hw-7-sec-11.4-solns

hw-7-sec-11.4-solns - REAS AND LENGTHS lN POLAR COORDINATES...

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Unformatted text preview: REAS AND LENGTHS lN POLAR COORDINATES 891 SECTION 11.4 A dy M—t tan¢'tan6 Ez‘i’mng dm/d" 2mg Mawmnw—Wmmzmz HL/dh 9 dm dcc/dé) an d?“ . dT ' i 2 it”: dflftane (@51119-1-T’C050)—tan0<EECOSQ-TSIDB> # rcosO+T~ 8:25: -’ dr dr sin29 i—k—y-tarfi <%%c059-rsin6>+tan0<§%sin6+rc050> 35c050+36~ €030 d6 6 _ T00826+rsin29 _ 7" it cos2 0 + £3;- sin2 6 dr/de 84. (a rzee :> dr/dG:ee,sobyExerCiseS3,tantb:r/ee:1 :> 1,!) : arctanl : a (b) The Cartesian equation of the tangent line at (1, 0) is y = m ~ 1, and that of the tangent line at (0, 6"”) is y = 6‘” ~ :17. t and radial lines, that ent of the angle between the tangen dr 1 is, a : tam/J. Then, by Exercise 83,11 : Fri/563 :> 21—5 : a?" :> (e) Let a be the tang r : Gee/a [by Theorem 10.4.2]. 11.4 Areas and Lengths in Polar Coordinates /4 /4 5 __L__7T5 7r/4 17/4 1r 7r 1.r:620363§. [12/0 mezfo §(62)2d0=/ le4de=tees1o =ala<1>=m4o 2.T‘—'—‘6 , f‘» 3 r:sm6,1‘3—§0§2—;— 2W/31 2 1 2M3 1 27(/3 1 2 1 4 1 2 Azfl/g —sm QdGZZL/a (1—c0520)d6:Z[0—-sm2(9]fl/3 =Z[—3’3~§sm—3l—§+§sm?fl] __ 71 f 11' f 7r 7r ~1[%-%(-—3)~-+1(: 1—1(§+—@)=fi+—@ 7r 2 7" (I) :x/sin0,0§0<7r A— 1(Vsm6> d9—/ i81n6d9—-[~—0089]g=%+%:1 o 0 2w 27r 2 211' 2“ 5 T-x/5,0§6<27r A- gr dQ—f MW? 1102/ 56d9:[:92]0 : 2 0 0 o 6.T:1+c059,0£0§fi. Azf %(1+c059)2d6=%/ (1+2c050+cos20)d6:%/ [1+2c0s6+%(1+c0520)]d9 0 0 0 [%9+2sin6+%sin29]g = %(%7r+0+0) —— : 1475 A +2c059+%cos20) d9:% 1 2 mmwew was €09 1!, 892 3 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 7. r=4+3sin9, —%§6§§. 7r/2 7r/2 7r/2 42/ %((4+331n6)2d6=§/ (16+24sin9+9sin26)d0=§/ (16+93in26)d6 [byTheorem5.5.6(b)] _.,r/2 , 7r/2 77r/2 7r/2 : g -2/ [16 + 9 . g1 -— cos20)] do [by Theorem 5.5.6(a)] O 7r/2 2/ (g_ 2 0 @TzsinZO, 0303? 7r/2 7r/2 142/0 55111229559ng gn—coswme:§[9—§sin4a]g/2:g(g)zg NRC cos26)d6 : [%9— game]? : (4% —0) — (0—0) 41" 9. The area above the polar axis is bounded by r = 3 cos 0 for 0 = 0 to 9 2 77/2 [not 7r]. By symmetry, A 2 2 f0“ 5,3 d6 : fO"/2(3 cose)2 d6 : 32 "/2 cos2 (we 0 =9f0“/2%<1+cos20)d0= gimme]? : grew) e(0+0)l = Also, note that this is a circle with radius %, so its area is 7r(%)2 2 37". 10. A = f021r%7'2 d0 2 02" %[3(1 + cos 6)]2d6 % 02711 + 2 cos9 + cos2 9) d0 3 02” [1 +2cos0+ g0 +cos29)] d0 = 38—0 + 2sin6+ 51112.9]? 2 gfi 11. The curve goes through the pole when 9 : 7r / 4, so we’ll find the area for 0 S 9 S 7r/4 and multiply it by 4. A : 4 [0“4 W d6 = 2 f5” (4 cos 29) d6 = 8fO"/4cos26d9 = 4[sin20]"/4 * 4 0 .4 12. To find the area that the curve encloses, we’ll double the area to the left of the vertical axis. A: Mfg/2 a2 — sin0)2d0 : [332(4 ~ 4sin6+sin29)d0 : 372/2[4 —4sin6+ %(1 — 00526)]d0 = fag/2 (% — 4sin0 — %cos29) d6 7r = [26+4c080—isin26117é2 : (277“) — <27") = 0r: We could have doubled the area to the right of the vertical axis and integrated from ~7r / 2 to 7r/2. 0r: We could have integrated from 0 to 27r [simpler arithmetic]. . 1- “uL'Szmu-amimgmstem, ,, r ‘ SECTION 11.4 AREAS AND LENGTHS IN POLAR COORDINATES C xx 13. One-sixth of the area lies above the polar axis and is bounded by the curve r=2cos39 r :2cos36’for0 : 0t00 = 7r/6. g A = sag/6 §(2cos39)2d9 = 12 f0“ cos? 30d6 B 2 6[6 + %sin69]g/6 = 6(%) = 7r 7r/6 0 (1 + cos 69) d6 ll 14. A (2 + cos 20)2 d6 : % f5" (4 + 4c0s 26 + cos2 26) d0 H f5" % 1 21r 20 ago + 2sin26 + gsinw] ll (4+4cos20+ ; +§cos46) d0 1 21r 0‘2 (971') = 921 H 15. A: [02“ $0 + 25111 66)2 d6 z; 02% + 4511166 + 4m2 60) d0 H 5.102" [1 + 4311160 + 4- gm — cos120)] d6 [027T %[30 — % cos 66 fl % sin 126] (3 + 4 sin 60 — 2 cos 129) d0 _l "2 ll 27r 0 WIN 0) (0 0)] — 37r holy- who [(6% 1s. A=f;%(2sin0+3sin99)2d9:2fg/2§(2sin9+3sin90)2d0 : (f/Z(4sin20+1zsin9sin90+9sin299)d6 [2(1 ~ cos 20) + 12- %(c0s(6 — 90) — cos(6 + 90)) + %(1 — cos 186)] d6 [integration by parts could be used for f sin 0 sin 90 d0] _ "/2 “0 #15 : fJ/2(2—2C0529+6COS86#60081004— % — %C03186)d9 : [}§G#Sin26+%sin80—gsin100— 5111189 13 V_ Tr _ 4 7r/2 0 17. The shaded loop is traced out from 6 z 0 to 0 : 7r / 2. A- fg/Zgrzde: gfg/2sin229d6 z; 0”” a1 -c0s40)d6 : fie— isin4GE/2 = fl?) = % 1s. A = [5/3 a4 sin 30)2 d6 = 8 fem sin2 39 d0 = 4f0"/3(1 - cos60) d6 ‘31 3 : 4[0 — ésin6OE/3 : 893 3.5 32 l / st 894 D CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES L fl @7‘20 :> 300850—0 56—g 0—173. A :fjgfo %(3c0856)2d6:f0"/109c0s2 56d9= gfo"/10(1+cos 109) d6: who [6 + % sin IOQJg/m = 3—3 (1 — 0051260030 : [a _ fisin12fljg/6 = g 21. This is a limaqon, with inner loop traced r = 1 + 2 sin (9 out between 6 : 161 and If?" [found by solving 1" = 0]. 7i _1 6 \‘ MW 6 0:? 37r/2 37r/2 37r/2 A:2/ §(1+2sin6)2d6=/ (1+4sin0+4sin26)d6:/ [1+4sin6+4-%(1~c0s26)]d6 7-rr/6 777/6 77r/6 = [6—4c050+20~sin29]::;:: (921)— +2\/§—¢7§) —7r_%§ 22. To determine when the strophoid r 2 2 cos 6 — sec 6 passes through the pole, we solve r=0 => 2cost9— :0 :> 2c0520—120 :> cosgfiz—l- :> c0st9 2 )2d6 = 0"/4(4cos26 — 4+se020)d0 I ‘ 0 [4. %(1+ c0826) — 4 +sec20] d0 = fO"/4(—2 + 2c0s20 +seC20) d6 =[—20+sin20+tan6]g/4=(—g+1+1)—0:2~g 23.2cos6':1 => cos6=l 2 0: A: 2g” %[(2C086)2 — 12] d6 : 0"/3(4cos26 —1)d49 : 0"” {4 [$0 + cos26)] — 1} d0 = o"/3(1 + 200829)!” I [9+sin26K/3 = §+§ A: ff” gm — sin0)2 *1]d6 = gffnsinze — 2sin6)d0 =iff"(1— cos26 —4sm0)d0 = fie — gsin2a+4cosqff =i7r+2 3 895 SECTION 11.4 AREAS AND LENGTHS IN POLAR COORDINATES miniscate r2 = 8 cos 29 and outside the circle 7" : 2, we first note that the two curves intersect when r2 = 8 cos 26 and 7‘ = 2, _1 -__ %.For —7r < 0 gmcosw— 5 <=> 26 # “71/3 that is, when cos 20 : / 6 or i57r / 6. The figure shows that the desired area is ori57r/3 <=> 0 = i7? mes the area between the curves from 0 to 7r/ 6. Thus, 4ti A: 4f"/6 [%(8c0526) # %(2)2] d6 = z 8 [sinza — an“; = 85/372 — 7r/6) : 4J5 # 47r/3 ’11 find the area A1 inside the curve "r = 2 + sin 0 26. To find the shaded area A, we and subtract «(g-)2 since 7’ : 3 sin 6 is a circle with radius l 2"(4 + 4sin9 + sin2 6) d6 A1: [02" %(2+sin9)2d0 = 2 O 1 2" [4+4sin6+ % - (1 — c0526)] d0 5 0 1 27' (9i +4sin9 — §cos26> d0 5 0 ago — 40059 — 51112613" : %[(97r * 4) e (-4)] = 2; H ll 27 30050:1+cos€ ¢> c050=% => ngor—g. A: 2L?” %[(3c059)2 — (1+c0s6)2]d0 =fgr/3(8cos26—2c056—1)d6=fg/3[4(1+cos20)—2€os0—1]d0 = fj‘/3(3+4cos20—2cos,9)de= [30+2sin29—2sino]g/3 :7T+\/§——\/§:7T 28.3sin622-sin0 :> 4sin6=2 :> sin9:% ——> 02%0r5?’r A = 2 jg; game)? — (2 — sin6)2] d6 = 7:762(9sin26r4+4sin6—sin29]d9 : ://62(851n26+4sin0—4)d0 :4f://62[2-%(1~c0s26)+sin6—1]d0 = 4f://62 (sin6 ~ cos 26) d0 = 4[—c056 — é-sin 29]:;: :4iro—oi-(-—ra-—ra>i=4ee>:m 898 37. 38. 39. 2sin26 = 1 => sin26 % 40. 41. 42. H. <L PARAMETRIC EQUATIONS AND POLAR COORDINATES Sec CHAPTER 11 The pole is a point of intersection. r=3sin6 1+sin6z3sin6 :> 1:2sin6 :> sin6:% :> 6=§0r%‘. The other two points of intersection are (g, g) and (g, r=1+sin6 > The pole is a point of intersection. l + sin 6 ,: 1—Cos6:1+sin6 —> cos6isin6 —1#tan6 r 37r 71r 6 — '4— or T. The other two points of intersection are (1 + , 3’41) and (1 — lg, “in r=1—cos6 7r 57r 137r 177r 26—6,6, 6,ore. By symmetry, the eight points of intersection are given by 5 137r 17 (1,6),where 6 = g, 1—", 72—, and 7;", and _ 77v 1171' 197r 231 (#1,0), where6 —— '15, T, E, and 3. [There are many ways to describe these points] Clearly the pole lies on both curves. sin 36 : cos 36 => tan 36 = 1 :> 36=%+mr[nanyinteger] => 6:%+§n => L 712a 6 : {L2, 51—”2', or 37", so the three remaining intersection points are ( sle ‘1‘ r=cos 36 t i The pole is a point ofintersection. sin 6 : sin 26 = 2 sin6 c056 <=> l sin6(l#2cos6)=0 <:> sin6:00rcos6=2 :> 2 the other intersection points are (fig—5, and (g, [by symmetry]. Clearly the pole is a point of intersection. sin 26 : cos 26 :> tan 26 = 1 => 26 = g + 21m [since sin 26 and cos 26 must be . . . . _ 7r _ 7r 97r pos1t1ve 1n the equat1ons] :> 6 — g + n77 :> 6 — g or —8—. 971' ‘ 1 7r So the curves also intersect at (7‘75, g) and ( 3+ ...
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This note was uploaded on 05/02/2011 for the course MATH 408D taught by Professor Chu during the Spring '09 term at University of Texas.

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hw-7-sec-11.4-solns - REAS AND LENGTHS lN POLAR COORDINATES...

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