) : 2, f{0,2) = f(0,—2) z —2. The critical points are (0,1),(0,2),(2,1),(2,2). f” = 6:1: — 6, fwy = 0 and fw = 123.} — 18 n— n— -—_-n (22) --u— OO 99 ...
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