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Math 101 Dec 91 Solutions

Math 101 Dec 91 Solutions - Math 101 Solutions Dee 1991 2 1...

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Unformatted text preview: Math 101 Solutions Dee. 1991 2 1. a)I=V/‘.1:3\/9-172 dz. 0 Use the substitution u = 9 — .172 du = —2;tdm (1:0) 5 (11:9) (1:2 E (11:5) Hence, 2 . 1 5 . / .17"\/9 —— :0" d2: = —3/ (9 —- u)u'/’du o - 9 1 " . . . = 3/ (911Vz —— 11"”)(111 9 = l 9(2113/2) — 3115/2 2 3 5 _ 9 = 19: — 10\/§ 5 Jr“ — 9 b) I = /md1. We use the method of partial fractions. \Vrite 124 _ 9 ‘- .1?“ __ 9 1:4 -— 91:2 —- 12(37 + 3)(1: - 3) A B C D 35+; z+3+:n-3 Then A(J: + 3)(1: — 3) + B2:(.r + 3)(.17 —- 3) + 0172(17 — 3) + Dz2(x + 3): .173 — 9 Substituting 2 = 3 : 54D = 18 => D = 1/3 17 = -3 => —-54C = -36 => C = 2/3 a: = 0 => —9.4 = *9 2 A = To find B, we substitute 1: = 1 and find that —8A-8B-—2C+4D=-—8 Now, using the values of A, C, and D from above we obtain B = 0. Hence, 373—9 1 2 l —— . = — d. [w—gm‘zd’ [(12+3(x+3)+3(1—3)) I — 1+21n|1+3|+lln|r 3|+C — at 3 I 3 ' c)I=/\/1T- _9d.1:. .1: Letting "2 = .172 —- 9 2" (In :: 217 (11: we obtain that /-'—"2_ 2 f I 9dr /—.—“-—du .17 II A H I 'L' + <9 V 3- II n — 3Arctan(13‘-) + C d) I = /cos3xe"i""d17. Let m 2: sin .1: dm = (3051: (11: so that /cosa.1:e"i'" d1: = [(1 — w")e""dm Integrate by parts; let (1 -- m") (In —2m (1m u e'“ (1m PM? u (it: = (1 — lll2)€m + 2me'" (1w \. Now, integrate by parts again; let 2. a) “it! first find Where the two curves 1' = 1 + sin!) and r = 33in!) 2m (in 2 dm v e'“ dm PM! u du H H ll (1 — w")e"’ + 2me'" — / 2e'" dw (1 - "’2)€m + 2me‘“ — 28'" + C (*1 + 210—m")e"’+ C = ——(1 _ Sin.1:)2e’"l"z + C II intersect. This occurs when 1+sin9=r=3sin0 => sinf) = 1/2 so that 0 = 7r/6 or 9 = 57r/6. Now, pictorially the region R is [£386 b) Now the area of R is given by 11/“ 1 . Fl“ 1 . area=2 / -(1 +sin0)‘ dbl—l -(3sin0)‘d9 _,_-/2 2 (I 2 3. a) Graphically the curve given by x = 8 cos” t, y = Ilsin‘1 t for 0 S t g 217 is \J v- b) Now, (8 cos“ t,4sin“ t) = (3J5, 1/2) implies that tan" t = 1/(3\/§) so that tant = l/\/§ Thus, as 0 S t S 217, we must have t = 77/6. Moreover, the given curve passes through the point (0, 4) when t = 1r/ 2. Therefore, the length of the curve from (0, 4) to (3J5, 1/2) lying in the first quadrant is given by arclength = 1r/2 = / \/(-24 cos2 tsin t)2 + (12 sinztcos t)2 dt 1r/6 1r/2 12/ sintcostt/4cos2 t + sin2 t dt W /e /G -4 Using the above picture surface area ll 5. Recall the following substituting x = —t2, we obtain 1r/2 12/ sintcos tv3coszt+ 1 dt , 1r 21+7ry‘/ (d—y) 41r/o ((2—ézd 1 -2: 1:21: 1‘2 k(1)t _o and so 1 0° _ k2-1/2 ——~fi(1+t’) +23 1)” This last series is valid whenever 0 < It] < 1. Now, integrating this series term-by—term 1/4 1 1/4 I: zk— —1/2 —-——dt = // —1 t (it 0 W0 +132) 1; (_ ) z * * , ”‘ = (_1) —t2 +1 2 k=o 4k+1 o k2; 1)k(4kl + 1)42'= k 243 1) (4]: + 11)42k + E" where 5,. is the error involved in approximating this series by its first n terms. Since we are dealing with an alternating series, the Alternating Series Test gives us that IA [(11 + 1)-st terml 1 (4n + 3)42n+2 ieni Choosing n = 1, we obtain 51 < 0.001. Hence, 1/4 1 d 1 1 ___.._ t” _ _ o «E (1 + 12) so with error less than 0.001 in absolute value. 6. Using the definition of the improper integral, we have M ed; d 1' c 8—“; d Azfiz‘ci'i‘oozfi” C = lim —e—‘/E 0—0” 0 _ ' __ -\/E - 25200 e ) = 1 b) 2 . Volume = / 21rz(2 — 2) dz using shells l 1 2 — 2 — 1 d ' d' k 7. a) The line y = 4x intersects the curve 3/ = l/z at the points (—1/2, -2) or ./d "K 11) l y usmg is s and (1/2, 2) while the line 3/ = 3/4 intersects this curve at the points (—2, ~1/2) and (2, 1/2). Thus, the region in the first quadrant bounded by these three curves is 9° 8) 2 . 5 _ 2 1 . 5 ‘ 1 O . 5 3X 0 - 5 1 1 - 5 2 2 ' 5 Since the density p = 6 is constant and the region is symmetric with respect to the line y = x, the center of mass (5,9') must lie on the line y = a: i.e., 37 = :2. Hence, we need only set up an integral for the moment M1: = 0 about the y-axis. Now, the mass of the region b) The area. of the region in part a), is given by area=/olfl(4x— E) dz+/l/2(% — E) dz m = p-area 91r 6--4- a7: 8. a) 2 The moment, M m = 0 about the y-axis is given by 3 M$=0 = /zp\/9-—$2dl 0 3 = 6/ xV9—32dz 0 Therefore, __ a i=Mx—0=i/ xVQ—zzdz area 91r o ...
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