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

# Calculus (3rd Edition)

This preview shows pages 1–6. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern