mst121 chapter a2

8/13/2019 Mst121 Chapter a2

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MST121Using Mathematics

Chapter A2

Lines and circles


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About this course

Thiscourse,MST121Using Mathematics,andthecoursesMU120OpenMathematics andMS221Exploring Mathematics provideaflexiblemeansofentrytouniversity-levelmathematics. Furtherdetailsmaybeobtainedfromtheaddressbelow.MST121usesthesoftwareprogramMathcad(MathSoft,Inc.)andothersoftwaretoinvestigatemathematicalandstatisticalconceptsandasatool inproblemsolving. Thissoftware isprovidedaspartofthecourse.

Thispublication formspartofanOpenUniversitycourse. DetailsofthisandotherOpenUniversitycoursescanbeobtainedfromtheStudentRegistrationandEnquiryService,TheOpenUniversity,POBox197,MiltonKeynesMK76BJ,UnitedKingdom: tel.+44(0)8453006090,[email protected],youmayvisittheOpenUniversitywebsiteathttp://www.open.ac.ukwhereyoucanlearnmoreaboutthewiderangeofcoursesandpacksofferedatall levelsbyTheOpenUniversity.TopurchaseaselectionofOpenUniversitycoursematerialsvisithttp://www.ouw.co.uk,orcontactOpenUniversityWorldwide,WaltonHall,MiltonKeynesMK76AA,UnitedKingdom,forabrochure: tel.+44(0)1908858793, fax+44(0)1908858787,[email protected]

TheOpenUniversity,WaltonHall,MiltonKeynes,MK76AA.Firstpublished1997. Secondedition2001. Thirdedition2008. Reprinted2008.Copyright1997,2001,2008TheOpenUniversitycAllrightsreserved. Nopartofthispublicationmaybereproduced,stored inaretrievalsystem,transmittedorutilised inanyformorbyanymeans,electronic,mechanical,photocopying,recordingorotherwise,withoutwrittenpermissionfromthepublisherora licencefromtheCopyrightLicensingAgencyLtd. Detailsofsuchlicences(forreprographicreproduction)maybeobtainedfromtheCopyrightLicensingAgencyLtd,SaffronHouse,610KirbyStreet,LondonEC1N8TS;websitehttp://www.cla.co.uk.Edited,designedandtypesetbyTheOpenUniversity,usingtheOpenUniversityTEX System. PrintedintheUnitedKingdombyCambrianPrinters,Aberystwyth.ISBN

978

07492

2938

2

3.2


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Contents

Study guide 4Introduction 51 Lines 6

1.1 Equations of lines 61.2 Some applications 16

2 Circles 202.1 Circles and their equations 202.2 Finding the circle through three points 232.3 Completing the square 262.4 Intersections of circles and lines 28

3 Trigonometry 333.1 Sine and cosine 333.2 Calculations with triangles 36

4 Parametric equations 404.1 Parametric equations of lines 404.2 Parametric equations of circles 43

5 Parametric equations by computer 46Summary of Chapter A2 47

Learning outcomes 47Appendix: Modelling the Earth 49Solutions to Activities 53Solutions to Exercises 58Index 60

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Study guide

Thischaptercontainsfivesections,whichare intendedtobestudiedconsecutively,andanappendix.Sections1,2and4areofaveragelength,andtheothertwoarerelativelyshort. TheAppendixcanberead,forinterest,atanytimeafterSection3.YouwillneedaccesstoyourDVDplayeratthestartofSubsection2.1. However,ifthis isnotconvenient,thencontinuewiththetextandviewthevideowhenthisbecomesfeasible. YouwillneedaccesstoyourcomputerandComputerBookA inordertostudySection5.Althoughthischaptermay look long, itshouldrequirenomorethanaveragestudytimesincesomeoftheideasincludedintheRevision Pack arerevisited inparticular,thetopicsofcoordinatesand lines,andtrigonometry.Thedivisionofyourtimeforthechaptermightbeasfollows.Studysession1: Section1.Studysession2: Section2.Studysession3: Section3.Studysession4: Section4andSection5.Sections4and5couldbesplit intotwostudysessions.Theorganisationofyourtimewithineachstudysession is likelytodependonyourexperienceandconfidenceinworkingwithalgebraicexpressions.Inadditiontothetopicsmentioned inthethirdparagraphabove,beforestudyingthischapteryoushouldalsobefamiliarwiththefollowingtopics fromtheRevision Pack : PythagorasTheorem; radianmeasureforangles; similartriangles.

TheoptionalVideoBandA(iv)Algebra workout Quadratic equations couldbeviewedatanystageduringyourstudyofthischapter.

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Introduction

Linesandcirclesareusedextensivelyasmodelsofobjectswithintherealworldandofpathswhichdescribethemotionsofobjects. Approximateoccurrencesofthesegeometricalshapesoccurinnature,andtheyhavebeenusedasabasisformanyhumanartefacts,bothancientandmodern.Thischapterconcentratesmainlyuponthemathematicaldescriptionandpropertiesof linesandcircles. While it isnaturaltothinkofthemingeometricterms, it isalsopossibletodescribethemalgebraically. Infact,it isoneofthetriumphsofmathematicstohavecreatedthe linkbetweengeometryandalgebra. Geometryreliesmoreuponourvisualsense,whereasalgebracanberegardedasameansofcodingvisual items inaformthatappearsmoreabstractbutrenderscalculationseasier. Inasense,algebracanbeusedtomodelgeometricstatements. It isoftenthecasethatageometricproblemmaybetranslated intoalgebraicterms,solvedusingalgebraicmanipulation,andtheninterpretedoncemoregeometrically. Thestudyhereof linesandcirclesillustratesthisprocess.Youwillalsoseethetopicoftrigonometry introducedhere,since itisconnectedbothwithcirclesandwithtriangles(whicharegeometricobjectsmadeupfromsegmentsof lines). Itoffersfurtherscopetodescribeand interpretageometricviewoftheworld inalgebraicterms.Section1showshow linescanberepresentedbyalgebraicequations,viaastatementofthegeometricalrestrictionwhich ineffectdefineswhata(straight) line is. Youwillthenseehowthepointatthe intersectionofa Inmathematics,thewordlinepairof linesmaybecalculatedbysolvingsimultaneouslytheequations usuallymeansstraight line.whichrepresentthe lines.Section2seekstocarryoutasimilarprogramme forcircles,whichreliesupon

the

use

of

Pythagoras Theorem.

An

important

algebraic

procedure,

knownascompleting the square foranexpressionoftheformx2 + 2px, is explainedandappliedhere.InSection3,thetrigonometricquantitiessine andcosine are introducedintermsofthecoordinatesofpointsonacircle. Theirexpressionasratioswithinaright-angledtriangle,withwhichyoumaybemorefamiliar, isalsomadeapparent.The ideas inthefirstthreesectionsaredevelopedinthe introductionofparametric equations inSection4. Foranobjecttravelling inspace,asingleequationthatrelatesthespatialcoordinatesoftheobjectcandescribe itsoveralltrajectory. Ontheotherhand,parametricequationscanbeusedtoexpresseachspatialcoordinateintermsoftheelapsedtime,whichpermitsthemotionalongthetrajectorytobedescribed. There iscomputerwork inSection5associatedwiththetopics inSection4.TheAppendix indicateshowsomeoftheideasofthechaptercanbeapplied intriangulation formappingpurposesand inaglobalpositioningsystem.

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1 Lines

TheadjectiveCartesiancomes

from

the

surname

of

thefamousFrenchmathematicianandphilosopherReneDescartes(15961650). He iscreditedwithbeingthefirsttorealisethatcurvesandsurfacescanbestudiednotjustusingideasofshape,butalsousingthetoolsofalgebra.

Thissectionshowshowcertainequationsdescribe lines,anddemonstratessomepracticalusesforthisalgebraicdescription. Subsection1.1introducesthegeneralformwhichtheequationofa linetakes,andshowshowtheappropriateequationmaybefound inspecificcases. Itgoesontoconsiderhowtheequationsoftwolinescanbeusedtofindtheirintersectionpoint. Subsection1.2 looksbrieflyatsomemodellingapplicationsofthismathematics.

1.1 Equations of lines

Points and coordinates

Asyouknow,points intheplanemaybespecifiedbypairsofcoordinates.Wegenerallyusearectangular orCartesian coordinatesystemtodothis,in

which

apair

of

straight

lines

at

right

angles

are

chosen

as

the

x-axisandy-axis ofthesystem. Thepointof intersectionoftheseaxes iscalled

theorigin,whichisusually labelledOanddescribedbythecoordinates(0, 0). Eachofthex-axisandy-axis isdrawnwithanarrowheadtoindicateapositivedirectionalongit,asshowninFigure1.1. Eachaxiscanberegardedasacopyoftherealnumber line,withauniformscalealong it. (Usually,thesamescaleisusedforeachaxis.)

Figure 1.1 CartesiancoordinatesFigure1.1showstwopoints,AandB. The point Ahascoordinates(1.5, 2.5),because itisreachedbymovingfromOby1.5unitsinthepositivex-directionand2.5units inthepositivey-direction. Similarly,pointB hascoordinates(3, 2),becauseit isreachedbymovingfromOby3units inthenegativex-directionand2units inthenegativey-direction.Moregenerally,apointwithcoordinates(x, y) is

totherightofthey-axis ifx > 0,tothe leftofthey-axis ifx < 0,

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SECTION 1 LINES

andabovethex-axis ify > 0,belowthex-axis ify < 0.

For a point (x, y) on the y-axis itself,wehavex=0. Similarly, if(x, y) lies onthex-axis,theny= 0.

Lines parallel to the coordinate axesYouprobablyhaveagood intuitivesenseofwhatthestraightnessassociatedwith linesmeans. Onewayofputtingit isthatasyoumovealongastraight line,youmaintainthesamedirection. However,thisobservation leavesussomewayshortofbeingabletodescribe linesbyequations.Theequationofastraight line,or indeedofanycurve,specifiesthepropertyorconditionthatissatisfiedbythecoordinates(x, y) of any point onthatlineorcurve. Forexample,thetwoaxesofaCartesiancoordinatesystemarethemselvesstraight lines. Allpointsonthex-axishavethepropertythattheiry-coordinate iszero,becausetoarriveatthesepointswedonotneedtomoveatall inthey-direction. Thex-coordinateofsuchapointdependsonhowfarfromtheoriginthepoint is,andwhetherweneedtomoveleft(x < 0)orright(x > 0)fromtheorigintoreachit. Thusthecoordinatesofallpointsonthex-axishavetheform(x, 0).Consequently,wedescribethex-axisasthelineconsistingofallpoints(x, y) for which y=0. Hencetheequationy=0 isanalgebraicrepresentationofthex-axis. Similarly,theequationx=0 isanalgebraicrepresentationofthey-axis,sincethepropertywhichdistinguishespointsonthey-axis isthattheirx-coordinate iszero;allpointsherehavecoordinatesoftheform(0, y).Thisconnectionbetweenlinesandequationsextendstoany linewhich isparallel toeitherthex- or y-axis. Consider,forexample,thelinewhich is3unitsverticallyabovethex-axis. Anypointonthis linehascoordinatesoftheform(x, 3). Wecanthereforeusetheequationy=3torepresentthe line. Similarly,theequationy=2representsthelinewhich isparalleltothex-axisbut2unitsverticallybelow it. The linesy= 3 and y=2areshowninFigure1.2.Ingeneral,anylineparalleltothex-axishasanequationoftheformy=c, where cisaconstant. Thevalueofc ispositiveforalineabovethex-axis,andnegativefora linebelow.

Activity 1.1 Parallel to the y-axis

(a) Aline isparalleltoand3unitstotheleftofthey-axis. Writedowntheequationwhichrepresentstheline.

(b) Writedownthegeneralformofequationforalineparalleltothey-axis. Explainhowtodistinguishbetweenthetwocases inwhichtheline istotheleftortotherightofthey-axis.

(c) Sketcheachofthe linesx=3 and x= 4. Solutionsaregivenonpage53.

Thelineorcurveissometimescalledthelocus oftheequation.

Figure 1.2Linesparalleltothex-axis

Thephrasethe liney= 3 is shorthandforthelinewhoseequationisy=3. Shorthandforms likethisareincommonuse.

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CHAPTER A2 LINES AND CIRCLES

Asyoufoundinthisactivity,anylineparalleltothey-axishasanequationoftheformx=d, where disaconstant. Itremainsnowtodeterminewhatequationscorrespondto linesthatarenotparalleltoacoordinateaxis.Equations of lines in general

TheshorthandA(2,0),for Considerfirstthelinewhichcutsthex-axisatthepointA(2,0)andtheexample,meansthatthe

y-axisat

the

point

B(0,

1).

This

line

is

shown

in

Figure

1.3.

pointlabelledAhascoordinates(2,0).

Figure 1.3 P liesonthe linethroughAandBInthefollowingdiscussion, ThepointP,withcoordinates(x,y), ischosento lieonthis line. Whatx > 0 and y > 1. algebraicconditiononxandy representsthefactthat(x,y) liesonthe

line?LetQbethepointonthesameverticallineasP and on the same horizontal lineasB seeFigure1.3. Thenthestraightnessofthe line

Similartriangleshavethe ABP meansthatthetrianglesABOandBPQaresimilar toeachother,sameshape. sothattheratiosof lengthsofcorrespondingsidesareequal. Forexample,

wehaveOB QP

= .AO BQ This leadstothealgebraicconditionwhichweseek. Thehorizontal lengthAOandvertical lengthOBarefound,fromthecoordinatesofthepointsA(2,0),B(0,1)andO(0,0),tobe

AO= 0 (2)=2 and OB= 1 0 = 1.ThepointQhasthesamex-coordinateasP andthesamey-coordinateasB,so itscoordinatesare(x,1). Itfollowsthat

BQ =x0 = x and QP =y1.Theconditionabovefortheratiosofcorrespondingsidesthereforeyieldstheequation

1 y1= .

2 xAftermultiplyingthroughbyxand then making ythesubjectoftheequation,thisbecomes

Infact,thisequationholdsfor y=allpoints(x,y)ontheline.

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x+ 1,whichistheequationofthelineshown inFigure1.3. Asacheck,thepairofvaluesx=2,y=0satisfiesthisequation(thepoint(2,0) liesontheline),asdoesthepairofvaluesx= 0, y=1(thepoint(0,1) liesonthe line).

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SECTION 1 LINES

It isworthanalysingthemanner inwhichthisequationwasderived,sincetheapproachcanbeappliedmoregenerally. ReferringbackoncemoretoFigure1.3,whatwedidwastoequatetheratiosOB/AOandQP/BQ,workingfromapairofsimilartriangles. Inthefirstoftheseratios,OB istheverticalchangefromAtoB,whileAO isthehorizontalchangefromAtoB. Thesecondratiosimilarly featurestheverticalchangeQP andhorizontalchangeBQ betweenthetwopointsB andP. Infact,whateverpairofpoints ischosenonthe line,

theratioofverticaltohorizontalchangebetweenthem isthesame.Thisratio isacharacteristicpropertyofthe line itself,called itsslope orgradient.Wenowconsiderthisideamoregenerally. SupposethatA(x1, y1) and B(x2, y2)areanytwopointsonagiven line,asshowninFigure1.4.

Figure 1.4 Riseandrun

xStartingfromA, we reach Bbymovinghorizontally(inthex-direction)by

2

x1 alongAC,andvertically(inthey-direction)byy2

y1 alongCB .

Wecalltheverticalchangey2 y1 therise inmovingfromAtoB,whilethehorizontalchangex2 x1 iscalledthecorrespondingrun. As drawn in Figure1.4,boththeriseandtherunfromAtoB arepositivequantities,sinceB istotherightofandaboveA. If,alternatively,B werebelowA(sothaty2 < y1),thentherisefromAtoB wouldbenegative. Infact,therise inthiscasewouldactuallydescribeafall. IfB weretothe leftofA(sothatx2 < x1),thentherunfromAtoB wouldbenegative.Aswasremarkedupon inaspecificcaseearlier,thestraightnessofthe lineinFigure1.4canbedescribedbytheobservationthat,howeverthetwopointsAandB arechosenonthe line,theratiooftherisetotherunfromAtoB isthesame. Thisratio iscalledtheslope orgradient ofthe line.

Rise, run and slope

Foranypairofpoints,A(x1, y1) and B(x2, y2),onagiven line,therise fromAtoB isy2 y1 andtherun fromAtoB isx2 x1. If therun isnotzero,thenthequantityriserun(calledtheslope ofthe line) is independentofthepairofpointschosen.

TheslopeofthelineinFigure1.3is 1 .2Notetheuseofsubscriptnotationhere,withx1 forx-coordinateoffirstpoint,x2 forx-coordinateofsecondpoint,andsoon.

Inthesedefinitions,theorderofthepointsissignificant.TherisefromB toAhastheoppositesigntotherisefromAtoB.

Thisfollowsfromthefactthat

all

right-angled

triangles

ABC,asinFigure1.4,whichcanbedrawnwithA,B onthe line,aresimilartooneanother.Thedefinitionofslopecanbesummedupinthephraseslopeequalsriseoverrun.

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Activity 1.2 Two special cases

(a) Thestatementaboveexcludesthecasewheretherunbetweenthetwopoints iszero(as itmust,toavoidtheundefinedoperationof divisionbyzero). Whattypeofstraight lineleadstoazerorunbetweentwopointsonthe line?

(b)

Whattype

of

line

has

zero

slope?

Solutionsaregivenonpage53.

Asyousaw inthesolutiontoActivity1.2(a),everylineoftheformx=d,whered isaconstant,hasinfinite slope.Wefoundtheequationforthe lineshowninFigure1.3byequatingtwoexpressionsfortheslopeoftheline: one intermsofthecoordinatesofthe

1twoknownpointsAandB,whichgavethenumericalvalue2

fortheslope,andtheother intermsofB andafurthergeneralpointP whosecoordinates(x,y)wereunspecified. Thisproceduregeneralisesasfollows.IfA(x1, y1) and B(x2, y2)areanytwopointswithknowncoordinatesonagivenline,thentheslopemofthe line isgivenby

risefromAtoB y2 y1m=

runfromAtoB = x2 x1 , (1.1)as illustrated inFigure1.5(a).

Figure 1.5 Riseandrun: particularandgeneralcasesNowletP(x,y)beageneralpointonthe line,asshown inFigure1.5(b).Theslope isexpressibleonceagainas

risefromAtoP yy1m= = . (1.2)runfromAtoP xx1

Equation(1.1)permitscalculationofthevalueofm,givenanytwopointsonthe line. Equation(1.2)thenprovidesanalgebraicconnectionbetweenthevariablesxandy,whichembodiestheconditionthatthepoint(x,y)liesonthe line. Equation(1.2)canberearrangedasfollows:

yy1 =m(xx1); that is, y=mx+ (y1 mx1).

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SECTION 1 LINES

Herem,x1 andy1 areconstants. Theequationofa linethereforehastheform

y=mx+c,wheremandcareconstantsforanyparticular line.Thesymbolmstandsfortheslope ofthe line. To interpretthesymbolc,notethatthepoint(0, c) liesonboththey-axisandonthe liney=mx+c. Hencecisthevalueofy wherethe liney=mx+ccutsthey-axis,whichiscalledthey-intercept oftheline. Similarly,thevalueofxwherethelinecutsthex-axisiscalledthex-intercept ofthe line. Theseinterceptsare illustrated inFigure1.6.

Figure 1.6 InterceptsTheslopemofa linegivesan indicationofhowsteeptheline is.Ifm iszero,thenthe line ishorizontal(paralleltothex-axis).Ifm ispositive,thenthelinerisesfrom lefttoright. Thesteepnessoftheline increasesforgreatervaluesofm. This is illustrated inFigure1.7(a)for linesthroughtheorigin, forwhichc=0andtheequationofthelinebecomesy=mx.Ifm isnegative,thenthelinefallsfrom lefttoright. The largerthemagnitude ofm,thesteeperisthecorresponding line(seeFigure1.7(b)).

Thiscoversallcasesexceptthatoflinesparalleltothey-axis,asconsidered inActivity1.2(a).

ThelineshowninFigure1.3hasy-intercept1andx-intercept2.

Themagnitude ofanumbera isaifa0,andaifa < 0. Forexample,themagnitudesof2and2areboth2.

Figure 1.7 Steepnessandslope

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Activity 1.3 Sketching lines a reminder

Tosketchaliney=mx +c, it issufficienttofindtwopointsontheline.Itisusuallyconvenienttochooseonepointtobe(0, c). Sketchthe linewhichcorrespondstoeachofthefollowingequations.

1(a) y=3

x (b) y=2x (c) y= 2x 1 (d) y=3x + 1 Solutionsaregivenonpage53.Activity1.3concernedlineswhoseequationsweregiven. It isoftenthecasethatweneedtofindtheequationofa line,startingfromotherinformationabout it. Thenextexampleshowshowtotacklesuchproblems,usingthe informationabouttheequationsof linesobtainedsofar,whichissummarisedbelow.

The equation of a line

Anylinewhichisnotparalleltothey-axishasanequationoftheformy=mx +c,

wheremistheslopeofthe lineandcis itsy-intercept. Theequationcanalsobeexpressedas

yy1 =m(x x1),Notethattheformulaforthe where(x1, y1)isanyonepointonthe line. Theslopemcanbeslopecanequallywellbe calculatedfromtheformulawrittenas y2 y1

m= ,y1 y2 x2 x1m= ,x1 x2

where(x1, y1) and (x2, y2)aretwopointsonthe line.showingthattheorder inwhichthetwopointsaretakendoesnotaffecttheoutcome,providedthattheorderisthesame inthedenominatorasinthe Example 1.1 Finding equations of linesnumerator.

(a) Findtheequationofthe linewhichhasslope3andpassesthroughthepoint(1, 2).

(b) Findtheequationofthe linewhichpassesthroughthepoints(1, 5)and(4, 7).

Solution

(a) Sinceboththeslopeofthe lineandonepointon itaregiven,wemayapplytheequationyy1 =m(x x1) in this case, with m= 3 and (x1, y1) = (1, 2). Thisgives

y(2)=3(x 1).Onmultiplyingoutthebracketsandthensubtracting2frombothsides,thisbecomes

y= 3x 5.Asacheck,theequation isindeedsatisfiedbythepairofvaluesx= 1 andy=2(as itmustbe ifthepoint(1, 2) istolieonthe lineasspecified).

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SECTION 1 LINES

(b) Heretheslopeofthe line isnotgiveninitially,butitcanbecalculatedfromtheriseandrun involved inmovingbetweenthegivenpoints.From (1, 5)to(4, 7),therun is41 = 3 and the rise is 75 = 12. Hencetheslopemisgivenby

rise 12m= = =4.

run 3Nowwecanproceedas inpart(a),applyingtheequationyy1 =m(x x1) with m=4 and either (x1, y1) = (1, 5)or (x1, y1) = (4, 7). Takingthefirstpossibility,wehave Youmayliketocheckthat

thesecondpossibilityleadstoy5 = 4(x 1).

thesameequationfortheline.Onmultiplyingoutthebracketsandthenadding5tobothsides,thisbecomes

y=4x + 9.Asacheck,theequation is indeedsatisfiedbythepairofvaluesx= 1 andy=5(as itmustbe ifthepoint(1, 5) isto lieonthe lineasspecified). Similarly,thepoint(4, 7)doeslieonthelineasrequired,sincetheequation issatisfiedbythepairofvaluesx= 4 and y=

7.

Activity 1.4 Finding equations of lines

(a) Findtheequationofthe linewhichhasslope2andpassesthroughthepoint(5, 3).

(b) Findtheequationofthe linewhichhasx-intercept3andy-intercept6(thatis,the linewhichpassesthroughthepoints(3, 0)and(0, 6)).


Parallel lines

Notethattwo(distinct) linesareparallelwhenevertheyhavethesameslope,sincetheyarethen equallysteep. Forexample,the liney= 3x + 2 isparalleltotheliney= 3x. The +2 inthefirstequation indicatesthatthefirst lineisaverticaldistance2unitsabovethesecond line. Similarly,the liney= 3x 1 isalsoparalleltotheliney= 3x,butaverticaldistance1unitbelow it. ThesethreelinesareshowninFigure1.8. Differentscaleshavebeen

usedfortheaxesinFigure1.8 inorderthatthefigureisnottoolarge. Thefactthatthelinesareparallelisnotaffected.

Figure 1.8 Parallellines

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Perpendicular lines

Weseenextthatthereisasimpleconnectionbetweentheslopesoftwolineswhichareperpendicular(atrightanglestoeachother). Firsttrythefollowingactivity.

Activity 1.5 Perpendicularity in a particular case

PlotthepointsA(2,3),B(1,1)andC(4,2),usingthesamescaleforeachaxis,andjoinAB andAC. Observethatthetwo linesegmentsAB andAC appeartobeperpendicular,andthencalculatetheirslopes.Asolution isgivenonpage54.

NowconsiderthegeneralsituationdepictedinFigure1.9,inwhichthelinesegmentAC hasbeenconstructedbyrotatinga linesegmentAB through

Congruenttriangleshavethe arightangleaboutA,sothatthetwotrianglesmarkedarecongruent.sameshapeandsize.

Thephraseequalinmagnitudemeans equalexceptforthepossibilitythatthey may be of opposite sign.

NotethattheparticularslopesfoundinActivity1.5satisfythisrelationship.

Alinewithslope0ishorizontal;a linewithinfiniteslopeisvertical.

Figure 1.9 CongruenttrianglesBecausethesetrianglesarecongruent,therisefromAtoB isequalinmagnitudetotherunfromAtoC,andtherunfromAtoB isequal inmagnitudetotherisefromAtoC. However,iftherisefromAtoB ispositive(as inFigure1.9),thentherunfromAtoC isnegative,andviceversa,whereastherunfromAtoB andtherisefromAtoC alwayshavethesamesign. Hencewehave

slopeofAB= risefromAtoBrunfromAtoB =

runfromAtoCrise fromAtoC =

1slopeofAC.

Anotherwayofexpressingthis istosaythat(slopeofAB)(slopeofAC) = 1.

Thiscalculation

depends

on

the

slope

of

AC

not

being

0or

infinite,

since

otherwisethedivision involvedwouldnotmakesense.Ingeneral, iftwo linesareperpendicular(butnotparalleltotheaxes),thentheproductoftheirslopesis1,andviceversa.

Perpendicularity condition

Iftwolinesareperpendicular,theneither theproductoftheirslopesis1or onehasslope0andtheotherhas infiniteslope.

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SECTION 1 LINES

Activity 1.6 Applying the perpendicularity condition

(a) ThepointAhascoordinates(1,4)andthepointB hascoordinates(5,2). FindtheslopeofthelineAB.

(b) What istheslopeofeachlineperpendiculartoAB?(c) Findtheequationofthe linewhichpassesthroughAand is

perpendiculartoAB.Solutionsaregivenonpage54.

Inter section of two lines

Youhaveseenhowanylinemayberepresentedbyanequationoftheformy=mx+cor(ifparalleltothey-axis)x=d, where m,canddareconstants. Itwaspointedoutthattwolineswhichhavethesameslopemareparalleltoeachother,andhenceneverintersect. Ontheotherhand, Twocurvesintersect, or meet,anytwolineswithdifferentslopesmust intersectatexactlyonepoint. iftheyhaveapointinThismaybevisible ifthetwo linesareplottedonagraph,eitherbyhand common.orbycomputer,andfromsuchaplotthepositionofthe intersectionpointcanbeestimated.The intersectionpointcanalsobefoundalgebraically. Itliesonbothofthe lines,whichmeansthatitscoordinates(x,y)satisfytheequationsofboth lines. Thevaluesofthesecoordinatescanthereforebefoundbysolvingthetwoequationssimultaneously. This isillustratedbelow.

Example 1.2 Finding points of intersection

Findthepointatwhichthe liney= 2x+3meetsthe liney=4x+ 1. Solution

Thecoordinates(x,y)ofthepointofintersectionAsatisfybothequations,y= 2x+ 3 and y=4x+ 1.

Hencethex-coordinateofAmustsatisfytheequation Thisapproachtosolving2x+ 3 = 4x+ 1; that is, 6x=2, theseequationsisequivalent

tosubstitutingtheexpressionwhosesolution isx=1

3. The y-coordinateofA isthenfoundby

substitutingthisvalueforx intoeitherofthetwooriginalequations.For

example,

putting

x

=

1

3 intothe

equation

y= 2x

+ 3

gives

fory fromthefirstequationinthesecondequation,asdiscussed inChapterA0.

y= 2 ( 13

) + 3 = 73

. Hencethepointof intersectionofthetwolineshascoordinates( 71

3,

3).

Asacheck,thesecoordinatesalsosatisfythesecondequation,y=4x+ 1, since we have 4( 1

3) + 1 = 7

3.

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Activity 1.7 Finding points of inter section

(a) Findthepointatwhichthe liney= 5x 7 intersectsthe liney=3x +1,bysolvingthetwoequationssimultaneously.

(b) Showthat it isnotpossibletofindthepointatwhichthe liney= 2x +3meetsthe liney= 2x 3,andexplainwhythis isso.


1.2 Some applications

Wenow lookbrieflyathowthemathematics inSubsection1.1canbeappliedintheprocessofmodellingrealproblems. Thetopicofmathematicalmodellingwasconsideredfirst inChapterA1,Section7,wherea looseframeworktodescribe itwas introduced. Thisframeworkencompassedfivekeystages,andinthissubsectionweareconcernedwithaspectsofthethreemiddlestages,namely, Createthemodel, Dothemathematicsand Interprettheresults.

modelCreate Thestageofcreatingthemodelinvolveschoosingvariables,stating

assumptionsandformulatingmathematicalrelationshipsbetweenthechosenvariables. InChapterA1therelationships involvedwererecurrence

Thevariablesaresaidtobe systems. Hereweare interestedinthepossibilityofformulating linearlinearly related insuchacase. relationships,that is,relationshipswhichwouldberepresentedbystraight

lines ifplottedonagraph. Youhavealreadyseenthealgebraic formy=mx +cthatsucharelationshiptakes,whenthevariablesarexandy.Youneedtobeabletorecognisethelinearformalsowhenothersymbolsarechosenforthevariables. Itwilloftenbethecasethattheconstantsmandcrepresentsomethingsignificant inthesituationbeingmodelled.

Herethesymbolstandstake For example, if trepresentstimeinseconds,andsmetresisthepositionoftheplacesofxandy, anobjectrelativetosomefixedpoint,thenthe linearrelationshiprespectively. It isusualto s= 20t +100givesthepositionoftheobjectatanytime,as indicated indenotetimebyt. thefollowingtable. Herem= 20 and c=100.

t 0 1 2 3 . . . s 100 120 140 160 . . .

Inthecontextofmotion, Theobjectchangespositionatasteadyvelocityof20ms1 (metrespersteadyisoftenusedtomean second),and isatpositions=100whent= 0. constant.

Sometimeswithinamodellingsituationtherearetwo linearrelationshipsof interestbetweenthesamepairofvariables. Forexample,twoobjectsinsteadymotionoverthesameperiod leadtotwodistancetimeequationsthatrepresentstraight lines. Itmightthenbeamatterof interesttodeterminewhere, ifatall,theobjectswillmeeteachother.Thiscorrespondstofindingthepointof intersectionoftwostraight lines,eitherbydrawingtheirgraphsorbyadoptingthealgebraicapproachoutlinedattheendofSubsection1.1. ThiscorrespondstotheDothemathematicsstageofmodelling.Domathematics

16


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SECTION 1 LINES

Example 1.3 Coinciding hands

Thepositionsoftheminuteandhourhandsofatraditionalclockcoincideat12noon. Theynextcoincideatsometimeafter1pm(seeFigure1.10).Byfollowingthestepsbelow,findthistime.

Figure 1.10 Coincidinghands(a) Takettobethetime,measuredinminutes,after1pm. Take(>0)to

betheangle indegreesthroughwhichahandhasrotated(clockwise!),measuredfromthe12oclockposition. Assumethatthehandsmovesteadily,ratherthan indiscretejumps. Finda linearrelationshipbetweentandforeachoftheminutehandandthehourhand.

(b) Findthepointof intersectionofthetwolineswhoseequationswerefound inpart(a).

(c) By interpretingtheanswertopart(b)appropriately,estimatewhenthehourandminutehandsnextcoincideontheclock-faceafter12noon.

Solution

(a) Thereare360inarevolutionofeitherhand. Theminutehand360undergoesonerevolutionperhour,or60

= 6perminute. Hencetheslopeofthe linerepresentingthe linearrelationshipbetweentand inthe(t,)-plane is6. Also,sincet= 0 at 1 pm, we have = 0 at t= 0. Hencethemotionoftheminutehand isdescribedbythe linewithequation

= 6t,asshowninFigure1.11.Similarly,thehourhandundergoesonerevolutionevery12hours,or

3601260 = 12 perminute. At1pm(t=0),thehourhandhasreached=30,whichis

1

12 ofacompleterevolution. Itsequationthereforetakestheform

Unlessthereisgoodreason,ashere, itisusualtomeasureanticlockwiserotationsbypositiveangles.

= 12

t+c, where= 30 when t= 0.Inotherwords,theline= 1

2t+cpassesthroughthepoint

Figure 1.11Theline= 6t

(t,) = (0,30),sothatc=30. The linearrelationship forthehourhand istherefore

= 12

t+ 30.

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Thesymbolisreadasisapproximatelyequalto.Inthischapter,theapproximationsarecorrecttothenumberofsignificantfiguresgiven.

Interpretresults


1(b) Thetwolines,= 6tand=2

t +30,meetwhen16t= t +30; that is, t= 60 5.4545.2 11

(c) Since0.4545minutes is600.454527seconds,thetwohandsnextcoincide(after12noon)atabout5minutesand27secondspast1pm.

Thelast

part

of

Example

1.3

corresponds

to

the

Interpret

the

results

stageofmodelling.

Activity 1.8 Modelling the motion of trains

ApassengertrainsetsofffromEdinburghWaverleytowardsLondonKingsCross(600kmaway)at7am. Ittravelsatanaveragespeedof100kmperhour. Onthesameday,afreighttrainstartsat8.30amfromLondontoEdinburgh,travellingatanaveragespeedof60kmperhour. Inthisactivityyouareaskedtoestimatethetimeandpositionatwhichthetwotrainspassoneanother.Letdbethedistance inkilometresfromEdinburgh,andtthetime inhourssince7am. Aspartofthemodellingprocess,youshouldassumethatthetwotrainstravelsteadilyattheaveragespeedsgiven.(a) Writedownanequationwhichrelatesdandtforthepassengertrain.(b) Forthefreighttrain,whatarethevaluesoftanddat8.30am? Find

anequationwhichrelatesdandtforthefreighttrain. (Notethattheslopeofthe line involvedherewillbenegative,sinceduringthetrainsjourneyddecreaseswhiletincreases.)

(c) Bysketchingthetwo linesonagraph,estimatewhenandwherethetwotrainspassoneanother.

(d) Usingalgebra,estimatethetimeanddistancefromEdinburghatwhichthetwotrainspassoneanother.


Summar y of Section 1

Thissectionhasreviewedor introduced:

x

theslopeofastraight line,whichis riserunforanypairofpointsA(x1, y1) and B(x2, y2)ontheline,wheretherunfromAtoB is

2 x1 andtherise isy2 y1; theequationy=mx +c,whichrepresentsastraight line inthe(x, y)-planethathasslopemandy-interceptc;

theequationyy1 =m(x x1),whichrepresentsastraight line inthe(x, y)-planethathasslopemandpassesthroughthepoint(x1, y1);

thefactthatparallel lineshavethesameslope; theconditionthat iftwo linesareperpendicular,theneither the

productoftheirslopes is1or onehasslope0andtheotherhasinfiniteslope;

themethodforfindingthepointof intersectionoftwolinesbysolvingthe

equations

of

the

lines

simultaneously.

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SECTION 1 LINES

Exercises for Section 1

Exercise 1.1

(a) Writedowntheriseandrunfrom(3,2)to(1,10).(b) Calculatetheslopeofthelinewhichpassesthrough(3,2)and(1,10).(c) Findtheequationofthe linedescribed inpart(b).(d)

Find

the

x-intercept

and

y-intercept

of

the

line

described

in

part

(b).

Exercise 1.2

Findtheequationofthe linethatpassesthroughthepoints(2,3)and(1,2).Exercise 1.3

Findtheequationofthe lineperpendiculartothatinExercise1.2whichpassesthroughthepoint(2,3).Exercise 1.4

Findthe

point

of

intersection

of

the

two

lines

y= 5x

7 and y=x+11.

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2 Circles

If it isinconvenienttowatchthevideonow,thencontinuewiththetextandviewthevideolater.

TostudySubsection2.1youwillneedaDVDplayerandDVD00104.Thevideoband indicatesvariouswaysof seeingcirclesand illustratessomeoftheirproperties. IntheremainderofSubsection2.1,youwillseehowPythagorasTheorem leadstoalgebraicequationsforcircles.Subsection2.2demonstratesamethodforfindingtheparticularcirclethatpassesthroughthreespecifiedpoints. Subsection2.3showsyouhowtotellwhetheragivenalgebraicequationdescribesacircle,and ifso,whichcircle itdescribes. InSubsection2.4youwillseehowtolocateanypointsatwhichagiven linecutsagivencircle.

2.1 Circles and their equations

Justasyouhavean intuitiveideaofwhatthe straightnessofastraightline involves,youprobablyhaveagoodunderstandingofwhatthecircularityofacircleentails. Thecircle is, inasense,the mostperfectofgeometricalfiguresinaplane. Mathematicallyspeaking,thisperfectioncanbetieddownasfollows.Associatedwithanycirclethere isaspecialpointcalled itscentre. If the circle iseitherrotated throughany angleabout itscentre,orreflectedacrossany linethrough itscentre,thentheresultingfigureoccupiespreciselythesameposition intheplaneastheoriginalcircle. Onlyacirclepossessesthesefeatures.Youwillalsobefamiliarwiththeideathatallpointsonacircleareatthesamedistancefromitscentre,andthatthisdistanceiscalledtheradius ofthecircle. Thevideobandreferstothisandtootherpropertiesofcircles.

Activity 2.1 Properties of circles

Thevideobandstartswithan introductionthatshowsvariousoccurrencesofnearlycircularshapes inthenaturalandmanufacturedworlds,andcontraststhesewiththeidealcirclewhichhumanssee intheirminds.This is followedbyfourshortepisodesthataddressdifferentpropertiesofcircles.Comment

Thefourepisodesinthevideobandarerelatedtothefollowingproperties.[1]Points equidistant from a fixed pointThis leadstothebest-knownwayofconstructingacircle.

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SECTION 2 CIRCLES

[2]Three points fix a circleArising fromthisfact,thecentreofthecirclethatpassesthroughthreegivenpointscanbefoundbylocatingthe intersectionoftheperpendicular Thereismoreaboutbisectorsofthe linesegmentsthatjoinpairsofpoints. Theradiusofthe thisconstruction incircle isthedistancefromthecentretoanyofthethreegivenpoints. Subsection2.2.Inthespecialcase inwhichthethreegivenpoints lieona line,thisconstructionbreaksdown,sincetheperpendicularbisectorsareparalleltooneanother. Inthiscasewecanthinkofthe centreasbeing infinitelyfaraway. Inotherwords,astraight linecanberegardedasa circleof infiniteradius.[3]Angles subtended by a fixed chordTheanglemadeatapointonthecircumferenceofacircle,bydrawinglinestothepointfrombothendsofafixedchord(whichiscalledtheanglesubtended bythechordatthepoint), isthesamewhereveronthecircumferencethepoint ischosen,providedthat it istakenonthesamesideofthechord.Ifthechordisnotadiameter,then itsubtendsanacute angle(ananglebetween0and90)onthelongerarc,andanobtuse angle(ananglebetween90and180)ontheshorterarc(seeFigure2.1). Thesetwoanglesaddupto180. Ifthechordisadiameter,thenbothanglesareexactly90. Figure 2.1[4]Horizontal and vertical components SubtendedanglesTheuniformmovementofapointaroundacircleinaverticalplanecanbeseenashavingtwocomponents: avertical(ory-)component,showingthevariableheightofthepoint,andahorizontal(orx-)component,showingtheside-to-sidemotion. Thegraphofeachcomponentwithtimehasasimilar wavyshape. Youwillsee laterthatforacirclewithunitradius,thesecomponentsarebydefinitionthesine andcosine oftheanglethroughwhichthepointhastravelledaroundthecentreofthecircle.

Afundamentalresultneededforthealgebraicdescriptionofcircles isPythagoras Theorem. SupposethatABC isaright-angledtriangle,withtherightangleatC,asshown inFigure2.2. Thelongestsideofsuchatriangle,which isthesideoppositetherightangle, iscalledthehypotenuse. We denote by AB the lengthofthehypotenuse. Similarly,BCandAC representthe lengthsoftheothertwosidesofthetriangle.PythagorasTheoremstatesthat inaright-angledtriangle,thesquareofthehypotenuseisequaltothesumofthesquaresoftheothertwosides;that is,

AB2 =AC2 +BC2.Weareinterestedhereinwhatthetheoremtellsusaboutthedistancebetweentwopoints. InFigure2.2,the lengthAB isthedistancefromAtoB,andaccordingtoPythagorasTheorem,thisdistancecanbeexpressed intermsofthelengthsAC andBC. SupposethatAhascoordinates(x1, y1) and that B hascoordinates(x2, y2). ThesepointsareshowninFigure2.3,overleaf,togetherwiththeriseandrunfromAtoB.

Figure 2.2Right-angledtriangleABC

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Figure 2.4 CirclewithcentreO andradius2

Figure 2.3 RiseandrunfromAtoBHerethelinesegmentAB can be considered as the hypotenuse of a right-angledtrianglewhoseothertwosideshave lengthsequaltothemagnitudesoftheriseandtherun,respectively. ApplyingPythagorasTheoremtothisright-angledtrianglegives

AB2 = run2 + rise2.SincetheriseandrunfromAtoB aregivenbyrun=x2 x1 andrise=y2 y1,wecandeducethefollowingresult.

Distance between two points

ThedistanceAB betweentwopointsA(x1, y1) and B(x2, y2) is givenby

AB2 = (x2 x1)2 + (y2 y1)2.

Activity 2.2 Finding distances between pairs of points

Findthedistancebetweeneachofthefollowingpairsofpoints.(a) (3,2)and(7,5) (b) (1,4)and(3,2)Solutionsaregivenonpage55.

Theformulaaboveforthedistancebetweentwopoints isthekeytowritingdownanequationforacircle,sinceadefiningpropertyofacircleisthatallpointsareatthesamedistance(theradius)fromthecentre.Suppose,forexample,thatweseekanequationforthecircleshown inFigure2.4,whichhascentreattheoriginO andradius2. SupposethatP(x,y) isanypointonthecircumferenceofthiscircle. ThenthedistancefromOtoP isOP =2. Accordingtotheboxedformulaabove,wehave

2OP2 = (x0)2 + (y0)2 =x2 +y .Wededucethatx2 +y2 =4. Hencetheequationx2 +y2 = 4 describes preciselythecirclewithcentreattheoriginandradius2,sinceapoint liesonthecircleonly if itscoordinates(x,y)satisfythisequation.

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SECTION 2 CIRCLES

Thisargumentgeneralisestoanycircle. Thesquareofthedistancefromafixedpoint(a,b) to any point (x,y) intheplane isgivenby(xa)2 + (yb)2. If (x,y) liesonthecirclewhosecentreis(a,b) and whoseradius isr,thenthecoordinates(x,y) must satisfy the equation

(xa)2 + (yb)2 =r2.

The equation of a circle

Thecirclewithcentre(a,b)andradiusrhasequation Foranyparticularcircle,a,b(xa)2 + (yb)2 =r2. andr areconstants.

Activity 2.3 Writing down equations of circles

Write down the equation of the circle in the (x,y)-planewhichsatisfieseachofthefollowingspecifications.(a) Centreat(0,0),radius3.

(b) Centreat(5,7),radius 2.(c) Centreat(3,1),radius1.Solutionsaregivenonpage55.

Activity 2.4 Writing down the centre and radius

Writedownthecentreandradiusofthecirclespecifiedbyeachofthefollowingequations.(a) (x1)2 + (y2)2 = 25 (b) (x+ 1)2 + (y+ 2)2 = 49 (c) (x)2 + (y+)2 =2(d) x2 + (y 3 )2 = 7 Solutionsaregivenonpage55.

2.2 Finding the circle through three points

Aswas

pointed

out

during

the

Video

Band,

if

three

points

are

given

which

donotall lieonasinglestraight line,thenthere isacircle(andonlyone)thatpassesthroughallthreepoints. Youwillnowseehowthecentreandradius(andhencetheequation)ofthiscirclemaybefound,giventhethreepoints incoordinate form.ThebasisofthemethodwasreferredtointheCommentforActivity2.1,and is illustrated inFigure2.5overleaf.

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Theline segment ABcomprises that part of the line throughAandB which liesbetweenAandB,includingAandB themselves.

Figure 2.5 A,B andC defineacircleIffixedpointsA,B andC defineacircle,thenitscentreDmustbeequidistantfromthemall.NowthepointswhichareequidistantfromAandB makeupa linewhichcutsthe linesegmentAB halfwayalong itslength(atM, say) and is at rightanglestoAB. This line, shown as MD inFigure2.5, iscalledtheperpendicular bisector ofAB.HencethecentreDmust lieontheperpendicularbisectorofeachofthelinesegmentsAB,BCandAC. Takinganytwooftheseperpendicularbisectors(asdrawninFigure2.5),thecentreD isattheirpointofintersection. OnceDhasbeenlocated,theradius isthedistancefromDtoanyoftheoriginalthreepoints.Thisprescription involvesfindingtheperpendicularbisectorofa linesegmentbetweentwogivenpoints,whichpassesthroughthemidpointofthelinesegment. Wethereforeneedtostartbyobtainingthepositionofthemidpoint,whosecoordinatesaregivenbytherulebelow.

Midpoint rule

ThemidpointofthelinesegmentbetweentwopointsA(x1, y1) and B(x2, y2)hascoordinates

1 12

(x1 +x2),2(y1 +y2) ;that is,itsx-coordinate isthemeanoftheoriginalx-coordinates,anditsy-coordinate isthemeanoftheoriginaly-coordinates.

Themethoddescribedaboveisapplied inaspecificcase inthefollowingexample.

Example 2.1 Finding the circle through three points

FindthecentreandradiusofthecirclethatpassesthroughthethreepointsA(4,8),B(1,1)andC(3,3).

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SECTION 2 CIRCLES

Solution

LetM,N bethemidpointsofthelinesegmentsAB,BC,respectively,andletDbethecentreofthecircle(seeFigure2.6).

Figure 2.6 FindingthecentreThe linesegmentAB hasslope(18)(14)=3andmidpointM at

7

1 1 52

(4+1),2

(8+(1)) =2

, .2

ItsperpendicularbisectorMD thereforehasslope13

(usingthe 7

5perpendicularityconditionfromSubsection1.1)andpassesthrough2

, .2

Itsequation isy7 =1x5 x+13; that is, y=1 .

2 3 2 3 3

Similarly,the linesegmentBChasslope(3(1))(31)= 1 and2

midpointN at1 1

2(1+(3)),

2(1 + (3)) = (1, 2).

ItsperpendicularbisectorND thereforehasslope2andpassesthrough(1, 2). Itsequation is

y(2)=2(x(1)); that is, y=2x4.Thetwoperpendicularbisectors,y=1x+13 (M D) and y=2x4

3 3

(N D),intersectatD, where x+131 =2x4; that is, x=5.

3 3

BysubstitutingthisvalueofxintotheequationforN D,wefindthatthecorrespondingvalueofy isy=2(5)4=6,sothecentreDofthecircle isat(5, 6).The

radius

ris

the

distance

between

D

and

A

(or

B, or

C),

so

it

is

given

by

r2 = (4 (5))2 + (8 6)2 =85;

thustheradius isr= 859.22.Theresultingequationofthecircle is(x+ 5)2 + (y6)2 =85. Asacheck, Another(rough)check istothisequationshouldbesatisfiedbythecoordinatesofeachofA,B andC. sketchthepositionsoftheFor example, for B wehave fourpointsA,B,C andD, as

inFigure2.6,andtoverify(1+5)2 + (16)2 = 36 + 49 = 85. thatthecoordinatesfoundfor

D andthecalculatedradiuslookapproximatelycorrect.

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Activity 2.5 Finding the circle through three points

Byfollowingthestepsbelow,findthecentre,radiusandequationofthecirclethatpassesthroughthethreepointsO(0,0),A(4,2)andB(8,6).(a) ShowthattheperpendicularbisectorofOAhasequationy= 2x+ 5. (b) ShowthattheperpendicularbisectorofOBhasequation

x+25y=4 .3 3(c) Bysolvingsimultaneouslytheequationsfoundinparts(a)and(b),

findthecentreofthecircle.(d) Useyouranswertopart(c)tofindtheradiusofthecircle.(e) Writedowntheequationofthecircle.Solutionsaregivenonpage55.

2.3 Completing the square

InActivity2.3youwereaskedtowritedowntheequationsofcircleswithgivencentreandradius. Althoughastandardformforsuchequationshasbeengiven,theymayalsobeexpressedindifferentbutequivalentways.Forexample,theequationx2 +y2 =9,foracirclewithcentre(0,0)andradius3, isequivalentto2x2 + 2y2 =18, inwhicheachtermofthefirstequationhasbeenmultipliedby2. Whenevertheequationofacircle(or indeedofa line) ismultipliedthroughbyanon-zeromultiple inthisway,theresultingequation isanequivalentrepresentationofthesamecircle(orline).Similarly,wecouldrearrangetheequation

(x5)2 + (y7)2 =2 (foracirclewithcentre(5,7)andradius 2)bymultiplyingoutthetwobrackets,toobtain

x2 10x+ 25 + y2 14y+ 49 = 2; thatis,

x2 10x+y2 14y+ 72 = 0.Thisequationdescribesthesamecircle,but it isno longerpossibletoreadoffdirectlythecoordinatesofthecentreandthevalueoftheradius.However,anequationmaybe initiallyobtained insuchaform,whichraisesthequestionofhowyouwouldthenfindthecentreandradiusfromtheequation. Thisquestion isnowaddressed.Themethodrequireddependsuponatechniqueknownascompleting

Thisidentityappearedin the square. This isbasedonthealgebraic identityChapterA0,Section4,inthe

x2 + 2px+p2 = (x+p)2.form

(a+b)2 =a2 + 2ab+b2. Onsubtractingthep2 frombothsidesoftheequation,wehave2x2 + 2px= (x+p)2 p .

Theright-handside iscalledthecompleted-square form oftheleft-handside,sincethevariablexappearsontherightonlywithina

Youwillseeanotheruseof squaredterm. Atermofthissquaredform isexactlywhatweseekincompletingthesquarein orderto identifythex-coordinateofacirclescentre.Subsection4.1.26


27/60

SECTION 2 CIRCLES

Example 2.2 Finding the centre and radius

Verifythattheequationx2 + 4x+y2 5y3 = 0

describesacircle,andfind itscentreandradius.SolutionIfwecanfindnumbersa,bandr forwhichtheequationgivenisequivalentto(xa)2 + (yb)2 =r2,thenweshallknowthattheequationdoes indeeddescribeacircle,withradiusrandcentreat(a,b).First,considertheexpressionx2 + 4x. Comparingthisexpressionwiththeidentity

2x2 + 2px= (x+p)2 p ,wecanmatchx2 + 4xwiththe left-handsidebyputting2p= 4; that is, p=2. Puttingthisvaluealso intotheright-handside,weobtainthecompleted-squareform

x2 + 4x= (x+ 2)2 22 = (x+ 2)2 4.Nowproceedsimilarlyfortheexpressiony2 5y. With y inplaceofx, the generalform is

2y2 + 2py= (y+p)2 p .We can match y2 5ywiththeleft-handsidebyputting2p=5,that is,p=5 . Puttingthisvaluealso intotheright-handside,weobtainthe

2

completed-squareform2

)2 252

)2 (5y2 5y= (y52

)2 = (y5 .4

Itremainstosubstitutebothofthecompleted-squareforms intotheequationgiven,namely

x2 + 4x+y2 5y3 = 0.Intermsofthecompleted-squareforms,thisbecomes

2)2 25(x+ 2)2 4 + (y5

43 = 0.

Oncollectingthenumbertermsandrearranging,wehave53

2)2 =(x+ 2)2 + (y5

4.

5Hencetheequation is indeedthatofacircle,whichhascentre(2,2

) and 1radius 533.64.2

Activity 2.6 Finding the centre and radius

(a) Findthecompleted-squareformoftheexpressionx2 4x.(b) Findthecompleted-squareformoftheexpressiony2 6y.(c) Usingyouranswerstoparts(a)and(b),verifythattheequation

x2 4x+y2 6y12=0describesacircle,andfinditscentreandradius.


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2

Hence3x2 + 2y2 = 1 is not theequationofacircle.

Theremainderofthissubsectionwillnotbeassessed.


x

Notethatnotallequationsinvolvingx2 +y2,x,yandnumbertermswillnecessarilydescribecircles. Tobeacircle,theequationmustbeexpressibleas(xa)2 + (yb)2 equaltoapositive quantity,namelyr ,thesquareoftheradiusofthecircle. Anexamplewhich isnotofthispattern isx2 +y2 =1. Sincex2 isnotnegativeforanyvalueofx, and similarly fory2,there isnopoint intheplanewhichsatisfiesthecondition

2 +y2 =1. Anothernon-circleequation isx2 +y2 = 0. This is satisfied bythesinglepoint(x,y) = (0,0),andasinglepoint isnotusuallyregardedasacircle.Tohaveachanceofdescribingacircle,theequationmusthavethesame

2coefficientforx2 andy . Intheexamplesconsideredsofar,thiscoefficienthasbeenchosentobe1; ifthis isnotinitiallythecase,thenitcanbeachievedbydividingthroughbythecoefficientconcerned.Forthisreason,theformulawhichwederivedearlierforcompletingthesquare issufficientforthepurposeof identifyingthecentreandradiusofacirclegiven ingeneralform. Italsosufficestoderivetheformulaforsolutionofageneralquadraticequation,as isnowshown.Ageneralquadraticequation inthevariablexhastheform

ax2 +bx+c= 0,wherea,bandcareconstants. Weassumethata=0,sinceotherwisethereisnoquadraticterm. Ondividingthroughbya, we obtain

x2 + b x+ c = 0. (2.1)a a

Thetermsx2 + (b/a)xcanbematchedwiththe left-handsideofourprevious completingthesquareequation,

2x2 + 2px= (x+p)2 p .Thematching isachievedwithp=b/(2a),sowehave

b b2 b2x2 + x= x+ .a 2a 4a2

Thusequation(2.1) isequivalenttob2 b2 c

x+ + = 0; 2a 4a2 a

thatis,b2 b2 c b2 4ac

x+ = =4a2

.2a 4a2 a

Finally,wetakethesquarerootandthensubtractb/(2a) from both sides. This

leads

to

the

familiar

formula

b b2 4ac b b2 4ac

x= = .2a 4a2 2a

2.4 Inter sections of circles and lines

YousawinSubsection1.1that iftwonon-parallel linesarerepresentedbyequations,thenthe intersectionpointofthe linesmaybefoundbysolvingtheirequationssimultaneously. Thesame istrueforotherpairsofcurvesforwhichtheequationsareknown.

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SECTION 2 CIRCLES

Inparticular,wecanfindwhereagiven line intersectsagivencircle. Aswithparallel lines,theremayturnouttobeno intersectionpoint insomecases. Fora lineandacircletherearethreepossibilities,asillustrated inFigure2.7.

Figure 2.7 Three possible cases The linemayintersectthecircle intwodistinctpoints,asshown in(a).Alternatively, itmaytouchthecircleatasinglepointas in(b), inwhichcasethe line isatangent tothecircle. Thirdly,thelinemaynot intersectthecircleatall,asshown in(c).Whentheequationsofa lineandacirclearesolvedsimultaneously,thesolutionprocessleadstoaquadratic equation,andthethreecasesthatarise insolvingsuchanequation(tworealroots,onerootornorealroot)correspondtothethreepossibleoutcomesshowninFigure2.7.

Example 2.3 Finding where line and circle meet

Findanypointsatwhichthecircle(x+ 7)2 + (y2)2 =80intersectseachofthefollowing lines. Notethatthethree lines(a) y= 2x4(b) y= 2x given

are

all

parallel,

since

eachhasslope2.

(c) y= 2x6Solution

(a) Usingtheequationofthe line,y= 2x4,tosubstitutefory intheequationofthecircle,wehave

(x+ 7)2 + (2x42)2 = 80.Onmultiplyingoutthebracketsandsimplifying,weobtainthesuccessive(equivalent)equations

(x+ 7)2 + (2x6)2 = 80,x2 + 14x+ 49 + 4x2 24x+ 36 = 80,5x2 10x+ 5 = 0,x2 2x+ 1 = 0.

The lastequationfactorisesas(x1)2 =0,whichhasthesinglesolutionx=1. Onfeedingthisvaluebackintotheequationoftheline,wefindy= 2 14 = 2.Hencethe line istangenttothecircleatthepoint(1,2).

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SECTION 2 CIRCLES

Activity 2.7 Finding where line and circle meet

Findanypointsatwhichthecircle(x3)2 + (y+ 4)2 =53intersectstheliney=x+ 2. Asolution isgivenonpage56.

Ifthelinegivenisparalleltothex-axis,thenthecalculationbecomessimpler. Forexample,the liney=3meetsthecircle(x2)2 + (y+ 1)2 = 25 where

(x2)2 + (3 + 1)2 =25; that is, (x2)2 = 25 16=9.Hencewehavex2 = 3,sox= 5 or x=1. Thepointsof intersectionare(5,3)and(1,3),sincethe line involved isy= 3. Similarsimplificationoccurs ifthegivenline isparalleltothey-axis,butherethequadraticequationtosolvewillbe intermsofy ratherthanx.Youmightwonderwhetherthesolvingsimultaneousequationsapproach Theremainderofthiscanalsobeappliedtotheproblemofdeterminingwhere(ifatall)two subsectionwillnotbecircles intersect. Itcan: themethod is illustratedbrieflybelow. Consider assessed.thecaseofthetwocircles

(x3)2 + (y+ 4)2 = 53 and (x+ 2)2 + (y1)2 = 13.Aftermultiplyingoutthebracketsandsimplifying,theseequationsbecome,respectively,

x2 6x+y2 + 8y28=0,x2 + 4x+y2 2y8 = 0.

Onsubtractingthesecondequationfromthefirst,wehave10x

+ 10y

20

=

0;

that

is,

y=

x

+ 2,

which istheequationofastraightline. This linehasthepropertythatifA isapointlyingonthe lineandoneofthecircles,thenAalso liesontheothercircle. The intersectionpointsofthetwocirclesarethereforefoundbylocatingwherethelinemeetseitherofthecircles.YoushowedinActivity2.7thatthislinemeetsthefirstofthecirclesgivenatthetwopoints(4,2)and(1,3). Hencethesearethetwointersectionpointsofthecircles.Ingeneral,aswithalineandacircle,twocirclesmaymeetattwopoints,onepointornopoints. Thesethreecasesare illustrated inFigure2.9. Ineachcase,thelinewhich isgeneratedfromthecircles inthemannerdescribedaboveisdrawn in. It isalwaysperpendiculartothe linejoiningthecirclecentres. Wherethecirclesmeetatasinglepoint(case(b)),theline isacommontangenttothecircles.

Figure 2.9 Three possible cases 31


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Thissectionhasintroduced: theformulaAB2 = (x2 x1)2 + (y2 y1)2 givingthedistanceAB

betweentwopoints A(x1, y1) and B(x2, y2); theequation(xa)2 + (yb)2 =r2 todescribeacirclewith

centre(a,b)andradiusr;1 1 thecoordinates 2(x1 +x2),2(y1 +y2) forthemidpointofthe linesegmentbetweentwopoints(x1, y1) and (x2, y2);

themethodforfindingacirclethroughthreegivenpoints; thecompleted-squareformforx2 + 2px, namely

2x2 + 2px= (x+p)2 p ,anditsuse,wherenecessary,infindingthecentreandradiusofacirclewhoseequationisgiven;

themethodforfindingthepointsof intersectionofalineandacircle,bysolvingtheirequationssimultaneously.


Exercise 2.1

(a) Writedownanequationwhichdescribesthecirclewithcentre(2,3)andradius4.

(b) Writedownthecentreandradiusofthecirclewithequation(x+ 5)2 + (y4)2 = 17.

Exercise 2.2

FindthecentreandradiusofthecirclewhichpassesthroughthethreepointsA(2,2),B(1,0)andC(1,6). Writedowntheequationofthecircle.Exercise 2.3

(a) Findthecompleted-squareformoftheexpressionx2 + 14x.(b) Findthecompleted-squareformoftheexpressiony2 24y.(c) Usingyouranswerstoparts(a)and(b),verifythattheequation

x2 + 14x+y2 24y96=0describesacircle,andfinditscentreandradius.

Exercise 2.4

Findanypointsatwhichthecircle(x3)2 + (y+ 7)2 =65 intersectsthe3liney=2

x+3 .2

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3 Trigonometry

Trigonometry isconcernedwithvariousratiosof lengthsthatareassociatedwithangles. InSubsection3.1youwillseedefinitionsoftwooftheseratios,thesine andcosine,togetherwithsomeoftheirbasicproperties. Subsection3.2thenexplainshowthesineandcosinemaybeputtousewithintriangles,forthepurposeofdeducing lengthsfromanglesizesorviceversa.It isassumedthatyouhavecomeacrossmostofthetopics inthissectionpreviously. Theywillbecoveredquiterapidly.

3.1 Sine and cosine

ConsiderthecircleshowninFigure3.1. Thiscirclehasradius1andcentreattheorigin,O. Itisoftencalledtheunit circle.

Figure 3.1 TheunitcircleConsiderthepointP onthecircumferenceoftheunitcircle,placedsothattheanglebetweenthepositivex-axis and the line segment OP, measured anticlockwise,is, where ispositive.SupposenowthatP hascoordinates(x,y). Thisisthesameassayingthatthevertical linethroughP meetsthex-axisat(x,0),andthehorizontallinethroughP meetsthey-axisat(0, y),asmarked. Thenthecosine andsine oftheanglearedefinedby

cos=x, sin=y.Theanglemaybemeasuredeither indegrees (360percompleterevolution)or inradians (2radianspercompleterevolution). Weshalluseradiansforthemoment,butswitchtodegreesinthecontextoftriangles inSubsection3.2.Asanexampleofapplyingthedefinitionsofcosineandsine,supposethat

If isnegative,thenP isplacedbyrotatingclockwisefromthepositivex-axisthroughtheangle.

Thesedefinitionsapplytoallvalues of negative,zeroandpositive.

. Then P hascoordinates(0,1),socos( )=0andsin( ) = 1. = 12 12 12

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Figure 3.2Signsinquadrants

These identitiesarevalidforallvaluesof.

Activity 3.1 Values of cos and sin

ByplacingP atappropriatepointsontheunitcircle,findthevaluesofthefollowing.(a) cos0andsin0 (b) cos(1) and sin(1) (c) cosandsin

2 2


Thevaluesofsineandcosinemaybepositive,negativeorzero,dependingwhereonthecircleP lies. Thesignsofthesetrigonometricquantitieswithinthefourquadrants oftheplane(thefourpartsseparatedbytheaxes)areasshown inFigure3.2. Thesefollowdirectlyfromthesignsofthex- and y-coordinates inthesequadrants.Certainformulaswhichrelatethesinesandcosinesofanglesmaybederivedby lookingatthegeometryofpointsontheunitcircle. NotefirstthatthepositionofP isunaffected iftheangle is increasedordecreasedbyacompleterevolution. Insymbols,thismaybeexpressedas

cos(+ 2) = cos and sin(+ 2) = sin ,whichsaysthatthevaluesofbothcosineandsinerepeatthemselvesevery2radians.Considernowtheeffectofreflectingpointsonthecircleacrossthex-axis,asshowninFigure3.3(a). IfthepointP(x,y)correspondstotheangle,andthepointQ isthereflectionofP inthex-axis,thenQhascoordinates(x,y)andcorrespondstotheangle. Hencewehave

cos() = cos and sin() = sin.

Figure 3.3 Reflections

HerethepointQcorrespondstotheangle 1.2

Notethatwewritecos2 for(cos)2 andsin2 for(sin)2.

Similarly,thecaseofreflection inthey-axis(seeFigure3.3(b),wherethepointQcorrespondstotheangle) gives

cos() = cos and sin() = sin .Furtherusefulidentitiesareobtainedbyconsideringreflection inthelinex=y (seeFigure3.3(c)). Theseare

cos(1) = sin and sin(1) = cos .2 2

Anotherkeypropertyofsineandcosinefollows fromthefactthatP liesontheunitcircle,soitscoordinatessatisfytheequationx2 +y2 = 1. Hence,foranyvalueof, we have

cos2

+ sin2

= 1.34


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SECTION 3 TRIGONOMETRY

Activity 3.2 More values of cos and sin

(a) Byapplyingtheformulacos(1

2) = sin ,1showthatcos = sin 1 . Then use the formula 4 4

cos2

+ sin2 = 1 tofindthecommonvalueofcos(1

4)andsin(14).(b) Byconsideringthegeometryofthediagram inFigure3.4, inwhich

OPQisanequilateraltrianglebisectedbythex-axis,findthevalues Thesymbol isreadasofcos(1

6)andsin(16). triangle.

Figure 3.4 TriangleOPQ(c) Whatarethevaluesofcos(1

3)andsin(13)?Solutionsaregivenonpage56.

Thevaluesofthesineandcosineofanyangle(expressedineitherdegreesorradians)maybeobtainedrapidlyfromacalculator,butvaluesfortheparticularangles inActivities3.1and3.2cropupoftenenoughtomake itworthcommittingthemtomemory ifpossible. Ifthis istoodauntinga

prospect,thenthesevaluesImaginenowthatthepointP inFigure3.1rotatesatasteadyratearound

can be looked up in the thecircle. Thecorrespondingvaluesofcosandsin(thecoordinates Handbook.ofP)oscillateupanddownbetween1and1. Yousaw intheVideoBand inSubsection2.1(fromwhichastill isreproduced inFigure3.5)the You will see more about these wavygraphsthatresult. graphsinChapterA3.

Figure 3.5 Wavygraphs35


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Wenowdefineathirdtrigonometricratio,thetangent ofanangle. Itisdefinedby

sintan= ,

coswhichmakessenseonly ifcos=0. IntermsofFigure3.1,this isequivalenttotan=y/x(x=0),sotan isundefinedifthepointP liesonthey-axis(that is,for=

12

,

32

,

52

,...). Notethat ifa linemakesananglemeasuredanticlockwisefromthepositivex-axis,thentan istheslope ofthe line,asdefinedinSubsection1.1.Thevalue= 12, for which

thetangentisundefined,correspondstolinesofinfiniteslope. Activity 3.3 Values of tan

Findthevalueofeachofthefollowing.(a) tan0 (b) tan (c) tan(1

4) (d) tan(1

6) (e) tan( 1

3)


Itfollows fromthecorrespondingformulasforcosandsinthatsintan() = sin() = =tan

cos() cosand

sintan() = sin() = =tan.

cos() cosThetangent ispositivewherebothcosineandsinearepositiveorwheretheyarebothnegative. Figure3.6 indicateswhichofsin,cosandtanare

Figure 3.6 positiveineachquadrant.Signsinquadrants

Activity 3.4 Values of sin, cos and tan

Usetheformulasforsin,cosandtanandthesolutionstoActivities3.2and3.3tofindthevalueofeachofthefollowing.(a) sin(5

6) (b) cos( 5

6) (c) tan( 2

3) (d) tan(1

3)


Toconcludethissubsection,threefurthertrigonometricratios,thesecant,For example, cosecantandcotangentofanangle,aredefinedby

1 23= 2/ 3 = 3, 1 1 1sec 6

sec= , cosec= , cot= .sin14 2, tancoscosec = Thesearesometimesusedwhenwewishtoavoidreciprocals.1 1= 1/ 3 = 3.cot 3 3

3.2 Calculations with triangles

Figure3.7showsanamendedcopyofthediagram(Figure3.1)referredtowhendefiningcosineandsine. ThelinesegmentOP isextendedtosomepointA,andaperpendicularisdroppedfromAtothex-axisatB. Here it isassumedthat isacute,thatis,between0and/2 radians (90).

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Figure 3.7 DefiningAandBSinceOABandOPX aresimilar, it followsthat

OB OX xOA = OP = 1 =x= cos ,

andsimilarlythatAB/OA= sin ,whateverthesizeoftheright-angledtriangleOAB.NowconsiderOABon itsown(Figure3.8(a)). ThesideOAisthehypotenuseofthetriangle. ThesideOB isadjacent totheangle, and thesideAB isopposite totheangle.

Figure 3.8 Adjacent,opposite,hypotenuseThisgives,forananglewithinaright-angledtriangle,theformulas Theseformulasareoftenused

adjacent opposite opposite tointroducethecosineandcos= . sine.

hypotenuse, sin= hypotenuse, tan= adjacentNotethatbothofthedescriptionsadjacentand oppositearerelativetotheanglecurrentlyunderconsideration. Indeed,ifisthethirdangle inthetriangle,asshowninFigure3.8(b),then Theseareinstancesofgeneral

adjacent(to)= opposite(to) = sin , rulesfromSubsection3.1,cos= namely

hypotenuse hypotenusecos(1

2

21

) = sin ,and,similarly,sin= cos .

sin( ) = cos ,Theserelationshipsmeanthat, inanyright-angled triangle,thespecificationofeither(a)oneotherangleandthe lengthofanyoneside,or(b)the lengthsofanytwosidesissufficienttoenableustodetermineallanglesandside lengthsofthetriangle. Thisprocess isoftencalledsolving the triangle .

since= 12.

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Activity 3.5 Solving triangles

Findtheunmarkedanglesand lengthsofsides ineachofthetwotrianglesTheinversesineofanumber inFigure3.9below. Inpart(b)youwillneedtoapplythe inversesineorgivesananglewhosesineis inversecosinefacilityonyourcalculator. (Makesurethatyourcalculatorthatnumber(wherethisis issettowork indegrees.)possible),andsimilarlyforthe inversecosine.

Figure 3.9 TwotrianglesSolutionsaregivenonpage57.

Theremainderofthis Toconcludethissubsection,therefollowsabrief indicationofhowsubsectionwillnotbe trigonometrycanalsobeappliedtosolvetrianglessuchasthatshown inassessed. Figure3.10(a),whichdonot containarightangle. This illustratesa

standardconventionforlabellingtheanglesandside lengthsofatriangle,withlengthaoppositeA, length boppositeB,and lengthcoppositeC.

Infull,thesineruleissinA sinB sinC

= = .a b c

Figure 3.10 AlabellingconventionTherules forsolvingsuchtrianglesdependessentiallyuponbeingabletoviewthemasbeingmadeupofright-angledtriangles. Forexample,bydroppingtheperpendicular fromC toAB inABC, we divide this triangle intotwosmalleroneswithrightangles,ADC andCDB, as showninFigure3.10(b). Denotingthe lengthofCDbyh, we have h=bsinA(inADC) and h=asinB (inCDB). Itfollowsthat

sinA sinB= ,

a bwhichisknownasthesine rule. Giventhe lengthofonesideforanytrianglewhoseanglesareknown,thisrulepermitsustodeducethelengthsoftheremainingsides.

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Thissectionhasreviewedorintroduced: thedefinitionscos=x, sin =y, where (x,y)arethecoordinatesof

a point P ontheunitcirclesuchthattheanglefromthepositivex-axistoOP is;

thefollowingpropertiesofcosandsin:

cos(+ 2) = cos , sin(+ 2) = sin ,cos() = cos , sin() = sin,cos() = cos, sin() = sin ,cos(1

2) = sin , sin(1

2) = cos ,

cos2 + sin2 = 1; thedefinitiontan= sin /(cos),wherecos=0,togetherwiththe

propertiestan() = tan, tan() = tan;

thedefinitionssec= 1/(cos),cosec= 1/(sin) and

cot= 1/(tan),whereineachcasethedenominatorsarenon-zero; withinaright-angledtrianglehavingacuteangle,theformulasadjacent opposite opposite

cos= tan=hypotenuse, sin= hypotenuse, adjacent.


Exercise 3.1

(a) Replacebyintheformulascos() = cos, sin() = sin .

Hencederiveformulasforcos(+) intermsof cos,sin(+) in terms of sin .

(b) Usetheseformulastofindthevalueofeachofthefollowing.(i) sin(7

6) (ii) cos(7

6) (iii) tan(7

6)

Exercise 3.2

Findtheunmarkedanglesand lengthsofsidesineachofthetwotrianglesinFigure3.11below.

Figure 3.11 Twotriangles

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4 Parametric equations

Unlesstisexplicitlylimitedinsomeway,it istobeassumed in such equations thatttakesallrealvalues.NotethattheuseofparameterhereisdifferentfromthatinChapterA1.

InSections1and2,yousawhowequationswhichrelatethex- and y-coordinatesofapointprovideanalgebraicdescriptionofa lineoracircle. Inthissectionyouwillseeanalternativewayofdoingthis. Inplaceofthesingleequationforalineorcircleemployedpreviously,weusetwoequations,whichexpresseachofxandy intermsofafurthervariable,sayt,calledtheparameter fortheequations.Youcanthinkofthetwoequationsastellingyoutheposition(x,y) of a smallobjectattimet,astheobjectmovesalonga lineorcircle. Theapparentextracomplexityofintroducingathirdvariable,t, is offset by additional information: youknownotonlythepathalongwhichtheobjectmoves,butalso itspositiononthepathatanyparticulartime.We look inSubsection4.1atparametricequationsfor linesand inSubsection4.2atparametricequationsforcircles.

4.1 Parametric equations of lines

Considertheequationy= 2x+3,whichdescribesaparticularstraightline. Thissaysthatthey-coordinateofanypointonthe linemustequaltwicethex-coordinateofthatpoint,plus3. Anotherwayofspecifyingthisistosaythatallpointsonthe linehavecoordinatesoftheform(x,2x+ 3), where x isanyrealnumber. Wecouldequallywelluseanotherletterhereinplaceofx,forexample,t. Thenthelineconsistsofallpointsoftheform(t,2t+3),whichinturnsaysthatthelineismadeupofpoints(x,y)suchthat

x=t, y= 2t+ 3.Theadditionalvariabletwhichhasbeen introducediscalledaparameter,andthetwoequationsarecalledparametric equations.Theprocessofdescribingxandy intermsoft iscalledparametrisationoftheoriginal line.Theequationsx=t,y= 2t+3representjustonewayofparametrisingtheliney= 2x+3. Forexample, ifwereturnedtothecoordinate form(x,2x+3),andthenreplacedxby4t+1,wewouldarriveatanothervalidparametrisation:

x= 4t+ 1, y = 2(4t+ 1) + 3 = 8t+ 5.Whyshouldwewanttodothis? Onthefaceof it,suchpairsofequationsareamorecomplicateddescriptionofthe linethanthesingleequation,y= 2x+3,fromwhichwestarted. However,wehavealsopotentiallygained informationthatcouldbeuseful inamodellingcontext. Iftrepresentstime,andx= 4t+ 1, y= 8t+5representthecoordinatesofamovingobject,thenwecansaythatwhent=0theobjectisat(1,5),whent= 1

2 it isat(3,9),andsoon. Foreachchosentime,thereisa

correspondingpositiononthe line.Figure4.1illustratestheabovetwoparametrisationsoftheliney= 2x+ 3.

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SECTION 4 PARAMETRIC EQUATIONS

Figure 4.1 Twoparametrisationsofoneline

InSubsection

1.1,

you

saw

how

to

obtain

the

equation

of

aline

specified

byeitherof(a) theslopeofthe lineandonepointon it;(b)twopointsontheline.Ineachcase,anappropriateparametrisationofthe line isasfollows. Youmayliketocheckthat(a) Ifa line inthe(x, y)-planehasslopemandpassesthroughthepoint thesetwoparametrisations

work,bysubstitutingthem(x1, y1),then itcanbeparametrisedas intotheappropriateformof

x=t +x1, y =mt +y1. theequationofa line.Here(x, y) = (x1, y1) when t=0. Theparametertistherunfrom(x1, y1) to (x, y).

(b)Ifa line inthe(x, y)-planepassesthroughthetwopoints(x1, y1) and (x2, y2),then itcanbeparametrisedas

x=x1 +t(x2 x1), y =y1 +t(y2 y1).Here(x, y) = (x1, y1) when t= 0, and (x, y) = (x2, y2) when t= 1. The magnitudeoftheparametertgivesthedistancefrom(x1, y1) to (x, y)asaproportionofthedistancefrom(x1, y1) to (x2, y2).Thisparametrisation is illustrated inFigure4.2. Figure 4.2 Case(b)Notethatthemidpointofthe linesegmentfromA(x1, y1) to B(x2, y2)

1correspondstotheparametervaluet=2

(seeFigure4.3). Putting1t=2

intotheparametricequationsabovegivesthemidpointformulastated inSubsection2.2,namely,

1 1(x, y) = 2

(x1 +x2),2(y1 +y2) .

Writedownparametricequationsfor:(a) the linewhichhasslope3andpassesthroughthepoint(1, 2);

Figure 4.3 Midpoint(b) the linewhichpassesthroughthetwopoints(1, 5)and(4, 7).Solutionsaregivenonpage57.

Activity 4.1 Parametric equations for specified lines

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SECTION 4 PARAMETRIC EQUATIONS

Theterm5(t 5)2 nevertakesavalue lessthan0,and itachievesthe Nosquareofarealnumberisvalue 0 when t=5. Atthistime,thevalueofd2 is5. Hencetheclosest negative.approachdistance is 52.24(km),andthisoccurswhent=5(hours).

Activity 4.3 Closest approach

Twoaircraftflyalongstraight-linecoursesatthesameheightandatsteadyspeeds. WithreferencetoagivenCartesiancoordinatesystem,thepositionofAircraft1 is(500t + 10, 312t 12)andthepositionofAircraft2 is(496t + 22, 310t 23),both intermsoftimet. Here distanceismeasured inkilometresandtime inhours. What istheclosestthattheaircraftapproacheachother,andatwhattimedoesthisclosestapproachoccur?Asolution isgivenonpage57.Ifwewanttospecifyonlypart ofa line,thenthiscanbeachievedbyrestrictingtheparametertoonlycertainrealvalues. Forexample, itwaspointedoutearlierthatthe linejoiningA(x

1, y

1) and B(x

2, y

2) can be

parametrisedasx=x1 +t(x2 x1), y =y1 +t(y2 y1), Notethat iftisreplacedhere

by1t,thenthenewwhere(x, y) = (x1, y1) at t= 0 and (x, y) = (x2, y2) at t= 1. As they parametrisationdescribesthestand,theseequationsapplyforallrealnumberst,andrepresenta line samelinebuttraversedinthewhichextendsinfinitely farineitherdirection. Todescribejustthe line oppositedirection,witht= 0 segment fromAtoB,wecouldaddtheconditionthattshouldbeno less at(x2, y2) and t= 1 at than0andnogreaterthan1. Insymbols,thiscanbewrittenas0t1, (x1, y1).which is read as 0 less than or equal to t, and tlessthanorequalto1.Theequationsthenread

x=x1 +t(x2

x1), y =y1 +t(y2

y1) (0

t

1).

4.2 Parametric equations of circles

Inthecaseofcircles,wehaveareadyparametrisationsuggestedbythevideoanimationwhichwas illustrated inFigure3.5. Theequations

x= cos , y= sin provideaparametrisationoftheunitcircle,withtheangleastheparameter(seeFigure3.1). Theangleherecanbereplacedbyt(time),butequally itcouldbereplacedby2t, 3toranynon-zeromultipleoft.Theseeachdescribeamovingpointwhichtracesoutthesamecircularpath,buttheydiffer intherateatwhichthepointrotates. Tobemorespecific,allparametrisationsoftheform

x= cos(kt), y = sin(kt),wherek isanon-zeroconstant,representtheunitcircle. Fork=1(thatis,=t),pointsonthecirclearereachedatthetimes indicated inFigure4.4(a). Themotion is intheanticlockwisedirection,sincewehaveassumedthatangles increase inthisdirection. Fork=2(that is,= 2t),

1 1 1motiontakesplaceattwicetherate,reaching=4

att=8

,= at2

1 1t= 1,andsoon(seeFigure4.4(b)). Fork=2

(that is,= t),therate4 2

ofmotion isonlyahalfthatwith=t(seeFigure4.4(c)). Ineachanticlockwiserevolution, increasesby2(radians),andhencethetimetincreasesby2/k.

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Figure 4.4 ThreeratesofmotionAsshown,themotion isanticlockwise ifk ispositive. Ifk isnegative(forexample,k=1 and =t),thenthemotionisintheclockwisedirectionaroundthecircle.

Activity 4.4 Parametrisation for a given rate of motion

Findaparametrisationoftheunitcirclewhichcorrespondstomotionanticlockwiseattherateofonerevolutionperunittime.Asolution isgivenonpage57.

Aparametrisationsuchasx= cos t,y= sin tdiffers inan importantrespectfromtheparametrisationsseenearlierforstraight lines. Themotionrepresentedbytheparametrisationhererepeats itselfeveryrevolution(2radians),since

cos(t+ 2) = cos t, sin(t+ 2) = sin t.Henceanysinglepointonthecirclecorrespondsto infinitelymanyvaluesoft. Ifdesired,thisrepetitioncanbeavoidedbyplacingarestrictiononthevaluesoft. Forexample,theequations

x= cos t, y= sin t (0t2)representjustonerevolutionaroundtheunitcircle,startingandfinishingat(x,y) = (cos 0,sin0)=(1,0). Forotherratesofmotionaroundthecircle,therestrictionontwillneedtobeadjustedtoachievejustonerevolution. Forexample,theparametrisation

Eventhought= 0 and t= 2givethesamepoint, itisconventionaltoincludebothvaluesintherangefort.

x= cos(3t), y = sin(3t) (0 t 23 )representsacompleteturnaroundthecircleatthreetimesthepreviousrate,whichiscompletedinonethirdofthetime.Iflessthanthewholecircle istobedescribed(acirculararc),thenthiscanbeachievedbyspecifyinganevensmallerrangeofvaluesfort. For example,

x= cos t, y= sin t (0t)istheupperunitsemi-circle(seeFigure4.5(a)),whereas

x= cos t, y= sin t (

t

2)isthelowerunitsemi-circle(Figure4.5(b)).

Alternatively,therangefortforthe lowersemi-circlecouldbetakenast0.44


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5 Parametric equations by computer

Tostudythissectionyouwillneedaccesstoyourcomputer,togetherwithComputerBookA.Inthissectionyouare invitedtouseyourcomputertoplot linesandcircleswhicharedescribedbyparametricequations. Youwillseealsothatthesameapproachcanbeadoptedwithothercurves.


Thissectionhasintroduced: theapproachneededtoplot,bycomputer,a line,circleorothercurve

givenbyparametricequations.

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Summar y of Chapter A2

Inthischapteryouhavestudiedthealgebraicdescriptionsof linesandcircles. Thesedescriptionspermitthesolutionofvariousproblemsstatedinitially ingeometricratherthanalgebraic form. The linksbetweengeometryandalgebraareofthegreatestmathematical importance.These linksextendtothealgebraicdescriptionofmotion,whichisprovidedbyparametricequationsformovingobjects. Trigonometryprovidesusefulparametrisationsforcircularmotion.

Lear ning outcomes

Youhavebeenworkingtowardsthefollowing learningoutcomes.Terms to know and use

Rectangularor

Cartesian

coordinates,

axis,

origin,

rise

and

run,

slopeorgradientofaline, infiniteslope, intercept,centreandradiusofacircle,subtendedangle,distancebetweentwopoints,linesegment,perpendicularbisector,midpoint,completingthesquare,completed-squareform,unitcircle,cosine,sine,degree,radian,quadrant,tangent,adjacent,opposite,hypotenuse,solvingatriangle,parameter,parametrisation,parametricequations.

Symbols and notation to know and use

(x1, y1) for first point, (x2, y2)forsecondpoint,A,ABC,sin, cos , tan , cosec , sec , cot , sin2 , etc., 0 t1.

Mathematical skills

Obtaintheequationofaline,eitherfromitsslopeandonepointonit,orfromtwopointson it.

Writedowntheslopeofa linethat isperpendiculartoa linewithgivenslope.

Evaluatethedistancebetweentwopointswithgivencoordinates. Writedowntheequationofacirclewithgivencentreandradius. Writedownthecoordinatesofthemidpointofthe linesegment

betweentwogivenpoints. Findtheequationofthecirclewhichpassesthroughthreegivenpoints,notallonthesamestraight line. Completethesquareforagivenexpressionoftheformx2 + 2px. Identifyfeaturesofalineorcirclefrom itsequation. Findthepointof intersectionoftwogivennon-parallel lines,orany

pointsof intersectionofagivenlineandcircle,bysolvingthecorrespondingequationssimultaneously.

Usetrigonometricratiostosolveright-angledtriangles. Writedownparametricequationsforagiven lineorcircle,orpart

thereof. Fromgivenparametricequationsforaline,derivethecorrespondingnon-parametricequation.

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Mathcad skills

Plotacurvewhich isspecifiedbyparametricequations.Modelling skills

Identifythevariables inasituation,andfindsuitableequationstorelatethem.

Appreciatethat, intheprocessofmathematicalmodelling,doingthemathematicscomesaftercreatingthemodelandis followedbyinterpretingtheresults.

Ideas to be aware of

Curves(including linesandcircles)mayberepresentedbyalgebraicequations.

Theslopeofaline(thequantityriserun) isapropertywhichisindependentofthetwopointsonthe linechosentocalculatetheriseandrun.

Theequationofastraight line isdeterminedby itsslopeandone

pointon it,oralternativelybytwopointson it. Theequationofacircle isdeterminedbyitsradiusandthepositionofitscentre.

Threepoints(notonthesamestraight line)determineauniquecircle. Thepointswhichareequidistantfromtwogivenpoints,AandB,

formtheperpendicularbisectorofthe linesegmentAB. Sineandcosinecanbeseenasthecomponentsofsteadycircular

motion. Theslopeofaline isequaltothetangentoftheanglewhichitmakes

withthepositivex-axis. Anon-parametricequationdescribesapath,whereasparametricequationscandescribethepositionatanygiventimeofanobject

movingalongthatpath. Completingthesquarepermitstheminimumormaximumvalueofa

quadraticexpressiontobefound.

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Appendix: Modelling the Earth

Thisappendixcontainsacoupleof illustrationsofhowthemathematicswhichhasbeenintroducedinthischaptercanbeapplied intherealworld.Ineachcase,mathematicalmodelling is involvedintranslatinga realproblem intomathematicalterms,solving itandtheninterpretingthesolutiononcemore,though it isthemathematical ideaswhichwedwellmostuponhere.Thefirstexampleconcernsthebasisofamajorground-surveyingtechniquewhichpredatedthemoresophisticatedmethodsnowavailable.Currentgroundsurveysmakesubstantialuseofthesatellite-basedglobalpositioningsystemdescribedinthesecondexample.

Applying trigonometr y

Considertheproblemofmappingacountry. Abasicrequirementofamapisthat itshouldrepresenttoscale,withreasonableaccuracy,therelativedistancesandorientationsofallplacesofsignificance. Onewayofdoingthisistodividethecountryup intoanetworkoftriangles. Ifthe lengthsofallsidesofeachtrianglearemeasured,thenitsshapeisalsodetermined,sothatwecanrepresentitcorrectlytoscaleonamap.Drawbacksofthisapproacharethe labourofcarryingoutallofthemeasurements,theproblemsposedinplacesbyharshterrain,andthedifficulty,beforethecomingofphotographyandradarechomethods,ofmaking lengthmeasurementsonthisscaleaccurately. Thesedrawbackswereovercomeduringtheeighteenthandearlynineteenthcenturies,byuseofthetheodolite(an instrumentcapableofmeasuringanglesaccurately)combinedwithapplicationofthesinerule,

sinA sinB= .

a bIftheanglesofatriangleandoneofitssideshavebeenmeasured,thenthesineruleprovidesasimplewayinwhichtocalculatethelengthsofthetworemainingsides.Supposenowthat, inthenetworkoftriangleswhichcoversthecountry,alltheangleshavebeenaccuratelyrecorded,andthatthelengthofasinglebaseline,which isthesideofoneofthetriangles,hasalsobeenmeasured(theside labelled1 inFigureA.1).

Thematerial inthisappendixwillnotbeassessed.

Asasimplifyingassumption,thedeparturesfromflatnessthatoccuronthesurfaceoftheEarthareignoredhere,thoughitis importanttotaketheseintoaccountwhendrawinganaccuratemap.

The sine rule was introduced attheendofSection3.

Figure A.1 Networkoftriangles

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ThefirstnationalmaptobebasedupontriangulationwastheAtlas National de France.Theworktookplaceovertheyears17331789. ThismapisalsoknownastheCartedeCassini,becausetheprojectinvolvedfoursuccessivegenerationsoftheCassinifamily.


Thenthe lengthsofthesides labelled2and3canbefoundbyapplyingthesinerule. Oncethishasbeendone,wehavetwomoreknownlengthsonthediagram,sothatothertriangleswhich includethesesidescanbesolvedinthesamemanner,byfurtherapplicationsofthesinerule.Continuinginthisway,the lengthsofallsideswithinthenetworkcanbeobtained. Oncetheinitialtriangulationofaregioniscomplete,moredetailcanbeobtainedusingsuccessivelyfinernetworksofsmallertriangles.Thefirst localtriangulationsofthistypetookplace inFrance inthelateseventeenthcentury. ThefirstTrigonometricalSurvey(latercalledOrdnanceSurvey)oftheBritishIsleswascompleted in1873. Itreliedupontheaccuratemeasurement(made in1784)ofabaselineofabout5miles in lengthonHounslowHeath.

Global positioning

SinceJuly1995,theGlobalPositioningSystem(GPS)developedandrunbytheUnitedStatesDepartmentofDefensehasbeenfullyoperational.Thissystem isavailable forcivilianaswellasformilitarypurposes,andhasalreadyfoundwideapplicationsforbothcommercialand leisureuse.Itenablesuserstoestablishtheirposition inspace(ontheEarthssurfaceoraboveit)withconsiderableaccuracy. However,thesystem issetupsothatthefullestaccuracy isavailableonlytotheUSmilitary.Thesystemhasthreecomponents.(a)Twenty-foursatellitesaremaintained inhighorbitsabovetheEarth,

eachwithanorbitalperiodof12hours. Sixseparateplanarorbitsareused,withfoursatellites ineachorbit. Thisconfigurationensuresthat,withanunobstructedviewskywards,aminimumofsixsatellitesarealwaysvisiblefromanypointontheEarth.

(b)Severalgroundstationsaroundtheworldmonitortheprecisepositionsofthesatellites,andalsotransmit informationtothem. Thesestationsare linkedtothecentreoftheoperationatColoradoSprings.

(c) Areceiverwithbuilt-inprocessoraccompanieseachuserofthesystem.Receiversvarywidely insophistication: sometypesarehand-heldandrelativelycheap;othersarebuilt intoshipsoraircraft.

Thesystemworksasfollows. Theusersreceiverlocksontoradiosignalsfromseveralsatellites. Assumingthataclockinthereceiverisaccuratelysynchronisedwiththose inthesatellites,thetime lagbetweenthetransmissionandreceptionofeachsignalcanbemeasured. Sinceradiosignalstravelatthespeedof light(about300000kms1),thetimelagmeasurementscanbetranslated intodistances. Eachsatellitetransmitsanaccurate

record

of

its

own

position

at

each

moment,

relative

to

a

three-dimensionalCartesiancoordinatesystemwhich isfixedrelativetotheEarth.Thereceiverthenhasestimatesofthedistancestoseveralsatellitesofknownpositionataparticulartime,andcancalculate itsownpositionatthattimerelativetothesamecoordinatesystem. Thiscanthenbetranslated,ifdesired, intothelongitude, latitudeandaltitudeofthereceiver. Repeatedcalculationsofthistypecanalsotelltheuserwhattheirspeedanddirectionofmotionare.

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APPENDIX: MODELLING THE EARTH

Tounderstandhowthepositionofthereceivercanbederivedfromthecalculateddistancesfromthesatellites, letussimplifymattersbyconsideringthecorrespondingsituation intwospacedimensionsratherthantheoriginalthree.FigureA.2showsthepositionoftheuseratapointP relativetotwotransmittingdevices(correspondingtosatellites)atpointsAandB. The coordinatesofAandB areknown,relativetoafixedoriginatOandCartesian(x,y)-axes.

Figure A.2 Points A,B andPThepointP isattheintersectionoftwocircles,withcentresatAandatB. Thecoordinatesofthesecentresareknown,andsoaretheradii(thedistancesfromthetransmitters),sothat ineachcasewecanwritedownanequationforthecircle. FindingthepositionofP thereforeamountstofindingapointwherethetwocircles intersect. ThemathematicsneededtolocatetheseintersectionpointswasdiscussedattheendofSubsection2.4.Hencemeasurementsfromjusttwotransmittersshouldsufficetofixthepositionoftheuserinoursimplifiedtwo-dimensionalsituation. Thisgeneralises

in

a

straightforward

manner

to

three

space

dimensions,

with

thecirclesreplacedbyspheres. Measurementsfromthreesatellitesfixthepositionoftheuser. Theequations involvedlookmorecomplicatedthanthoseforcircles,buttheycanbesolvedsimultaneouslyforthe intersectionpoint inaverysimilarway. However,there isasnag!Itwasassumed,atthestartofdescribinghowthesystemworks,thattheclock inthereceiverwasaccuratelysynchronisedwiththose inthesatellites. Thesatellitesdoindeedincorporatehighlyaccurateclocks.Receivers,ontheotherhand,wouldbeprohibitivelyexpensiveiftheyhadclocksofcomparableaccuracy. Thereforetheirtimingdevices,althoughgoodbyeverydaystandards,arenotadequatelyaccurateforthepurposedescribedhere. Toappreciatewhat is involved,notethatwiththetransmittedsignalstravellingatthespeedof light,atimingerrorofamillisecondcorrespondstoadistanceerrorof300km!Luckily,there isasimplewayaroundthisproblem,which istouse data

from an additional satel lite. Toseehowthisworks intermsofourtwo-dimensionalsituation, lookatFigureA.3.

Infact,therewillbetwosuchintersectionpoints,butinpractice itwillbepossibletodiscountoneasgivinganabsurdresult.

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Figure A.3 EffectsoftimeerrorinreceiverAsbefore,therearetransmittersatknownpositionsAandB,butthere isalsoathirdtransmitteratpositionC. Ifthereceiversclockwereaccurate,thenthethreecalculateddistanceswouldbeequaltotheradiiofthreecircleswhichintersectatasinglepoint. Duetotimeerror inthereceiver,thisdoesnotoccur. Instead,thecalculateddistances leadtothecirclesshownsolid inFigureA.3,whichdonot intersectmutuallyatapoint.Thisshowsupthetimeerror inthereceiver. Sincethesecalculationsareclearly inaccurate,thethreedistancesarecalledpseudoranges fromthetransmitters.Thesavinggracehereisthatthetimeerror isthesameforeachofthe

Thesymbolsandarethe threetimemeasurements. Hencewemayseekavalueof,thetimeerror,Greekletters tauandrho, suchthatremovingadistance=c(wherec isthespeedof light) fromrespectively. eachpseudorangewillproducethreecircleswhichdo intersectinasingle

point(thebrokencircles inFigureA.3). This intersectionpoint isthepositionP oftheuser. Asaby-product,thevaluefoundfor isthe

Theuserthenhasa receiverstimeerror,whichcanbeusedtocorrectthetimeregisteredbycalculated

time

which

is

as

thereceiver.accurateasthatonthe

satellites! Hence ittakesthreetransmittersto locateaccuratelyintwospacedimensionsareceiverwithaslightly inaccurateclock. Inthreespacedimensions,asimilarapproachsucceedswithdatafromfoursatellites.

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Solutions to Activities

Solution 1.1

(a) Anypointonthe lineparalleltoand3unitstothe leftofthey-axishascoordinatesoftheform(3, y),sothesecoordinatessatisfytheconditionx=3. Thisisthereforetherequiredequationofthe line.

(b) Ingeneral,anylineparalleltothey-axishasanequationoftheformx=d, where d isaconstant. Thevalueofd isnegativefora linetothe leftofthey-axis,andpositivefora linetotheright.

(c)

Figure S.1

Solution 1.2

(a) Iftheruniszero,thenx1 =x2. One of the pointsisverticallyabovetheother,sotheline isparalleltothey-axis. Althoughtheslopeisnotthendefinedbytheexpressionriserun,wesaythatsuchalinehasinfinite slope.(AsyoufoundinActivity1.1,suchalinehasanequationoftheformx=d, where d isaconstant. Herex1 =x2 =d.)

(b) Iftheslopeiszero,thentherisey2 y1 mustbezero,soy2 =y1. Bothpointsareatthesamehorizontallevel,sotheline isparalleltothex-axis. (Aspointedoutearlier,suchalinehasanequationoftheformy=c, where cisacons

mst121 chapter a2

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