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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. lncosy§ = § —ln\/§. (b) The area of the region is A = fgﬂgi—If = sin—1(§)lg = sin‘l(%) = g. The ycoordinate 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 crosssections 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+452+262+452+242+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,;(11k—oolk) =
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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 Spring '08
 Broughton
 Math

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