1Preliminr topics jectives To revise the properties of sine, cosine nd tngent. To revise the sine rule nd the cosine rule. To revise geometr in the plne, including prllel lines, tringles nd circles. To revise rithmetic nd geometric sequences. To revise rithmetic nd geometric series. To revise infinite geometric series. To work with the modulus function. To revise Crtesin equtions for circles. To sketch grphs of ellipses from their Crtesin equtions. To sketch grphs of hperols from their Crtesin equtions. To consider smptotic ehviour of hperols. To use prmetric equtions to descrie curves in the plne. To investigte the distriution of smple mens using simultion. The first si sections of this chpter revise knowledge nd skills from Specilist Mthemtics Units 1 & tht re required in this course. For further detils, we refer ou to the relevnt chpters of Specilist Mthemtics Units 1 &. This chpter lso introduces prmetric equtions, which re used in Chpters 6 nd 1. The finl section of this chpter gives rief empiricl introduction to the stud of the distriution of smple mens, which will e investigted further in Chpter 15. Chpter 1
Chpter 1: Preliminr topics 1 Circulr functions Defining sine, cosine nd tngent The unit circle is circle of rdius 1 with centre t the origin. It is the grph of the reltion + = 1. We cn define the sine nd cosine of n ngle using the unit circle. Definition of sine nd cosine For ech ngle θ, there is point P on the unit circle s shown. The ngle is mesured nticlockwise from the positive direction of the -is. cos(θ ) is defined s the -coordinte of the point P sin(θ ) is defined s the -coordinte of the point P For emple: (0.8660, 0.5) ( 0.7071, 0.7071) ( 1, 0) ( 0.176, 0.988) θ (0, 1) (0, 1) (1, 0) P(cos(θ ), sin (θ )) 15 100 0 sin 0 = 0.5 (ect vlue) sin 15 = 1 0.7071 sin 100 0.988 cos 0 = 0.8660 cos 15 = 1 0.7071 cos 100 0.176 Definition of tngent tn(θ ) = sin(θ ) cos(θ ) The vlue of tn(θ ) cn e illustrted geometricll through the unit circle. considering similr tringles PP nd TT, it cn e seen tht TT T = PP P i.e. TT = sin(θ ) cos(θ ) = tn(θ ) P θ P sin (θ ) = PP T(1, tn (θ )) T
The trigonometric rtios For right-ngled tringle C, we cn construct similr tringle C tht lies in the unit circle. From the digrm: C = sin(θ ) nd C = cos(θ ) The similrit fctor is the length, giving C = sin(θ ) nd C = cos(θ ) C = sin(θ ) nd C = cos(θ ) This gives the rtio definition of sine nd cosine for right-ngled tringle. The nming of sides with respect to n ngle θ is s shown. sin(θ ) = opposite hpotenuse cos(θ ) = djcent hpotenuse tn(θ ) = opposite djcent Definition of rdin In moving round the unit circle distnce of 1 unit from to P, the ngle P is defined. The mesure of this ngle is 1 rdin. ne rdin (written 1 c ) is the ngle sutended t the centre of the unit circle n rc of length 1 unit. Note: ngles formed moving nticlockwise round the unit circle re defined s positive; those formed moving clockwise re defined s negtive. Degrees nd rdins The ngle, in rdins, swept out in one revolution of circle is π c. π c = 60 π c = 180 1 Circulr functions 1 hpotenuse θ 1 θ djcent 1 1 C C P 1 c C opposite 1 unit 1 1 c = 180 π or 1 = πc 180 Usull the smol for rdins, c, is omitted. n ngle is ssumed to e mesured in rdins unless indicted otherwise.
Chpter 1: Preliminr topics The following tle displs the conversions of some specil ngles from degrees to rdins. ngle in degrees 0 0 5 60 90 180 60 ngle in rdins 0 Some vlues for the trigonometric functions re given in the following tle. π 6 sin cos tn 0 0 1 0 π 1 1 6 π π π 1 1 1 1 1 0 undefined The grphs of sine nd cosine sketch of the grph of f : R R, f () = sin is shown opposite. s sin( + π) = sin for ll R, the sine function is periodic. The period is π. The mplitude is 1. sketch of the grph of f : R R, f () = cos is shown opposite. The period of the cosine function is π. The mplitude is 1. π π 1 π π π f() = sin π π π π π π 1 1 f () = cos π π π π π π 1 For the grphs of = cos(n) nd = sin(n), where > 0 nd n > 0: Period = π n mplitude = Rnge = [, ]
Smmetr properties of sine nd cosine 1 Circulr functions 5 The following results m e otined from the grphs of the functions or from the unit-circle definitions: sin(π θ) = sin θ sin(π + θ) = sin θ sin(π θ) = sin θ sin( θ) = sin θ cos(π θ) = cos θ cos(π + θ) = cos θ cos(π θ) = cos θ cos( θ) = cos θ sin(θ + nπ) = sin θ cos(θ + nπ) = cos θ for n Z ( π ) sin θ = cos θ ( π ) cos θ = sin θ Emple 1 Convert 15 to rdins. 15 15 πc = = πc 180 Emple Find the ect vlue of: Convert 1.5 c to degrees, correct to two deciml plces. 1.5 c 1.5 180 = = 85.9 to two deciml plces π sin 150 cos( 585 ) sin 150 = sin(180 150 ) Emple = sin 0 = 1 Find the ect vlue of: ( 11π ) sin 6 ( 11π ) sin 6 ( = sin π π 6 ( π ) = sin 6 ) cos( 585 ) = cos 585 ( 5π ) cos 6 = cos(585 60 ) = cos 5 = cos 5 = 1 ( 5π ) cos = cos( 7 1 6 π) ( π ) = cos = 1 = 0
6 Chpter 1: Preliminr topics The Pthgoren identit For n vlue of θ: cos θ + sin θ = 1 Emple If sin( ) = 0. nd 0 < < 90, find: cos( ) tn( ) sin ( ) + cos ( ) = 1 tn( ) = sin( ) cos( ) = 0. 0.91 0.09 + cos ( ) = 1 cos ( ) = 0.91 = 91 cos( ) = ± 0.91 Since 0 < < 90, this gives cos( ) = 91 91 0.91 = 100 = 10 of equtions involving sine nd cosine = 91 91 If trigonometric eqution hs solution, then it will hve corresponding solution in ech ccle of its domin. Such n eqution is solved using the smmetr of the grph to otin solutions within one ccle of the function. ther solutions m e otined dding multiples of the period to these solutions. Emple 5 The grph of = f () for f : [0, π] R, f () = sin is shown. For ech pronumerl mrked on the -is, find the other -vlue which hs the sme -vlue. For =, the other vlue is π. For =, the other vlue is π. For = c, the other vlue is π (c π) = π c. For = d, the other vlue is π + (π d) = π d. 1 1 π c d π
1 Circulr functions 7 Emple 6 ( Solve the eqution sin + π ) = 1 for [0, π]. Let θ = + π. Note tht 0 π 0 π π + π 1π π θ 1π ( To solve sin + π ) = 1 for [0, π], we first solve sin θ = 1 for π θ 1π. Consider sin θ = 1. θ = π 6 or 5π 6 or π + π 6 or π + 5π 6 or π + π 6 or π + 5π 6 or... The solutions π 9π nd re not required, s the lie outside the restricted domin for θ. 6 6 For π θ 1π : θ = 5π 6 + π 6 = 5π 6 = π 6 = π Using the TI-Nspire 1π or or 17π or 5π 6 6 6 1π or or 17π or 5π 6 6 6 11π or or 15π or π 6 6 6 or 11π 1 or 5π or π 1 Ensure our clcultor is in rdin mode. (To chnge the mode, go to con > Settings > Document Settings.) Complete s shown. Note: The Grph ppliction hs its own settings, which re ccessed from Grph pge using menu > Settings.
8 Chpter 1: Preliminr topics Using the Csio ClssPd pen themppliction. Ensure our clcultor is in rdin mode (with Rd in the sttus r t the ottom of the min screen). Enter nd highlight ( sin + π ) = 1 0 π Select Interctive > Eqution/Inequlit > solve. Trnsformtions of the grphs of sine nd cosine The grphs of functions with rules of the form f () = sin(n + ε) + nd f () = cos(n + ε) + cn e otined from the grphs of = sin nd = cos trnsformtions. Emple 7 Sketch the grph of the function ( h: [0, π] R, h() = cos + π ) + 1 ( We cn write h() = cos ( + π )) + 1. 6 The grph of = h() is otined from the grph of = cos : diltion of fctor 1 from the -is diltion of fctor from the -is trnsltion of π units in the negtive direction of the -is 6 trnsltion of 1 unit in the positive direction of the -is. First ppl the two diltions to the grph of = cos. π = cos() π π π 5π π 7π π
1 Circulr functions 9 Net ppl the trnsltion π units in 6 the negtive direction of the -is. π 6 π 1 ppl the finl trnsltion nd restrict the grph to the required domin. The grph of tn π 7π 1 sketch of the grph of = tn θ is shown elow. π π π π 5 = cos + 5π 6 π 1π 1 π 5π 6 π 6 19π 1 π π 5π π π 11π 6 5π 1 11π 6 π, 5 π, Notes: The domin of tn is R \ { (k + 1) π : k Z }. The rnge of tn is R. The grph repets itself ever π units, i.e. the period of tn is π. The verticl smptotes hve equtions θ = (k + 1) π, where k Z. θ
10 Chpter 1: Preliminr topics Using the TI-Nspire pen Grphs ppliction nd define f 1() = tn(). Using the Csio ClssPd pen the menum; select Grph & Tle. Enter tn() in 1, tick the o nd tp $. If necessr, select Zoom > Quick > Quick Trig or tp 6 to mnull djust the window. Smmetr properties of tn The following results re otined from the definition of tn: tn(π θ) = tn θ tn(π + θ) = tn θ Emple 8 Find the ect vlue of: tn 0 tn 0 = tn(60 0 ) = tn 0 tn(π θ) = tn θ tn( θ) = tn θ ( π ) tn ( π ) tn ( = tn π + π ) ( π ) = tn = 1 =
1 1 Circulr functions 11 of equtions involving tn The procedure here is similr to tht used for solving equtions involving sin nd cos, ecept tht onl one solution needs to e selected then ll other solutions re one period length prt. Skillsheet Emple 1 Emple 9 Solve the following equtions: tn = 1 for [0, π] tn( π) = for [ π, π] tn = 1 ( π ) Now tn = 1 = π = π or 7π Let θ = π. Then or π + π or π + π or π + π or 11π π π π π π π π π θ π or 15π To solve tn( π) =, we first solve tn θ =. θ = π θ = π π = π Eercise 1 1 or π π or π π or π π or π or π = π or π = π or π 6 or 5π or 5π or π or π or 8π or 8π or 5π or 5π 6 Convert the following ngles from degrees to ect vlues in rdins: i 70 ii 50 iii 50 iv 15 Convert the following ngles from rdins to degrees: 5π i π 7π ii iii iv 11π 1 6 v v 10 1π 9 vi vi 15 11π 1
1 Chpter 1: Preliminr topics 1 Emple Emple Emple Emple 5 Perform the correct conversion on ech of the following ngles, giving the nswer correct to two deciml plces. 5 Convert from degrees to rdins: i 7 ii 100 iii Convert from rdins to degrees: i 1.7 c ii 0.87 c iii 5.8 c Find the ect vlue of ech of the following: iv iv 51 0.1 c v v 06 c sin(15 ) cos( 00 ) c sin(80 ) d cos(0 ) e sin( 5 ) f sin(0 ) Find the ect vlue of ech of the following: ( π ) ( π ) sin cos ( 5π ) ( 9π ) d cos e cos ( 1π ) ( 9π ) g cos h cos 6 6 If sin( ) = 0.5 nd 90 < < 180, find: cos( ) tn( ) 6 If cos( ) = 0.7 nd 180 < < 70, find: sin( ) tn( ) 7 If sin = 0.5 nd π < < π, find: cos 8 If sin = 0. nd π 9 cos The grph of = f () for f : [0, π] R, f () = cos is shown. tn < < π, find: tn For ech pronumerl mrked on the -is, find the other -vlue which hs the sme -vlue. 1 ( cos π ) c ( 11π ) f sin ( sin π ) i 6 vi vi 10 8.9 c c d π π 1
1 1 Circulr functions 1 Emple 6 Emple 7 Emple 8 Emple 9 10 11 1 Solve ech of the following for [0, π]: sin = sin() = ( c cos() = 1 sin + π ) d = 1 ( cos ( + π )) ( e = 1 sin + π ) f = Sketch the grph of ech of the following for the stted domin: ( f () = sin(), [0, π] f () = cos + π ) [ π ],, π ( f () = cos ( + π )) c, [0, π] d f () = sin() + 1, [0, π] ( f () = sin π ) e +, [0, π] Find the ect vlue of ech of the following: ( 5π ) ( tn tn π ) ( tn 9π ) c 6 d tn 0 1 If tn = 1 nd π π, find the ect vlue of: sin cos c tn( ) d tn(π ) 1 If tn = nd π π, find the ect vlue of: sin cos c tn( ) d tn( π) 15 Solve ech of the following for [0, π]: tn = ( tn π ) = 6 ( ( π ) c tn + = 0 d tn ) + = 16 Sketch the grph of ech of the following for [0, π], clerl lelling ll intercepts with the es nd ll smptotes: ( f () = tn() f () = tn π ) ( f () = tn + π ) ( c f () = tn + π ) d
1 Chpter 1: Preliminr topics 1 The sine nd cosine rules In this section, we revise methods for finding unknown quntities (side lengths or ngles) in non-right-ngled tringle. Lelling tringles The following convention is used in the reminder of this chpter: Interior ngles re denoted uppercse letters. The length of the side opposite n ngle is denoted the corresponding lowercse letter. For emple, the mgnitude of ngle C is denoted, nd the length of side C is denoted. The sine rule The sine rule is used to find unknown quntities in tringle in the following two situtions: 1 one side nd two ngles re given two sides nd non-included ngle re given. In the first cse, the tringle is uniquel defined up to congruence. In the second cse, there m e two tringles. Sine rule For tringle C: sin = sin = c sin C Proof We will give proof for cute-ngled tringles. The proof for otuse-ngled tringles is similr. In tringle CD: sin = h h = sin In tringle CD: i.e. sin = h sin = sin sin = sin Similrl, strting with perpendiculr from to C would give sin = c sin C c c C D h C C
1 The sine nd cosine rules 15 Emple 10 Use the sine rule to find the length of. c sin 1 = 10 sin 70 10 sin 1 c = sin 70 = 5.809... The length of is 5.8 cm, correct to two deciml plces. Emple 11 Use the sine rule to find the mgnitude of ngle XZY, given tht Y = 5, = 5 nd z = 6. 5 sin 5 = 6 sin Z sin Z sin 5 = 6 5 6 sin 5 sin Z = 5 = 0.5071... 5 cm 0.7 Z = (0.7... ) or Z = (180 0.7... ) X c 70 10 cm 5 cm 6 cm 19.5 5 cm Hence Z = 0.7 or Z = 19.5, correct to two deciml plces. Notes: Rememer tht sin(180 θ) = sin(θ ). 6 cm When ou re given two sides nd non-included ngle, ou must consider the possiilit tht there re two such tringles. n ngle found using the sine rule is possile if the sum of the given ngle nd the found ngle is less thn 180. Z 1 X Z 1 Z 5 5 C Y Y
16 Chpter 1: Preliminr topics The cosine rule The cosine rule cn e used to find unknown quntities in tringle in the following two situtions: 1 two sides nd the included ngle re given three sides re given. In ech cse, the tringle is uniquel defined up to congruence. Cosine rule For tringle C: or equivlentl = + c c cos cos = + c c The smmetricl results lso hold: = + c c cos c = + cos C Proof We will give proof for cute-ngled tringles. The proof for otuse-ngled tringles is similr. In tringle CD: cos = = cos Using Pthgors theorem in tringles CD nd CD: = + h = (c ) + h Epnding gives = c c + + h Emple 1 = c c + (s = + h ) = + c c cos (s = cos ) For tringle C, find the length of in centimetres correct to two deciml plces. c C D h c c 10 cm 67 C 5 cm C
1 The sine nd cosine rules 17 c = 5 + 10 5 10 cos 67 = 85.968... c = 9.696... The length of is 9.7 cm, correct to two deciml plces. Emple 1 For tringle C, find the mgnitude of ngle C correct to two deciml plces. cos = + c c = 1 + 6 15 1 6 = 0.15 = (108.099... ) 6 cm The mgnitude of ngle C is 108.1, correct to two deciml plces. Emple 1 In C, C = 8, C = 1 cm nd = 15 cm. Find correct to two deciml plces: C C Find C using the cosine rule: = + c c cos = 1 + 15 1 15 cos 8 = 1 + 5 60 cos 8 = 18.8976... = 17.8577... C 1 cm 15 cm cm 1 cm 8 Find C using the sine rule: sin = c sin C sin C = c sin = 15 sin 8 17.8577 C 15 cm Thus C = = 17.86 cm, correct to two deciml plces. Thus C = 56.8, correct to two deciml plces. Note: In prt, the ngle C = 1.7 is lso solution to the eqution, ut it is discrded s possile nswer ecuse it is inconsistent with the given ngle = 8.
18 Chpter 1: Preliminr topics 1 Eercise 1 Skillsheet Emple 10 Emple 11 Emple 1 Emple 1, 1 Emple 1 1 In tringle C, C = 7, C = 55 nd = 10 cm. Find correct to two deciml plces: C C In C, C =, C = 8.5 cm nd = 5.6 cm. Find correct to two deciml plces: the two possile vlues of C (one cute nd one otuse) C in ech cse. In tringle C, C = 58, = 6.5 cm nd C = 8 cm. Find correct to two deciml plces: C C In C, = 5 cm, C = 1 cm nd C = 10 cm. Find: the mgnitude of C, correct to two deciml plces the mgnitude of C, correct to two deciml plces. 5 The djcent sides of prllelogrm re 9 cm nd 11 cm. ne of its ngles is 67. Find the length of the longer digonl, correct to two deciml plces. 6 In C, C = 5, = 10 cm nd C =.7 cm. Find correct to two deciml plces: C C 7 In C, C = 5, C = 60 nd C = 1 cm. Find. 8 In PQR, QPR = 60, PQ = cm nd PR = cm. Find QR. 9 In C, the ngle C hs mgnitude 0, C = 0 cm nd = 18 cm. Find the distnce C correct to two deciml plces. 10 In C, the ngle C hs mgnitude 0, C = 10 cm nd = 8 cm. Find the distnce C using the cosine rule.
1C Geometr prerequisites 19 1C Geometr prerequisites This section lists some geometric results tht ou should e fmilir with nd e le to ppl in emples. Prllel lines If two prllel lines re crossed trnsversl, then: lternte ngles re equl corresponding ngles re equl co-interior ngles re supplementr. If two lines crossed trnsversl form n equl pir of lternte ngles, then the two lines re prllel. ngle sum of polgon The sum of the interior ngles of n n-sided polgon is (n )180. Tringles Tringle inequlit In C: < + c, < c + nd c < +. Pthgors theorem nd its converse In C: If C is right ngle, then + = c. If + = c, then C is right ngle. Properties of isosceles tringles The se ngles of n isosceles tringle re equl. The line joining the verte to the midpoint of the se of n isosceles tringle is perpendiculr to the se. The perpendiculr isector of the se of n isosceles tringle psses through the opposite verte. c c + = 180 Circle geometr The ngle t the centre of circle is twice the ngle t the circumference sutended the sme rc. θ θ C C
0 Chpter 1: Preliminr topics The ngle in semicircle is right ngle. ngles in the sme segment of circle re equl. qudrilterl is cclic if nd onl if the sum of ech pir of opposite ngles is 180. tngent to circle is perpendiculr to the rdius drwn from the point of contct. The two tngents drwn from n eternl point re the sme length, i.e. PT = PT. lternte segment theorem The ngle etween tngent nd chord drwn from the point of contct is equl to n ngle in the lternte segment. z w P T T θ θ θ + = 180 z + w = 180 = If nd CD re two chords of circle tht cut t point P, then P P = PC PD C P D C θ T D θ P P
1C 1C Geometr prerequisites 1 Emple 15 Emple 15 Find the mgnitude of ech of the following ngles: C DC c CD d CD e D C = 9 (verticll opposite) DC = 87 (opposite ngle of cclic qudrilterl) 9 c CP = 85 (verticll opposite) CD = ( 180 (60 + 85) ) = 5 (ngle sum of tringle, CP) d CD = 5 (ngle sutended the rc CD) DC = 87 (from ) CD = ( 180 (87 + 5) ) = 58 (ngle sum of tringle, CD) P 60 e D = ( 180 (60 + 58) ) = 6 (opposite ngle of cclic qudrilterl) Eercise 1C 1 Find the vlues of,, z nd. Find the mgnitude of ech of the following: RTW TS W c d TRS RWT is tngent to the circle t C. Find the vlues of, nd c. T R C 68 6 0 z 50 85 150 105 7 c C S D W
Chpter 1: Preliminr topics 1C CD is squre nd X is n equilterl tringle. Find the mgnitude of: DXC XDC 5 Find the vlues of,, c, d nd e. 6 Find in terms of, nd c. W X e d 69 7 105 c Z Y 7 Find the vlues of nd, given tht 8 Find the vlues of,, c nd d. is the centre of the circle. 0 c 70 D 50 c d 9 Find the vlues of nd. 10 Find the vlues of nd, given tht is the centre of the circle. X 0 11 For ech of the following, find the vlue of : D 6 P 8 C C P 1 0 D 50 C c J L M 8 K X N C
1D Sequences nd series 1D Sequences nd series The following re emples of sequences of numers: 1,, 5, 7, 9,... 0.1, 0.11, 0.111, 0.1111, 0.11111,... 1, 1 9, 1 7, 1 c 81,... d 10, 7,, 1,,... e 0.6, 1.7,.8,..., 9. Ech sequence is list of numers, with order eing importnt. The numers of sequence re clled its terms. The nth term of sequence is denoted the smol t n. So the first term is t 1, the 1th term is t 1, nd so on. sequence m e defined rule which enles ech susequent term to e found from the previous term. This tpe of rule is clled recurrence reltion, recursive formul or n itertive rule. For emple: The sequence 1,, 5, 7, 9,... m e defined t 1 = 1 nd t n = t n 1 +. The sequence 1, 1 9, 1 7, 1 81,... m e defined t 1 = 1 nd t n = 1 t n 1. Emple 16 Use the recurrence reltion to find the first four terms of the sequence t 1 =, t n = t n 1 + 5 t 1 = t = t 1 + 5 = 8 t = t + 5 = 1 t = t + 5 = 18 The first four terms re, 8, 1, 18. Emple 17 Find the recurrence reltion for the following sequence: 9,, 1, 1,... = 1 9 i.e. t = 1 t 1 1 = 1 i.e. t = 1 t The sequence is defined t 1 = 9 nd t n = 1 t n 1.
Chpter 1: Preliminr topics sequence m lso e defined eplicitl rule tht is stted in terms of n. For emple: The rule t n = n defines the sequence t 1 =, t =, t = 6, t = 8,... The rule t n = n 1 defines the sequence t 1 = 1, t =, t =, t = 8,... The sequence 1,, 5, 7, 9,... cn e defined t n = n 1. The sequence t 1 = 1, t n = 1 t n 1 cn e defined t n = 1 n. Emple 18 Find the first four terms of the sequence defined the rule t n = n +. t 1 = (1) + = 5 t = () + = 7 t = () + = 9 t = () + = 11 The first four terms re 5, 7, 9, 11. rithmetic sequences sequence in which ech successive term is found dding fied mount to the previous term is clled n rithmetic sequence. Tht is, n rithmetic sequence hs recurrence reltion of the form t n = t n 1 + d, where d is constnt. For emple:, 5, 8, 11, 1, 17,... is n rithmetic sequence. The nth term of n rithmetic sequence is given t n = + (n 1)d where is the first term nd d is the common difference etween successive terms, tht is, d = t k t k 1, for ll k > 1. Emple 19 Find the 10th term of the rithmetic sequence, 1,, 5,.... =, d = t n = + (n 1)d t 10 = + (10 1) =
rithmetic series 1D Sequences nd series 5 The sum of the terms in sequence is clled series. If the sequence is rithmetic, then the series is clled n rithmetic series. The smol S n is used to denote the sum of the first n terms of sequence. Tht is, S n = + ( + d) + ( + d) + + ( + (n 1)d ) Writing this sum in reverse order, we hve S n = ( + (n 1)d ) + ( + (n )d ) + + ( + d) + dding these two epressions together gives S n = n ( + (n 1)d ) Therefore S n = n ( + (n 1)d ) Since the lst term l = t n = + (n 1)d, we cn lso write S n = n ( ) + l Geometric sequences sequence in which ech successive term is found multipling the previous term fied mount is clled geometric sequence. Tht is, geometric sequence hs recurrence reltion of the form t n = rt n 1, where r is constnt. For emple:, 6, 18, 5,... is geometric sequence. The nth term of geometric sequence is given t n = r n 1 where is the first term nd r is the common rtio of successive terms, tht is, r =, t k 1 for ll k > 1. Emple 0 Find the 10th term of the sequence, 6, 18,.... =, r = t n = r n 1 t 10 = 10 1 = 9 66 t k
6 Chpter 1: Preliminr topics Geometric series The sum of the terms in geometric sequence is clled geometric series. n epression for S n, the sum of the first n terms of geometric sequence, cn e found using similr method to tht used for rithmetic series. Let Then Sutrct (1) from (): Therefore S n = + r + r + + r n 1 (1) rs n = r + r + r + + r n () rs n S n = r n S n (r 1) = (r n 1) S n = (rn 1) r 1 For vlues of r such tht 1 < r < 1, it is often more convenient to use the equivlent formul S n = (1 rn ) 1 r which is otined multipling oth the numertor nd the denomintor 1. Emple 1 Find the sum of the first nine terms of the sequence 1, 1 9, 1 7, 1 81,.... = 1, r = 1, n = 9 1 ( ( 1 1 ) 9 ) S 9 = 1 1 = 1 ( ( 1 9 ) 1 ) 0.99975 Infinite geometric series If geometric sequence hs common rtio with mgnitude less thn 1, tht is, if 1 < r < 1, then ech successive term is closer to zero. For emple, consider the sequence 1, 1 9, 1 7, 1 81,... In Emple 1 we found tht the sum of the first 9 terms is S 9 0.99975. The sum of the first 0 terms is S 0 0.9999999986. We might conjecture tht, s we dd more nd more terms of the sequence, the sum will get closer nd closer to 0.5, tht is, S n 0.5 s n.
1D Sequences nd series 7 n infinite series t 1 + t + t + is sid to e convergent if the sum of the first n terms, S n, pproches limiting vlue s n. This limit is clled the sum to infinit of the series. If 1 < r < 1, then the infinite geometric series + r + r + is convergent nd the sum to infinit is given S = 1 r Proof We know tht S n = (1 rn ) 1 r = 1 r rn 1 r s n, we hve r n 0 nd so rn 1 r 0. Hence S n s n. 1 r Emple Find the sum to infinit of the series 1 + 1 + 1 8 +. = 1, r = 1 nd therefore S = 1 1 r = = 1 1 1 Using CS clcultor with sequences Emple Use clcultor to generte terms of the geometric sequence defined t n = 51(0.5) n 1 for n = 1,,,... Using the TI-Nspire Sequences defined in terms of n cn e investigted in Clcultor ppliction. To generte the first 15 terms of the sequence defined the rule t n = 51(0.5) n 1, complete s shown. Note: ssigning (storing) the resulting list s tn enles the sequence to e grphed. The lists n nd tn cn lso e creted in Lists & Spredsheet ppliction.
8 Chpter 1: Preliminr topics Using the Csio ClssPd pen the menum; select Sequence. Ensure tht the Eplicit window is ctivted. Tp the cursor net to n E nd enter 51 0.5 n 1. (To enter n 1, select the eponent utton in the Mth1 keord.) Tick the o or tp EXE. Tp # to view the sequence vlues. Tp 8 to open the Sequence Tle Input window nd complete s shown elow; tp K. Emple Use CS clcultor to plot the grph of the rithmetic sequence defined the recurrence reltion t n = t n 1 + nd t 1 = 8. Using the TI-Nspire In Lists & Spredsheet pge, nme the first two columns n nd tn respectivel. Enter 1 in cell 1 nd enter 8 in cell 1. Enter = 1 + 1 in cell nd enter = 1 + in cell. Highlight the cells nd using shift nd the rrows. Use menu > Dt > Fill to generte the sequence of numers. To grph the sequence, open Grphs ppliction ( ctrl I > dd Grphs). Crete sctter plot using menu > Grph Entr/Edit > Sctter Plot. Enter the list vriles s n nd tn in their respective fields. Set n pproprite window using menu > Window/Zoom > Zoom Dt. Note: It is possile to see the coordintes of the points: menu > Trce > Grph Trce. The sctter plot cn lso e grphed in Dt & Sttistics pge.
1D Sequences nd series 9 lterntivel, the sequence cn e grphed Using the Csio ClssPd pen the menu m; select Sequence P G. Ensure tht the Recursive window is ctivted. Select the setting & s shown elow. ES directl in the sequence plotter ( menu > Grph Entr/Edit > Sequence > Sequence) with initil vlue 8. Tp the cursor net to n+1 nd enter n +. Note: The smol n cn e found in the dropdown menu n, n. L Enter 8 for the vlue of the first term, 1. Tick the o. Tp # to view the sequence vlues. Tp $ to view the grph. Tp 6 nd djust the window setting for the first FI N 15 terms s shown elow. Select nlsis > Trce nd use the cursor I to view ech vlue in the sequence.
0 Chpter 1: Preliminr topics 1D Eercise 1D Emple 16 1 Use the recurrence reltion to find the first four terms of the sequence t 1 =, t n = t n 1. sequence is defined recursivel t 1 = 6, t n+1 = t n 1. Find t nd t. Use CS clcultor to find t 8. Emple 17 Find possile recurrence reltion for the sequence, 6, 18,.... Emple 18 Find the first four terms of the sequence defined t n = n for n N. 5 sequence is defined recursivel 1 = 5, n+1 = n + 6. Find nd. Use CS clcultor to find 10 nd to plot grph showing the first 10 terms. 6 The Fioncci sequence is given the recurrence reltion t n+ = t n+1 + t n, where t 1 = t = 1. Find the first 10 terms of the Fioncci sequence. Emple 19 7 Find the 10th term of the rithmetic sequence, 7, 10,.... Emple 0 8 Clculte the 10th term of the geometric sequence, 6, 18,.... Emple 1 9 Find the sum of the first 10 terms of n rithmetic sequence with first term nd common difference. 10 Find the sum of the first eight terms of geometric sequence with first term 6 nd common rtio. Emple 11 Find the sum to infinit of 1 1 + 1 9 1 7 +. 1 The first, second nd third terms of geometric sequence re + 5, nd respectivel. Find: the vlue of the common rtio c the difference etween the sum to infinit nd the sum of the first 10 terms. 1 Find the sum to infinit of the geometric sequence,,,,... in terms of. 1 Consider the sum S n = 1 + + + + n 1. n 1 Clculte S 10 when = 1.5. i Find the possile vlues of for which the sum to infinit S eists. ii Find the vlues of for which S = S 10. 15 Find n epression for the sum to infinit of the infinite geometric series 1 + sin θ + sin θ + Find the vlues of θ for which the sum to infinit is.
1E The modulus function 1 1E The modulus function The modulus or solute vlue of rel numer is denoted nd is defined if 0 = if < 0 It m lso e defined s =. For emple: 5 = 5 nd 5 = 5. Emple 5 Evlute ech of the following: i ii i ii c i 6 + ii 6 + i = 6 = 6 ii = = 6 Note: = i = = ii = = Note: = c i 6 + = = ii 6 + = 6 + = 8 Note: 6 + 6 + Properties of the modulus function = nd = = implies = or = + + If nd re oth positive or oth negtive, then + = +. If 0, then is equivlent to. If 0, then k is equivlent to k k +. The modulus function s mesure of distnce Consider two points nd on numer line: n numer line, the distnce etween points nd is =. Thus cn e red s on the numer line, the distnce of from is less thn or equl to, nd cn e red s on the numer line, the distnce of from the origin is less thn or equl to. Note tht is equivlent to or [, ].
Chpter 1: Preliminr topics Emple 6 Illustrte ech of the following sets on numer line nd represent the sets using intervl nottion: { : < } { : } c { : 1 } (, ) (, ] [, ) c [, 5] The grph of = The grph of the function f : R R, f () = is shown here. This grph is smmetric out the -is, since =. Emple 7 1 0 1 0 1 0 1 5 ( 1, 1) For ech of the following functions, sketch the grph nd stte the rnge: f () = + 1 f () = + 1 Note tht = if, nd = if. + 1 if f () = + 1 = + 1 if < if = if < Rnge = [1, ) ( ) + 1 if f () = + 1 = ( ) + 1 if < + if = + if < Rnge = (, 1] (0, ) (1, 1) (, 1) (0, ) (, 1)
1E The modulus function Using the TI-Nspire Use menu > ctions > Define to define the function f () = s( ) + 1. Note: The solute vlue function cn e otined tping s( ) or using the D-templte plette t. pen Grphs ppliction ( ctrl I > Grphs) nd let f 1() = f (). Press enter to otin the grph. Note: The epression s( ) + 1 could hve een entered directl for f 1(). Using the Csio ClssPd InM, enter the epression + 1. Note: To otin the solute vlue function, either choose s( from the ctlog (s shown elow) or selectfrom the Mth1 keord. Tp $ to open the grph window. Highlight + 1 nd drg into the grph window. Select Zoom > Initilize or use 6 to djust the window mnull. Note: lterntivel, the function cn e grphed using the Grph & Tle ppliction. Enter the epression in 1, tick the o nd tp $.
Chpter 1: Preliminr topics Functions with rules of the form = f() nd = f( ) If the grph of = f () is known, then we cn sketch the grph of = f () using the following oservtion: f () = f () if f () 0 nd f () = f () if f () < 0 Emple 8 Sketch the grph of ech of the following: = = 1 The grph of = is drwn nd the negtive prt reflected in the -is. = 1 = 1 The grph of = 1 is drwn nd the negtive prt reflected in the -is. The grph of = f ( ), for R, is sketched reflecting the grph of = f (), for 0, in the -is. Emple 9 Sketch the grph of ech of the following: = = 1 The grph of =, 0, is reflected in the -is. The grph of =, 0, is reflected in the -is.
1E 1E The modulus function 5 Skillsheet Eercise 1E 1 Emple 5 5 + Emple 6 Emple 7 Emple 8 Emple 9 Evlute ech of the following: d 5 e 5 + 5 Solve ech of the following equtions for : c f 5 5 + 1 = = c 5 = 9 d 9 = 0 e = f + = 8 g 5 + 11 = 9 For ech of the following, illustrte the set on numer line nd represent the set using intervl nottion: { : < } { : 5 } c { : 1 } d { : < } e { : + 5 } f { : + 1 } For ech of the following functions, sketch the grph nd stte the rnge: f () = + 1 f () = + + c f () = + 1 d f () = 1 5 Solve ech of the following inequlities, giving our nswer using set nottion: { : 5 } { : } c { : 1 } d { : 5 < } e { : + 7 } f { : + 1 } 6 Solve ech of the following for : 7 8 + = 6 5 = 10 c 1 + = 10 Sketch the grph of ech of the following: = 9 = c = 1 d = 0 e = 8 f = Sketch the grph of ech of the following: = = c = 7 + 1 d = 1 e = + 1 f = + 1 9 If f () = + with f () = nd f ( 1) =, find the vlues of nd. 10 Prove tht +. 11 Prove tht. 1 Prove tht + + z + + z.
6 Chpter 1: Preliminr topics 1F Circles Consider circle with centre t the origin nd rdius r. If point with coordintes (, ) lies on the circle, then Pthgors theorem gives + = r The converse is lso true. Tht is, point with coordintes (, ) such tht + = r lies on the circle. Crtesin eqution of circle The circle with centre (h, k) nd rdius r is the grph of the eqution ( h) + ( k) = r r P(, ) Note: This circle is otined from the circle with eqution + = r the trnsltion defined (, ) ( + h, + k). Emple 0 Sketch the grph of the circle with centre (, 5) nd rdius, nd stte the Crtesin eqution for this circle. The eqution is ( + ) + ( 5) = which m lso e written s + + 10 + 5 = 0 The eqution + + 10 + 5 = 0 cn e unsimplified completing the squre: + + 10 + 5 = 0 + + + 10 + 5 + 5 = 9 ( + ) + ( 5) = This suggests generl form of the eqution of circle: + + D + E + F = 0 Completing the squre gives i.e. + D + D + + E + E + F = D + E ( + D ) ( + + E ) = D + E F 7 5
1F 1F Circles 7 If D + E F > 0, then this eqution represents circle. ( If D + E F = 0, then this eqution represents one point D, E ). If D + E F < 0, then this eqution hs no representtion in the Crtesin plne. Emple 0 Emple 1 Sketch the grph of + + + 6 1 = 0. Stte the coordintes of the centre nd the rdius. Complete the squre in oth nd : + + + 6 1 = 0 + + + + 6 + 9 1 = 1 ( + ) + ( + ) = 5 The circle hs centre (, ) nd rdius 5. Emple + 1 6 (, ) Sketch grph of the region of the plne such tht + < 9 nd 1. Eercise 1F 1 = 1 required region For ech of the following, find the eqution of the circle with the given centre nd rdius: 1 centre (, ); rdius 1 centre (, ); rdius 5 c centre (0, 5); rdius 5 d centre (, 0); rdius Emple 1 Find the rdius nd the coordintes of the centre of the circle with eqution: + + 6 + 1 = 0 + + 1 = 0 c + = 0 d + + 10 + 5 = 0
8 Chpter 1: Preliminr topics 1F Emple Sketch the grph of ech of the following: + + + = 0 + + = 6 c + + 8 10 + 16 = 0 d + 8 10 + 16 = 0 e + 8 + 5 + 10 = 0 f + + 6 9 = 100 For ech of the following, sketch the grph of the specified region of the plne: + 16 + 9 c ( ) + ( ) < d ( ) + ( + ) > 16 e + 16 nd f + 9 nd 1 5 The points (8, ) nd (, ) re the ends of dimeter of circle. Find the coordintes of the centre nd the rdius of the circle. 6 Find the eqution of the circle with centre (, ) tht touches the -is. 7 Find the eqution of the circle tht psses through (, 1), (8, ) nd (, 6). 8 Consider the circles with equtions + 60 76 + 56 = 0 nd + 10 1 + 9 = 0 Find the rdius nd the coordintes of the centre of ech circle. Find the coordintes of the points of intersection of the two circles. 9 Find the coordintes of the points of intersection of the circle with eqution + = 5 nd the line with eqution: = 1G Ellipses nd hperols = lthough the Crtesin equtions of ellipses nd hperols re not included in the Specilist Mthemtics stud design, the re mentioned in the contet of vector clculus. Completing this section is not essentil, ut will help ou when working with ellipses nd hperols in Chpter 1. Ellipses For positive constnts nd, the curve with eqution + = 1 is otined from the unit circle + = 1 ppling the following diltions: diltion of fctor from the -is, i.e. (, ) (, ) diltion of fctor from the -is, i.e. (, ) (, ). The result is the trnsformtion (, ) (, ).
1G Ellipses nd hperols 9 1 (, ) (, ) (, ) (, ) 1 1 1 1 The curve with eqution + = 1 1 is n ellipse centred t the origin with -is intercepts t (, 0) nd (, 0) nd with -is intercepts t (0, ) nd (0, ). If =, then the ellipse is circle centred t the origin with rdius. Ellipse + = 1 where > is the mjor is is the minor is Crtesin eqution of n ellipse The grph of the eqution ( h) ( k) + = 1 Ellipse + = 1 where > is the minor is is the mjor is is n ellipse with centre (h, k). It is otined from the ellipse + = 1 the trnsltion (, ) ( + h, + k). (h, k + ) (h, k) (h, k) (h, k ) (h +, k)
0 Chpter 1: Preliminr topics Emple Sketch the grph of ech of the following ellipses. Give the coordintes of the centre nd the is intercepts. 9 + = 1 + 9 = 1 ( ) ( ) c + = 1 d + + + 6 = 0 9 16 Centre (0, 0) Centre (0, 0) is intercepts (±, 0) nd (0, ±) is intercepts (±, 0) nd (0, ±) c Centre (, ) -is intercepts When = 0: -is intercepts When = 0: ( ) + = 1 9 16 ( ) 16 = 5 9 ( ) = 16 5 9 = ± 5 ( ) + 9 9 16 = 1 ( ) 9 = 7 16 ( ) = 9 7 16 = ± 7 + 5 ( 1, ) 5 7 (, 7) (, ) (, 1) (5, ) 7 +
1G Ellipses nd hperols 1 d Completing the squre: i.e. + + + 6 = 0 ( + 8 + 16) + + 6 8 = 0 ( + ) + = 1 ( + ) + Centre (, 0) is intercepts ( 6, 0) nd (, 0) Given n eqution of the form + + C + E + F = 0 1 = 1 (, ) ( 6, 0) (, 0) (, 0) (, ) where oth nd re positive, the corresponding grph is n ellipse or point. If =, then the grph is circle. In some cses, s for the circle, no pirs (, ) will stisf the eqution. Hperols The curve with eqution = 1 is hperol centred t the origin with is intercepts (, 0) nd (, 0). The hperol hs smptotes = nd =. To see wh this should e the cse, we rerrnge the eqution of the hperol s follows: = 1 = 1 = ( ) 1 s ±, we hve 0 nd therefore i.e. ± (, 0) = (, 0) =
Chpter 1: Preliminr topics Crtesin eqution of hperol The grph of the eqution ( h) ( k) = 1 is hperol with centre (h, k). The smptotes re k = ± ( ) h Note: This hperol is otined from the hperol with eqution = 1 the trnsltion defined (, ) ( + h, + k). Emple For ech of the following equtions, sketch the grph of the corresponding hperol. Give the coordintes of the centre, the is intercepts nd the equtions of the smptotes. 9 = 1 9 = 1 c ( 1) ( + ) = 1 Since 9 = 1, we hve = (1 9 ) 9 Thus the equtions of the smptotes re = ±. If = 0, then = 9 nd so = ±. The -is intercepts re (, 0) nd (, 0). The centre is (0, 0). Since 9 = 1, we hve = (1 9 + ) Thus the equtions of the smptotes re = ±. The -is intercepts re (0, ) nd (0, ). The centre is (0, 0). ( 1) ( + ) d = 1 9 = = = (, 0) (, 0) (0, ) (0, ) =
1G Ellipses nd hperols c First sketch the grph of = 1. The smptotes re = nd =. The centre is (0, 0) nd the is intercepts re (1, 0) nd ( 1, 0). Note: This is clled rectngulr hperol, s its smptotes re perpendiculr. Now to sketch the grph of ( 1) ( + ) = 1 we ppl the trnsltion (, ) ( + 1, ). The new centre is (1, ) nd the smptotes hve equtions + = ±( 1). Tht is, = nd = 1. is intercepts If = 0, then =. If = 0, then ( 1) = 5 nd so = 1 ± 5. Therefore the is intercepts re (0, ) nd (1 ± 5, 0). = ( 1, 0) (1, 0) = = 1 = (1 5, 0) (1, ) (1 + 5, 0) (0, ) (, ) ( 1) ( + ) d The grph of = 1 is otined from the hperol 9 9 = 1 through the trnsltion (, ) (, + 1). Its centre will e (, 1). = 1 9 (0, ) (0, _ ) = = ( The is intercepts re 0, 1 ± 1 ). (, 1) (, ) (, 1) _ = + 7_ ( 1) ( + ) = 1 9 = _ _ 1 Note: The hperols = 1 nd 9 9 = 1 hve the sme smptotes; the re clled conjugte hperols.
Chpter 1: Preliminr topics 1G Eercise 1G Skillsheet Emple Emple 1 Sketch the grph of ech of the following. Lel the is intercepts nd stte the coordintes of the centre. 9 + 16 = 1 5 + 16 = 00 ( ) ( 1) c + = 1 ( ) d + = 1 9 16 9 e 9 + 5 5 100 = f 9 + 5 = 5 g 5 + 9 + 0 18 16 = 0 h 16 + 5 + 100 8 = 0 ( ) ( ) i + = 1 j ( ) + ( 1) = 16 9 Sketch the grph of ech of the following. Lel the is intercepts nd give the equtions of the smptotes. 16 9 = 1 16 9 = 1 c = d = e 8 16 = 0 f 9 5 90 + 150 = 5 ( ) ( ) g = 1 h 8 + = 0 9 i 9 16 18 + 151 = 0 j 5 16 = 00 Find the coordintes of the points of intersection of = 1 with: = 1 + = 1 Show tht there is no intersection point of the line = +5 with the ellipse + = 1. 5 Find the points of intersection of the curves + = 1 nd 9 9 + = 1. Show tht the points of intersection re the vertices of squre. 6 Find the coordintes of the points of intersection of 16 + = 1 nd the line with 5 eqution 5 =. 7 n the one set of es, sketch the grphs of + = 9 nd = 9.
1H Prmetric equtions 5 1H Prmetric equtions In Chpter 1, we will stud motion long curve. prmeter (usull t representing time) will e used to help descrie these curves. In this section, we give n introduction to prmetric equtions of curves in the plne. The unit circle The unit circle cn e epressed in Crtesin form s { (, ) : + = 1 }. We hve seen in Section 1 tht the unit circle cn lso e epressed s { } (, ) : = cos t nd = sin t, for some t R The set nottion is often omitted, nd we cn descrie the unit circle the equtions = cos t nd = sin t for t R These re the prmetric equtions for the unit circle. We still otin the entire unit circle if we restrict the vlues of t to the intervl [0, π]. The following three digrms illustrte the grphs otined from the prmetric equtions = cos t nd = sin t for three different sets of vlues of t. t [0, π] t [0, π] t [ 0, π 1 Circles 1 1 1 Prmetric equtions for circle centred t the origin 1 1 The circle with centre the origin nd rdius is descried the prmetric equtions = cos t nd = sin t The entire circle is otined tking t [0, π]. 1 Note: To otin the Crtesin eqution, first rerrnge the prmetric equtions s = cos t nd = sin t Squre nd dd these equtions to otin + = cos t + sin t = 1 This eqution cn e written s + =, which is the Crtesin eqution of the circle with centre the origin nd rdius. ] 1 1
6 Chpter 1: Preliminr topics The domin nd rnge of the circle cn e found from the prmetric equtions: Domin The rnge of the function with rule = cos t is [, ]. Rnge Emple 5 Hence the domin of the reltion + = is [, ]. The rnge of the function with rule = sin t is [, ]. Hence the rnge of the reltion + = is [, ]. circle is defined the prmetric equtions = + cos θ nd = 1 + sin θ for θ [0, π] Find the Crtesin eqution of the circle, nd stte the domin nd rnge of this reltion. Domin The rnge of the function with rule = + cos θ is [ 1, 5]. Hence the domin of the corresponding Crtesin reltion is [ 1, 5]. Rnge The rnge of the function with rule = 1 + sin θ is [, ]. Hence the rnge of the corresponding Crtesin reltion is [, ]. Crtesin eqution Rewrite the prmetric equtions s = cos θ nd 1 = sin θ Squre oth sides of ech of these equtions nd dd: i.e. ( ) ( 1) + = cos θ + sin θ = 1 9 9 ( ) + ( 1) = 9 Prmetric equtions for circle The circle with centre (h, k) nd rdius is descried the prmetric equtions = h + cos t nd = k + sin t The entire circle is otined tking t [0, π]. Prmetric equtions in generl prmetric curve in the plne is defined pir of functions = f (t) nd = g(t) The vrile t is clled the prmeter. Ech vlue of t gives point ( f (t), g(t) ) in the plne. The set of ll such points will e curve in the plne.
1H Prmetric equtions 7 Note: If = f (t) nd = g(t) re prmetric equtions for curve C nd ou eliminte the prmeter t etween the two equtions, then ech point of the curve C lies on the curve represented the resulting Crtesin eqution. Emple 6 curve is defined prmetricll the equtions = t nd = t for t R where is positive constnt. Find: the Crtesin eqution of the curve the eqution of the line pssing through the points where t = 1 nd t = c the length of the chord joining the points where t = 1 nd t =. The second eqution gives t =. Sustitute this into the first eqution: ( ) = t = ( ) = = This cn e written s =. t t = 1, = nd =. This is the point (, ). t t =, = nd =. This is the point (, ). The grdient of the line is (t, t) ( ) m = = 6 = Therefore the eqution of the line is = ( ) which simplifies to = +. c The chord joining (, ) nd (, ) hs length ( ) + ( ( )) = 9 + 6 = 5 = 5 (since > 0)
8 Chpter 1: Preliminr topics Ellipses Prmetric equtions for n ellipse The ellipse with the Crtesin eqution + = 1 cn e descried the prmetric equtions = cos t nd = sin t The entire ellipse is otined tking t [0, π]. Note: We cn rerrnge these prmetric equtions s = cos t nd = sin t Squre nd dd these equtions to otin + = cos t + sin t = 1 The domin nd rnge of the ellipse cn e found from the prmetric equtions: Domin The rnge of the function with rule = cos t is [, ]. Rnge Emple 7 Hence the domin of the reltion + = 1 is [, ]. The rnge of the function with rule = sin t is [, ]. Hence the rnge of the reltion + = 1 is [, ]. Find the Crtesin eqution of the curve with prmetric equtions = + sin t nd = cos t for t R nd descrie the grph. We cn rerrnge the two equtions s = sin t nd = cos t Now squre oth sides of ech eqution nd dd: ( ) ( ) + = sin t + cos t = 1 9 Since ( ) = ( ), this eqution cn e written more netl s ( ) + 9 ( ) = 1 This is the eqution of n ellipse with centre (, ) nd is intercepts t (, 0) nd (0, ).
Hperols 1H Prmetric equtions 9 In order to give prmetric equtions for hperols, we will e using the secnt function, which is defined sec θ = 1 cos θ if cos θ 0 The grphs of = sec θ nd = cos θ re shown here on the sme set of es. The secnt function is studied further in Chpter. We will lso use n lterntive form of the Pthgoren identit cos θ + sin θ = 1 Dividing oth sides cos θ gives 1 + tn θ = sec θ We will use this identit in the form sec θ tn θ = 1 1 π π π π π π 1 Prmetric equtions for hperol The hperol with the Crtesin eqution = 1 cn e descried the prmetric equtions = sec t nd = tn t for t ( π, π ) ( π, π ) Note: We cn rerrnge these prmetric equtions s = sec t nd = tn t Squre nd sutrct these equtions to otin = sec t tn t = 1 = sec θ = cos θ The domin nd rnge of the hperol cn e determined from the prmetric equtions. Domin There re two cses, giving the left nd right rnches of the hperol: For t ( π, π ), the rnge of the function with rule = sec t is [, ). The domin [, ) gives the right rnch of the hperol. For t ( π, π ), the rnge of the function with rule = sec t is (, ]. The domin (, ] gives the left rnch of the hperol. Rnge For oth sections of the domin, the rnge of the function with rule = tn t is R. θ
50 Chpter 1: Preliminr topics Emple 8 Find the Crtesin eqution of the curve with prmetric equtions = sec t nd = tn t for t ( π, π ) Descrie the curve. Rerrnge the two equtions: = sec t nd = tn t Squre oth sides of ech eqution nd sutrct: 9 16 = sec t tn t = 1 The Crtesin eqution of the curve is 9 16 = 1. The rnge of the function with rule = sec t for t ( π, π ) is (, ]. Hence the domin for the grph is (, ]. The curve is the left rnch of hperol centred t the origin with -is intercept t (, 0). The equtions of the smptotes re = nd =. Finding prmetric equtions for curve When converting from Crtesin eqution to pir of prmetric equtions, there re mn different possile choices. Emple 9 Give prmetric equtions for ech of the following: + = 9 c 16 + = 1 ( 1) ( + 1) = 1 9 ne possile solution is = cos t nd = sin t for t [0, π]. nother solution is = cos(t) nd = sin(t) for t [0, π]. Yet nother solution is = sin t nd = cos t for t R. ne solution is = cos t nd = sin t. c ne solution is 1 = sec t nd + 1 = tn t.
1H 1H Prmetric equtions 51 Skillsheet Emple 5 Using the TI-Nspire pen Grphs ppliction ( con > New Document > dd Grphs). Use menu > Grph Entr/Edit > Prmetric to show the entr line for prmetric equtions. Enter 1(t) = cos(t) nd 1(t) = sin(t) s shown. Using the Csio ClssPd pen the Grph & Tle ppliction. From the toolr, select Tpe > PrmTpe. Use the Trig keord to enter the equtions s shown on the right. Tick the o nd tp $. Use 6 to djust the window. Eercise 1H 1 Find the Crtesin eqution of the curve with prmetric equtions = cos(t) nd = sin(t), nd determine the domin nd rnge of the corresponding reltion. Emple 6 curve is defined prmetricll the equtions = t nd = 8t for t R. Find: the Crtesin eqution of the curve the eqution of the line pssing through the points where t = 1 nd t = 1 c the length of the chord joining the points where t = 1 nd t =.
5 Chpter 1: Preliminr topics 1H Emple 7 Find the Crtesin eqution of the curve with prmetric equtions = + sin t nd = cos t for t R, nd descrie the grph. Emple 8 Emple 9 Find the Crtesin eqution of the curve with prmetric equtions = sec t nd = tn t for t ( π, π ), nd descrie the curve. 5 Find the corresponding Crtesin eqution for ech pir of prmetric equtions: c = cos(t) nd = sin(t) = cos t nd = sin t d = sin(t) nd = cos(t) = sin t nd = cos t e = tn(t) nd = sec(t) f = 1 t nd = t g = t + nd = 1 h = t 1 nd = t + 1 t = t 1 ( nd = t + 1 ) i t t 6 For ech of the following pirs of prmetric equtions, determine the Crtesin eqution of the curve nd sketch its grph: = sec t, = tn t, t ( π, π ) = cos(t), = sin(t) c = cos t, = + sin t d = sin t, = cos t, t [ π, π ] e = sec t, = tn t, t ( π, π ) f = 1 sec(t), = 1 + tn(t), t ( π, π ) 7 circle is defined the prmetric equtions 8 = cos(t) nd = sin(t) for t R Find the coordintes of the point P on the circle where t = π. Find the eqution of the tngent to the circle t P. Give prmetric equtions corresponding to ech of the following: c + = 16 ( 1) + ( + ) = 9 9 = 1 ( 1) ( + ) d + = 9 9 9 circle hs centre (1, ) nd rdius. If prmetric equtions for this circle re = + cos(πt) nd = c + d sin(πt), where,, c nd d re positive constnts, stte the vlues of,, c nd d. 10 n ellipse hs -is intercepts (, 0) nd (, 0) nd -is intercepts (0, ) nd (0, ). Stte possile pir of prmetric equtions for this ellipse. 11 The circle with prmetric equtions = cos(t) nd = sin(t) is dilted fctor of from the -is. For the imge curve, stte: possile pir of prmetric equtions the Crtesin eqution.
1H 1I Distriution of smple mens 5 ( t ( t 1 The ellipse with prmetric equtions = cos nd = + sin is trnslted ) ) units in the negtive direction of the -is nd units in the negtive direction of the -is. For the imge curve, stte: possile pir of prmetric equtions the Crtesin eqution. 1 Sketch the grph of the curve with prmetric equtions = + sin(πt) nd = + cos(πt) for: t [ 0, 1 ] t [ 0, 1 ] t [ 0, ] c For ech of these grphs, stte the domin nd rnge. 1I Distriution of smple mens In Specilist Mthemtics Units 1 &, ou m hve investigted the smpling distriution of smple mens. This topic will e covered more formll in Chpter 15. In this section, we revise some of the ides from Units 1 &. Summr of concepts popultion is the set of ll eligile memers of group which we intend to stud. popultion does not hve to e group of people. For emple, it could consist of ll pples produced in prticulr re, or ll components produced fctor. smple is suset of the popultion which we select in order to mke inferences out the popultion. Generlising from the smple to the popultion will not e useful unless the smple is representtive of the popultion. The simplest w to otin vlid smple is to choose rndom smple, where ever memer of the popultion hs n equl chnce of eing included in the smple. The popultion men µ is the men of ll vlues of mesure in the entire popultion; the smple men is the men of these vlues in prticulr smple. The popultion men µ is popultion prmeter; its vlue is constnt for given popultion. The smple men is smple sttistic; its vlue is not constnt, ut vries from smple to smple. The smple men X cn e viewed s rndom vrile, nd its distriution is clled smpling distriution. The vrition in the smpling distriution decreses s the size of the smple increses. When the popultion men µ is not known, we cn use the smple men s n estimte of this prmeter. The lrger the smple size, the more confident we cn e tht the smple sttistic gives good estimte of the popultion prmeter.
5 Chpter 1: Preliminr topics n emple Suppose tht one million people live in prticulr cit nd we know tht the men IQ for this popultion is 100 nd the stndrd devition is 15. This emple illustrtes the ides listed in the summr: Popultion The popultion is the one million people living in the prticulr cit. Smple We will tke rndom smple of 00 people from the popultion. Popultion men µ We re considering IQ nd the popultion men µ is 100. Smple men The smple men is otined determining the men IQ of the 00 people in the smple. Rndom vrile X If we tke numer of smples of size 00 from the sme popultion nd determine the men IQ for ech of these smples, we otin distriution of smple mens. The mens of these smples re the vlues of the rndom vrile X. To use technolog to investigte the rndom vrile IQ, we use the norml distriution. You will stud this distriution in Mthemticl Methods Units &. For now it is enough to know tht mn commonl occurring rndom vriles such s height, weight nd IQ follow this distriution. This histogrm shows the distriution of the IQ scores of 1000 people rndoml drwn from the popultion. You cn see tht the distriution is smmetric nd ell-shped, with its centre of smmetr t the popultion men. Frequenc 160 10 10 100 80 60 0 0 0 50 60 70 80 90 100 110 10 10 10 150 The norml distriution is full defined its men nd stndrd devition. If we know these vlues, then we cn use technolog to generte rndom smples. We will use TI clcultor, ut the tsk m e crried out in similr w with other technolog. Using the TI-Nspire To generte rndom smple of size 00 from norml popultion with men 100 nd stndrd devition 15: Strt from Lists & Spredsheet pge. Nme the list iq in Column. In the formul cell of Column, enter the formul using Menu > Dt > Rndom > Norml nd complete s: = rndnorm(100, 15, 00) Note: The snt is: rndnorm(men, stndrd devition, smple size)