Math 101 Apr 89 Solutions - Math 101 April 1989 = % /(Sinu...

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Unformatted text preview: Math 101 April 1989 = % /(Sinu _ c082 u sin u) du 1 1 , = 5(5cos 2z-cos2z) +0 1 1 2 1. a) (substitution) Let u = 4 + cos 2;, du = —2 sin 2:: dz. Get = ‘2' c°s 2” cos 2‘” - 1) + /\/4+c032:c sianda: =—%/u1/2du = __:- I gas/2 + C d) (Complete the square; trig substitution) 1 3/2 63—32 = —(zz—62+9-9) = —— 2 3(4+cos z) +C = _((z-3)2_9) = 9 — (a: — 3)2 b) (Integration by parts) Let u = lnz dv = 2z+1dz ' -— _ du = Ad; 0 = L2H”) 3 z_3 3sin0 _ z 3 a 3c030d0 = dz The“ ,___9 ( 3y 3c030 = 9 — (a: - 3)2 — z — f(2:z:+ l)ln1:d:c = (:1:2 +z)1nz -- ‘/(:::2 +z)%da: _ 2 _ — (x +z)lnx /(x+1)da: so 1 . = (32+x)lnz—(—32+3)+C / 4+2: _ /4+(3sm0+3). ado 2 ‘ -—m dz _ —————3 cos 0 3cos = /(38in0+7)d0 c) 2(Substitution, and trig identity) Let u = 22:, du = 2 dz, write = _3 case + 70 + C u = 1 — cos2 u. We obtain /sin3 2min: = é/sinaudu 1 i/sinuu — cos2 n) du sin 7arcsin(z;3) — V6z- 1:2 +C 9) (Trig- S‘letitUtion) iarctang- in second: ‘ 1 1 tin d2: 2 - 5‘“'*"z/7+/12+22 ’3 2 Si” = ‘ 1 1 1 a: = -ln|:1:|——ln(zz+4)+-arctan—+C 2 2 4 2 2 cosflda = ——2d:c 132-4 1" a _ 32—4 cos — a: 2. First find intersections. Find 1:2 = 3+6 => 22—2—6 = 0 => (1:- 3)(z+2)=0=>z=3andz=—2.. So 15 / dz __ fsin0 —c030d0 zZVz2—4 _ 2cos9 2 12-5 1 10 = —- sin0d0 4 7.5 - ~1-cos9+C ' 4 2.. = z 4+C 4a: -4 4 2 4 Hence, 1') (Partial fractions) 3 z+2 A Bx+C A = /((z+6)—:cz)dz ——-—2 =——+ 2 -2 2(11: +4) 2: a; +4 1:3 3:2 3 A+B = 0 1 1 = (-g+7+6$)l_2 => C = 1 =>A=—,B=——,C=1.Sotheintegralis 2 2 125 6 1 1 —%z+l 1 1 zda: d2: $2+4 )dz _ 2lnlzl_2/zz+4+ x2+4 Let u = 2:2 + 4 du = 21:11:: in first integral' note / dz = a ! $2+a2 3. Note the symmetry of the region about the :c-axis: [/21 A = / -r2d0 0 1 2 = 21r (1-y‘)dy l [/2 ° = —/ (l-sin0)2d0 gs 1 A 2 0 = 27r(y—? lo 1 '12 4 87r = 5/ (l—23in0+sin29)d0 = 21r-§=—5— 0 1 r/2 1 */2 . 2 Alternative calculation: eliminate 3/, so = 5(9+2c°39) o + 5 j; 31“ ado ‘ 2 ‘ 81r In 1 “"1 V=2 2 -d=4-—‘/2=— =_-_ _ -_ "/0 fix: 1r 52 o 5 2(2 2)+2 0 2(1 cos29)do 1r 1 1 . II? — Z—1+Z(0—§sm20)o 4.r=1—sin0: 1r 1r 31r = --1+-=——1 4 8 8 0 ‘- o 1 2 d1: 2 dy 2 7r = — _ 3" 5. Speedv (dt) +(dt) . We have 2 a: = t2 y = t3 dz _ dy __ 2 a? — 2t dt — 3‘ so 1) = V49 + 9t‘. . . oo 2 The distance travelled is S / 1 d3 since m 20".“ an x 2 1 1 37/2 1:2 00 a = -/t vdt = /z_3/2dx=—2z’1/2l:° 1 = 2<oo II o\, n on ‘O 6 N I A a. 95 Since A and B converge, I = A + B also converges. Let 18t dt 2 2 7. 4 = 4: (u = 40) x + y Y 40 ._ 1 1/2 X ll oh :1 3 : Q- A es H M V III N Then _ 1 2 a/zl40 __ 3 _ . I — 00 x2 dz - . . . . . fl 2 d h o m - The area. of an equilateral tnanzgle With sxdes s is A = 2 s an t e _ /1 9:2 ‘11! + /°° 1:2 (13 volume of the region is V = / «75.92 dz. But :02 + 43/2 = 4 => 1/ = o «4 + x7 o 4 + z 2 :i:%\/4 - 1:2, and so 3 = 2|y| = 4 — 1:”. Hence 1 2:2 a: 215 __.,,2 2 z A=/o‘/%<oo, >< V 12¢:(jr)d 2-7 /0 (4-32)“- (by symmetry) interval. 1 3 = «Eon—q )l B _ /°° zzdx 3 o - 1 V4+z7 % 3 Write this as I = A + B, where since A is the integral of a continuous function over a finiteBclosed 2 7| 2( k=1 8. Sn: 1 (k + 2)(k + 3) +(§)""1) = U" + V", where " 1 " 2 U" = —————, d V" = — k". §(k+2)(k+3) a" g9) Vn is geometric, so 1—(2)n 2 V,,= 13% =3(1—(§)") " 1 U" = —-—- I§(lc+2)(k+3) _ "(_1__;) k=l k+2 k+3 1 1 1 1 1 1 - ((m‘m)+(m‘m)+“+(n+2 n+3 1 I _ (ii—n+3) _ _ 1 1 2,, . _10 So Sn—Un-i-V -— 5—n+3)+3(1—(§) ),andhencenlergoS —?. 9. 6" = k=0 00 1: —2t’ _ (-2”) e — Z kl k=0 _ 2 (2‘2)2 (2‘2? -1 213+ 2! — 3! +-- 0.1 2 I = f 6-21 dt 0 10. n WMs v: N 73' + g—d u [‘18 II [V]: where 6,, is the error from truncating the series at in terms. For an alternating series, we have lenl < |(n + 1)th terml. We require that _2 m -(2m+l) m —(2m+1) ( ) 10 _2 10 (104’ m! 2m + 1 — 2m + 1 . . 23 10-7 8 _, where m E n+1. With m = 3, this evaluates to y 7 , or 5-10 < 10'7. Hence, . 10'3 22 10"5 -1 -7 13-810 ~2T+§T+€m IEnI<10 a) Let 1(1) be the Taylor polynomial to three terms: 1(2) = m) + News - a) + ’"(f’e ~ ar 2 1r t =--. Sea 4 1 m) = ln(sec 1:) => fig) = 51112 1 1r I _ , = ' — = f (z) — sect secztanz tana: => f 1 f”(z) = seczz => f”(z) = 2 -1 I I 2 SoI(a:)— 21n2+(a: 4)+(z 4) b) Taylor’s theorem says f (z) = I (z) + 6 where E is the error. For a: = §+ 0.01, we have M; + 0.01) = mg + 0.01) + e and 1 1G + 0.01) = 511. 2 + 0.01 + (0.01)2 SO “E + 0.01) = $111 2 + 0.01 + (0.01)2 + 6 Now estimate error a. For some t E [0, 0.01] we have fIII(_:_ + t) 3! (0.01)3 5 :: sinz III __ But f (1:)— 2cosaz, so m 7" If (2' + t)l IA N ll 3 So finally get 8J5 Isl < v x 10‘6 < 3 x10"° ...
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This note was uploaded on 01/30/2011 for the course MATH 101 taught by Professor Broughton during the Spring '08 term at The University of British Columbia.

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Math 101 Apr 89 Solutions - Math 101 April 1989 = % /(Sinu...

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