Calculus Third Editon By Strauss, Bradley and Smith sec6.3

Calculus (3rd Edition)

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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- rena-e [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:. lining-an 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 ' ' _' crime-section. 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 5m36-2m6—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“? — ecu-3a = 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 zliJIQ-mafi : 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 Int-93m! 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 reap-ect- 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)I|fl-; = 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: =4llaw520-2cm0—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 lit-em. 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's-M 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] = —30-in 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 : film-ct? 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 — tan-5' _ 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'tancr-1''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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