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102-2007&2008-3-F10-August2008

102-2007&2008-3-F10-August2008 - KUWAIT UNIVERSITY...

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Unformatted text preview: KUWAIT UNIVERSITY Department of Mathematics 3; Computer Science Math 102 Aug 2, 2008 Calculus [1 Final Exam Time: 120 minutes 10. No calculators or cell phones are permitted during the exam . Given that flat) = \/$3 — 3:2 + 3:. [4 pts} (a) Sh0w that f is one-to—one in the interval [0, 00). ('3) Find (FUTU- _ Find the derivative of y = tan—1&0 5—2:) + ln [(3x + 1) (2 +cosss)]. 4 pts] 1 . Evaluate ] ————- tint. 3 pts] coshz i 1 E aluat f 1 d 3 ts] . V e 3: . ‘ $5 + 3: p 2 . Find lim (“3 + )3. 3 ptsl taco x i 1 0° :2: . Evaluate the improper integral / j d9: if it converges. '3 pts] 0 8 $2 In a _ ' . . I . The curve y z — — T, 2 S I} S 4, is rotated about the :E-aJClS. Find the area of the resulting surface. i4 pts] . Find the centroid of the region bounded by the curves y : 3:3 , x + y = 2.. and x z 0. [4 PtSl Sketch the graphs of the polar equations 1” = 3 e 3 sint1 and r = l + sinfl , and find the area of the region that lies inside both graphs. [6 ptsi A curve C is given by the parametric equations a: : sint + cost and y 2 sint — cost, where —1r 3 t g 77‘ [6 pts] (a) Find the points P[;r,y} on C where it has horizontal or vertical tangent lines. (b) Find the length of the curve C“ CO SOLUTIONS Calculus B Final Exam Summer 2008 1. Given that fix) 2 1/ m3 7 m2 + m. [4 pts] (:1) Show that f is one—to—one in the interval [0, 00). 3m2—2m+1 (x—1/3)2+2/9> ff(ml=W=g m 0, cc>0 (b) Find (Jr—Ira). By inspection, f(1) =1 so that f’(1): 1 and (f'1)’(1}= 4 =1 4WD 2. Find the derivative of y = tan-10: 8'3} + In [(3m + 1) (2 + cos 23)] [4 pte] e’ac A we" 3”“ 1113 sing: = -1 -w 1 T 1 12 1, :1; tan (5'55 )+“(3 ‘3' )+n( “Wm“? 1+x2e—2x+3$+1 2+cosx 1 3. Evaluate f make. [3 pics} [0032—wa=[de=/(c0thmcschm+cach2:r)dm=v(cach3:+cotha;)+c cosh :r— 1 smh .7: OR 1 2 26‘” 2e“: 72 ~————d.1:= — = — = we; : [coshx—l fex+e"°—2d$ [823+1—2e1dx (ex—1PM: (ex—1+.3 4. Evaluate f 51 dm. Putu::c4 [3 pts] m +m 1 1 4m3 1 1111 1 1 1 1 u —d = — d d : "l fm(m4+1}‘” 4/m4{a:4+1) “r” 4fu(u+l) 4/[11 n+1] “ 4 I1u+1+c _ _ 3+2 1 ar+2 I 5- 111mm 13 ms] hay =x(ln(z+2)—ln(z—1)) = w A» limlny:lim 1/(m+2)— 1/(1— 1) 1/9: mm—2 2 limlny:lim($+—23)x(-x-:~fi=3 —1- limy=e3 6. Evaluate the improper integral ] grab: if it converges. Integrate by parts: [3 pts] 0 3 3. lim we’zdmzlim —e’z(a:+1) : lim-t+l+1=1 t—nac 0 t—aoo 0 t—too et 99 7. 10. 2 l The curve y : It? — % , 2 g a: 5 4, is rotated about the r-axis. Find the area of the resulting surface. [4 pts] 1 1 1 1 l 2 I r 2 2 2 : _ _ 1 = _ _ 1 z _ = ( no 3’ x 4m _’ (y) + “”3 2+16m2+ +2+lfim2 “Heel 31/242173; 1+(yrlidm2fil4(%2_¥) ($+$)dx=QgT/:(%+gwf_gjfii%)dm 2 4 82g[m4+x2—m2lna:—(1nfj]2=m . Find the centroid of the region bounded by the curves 9 = 2:3 , a: + y : 2 , and a: = 0. [4 pts] Intersection point: 3:3 = 2 - m gives a: : 1. The line y + x = 2 lies above the curve 3; = $3 in the first quadrant between the lines a: = Cl and x = 1. The area. of the region is 1 5 A=](2—m—xg]dr=— 9 4 and the moments are: 23 11 1 szp] §{(2—x)2—m5)dm=fip, Myzp/ 33(27927x3)d:c 7 U D E? which give the centroid: My 28 Mr, 92 9‘ : p—A : e y = p— = m Sketch the graphs of the polar equations 7‘ z 3 — 3 sinfl and 7' t 1 + sin 9, and find the area of the region that lies inside both graphs. [6 pts] Due to symmetry, we need one intersection point in the first quadrant: 1 + sin 9 : 3 — 3 sinfl which gives sin 9 = 1/2, so that B : 7r/6. The area becomes "/51 ”/21 11 A=2 / —(l+sin6)2d6+f —(3—35in6)2d9 :.-.:—n—9~/§ .—n/22 we 2 2 A curve C is given by the parametric equations x = sint + cost and y = sint — cost, where ingtgrr. [film] (a) Find the points POE, y] on C where it has horizontal or vertical tangent lines. dy __ dy/dt _ cost+sint _ 1 +tant _d.r _. elm/alt _ cost—sint _ 1 etant Horizontal tangent line: 1 + tent = 0 occurs at t = —7r/4 ,37r/4, that is at i"-“(01 313%?) Vertical tangent line: 1 — tent = 0 occurs at t = 1r/4 , 737174, that is at P(i\/§,U). (b) Find the length of the curve C. d9: 3 dy 2_ _ 2 _ 2_ (a) +(E) —(costis1nt) +(cost+51nt) _2 sz (3—1:)2+(%)2dt=/j mmm own m ...
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