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Math 101 Apr 96 Solutions

# Math 101 Apr 96 Solutions - Math 101 Solutions April 1996(a...

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Unformatted text preview: Math 101 Solutions April 1996 (a) Let V be the volume of the solid. Then V = f011rR(:r)2d1: = fol7r{1:3}d1: = 1rf011:6d1: = \$257”) =% (b) Let ds = 1/1 + (£51)2 d1: be an element of the surface. Then =21r foy(1:)d1: 13 the total area of the surface of the solid. Since (13— — ‘A/1+(31:2)2d1:— - v1+91:4d1:,A= 9mm = ;—7(10\/T6 — 1). 27rf01:x3\/1+91:4d1:=1"—8-§(1+ 2. Let d3 = ‘/1+(§—§)2d1: be a length element of the curve. Then L \/1 +sin2 1:111; 11%!) = 11%") = A?) = 3. (a) = f:"ds 2n . = 0Wd1= The Trapezoidal Rule states that L z a—[M f("0)+f( )+f(2")+f(3—")+f(“—")+f(5—”)+%f(“—")] where f(:r) = v1+sin21:. Since f(0)-- — 4, then L z f(7r)=f(21r)=1and f(%) = (2 + 2J7) x 7.6356. ln|cosy||§ = § —-ln\/§. (b) The area of the region is A = fgﬂgi—If = sin—1(§)lg = sin‘l(%) = g. The y-coordinate of the centroid of this region is ‘ = %£ydA = ﬁfﬂﬂx )]2d1: = 3f: '1?!va which can be re- 4costdt to 17 = ﬁfi sectdt = duced via :1: = 4sint and d1: 3371n(sect + tan t)0F = 23; In ﬂ. an (a) The area of the region is A = log“ — tan y)dy ll ylo + (b) This volume is equivalent to removing a volume of radius R(y) = 2 tany from a cylinder of radius 1 and height!E . Thus, V = "T — «2—21 f0? 7r(tany)2 dy— _. ——-—1rf07tan2ydy= ——7r tan yl0 +11ng = 2 . Rearranging‘i — ——k\/P Pgives P 2 dP— - —kdt. Integrating both sides yields 2P5— — —kt + C, or P(t)- — Z(—- —-kt + C)2. To determine k and C, the conditions P(O )— - 90, 000 and P(6)= 40, 000 msut be used. 19(0): — =90, 000 givesC: 600 and P(6)= ;(- kt+600)2— ~ 40, 000 gives 11— - 1—00. Thus, P(t)— - l(600 — l—mt t)2 , and so P(t0)— — 10, 000 occurs when to = 12 weeks. (a)Sinceh+z=1—z+1:2—13+1:4—... ,thenTl;=1—t“+ t8 — t” + t16 — .... Integrating this power series gives [(12), 1e I(z)= f0’(1—t4+t8—t12+t16—...)dt = (t—%t5+%t9——11§t13+1‘—7t”—...)|3 = x— gzr’ + \$39 — T153313 + 11.72:" —.... (b) Noting that 11—3213 = Eli” < 0.0001 and that the series is alter- nating, IS —— Sal < 0.0001, where 33 is the series truncated after three _ 1 1 1 ~ ~ 1 (c) Since the next term after 83 is negative and thus overcompensates (since I (2:) is an alternating series), the value computed in (b) is larger than the true value of I (%) 7. (a) (b) Using symmetry, %A = %f0% r09)2 (10, and so A tiff sin4(30) (3 d0). Using the identity sinzt = W, sin‘t = % — % cos 2t + % cos2 2t. This can be reduced further using coszt = W to sin‘lt = g- — %cos2t + %cos4t. Thus, A = 4fo%(% — §c0s(60) + 303020)) = (g0 — sin(66) + §sin(120))|§ = 37"- . Since the vertical cross-sections of the pool are the bottom halves of circular disks, V = 325 016 R(2:)2 dz, where R(:c) is the radius of the pool a distance a: from one end. Simpson’s Rule states, then, that V e %§(R(0)2 + 412(2)2 + 212(4)? + 4R(6)2 + 212(8)2 + 412(10)2 + 9. 10. 11. 2R(12)2+4R(14)2+R(16)2 = g—(0+4-52+2-62+4-52+2-42+4- 32 + 2 . 42 + 4 . 52 +0) = 4732" 3 494.277. (a) If (13 = y] 13%)? + (1%)2 dt is a small length of the curve, then L = flmds. Since ‘15 = (%?)+(S—T§:£)dt = ,lﬁgdt = 59, then szioo i175" (b) 1m < k < 1,thenL = f° g; = ﬁ—kzhkﬁo = 1—1,;(11-k—ool-k) = 00, since I: < 1 means (l—k) > 0. Ifk = 1, then L =1°°£t£ = int]? = lnoo = 00. If k > 1, then L = 100%; : i—l—ktl—kh” = ﬁnk" -— 00“") = i—l—kll‘k, since k > 1 means that (1 — k) < 0. Thus, the integral is finite for k > 1. Rewriting m% = —(mg + Irv?) gives 1.519%; = —%dt. Integrat— ing both sides yields mistan”1(‘/;lk—gv) = —%t + C, or v(t) = ﬂtaM—‘lénlt + C), where C can be determined from v(0) = v0. Thus, C = tan“(‘/;—"’55vo). The highest point will be reached when the velocity is zero, i.e. v(th) = 0. This occurs when t}, = —— I k ﬁtan 1( h—gvo). Using integration by parts with u = f(:r), du = f’(1:)d:c, dz) = cos(n:c) dz, and v = ésin(na:), allows us to write f0“ f (1:) cos(n:c) d1: as M f (:1:)|3’r — l 2" f’(:1:) sin(nx) dx. Since f (x) is continuous and n n 0 the interval of integration is bounded, f0?" f’ (x) sin(na:) dz = k, where k: is some ﬁnite constant. Also, sin(m:) 5 1 means that the first term is bounded by ﬂéﬂ = 9f, where M = f (2n) is another ﬁnite constant. Thus the limit becomes limn_,°° ”4”" = . n ...
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