PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS

Transcription

1 PARAMETRIC EQUATIONS, POLAR COORINATES, AN VECTOR FUNCTIONS. Parameric Equaions Preliminar Quesions. escribe he shae of he curve cos ; sin. For all, C. cos / C. sin / 9.cos C sin / 9 9; herefore he curve is on he circle C 9. Also,eachoinonhecircle C 9 can be reresened in he form. cos ;sin / for some value of. Weconcludehahecurve cos, sin is he circle of radius cenered a he origin.. How does C cos ; 5 C sin differ from he curve in he revious quesion? In his case we have. / C. 5/. cos / C. sin / 9.cos C sin / 9 9 Therefore, he given equaions aramerize he circle of radius cenered a he oin.; 5/.. Wha is he maimum heigh of a aricle whose ah has arameric equaions 9,? The aricle s heigh is.tofindhemaimumheighwesehederivaiveequalozeroandsolve: The maimum heigh is.0/ 0. d d d d. / 0 or 0. Can he arameric curve.; sin / be reresened as a grah f./?whaabou.sin ;/? In he arameric curve.; sin / we have and sin, herefore, sin. Thais,hecurvecanbe reresened as a grah of a funcion. In he arameric curve.sin ;/ we have sin,, herefore sin. This equaion does no define as a funcion of,hereforehearamericcurve.sin ;/ canno be reresened as a grah of a funcion f./. 5. Mach he derivaives wih a verbal descriion: (a) d (b) d d d (i) Sloe of he angen line o he curve (ii) Verical rae of change wih resec o ime (iii) Horizonal rae of change wih resec o ime (a) The derivaive d d ishehorizonalraeofchangewihresecoime. (b) The derivaive d d ishevericalraeofchangewihresecoime. (c) The derivaive d d ishesloeofheangenlineohecurve. Hence, (a) $ (iii), (b) $ (ii), (c) $ (i) (c) d d

2 8 C H A P TE R PARAMETRIC EQUATIONS, POLAR COORINATES, AN VECTOR FUNCTIONS Eercises. Find he coordinaes a imes 0,,ofaariclefollowingheah C, 9. Subsiuing 0,, and ino C, 9 gives he coordinaes of he aricle a hese imes resecivel. Tha is,. 0/ C 0 ; ).; 9/. / C 9; 9 ).9; /. / C 65; 9 9 ).65; 9/:. Find he coordinaes a 0; ; of a aricle moving along he ah c./.cos ; sin /. Seing 0,, and in c./.cos ; sin / we obain he following coordinaes of he aricle: 0W.cos 0; sin 0/.; 0/ W.cos ; sin /.0; / W.cos ; sin /.; 0/. Show ha he ah raced b he bulle in Eamle is a arabola b eliminaing he arameer. The ah raced b he bulle is given b he following arameric equaions: 80; 00 :9 We eliminae he arameer. Since 80, wehave 80. Subsiuing ino he equaion for we obain: 00 : :9 5 : The equaion 5 :9 600 isheequaionofaarabola.. Use he able of values o skech he arameric curve../;.//,indicainghedirecionofmoion We markhegivenoinsonhe-lane and connec he oins corresonding o successive values of in he direcion of increasing. Wegehefollowingrajecor(hereareohercorrecanswers): = 6 = = = = = 0 5. Grah he arameric curves. Include arrows indicaing he direcion of moion. (a).; /, << (b).sin ;sin /, 0 (c).e ;e /, << (d). ; /, (a) For he rajecor c./.; /, << we have. Alsohewocoordinaesendo and as!and! resecivel. The grah is shown ne: =

3 SECTION. Parameric Equaions 9 (b) For he curve c./.sin ;sin /, 0, wehave.sin is increasing for 0, decreasing for and increasing again for. Henceheariclemovesfromc.0/.0; 0/ o c. /.; /, henmovesbacko c. /. ; / and hen reurns o c./.0; 0/.Weobainhefollowingrajecor: π = (,) π = (,) = 0 0< = π (, ) = π (, ) < These hree ars of he rajecor are shown ogeher in he ne figure: π = (,) = 0 = π = π (, ) (c) For he rajecor c./.e ;e /, <<, wehave. Howeversince rajecor is he ar of he line, 0<. lim! e 0 and lim! e,he (d) For he rajecor c./. ; /,, wehaveagain. Sincehefuncion is increasing he aricle moves in one direcion saring a.. / ;. / /. ; / and ending a. ; /.; /. Therajecorisshownne: = (,) = (, ) 6. Give wo differen aramerizaions of he line hrough.; / wih sloe. The equaion of he line hrough.; / wih sloe is. / or 7. Onearamerizaionis obained b choosing he coordinae as he arameer. Tha is,. Hence 7 and we ge, 7, <<.Anoheraramerizaionisgivenb, 7, <<. In Eercises 7, eress in he form f./b eliminaing he arameer. 7. C, We eliminae he arameer. Since C, wehave. Subsiuingino we obain. / )

4 50 C H A P TE R PARAMETRIC EQUATIONS, POLAR COORINATES, AN VECTOR FUNCTIONS 8., From,wehave.Subsiuingin we obain 9., an. C e /. / ) ; 0: Relacing b in he equaion for we obain an. C e /. 0., C From we ge. Subsiuing ino C we obain C. / C C ; 0: Since we mus have afuncionof,weshouldrobablchooseeiherheosiiveornegaiveroo.. e, 6e. C, We eliminae he arameer. Since e,wehave ln or ln.subsiuingin 6e we ge 6e 6e. ln / 6e ln 6e ln 6 ) 6 ; >0: From C,wege or. We now subsiue in o obain ). / ; :. ln, Since ln we have e.subsiuingin we obain e.. cos, an We use he rigonomeric ideni sin cos o wrie We noweress in erms of : an sin cos cos : cos an ) ; 0: Since we mus have afuncionof,weshouldrobablchooseeiherheosiiveornegaiveroo. In Eercises 5 8, grah he curve and draw an arrow secifing he direcion corresonding o moion. 5., Le c./../;.//. ; /.Thenc. /../;.// so he curve is smmeric wih resec o he -ais. Also, he funcion is increasing.hencehereisonlonedirecionofmoiononhecurve.thecorresondingfuncion is he arabola./ 8.Weobainhefollowingrajecor: = 0 6. C, C

5 SECTION. Parameric Equaions 5 We find he funcion b eliminaing he arameer. Since C we have, hence C. / or C. Also, since C and C are increasing funcions, he direcion of moion is he direcion of increasing. We obain he following curve: 6 = 0 = (6, 5) (, ) , sin We find he funcion b eliminaing. Since,wehave. Subsiuing sin. We obain he following curve: ino sin we ge ( π,0) (π,0) 8., From we have =.Hence, =.Sincehefuncions and are increasing, here is onl one direcion of moion, which is he direcion of increasing. Noicehaforc./. ; / we have c. /. ; /../;.//.Hencehecurveissmmericwihresecohe ais. We obain he following curve: 9. Mach he aramerizaions (a) (d) below wih heir los in Figure, and draw an arrow indicaing he direcion of moion π (I) (II) FIGURE (III) (IV) (a) c./.sin ; / (b) c./. 9; 8 / (c) c./. ; 9/ (d) c./. C ; 5 / (a) In he curve c./.sin ; /he -coordinae is varing beween and so his curve corresonds o lo IV. As increases, he -coordinae is decreasing so he direcion of moion is downward.

6 5 C H A P TE R PARAMETRIC EQUATIONS, POLAR COORINATES, AN VECTOR FUNCTIONS π π (IV) c./.sin ; / (b) The curve c./. 9; 8 / inersecs he -ais where 8 0; or ; 0, and. These are he oins. ; 0/ and. 9; 0/.The-inerces are obained where 9; or : The -inerces are.0; / and.0; /.As increases from o 0, decreases, and as increases from 0 o, increases. We obain he following rajecor: (c) The curve c./. ; 9/ inersecs he -ais where 0, or. The-inerce is.0; 8/. The -inerces are obained where 9 0 or. Theseareheoins. ; 0/ and.; 0/.Seing we ge 9. / 9 8: As increases he coordinae decreases and we obain he following rajecor: 0 5 (III) (d) The curve c./. C ; 5 / is a sraigh line, since eliminaing in C and subsiuing in 5 gives 5 C which is he equaion of a line. As increases, he coordinae C increases and he -coordinae 5 decreases. We obain he following rajecor: 5 5 (I)

7 SECTION. Parameric Equaions 5 0. Aariclefollowsherajecor./ C ;./ 0 wih in seconds and disance in cenimeers. (a) Wha is he aricle s maimum heigh? (b) When does he aricle hi he ground and how far from he origin does i land? (a) To find he maimum heigh./,wesehederivaiveof./ equal o zero and solve: d d d d.0 / 0 0 ) 0: The maimum heigh is.0/ cm. (b) The objec his he ground when is heigh is zero. Tha is, when./ 0. Solvingfor we ge 0.0 / 0 ) 0; 0: 0 is he iniial ime, so he soluion is 0.Ahaime,heobjec s coordinae is.0/ 0 C 0 00.Thus, when i his he ground, he objec is 00 cm awa from he origin.. Find an inerval of -values such ha c./.cos ;sin / races he lower half of he uni circle. For, wehavec./. ; 0/. As increases from o, he-coordinae of c./ increases from o, and he -coordinae decreases from 0 o (a =) andhenreurnso0.thus,for in Œ;, heequaionraceshe lower ar of he circle.. Find an inerval of -values such ha c./. C ; 5/ aramerizes he segmen from.0; 7/ o.7; 7/. Noe ha C 0 a =, and C 7 a. Also, 5 akes on he values of 7 and 7 a = and.thus,heinervalisœ =;. In Eercises 8, find arameric equaions for he given curve.. 9 This is a line hrough P.0; 9/ wih sloe m. Usinghearamericreresenaionofaline,asgivenin Eamle, we obain c./.; 9 / C 9 Leing ields he arameric reresenaion c./.; 8 /. We define he arameer. Then, 5 C 5 C, giving us he aramerizaion c./ 5 C The curve C 9 is a circle of radius 7 cenered a he origin. We use he arameric reresenaion of a circle o obain he reresenaion c./.7 cos ;7sin /. 7.. C 9/ C. / 9 This is a circle of radius 7 cenered a. 9; /. Usinghearamericreresenaionofacirclewegec./. 9 C 7 cos ; C 7 sin /. 8. C 5 This is an ellise cenered a he origin wih a 5 and b. Usinghearamericreresenaionofanellisewe ge c./.5 cos ;sin / for. 9. Line of sloe 8 hrough. ; 9/ Using he arameric reresenaion of a line given in Equaion, we ge he aramerizaion c./. C ;9C 8/. 0. Line hrough.; 5/ erendicular o The line erendicular o has sloe m. We use he arameric reresenaion of a line given in Equaion oobainhearamerizaionc./. C ;5 /. ;.

8 5 C H A P TE R PARAMETRIC EQUATIONS, POLAR COORINATES, AN VECTOR FUNCTIONS. Line hrough.; / and. 5; / We use he wo-oin aramerizaion of a line wih P.a; b/.; / and Q.c; d/. 5; /. Thenc./. 8; C / for <<.. Line hrough ; 6 and 7 6 ; 5 We use he wo-oin aramerizaion of a line wih P.a; b/ ; 6 and Q.c; d/ 7 6 ; 5. Then for <<.. Segmen joining.; / and.; / c./ ; 6 C We use he wo-oin aramerizaion of a line wih P.a; b/.; / and Q.c; d/.; /. Thenc./. C ; C /;sincewewanonlhesegmenjoininghewooins,wewan0.. Segmen joining. ; 0/ and.0; / We use he wo-oin aramerizaion of a line wih P.a; b/. ; 0/ and Q.c; d/.0; /. Thenc./. C ; /; sincewewanonlhesegmenjoininghewooins,wewan0. 5. Circle of radius wih cener.; 9/ Subsiuing.a; b/.; 9/ and R in he arameric equaion of he circle we ge c./. C cos ;9C sin /. 6. Ellise of Eercise 8, wih is cener ranslaed o.7; / Since he cener is ranslaed b.7; /, soiseveroin.thusheoriginalaramerizaionbecomesc./.7 C 5 cos ; C sin / for. 7.,ranslaedsohaheminimumoccursa. ; 8/ We ma aramerize b.; / for <<. Theminimumof occurs a.0; 0/, sohedesired curve is ranslaed b. ; 8/ from.thusaaramerizaionofhedesiredcurveisc./. C ; 8 C /. 8. cos ranslaed so ha a maimum occurs a.; 5/ Amaimumvalueof cos occurs a 0. Hence,hecurve cos. /, or C cos. / has a maimum a he oin.; 5/. Wele, hen C and C cos. Weobainhereresenaionc./. C ; C cos /. In Eercises 9, find a aramerizaion c./ of he curve saisfing he given condiion. 9., c.0/.; / Le./ C a and./. C a/. Wewan.0/, huswemususea. Ourlineis c./../;.//. C ;. C / /. C ; C /. 0., c./.; / Le./ C a; since./ we have C a so ha a. Then. / 7, so ha he line is c./. ; 7/ for <<.., c.0/.; 9/ Le./ C a and./. C a/.wewan.0/, huswemususea. Ourcurveis c./../;.//. C ;. C / /. C ; C 6 C 9/.. C, c.0/.; / This is a circle of radius cenered a he origin, so we are looking for a aramerizaion of ha circle ha sars a a differen oin. Thus insead of he sandard aramerizaion. cos ;sin /, 0 mus corresond o some oher angle!. We choose he aramerizaion. cos. C!/;sin. C!// and mus deermine he value of!. Now, Since.0/ ; so cos.0 C!/ cos! and! cos or 5.0/ ; we have sin.0 C!/ sin! and! sin or Comaring hese resuls we see ha we mus have! sohahearamerizaionis c./ cos C ; sin C

9 SECTION. Parameric Equaions 55. escribe c./.sec ;an / for 0 < inheform f./.secifhedomainof. The funcion sec has eriod and an has eriod. Thegrahsofhesefuncionsinheinerval, areshownbelow: sec an sec ) sec an ) an sin cos cos cos sec Hence he grah of he curve is he herbola. Thefuncion sec is an even funcion while an is odd. Also has eriod and has eriod.ifollowshaheinervals <, < < and < <race he curve eacl once. The corresonding curve is shown ne: = = = (, 0) (, 0) = 0 = + = + c./.sec ;an /. Find a aramerizaion of he righ branch ( >0)ofheherbola a b using he funcions cosh and sinh.how can ou aramerize he branch < 0? We show firs ha cosh, sinh aramerizes he herbola when a b : hen.cosh /.sinh / : using he ideni cosh sinh. Generalizehisaramerizaionogeaaramerizaionforhegeneralherbola. a /. b / : a cosh ; b sinh : We mus of course check ha his aramerizaion indeed aramerizes he curve, i.e. ha a cosh and b sin saisf he equaion. a /. b / : a cosh b sinh.cosh /.sinh / : a b a b The lef branch of he herbola is he reflecion of he righ branch around he line 0, soiclearlhashearamerizaion a cosh ; b sinh : 5. The grahs of./ and./ as funcions of are shown in Figure (A). Which of (I) (III) is he lo of c./../;.//? Elain. () () (A) (I) FIGURE (II) (III)

10 56 C H A P TE R PARAMETRIC EQUATIONS, POLAR COORINATES, AN VECTOR FUNCTIONS As seen in Figure (A), he -coordinae is an increasing funcion of,while./ is firs increasing and hen decreasing. In Figure I, and are boh increasing or boh decreasing (deending on he direcion on he curve). In Figure II, does no mainain one endenc, raher, i is decreasing and increasing for cerain values of. Theloc./../;.// is lo III. 6. Which grah, (I) or (II), is he grah of./ and which is he grah of./ for he arameric curve in Figure (A)? (A) (I) FIGURE (II) As indicaed b Figure (A), he -coordinae is decreasing and hen increasing, so lo I is he grah of. Figure (A) also shows ha he -coordinae is increasing, decreasing and hen increasing, so lo II is he grah for. 7. Skech c./. ; / following he ses in Eamle 7. We noe ha./ is odd and./ is even, hence c. /.. /;. //../;.//.I follows ha c. / is he reflecion of c./ across -ais. Tha is, c. / and c./ are smmeric wih resec o he -ais; hus, i suffices o grah he curve for 0. For 0, wehavec.0/.0; 0/ and he -coordinae./ ends o as!.to analze he -coordinae, we grah./ for 0: 8 6 We see ha./ < 0 and decreasing for 0<<=,./ < 0 and increasing for = <<and./ > 0 and increasing for >.Also./ ends o as!.therefore,saringaheorigin,hecurvefirsdirecsohelefofhe-ais, hen a = i urnsoherigh,alwaskeeinganuwarddirecion.thearofheahfor 0 is obained b reflecing across he -ais. We also use he oins c.0/.0; 0/, c./. ; /, c./.0; / o obain he following grah for c./: = (0, ) = = = (, ) = 0 = = 0 = Grah of c./ for 0. Grahof c./ for all. 8. Skech c./. ; 9 / for 0. The grahs of./ and./ 9 for 0 are shown in he following figures: /./ 9

11 SECTION. Parameric Equaions 57 The curve sars a c. /.; 7/. For <<0,./ is decreasing and./ is increasing, so he grah urns o he lef and uwards o c.0/.0; 9/. Thenfor0<<,./ is decreasing and so is./,hencehegrahurnsohelefanddownwards owards c./. ; 5/. For <<0,./ is increasing and./ is decreasing, hence he grah urns o he righ and downwards, ending a c.0/.60; 9/. Theinerceareheoinswhere. / 0 or 9 0, hais 0; ;. Thesearehe oins c.0/.0; 9/, c./.0; 7/, c./. ; 0/, c. /.; 0/.Theseroeriesleadohefollowingah: = 0 (0, 9) =, (, 5) (, 0) =, (, 0) = = (0, 7) = (, 7) In Eercises 9 5, use Eq. (7) o find d=d a he given oin. 9.. ; /, B Eq. (7) we have Subsiuing we ge 50.. C 9; 7 9/, We find d d : 5..s s; s /, s Using Eq. (7) we ge Subsiuing s we obain d d 0./ 0./. 0 /. / 0 d d ˇ ˇ. / 6 : d.7 9/0 d. C 9/ 0 7 ) d ˇ 7 d ˇ : d d 0.s/ 0.s/.s 0 /.s s/ 0 s d d s ˇ s ˇs s s s. /. / : 5..sin ; cos /, 6 Using Eq. (7) we ge d d 0./ sin 0./ cos Subsiuing 6 wege d d sin cos ˇ sin ˇ=6 cos

12 58 C H A P TE R PARAMETRIC EQUATIONS, POLAR COORINATES, AN VECTOR FUNCTIONS In Eercises 5 56, find an equaion f./for he arameric curve and comue d=d in wo was: using Eq. (7) and b differeniaing f./. 5. c./. C ; 9/ Since C,wehave. Subsiuing in 9 we have 9 9 C iffereniaing 9 C 5. c./ ; d gives d 9 d. We now find using Eq. (7): d d d 0./. 9/0 0./. C / 0 9 Since we have. Subsiuingin ields./ : We differeniae : Now, we find d using Eq. (7). Thus, d d d 0./ 0./ d d 0 0 : Since,henhis is he same as. 55. s, s 6 C s We find as a funcion of : We now differeniae C.Thisgives s 6 C s s C s C : d d : Alernaivel, we can use Eq. (7) o obain he following derivaive: d d 0.s/ s 6 0.s/ C s 0 s 0 6s5 s s s s 6 : Hence, since s, 56. cos, cos C sin d d : To find as a funcion of,wefirsuseherigonomericidenisin cos o wrie cos C cos : We subsiue cos o obain C.iffereniainghisfuncionields Alernaivel, we can comue d using Eq. (7). Tha is, d Hence, since cos, d d 0./ 0./ cos C sin 0.cos / 0 d : d d : d sin C sin cos sin cos :

13 SECTION. Parameric Equaions Find he oins on he curve c./. ; 6/ where he angen line has sloe. We solve d d 6 6 or 6 8 6, or 6 0, sohesloeis a 0; 6 and he oins are.0; 0/ and.96; 80/ 58. Find he equaion of he angen line o he ccloid generaed b a circle of radius a. The ccloid generaed b a circle of radius can be aramerized b c./. sin ; cos / Then we comue d ˇ sin ˇ d ˇ= cos ˇ= sohahesloeofheangenlineis and he equaion of he angen line is cos sin or C 8 In Eercises 59 6, le c./. 9; 8/ (see Figure ) FIGURE Plo of c./. 9; 8/. 59. raw an arrow indicaing he direcion of moion, and deermine he inerval of -values corresonding o he orion of he curve in each of he four quadrans. We lo he funcions./ 9 and./ 8: We noe carefull where each of hese grahs are osiive or negaive, increasing or decreasing. In aricular,./ is decreasing for <0,increasingfor>0,osiiveforjj >,andnegaiveforjj <.Likewise,./ is decreasing for <,increasingfor >,osiivefor>8or <0,andnegaivefor0<<8.Wenowdrawarrowsonheahfollowinghedecreasing/increasing behavior of he coordinaes as indicaed above. We obain: 60 = 0 ( 9,0) 0 0 = (0, 5) 0 = (0,) 0 0 = (7, 6) = 8 (55,0) 0 60

14 60 C H A P TE R PARAMETRIC EQUATIONS, POLAR COORINATES, AN VECTOR FUNCTIONS This lo also shows ha: The grah is in he firs quadran for < or >8. The grah is in he second quadran for <<0. The grah is in he hird quadran for 0<<. The grah is in he fourh quadran for << Find he equaion of he angen line a. Using he formula for he sloe m of he angen line we have: m d ˇ 8 0 d ˇ 9 ˇ 0 ˇ 8 j j 0: Since he sloe is zero, he angen line is horizonal. The -coordinae corresonding o is 8 6.Hencehe equaion of he angen line is Find he oins where he angen has sloe. The sloe of he angen a is d d The oin where he angen has sloe corresonds o he value of ha saisfies d d ) ) 8: We subsiue 8 in./ 9 and./ 8 o obain he following oin:.8/ / Find he oins where he angen is horizonal or verical. In Eercise 6 we found ha he sloe of he angen a is d d ).55; 0/ The angen is horizonal where is sloe is zero. We se he sloe equal o zero and solve for. Thisgives The corresonding oin is 0 ) :../;.//. 9; 8 /.7; 6/: The angen is verical where i has infinie sloe; ha is, a 0. Thecorresondingoinis..0/;.0//.0 9; 0 8 0/. 9; 0/: 6. Le A and B be he oins where he ra of angle inersecs he wo concenric circles of radii r < R cenered a he origin (Figure 5). Le P be he oin of inersecion of he horizonal line hrough A and he verical line hrough B. Eresshe coordinaes of P as a funcion of and describe he curve raced b P for 0. B A r P R FIGURE 5

15 SECTION. Parameric Equaions 6 We use he arameric reresenaion of a circle o deermine he coordinaes of he oins A and B. Thais, A.r cos ;r sin /; B.R cos ;R sin / The coordinaes of P are herefore P.R cos ;r sin / In order o idenif he curve raced b P,wenoicehahe and coordinaes of P saisf R cos and r C cos C sin : R r sin. Hence The equaion R C r is heequaionofellise.hence,hecoordinaesofp,.r cos ;r sin / describe an ellise for A0-fladderslidesdownawallasisboomB is ulled awa from he wall (Figure 6). Using he angle as arameer, find he arameric equaions for he ah followed b (a) he o of he ladder A, (b)heboomofheladderb, and(c)heoinp locaed ffrom he o of he ladder. Show ha P describes an ellise. A P = (, ) 6 FIGURE 6 B (a) We define he -coordinae ssem as shown in he figure: A 0 0 B As he ladder slides down he wall, he -coordinae of A is alwas zero and he -coordinae is 0 sin. Thearameric equaions for he ah followed b A are hus 0; 0 sin ; is beween and 0. The ah described b A is he segmen Œ0; 0 on he -ais. 0 0

16 6 C H A P TE R PARAMETRIC EQUATIONS, POLAR COORINATES, AN VECTOR FUNCTIONS (b) As he ladder slides down he wall, he -coordinae of B is alwas zero and he -coordinae is 0 cos. Thearameric equaions for he ah followed b B are herefore 0 cos ; 0; is beween and 0. The ah is he segmen Œ0; 0 on he -ais. 0 0 (c) The and coordinaes of P are cos, 6 sin. The ah followed b P has he following aramerizaion: c./. cos ;6sin /; is beween and 0. P(, ) 6 0 As shown in Eamle, he corresonding ah is a ar of an ellise. Since is varing beween he ellise in he firs quadran. and 0, weobainhearof 6 0 In Eercises 65 68, refer o he Bézier curve defined b Eqs. (8) and (9). 65. Show ha he Bézier curve wih conrol oins has aramerizaion P 0.; /; P.; /; P.6; 5/; P.7; / c./. C 6 C ;C 5 9 / Verif ha he sloe a 0 is equal o he sloe of he segmen P 0 P. For he given Bézier curve we have a 0, a, a 6, a 7, andb 0, b, b 5, b. Subsiuing hese values in Eq. (8) (9) and simlifing gives./. / C 9. / C 8. /C 7 C C 9. C / C 8 8 C 7 C C 9 8 C 9 C 8 8 C 7 C C 6 C

17 SECTION. Parameric Equaions 6./. / C 6. / C 5. /C. C / C 6. C / C 5 5 C C C 6 7 C 6 C 5 5 C C 5 9 Then c./. C 6 C ;C 5 9 /; 0 : We find he sloe a 0. Usingheformulaforsloeofheangenlinewege d d. C 5 9 / C 6 C / 0 6 C 6 9 ) d ˇ d ˇ0 6 : The sloe of he segmen P 0 P ishesloeofhelinedeerminedbheoinsp 0.; / and P.; /. Thais, 8. We see ha he sloe of he angen line a 0 is equal o he sloe of he segmen P 0P, as eeced. 66. Find an equaion of he angen line o he Bézier curve in Eercise 65 a. so ha a, and We have Thus he angen line is d d./0 0 7./ 0 6 C 6 9 d ˇ d ˇ= C 6 9 ˇ ˇ= ; or 7 C Find and lo he Bézier curve c./ assing hrough he conrol oins P 0.; /; P.0; /; P.5; /; P.; / gives Seing a 0, a 0, a 5, a, andb 0, b, b, b ino Eq. (8) (9) and simlifing./. / C 0 C 5. /C. C / C 5 5 C 9 C 6./. / C 6. / C. / C. C / C 6. C / C C 6 C 6 C 6 C 6 C C C 6 We obain he following equaion The grah of he Bézier curve is shown in he following figure: c./. 9 C 6 ;C 6 /; 0 :

18 6 C H A P TE R PARAMETRIC EQUATIONS, POLAR COORINATES, AN VECTOR FUNCTIONS 68. Show ha a cubic Bézier curve is angen o he segmen P P a P. The equaions of he cubic Bézier curve are./ a 0. / C a. / C a. / C a./ b 0. / C b. / C b. / C b We use he formula for he sloe of he angen line o find he sloe of he angen line a P.Weobain d d 0./ 0./ b 0. / C b.. /. //C b.. / / C b a 0. / C a.. /. //C a.. / / C a () The sloe of he angen line a P is obained b seing in (). Tha is, m 0 C 0 b C b 0 C 0 a C a b b a a () We comue he sloe of he segmen P P for P.a ;b / and P.a ;b /.Wege m b b a a Since he wo sloes are equal, we conclude ha he angen line o he curve a he oin P is he segmen P P. 69. Abullefiredfromagunfollowsherajecor a; b 6.a; b > 0/ Show ha he bulle leaves he gun a an angle an b a and lands a a disance ab=6 from he origin. The heigh of he bulle equals he value of he -coordinae. When he bulle leaves he gun,./.b 6/ 0. Thesoluionsohisequaionare 0 and 6 b, wih 0 corresonding o he momen he bulle leaves he gun. We find he sloe m of he angen line a 0: d d 0./ b 0./ a ) m b a ˇ b ˇ0 a I followshaan b a or an b a. The bulle lands a 6 b. We find he disance of he bulle from he origin a his ime, b subsiuing 6 b in./ a. Thisgives b ab Plo c./. ; C 8/ for.findheoinswhereheangenlineishorizonalorverical. The grah of c./. ; C 8/, is shown in he following figure: = 0 (0, 8) =.5 (.,.8) = ( 5, ) =.5 (.,.8) =, (5, ) =.5 0 (.9, ) =.5 (.9, ) We find he sloe of he angen line a : d d 0./ 0./. 0 C 8/. / 0 () The angen line is horizonal where d 0. Tha is, d 0 ). 6/ 0 ) 0; 6; 6:

19 SECTION. Parameric Equaions 65 We findhecorresondingoinsbsubsiuinghesevaluesof in c./.weobain: c.0/.0; 8/; c. 6/. :9; /; c. 6/.:9; /: The angen line is verical where he sloe in () is infinie, ha is, where 0 or :5. Wefindheoins b seing in c./.wege c. :; :8/; c.:; :8/: 7. Plo he asroid cos, sin and find he equaion of he angen line a. The grah of he asroid cos, sin is shown in he following figure: π = (0, ) = π (, 0) = 0 (, 0) π = (0, ) The sloe of he angen line a is m d ˇ.sin / 0 ˇ d ˇ=.cos / 0 sin cos ˇ ˇ= cos sin ˇ an ˇ. sin / ˇ= cos ˇ= ˇ= We findheoinofangenc:! ; cos ; sin 8 ; 8 The equaion of he angen line a is, hus, 8 ) C 8 7. Find he equaion of he angen line a oheccloidgeneraedbheunicirclewiharamericequaion(5). We find he equaion of he angen line a oheccloid sin, cos.wefirsfindhederiva- ive d d : The sloe of he angen line a isherefore: We findheoinofangenc: The equaion of he angen line is, hus,! d d 0./. cos /0 0./. sin / 0 sin cos m d ˇ sin d ˇ= cos ; sin ; cos ;!! )! C 7. Find he oins wih horizonal angen line on he ccloid wih arameric equaion (5).

20 66 C H A P TE R PARAMETRIC EQUATIONS, POLAR COORINATES, AN VECTOR FUNCTIONS The arameric equaions of he ccloid are sin ; cos We find he sloe of he angen line a : The angen line is horizonal where i has sloe zero. Tha is, d. cos /0 d. sin / 0 sin cos d d sin cos 0 ) sin 0 cos ).k /; k 0; ; ; : : : We find he coordinaes of he oins wih horizonal angen line, b subsiuing.k / in./ and./.thisgives The required oins are.k / sin.k /.k / cos..k //. /..k /; /; k 0; ; ; : :: 7. Proer of he Ccloid Prove ha he angen line a a oin P on he ccloid alwas asses hrough he o oin on he rolling circle as indicaed in Figure 7. Assume he generaing circle of he ccloid has radius. Tangen line Ccloid FIGURE 7 The definiion of he ccloid is such ha a ime, heoofhecirclehascoordinaesq.; / (since a ime he circle has roaed eacl once, and is circumference is ). Le L be he line hrough P and Q. ToshowhaL is angen o he ccloid a P i suffices o show ha he sloe of L equals he sloe of he angen a P.Recallhaheccloidis aramerized b c./. sin ; cos /.ThenhesloeofL is and he sloe of he angen line is and he wo are equal. 0./. cos /0 0./. sin / 0 sin cos. cos /. sin / C cos sin sin. C cos / cos sin. C cos / sin C cos sin 75. A curae ccloid (Figure 8) is he curve raced b a oin a a disance h from he cener of a circle of radius R rolling along he -ais where h<r.showhahiscurvehasaramericequaions R h sin, R h cos. h R π π FIGURE 8 Curae ccloid. Le P be a oin a a disance h from he cener C of he circle. Assume ha a 0, helineofcp is assing hrough he origin. When he circle rolls a disance R along he -ais, he lengh of he arc c SQ (see figure) is also R and he angle SCQ has radian measure.wecomuehecoordinaes and of P. C S h P A C R 0 R Q

21 SECTION. Parameric Equaions 67 We obain he following aramerizaion: R PA R h sin. / R h sin R C AC R C h cos. / R h cos R h sin ; R h cos : 76. Use a comuer algebra ssem o elore wha haens when h >Rin he arameric equaions of Eercise 75. escribe he resul. Look firs a he arameric equaions hsin, hcos.thesedescribeacircleofradiush.seeforinsance he grahs below obained for h and h c() = ( h*sin(), h*cos()) h =, 5 Adding R o he coordinae o obain he arameric equaions h sin, R h cos, ieldsacirclewihiscener moved u b R unis: c() = ( h*sin(), R h*cos()) R =, 5 h = 5 Now, we add R o he coordinae o obain he given arameric equaion; he curve becomes a sring. The figure below shows he grahs obained for R and various values of h. We see he inner loo formed for h > R Show ha he line of sloe hrough. ; 0/ inersecs he uni circle in he oin wih coordinaes C ; C Conclude ha hese equaions aramerize he uni circle wih he oin. ; 0/ ecluded (Figure 9). Show furher ha =. C /. 0

22 68 C H A P TE R PARAMETRIC EQUATIONS, POLAR COORINATES, AN VECTOR FUNCTIONS (, ) Sloe (, 0) FIGURE 9 Uni circle. The equaion of he line of sloe hrough. ; 0/ is. C /. Theequaionofheunicircleis C. Hence, he line inersecs he uni circle a he oins.; / ha saisf he equaions: Subsiuing from equaion () ino equaion () and solving for we obain This gives. C / () C () C. C / C C C. C / C C. / 0 ;. C /. /. C /. C / C So and. The soluion corresonds o he oin. ; 0/. Weareineresedinhesecondoinof C inersecion ha is varing as varies. Hence he aroriae soluion is C We findhe-coordinae b subsiuing in equaion (). This gives!. C / C C C C C C We concludehahelineandheunicircleinersec,besidesa. ; 0/, aheoinwihhefollowingcoordinaes: C ; () C Since hese oins deermine all he oins on he uni circle ece for. ; 0/ and no oher oins, he equaions in () aramerize he uni circle wih he oin. ; 0/ ecluded. We show ha. Using () we have C C C C C C C C C C C : 78. The folium of escares is he curve wih equaion C a, wherea 0 is a consan (Figure 0). (a) Show ha he line inersecs he folium a he origin and a one oher oin P for all ; 0.Eresshecoordinaes of P in erms of o obain a aramerizaion of he folium. Indicae he direcion of he aramerizaion on he grah. (b) escribe he inerval of -values aramerizing he ars of he curve in quadrans I, II, and IV. Noe ha is a oin of disconinui of he aramerizaion. (c) Calculae d=d as a funcion of and find he oins wih horizonal or verical angen. II I III IV FIGURE 0 Folium C a.

23 SECTION. Parameric Equaions 69 (a) We find he oins where he line ( ; 0) andhefoliuminersec,bsolvinghefollowingequaions: Subsiuing from () in () and solving for we ge () C a () C a. C / a 0.. C / a/ 0 ) 0; Subsiuing in () we find he corresonding -coordinaes. Tha is, 0 0; a C a C a C Weconcludehaheline, 0; inersecs he folium in a unique oin P besides he origin. The coordinaes of P are: a C ; a ; 0; C The coordinaes of P deermine a aramerizaion for he folium. We add he origin so 0 mus be included in he inerval of. We ge! a c./ C ; a C ; To indicae he direcion on he curve (for a>0), we firs consider he following limis: lim./ lim!./! lim./ lim./ 0 lim./ lim./ 0!!!! lim./ lim! C./! C lim./ 0 lim./ 0!0!0 These limis deermine he direcions of he wo ars of he folium in he second and fourh quadran. The loo in he firs quadran, corresonds o he values 0 <, andiisdirecedfromc./. a ; a / o c./. a ; a / where and are wo chosen values in he inerval 0 <.Thefollowinggrahshowshedirecedfolium: < < 0 < < 0 < = + = 0 = = = (b) The limis comued in ar (a) indicae ha he ars of he curve in he second and fourh quadrans corresond o he values <<0and << resecivel. The loo in he firs quadran corresonds o he remaining inerval 0 <. (c) We find he derivaive d, using he Formula for he Sloe of he Tangen Line. We ge d a 0 d d 0./ 0./ C 0 a C Horizonal angen occurs when d 0. Tha is, d 6a.C / a.c / a.c / a.c / 6a a a 6a. /. / 0 ). / 0; 0 ) 0; :

24 70 C H A P TE R PARAMETRIC EQUATIONS, POLAR COORINATES, AN VECTOR FUNCTIONS The corresonding oins are: c.0/..0/;.0//.0; 0/ c ; a C ; a! C Verical angen line occurs when d d isinfinie.thais, The corresonding oin is 0 ) c ; 0 a C ; A C a ; a a; a : 79. Use he resuls of Eercise 78 o show ha he asmoe of he folium is he line C a. Hin: Show ha lim. C /! a. We mus show ha as!or! he grah of he folium is geing arbiraril close o he line C a, and he derivaive d isaroachinghesloe of he line. d In Eercise 78 we showed ha!when!. / and! when!. C /.Wefirsshowhahegrahis aroaching he line C a as!or!,bshowingha lim C lim C a.!! C For./ a a,./ C C, a >0,calculaedinEercise78,weobainusingL Hôial srule: We now show ha d d obain lim. C / lim! lim. C / lim! C!! C a C a C lim! a C a C lim! C a C 6a a C 6a a 6a a 6a a a d 6a a isaroaching as! and as! C. Weuse d a 6a comued in Eercise 78 o d lim! d lim 6a a! a 6a 9a 9a d lim! C d lim 6a a! C a 6a 9a 9a We conclude ha he line C a is an asmoe of he folium as!and as!. 80. Find a aramerizaion of nc C nc a n n,wherea and n are consans. Following he mehod in Eercise 78, we firs find he coordinaes of he oin P where he curve and he line inersec. We solve he following equaions: nc C nc a n n Subsiuing in he second equaion and solving for ields nc C nc nc a n n n. C nc / nc a n n 0 n.. C nc / a n / 0 ) 0; a n C nc We assume ha (so C nc 0) andobainonesoluionbesidesheorigin.thecorresonding coordinaes are Hence, he oins aramerizaion: a n anc C nc C nc a n anc,,, areeaclheoinsonhecurve.weobainhefollowing C nc C nc a n anc ; ; : C nc C nc

25 SECTION. Parameric Equaions 7 8. Second erivaive for a Paramerized Curve Given a aramerized curve c./../;.//,showha Use his o rove he formula d d d 0./ 00./ 0./ 00./ d 0./ d d 0./ 00./ 0./ 00./ 0./ Bheformulaforhesloeofheangenlinewehave d d 0./ 0./ iffereniaing wih resec o,usinghequoienrule,gives d d d 0./ d d d 0 0./ 00./ 0./ 00././ 0./ B he Chain Rule we have d d d d d d d d d d Subsiuing ino he above equaion and using d d d=d 0 gives./ d d d d 0./ 00./ 0./ 00./ 0./ 0./ 0./ 00./ 0./ 00./ 0./ 8. The second derivaive of is d =d. VerifhaEq.()aliedoc./.; / ields d =d.infac, an aramerizaion ma be used. Check ha c./. ; 6 / and c./.an ;an / also ield d =d. so ha indeed For he aramerizaion c./.; /,wehave For c./. ; 6 /,wehave so ha again Finall, for c./.an ;an /, and 0./ ; 00./ 0; 0./ ; 00./ 0./ 00./ 0./ 00./ 0 0./ 0./ ; 00./ 6; 0./ 6 5 ; 00./ 0 0./ 00./ 0./ 00./ 0./ / / sec ; 00./ an sec ; 0./ an sec ; 00./ 6 sec sec 0./ 00./ 0./ 00./ 0./ sec.6sec sec / an sec.an sec / sec 6 6 sec6 sec sec an sec 6 sec6 sec 6 6 sec6 sec.. C sec /// sec 6

26 7 C H A P TE R PARAMETRIC EQUATIONS, POLAR COORINATES, AN VECTOR FUNCTIONS In Eercises 8 86, use Eq. () o find d =d. 8. C, 7, Using Eq. () we ge We find he firs and second derivaives of./ and./: 8. s C s, s, s 0./ C ) 0./ C 6 00./ 6 C ) 00./ 6 C 0./ ) 0./ 8 00./ ) 00./ d ˇ d 0./ 00./ 0./ 00./ ˇ ˇ 0./ ˇ Since 0.s/ s C s, we have 0./ 0.Hence,Eq.()cannobeusedocomue d a s. d C 9,, We comue he firs and second derivaives of./ and./: 0./ 8 ) 0. / 8 00./ 0 ) 00. / 0 0./ ) 0. / 00./ 0 ) 00. / 0 Using Eq. () we ge d ˇ d 0. / 00. / 0. / 00. / ˇ 0. / 8 0. / cos, sin, Using Eq. () we ge We find he firs and second derivaives of./ and./: d ˇ d ˇ 0./ sin ) 0 00./ cos ) 00 0./ cos ) 0 00./ sin ) Use Eq. () o find he -inervals on which c./. ; / is concave u. The curve is concave u where d d > 0.Thus, We comue he firs and second derivaives: 0./ 00./ 0./ 00./ 0./ >0 () 0./ ; 00./ 0./ ; 00./ 6

27 SECTION. Parameric Equaions 7 Subsiuing in () and solving for gives.6 8/ 8 6 C 8 8 Since 6 C 8>0for all,hequoienisosiiveif8 >0.Weconcludehahecurveisconcaveufor> Use Eq. () o find he -inervals on which c./. ; / is concave u. The curve is concave u where d d > 0.Thais, We comue he firs and second derivaives: Subsiuing in () and solving for gives 0./ 00./ 0./ 00./ 0./ >0 () 0./ ; 00./ 0./ ; 00./.8 8/ 8 6 C 8 8 C This is clearl osiive for >0.For<0,wewanC > 0,whichmeans >, so < (b aking he recirocal of boh sides), so <. Thus, we see ha our curve is concave u for < and for > Area Under a Paramerized Curve Le c./../;.//, where./ > 0 and 0./ > 0 (Figure ). Show ha he area A under c./ for 0 is Z A./ 0./ d 0 Hin: Because i is increasing, he funcion./ has an inverse g./ and c./ is he grah of.g.//. Al he change-of-variables formula o A R. /. 0 /.g.//d. c() ( 0 ) FIGURE ( ) Le 0. 0 / and. /.Wearegivenha 0./ > 0,hence./ is an increasing funcion of,soihas.g.//d.recallha is a funcion of and g./, soheheigh an inverse funcion g./. TheareaA is given b R 0 a an oin is given b.g.//.wefindhenewlimisofinegraion.since 0. 0 / and. /,helimisfor are 0 and,resecivel.alsosince 0./ d d, we have d 0./d.Performinghissubsiuiongives Since g./,wehavea Z 0 Z A.g.//d 0./ 0./ d. Z 0.g.// 0./ d: 90. Calculae he area under over Œ0; using Eq. () wih he aramerizaions. ; 6 / and. ; /. The area A under on Œ0; is given b he inegral Z A./ 0./ d 0 where. 0 / 0 and. /.Wefirsusehearamerizaion. ; 6 /.Wehave./,./ 6.Hence, 0. 0 / 0 ) 0 0

28 7 C H A P TE R PARAMETRIC EQUATIONS, POLAR COORINATES, AN VECTOR FUNCTIONS. / ) Also 0./.SubsiuinghesevaluesinEq.()weobain Z Z A 6 d 8 d ˇˇˇˇ 0 9 Using he aramerizaion./,./,wehave 0./.Wefind 0 and : 0. 0 / 0 ) 0 0. / ) or : Equaion () is valid if 0./ > 0, haisif>0.hencewechooseheosiivevalue,. WenowuseEq.()oobain Z Z A d 5 d ˇˇˇˇ 0 6 Boh answers agree, as eeced. 9. Wha does Eq. () sa if c./.; f.//? In he aramerizaion./,./ f./we have 0./, 0. 0 /,. /.HenceEq.()becomes A Z 0 Z../ 0 /./ d f./d. 0 / We see ha in his aramerizaion Eq. () is he familiar formula for he area under he grah of a osiive funcion. 9. Skech he grah of c./.ln ; / for and comue he area under he grah using Eq. (). We use he following grahs of./ ln and./ for :./ ln./ We see ha for <<,./ is osiive and increasing and./ is osiive and decreasing. Also c./.ln ; /.0; / and c./.ln ; /.ln ; 0/. Addiionalinformaionisobainedfromhederivaive d. /0 d.ln / 0 = ; ielding d ˇ d ˇ and d ˇ : d ˇ We obain he following grah: (0, ) = = (ln, 0)

29 SECTION. Parameric Equaions 75 WenowuseEq.()ocomueheareaA under he grah. We have./ ln, 0./,./, 0,.Hence, Z Z A./ 0./ d. / Z 0 d d ln ˇ. ln /. ln / ln 0:86 9. Galileo ried unsuccessfull o find he area under a ccloid. Around 60, Gilles de Roberval roved ha he area under one arch of he ccloid c./.r R sin ;R R cos / generaed b a circle of radius R is equal o hree imes he area of he circle (Figure ). Verif Roberval s resul using Eq. (). R πr πr FIGURE The area of one arch of he ccloid equals hree imes he area of he generaing circle. This reduces o Z 0 Z.R R cos /.R R sin / 0 d R. cos / d R : 0 Furher Insighs and Challenges 9. Prove he following generalizaion of Eercise 9: For all >0,heareaofheccloidalsecorOP C is equal o hree imes he area of he circular segmen cu b he chord PC in Figure. P R P R O C = (R, 0) O C = (R, 0) (A) Ccloidal secor OPC FIGURE (B) Circular segmen cu b he chord PC ro a erendicular from oin P o he -ais and label he oin of inersecion T,anddenoeb he cener of he circle. Then he area of he ccloidal secor is equal o he area of OP T lus he area of PTC.Thelaerisarianglewih heigh./ R R cos and base R.R R sin / R sin, soisareais R sin. cos /.TheareaofOP T,using Eq. (), is Z Z Z.u/ 0.u/ du.r R cos u/.ru R sin u/ 0 du R. cos u/ du R sin C sin cos so ha he oal area of he ccloidal secor is R sin C sin cos C R sin. cos / R R sin R. sin / The area of he circular segmen is he area of he circular secor P C subended b he angle less he area of he riangle P C. The riangle P C has heigh R cos and base R sin sohaisareaisr cos sin R sin,andheareaofhecircular secor is R R. Thus he area of he circular segmen is which is one hird he area of he ccloidal secor. R. sin /

30 76 C H A P TE R PARAMETRIC EQUATIONS, POLAR COORINATES, AN VECTOR FUNCTIONS 95. erive he formula for he sloe of he angen line o a arameric curve c./../;.// using a mehod differen from ha resened in he e. Assume ha 0. 0 / and 0. 0 / eis and ha 0. 0 / 0. Showha. 0 C h/. 0 / lim h!0. 0 C h/. 0 / 0. 0 / 0. 0 / Then elain wh his limi is equal o he sloe d=d. rawadiagramshowinghaheraioinhelimiishesloeofasecan line. Since 0. 0 / and 0. 0 / eis, we have he following limis:. 0 C h/. 0 / lim 0. 0 C h/. 0 /. 0 /; lim 0. 0 / () h!0 h h!0 h We use Basic Limi Laws, he limis in () and he given daa 0. 0 / 0,owrie.. 0 C h/. 0 / 0 Ch/. 0 / lim h!0. 0 C h/. 0 / lim h h!0. 0 Ch/. 0 / h lim h!0. 0Ch/. 0 / h. lim 0 Ch/. 0 / h!0 h 0. 0 / 0. 0 /. 0 C h/. 0 / Noice ha he quoien. 0 C h/. 0 / is he sloe of he secan line deermined b he oins P.. 0/;. 0 // and Q.. 0 C h/;. 0 C h//. Hence,helimiofhequoienash! 0 is he sloe of he angen line a P,haishederivaive d d. ( 0, h) Q ( 0 ) P ( 0 ) ( 0 + h) 96. Verif ha he racri curve (` >0) c./ ` anh ; ` sech ` ` has he following roer: For all, hesegmenfromc./ o.; 0/ is angen o he curve and has lengh ` (Figure ). c() FIGURE The racri c./ ` anh `; ` sech. ` Le P c./ and Q.; 0/. P((), ()) Q = (, 0) The sloe of he segmen PQ is./ 0 m./ ` sech ` ` anh sinh ` `

31 SECTION. Parameric Equaions 77 We comuehesloeofheangenlineap : m d 0 d 0./ ` sech ` 0./ 0 ` ` sech ` anh ` ` anh ` ` ` sech ` sech anh ` ` sech sech anh ` ` anh sech ` anh sinh ` ` ` ` Since m m,weconcludehahesegmenfromc./ o.; 0/ is angen o he curve. We now show ha jpqj `: s q jpqj../ / C../ 0/ ` anh s ` ` sech C ` sech ` ` s anh ` C sech ` ` sech ` sinh ` C sech ` ` s sinh C ` sech ` cosh ` 97. In Eercise 58 of Secion 9., we described he racri b he differenial equaion d d ` ` ` ` Show ha he curve c./ idenified as he racri in Eercise 96 saisfies his differenial equaion. Noe ha he derivaive on he lef is aken wih resec o,no. Noe ha d=d sech.=`/ anh.=`/ and d=d sech.=`/ anh.=`/.thus, Muliling o and boom b `=` gives d d d=d d=d sech.=`/ anh.=`/ d d ` =` =` InEercises98and99,referoFigure In he aramerizaion c./.a cos ;b sin / of an ellise, is no an angular arameer unless a b (in which case he ellise is a circle). However, can be inerreed in erms of area: Show ha if c./.; /, hen.=ab/a, wherea is he area of he shaded region in Figure 5. Hin: Use Eq. (). (, ) FIGURE 5 The arameer on he ellise C. a b We comue he area A of he shaded region as he sum of he area S of he riangle and he area S of he region under he curve. The area of he riangle is S.a cos /.bsin / ab sin () (, ) S S

32 78 C H A P TE R PARAMETRIC EQUATIONS, POLAR COORINATES, AN VECTOR FUNCTIONS The area S under he curve can be comued using Eq. (). The lower limi of he inegraion is 0 0 (corresonds o.a; 0/) and he uer limi is (corresonds o../;.//). Also./ b sin and 0./ asin. Since 0./ < 0 on he inerval 0<< (which reresens he ellise on he firs quadran), we use he osiive value a sin o obain a osiive value for he area. This gives Z S 0 Z b sin u a sin udu ab 0 cos u du ab Z ab 0 sin ab 0 ab sin udu u ab sin ˇˇˇˇ sin u 0 () Combining () and () we obain A S C S ab sin Hence, A ab. 99. Show ha he aramerizaion of he ellise b he angle is C ab ab sin ab ab cos a sin C b cos ab sin a sin C b cos We consider he ellise a C b : For he angle we have an, hence, Subsiuing in he equaion of he ellise and solving for we obain a C an b b C a an a b.a an C b / a b a b a an C b an () a b cos a sin C b cos We now ake he square roo. Since he sign of he -coordinae is he same as he sign of cos,weakeheosiiveroo,obaining Hence b (), he -coordinae is ab cos a sin C b cos () ab cos an an a sin C b cos Equaliies () and () give he following aramerizaion for he ellise: c./ ab cos ; a sin C b cos ab sin a sin C b cos ab sin a sin C b cos! ()