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Unformatted text preview: Math 101 Solutions Dee. 1991 2
1. a)I=V/‘.1:3\/9172 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+:n3 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 —8A8B—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+21n1+3+llnr 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° _ k21/2
——~ﬁ(1+t’) +23 1)” This last series is valid whenever 0 < It] < 1. Now, integrating this series termby—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 yaxis. 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 = parea
91r 64 a7: 8. a) 2 The moment, M m = 0 about the yaxis 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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 Spring '08
 Broughton
 Math

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