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102-2005&2006-2-F10-July2006

102-2005&2006-2-F10-July2006 - m w’>W 4t...

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Unformatted text preview: m w’ >W 4t: _ filth/1931115! 27, 2006 Ktinjajt Univei‘SI y .- Math 102 Jeptof Math Sc Comp Sci Final Examination Duration: 2 hours The use of calculator is not allowed. Turn off your mobile phone or pager ‘51 - = * >‘ "f"? y 1. Let fir) = tan"I I +tan_1(ln;r) "12, I > D. ' “I ' V .5- (a) Show that f is one—to-one Over the interval (0,00). («5' ‘ 4- _r 39-“ l (1 point) (b) Find the domain of 1'“. (1 point) (c) Show that P({], 1) is on the graph of f ‘1 and find the slope of the tangent line at P. (2 points) 1 ' 2. Evaluate lim+ (sin rjlnx , if it exists. (3 points) :—‘0 3. Evaluate the following integrals. (3 points each) ' sin 23: x/zt:i - 4 . 2 c053 :r. ' —— a b a 2 3 sec :33 dc: d #— dx (a) f1+sin2r x U f a? a: (c) /$( ) [) V1+sinz ‘ I 4 1 ' . . f Cl. Determine if the im re er inte ral f —-4~u—-—— tie: is conver ent or diver ant, and find its value i p p . g 1 #8 + 21' u 1:. g g it is convergent. - , , (3 points) 5. Find the length of the cardioid r : 1 + sin 6. I I (4 points) r1“; El ll? _ ”iii 6. Find the equation, inqectangular coordinates, of the tangent line to the graph'of the polar equation r 2 2 + aim?‘ at the poinmfimo 9 =_ O.__ (‘1 l Vi) _ (4 points) 7. Sketch the graphs of the polar equations r 2 cost} and r = 1 facet}1 and find the area that lies inside . both graphs. . . y 3,} ”.53 .( ”J h {ng5? (4 points) a» 57.1 _ .--"..~--- . ~._)'s.:a fl \ 8. Consider the curve C given by the parametric equations a” fl». m r: .. .. a; l. I. _ i zzeicost yzeisint, 0315231" (If/CD ‘ v (a) Show that (2 points) ema— er ” ‘1 ‘* (b) The curve C is revolved about the I—ELXlS‘ Find the area of the resulting surface. 1‘ (4 points) vi“ (I skin 00 CO rij f’($)= my-F 71—5—27 > 0 for all I > 0. So f is increasing on (0,00) and therefore is (1—1) there. :55”) D,-1=R,=(1im f(2 2,) limoof(z))=(_3_47£,§41£) x~40+ if (C) f(1)=0 soP(0 1)isonthegrapho{f 1 Theslope:%92— WWW—177:;— 4.1:? 1 A ' __1_ 2; 2. — sinz In: In? : L—“S‘”. Iim 1m _ lim ——-——-‘“5‘” =I1irn cosrzl Iim sinx 1M =e. 2 y_ ( )— In: 2-»04‘ J 1‘04. In! 0+ 5””: 2—.o+ ( ) 3. sin 2: __ 25in 1 cos: “=5_‘“ 3 2n _ . 2 (a) /—~r1+5in:’d1 _ /——,— l+—_—sn—in I dz: _ /1+u du ln(1+sm z)+C (b)/ @311: =2“ / Zggg—giazsecetanade = / :2—3‘d0: / (sec9—cos)0d0 =1nlsec9+tam9| —sin9+C:ln!z+ m. ~3EIE+C (c) /23:3 sec2 z2drc : [$2 (2msec2 r2) dz = /:t2 (can :2),dz 2 .12 (tan 2:2) — /taan1:2d$ ' = 3:2 (tan 2:2) + in Icos I2| + C ‘ [TigjéfideI = f (1_——sin2:z:) (1 +sin 1:)—% (:05sz: = f%du 2 2 mu mm —%‘(1+u)5+§- (1+u) —§(l+sinz)+ +§(1+sinr) »L 4. fi'fl = lim dz = lim sin—l E L =sin“! : fl . l 8+21—x “.4- . 1 79~(I_1)2 Laq‘ ( 3 )l] L Run 2%? -1! _1 2 2 _ _ cosflsin 9+ 2+sin 0 cos 0 _. cos 9 cos I}— 2+sin 9 sin 9 The point of tangency is P(2,0)v The equation is y = 2(x _ 2) U‘ [0 Lfinl: A v u + ~1 as Q. Cb i N 3.1121 N A ,.‘ + S’.‘ :1 2 % Ii c059 d9=—4\/‘ 2ml§% =8' l—sin H ”H = 2. 9:0 6. fi = cos 9‘ The slope— — w :5 cosO—rsin 9 § ' 5 % Area: [(1—c056))2_r16+:{8052 9=_[(1—2c059)d9+_cos[ zflzfgw—fl 0 g 0 0 2 8. (a) ($132+ ('1‘): @cozt—sinty +e‘(25int+cott)2= 8‘6 +1): 38‘ 27v 2w \/5 (b) Area: /y (2in )dt: if e (sint)dt t: T (I — 82 7r) by using integration by parts ...
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