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Unformatted text preview: 5.: Pole: Fauna and Areas. Page: 334—335
1. Step 1. Find the simultaneous solution of the given system of equations. Step ‘2. Determine
whether the pole lies on the two graphs.
Step 3. Grepit the curves to loot: for other
points of intersection. 1. Sketch the graph of the polar region whose
urea. in to be found. Determine rays r3 = n
and E = 15' that bound the region. Compute the area by the formula. 13
A = Ema]? re
ﬂ'
3. I. lﬂnnierate b. eirele
e. tone {3 petals] :2]. none {spiral}
e. wdioid 1'. line
5. lemnimte h.11ma.eon
# I renae [4 petals] b. lemnisente
c. circle d. rose {iii petals]
1:. none {spiral} f. lemnieente
5. none [3 petals} h. enrdinid
5. I. r4366 {4 petnie] h. circle
1:. liningan ti. enrdinid
f. lime :. line
5. rm [5 petals} 11. line . E, Addihbnaf Appffcﬂims of the Int _ 1?. 19. 21. . The pole. P1, in a paint
' ' _' crimesection. r1=f2
4mﬂ=4aluﬂ
“113:1
«=2! Pita ﬂ! 1E] 23. The pole is not a point. ufintcmmtlon.
r1 — 1":
4 sin 6‘ =
am 3 = 1}
a  g a; Pin. g} and Pam. 5") Hon {5.3, Polar Forms and Areas Page 265 Ill. The pole in not. a point 1 — 2 M6 +098” = 4 aingﬁ'
of Inbrmt1:n. 1 _ 20055 + m!” = 4(1 _ 00313)
*‘1='"‘2 5m362m6—3=ﬂ
93'“29=9 (5mﬂ+3)(msﬂ—1)=J
smEE=l m3=IIﬂ§
29:11. MT 3
FT msﬂ=l 13085: —5
:1r MT _
TITS“ 9:0 3:605 1(#§] :5 2.214
‘II'
1W. T). P2113. T} P,(3.2. 2.2; 25. The W131 P1, is a point
of intersection.
I"1 = r3 l+uin6=1— 31110 sin I9 = I}
a = 0‘ r 29. The pole is not is a point ul’interaectiou.
Pal}, {I}, P591 11'} r1; = r2
2 an a : 175$, . The pen, PI, is a point. 200“? — ecu3a = 4 n
of intersection. 1mg“ _ 2mg + 4 = D ‘
For 3 = a! + my The“ are no
r = 36 = :r + 3:11!“ intmvectiun point; P4: + :lrrrr. @1134 we
even :1: so. A = H aim an = ﬁa— yum}?
Pnizfl' + 3r”, u
for odd :1. = ﬁ g :3 0.0226
:15 17. The pole, P1. is a point
of intersection. :1. A = M29 a = kﬂ+£5ain mfg“
D 2_ 2
1": — '2
. . 2 r «1'5
5m ’23 = 2m] 6' :31?th :5 0.2392
zliJIQmaﬁ : 23min? nu
5:0 m59=sinﬂ 32.A=£rl. mzﬂdﬂ=ﬂnﬂlﬂéu=l
9:0,: kg}; —m
P2: (1.%] '1’: fr”
:3. The pain. P1, is a. paint ofintersection. 33' A =%‘[ sing a = “g'mﬂlrjs
 w s
'"1 = “2 vii
1—mae;=4siua =T"°*“m 1 wwuﬂzﬁsinﬂ P15: 266 Chapter 6, Addition! Appﬁcaﬂbns of the Int93m! as. ,d. :%I;a=2—£,§I’:=T a: 4.1335 and = J HM“ .29]: d9 __.%Tl — c0343 a U I qr!!! U
=[a — }sm 4ml1 =g as 1,5703 43. We will ﬁnd the arm from R = {I to L1 = {
and multiply by 3. “3 1r}:
A = (3)34 a2 air.“ 33 d0 = 3,12! (1%Ogﬁﬂ) '
'J' a 2 _ 1:13 2
=§;_(e%amsag‘ u =% 44. The curves inberaect when 4 cos 1‘? = 2 or w
= 7,1'3, 11'f3. Both curves are
symmetric with reapect to the m—Miﬂ, so m
A=2J go: ma}?  2‘] d3
0
_ r “fa—49f
_.(Sﬂ+4sm2fi — «mu “Tux/5 ‘15. The curves inmmt when we {3 = I]; [i = :12.
iiirﬂ. Using symmetry.
«12 A = [12 — c1 4 mm dz:
0
2‘
= a°[2nuiI16%{G' +%5iI129)Iﬂ; = 2(2 — a
46. The curves intcrncct when r = I] or when 1— coeﬂzsin3m=irfl
If: .4 =H [sinia — (1 — cos 93131.10
U
1U?
:,}I(_1+2me — c0329}dﬂ
U are
[:1
u g—ﬂ+ﬁsinﬂ—%sin2ﬂ] —% Section 5.3, Poiar Farms and Areas ‘7. Solving simultaneously:
6 cos ti = 2 + 2 cm ti
1 cos 6 = 2 Also note that the pole, [I], ll). is a solution.
aria
.4 =(2}1§I [00 cash? — 4(1 + one 01F] 00 I:
:13 dI[9weaﬁ—1r2weﬂ—mazﬁ]dﬁ D
If: =4llaw5202cm0—1103
0
=4(36+23in29—28in€) at}: = 4x
0 43. The curves r2 = lines 26' and r = 2 intersect
when 4 = 8 (:00 ‘25”; I9 = :lz'irfﬁ, £51115.
Using symmetry, IIE
A=4I—;—[3mse0 — 29100
0
= [s sin 20 — 30H”? =4(V’§ — g] 49. The inner loop is treated out for «f6 5 ﬂ 5 5rfﬁ
and the outer loop [or 5:16 5 9 5 litem. Thus. the area. between the loops is
“we. 01:} he —4einﬁ}2dﬂ 5115
Mr] —[ g2  43in 0}2 00
0110
13:70
Inn‘s
a [00 + a (.059 — 23in 20]
=(s«+w’i} # Hr — WE)
= 4x + 12¢? :2: 33.3510 50. To ﬁnd where the line intersects the
lemniaeate, solve: = [00 + Eco: 0 — 2m 20]
Sefﬁ
we 1
= '2 23
—'—i[;c C03 0' A: ilJAlSEIl 51. 6‘2. 53. Pig: 26? Thus. tiling symmetry, the area is
Imam A=2 1 20—71%; = [sin 20 — tan 0] “:33 s: 0.3003 The line and the lemniscete intersect where: m = 2sin29 i9 :25 0.2374,!)3854 Thin!T the area is ad'sM
A: 1  _ 1 J. Emsmﬂrﬁ m2 M
0.23% m“ :5 0.0074
8.251"! _[—cm 29 __ ten 9] At each point P on the limegon
r = '2 + 31:05 9. the y—coordinete in 1r = mint? = (2 + sens ﬂisinﬂ
The muhnum value of 1; occurs when y'iﬂ] = for 0 5 0 5m—1(—§
We have. 010) : ecosa + 30:09:? — sin20} ﬁe) = 0 when 0 9: 0.9?06 50 the maximum
value of 3: is 3.0439. At each point For: r: 3 + 30in H, the :
ewrdinate in z: renal? = (3 + taintijmaﬂ The maximum value of reecurs when
19(19): I] for 1r_f2 g ti 5 xiii. Since I'm] = —30in H + 3(cou23  siniﬁ]
09(9) = I] when. t? = €— ee the maximum value of
ale MEN 9:: 3.3011. 51. a. Point Hr, ﬂ] is at a distance r = r. from the pole. 0P makes an angle ti with respect
to the original pole: exie. Let that axle be rotated through an angle ct. Then G —r.'r
is the angle between the rotated exie and 6?. Sieee r: r'. {(0) = {[0  a). Page 268 b. According to part .
the graph of r = '2 sedti — g
is obtained by
rotating r : ﬁlmct? through an angle
«[3 about the
origin. Since r = 29cc t? in the line runs 6 = 2 or r = 2, the graph is a. rotated line. . _ 3 .
5!}. Converting r _ m to Carlenmn coordinates, we obtain rmti=ti _ __ —19' 11'
r—LBII f 0534:? y“ = tan :+ means: y“ z n for all u 5 a: q %. Thus. the graph is = always rising and y _. +0.3. as I _, Gin] — ' so there is a vertical asymptote at z = 5?. Banana r: {[9} and z = rcns t'. 1;: rain 3.
we have a: = Halt.0: ti and y = f{ﬂ}«inﬁ.
Using the chain rule, we ﬁnd that ﬂy  (ﬂy dz) or, equivalently. 33' EE BE
in dy__21§
iii—"71'? t =ﬂ+ﬂ+ﬁ°m"=W Chapter 6, Additionai Appiications of the in: Because a: = {(Ekw i] and y = ﬂtﬂa‘m ﬂ,
follow that $=;(933%{ma]+cma% = —]'[Ei]nin 5' + {flimm ti :3? = my) ﬁsh 91+ sin a % = ﬂﬂkoa t' + I‘llﬂjnsin H
and the slope of the tangent line is given if
dy _ 3% _ {mm :3 + .r‘szina 58. Let gt be the angle betwccn the tangent ‘
through P and the positive m—axis. Since
51 = {[zjn‘mﬁ' and 1'. = con ti. _ _dyF Itﬂlcma+rmm
m'mni’ﬁ'ﬁ' —f(e]sinﬂ+1'(ﬂ)nu
taugt — tan5' _ ain't?
ma {{3} cos 6' + {'03) sin ti  fill?) sin I? «1— f’lﬁli'} cos :9
fit?) cos ii + {‘01) sin ti sin 3
~—ﬂﬁ) sin 6 + ITS) cost? 60% 5’ ﬂﬂ)ms‘ﬂ+f’{ﬂ}ainﬂcnaﬂ+ﬂﬂ)sin2ﬁ — I'Eﬂjain
— ﬂinginﬂcosti + I'(5)coa ﬂ‘+_f[ﬂ]ain9casﬂ+f'( 1+ fig}
rm
5!]. I. tana:ﬁ9£ui§=—wt3
b dI'v—Eiainvlii'tancr1''m'iﬁ'
’33— ‘ _ sing
39
c. ﬁ=ﬁ¢3ﬂnaua=§5E=§
IE lit]. The solution to this problem is found in V
34. 1984 issue of Schaai Science and Mathem with. p. 265. 6.4 An: Imgth and Surface Area. Fags 393—395 1. 131+[ﬂx]. =3f1+31= JITI 2
a: Iﬂdx=3m
—I ...
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 Addihbnaf, Schaai Science, Pnizfl, 32.A=£rl. mzﬂdﬂ=ﬂnﬂlﬂéu=l

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