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Introductionandoverview
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Acknowledgements
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Abouttheauthors
DavidSadler isaretiredMathematicsteacher.Hetaughtfor36yearsatSydneyGrammarSchoolandwas HeadofMathematicsfor7years.HealsotaughtatUNSWforoneyear.HewasanHSCmarkerformanyyears andhasbeenapresenteratvariousconferencesandprofessionaldevelopmentcourses.Hehasastrongpassion forexcellenceinmathematicseducationandhaspreviouslyco-authoredseveralseniortextsforCambridge Universitypress.
DerekWard taughtMathematicsatSydneyGrammarSchoolfor31years,andwasMasterinChargeof ExaminationStatisticsfor20years.HehasanMScinAppliedMathematicsandaBScDipEd,bothfrom theUniversityofNSW,wherehewassubsequentlySeniorTutorforthreeyears.Hehasakeeninterestin mathematicseducationandhasco-authoredseveralothertextsforCambridgeUniversitypress,aswellasinternal publicationsforSydneyGrammarSchool.DerekhasanAMusAinFlute,andhasanextensivebackgroundin singing.HesingsinvariouschoirsandconductsachoirinNewZealand,wherehenowlives.
WilliamMcArthur hasbeenaMathematicsteacherfor15years,andhastaughtatSydneyGrammarSchool since2016.HehasaBScinMathematics,aBEd,andanMEd(AssessmentandEvaluation),allfromthe UniversityofNSW.HehaspresentedatHSCMathematicsrevisionseminars,andworkedwithpre-service MathematicsteachersattheUniversityofNotreDame.WilliamhasexperienceinHSCmarking,andhasbeen involvedwiththeNSWStage6Mathematicscurriculumreview.
1 ComplexNumbersI
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Chapterintroduction
Oneofthesignificantpropertiesoftherealnumbersisthatanyofthefourarithmeticoperationsofaddition, subtraction,multiplicationanddivisioncanbeappliedtoanypairofrealnumbers,withtheexceptionthat divisionbyzeroisundefined.Asaresult,everylinearequation
ax + b = 0where a 0 canbesolved.
Thesituationisnotsosatisfactorywhenquadraticequationsareconsidered.Therearesomequadratic equationsthatcanbesolved,butothers,like
x 2 + 2x + 3 = 0, havenorealsolution.Thisapparentinconsistency,thatsomequadraticshaveasolutionwhilstothersdonot, canberesolvedbytheintroductionofanewtypeofnumber,thecomplexnumber.
Butthereismoretocomplexnumbersthanjustsolvingquadraticequations.Inthischapterthereaderisshown anapplicationtogeometryand,laterinthecourse,complexnumberswillbeusedinthestudyofpolynomials. Thesenewnumbershavemanyapplicationsbeyondthiscourse,suchasinevaluatingcertainintegralsandin solvingproblemsinelectricalengineering.Complexnumbersalsoprovidelinksbetweenseeminglyunrelated quantitiesandareasofmathematics.Hereisastunningexample.Thefourmostsignificantrealnumbers encounteredsofarare0,1, e and π.Althoughtheproofisbeyondthescopeofthiscourse,thesefourare connectedinaremarkablysimpleequationinvolvingthespecialcomplexnumber i,namely
e iπ + 1 = 0.
1A TheArithmeticofComplexNumbers
IntroducingANewTypeofNumber
Theinvestigationisbegunbyexaminingtherootsofvariousquadraticequations.Forconvenienceinpresenting thenewwork,themethodofcompletingthesquareisusedexclusively.
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Firstconsideronlythosequadraticequationswithrationalsolutions,suchastheequation x2 4x 12 = 0. Completingthesquare:
(x 2)2 = 16
so x 2 = 4or 4 whichleadstothetworoots
α = 6and β = 2.
Notethat α + β = 4and αβ = 12.
Repeatingthisprocessforanumberofquadraticswithrationalsolutions,itsoonbecomesevidentthatif
ax2 + bx + c = 0hassolutions α and β then
α + β = b a and αβ = c a
Furtherinvestigationrevealsthattherearesomequadraticequationswhichdonothaverationalsolutions,suchas x2 4x 1 = 0.Completingthesquare:
(x 2)2 = 5.
Hereinliesaproblemsincethereisnorationalnumberwhichwhensquaredequals5.Thisproblemisovercome byintroducinganewtypeofnumber,inthiscasetheirrationalnumber √5whichhasthepropertythat
Assumingthatthenormalrulesofalgebraapplytothisnewnumber,itfollowsthat
5,so that5hastwosquareroots,namely √5and √5.Iftheintroductionofthisnewtypeofnumberisvalidthenthe solutionmayproceed.Thus
x 2 = √5or √5 whichleadstothetworoots
α = 2 + √5and β = 2 √5.
Notethat α + β = 4and αβ = 1.
Repeatingthisprocessforanumberofquadraticswithirrationalsolutions,itsoonbecomesevidentthatif ax2 + bx + c = 0hasirrationalroots α and β then
α + β = b a and αβ = c a
Sincethisisconsistentwiththequadraticequationswithrationalsolutions,itseemsthattheintroductionofsurds intothenumbersystemisvalid.IndeedsurdshavebeenusedsinceYear8andstudentswillbeproficientintheir use.
Yetfurtherinvestigationrevealsthattherearesomequadraticequationswhichhaveneitherrationalnorirrational solutions,suchas x2 4x + 5 = 0.Completingthesquareyields: (x 2)2 = 1.
Againthereisaproblemsincethereisnoknownnumberwhichwhensquaredequals 1.Justasbefore,this problemisovercomebyintroducinganewtypeofnumber.Inthiscasethesocalledimaginarynumber i is introducedwhichhasthepropertythat i2 = 1.Assumingthatthenormalrulesofalgebraapplytothisnew number,itfollowsthat( i)2 = i2 = 1,sothat 1hastwosquareroots,namely i and i .Iftheintroductionof thisnewtypeofnumberisvalidthenthesolutionmayproceed.
Thus
x 2 = i or i whichleadstothetworoots
α = 2 + i and β = 2 i
Notethat α + β = 4and αβ = (2 i)(2 + i)
= 22 i2 (differenceoftwosquares)
= 4 + 1(since i2 = 1) = 5.
Repeatingthisprocessforanumberofquadraticswithsolutionswhichinvolvetheimaginarynumber i,itsoon becomesevidentthatif ax2 + bx + c = 0hassolutions α and β then
α + β = b a and αβ = c a
Sincethisisconsistentwithallpreviouslyencounteredquadraticequations,itseemsreasonabletoincludethe imaginarynumber i inthenumbersystem.
ANewNumberinArithmetic
Theimaginarynumber i isnowformallyincludedintothesystemofnumbers.Ithasthespecialpropertythat i2 = 1.Thisnewnumber i willbetreatedasifitwereanalgebraicpronumeralwhenitiscombinedwithreal numbersusingthefourarithmeticoperationsofaddition,subtraction,multiplicationanddivision.
1Anewnumber
Thenewnumber i hasthespecialpropertythat i2 = 1.
Itmaybeusedlikeapronumeralwithrealnumbersinaddition,subtraction,multiplicationanddivision.
Itisinstructivetowriteoutthefirstfourpositivepowersof i.Theyare:
(bydefinition)
Writingoutthenextfourpowersof i,itisfoundthatthissequencerepeats.
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Itshouldbeclearfromthesecalculationsthatthesequencecontinuestocycle. Ingeneral,theresultcanbedeterminedfromtheremainderwhentheindexisdividedby4.
2Powersoftheimaginarynumber
Apowerof i maytakeonlyoneoffourpossiblevalues.If k isaninteger,thenthesevaluesare:
Example1
Simplify: i23 a i7 + i9 b Solution i23 = i4×5+3 = i a i7 + i9 = i + i = 0 b
ComplexNumbers
Since i hasbeenincludedinthenumbersystemandsinceitistobetreatedasapronumeral,thenumbersystem mustnowincludetherealnumbersplusnewquantitieslike
2i , 7i ,5 + 4i and √6 3i .
Thesetwhichincludesallsuchquantitiesaswellastherealnumbersisgiventhesymbol C.Eachquantityin C iscalleda complexnumber.Thus5,2i and √6 3i areallexamplesofcomplexnumbers.Inthefirstcase,5 isalsoarealnumber,andtherealnumbersformaspecialsubsetofthecomplexnumbers.Thenumber2i isan exampleofanotherspecialsubsetofthecomplexnumbers.Thissetconsistsofalltherealmultiplesof i ,which arecalled imaginarynumbers.Thus 7i isanotherexampleofanimaginarynumber.
3Twonewtypesofnumbers
Let a and b berealnumbers.
Complex Numbers:Numbersoftheform a + ib arecalled complexnumbers.Thesetofallcomplex numbersisgiventhesymbol C
Imaginary Numbers:Numbersoftheform ib,thatisthecomplexnumbersforwhich a = 0,arecalled imaginarynumbers
Againnotingthat i istreatedasapronumeral,theaddition,subtractionandmultiplicationofcomplexnumbers presentsnoproblem.
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Thefollowingexamplesofmultiplicationinvolvebinomialexpansionsandthepropertythat i2 = 1.
Example2
Simplify:
(4 + 3i)2 = 16 + 24i + 9i2 = 16 + 24i 9 = 7 + 24i c (2 + 5i)(2 5i) = 4 25i2 = 4 + 25 = 29 d
Thelastthreeexamplesabovedemonstratetheexpansionsof(x + iy)2,(x iy)2 and(x + iy)(x iy)forrealvalues of x and y .Notethatinthefinalexample,theresultisthesumoftwosquaresandisarealnumber.Thiswill alwaysbethecase,regardlessofthevaluesof x and y
4Thesumoftwosquares
Let x and y berealnumbers,then
(x + iy)(x iy) = x 2 + y 2 whichisalwaysarealnumber.
ComplexConjugates
Thelastresultissignificantandwillbeusedfrequently.Clearlythepairofnumbers x + iy and x iy have aspecialrelationship.Theyarecalled complexconjugates.Thusthecomplexconjugateof3 + 2i is3 2i. Similarlytheconjugateof7 5i is7 + 5i
Whentheconjugateisrequired,thecomplexnumberiswrittenwithabaraboveit.Thus:
5Complexconjugates
Let x and y berealnumbers,thenthetwocomplexnumbers x + iy and x iy arecalledcomplex conjugates. A Theconjugateof x + iy is x + iy = x iy
B Theconjugateof x iy is x iy = x + iy .
Division
Justlikerealnumbers,divisionbyzeroisundefined.Dividingacomplexnumberbyanyotherrealnumber presentsnoproblem.Aswithrationalnumbers,fractionsshouldbesimplifiedwhereverpossiblebycancelling outcommonfactors.
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Thereisapotentialproblemifonecomplexnumberisdividedbyanother,suchasin 2+i 3 i .Asitstands,itisnot clearthatthissortofquantityisevenallowedinthenewnumbersystem,sinceitisnotinthestandardform x + iy.
Theproblemisresolvedbytakingasimilarapproachtothatusedtodealwithsurdsinthedenominator.The processhereiscalled realisingthedenominator.Thusifthedivisorisanimaginarynumberthensimplymultiply thefractionby i/i,asinthefollowingtwoexamples.
Example3
Realisethedenominatorsof:
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Ifontheotherhandthedenominatorisacomplexnumberthenthemethodistomultiplytopandbottombyits conjugate,asdemonstratedhere.
Example4
Realisethedenominators:
6Realisingthedenominator
Therearetwocases.
A Ifthedenominatorisanimaginarynumber,multiplytopandbottomby i
B Ifthedenominatoriscomplex,multiplytopandbottombyitsconjugate.
Thereisnowsignificantevidencethatthecomplexnumbersformavalidnumbersystem.Ithasbeenseenonthe previouspagesthatthefourbasicarithmeticoperationsofaddition,subtraction,multiplicationanddivisionall behaveinasensibleway,consistentwithrealarithmetic.
AConventionforPronumerals
Itisoftennecessaryindevelopingthetheoryofcomplexnumberstoperformalgebraicmanipulationswith unknowncomplexnumbers.Inordertohelpdistinguishbetweenrealandcomplexvariables,theconventionthat willbeusedinthistextisthatthepronumerals x, y, a and b willrepresentrealnumbersandthepronumerals z and w willrepresentcomplexnumbers.Thusinastatementlike“Let z = x + iy”itisautomaticallyunderstood that x and y arerealwhilst z iscomplex.
RealandImaginaryParts
Giventhecomplexnumber z = x + iy,therealpartof z istherealnumber x,andtheimaginarypartof z isthereal number y.Itisconvenienttodefinetwonewfunctionsofthecomplexvariable z forthesetwoquantities.Thus
Re(z) = x and Im(z) = y fromwhichitisclearthat
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z = Re(z) + i Im(z).
Example5
DetermineRe(z2 iz)when z = 3 i
Solution
Expandingthequadraticin z first,
z2 iz = (3 i)2 i(3 i)
= 8 6i 3i 1
= 7 9i ,
soRe(z2 iz) = 7.
Iftwocomplexnumbers z and w areequal,byanalogywithsurds,itisnaturaltoexpectthatRe(z) = Re(w)and Im(z) = Im(w).Thisisinfactthecase.
7Equalityofcomplexnumbers
Iftwocomplexnumbers z and w areequalthen
Re(z) = Re(w)andIm(z) = Im(w).
Proof Let z = x + iy and w = a + ib,andsupposethat z = w.Then
x + iy = a + ib .
Rearranging i(y b) = a x [∗∗]
Bywayofcontradiction,supposethat y b 0,then i = a x y b ,whichisarealnumber.
But i isanimaginarynumberandsothereisacontradiction.Thus y b = 0andhence y = b .Itfollows fromequation[∗∗]that x = a ,andtheproofiscomplete.
ThecarefulreaderwillhavenoticedthatthedefinitionsofRe(z)andIm(z)givenabovearenotintermsofthe variable z.Bothofthesefunctionscanbeexpressedintermsof z byfirstwritingdown z anditsconjugate:
z = x + iy ,
z = x iy .
Thispairofsimultaneousequationscanbesolvedfor x and y toobtain:
Re(z) = 1 2 (z + z)andIm(z) = 1 2i (z z).
8Realandimaginaryparts
Thesecanbewrittenasfunctionsof z
Re(z) = 1 2 (z + z)andIm(z) = 1 2i (z z).
TheArithmeticofConjugates
Sincetakingthecomplexconjugateof z simplychangesthesignoftheimaginarypart,whenitisappliedtwicein successiontheendresultleaves z unchanged.Thus
(z) = x + iy = x iy = x + iy = z .
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Anotherimportantpropertyoftakingconjugatesisthatitcommuteswiththefourbasicarithmeticoperations.For example,withaddition,
(3 + i) + (2 4i) = 5 3i = 5 + 3i , and 3 + i + 2 4i = 3 i + 2 + 4i = 5 + 3i .
Thus (3 + i) + (2 4i) = 3 + i + 2 4i
Noticethatitdoesnotmatterwhethertheadditionisdonebeforeoraftertakingtheconjugate,theresultisthe same.Hereisanexamplewithmultiplication.
(3 + i)(2 4i) = 10 10i = 10 + 10i , and 3 + i × 2 4i = (3 i)(2 + 4i) = 10 + 10i .
Thus (3 + i)(2 4i) = 3 + i × 2 4i
Againnoticethatitdoesnotmatterwhetherthemultiplicationisdonebeforeoraftertakingtheconjugate,the resultisthesame.Thisisalwaysthecaseforaddition,subtraction,multiplicationanddivision.
9Thearithmeticofconjugates
Thetakingofcomplexconjugatesiscommutativewithaddition,subtraction,multiplicationanddivision. w + z = w + z a w z = w z b wz = w × z c w ÷ z = w ÷ z d
Theproofoftheseresultsisleftasaquestionintheexercise.Therearetwospecialcasesoftheseresults.Toget theconjugateofanegative,put w = 0into b
( z) = 0 z = 0 z thus ( z) = z
Fortheconjugateofareciprocal,put w = 1in d toget
z 1 = 1 ÷ z = 1 ÷ z = 1 ÷ z thus z 1 = (z) 1
IntegerPowers
Thecarefulreaderwillhavenotedthatseveraloftheexamplesusedaboveinvolvepowersofacomplexnumber despitethefactthatthemeaningof zn hasnotyetbeenproperlydefined.Iftheindex n isapositiveintegerthen themeaningof zn isanalogoustotherealnumberdefinition.Thus
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z n = z × z × ... × z n factors or,therecursivedefinitionmaybeused:
z 1 = z ,
z n = z × z n 1 for n > 1.
Justliketherealnumbers,if z = 0then z0 isundefined.Forallothercomplexnumbers, z0 = 1.Againcontinuing theanalogywiththerealnumbers,anegativeintegerpoweryieldsareciprocal.Thusif n isapositiveintegerthen
z n = 1 zn , z 0.
Aswithotherdivisionbycomplexnumbers,thedenominatorisusuallyrealisedbymultiplyingbytheconjugate. Thecasewhen n = 1occursfrequentlyandshouldbelearnt.
z 1 = 1 z = z zz
Indiceswhicharenotintegerswillnotbeconsideredinthistext.
TheStructureofNumbers
Animportantquestiontoaskatthisstageis:howdocomplexnumbersfitintothestructureofnumbers?Thisis bestunderstoodbyreviewinghownumbershavebeendeveloped.Thenaturalnumbers N enableadditionand multiplicationtobecarriedoutbutnotasubtractionlike2 5.Theinclusionofthenegativenumberstocreate theintegers Z allowstheanswer2 5 = 3,anditisclearthat N ⊂ Z .Adivisionlike2 ÷ 5isnotpossiblewith integersandsofractionsareincludedtocreatethelargersetofrationals Q,with2 ÷ 5 = 2 5 and N ⊂ Z ⊂ Q
Thereisaspecialconnectionbetween Q andlinearequations.Therearelinearequationswithnaturalcoefficients thatdonothavenaturalsolutions.Bywayofexample2x + 6 = 0hassolution x = 3whichisaninteger.Ina similarway,therearelinearequationswithintegercoefficientsthatdonothaveintegersolutions.Forexample 6x 2 = 0hassolution x = 1 3 whichisarationalnumber.Incontrast,everylinearequation ax + b = 0with rationalcoefficients a and b hasasolutionwithinthesetofrationalnumbers. Asmentionedearlier,surdslike √3arerequiredtosolvecertainquadraticequations. Otherirrationalnumberslike π and e havealsobeenmet.Combiningtheirrational numberswiththerationalnumbersproducesthereals R,with N ⊂ Z ⊂ Q ⊂ R .Inthis section,numbersoftheform x + iy havebeenincludedwiththerealstocreatethe complexnumbers C.ThesituationisshownintheEulerdiagramontheright,where N ⊂ Z ⊂ Q ⊂ R ⊂ C N Z R C Q 3 i 3
Thereisaspecialconnectionbetween C andquadraticequations.Therearequadraticequationswithrational coefficientsthatdonothaverationalsolutions.Bywayofexample x2 3 = 0hassolutions x = √3or √3, neitherofwhicharerational.Inasimilarway,therearequadraticequationswithrealcoefficientsthatdonot haverealsolutions.Forexample z2 + 9 = 0hassolutions z = 3i or 3i,whicharecomplexnumbers.Incontrast, everyquadraticequation ax2 + bx + c = 0withcomplexcoefficients a, b and c hassolutionsinthesetofcomplex numbers.ThiswillbedemonstratedinSection 1B.
Exercise1A
1 ClassifyeachofthesenumbersusingthesmallestpossiblesubsetintheEuler diagramontheright.
a 7 b 13 c 1 2 3 d √37
2 UsetherulegiveninBox 2 tosimplify:
FOUNDATION
3 Evaluate:
4 Expressintheform a + ib,where a and b arereal. (7 + 3i) + (5 5i) a
5 Expressintheform x + iy,where x and y arereal. (4 + 5i)i a (1 + 2i)(3 i) b (3 + 2i)(4 i) c
( 7 + 5i)(8 6i) d (5 + i)2 e (2 3i)2 f (2 + i)3 g (1 i)4 h (3 i)4 i
6 UsetheruleforthesumoftwosquaresgiveninBox 4 tosimplify:
(1 + 2i)(1 2i) a (4 + i)(4 i) b (5 + 2i)(5 2i) c ( 4 7i)( 4 + 7i) d
7 Expressintheform x + iy,where x and y arereal. 1 i a 2 + i i b 5 i 1 i c 6 7i 4 + i d 11 + 13i 5 + 2i e (1 + i)2 3 i f
8 Let z = 1 + 2i and w = 3 i.Find,intheform x + iy : (iz) a w + z b 2z + iw c Im(5i z) d z2 e
9 Let z = 8 + i and w = 2 3i.Find,intheform x + iy : z w a
10 Let z = 2 i and w = 5 12i.Find,intheform x + iy : zw a (1 + i)z w b 10 z c w 2 3i d Re (1 + 4i)z e
11 Byequatingrealandimaginaryparts,findtherealvaluesof x and y giventhat: (x + iy)(2 3i) = 13i a
+ 4i)(x + iy) = 6 + 7i b (1 + i)x + (2 3i)y = 10 c
DEVELOPMENT
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12
Expressintheform x + iy,where x and y arereal.
13
Giventhat z = x + iy and w = a + ib,where a, b, x and y arereal,provethat: z + w = z + w a z w = z w b zw = z w c z2 = (z)2 d 1 z = 1 z , z 0 e z w = z w , w 0 f
14 Let z = a + ib,where a and b arerealandnon-zero.Provethat:
z + z isreal, a z z isimaginary, b z2 + (z)2 isreal, c zz isrealandpositive. d
15 Let z = a + ib,where a and b arereal.If z z i isreal,showthat z isimaginaryor0. 16 Provethatif z2 = (z)2 then z canonlybepurelyrealorpurelyimaginary.
17 If z = x + iy,where x and y arereal,expressintheform a + ib,where a and b arewrittenintermsof x and y. z 1 a z 2 b z 1 z + 1 c
CHALLENGE
18 Ifboth z + w and zw arereal,provethateither z = w orIm(z) = Im(w) = 0.
cos
,showthat
1B
QuadraticEquations
Nowthatthearithmeticofcomplexnumbershasbeensatisfactorilydeveloped,itisappropriatetoreturnto theoriginalproblemofsolvingquadraticequations.Toreflectthefactthatthesolutionsmaybecomplex,the variable z willbeused.
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QuadraticEquationswithRealCoefficients
Thesimplestquadraticequationsaretheperfectsquare
(z λ)2 = 0
forwhich z = λ , andthedifferenceoftwosquares
z2 λ2 = 0
forwhich z = λ or λ
Itisnowalsopossibletosolveequationsinvolvingthesumoftwosquares,usingtheresultofBox 4 in Section 1A.
Given z2 + λ2 = 0
(z + iλ)(z iλ) = 0(thesumoftwosquares)
so z = iλ or iλ
Thustherearethreepossiblecasesforasimplequadraticequation:aperfectsquare,thedifferenceoftwo squares,orthesumoftwosquares.
Example6
Findthetwoimaginarysolutionsof z2 + 10 = 0.
Solution
Fromthesumoftwosquares
(z + i√10)(z i√10) = 0 so z = i√10or i√10.
Formoregeneralquadraticequations,itissimplyamatterofcompletingthesquarein z toobtainoneofthesame threesituations:aperfectsquare,thedifferenceoftwosquares,orthesumoftwosquares.
Example7
Findthecomplexsolutionsof z2 + 6z + 25 = 0.
Solution
Completingthesquare:
(z + 3)2 + 16 = 0
so(z + 3 + 4i)(z + 3 4i) = 0(sumoftwosquares) thus z = 3 4i or 3 + 4i
Noticethatthesumoftwosquaressituationalwaysyieldstworootswhicharecomplexconjugates.
10Quadraticequationswithrealcoefficients
Completethesquarein z toobtainoneofthefollowingsituations:
A A perfectsquare:Thereisonlyonerealroot.
B Thedifferenceoftwosquares:Therearetworealroots.
C Thesumoftwosquares:Therearetwoconjugatecomplexroots.
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Thereareseveralwaysofprovingtheassertionthatcomplexsolutionstoquadraticequationswithreal coefficientsmustoccurasconjugatepairs.Theapproachpresentedherewilllaterbeextendedtoencompassall polynomialswithrealcoefficients.
Proof Let Q(z) = az2 + bz + c,where a, b and c arerealnumbers.Supposethattheequation Q(z) = 0 hasa complexsolution z = w.Itfollowsthat
aw 2 + bw + c = 0.
Taketheconjugateofbothsidesofthisequationtoget
aw2 + bw + c = 0.
Nowtheconjugateofarealnumberisthesamerealnumber.Further,asnotedinBox 9,takinga conjugateiscommutativewithadditionandmultiplication.Thusthelastequationbecomes
a(w)2 + b(w) + c = 0, thatis, Q(w) = 0.
Hence z = w isalsoacomplexrootof Q(z) = 0. Thatis, Q(z) = 0 musthavetwoconjugatecomplexroots, z = w and z = w,andtheproofiscomplete.
Example8
Findaquadraticequationwithrealcoefficientsgiventhatoneoftherootsis w = 5 i
Solution
Thecoefficientsarerealsotherootsoccurinconjugatepairs.Hencetheotherrootis w = 5 + i.Thusthe monicquadraticequationis:
z (5 i) z (5 + i) = 0 or (z 5) + i (z 5) i = 0 thus (z 5)2 + 1 = 0. Finally z2 10z + 26 = 0.
Ingeneral,aquadraticequationwithrealcoefficientswhichhasacomplexroot z = α is z 2 2Re(α) z + α α = 0, ascanbeobservedinthethreeexamplesabove.Theproofisquitestraightforward,andisoneofthequestionsin theexercise.
11Realquadraticequationswithcomplexroots
Aquadraticequationwithrealcoefficientswhichhasacomplexroot z = α is z 2 2Re(α) z + α α = 0.
TheQuadraticMethod
Manyreaderswillknowthequadraticformulaas
x = b ± √b2 4ac 2a
Thereisaproblemwiththisformulawhencomplexnumbersareinvolved.Whenappliedtorealnumbers,the symbol √ meansthepositivesquareroot,butitisunclearwhat“positive”meanswhenappliedtocomplex numbers.Itmightbetemptingtosaythat i ispositiveand i isnegative,butthenwhatistobesaidabout numberslike( 1 + i)or(1 i)?Inshort,itdoesnotmakesensetospeakofpositiveandnegativecomplex numbers,andsothepositivesquareroothasnomeaning.Thusitisnotappropriatetoblindlyusethequadratic formulatosolveanequationwithcomplexroots.
Recallthatthequadraticformulaarosefromapplyingthemethodofcompletingthesquare.Herethatprocessis reviewed.
Given az2 + bz + c = 0,
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Nowsupposethereexistsanumber λ,possiblycomplex,suchthat ∆= λ2 .
Then
whence
fferenceoftwosquares) andso
Thusifthereisanumber λ,possiblycomplex,forwhich λ2 =∆,thenthesolutiontothequadraticequationcan bewrittenusingthelastlineabove.Ifthequadraticformulaistobeappliedthenthismethodshouldalwaysbe followed.
12Thequadraticmethod
Usethefollowingstepstosolve az2 + bz + c = 0.
1 Firstfind ∆= b2 4ac
2 Nextfindanumber λ,possiblycomplex,suchthat λ2 =∆ .
3 Finally,therootsare
Example9
Solve z2 + 2z + 6 = 0. Solution ∆= 22 4 × 1 × 6 = 20 = 2i√5 2 , hence z = 2 2i√5 2 or 2 + 2i√5 2 = 1 i√5or 1 + i√5.
ComplexSquareRoots
Beforeextendingtheaboveworktothecaseofaquadraticequationwithcomplexcoefficients,itisnecessaryto developmethodsforfindingthesquarerootsofcomplexnumbers.
Thefirstthingtonoticeisthat,justlikerealnumbers,everycomplexnumberhastwosquareroots.Theproofis quitestraightforward.Supposethatthecomplexnumber z isasquarerootofanothercomplexnumber w then z2 = w .
Further( z)2 = z2 = w .
Hence w hasasecondsquarerootwhichistheoppositeofthefirst,namely( z).
Thusforexample 2i hastwooppositesquareroots,(1 i)and( 1 + i).Thisisnotreallyverysurprisingsince allrealnumbers(otherthanzero)havetwooppositesquareroots.Forexample,9hassquareroots3and 3, whilst 5hassquareroots i√5and i√5.Theproofthattherearenomorethantwosquarerootsisleftasan exercise.
ComplexSquareRootsandPythagoras
Asimplewaytofindcomplexsquarerootsistoequatetherealandimaginarypartsof z2 = w inordertoobtaina pairofsimultaneousequations.
Given(x + iy)2 = a + ib ,where x, y, a and b arereal,
x2 y2 + 2ixy = a + ib
Equatingrealandimaginarypartsyields
x2 y2 = a and xy = 1 2 b .
Inmanycasesthispairofequationscanbeeasilysolvedbyinspectingthefactorsof 1 2 b,asinthefollowing example.
Example10
Findthesquarerootsof7 + 24i.
Solution
Let(x + iy)2 = 7 + 24i ,where x and y arereal, then(x2 y2) + 2ixy = 7 + 24i
Equatingrealandimaginarypartsyieldsthesimultaneousequations
x2 y2 = 7 and xy = 12.
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Inspectingthefactorsof12,itisclearthat x = 4and y = 3,or x = 4and y = 3.Hencethesquarerootsof 7 + 24i aretheopposites
4 + 3i and 4 3i
Somereaderswillhavenoticedinthisexamplethat7and24arethefirsttwonumbersofthePythagorean triad7,24,25.Thisisnocoincidence.Itisoftenthecasethatif b isevenandthenumbers |a|, |b| and √a2 + b2 formaPythagoreantriadthentheresultingequationsfor x and y canbesimplysolvedbyinspectingthefactors of 1 2 b
13ComplexsquarerootsandPythagoras
Given(x + iy)2 = a + ib,equatetherealandimaginarypartstogetthesimultaneousequations x 2 y 2 = a xy = 1 2 b
If b isevenandthenumbers |a|, |b| and √a2 + b2 formaPythagoreantriadthentheseequationscanoften besolvedbyinspectingthefactorsof 1 2 b .
QuadraticEquationswithComplexCoefficients
Simplequadraticequationswithcomplexcoefficientscannowbesolved.Allthatisneededistocombinethe abovemethodforfindingtherootsofacomplexnumberwitheitherthemethodofcompletingthesquareorthe quadraticmethodinBox 12
Example11
Completethesquaretosolve
Solution
Rearranging
Let (x + iy)2 = 3 + 4i then x2 y2 = 3 and xy = 2 sobyinspectiononesolutionis x = 1and y = 2.
Hence(z (1 + 3i))2 = (1 + 2i)2 andthus z = (1 + 3i) + (1 + 2i)or(1 + 3i) (1 + 2i) thatis z = 2 + 5i or i .
Whilstthefocushereisonthequadraticmethodandcompletingthesquare,thosetwomethodsshouldonlybe usedwhenrequired.Itisalwayspreferabletosolveaquadraticequationbyfactorswhenevertheycanbeeasily identified.
Example12
Solve z2 + 4iz 3 = 0.
Solution
Notingthat 3 = i × 3i andthat4i = i + 3i , (z + i)(z + 3i) = 0 hence z = i or 3i .
HarderComplexSquareRoots
OftenthesimultaneousequationsgiveninBox 13 cannotbesolvedbyinspection.Fortunatelythereisanidentity thatcanbeusedtohelpsolvetheseequations.Recallthatif
(x + iy)2 = a + ib then x2 y2 = a (1) and2xy = b (2)
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Squaringtheseandadding:
a2 + b2 = (x2 y2)2 + (2xy)2
= (x2)2 + 2x2y2 + (y2)2
= (x2 + y2)2
Hence x2 + y2 = √a2 + b2 (3)
Equations(1)and(3)nowformaverysimplepairofsimultaneousequationstosolve.Equation(2)isusedto determinewhether x and y havethesamesign,when b > 0,oroppositesign,when b < 0.Alsonoticethenow obviouslinkwithPythagorasthroughtherighthandsidesoftheseequations: a, b and √a2 + b2
14Squarerootsofacomplexnumber
Given(x + iy)2 = a + ib then x and y aresolutionsofthepairofsimultaneousequations x 2 y 2 = a x 2 + y 2 = √a2 + b2
withthesamesignif b ispositive,andoppositesignif b isnegative.
Example13
Determinethetwosquarerootsof 4 + 2i
Solution
Let(x + iy)2 = 4 + 2i .SinceIm( 4 + 2i) > 0, x and y havethesamesign.Further,( 4)2 + 22 = 20,sosolve x2 y2 = 4 (1) and x2 + y2 = √20. (2)
Adding(1)and(2)yields
2x2 = 4 + 2√5
so x = 2 + √5or 2 + √5.
Subtracting(1)from(2)yields
2y2 = 4 + 2√5
so y = 2 + √5or 2 + √5.
Hence x + iy = 2 + √5 i 2 + √5or 2 + √5 + i 2 + √5.
Infact,theresultinBox 14 canbeusedtodevelopaformulaforthesquarerootsofanycomplexnumber,which isderivedinoneoftheexercisequestions.Howeverthatformulaisnotpartofthecourseandshouldnotbe memorised.
HarderQuadraticEquations
Anyquadraticequationcannowbesolved,includingthosewithcomplexdiscriminants.Box 14 isusedtofind thesquarerootsofdiscriminantsthatcannotbefoundbyinspection.
Example14
Solve z2 + (4 2i)z + 1 = 0byusingthequadraticmethod.
Solution
∆= (4 2i)2 4 = 8 16i .
Let(x + iy)2 = 8 16i .NowIm(8 16i) < 0so x and y haveoppositesign.
Further, x2 + y2 = √320so
x2 y2 = 8 (1) and x2 + y2 = 8√5 (2)
Addingandsubtractingequations(1)and(2)yields
2x 2 = 8 + 8√5 x 2 = 4(1 + √5) and2
Exercise1B
1 Solvefor z.
=
2 Writeasaproductoftwocomplexlinearfactors.
3 Formaquadraticequationwithrealcoefficientsgiventhatonerootis:
4 Ineachcase,findthetwosquarerootsofthegivennumberbytheinspectionmethod.
5a Findthetwosquarerootsof 3 4i.
b Hencesolve z2 3z + (3 + i) = 0.
6a Findthetwosquarerootsof 8 + 6i
b Hencesolve z2 (7 i)z + (14 5i) = 0.
7 UsethemethodoutlinedinBox 12 tosolvefor z
z2 z + (1 + i) = 0 a z2 + 3z + (4 + 6i) = 0 b
z2 6z + (9 2i) = 0 c (1 + i)z2 + z 5 = 0 d
z2 + (2 + i)z 13(1 i) = 0 e iz2 2(1 + i)z + 10 = 0 f
8a Findthevalueof w if i isarootoftheequation z2 + wz + (1 + i) = 0.
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b Findtherealnumbers a and b giventhat3 2i isarootoftheequation z2 + az + b = 0.
c Giventhat1 2i isarootoftheequation z2 (3 + i)z + k = 0,find k andtheotherrootoftheequation.
9 Findthetwocomplexnumbers z satisfying zz = 5and z z = 1 5 (3 + 4i).
10a Solve z2 2z cos θ + 1 = 0for z bycompletingthesquare.
b Rearrangingtheequationinpart a givescos θ = 1 2 z + 1 z .Confirmthisresultforeachofthesolutions topart a bysubstitution.
11a Showthat A3 B3 = (A B)(A2 + AB + B2).
b Usepart a tohelpsolve: z3 = 1, i z3 + i = 0. ii
12 Considerthequadraticequation az2 + bz + c = 0,where a, b and c arerealand b2 4ac < 0.Supposethat ω isoneofthecomplexrootsoftheequation.
a Explainwhy aω2 + bω + c = 0.
b Bytakingtheconjugateofbothsidesoftheresultin a,andusingthepropertiesofconjugates,showthat a ω 2 + b ω + c = 0.
c Whathaveyoujustprovedaboutthetwocomplexrootsoftheequation?
13 Supposethat z = α isacomplexsolutiontoaquadraticequationwithrealcoefficients.
a Whichothernumberisalsoasolutionofthisquadraticequation?
b Henceprovethatonesuchquadraticequationis z2 2Re(α)z + α α = 0.
14 Let(x + iy)2 = a + ib,thenwehave x2 y2 = a and2xy = b
a Repeatthefollowingforseveraldifferentpositivevaluesof a and b
i Graphtheequationsonthesamenumberplaneusingsuitablesoftware.
ii Whatfeatureofthegraphindicatesthattherearetwosquarerootsof a + ib?
b Investigatehowthegraphchangeswheneither a or b orbotharenegativeorzero.
15 UsetheresultsofBox 14 tofindthetwosquarerootsof: i a 6 + 8i b 2 + 2i√3 c 10 24i d 2 4i e
16 Findthediscriminantanditssquareroots,andhencesolve:
z2 + (4 + 2i)z + (1 + 2i) = 0 a z2 2(1 + i)z + (2 + 6i) = 0 b
z2 + 2(1 i√3)z + 2(1 + i√3) = 0 c z2 + (1 i)z + (i 1) = 0 d
17 Let α and β bethetwocomplexrootsof z3 = 1.Showthat:
CHALLENGE
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β = α, a α2 = β and β2 = α, b 1 + α + α2 = 0, c d thesumofthefirst n termsoftheseries1 + α + α2 + α3 + iseither0,1or α2,dependingonthe remainderwhen n isdividedby3.
18 Let a, b and c berealwith b2 4ac < 0,andsupposethatthequadraticequation az2 + bz + c = 0has complexsolutions α = x + iy and β = u + iv .
a Byconsideringthesumandproductoftheroots,showthat Im(α + β) = 0andIm(αβ) = 0.
b Henceshowthat α = β .
19 Let(x + iy)2 = a + ib,where b 0.UsetheresultofBox 14 toprovetheformula:
Explainthesignificanceoftheterm b/|b| inthisformula.
1C TheArgandDiagram
Mathematicsrequiresaknowledgeofnumbers,andthroughouthighschoolthatunderstandingofnumbers hasbeenenhancedbybeingabletoplotthemonanumberline,tovisualisetheirpropertiesandrelationships. Initiallytherewerethenaturalnumbers,shownatdiscreteintervalsonthenumberline.
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Whennegativenumberswereincludedtocreatetheintegers,thenumberlinewasextendedtotheleftofthe origintoshowthesenewnumbers.
Nextcametherationals,thefractionswhichexistinthespacesbetweenintegers.Eventuallytheirrationalswere discoveredandincluded,whichfitinthegapsthataresomehowleftbetweenrationals.Someirrationalnumbers like √2canbeconstructedgeometrically,butotherslike e and π canonlybeapproximatedtosomanydecimal places.Theconstructionfor √2isshownherealongwiththepositionsof
Thenumberlineisnowfull,therealshavefilleditup,andthereisnospaceleftforanynewobjectslikecomplex numbers.Further,sincecomplexnumberscomeintwoparts,realandimaginary,thereisnosatisfactorywayof representingtheminaonedimensionalnumberline.Atwodimensionalrepresentationisneeded.
TheComplexNumberPlane
Keepingtothingsthatarefamiliar,thenumberplanewouldseemtobeaconvenientwaytorepresentcomplex numbers.Moreformally,foreachcomplexnumber z = x + iy therecorrespondsapoint Z(x, y)intheCartesian plane.Equally,givenanypoint W(a, b)intherealnumberplane,theassociatedcomplexnumberis w = a + ib.
Thusinthediagramontheright,thecomplexnumbers4 + i and 3 + 3i arerepresented bythepoints A and B respectively.Thepoints C and D representthecomplexnumbers 1 2i and2 4i.Severaldifferentnamesareusedtodescribeacoordinateplane thatisusedtorepresentcomplexnumbers.Onenameisthe Arganddiagram,after theFrenchmathematicianJean-RobertArgand,borninGenevain1768.Theterms complexnumberplane or z-planearealsoused.
15TheArganddiagram
Thecomplexnumber z = x + iy isassociatedwiththepoint Z(x, y)intherealnumberplane.Acomplex numbermayberepresentedbyapoint,andapointmayberepresentedbyacomplexnumber.
Asaconvenientabbreviation,thepoint Z(x, y)willsometimesbesimplyreferredtoasthepoint z intheArgand diagram.Itisimportanttorememberthatthecomplexnumberplaneisjustarealnumberplanewhichisused todisplaycomplexnumbers.Bythenatureofthisrepresentation,iftwocomplexnumbersareequalthenthey representthesamepoint.Theconverseisalsotrue.Iftwopointscoincidethentheyrepresentthesamecomplex number.
TheRealandImaginaryAxes
IfIm(z) = 0,thatis z = x + 0i,then z isarealnumberandthecorrespondingpoint Z(x,0)intheArganddiagram liesonthehorizontalaxis.Thusthehorizontalaxisiscalledthe realaxis
Likewise,ifRe(z) = 0,thatis z = 0 + iy,then z isanimaginarynumberandthecorrespondingpoint Z(0, y)inthe Arganddiagramliesontheverticalaxis.Thustheverticalaxisiscalledthe imaginaryaxis
SomeSimpleGeometry
Nowthatthecomplexplanehasbeenintroduced,itisimmediatelypossibletoobservethegeometryofsome simplecomplexnumberoperations.Thesimplestofthesearethegeometriesofconjugates,opposites,and multiplicationby i,whicharenowinvestigated.
Let z = x + iy ,thentheconjugateis
z = x iy , thatis, y hasbeenreplacedby y.Thiswasencounteredintheworkongraphsand isknowntobeareflectionintherealaxis.Thisisclearlyevidentintheexampleof z1 = 3 + 2i and z2 = 3 2i = z1 shownontheright.
16Thegeometryofconjugates
Thepoints z and z intheArganddiagramarereflectionsofeachotherintherealaxis.
Next,let z = x + iy ,thentheoppositeis
z = x iy
Inthiscase, x and y havebeenreplacedby x and y respectively.Thustheresultis obtainedbyreflectingthepointinbothaxesinsuccession.Alternatively,itisarotation by π abouttheorigin.Thediagramontherightwith z1 = 1 2i and z
demonstratesthisrotation.
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17Thegeometryofopposites
Thepoints z and z intheArganddiagramarerotationsofeachotherby π abouttheorigin.
Nowlet z1 = a + ib ,then z2 = iz1 isgivenby z2 = b + ia
Considerthecorrespondingpoints Z1 and Z2 shownintheArganddiagramontheright, whereneither a nor b iszero.Theproductofthegradientsof OZ1 and OZ2 is b a × a b = 1.
Hence OZ2 isperpendicularto OZ1.Thatis,multiplicationby i isequivalenttoananticlockwiserotationby π 2 abouttheorigin.Thesituationisthesameevenwhen z1 isrealorimaginary,butnotzero,andtheproofisleftas anexercise.
18Thegeometryofmultiplicationby i
Thepoint iz intheArgandDiagramistheresultofrotatingthepoint z by π 2 anticlockwiseaboutthe origin.
Za,b1()
Zb,a2() x y O
Notethatmultiplicationby i twiceinsuccessionyieldsarotationof2 × π 2 = π.Thisisconsistentwiththe geometryofopposites,since i(iz) = i2z = z
Example15
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Let z = x + iy.Determine i z andhencegiveageometricinterpretationoftheresult.
Solution
i z = i(x iy) = y + ix
Thisisjust z with x and y swapped.Thusitisareflectionintheline y = x,asencounteredwithinverse relations.
CurvesintheArgandDiagram
Sofarattentionhasbeengiventoindividualpointsinthecomplexplane.Oftenanequationin z willcorrespond toawellknownlineorcurveintheArganddiagram.Inthesimplecasesdealtwithhere,theequationcanbe foundbyputting z = x + iy.
Example16
Graphthefollowing:
Re(z) = 2, a Im(z) = 1. b
Solution
Theequationsgive:
a theverticalline x = 2and b thehorizontalline y = 1, asshowninthediagramontheright.
Notethatthesetwolinesintersectat z = 2 i.
19Verticalandhorizontallines
IntheArganddiagram:
• theequationRe(z) = a is theverticalline x = a ,
• theequationIm(z) = b is thehorizontalline y = b ,
• thesetwolinesintersectat z = a + ib .
Example17
Letthepoint P inthecomplexplanerepresentthenumber z = x + iy.Giventhat z z = 9,findthecurvethat P ison,andsketchit.
Solution
Thegivenequationbecomes
(x + iy)(x iy) = 9
so x2 + y2 = 32 thatis,acirclewithcentretheoriginandradius3.
Insomeexamplesitisbesttomanipulatethegivenequationin z first,andthensubstitute x + iy.Itisalso importanttonoteanyrestrictionson z beforestarting.Bothofthesepointsfeatureinthefollowingexample.
Example18
Findanddescribethecurvespecifiedby 1 z + 1 z = 1.
Solution
Notethatinthegivenequation z 0,sincetheLHSisundefinedthere.Multiply bothsidesbythelowestcommondenominatortoget
z + z = z z
so2x = x2 + y2
or0 = x2 2x + y2 thus1 = (x 1)2 + y2 thatis,thecirclewithradius1andcentre(1,0),excludingtheorigin. x y
Exercise1C
1 Writedownthecoordinatesofthepointinthecomplexplanethatrepresents:
FOUNDATION
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2 Writedownthecomplexnumberthatisrepresentedbythepoint: ( 3,0) a (0,3) b (7, 5) c (a, b) d
3 Let z = 1 + 3i,andlet A, B, C and D bethepointsrepresenting z, iz, i2z and i3z respectively.
a Plotthepoints A, B, C and D inthecomplexplane.
b Whattypeofspecialquadrilateralis ABCD?
c Whatappearstobethegeometriceffectofmultiplyingacomplexnumberby i?
4 Let z = 3 + i and w = 1 + 2i.Plotthepointsrepresentingeachgroupofcomplexnumbersonseparate Arganddiagrams.LookforthegeometryofBoxes 16 to 18 inparts a to c,anddescribetheshapesformed inparts d to f
z, iz, z, iz a w, iw, w, iw b z, z, w, w c
5 GraphthefollowingsetsofpointsintheArganddiagram. Re(z) = 3 a Im(z) = 2 b
z) = Im(z) c 2 Re(z) = Im(z) d
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6
Letthepoint P representthecomplexnumber z = √3 + i,andletthepoints Q, R, S and T ,represent z, z, iz and 1 z respectively.PlotallthesepointsonanArganddiagram.
7 Showthatthepointrepresenting z isareflectionofthepointrepresenting z inthe y-axis.
8 Considerthepointsrepresentedbythenon-zerocomplexnumbers z, z, z and z
Showthatthesepointsformarectanglebyusing:
a coordinategeometrytoshowthatthediagonalsareequalandbisecteachother, b thegeometryofconjugatesandopposites.
9 Inthetextitwasproventhatwhen z iscomplex, iz isarotationby π 2 abouttheorigin.Provethesameresult when z is: real, a imaginary. b
10 Thenumbers z = a + ib and w = iz areplottedinthecomplexplaneat A and B respectively.Inthe following,assumethat a and b arenon-zero.
a Byconsideringgradients,showthat OA ⊥ OB
b Usethedistanceformulatoshowthat OA = OB
c Whattypeoftriangleis △OAB?
11
Thepoint P inthecomplexplanerepresentsthenumber z.Findanddescribethecurvethat P isongiven that
12 Thecomplexnumber z isrepresentedbythepoint C intheArganddiagram.Findanddescribethecurve that C isonif Re z 6 z = 0. 13 Showthat(z 2) (z 2) = 9representsacircleintheArganddiagram. 14 FindanddescribethecurveintheArganddiagramspecifiedby zz = Re(z 1 + 3i) 2
CHALLENGE 15 Sketchthecurveinthecomplexplanespecifiedbythegivenequationforrealvaluesof c.
16 Showthatthepointrepresenting iz isareflectionofthepointrepresenting z in y = x 17 Showthat 1 z isareflectionandenlargementof z
1D Modulus-ArgumentForm
Thelocationofapoint P isdeterminedbyitshorizontalandverticalcoordinates, x and y.Recallfromthestudyoftrigonometrythatthelocationcanalsobedetermined bythedistance OP = r fromtheoriginandtheangle θ throughwhichtheray OA is rotatedtocoincidewithray OP.Thesituationisshowninthenumberplaneonthe right.
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TheModulusandArgumentofaComplexNumber
IntheArganddiagramthedistance r iscalledthe modulus of z,andowingtoitsgeometricdefinitionasa distanceitiswrittenas |z|.Onsquaring:
|z|2 = r2 = x2 + y2 = (x + iy)(x iy)(sumoftwosquares) hence |z|2 = zz .
Theangle θ iscalledthe argument of z,andiswritten θ = arg(z).Justaswithtrigonometry, θ cantakeinfinitely manyvaluesforthesamepoint P.Itisoftennecessarytorestricttheangletojustonevaluecalledthe principal argument,whichiswrittenArg(z).Thevalueoftheprincipalargumentisalwaysintherange π < Arg(z) ≤ π Notethestrictinequalityonthelefthandside,andtheuseofradianmeasure.
Thereisaproblemwithmeasuring θ when z = 0becausethen P coincideswiththeoriginandthereisnoangleto measure.Forthisreason,botharg(0)andArg(0)areundefined.However,itshouldbeclearthat |0| isdefinedand that |0| = 0.
20Modulusandargument
Let P representthecomplexnumber z = x + iy intheArganddiagram,withorigin O
• The modulus of z isthedistance |z| = r = OP.Notethat |z|2 = zz.
• The argument of z 0isanyanglearg(z) = θ thattheray OP canmakewiththepositivedirectionof therealaxis.Incontrast,arg(0)isundefined.
• The principalargument of z 0istheuniqueangleArg(z) = θ whichisintherange π < θ ≤ π.Like arg(0),Arg(0)isundefined.
Fromthetrigonometricdefinitionsitisclearthat
x = r cos θ
and y = r sin θ ,
fromwhichitfollowsthat z = r cos θ + ir sin θ
Noticethatthemodulus r isacommonfactorinthislastexpressionanditismorecommonlywrittenas
z = r(cos θ + i sin θ) or z = r cis θ forshort.
Inordertocontrastthetwowaysofwritingacomplexnumber, z = x + iy iscalled real-imaginary or Cartesian formwhilst z = r(cos θ + i sin θ)iscalled modulus-argument form,or mod-arg formforshort.Anothernamefor mod-argformis polarform,becausetheradiusismeasuredfromtheoriginwhichactsasapole.Equations(1) and(2)aboveservetolinkthetwoforms.
(1)
(2)
Example19
Expresseachcomplexnumberinreal-imaginaryform.
z = 4cis π a z = 2cis π 6 b
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Solution
z = 4cos π + 4i sin π = 4 a
Example20
Expresseachcomplexnumberinmod-argformusingtheprincipalargument.Inpart c giveArg(z)correctto twodecimalplaces.
z = 5i a z = 3 3i b z = 4 3i c
Solution
Ineachcaselet z = r cis θ,with Z thepointintheArganddiagram.
a Inthiscase Z isonthepositiveimaginaryaxisso r = 5 and θ = π 2 , hence z = 5cis π 2 .
b Now Z isinthefourthquadrantwith r2 = 32 + 32 or r = 3√2. Thuscos θ = 1 √2 and θ = π 4 , hence z = 3√2cis π 4 .
c InthiscasethereisaPythagoreantriadso r = 5.
Nowcos θ = 4 5 and Z isinthethirdquadrant so θ = π + cos 1 4 5 ≑ 2.50radians, hence z ≑ 5cis( 2.50).
21Formsofacomplexnumber
• x + iy iscalledthe real-imaginary formor Cartesian formof z.
• r(cos θ + i sin θ) = r cis θ iscalledthe modulus-argument formof z
• Theequationsrelatingthetwoformsare:
x = r cos θ and y = r sin θ
Notethatsomepeopleprefertowrite
cos θ = x |z| andsin θ = y |z| whenfinding θ.Thisisobviouslyobtainedbyputting r = |z| inthelastdotpointofBox 21 andthendividing.
SomeSimpleAlgebra
Aswillberevealedovertheremainderofthischapter,theuseofmod-argformisapowerfultool,bothin simplifyingmuchalgebraandinprovidinggeometricinterpretations.
Perhapsthemostobviousthingtonoticeisthat | cis θ| = 1.Thegeometryofthesituationmakestheresult obvioussinceif z = cos θ + i sin θ thenthepoint Z(cos θ,sin θ)liesontheunitcircle.Hence |z| = OZ = 1.Hereis analgebraicderivationofthesameresult.
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| cos θ + i sin θ|2 = cos2 θ + sin2 θ = 1, hence | cis θ| = 1.
Thisidentityhasimmediateapplicationsinquadraticequations.Ifcis θ isoneoftherootsofamonicquadratic thentheconstanttermmustbe1,asthefollowingexampledemonstrates.
Example21
Findaquadraticequationwithrealcoefficientsgiventhatonerootis z = cis θ
Solution
Thecoefficientsarerealsotheotherrootmustbe cis θ.Thus: (z cis θ)(z cis θ) = 0 or z2 (cis θ + cis θ)z + cis θ × cis θ = 0 thatis z2 2Re(cis θ)z + | cis θ|2 = 0 thus z2 2z cos θ + 1 = 0.
TheProductofTwoComplexNumbers
Themodulus-argumentformoftheproductoftwonumbersisaparticularlyimportantresult.Let w = a cis θ and z = b cis φ,with a > 0and b > 0.Then: wz = a(cos θ + i sin θ) × b(cos φ + i sin φ) = ab (cos θ cos φ sin θ sin φ) + i(cos
) = ab cos(θ + φ) + i sin(θ + φ) = ab cis(θ + φ). Thus |wz| = ab andarg(wz) = θ + φ . Thisyieldsthefollowingtwosignificantresults:
|wz| = |w||z| andarg(wz) = arg(w) + arg(z).
Inwords;themodulusoftheproductistheproductofthemoduli,andtheargumentoftheproductisthesumof thearguments.
Example22
Let w = √3 + i and z = 1 + i
a Evaluate wz inreal-imaginaryform.
b Express w and z inpolarformandhenceevaluate wz inpolarform.
c Hencefindtheexactvalueofcos 5π 12
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Solution
a wz = (√3 + i)(1 + i)
= (√3 1) + i(√3 + 1).
b Now w = 2cis π 6 z = √2cis π 4 , hence wz = 2√2cis( π 6 + π 4 ) = 2√2cis 5π 12 .
c Equatingtherealpartsofparts a and b yields 2√2cos 5π 12 = √3 1 hencecos 5π 12 = √3 1 2
2 or
22Theproductoftwocomplexnumbers
Let w and z betwocomplexnumbers.
• Themodulusoftheproductistheproductofthemoduli,thatis:
|wz| = |w||z|
• Theargumentoftheproductisthesumofthearguments,thatis: arg(wz) = arg(w) + arg(z)(provided w 0and z 0).
Giventheconnectionbetweenmultiplicationanddivision,itmightbeexpectedthatthemodulusofaquotientis thequotientofthemoduli,andthisisindeedthecase.Thereadermayalsoobservethattheargfunctionseemsto behavelikethelogfunctionsothattheargumentofaquotientisthedifferenceofthearguments.Thisisalsothe case.Aproofofboththeseresultsisaskedforinaquestionoftheexercise.
Example23
Let w = √3 + i and z = 1 + i
a Evaluate w ÷ z inreal-imaginaryform.
b Usethemod-argformsof w and z tofindtheexactvalueofcos π 12
Solution
a w z = √3 + i 1 + i = 1 2 (√3 + i)(1 i) = 1 2 (1 + √3) + 1 2 i (1 √3).
b Themoduliandargumentsof w and z aregiveninthepreviousexample.Usingthequotientofmoduliand thedifferenceofarguments, w z = 2 √2 × cis( π 6 π 4 ) = √2cis( π 12 ).
Equatingtherealpartwithpart a,andusingthefactthatcosiseven, √2cos π 12 = 1 2 (1 + √3), thus,cos π 12 = 1 + √3 2√2 or
SomeSimpleGeometryAgain
Itisinstructivetore-examinethegeometryofconjugates,oppositesandmultiplicationby i usingmod-argform. Beginningwiththeconjugate,recallthattheresultisareflectionintherealaxis.Thusthemodulusshouldbe unchanged,andtheargumentshouldbeopposite.
Let z = r cis θ then
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= r cos θ
sin θ
r cos( θ) + ir sin( θ)
r cis( θ) hence |z| = |z| and arg(z) = arg(z)
thatis,themodulusisunchangedandtheangleisopposite,asexpected.
Thecasesofoppositesandmultiplicationby i aremoresimplydealtwith.Recallthattheseoperations representedrotationsinthecomplexplaneby π and π 2 respectively.Thus,again,themodulusshouldbethesame, andtheargumentshouldbeincreasedappropriately.Lookingatoppositesfirst:
andarg( z) = arg( 1 × z) = arg( 1) + arg(z) = π + arg(z).
Thatis,themoduliofoppositesareequalandtheargumentsdifferby π Similarly |iz| = |i||z| = |z| , andarg(iz) = arg(i) + arg(z) = π 2 + arg(z).
Thatis,themoduliof z and iz areequalandtheargumentsdifferby π 2 .Inbothcasestheresultsareexactlyas expected.Alsonoticethattheprincipalargumentsof 1and i havebeenused.Asanexercise,justifywhythisis correct.
TheGeometryofMultiplication
ThegeometryofmultiplicationisevidentintheresultsofBox 22.Theproductofthemoduliindicatesan enlargementwithcentretheorigin,andthesumoftheargumentsrepresentsananticlockwiserotationaboutthe origin.
Considerthesetwotransformationsindividuallyandlet w = r cis θ.When θ = 0theproduct wz reducesto wz = rz,whichisanenlargementwithoutanyrotation.Thusboth z and rz lieonthesameray.
When |w| = r = 1theproduct wz becomes wz = z cis θ.UsingBox 22: |wz| = |w||z| = |z|
andarg(wz) = arg w + arg z = θ + arg z
Thisissimplyarotationwithoutanyenlargement.Thus z and z cis θ bothlieonacircleofradius |z|.The followingexampleservestodemonstratethesituation.
Example24
Let z = 1 + i.
a Find,inCartesianform,thecomplexnumber w suchthat wz isarotationof z by π 3 abouttheorigin.
b Evaluate wz inCartesianform.
c Verifythat |wz| = |z|,thenplot z and wz onanArganddiagram.
Solution
a Clearly w = cis π 3 = 1 2 (1 + i√3).
b wz =
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c Now |wz|
= 2. Hence |wz| = √2 = |z|
23Thegeometryofmultiplication
Let w = r cis θ.Thenthecomplexnumber wz istheresultofarotationof z by θ anticlockwiseaboutthe originandanenlargementof z byfactor r withcentretheorigin.
Thecorrespondinggeometryforthedivisionofcomplexnumbersissimilarandisdealtwithinoneofthe exercisequestions.
TheresultsinBoxes 22 and 23 cannowbeusedtohelpsolveproblemsinvolvingrotationsandenlargementsin theCartesianplane.
Example25
Rectangle OABC hastwoofitsverticesat A(2,0)and C(0,1).Thisrectangleis rotatedabouttheoriginandenlargedtoobtainrectangle OBDE.Thesituationis showninthediagramontheright.
a Findtheenlargementfactorandfindtheangleofrotationcorrecttothenearest minute.
b Findthecoordinatesof D and E.
Solution
Thevertices A, B and C correspondtothecomplexnumbers a = 2, b = 2 + i and c = i,respectively.Let w be thecomplexnumbercorrespondingtotherotationandenlargement.
a Then b = wa andhence w = b a = 1 + 1 2 i
Theenlargementfactoris |w| = 1 + 1 4 =
Let θ betheangleofrotation,sothat θ = arg(w).
Clearly, θ isacutewithcos
b Let D and E havecorrespondingcomplexnumbers d and e,then d = wb
= (1 + 1 2 i)(2 + i) = 3 2 + 2i and e = wc = 1 2 + i .
Hence D isat( 3 2 ,2)and E isat( 1 2 ,1).
ShiftingintheComplexPlane
Recallthat,forarealnumber x,thevalueof |x| isthedistancefromtheoriginto x. Shiftingthis, |x a| isthedistancefrom a to x.TheresultsforshiftingintheCartesian planecanalsobeappliedtotheArganddiagram.Thussince |z| isthedistancefromthe originto z,itfollowsthat |z w| isthedistancefrom w to z.Analgebraicproofofthis resultisthesubjectofaquestionintheexercise. x y w z
Likewise,sincearg(z)istheangleattheoriginbetween z andthepositiverealaxis,thevalueofarg(z w)isthe angleatthevertex w between z andtherighthalfofthehorizontallinethrough w.Thesituationforboth |z w| andarg(z w)isshowninthediagramaboveontheright.
Example26
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Letpoints W and Z represent w = 2 + 2i and z = 1 + 5i respectively.Find: a thelengthof WZ, b theangle θ between WZ andthepositivedirectionofthe x-axis.
Solution
Firstnotethat z w = 3 + 3i
a WZ = |z w| = 3√2.
b Bycorrespondinganglesonparallellines,
θ = Arg(z w)
= Arg( 3 + 3i)
= 3π 4 . arg zw x y Z W
24Shiftinginthecomplexplane
Functionsofcomplexnumberscanbeshiftedinasimilarwaytofunctionsofrealnumbers.Inparticular:
• |z w| isthedistancefrom w to z
• arg(z w)istheangleatthevertex w between z andtherighthalfofthehorizontallinethrough w
Exercise1D
1 Find |z| given:
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FOUNDATION
2 FindArg(z)given:
3 Expresseachcomplexnumberintheform
4 Repeatthepreviousquestionforeachofthesecomplexnumbers,writing θ inradianscorrecttotwodecimal places.
5 Expressintheform a + ib,where a and b arereal.
6 Giventhat z = 1 i,expressinmodulus-argumentform(thatis,inpolarform):
7 Simplifyeachexpression,leavingyouranswerinmodulus-argumentform.
8 Findthedistance |z w| betweenthefollowingpairsofnumbersinthecomplexplane.
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9
Supposethatmultiplyingacomplexnumberby w producesarotationof θ radiansabouttheorigin.Find w inCartesianformforthegivenvaluesof θ
10 Let z beanon-zerocomplexnumbersuchthat0 < arg z < π 2 .Indicatepoints A, B, C and D inthecomplex planerepresentingthecomplexnumbers z, iz,(2cis π 3 )z and 1 2 cis( π 4 ) z
11 Statethedilationfactor(centredattheorigin)andtherotationangle(abouttheorigin)producedwhena complexnumberismultipliedby:
12 Isoscelestriangle ABC hasverticesat A(1,0), B(3,0)and C(2,2).Thistriangleisenlargedandrotated,in bothcaseswithcentretheorigin.Theresultis △PQR ∼ △ABC withvertex Q at C.Useasimilarmethodto Example 25 todothefollowing.
a Drawthesituationinthecomplexplane.
b Findtheenlargementfactorandangleofrotation.
c Findthecoordinatesof P and R
13 Replace z with z ÷ w inBox 22 toprovethatfor z 0and w 0: z w = |z| |w| a arg z w = arg z arg w b
14 Supposethat z1 = √3 + i and z2 = 2√2 + 2√2 i. Write z1 and z2 inmod-argform. a Hencewrite z1z2 and z2 z1 inmod-argform. b
15 Repeatthepreviousquestionfor z
16a Express 1 + i√3 1 + i inCartesianform.
b Write1 + i and1 + i√3inpolarformandhenceexpress 1 + i√3 1 + i inpolarform.
c Hencefindcos π 12 insurdform.
17 Let z1 = 1 + 5i and z2 = 3 + 2i,andlet z = z1 z2
a Find |z| withoutfinding z.
b Findtan(tan 1 5 tan 1 2 3 ),andhencefindarg z withoutfinding z
c Hencewrite z intheform x + iy,where x and y arereal.
18 Showthatforanynon-zerocomplexnumber z = r cis θ: z z = |z|2 , a arg(z2) = 2arg(z), b if |z| = 1then z = z 1 . c
19 Let z beanynon-zerocomplexnumber.Byconsideringarg(|z|2),usetheresultinpart a oftheprevious questiontoprovethatarg z = arg z
20 Let z = cos θ + i sin θ.Determine z2 intwodifferentwaysandhenceshowthat: cos2θ = cos2 θ sin2 θ a sin2θ = 2sin θ cos θ b
21 Thecomplexnumber z satisfiestheequation |z 1| = 1.Provethat |z|2 = 2Re(z)by:
a letting z = x + iy, b squaringtheequationandthenusingtheresult |z|2 = zz
22 Giventhat z isacomplexnumbersatisfying |2z 1| = |z 2|,provethat |z| = 1by:
a letting z = x + iy, b squaringtheequationandthenusingtheresult |z|2 = zz.
23 Let z = 1 + cos θ + i sin θ,where
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CHALLENGE
<
Showthat |z| = 2cos θ 2 andarg z = θ 2 a Henceshowthat z 1 = 1
24 Let w = x1 + iy1 and z = x2 + iy2.
a Showthatthedistance WZ betweenthepoints w and z is |z w|.
b Showthattheanglethattheline WZ makeswiththepositiverealaxisisarg(z w).
25 Let z = cis θ and w = cis φ,notingthat |z| = |w| = 1.Evaluate z + w inmod-argformandhenceshowthat arg(z + w) = 1 2 (arg z + arg w).[Hint:Usethesumtoproductidentitiescos
andsin θ + sin φ =
26a ProvethatRe(z) ≤|z|.Underwhatcircumstancesaretheyequal? b Provethat |z + w|≤|z| + |w|.Beginbywriting |z + w|2 = (z + w)(z + w).
1E
VectorsandtheComplexPlane
Thegeometryofmultiplicationanddivisionbecameevidentwiththeintroductionofthemodulus-argumentform intheprevioussection.Sincetheargumentsareaddedorsubtracted,itisclearthatarotationisinvolved.Since themoduliaremultipliedordivided,itisclearthatanenlargementisinvolved.
Sofar,theobservedgeometryofadditionandsubtractionhasbeenlimited.Abetterunderstandingofthese twooperationsisdesirableandcanbeachievedbyyetanotherrepresentationofcomplexnumbers,thistime asvectors.Inthissection,afewvectortoolswillbeintroducedtohelpbetterunderstandcomplexnumbers.A detailedstudyofvectorsisgivenlaterinbothExtensioncourses.
Vectors
Inthesimplegeometricdefinitionusedinthissection,avectorhastwocharacteristics,amagnitudeanda direction.Thustheinstructiononapiratetreasuremap“walk40paceseast”isanexampleofadisplacement vector.Themagnitudeis“40paces”andthedirectionis“east”.AtraintravellingfromSydneytoPerthacrossthe Nullarborat120km/hisanexampleofavelocityvector.Themagnitudeis120km/handthedirectioniswest.
InthenumberplaneandArganddiagram,avectorisrepresentedbyanarrow,whichismoreproperlycalleda directedlinesegment.Thelengthofthearrowisthemagnitudeofthevectorandthedirectionofthearrowis thedirectionofthevector.Whennamingavector,thetwoletternameofthelinesegmentisusedwithanarrow aboveittoindicatethedirection,asshownintheseexamples.
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VectorsandComplexNumbers
Lookfurtherattheexamplesabove.Inthefirst,itisnaturaltoassumethatthevector −−→ OZ representsthecomplex number z.Butwhatcomplexnumberdoesthevector −−→ WZ representinthesecondexample?
Themagnitudeofavectoristhedistancebetweentheend-pointsofitsarrow.Thatis,themagnitudeof −−→ WZ is |z w|.Thedirectionofavectormaybespecifiedbytheangleitsarrowmakeswiththehorizontal.Thatis,the directionof −−→ WZ isgivenbyArg(z w).Sinceanycomplexnumberiscompletelydeterminedbyitsmodulusand argument,itfollowsthatthe −−→ WZ mustrepresentthecomplexnumber(z w).Thisisalwaysthecase,regardless ofthevaluesof z and w
25Vectorsandcomplexnumbers
Letthevectorfrom w to z intheArganddiagrambe −−→ WZ.Thenthevector −−→ WZ representsthecomplex number(z w).
EqualVectors
Iftwovectors −−→ AB and −−→ PQ representthesamecomplexnumberthentheyhavethesamemagnitudeanddirection. Thusitmakessensetowrite
−−→ AB = −−→ PQ sincethereisnothingtodistinguishbetweenthem.
Bywayofexample,let Z0, Z1 and Z2 representthecomplexnumbers1 + 2i,2 + i and3 + 3i respectively.Thenby Box 25, −−−→
OZ0 = (1 + 2i) 0 = 1 + 2i and −−−→ Z1Z2 = (3 + 3i) (2 + i) = 1 + 2i
Thatis,thesamecomplexnumberisrepresentedbybothvectors,andhence
26Equalvectors
Vectorswiththesamemagnitudeanddirectionareequal.Equalvectorsrepresentthesamecomplex number.
AdditionandSubtraction
Considerthethreepoints A, B and C whichrepresentthecomplexnumbers w, z + w and z.Now −−→ OB = (z + w)and,byBox 25, −−→ AC = (z w).Sothediagonalsof OABC havespecialsignificance.Doesthisquadrilateralhaveanyotherspecialcharacteristics?
Themagnitudeof −−→ AB is
|(z + w) w| = |z| , thus AB = OC.Likewise,themagnitudeof −−→ CB is
|(z + w) z| = |w| ,
thus CB = OA.Nowsince OABC hasoppositesidesofequallength,itmustbeaparallelogram.Consequently theoppositesidesarealsoparallel.
Hence −−→ OC = −−→ AB and −−→ OA = −−→ CB
Thusinordertoaddorsubtracttwocomplexnumbersgeometrically,simplyconstructtheparallelogram OABC fromthevectors −−→ OA and −−→ OC.Thenthesumisthevector −−→ OB andthedifferenceis −−→ AC
Example27
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Giventwonon-zerocomplexnumbers w and z withequalmoduliandacutearguments,showthat Arg(w + z) = 1 2 Arg(w) + Arg(z) .
Solution
Considerthepoints OABC inthe z-planerepresentingthecomplexnumbers0, w, w + z and z respectively.Now OABC isaparallelogram.Further OA = OC ,since |w| = |z| .Thusinfact OABC isarhombus.Sincethediagonal OB oftherhombus bisectstheangleatthevertex O,itfollowsthat
Arg(w + z) = 1 2 Arg(w) + Arg(z) .
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27Thegeometryofadditionandsubtraction
Letthepoints O, A and C representthecomplexnumbers0, w and z.Constructtheparallelogram OABC.
• Thediagonalvector −−→ OB representsthecomplexnumber(z + w).
• Thediagonalvector −−→ AC representsthecomplexnumber(z w).
Example28
Thepoints OABC representthecomplexnumbers0, w, w + z and z.Giventhat z w = i(z + w),explainwhy OABC isasquare.
Solution
Firstly OABC isaparallelogram,where −−→ OB represents z + w and −−→ AC represents z w.Since z w = i(z + w)it followsthat
arg(z w) = arg(i(z + w)) = arg(i) + arg(z + w) = π 2 + arg(z + w).
Italsofollowsthat |z w| = |i(z + w)| = |i|×|z + w|
Thusthediagonals OB and AC areatrightanglesandhavethesamelength.Hence OABC isasquare.
TheTriangleInequality
Animportantidentityencounteredwiththeabsolutevalueofrealnumbersisthetriangleinequality
|x|−|y| ≤|x + y|≤|x| + |y| .
Giventhattheabsolutevalueofarealnumberisanalogoustothemodulusofacomplexnumber,itisnot surprisingthatthesameresultholdsforthemodulusofcomplexnumbers,thatis |z|−|w| ≤|z + w|≤|z| + |w|
Thisresultcanbeexplainedintermsofthegeometryoftheadditionofcomplex numbers.Consideronlythepoints O, A and B asdefinedpreviouslyandshownin thediagramontheright.Recallthatthethreevectors −−→ OA, −−→ AB and −−→ OB representthe complexnumbers w, z and z + w respectively.Hencethethreemoduli |w|, |z| and |z + w| arethelengthsofthesidesof △OAB. x y O A B zw w z
ItisawellknownresultofEuclideangeometrythatthelengthofonesideofatrianglemustbelessthanorequal tothesumoftheothertwo,thus
|z + w|≤|z| + |w| , withequalitywhen O, A and B arecollinear.
Similarlythelengthofonesideisgreaterthanorequaltothedifferenceoftheothertwo.Thus,
|z|−|w| ≤|z + w| , withequalityagainwhenthepointsarecollinear.
Combiningthesetwoyields
|z|−|w| ≤|z + w|≤|z| + |w| , andreplacing w with w throughoutgives
|z|−|w| ≤|z w|≤|z| + |w|
Theseinequalitiesarecalledthe triangleinequalities,aftertheirgeometricorigins.
28Thetriangleinequalities
Forallcomplexnumbers z and w :
• |z|−|w| ≤|z + w|≤|z| + |w| • |z|−|w| ≤|z w|≤|z| + |w|
TwoSpecialCases
Avectorapproachisveryhelpfulinanalysingthegeometryintwospecialcasesofdivision.Let z1, z2, z3 and z4 befourcomplexnumberscorrespondingtothepoints A, B, C and D,andlet z2 z1 z3 z4 = λ , sothat z2 z1 = λ(z3 z4).
Supposethat λ isrealandnon-zero,thenonevectorisamultipleoftheother.Hencebothvectorshavethesame directionif λ > 0(butmaydifferinlength)andoppositedirectionif λ < 0.Ineithercasethelines AB and CD are parallel.
Inthecasewhere λ isimaginary,thetwovectorsmustbeperpendicular,sincemultiplicationby i isequivalent toarotationby π 2 .Hence AB ⊥ CD.ThesignofIm(λ)determineswhethertherotationisanticlockwiseor clockwise.
Example29
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Let z1 and z2 beanytwocomplexnumbersrepresentingthepoints A and B inthecomplexplane.Considerthe complexnumber z givenbytheequation z z1 z2 z1 = t where t isreal.Letthepoint C represent z.
a Showthat A, B and C arecollinear.
b Henceshowthat AC : BC = |t| : |1 t|
Solution
a Firstnotethat −−→ AB represents z2 z1 andthat −−→ AC represents z z1.Since z z1 z2 z1 isrealitfollowsthat AB and AC areparallel.Furthersince A iscommontobothlines,itfollowsthat A, B and C arecollinear.
b If t < 0then −−→ AC hastheoppositedirectionto −−→ AB andtheorderofthepoints is CAB.
If t = 0then A and C coincide.
If0 < t < 1thenboth −−→ AC and −−→ AB havethesamedirectionand −−→ AB hasthegreater magnitude.Hencetheorderofthepointsis ACB
If t = 1then C and B coincide.
If t > 1thenthevectorshavethesamedirectionbut −−→ AC hasgreatermagnitude,so theorderis ABC.
Ineachcaselet |z2 z1| = r.Then: AC : AB = |z z1| : |z2 z1| = |t|r : r = |t| :1.
Henceinallcases AC : BC = |t| : |1 t|.Thethreenon-zerocasesareshownon theright.
Exercise1E
1 Inthediagramontheright, OABC isaparallelogram.Thepoints A and C represent5 + i and2 + 3i respectively.Findthecomplexnumbersrepresentedby:
a thevector OB, b thevector AC, c thevector CA.
2 Inthediagramontheright, OPQR isasquare.Thepoint P represents4 + 3i. Findthecomplexnumbersrepresentedby:
a thepoint R, b thepoint Q, c thevector QR, d thevector PR.
3 Inthediagramontheright,intervals AB, OP and OQ areequalinlength, OP is parallelto AB and ∠POQ = π 2 .If A and B representthecomplexnumbers3 + 5i and9 + 8i respectively,findthecomplexnumberwhichisrepresentedby Q.
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4 Inthediagramontheright, OABC isasquare.Thepoint A representsthe complexnumber2 + i
a Findthenumbersrepresentedby B and C
b Ifthesquareisrotated45◦ anticlockwiseabout O togive OA′ B′C′,findthe numberrepresentedby B′ .
5 Inthediagramontheright, AB = BC and ∠ABC = 90◦.Thepoints B and C represent5 + 3i and9 + 6i respectively.Findthecomplexnumbersrepresented by:
a thevector BC, b thevector BA, c thepoint A.
6 Thediagramontherightshowsasquare ABCD inthecomplexplane. Thevertices A and B representthecomplexnumbers9 + i and4 + 13i respectively.Findthecomplexnumbersthatcorrespondto:
a thevector AB, b thevertex D
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7 Inthediagramontheright,thepoints P and Q correspondtothecomplex numbers z and w respectively.Thetriangle OPQ isisoscelesandtheangle POQ is arightangle.
Provethat z2 + w2 = 0.
8 IntheArganddiagramontheright, ABCD isasquare,and OE and OF are parallelandequalinlengthto AB and AD respectively.Thevertices A and B correspondtothecomplexnumbers w1 and w2 respectively.Whatcomplex numberscorrespondtothepoints E, F, C and D?
9 Inthediagramontheright,theverticesofatriangle ABC arerepresentedby thecomplexnumbers z1, z2 and z3 respectively.Thetriangleisisosceles,and right-angledat B.
a Explainwhy(z1 z2)2 + (z3 z2)2 = 0.
b Supposethat D isthepointsuchthat ABCD isasquare.Find,intermsof z1, z2 and z3,thecomplexnumberthatthepoint D represents.
10 IntheArganddiagramontheright, OABC isarectangle,with OC = 2OA.The vertex A correspondstothecomplexnumber ω.
a Whatcomplexnumbercorrespondstothevertex C?
b Whatcomplexnumbercorrespondstothepointofintersection D ofthe diagonals OB and AC?
11 Theverticesofanequilateraltriangleareequidistantfromtheorigin.Oneof itsverticesisat1 + √3i.Findthecomplexnumbersrepresentedbytheother twovertices.
[Hint:Whatistheanglesubtendedbytheverticesattheorigin?]
12
Given z = 3 + 4i,findthetwopossiblevaluesof w sothatthepointsrepresenting0, z and w forma right-angledisoscelestrianglewithright-angleatthepointrepresenting: 0 a z b w c
13 Giventhat z1 = 1 + i, z2 = 2 + 6i and z3 = 1 + 7i,findthethreepossiblevaluesof z4 sothatthepoints representing z1, z2, z3 and z4 formaparallelogram.
14 Supposethatthecomplexnumber z hasmodulusone,andthat0 < Arg z < π 2 Provethat2Arg(z + 1) = Arg z
15 Theverticesofthequadrilateral ABCD inthecomplexplanerepresentthecomplexnumbers z1, z2, z3 and z4 respectively.
a If z1 z2 = z4 z3 thenshowthatthequadrilateral ABCD isaparallelogram.
b If z1 z2 = z4 z3 and z1 z3 = i(z4 z2)thenshowthat ABCD isasquare.
16 AtrianglehasverticesatpointsintheArganddiagramwhichrepresentthe complexnumbers z1, z2 and z3. If z2 z1 z3 z1 = cos π 3 + i sin π 3 ,showthatthetriangleisequilateral.
17 Inthediagramontheright,thepoints P1, P2 and P3 representthecomplex numbers z1, z2 and z3 respectively.If
thenshowthat
18 Let z1 = 2i and z2 = 1 + √3 i
a Express z1 and z2 inmod-argform.
b Plotinthecomplexplanethepoints P, Q, R and S representing z1, z2, z1 + z2 and z1 z2 respectively.
c Findtheexactvaluesof: Arg(z1 + z2) i Arg(z1 z
19a Provethatforanycomplexnumber
b Henceprovethatforanycomplexnumbers z
and
c Explainthisresultgeometrically.
20 Byputting z1 = r
21 Inthediagramontheright,thepoints P and Q representthecomplexnumbers z and w respectively.
a Explainwhy |z w|≤|z| + |w|
b Indicateonthediagramthepoint R representing z + w
c Whattypeofquadrilateralis OPRQ?
d If |z w| = |z + w|,whatcanbesaidaboutthecomplexnumber w z ?
22a Provethatthepoints z1, z2 and z3 arecollinearif z3 z1 z2 z1 isreal. b Henceshowthatthepointsrepresenting5 + 8i,13 + 20i and19 + 29i arecollinear.
CHALLENGE
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23 Thecomplexnumbers ω1 and ω2 havemodulus1,andarguments α1 and α2 respectively,where0 < α1 < α2 < π 2
ShowthatArg (ω1 ω2) = 1 2 (α1 + α2 π).
24 [Circle Geometry]Itisknownthatarg z4 z1 z2 z1 + arg z2 z3 z4 z3 = π.Explainwhythepointsrepresenting thesecomplexnumbersareconcyclic.
25 [Circle Geometry]Thepointsrepresentingthecomplexnumbers0, z1, z2 and z3 areconcyclicandin anticlockwiseorder.Provethatthepointsrepresenting 1 z1 , 1 z2 and 1 z3 arecollinear.
[Hint:Showthat z 1 1 z 1 2 z 1 1 z 1 3 isreal.]
1F CurvesandRegionsintheArgandDiagram
Inmanysituationsasetofequationsorconditionsonavariablecomplexnumber z yieldsasetofpointsinthe Arganddiagramwhichisafamiliargeometricobject,suchasalineoracircle.Themainaimofthissectionisto provideageometricdescriptionforacurveorregionspecifiedalgebraically.Thereforetheexamplesinthistext havebeengroupedbythevariousgeometries.
Therearetwobasicapproachesusedinthissection,algebraicorgeometric.Theadvantageofthealgebraic approachisthatmostreaderswillalreadybeproficientatmanipulatingequationsin x and y.Unfortunatelythe geometryofthesituationmaybeobscuredbythealgebra.Theadvantageofthegeometricapproachisthatitwill oftenprovideaveryelegantsolutiontotheproblem,butmayalsorequireakeeninsight.Bothmethodsshouldbe practised,withtheaimtobecomeproficientatthegeometricapproach.
StraightLines
SomesimplestraightlineshavealreadybeenencounteredinSection 1C,suchastheverticallineRe(z) = a.Here aresomeotherexamplesandtheirgeometricinterpretations.
Incoordinategeometry,giventhecoordinatesoftwopoints A and B,thetaskoffindingtheequationofthe perpendicularbisectorof AB isalengthyone.Theequivalentcomplexequationisremarkablysimple.
Example30
Let z1 = 4and z2 = 2i,andletthevariablepoint z satisfytheequation |z
a Put z = x + iy andhenceshowthat z liesonthestraightline y + 2x 3 = 0.
b Describethislinegeometricallyintermsof z1 and z2
Solution
a Substitutethevaluesof z1 and z2,thensquaretoget (x 4)2 + y2 = x2 + (y + 2)2 or(x 4)2 x2 = (y + 2)2 y2
thus 4(2x 4) = 2(2y + 2)(differenceofsquares)
so4 2x = y + 1 hence y + 2x 3 = 0.
b Let z, z1 and z2 bethepoints P, A and B intheArganddiagram.Sincethe modulusisadistance,thegivenequationyields
PA = PB
Thuseither △APB isalwaysisosceles,or P bisects AB.Hence P isonthe perpendicularbisectorof AB. A
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29Theperpendicularbisectorofaninterval
Let z1 and z2 bethefixedpoints A and B intheArganddiagram,andlet z beavariablepoint P.If |z z1| = |z z2|
then P isontheperpendicularbisectorof AB
Example31
Let z0 = a + ib bethefixedpoint T andlet z = x + iy beavariablepoint P inthecomplexplane.Itisknown that
z z0 = ikz0 , where k isarealnumber.Itisalsoknownthat P isonastraightline.
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a Findtheequationofthatstraightlineintermsof a, b, x and y.
b Whatisthegeometryofthesituation?
Solution
a Thegivenequationexpandsto
x + iy (a + ib) = ik(a + ib).
Equatingrealandimaginarypartsyields
x a = kb and y b = ka . Eliminating k fromthispairofequationsgives
b(y b) = a(x a) or ax + by = a2 + b2
b Somereaderswillrecognisethisequationasthetangenttoacircle.Thisgeometryisconfirmedby examiningthegivenequationmoreclosely.
c Sincemultiplicationby i representsarotationof π 2 ,itfollowsthatfor k 0thevector −−→ TP isperpendicular to −−→ OT .Thatis, P liesonalineperpendicularto OT .Further,when k = 0, z = z0,sothislinepasses through T .Thatis, PT isthetangenttothecirclewithradius OT ,asshownabove.
Rays
ThehorizontalandverticallinesinSection 1C andthefirstexampleabovedemonstratesomeofthegeometryof therecentlyintroducedfunctionsRe(z),Im(z)and |z|.ThenewfunctionArg(z)describesarayinthe z-plane.
Example32
Thecomplexnumber z satisfiesArg(z) = π 3
a Let |z| = r.Write z inmodulus-argumentform.
b Plot z when r = 1,2,3,4,andobservethat z liesonaray.
c Explainwhytheoriginmustbeexcluded.
d UseshiftingtosketchArg(z 2 i) = π 3 .
Solution
a z = r(cos π 3 + i sin π 3 ).
b Seethefirstgraphontheright.
c Arg(0)isundefinedsotheoriginisnotincluded.
d Arg(z 2 i) = Arg(z (2 + i))sotherayhasbeen shiftedtothepoint2 + i,asshownontheright.
30RaysintheArganddiagram
• TheequationArg(z) = θ representstheraywhichmakesanangle θ withthepositiverealaxis,omitting theorigin.
• ThegraphArg(z z0) = θ istheresultofshiftingtheaboverayfromtheorigintothepoint z0 .
CirclesandParabolas
Thecirclecaneasilybewrittenasanequationofacomplexvariable.Theparabolacanalsobewrittenas anequationin z,thoughitissomewhatcontrivedandisnotasignificantresultinthiscourse.Thegeometric definitionsofeachintermsofdistancesisthekey.
Example33
Considertheequation |z z0| = r forsomefixedcomplexnumber z0 = a + ib andpositiverealnumber r
a Explainwhythisrepresentsacircle.Statethecentreandradius.
b Confirmtheresultbyputting z = x + iy andfindingtheCartesianequation.
c Expand |z 1|2 intermsof z and z.Hencedeterminethecurvespecifiedby |z|2 = z + z
Solution
a Theequationspecifiesthatthedistancebetween z and z0 isfixed.Thisisthe geometricdefinitionofacircle.Thecentreis z0 andtheradiusis r
b Beginbysquaringbothsides:
c From |w|2 = w w itfollowsthat |z 1|2 = (z 1) (z 1)
= (z 1)(z 1)
= |z|2 (z + z) + 1.
Since |z|2 = z + z ,italsofollowsthat
|z 1|2 = 1, thatis,thecirclewithcentre z = 1andradius1.
31CirclesintheArganddiagram
Let z0 bethefixedpoint C intheArganddiagram,andlet z beavariablepoint P.If |z z0| = r
thenthepoint P isonthecirclewithcentre C andradius r
Example34
Let S bethefixedpoint1 + 2i and z bethevariablepoint P intheArganddiagram.Itisknownthat |z (1 + 2i)| = Im(z).Showalgebraicallythat P lies onaparabolabyputting z = x + iy
Solution
Squaringbothsidesofthegivenequation: (x 1)2 + (y 2)2 = y2 so (x 1)2 = y2 (y 2)2 = 4y 4 thus (x 1)2 = 4(y 1).
Thisistheequationofaparabolawithvertexat1 + i,asshownontheright. x y
Thislastexamplemayseemalittlecontrived,butitdemonstratesthattherearemanysituationswherefurther insightintoaproblemmaybeachievedbyconsideringacorrespondingprobleminthecomplexplane.
Regions
Inmanyinstancesacurvedividestheplaneintotwoormoreregions.Insimplecasesaregionisdefinedby thecorrespondinginequation.Whenmoreintricateregionsarerequired,twoormoresimpleregionsmaybe combinedbytakingtheunionorintersectionofthecorrespondinginequations.Somecommonexamplesfollow.
Example35
Sketchthefollowingregionsinthecomplexplane.
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Solution
Thefirstthreecanbeeasilyexplainedgeometrically.
Thisis1 ≤ x ≤ 3,theverticalstrip between x = 1and x = 3.
Put z = r cis θ toget0 ≤ θ ≤ π 4 , whichdefinesawedgeexcludingthe origin,sinceArg(0)isundefined.
Theboundarycurveisthecircle withradius1andcentre2 i,andis notincluded.Theregionincludes thecentreofthecirclesincethe LHSoftheinequalityiszerothere. c x y i
Theperpendicularbisectorofthesegment from0to 2 + 2i istheboundary, andisnotincluded.Theregion includesthepoint 2 + 2i,since theRHSoftheinequalityiszerothere.
Example36
a Sketchtheregions |z i|≤ 1and i
b Hencesketch theunionand i theintersectionoftheseregions. ii
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b Theboundariesintersectat 1 + i and,fromtrigonometry,theyintersectagainat
)and 1 2 + i (1 √3 2 ).Herearethegraphs.
CircleGeometry
ManyofthecirclegeometrytheoremsencounteredinYear10maybeexpressedintermsofacomplexnumber. Onesignificantresultisincludedhere,withotherexamplestobefoundintheexercise.
Example37
Let z1 = 3 + i and z2 = 1 i.
Describeandsketchthesetofpoints z,wherearg z z1 z z2 = π 4 .
Solution
Let z1, z2 and z representthepoints A, B and P respectively.Firstnotethattheequationcanbewrittenas arg(z z1) arg(z z2) = π 4
Recallthat z z1 isthevector −−→ AP,soarg(z z1)isthedirectionofthisvector.Likewisearg(z z2)isthe directionofvector −−→ BP.Thusthedifferenceistheanglebetweenthemandisalways π 4 .Usingtheconverse oftheanglesinthesamesegmenttheorem,itfollowsthat P mustlieonthearcofacirclewithchord AB Further,since
∠APB = π 4 < π 2
itisamajorarc.As P movesalongthearcfrom A to B,itmovesanticlockwise aboutthecentre,becauseanglesaremeasuredanticlockwise.Lastly,sincearg(0)is undefined,theendpointsofthearcarenotincluded.Itsimplyremainstofindthe centreandradiusofthiscircle.Let C bethecentre,then
∠ACB = π 2 (anglesatthecentreandcircumference) hence ∠CAB = π 4 (baseanglesofisoscelestriangle).
SinceArg(z1 z2) = π 4
itfollowsthat AC ishorizontal,asthealternateanglesareequal.Thus BC isvertical.Hence C = 1 + i isthe centreofthecircleand AC = 2isitsradius.
32Thearcofacircle
Letpoints A and B representthecomplexnumbers z1 and z2.Letthevariablepoint P represent z.The equation Arg z z1 z z2 = α where0 < α < π , isthearc AB ofacirclewiththeendpointsexcluded.As P movesalongthearcfrom A to B itsmotionis anticlockwiseaboutthecentre.
Exercise1F
1 SketcheachstraightlinebyusingtheresultofBox 29,thenfinditsCartesianequation.
3
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2 Sketchtheraysspecifiedbythefollowingequations.Box 30 maybeofhelp.
3 UseBox 31 tosketchthesecircles.
4 GraphthefollowingregionsintheArganddiagram.
5 In eachcasesketchtheboundaryorboundariesoftheregionandthenshadetheregion.
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DEVELOPMENT
6 Ineachcasesketch i theintersectionand ii theunionofthegivenpairofregions.
7 Put z = x + iy tohelpsketchthesehyperbolas. z2 (z)2 = 16i a z2 (z)2 = 12i
8 Ineachcasethegivenequationrepresentsaparabola.FindtheCartesianequationbyputting z = x + iy,and hencesketchtheparabola.
9 Byputting z = x + iy,orotherwise,sketchthegraphdefinedbytheequation:
z) = |z| a
10 Sketchthearcsofcirclesspecifiedbythefollowingequations,showingthecentreandradiusineachcase.
11 Acomplexnumber z satisfiesarg z = π 3 .
a Useadiagramtoshowthat |z 2i|≥ 1.
b Forwhichvalueof z is |z 2i| = 1?
12 ConsiderthegraphintheArganddiagramontheright.
a Writedownanequationforthisgraphintermsof z
b Findthemodulusandargumentof z atthepointwhere |z| takesitsminimum value.
c Hencefind z inCartesianformwhen |z| takesitsleastvalue.
13a Acomplexnumber z satisfies |z 1| = 2.Drawadiagramandhencefindthegreatestandleastpossible valuesof |z|.
b If z isacomplexnumbersuchthatRe(z) ≤ 2 and |z 3| = 2,showwiththeaidofadiagramthat 1 ≤|z|≤ √7.
14a Acomplexnumber z satisfies |z 2| = 1.
Sketchthegraphof |z 2| = 1. i
Showthat 1π 6 ≤ Arg z ≤ 1π 6 ii
b Thecomplexnumber z issuchthat |z| = 1.Useyouranswerstopart a toexplainwhy 1π 6 ≤ Arg(z + 2) ≤ 1π 6
15 Thecomplexnumber w satisfies |w| = 10and0 ≤ Arg w ≤ π 2 ,andthecomplexnumber z isspecifiedby z = 3 + 4i + w
a Sketchthegraphof z = 3 + 4i + w
b Useyoursketchtodeterminethemaximumvalueof |z| .
c Whatisthevalueof z forwhichthismaximumoccurs?
16a Showthatthecircleequation |z z0| = r isequivalentto z z (z z0 + zz0) + z0 z0 r 2 = 0.
[Hint:Squarebothsidesof |z z0| = r andusetheresult |w|2 = w w .]
Usepart a towritetheseequationsintheform |z z0| = r,andstatethecentreandradiusofeachcircle.
z z + 2(z + z) = 0 i
z z (1 + i)z (1 i)z + 1 = 0 ii 1 z + 1 z = 1 iii
17 Thevariablepoint P correspondstothecomplexnumber z = √2(p + 1 p i).Thispointtracesoutacurveas theparameter p changes.
a FindtheCartesianequationofthiscurve.Beginbyputting z = x + iy.Equaterealandimaginaryparts, theneliminatetheparameter p
b Thepoint P isrotatedclockwiseby π 4 abouttheoriginto Q.Let Q correspondtothecomplexnumber w. Usemultiplicationbyanappropriatecomplexnumbertoexpress w intermsoftheparameter p
c FindtheCartesianequationofthecurvethat Q tracesoutas p varies.
18 Sketchthegraphofthepoints z forwhich z 1 z i is: real, a imaginary(non-zero). b
19
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Sketchthegraphof: arg(z + i) = arg(z 1) a arg(z + i) = arg(z 1) + π b
20a Thevariablecomplexnumber z satisfies |z 2 i| = 1.Useadiagramtofindthemaximumandminimum valuesof:
|z| i |z 3i| ii
b Acomplexnumber z satisfies |z| = 3.Useasketchtofindthegreatestandleastvaluesof |z + 5 i|
c Thevariablecomplexnumber z satisfies |z z0| = r.Useasimilarapproachtoparts a and b tofindthe maximumandminimumvaluesof:
|z| i |z z1| ii
d Confirmyouranswerstothepreviouspartsbyusingthetriangleinequality |z|−|w| ≤|z + w|≤|z| + |w| .
CHALLENGE
21 Describethegraphofarg z z1 z z2 = α,where α isconstant,if:
= 0
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22 [Pythagoras]Describethegraphdefinedbytheequation |
23 Thevariablepoint P correspondstothecomplexnumber z = p + p2i.Itshouldbeobviousthat P lieson theparabola y = x2.Thisparabolaisrotatedthroughanacuteanglesothatitsaxisofsymmetryistheline y = x.UsemultiplicationofcomplexnumberstodeterminetheCartesianequationofthisnewparabola.
24 Supposethat k|z z1| = ℓ|z z2|,where k ℓ andbotharepositiverealnumbers.
a Showthatthegraphspecifiedbytheequationisacirclewithcentre
byletting z = x + iy, i bygeometricmethods. ii
b Whathappensinthelimitas k approaches ℓ?
Chapter1Review
Reviewactivity
• Createyourownsummaryofthischapteronpaperorinadigitaldocument.
Chapter1Multiple-choicequiz
• Thisautomatically-markedquizisaccessedintheInteractiveTextbook.AprintablePDFWorksheetversionis alsoavailablethere.
ChapterReviewExercise
1 If z = 3 i and w = 17 + i,find:
2 Writeasaproductoftwocomplexlinearfactors.
3 Solveeachquadraticequationfor z.
4 Findthesquarerootsof:
5 Solvefor z:
6 Expresseachcomplexnumberinmodulus-argumentform.
7 ExpresseachcomplexnumberinCartesianform.
8 Simplify:
9 Findtheresultwhen z = 2 + i isrotatedabouttheoriginby:
10 Sketchthegraphinthecomplexplanerepresentedbytheequation:
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11 Shadetheregioninthecomplexplanethatsimultaneouslysatisfies |z|≥ 1,Re(z) ≤ 2 and
12
Supposethat z = 1 + √3 i and w = 1 + i
a Find z w intheform a + ib,where a and b arereal.
b Write z and w inmodulus-argumentform.
c Hencewrite z w inmodulus-argumentform.
d Deducethatcos 5π 12 = √6 √2 4
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13
Sketchthegraphspecifiedbytheequation:
14 Atriangle PQR inthecomplexplaneisisosceles,with ∠P = 90◦.Thepoints P and Q representthecomplex numbers4 2i and7 + 3i respectively.Itisalsoknownthatthepoints P, Q and R areinanticlockwiseorder. Findthecomplexnumbersrepresentedby: thevector PQ, a thevector PR, b thepoint R c
15 If z1 = 4 i and z2 = 2i,findineachcasethetwopossiblevaluesof z3 sothatthepointsrepresenting z1, z2 and z3 formanisoscelesright-angledtrianglewiththeright-angleat: z1 a z2 b
16 InanArganddiagram, O istheoriginandthepoints P and Q representthecomplexnumbers z1 and z2 respectively.
Iftriangle OPQ isequilateral,provethat z1 2 + z2 2 = z1z2
17 If z1 = 2cis π 12 and z2 = 2i,find: arg(z1 + z2) a arg(z2 z1) b
18 If z1 and z2 arecomplexnumberssuchthat |z1| = |z2|,provethat arg(z1z2) = arg (z1 + z2)2 .
19 If z = cis θ,showthat z2 1 z2 + 1 = i tan θ
20 Thepoints A, B, C and O representthenumbers z, 1 z ,1and0respectivelyinthecomplexplane.Giventhat 0 < arg z < π 2 ,provethat ∠OAC = ∠OCB
21a Bydrawingasuitablediagram,provethetriangleinequality |z1 z2|≥|z1|−|z2| b Hencefindthemaximumvalueof |z| giventhat z 4 z = 2.
2 Proof
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Chapterintroduction
TherearemanyopportunitiesforpresentingorreadingproofsintheExtension1andExtension2courses.In thischapter,thebasicconcepts,terminologyandnotationarediscussed,andafewcommonmethodsofproof aredemonstrated.
Logicunderpinsproofandsothechapterbeginswithtwosectionsonbasiclogicprinciples.Thesesectionsare necessarilylongasthereisalotofmaterialtoinclude.Theyarealsodifficult,insomeinstancesbecausethe workseemsnottofollowintuition.Whilstthesesectionsareimportant,thetimedevotedtothemshouldnotbe tothedetrimentoftherestofthechapter.
ThetwokeyconceptsinSections 2A and 2B areprovenimplications,whicharealwaystrue,andlogically equivalentstatements.Theseareappliedinthevarioustypesofproofpresentedintheremainderofthechapter.
Section 2C investigatesafewsimpleproblemsinnumbertheory,whichisastudyofthestructuresofthe integers.ProofsbycontradictionandbycontrapositionarepresentedinSection 2D.Algebraicinequalities areproveninSection 2E,whilstSection 2F introduceshardertypesofinduction.Thefinalsectionconsiders inequalitiesincalculusandgeometry,andrequiresintegration.ItmaybeappropriatetopostponeSection 2G untilafterChapter6.
2A LogicandTheLanguageofProof
Thissectionandthenextintroducebasicconcepts,terminologyandnotationthatarefundamentaltotheproofs demonstratedthroughouttheremainderofthechapter.
Statement
Statementsarethebasicbuildingblocksofproof.Despitebeingsofundamental,itisratherdifficulttodefine whatastatementis.TwoofthedefinitionsgivenintheMacquarieDictionaryareasfollows:
1 somethingstated
2 acommunicationordeclarationinspeechorwritingsettingforthfacts,particulars,etc.
Thelatterisperhapsmorerelevanttomathematicalproof.Thusastatementcanbeasimplesentence,suchas: n isamultipleof3.
Itmayalsobeamathematicaldeclaration,suchas: let p beaprimenumber.
Astatementmaytaketheformofanassertionordefinition,suchas: evennumbersaredivisibleby2, oritmaybeaclaimtobeprovenoradeductionwhichhasalreadybeenproven,suchas: everymultipleof6isalsoamultipleof3.
Althoughthesefourexamplesdonotsaypreciselywhatastatementis,theywillenablediscussionsabout statementswithsomeagreementaboutwhatismeant.
PropositionalSymbol
Itissometimesconvenienttomanipulatestatementsinanalgebraicway.Acapitallettercalleda propositional symbol willbeusedinthiscoursetorepresentastatement.Hereisanexamplefromthepreviousstatements.
P : n isamultipleof3.
Thismaybereadas
Let P bethestatement nisamultipleof3
Theletter P isusedasitisthefirstletteroftheword proposition,asynonymforastatement.
Whereaprobleminvolvesmultiplestatementsitisusualtoassigncapitalletterssequentiallystartingat P.Here arethreerelatedexamples.
P : n isamultipleof3
Q : n isamultipleof2
R : n isamultipleof6.
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1Propositionalsymbol
Astatementmayberepresentedbyacapitallettercalledapropositionalsymbol.Whereaproblem involvesmultiplestatementsitisusualtoassigncapitalletterssequentiallystartingat P
LogicalValues
Everystatementmusttakeoneoftwologicalvalues:trueorfalse.Usingproposition P above, P istruewhen n = 6,but P isfalsewhen n = 7.
Noticeinthisexamplethatthelogicalvalueofastatementmaychangewithcircumstance,butitcannotbe simultaneouslybothtrueandfalse.Thereisnointeger n whichisbothamultipleof3andnotamultipleof3. Likewiseastatementcannotbeneithertruenorfalse.Givenaninteger n,itmustalwaysfallintooneoftwo categories:amultipleof3,ornotamultipleof3.
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2Logicalvalues
Astatementmusttakeoneoftwologicalvalues:trueorfalse.
AStatementthatisAlwaysTrue
Whilstsomestatementsmaychangelogicalvalueduetocircumstance,otherstatementsneverchangetheir logicalvalue.Considerthesestatements:
S :16isanoddnumber, T :17isaprimenumber. Clearly S isalwaysfalseand T isalwaystrue.Notmuchmorecanbesaidatthispoint.Later,however,itwillbe seenthatstatementsthatarealwaystruearefundamentaltothewayproofswork.
AProvenStatement
Astatementthatcanbeshownbysomemeanstobealwaystrueissaidtobeproven.Forexample, P :34isaFibonaccinumber isalwaystrue,asisdeterminedbysimplywritingoutthefirstfewterms: 1,1,2,3,5,8,13,21,34, ...
Astatementwhichisshowntobetrueissaidtobe proven,andtheevidenceusedtoestablishthetruthiscalled the proof.Thus,inthiscase P isproven,andtheproofisthelistingofthefirstfewtermsofthesequence.
3Aprovenstatement
Astatementwhichisshowntobetrueissaidtobeproven,andtheevidenceusedtoestablishthetruthis calledtheproof.
ExamplesandCounterexamples
Whilstitisrarelyfeasible,astatementcanbeshowntobeprovenbyexample,butcaremustbetakentoexamine everypossiblecase.Onestatementwherethisisfeasibleis: therearethreeprimesonaregulardie.
Theprimesonadieare2,3and5,sothestatementisproven.Clearlythisisacontrivedsituation,butitservesas areminderthatproofbyexhaustiveexamplesissometimespossible.
Incontrast,thestatement allprimesareodd isclearlyfalse,since2isbothprimeandeven.Whenastatementisshowntobefalsebyexampleinthisway, thatexampleiscalleda counterexample.Thestatementissaidtobe disproved,meaningthatitisnotalwaystrue. Itisimportanttonotethatonlyonecounterexampleisneededinordertoshowthatastatementisdisproved.
4Acounterexample
Anexamplethatshowsthatastatementisnotalwaystrueiscalledacounterexampleandthestatementis saidtobedisproved.
Negation
The negation ofastatementchangesitslogicalvalue.InEnglish,negationiscommonlyassociatedwiththeword not .Thus,whilstthestatement allMondaysareschooldays isclearlyfalse,itsnegation notallMondaysareschooldays isequallyclearlytrue.InEnglish,thislaststatementmightbeconsideredclumsy,andsomaybereplacedwith someMondaysareholidays.
Thus,whilstthepresenceoftheword not clearlyindicatesanegation,otherwordscanbeusedtoproducea negation.
Thenotation ¬P isusedforthenegationofproposition P inthesenotes.Othernotationsinclude ∼P and!P,the latterbeingcommonincomputing.When P istruethevalueof ¬P isfalse.Likewise,when P isfalsethevalue of ¬P istrue.
5Negation
Thenegationofastatement P is ¬P,readas notP.When P istruethevalueof ¬P isfalse.When P is falsethevalueof ¬P istrue.
TruthTables
Firstly,itmustbeemphasisedthattruthtablesarenotpartofthecourse.Nevertheless,theyhavebeenincluded inthistextbook,asatruthtableisavaluabletoolinpropositionallogic.Itisoftenthecasethatatruthtablecan greatlysimplifyaverydifficultproblem.
Whenaprobleminvolvesmultiplestatements,thedifferentlogicalvaluesofthestatementsmaybedisplayed inatruthtable.Theletters T or F cannotbeusedfortrueorfalsebecauseuppercaselettersarereservedfor propositionalsymbols.Inthisbook,1representstrueand0representsfalse,owingtotheiruseincomputing. Othersymbolsusedfortrueinclude t or ⊤ andsymbolsforfalseinclude f or ⊥.Bywayofexample,hereisthe truth tableforthetwostatements P : nisamultipleof3 and Q : nisamultipleof2.
P Q Example
1 1 n = 6
1 0 n = 3
0 1 n = 4
0 0 n = 5
Examplesarenotnormallyincludedinatruthtable.Theyhavebeenincludedheretoshowhoweachcombinationoflogicalvaluesispossible.
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Equivalence
Introducingsomemorenotation,thesymbol ≃ isusedtoindicatethattwoexpressionsarelogicallyequivalent. Thatis,theyhavethesamelogicalvaluewithineachrowofthetruthtable.Forexample,letpropositions Q and R be
Q :6 = 2 × 3 andlet
R :ChristmasDayisapublicholiday.
Thepropositionssaydifferentthingssotheyarenotequal.Nevertheless,botharetrue,so Q ≃ R.Thisisrather likeusing ∼ forsimilartriangles:thetwoobjectsarenotthesamebuttheysharecharacteristics.Unfortunately, somesourcesuse = insteadof ≃ forequivalentstatements.
Thepreviousexampleiscontrived.Hereisamorerealisticproblem. Experiencesuggeststhatadoublenegationeffectivelycancelsout.Indeeditis correctthat
¬¬P ≃ P,
asseeninthetruthtableontheright.When P istrue,sotoois ¬¬P andwhen P isfalse,sotoois ¬¬P
Further,itisseeninthisequivalencethatnegationbehaveslikethenegativesignofarithmetic:twonegatives essentiallycancelout.
TheLogicalConnectorsAnd&Or
Twoormorepropositionscanbecombinedusingthelogicalconnectors:
∧ whichisreadas and
∨ whichisreadas or
Thelogicalexpression P ∧ Q istruewhen P and Q arebothtrue,otherwiseitisfalse.Incontrast, P ∨ Q istrue whenatleastoneof P or Q istrue,otherwiseitisfalse.Thismaybesummarisedinatruthtable.Therearefour rowsbecauseeachof P and Q canbetrueorfalse.
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Example1
Let F = {1,2,3,5} bethesetofFibonaccinumbersonaregulardieand T = {1,3,6} bethesetoftriangular numbers.
a Write F ∪ T inwordsthenwritedowntheelementsoftheset.
b Let P : x ∈ F andlet Q : x ∈ T betwostatements. Write P ∨ Q inwordsthenlistthevaluesof x when P ∨ Q istrue.
Solution AVenndiagramofthesituationisshownbelowontheright.
a Theunionofthesetsis
F ∪ T = thenumbersin F or T = {1,2,3,5,6}
b Combiningthestatementswith ∨ gives P ∨ Q istruewhen x ∈ F or x ∈ T otherwiseitisfalse,so
P ∨ Q istruewhen x ∈{1,2,3,5,6}
Continuedonthenextpage
Noticethattheanswerstoparts a and b arealmostidentical.Thisishardlysurprising.TheEnglish mathematicianJohnVennoriginallycreatedhisdiagramstosolveproblemsinlogic.Hisuseofshading, however,differsfrommodernuse.Keenreadersmaycaretoinvestigatethisfurther.
6ThelogicalconnectorsAnd&Or
Let P and Q betwostatements:
• P ∧ Q istruewhen P istrueand Q istrue,otherwise PandQ isfalse.
• P ∨ Q istruewhen P istrueor Q istrue,otherwise PorQ isfalse.
OrderofPrecedence
Thevalueofanarithmeticexpressiondependsentirelyontheorderofoperations.Thus
2 × 3 + 4 = 10.
Iftheorderneedstobechangedthenbracketsareused,suchasin
2 × (3 + 4) = 14.
Likewise,thevalueofalogicalexpressiondependsentirelyontheorderinwhichthelogicalconnectorsare applied,calledthe orderofprecedence.Theconnectorsmetsofarhavebeenintroducedintheirorderof precedence.Thatis,apply
¬ before ∧ then ∨ Bywayofexample,let P : 8iseven and Q : 9isprime.Clearly P istrueand Q isfalse.Theexpression ¬P ∧ Q thenbecomes
¬P ∧ Q ≃¬1 ∧ 0
0 ∧ 0
0.
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8isnotevenand9isprime, botharefalse, sothestatementisfalse.
Thus,thevalueisfalsebecausethe not isappliedbeforethe and.Further,likearithmetic,bracketscanbeusedto changetheorderofprecedence.Forexample,withthesame P and Q,
¬(P ∧ Q) ≃¬(1 ∧ 0) ≃¬0 ≃ 1. notboth8isevenand9isprime, whichisnotfalse, sothestatementistrue. Thatis,theanswerisnowtruebecausethe and isappliedbeforethe not.
7Orderofprecedence
Inlogicalexpressions, not isappliedbefore and whichisappliedbefore or.Bracketsareusedtochange theorderofprecedence.
And,Or&Negation
Inarithmetic,bracketsinanexpressioncanbeexpandedbyusingthedistributivelaw.Thus 2 × (3 + 4) = 2 × 3 + 2 × 4.
Giventheprevioussimilaritiesbetweenarithmeticandlogic,itseemsreasonabletoask:istheresomethinglike thedistributivelawinlogic?Theansweris: yes.Herearetherulesforthenegationof and and or:
¬(P ∧ Q) ≃¬P ∨¬Q,
¬(P ∨ Q) ≃¬P ∧¬Q
Noticethatineachcasethe not isdistributedtoeachpropositionandthelogicalconnectorisswapped.Theseare knownas deMorgan’slaws aftertheBritishmathematicianAugustusdeMorgan.
Example2
Supposethataschoolofferstwolanguages:FrenchandJapanese.Studentsmaychoosetostudyonesubject ortheother,ormaystudybothsubjects.Foraparticularstudent x,considerthepropositions P : x ∈ F,the studentstudiesFrench,and Q : x ∈ J,thestudentstudiesJapanese.
a Suppose ¬(P ∨ Q)istrue.Writethisinwords.
b Explainwhy ¬P ∧¬Q isalsotrue,thenwritethisinwords.
Solution
a ThestudentdoesnotstudyFrenchorJapanese.
b ¬(P ∨ Q) ≃¬P ∧¬Q,soifoneistruethensomusttheother.ThestudentdoesnotstudyFrenchanddoes notstudyJapanese.
8And,Or&Negation
Therulesfornegationof and and or statementsare:
• ¬(P ∧ Q) ≃¬P ∨¬Q
• ¬(P ∨ Q) ≃¬P ∧¬Q
And,Or&Negation—Enrichment
TheequivalenceoftheexpressionsinBox 8 iseasilydemonstratedintruthtables.Hereisthetableforthefirst law.
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Althoughstudentsarenotexpectedtoconstructthistable,theequivalenceofthetwoexpressionsisclearly evidentinthe4thand7thcolumns.Thetablefor ¬(P ∨ Q) ≃¬P ∧¬Q isleftasanenrichmentexercise.
Implication
AnexpressioninEnglishlike ifitisrainingthenIamwet isreferredtoinlogicasan implication.Anarrowindicatesanimplication.Thus,putting P : itisraining and Q : Iamwet,theaboveimplicationiswrittenas P → Q or P =⇒ Q
It maysurprisesomethatthelogicalvalueofanimplicationmaybeeithertrueorfalse.Perhapsitisraining whichiswhyIamwetsothattheimplicationistrue.OntheotherhanditmayberainingbutIamdrybecauseI aminsidemyhouseinwhichcasetheimplicationisfalse.Anotherpossibilityisthatitisnotraining.Inthatcase theimplicationistrueregardlessofwhetherornotIamwet.Thereasoninghereisthatanythingcanbededuced fromafalsehood,andtheimplicationmaybedescribedas vacuously true.
Hereisthetruthtablesummarisingallthis.
P Q P → Q
1 1 1
1 0 0
0 1 1
0 0 1
Hereisamoremathematicalexampleshowinghowitworks.
Example3
Let P : xisamultipleof6 andlet Q : xisdivisibleby9.Completeatruthtablefor P → Q whichincludesan exampleof x foreachrow.
Solution
TheimplicationinEnglishis: if x isamultipleof6thenitisdivisibleby9.
Therearesomevaluesof x whichmaketheimplicationtrue,suchas x = 18.Somereadersmayinstinctively thinkthattheimplicationisfalsebecause24isdivisibleby6butnotby9.Moreprecisely, x = 24isa counterexamplewhichshowsthattheimplicationisdisproved(notalwaystrue).Hereisthetable.
P Q P → Q x 1 1 1 18 1 0 0 24
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Noticethat P → Q isvacuouslytrueinthelasttworowsofthetable.
9Implication
Asentenceoftheform ifPthenQ iscalledanimplication.Itiswritteninsymbolsas
P → Q or P =⇒ Q
Theimplicationisfalsewhen P istrueand Q isfalse,otherwiseitistrue.
Itisinstructivetoexpressanimplicationintermsofsets.So,put
A = {themultiplesof6} and B = {numbersdivisibleby9} sothatthepropositionsinthelastexampleare
P : x ∈ A and Q : x ∈ B.
IntheVenndiagramshownontheright,thethreeshadedregionscorrespondtorows1, 3and4ofthetruthtableabovewhere P → Q istrue.Lookingcarefully,theshaded partsareequivalentto A ∪ B.Thissuggeststhat
P → Q ≃¬P ∨ Q.
Thisisindeedalwaysthecase.Aproofmightuseatruthtableandthisisleftasanenrichmentexercise.
Itisnoweasytoshowthat P → Q isalsologicallyequivalentto ¬(P ∧¬Q).
¬(P ∧¬Q) ≃¬P ∨¬¬Q (bydeMorgan’slaw)
≃¬P ∨ Q (doublenegation)
≃ P → Q (shownabove).
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10Otherformsofimplication
Theimplication P → Q islogicallyequivalentto ¬P ∨ Q or ¬(P ∧¬Q).
Animplicationcomesbelow or intheorderofprecedence.Thusintheexpression
P →¬Q ∨ P
the not isevaluatedfirst,thenthe or andtheimplicationisevaluatedlast.
TheNegationofanImplication
Itisoccasionallyusefultoconsiderthenegationofanimplication,viz.
¬(P → Q).
Notethatthebracketsarenecessaryherebecauseoftheorderofprecedenceofthelogicalconnectors.An equivalentstatementinEnglishmightbe
P doesnotimply Q, butthiscannotbere–wordedasasimple if...then statementinanypracticalway.Nevertheless, ¬(P → Q)canbe expressedinasimplemannerbyapplyingthesecondformulainBox 10.Since
P → Q ≃¬(P ∧¬Q), negatingbothsidesgives
¬(P → Q) ≃¬¬(P ∧¬Q)
≃ P ∧¬Q (doublenegation)
Example4
Considerthestatement: ifthisisanowlthenitisabird.
a Writeanequivalentstatementusing or.
b Writethenegationasanequivalentstatementusing and Solution
a Thisisnotanowloritisabird.
b Thisisanowlanditisnotabird.
ProvenImplications
Asageneralrule,mathematiciansareonlyinterestedinthingsthatareknowntobealwaystrue,thatis,proven. Animplicationthatisalwaystrueiscalleda provenimplication.Mostoftheproofsintheremainderofthis chapterutiliseprovenimplications.
Bywayofexample,considersolving2x + 4 = 10.Thefirsttwolinesmightbe
2x + 4 = 10, 2x = 6.
Theselinesofworkingtogetherformanimplication,butwithoutthelogicalsymbols.Whentheworkingis explained,thelogicisrevealed.
If x isasolutionof2x + 4 = 10then x isasolutionof2x = 6. Thustheworkingcanbewritteninlogicalsymbolsas:
P : x isasolutionof2x + 4 = 10
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Q : x isasolutionof2x = 6
P → Q
Theimplicationisprovenasfollows.
When x = 3both P and Q aretruesotheimplicationistrue. Amistakeinthecalculationsmightgiveanyanswersuchas x = 4.
Oddly,when x = 4theimplicationistrue.
Thisisbecause P isfalsesotheimplicationisvacuouslytrue.
Example5
Astudentwrotethefollowingwhenansweringaquestion.
If x isamultipleof6then x iseven.
Rewritetheimplicationinsymbolsthenprovetheimplication.
Solution
Let P : xisamultipleof6 and Q : xiseven,thenthestatementcorrespondswith P → Q.
When P istrue,let x = 6m forsomeinteger m.Then
x = 6m = 2 × (3m)
Hence x isamultipleof2.Thatis, x isevenandtheimplicationistrue.
Thisisprobablyallthatwouldberequiredinatest.Forthesakeofcompleteness,however,itshouldalsobe statedthatwhen P isfalsetheimplicationisvacuouslytrue.Thus, P → Q isprovenregardlessofwhether P is trueorfalse.
When P → Q isaprovenimplication,itissometimesstatedthat Q isalogicalconsequenceof P.Thus: x isevenasalogicalconsequenceofitbeingamultipleof6.
OtherwordsusedinEnglishforsuchlogicaldeductionsinclude:hence,thus,so,sothat,consequently, andtherefore.ThesewordshavebeenregularlyseenintheproofsofvarioustheoremsgiveninYear11 mathematics.
SometimestheorderofthetwostatementsisreversedinEnglish,socaremustbetakentodeterminewhich statementisthelogicalconclusionoftheother.Forexample,thelogicalorderofthededuction a2 isoddbecause a isodd ismadefarclearerwhenitisre-writtenas:if a isoddthen a2 isodd.Itisgoodpracticewhenwriting mathematicstoputtheconclusionsecondinthisway.
Continuedonthenextpage
Twootherwordscommonlyassociatedwithprovenimplicationsare each and all .Hereisasimilarexample writtenallthreeways.
Alloddsquaresareonemorethanamultipleof4. Eachoddsquareisonemorethanamultipleof4.
Ifanumberisoddthenitssquareisonemorethanamultipleof4.
11Provenimplications
Aprovenimplicationisalwaystrue.
Exercise2A
1 Identifyeachofthesesymbols.
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FOUNDATION
2 Indicatewhethereachstatementistrueorfalse.
a Onecounterexampleisnotsufficienttodisproveastatement.
b Negationalwayschangesthelogicalvalueofastatement.
c A =⇒ B meansthat A isalogicalconsequenceof B
d Thenegationofan and statementisan or statement.
e Animplicationcanbewrittenasan if...then... statement.
f Theword or isassociatedwiththeintersectionoftwosets.
3a Whichstatementdoesnothavealogical(ortruth)value?
Someintegersareprimenumbers. A Australiadoesnothaveaprimeminister. B HistoryismoreinterestingthanScience. C Allcarsarered. D
b Whichstatementhasthesamemeaningas‘P istruewhenever Q istrue’? If P istruethen Q istrue. A If Q istruethen P istrue. B P and Q arebothtrue. C Onlyoneof P and Q istrue. D
c Supposethat P istrueand Q isfalse.Whatisthelogicalvalueof P =⇒ Q? True. A False. B Itcouldbetrueorfalse. C Itcannotbedetermined. D
d Whatisthenegationof‘P istrueand Q istrue’? P isfalseor Q isfalse. A P istrueor Q istrue. B P isfalseand Q isfalse. C Either P isfalseor Q isfalse. D
e Whatisthenegationof‘P istrueor Q isfalse’?
P and Q arebothfalse. A
P istrueand Q isfalse. B
P isfalseor Q istrue. C
P isfalseand Q istrue. D
f Whatisthenegationof‘if P istruethen Q istrue’?
P istrueand Q isfalse. A
P isfalseand Q istrue. B
If P isfalsethen Q istrue. C
If P isfalsethen Q isfalse. D
4 Writedownthenegationofeachstatement.
Allcarsarered. a a > b b Hillarylikessteakandpizza. c BilliscorrectorDaveiscorrect. d
IfIliveinTasmaniathenIliveinAustralia. e IfNikhildoesn’tstudythenhewillfail. f 3 ≤ x ≤ 8 g x < 5or x ≥ 0 h
5 Aten-sideddiehasthenumbers1to10onitsfaces.Let F bethesetofFibonaccinumbersand P thesetof primenumbersonthedie.
a Listtheelementsbelongingto:
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b Confirmthat:
6 Statewhethereachstatementistrueforall x ∈ R.Ifitisfalse,provideacounterexample.
7 Statewhethereachstatementistrueforall a, b ∈ R.Iffalse,provideacounterexample. If a > b then a2 > b
8 Giventhat ∧ isthelogicsymbolfor‘and’and ∨ isthelogicsymbolfor‘or’,rewriteeachexpressioninan equivalentformwithoutbrackets.
¬(¬P) a ¬(P ∧ Q) b
¬(P ∨ Q) c ¬(¬P ∧ Q) d
¬(P ∨¬Q) e ¬(¬P ∧¬Q ∧¬R) f
¬(P =⇒ Q) g ¬(P =⇒¬Q) h
9
[Thisproofaimstoestablishthat2 = 3.]
Let x = 2and y = 3. =⇒ x2 5x = 6and y2 5y = 6 (Line1)
(Line2)
(Line3)
(Line4)
(Line5)
(Line6) =⇒ 2 = 3 (Line7)
Inwhichlineistheimplicationnottrue?
10 Supposethat P isthestatement‘JackdoesMathematicsExtension2’andthat Q isthestatement‘Jackis crazy’.WriteeachofthefollowingasEnglishsentences.
11 Considerthisstatement: ‘IfeitherAnnaorBryanpassedtheexam,theneitherAnnaandChrisbothpassedorBryanandChrisboth passed.’
Thestatementisknowntobefalse.DeterminewhetherChrispassedorfailed.
12 Considerthecompoundstatement(P ∨ Q) ∧ (¬P).Giventhatthisstatementistruecommentonthelogical (ortruth)valuesof P and Q
13 Animplicationisfalseifthepremiseistruebuttheconclusionisfalse.Otherwiseitistrue.Statewhether eachimplicationistrueorfalse.
a If27isaprimenumberthen27isanoddnumber.
b IftheprimeministerofAustraliaisarobotthentheprimeministerofNewZealandisarobot.
c Ifanumberisoddthenallitsprimefactorsareodd.
d If p isthelargestknownprimenumberthenitcouldbethelargestprimenumberthatexists.
14 Explainwhy P =⇒ Q isequivalentto ¬P ∨ Q. [Hint:When P =⇒ Q istrueitisnotpossiblefor P tobetrueand Q tobefalse.]
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CHALLENGE
15 Onatrain,Pender,SadlerandWardarethefireman,guardanddriver,butNOTrespectively.Alsoaboard thetrainarethreepassengerswhohavethesamelastnames:DrPender,MrSadlerandMrWard.The followingfactsareknown.
1 MrSadlerlivesinSydney.
2 TheguardlivesexactlyhalfwaybetweenMelbourneandSydney.
3 MrWardearnsexactly$100000peryear.
4 Theguard’snearestneighbour,oneofthepassengers,earnsexactlythreetimesasmuchastheguard.
5 Penderbeatsthefiremanatpool.
6 Thepassengerwhosenameisthesameastheguard’slivesinMelbourne. Whoisthedriver?Clearlyexplainyourlogic.
16 Themethodof modusponens assertsthatif P istrueand P → Q istruethen Q mustbetrue.Insymbols, P ∧ (P → Q) → Q isalwaystrue.Useatruthtabletoconfirmthis.
2B
FurtherLogicandLanguageofProof
Section 2A introducedmostofthelogicrequiredforconstructingsoundproofs.Thissectionpresentsthe remaininglogicandterminologyforthiscourse.
Quantifiers
Thewords all and some wereusedinthediscussionsofSection 2A.Thesewords,andtheirsynonyms,arecalled quantifiers becausetheyindicateaquantity.Bywayofexample,considertheodds D,primes E andFibonacci numbers F onadieasshowninthenexttwoVenndiagrams.
Thefirstdiagrammakesitclearthat alloftheprimesonadieareFibonaccinumbers, Itisalsoseenthattheword all indicatesthatasubsetisinvolved.Inthiscase, E ⊆ F
Incontrast,theseconddiagrammakesitclearthat someoftheprimesonadieareodd.
Further,theword some indicatesthattheintersectionofsetsisinvolved.Inthiscase, D ∩ E isnotempty.
Inmathematics,theword all isfrequentlyreplacedwiththesymbol ∀,meaning forall.Thestatementabout primesandFibonaccinumberscanthenbewrittensuccinctlyas
∀x ∈ E, x ∈ F
Thatis,forallnumbers x inset E, x isalsoinset F.Itisworthnotingthatthisisequivalentto ifx ∈ Ethen x ∈ F. ThisequivalencewasfirstseeninthethreestatementsaboveBox 11 intheprevioussection.
Animportantsynonymfor some is thereexists whichisgiventhesymbol ∃ .Thus,continuingwiththeoddsand primesonadie,therelationshipcanbestatedas:
∃ x ∈ E suchthat x ∈ D
Thatis,thereexistsaprimeonadiethatisodd.Itisalsotruethat:
∃ x ∈ E suchthat x D
Thatis,thereexistsaprimeonadiethatisnotodd.
Example6
WritethefollowingmathematicalstatementinEnglish.
Solution
[Thiswillbecomeeasywithexperience,butinitiallyitmayhelptowriteastatementoutexactlyasitappears, thenrewordittomakeitclearer.]
Thestatementtranslatesas:
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forall m intheintegers,thereexists k intheintegerssuchthat(2m + 1)2 = 4k + 1.
TheLHSoftheequationisthesquareofanoddnumberandtheRHSisonemorethanamultipleof4.Thus, inEnglish,thestatementmaybesimplifiedto thesquareofanyoddnumberisalwaysonemorethanamultipleof4.
Continuedonthenextpage
Althoughitwasnotaskedfor,itissimpletoprovethisbyexpandingtheLHS: (2m + 1)2 = 4m 2 + 4m + 1 = 4m(m + 1) + 1.
Now4m(m + 1)isamultipleof4,sothattheRHSisonemorethanamultipleof4,asclaimedinthe statement,whichisnowproven.
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12Quantifiers
Thewords all and some,andtheirsynonyms,arecalledquantifiers.
• Thesymbol ∀ means forall andisassociatedwithsubsets.
• Thesymbol ∃ means thereexists andisassociatedwithintersections.
Studentsareexpectedtounderstandlogicalsymbolswhenevertheyareusedinquestions.Studentsarealso expectedtobeabletousesymbolsaspartofananswer,however,theyshouldavoidsituationswherethesymbols becomeadistraction.Bywayofexample,considerthefollowingtwostatementsinvolvingthesetofprimes E
Notallprimesareodd.
∃ x ∈ E suchthat x 2m + 1, ∀ m ∈ Z.
Thestatementssaythesamething,butinthiscasetheEnglishisfarclearer,succinctandthereforepreferable.
NegationandQuantifiers
Itisclearthatthestatement allprimesareodd isfalse,because2isaprimewhichisnotodd.Negatingthisstatement, notallprimesareodd isthenclearlytrue.Thistruestatementcanbere-wordedas thereexistsaprimethatisnotodd.
Observethattheresultofnegationistochangethequantifierfrom all to thereexists andtochangethe characteristicfrom odd to notodd.
Asimilarsituationariseswithnegatingastatementinvolving thereexists.Bywayofexample, thereexistsasquarewithunequalsides isfalse.Negatingthisstatement, theredoesnotexistasquarewithunequalsides isclearlytrue.Thistruestatementcanbere-wordedas allsquareshaveequalsides. Onceagain,thequantifierisswappedandthecharacteristicisnegated.
13Negationandquantifiers
• Thenegationof allAareB is thereexistsAthatisnotB
• Thenegationof thereexistsAthatisB is allAarenotB
SufficientandNecessary
Thesetwotermsareassociatedwithimplicationsthatarealwaystrue.Theyareoften usedinmathematicalproofsordiscussionsaboutproofs.Onceagainlet E = {2,3,5} be theprimenumbersandlet F = {1,2,3,5} betheFibonaccinumbersonadie.Itcanbe seenintheEulerdiagramthattheimplication
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x ∈ E → x ∈ F, thatistosay
ifanumberonadieisprimethenitisaFibonaccinumber,
isalwaystrue.Thefirstpropositionofaprovenimplicationiscalleda sufficient condition.Thatis,itissufficient toknowthatanumberonadieisaprimetoguaranteethatitisaFibonaccinumber.Thesecondpartofthe if...then... statementiscalleda necessary condition.Thatis,itisnecessarytoknowthatanumberonadieisa Fibonaccinumberinorderforittobeprime.However,itdoesnotguaranteethatitisaprime.Ascanbeseenin thesetdiagram,itcouldbethenumber1.
Example7
Theimplication ifm > 5 thenm > 3isalwaystrue.
a Explainwhy m > 5issufficientfor m > 3.
b Explain,withanexample,why m > 3isnotsufficientfor m > 5.
Solution
a Anynumberthatisgreaterthan5isautomaticallygreaterthan3.
b When m = 4(or5)thestatement m > 3istruebut m > 5isfalse.
14Sufficientandnecessary
Intheprovenimplication P → Q,
• itissufficientthat P istruetoguaranteethat Q istrue,
• itisnecessarythat Q istrueinorderthat P istrue.
ConverseStatements
Considerthefollowingtwostatements.
If x isamultipleof4then x iseven.
If x iseventhen x isamultipleof4.
Whenthetwopartsofan if...then... statementareswappedinthisway,theresultiscalledthe converse.Using logicnotation,let P : xisamultipleof4 andlet Q : xiseven.Theconverseof P → Q
is Q → P
Inthiscase, P → Q isalwaystrue,buttheconverse Q → P isfalsewhen x = 6,forexample.Inotherwords,the converseofastatementisnotlogicallyequivalenttotheoriginalstatement.
Incontrast,herearetwoconversestatementsfromYear9geometry.
If ABCD isarhombusthenitsdiagonalsbisectthevertexangles. Ifthediagonalsof ABCD bisectthevertexanglesthenitisarhombus. Inthiscasebothstatementsarealwaystrue,butitisimportanttorealisethattheconversestatementisstillnot thesameastheoriginal.Inthefirststatement,itisgiventhatthequadrilateralisarhombus.Thisiscalledthe premise,andthe conclusion isthatthediagonalsbisectthevertexangles.Thesecondstatementdiffersbecause thepremiseisnowthatthediagonalsbisectthevertexangles,andtheconclusionisthatitisarhombus.
Example8
Let P : xisaprime andlet Q : xisodd
a Findavalueof x forwhich P → Q isfalse,
b Whatisthelogicalvalueof Q → P when x isthevalueinpart a?
Solution
a When x = 2, P → Q isfalse:2isprimebutnotodd.
b When x = 2, Q → P istruebecause2isnotodd,sotheimplicationisautomaticallytrue.
15Conversestatements
Everyimplicationhasacorrespondingconverse.
Theconverseofthestatement ifPthenQ isthestatement ifQthenP. Ingeneral,animplicationisnotlogicallyequivalenttoitsconverse.
IfandOnlyIf
Herearetwoimplicationsfromgeometry.Eachistheconverseoftheotherandbotharealwaystrue. Ifatriangleisisoscelesthenitsbaseanglesareequal and ifthebaseanglesofatriangleareequalthenitisisosceles.
Thistypeofpairingofconverseimplicationsissocommoninmathematicsthatitiscontractedandwrittenas: atriangleisisosceles ifandonlyif itsbaseanglesareequal.
Whenitiswrittendown,itisoftenabbreviatedfurtherbyusing iff insteadof ifandonlyif,however,itisstillread outas ifandonlyif. Thus, atriangleisisoscelesiff itsbaseanglesareequal.
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Thenotationfor ifandonlyif is ↔ or ⇐⇒.Putting P : thetriangleisisosceles and Q : thebaseanglesareequal theabovestatementiswrittenas
P ↔ Q or P ⇐⇒ Q. Anyiff shouldbeevaluatedlastwhenevaluatingthetruthofanexpression.
Example9
Let P : thesumofthedigitsofnisdivisibleby3 andlet Q : nisdivisibleby3.Consider P ↔ Q. a WritethisinEnglishusing ifandonlyif. b Writethetwoimplicationsthatthisrepresents; inEnglish, i usinglogicnotation. ii
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Solution
a Thesumofthedigitsof n isdivisibleby3ifandonlyif n isdivisibleby3. bi Ifthesumofthedigitsof n isdivisibleby3then n isdivisibleby3, and if n isdivisibleby3thenthesumofitsdigitsisdivisibleby3. ii Writingthetwoimplicationsandjoiningthemwith and gives (P → Q) ∧ (Q → P).
Inboththisexampleandtheintroductoryexample,bothimplicationsarealwaystrueandhencetheresultof theiff isalsoalwaystrue.AsstatedinSection 2A,mathematicsisgenerallyabouttruestatements,somost examplesofiff inthischapterarealwaystrue.
Nevertheless,aniff maynotalwaysbetrue.Considerthestatement x isamultipleof4iff x iseven.
Itistruewhen x = 8butfalsewhen x = 6.Thisisbecauseonestatementistrue(6isindeedeven)buttheother isfalse(6isnotamultipleof4).
16Ifandonlyifstatements
Let P and Q betwopropositions.Thecompoundstatement ifPthenQandifQthenP iswrittenas Pif andonlyifQ whichisabbreviatedas Piff Q.Thenotationforthisis
P ↔ Q or P ⇐⇒ Q.
Itsvalueistruewhenboth P and Q aretrueorwhenboth P and Q arefalse,otherwisethestatementis false.
Anotherimportantobservationcanbemadefromallthis.Aniff thatisalwaystrueindicatesthat P isequivalent to Q.Thisissimplyanotherwayofsayingthateitherbotharetrueorbotharefalse.Returningtothegeometry above, atriangleisisoscelesiff itsbaseanglesareequal isalwaystrue.Thatis, atriangleisisosceles isequivalenttothestatement thebaseanglesofatriangleare equal. Givenanytriangle,eitherbothstatementsaretrueorbothstatementsarefalse.
17Equivalentstatementsandiff
• If P ↔ Q isalwaystruethen P ≃ Q
• If P ≃ Q thenthevalueof P ↔ Q isalwaystrue.
TheContrapositive
Therearemanyexamplesofpairsofequivalentstatements,butoneparticularsituationplaysanimportantpartin manyproofs.Bywayofexample,recallthat if x isaprimeonadiethenitisaFibonaccinumber.
Thisislogicallyequivalenttothe contrapositive statement if x isnotaFibonaccinumberthenitisnotaprimeonadie.
Writingthisinlogicnotation,thecontrapositiveof P → Q is ¬Q →¬P
Noticethatthetwopartsofthe if...then... statementhavebeennegatedandtheorderswapped.Thisisalways thecasewiththecontrapositivestatement.Itisalsoalwaysthecasethatanimplicationisequivalenttoits contrapositive,thatis
P → Q ≃¬Q →¬P.
Theequivalenceofanimplicationanditscontrapositivecanbeprovenbyatruthtable,andthisisleftasan enrichmentexercise.Apracticalexamplewilldemonstratethisequivalence.Thus,thestatement ifIownadogthenIhaveapet islogicallyequivalenttoitscontrapositive ifIdonothaveapetthenIdonotownadog.
Thismeansthatanimplicationanditscontrapositivecanbewrittenintheformofan ifandonlyif statementthat isalwaystrue.
Itisalwaystruethat: P implies Q ifandonlyifnot Q impliesnot P. or,writingthisentirelyinsymbols:
18Thecontrapositive
Animplication,if P then Q,islogicallyequivalenttoitscontrapositive,ifnot Q thennot P.Thatis, P → Q ≃¬Q →¬P
Exercise2B
1 Identifyeachofthesesymbols.
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FOUNDATION
2 Writedowntheconverseofeachstatement,andstatewhethertheconverseistrueorfalse.
a Ifatrianglehastwoequalsides,thenithastwoequalangles.
b Ifanumberisodd,thenitssquareisodd.
c IfIamahorse,thenIhavefourlegs.
d Ifanumberendswiththedigit6,thenitiseven.
e Everysquareisarhombus.
f If √n ∈ R,then n ≥ 0.
3 Indicatewhethereachstatementistrueorfalse.
a Havingfourlegsisanecessaryconditionforbeingacat.
b Havingfourlegsisasufficientconditionforbeingacat.
c Owningacarisanecessaryconditionforholdingadriver’slicence.
d Owningacarisasufficientconditionforholdingadriver’slicence.
e Iftwostatementsareequivalenttheneachisanecessaryconditionfortheothertobetrue.
f Iftwostatementsareequivalenttheneachisasufficientconditionfortheothertobetrue.
4 Writedown(in‘if ... then ...’form)thetwoconversestatementsequivalenttoeach‘ifandonlyif’ statement.
a Anumberisdivisibleby15ifandonlyifitisdivisiblebyboth3and5.
b Atrianglehastwoequalanglesifandonlyifithastwoequalsides.
c Aninteger n greaterthanoneisprimeifandonlyifitsonlydivisorsare1and n
d Aquadrilateralisaparallelogramifandonlyifithasapairofoppositesidesthatareequalandparallel.
5 Writedownthecontrapositiveofeachstatement.
a IfIwatermyplants,thentheywillgrow.
b IfyoudonotliveinAustralia,thenyoudonotliveinMelbourne.
c Ifatrianglehasthreeequalsides,thenithasthreeequalangles.
d IfIlikecycling,thenIdonotlikemotorists.
e Ifanumberiseven,thenthenextnumberisodd.
f If a and b arebothpositiveand a > b,then 1 a < 1 b
6 Considerthecontrapositiveofthestatement:
“Ifit’smybirthday,thenIwillreceiveapresent".
a WhatcanbededucedifIdonotreceiveapresent?
b Whatcanbededucedifthestatementisfalse?
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DEVELOPMENT
7a Whichofthefollowingbestdescribestheequivalenceoftwostatements?
A Oneofthestatementsimpliestheother.
B Thetwostatementsusethesamelogic.
C Eachstatementimpliestheother.
D Thetwostatementshavethesamemeaning.
b Whatistheconverseof ¬P =⇒ Q?
A Q =⇒¬P
B ¬Q =⇒ P
C P =⇒ Q
D P =⇒¬Q
c Whatisthecontrapositiveof ¬P =⇒ Q?
A Q =⇒¬P
B ¬Q =⇒ P
C P =⇒ Q
D P =⇒¬Q
d Whichstatementhasthesamemeaningas‘P if Q’?
A Q if P
B P ifandonlyif Q.
C If Q then P
D If P then Q.
e Whichstatementhasthesamemeaningas‘P onlyif Q’?
A P if Q.
B Q onlyif P
C If Q then P.
D If P then Q
f Whichstatementisequivalentto‘∀ k ∈ Z ∃ ℓ ∈ Z suchthat k <ℓ’?
A Givenanyinteger,thereexistsalargerinteger.
B Givenanyinteger,thereexistsasmallerinteger.
C Givenanytwointegers,oneisalwayssmallerthantheother.
D Thereexistintegers k and ℓ suchthat k <ℓ.
g Whatisthenegationof‘∀ k ∈ Z ∃ ℓ ∈ Z suchthat k <ℓ’?
A ∀ ℓ ∈ Z ∃ k ∈ Z suchthat k <ℓ
B ∃ k ∈ Z suchthat ∀ ℓ ∈ Z k <ℓ
C ∀ k ∈ Z ∃ ℓ ∈ Z suchthat k ≥ ℓ
D ∃ k ∈ Z suchthat ∀ ℓ ∈ Z k ≥ ℓ
8 Writedownthenegationofeachstatement.
∀ x ∈ Z 6x iseven. a
∃ y ∈ R suchthat y2 < 0. b
∀ m ∈ Z ∃ n ∈ Z suchthat m + n = 20. c
∃ θ ∈ R suchthat ∀k ∈ Z sin θ ≤ 1 + k2 . d
9 Whichofthesestatementsarealwaystrue?Assumethatallvariablesarereal. itisraining ⇐⇒ therearecloudsinthesky a 3a = 6 ⇐⇒ 5a = 10 b a ≥ b ⇐⇒−b ≥−a c x = 5 ⇐⇒ x2 = 25 d x = 5 ⇐⇒ x3 = 125 e a isaninteger ⇐⇒ a2 isaninteger f θ = 1π 6 ⇐⇒ sin θ = 1 2 g sin θ = 1 √2 ⇐⇒ tan θ = ±1 h
10 WriteeachstatementasanEnglishsentence,withoutanyuseofsymbols.
∀ n ∈ Z ∃ m ∈ Z suchthat m > n a a ∈ R and a > 0 =⇒ a + 1 a ≥ 2 b
11 Considerthefollowingstatements:
1 IfIdonotdomyhomeworkthenIwillfail.
2 IfIstudyhardthenIwillpass.
Ifstatements 1 and 2 arebothtrueandIpassed,then: DidIdomyhomework? a DidIstudyhard? b
12 Considerthesetwostatements:
P:IfRodisAmy’sfatherthenAmyisNeil’sniece.
Q:NeilisRod’sbrother.
Whichofthefollowingisequivalenttothestatement P =⇒ Q?
A IfNeilisRod’sbrotherthenRodisAmy’sfatherandAmyisnotNeil’sniece.
B IfNeilisnotRod’sbrotherthenRodisAmy’sfatherandAmyisnotNeil’sniece.
C IfNeilisRod’sbrotherthenRodisAmy’sfatherandAmyisNeil’sniece.
D IfNeilisnotRod’sbrotherthenRodisnotAmy’sfatherandAmyisNeil’sniece.
13 Provebywayoftruthtableorotherwisethat P ↔ Q isequivalentto P ∧ Q ∨¬P ∧¬Q.
14 Theformal‘epsilon-delta’definitionofcontinuity,limx→a f (x) = f (a),isasfollows:
0suchthat
a Given f (x) = 4x 3,writean‘epsilon-delta’proofofthefactthatlim
[Hint:Firstfind δ intermsof ε.]
b Writedownthenegationofthe‘epsilon-delta’definitiongivenabove.
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CHALLENGE
2C NumberProofs
Nowthatthebasicconcepts,terminologyandnotationofproofareunderstood,theycanbeputtogethertowrite proofsofsimpleresults.ThegeometrystudiedinYears7to10involvedmanysuchproofs,anditwouldbeworth reviewingsomeofthoseinlightofthisnewunderstanding.InthissectionandinSection 2D,however,attention isfocusedonsomebasicresultsin numbertheory,thatis,thestudyofthestructuresandnumberpatternsin:
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Z theintegers 3, 2, 1,0,1,2,3,
Z+ thepositiveintegers1,2,3, ...
N thenaturalnumbers0,1,2,3,
Aproperinvestigationofnumbertheorywouldoccupyanentirebook,soonlyafewsimpleresultsindivisibility areprovenhere.Theproofsrelyonimplicationsthatarealwaystrueandiffsthatarealwaystrue.
Divisibility
Itisimportanttobeginwithacleardefinitionofwhatdivisibilitymeans.Let a, b, m ∈ Z andsupposethat b = am. Thenumbers a and m arecalledfactorsof b.Furthermore b issaidtobedivisibleby a.(Equally, b isdivisibleby m.)Toputitanotherway,if b and a areintegersand b isdivisibleby a thenthereexistsaninteger m suchthat b = am
19Divisibility
If a, b ∈ Z and b isdivisibleby a then ∃ m ∈ Z suchthat b = am
Thisdefinitionwillbeessentialinthefollowingexamples,andintheexercisequestions.Althoughthewording ofthedefinitionmayseemobscure,inpracticeitisquiteobvious.Forexample,12isdivisibleby4because 12 = 4 × 3.
Example10
Let a and b betwointegersdivisibleby3.
a Provethat(a + b)isdivisibleby3.
b Provethat(ax + by)isdivisibleby3, ∀ x, y ∈ Z
Solution
a Bythedefinitionofdivisibility, ∃ m, n ∈ Z suchthat
a = 3m and b = 3n.
Hence(a + b) = (3m + 3n)
= 3(m + n).
Thatis,(a + b)isdivisibleby3.
b Likewise, ∀ x, y ∈ Z, (ax + by) = (3mx + 3ny)
= 3(mx + ny)
Thatis,(ax + by)isdivisibleby3.
Example11
Provethat a2 a isevenforall a ∈ Z.
Solution
Factoring, a2 a = a(a 1).
Now,if a isodd,then(a 1)iseven,andif a iseven,then(a 1)isodd. Ineithercase,theproductofanoddandeveniseven.
Hence a2 a iseven.
Note: Thisisessentiallyproofbyexample,witheverycaseexamined.
Example12
Astudentclaimsthatifaninteger n isdivisiblebyboth4and6thenthenumberisdivisibleby4 × 6 = 24.
a Disprovethisclaimbyfindingacounterexample.
b Explainwhathasgonewrong,anddeterminethecorrectconclusion.
Solution
a Clearly12 = 4 × 3and12 = 6 × 2,hence12isdivisiblebyboth4and6,butitisnotdivisibleby24.
b Theproblemisthat4and6havecommonfactor2.
Since n isdivisibleby4and6,thereexistintegers k and ℓ suchthat
n = 4k and n = 6ℓ
so4k = 6ℓ
or2k = 3ℓ.
Since2and3areprimes, k isdivisibleby3(and ℓ iseven),sothereexistsaninteger m suchthat k = 3m
Hence n = 4 × 3m = 12m.
Thatis,if n isdivisiblebyboth4and6thenitisdivisibleby12.
Example13
Let n = 10x + y,where n, x, y ∈ Z+ (Thatis,thenumbersareallpositiveintegers.)
a Provethatif n isdivisibleby7then(x 2y)isalsodivisibleby7.
b Further,provethattheconverseistrue.
c Writetheresultasaniff statement.
d Hencedeterminewhetherornot3871isdivisibleby7.
Solution
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a Since n isdivisibleby7, ∃ m ∈ Z+ suchthat 10x + y = 7m
so x 2y = 7m 9x 3y (subtract9x + 3y frombothsides)
= 7m 9x 3(7m 10x)(since y = 7m 10x)
= 21x 14m
or x 2y = 7(3x 2m).
Hence(x 2y)isdivisibleby7.
b Sincetheequationsinpart a areequivalent,itfollowsthat
10x + y = 7m ↔ x 2y = 7(3x 2m) isalwaystrue.Thustheimplication
if(x 2y)isdivisibleby7then n = 10x + y isdivisibleby7 isalsoalwaystrue.
c Thatis,combiningparts a and b,itisproventhat
n = 10x + y isdivisibleby7ifandonlyif(x 2y)isdivisibleby7.
d Thedivisibilitytestinpart c canbeappliedrecursively.Thus: 3871isdivisibleby7iff 387 2 × 1 = 385isdivisibleby7. 385isdivisibleby7iff 38 2 × 5 = 28isdivisibleby7. Now28 = 7 × 4isclearlydivisibleby7. Hence3871isdivisibleby7.
Exercise2C FOUNDATION
Note: Inthisexerciseyoumayassumethatallpronumeralsrepresentintegers.
1a If a and b areeven,provethat a + b iseven.
b If a and b areodd,provethat a + b iseven.
c If a isevenand b isodd,provethat a + b isodd.
2a If a and b areeven,provethat ab iseven.
b If a and b areodd,provethat ab isodd.
c If a isevenand b isodd,provethat ab iseven.
3 Provethat:
a thesquareofanevennumberiseven,
b thesquareofanoddnumberisodd.
4 Provethat:
a if b and b + c arebothdivisibleby a then c isdivisibleby a, b if b and b c arebothdivisibleby a then c isdivisibleby a
5 If b and c arebothdivisibleby a,provethat bx + cy isdivisibleby a forany x, y ∈ Z
6 If a, b, c and d areconsecutiveintegers,provethat:
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7 Provethatif a b iseventhen a2 b2 isdivisibleby4.
8 Supposethat2a + b and3a + 2b arebothdivisibleby n.Provethat a and b arebothdivisibleby n.
9 Supposethat a2 + a and a2 a arebothdivisibleby4.Provethat a iseven.
10 Provethat a3 a isdivisibleby6 ∀ a ∈ Z
11
12
13
14
Provethattheremainderis1whenthesquareofanoddnumberisdividedby8.
Provethateveryprimenumbergreaterthan3iseitheronemoreoronelessthanamultipleof6.
Provethatanumberisdivisibleby6ifandonlyifitisdivisiblebyboth2and3. (Rememberthattoprove A ⇐⇒ B,youmustproveboth B =⇒ A and A =⇒ B.)
Provethatanintegeristhesumof7consecutiveintegersifandonlyifitisdivisibleby7.
15a If n isodd,provethatthesumof n consecutivenumbersisdivisibleby n.
b If n iseven,isthesumof n consecutivenumbersdivisibleby n?Explainyouranswer.
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16
Provethata4-digitnumberisdivisibleby3ifandonlyifthesumofitsdigitsisdivisibleby3.
17 Let n = 10x + y,where n, x, y ∈ Z+ .
a Provethatif n isdivisibleby13,then x + 4y isalsodivisibleby13.
b Provethattheconverseofpart a istrue.
c Combine a and b intoan‘ifandonlyif’statementusingmathematicalsymbols.
d Usepart a recursivelytoshowthat8112isdivisibleby13.
18 Ifthepolynomial P(x) = a0 + a1 x + a2 x2 + ···
,withintegercoefficients
,hasan integerzero x = k, k 0,provethattheconstantterm a0 isdivisibleby k
19a Showthat xn 1 = (x 1)(
b Henceprovethat: i 7n 1isdivisibleby6 ∀ n ∈ Z+ , ii if an 1isprime,then a = 2for a > 0and n ∈ Z+
2 + x + 1).
20 Supposethat n = paqb,where p and q areprimesand p q.
a Usecombinatorics(acountingargument)toexplainwhy n has(a + 1)(b + 1)factors.
b Hencedeterminethenumberoffactorsof80000.
21 Provethestatement: a c isadivisorof ab + cd =⇒ a c isadivisorof ad + bc CHALLENGE
22 Supposethat p isaprimenumbergreaterthan30.Provethatwhen p isdividedby30,theremainderis either1orprime.
23 Numberssuchas6and28areknownas perfect numbersbecausetheyareequaltothesumoftheirfactors, excludingthenumberitself.
a Confirmthat6and28areperfectnumbers.
b Provethatif2n 1isprime,then2n 1(2n 1)isaperfectnumber.
[Hint:Separatethefactorsof2n 1(2n 1)intotwogroups:thosethatarestrictlypowersof2andthose thathavethefactor2n 1.]
2D
ProofbyContrapositionandbyContradiction
Twomethodsofproofcommonlyusedinmathematicsinvolvenegation.Thesearecalled proofbycontraposition and proofbycontradiction.ThelatterisalsogiventheLatinname reductioadabsurdum,whichliterallymeans reducetoabsurdity .Thereasonforthiswillbeexplainedlater.
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ProofbyContraposition
Thisstyleofprooftakesadvantageofthefactthatanimplicationisequivalenttoitscontrapositive.Thuswhenan implicationisnoteasytoprovedirectly,itmaybesuitabletouseproofbycontrapositioninstead.Itisimportant toclearlystatewhatisbeingdoneattheoutset.
Example14
Provethatif n2 iseventhen n iseven.
Solution
Thecontrapositiveisproveninstead.Thatis:
if n isnoteventhen n2 isnoteven, or,morenaturally:
if n isoddthen n2 isodd.
Let n = 2m + 1,then n isoddand
n 2 = (2m + 1)2
= 4m 2 + 4m + 1
= 2(2m 2 + 2m) + 1
whichisonemorethanamultipleof2,andisthusodd.
Hence,bythecontrapositive,if n2 iseventhen n iseven.
20Proofbycontraposition
Animplicationanditscontrapositiveareequivalent.Henceanimplicationmaybeprovenbyprovingits contrapositiveinstead.
ProofbyContradiction
Thisstyleofproofisbasedonthefactthatastatementcanonlyhaveoneoftwovalues,trueorfalse,andthe negatedstatementmusthavetheoppositevalue.Thatis,ifastatementistruethenitsnegationmustbefalse. Henceprovingastatementisequivalenttoshowingthatitsnegationisfalse.Thuswhenastatementisnoteasyto provedirectly,itmaybesuitabletoshowinsteadthatthenegatedstatementisfalse.
Thefirststepistowritedownthenegationasanassumption.Itisthenshownthatthisleadstoanabsurd statement,like 1 > 2,oritleadstoacontradictionoftheinitialassumption.Thisiswhythemethodisalso called reductioadabsurdum .Sincethenegationisequivalenttoafalsestatement,thenegationisalsofalse.It mustthereforeholdthatthepropositionistrue.
Example15
Provethat √2isirrational.
Solution
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Bywayofcontradiction,assumethat √2isrational.
Thatis,assumethereexist m, n ∈ N (naturalnumbers)suchthat √2 = m n where n ≥ 1andtheHCF(highestcommonfactor)of m and n is1. Thatis,thefractionhasbeenreducedtolowestterms.
Squaringandre-arranginggives
2n 2 = m 2 .
Thus m2 isdivisibleby2.
Nowif m werenotdivisibleby2then m2 wouldnotbedivisibleby2. Hence m isalsodivisibleby2.Solet m = 2p andwrite
2n2 = 4p2 or n2 = 2p2
Thus n2 isdivisibleby2.
Nowif n werenotdivisibleby2then n2 wouldnotbedivisibleby2. Hence n isalsodivisibleby2. Thatis,2isacommonfactorof m and n ButtheHCFis1,sothereisacontradiction.
Hence √2isirrational.
21Proofbycontradiction
Provingastatementisequivalenttoshowingthatitsnegationisfalse.Beginbywritingdownthenegation andshowthisleadstoacontradiction.
Noticethecarefuluseoftwocontrapositivestatementsintheaboveproof.Eachisarestatementofwhatwas proveninExample 14.Theproofthat √3isirrationalrequiressimilarsteps.Thatis: if m werenotdivisibleby3then m2 wouldnotbedivisibleby3.
Aproblemarisesatthispointbecausethatclaimhasnotbeenproven.Likewise,theproofthat √7isirrational leadstotheclaimthat
if m werenotdivisibleby7then m2 wouldnotbedivisibleby7 whichalsohasnotbeenproven.Inthiscourse,allsuchclaimswillbetakentobeintuitivelyobviousunlessa proofisspecificallyrequiredbythequestion.
Thenextexamplefollowsaslightlydifferentargumenttoshowthatlog2 3isirrational.Thisresultisextremely importantinmusic.Musicianswillknowthatwhenafifthisplayedonapiano,sayfromAtoE,itisnotapure fifthbutslightlyshort.Apurefifthhasfrequenciesintheratio2:3.ThususingA440shouldgiveE660,butona pianoitisapproximatelyE659.Thisisdonesothatthecycleoffifthsworks,anditcomesaboutbecauselog2 3is irrational.
Example16
Provethatlog2 3isirrational.
Solution
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Bywayofcontradiction,assumethatlog2 3isrational.
Thatis,assumethereexist m, n ∈ N suchthat
log2 3 = m n where n ≥ 1andtheHCFof m and n is1. Thatis,thefractionhasbeenreducedtolowestterms. Fromthedefinitionoflogs,thiscanbere-writtenas
3 = 2 m n
Takethe nth powerofbothsidestoget
3n = 2m
Nowsince n ≥ 1itfollowsthat3isafactoroftheLHS. Butclearly3isnotafactoroftheRHS,sothereisacontradiction. Hencelog2 3isirrational.
TheFundamentalTheoremofArithmetic
Inthelastproofitwasassumedthatapositivepowerof3cannotequalapositivepowerof2.Intuitionand experiencecertainlyseemtosuggestthisistrue,butithasnotbeenproven.Theresultisaspecificcaseofamore generaltheorem.Thetheoremeffectivelystatesthatthereisonlyonewaytowriteanumberinprimefactored form.Thisissuchanimportantresultinnumbertheorythatitisgiventhename thefundamentaltheoremof arithmetic .Althoughanyproofofthisresultisbeyondthescopeofthiscourse,itissuchanimportanttheorem thataproofhasbeenincludedintheappendixtothischapter.
JustifyingTheseProofMethods—Enrichment
RecallthatinSection 2B itwasshownthatanimplicationisequivalenttoitscontrapositive.Thatis, P → Q ≃¬Q →¬P.
Inthemethodofproofbycontraposition,itisshownthat ¬Q →¬P istrue.Hence,bytheaboveequivalenceit followsthat P → Q isalsotrue.
Thesituationismorecomplicatedforproofbycontradiction.ThewordcomesfromtheLatin contradico meaning tospeakagainst.Inthecontextoflogic,itistheresultofapropositionanditsnegation.Thatis, Q ∧¬Q whichisclearlyfalse.
Nowsupposethat ¬P leadstoacontradiction.Inotherwords,supposethat
¬P → Q ∧¬Q isatrueimplication.Thatmeansthat ¬P → 0istrue.Thislastexpressionhasaverysimpletruthtable.
P ¬P ¬P → 0
1 0 1
0 1 0
Theonlyrowwheretheimplicationistruecoincideswith P beingtrue.Thus,ifitistruethat ¬P impliesa contradictionthen P mustbetrue.Thiscanbewritteninsymbolsas
(¬P → Q ∧¬Q) → P andthislastimplicationisalwaystrue.
Toputthisintocontext,inExample 16,proposition P isthatlog2 3isirrational.Thetrueimplicationofthe negation ¬P isthecontradiction
Q :3isafactorof2m and ¬Q :3isnotafactorof2m
Hence P istrueandlog2 3isirrational.
Studentsmayalsoliketoinvestigatethestatement(¬P → Q ∧¬Q) → P whenproposition P isfalse.
Exercise2D
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FOUNDATION
1 Provebycontradictionthatlog7 13isirrational. Startbyassumingthatlog7 13isrational.Letlog7 13 = m n ,where m, n ∈ Z+ and m and n havehighest commonfactor1.
2 Provebycontradictionthat √5isirrational. Startbyassumingthat √5isrational.Let √5 = m n ,where m, n ∈ Z+ and m and n havehighestcommon factor1.
3 Considerthisstatementfor a ∈ N:‘If a2 isoddthen a isodd.’
a Writedownthecontrapositiveofthestatement.
b Provetheoriginalstatementbyprovingitscontrapositive.
4 Byprovingthecontrapositive,provethatif m2 + 4m + 7iseven,then m isodd.
5 Supposethat a, b ∈ Z+.Byprovingthecontrapositive,provethatif ab iseven,then a isevenor b iseven. DEVELOPMENT
6a Provebythecontrapositivethatif r isirrationalthen1 + r isalsoirrational.
b Provethatlog3 5isirrational.
c Henceprovethatlog3 15isirrational.
7a Provebythecontrapositivethatif r isirrationalthen2r isalsoirrational.
b Provethat √11isirrational.
c Henceprovethat √44isirrational.
8 Byprovingthecontrapositive,provethatif m isodd,thentheequation x2 + x = m hasnointegersolutions.
9 Supposethatthenumber n iscompositeandhastwodistinctprimefactors p1 and p2.Usecontradictionto provethatatleastoneof p1 and p2 islessthan √n.
10 Provebycontradictionthatif n ∈ Z+ then √4n 2isirrational.
11 Provebycontradictionthat √3 + 1isirrational.
12a Provethat √6isirrational.
b Henceprovethat √3 + √2isirrational.
13
Supposethat ∃ n ∈ N suchthat n2 + 2isdivisibleby4.
a Deducethat n iseven.
b Henceprovebycontradictionthatnosuch n exists.
14a Explainwhyeveryoddnumberisonemoreoronelessthanamultipleof4.
b Provethattheproductofanytwopositiveintegersoftheform4n + 1,where n isapositiveinteger,is alsooftheform4n + 1.
c Henceprovebycontradictionthatanycompositenumberoftheform4n 1musthaveatleastoneprime factoroftheform4n 1.
15 Provebycontradictionthatthereareinfinitelymanyprimenumbers.
[Hint:Assumethat p isthelargestprimenumberandconsiderthenumber(p! + 1).]
16 Provebycontradictionthatif a2 + b2 = c2,for a, b, c ∈ Z+,then a isevenor b iseven.
17 Supposethat a, b, c and d arepositiveintegersand c isnotasquare.
a Provethatif a b + √c + d √c isrational,then b2d = c(a + d).
b Henceprovebycontradictionthat a 1 + √c + d √c isirrational.
18 Supposethat n ∈ Z+ and p isprime.
Provethattherecannotexist k ∈ Z+ suchthat(3k + 2)2 = n2 + p
19 Let ω beanon-realcuberootof1.
a Showthat ω2 + ω + 1 = 0.
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CHALLENGE
b Provebycontradictionthatif n isapositivemultipleof3,then x2 + x + 1isnotafactorof(
20 Considerthefollowingtheoremaboutprimenumbers:
If ∀ a, b ∈ Z+ , p|ab =⇒ p|a or p|b then p isprime.
Note: that p|a means p divides a,or p isafactorof a
a Statethecontrapositiveofthetheorem.
b Provethetheorembyprovingthecontrapositive.
[Hint:Put a = p1,where p1 isaprimedivisorof p,and b = p p1 .]
21 Supposethat p isaprimenumbergreaterthan3andthatforsome n ∈ N, pn isa20-digitnumber.Provethat amongthese20digits,thereareatleastthreethatareequal.
[Hint:Useproofbycontradictionandthetestfordivisibilityby3.]
2E AlgebraicInequalities
Theinequality a > b canbedefinedforrealnumbers a and b geometricallyas a istotherightof b onthenumber line.Mostoftheinequalitiesinthissectionaremoreeasilysolvedusingthefollowingalgebraicdefinition:
If a b > 0then a > b.
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Inwords: a isgreaterthan b when a b ispositive.Anumberofusefulresultsfollowfromthisdefinition,for realnumbers a, b, c and d.Theyinclude:
if a > b then a ± c > b ± c,
if a > b > 0then 1 b > 1 a > 0anditsconverse, if |a| > |b| then a2 > b2 anditsconverse,
if a > b and b > c then a > c,
if a > b and c > d then a + c > b + d,
if a > b and c > 0then ac > bc,
if a > b and c < 0then ac < bc
Threeoftheseareprovedhereandtheremainderarelefttotheexercise.
Example17
Proveforallrealnumbers a, b and c thatif a > b then a + c > b + c
Solution
Let a > b then a b > 0.Nowconsider
LHS RHS = (a + c) (b + c)
= a b (expandingbrackets)
> 0 (since a b > 0).
Thatis(a + c) (b + c) > 0, hence (a + c) > (b + c) (bydefinition).
Notethatthiscanalsobeshowngeometrically.Thenumber a istotherightof b.Shiftingbothbythesame amount c meanstheykeepthesamerelativeposition.Hence a + c > b + c
Example18
Proveforallrealnumbers a, b and c thatif a > b and c > 0then ac > bc.
Solution
Let a > b then a b > 0.Nowconsider
LHS RHS = ac bc
= c(a b)(factoring)
> 0 × 0(sincebotharepositive). Thatis ac bc > 0, hence ac > bc (bydefinition).
Notethatthiscanalsobeshowngeometrically.Thenumber a istotherightof b.Stretchingbothbythesame amount c > 0meanstheykeepthesamerelativeposition.Hence ac > bc.
Noticethatbothexamplesabovebeginbyputtingeverythingononeside.Thisisoneofseveralmethodsfor provinginequalities.Theseare:
• puteverythingononeside,
• squaresofrealscannotbenegative,
• combinationsofinequalities,
• beginwithaknownresult.
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Examplesarenowgivenforeachofthesecategories.
PutEverythingonOneSide
Inafewinstancestheinequalitycaneasilybeprovedbymovingalltermstoonesideandconsideringthesignof thatexpression.Twoexamplesfollow,thefirstofwhichproveswhatisintuitivelyobvious.
Example19
Lettherealnumbers a, b and c berelatedby a > b and b > c.Provethat a > c.
Solution
Fromthegiveninformation, a b > 0and b c > 0.Itisrequiredtoshowthat a c > 0.
Now a c = (a b) + (b c)(subtractingthenadding b) > 0 + 0 (since a b > 0and b c > 0).
Thatis a c > 0, hence a > c (bydefinition).
Thisisgeometricallyobviousonanumberline.Thenumber b istotherightof c and a istotherightof b Hence a istotherightof c
Example20
Let a < b berealnumbers.Provethattheaverageofthesquaresof a and b isgreaterthanthesquareofthe average.
Solution
Thecorrespondinginequalitytoproveis
SquaresCannotbeNegative
Thecrucialstepinthepreviousexamplewasthatthesquareofarealnumbercannotbenegative.Thatfactcanbe deducedfromtheassumptionsatthestartofthissection,andtheproofisleftasanexercise.Thisimportantresult canbeusedtosolvenumerousotherinequalities.
ArithmeticandGeometricMeans
Recallfromsequencesandseriesthatif a and b arepositiverealnumbersthenthesequence a, x, b willbe arithmeticif
x = a + b 2
Thevalueof x iscalledthe arithmeticmean of a and b.
Likewise,thesequence a, y, b willbegeometricif y = √ab.
Thevalueof y iscalledthe geometricmean of a and b.
22Arithmeticandgeometricmeans
Supposethat a and b arepositive.
• Thearithmeticmeanof a and b is a + b 2
• Thegeometricmeanof a and b is √ab
TheAM/GMInequality
Oneveryimportantresultisthatthearithmeticmeanisatleastaslargeasthegeometricmean.Thisissometimes calledthe AM/GMinequality.Itisprovedbynotingthatsquaresofrealnumberscannotbenegative.
Example21
ProvetheAM/GMinequality.
Solution
Let a and b betwopositiverealnumbers. Sincesquarescannotbenegative,
Expanding
or
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23TheAM/GMinequality
Thearithmeticmeanandgeometricmeanoftwopositivenumbers a and b arerelatedbytheinequality a + b 2 ≥ √ab.
TheAM/GMinequalitycanalsobeprovenusingcirclegeometryandthatisdoneinanexampleinSection 2G.
CombinationsofInequalities
Havingestablishedaninequality,itcanberestatedusingothervariablesandtheresultscombinedtoformanew inequality.
Example22
a Supposethat1 ≤ k ≤ n
i Showthat n ≤ k(n k + 1).
ii Explainwhy
b Henceprovethat,forallpositiveintegers n
Solution
ai ItiseasytofactoriseLHS RHS,thenshowthatthisisneverpositive,andhenceLHS ≤ RHS.Thatis leftasanexercise.Alternatively,therighthandsideisaquadraticin k withnegativeleadingcoefficient:
k(n + 1) k2
Henceitsgraphisconcavedown,andanyminimumwilloccuratanendpointofthedomain.Direct substitutionofthetwoendpointsgives 1
ii Thelefthandinequalityfollowsdirectlyfrompart i. BytheAM/GMinequality,
b Bypart a itfollowsthat
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Nowmultiplyalltheseresultstogethertoget
Thisexamplehappenedtousemultiplicationofinequalities.Otherexamplesmayrequireaddition,subtractionor division.Multiplicationanddivisionshouldbeavoidedunlessitisguaranteedthatthequantitiesinvolveddonot changesign.Otherwiseitisnotknownwhetherthedirectionoftheinequalityisaffected.
BeginwithaKnownResult
Manyproblemsbeginwithaknownresultfromwhichanotherinequalityistobeobtained.Animportant exampleofthisistherelation
|x| + |y|≥|x + y|≥ |x|−|y|
whichisknownasthetriangleinequality.Thiswasproveninthechapteroncomplexnumbersbyusing geometry.Inthefollowingexamplethelefthandinequalityisprovenalgebraically.
Example23
Usethefactthat |a|≥ a toprovethat |x| + |y|≥|x + y|
Solution
Squarethelefthandsidetoget (|x| + |y|)2 = x2 + 2|x||y| + y2
x2 + 2xy + y2 (since |a|≥ a) so(|x| + |y|)2 ≥ (x + y)2 .
Nowtakethesquarerootofbothsidestoget |x| + |y|≥|x + y| (since √a2) = |a| ontheRHS.
Theproofoftheotherpartofthetriangleinequalityisleftasanexercise. Inotherproblemsrestrictionsareplacedonthevariables,suchas a + b + c = 1,fromwhichitcanbeshown that ab + ac + bc < 1 3 .Theusualapproachistobeginmanipulatingtheexpressionandatcrucialstepsapply thegivenrestrictions.
Insomeinstancesthesolutionalsoinvolveschangingthevalueofafractionbyalteringthenumeratoror denominator.Forexample,decreasingthenumeratororincreasingthedenominatorwillreducethevalueof thefraction.Thisisclearlyevidentinthefollowingnumericalexample.
3 5 > 2 5 > 2 7
Thestrategyisusedtwiceinthenextexample,alongwitharestriction.
Example24
Let a, b and c bethreepositiverealnumbers.Itisknownthat a +
Showthat
b Henceshowthat
Solution
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a LHS RHS = (a + b)(1 +
(1 + a + b)(1 + c) = (a + b) c (1 + a + b)(1 + c) ≥ c c (1 + a + b)(1 + c) (since a + b ≥ c.)
ThusLHS RHS ≥ 0,andhenceLHS ≥ RHS.
b a 1 + a + b
(bypart a)
sinceincreasingthedenominatorsdecreasesthefirsttwofractions. Hence
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Exercise2E FOUNDATION
1 IneachpartprovetheresultbyconsideringLHS RHS.
a If a > b and c > d then a + c > b + d.
b If a > b and c < 0then ac < bc
c If a > b > 0then 1 b > 1 a .
2 Provethatif |a| > |b| then a2 > b2.[Hint:Usethefactthat |a| = √a2.]
3 Giventhat a > 1,provethat a2 > 1byprovingthatLHS RHS > 0.
4 Provetheseinequalitiesfor a, b ∈ R.[Hint:BeginwithLHS RHS.]
5 Provethat a + 1 a ≥ 2for a > 0.
6 Prove,for a, b > 0,that:
7 If a > b > 0,provethat:
8a Giventhat x and y arenon-negative,provethat x + y ≥ 2√xy. b Henceprovethat(x + y)(x + z)(y + z) ≥ 8xyz,where z isalsonon-negative.
9 Supposethat p, q and r arerealanddistinct.
a Provethat p2 + q2 > 2pq
b Usepart a threetimestoprovethat p
c Giventhat p + q + r = 1,provethat pq + qr + rp < 1 3 [Hint:Beginwith(p + q + r)2 = 1andusepart b.]
10 Supposethat a, b and c arepositive.
a Provethat a2 + b2 ≥ 2ab
b Henceprovethat
c Giventhat
d If x, y and z arepositive,deducethat x + y + z ≥ 3 (xyz) 1 3
11a Showthat 1 k2 < 1 k(k 1) for k > 1.
b Showthat 1 k(k 1) = 1 k 1 1 k .
c Henceshowthat 1 12 + 1 22 + 1 32 + ··· + 1 n2 < 2.
12 Supposethat a, b and c arerealnumbers.
a Provethat
b Henceprovethat
c Usingsuitablesubstitutionsinpart b,orotherwise,provethat
d Deducethatif a + b + c = d,then a4 +
13a Supposethat a, b, c and d arepositive.
Usethefactthat
b Henceshowthat
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abcd
14a Showthat a2 + b2 ≥ 2ab forallrealnumbers a and b.
b Henceshowthat a2 + b2 + c2 ab bc ca ≥ 0.
c Deducethatforallpositiverealnumbers a, b and c: (a + b + c)2 ≥ 3(ab + bc + ca).
d Supposethat a, b and c arethesidelengthsofatriangle. Explainwhy(b c)2 ≤ a2 . i
Deducethat(a + b + c)2 ≤ 4(ab + bc + ca). ii
15 Supposethat a, b and c arepositive.
a Provethat a b + b a ≥ 2.
b Henceshowthat (a + b + c) 1 a + 1 b + 1 c ≥ 9.
ci Provethat a3 + b3 ≥ a c + b c abc,andwritedownsimilarinequalitiesfor b3 + c3 and c3 + a3 ii Henceprovethat a3 + b3 + c3 ≥ 3abc. iii Deducethat a b + b c + c a ≥ 3.
16 Supposethat x and y arepositive.
a Provethat 1 x + 1 y ≥ 4 x + y
bi Provethat 1 x2 + 1 y2 ≥ 2 xy
ii Usepart i andthesquareofpart a toprovethat 1 x2 + 1 y2 ≥ 8 (x + y)2
17 Supposethat a, b, c > 0.
a UseasuitableAM/GMinequalitytoprovethat a3 b + b3 c + bc ≥ 3ab
b Henceprovethat a3 b + b3 c + c3 a ≥ ab + bc + ca
18 Itisknownthat |a|2 = a2 and |a|≥ a foranyrealnumber a Let x and y beanytworealnumbers.Usetheaboveresultstoprovethetriangleinequality |x y|≥ |x|−|y|
[Hint:BeginwithLHS2 RHS2.]
19 [Triangleinequalitywithcomplexnumbers]
a Let z = x + iy beacomplexnumber.ProvealgebraicallythatRe(z) ≤|z|
b Let z and w betwocomplexnumbers.Provethat |z + w|≤|z| + |w|. Beginbywriting |z + w|2 = (z + w)(z + w).
c Underwhatcircumstancesis |z + w| = |z| + |w|?
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CHALLENGE
20 Giventhat a > b > c > 0,provethat a
(abc)
21 Given a, b, c > 0and abc = 1,usetheAM/GMinequalitywiththreetermstoprovethat 1 + ab 1 + a + 1 + bc 1 + b + 1 + ca 1 + c ≥ 3.
22a Provethat
b Supposethat a1, a2, , an 1 arepositiveintegers(notnecessarilyconsecutive)suchthat1 < a1 < a2 <
.Explainwhy
c Henceprovethat
23a Supposethat a,
areallpositive. i Provethat(a +
Henceprovethat(
b Nowconsideratriangle ABC withsidelengths a, b and c i Provethatsin
24 If a, b, c > 0,provethat
[Hint:BeginbyaddingonetoeachfractionontheLHS.]
2F FurtherProofbyInduction
InductionisanimportanttopicintheMathematicsExtension1course,andcandidatesinExtension2are expectedtobeproficientatthetwomainstylesalreadymet,sumsanddivisibility.Theexercisequestionsinclude problemswhichreviewthiswork.Inductionwillfurtherbeappliedtoinequalitiesandverifyingformulaefor sequencesspecifiedrecursively.
ReviewofInduction
Recallthattherearethreemainpartstoaninductionproof.First,thestatementisverifiedforanyinitialterms. Thesecondstepprovestheimplicationthatifthestatementistrueforsomeinteger k thenitistrueforthenext integer(k + 1).Eachproofthenconcludeswithanappealtotheprincipleofmathematicalinduction.
Example25
Provebymathematicalinductionthat
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forallpositiveintegers n.
Solution
Thisisjustthe
Alsorecallthatforthepartialsumsofanysequence
A When n = 1,LHS = ( 1)1 × 12 = 1 andRHS = 1 2 ( 1)1 × 1 × 2 = 1 = LHS sotheresultistruefor n = 1.
B Assumethestatementistrueforthepositiveinteger n = k.
Thatis,assumethat
Nowprovethestatementfor n = k + 1.Thatis,provethat
But LHS = S
sobytheinductionhypothesis(†):
= RHS.
C Itfollowsfromparts A and B bymathematicalinductionthattheresultistrueforallintegers n ≥ 1.
(†)
Harderquestionsmayinvolvedivisibilityproblems.Inothercasesthestatementmayonlybetrueforvaluesof n insomespecifiedsequence.Theeasiestwaytomanagethisistoreplace n withtheformulaforthe mth termof thesequence.Thenextexampledemonstratesbothsituations.
Example26
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Provebymathematicalinductionthat3n 2n isdivisibleby5when n isapositiveeveninteger.
Solution
Since n iseven,put n = 2m
Nowshowthat32m 22m isdivisibleby5for m ∈ Z+
A When m = 1,32m 22m = 9 4 = 5 × 1 sotheresultistruefor m = 1.
B Assumethestatementistrueforthepositiveinteger m = k Thatis,assumethat32k 22k = 5p,forsomeinteger p (†)
Nowprovethestatementfor m = k + 1.
Thatis,provethat32(k+1) 22(k+1) isdivisibleby5.
Now32(k+1) 22(k+1) = 32k × 9 22k × 4
sobytheinductionhypothesis(†): 3
= 5(9p + 22k ) whichisdivisibleby5.
C Itfollowsfromparts A and B bymathematicalinductionthattheresultistrueforallintegers m ≥ 1,and henceistrueforallpositiveevenintegers n
ProvingInequalities
Oftenitisnecessarytoproveaninequalitybyinduction,thuscombiningthetechniquesofthissectionwiththose presentedinSection 2E.Inmanycases,thebestapproachfortheinequalityistoputeverythingononeside.
Example27
Provebymathematicalinductionthat2n > n3 forallintegers n ≥ 10.
Solution
A When n = 10,LHS = 210 = 1024 and RHS = 103 = 1000 sotheresultistruefor n = 10.
B Assumethestatementistrueforthepositiveinteger n = k,where k ≥ 10. Thatis,assumethat2k > k3 (†)
Nowprovethestatementfor n = k + 1.Thatis,provethat
2k+1 > (k + 1)3 .
Continuedonthenextpage
ThiswillbedonebyprovingLHS RHS > 0.
LHS RHS = 2k+1 (k + 1)3
= 2 × 2k (k3 + 3k2 + 3k + 1),
sobytheinductionhypothesis(†):
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LHS RHS > 2k3 (k3 + 3k2 + 3k + 1)
= k3 (3k2 + 3k + 1)
> k3 (3k2 + 3k2 + 3k2)(as3k < 3k2 and1 < 3k2)
= k2(k 9) > 0 (since k ≥ 10).
C Itfollowsfromparts A and B bymathematicalinductionthattheresultistrueforallintegers n ≥ 10.
ProvingRecursiveFormulae
Recallthatsequencescanbedefinedusinganinitialvalueandarecursiveformula.Forexample,considerthe sequencedefinedby
Tn = Tn 1 + 2n,for n > 1,where T1 = 2.
Thefirstfewtermsofthatsequenceare2,6,12,20,30,42, Itappearsthatasimplerformulaforthissequence is
Tn = n 2 + n andthiscanbeprovedbymathematicalinduction.
Example28
Forthesequencedefinedrecursivelyby
Tn = Tn 1 + 2n,for n > 1,where T1 = 2, provebymathematicalinductionthat
Tn = n 2 + n,forall n ≥ 1.
Solution
A When n = 1,RHS = 12 + 1 = 2
sotheresultistruefor n = 1.
B Assumethestatementistrueforthepositiveinteger n = k
Thatis,assumethat Tk = k2 + k (†) Nowprovethestatementfor n = k + 1.Thatis,provethat
Tk+1 = (k + 1)2 + (k + 1)
= k2 + 3k + 2.
Fromthegivenrecursionformula:
LHS = Tk + 2(k + 1) sobytheinductionhypothesis(†):
LHS = k2 + k + 2(k + 1)
= k2 + 3k + 2 = RHS.
C Itfollowsfromparts A and B bymathematicalinductionthattheresultistrueforallintegers n ≥ 1.
Noticehoweasythislastinductionproofwas.Itisgenerallythecasethatrecursiveformulaeareeasytoproveby induction,butthereareexceptions.
Exercise2F
1 Provebyinductionthatforallpositiveintegervaluesof n:
2 Provebyinductionthatforallpositiveintegervaluesof n: 5n + 3isdivisibleby4 a
n + 2n+1 isdivisibleby3 c
n 5n + 4isdivisibleby5 e
3 Provebyinductionthatforallpositiveintegervaluesof n:
+
4 Provetheseinequalitiesbymathematicalinduction: n2 ≥ 3n 2for n ≥ 1 a 2n ≥ 1 + 3n for n ≥ 4 b
FOUNDATION
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DEVELOPMENT
5a Provebyinductionthat(1 + c)n > 1 + cn forallintegers n ≥ 2,where c isanon-zeroconstantgreater than 1.
b Henceshowthat 1 1 2n n > 1 2 forallintegers n ≥ 2.
6a Solvetheinequation x2 > 2x + 1.
b Henceprovebyinductionthat2n > n2 forallintegers n ≥ 5.
7 Ineachcaseasequencehasbeendefinedrecursivelyandthenaformulagivenforthe nthterm.Use mathematicalinductiontoproveeachformula.
a If T1 = 1and Tn = Tn 1 + n for n ≥ 2,then Tn = 1 2 n(n + 1)for n ≥ 1.
b If T1 = 1and Tn = 2Tn 1 + 1for n ≥ 2,then Tn = 2n 1for n ≥ 1.
c If T1 = 5and Tn = 2Tn 1 + 1for n ≥ 2,then Tn = 6 × 2n 1 1for n ≥ 1.
d If T1 = 1and Tn = 3Tn 1 1 4Tn 1 1 for n ≥ 2,then Tn = n 2n 1 for n ≥ 1.
8 [FormulaeforAPsandGPs]Usemathematicalinductiontoproveeachresult.
a If T1 = a and Tn = Tn 1 + d for n ≥ 2,then Tn = a + (n 1)d for n ≥ 1.
b If T1 = a and Tn = r × Tn 1 for n ≥ 2,then Tn = arn 1 for n ≥ 1.
c If S 1 = a and S n S n 1 = a + (n 1)d for n ≥ 2then
S n = 1 2 n 2a + (n 1)d for n ≥ 1.
d If S 1 = a and S n S n 1 = arn 1 for n ≥ 2and r 1then
S n = a (rn 1) r 1 for n ≥ 1.
9
Provebyinductionthatforoddpositivevaluesof n:
a 7n + 13n + 19n isdivisibleby13, b n3 n isdivisibleby24.
10
Proveeachdifferentiationresultbyinductionforallintegers n ≥ 1.
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11
12
a dn dxn 1 x = ( 1)nn! xn+1
b dn dxn (sin ax) = an sin ax + nπ 2
Provebyinductionthattheinterioranglesumofapolygonwith n sidesis(n 2) × 180◦.[Hint:Dissectthe (k + 1)-gonintoa k-gonandatriangle.]
Provebyinductionthatapolygonwith n sideshas 1 2 n(n 3)diagonals.
13 Provebyinductionthat n linesintheplane,notwobeingparallelandnothreeconcurrent,dividetheplane into 1 2 (n2 + n + 2)regions.[Hint:The(k + 1)thlinewillcrosstheother k linesin k distinctpoints,andso willadd k + 1regions.]
14 Provebyinductionthatforallpositiveintegervaluesof n:
a 1 12 + 1 22 + 1 32
15 Supposethat a, b > 0and n isapositiveinteger.
a Provetheinequality
b Henceprovebyinductionthat
16a Byrationalisingthenumerator,provethat
b Henceprovebyinductionthat1
17a Provetheidentity2cos A sin B = sin(A + B) sin(A B).
b Henceprovebyinductionthatforallpositiveintegers n,
18a Provethatforpositivevaluesof x and y, x y + y x ≥ 2.
b Henceprovebyinductionthatforpositivevaluesof
c Deducethatcosec2 θ + sec2 θ + cot2 θ ≥ 9cos2 θ
CHALLENGE
19 Ineachcaseasequencehasbeendefinedrecursivelyandthenaformulagivenforthe nthterm.Usea strongerformofinductiontoproveeachformula.
a If T1 = 3, T2 = 6,and Tn = 3 Tn 1 2 Tn 2 1for n ≥ 3then Tn = n + 2n for n ≥ 1.
b If T1 = 8, T2 = 34,and Tn = 8 Tn 1 15 Tn 2 for n ≥ 3then Tn = 5n + 3n for n ≥ 1.
c If T1 = 12, T2 = 30,and Tn = 5 Tn 1 6 Tn 2 for n ≥ 3then Tn = 3 × 2n + 2 × 3n for n ≥ 1.
d If T0 = T1 = 2,and Tn = 2 Tn 1 + Tn 2 for n ≥ 2then
20a Provetheinequality
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b Henceprovebyinductionthatforall n ≥ 2,
21 Define S n asthesumoftheproductsofallthepairsofdistinctintegersthatcanbeformedfromthefirst n positiveintegers.
So,forexample, S 3 = 1 × 2 + 1 × 3 + 2 × 3.
Provebymathematicalinductionthat S n = 1 24 (n 1)(n)(n + 1)(3n + 2)for n ≥ 2.
22 Asquadof n footballersputstheirtrainingtopsouttowash.Whenthewashinghasfinisheddrying,each playertakesatrainingtop,butitisfoundthatno-onehastheirown.Thissituationiscalledaderangement.
Let Dn bethenumberofderangements.
a Insomederangements,Benandanotherplayerhaveeachother’stop.Explainwhythenumberofthese derangementsis(n 1)Dn 2,for n > 2.
b Findasimilarformulafortheremainingderangementsandhenceshowthat
n = (n 1)Dn 1 + (n 1)Dn 2,for n > 2.
c Usetheaboveresulttoshowthat
d Find D1 and D2,thenprovebyinductionthat
e Henceprovebyinductionthat
2G InequalitiesinGeometryandCalculus
Thequestionsinthissectionoftenrequirethecalculusofintegrationorcurvesketching.Consequentlyitmaybe appropriatetodelaythissectiontolaterinthecourse,suchasattheendofChapter6.
AnInequalityUsingGeometry
TheAM/GMinequalitywasprovenalgebraicallyinSection 2E.Hereisageometricproofofthisimportant result.Ithastheadvantageofallowingreaderstoseehowtheinequalityworks,andidentifythespecialcaseof equality.
Example29
ProvetheAM/GMinequalityforpositiverealnumbers a and b usingcirclegeometry.Thatis,provethat a + b 2 ≥ √ab andidentifywhenthetwosidesareequal.
Solution
Constructlinesegment PQ with |PQ| = a + b
Constructthecirclewithdiameter |PQ| andcentre O
Theradiusofthiscircleisclearly
r = a + b 2 (thearithmeticmean.)
Let V bethepointthatdivides PQ intheratio a : b
Constructchord UW perpendicularto PQ at V.
Hence V bisects UW,solet UV = VW = x
Now x2 = ab (productofinterceptsofintersectingchords) so x = √ab (thegeometricmean.)
Finally,thelongestchordinacircleisadiameter,hence r ≥ x thatis a + b 2 ≥ √ab withequalitywhen UW isalsoadiameter.Thatis, a = b.
AnInequalityUsingCalculus
Manyinequalitiesarisethroughthestudyofcalculusandgeometry.Asaverysimple example,usinganupperandalowerrectangletoapproximatetheareabetweenthe hyperbola y = 1 x andthe x-axisfor1 < x < 2gives
1 2 < log2 < 1.
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Oftenknowledgeofothertopicsisalsorequired.Thefollowingworkedexamplemakes useofcalculusandageometricseriestofindanapproximationforlog 3 2 whichis accuratetotwodecimalplaces. x y yx 1 2
Example30
Considerthegeometricseries
S 2n = 1 h + h2 ... + h2n , where0 < h < 1.
a Showthat S 2n 1 < 1 1 + h < S 2n.
b Integratethepreviousresultbetween h = 0and h = x,where0 < x < 1,andhencewritedownapolynomial inequalityforlog(1 + x).
c Use n = 3toestimatethevalueoflog 3 2 .
Solution
a Now S 2n
(byGPtheory) = 1 h2n 1 + h < 1 1 + h (increasenumerator) and S 2n = 1 h + h2 ...
(byGPtheory)
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c Put n = 3and x = 1 2 toget
Thesecalculationssuggestthatlog 3 2 ≑ 0.405,correcttothreedecimalplaces.Theactualvalueis0.40547, correcttofivedecimalplaces.
TheSqueezeTheorem
Limitsarerequiredforcalculusinbothintegrationanddifferentiation.Aproperstudyoflimitsmaybea componentofuniversitystudy,butathighschooltheevaluationofalimithasreliedonintuition.Theresultis thatalimitlike
lim x→0 sin x x cannotbedeterminedeasily.The squeezetheorem,alsoknownasthe sandwichtheorem,allowssuchalimit tobedeterminedbysqueezingitbetweentwoknownlimits.Thefollowingstatementofthetheoremmaybe consideredimprecise,butitisallthatisneededfortheproblemsinthiscourse.
Ifthreefunctionsarerelatedby
f (x) ≤ g(x) ≤ h(x)
forallvaluesof x near x = k,butnotnecessarilyat x = k,andif
limx→k f (x) = limx→k h(x) = L
thenlimx→k g(x) = L.
Thediagramontherightshowsonepossiblesituation.Itmayhelptounderstandwhat isgoingon.Inthisparticularcase,allthreefunctionsaredefinedat x = 2.Thegraph of y = g(x)remainsbetween y = f (x)and y = h(x),soas x → 2thevalueof g(x)is squeezedbetweenthevaluesof f (x)and h(x).Sinceboth f (x)and h(x)approach y = 1, sotoomust g(x).
Unfortunately,theproofofthesqueezetheoremrequiresmathematicsbeyondthescopeofthiscourse. Nevertheless,itmaybeusedtohelpevaluatelimits.
Example31
Inthediagramontheright,thetangent AB subtendsanacuteangle x atthecentre O ofacircle(notdrawn)withradius OA = 1.Line OB intersectsthecircleat C.It shouldbeclearthat areaof △OAC ≤ areaofsector OAC ≤ areaof △OAB
a Usethisresulttoshowthat1 ≤ x sin x ≤ 1 cos x ,for0 < x < π 2
b Writedowntheresultofthereciprocaloftheinequalityinpart a andexplainwhy itisalsovalidfor π 2 < x < 0.
c Usethesqueezetheoremtodeterminelimx→0 sin x x .
d Henceprovethat d dx sin x = cos x
Solution
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a Usingthegiveninequalityandevaluatingtheareas, 1 2 sin x ≤ 1 2 x ≤ 1 2 tan x.
Doublinganddividingbysin x,sincesin x 0for0 < x < π 2 , 1 ≤ x sin x ≤ 1 cos x
Continuedonthenextpage
b Thereciprocalreversesthisorder,giving cos x ≤ sin x x ≤ 1
Nowboth x andsin x areoddfunctionshence sin x x iseven.Sinceallthreefunctionsareeven,itfollowsthat theinequationisalsovalidfor π 2 < x < 0.
c Fromparts a and b,forall x near x = 0, cos x ≤ sin x x ≤ 1
thuslimx→0 cos x ≤ limx→0 sin x x ≤ limx→0 1
so 1 ≤ limx→0 sin x x ≤ 1
hence,bythesqueezetheorem
limx→0 sin x x = 1.
d d dx sin x = limh→0 sin(x + h) sin x h (bydefinition) = limh→0 sin x cos h + cos x sin h sin x h (expanding) = limh→0 cos h 1 h sin x + sin h h cos x (liketerms) = limh→0 cos2 h 1 h(cos h + 1) sin x +
h h cos x = limh→0 sin2 h h(cos h + 1) sin x + sin h h cos x (byPythagoras) = limh→0 sin h h cos x sin h h sin
1 × cos x 1 × 0 × sin x.(from c)
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Thus d dx sin x = cos x.
Notethatsin(x + h)isexpandedonthesecondlineofthesolutiontopart d.Analternativemethodusesthe sumstoproductstrigonometricformulaewhicharestudiedinChapter4.Thisproofmightbereviewedatthat stage.
24Thesqueezetheorem
Ifthreefunctionsarerelatedby f (x) ≤ g(x) ≤ h(x)forallvaluesof x near x = k,butnotnecessarilyat x = k,andif limx→k f (x) = limx→k h(x) = L thenlimx→k g(x) = L.
Exercise2G FOUNDATION
1a Aregulardodecagonisdrawninsideacircleofradius1cmandcentre O sothatitsverticeslieonthecircumference,asshowninthefirstdiagram. Determinetheareaof △OAB,andhencefindtheexactareaoftheinscribed dodecagon.
bi Usetheformulafortan2θ toshowthattan15◦ = 2 √3.
ii Anotherregulardodecagonisdrawnwithcentre O,sothateachsideis tangenttothecircle,asshownintheseconddiagram.Findtheareaof △OGH andhencefindtheexactareaofthecircumscribeddodecagon.
c Byconsideringtheresultsinparts a and b,showthat
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2 Thepoints A, P and B onthecurve y = 1 x have x-coordinates1,1 1 2 and2 respectively.Thepoints C and D arethefeetoftheperpendicularsdrawnfrom A and B tothe x-axis.Thetangenttothecurveat P cuts AC and BD at M and N respectively.
a Findtheareaoftrapezium ABDC
b Byfindingtheareaofarectangle,orotherwise,findtheareaoftrapezium MNDC
c Henceshowthat 2
3 Supposethat m ≤ f (x) ≤ M intheinterval a ≤ x ≤ b.Useadiagramtohelpprovethat
4 Itisknownthat 1 4 x
+ x 1 ≤ f (x) ≤ 2 3 x near x = 2.Findlimx→2 f (x). DEVELOPMENT
5 Thediagramshowsupperrectanglesforthegraphof y = 1 x
a Byconsideringappropriateareas,showthat 1 + 1 2 + 1 3 + + 1 n ≥ loge(n + 1).
b Whatdoyouconcludeabouttheinfiniteseries
6a Usecalculustoshowthatthegraphof y = ln x isconcavedownthroughoutitsdomain.
b Sketchthegraphof y = ln x,andmarktwopoints A(a,ln a)and B(b,ln b)onthecurve,where0 < a < b.
c Findthecoordinatesofthepoint P thatdividestheinterval AB intheratio2:1.
d Usingparts a and c,deducethat
).
7 Thediagramshowsthepoints A(0,1)and B( 1, e 1)onthecurve y = ex,and thepoint C( 1,0)onthe x-axis.Thetangentat P( 1 2 , e 1 2 )intersects OA at M and BC at N.
a Determinetheexactareaoftheregionboundedbythecurve, BC, CO and OA.
b Findtheareaoftrapezium OABC
c Showthattheareaof OMNC isequalto CO × PQ,andhencefindtheareaof thetrapezium.
d Henceshowthat 1 2 (3 + √5) < e < 3.
8 Thefunction f (x)isdefinedby f (x) = x loge(1 + x2).
a Showthat f ′(x)isnevernegative.
b Explainwhythegraphof y = f (x)liescompletelyabovethe x-axisfor x > 0.
c Henceprovethat ex > 1 + x2,forallpositivevaluesof x
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9 Let f (x) = xn e x,where n > 1.
a Showthat f ′(x) = xn 1e x(n x).
b Showthat(n, nne n)isamaximumturningpointofthegraphof f (x),andhencesketchthegraphfor x ≥ 0.(Don’tattempttofindpointsofinflection.)
c Explainwhy xne x < nne n for x > n (Beginbyconsideringthegraphof f (x)for x > n.)
d Deducefrompart c that(1 + 1 n )n < e
10 Considerthefunction y = ex 1 x 10 10
a Findthetwoturningpointsofthegraphofthefunction.
b Discussthebehaviourofthefunctionas x →∞ andas x →−∞
c Sketchthegraphofthefunction.
d Fromyourgraph,deducethat ex ≤ 1 x 10 10 ,for x < 10.
e Henceshowthat 11 10 10 ≤ e ≤ 10 9 10 .
11a Let a, b and c bethelengthsofthesidesofatriangleandlet ∠A beoppositeside a.Usethecosineruleto helpprovethat |b c|≤ a ≤ b + c Thatis,provethatonesideofatriangleislongerthanthedifferencebetweentheothertwosidesand shorterthanthesumoftheothertwosides.
b Henceproveforanytwocomplexnumbers z and w that
c Underwhatcircumstancesis |z|−|w| = |z + w|?
12 Inthisquestionyoumayassumethatsimpleexponentialcurvesareconcaveup.
a Showbydirectcalculationthat: 66 < 3 × 56 , i
b Thepoints A 1 6 ,3 1 6 and B(0,1)lieontheexponentialcurve y = 3x.Thepoints B and C 1
ontheexponentialcurve y = 5 2 x
i Usepart a toshowthatthegradientofchord AB isgreaterthan1andthegradientofchord BC isless than1.
ii Henceshowthat 5 2 < e < 3.
13
Let |t| < 1andlet N beapositiveinteger.
a Showthat1 +
b Showthatthedifferencebetweenthetwois
c Integratetheresultinpart a between0and x,where0 < x < 1.Henceshowthat:
[Hint:Write
d Explainwhy ∫
e Useparts b to d toshowthat
f Hencefindln2correcttothreedecimalplaces.
14 Thegraphof y = 1 t ,for t > 0,isshown.
Let x > 1and α > 0.
a Byconsideringupperandlowerrectangles,showthat
b Usethesqueezetheoremtoshowlim
15a Let n > 1and k bepositiveintegers.Uselowerrectanglestoprovethat 1 1 n
b Henceprovethat
16 Considertheintegral ∫ n+x n 1 t dt.
a Useupperandlowerrectanglestoshowthat
b Usethesqueezetheoremtoshowthatlimn→∞
= ex foranygivenvalueof x
c Usetrialanderrortodeterminehowbig n needstobesothat 1 + x n n ≑ ex correcttothreedecimal placeswhen x = 0.1.
17 Considertheintegral ∫ a 0 x2 dx.
a Use n lowerand n upperrectanglesofwidth a n toshowthat:
b Provebyinductionthat12 + 22 + 32 + + n2 = 1 6 n(n + 1)(2n + 1)forallpositiveintegers n
c Bytakinglimitsas n →∞,usethesqueezetheoremtoshowthat
18 Thediagramontherightshowsthecurve y = f (x)intheinterval a ≤ x ≤ b where f ′′(x) > 0.Thecorrespondingchordis CD and MN istangentto y = f (x)at P where x = a+b 2 .
a Useareastobrieflyexplainwhy (
b Byputting f (x) = 1 x2 in a,showthatfor n = 2, 3,4, , 4 (2n 1)2 <
c Deducethat 4
d Showthat
CHALLENGE
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Henceshowthat
19a Find d dx (x ln x) andhencefind ∫ ln xdx
b Showthat ∫ n 1 ln xdx = n ln n n + 1.
c Usethetrapezoidalruleontheintervalswithendpoints1,2,3,..., n toshowthat ∫ n 1 ln xdx > 1 2 ln n + ln(n 1)!
d Henceshowthat n! < n n+ 1 2 e1 n
Note: ThisisapreparatorylemmaintheproofofStirling’sformula n! ≑
2π n n+ 1 2 e n,whichgivesan approximationfor n!whosepercentageerrorconvergesto0forlargeintegers n
20a Provethatloge x ≤ x 1for x > 0.
b Supposethat p1, p2, p3,..., pn arepositiverealnumberswhosesumis1. Provethat n r=1 loge(npr ) ≤ 0.
c Let x1, x2, x3,..., xn bepositiverealnumbers.Provethat x
21 [Aproofthat e isirrational]Forpositiveintegervaluesof n,let S n = 1 + n r=1 1 r! .
a Provebyinductionthat e S n = e ∫ 1 0 xn n! e x dx forallpositiveintegervaluesof n. [Hint:Youwillneedtouseintegrationbyparts,whichiscoveredinSection 6F.]
b From a deducethat0 < e S n < 3 (n + 1)! forallpositiveintegervaluesof n.
c Use b todeducethat (e S n) n!isNOTanintegerfor n = 2,3,4,
d Showthattherecannotexistpositiveintegers p and q suchthat e = p q
Chapter2Review Reviewactivity
• Createyourownsummaryofthischapteronpaperorinadigitaldocument.
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Chapter2Multiple-choicequiz
• Thisautomatically-markedquizisaccessedintheInteractiveTextbook.AprintablePDFWorksheetversionis alsoavailablethere.
ChapterReviewExercise
1 Answertrueorfalseineachpart.
a Inmathematics,astatementisasentencethatcanbebothtrueandfalse.
b Thestatement‘P or Q’canbedenotedby P ∨ Q.
c Thenegationof P =⇒ Q is(P and ¬Q).
d Theconverseandthecontrapositiveofastatementareequivalentstatements.
e ¬(P and Q) ⇔ (¬P and ¬Q).
f If P =⇒ Q and Q =⇒ P,then P and Q arethesamestatement.
g ‘P implies Q’isequivalentto‘¬Q implies ¬P’.
h (P =⇒ Q) ⇔¬(P and ¬Q).
2 Writedowntheconverseofeachstatement,andstatewhethertheconverseistrueorfalse.
a Ifaquadrilateraliscyclic,thenitsoppositeanglesaresupplementary.
b Iftwonumbersarebothodd,thentheirsumiseven.
c Everyrhombusisaparallelogram.
3 Writedownthenegationofeachstatement.
a Allmathematiciansareintelligent.
b SuzielikesPhysicsandChemistry.
c IfIamonvacation,thenIamnotworking.
4 Writedownthecontrapositiveofeachstatement.
a IfIamabicycle,thenIhavetwowheels.
b Ifanumberisodd,thenitslastdigitisnot6.
c Asquarehasfourequalsides.
5 Writedown(in if then form)thetwoconversestatementsequivalenttoeach ifandonlyif statement.
a Anumberisevenifandonlyifitisdivisibleby2.
b Aquadrilateralisaparallelogramifandonlyifitsdiagonalsbisecteachother.
c a isdivisibleby b ifandonlyif ∃ c ∈ Z suchthat a = bc
6 Writedown:
a thenegationof ¬A and ¬B,
b thecontrapositiveof A =⇒ B,
c theconverseof A =⇒¬B,
d thenegationof A =⇒¬B.
7 Supposethatthestatement ¬P ∧ (Q =⇒ P)istrue.Whatcanweconcludeabout P and Q?
8 Provethat:
a thesumofthreeconsecutiveintegersisdivisibleby3, b theproductofthreeconsecutiveevennumbersisdivisibleby8, c theproductoftwoconsecutiveevennumbersisdivisibleby8.
9 Provethattheremainderis1whenanoddsquarenumberisdividedby4.
10 Provethatanintegerisdivisiblebyboth3and5ifitisdivisibleby15.
11 If n isodd,usedivisibilityargumentstoprovethat n3 n isdivisibleby24.
12 Iftheinteger n is not divisibleby3,provethat n2 + 2mustbedivisibleby3.
13a Showbyexpandingthatif n isodd, x
+ 1 = (x + 1)(
b Henceprovethatif n isodd: i 2n + 1isdivisibleby3, ii 2mn + 1isdivisibleby2m + 1.
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1).
14 Supposethat abc isa3-digitnumber,with a c > 1.Ifwesubtract cba from abc andthenaddtheresultof thissubtractiontothenumberobtainedbyreversingitsdigits,provethattheansweris1089.
15 Provebycontradictionthat: √7isirrational, a log3 7isirrational. b
16 Byprovingthecontrapositive,provethatif a2 iseventhen a iseven.
17a Provethatif √5 + √3isrational,thensois √5 √3.[Hint:Consider 1 √5+√3 .] b Henceprovebycontradictionthat √5 + √3isirrational. (Youmayusethefactthat √5isirrational.)
18 Provethatif2n 1isprimethen n isprimebyprovingthecontrapositive.
19a Provethat x + y ≥ 2√xy for x, y > 0.
b Henceprovethat(a + b)(1 + ab) ≥ 4ab for a, b > 0.
20a Provethat a b + b a ≥ 2for a, b > 0.
b Henceprovethat ab(a + b) + bc(b + c) + ca(c + a) ≥ 6abc for a, b, c > 0.
21a Usethefactthat x + y 2 ≥ √xy,for x, y > 0,toprovethat
bi Showthat
ii Henceprovethat a + b + c 3 ≥ 3 √abc.
22 Itisknownthat x 2 < g(x) < x2 x 1near x = 1.Findlimx→1 g(x).
23 Provebyinductionthatforallpositiveintegervaluesof n:
a n r=1(r + 1) × 2r = n × 2n+1
b n r=1 r2(r + 1) = 1 12 n(n + 1)(n + 2)(3n + 1)
24 Provebyinductionthatforallpositiveintegervaluesof n:
a 6n + 4isdivisibleby5,
b n3 + 2n isdivisibleby3.
25a Provebyinductionthatforallpositiveintegervaluesof n:
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b Whatisthevalueoflimn→∞
26 Provebymathematicalinductionthat:
a n(n + 2)isdivisibleby8forevenpositivevaluesof n,
b 3n + 7n isdivisibleby10foroddpositivevaluesof n,
c 4n + 5n + 6n isdivisibleby15foroddpositivevaluesof n.
27 Usemathematicalinductiontoprovethatif T1 = 3and Tn = Tn 1 + 4n for n ≥ 2,then Tn
n ≥ 1.
28 Foracertainsequence, a1 = 1and an+1 = √2an + 1for n ≥ 1. Provebyinductionthat an < 3for n ≥ 1.
29 Provebyinductionthat n! > 3n for n ≥ 7.
30 Provebyinductionthat,for n ≥ 1,the nthderivativeof xe x is( 1)n(x n)e x
31 Prove,for n ≥ 3,thattheexterioranglesumofaconvex n-sidedpolygonis360◦
32a Provebyinductionthat2n > n,forallpositiveintegers n.
b Henceshowthat1 < n √n < 2for n ≥ 2.
c Supposethat a and n arepositiveintegers.Itisknownthatif n √a isrationalthenitisaninteger.Whatcan wededuceabout n √n,where n isapositiveintegergreaterthan1?
Appendix:TheFundamentalTheoremofArithmetic
ThefundamentaltheoremofarithmeticisnotpartoftheExtension2course.Itsproofisincludedherefirstly becausethetheoremissoimportantinthestudyofnumbertheoryandsecondlytojustifyitsuseinseveralproofs earlierinthischapter,mostnotablyinprovingthatlog2 3isirrational.Theproofofthetheoremreliesonseveral otherresults,whichalsoneedexplanationandproof.
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Theorem. Supposethattwointegersaandbhave HCF = d.
Then (a bq) isdivisiblebydforanyintegerq.
Proof Both a and b havefactor d,sothereexistintegers m and n suchthat
a = md and b = nd.
Thus a bq = md ndq = d(m nq).
Hence(a bq)isdivisibleby d.
Usingthedivisionalgorithm,forintegers a and b itisalwayspossibletowrite
a = bq1 + r1 where0 ≤ r1 < b, andwhere q1 and r1 areintegers.Itfollowsfromtheaboveproofthatboth b and r1 aredivisibleby d,theHCF of a and b.Nowrepeatthedivisionalgorithmtoget
b = r1q2 + r2 where0 ≤ r2 < r1, andwhere q2 and r2 areintegers.Onceagainitfollowsthatboth r1 and r2 aredivisibleby d.Nowcontinuethe processtogetadecreasingsequenceofpositiveremainders r1, r2, whichmustthereforeterminate.Further, everyterminthissequenceisdivisibleby d.Consequentlythelastterminthissequenceofremaindersis d itself, theHCFof a and b.Anexampleshouldhelpconvincethereaderofthis.
FindingtheHCFof81and66:
81 = 66 × 1 + 15
66 = 15 × 4 + 6
15 = 6 × 2 + 3
6 = 3 × 2.
ThustheHCFof81and66is3.
Moresignificantly,whenthedivisionalgorithmiswrittenoutlikethis,theresultisasetofequationsthatcanbe usedtowritetheHCFasasumofmultiplesoftheoriginalnumbers.Itissimplyamatterofworkingbackwards throughtheequations.Inthiscase:
3 = 15 6 × 2(from[3])
= 15 (66 15 × 4) × 2(from[2])
= 15 × 9 66 × 2
= (81 66) × 9 66 × 2(from[1])
= 81 × 9 66 × 11.
Inotherwords,usingthesetwoalgorithms,itisalwayspossibletofindthevaluesof x and y suchthat ax + by = d.Inthiscase, x = 9and y = 11.
Itisappropriatetopauseatthispointtogivearemarkableproofthat √n iseitheranintegerorirrational whenever n isapositiveinteger.
Theorem. Ifn ∈ Z+ theneither √n ∈ Z or √nisirrational.
Proof Supposethat √n isrational.Then ∃ a, b ∈ Z+ suchthat
√n = a b or b√n = a (∗∗) where b ≥ 1andtheHCFof a and b is1.Thus,fromabove,thereexistintegers x and y suchthat ax + by = 1.
Multiplythisequationby √n toget
√n = (a√n)x + (b√n)y
Nowusetheresultofequation(∗∗)toget:
√n = (bn)x + (a)y
TheRHSofthislastequationisasumofproductsofintegers,andmustthereforebeaninteger.Henceif √n isrationalthenitmustbeaninteger.Otherwise √n isirrational. Hereisanotherusefultheoremthatwillbeneeded.
Theorem. IftheHCFofprimepandintegerais1,andifintegerabisdivisiblebyp,thenbisdivisiblebyp.
Proof SincetheHCFof p and a is1, ∃ x, y ∈ Z suchthat ax + py = 1 thus abx + pby = b.
But ab isdivisibleby p,soput ab = mp toget mpx + bpy = b thatis p(mx + by) = b, bywhich b isdivisibleby p
Asaconsequenceofthisproof,ifaninteger n isdivisiblebyprime p then p dividesatleastonefactorof n Theproofofthisstrongerresultcanbedonebyinduction,andisleftasanexercise.Withthesetools,itisnow possibletoprovethefundamentaltheoremofarithmeticforpositiveintegers.
Theorem. Ifapositiveintegerncanbewrittenastheproductofitsprimesintwoways,thenoneisa rearrangementoftheother.
Proof Suppose,bywayofcontradictionthereexistsapositiveinteger n whichcanbewrittenastheproductof itsprimefactorsintwodifferentways.Thenletthoseproductsbe n = p1 × p2 × ... × pk = q1 × q2 × ...
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qℓ, where,forthesakeofsimplicity,itisassumed k <ℓ.Then,bydefinition, q1 × q2 × × qℓ isdivisibleby p1
Itfollowsfromabovethat p1 mustdivideoneoftheseprimefactors.Letthisbe q j.Butifoneprime dividesanotherthentheymustbeequal.Thatis p1 = q j
Nowre-orderandre-labeltheprimessothat p1 = q1.Thatis: n = p
with p1 = q1.
Cancelthisprime p1 sothat
n p1
Nowrepeattheaboveargumentwith p2, p
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... p
,(orwriteitoutproperlyusinginduction)toget n p
Butaproductofprimescannotequal1.Hence k = ℓ andeachprime pi isequaltoacorresponding prime qi
25Thefundamentaltheoremofarithmetic
Ifapositiveinteger n canbewrittenastwodifferentproductsofprimes
then k = ℓ andtheprimes p
areare-arrangementoftheprimes q1, q2, , qℓ
Notethattheaboveproofcanbeadaptedtoincludenegativeintegervaluesof n,andthisisleftasanexercise.
3 ComplexNumbersII: Polynomials&deMoivre
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Chapterintroduction
ThischaptercontinuesthestudyofcomplexnumbersbeguninChapter1wheretheywereappliedtosolve quadraticequations.InSection 3A,theExtension1theoryforotherpolynomialequationsissimilarly broadenedtoincludecomplexnumbers.
Complexpolynomialequationsofthetype zn 1 = 0aresignificant.Thekeytosolvingtheseequationsis deMoivre’stheorem,whichispresentedinSection 3B.Oneconsequenceofthistheoremisthattrigonometric identitiescanbequicklydeveloped,andsomecommonidentitiesareinvestigatedinSection 3C.Finally, deMoivre’stheoremisusedinSection 3D tofindthecomplexrootsofnumbers,andtodeducerelationships betweentheroots.
3A PolynomialsandComplexNumbers
Asignificantapplicationofcomplexnumbersisinthestudyofpolynomials.Thissectionfurtherdevelopsthe workonpolynomialsalreadydoneintheMathematicsExtension1course.Thatworkisassumedknowledge thoughsomepartsofthetheoryarerepeatedhereforthesakeofconvenience.Thefocusisonpolynomialswith realcoefficientsandtherelationshipswiththezeroes,particularlywhentheyareeithercomplex,orrealand repeated.
ThecruxofthetheoryislaterinthissectionwheretheFundamentalTheoremofAlgebraispresentedalongwith someofitsconsequences.Thetheoremisleftunprovenasanyproofisbeyondthescopeofthecourse.
PolynomialswithIntegerCoefficients
Ifapolynomialwithintegercoefficientshasanintegerzero x = k,then k isafactoroftheconstantterm.Thisisa significantaidinfactorisingapolynomial.
Example1
Itisknownthatthepolynomial P(x) = x3 x2 8x 6hasonlyoneintegerzero.Finditandhencefactorise P(x)completely.
Solution
Thezeroisafactorof6,sothepossiblevaluesare: ±1, ±2, ±3, ±6.Testingtheseonebyone: P(1) = 14, P( 1) = 0, andthereisnoneedtocontinuefurther.Bythefactortheorem,(x + 1)isafactorof P(x).Performingthelong division:
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Thus P(x) = (x + 1)(x2 2x 6) = (x + 1) (x 1)2 7 (completingthesquare) = (x + 1)(x 1 √7)(x 1 + √7)(differenceoftwosquares.)
1Integercoefficientsandzeroes
Ifthepolynomial
(
) =
withintegercoefficients a0, a1, a2, ... , an,hasanintegerzero x = k,then k isafactoroftheconstantterm a0.
Proof Since P(k) = 0,itfollowsthat
a0 + a1k + a2k2 + + ankn = 0
so a1k + a2k2 + ... + ankn = a0
thus k × (a1 + a2k + + ankn 1) = a0
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Sinceallthetermsinthebracketsareintegers,theresultisalsoaninteger.Thus,thelefthandsideistheproduct oftwointegers.Hence k isafactorof a0
PolynomialsandComplexNumbers
Considerthegeneralpolynomial
P(x) = a0 + a1 x + a2 x 2 + ... + an xn .
Eachterminthisexpressioninvolvesanintegerpowerandmultiplicationbyaconstant.Thetermsarethen simplyadded.Sinceintegerpowers,multiplicationandadditionareallnaturaloperationswithcomplexnumbers, itfollowsthatthepolynomialcanbeevaluatedwhen x isacomplexnumber.Forexampleif P(x) = x2 2x + 4 thenat x = i itsvalueis
P(i) = i2 2i + 4 = 3 2i .
Insomeexamplesthepolynomialwillbewrittenasafunctionof z inordertoemphasisethefactthatcomplex numbersmaybesubstituted.
RemaindersandFactors
HereisaquicksummaryofcertainimportantresultsfromtheMathematicsExtension1course.Inthe usualnotation,let P(x)and D(x)beanypairofpolynomials,where D(x) 0.Thereisauniquepairof polynomials Q(x)and R(x),suchthat
P(x) = D(x) × Q(x) + R(x), andwhereeither
deg(D) > deg(R)or R(x) = 0.
Thisisknownasthedivisiontheorem.Asaconsequence,if D(x) = (x α)then R(x)mustbeaconstant,either zeroornon-zero.Letthisconstantbe r.Re-writingthedivisiontheorem:
P(x) = (x α) × Q(x) + r , whence P(α) = r , whichisknownastheremaindertheorem.
If R(x) = 0thenfromthedivisiontheorem
P(x) = D(x) × Q(x),
sothat P(x)isaproductofthefactors D(x)and Q(x).Inparticular, x α isafactorof P(x)ifandonlyif P(α) = 0.Thisisknownasthefactortheorem.
Thedivisiontheorem,theremaindertheoremandthefactortheoremarevalidforcomplexnumbersaswellas realnumbers.Thoughtheseclaimswillnotbeprovenhere,theresultsmaybefreelyappliedtosolveproblems.
Example2
Let P(x) = x3 2x2 x + k ,where k isreal.
a Showthat P(i) = (2 + k) 2i .
b When P(x)isdividedby x2 + 1theremainderis4 2x.Findthevalueof k
Solution
a P(i) = i3 2i2 i + k
= i + 2 i + k
= (2 + k) 2i
b Bythedivisiontheorem,
Thus P(i) = 4 2i hence(2 + k) 2i = 4 2i
Equatingtherealpartsgives k = 2.
RealCoefficientsandRemainders
Supposethatthepolynomial P(z)hasrealcoefficients.Iftheremainderwhen P(z)isdividedby(z α)is β thentheremainderwhen P(z)isdividedby(z α)is β .Usingtheremaindertheorem,thisisequivalenttothe statementthatif P(α) = β then P(α) = β
Example3
a Usetheremaindertheoremtofindtheremainderwhen P(z) = z3 2z2 + 3z 1isdividedby(z i).
b Hencefindtheremainderwhen P(z)isdividedby(z + i).
Solution
Theremainderis: P(i) = i3 2i2 + 3i 1
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2Realcoefficientsandremainders
Ifthepolynomial P(z)hasrealcoefficientsandif P(α) = β then P(α) = β
Theproofisnottoodifficultandisdealtwithinaquestionoftheexercise.
RealCoefficientsandComplexzeroes
Supposethatthepolynomial P(z)hasrealcoefficients.If P(z)hasacomplexzero z = α thenitisguaranteedto haveasecondcomplexzero z = α .Further,bythefactortheorem,thereexistsanotherpolynomial Q(z)suchthat: P(z) = (z α)(z α) × Q(z) = (z 2 (α + α)z + αα) × Q(z)
= (z 2 2Re(α)z + |α|2) × Q(z).
Thus P(z)hasaquadraticfactorwithrealcoefficients:(z2 2Re(α)z + |α|2).
Example4
Considerthepolynomial P(z) = 2z3 3z2 + 18z + 10.
a Giventhat1 3i isazeroof P(z),explainwhy1 + 3i isanotherzero.
b Findthethirdzeroofthepolynomial.
c Hencewrite P(z)asaproductof: i linearfactors, ii alinearfactorandaquadraticfactor,bothwithrealcoefficients.
Solution
a Since P(z)hasrealcoefficients, (1 3i) = 1 + 3i isalsoazero.
b Letthethirdzerobe a ,thenbythesumoftheroots:
a + (1 3i) + (1 + 3i) = 3 2 so a + 2 = 3 2 and a = 1 2 . ci Bythefactortheorem:
P(z) = 2(z + 1 2 )(z 1 + 3i)(z 1 3i) = (2z + 1)(z 1 + 3i)(z 1 3i).
ii P(z) = (2z + 1)(z2 2z + 10).
3Realcoefficientsandzeroes
Ifthepolynomial P(z)hasrealcoefficientsandacomplexzero z = α thenitisguaranteedtohavea secondcomplexzero z = α
Consequently P(z)has(z2 2Re(α)z + |α|2)asafactor,whichisaquadraticwithrealcoefficients.
Proof Supposethatthecomplexnumber z = α isazeroofthepolynomial
P(z) = a0 + a1z + a2z 2 + + anz n , wherethecoefficients a0, a1, , an areallreal.Thatis P(α) = 0.Then P(
) =
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(since z )n =
(since c
)
Thus, P(α) = P(α) = 0 = 0.
Hence z = α isalsoazeroofthepolynomial P(z).Further,asshownabove: P(z) = (z2 2Re(α)z + |
|2) × Q(z).
The FundamentalTheoremofAlgebra
Alltheworkencounteredsofarinthissectiondealswithfindingthezeroesofvariouspolynomials.Uptothis pointithasbeenpossibletosidestepanimportantquestion:doeseverypolynomialhaveazero?Forthereisno pointinsearchingforoneifnoneexists.
Inordertoemphasisethispoint,considerthepolynomial P(x) = x2 + 1.Clearlythisfunctionhasnorealzero, andthereisnopointinsearchingforone.Yetthepolynomialdoesindeedhavetwozeroes,bothofwhichhappen tobecomplexnumbers:namely i and i.Coulditbethatthereisanotherpolynomialwhichhasneitherrealnor complexzeroes?
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Theanswertothisquestionis:everypolynomialwithdegree ≥ 1hasatleastonezero,thoughthatzeromay becomplex.Thisissuchanimportantandbasicfactinthestudyofmathematicsthatitisgivenatitle— The FundamentalTheoremofAlgebra.
4Thefundamentaltheoremofalgebra
Everypolynomialwithdegree ≥ 1hasatleastonezero,thoughthatzeromaybecomplex.
SeveraleminentmathematiciansworkedonthistheoremincludingLeibniz,EulerandArgand.Butcredit isusuallygiventoGaussforthefirstproof,whichhepresentedinhisdoctoralthesisin1799.This,orany otherproofofthetheorem,isbeyondthescopeofthiscourse.Althoughthewordinggivenintheboxaboveis imprecise,itisusuallysufficientfortheproblemsencounteredatthislevel.
TheDegreeandtheNumberofZeroes
AlthoughtheFundamentalTheoremofAlgebracannotbeprovenhere,itispossibletoprovetwosignificant consequencesofthetheorem.Thefirstisthateverypolynomialofdegree n ≥ 1withcomplexcoefficientshas precisely n zeroes,ascountedbytheirmultiplicities.
Thisisalsotrueforpolynomialswithrealcoefficients.Todemonstratetheresult,thecubic P(x) = x3 3x2 + 4 = (x 2)2(x + 1)hasthreezeroes:thesimplezero x = 1andthedoublezero x = 2.Sometimesthisiswrittenas x = 1,2,2.
5Thedegreeandthenumberofzeroes
Everypolynomialofdegree n ≥ 1withcomplexcoefficientshasprecisely n zeroes,ascountedbytheir multiplicities.
Proof Thisproofusesinductiononthedegreeofapolynomial.
A Considerthegeneralpolynomialofdegreeonewithcomplexcoefficients:
P1(x) = a0 + a1 x ,where a1 0.
Clearlythispolynomialhasonezero x = α1,where α1 = a0 a1 |a1|2 . Thustheresultistruefor n = 1.
B Supposethattheresultistrueforsomeinteger k ≥ 1.Thatis,supposethateverypolynomial Pk (x)of degree k withcomplexcoefficients,
Pk (x) = a0 + a1 x + + ak x k ,where ak 0, has k zeroes, x = α1, ... , αk ,ascountedbytheirmultiplicities. [**] Next,provethestatementistruefor n = k + 1.Thatis,provethateverypolynomial Pk+1(x)ofdegree k + 1withcomplexcoefficients,
Pk+1(x) = a0 + a1 x + ... + ak+1 x k+1 ,where ak+1 0, has k + 1zeroesascountedbytheirmultiplicities.
Nowforanyparticularpolynomial Pk+1(x),thatpolynomialhasatleastonezerobytheFundamental TheoremofAlgebra.Letthiszerobe x = αk+1.Then,bythefactortheorem,itfollowsthat Pk+1(x) = (x αk+1)Qk (x)
forsomepolynomial Qk (x)ofdegree k.Butbytheinductionhypothesisabove[∗∗], Qk (x)has k zeroes, allofwhicharethusinheritedby Pk+1(x).
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Hence Pk+1(x)has k + 1zeroes, x = α1, , αk , αk+1,ascountedbytheirmultiplicities.Clearlythis followsforeachandeverypolynomial Pk+1(x).
C Itfollowsfromparts A and B bymathematicalinductionthatthestatementistrueforallintegers n ≥ 1.
RealLinearandQuadraticFactors
ThesecondsignificantconsequenceoftheFundamentalTheoremofAlgebraisthateverypolynomialofdegree n ≥ 1withrealcoefficientscanbewrittenasaproductoffactorswhichareeitherlinearor irreducible quadratics, eachwithrealcoefficients.Inthiscontexttheword irreducible isusedtoindicatethatthequadratichasnoreal zero.
Inordertodemonstratetheresult,noticethatthepolynomial P(x) = x3 1canbewrittenastheproduct
P(x) = (x 1)(x 2 + x + 1).
Thequadraticfactor(x2 + x + 1)isirreduciblesinceithasnorealzero.
6Reallinearandquadraticfactors
Everypolynomialofdegree n ≥ 1whichhasrealcoefficientscanbewrittenasaproductoffactorswhich areeitherlinearorirreduciblequadratics,eachwithrealcoefficients.
Proof Let Pn(x) = a0 + a1 x + ... + an xn beapolynomialwithdegree n ≥ 1whichhasrealcoefficients.Bythe previousresult,thispolynomialhas n zeroes.Letthesezeroesbe x = α1, ... , αn.Ifnoneofthesezeroes arecomplexthentheresultisobviouslytruesincethen,usingproductnotation,
Pn(x) = an n k=1 (x αk ).
Ifallofthezeroesarecomplexthentheresultisagainobviouslytrue.Since Pn(x)hasrealcoefficients, therootscomeinconjugatepairs,bywhichitisknownthat n iseven.Thus
Pn(x) = an n/2 k=1 (x αk ) × (x αk ) = an n/2 k=1 x 2 2Re(αk )x + |αk |2
Lastly,ifsomeofthezeroesarecomplexnumbersthenonceagaintheyoccurasconjugatepairs,sincethe coefficientsof Pn(x)arereal.Letthenumberofconjugatepairsbe j,where1 < 2 j < n.Nowre-orderand re-labelthezeroeswiththeconjugatepairslistedfirst.Thusthefirstconjugatepairis x = α1, α1 ,andthe lastconjugatepairis x = α j, α j .
Ifthereareanyotherzeroesthentheyarereal.Thefirstoftheseis x = α2 j+1 andthelastis x = αn .Soby thefactortheorem,andusingproductnotation:
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Ineachofthethreecasestheresultisaproductoffactorswithrealcoefficients,whichareeitherlinearor irreduciblequadraticfactors.Putmoresimply,multiplyallthecomplexfactorstogetherinconjugatepairs togetirreduciblequadraticfactorswithrealcoefficients,andanyremainingfactorsarebothlinearand real.
Exercise3A FOUNDATION
1 Itisknownthatineachcasethegivenpolynomial P(x)hasonlyoneintegerzero.Finditandhence factorise P(x)completely.
(x) = x3 6x
2 Itisknownthat1 + i isazeroofthepolynomial P(x) = x3 8x2 + 14x 12.
a Whyis1 i alsoazeroof P(x)?
b Usethesumofthezeroestofindthethirdzeroof P(x).
3 Itisknownthat1 2i isazeroofthepolynomial P(x) = x3 + x + 10.
a Writedownanothercomplexzeroof P(x),andgiveareasonforyouranswer.
b Henceshowthat x2 2x + 5isafactorof P(x).
c Findthethirdzero,andhencewrite P(x)asaproductoffactorswithrealcoefficients.
4 Itisknownthat 3i isazeroofthepolynomial P(z) = 2z3 + 3z2 + 18z + 27.
a Writedownanothercomplexzeroof P(z).Justifyyouranswer.
b Hencewritedownaquadraticfactorof P(z)withrealcoefficients.
c Write P(z)asaproductoffactorswithrealcoefficients.
5 Let P(z) = 2z3 13z2 + 26z 10.
a Showthat P(3 + i) = 0.
b Statethevalueof P(3 i),andgiveareasonforyouranswer.
c Hencewrite P(z)asaproductof: i linearfactors, ii alinearfactorandaquadraticfactor,bothwithrealcoefficients.
DEVELOPMENT
6 Considerthepolynomial Q(x) = x4 6x3 + 8x2 24x + 16.
a Itisknownthat Q(2i) = 0.Whydoesitfollowimmediatelythat Q( 2i) = 0?
b Byusingthesumandtheproductofthezeroesof Q(x),orotherwise,findtheothertwozeroesof Q(x).
c Hencewrite Q(x)asaproductof: fourlinearfactors, i threefactorswithrealcoefficients, ii twofactorswithintegercoefficients. iii
7a Solvetheequation x4 3x3 + 6x2 + 2x 60 = 0giventhat x = 1 + 3i isaroot.
b Solvetheequation x4 6x3 + 15x2 18x + 10 = 0giventhat x = 1 i isaroot.
8 Twoofthezeroesof P(z) = z4 12z3 + 59z2 138z + 130are a + ib and a + 2ib,where a and b arerealand b > 0.
a Findthevalueof a byconsideringthesumofthezeroes.
b Usetheproductofthezeroestoshowthat4b4 + 45b2 49 = 0,andhencefind b.
c Henceexpress P(z)astheproductofquadraticfactorswithrealcoefficients.
9 Supposethat P(x) = x3 + kx2 + 6,where k isreal.
a Showthat P(2i) = (6 4k) 8i .
b When P(x)isdividedby x2 + 4theremainderis 4x 6.Findthevalueof k
10 Let P(x) = x3 x2 + mx + n,whereboth m and n areintegers.
a Showthat P( i) = (1 + n) + i(1 m).
b When P(x)isdividedby x2 + 1theremainderis6x 3.Findthevaluesof m and n .
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11
Supposethat P(x) = x3 + x2 + 6x 3.
a Usetheremaindertheoremtofindtheremainderwhen P(x)isdividedby x + 2i
b Hencefindtheremainderwhen P(x)isdividedby: x 2i , i x2 + 4. ii
12 Let P(z) = z8 5 2 z4 + 1.Supposethat w isarootof P(z) = 0.
a Showthat iw and 1 w arealsorootsof P(z) = 0.
b Findoneoftherootsof P(z) = 0inexactform.
c Hencefindalltherootsof P(z) = 0.
13
Supposethat P(x) = x4 + Ax2 + B,where A and B arepositiverealnumbers.
a Explainwhy P(x)hasnorealzeroes.
b Giventhat ic and id arezeroesof P(x),where c and d arerealand c d,writedowntheothertwo zeroesof P(x),andgiveareason.
c Provethat c4 + d4 = A2 2B.
14 Thepolynomial P(x) = x3 + cx + d,where c and d arerealandnon-zero,hasanegativerealzero k,andtwo complexzeroes.Thegraphof y = P(x)hastwoturningpoints.
a Whatcanbesaidaboutthetwocomplexzeroesof P(x),andwhy?
b Byconsidering P′(x),showthat c < 0.
c Sketchthegraphof y = P(x).
d If a ± ib,where a and b arereal,arethecomplexzeroesof P(x),deducethat a > 0.
e Provethat d = 8a3 + 2ac.
15 Considerthepolynomialfunction f (x) = x3 3x + k,where k isanintegergreaterthan2.
a Showthat f (x)hasexactlyonerealzero r,andexplainwhy r < 1.
b Giveareasonwhythetwocomplexzeroesof f (x)formaconjugatepair.
c Ifthecomplexzeroesare a + ib and a ib,usetheresultforthesumoftherootstwoatatimetoshow that b2 = 3(a2 1).
d Findthethreezeroesof f (x)giventhat k = 2702,andthat a and b areintegers.
16 Considerthepolynomial P(z) = z4 +
2 + 20z + 25.Itisknownthat z = 1 + 2i isadoublezeroof P(z).
a Explainwhy z = 1 2i isalsoadoublezeroof P(z).
b Explainwhy P(z)hasnootherzeroes.
c Hencefactorise P(z)overthecomplexnumbersandthenovertherealnumbers.
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CHALLENGE
17 Inthetextitwasproventhatif P(z)isapolynomialwithrealcoefficientsandif P(α) = 0then P(α) = 0. Useasimilarapproachtoprovethatif P(α) = β then P(α) = β .
18 Let P(x) = a0 + a1 x + a2 x2 + + an xn beapolynomialwithintegercoefficients.Supposethat P(x)hasa rationalzero x = p q where p and q havehighestcommonfactor1.Showthat p isafactorof a0 andthat q is afactorof an .
19 UsetheFundamentalTheoremofAlgebratocarefullyexplainwhyeverypolynomialofodddegreewith realcoefficientshasatleastonerealzero.
20a Thepolynomial P(z)hasrealcoefficientsandadoublecomplexzero z = α.
i Explainwhy z = α isalsoadoublezero.
ii Showthat z2 2Re(α)z + |α|2 2 isafactorof P(z).
iii Henceprovethat P′(z)hassimplecomplexzero α
b Followsimilarworkingtoshowthatif P(z)hasrealcoefficientsandacomplexzero α withmultiplicity m then P′(z)hascomplexzero α with multiplicity(m 1).
21a Let u and v betwonumbersoftheform u = a + b√c,where a, b and c arerationalnumbers,with √c anirrationalconstant.Letthenotation u∗ indicatethevalueof u whenthesignof b isreversed.Thatis, u∗ = a b√c
Showthat u∗ + v∗ = (u + v)∗ i
Showthat λu∗ = (λu)∗ whenever λ isarationalnumber. ii Provebyinductionthat(un)∗ = (u∗)n forpositiveintegers n iii
b Supposethat u = a + b√c isazeroofacertainpolynomialwithrationalcoefficients.Usetheresultsof part a toshowthat u∗ = a b√c isalsoazeroofthispolynomial.
3B PowersofComplexNumbers
Recallthatwhencomplexnumbersaremultipliedtheargumentsareadded,asfollows:
arg(wz) = arg(w) + arg(z).
Nowput w = z = cis θ.Thatis,bothareequalandhavemodulus1.Then:
z2 = z × z
= cis(θ + θ)(addingarguments)
= cis2θ
Nextput w = z2,sothat
z3 = z2 × z
= cis(2θ + θ)(againbyaddingarguments)
= cis3θ
Theseinitialcalculationssuggestthesimplerelationship
z n = cis nθ , whenever |z| = 1,atleastforpositiveintegers n .Infacttheresultistrueforallintegers,whichisnowproven.
deMoivre’sTheorem
Let z = cos θ + i sin θ .Itcanbeproventhat
z n = cos nθ + i sin nθ forallintegers n.Theproofisintwoparts,beginningwithaproofbyinductionfor n ≥ 0.Conjugatesarethen usedtoextendtheprooftonegativeintegers.
Proof Asalwayswithproofbyinduction,firstprovetheresulttrueforthestartingvalue.
A When n = 0
LHS = z0 = 1, since z 0.
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RHS = cos0 + i sin0 = 1 + 0i = LHS.
Hencethestatementistruefor n = 0.
B Supposethattheresultistrueforsomeinteger k ≥ 0,thatis
zk = cos kθ + i sin kθ [†] Nowprovethestatementfor n = k + 1.Thatis,provethat
zk+1 = cos (k + 1)θ + i sin (k + 1)θ
LHS = zk × z = (cos kθ + i sin kθ) × (cos θ + i sin θ)(bythehypothesis[†]) = cos (k + 1)θ + i sin (k + 1)θ (bythesumofarguments) = RHS.
Hencetheresultistruefor n = k + 1.
C Itfollowsfromparts A and B bymathematicalinductionthatthestatementistrueforallintegers n ≥ 0.
D Finally,extendtheresulttothenegativeintegers. Todothis,considerthevalueof z n when n isapositiveinteger.
z n = z 1 n
= ( z )n
= cos( θ) + i sin( θ) n
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(since |z| = 1)
= cos( nθ) + i sin( nθ)(bypart C,since n ispositive,) andtheproofiscomplete.
7deMoivre’stheorem
Let z = cos θ + i sin θ beacomplexnumberwithmodulus1.Thenforallintegers n, z n = cos nθ + i sin nθ
Oneimmediateconsequenceoftheabovetheoremisthatif z = r cis θ then zn = rn cis nθ.Thusif r > 1and θ > 0thenas n increasessotoodoesthemodulusandargumentof zn.Thatis,thepointsrepresenting zn lieonan anticlockwisespiral.
Bywayofexample,thetablebelowshowsthevaluesof zn inthecasewhen z = i√2for integervaluesof n between 4and4.
Thecorrespondingpointsareshowninthegraphaboveandarejoinedwithasmoothcurve.Noticethatthespiral doesnotcuttheaxesatrightangles.
Amorepracticalapplicationistoquicklysimplifyintegerpowersofcomplexnumbers,asinthefollowing example.
Example5
a Write z = √3 + i inmodulus-argumentform.
b Henceexpress z7 infactoredreal-imaginaryform.
Solution
a Itshouldbeclearthat z = 2(cos 5π 6 + i sin 5π 6 ).
b UsingdeMoivre’stheorem,
z 7 = 27(cos 5π 6 + i sin 5π 6 )7
= 128(cos 35π 6 + i sin 35π 6 )
= 128(cos π 6 + i sin π 6 )
= 64(√3 i).
Example6
Forwhichintegervaluesof k is(1 + i)k imaginary?
Solution
Now(1 + i) = √2(cos π 4 + i sin π 4 ) so(1 + i)k = √2k (cos kπ 4 + i sin kπ 4 )(bydeMoivre) whichisimaginarywhen kπ 4 isanoddmultipleof π 2 .
Thus kπ 4 = (2n+1)π 2 where n isaninteger, thatis k = 4n + 2, hence k = , 6, 2,2,6,10,
Exercise3B
1 Writeeachexpressionintheformcis nθ: (cos θ + i sin θ)5 a (cos θ + i sin θ) 3 b (cos2θ + i
FOUNDATION
θ)4 c
2 Simplifyasfullyaspossible: (cos θ + i sin θ)6(cos θ + i sin θ) 3 (cos θ i sin θ)4 a (cos3θ + i sin3θ)5(cos2θ i sin2θ) 4 (cos4θ i sin4θ) 7 b
3 Writeeachexpressionintheform a + ib,where a and b arereal: (cos π 4 +
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6 )5
(cos 2 3 + i sin 2 3 ) 2 d (cos 3 8 i sin 3 8 ) 6 e (cos 5π 12 i sin 5π 12 )4 f
4a Write1 + i intheform r(cos θ + i sin θ).
b Hence,orotherwise,find(1 + i)17 intheform a + ib,where a and b areintegers.
5 Let z = 1 + i√3.
Express z inmod-argform. a
Express z11 intheform a + ib,where a and b arereal. b
6 Let z = √3 + i.
Findthevaluesof |z| andarg z. a
Hence,orotherwise,showthat z7 + 64z = 0. b
7a Express √3 i inmod-argform.
b Express(√3 i)7 inmod-argform.
c Henceexpress(√3 i)7 intheform x + iy,where x and y arereal.
8a Express 1 i√3inmod-argform.
b Express( 1 i√3)5 inmod-argform.
c Henceexpress( 1 i√3)5 intheform x + iy,where x and y arereal.
9a Express z = √2 i√2inmod-argform.
b Hencewrite z22 intheform a + ib,where a and b arereal.
10 Showthat:
(1 + i)10 ispurelyimaginary a (1 i√3)9 isreal b 1 + i isafourthrootof 4 c √3 i isasixthrootof 64 d
11 If k isamultipleof4,provethat( 1 + i)k isreal.
12a Findtheminimumvalueofthepositiveinteger m forwhich(√3 + i)m is: real, i purelyimaginary. ii
b Evaluate(√3 + i)m foreachoftheabovevaluesof m.
13a Provethat(1 + i)n + (1 i)n isrealforallpositiveintegervaluesof n
b Determinethepositivevaluesof n forwhich(1 + i)n + (1 i)n = 0.
14 UsedeMoivre’stheoremtoprovethat
15a Showthatif n isdivisibleby3then(1
b Simplifytheexpressionif n isnotdivisibleby3.
16 Showthat 1
17 Provethat(1 +
18 Let z = cis π n ,where n isapositiveinteger.Showthat:
19a Usegeometrytoexpress1 + cos θ + i sin θ inmodulus-argumentform.
b Henceshowthat1 + 4cos θ + 6cos2θ + 4cos3θ + cos4θ = 16cos4( θ 2 )cos2θ
DEVELOPMENT
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3C TrigonometricIdentities
AusefulapplicationofdeMoivre’stheoreminvolvescombiningitwithbinomialexpansionstoobtainvarious trigonometricidentities.
Example7
a Expresscos3θ intermsofpowersofcos θ.
b Henceshowthat x = cos π 9 isasolutionof8x3 6x 1 = 0.
c Findthevalueofcos π 9 cos 5π 9 cos 7π 9 .
Solution
a Let z = cos θ + i sin θ,then z3 = (cos θ + i sin θ)3 sobydeMoivre’stheorem cos3θ + i sin3θ = cos3 θ + 3i cos2 θ
Taketherealparttoget
=
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b Let x = cos θ,sothat 4x3 3x = cos3θ . Thus4x3 3x = 1 2 when θ =
Hence8x3 6x 1 = 0hassolution x = cos π 9 .
c Therootsofthecubicin b are x = cos θ wherecos3θ = 1 2 .Thesolutionsofcos3θ = 1 2 whichgivedistinct valuesofcos θ are θ = π 9 , 5π 9 and 7π 9 .Hence,bythetheproductoftheroots,
ApplicationsinCalculus
Thecalculusofcomplexvariablesisbeyondthiscourse.Nevertheless,atrigonometricidentitythathasbeen derivedfromeithertherealpartortheimaginarypartofdeMoivre’stheoremcansafelybeusedtosolve problemsincalculus.
Example8
a Let z = cos θ + i sin θ.Showthat zn z n = 2i sin nθ
b Expand(z z 1)5 .
c Useparts a and b toshowthat16sin5 θ = sin5θ 5sin3θ + 10sin θ
d Hencefind ∫ π 2 π 4 sin5 θ dθ
Solution
a z n z n = z n zn (since |z| = 1)
= 2i Im(z n)
= 2i Im(cos nθ + i sin nθ)(bydeMoivre)
= 2i sin nθ
b (z z 1)5 = z5 5
c Rearrangingpart b,
so(2i sin θ)
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θ (bypart a)
thus16sin5 θ = sin5θ 5sin3θ + 10sin θ (equatingimaginaryparts.)
d Dividingby16andintegratingyields
Exercise3C
2 Usesimilarmethodstothepreviousquestiontoshowthat:
3 Let z =
a UsedeMoivre’stheoremtoshowthat
Henceshowthatcos
4 Repeatthemethodsofthepreviousquestiontoshowthat
FOUNDATION
5a UsethemethodsofQuestion 3 toshowthat
cos5 θ = 1 16 (cos5θ + 5cos3θ + 10cos θ).
b Henceevaluate ∫ π 2 0 cos5 θ dθ.
6a UsedeMoivre’stheoremtoprovetheidentity
cos6α = 32cos6 α 48cos4 α + 18cos2 α 1.
b Henceshowthatthepolynomialequation32
5,7,9,11.
c Usetheproductofthesesixrootstodeducethatcos π 12 cos 5π 12 = 1 4
7 Theidentitytan4θ = 4tan θ
a Usethisidentitytoshowthattheequation
tan 9π 16 ,tan 13π 16
b Henceshowthattan
8a Let z = cos θ + i sin θ.Showthat zn z
b Expand z z 1 6 thenusepart a toshowthat
2
DEVELOPMENT
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c Hencefindthestationarypointsof y = cos6θ 6cos4θ + 15cos2θ 20.
9a UsedeMoivre’stheoremtoshowthat
b Henceshowthattheequation16x5 20x3 + 5x 1 = 0hasroots x = 1,sin π 10 ,sin
c Byequatingcoefficients,orotherwise,findthevaluesof b and c forwhich 16x 4 + 16x 3 4
d Henceexplainwhytheequation16
= 0hastwodoubleroots.
ei Usepart b toshowthattheequation16x4 + 16x3 4x2 4x + 1
sin 13π 10 ,sin 17π 10
ii Doesthiscontradictpart c whichassertsthattheequationhastwodoubleroots?
f Hencefindexactvaluesforsin π 10 andsin 3π 10 10a Expresscos7θ intheform A
b Usethesubstitution x = 4cos2 θ toshowthattheequation x
c Henceevaluate:
11a Showthatsin5 θ = 1 16 (sin5θ 5sin3θ + 10sin θ).
b Hencesolvetheequation16sin5 θ = sin5θ for0 ≤ θ < 2π.
12a UsedeMoivre’stheoremtoshowthat
b Henceshowthattheequation
c Deducethattan π 5 tan 2 5 = √5andthattan2 π 5 +
13 Let z = cos θ + i sin θ.
a Showthat2cos
b Henceshowthat
c Deducethatcos3 θ
14a UsedeMoivre’stheoremtoexpress
b Hencesolve x6 6x4 + 10x2 4 = 0,leavingtherootsintrigonometricform.
15 Let n beapositiveinteger.
a UsedeMoivre’stheoremtoshowthat
b Henceshowthatthepolynomial
d Usethefactthatcot
CHALLENGE
haszeroesoftheform
3D
RootsofComplexNumbers
RecallfromapreviousexamplethatthepointsintheArganddiagramwhichrepresent zn,where n isaninteger, lieonaspiralwhenever |z| 1.When |z| = 1,itshouldbeclearthatthepointslieontheunitcircle.Further,if z = cos θ + i sin θ thentheangleattheoriginsubtendedbysuccessivepointsis
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arg(z n) arg(z n 1) = arg zn zn 1 = arg(z) = θ
Thatis,theangleisconstant.Thussuccessivepointsareregularlyspacedabouttheunitcircle.
Forexample,thesketchontherightshowsthepoints zn for n = 1,2,3,4,5,6,where z = cos 1 2 + i sin 1 2 .Notethat
arg(z) = 1 2 ≑ 28◦39′ , whichistheanglesubtendedattheoriginbyanypairofsuccessivepoints.Itshould beclearthat2π ÷ 1 2 = 4π isirrational,andhencenoneofthepointscoincide,evenfor largervaluesof n.Inthatsense,thisisnotaveryinterestingexample.
RootsofUnity
Significantconfigurationsofpointsarisewhenequationsoftheform
z n = 1
aresolved.Therearealways n solutionsandthepointsareequallyspacedabouttheunitcircleinthecomplex plane.Thesenumbersarecalledthe rootsofunity,forobviousreasons.
8Rootsofunity
Thesearetherealandcomplexsolutionsoftheequation
z n = 1
where n isapositiveinteger.WhenthesolutionsareplottedintheArganddiagram,the n pointsare equallyspacedaroundtheunitcircle.
Thesituationisbestdemonstratedbyexample.
Example9
a Solve z6 = 1.
b Plotthesolutionsontheunitcircleinthecomplexplane.
i Whatistheanglesubtendedattheoriginbysuccessiveroots?
ii Whatregularpolygonhasthesepointsasvertices?
c Let α = cis( π 3 ).Showthatthelist1, α, α2 , α3 , α4 and α5 includesallsixrootsof z6 = 1.
d Let β = cis 2π 3 .Whichrootsof z6 = 1canbewrittenintheform βm,where m isaninteger?Whatpolygon hasthesepointsasvertices?
Solution
a Let z = cis θ andnotethat1 = cis2kπ,where k isaninteger.Thus cis6θ = cis2kπ (bydeMoivre)
so6θ = 2kπ
hence θ = kπ 3
Applytherestriction π < θ ≤ π toobtainallthedistinctsolutions.Thus
π < kπ 3 ≤ π
so 3 < k ≤ 3.
Hencethesixrootsof z6 = 1are
cis( 2π 3 ),cis( π 3 ),1,cis π 3 ,cis 2π 3 and 1.
b Thegraphontherightshowsthesesixroots. i Clearlytheangleatthecentreis π 3 . ii Thesearetheverticesofaregularhexagon.
c UsingdeMoivre’stheorem,thegivenlistis: 1,cis( π 3 ),cis( 2π 3 ),cis( 3π 3 ) = 1, cis( 4π 3 ) = cis 2π 3 andcis( 5
Thisisthesamelistasgivenintheanswertopart a,butsimplyinadifferentorder.
d Now βm = cis 2mπ 3 bydeMoivre’stheorem.Hencearg(βm) isamultipleof 2π 3 .Thustheonlypossiblevalues that βm maytakeare: cis( 2π 3 ),1andcis 2π 3
Thatis,onlythesethreerootscanbewrittenasapowerof β.ThepointsintheArganddiagramformthe verticesofanequilateraltriangle.
Thereareseveralimportantfeaturestoobserveintheexampleabove.First,therearesixsolutions,asassuredby thefundamentaltheoremofalgebra.Secondly,arg(1) = 2kπ isusedinsteadoftheprincipalargumentArg(1) = 0. Thereasoningisthat,evenifArg(z) = θ,thevalueof nθ maynotbetheprincipalargumentof zn.Thisisseen above:whenArg z = θ = 2π 3 ,thevalueof6θ = 4π whichisclearlynotaprincipalvalue.Thirdly,thevaluesof therootsinthesolutionarewrittenwiththeirprincipalargumentsbyrestricting k,asshowninpart a
9Findingtherootsofunity
Tosolve zn = 1,beginbyletting z = cos θ + i sin θ andbynotingthat1 = cos2kπ + i sin2kπ.Thus,by deMoivre’stheorem: cos nθ + i sin nθ = cos2kπ + i sin2
sothat nθ = 2kπ.Thenapply π < θ ≤
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tofindtherestrictionson k
Further,anypowerofarootwillalwayscoincidewithoneorotheroftheroots.Inthecaseof αm intheexample, eachoftheotherrootswasvisitedonebyoneas m increased.Inthecaseof βm,onlysomeoftheotherrootswere obtained.Noticethelinkherewiththefactorsof6.
RootsofComplexNumbers
Thetechniqueforfindingrootsofunitycanbeadaptedtofindtherootsofanycomplexnumber.Thetrickisto allowboththemodulusandtheargumenttobeunknown.Itisthenamatterofequatingmoduliandarguments. Onceagain,therootsareequallyspacedaroundacircle.
Example10
Findthecuberootsof8i.
Solution
Thatis,solvetheequation z3 = 8i
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Beginbyletting z = r cis θ andputting8i = 8cis( π 2 + 2kπ),then:
r3 cis3θ = 8cis( π 2 + 2kπ)(bydeMoivre.)
Equatethemodulitoget r = 2,andfromthearguments
θ = (4k+1)π 6
Fortheprincipalargumentinthesolution
π < (4k+1)π 6 ≤ π
so 6 < 4k + 1 ≤ 6 thus k = 1,0,1
andthethreerootsare:2cis( π 2 ),2cis
Thatis 2i, √3 + i and √3 + i ,whichlieonthecirclewithradius2.
Thetechniquescannowbeusedtofindtheexactvaluesofthetrigonometricfunctionsatcertainrational multiplesof π,asinthefollowingexample.
Example11
Considertheequation z5 + 1 = 0.
a FindtherootsofthisequationandshowthemontheArganddiagram.
b Factorise z5 + 1:
i asaproductoflinearfactors, ii asaproductoflinearandquadraticfactorswithrealcoefficients.
c Evaluatecos π 5 + cos 3π 5 .
d Let α beacomplexrootof z5 + 1 = 0,thatis α 1.
i Showthat1 α + α2 α3 + α4 = 0.
ii Findaquadraticequationwithroots(α4 α)and(α2 α3).
e Put α = cis π 5 inpart d,andhenceevaluatecos π 5 .
Solution
a Let z = cis θ andnotethat 1 = cis(2k + 1)π,where k isaninteger.Thus cis5θ = cis(2k + 1)π (bydeMoivre)
so5θ = (2k + 1)π
hence θ = (2k+1)π 5 .
Continuedonthenextpage
Applytherestriction π < θ ≤ π toobtainallthedistinctsolutions.Thus
π < (2k+1)π 5 ≤ π
so 5 < (2k + 1) ≤ 5 or 3 < k ≤ 2.
Hencethefiverootsare:
cis( 3π 5 ),cis( π 5 ),cis
,cis
and 1, orinconjugatepairs,
b Usingtherootsofthegivenequation,
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c Bythesumoftheroots
di Since α isacomplexroot,
so(
+ 1)(1
0(fromGPtheory) thus1
= 0(since α 1) ii Thesumoftherootsis
Theproductoftherootsis
Hencetherequiredquadraticis
e With
Exercise3D FOUNDATION
1a Findthethreecuberootsofunity,expressingthecomplexrootsinbothpolarandCartesianform.Use therestriction π < θ ≤ π
b Showthatthepointsinthecomplexplanerepresentingthesethreerootsformanequilateraltriangle.
c If ω isoneofthecomplexroots,showthattheothercomplexrootis ω2
d Writedownthevaluesof:
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e Showthat: i (1 + ω2)3 = 1 ii (1 ω ω2)(1 ω + ω2)(1 + ω ω2) = 8 iii (1 ω)(1 ω2)(1 ω4)(1 ω5) = 9
2 Considertheequation z3 = i,whoserootsarethethreecuberootsof i.
a Write i inmod-argformwithprincipalargument.
b Hencewrite i inmod-argformwithageneralargument.
c If z = cos θ + i sin θ,showthat θ = (4k 1)π 6 .
d Hencefindthethreerootsoftheequationinmod-argformwithprincipalarguments.
e ExpressthethreerootsinCartesianform.
3a Solvetheequation z6 = 1,expressingthecomplexrootsintheform a + ib, where a and b arereal.
b PlottheserootsonanArganddiagramandshowthattheyformaregularhexagon.
c If α isthecomplexrootwithsmallestpositiveprincipalargument,showthattheotherthreecomplex rootsare α2 , α 1 and α 2 .
d Showthat z6 1 = (z2 1)(z4 + z2 + 1).
e Hencewrite z4 + z2 + 1asaproductofquadraticfactorswithrealcoefficients.
4a Find,intheform a + ib,thefourfourthrootsof 1.
b Hencewrite z4 + 1asaproductoftwoquadraticfactorswithrealcoefficients.
5a Find,intheform a + ib,thesixrootsoftheequation z6 + 1 = 0.
b Henceshowthat z6 + 1 = (z2 + 1)(z2 √3 z + 1)(z2 + √3 z + 1).
c Dividebothsidesofthisidentityby z3 andthenlet z = cos θ + i sin θ toshowthat
6a Find,inpolarform,thefivefifthrootsof i.
b Find,inpolarform,thefourfourthrootsof i
c Find,intheform a + ib,thefourfourthrootsof 8 8√3 i
d Find,inpolarform,thefivefifthrootsof16√2 16√2 i.
7a Findthefivefifthrootsof 1,writingthecomplexrootsinmod-argform.
DEVELOPMENT
b If α isthecomplexrootwithleastpositiveprincipalargument,showthat α3 , α7 and α9 aretheotherthree complexroots.
c Showthat α7 = α2 andthat α9 = α4
d Usethesumoftherootstoshowthat α
8a Findthesevenseventhrootsofunity.
b Byconsideringthesumoftheroots,showthatcos
c Write z7 1asaproductofonelinearandthreequadraticfactors,allwithrealcoefficients.
d If α isthecomplexseventhrootofunitywiththeleastpositiveprincipalargument,showthat
α5 and α6 aretheotherfivecomplexroots.
e Acertaincubicequationhasroots
4.Usetherelationshipsbetweentheroots andcoefficientstoshowthattheequationis x
9ai Findthefivefifthrootsofunity,writingthecomplexrootsinmod-argform.
ii Showthatthepointsinthecomplexplanerepresentingtheserootsformaregularpentagon.
iii Byconsideringthesumofthesefiveroots,showthatcos 2 5 +
bi Showthat
ii Henceshowthat
iii Byequatingthecoefficientsof z inthisidentity,showthatcos π 5 = 1+√5 4
ci Usethesubstitution x = u + 1 u toshowthattheequation x2
10a Findtheninthrootsofunity.
b Henceshowthat
c Deducethat
11
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and
=
3
5 + ρ6 aretherootsofthe quadraticequation x2 + ax + b = 0,where a and b arereal.
a Showthat ρ + ρ2 + + ρ6 = 1.
b Findthevaluesofthecoefficients a and b
c Deducethat
12 Let ω = cis 2 9
a Showthat ωk ,where k isaninteger,isasolutionoftheequation z9 = 1.
b Showthat ω + ω2 + ω3 + ω4 + ω5 + ω6 + ω7 + ω8 = 1.
c Henceshowthatcos 2 9 + cos 4 9 = cos π 9
d Deducethatcos π 9 cos 2 9 cos 4 9 = 1 8
13ai UsedeMoivre’stheoremtoexpresscos4θ andsin4θ intermsofcos θ andsin θ
ii Henceprovethattan4θ = 4tan θ 4tan3 θ 1 6tan2 θ + tan4 θ
iii Deducethattan 1 24 7 = 4tan 1 1 3 .
b Hencefindthefourfourthrootsof7 + 24i inCartesianform.
14 Supposethat ω3 = 1and ω 1.Let k beapositiveinteger.
a Whatarethetwopossiblevaluesof1 + ωk + ω2k ?
b Usethebinomialtheoremtoexpand(1 + ω)n and(1 + ω2)n,where n isapositiveinteger.
c Let ℓ bethelargestintegerforwhich3ℓ ≤ n.Showthat
d If n isamultipleof6,showthat
+ n 3 +
15a Showthattheequation(
for k = 1,2,3.
b Henceshowthat
c Bymakingasuitablesubstitutionintothisidentity,deducethat
16 Considertheequation(
CHALLENGE
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n isapositiveinteger.
a Showthateveryrootoftheequationispurelyimaginary.
b Lettherootsberepresentedbythepoints P1, P2, ... ,P2n intheArganddiagram,andlet O betheorigin. Showthat OP1 2 + OP2 2 + ··· + OP2n 2 = 2n(2n 1).
Chapter3Review
Reviewactivity
• Createyourownsummaryofthischapteronpaperorinadigitaldocument.
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Chapter3Multiple-choicequiz
• Thisautomatically-markedquizisaccessedintheInteractiveTextbook.AprintablePDFWorksheetversionis alsoavailablethere.
ChapterReviewExercise
1 If3i isazeroofapolynomial P(z)withrealcoefficients,explainwhy z2 + 9isafactorof P(z).
2 Itisknownthat2 + 5i isazeroofthepolynomial P(z) = z3 8z2 + 45z 116.
a Whyis2 5i alsoazeroof P(z)?
b Usethesumofthezeroestofindthethirdzeroof P(z).
c Hencewrite P(z)asaproductoftwofactorswithrealcoefficients.
3 Solvetheequation x4 6x3 + 15x2 18x + 10 = 0giventhat x = 1 i isaroot.
4 Simplify: (cos θ
5 Simplify cos(
6a Write1 i inmod-argform.
b Hencefind (1 i)13 inCartesianform.
7 Find 1 2 √3 2 i 8 inCartesianform.
8 Supposethat z = 1 √2 (1 + i).
Usethemodulus-argumentformof z toshowthat1 + z + z2 + z3 + z4 = √2 + 1 i
9a UsedeMoivre’stheoremtoevaluate(√3 + i)12 + (√3 i)12 b If n isapositiveinteger: i provethat(√3 + i)n + (√3 i)n isreal, ii determinethevaluesof n forwhich(√3 + i)n + (√3 i)n isrational.
10 If z = cos θ + i sin θ,showthat1 + z4 = 2cos2θ (cos2θ + i sin2θ).
11a UsedeMoivre’stheoremtofindcos6θ andsin6θ intermsofpowersofcos θ andsin θ.
b Henceshowthattan6θ = 2t 3 10t2 + 3t4 1 15t2 + 15t4 t6 ,where t = tan θ
12a Expand z + 1 z 4 and z 1 z 4
b Byletting z = cos θ + i sin θ,provethatcos4 θ + sin4 θ = 1 4 (cos4θ + 3).
13a Showthat(z z 1)7 = (z7 z 7) 7(z5 z 5) + 21(z3 z 3) 35(z z 1).
b If z = cos θ + i sin θ,showthat z z 1 = 2i sin θ andthat(zn z n) = 2i sin nθ.
c Henceprovethatsin7 θ = 1 64 (35sin θ 21sin3θ + 7sin5θ sin7θ)
d Find ∫ 35sin θ 64sin7 θ dθ.
14a UsedeMoivre’stheoremtoprovethatcos5θ = 16cos5 θ 20cos3 θ + 5cos θ.
b Hencesolvetheequation16x4 20x2 + 5 = 0,givingtherootsintrigonometricform.
c Showthatcos π 10 cos 3π 10 = √5 4 .
d If u = 2x2 1,showthat4u2 2u 1 = 0.
e Deducetheexactvalueofcos π 5 .
15 Let z = cos θ + i sin θ andsupposethat n isapositiveinteger.
a Showthat zn + z n = 2cos nθ
b Provethat2cos A sin B = sin (A + B) sin (A B)
c Henceshowthat z2n + z2n
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2
sin θ = sin (2n + 1) θ.
d Usethepreviouspartandtheresultcos3A = 4cos3 A 3cos A toprovetheidentity
16 Supposethat ω isacomplexcuberootof 1.
a Showthattheothercomplexrootis ω2 .
b Evaluate(6ω + 1)(6ω2 1).
17 Solvetheequation z3 8i = 0,expressingtherootsinCartesianform.
18 Find,inpolarform: thethreecuberootsof2 + 2i, a thesixsixthrootsof i. b
19 Supposethat z = 4√3
a Simplify z,writingyouranswerinpolarform.
b Showthat z 8 + i z 8 2 + z 8 3 = 2i.
c Findthethreecuberootsof z inpolarform.
20a Solvetheequation z5 + 1 = 0,expressingtherootsinmod-argform.
b Henceshowthat z5 + 1 = (z + 1) z2 2cos π 5 z + 1 z2 + 2cos 2 5 z + 1
c Deducethat2cos 2 5 2cos π 5 + 1 = 0,andhencefindtheexactvaluesofcos π 5 andcos 2 5 .
21a Findthefifthrootsofunityinpolarform.
b Let α bethecomplexfifthrootofunitywiththesmallestpositiveargument,andsupposethat u = α + α4 and v = α2 + α3
c Showthat u and v aretherootsofthequadraticequation x2 + x 1 = 0.
d Showthat u v = √5.
e Usethevaluesof u + v and u v todeducethatcos 2 5 = 1 4 √5 1 .
22a Findthesevenseventhrootsof 1inmod-argform.
b Henceshowthat:
cos π 7 + cos 3 7 +
5 7 = 1 2
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c Dividebothsidesoftheidentityin biii by z3 andlet z = cos θ + i sin θ toshowthat
23 Supposethat n isanintegergreaterthan2and ω isan nthrootofunity,where ω 1.
a Byexpandingtheleft-handside,showthat
b Usingtheidentity 1 z2 1 = z 1 z z 1 ,orotherwise,provethat
c Hence,if ω = cos 2
2π n ,findtherealpartof
ω 1 . d Deducethat1 + 2cos
e Byexpressingtheleft-handsideoftheresultinpart d intermsofcos π 5 andcos 2 5 ,findtheexactvalue ofcos π 5
24 Considertheequation z2n+1 = 1,where n isapositiveinteger.
a Findtherootsoftheequation,expressingtheminpolarform.
b Henceshowthat z
c Deducethat2n
25 Supposethat z1, z2, , zn arethe nthrootsof1.
Let Pk bethepointrepresenting zk for k = 1,2, ... , n,andlet A bethepointrepresentingthecomplex number ω,where |ω| = 1.
a Showthat APk 2 = (ω zk )(ω zk ).
b Henceshowthat n k=1 APk 2 = 2n
c Supposeinsteadthat A representstherealnumber x,where |x| < 1. Showthat AP1 × AP2 ×···× APn = 1 xn
4
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Vectors
Chapterintroduction
TheworkonvectorsintheMathematicsExtension1courseisnowextended.Thenewmaterialrequiresa thoroughknowledgeoftheExtension1topic,andstudentsmaybenefitfromareviewofthattheorybefore beginningthischapter.Agoodunderstandingofthreedimensionalgeometryisalsoneeded.Therelevanttheory isincludedintheappendixtothischapter.
Thefocusinthefirstpartofthechapterisonthevectorequationsoflinesandothersimplecurves,inbothtwo andthreedimensions.Section 4A introducesthevectorequationofaline.Intersectingandnon-intersecting linesarediscussedinSection 4B,withthespecialcaseofperpendicularlinesdealtwithinSection 4C.The vectorequationssatisfiedbycirclesandspheresareinvestigatedinSection 4D.Linesandcirclesbothuse parameters.SomeothercurvesthatuseparametersarepresentedinSection 4E.
Thefinalpartofthechapterismoretheoretical.Furthercharacteristicsofthedotproductareinvestigatedin Section 4F.Thechapterconcludeswithvectorapproachestogeometry.Twodimensionalproblemsaredealt withinSection 4G andthreedimensionalproblemsinSection 4H.
Avectormaybetypesetas r ∼ or r.Whilstsomeexercisequestionsandworkedexamplesusetheboldnotation, itisunsuitableforhandwrittenwork,soithasbeenmostlyavoided.Thetheoryispresentedusingthe‘tilde’ notation, r.
4A TheVectorEquationofaLine
Oneoftheadvantagesofvectorsisthatthederivationofcertainresultsandtheequationsofcertainobjectsare thesamebothintwodimensionsandinthreedimensions.Forexample,theformulafortheprojectionofone vectorontoanotherisalways
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projb a = b ∼ a ∼ b b b.
Thefocusofthissectionisonthevectorequationsoflines,astheytoohavethesameforminbothtwo dimensionsandthreedimensions.
LinesThroughtheOrigin
Let O betheoriginandlet B beanotherpointwithpositionvector b.Considerthe line OB.Let R beavariablepointin OB withpositionvector r.Itfollowsthat OR||OB andhence
r ∼ = λb ∼
Thusthepositionvectorofeverypointin OB isobtainedas λ varies,asshowninthediagramabove,andthisis thevectorequationoftheline OB
Thepropernamefortheequationisa parametricvectorequation ,sinceitinvolvesa parameter λ.Also,the vector b ∼ iscalleda directionvector sinceitspecifiesthedirectionoftheline.
Example1
Findthevectorequationofthelinethroughtheoriginandthepoint B(2,3).
Solution
Thepoint B haspositionvector b = 2i + 3j.Let r = xi + yj,thentheequationof OB is r = λb, thatis
1Linesthroughtheorigin
Thelinethroughtheoriginandanotherpoint B withpositionvector b hasvectorequation r = λb.
Thescalar λ iscalledaparameterand b iscalledthedirectionvector.
TheParametricEquations
Thecomponentsofthevectorequationinthelastexampleare
x = 2λ
y = 3λ
Thesearecalledthe parametricequations ofthestraightline.Theyformapairofsimultaneousequationswith parameter λ
TheCartesianequationofalineintwodimensionscanalwaysbefoundbyeliminatingtheparameter.Inthiscase thatequationis
y = 3 2 x
Thereisnorestrictiononthedomainoftheparameterintheworkedexample,andsothewholelineisobtained. Incontrast,setting0 ≤ λ ≤ 4wouldyield
y = 3 2 x with0 ≤ x ≤ 8(since x = 2λ.)
Thatis,theresultisthelinesegmentwithendsattheoriginand(8,12).
Therearemanytwodimensionalcurveswhichcanbedefinedthroughparametricequations.Someofthesewill beencounteredinalatersection.Ineachcase,theparameteriseliminatedandanyrestrictionsnotedtofindthe Cartesianequation.
Example2
DeterminetheCartesianequationofthelinesegment
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Solution
Theparametricequationsare
x = 2µ with 2 ≤ x ≤ 2 and y = 4µ = 2(2µ).
Hencetheresultisthelinesegment
y = 2x, 2 ≤ x ≤ 2.
Note: Thatinthisexampletheparameterwaslabelled µ.Anysymbolcanbeused.TheGreekletter λ isoften usedforproblemsinvolvinglines,however,becauseitisequivalenttotheletter ℓ,thefirstletterofthe word line.
TheDirectionVectorandtheGradient
Intwodimensions,thedirectionvector b ∼ ofthevectorequation r ∼ = λb ∼ andthegradient m oftheCartesian equation y = mx arecloselyrelated.Thevector b = b1 i + b2 j specifieshow x and y changeas λ varies.In particular,thecoordinatesof R for λ = 0and1are:
λ = 0, R = (0,0)and λ = 1, R = (b1, b2).
Fromthisitisclearthat b2 istheriseand b1 istherun.Hencethegradientis
m = b2 b1 provided b1 0.
Therestriction b1 0meansthatthelinecannotbevertical.
Thusitissometimesconvenienttoput b1 = 1sothat b2 = m andthegradientappearsexplicitlyinthevector equation.Thisgives
Example3
Writetheequationoftheline y = 2x invectorform.
Solution
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2Thedirectionvectorandthegradient
Thetwoequationsofanon-verticallinethroughtheoriginintwodimensionsare r ∼ = λb ∼ and y = mx
Fromthis
FindingaVectorEquation
WhilsttheformulainBox 2 isuseful,itissometimesbettertofindavectorequationbycarefullychoosinga parameter.
Example4
Findthevectorequationoftheline y = 2 3 x
Solution Let
or
Noticethatbyputting x = 3λ therearenofractionsinthefinalresult.Anothercarefulchoiceofparameteris neededwhencirclesarestudiedlaterinthischapter.
TheLineThroughaGivenPoint
Let O betheoriginandlet A and B betwootherpointswithpositionvectors a ∼ and b ∼ respectively.Considertheline ℓ through A parallelwith OB.Let R beavariablepointin ℓ withpositionvector r.Itfollowsthat AR||OB andhence r a = λb. O B R r a b A
Thusthepositionvectorofeverypointin ℓ isobtainedas λ varies,asshowninthediagramabove,andthisisthe parametricvectorequationoftheline.
Itisimportanttoobservethattheequation r ∼ a ∼ = λb ∼ isconsistentwiththewayafunctionchangeswhenits graphisshifted.Theline OB withequation r ∼ = λb ∼ hasbeenshiftedsothatitpassesthroughthepoint A with positionvector a.Consequentlythevariablevector r hasbeenreplacedwith r a.Despitethissignificantresult, itismorecommontowritetheequationas
Example5
Writedownthevectorequationofthelinethrough A withdirection −−→ OB,where A = ( 2, 1,3)and B = (1,0,1).Thendeterminewhetherornot C = (0, 1,4)isonthisline.
Solution
Thevectorequationissimply
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Thepoint C willbeonthelineifthereisasolutionto
Fromthecomponents,both λ = 2and λ = 1.Clearlytheseareinconsistentandhence C doesnotlieonthe line.
Note: Thatitisusuallyeasiertoobserveaninconsistencylikethiswhencolumnvectorsareused,ratherthan rowvectors.
3Thelinethroughagivenpoint
Thevectorequationofthelinewithdirection b whichpassesthroughapointwithpositionvector a is
Example6
Showthat
isequivalenttothegradient–interceptformulaintwodimensionalcoordinate geometry.
Solution
Theparametricequationsare
x = λ
y = λm + c thusbysubstitutionitisclearthat y = mx + c,whichisthegradient–interceptformulaincoordinategeometry.
VerticalLinesinTwoDimensions
Theadvantageofusingavectorequationlike r ∼ = a ∼ + λb ∼ ratherthanaCartesianequationlike y = mx + c iswhen dealingwithverticallinesintwodimensions.Essentially,thevectorequationdoesnotchange,onlyitsdirection vector.Thustheequationofaverticallineis
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or,usingaunitdirectionvector, b ∼ = j.Thusthevectorequationofaverticallinethrough A withpositionvector a ∼ canbewrittendownwithoutanyworking.
Example7
Writedownthevectorequationoftheverticallineintwodimensionsthatpassesthrough A( 2,1).
Solution
Thelineis r = 2i + j + λj.
TheLineThroughTwoGivenPoints
Inthetwopointmethodofcoordinategeometry,thefirststepistousethegivenpoints todeterminethegradientoftheline.Likewise,forthevectorequation,thefirststepis tofindthedirectionvectoroftheline.Letthepoints A and B havepositionvectors a and b.Thenadirectionvectorofline ℓ isgivenby
AB = (b a).
Itisnowatrivialmattertowritedownthevectorequationof AB,whichis
r ∼ = a ∼ + λ(b ∼ a ∼ ).
Example8
Considerthepoints A( 1, 2,3)and B( 2,1,0).
a Evaluate −−→ AB
b Hencedeterminethevectorequationof AB
Solution
a Usingrowvectors,fromthepositionvectorsof A and B: −−→ AB = ( 2,1,0) ( 1, 2,3) = ( 1,3, 3).
b Thusthelineinrowvectorformis
(x, y, z) = ( 1, 2,3) + λ( 1,3, 3).
4Thelinethroughtwogivenpoints
Supposethat A haspositionvector a ∼ and B haspositionvector b ∼.Thevectorequationoftheline AB is r ∼ = a ∼ + λ(b ∼ a ∼ ).
TheEquationofaLineSegment
Thelinesegment AB issimplytheresultofrestrictingtheparameterinthevector equationofline AB.Theequationofthewholelineis,fromabove, r ∼ = a ∼ + λ(b ∼ a ∼ ).
Itshouldbeobviousthat r ∼ = a ∼ when λ = 0andthat
When λ takesvaluesbetween0and1,thevariouspointsalongthelinesegmentareobtained.Forexample,the diagramshowsthepoint R when λ = 1 3 .Thatis,point R is 1 3 ofthewayfrom A to B.Thus,everypointintheline segmentcanbereachedif0 ≤ λ ≤ 1.
Example9
Findthevectorequationofthelinesegment AB where A = ( 1,2)and B = (3,3).
Solution
Inthiscase
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5Theequationofalinesegment
Supposethat A haspositionvector a and B haspositionvector b.Thevectorequationofthelinesegment AB is
RatioDivision
Inthediscussionabove,itwasstatedthat λ = 1 3 inthediagram.Thevectorequationofalinecanbeusedto deriveaformulaforthecoordinatesofapointthatdividesagivenintervalintoanyspecifiedratio.Asusual, supposethatthepositionvectorsofthepoints A and B are a and b respectively.Furthersupposethat AB is dividedintotheratio k : ℓ bythefixedpoint P withpositionvector p ∼ .Thediagrambelowshowsthesituation.
Fromthegivenratioitfollowsthat
(b a).
Hence,byvectoraddition,
andso
Thatis p
Itmaybemoreconvenienttoput
+ℓ inthesecondlastlinetoget: p ∼ = (1 λ)a + λb
Whicheverformulaisused,oncethecomponentsof p ∼ havebeencalculated,thecoordinatesof P canimmediately bewrittendown.
Example10
a Findthecoordinatesof P thatdividesinterval AB intheratio1:2,where A = (1, 3,2)and B = ( 5,6, 1).
b Checktheresultbycalculating −−→ AB and −−→ AP
Solution
a Inthiscase k = 1and
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Hence P = ( 1,0,1).
b First,usingsubtractionofvectors
,giving
so AP = √14 = 1 3 AB (asexpected.)
6Ratiodivision
Lettheinterval AB bedividedintotheratio k : ℓ bypoint P.Letthecorrespondingpositionvectorsbe a ∼ , b and p ∼ .Thecoordinatesof P canbefoundfromthevectorequation p ∼ = (1 λ)a + λb where λ = k k+ℓ .
When P liesbetween A and B,asintheabovederivationoftheformula,itissaidthat P divides AB internally. When P liesoutsidetheinterval AB itissaidthat P divides AB externally.Theonlychangetotheformulainthis caseisthatoneof k or ℓ ismadenegative.Theproofofthisresultisleftasanexercise.
Exercise4A FOUNDATION
Note: Throughoutthisexercise λ isarealparameter.
1 Writedownavectorequationofthelinethat:
a passesthroughtheoriginandhasdirectionvector
b passesthroughthepoint( 3, 1)andisparalleltothevector5 i + 5 j
2 Writedownavectorequationofthelinethat:
a passesthroughthepoint(7,0, 5)andhasdirectionvector 4i 6j + 9k,
b passesthroughthepoint(3,4,5)andisparalleltothevector( 6, 7, 8).
3 Aline ℓ passesthroughthepoint( 1,3)andhasdirectionvector2 i j ∼ .
a Writedownavectorequationoftheline.
b Whatisthegradientoftheline?
c WhatistheCartesianequationoftheline?
4 AlinehasCartesianequation y = 2 3 x 4.
a Findthepositionvectorofthepointonthelinewhere x = 3.
b Usethegradientofthelinetofindadirectionvectoroftheline.
c Hencewritedownavectorequationoftheline.
5a ConsiderthelinewithCartesianequation3x 5y + 30 = 0.
i Findadirectionvectoroftheline.
ii Writedownthepositionvectorofapointontheline. (The y-interceptisagoodchoice.)
iii Hencewritedownavectorequationoftheline.
b Useasimilarapproachtofindavectorequationofeachoftheselines.
6 Considerthelinewithvectorequation
a Writedownapairofparametricequationsoftheline.
b HencefindtheCartesianequationoftheline.
7 FindtheCartesianequationofeachline.
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8 Findavectorequationoftheline AB given: A(4,3)and B(6,0) a A( 7,5)and B( 13, 8) b
9 Findavectorequationoftheline PQ given: P( 1,3,1)and Q(2,4,5) a P(7, 11,14)and Q(17,9, 16) b
(8, 13,11) a
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,showthatthevectorequation
isequivalenttothe
point–gradientformula
13 Thepoints A, B and C havepositionvectors2i 3j + k, 4i + 6k and14i 9j 9k respectively.
a Findavectorequationoftheline AB.
b Determinewhetherornotthepoint C liesontheline AB
14 Thepoints A, B and C havepositionvectors
Findthevalueof α forwhichthepoint C liesontheline AB
15 Thepoints A, B, C and D haverespectivepositionvectors(0,1,2),( 1, 1,0),(4,0,1)and(3,1,2)relative toanorigin O.Let M bethemidpointof AC
a Findthepositionvectorof M
b Findavectorequationoftheline BD.
c Showthat M liesontheline BD
d Findtheratio BM : MD.
16 Findthecoordinatesofthepoint P that:
a dividestheinterval AB intheratio2:3given A( 2,2)and B(4,7),
b dividestheinterval CD intheratio1:2given C(3,2, 4)and D(6, 1,2).
17 Findavectorequationofthelinesegment AB andstatetherestrictionontheparameter,given A(1, 2)and B(5,4).
18 Findavectorequationofthelinesegment PQ andstatetherestrictionontheparameter,given P( 1,1, 2) and Q(2,3, 1).
19 Thepoints A and B havepositionvectors a and b respectively.Describethepartoftheline AB thatis representedbythevectorequation r ∼ = a + λ(b a)if:
20 Alinehasvectorequation
a Showthat −−→ AP =
b Henceshowthat
AP
c Writedowntheminimumvalueof
CHALLENGE
.Thevariablepoint P isonthelineand A isthefixedpoint(3,7).
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AP| andstatethevalueof λ forwhichitoccurs.
d Statethecoordinatesof P forwhich | −−→ AP| isminimised.
e Confirmthat −−→ AP isperpendiculartothelinewhen | −−→ AP| isminimised. 21 Aline ℓ hasvectorequation
.Thevariablepoint P ison ℓ and A isthepoint(1,1,1).
a Find | −−→ AP|2 intermsof λ.
b Hencefindtheminimumdistancefrom A to ℓ
c Showthattheminimumdistancefoundin b istheperpendiculardistance.
d Repeat b,thistimeusingthescalarproductof −−→ AP andthedirectionvectorof ℓ
4B Parallel,IntersectingandSkewLines
Twolinesintwodimensionsmusteitherbeparallelortheyintersect.Inthree dimensionsitispossiblethattwolinesarenotparallelanddonotmeet.Suchpairsof linesaredescribedasbeing skew.Forexample,theoppositeedges AB and CD ofthe tetrahedronontherightareskewlines.Inthefirstpartofthissection,however,the focuswillbeonparallellinesandintersectinglinesintwoandthreedimensions.The sectionconcludeswithaninvestigationofskewlinesinthreedimensions.
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ALineParalleltoAnotherLine
Twolinesareparalleliftheyhavethesamedirection.Thusasimplewaytofindtheequationofalineparallelto anotherlineistousethesamedirectionvector.
Example11
Findthelinethrough(2,3, 1)thatisparallelto r = (1,2,0) + λ(1, 1,2).
Solution
Usingthepositionvectorofthegivenpointandthedirectionvectorofthegivenline,theequationis r = (2,3, 1) + λ(1, 1,2).
7Alineparalleltoanotherline
Thelinethrough P withpositionvector p thatisparallelwiththeline
ParallelVectors
Twovectors u ∼ and v ∼ areparallelifthereisaconstant k forwhich u ∼ = kv ∼
Thevalueof k iseasilyfoundbytakingtheratiosofmatchingcomponents.
Example12
Showthat u ∼ = 6 i ∼ 8 j ∼ isparallelto v = 9 i ∼ + 12 j ∼
Solution
Itisclearthat u ∼ = 2 3 v ∼ sotheyareparallel.
Thereisanothertestforparallellineswhichtakesmoresteps,butwhichisimportantinprogramming,soit isincludedhere.Recallthatifvectors u ∼ and v ∼ includetheangle θ then u ∼ v ∼ = |u ∼ ||v ∼ | cos θ.Forparallelvectors θ = 0◦ or180◦ so | cos θ | = 1.Hencetwodirectionvectorswillbeparallelif |u v ∼ | = |u||v ∼ |.
Example13
RepeatExample 12 usingthedotproducttest.
Solution
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Evaluatingthevariousdotproducts:
|u ∼ v| = |− 54 96| = 150
u u = 36 + 64 = 100 v v = 81 + 144 = 225 so |u||v| = 10 × 15 = 150 = |u ∼ v ∼ | andhencethevectorsareparallel.
8Parallelvectors
Vectors u ∼ and v areparallelifandonlyif
CollinearPoints
Ifthepoints A, B and C arecollinear,thenthevectors −−→ AB, −−→ AC and −−→ BC areparalleltothe directionvectoroftheline.Thustheyareparalleltoeachother.Thisprovidesatestfor collinearpoints.Thepoints A, B and C arecollinearifandonlyif
Example14
Determinewhetherornotthepoints A(2,1,4), B(1,3,2)and C(0,5,0)arecollinear.
Solution Fromthecoordinates −−→ AB = i + 2 j ∼ 2k and −−→ AC = 2 i + 4 j ∼ 4k = 2 −−→ AB, hencethepointsarecollinear.
9Collinearpoints
Thepoints A, B and C arecollinearifandonlyif
Notethatthetest −−→ AB ∥ −−→ AC canbereplacedwithanyconvenientpairingofthelettersintovectors,suchas −−→ BA ∥
ParallelLines
Linesthatareparallelhavedirectionvectorsthatareparallel.
Example15
Thevectorequationsoftwolinesaregivenbelow.Showthattheselinesareparallel.
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thedirectionvectorsareparallel,andhencethelinesareparallel.
Takecarewiththelogichere.Twolinescanonlybeparalleliftheirdirectionvectorsareparallel.Incontrast, twoequationswithparalleldirectionvectorsmayrepresenteitherparallellinesorthesameline.
Example16
Showthattheequationsbelowrepresentthesameline.
Thedirectionvectorsareequal,butputting
+ 2inthesecondequationgives
Thus,theequationsrepresentthesameline.
10Parallellines
Twostraightlineequationswithparalleldirectionvectorsmayrepresenteitherparallellinesorthesame line.
Non-parallelLines
TheconditionsinBox 8 forparallelvectorscanalsobeusedinthenegativetoshowthattwolinesarenot parallel.
Example17
Provethatthelinesbelowmustintersect.
Solution
Twolinesintwodimensionswillintersectiftheyarenotparallel.Takingthedotproductofthedirection vectors:
Clearlythisisnottheproductofthevectorlengthssothelinesarenotparallel. Hencethelinesintersect.(Infact,theyareperpendicular.Why?)
11Non-parallellines
Twostraightlineswillnotbeparalleliftheirdirectionvectorsarenotparallel.
IntersectingLines
Analternatewaytosolvethelastexampleistotrytofindthepointofintersection.Iftwolinesareknown tointersectthenthepointofintersectioncanbefoundbyequatingtheircomponents.Thisyieldsasetof simultaneousequations.
Example18
Findthepointofintersectionofthelines
Solution
Equatingtherighthandsides:
andrearrangingthenyields
Equatingthecomponents:
Thatis,thelinesintersectatthepoint(2,1).
Note: Thatthesamepointisobtainedwhen µ = 2issubstitutedinto
Whenthismethodisappliedtointersectinglinesinthreedimensions,therearethreeequationstosolve simultaneously.Thebestwaytomanagethisistosolvetwoofthemandthenusethethirdequationtocheckthe solution.
Example19
Findthepointofintersectionofthelines
Solution
Equatingtherighthandsides
Equatingthefirsttwocomponents:
µ = 1 and
µ = 5, whichhavesolution λ = 2and µ = 3.
Checkingthe k ∼ component, 2 × 2 + 3 × 1 = 1(whichisconsistent.)
Using r ∼ 1 thepointofintersectionis
Thatis,thelinesintersectat(3,0, 1).
Note: Thatcheckingthesolutionintheremainingcomponentisimportant.Donotskipthisstep,althoughit mayoftenbedonementally.
12Intersectinglines
Iftwolinesareknowntointersectthenthepointofintersectioncanbefoundbyequatingtheir components.Thisyieldsasetofsimultaneousequations.Allthreeequationsmustbecheckedinthree dimensions.
SkewLinesandInconsistentEquations
Skewlinesarenotparallelandsothefirststepintestingforskewlinesistodeterminewhetherthedirection vectorsareparallel.Also,skewlinesdonotintersect.Tryingtofindapointofintersectionbyequating componentsyieldsasetofinconsistentsimultaneousequations.
Example20
Determinewhetherornotthefollowingareskewlines.
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Solution
Itshouldbeclearthatthereisnosolutionfor k totheequation
hencethelinesarenotparallel.Equatingthefirsttwocomponents:
and
Addingthesegives 1 + 3λ = 4 bywhich λ = 1,andso µ = 1.Substitutingtheseintothe z-componentsgives 2 1 1 2.
Theequationsareinconsistentandthelinesarenotparallel,hencetheseareskewlines.
Noticeinthisexamplethattheinconsistencywasnotfounduntilthe z-componentwaschecked.Thisiswhy itissoimportanttocheckallthreecomponentswhendeterminingwhetherornottwolinesintersectinthree dimensions.
13Skewlinesandinconsistentequations
Toshowthattwolinesareskew,firstshowthattheyarenotparallel.Thenshowthatequatingthe componentsyieldsasetofinconsistentsimultaneousequations.
Exercise4B FOUNDATION
Note: Throughoutthisexercise λ and µ arerealparameters.
1 Writedownavectorequationofthelinethat:
a passesthroughthepoint(3, 2, 4)andisparalleltotheline
r = 2i 2j + k +
(5i 3j k),
b passesthroughthepoint(2,1, 1)andisparalleltothe z-axis.
2 Showthatthetwolinesineachpartareparallel.
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3 Considerthelineswithvectorequations r = (0, 5) + λ(2,1)and r = µ(3, 1).
a Findtheirpointofintersection.Beginbysettingupapairofsimultaneousequationsin λ and µ
b Findtheacuteanglebetweenthelinesbyusingtheirdirectionvectors.
4 Findthepointofintersectionofthelines
5 Findthepointofintersectionofeachpairoflines.
6 Twoships A and B startsailingfromdifferentportsat9am.Theirrespectivepositionvectors,with componentsinkilometresafter t hours,are:
Showthattheshipswillcollide,andstatewhenandwherethiswillhappen.
7 Boat P leavesaharbourat11am.After t hoursitspositionis
.Anotherboat Q leavesabay at11amandafter t hoursitspositionvectoris
.Thecomponentunitsforbothposition vectorsarekilometres.
Showthattheboatsdonotcollide,butthattheirpathscross.
8 Showthatthethreepointsineachpartarecollinear. ( 4,15),(2,3),(13, 19) a ( 2, 3,7),( 5,12, 2),(1, 18,16) b
9 Showthatthetwovectorequations
[Hint:Onepossibleapproachistoshowthatthelinesareparallelandpassthroughacommonpoint.]
10 Showthatthelines
)areskew.(Youmustshow thattheyarenotparallelanddonotintersect.)
11 Ineachcaseeithershowthatthegivenlinesareskeworfindtheirpointofintersection.
12 Findthevalueof
)intersect.
13 Theline ℓ1 passesthroughthepoints(2,0,1)and( 1,3,4),whiletheline ℓ2 passesthrough( 1,3,0)and (4, 2,5).
Findthepointofintersectionof ℓ1 and ℓ2. a
b
Find,tothenearesttenthofadegree,theacuteanglebetween
14 Showthatthelines
R,areconcurrent.
15 Showthatthelines
concurrent.
16 Considerthelineswithequations
a Findtheirpointofintersection.
b Findtheacuteanglebetweenthelines.[H
17 Thelines
haverespectiveequations
a Showthatthelinesformarhombusbyshowingthatoppositesidesareparallelandapairofadjacent sidesareequal.
b Findvectorequationsoftheanglebisectorsoftherhombus.
[Hint:Thediagonalsofarhombusbisecttheanglesoftherhombus.]
Showthatthelines
a Findtheirpointofintersection P
b Findvectorequationsofthebisectorsoftheanglesformedbythelinesat P
20 Twoplanes A and B arebeingobservedfromacontroltoweratanorigin O Theplaneshavebeenflyingfor t1 minutesand t2 minutesrespectively.
Theirrespectivepositionvectors(inkm)withrespecttotimearegivenby
a Showthattheirflightpathsintersect.
b Ifplane B tookoff T minutesafterplane A,showthatthedisplacementvector d ∼ (t)fromplane B to plane A isgivenby d
,where t minutesisthetimeelapsedsinceplane B tookoff
c Forsafetyreasons,thetwoplanesmustmaintainaminimumseparationof10km.Determinethe minimumpossiblevalueof T ,correcttothreesignificantfigures,sothattheplanesmaintainthe minimumseparation.
4C
PerpendicularLines
Perpendicularlinesintersectatrightangles.ThustheworkinSection 4B onintersectinglineswillbeneeded here.Thegeometryofthesituationissimplerintwodimensionsandthatwillbeconsideredfirst.Perpendicular linesinthreedimensionsareconsideredinthesecondpartofthissection.
PerpendicularLinesinTwoDimensions
Linesthatareperpendicularintersectatrightangles.Intwodimensions,linesthatareatrightanglesmust intersectsoitissufficienttoshowthattheirdirectionvectorsareperpendicular.Thus,ifthedirectionvectors are u and v ∼ thenshowthat u v ∼ = 0.
Example21
Showwhetherornotthelinesbelowareperpendicular.
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Solution
Thedotproductofthedirectionvectorsis
Hencethelinesareperpendicular.
14Perpendicularlinesintwodimensions
Thelines r
and
intwodimensionsareperpendicularifandonlyif u ∼ v ∼ = 0.
PerpendicularVectorsinTwoDimensions
Foragivenvectorintwodimensionsitiseasytofindaperpendicularvector.
Suppose a ∼ = xi ∼ + yj isthepositionvectorofthepoint A intheArganddiagram. Thus A representsthecomplexnumber α = x + iy.Theresultofrotating A by90◦ abouttheoriginis β = iα,and β = y + ix.Letthisrepresent B withpositionvector b = yi + xj ∼ .Itfollowsthat a and b areperpendicular,ascanbecheckedbythedot product. Re Im Ax,y() By,x()
Continuingthisprocess,thecomplexnumbers
, i2α and i3α correspondtothevectors a, b, a and b.That is,thevectors
formasequencewhereeachsubsequentvectoristheresultofarotationby90◦.Further,everyvectorinthis sequencehasthesamemagnitude.Thisfeatureprovesusefulincertainproblems.
Example22
Acertainlineintwodimensionshasgradient 3 5 .Writedownadirectionvectorofalineperpendiculartothis.
Solution
Thegivenlinehasdirectionvector5 i ∼ 3 j.Henceadirectionvectorofaperpendicularlineis3 i ∼ + 5 j
15Perpendicularvectorsintwodimensions
Inthesequenceofvectors
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and
eachsubsequentvectoristheresultofarotationby90◦.Further,everyvectorinthissequencehasthe samemagnitude.
Asimplewaytorememberthisresultistoswapthecomponentsandreversethesignofthenewfirstcomponent.
ALinePerpendiculartoAnotherinTwoDimensions
Itisnowsimpletodeterminethevectorequationofalinethroughagivenpointthatisperpendiculartoagiven lineintwodimensions.UsethepositionvectorofthepointanduseBox 15 tofindtherequireddirectionvector.
Example23
Whatisthevectorequationofthelineperpendicularto2x 3y + 4 = 0whichpassesthroughthepoint( 5,6)?
Solution
Thegradientofthegivenlineis 2 3 soithasdirectionvector
Thustheperpendicularlinehasdirectionvector
Hencetherequiredlineis
16Alineperpendiculartoanotherintwodimensions
Thelineintwodimensionsthrough P withpositionvector p thatisperpendicularto
PerpendicularLinesinThreeDimensions
RecallfromSection 4B thatiftwolinesarenotparallelthentheymayeitherintersectorbeskew.Thus,inorder toshowthattwolinesinthreedimensionsareperpendicular,itisnecessarytoshowboththatthelinesintersect andthattheangleformedisarightangle.
Supposethatthelinesare:
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Firstcheckthatthedirectionsareperpendicularbyshowingthat
u · v = 0.
Thesecondstepistoshowthatthelinesintersectbysolving a + λu = b + µv.
Example24
Testwhetherornottheselinesareperpendicular. r = 2 i j k + λ i + 2k r ∼ = i ∼ + 2
Solution
Thedotproductofthedirectionvectorsis
1 × 2 + 0 × 1 + 2 × ( 1) = 0 sothedirectionvectorsareperpendicular.
Nowtrytosolve 2
Equatingthe j ∼ componentsgives
1 = 2 + µ
so µ = 3.Thenfromthe i components
2 + λ = 1 6
so λ = 9.Checkingthe k components,
1 18 = 1 + 3
whichisclearlyinconsistent,sothereisnosolution. Thusthelinesdonotintersectandhencetheyarenotperpendicular.
17Perpendicularlinesinthreedimensions
Thelines r ∼ = a ∼ + λu ∼ and r ∼ = b ∼ + µv areperpendicularprovidedthatboth
• u v ∼ = 0,
• thereisasolutionto a ∼ + λu ∼ = b ∼ + µv
ALinePerpendiculartoAnotherinThreeDimensions
Theproblemoffindingthelineperpendiculartoagivenline ℓ andwhichpassesthroughagivenpoint P canbe solvedviaaprojection.
Letline ℓ havevectorequation r ∼ = a + λu where A isapointon ℓ withpositionvector a Letpoint P havepositionvector p.Thediagramontherightshowswhatisseeninthe planedefinedbyline ℓ andpoint P.Thepoint Q isthefootoftheperpendicularfrom P to ℓ. A Q uP
Thus −−→ AQ = proju ∼ −−→ AP = (p a ∼ ) u ∼ u
Henceadirectionvectoroftherequiredlineis
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Finally,theequationofthelineperpendicularto ℓ andthrough P is
Thiscomplicatedformulaisnotpartofthecoursesothereisnoneedtomemoriseit.Themethodisstraightforward,however,andiseasytoapply.Summarising:
• drawadiagramofthetwodimensionalsituation
• project −−→ AP ontothedirectionvectorofthelinetoget −−→ AQ
• calculate −−→ QP = −−→ AP −−→ AQ,thedirectionvectoroftheperpendicularline
• writedowntheequationoftheperpendicularline,viz. r ∼ = p + λ −−→ QP
Example25
Findtheequationofthelinethrough P(1,2,3)whichisperpendiculartotheline
and u = 2 i j ∼ .
Solution
Let A havepositionvector a andlet Q bethepointonthelinenearest P.Thesituationisshowninthediagram.
AP = 2 i ∼ + 2 j + k ∼ so −−→ AP u = 4 2 = 2 and u ∼ u ∼ = 4 + 1 = 5. Thus −−→ AQ = proju −−→ AP = 2 5 u so −−→ QP = 2 i + 2 j ∼ + k 2 5 (2 i j ∼ ) = 6 5 i + 12 5 j + k andhencetheperpendicularlinehasequation r = ( i + 2 j + 3k) + λ( 6 5 i + 12 5 j + k).
Notethatthelinecanbewrittenwithoutfractionsbyputting λ = 5µ toget
r = ( i + 2 j + 3k) + µ(6 i + 12 j + 5k).
Further,thepointofintersection Q canbefound.Ithaspositionvector q = a + 2 5 u
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Note: ThatthismethodalsoworksintwodimensionsbutitisfastertousetheresultinBox 16.
18Alineperpendiculartoanotherinthreedimensions
Thestepstofindthelinethroughagivenpoint P thatisperpendiculartoagivenlinethrough A are:
• drawadiagramofthetwodimensionalsituation
• project −−→ AP ontothedirectionvectorofthegivenlinetoget −−→ AQ
• calculate −−→ QP = −−→ AP −−→ AQ,thedirectionvectoroftheperpendicularline
• writedowntheequationoftheperpendicularline,viz. r
TheDistancefromaPointtoaLine
Thediagramusedtofindthelinethroughapoint P whichisperpendiculartoagivenline ℓ canalsobeused toquicklyandeasilyfindthedistancefromthepointtotheline.ThetrickistousePythagoras’theoreminthe associatedrightangletriangle.Itishelpfultopractisedrawingthediagram,soherearethedetailsagain.
Letline ℓ havevectorequation r = a + λu where A isapointon ℓ withpositionvector a
Letpoint P havepositionvector p ∼ .Thediagramontherightshowswhatisseeninthe planedefinedbyline ℓ andpoint P.Thepoint Q isthefootoftheperpendicularfrom P to ℓ
Thus
AQ = proju
Pythagoras’theoremcannowbeappliedto △APQ tofind
Thiscanbewrittenusingthedotproductas
Thiscomplicatedformulaisnotpartofthecoursesothereisnoneedtomemoriseit.Themethodisstraightforward,however,andiseasytoapply.Summarising:
• drawadiagramofthesituation
• project −−→ AP ontothedirectionvectorofthelinetoget −−→ AQ
• usePythagoras’theoremin △APQ tofindthelengthof QP
Example26
Findthedistancefrom P(1,2,3)totheline r ∼ = a ∼ + λu ∼ ,where a ∼ = i ∼ + 2k ∼ and u ∼ = 2 i ∼ j
Solution
Let A havepositionvector a ∼ andlet Q bethepointonthelinenearest P.Thesituationisshowninthediagram.
Asbefore,
and
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Thatis QP = 41 5 units(or 1 5 √205withrationaldenominator.)
Note: Thatthismethodworksinbothtwoandthreedimensions.Alsonotethatitwasnotnecessarytofind
19Thedistancefromapointtoaline
Thestepstofindtheperpendiculardistancefromapoint P toalinethrough A are:
• drawadiagramofthesituation
• project −−→ AP ontothedirectionvectorofthelinetoget −−→ AQ
• usePythagoras’theoremin △APQ tofindthelengthof QP.
Arelatedproblemistofindthedistancefromagivenpoint P toalinethatpassesthroughtwogivenpoints A and B.Insuchcases,use −−→ AB asthedirectionvectorofthelineandproceedasbefore.
Example27
Findtheperpendiculardistancefrom P = (2,1,0)tothelinethrough A = ( 1,0,2)and B = (1,1,3).
Solution
Thesituationisdrawnontheright.Thelinehasdirectionvector
√354.)
Note: Throughoutthisexercise λ and µ arerealparameters.
1 Theline ℓ hasvectorequation r =
a Writedownadirectionvectorofalineperpendicularto ℓ
b Writedownavectorequationofthelinethrough(5,3)thatisperpendicularto ℓ.
2 Writedownavectorequationofthelinethrough( 2,3)thatisperpendiculartotheline r = i 2j + λ(4i + 5j).
3a Considertheline ℓ withCartesianequation3x y 4 = 0.
i Writedownadirectionvectorof ℓ
ii Writedownavectorequationofthelinethrough(6, 4)perpendicularto ℓ
b Likewisefindthelinethrough( 2, 2)perpendicularto x + 5y + 3 = 0invectorform.
4 Showthatthelines
5 Determinethevalueof k forwhichthelines
(k,4)are: parallel, a perpendicular. b
6 Findthevalueof k forwhichthelines
7 Thepoints P and Q havepositionvectors 3 i +
and
areperpendicular.
j and i + 8 j respectively.Findavectorequationoftheline through Q thatisperpendiculartotheline PQ.
8a Determinewhetherthefollowingpairsoflinesintwodimensionsareperpendicular.
b Determinewhetherthefollowingpairsoflinesinthreedimensionsareperpendicular.
9 Supposethat A isthepoint(1,2,0)ontheline ℓ withdirectionvector
Let Q bethepointon ℓ suchthat PQ ⊥ ℓ.
a Find −−→ AP.
b Find −−→ AQ byfindingtheprojectionof −−→ AP ontothedirectionvectorof ℓ
c Find −−→ QP.
DEVELOPMENT
andlet P bethepoint(0,0,2).
d Hencewritedownavectorequationofthelinethrough P perpendicularto ℓ
e Findthecoordinatesof Q.
10 Usetheapproachinthepreviousquestiontofindavectorequationofthelinethroughthepoint(5,2, 3) thatisperpendiculartotheline r = 2 i k + λ( i + 2 j ∼ + k).
11
Supposethat A isthepoint(1,5)ontheline ℓ withdirectionvector
Let Q bethepointon ℓ suchthat PQ ⊥ ℓ
a Find −−→ AP
b Find −−→ AQ byfindingtheprojectionof −−→ AP ontothedirectionvectorof ℓ.
andlet P bethepoint(2, 2).
c HenceusePythagoras’theoremtofindtheperpendiculardistancefrom P to ℓ
12 Ineachpartusetheapproachinthepreviousquestiontofindtheperpendiculardistancefromthegiven pointtothegivenline.
(6, 3),
( 2, 2,9),
13 Findtheperpendiculardistancefromthepoint P(1, 2,0)totheline AB where A and B arethepoints (2,0, 1)and(0,0,2)respectively.
14 Findavectorequationofthelinethroughthepoint(1, 1,1)thatisperpendiculartotheline r = 8 i ∼ 2 j 2k ∼ +
(2 i ∼ j k ∼)andhenceshowitisparalleltothe xy-plane.
15 Findtheperpendiculardistancebetweentheparallellines
CHALLENGE
16 Considerthelineswithequations
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a Showthatthelinesareskew.
b Thepoint A lieson r2 withparameter µ = k,where k isaconstant.
i Write −−→ OA asasinglecolumnvector.
ii Find −−→ OB,where −−→ OB istheprojectionof −−→ OA ontotheline r ∼ 1
iii Hencefindthevector −−→ BA.
iv Whatistheminimumvalueof | −−→ BA|?Whatdoesthisvaluerepresentinrelationtothelines r ∼ 1 and r ∼ 2?
17 Sanjaywalksinastraightlineataconstantspeedfromthepoint P( 4,0)tothepoint Q(11,10).His positionvector r ∼ canbeexpressedintheform r ∼ = a ∼ + tb ∼ ,where t minutesisthetimeforwhichhehasbeen walking.Hearrivesat Q after5minutes.
a Find a ∼ and b ∼ .
b SupposethatArjunisatthepoint A(7,3)andheintendstotossaballtoSanjaywhenSanjayreachesthe pointonhiswalkthatisclosestto A.Findthevalueof t whenArjuntossestheballtoSanjay.
18 Usevectorstoshowthattheperpendiculardistance d fromthepoint(a, b)totheline y = mx + c is d = |ma + c b|
m2 + 1
4D VectorEquationsofCirclesandSpheres
Thevectorequationofacircleorsphereisbasedonthegeometricdefinition.Thatis,asetofpointsthatarea givendistancecalledtheradiusfromagivenpointcalledthecentre.Theresultisthatthevectorequationsfor bothobjectslooksimilar.Likestraightlines,thevectorequationsofcirclescanbewritteninparametricform. Thatwillbethefocusofthelatterpartofthissection.
CirclesinTwoDimensions
Letthevariablepoint R withpositionvector r ∼ lieonthecirclewithradius r andcentretheorigin.Itfollowsthat OR = r.Thusthevectorequationofthecircleis
|r| = r.
NoticethatthisequationisconsistentwiththeequationofacircleintheArganddiagram,viz. |z| = r.This ishardlysurprising,sincevectorshavealreadybeenusedinconjunctionwithcomplexnumbersinanearlier chapter.Thusthevectorequationofacircleintwodimensionscanbewrittendownwithoutworking.
Example28
Writedownthevectorequationofthecirclewithcentretheoriginandradius5.
Solution
If r isthepositionvectorofanypointonthecirclethen |r| = 5.
Whilstitistruethatacirclewithradius r andcentretheoriginhasvectorequation |r ∼ | = r,theconverseisnottrue inthesensethatthingsthatarenotcirclesalsosatisfy |r ∼ | = r.Forexample,thevectorequation
r ∼ = λ( i ∼ + j)for |λ| = 1
representsthetwoisolatedpoints(1,1)and( 1, 1),yetthisequationalsosatisfies |
= √2.Noticethatthese twopointslieonthecirclewithradius √2.Ingeneral,ifanobjectsatisfies |
r thenitiseitheracircleorsome partofthecircumference,asaboveorasinthefollowingexample.
Example29
Determinethepointonthecirclewithcentretheoriginandradius2whichisclosesttotheline 2x + 4y 15 = 0.
Solution
Fromthediagramontheright,thepoint P liesinthefirstquadrant,andliesinthe linethroughtheoriginperpendiculartothegivenline.
Adirectionvectorofthegivenlineis
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Thus,aperpendicularvectoris
CircleswithOtherCentres
Nowsupposethatthecircle |r ∼ | = r isshiftedsothatthecentreisat C withpositionvector c ∼ .Then,fromthe shiftingresults, |r ∼ c ∼ | = r
Onceagain,theequationisconsistentwith |z α| = r forcomplexnumbers,whichisacirclewithcentre α and radius r intheArganddiagram.
Example30
Theline r
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intersectsthecirclewithcentre
andradius3at P and Q.Themidpointof chord PQ is M.Findthecoordinatesof M
Solution
Thecirclehasequation
9aftersquaring.Substitutetheequationofthelinetoget
Expandingthebracketsandcollectingterms:
Now P and Q correspondtotherootsofthisequation.
Thus,themidpointcorrespondstotheaverageoftheroots,whichis
Hencethemidpointhaspositionvector
thatis, M
20Circlesintwodimensions
Let r bethepositionvectorofavariablepointonthecirclewithcentre c andradius r.Then
Spheres
Theformofthevectorequationofasphereappearsidenticaltothatofacircleintwodimensions.Thusthe spherewithcentretheoriginandradius r is |r ∼ | = r
Likewise,thespherewithcentre c ∼ is |r ∼ c ∼ | = r, where,ofcourse,eachvectorhasthreecomponentssincethesituationisnowthreedimensional.
Example31
FindtheCartesianequationofthespherewithcentre c = i j k whichpassesthrough a = 2 i + 1 j + 5k. Solution
Now a ∼ c ∼ = 3 i ∼ + 2 j + 6k ∼,sotheradiusisgivenby r 2 = a ∼ c ∼ 2 = 32 + 22 + 62 = 49.
Thustheequationofthesphereis |r ∼ c ∼ |2 = 49,soinCartesianformthisgives (x + 1)2 + (y + 1)2 + (z + 1)2 = 49
21Spheres
Inthreedimensions,let r bethepositionvectorofavariablepointonthespherewithcentre c ∼ and radius r.Then
r c| = r.
Recallthatapartofacircle,suchasanarc,satisfiesthevectorequationofacircle.Likewise,apartofasphere, suchasagreatcircle,satisfiesthevectorequationofasphere.
TheStandardParametricVectorEquationofaCircle
Thestandardparametricvectorequationofacircleintwodimensionswithcentrethe originandradius r isbasedonthetrigonometricdefinitions:
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Thusthevectorequationforthevariablepoint R is
Theparameteris θ and,unliketheequationofastraightline,itappearswithinthevectorandnotasascalar multiple.Thevalueof θ isoftenrestricted.Forexample,therestriction π < θ ≤ π correspondstoapoint movingoncearoundthecircleas θ increasesfrom π to π
Example32
Drawagraphofthepointswithpositionvector
Solution
Thisistheminorarcofthecirclewithcentretheoriginandradius2, from(2,0)to(1, √3)asshown.
Whenthecircleisshiftedtohavecentrewithpositionvector c ∼ theresultis
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Itissometimesconvenienttowritethisas
+ yj
where c = xc i + yc j ∼ .TheequivalentCartesianequationcanbefoundbyapplyingtheresultofBox 20.
Example33
DeterminetheCartesianequationofthecurveintwodimensionswithvectorequation
Solution
Firstrewritetheequationas
22Thestandardparametricvectorequationofacircle
Thestandardparametricvectorequationofacircleintwodimensionswithradius r andcentrewith positionvector c is
CirclesinThreeDimensions
Therearemanyotherwaystoparameteriseacircletoobtainitsvectorequation.Theonegivenbelowisbased onBox 22 andcanbeusedinbothtwoandthreedimensions.Itissimplyamatterofreplacing i and j ∼ withtwo perpendicularunitvectors.Thusthegeneralequationofacircleinbothtwoandthreedimensionsis
Whentheunitvectorsˆa and ˆ b arereplacedwithtwoofthebasisvectors i , j ∼ or k thenthecircleliesinaplane parallelwiththecorrespondingcoordinateplane.
Example34
Describethecirclewithequation
r = 3 i + 4(cos θ j ∼ + sin θ k).
Solution
Thebasisvectors j and k ∼ correspondtothe yz-plane.Thusitisthecirclewithcentre(3,0,0)andradius4, parallelwiththe yz-plane.
Thereareanynumberofspheresthatintersectwiththecircleintheaboveexample.Theobviousoneisthe spherewithcentre(3,0,0)andradius4.Thisisbecausetheintersectionofaspherewithaplanethroughits centreisacirclewiththesameradius.Iftheplanedoesnotpassthroughthecentre,thentheradiusofthe circleissmallerthanthesphere,asinthenextexample.
Example35
Showthatthecircleinthepreviousexampleliesonthesurfaceofaspherewithcentretheorigin,andstateits radius.
Solution
Itisrequiredtoshowthat |r| = r andtofind r.Now
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= 25 thus |r| = 5.
Thatis,thecircleliesonthespherewithcentretheoriginandradius5.
Note: Thatmostchoicesofˆa and ˆ b willproducecirclesthatareneitherhorizontalnorvertical.Theangles thatsuchcirclesmakewiththecoordinateaxesareoftenwrittenintermsoftheanglesusedto parameteriseasphere.
Enrichment—SpheresandParameters
Thestandardparametersforaspherewithradius r andcentreattheoriginusetwo angles, θ and φ,tolocateaspecificpoint R onthesurfaceofthesphere.Theangle θ is thefamiliarangleinthe xy-plane.Itiscalledthe azimuthangle.Theangle φ isformed bytheradius OR andthe z-axisattheorigin.Itiscalledthe polarangle or zenithangle Thesituationisshownontheright.
Asanexercise,showthatthesedefinitionsfortheanglesleadtothefollowingparametricvectorequationofthe spherewithradius r andcentreattheorigin:
Thisparameterisationofthesphereiscloselylinkedwiththelinesoflatitudeandlongitudeusedingeography andnavigation.Todemonstratethis,andasafurtherexercise,usesuitable3Dgraphingsoftwaretoshowasphere ofsuitableradius,displayingthelineswhere θ and φ aremultiplesof π 6
Exercise4D FOUNDATION
Note: Throughoutthisexercise λ and µ arerealparameters.
1 Acirclehascentre(6, 9)andradius2√7.Writedown: a aCartesianequationforthecircle, b avectorequationforthecircle, c apairofparametricequationsforthecircle.
2 Aspherehascentre( 2,7, 4)andradius9.Writedown: aCartesianequationforthesphere, a avectorequationforthesphere. b
3 WritedownaCartesianequationfor:
4 WritedownavectorequationandaCartesianequationforthecirclewithparametricequations
5 Findavectorequationfor:
6 Showthatthepoint P(8, 5,2)liesonthesurfaceofthesphere
7 Determinewhetherthepoint A( 4, 5,6)liesinsideoroutsidethesphere
8 Aspherehasvectorequation r (2 i + j ∼ k) · r (2
) = 20.Writedownthecentreandradiusof thesphere.
9 Supposethatthepoint A haspositionvector a =
a Writedownavectorequationforthecirclepassingthrough A withcentreattheorigin.
b Explain,withtheaidofadiagram,whythetangenttothecircleat A hasvectorequation r (3 i j) (3 i j) = 0.
c HencefindtheCartesianequationofthetangent.
10 Alinehasequation
a Writedownapairofparametricequationsrepresentingtheline.
b Hencefindthepointsofintersectionofthelineandthecircle.
a Explainwhythecircleofintersectionofthetwospheresisparalleltothe xy-plane.
b Determinethecentreandradiusofthecircleofintersection.
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Findthepointswhere ℓ intersects S
14 Considertheline
a Findthepointsofintersectionofthelineandthesphere.
b Showthatthelinepassesdirectlyabovethecentreofthesphere,andwritedownthecoordinatesofthe pointatwhichthishappens.
c Let C bethecentreofthesphere,letthepointsinpart a be P and Q andlet R bethepointinpart b.By consideringasuitabledotproduct,usesimplegeometrytoexplainwhy R isnotthemid-pointof PQ.
15 Supposethat,inthreedimensions,theline ℓ passesthroughthepoint A.Letthevectorequationof ℓ be r = a + λv ∼ where a 0 ∼ isthepositionvectorof A and v ∼ isavectorparallelwith ℓ.Inthisquestionyoumay assumethatthereisonlyoneplane P thatcontainsboth ℓ andtheorigin O.
a Useasuitableprojectiontohelpshowthat L,thedistancefrom O to ℓ,isgivenby
b Whatshapeisformedwhentheplane P intersectsthespheregivenbythevectorequation |r| = R ?
c Inthispart,assumethat R < L
i Drawatwodimensionaldiagramoftheshapeinpart b.Includeinyourdiagramtheorigin,the point A andtheline ℓ
ii Useyourdiagramtohelpfindtheminimumdistancefromthespheretotheline.
d Howmustyouranswertopart cii changeif R ≥ L ? CHALLENGE
16 Showthat
istheequationofacircle.
17 Considerthelinewithvectorequation r ∼ = b ∼ + λm ∼ ,where m ∼ isaunitvector,andthesphere |r ∼ | = a
a Ifthelineisatangenttothesphere,provethatthepointofcontactcorrespondstothepointontheline where λ = (b ∼ m ∼ ).
b Henceprovethatatangenttothesphereisperpendiculartotheradiusdrawntothepointofcontact.
4E
OtherParametricCurves
Sections 4A to 4D presentedthestandardparametricvectorequationsforlinesandcircles.Theconceptofusing parameterstodefineacurveisnowextendedtootherfamiliarcurves.Firstly,however,thecircleisrevisited.
OtherParametersforCircles
Forsomeproblemsitisconvenienttospecifyacircleinawaythatdiffersfromthestandardparametersgivenin
Box 22.Thevectorequation |r ∼ c ∼ | = r maystillbeusedtofinditscentreandradius.
Example36
Considerthevectorequation r ∼ = c + a cos
a ShowthatthisisacirclebyfindingitsCartesianequation.
b Whereonthecircleis θ = 0andinwhichdirectionisthecircletraversedas θ increases?
Solution
a Beginningbyshowingthat
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Thisistheequationofacirclewithcentre(0,2)andradius
b When θ = 0,
and when
sothecircleisbeingtraversedclockwise,asseeninthediagramontheright.
FindingtheCartesianEquation
RecallthattheCartesianequationofalinewasobtainedfromitsvectorequationbysolvingsimultaneous equations.Thatapproachisoftenusedtoidentifyacurvewhentheparametersareunfamiliar.Iftheparameters involvetrigonometrythenlookforthesumsanddifferencesofsquares,asusual.
Example37
Thevariablepoint R onacertaincurvehaspositionvector r ∼ = (sec t tan t) i + (sec t + tan t) j ∼ .Identifythis curvebyfindingitsCartesianequation.
Solution
Writingdownthecomponentsofthiscurve:
x = sec t tan t
y = sec t + tan t
Multiplyingthesegives
xy = sec2 t tan2 t = 1.
Thisisthefamiliarhyperbolainthe1stand3rdquadrants:
y = 1 x x y
ABézierCurve
Anamazingapplicationofparametersisa Béziercurve .Thisistrulymodernmathematics,beingdeveloped independentlybymathematiciansattwoFrenchcarmanufacturersin1959and1960.Originallydevelopedfor technicaldrawingsofautomotiveparts,theyarenowextensivelyusedinmanyotherapplications,particularly thoseinvolvingcomputergraphics.Thecurvesareremarkablysimpletoconstruct,beingbasedontheratio divisionofintervals.
ThefollowingexampleusesasimplequadraticBéziercurve.Giventhreepointsinvectorform,theBéziercurve joinsthefirstandlastwithasmoothcurvebycleverlyusingthemiddlepoint.Inthiscase,thecurveisaparabola.
Example38
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a Findvectorequationsofthepoint B0 whichdivides A0 A1 intheratio t :1 t,andthepoint B1 which divides A1 A2 inthesameratio.
b Hencefindthevectorequationofthepoint P whichdivides B0 B1 intheratio t :1 t,for0 ≤ t ≤ 1.
c Showthat P = A0 when t = 0,andthat P = A1 when t = 1.
d FindtheCartesianequationof P. Solution
a UsingtheformulaforratiodivisioninBox 6:
b ∼ 0 = (1 t)a ∼ 0 + ta ∼ 1 b1 = (1 t)a1 + ta2
b Likewise:
thatis p = (1
ThisisthequadraticBézierCurvefrom A0 to A2 using A1
c At t = 0,thesecondandthirdtermsareclearlyzero,leaving p ∼ = a0 At t = 1,thefirstandsecondtermsareclearlyzero,leaving p = a ∼ 2
d Firstwritethevectorequationincomponentform.
Nowwritingoutthecomponentsassimultaneousequations:
Expandingandsimplifyingtheseequationsgives:
x = 3t 1
2y = 12t 9t2 = 3t(4 3t)
so2y = (x + 1)(4 (x + 1))(from A0 to A2) thatis y = 1 2 (x + 1)(3 x)for 1 ≤ x ≤ 2.
Thecurveisshownasthesolidcurveontherightbetween A0 and A2.Thedashedpartofthecurveshows therestoftheparabola.Noticethattheintervals A0 A1 and A1 A2 aretangenttotheparabola.Thisisalways thecasewithaquadraticBéziercurve.
Asafinalnote,thediagramsinthisseriesoftextbooksweredrawnusingvariousBéziercurves,includinglinear, quadraticandcubiccurves.Further,theratiodivisionformulainBox 6 istheformulausedforlinearBézier curves.
ThreeDimensionalCurves
Curvesinthreedimensionscanbeextremelycomplicatedandmostarebeyondthe scopeofthecourse.Inafewinstancesthecurvecanbeunderstood,suchasthehelix
∼ = 2cos θ i + 2sin θ j ∼ + θ 2
k shownontheright.Thehelixislikeacircularstaircase,increasinginheightasitwinds aroundthe z-axis.
Insomecases,thecurveoraprojectionofitcanberepresentedasafamiliarfunctionintwodimensions.For example,whenthehelixaboveisprojectedintothe xy-planetheresultisacirclewithcentretheoriginand radius2.
Example39
Considerthevariablepoint R withpositionvector
a Thepoint R tracesoutacircleinthreedimensionsastheparameter t varies.Confirmthisbyshowingthatit satisfies |r ∼ c| = r
b Thevector r isprojectedintothe xy-planetoobtain
s ∼ = proj i r + proj j ∼ r, thepositionvectorofpoint S .Find s ∼ intermsoftheparameter t.
c DeterminetheCartesianequationofthecurvetracedoutby S as t varies.
d Explainhowthiscurvecanbeobtainedbytransforming x2 + y2 = 1.
Solution
a Let c ∼ = 4 i ∼ + 2 j + 3k ∼ .Then |r c|2 = (3cos ti + 5sin tj ∼ + 4cos tk) · (3cos ti + 5sin tj ∼ + 4cos tk) = 9cos2 t + 25sin2 t + 16cos2 t = 25 hence |r ∼ c ∼ | = 5. Thusthecirclehascentre(4,2,3)andradius5.
b Thecomponentsof s aretheprojectionsof r inthedirectionsof i and j.
Thus x = r · i = (4 + 3cos t) and y = r · j ∼ = (2 + 5sin t).
Hence s = (4 + 3cos t) i + (2 + 5sin t) j ∼ .
c Next,makethetrigonometricfunctionthesubjectofeachcomponent:
cos t = x 4 3 , sin t = y 2 5 .
Thus,squaringandaddingintheusualwaygives (x 4)2 9 + (y 2)2 25 = 1.
d Thisistheunitcircle x2 + y2 = 1stretchedhorizontallybyfactor3andverticallybyfactor5tobecomean ellipsewithcentreattheorigin.Finallythisellipseistranslatedtohavecentreat(4,2). Thereasontheprojectionisanellipseisthattheoriginalcircleispositionedatanangletothe xy-plane. Itmayhelptoimaginetheellipseastheshadowofthetiltedcirclewhenalightisshonefromdirectly overhead.
Enrichment—RotationsinTwoDimensions
Complexnumberscanbeappliedtotheparametricequationofacurvetorotateitandsocreateanewcurve.
Example40
ThehyperbolainExample 37 isrotatedabouttheoriginby π 4 .Usecomplexnumbersandthegivenparameters tofindtheCartesianequationofthisnewcurve,thensketchit.
Solution
Thepoint Z onthehyperbolaisrepresentedbythecomplexnumber a + ib = (sec t tan t) + (sec t + tan t) i.
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Multiplythisbycis π 4 torotateitabouttheorigin.
(a + ib)cis π 4 = (a + ib) × 1 √2 (1 + i) = 1 √2 (a b) + 1 √2 (a + b) i = √2tan t + √2sec ti
Continuedonthenextpage
Thustheparametricvectorequationforthenewcurveis r = √2( tan ti + sec tj ∼ )
Squarethensubtractthecomponentstoget
y2 x2 = 2(sec2 t tan2 t), thus y2 x2 = 2.
Thesketchisontheright,withinterceptsat y = ±√2.
Theoriginalasymptotes y = 0and x = 0havealsobeenrotatedby π 4 .Thusthe asymptotesforthishyperbolaarethelines y = x and y = x
Exercise4E
1 Alinehasvectorequation
a Writedownapairofparametricequationsoftheline.
b FindtheCartesianequationoftheline.
2 Acirclehasvectorequation r
a Writedownapairofparametricequationsofthecircle.
b FindtheCartesianequationofthecircle.
3 Afunction y = f (x)isdefinedbythevectorequation r ∼ (t) = (t 2) i
a Statethedomainofthefunction f (x).
b FindtheCartesianequationofthefunction.
c Sketchthegraphofthefunction.
4 Repeatthepreviousquestiongiven r(t) = 2ti +
5 Afunction y = g(x)hasvectorequation
a Statethedomainandrangeofthefunction g(x).
b FindtheCartesianequationofthefunction.
c Sketchthegraphofthefunction.
6 Sketchthecurvewithvectorequation:
7 Describegeometricallythegraphof
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8 Writedownapossibleparametricvectorequationfor:
a thelinethrough(1,2,3)thatisparalleltothe z-axis, b thelinethrough(4,5,6)thatisparalleltothe y-axis,
c thecirclewithcentre(3,0,0)andradius2unitsthatisparalleltothe yz-plane, d thecirclewithcentre(4,5,6)andradius7unitsthatisparalleltothe xz-plane.
9
Thesquarepyramid ABCDE isregular,meaningthattheedgeshaveequal length.Twoverticesareat A(2,2,0)and B( 2,2,0).
a UsePythagoras’theoremin △OAE tofindthecoordinatesoftheapexat E
b Hencefindthevectorequationofthedirectedlinesegmentfrom E to A.
Sketchthecurvespecifiedbytheparametricvectorequation:
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Describethegraphof
13 Thepositionvectorofapoint P onacurveisgivenby
a Showthateverypointonthiscurveliesonacirclewithcentreattheorigin O
b Whichpointonthiscircleisexcludedfromthecurve?
c Letthecurveintersectthe x-axisat A andlet ∠AOP = θ.Whatistherelationshipbetween t and θ?
14 FindtheCartesianequationcorrespondingtoeachofthefollowingvectorequations.
15 [TheInverseofaPoint]Theinverseofthepoint P withpositionvector p ∼ 0 isthepoint Q withposition vector q ∼ suchthat
and
Inthisquestion,let p ∼ = 2 j ∼ + λ i ∼
1.
a Describetheshapethat P tracesoutintheCartesianplaneas λ varies.
b Showthat q = 1 | p|2 p andhencefind q.
c Plot Q intheCartesianplanefor λ = 2,0,2andinthelimitas λ →∞
d Whatcircledothesepointslieon?Giveyouranswerintheform |q c| = R.
e Provethat Q liesonthiscircleforallvaluesof λ
16 Eachpoint R onahelixhaspositionvector:
2cos
+ 2sin
k for θ ≥ 0.
a Thevector r ∼ isprojectedintothe xy-planetoobtain s = proj i r ∼ + proj j r ∼ thepositionvectorof S .Write s intermsoftheparameter θ
b DeterminetheCartesianequationofthecurvetracedoutby S for0 ≤ θ ≤ π. 2
17 Acircularhelixspiralsaroundthe x-axisasshowninthediagramontheright. Everypointonthehelixis4unitsfromthe x-axis.Writedownaparametric equationforthishelix.
18 Agraphof h = cos2θ for0 ≤ θ ≤ 2π isdrawnonarectangularsheetofpaper.The endsarethengluedtogethersoastoformacylinderwithonlythecurveshowing. Itisfoundthatthecylinderhasradius2cmandheight2cm.Thediagramon therightshowsthesituation,alongwiththeusualCartesianaxes.The xy-plane correspondswith h = 0.Thepoint(2,0,0)correspondswith θ = 0and(0,2,0) correspondswith θ = π 2 onthegraphpaper.
a Suggestapossibleparametricvectorequation r(θ)forthiscurve.Itmayhelptoconsiderthepoint P on thecylinderthatcorrespondstoagivenvalueof θ
b Re-writetheequationofthecurveintermsofthenewparameter ℓ thatisthearclengtharoundthecircle x2 + y2 = 4measuredanticlockwisefrom x = 2.
c Amediacompanyhasalogobasedonthisidea,butusesthegraph h = 2sin3θ instead.Thisisthen projectedontothe xz-planetoget s ∼ (θ) = proj i r + proj k r.Usesuitablesoftwaretographwhatisseenin the xz-plane.
CHALLENGE
19 Sketchthecurvedefinedparametricallybythevectorequation:
20 Acurvespiralstwicefromtheorigintothepoint(2,0,4)aroundtheparaboloid z = x2 + y2,asshowninthediagram.Determineapossibleparametricvector equationforthecurve,andstatetherestrictionontheparameter.
4F TheAlgebraoftheDotProduct
Thissectionfocusesonthewayvectorscanbemanipulatedalgebraically.Thosemanipulationsleadtofour significantresults:theCauchy-Schwarzinequality,thetriangleinequality,theareaofatriangleanddirection cosines.
Thealgebraofthedotproduct
Inthisseriesofbooks,thedotproductisdefinedgeometricallyas a
where θ istheanglebetweenthetwovectors.Severalusefulalgebraicresultsfollowfromthisdefinition:
DistributiveLaw:
Studentsareexpectedtobeabletoprovethesealgebraicresultsbyusingthegeometricdefinition,possiblywith theaidofadiagram.Thesearemostlystraightforward.Theexceptionisthegeometricproofofthedistributive lawinthreedimensions.Itisverydifficult,soitisdealtwithlaterinthissection.
Example41
Usethegeometricdefinitionofthedotproductandasuitablediagramtoprovethat
Solution
Let
AOB =
in
and
.Itfollowsthat
.The situationisshowninthediagramontheright.Fromthecosinerule:
hence,fromthedefinitionofthedotproduct
Notethatthesameresultcanalsobeachievedquicklybyfirstassumingthevalidityofthemagnitude,viz. |a b|2 = (a b) (a b).Thiscanthenbeexpandedbyassumingthevalidityofthedistributiveandcommutative laws.Theadvantageoftheaboveproofisthatitdoesnotrelyonsuchassumptions.
Thecosinerulecanalsobeusedtoderivethecomponentformofthedotproduct,whichinthreedimensionsis
Studentsareexpectedtobeabletoverifythealgebraicresultsforthedotproductusingthecomponentformin eithertwoorthreedimensions.Thesearequiteeasyandsoareleftasanexercise.
TheDistributiveLawinThreeDimensions
Theresultispatentlytruewhen a ∼ = 0.So,itisassumedthat a ∼ 0 inwhatfollows. Thedifficultywithgeometricallyprovingthedistributivelawinthreedimensionsisinconstructingasuitable diagramtoworkwith.Thediagramusedhereisobtainedinthreesteps.
Considerthevectors a = −−→ OA and b = −−→ OB.Thevector a hasbeenmadevery longhereasitmakesthefinalpictureeasiertofollow,butitsactuallength isirrelevanttotheproof.Asdiscussedintheappendix,thereisonlyone planethatpassesthrough B andisperpendicularto OA,whichislabelled π1 Let π1 intersect OA at P
Letthepoint C havepositionvector(b + c)sothat −−→ BC = c.Thereisonly oneplanethrough C thatisperpendicularto OA,whichislabelled π2.Note thatbothplanesareparallelsincetheyarebothperpendicularto OA.Let π2 intersect OA at R.Thisisshownintheseconddiagramontheright.
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Thethirddiagramshowsthepoint Q inplane π2.Itispositionedsothat −−→ CQ = −−→ BP.Thisisalwayspossiblesincetheplanesareparallel.Theproofof thedistributivelawcannowbegin.
Inquadrilateral BCQP |BP| = |CQ| (equalvectors) BP ∥ CQ (equalvectors) hence BCQP isaparallelogram. Thus −−→ PQ = −−→
Now −−→ OP istheprojectionof b onto a sohaslength
and
PR istheprojectionof −−→ PQ = c onto a sohaslength
istheprojectionof
sohaslength
Clearlythesumofthefirsttwolengthsequalsthethirdso
Finally,multiplythroughbythecommondenominatortoget
+
TheCauchy-SchwarzInequality
Recallthattherangeofcos θ is 1 ≤ cos θ ≤ 1.
Multiplyingthroughoutby |a||b| fornon-zerovectors a and b yields
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andhence
whichiscalledthe Cauchy-Schwarzinequality. TheCauchy-Schwarzinequalityisclearlyalsotrueif a ∼ = 0 ∼ or b = 0 sincethenallthreepartsoftheinequalityarezero.Thus,theinequalityholdsforallvectors a and b.
Observeinthisinequalitythatthedotproductandmagnitudeofavectorareconsistentwiththeabsolutevalue functionforrealnumbers,forwhich
Thevectorinequalitycanalsobederivedfornon-zerovectorsbyusingthedotproductandthefactthatthe magnitudesmustbepositive.
Considerthequadraticfunction
Thecoefficientof t2 ispositiveso Q(t)hasaminimumvalueat
Example42
Verifytheinequality
Asimplecalculationgives
Thus 14 ≤−1 ≤ 14 thatis, −|a||b|≤ a · b ≤|a||b|,asexpected.
23TheCauchy-Schwarzinequality
Forallvectors a and b, −|a||b|≤ a b ≤|a||b|.
TheTriangleInequalityInVectorForm
Thetriangleinequalityhasbeenderivedforrealnumbersandforcomplexnumbers.Invectorform,this inequalityis
Hereisaproofoftherighthandinequality.Beginbysquaringthemiddleterm.
Andso,fromtheCauchy-Schwarzinequalitydevelopedabove,
Theproofofthelefthandinequalityisaquestionintheexercise.
Therighthandequalityoccurswhen
sincethen
Thereaderislefttoshowthatthelefthandequalityoccurswhen a = b.
Example43
Verifytheinequality
Solution
Itshouldbeclearinthiscasethat
forthevectors
TheAreaofaTriangle
Itiseasytodeterminetheareaofatriangleusingthedotproduct.Supposethatin △OAB,vertex A hasposition vector a ∼ andvertex B haspositionvector b ∼ .Let θ betheanglebetweenthesetwovectors.
Then a
Itfollowsthat
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Re-arrangingthisthengives
Thislastresultissometimescalledthe perpendiculardotproduct.Itcanbeusedtofindtheareaof △OAB.
Notethatthissimpleformulaisnotpartofthecourse.Ifitistobeusedtosolveaproblemthenitmustbe derivedaspartoftheanswer.
Example44
Theverticesof △PQR areat P( 1,0,2), Q(1,1,1)and R(0, 2,3).Usethedotproducttohelpfindtheareaof thistriangle.
Solution
Let −−→ PQ = a ∼ and
PR = b ∼ ,thentheareaof △PQR isthesameastheareaof △OAB formedbytheposition vectors −−→ OA = a ∼ and −−→ OB = b ∼ .Let θ betheangleformedattheorigin.
Then
so
and
with
Itfollowsthat
and
Hence |△PQR| = 1 2 √35squareunits.
Theperpendiculardotproducthasotheruses.Forexample,itcanbeadaptedinordertofindthedistancefrom agivenpointtoagivenline.Keenreadersmayliketoinvestigatethisandthenredotherelevantquestionsfrom Exercise 4C
DirectionCosines
Thedirectionofalineisgivenbyitsdirectionvector.Itissometimesconvenienttospecifythedirectionvector intermsoftheanglesitmakeswiththecoordinateaxes.Intwodimensions,thegradient m = tan θ isallthatis needed.Thesituationinthreedimensionsisnowdealtwith.
Onewaytodealwiththoseanglesistousethedotproduct.Supposethattheangle θ between a ∼ andtheunit vectorˆu ∼ isrequired,whereˆu ∼ isalignedwithoneofthecoordinateaxes.Then
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But 1 |a ∼ | a issimplytheunitvectorˆa so cos θ =
Thusreplacingtheunitvectorˆu witheachbasisvectorgivesthecosineoftheanglebetweenˆa andthe correspondingcoordinateaxis.Hencetheyarecalled directioncosines .Thatis,thedirectioncosinesinthree dimensionsare
Inotherwords,thedirectioncosinesofavectoraresimplythecomponentsofthecorrespondingunitvector.It isthereforenotsurprisingthatthedirectionofavectorinthreedimensionsiscompletelydeterminedifthethree directioncosinesareknown.
Example45
Evaluatethedirectioncosinesfor a
= −−→ OA,where A isthepoint(2,3,4).Hencegivetheangletoeachaxis correcttothenearestdegree.
Solution
Firstcalculate |a ∼ | |a ∼ |2 = 4 + 9 + 16 = 29.
Hencethedirectioncosinesof a are:
Thecorrespondingangleswiththeaxesare:
25Directioncosines
Thedirectioncosinesofavector a inthreedimensionsarethecomponentsoftheunitvectorˆa,viz:
1 Assumingthat i ∼ , j and k ∼ arethestandardunitvectorsinthepositive x, y and z directions,evaluate:
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2 Let a
and b
3
ConfirmthattheCauchy-Schwarzinequalityissatisfiedineachcasebelow.
6 Given
,usethealgebraicformulaforthedot producttoprovethedistributivelaw
7 Giventhreenon-zerovectors
then c ∼ (a ∼ + b ∼ ) = 0.
8a Usethedefinition a ∼ b
b Henceshowthatforvectors
, b
and c ∼,provethatif
9 Point P haspositionvector p andline ℓ hasvectorequation r = a + λv ∼.Whatcanyouconcludeabout P and ℓ if |(p ∼ a) · v| = |(p ∼ a)||v
10 Giventwonon-zerovectors a ∼ and b ∼,usescalarproductstoshowthat: a if a ∼ + b ∼ and a ∼ b
areperpendicularthen |
|, b if |a + b| = |a b| then a and b areperpendicular.
11 Thepoints A, B and C havepositionvectors a, b and c respectivelyrelativetotheorigin O.If −−→ AB ⊥ −−→ OC and −−→ BC ⊥ −−→ OA,provethat −−→ AC ⊥ −−→ OB.(Youareprovingthat O isthe orthocentre of △ABC.)
12 Theadjacentsidesofaparallelogramarerepresentedbythevectors a
Findtheexactareaoftheparallelogram.
and
13a Twovectors a and b aresuchthat a + b isperpendicularto a and |b| |a| = √2. Showthat2a + b isperpendicularto b. b Whatspecialquadrilateralisformedbythepointswithpositionvectors0 ∼ , a ∼ ,2a ∼ + b ∼ and a ∼ + b ∼ ?
14
Supposethat a ∼ , b ∼ and c ∼ arenon-zerovectorsinthreedimensionssuchthateachvectorisperpendicularto theothertwo.(Suchvectorsarecalled mutuallyorthogonal.)
Considertheequation
a Byconsidering a ∼ (
b Similarly,provethat λ2 = λ3 = 0.
,where
),provethat λ1 = 0.
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15 [TheDistributiveLawinTwoDimensions]Thedistinctpoints O, A, B and C lie inaplane.Apossibleconfigurationisshownontheright.Let
and −−→ BC = c ∼.Theprojectionof b ∼ onto a ∼ is −−→ OP and Q ischosensothat −−→ PQ =
Line OA intersects CQ at R
a Provethat ∠ORC isarightangle.
b Hencecompletethegeometricproofofthedistributivelawforthedotproduct intwodimensions.Thatis,provethat
[Hint:Beginbyconsideringtheprojectionsof b ∼ and c ∼ onto a ∼ .]
16 Inthenotes,therighthandinequalityofthetriangleinequalityisproven.Adoptasimilarapproachtoprove thelefthandinequality,thatis, ||
[Hint:UsethelefthandpartoftheCauchy-Schwarzinequality
CHALLENGE
17 Apositionvector v in3-dimensionalspacemakesanglesof α, β and γ withthe x, y and z axesrespectively.
a Byconsideringthedotproductof v withthestandardunitvectors i , j ∼ and k,provethat cos2 α + cos2 β + cos2 γ = 1.
b If α + β = π 2 ,whatcanbededucedaboutthevector v?
18a Byconsideringthedotproductoftwovectors,showthatforany
b Giventhat x, y,and z arenon-negativeand x + y + z = 3,findtheminimumpossiblevalueof
19 Supposethat a ∼ and b ∼ areperpendicularunitvectorsinthreedimensions.Let v ∼ beanyvectorinthree dimensionsandwrite
v ∼ = (v
where c isaunitvectorand λ isaconstantscalar.
a Showthat c ∼ isperpendiculartoboth a ∼ and b ∼ .
b Henceshowthat λ = v · c
c Deducethat |v ∼ |
20 Given r ∼ (t) = x(t) i ∼ + y(t) j + z(t)k ∼ ,avectorfunctionoftheparameter t,thederivativewithrespectto t is r ∼ ′(t) = x′(t) i ∼ + y′(t) j + z′(t)k ∼
a Showthatif f (t)and g(t)havederivatives f ′(t)and g′(t)then
d
dt ( f ∼ (t) g ∼ (t)) = f ∼ ′(t) g ∼ (t) + f ∼ (t) g ∼ ′(t).
b Henceshowthatif r ∼ isacirclewithcentretheoriginthen r ∼ and r ∼ ′ areperpendicular.
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21 [VectorsinStatistics]Avectorintwodimensionshastwocomponentswhereasavectorinthreedimensions hasthreecomponents.Extendingthispattern,an n-dimensionalvectorhas n components.Thedotproduct maybeextendedinasimilarwaytomean
a ∼ b ∼ = a1 b1 + a2 b2 + + an bn
Letthe n datavalues {x1, x2, , xn} berepresentedbythe n-dimensionalvector x ∼ = (x1, x2, , xn).
Letthevector u ∼ haveall n componentsequalto1.Itshouldbeclearthat u x = 1 × x1 + 1 × x2 + ... + 1
Foreachscore,thequantity(t xi)iscalledthedeviationfrom t.Thesumofthesquaredeviationsisthus S (t) = (t xi)2 .
a Showthat S (t) = (tu ∼ x ∼ ) (tu ∼ x ∼ ).
b Expandtheresultinpart a andsowrite S (t)asaquadraticfunctionwithcoefficientsthataredot products.
c Writethevalueof t thatminimises S (t)indotproductform,thenshowthatthisvalueis t = x.
d Findproju x ∼ usingtheusualdotproductformula.Whatdoyounotice?
e Whatfamiliarstatisticalvalueis S (x) u · u ?
4G VectorsandGeometryinTwoDimensions
Whilstvectorsprovideapowerfultoolfortacklingproblems,inmanyinstancesingeometrytheEuclideanproofs studiedinYears7to10arebyfarthesimplestandbest.Thereareafewcases,however,wherevectorproofs arebothefficientandelegant.Thisshortsectionconsiderssomeofthoseproofs.Asageneralrule,theproblems involvelengthsorrightangles.
Example46
Point C isoutsideacirclewithcentre O.Thepointsofcontactofthetwotangentsfrom C tothecircleare A and B.Let −−→ OA = a, −−→ OB = b and −−→ OC = c.Provethefollowingresults.
a Tangents CA and CB subtendequalanglesatthecentre O
b CA = CB.
Solution
Lettheradiusofthecirclebe r.
a Theanglebetweenaradiusandtangentatthepointofcontactis90◦ .
Hence(c a)
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or
so
b Nextconsiderthesquareofthelengthof AC
, thatis, AC = BC.
Thisisastandardresultfromcirclegeometry:thatthetwotangentstoacirclefromanexternalpointareequal. ItisusuallyprovedusingtheRHScongruencetest.Inthenextexample,vectorsareusedinsteadoftheAA similaritytest.
Example47
Thepoint P haspositionvector p ∼ with |p ∼ | > r.Thepoint U haspositionvector u andliesonthecirclewith radius r andcentre O.Itisknownthat ∠OUP = 90◦.Thepoint Q haspositionvector q ∼ whichistheprojection of u onto p ∼
a Drawadiagramofthesituation.
b Usevectorstoprovethat OP × OQ = r2 .
Solution
a Thediagramisshownbelowontheright.
b Now u · (p ∼ u) = 0(∠OUP = 90◦)
so u ∼ p = u ∼ u ∼ = r2 .
Sincethepoints O, Q and P arecollinear, OP × OQ = p ∼ · q ∼ = p proj p u
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Exercise4G
1 Inthediagramtotheright, OACB isarhombus.
Let −−→ OA = a and −−→ OB = b.
a Whyis |a| = |b|?
b Bysquaringtheresultinpart a,showthat a ∼ a ∼ = b ∼ b ∼
c Express −→ OC and −−→ BA intermsof a and b.
d Henceshowthat −−→ OC −−→ BA = 0.
e Whathaveyoujustprovenaboutthediagonalsofarhombus?
2 Inthediagramtotheright, ABCD isaquadrilateral.Points P, Q, R and S arethe midpointsof AB, BC, CD and DA respectively.Let −−→ AB = a, −−→ BC = b, −−→ AD = d and −−→ DC = c ∼
a Explainwhy a ∼ + b ∼ = d + c ∼
b Express −−→ PQ intermsof a and b.
c Express −−→ SR intermsof d ∼ and c ∼
d Henceshowthat −−→ PQ = −−→ SR.
e Deducethatthequadrilateral PQRS isaparallelogram.
3 Inthediagramtotheright, O isthecentreofasemi-circle APB withdiameter AB Let −−→ OA = a ∼ and −−→ OP = p
a Write −−→ OB intermsof a ∼
b Express −−→ AP and −−→ BP intermsof a and p ∼ .
c Henceprovethat ∠APB = 90◦
d Whatcirclegeometrytheoremhaveyoujustproven?
4 Inthediagramtotheright, OABC isaparallelogramwithequaldiagonals OB and AC.Points A and C haverespectivepositionvectors a ∼ and c ∼ relativeto O
a Explainwhy −−→ CB = a ∼
b Write −−→ OB intermsof c and a.
c Write −−→ AC intermsof c ∼ and a ∼
d Explainwhy |c + a| = |c a|
e Usetheresultinpart d,andthefactthat |v|2 = v · v,toshowthat a · c = 0.
f Whatcanyounowsayaboutaparallelogramwithequaldiagonals?
5 Usevectorstoprovethatthesumofthesquaresofthelengthsofthetwodiagonalsofaparallelogramis equaltothesumofthesquaresofthelengthsofthefoursides.
DEVELOPMENT
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6 Prove,usingvectors,thatthediagonalsofaparallelogrambisecteachother.
7 Inthediagramontheright, AD and BE arealtitudesof △ABC intersectingat P Let a ∼ , b ∼ , c ∼ and p betherespectivepositionvectorsof A, B, C and P relativetoan origin O.
a Explainwhy(p a ∼ ) (c ∼ b ∼ ) = 0.
b Explainwhy(p ∼ b) · (a c) = 0.
c Henceshowthat(p c) (a b) = 0.
d Deducethatthethreealtitudesofatriangleareconcurrent.
8 Inthediagramontheright, PL and PM aretheperpendicularbisectorsof sides AB and BC of △ABC,and N isthemidpointof AC.Let −−→ AB = u ∼ , −−→
and −→ PL = w ∼
a Find −−→ PN intermsof v ∼ and w ∼
b Henceprovethatthethreeperpendicularbisectorsofthesidesofatriangleare concurrentbyprovingthat −−→ PN · −−→ AC = 0.
9 Thediagramontherightshowstrapezium OABC with OA ∥ CB.Themidpoints ofthenon-parallelsides OC and AB are M and N respectively.Let −−→ OA = a, −−→ OB = b ∼ and −−→ OC = c ∼
a Showthat −−−→ MN = 1 2 (a ∼ + b ∼ c ∼ )
b Explainwhy b c = ka,where k isaconstant.
c Henceshowthat MN isparallelto OA and CB
10 Inthediagramontheright, PABC and PUVW aresquares. Let −−→ PA = a, −−→ PC = c, −−→ PU = u and −−→ PW = w
a Provethat a · u + c · w = 0.
b Henceprovethat AW ⊥ UC.
11 [ATheoremofEuler]Thediagramshowsconvexquadrilateral ABCD.The midpointofdiagonal AC is M1 andthemidpointofdiagonal BD is M2.Itisa theoremthat
AB2 + BC2 + CD2 + DA2 = AC2 + BD2 + 4(M1 M2)2 .
Thatis,thesumofsquaresofthesidesisequaltothesumofsquaresofthe diagonalsplus4timesthesquareofthedistancebetweentheirmidpoints.
Usetheresult |v|2 = v · v towriteavectorproofofthistheorem.
Beginbyletting −−→ AB = b ∼ , −−→ AC = c ∼ and −−→ AD = d ∼
12 Supposethat OABC isaparallelogram.Let M bethemidpointof OA andlet P bethepointofintersection of MC and OB.Prove,usingvectors,that OP = 1 3 OB [Hint:Vectors −−→ OP and −−→ MP arescalarmultiplesof −−→ OB and −−→ MC respectively.]
13 Supposethat OABC isatrapeziumwithparallelsides OA and CB suchthat OA = 3 CB.If E and F arethe respectivemidpointsofthediagonals OB and AC,usevectorstoprovethat EFBC isaparallelogram.
14 Inthediagram, AX = 1 2 XB, OY = YX and OA = 3AP.Thelines YQ and PQ are parallelto OA and OB respectively.Let −−→ OA = a ∼ and −−→ OB = b ∼
a Express −−→ OP and −−→ OY intermsof a ∼ and b ∼
b Findthevaluesof m and n suchthat −−→ YQ = ma and −−→ PQ = nb.
c Henceexplainwhy OAQY isaparallelogram.
15 Thepositionvectorsoftwopoints A and B are a and b respectively.
a Showthatthepoint P withpositionvector p ∼ = λa + (1 λ)b,where λ ∈ R,iscollinearwith A and B.
b Showthatthepoint Q withpositionvector q = 3a ∼ 2b ∼ liesontheline BA
c Whatistheratio QA : QB?
d Thepoint R haspositionvector r = 3a ∼ + 5b ∼ andisnotontheline AB.Find,intermsof a ∼ and b ∼ ,the positionvectorofthepointwhereline OR meetsline AB.
e Showthatthevalueof λ forwhich OP and AB areperpendicularis
a
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CHALLENGE
16 Consideratriangle OAB,with P, Q and R themidpointsof OB, OA and AB respectively.Supposethatthe medians AP and BQ intersectat C,andlet −−→ OA = a and
OB = b.
a Explainwhy −−→ AC = λ1 1 2 b a and −−→ BC = λ2 1 2 a b ,where λ1 and λ2 bothliebetween0and1.
b Findthevaluesof λ1 and λ2
c Henceprovethatthethreemediansofatriangleareconcurrent,andstatetheratioinwhich C divides eachmedian.
17 Supposethat C dividestheinterval AB internallyintheratio m : n.Let −−→ OA = a, −−→ OB = b and −−→ OC = c.
a Showthat c = mb + na m + n .
b If m : n = |a| : |b|,deducethat OC bisects ∠AOB.
4H
VectorsandGeometryinThreeDimensions
Manyoftheproblemsinthissectioninvolvesolidssuchasthetetrahedronontheright. Therearethreeanglesthatarecommonlyrequired:
• theangleformedbytwolines,suchasedges AB and BC,
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• theangleformedbyalineandaplane,suchasedge BD andbase △ABC,
• theangleformedbytwoplanes,suchas △ABC and △ABD
Thelasttwodotpointsrequirefurtherexplanation.Anangleisformedbytwointersectinglines,yetthesedot pointsinvolveplanes.
TheZenithAngle
Thediscussionintheappendixstatesthatalineperpendiculartoaplaneiscalleda normal totheplane.Anormal isneededinordertofindtheanglebetweenalineandaplane.
Letline AB intersectplane P at A.Draw AC,thenormaltotheplaneat A.Thesituation isshownontheright.Thepoint C maybedescribedasbeingatthe zenith of A.The word zenith isderivedfromtheArabic samtal-ra’s,meaning directlyoverhead.Thus, ∠BAC iscalledthe angletothezenith,ormoresimplythe zenithangle.
Theanglebetweenthelineandtheplaneisdefinedtobethecomplementofthezenithangle.Ifthedirection vectorsfor AB and AC areknownthenthezenithanglecanbefoundbythedotproduct.Itisthensimplyamatter offindingthecomplementtoobtaintheanglebetweenthelineandtheplane.
Example48
Arectangularprismhasthreeedgesonthecoordinateaxes,withonecornerattheorigin O andtheopposite cornerat P(4,2,3).Findtheanglebetweenthespacediagonal OP andthe xy-plane.Givetheanswercorrectto thenearestdegree.
Solution
Let Q bethecorneroftheprismbeneath P.Lettheanglesubtendedattheorigin by PQ be θ.Thesituationisshownontheright.The z-axisisnormaltothe xy-plane withdirectionvector k.Line OP hasdirectionvector
ˆ p ∼ = 1 √29 (4 i + 2 j ∼ + 3k).
Thusˆp ∼ · k = 3 √29 andthezenithangleat O is cos 1( 3 √29 ) = 56◦ hence θ = 34◦ (correcttothenearestdegree.)
Insuchasimpleproblem,itiseasiertousePythagoras’theoremin △POQ.Readersareencouragedtouse vectormethodshere,however,astherearecertainproblemswherevectorswillberequired.Noticethat QP isthenormaltothe xy-planethrough P.Thatis,usingPythagoras’theoremstillmakesuseofanormaltothe plane.
TheDihedralAngle
Theanglebetweentwoplanesiscalledthe dihedralangle.Theword dihedral comes fromtheAncientGreekprefix di- meaning two andtheword hedron meaning base, whichinfersaplane.
Thediagramaboveontherightsuggeststwopossibleanglesatthelineofintersection, oneacuteandtheotherobtuse.Obtuseanglesarenotconsidered,however,and sothedihedralangleisalwayseitheracuteorarightangle.Thediagramonthe immediaterightmayhelptounderstandthedefinitionthatfollows.Chooseathirdplane perpendiculartothelineofintersection.Thenewlinesofintersectionbetweenthefirst twoplanesandthethirdformthedihedralangle θ.
Thefollowingexampledemonstratesthatifthedihedralangleisnotusedthentheanglebetweentwoplanescan vary.Thefinalansweristhemaximumoftheanglesinthisparticularcase,whichinvolvesthedihedralangle.
Example49
Thetwofixedpoints P(0, a, b)and Q(0, a, b)lieonplanesthroughthe x-axisthat areequallyinclinedtothe xy-plane.Thevariablepoint R(λ,0,0)liesonthe x-axis with ∠PRQ = θ.Let p
and
withcorrespondingunitvectorsˆp
andˆq
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a Showthatˆ
b Findthevalueof λ thatmaximises θ.Thendescribethecorresponding geometricalsituation.
Solution
a Itshouldbeclearthat
Hence,ˆ
b Thevalueof θ ismaximisedwhenˆ
isminimised.Thathappenswhenthefractionismaximised. Thus,put λ = 0.Inthatcase P, Q and R lieinthe yz-planewith R attheorigin.Thatis,thedihedralangle (when a ≤ b)oritssupplement(when a > b)isthemaximumoftheseangles.
Asafinalnote,thedihedralanglefortwoplanesisequaltotheacuteanglebetweenthenormals.Thus,ifnormal vectorsareknownorcaneasilybefound,thedotproductcanbeusedtoobtainthedihedralangle.Thisresult canbeprovenbydrawingthesituationintheplaneofthedihedralangle,andisleftasanexerciseinEuclidean geometry.
Exercise4H FOUNDATION
1 Acubeofsidelengthoneunithasverticesat O(0,0,0), A(1,0,0), B(1,1,0), C(0,1,0), D(0,0,1), E(1,0,1), F(1,1,1)and G(0,1,1).Usevectormethodstoshowthattheacuteanglebetweenthediagonals AG and CE isarccos 1 3 .
2 Arectangularprismis1unitby2unitsby3units.Bygivingitsverticesappropriatecoordinates,findthe threepossiblevaluesofcos θ where θ istheacuteanglebetweenapairofspacediagonalsoftheprism.
3 Thediagramshowsarectangularprism.Thevertices A, B and C have coordinates(4, 8,4),(1, a,1)and(b, c,2)respectively.
a Usethefactthat −−→ OA isperpendicularto −−→ OB tofindthevalueof a
b Findthevaluesof b and c
c Findthecoordinatesof D, X and Y
4 Thediagramshowsatriangularpyramid.Itsbaseisaright-angledisosceles triangle,anditsperpendicularheightisequaltothelengthoftheequalsidesofthe base.
a Let M bethemidpointof BC.Explainwhy MD and MA arebothperpendicular to BC
b Bygivingtheverticesofthepyramidappropriatecoordinates,usevector methodstoshowthattheacuteanglebetweenthefrontface ABC andthe base BCD iscos 1 1 √3
5 Thepositionvectorsoftheverticesofatetrahedron(thatis,atriangularpyramid) ABCD
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a Whatisthesizeof ∠CBD?
b Showthat AB isperpendiculartoboth BC and BD
c Calculatethevolumeofthetetrahedron.
6 Thecube OABCDEFG hasvertices O(0,0,0), A(a,0,0), B(a, a,0), C(0, a,0), D(0, a, a), E(0,0, a), F(a,0, a)and G(a, a, a),where a > 0.
a Useascalarproducttofindtheanglebetweenthetwofacediagonals OF and OD.
b Whattypeofspecialtriangleis △OFD?
c Whatspecialnameisgiventothetriangularpyramid OBDF?
d Find ∠FXD,where X isthecentreofthecube.(Thisangleisthebondingangleofcarbontetrachloride.)
7 Thepoints A and B havepositionvectors
linesegment AB suchthat AP : PB = λ :1 λ
a Showthat −−→ OP = (1 + λ) i ∼ + (2
b Hencefindthevalueof λ forwhich: OP isperpendicularto AB, i
8 Thepoints A, B and C haverespectivepositionvectors3 i ,2 j ∼ and9k
Showthatcos ∠ACB = 27 √850
.Thepoint P liesonthe
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a Hencefindtheareaof △ABC
b
Findthevolumeofthetriangularpyramid OABC,andhencefindtheperpendiculardistancefromthe origintotheplane ABC c
9 Thepoints A, B and C havepositionvectors
a Findcos ∠BAC,andhenceshowthat △ABC hasarea12√10squareunits.
b Findthepositionvectorsof D and E,where D isthepointthatdivides AB internallyintheratio1:2 and E isthemidpointof CD.
c Showthat OE isperpendiculartoboth AB and AC,andhenceperpendiculartotheplane ABC
d Findthevolumeofthetriangularpyramid OABC.
10 Aquadrilateralisafoursidedfigureinaplane.Inthisquestion,supposethatthedefinitionofaquadrilateralisextendedtoincludethecaseinthreedimensionswherethefourverticesdonotalllieinaplane.Let OABC besuchaquadrilateralwithverticesat O(0,0,0), A(a1, a2, a3), B(b1, b2, b3)and C(c
).
Giventhat M, N, P and R aretherespectivemidpointsof OA, AB, BC and CO,usevectorstoprove that MNPR isaparallelogram.
11 Thediagramshowsaparallelepiped,asolidinwhicheveryfaceisa parallelogram.Vertices A, B,and C havepositionvectors −−→ OA = 5 i ∼ , −−→ OB = i ∼ + 3k ∼ and −−→ OC = i ∼ + 4 j
a Writedownthepositionvectorsof X, Y, Z and D
b Calculatethesizeof ∠OZY tothenearesttenthofadegree.
c Thepoint P divides CZ intheratio λ :1.Findthevalueof λ forwhich OP isperpendicularto CZ.
12 Asquarepyramidhasbase OABC,where O istheorigin,andapex V.Thepoints A, B, C and V have respectivepositionvectors2 i ,2 i + 2 j,2 j and i + j + 3k.
a Find −−→ BV.
b Findavector x thatisperpendiculartoboth −−→ OA and −−→ OV,andhenceisperpendiculartotheplane OAV
c Calculate,tothenearestdegree,theacuteanglebetweenthevectors x and −−→ BV.
d Hencedetermine,tothenearestdegree,theacuteanglebetweentheedge VB ofthepyramidandtheside face OAV
13
Thediagramshowsasquarepyramid VABCD.Thebasehassidelength4cm andtheperpendicularheight OV is2h cm.Themidpointsof AB, BC, VC and VA are P, Q, M and N respectively.
a Findthepositionvectors −−→ OA, −−→ OB, −−→ OC and −−→ OD.
b Findthevectors −−→ PM and −−→ QN intermsof h
c Findthepositionvector −−→ OX,where X isthepointofintersectionof PM and QN.
d Giventhat OX isperpendicularto VB,findthevalueof h
e Showthat PQMN isarectangle.
14 Thediagramshowsatriangularpyramid ABCD.Supposethatthepoint P issuch that −−→ DP = 1 4 −−→ DB + 1 3 −−→ DC and Q isthepointofintersectionof DP producedand BC
Let −−→ DB = u ∼ and −−→ DC = v ∼ .
a Findtwodifferentexpressionsfor −−→ BQ andhenceshowthat −−→ BQ = 4 7 −−→ BC
b Let −−→ AB = b ∼ and −−→ AC = c ∼.Supposethatthepoint R issuchthat −−→ AR = 3 −−→ AB + 4 −−→ AC
Provethat R liesontheline AQ.
15 Thediagramontherightshowsatriangularpyramid OABC.Letthenotation
|XYZ| indicatetheareaof △XYZ
Provethat |AOB|2 + |BOC|2 + |COA|2 = |ABC|2
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CHALLENGE
16 LikeQuestion 10,supposethatthedefinitionofaquadrilateralisextendedtoincludethecaseinthree dimensionswherethefourverticesdonotalllieinaplane.Useavectorapproachtoprovethatthesumof thesquaresofthesidesofsuchaquadrilateralisequaltothesumofthesquaresofthediagonalsplusfour timesthesquareofthedistancebetweenthemidpointsofthediagonals.Thatis,repeatQuestion 11 from Exercise 4G butnowinthreedimensions.
[Hint:Taketheoriginasoneoftheverticesandthenspecifygeneralcoordinatesfortheotherthree vertices.]
Chapter4Review
Reviewactivity
• Createyourownsummaryofthischapteronpaperorinadigitaldocument.
Chapter4Multiple-choicequiz
• Thisautomatically-markedquizisaccessedintheInteractiveTextbook.AprintablePDFWorksheetversionis alsoavailablethere.
ChapterReviewExercise
Note: Throughoutthisexercise λ and µ arerealparameters.
1 Alinehasvectorequation r ∼ = 2
j + k ∼ ).Writedown: a thepositionvectorofapointontheline, b adirectionvectoroftheline.
2 Writethelinearequation y = 2x + 3inparametricvectorform.
3 FindtheCartesianequationoftheline r = 2i 4j + λ(i + 3j).
4 Findaparametricvectorequationoftheline PQ given P(1,1, 1)and Q(2, 1,2).
5 Determinewhetherornoteachpointliesontheline r ∼ = (6, 4, 3) + λ( 5,2,7). ( 4,0,13) a (16, 8, 17) b
6 Aline ℓ hasvectorequation r = 6i + 4j +
(2i + j 2k).Findthepointsofintersectionof ℓ with the yz, xz and xy-planes.
7 Findthepointofintersectionofeachpairoflines.
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9 Findthevalueof a forwhichthelines
aj 8k ∼)areperpendicular.
10 Supposethat A isthepoint(2, 1)ontheline ℓ withdirectionvector
Let Q bethepointon ℓ suchthat PQ ⊥ ℓ.
a Find −−→ AP.
b Find −−→ AQ byfindingtheprojectionof −−→ AP ontothedirectionvectorof ℓ
andlet P bethepoint( 1,5).
c HenceusePythagoras’theoremtofindtheperpendiculardistancefrom P to ℓ.
11 Supposethat A isthepoint(1,1,0)ontheline ℓ withdirectionvector
Let Q bethepointon ℓ suchthat PQ ⊥ ℓ
a Find −−→ AP
b Find −−→ AQ byfindingtheprojectionof −−→ AP ontothedirectionvectorof ℓ.
c Find −−→ QP
andlet P bethepoint(1,0, 1).
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d Hencewritedownavectorequationofthelinethrough P perpendicularto ℓ.
e Findthecoordinatesof Q
f Findtheperpendiculardistancefrom P to ℓ
12 Given a
and
R,provethat(
13 Foranytwonon-zerovectors
14 Aspherehascentre(3, 4,2)andradius √7.Writedown:
a aCartesianequationforthesphere,
b avectorequationforthesphere.
15 Showthatthepoint P(5, 1,4)liesoutsidethesphere
16 Findavectorequationforthesphere x2 + y2 + z2
17 Findthepointsofintersectionoftheline
)andthesphere (x + 2)2 + (y 3)2 + (z + 4)2 =
18 Sketchthegraphof r
19 FindtheCartesianequationcorrespondingtoeachofthefollowingvectorequations.
20 Thediagramontherightshowsrectangle OABC withdiagonalsthatare perpendicular.Let
and
a Whatisthevalueof a ∼ · c ∼ ?
b Write −−→ OB and −−→ AC intermsof a and c.
c Henceprovethat OABC isasquare.(Recallthat
21 Thesquare ABCD shownhassidelength ℓ.Let
a Express −−→ AC and −−→ BD intermsof a and b
b Whatisthevalueof a b ?
c Write a ∼ · a ∼ intermsof ℓ
d Find(a + b) (b a).
e Whathaveweproveninpart d aboutthediagonalsofasquare? AB
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22 Usevectorstoproveeachofthesetheorems.
a Themidpointsofthesidesofarhombusaretheverticesofarectangle.
b Themidpointsofthesidesofarectanglearetheverticesofarhombus.
23 Usevectorstoprovethataquadrilateralisaparallelogramifandonlyifitsdiagonalsbisecteachother.
24 Atriangularpyramidhasvertices O(0,0,0), A(2a,0,0), B(0,2a,0)and C(0,0,3a).Usevectorstofind, correcttothenearestdegree,theanglebetweenthefrontface ABC andthebase AOB
25 Thepoints A, B, C and D havepositionvectors
3 i ∼ + j + 2k ∼
a Showthattriangle ABC isright-angled.
b Showthattriangle ABD isisosceles.
c Showthatthevector j k ∼ isperpendiculartotheplanecontainingtriangle ABC byshowingthatitis perpendiculartotwonon-parallelvectorsintheplane.
d Showthat BD passesthroughthemidpoint M of AC
e Findtheratio BM : MD.
26 Thepoints A, B, C and D havepositionvectors
+ 7k.Thepoint P lieson BC suchthat BP = 1 3 PC,andthepoint Q lieson AP producedsuch that AP = 1 4 PQ.
a Findthepositionvectorsof P and Q.
b Determinethepositionvectorofthepoint R on PQ suchthat DR isperpendicularto PQ
c Findtheratio PR : RQ.
Appendix:SomeGeometryin3D
Thethreefundamentalobjectsofgeometryarepoints,linesandplanes.Becausetheyaresofundamental,it isdifficulttogiveprecisedefinitionsofthem,justasitisdifficulttodefinepreciselywhatanumberis.The followingdescriptionsofpoints,linesandplanesdon’treallysaywhattheyare,butwillenablediscussionsabout themwithsomeagreementaboutwhatismeant.
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Point
A point marksaposition.Asitisalocation,ithasnosize.Indiagrams,pointsaremarkedwith adotorcross-hair,likethoseontheright.Italiccapitallettersareusedtolabelpoints.
Line
A line hasnobreadth,butextendsinfinitelyinoppositedirections.Indiagrams,lines aredrawnwitharuler,liketheoneontheright.Italiclowercaselettersareusedtolabel lines.
Theword line alwaysmeans straightline,anddoesnotincludecurves.Alinecanalsobethoughtofasthepath ofapointmovinginafixeddirection,likethepathofathinbeamoflight.Itisassumedthatastraightlineisthe shortestpathbetweentwopoints.
Plane
A plane hasnothickness,butextendsinfinitelyinalldirections.Indiagrams,aplane isdrawnasthoughitisasheetofpapershowninperspective,likeontheright.Inthis text,aregionwithinaplanewillusuallybelabelledbyitsvertices.
Inthesamewaythatalineisthepathtracedbyapointmovinginafixeddirection,aplaneisthepathtracedby thelinemovinginasecondfixeddirection.Apracticalexampleistheedgeofaruler,representingtheline,asit isdraggedacrossadesk,representingtheplanethatistracedout.
Theword plane alwaysmeans flatplane,anddoesnotincludecurvedsurfaceslikecylinders,conesorspheres.A linejoininganytwodistinctpointsinaplaneliesentirelywithinit.
DefiningaPlane
Itwillbeassumedthatthereisonlyoneplanewhichpassesthroughthreedistinctfixed points.Thethreepointsaresaidto definetheplane. Itiswhyatripodneverwobbles, regardlessofhowuneventhegrounditisplacedon.
Nowconsidertheline AB andthepoint C.Sincethesamethreepointshavebeenused, itisclearthatthereisonlyoneplanewhichpassesthroughalineandapointwhichis notontheline.Thusaplanecanalsobedefinedinthisway.
Further,considerthetwolines AB and AC.Sincethesamethreepointshavebeenused, itisclearthatthereisonlyoneplanewhichpassesthroughtwolinesthatintersectina singlepoint.Thusaplanecanalsobedefinedinthisway.
Finallyconsidertheline AB andthelineparallelwith AB through C.Onceagain,the samethreepointshavebeenusedsoitisclearthatthereisonlyoneplanewhichpasses throughtwoparallellines.Thusaplanecanalsobedefinedinthisway.
ArrangementsofLinesinaPlane
Therearepreciselytwoarrangementsoflinesinplanes.Eithertheyareparallelortheyintersect.Thisis demonstratedinthediagramsaboveforthelasttwowaystodefineaplane.Therearefourarrangementswhen therearethreelines,andtheseareshownbelow.
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ArrangementsofPlanesinSpace
Thereisasimilararrangementofplanesinspacecorrespondingtoeachofthearrangementsoflinesinaplane. Twoplanesinspacecaneitherbeparallelorintersectinaline,asshownbelow.
Andherearethearrangementsofthreeplanesinspacecorrespondingtothearrangementsshownaboveofthree linesinaplane.
Inaddition,threeplanesinspacecanintersectinasinglepoint,asshownontheright. Thisdiagramshowsthespecialcasewheneachplaneisatrightanglestotheothertwo. Threeplanesinspaceintersectingatapointdonothavetobeatrightanglestoeach other.Asimpleexampleisatetrahedron,wherepairsoftriangularfacesareat71◦ to eachother,yeteachgroupofthreemeetsatacommonvertex.
ArrangementsofTwoLinesinSpace
Asnotedabove,iftwodistinctlinesareparallelthentheydefineaplane,andtwolineswhichintersectina pointalsodefineaplane.Thereisathirdpossibleconfigurationoftwolinesinspacewherethelinesareneither parallelnorintersecting.
Considertwoparallelplanes.Foranylinedrawninoneplane,itisalwayspossibleto drawalineintheotherwhichisnotparallelwiththegivenline.Anexampleofthe situationisshownontheright.Pairsoflineslikethesewhicharenotparallelanddo notintersectarecalled skewlines .Further,thereisonlyonepairofparallelplanesthat containanygivenpairofskewlines.
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Onceagain,atetrahedronprovidesapracticalexampleofthesituation.Theopposite edges AB and CD areclearlynotparallelandclearlydonotintersect,andhencethey areskewlines.Asanexercise,trytoidentifyanypairofparallellines,eachpairof intersectinglines,andeachpairofskewlines.Thenextendtheexercisetoeachofthe Platonicsolids.
LinesandPlanesthatarePerpendicular
Anobjectthatisperpendiculartoalineinthreedimensionsisaplane.Further,thereisonlyoneplanethatpasses throughagivenpointandisperpendiculartoagivenline.
Inparticular,ifplane P isperpendiculartoline ℓ andiftheseintersectat A thenevery linein P thatpassesthrough A isperpendicularto ℓ.Consequently,ifthenon-parallel lines AB and AC arebothperpendicularto ℓ thenthepoints B and C alsolieinplane P Thesituationisshownontheright.
Conversely,anobjectthatisperpendiculartoaplaneinthreedimensionsisaline. Further,thereisonlyonelinethatpassesthroughagivenpointandisperpendicular toagivenplane.Consequently,ifpoint A isinplane P andifboth AB and AC are perpendicularto P then A, B and C arecollinearasshownontheright.Thisline ℓ is calledthe normal to P at A.
FurtherGeometryinThreeDimensions
Thereismuchmorethatcouldbesaidaboutgeometryinthreedimensions.Whathasbeenincludedherearethe essentialconceptsandconfigurationsthatarelikelytobeneededinthestudyofpoints,linesandplanesthrough vectors.Furtherstudyofgeometryinthreedimensionsmightbesuitableasanenrichmentproject.
5
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TrigonometricIdentities
Chapterintroduction
Supposethattwofunctions f (x)and g(x)haveprimitives F(x)and G(x),andhavederivatives f ′(x)and g′(x). Itshouldbeobviousthatthesum y = f (x) + g(x)hasthesimplederivative y′ = f ′(x) + g′(x).Equally,itshould beobviousthattheindefiniteintegralis ∫ ydx = F(x) + G(x) + C,forsomeconstant C.Incontrast,theproduct y = f (x) × g(x)hasthemorecomplicatedderivative y′ = f ′(x) × g(x) + f (x) × g′(x).Further,theprimitiveofthe productisunknownatthisstage.Thus,whenevercalculusisinvolveditisfareasiertodealwithasumthana product.
Theaboveadvantagesanddisadvantagesareamplifiedwhentrigonometryisalsoinvolved.So,itisusefultobe abletoconvertproductsoftrigonometricfunctionstosumsinordertoapplycalculus.Hencethischapterbegins withtherequiredtrigonometricidentitiesinSection 5A.
Section 5B introducesasubstitutionthatconvertstrigonometricfunctionsintoratiosofpolynomials.Applying calculustopolynomialsandsolvingpolynomialequationshavebothbeencloselystudiedinYears11and12, sothisidentitywillprovemostuseful.Thereareotherapplicationsoftheseidentitiesthatwillbemetinalater chapter.
5A ProductstoSums
Itwillbeshowninthissectionthatproductsofsineandcosinecanbeexpressedasasumordifference.For example,
2sin 3 2 x cos 1 2 x = sin2x + sin x.
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Eachexpressionhasitsadvantages.Itiseasiertofindthe x-interceptsoftheproduct y = 2sin 3 2 x cos 1 2 x thanthe sum.Incontrast,itiseasiertodifferentiatethesum y = sin2x + sin x thantheproduct.
ProductstoSums
Thederivationoftheseformulaebeginswiththecompoundangleformulaeforcosineandsine.
cos(A + B) = cos A cos B sin A sin B
cos(A B) = cos A cos B + sin A sin B
sin(A + B) = sin A cos B + cos A sin B
sin(A B) = sin A cos B cos A sin B
Noticethat[2] + [1]cancelsoutsin A sin B,leaving
2cos A cos B = cos(A B) + cos(A + B),
andlikewise[2] [1]gives
2sin A sin B = cos(A B) cos(A + B).
Adding[3] + [4]cancelsoutcos A sin B,leaving
2sin A cos B = sin(A + B) + sin(A B),
andlikewise[3] [4]gives
2cos A sin B = sin(A + B) sin(A B). [8]
Notethatswappingangles A and B in[7]alsogives[8].
Equations[5]to[8]arecollectivelyknownasthe productstosums formulae.Noticethattherighthandsidesof equations[6]and[8]areeachdifferences.Nevertheless,theword sum isusedtodescribethemsinceeachcanbe writtenasasuminvolvinganegative.Forexample,
2sin A sin B = cos(A B) cos(A + B) = cos(A B) + cos(A + B)
Example1
Acertaincurvehastheequation y = 2sin 1 2 x cos 3 2 x.Findthecoordinatesofthetwostationarypointsinthe interval0 ≤ x ≤ π.Approximateeachvalueto2decimalplaces.
Solution
Firstapplyequation[7]with A = 1 2 x and B = 3 2 x toget
y = sin2x sin x (sincesineisodd).
Hence y′ = 2cos2x cos x = 2(2cos2 x 1) cos x (doubleangleformula)
= 4cos2 x cos x 2.
Continuedonthenextpage
Thus,thestationarypointsoccurwhen
4cos2 x cos x 2 = 0.
Solvingthisquadraticincos x forall x gives x y
cos x = 1+√33 8 or 1 √33 8 . Henceonevalueof x inthedomainisgivenby
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cos x = 1+√33 8
so x = cos 1 1+√33 8 ≑ 0.57(correctto2decimalplaces)
forwhich y ≑ 0.37.Thesecondvalueis
x = cos 1 1 √33 8 ≑ 2.21(correctto2decimalplaces)
forwhich y ≑ 1.76.Thesepointsareshownintheabovegraph.
Whilstequations[5]to[8]areeasytouse,theformulaeareoftenwrittenwiththefactorof2dividedout.
1Productstosums
cos A cos B = 1 2 cos(A B) + cos(A + B)
sin A sin B = 1 2 cos(A B) cos(A + B)
sin A cos B = 1 2 sin(A + B) + sin(A B)
cos A sin B = 1 2 sin(A + B) sin(A B)
ExactValues
Theproductstosumsformulaecanbeusedinafewcasestoobtainexactvaluesforthetrigonometricfunctions ofcertainangles.
Example2
a Express2sin54◦ cos18◦ asasumofsinefunctions.
b Writethisexpressiononlyintermsoftheangles72◦ and36◦
c Applydoubleangleformulaetoshowthat4cos2 36◦ 2cos36◦ 1 = 0.
d Hencefindtheexactvalueofcos36◦
Solution
a 2sin54◦ cos18◦ = sin72◦ + sin36◦
b Nowapplythecomplementaryangleresultstoget 2cos36◦ sin72◦ = sin72◦ + sin36◦
c Applytheresultsforsin2θ toget 4cos2 36◦ sin36◦ = 2cos36◦ sin36◦ + sin36◦ andsincesin36◦ 0thisgives
4cos2 36◦ = 2cos36◦ + 1 or4cos2 36◦ 2cos36◦ 1 = 0.
d Nowthequadraticformulagivesthevalues 2+√20 8 or 2 √20 8 Since36◦ isacute,takethepositiveroottoget cos36◦ = 1+√5 4 .
SumstoProducts
Theprocessofexpressingaproductoftrigonometricfunctionsasasummustbereversedinordertosolve equationslike
sin5x + sin x = 0.
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Althoughtherearesimpleformulaetodothis,theyarenotpartofthecourse.Instead,asimplepairof simultaneousequationsissolved.
Noticethat,intheformulaeofBox 1,thetwoanglesinthesumsare(A + B)and(A B).Thus,tosolvean equationinvolvingasum,firstidentifywhichoftheproductstosumsequationsisrequired,thenput(A + B)and (A B)equaltothetwoanglesinthesum.
Example3
Writetheequationsin5x + sin x = 0intermsofaproductandhencesolvetheequationinthedomain 0 ≤ x ≤ 2π
Solution
Therelevantproducttosumsformulais
sin(A + B) + sin(A B) = 2sin A cos B, soput A + B = 5x and A B = x
Addingtheseanddividingtheresultby2gives A = 3x forwhich B = 2x
Thus2sin3x cos2x = 0.
Thiscanbesolvedmentallytoget,forintegervaluesof n, x = nπ 3 (fromsin3x = 0) or x = (2n+1)π 4 (fromcos2x = 0).
Thusintherequireddomainthesolutionsare
Calculus
Oneelegantapplicationofthisapproachisinprovingthatthederivativeof y = sin x is y′ = cos x
Example4
a Showthatsin u sin x = 2cos 1 2 (u + x)sin 1 2 (u x).
b Thederivativeof y = f (x)maybedefinedas: dy dx = lim u→x f (u) f (x) u x
Usethisdefinitionandtheresultofpart a todifferentiate y = sin x.
Solution
a Put A + B = u and A B = x toget A = 1 2 (u + x) and B = 1 2 (u x).
Substitutetheseintotheequationforcos A sin B toget sin u sin x = 2cos 1 2 (u + x)sin 1 2 (u x).
b dy dx = lim u→x sin u sin x u x = lim u→x 2cos
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× 1
cos x
Anexercisequestionusesasimilarapproachforthederivativeof y = cos x.
Enrichment—TheSumstoProductsFormulae
Asmentionedabove,thereareformulaetoconvertsumstoproducts.Theyareeasilyobtainedbytwosubstitutionsappliedtotheproductstosumsformulae.Beginbyletting
A + B = C
A B = D.
Thispairofsimultaneousequationsiseasilysolvedfor A and B toget
A = 1 2 (C + D) B = 1 2 (C D).
Thus,theproductstosumsformulaefor A and B becomethesumstoproductsformulaefor C and D below. cos C + cos D = 2cos 1
Whenapplyingtheseformulaeinaquestion,itmayhelptothinkof 1 2 (C + D)astheaverageoftheangles.Also notethat 1 2 (C D) = C 1 2 (C + D).
Thatis,thesecondangleusedintheproductis C minustheaverage.
Applyingthethirdformulaabovetosin5x + sin x = 0,theequationinExample 3,givestheaverage3x and C minustheaverage2x.Thus
2sin3x cos2x = 0 exactlyasbefore.Asanexercise,checkthesolutiontoExample 4a usingthelastsumstoproductsformula.
Exercise5A
1 Writeasasumordifferenceoftwotrigonometricfunctions.
2cos35◦ cos15◦ a
◦ sin32◦ b sin49◦ sin22◦ c cos61◦ sin25◦ d
2sin3α cos α e cos2θ cos θ f
2sin(x + y)sin(x y) g
2 Show,withoutacalculator,that: 2cos75◦ cos15◦ = 1 2 a
FOUNDATION
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A + B)cos(2A B) h
◦ sin35◦ = cos10◦ 1 2 b 4sin40◦ cos10◦ = 1 + 2sin50◦ c
◦ cos35◦ = cos10◦ + sin10◦ d 2sin54◦ sin18◦ = sin54◦ sin18◦ e
3 Usetheproductstosumsidentitiestoprovethat: 2sin3θ cos2θ + 2cos6θ sin θ = sin7θ + sin θ a
α sin3α + 2cos5α cos2
4 Show,withoutacalculator,that:
5a Writesin45◦ sin15◦ asasumandhencefindtheexactvalueofsin15◦ . b Useasimilartricktofindtheexactvalueofsin75◦ DEVELOPMENT
6ai Showthat2sin2x cos x = sin3x + sin x. ii Hencesolvetheequationsin3x + sin x = 0,for0 ≤ x
b Usingasimilarmethod,solvecos3x + cos x = 0,for0 ≤ x ≤ π
7 Useaproductstosumsidentitytosolvesin x + 2cos2x sin x = 1 2 ,for0 ≤ x ≤ 2π.
8a Expresssin3x + sin x asaproductusingaproductstosumsidentity. b Hencesolvetheequationsin3x + sin2x + sin
9 Provetheseidentities.
4cos4x cos2x cos x = cos7x + cos5x + cos3x + cos x a 4sin5α cos3
b
10 Show,withoutacalculator,that: 8sin 4π 9 sin 2π 9 sin π 9 = √3 a cos 4π 9 cos 2π 9 cos π 9 = 1 8 b
11 Let y = 20cos θ sin(θ + 30◦)for0◦ ≤ θ < 180◦.Withouttheaidofcalculus,findthemaximumvalueof y andthecorrespondingvalueof θ.
12a Expresscos20◦ cos10◦ intermsofcos10◦ using: aproducttosumidentity, i adoubleangleidentity. ii b Henceshowthatcos10◦ isarootofthecubicequation8x3 6x √3 = 0.
13a Useaproductstosumsidentitytoprovethat:
2sin x(cos2x + cos4x + cos6x) = sin7x sin x
bi Deducethatcos 2π 7 + cos 4π 7 + cos 6π 7 = 1 2 ii Henceevaluatecos π 7 +
14 Useaproductstosumsidentitytoprovethat: sin x + sin3x + sin5x + + sin(2n 1)x =
15 Solveeachofthefollowingequations,for0 ≤ x ≤ π.
cos5x + cos x = 0 a sin4x sin x = 0 b
cos3x + cos5x = cos4x c
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cos4x + cos2x = cos3x + cos x d
sin x + sin2x + sin3x + sin4x = 0 e sin5x cos4x = sin3x cos2x f
16a Provethattan 1 2 (A + B) + tan 1 2 (A B) = 2sin A cos A + cos B b Hencefindtheexactvalueoftan15◦
17 Provethatsin(B C)cos(A D) + sin(C A)cos(B D) + sin(A B)cos(C D) = 0.
18 Substitute P = A + B and Q = A B intotheproductstosumsidentitiestoestablishthesesumstoproducts identities.
sin P + sin Q = 2sin 1 2 (P + Q)cos 1 2 (P Q) a sin P sin Q = 2cos 1 2 (P + Q)sin 1 2 (P Q) b cos P + cos Q = 2cos 1 2 (P + Q)cos 1 2 (P Q) c cos P
19 Usethesumstoproductsidentitiestoprovethat:
sin35◦ + sin25◦ = sin85◦ a cos36◦ cos84◦ = √3sin24◦ b
sin3θ + sin θ cos3θ + cos θ = tan2θ c sin6α sin4α + sin2α cos6α cos4α + cos2α = tan4α d
20a Showthatcos u cos x = 2sin 1 2 (u + x)sin 1 2 (u x). b Thederivativeof y = f (x)maybedefinedas: dy dx = lim u→x f (u) f (x) u x
Usethisdefinitionandtheresultofpart a todifferentiate y = cos x. CHALLENGE
21 Solvetheequationcos5x = sin x,for0 ≤ x ≤ π.[Hint:Writesin x ascos( π 2 x).]
22 If A, B and C arethethreeanglesofatriangle,provethat:
sin A + sin B + sin C = 4cos 1 2 A cos 1 2 B cos 1 2 C a
sin2A + sin2B + sin2C = 4sin A sin B sin C b
5B The t -Formulae
Theotheridentitiesforthechapterarisefromasubstitutionthattransformstrigonometricfunctionsintorational functions,thatis,ratiosoftwopolynomials.Theseidentitiesarefrequentlyreferredtoasthe t-formulae,because theletter t isoftenusedforthissubstitution.
The t -formulae
Hereisaderivationoftheidentities.Aswithmuchoftrigonometry,agoodplacetostartisthePythagorean identity,
cos2 θ 2 + sin2 θ 2 = 1.
Forcos θ 2 0,thelefthandsidecanbere-writtenas
cos2 θ 2 (1 + tan2 θ 2 ) = 1.
Solet t = tan θ 2 toget
cos2 θ 2 = 1 1 + t2 .
Thedoubleangleformulaforcosinegives
cos
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Thedoubleangleformulaforsinegives
thussin θ = 2t 1 +
Dividingtheresultforsin θ bycos θ thengives
Theserationalequivalentsforsin θ,cos θ andtan θ canbeusedtoproveotheridentitiesandsolvevarious problems.Thesolutionshouldalwaysbeginwithastatementofthesubstitution.
Example5
Provetheidentitysec
Hencefindtheacuteangle x suchthatsec
b Fromthepreviouspart tan( π 4 + x 2 ) = √3 andsince x isacute,itisguaranteedthat π 4 + x 2 = π 3 so x = 2π 3 π 2 = π 6 .
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Strictlyspeakingthereisaproblemwithpart a ofthelastexample.Theidentityisnotvalidforallvaluesof x. Indeedtheresultisonlytruewhen x (2n+1)π 2 forinteger n,thatis,when x isnotanoddmultipleof π 2 .Itisoften thecasethatsuchidentitiesarestatedwithoutanyrestrictions.Itislefttothereadertoworkoutwhattheseare whennecessary.
2 t -formulae
Let t = tan θ 2 sothat,for θ notanoddmultipleof π,
Asinthelastexample,the t-formulaeareparticularlyusefulwhenaprobleminvolvestwoangles,oneofwhich ishalftheother.Forthatreason,theidentitiesaresometimesreferredtoasthe half-angleformulae orthe double-angleformulae.Thesubstitutionshouldalwaysbeforthetangentofthesmallerangle.
Example6
a Provetheidentitycot θ = 1 + cos2θ sin2θ
b Hencedeterminetheexactvalueofcot π 8 .
Solution
a Let t = tan θ sothatcos2θ
b Put θ = π 8 intotheresultinpart a toget cot π 8 = 1 + cos π 4 sin π 4 = 1 + 1 √2 1 √2 × √2 √2 = √2 + 1.
LikeExample 5,theidentityin a isnotalwaysvalid.Asanexercise,determinetheimpliedrestrictionon θ.
RememberingtheFormulae
Thediagramontherightisoftenusedtorememberthe t-substitutionformulae.Itis basedonanalgebraicidentity.
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Thatis,thequantities(1 t2),2t and(1 + t2)formaPythagoreantriadfor0 < t < 1.Hencetheyformthesidesof arighttriangle,asshown.
Thiscanbeextendedinasimilarwaytothemannerinwhichtrigonometrywasextendedtoanglesofany magnitude.Thus,forallvaluesof t,thepointwithcoordinates x = (1 t2)and y = 2t isguaranteedtolieonthe circle x2 + y2 = r2,where r = (1 + t2).Hence,bythestandardtrigonometricresults,
,sin
Finally,noticethat θ isinthefirstquadrantwhen0 < t < 1,thesecondquadrantwhen t > 1,thethirdquadrant when t < 1andthefourthquadrantwhen 1 < t < 0.Theboundariesbetweenthequadrantsareat t = 1,0,1 andinthelimitas |t|→∞.Asanexercise,drawagraphofthepoints
forvariousvaluesof t toconfirm theseobservations.
SolvingEquations
The t-substitutionoftenmakessolvingatrigonometricequationsimpler,turningitintoapolynomialequation. Someequationsonlyinvolveasingleangle.Inthosecases,makethesubstitutionforhalfthisangle.
Example7
Solve3cos x 2sin x = 1for π < x ≤ π
Thenapproximatetheanswerscorrecttoonedecimalplace.
Solution
Let
Thegivenequationbecomes
CollectingtermsontheRHSgives
Thediscriminantis12,sothequadraticformulagives
Thegraphontherightshowsthesituation.
DegenerateEquations
Whenthe t-formulaeareusedtosolvecertainequationstheremaybepowersof t thatcancelout.Itispossible inthosecasesthatvalidsolutionsarelost.Ifpowersof t canceloutthencheckforvaluesoftheanglewhere t is undefined.Thiswillcorrespondtooddmultiplesof π 2 wherethetangentfunctionisundefined.
Example8
Solvecos x + sin x = 1for0 ≤ x < 2π.
Solution
Let t = tan x 2 sothatcos x =
Thegivenequationbecomes
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andsin
Collectingliketerms,the t2 termscancel,leaving
2t + 2 = 0 thus t = 1 thatis tan x 2 = 1.
Thiscorrespondsto x = 3π 2 forthegivendomain.
Thereisanothersolution,however,whichis x = π (wheretan x 2 isundefined).
Actually,thisequationiseasytosolvewithoutthe t-substitution:sin x = 1at x =
wherecos x = 0,and cos x = 1at x = π wheresin x = 0.Thisisalsoevidentinthegraphof y = cos x + sin x above.Nevertheless,the exampleshowstheimportanceofcheckingforothersolutionswhenpowersof t cancelout.
Exercise5B
1 Writeintermsof
2 Writeintermsof t,where t = tan θ.
3 Usethe t = tan 1 2 θ resultstofindtheexactvalueof:
4 Proveeachidentityusingthe t = tan 1 2 θ results.
cos θ(tan θ tan 1 2 θ) = tan 1 2 θ a 1 cos2x sin2x = tan x b 1 cos θ 1 + cos θ = tan2 1 2 θ c cos A + sin A 1 cos A sin A + 1 = tan 1 2 A d
5 Considertheequationcos x sin x = 1,where0 ≤ x ≤ 2π.
Usethesubstitution t = tan 1 2 x toshowthattheequationcanbewrittenas t2 + t = 0. a
Henceshowthattan 1 2 x = 0or 1,where0 ≤ 1 2 x ≤ π b
Hencesolvetheequationfor x. c
6 Considertheequation √3sin x + cos x = 1.
Showthattheequationcanbewrittenas t2 = √3 t,where t = tan 1 2 x a
Hencesolvetheequation,for0 ≤ x ≤ 2π. b
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DEVELOPMENT
7a Showthattheequation4cos x + sin x = 1canbewrittenas(5t + 3)(t 1) = 0,where t = tan 1 2 x.
b Hencesolvetheequation,for0◦ ≤ x ≤ 360◦.Givethesolutionscorrecttothenearestminutewhere necessary.
8a Showthattheequation3sin x 2cos x = 2canbewrittenas3t 2 = 0,where t = tan 1 2 x
b Hencesolvetheequationfor0◦ ≤ x ≤ 360◦,givingsolutionscorrecttothenearestminutewhere necessary.(Remembertocheck x = 180◦ asapossiblesolution,giventhattheresultingequationin t is linear.)
9a Showthattheequation6sin x 4cos x = 5canbewrittenas t2 12t + 9 = 0,where t = tan 1 2 x
b Showthattan 1 2 x = 6 + 3√3or6 3√3.
c Henceshowthat77◦35′ and169◦48′ arethesolutions(tothenearestminute)oftheequationoverthe domain0◦ ≤ x ≤ 360◦
10a If t = tan22 1 2 ◦,showthat 2t 1 t2 = 1.
bi Henceshowthattan22 1 2 ◦ = √2 1.
ii Whatdoestheotherrootoftheequationrepresent?
11 Usethemethodsofthepreviousquestiontoshowthat: tan67 1 2 ◦ = √2 + 1 a
12 Solveeachequation,for0◦ ≤ x ≤ 360◦,usingthe t-formulae.Givesolutionscorrecttothenearestminute wherenecessary. 5sin x + 4cos x = 5 a 7cos x 6sin x = 2 b 3sin x 2cos x = 1 c 5cos x + 6sin x = 5 d
13 Use t-formulaetoprovethattheequation8tan θ 4sec θ = 1hassolutions θ = 2tan 1 1 3 and θ = 2tan 1 5 inthedomain0 ≤ θ ≤ 2π.
14 Let θ1 and θ2 bethesolutionsoftheequation6cos θ + 17sin θ = 18inthedomain0 ≤ θ ≤ 2π.Giventhat θ1 > θ2,showthattan θ1 θ2 2 = 1 18 .
15 Considertheequation a cos x + b sin x = c,where a, b and c areconstantsand a + c 0.
CHALLENGE
a Showthattheequationcanbewrittenintheform(a + c)t2 2bt (a c) = 0,where t = tan 1 2 x
b Showthattherootsoftheequationarerealif c2 ≤ a2 + b2
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c Supposethattan 1 2 α andtan 1 2 β aredistinctrealrootsofthequadraticequationinpart a.Provethat tan 1 2 (α + β) = b a ,assumingthat a 0.
16 Giventheequation(2k 1)cos θ + (k + 2)sin θ = 2k + 1,where k isaconstant,use t-formulaetoprovethat tan θ = 4 3 or 2k k2 1 .
17 Let θ1, θ2, θ3 and θ4 besolutionsoftheequation a cos4θ + b sin4θ = c suchthattan θ1,tan θ2,tan θ3 and tan θ4 aredistinct.Usetheproductoftherootsofaquarticequationtoprovethat tan θ1 tan θ2 tan θ3 tan θ4 = 1.
Chapter5Review
Reviewactivity
• Createyourownsummaryofthischapteronpaperorinadigitaldocument.
Chapter5Multiple-choicequiz
• Thisautomatically-markedquizisaccessedintheInteractiveTextbook.AprintablePDFWorksheetversionis alsoavailablethere.
ChapterReviewExercise
1 Writeasasumordifferenceoftwotrigonometricfunctions.
2sin50◦ cos15◦ a
◦ sin26◦ b 2cos4A sin2A c
P + 2Q)cos(3P 2Q) d
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2 Show,withoutacalculator,that:
◦ cos16◦ = 1 + 2cos28◦ a
3 Useproductstosumsidentitiestoprovethat
4 Provetheidentity4cos6
5 Usesumstoproductsidentitiestoprovethat sin6A + sin2A cos6A + cos2A =
6 Usesumstoproductsidentitiestosolvetheseequations,for0
7 Writeintermsof t,where t = tan 1 2 θ
8 Usethe t = tan 1 2 θ resultstofindtheexactvalueof:
Proveeachidentityusingthe
10 Considertheequationsin
Usethesubstitution
Hencesolvetheequationfor
Solvetheequation
3cos
12a Showthattheequation2cos x + sin x = 1canbewrittenas(3t + 1)(t 1) = 0,where t = tan 1 2 x
b Hencesolvetheequation,for0◦ ≤ x ≤ 360◦.Givethesolutionscorrecttothenearestminutewhere necessary.
13a Showthattheequation4sin x cos x = 1canbewrittenas4t 1 = 0,where t = tan 1 2 x.
b Hencesolvetheequationfor0◦ ≤ x ≤ 360◦,givingsolutionscorrecttothenearestminutewhere necessary.(Remembertocheck x = 180◦ asapossiblesolution,giventhattheresultingequationin t is linear.)
14a Showthattheequation5sin x 3cos x = 4canbewrittenas t2 10t + 7 = 0,where t = tan 1 2 x
b Showthattan 1 2 x = 5 + 3√2or5 3√2.
c Hencesolvetheequationoverthedomain0◦ ≤ x ≤ 360◦,givingsolutionscorrecttothenearestminute.
15 Solveeachequation,for0 ≤ x ≤ 2π,usingthe t-formulae.Givesolutionscorrecttofourdecimalplaces.
4cos x + 3sin x = 1 a
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7sin x 5cos x = 4 b
6 Integration
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Chapterintroduction
Theartofintegrationisaskillthatallmathematiciansmustpossess,asintegralsariseinallareasofmathematics.Forexample,intheseeminglyunrelatedtopicofprimenumberstheintegral ∫ dx log x appears.
Asintegrationisanartform,itrequiresplentyofpracticetobecomeproficient.Thusstudentsareencouraged toattemptasmanyoftheexercisequestionsaspossibleinthetimetheyhaveavailable.
TheworkinthischapterbuildsonthecontentoftheMathematicsExtension1course.Amethodicalapproach isneededtostudythematerial.Inparticular,itisimportanttobeabletorecognisethedifferentformsof integrals,andtoquicklydeterminewhichmethodisbestused.
Thefirstfivesectionsarerelativelystraightforward,beingbasedonalgebraicmanipulation.InSection 6F the newmethodofintegrationbypartsisintroduced,whichisbasedontheproductrulefordifferentiation.Section 6G coversvarioushardertypesoftrigonometricintegrals.Section 6H introducestheconceptofintegralsthat canbereferencedbyanindex,andthecorrespondingreductionformulae.Thechapterconcludeswithasetof miscellaneousquestions.Theseprovideanopportunitytopractisechoosingthemostefficientmethodtoapply.
6A TheStandardIntegrals
StudentswillknowthateachexaminationisaccompaniedbyaHSCReferenceSheet,whichincludesvarious integrals.MostofthesehavealreadybeenencounteredintheMathematicsExtension1course.Theappendixto thischapterincludesatableofsimilarintegralsaswellassomeothercommonintegralsthatmaybeencountered inthiscourse.Theabilitytomanipulateanyintegralformulainsimplewaysisexpectedofallstudents.
Example1
Evaluate
Solution
Takeoutafactorof
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Exercise6A
FOUNDATION
1 UsetheIntegralCalculussectionsoftheHSCReferenceSheettodetermineeachoftheseindefinite integrals.
2 UsetheIntegralCalculussectionsoftheReferenceSheettofind:
3 UsetheIntegralCalculussectionsoftheHSCReferenceSheettoevaluateeachofthesedefiniteintegrals.
4 Find:
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5 Evaluate:
6 Evaluate
6B AlgebraicManipulation
Manyoftheintegralsencounteredcontainfractionswhichrequiresomesortofrearrangementbeforeproceeding. Inthefirstexamplethatfollowsthenumeratorisalmostidenticaltothedenominator.
Example2
Determine ∫ x2 1 x2 + 1 dx.
Notingthat x2 1 = (x2 + 1) 2thefractionmaybeseparated.
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Inharderproblemslongdivisionmayberequired.Insomecasesthenumeratorisclosetoamultipleofthe denominator,asinthenextworkedexample.
Example3
Find
UsingaCommonFactor
Somerationalfunctionsareeasiertodealwithaftermultiplicationordivisionbyacommonfactor.Theresultisa numeratorwhichisthederivativeofthedenominator.
Example4
Evaluate
Solution
Multiplynumeratoranddenominatorby e x toget:
∫ 1 1 e2x 1 e2x + 1 dx = ∫ 1 1 ex e x ex + e x dx = log(ex + e x) 1 1 = log(e + e 1) log(e 1 + e) = 0.(Whyisthisobviousfromthegraph?) x y
Noticeinthesolutionthattheprimitivelog(ex + e x)iswrittenwithoutanabsolutevalue.Thisisbecause ex + e x isalwayspositive,andhencetheabsolutevaluefunctionisredundant.Inthistext,theabsolutevaluewill normallybeomittedwheneveritisredundant.
Exercise6B FOUNDATION
1 Determinethefollowingbyrewritingthenumeratorintermsofthedenominator.
2 Evaluatethefollowing.
3 Evaluatethefollowing.Ineachcase,beginbyrewritingthegivenfractionastwofractionsbyseparatingthe termsinthenumerator.
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4a Let y = ln(x + √x2 + a2).Findandsimplify dy dx
b Hencefindaformulafor
c Usethisformulatodetermine:
5a Giventhat x3
b Giventhat x3 = (x3 + 1) 1andthat x
determine ∫ x3 x + 1 dx
c Usesimilarapproachestothoseshowninparts a and b todeterminethefollowing.
6 Evaluatethesebyfirstmultiplyingordividingbyanappropriatefactor.
7 Byusinglongdivisionorotherwise,determine:
8a Let y = ln(x + √x2 a2),where x > |a|.Findandsimplify dy dx .
b Hencefindaformulafor
c Usethisformulatodetermine:
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9 Dividenumeratoranddenominatorbyanappropriatefactortohelpdetermine
b Henceorotherwiseshowthat
c Explainwhytheabsolutevalueintheprimitiveissuperfluous.
6C Substitution
Manyofthetechniquesusedinintegrationarederivedfromdifferentiation.Thisisnotsosurprisingsincethe twoprocessesareessentiallymutuallyinverse.Oneparticularlyusefultechniqueissubstitutionwhichisthe integrationequivalentofthechainrulefordifferentiation,andissometimescalledthereversechainrule.
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TheIntegralFormoftheChainRule
Supposethat F isafunctionof u,whichisinturnafunctionof x.Furthersupposethat F(u)isaprimitiveof f (u). Differentiating F withrespectto x andapplyingthechainrulegives: d dx F(u) = dF du × du dx so d dx F(u) = f (u) × u′ .
Swappingsidesandintegratingthisresult
∫ f (u) × u′ dx = ∫ d dx F(u) dx or ∫ f (u) × u′ dx = F(u) + C
Itisthislastresultwhichprovesmostusefulforintegration.Thusifanintegrandcanbeexpressedasaproduct, whereonefactorisachainoffunctions f (u)andtheotherfactoris u′ thentheprimitivecanimmediatelybe writtendown.
1Theintegralformofthechainrule
If F(u)isaprimitiveof f (u)then ∫ f (u) × u′ dx = F(u) + C
Substitutions
Inthesimplestexamples,theprimitivecanbedeterminedmentally.Forexample,aneasyintegralinvolvingthe exponentialfunctionis ∫ 2xex2 dx = ex2 + C.
Inharderexamplesaformalprocedureshouldbefollowed.
Example5
Determine ∫ x2 √x3 + 1 dx byasuitablesubstitution.
Solution
Let I = ∫ x2 √x3 + 1 dx andput u = x3 + 1,then du dx = 3x2 or 1 3 du = x2 dx (treatingthederivativelikeafraction.)
Thus I = ∫ 1 3√u du = 2 3 √u + C. Hence I = 2 3 √x3 + 1 + C
Noticethatthefinalstepofthesolutionisabacksubstitutiontogettheintegral I intermsof x.Itisimportant toremembertodothis.
SubstitutionsandDefiniteIntegrals
Itisequallyimportanttofollowthisformalprocedurewhendefiniteintegralsareinvolved,payingparticular attentiontothelimitsofintegration.However,thebacksubstitutionstepisnotneeded.
Example6
Useasuitablesubstitutiontofind ∫ π 2 0 sin x (1 + cos x)3 dx.
Solution
Let I = ∫ π 2 0 sin x (1 + cos x)3 dx andput u = 1 + cos x toget du dx = sin x so du = sin xdx. When x = 0, u = 2, andwhen x = π 2 , u = 1, thus I =
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Thestepwherethelimitsareexpressedintermsofthesubstitutevariableisimportant.Hadthisstepnotbeen donethenthewronganswerwouldhavebeenobtainedsince
whichisundefinedatthelowerlimit.Againnoticethatthederivativeistreatedlikeafractioninthethirdline ofthesolution.
HarderExamples
Inexampleslikethoseabove,candidatesareexpectedtodeterminetheappropriatesubstitutionforthemselves.In harderproblemsthesubstitutionwillbegiven.Thechainrulemayhelpwhenfindingitsderivative.
Example7
Usethesubstitution u = √x tofind I = ∫ 1 x + √x dx
Solution
Write u2 = x.Then,differentiatingbothsides, d dx (u2) = d dx (x)
so2u du dx = 1(bythechainruleontheLHS) or2udu = dx.
Continuedonthenextpage
Hence I = ∫ 2udu
u2 + u = ∫ 2 du u + 1 = 2ln(u + 1) + C = 2ln(√x + 1) + C
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TakeCarewithSubstitutions
Therearemanyintegralswhichrequireacarefulchoiceofsubstitutionsoastoavoidsubsequentdifficulties.For example,thecorrectchoiceofsubstitutioninthepreviousexampleis u = √x Onfirstinspection,itwouldseemtomakenodifferencetomakethealternatesubstitution u2 = x,however observewhathappensinthedenominator. x + √x = u 2 + √u2 = u 2 + |u|.
Thusinthiscaseanewcomplicationhasbeenintroduced,namelytheabsolutevaluefunction.Ingeneral,the bestchoiceofsubstitutionisoftheform u = f (x).
Insomeinstancesitismorenaturaltouseasubstitutionoftheform x = g(u).Thenthedomainof u isrestricted toavoidanylatercomplication.Thisisoftenthecasewithtrigonometricsubstitutions.
Example8
Evaluate ∫ 1 0 √4 x2 dx usingasuitablesubstitution.
Solution
Theintegrandistheuppersemi-circlewithradius2,soput x = 2sin θ with
Thismeansthatcos θ ≥ 0.
Differentiating dx = 2cos
When x = 0, θ = 0, andwhen x = 1, θ =
Thus I =
(bythePythagoreanidentity)
(sincecos
(bythedouble-angleformula)
Noticethatinthesolution √cos2 θ = | cos θ|,butthisisfurthersimplifiedtocos θ sinceitisknownthat cos θ ≥ 0inthespecifieddomain.Furtherobservethatthesameoutcomewouldhavebeenachievedby substituting θ = sin 1 x 2 ,sincetherangeofinversesineisalso π 2 ≤ θ ≤ π 2 .Atrigonometricsubstitution shouldalwaysbemadeinthismanner,withtheinversefunctioninmind.
TwoGuidelinesforSubstitutions
Theinfinitevarietyofintegralsthatmaybeencounteredmakesitimpracticaltogiveaspecificrecipefor choosingthecorrectsubstitution.Howeverthefollowingtwoguidelinesmayhelp,andcanbeobservedin practiceinthepreviousexamples.
• Trytoreplacethepartoftheintegralwhichcausesdifficulty,suchastheinnermostfunctioninachain offunctions.Inparticular,iftheintegralinvolvessquare-rootsofsumsordifferencesofsquaresthena trigonometricsubstitutionislikelytowork.
• Itisbettertouseasubstitutionwhichisafunction u = f (x)ratherthanarelation x = g(u).Ifarelationmust beusedthenitisoftenbesttorestrictthedomaininasimilarmannertoaninversefunction.
Exercise6C FOUNDATION
1 Findtheseintegralsbythereversechainrule,thendothemagainusingasuitablesubstitution.
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2 Useasuitablesubstitutionwherenecessarytofind:
3 Useasuitablesubstitutionwherenecessarytoevaluate:
4a Useasuitablesubstitutiontohelpevaluate
b Howcouldthisintegralhavebeenevaluatedusingjustalgebraicmanipulation?
5 Usethegivensubstitutiontofind:
6 Ineachcase,usethegivensubstitutiontoevaluatetheintegral.
7 Ineachcase,usethegivensubstitutiontodeterminetheprimitive.
8 Ineachcaseusethegiventrigonometricsubstitutiontoevaluatetheintegral.Youmayassumethat 0 < θ < π 2
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9 Ineachcaseusethegiventrigonometricsubstitutiontoevaluatetheintegral.Youwillneedtheresult
10a Useasuitablesubstitutiontohelpevaluate
b Howcouldthisintegralhavebeenevaluatedusingjustalgebraicmanipulation? 11a Useasuitablesubstitutiontoshowthat
b Redothisproblembygeometricmeans.
12 Itisgiventhat ∫ π 0 cos x x + 1 dx = A,where A isaconstant.
13 Let I = ∫ π 2 0 sin x sin x + cos x dx
Usethesubstitution u = π 2 x
a Byaddingthetwoequalintegrals,findthevalueof I b
14a Showthat
b Let I = ∫
I i Henceevaluate I. ii
15a Useatrigonometricsubstitutiontoshowthat
b Howcouldthisintegralhavebeenevaluatedusingalgebrathengeometry?
CHALLENGE
16 Considertheindefiniteintegral I = ∫ dx x√x2 1 .Clearlythedomainoftheintegrandisdisjoint,being x > 1 or x < 1.Thusitseemsappropriatetouseadifferentsubstitutionineachpartofthedomain.
Find I for x > 1byusingthesubstitution u = √x2 1. a
Find I for x < 1byusingthesubstitution u = √x2 1. b
17a Useasuitablesubstitutiontodetermine ∫ 4 2+ε dx x2 √x2 4 ,where ε > 0. b Takethelimitofthisresultas ε → 0+ andhencefind ∫ 4 2 dx x2 √x2 4
6D PartialFractions
Inarithmetic,whengiventhesumoftwofractions,thenormalprocedureistocombinethemintoasingle fractionusingthelowestcommondenominator.Thus
1 3 + 1 2 = 5 6 .
Unfortunatelywhenthefractionsarefunctionsandintegrationisinvolved,thisisexactlythewrongthingtodo. Whilstitistruethat
3 x + 2 + 2 x 1 = 5x + 1 x2 + x 2 , whenconsideringthecorrespondingintegrals,
itshouldbeclearthatthelefthandsideisfarsimplertodeterminethantherighthandintegral.So:
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Thisexampleistypicalofintegralsofrationalfunctions.Itiseasiesttofirstsplitthefractionintoitssimpler components.Inmathematicalterminology,thefractionis decomposedintoitspartialfractions
IntegralsLeadingtoLogarithms
Manyoftheintegralsinthissection,andindeedthewholechapter,resultinthelogarithmofanabsolutevalue,as intheaboveexample.Thecarefulreaderwillnoticeaproblemwithwritingthingslike
1 x 3 dx = log |x 3| + C sinceitispossiblethatthevalueof C differsbetween x < 3and x > 3.Suchpeculiaritieswillnotbeconsidered here.
ATheoremAboutPartialFractions
Considertherationalfunction
P(x)
A(x) × B(x) , where P, A and B arepolynomials,withnocommonfactorsbetweenanypair,andwheredeg P < deg A + deg B Itisalwayspossibletowrite
P(x)
A(x) × B(x) = R
B
x) , wheretheremainders RA and RB arepolynomialswithdeg RA < deg A anddeg RB < deg B.However,theproofis beyondthescopeofthiscourse.
LinearFactors
Inthesimplestexamples, A(x)and B(x)arelinear.Sincethedegreesof RA and RB areless,theymustbe constants,yettobefound.
Example9
a Decompose x + 1 (x 1)(x + 3) intoitspartialfractions.
b Henceevaluate ∫ 6 2 x + 1 (x 1)(x + 3) dx
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Solution
a Let x + 1 (x 1)(x + 3) = A x 1 + B x + 3 ,where A and B areunknownconstants.Multiplythisequationby (x 1)(x + 3)toget: x + 1 = A(x + 3) + B(x 1)
or x + 1 = (A + B)x + (3A B).
Equatingcoefficientsoflikepowersof x yieldsthesimultaneousequations A + B = 1 3A B = 1.
Thesecanbesolvedmentallytoget A = 1 2 and B = 1 2 .Thus x + 1 (x 1)(x + 3) = ( 1 2 ) x 1 + ( 1 2 ) x + 3
b Hence ∫ 6 2 x + 1 (x 1)(x + 3) dx = 1 2 ∫ 6 2 1 x 1 + 1 x + 3 dx
= 1 2 ln(x 1) + ln(x + 3) 6 2 = 1 2 (ln5 + ln9) (ln1 + ln5) = ln3. y x
Thismethodofequatingcoefficientsoflikepowersof x isusuallyonlyconvenientinstraightforward examples.
FindingtheConstantsbySubstitution
Anothermethodoffindingtheunknownconstantsusessubstitution.Inmanycasesitisalsoaquickermethod.
Example10
Decompose 3x 5 (x 3)(x + 1) intopartialfractions.
Solution
Let 3x 5 (x 3)(x + 1) = A x 3 + B x + 1 ,where A and B areconstants, then3x 5 = A(x + 1) + B(x 3)(multiplyby(x 3)(x + 1)).
When x = 3,4 = 4A
so A = 1.
When x = 1, 8 = 4B
so B = 2.
Continuedonthenextpage
Thus 3x 5 (x 3)(x + 1) = 1 x 3 + 2 x + 1
Thecarefulreaderwillhavenoticedapointofcontentionwiththesolution.Thefractionisundefinedwhen x = 3andwhen x = 1,yetthesevalueswereusedinthesubstitutionsteps.Howcanthisbevalid?The answeristhatsomeofthedetailofthesolutionhasbeenomitted.Hereisamorecompleteexplanation.
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Since 3x 5 (x 3)(x + 1) = A x 3 + B x + 1 where x 1,3, itfollowsthat 3x 5 = A(x + 1) + B(x 3)where x 1,3.
Nowthislastequationistruewhenever x 1,3.Thatis,itisalinearequationwhichistrueforatleasttwo othervaluesof x.Henceitisanidentity,andsoitistrueforall x,including x = 1and x = 3.Thusthese valuescanbesubstitutedtodetermine A and B.Itisnotnecessarytogivethiscompleteexplanationaspartof asolution,butstudentsshouldbeawareofit.
NumeratorswithHigherDegree
Inslightlyharderproblems,thedegreeofthenumeratorisgreaterthanorequaltothedegreeofthedenominator. Insuchcases,thefractionshouldbeexpressedasasumofapolynomialandthepartialfractions.Longdivision maybeusedatthisstep,butitisofteneasiertouseapolynomialwithunknowncoefficients,asinthefollowing example.
Example11
Determine
Solution
Firstnotethat
TheCover-upRule
Thereisanevenquickermethodtodeterminetheconstantsofthepartialfractions,providedthattheoriginal denominatorisaproductofdistinctlinearfactors,andprovidedthatthedegreeofthenumeratorislessthanthe degreeofthedenominator.Thetrickistomultiplybyjustonelinearfactoratatime.
Example12
Express 7 5x (x + 1)(x 2)(x 3) inpartialfractionsform.
Solution
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+ 1)gives
Thismethodoffindingtheconstantsissometimescalledthe cover-uprule.Lookcarefullyathowthethree constantsaredetermined.Foreachconstant,thematchinglinearfactoriseffectivelyomitted,or“coveredup”. Thusfor C1,(x + 1)isleftoutoftheoriginalfraction.For C2,(x 2)isexcluded,andfor C3,(x 3)isomitted fromtheoriginalfraction.Ineachcase,theresultingrationalfunctionisthenevaluatedatthecorresponding valueof x.Withpractice,moststudentsshouldbeabletodeterminetheconstantsmentallyusingthismethod.
Enrichment—ProofoftheCover-upRule
Hereisaproofforthegeneralcase.
Proof Considertherationalfunction P(x) Q(x) wheredeg P < deg Q,andwhere Q(x)isaproductofdistinctlinear factors,thatis
notetheuseofproductnotation, . Let
Multiplythislastequationby(x ak )toget
Nowtakethelimitas x → ak .Alltermsexcept Ck ontherighthandsidearezeroandso: Ck = limx→ak P(x)(x ak ) Q(x) = limx→ak P(x)
n i=1 i k (x ai) (thatis,cancelthe kthlinearfactor) hence Ck = P(ak )
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Themathematicalnotationmayseemdifficult,buttheresultisexactlyasbefore.Togetthe kth coefficient Ck ,omitthe kthlinearfactorfromthedenominatorandevaluatetherestofthefractionat x = ak .
QuadraticFactors
Incertaininstances,thedenominatoroftherationalfunctionbeingconsideredwillhaveaquadraticfactorwith norealzero.Forexample,in
3x + 10
(x 2)(x2 + 4)
thequadraticfactor(x2 + 4)hasnorealzero.Thusthedenominatoroftherationalfunctioncannotbeexpressed asaproductofreallinearfactors.
Nevertheless,themethodforfindingthepartialfractiondecompositionremainsessentiallythesame.Theonly requirementisthatthedegreeofthenumeratorislessthanthedegreeofthedenominator.Thus,itfollowsthatfor anyquadraticfactorthenumeratorisanunknownlinearpolynomial.
Example13
a Rewrite 3x + 10 (x 2)(x2 + 4) initspartialfractions.
b Hencedetermine ∫ 3x + 10 (x 2)(x2 + 4) dx
Solution
a Let 3x + 10 (x 2)(x2 + 4) = A x 2 + Bx + C x2 + 4 ,where A, B and C areunknownconstants.Then 3x + 10 = A(x2 + 4) + (Bx + C)(x 2)
At x = 2 16 = 8A so A = 2.
Equatingcoefficientsof x2 yields 0 = 2 + B so B = 2.
At x = 0 10 = 8 2C so C = 1.
Thus 3x + 10 (x 2)(x2 + 4) = 2 x 2 2x + 1 x2 + 4 .
b Hence ∫ 3x + 10 (x 2)(x2 + 4) dx = ∫ 2 x 2 2x x2 + 4 1 x2 + 4 dx = 2ln |x 2|− ln |x 2 + 4|− 1 2 tan 1( x 2 ) + C
Hereisasummaryofthetechniquesmetsofarinthissection.
2Partialfractions
• Eachnumeratorshouldhavedegreeonelessthanitsdenominator.
• Usethecoveruprulewhentherearedistinctlinearfactors.
• Substitution,simultaneousequationsorequatingcoefficientsmayalsobeusedtodetermineunknown constants.
• Useapolynomialwithunknowncoefficientsorlongdivisionfornumeratorswithhigherdegree.
DenominatorswithRepeatedFactors
Whenafactorofapolynomialhasindexgreaterthanoneitiscalledarepeatedfactor.Forexample,inthe fraction
x2 + 5x + 4 (x 3)2(x + 1) , thefactor(x 3)isarepeatedfactorsinceithasindex2. WritingaseparatefractionforeachdistinctfactorandapplyingtheguidelinesinBox 2 wouldgivethefollowing partialfractiondecomposition.
x2 + 5x + 4 (x 3)2(x + 1) = ax + b (x 3)2 + c x + 1
Thelinearnumerator ax + b occursbecausethedenominatorisaquadratic.Thisexpressioncanbere-arrangedas follows.
x2 + 5x + 4 (x 3)2(x + 1) = a(x 3) + (b + 3a) (x 3)2 + c x + 1 = a x 3 + (b + 3a) (x 3)2 + c x + 1 = A x
Noticethatthefinallineincludesafractionforeachofthedenominators(x 3)and(x 3)2.Thatis,for increasingpowersof(x 3)beginningat1andendingat2,theindexforthegivenfraction.Eachnumeratorisan unknownconstant.
Itisbettertowritethedecompositionforrepeatedfactorsinthiswaywhendoingintegration.Noexplanationis required.Thus,ifthefactor(x a)isrepeatedthenincludeseparatefractionsforincreasingpowersofthatfactor, beginningat1andendingattheindexforthegivenfraction.Eachnumeratorisanunknownconstant.Then proceedintheusualway.
Example14
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a Findthepartialfractiondecompositionfor 8 x (x 2)2(x + 1) .
b Henceevaluate ∫ 1 0 8 x (x 2)2(x + 1) dx
Solution a Let
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Coefficientof x2:0 = A + C so A =
b Hence
3Denominatorswithrepeatedfactors
Ifthefactor(x a)isrepeatedthenincludeseparatefractionsforincreasingpowersofthatfactor, beginningat1andendingattheindexforthegivenfraction.Eachnumeratorisanunknownconstant. Thenproceedintheusualway.
Exercise6D FOUNDATION
1 Decomposethefollowingfractionsintopartialfractions.
2 Find:
3 Find:
4 Find:
5 Find:
6 Findtheexactvalueof:
7 Findtheexactvalueof:
8 Findtheexactvalueof:
9 Findtheexactvalueof:
10 Showthat:
Showthat
12 Usepartialfractionstofind:
13ai Let 2x2 + 1 (x 1)(x + 2) = A + B x 1 + C x + 2 .Findthevaluesof A, B and C. ii Hencefind ∫ 2x2 + 1 (x 1)(x + 2) dx
b Useasimilartechniquetopart a inordertofind: ∫ x2 2x + 3 (x + 1)(x 2) dx i
14a Findthevaluesof A, B and C suchthat 3x2 10 x2 4x + 4 = A + B x 2 + C (x 2)2
b Hencefind
3x2 66 (x + 4)(x 5) dx ii
DEVELOPMENT
3x2 10 x2 4x + 4 dx.
15ai Findthevaluesof A, B, C and D suchthat x3 3x2 4 (x + 1)(x 3) = Ax + B + C x + 1 + D x 3
ii Henceevaluate ∫ 1 0 x3 3x2 4 (x + 1)(x 3) dx.
b Useasimilarmethodtoevaluate ∫ 4 2 x3 + 4x2 + x 3 (x + 2)(x 1) dx
16a Findthevaluesof A, B, C and D suchthat 2x3 + x2 + 3x + 1 (x2 + 1)2 = Ax + B x2 + 1 + Cx + D (x2 + 1)2
b Hencefind ∫ 2x3 + x2 + 3x + 1 (x2 + 1)2 dx
17 Useappropriatemethodstofind:
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18 Find ∫ 8 (x + 1)3(x + 3) dx.
19a Inthenotationofthetext,if Q(x)isaproductofdistinctlinearfactorsthen:
Usethistoprovetheformula
b UsethisformulatoredoQuestion 6b and 7b.
:Whatisthevalueof Q(ak )?]
6E DenominatorswithQuadratics
Manypracticalapplicationsyieldintegralswithaquadraticinthedenominator.Itisoftensimplyamatterof applyingthefollowingtworesults:
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Anothercommonintegralis ∫ dx x2 a2 .Althoughaformulaexistsforthis,itisnotpartofthecourse.Itis expectedthatcandidatesdeterminetheprimitivebyuseofpartialfractionswheneverthistypeofintegralis encountered.
Example15
Evaluate ∫ 1 1 4 x2 4 dx.
Now 4 x2 4 = 4 (x 2)(x + 2) , solet 4 (x 2)(x + 2) = A x 2 +
QuadraticswithLinearTerms
Frequentlythequadraticwillhavealinearterm,suchasin3 + 2x x2.Intheseinstancesthemethodisto completethesquaretoobtainthedifferenceoftwosquares.
Example16
Find
Completingthesquareinthedenominator:
Thesolutionusesasubstitution.ThisstepmaybeomittedbyusingaresultfromtheExtension1course. Continuedonthenextpage
Recallthatif F(x)isaprimitiveof f (x)then
∫ f (ax + b) dx = 1 a F(ax + b) + C.
Theresultisacombinationofshiftingandstretchingalongthe x-axis.Ashiftdoesnotaffecttheareaundera graph,butastretchdoes,hencethefactor 1 a .
Inthisexample, f (x) = 1 √4 x2 ,theprimitiveis F(x) = sin 1 x 2 ,with a = 1and b = 1.Thusitispermissible towrite
1 4 (x 1)2 dx = sin 1 x 1 2 + C, withoutshowinganyworking.Example 17 issimilar.
Example17
Findthevalueof ∫ 1 1 9 7 + 4x + x2 dx
Solution
Completingthesquareinthedenominator:
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4Quadraticswithlinearterms
Completethesquare,thenusetheresult
)istheprimitiveof f (x).
LinearNumerators
Sofarinalltheexamplesthenumeratorhasbeenaconstant.Whenthenumeratorislinearitisbesttocarefully splititintotwoparts.Thefirsttermshouldbeamultipleofthederivativeofthequadraticinthedenominator. Thesecondtermwillthenbeaconstant.
Example18
Determine ∫ 4x + 3 x2 + 9 dx.
Inharderexamplesthequadraticwillalsocontainalinearterm.
Example19
Evaluate
5Linearnumerators
Whenthenumeratorislinearitisbesttosplititintoamultipleofthederivativeofthequadraticinthe denominatorplusaconstant.
RationalisingtheNumerator
Inpreviousworkithasbeenconvenienttorationalisethedenominatorwhenasurdappears.Incontrast,when calculusisinvolveditisoftenmoreconvenienttorationalisethenumeratorinstead.
Example20 Find
Rationalisingthenumerator
Noticeinthefirstlineofworking,byrationalising,thenumeratorhasbecomelinear.Thisistypicalofthe questionsdoneinthissection.
6Rationalisingthenumerator
Whencalculusisinvolveditisoftenconvenienttorationalisethenumerator.
Exercise6E
1 Findtheseintegrals.
2 Determinethefollowing.
3 Evaluatethefollowing.
4 Find:
5 Findthevalueof:
6 Determineeachprimitive.
7 Evaluate:
8a Whyisitnotvalidtoevaluate
Nevertheless,showthatitsvalueislim
9a Showthat3
Henceorotherwiseshowthat
usingthetechniquesofthissection?
FOUNDATION
CHALLENGE
6F IntegrationbyParts
Whilsttherearewellknownandrelativelysimpleformulaeforthederivativesofproductsandquotientsof functions,therearenosuchgeneralformulaefortheintegralsofproductsandquotients.Nevertheless,aswas foundintheprevioustwosections,certainquotientscanbeintegratedrelativelyeasily.Inthissection,amethod ofintegrationisdevelopedthatcanbeappliedtocertaintypesofproducts.Itbeginswiththeproductrulefor differentiation.
Now d dx (uv) = u′v + uv′ .
Swappingsidesandintegratingyields
∫ u′vdx + ∫ uv′ dx = uv, hence ∫ uv′ dx = uv ∫ u′vdx
Thislastequationprovidesawaytorearrangeanintegralofoneproductintoanintegralofadifferentproduct. Theformulaisappliedwiththeaimthatthenewintegralisinsomewaysimpler.Theprocessiscalled integration byparts
Example21
Useintegrationbypartstofind ∫ xex dx
Solution
Let I = ∫ xex dx = ∫ uv′ dx, where u = x and v′
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=
x ex + C or I = ex(x 1) + C.
Noticethelackofanyconstantofintegrationuntiltheprocessisfinished.
7Integrationbyparts
Theintegraloftheproduct uv′ canberearrangedtoget
whichiscalledintegrationbyparts.
ReducingPolynomials
Whenoneofthefactorsoftheintegrandisapolynomial,itiscommontolet u bethatpolynomial.Inthatwaythe newintegral,whichdependson u′,willcontainapolynomialoflesserdegree.Thatis,theaimistoreducethe degreeofthepolynomial.
Example22
Evaluate ∫ π 0 (x + 1)sin xdx
Solution
Let I = ∫ π 0 (x + 1)sin xdx = ∫ π 0 uv′ dx,
where u = (x + 1)and v′ = sin x
so u′ = 1and v = cos x
Thus I = uv π 0 ∫ π 0 u′vdx
= (x + 1)cos x π 0 + ∫ π 0 cos xdx
= (π + 1) + 1 + sin x π 0 ,
hence I = π + 2. y x
8Reducingpolynomials
Integrationbypartsmaybeusedtoreducethedegreeofapolynomial.Let u bethatpolynomialand v′ the otherfactor.
RepeatedApplications
Itmaybenecessarytoapplyintegrationbypartsmorethanonceinordertocompletetheprocessofintegration. Insimplerexamplesitmaybepossibletodosomeofthestepsmentally.
Example23
Evaluate ∫ 1 0 x2e x dx.
Solution
Let I = ∫ 1 0 x2e x dx andput u = x2 and v′ = e x
so u′ = 2x and v = e x
Then I = x2e x 1 0 + ∫ 1 0 2xe x dx (byparts.)
Thesecondtermisanotherintegralofaproduct.
Soput u = 2x and v′ = e x with u′ = 2and v = e x .
Thus I = e 1 + 2
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(bypartsagain)
y x e
ExceptionswithPolynomials
Althoughitiscommontoreducethedegreeofapolynomialusingintegrationbyparts,therearemany exceptions.Inthiscoursetheseexceptionstypicallyinvolvethelogarithmfunction.
Example24
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Determine ∫ x log xdx. Solution
Let I = ∫ x log xdx andput u = log x and v′ = x
Thus I = 1 2 x2 log x
(byparts) = 1 2 x2 log x ∫ 1 2 xdx = 1 2 x2 log x 1 4 x2 + C or I = 1 4 x2(2log x 1) + C
Integrandswhere v ′ = 1
Theprimenumber5hasonlytwodistinctfactors,namely1 × 5.Afunctionmaybetreatedlikeaprimeina similarway:
Thissomewhatartificialformoffactoringisappliedtofacilitateintegrationbyparts.Itisthenusualtoput u equaltothefunctionand v′ = 1.
Example25
Findthevalueof ∫ 1 2 0 sin 1 xdx. Solution
Let I = ∫ 1 2 0 sin 1 xdx = ∫ 1 2 0 1 × sin 1 xdx Put u = sin 1 x and v′ = 1 so u′ = 1 √1 x2 and v = x. Thus
TheImportanceofaDiagram
Thecarefulreaderwillhaveseenthatthereisasimplerwaytodothelastexample.Thekeyisinthediagram. Theareashadedisthedifferencebetweentheareasoftherectangle,width 1 2 andheight π 6 ,andtheunshaded portion.Thatportioninvolvesasimpleintegralalongthe y-axis.
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Thisexamplehighlightstheimportanceofadiagramtoseethemostefficientsolution.Inthiscase,itwouldbe bettertointegratealongthe y-axisthantouseintegrationbyparts.Itisleftasanexercisetoshowthattheresult isthesame.
ARecurrenceoftheIntegral
Integrationbypartsmayleadtoarecurrenceoftheoriginalintegral.Itisthensimplyamatterofcollectinglike terms.
Example26
Findaprimitiveof ex sin x
Solution
Let I = ∫ ex sin xdx andput u = sin x and v′ = ex
so u′ = cos x and v = ex .
Then I = ex sin x ∫ ex cos xdx (byparts.)
Nowput u = cos x and v′ = ex so u′ = sin x and v = ex
Thus I = ex sin x ex cos x + ∫ ex sin xdx (bypartsagain)
= ex(sin x cos x) I
So2I = ex(sin x cos x) givesonepossibleprimitive, hence I = 1 2 ex(sin x cos x) + C isthegeneralprimitive.
Inthisexampleitwasimportanttoapplythemethodconsistently.Noticethat u wasalwaysthetrigonometric functionand v′ wasalwaystheexponentialfunction.Asanexercisetohighlightthesignificanceofthese choices,repeattheworkedexercisebutput u = ex and v′ = cos x atthesecondintegrationbyparts.
Asafinalnote,thereisnoconstantofintegrationinthesecondlastlineofthesolutiontoExample 26,yetthe lastlineincludesaconstant.Thereisnothingtobealarmedabouthere.Theprocesshasledtoaspecificprimitive I = 1 2 ex(sin x cos x)and,aswithanyindefiniteintegral,thegeneralprimitiveisthenobtainedbysimplyadding aconstantatthelaststep.
Exercise6F FOUNDATION
1 Find:
2 Evaluate:
3 Useintegrationbypartswith v′ = 1tofind:
4 Findthevalueof:
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5 Ineachcaseuseintegrationbypartstoincreasethepowerof x.
DEVELOPMENT
6 Userepeatedapplicationsofintegrationbypartsinordertofind:
7 Theseintegralsaremorenaturallydonebysubstitution.However,theycanalsobedonebyparts.Use integrationbypartsandthenredoeachintegralusingasuitablesubstitution,inordertocomparethe efficiencyandeaseofeachmethod.
8 Determine:
9 Evaluate:
11 Showthat:
13 Usetrigonometricidentitiesandthenintegrationbypartstoshowthat:
14 Determineformulaeforthefollowing:
15 Evaluate:
6G TrigonometricIntegrals
Trigonometricintegralsarisefrequentlyinpracticalapplications.Thissectioncontainsthoseintegralsmore commonlyencountered,andisgroupedbytype.
PowersofCosineandSine
Therearetwomethodsfortheintegral
∫ cosm x sinn xdx (where m ≥ 0and n ≥ 0areintegers) dependingonwhethertheconstants m and n areoddoreven.Ifbothareeventhenitisbesttousethedouble angleidentities.
Example27
Evaluate ∫ π 2 0 4cos2 x sin2 xdx
Solution
Applythedoubleangleformulaforsinetoget:
π 2 0 4cos2 x sin2 xdx = ∫ π 2 0 sin2 2xdx = 1 2 ∫ π 2 0 (1 cos4x) dx (doubleangleformula) = 1 2 x 1 4 sin4x π 2 0
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Inthesecondmethodoneorbothof m and n isodd.Workwithcosineif m isodd,otherwiseworkwith sine.Theoddindexofthechosentrigonometricfunctioncanbereducedto1viathePythagoreanidentity, cos2 x + sin2 x = 1.Itisthenamatterofmakingasubstitutionfortheothertrigonometricfunction.Theresultisa polynomialintegral.
Example28
Determine ∫ cos3 x sin2 xdx
Solution
Let I = ∫ cos3 x sin2 xdx
= ∫ cos x(1 sin2 x)sin2 xdx (byPythagoras).
Put u = sin x, sothat du = cos xdx, then I = ∫ (1 u2)u2 du
= ∫ u2 u4 du
= 1 3 u3 1 5 u5 + C
= 1 3 sin3 x 1 5 sin5 x + C
9Powersofcosineandsine
Givenanintegraloftheform ∫ cosm x sinn xdx:
• if m and n arebotheventhenusethedoubleangleidentities,
• ifeither m or n isoddthenusethePythagoreanidentityandasubstitution.
PowersofSecantandTangent
Therearethreegeneralmethodsfortheintegral
∫ secm x tann xdx (where m ≥ 0and n ≥ 0areintegers), againdependingonwhethertheconstants m and n areoddoreven.Therearealsotwospecialcaseswhichshould bedealtwithfirst.
When m = 0and n = 1thesituationistrivial,viz:
∫ tan xdx = ∫ sin x cos x dx = ln | cos x| + C.
Averyclevertrickisrequiredfortheotherspecialcasewhen m = 1and n = 0.
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Noticethatinbothspecialcasestheresultisalogarithmicfunctionsincethenumeratoroftheintegrandisthe derivativeofthedenominator.
10Theintegralsofthetangentandsecantfunctions
∫ tan xdx = ln | cos x| + C
∫ sec xdx = ln | sec x + tan x| + C
Alternatively,the t-substitution,whichisreviewedlaterinthissection,canbeusedtofindtheprimitiveofsec x. Aquestionintheexercisedealswiththis.Theprimitiveoftan x canalsobefoundbythe t-substitution,butitis notrecommendedasitisinefficient.
Nowconsidertheremainingcasesofsecm x tann x.If m and n arebotheventhenseparateoutafactorofsec2 x andsubstitute u = tan x.Theresultisapolynomialintegral.ThePythagoreanidentity1 + tan2 x = sec2 x maybe requiredbeforeapplyingthesubstitution.Thefollowingtwoexamplesdemonstrateallthis,with m = 0inthefirst instance.
Example29
Find ∫ tan4 xdx
Solution ∫ tan4 xdx
= ∫ tan2 x sec2 xdx ∫ tan2 xdx (byPythagoras)
= ∫ tan2 x sec2 xdx ∫ sec2 x 1 dx (byPythagorasagain)
= ∫ u2 du ∫ sec2 xdx + ∫ 1 dx (where u = tan x)
= 1 3 u3 tan x + x + C = 1 3 tan3 x tan x + x + C.
Example30
Showthat ∫ π 4 0 sec4 x tan2 xdx = 8 15
Solution
Let I = ∫ π 4 0 sec4 x tan2 xdx
so I = ∫ π 4 0 sec2 x (tan2 x + 1)tan2 xdx (byPythagoras.)
Nowput u = tan x toget, I = ∫ 1 0 (u2 + 1)u2 du
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If n isoddthenfactoroutthetermsec x tan x andsubstitute u = sec x toobtainapolynomialintegral.The Pythagoreanidentitymayberequired.
Example31
Determinethevalueof ∫ π 3 0 sec3 x tan xdx
Solution
Let I = ∫ π 3 0 sec3 x tan xdx, so I = ∫ π 3 0 sec2 x × sec x tan xdx Put u =
Whenever m isoddand n isevenitisbesttointegratebyparts.OnceagainthePythagoreanidentitymaybe required.Theoriginalintegralmayalsore-appear,asinthefollowingexample.
Example32
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Find ∫ sec3 xdx. Solution
Let I = ∫ sec3 xdx = ∫ sec2 x × sec xdx.
Put u = sec x and v′ = sec2 x
so u′ = sec x tan x and v = tan x
Then I = sec x tan x ∫ sec x tan2 xdx (byparts)
= sec x tan x ∫ sec x(sec2 x 1) dx (byPythagoras.)
Thus I = sec x tan x ∫ sec3 xdx + ∫ sec xdx,
so I = sec x tan x I + log | sec x + tan x| (fromthespecialcase)
or2I = sec x tan x + log | sec x + tan x|, hence I = 1 2 sec x tan x + log | sec x + tan x| + C
11Powersofsecantandtangent
Givenanintegraloftheform ∫ secm x tann xdx whichisnotoneofthespecialcases:
• if m and n arebotheventhenfactoroutsec2 x andsubstitute u = tan x
• if n isoddthenfactoroutthetermsec x tan x andsubstitute u = sec x
• if m isoddand n iseventhenuseintegrationbyparts
ProductstoSums
Theformulaeforexpressingtheproductsofsineorcosineassumsweredevelopedinthepreviouschapter.Here theyareagainforconvenience.
12Productstosums
cos A cos B = 1 2 cos(A B) + cos(A + B) sin A sin B = 1 2 cos(A B) cos(A + B) sin A cos B = 1 2 sin(A + B) + sin(A B)
cos A sin B = 1 2 sin(A + B) sin(A B)
Theseformulaecanbeappliedtosimplifyanintegralinvolvingproducts.
Example33
Find ∫ cos3x cos2xdx.
Solution
∫ cos3x cos2xdx = 1 2 ∫ (cos x + cos5x) dx (productstosums) = 1 2 sin x + 1 10 sin5x + C.
Theintegralinthisexamplecanalsobedoneusingintegrationbyparts.Itismuchlessefficientasthemethod mustbeappliedtwice.Itisleftasanexerciseforthereadertoshowthattheresultis ∫ cos3x cos2xdx = 1 5 (3sin3x cos2x 2cos3x sin2x) + C
The t -substitution
The t-substitutionwasalsoderivedinthelastchapter.Ittoocanbeappliedtointegration,wheretheaimisto transformatrigonometricintegrandintoarationalfunction.Consequently,themethodofpartialfractionsisoften required.Asalways,thesolutionshouldbeginwithastatementoftheparticularsubstitutionbeingused.
Example34
Showthat ∫ π 2 0 4 3 + 5cos x dx = ln3.
Solution
Let I = ∫ π 2 0 4 3 + 5cos x dx.
Put t = tan x 2 then dt dx = 1 2 sec2 x 2 = 1 2 (1 + t2). So
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Let 4 (2 + t)(2 t) = A 2 + t + B 2 t (partialfractions) then A = 1and B = 1(bythecover-upmethod.)
Hence I =
13The t -substitution
Usethistotransformatrigonometricintegrandintoarationalfunction.Let t = tan x 2 ,then:
TrigonometricSubstitutions
ThePythagoreantrigonometricidentitiesinvolvethesumsanddifferencesofsquares.Inthecontextof integration,themostusefulidentitiesare: cos2 θ = 1 sin2 θ adifferenceofsquares, sec2 θ = tan2 θ + 1 asumofsquares,
adifferenceofsquares.
Integralsthatinvolveasumordifferenceofsquaresareoftensimplifiedwithatrigonometricsubstitution correspondingtoanequationinthislist.SomeexamplesofthesesubstitutionsweremetinExercise 6C.This exerciseincludessomefurtherexamples.
Example35
Useasuitabletrigonometricsubstitutiontodeterminetheindefiniteintegral
Solution
Put
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(sincesec θ > 0inthedomain)
(fromBox 10.)
1 Find:
2 Findeachofthefollowingintegralsbysubstitutingeither u = sin x or u = cos x.Youmayalsoneedtoapply thePythagoreanidentitycos2 x + sin2 x = 1.
3 Usethedoubleangleformulaetoevaluate:
4 Usethesubstitution u = tan x tofindthefollowing.YoumayalsoneedtoapplythePythagoreanidentity 1 + tan2 x = sec2 x.
5 Evaluate:
6 Determine:
7 Showthat:
8 Ineachcaseuseasuitabletrigonometricsubstitutiontoevaluatetheintegral.
9 Let I = ∫
Find I usingasuitablesubstitution. a Find I bythedoubleangleformulae. b Showthattheanswerstoparts a and b areequivalent. c
10 Evaluate:
11 Evaluate:
12 Findtheseintegralsbyfirstconvertingtheproductstosums.
13 Evaluatethesedefiniteintegrals.
14 Usethesubstitution
15 Usethesubstitution
DEVELOPMENT
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16a Usethe t-substitutiontoshowthat
b Showthattheprimitivein a isequivalenttoln
17a Let n beapositiveinteger.Useintegrationbypartstwicetoshowthat
b Henceevaluate ∫ 2π 0 ex cos2x cos xdx
18 Determine:
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CHALLENGE
19 Usethesubstitution
20 Find
21 FromdeMoivre’stheorem,itisknownthat(cos θ + i sin θ)3 = cos3θ + i sin3θ.Usethisresulttohelp determine ∫ cos3 θ dθ
22 Useintegrationbypartsandthefactthat
6H ReductionFormulae
Readerswillbefamiliarwithsequencesandseries,suchastheoddnumbers, 1,3,5,7, ... thatis un = 2n 1, orthepowersof2, 1,2,4,8, ... thatis un = 2n 1 .
Inthissection,sequencesofintegralsareconsidered,suchasthesequence
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Ofparticularinterestaretheequationswhichrelatethetermsofthesequence.
Continuingwiththisexample,itcanbeshownthat In = n 1 n × In 2 for n ≥ 2.
Suchequationsarecalled reductionformulae,becausetheyenabletheindextobereduced,inthiscasefrom n to n 2.Inpracticalterms,thismeansthatifoneoftheintegralsinthesequenceisknownthenothertermscanbe simplycalculatedfromitwithouttheneedforfurtherintegration.Returningtotheexampleabove,since
itfollowsthat
and
Thisisobviouslyasignificantsavingofeffortsinceitwasnotnecessarytofindtheprimitivesofsin3 x andsin5 x inordertoevaluate I3 and I5.Itshouldbeclearfromthisthatreductionformulaeareofparticularimportance.
Notethattheconventionistoevaluatethesequenceindex before theintegralisevaluated.Thus,onceagainusing thesameexample,
Identities
Inafewcasesthereductionformulacanbegeneratedbyuseofanidentity,asinthefollowingexample.
Example36
Let In = ∫ π 4 0 tann xdx
a Showthat In = 1 n 1 In 2 for n ≥ 2.
b Evaluate I1 andhencefind I5
Solution
a In = ∫ π 4 0 tann 2 x (sec2 x 1) dx (byPythagoras,for n ≥ 2)
= ∫ π 4 0 tann 2 x sec2 xdx ∫ π 4 0 tann 2 xdx
thus In = tann 1 x n 1 π 4 0 In 2
= 1 n 1 In 2 for n ≥ 2.
b I1 = ∫ π 4 0 tan xdx
= ln(cos x) π 4 0
Thus I3 = 1 2 I1 = 1 2 1 2 ln2, and I5 = 1 4 I3 = 1 2 log2 1 4 . y (tan x)5 y x 2 4 2
= 1 2 ln2.
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ByParts
Manyexamplesofreductionformulaeuseintegrationbyparts.
Example37
Let In = ∫ e 1 (log x)n dx
a Showthat In = e nIn 1 for n ≥ 1.
b Henceshowthat I3 = 6 2e.
Solution
a In = ∫ e 1 1 × (log x)n dx
= x(log x)n e 1 ∫ e 1 x × n x (log x)n 1 dx (byparts)
= (e 0) n ∫ e 1 (log x)n 1 dx
= e nIn 1
b I0 = ∫ e 1 1 dx = e 1.
Thus I1 = e I0 = 1, I2 = e 2I1 = e 2, and I3 = e 3I2
= 6 2e y x e y (log x)3
ByPartswithanIdentity
Someexamplesuseintegrationbypartsandanidentity.
Example38
Let In = ∫ 1 0 x2(1 x2)n dx
a Usetheidentity x2 ≡ 1 (1 x2)toshowthat
b Evaluate I0 andhencefind I3
Solution
a Applyintegrationbypartsfirsttoget:
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Exercise6H
1a Giventhat In = ∫ tann xdx,provethat In =
Henceshowthat
2a If In = ∫ xnex dx,showthat In = xnex nIn 1 for n ≥ 1.
b Henceshowthat ∫ x3ex dx = (x3 3x2 + 6x 6)ex + C.
3a If In = ∫ e 1 x(ln x)n dx,showthat In = 1 2 e2 1 2 nIn 1 for n ≥ 1.
b Find I0 andhenceshowthat I4 = 1 4 (e2 3).
4 Let un = ∫ π 2 0 cosn xdx
a Useintegrationbypartswith v′ = cos x toshowthat un = (n 1) ∫ π 2 0 sin2 x cosn 2 xdx for n ≥ 2.
b Henceshowthat un = (n 1)(un 2 un).
c Deducethat un = n 1 n un 2 for n ≥ 2,andhenceevaluate u5
FOUNDATION
5 Let Tn = ∫ π 4 0 secn xdx
a Showthat Tn = (√2)n 2 n 1 + n 2 n 1 Tn 2 for n ≥ 2. Youwillneedtouseintegrationbypartsandatrigonometricidentity.
b Deducethat T6 = 28 15
6 Let Cn = ∫ π 2 0 xn cos xdx,where n ≥ 0.
Provethat Cn = ( π 2 )n n(n 1)Cn 2,for n ≥ 2. a Henceevaluate C6. b
7 Supposethat In = ∫ xn 1 + x2 dx
Usealgebraicmanipulationtoshowthat In = xn 1 n 1 In 2. a Hencefind ∫ x5 1 + x2 dx b
8a If In = ∫ 1 0 (1 x2)n dx,showthat In = 2n 2
b Evaluate I0 andhencefind I4.
9a If un = ∫ 1 0 x(1 x3)n dx,showthat un = 3n 3n + 2 un 1 for n ≥ 1.
b Showthat u0 = 1 2 andhenceevaluate
11 Let un = ∫ π 2 0 sinn x cos2 xdx
a
12 Considertheintegral I
a Showthat I
DEVELOPMENT
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c Useintegrationbypartstoshowthat I
d Henceevaluate I2
13a Showthat(1 + t2)n 1 + t2(1 + t2)n 1 = (1 + t2)n
b Put Pn = ∫ x 0 1 + t2 n dt.Useintegrationbypartsandpart a toshowthat Pn = 1 2n+1 1 +
c Hencedetermine P4: bythereductionformula, i byusingthebinomialtheorem. ii
d Hencewrite1 + 4 3 x2 +
14 Let Tn = ∫ 1 0 xn √1 xdx.
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inpowersof(1 + x2).
a Deducethereductionformula Tn = 2n 2n + 3 Tn 1 for n ≥ 1.
b Showthat T3 = 32 315
c Usethereductionformulatohelpprovebyinductionthat Tn = n!(n + 1)! (2n + 3)! 4n+1
15 Let In = ∫ 1 0 (1 x2)n dx and Jn
a Applyintegrationbypartsto In toshowthat In = 2nJn 1 for n ≥ 1.
b Henceshowthat In = 2n 2n + 1 In 1 for n ≥ 1.
c Showthat Jn = In In+1,andhencededucethat Jn = 1 2n + 3 In
d Hencewritedownareductionformulafor
a Showthat I1 = 1 2 ln2.
b Showthat,for n ≥ 2, In + In 2 = 1 n 1
c For n ≥ 2,explainwhy In < In 2,anddeducethat 1 2(n + 1) < In < 1 2(n 1) .
d Usethereductionformulainpart b tofind I5,andhencededucethat 2 3 < ln2 < 3 4 .
17 Let In = ∫ π 2 0 sinn xdx andsupposethat Jn = nIn In 1,where n isapositiveinteger.
a Showthat In = n 1 n In 2 for n ≥ 2.
b Showthat Jn = Jn 1
c Henceshowthat Jn = π 2 forallpositiveintegervaluesof n.
d Explainwhy In < In 1 forallpositiveintegervaluesof n
e Deducethat π 2(n + 1) < In < π 2n
18 Supposethat In = ∫ π 3 0 cos nx sec xdx.
a Provethat In = 2 n 1 sin (n 1)π 3 In 2 for n ≥ 2. [Hint:Write nx as(n 2)x + 2x.]
b Evaluate ∫ π 3 0 cos5x sec xdx
CHALLENGE
6I MiscellaneousIntegrals
Aswasstatedinthechapterintroduction,integrationisanartformandrequiresmuchpractice.Inparticular,it isimportanttobeabletorecognisethedifferentformsofintegrals,andtoquicklydeterminewhichmethodis bestused.Tothatend,thissectionhasbeenincluded.Theexercisecontainsamixtureofallintegraltypes.Some questionscanbedonebymorethanonemethod.Itisuptothereadertodeterminewhichmethodisthemost efficient.
Exercise6I
1 Evaluate:
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2 Find:
3 Showthat:
4a Findtherationalnumbers A, B and C suchthat
b Henceshowthat
5 Useintegrationbypartstoshowthat
6a Evaluate ∫ π 3 0 sec4 xdx
b Henceevaluate ∫ π 3 0 sec6 xdx
7 Ineachcaselet t = tan x 2 inordertoshowthat:
2 0
3 +
FOUNDATION
DEVELOPMENT
8a Findthevaluesof A, B, C and D suchthat 4t (1 + t)2(1 + t2) = A 1 + t + B (1 + t)2 + Ct + D 1 + t2 .
b Henceusethe t-substitutiontoevaluate ∫ π 2 0 sin x 1 + sin x dx.
9 Usethesubstitution u = 6 √x toshowthat
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10
Find ∫ √a2 x2 dx using: thesubstitution θ = sin 1 x a , a integrationbyparts. b
11a Showthat ∫ 1 0 5 5x2 (1 + 2x)(1 + x2) dx = 1 2 (π + ln 27 16 ).
b Hencefind ∫ π 2 0 cos x 1 + cos x + 2sin x dx usingthesubstitution t = tan x 2
12a Findintegers P and Q suchthat 8sin x + cos x 2 = P(3sin x + 2cos x 1) + Q(3cos x 2sin x).
b Hencefind ∫ 8sin x + cos x 2 3sin x + 2cos x 1 dx. 13 Useasubstitutiontofind:
e√x dx a
√x 1 x 2 dx b
14a Given In = ∫ e 1 (ln x)n dx,showthat In = e nIn 1 for n ≥ 1.
b Henceevaluate I3
15a If Tn = ∫ π 0 sinn xdx,showthat Tn = n 1 n Tn 2 for n ≥ 2.
b Henceshowthat T5T6 = 1π 3 .
16a Let In = ∫ 1 0 1 √x n dx forintegers n ≥ 0.
i Showthat In = n n+2 In 1 for n ≥ 1.
ii Henceshowthat In = 2 (n + 2)(n + 1) .
b Thediagramontherightshowsthecurve y = 1 √x n for0 ≤ x ≤ 1andfor somepositiveinteger n.Theshadedregionboundedbythiscurveandtheaxes islabelled A,andtheremainderoftheunitsquareislabelled B.Bothregions arerotatedaboutthe x-axistocreatetwosolidswithvolumes VA and VB respectively. Itisfoundthat VA : VB = 1:44.Whatisthevalueof n?
17 Itisgiventhat ∫ 1 0 e2x x + 1 dx = M,where M isaconstant.
a Showthat ∫ k k 1 e 2x k + 1 x dx = Me 2k .
b Usepart a andintegrationbypartstoshowthat
∫ 1 0 ln(2 x) e2x dx = e2 ln2 M 2e2
CHALLENGE
18a Supposethat ∫ 1 xf (x) dx = g(x) + C.Useasuitablesubstitutiontoshowthat
b Hencefind ∫ 1 x(xn 1) dx
Usethesubstitution
b Alternatively,usetheresult(1 +
andpartialfractionstoshowthat
c Theformulaeinparts a and b agreewhenappliedtotheintegral
provided pq > 0.When pq ≤ 0theformulainpart a isincorrect.Whymightthatbe? 20a Usethesubstitution x = u toshowthat
Useasuitablesubstitutiontoshowthat
b Henceevaluate
b Henceevaluate
Supposethat
a Usethesubstitution x = tan θ toshowthat I =
b Next,usethesubstitution
c Finally,deducethat I = 1 8 π ln2.
24 Let In = ∫ 1 0 xn 1 (x + 1)n dx,for n = 1,2,3,
a Showthat I1 = ln2.
b Useintegrationbypartstoshowthat
c Themaximumvalueof
Usethisfacttoshowthat
d Deducethat In ≤ 1 n 2n 1 .
e Usethereductionformulainpart b andtheinequalityinpart d toshowthat
2 3 < ln2 < 17 24 .
25 Find ∫ ln x (1 + ln x)2 dx.
Chapter6Review
Reviewactivity
• Createyourownsummaryofthischapteronpaperorinadigitaldocument.
Chapter6Multiple-choicequiz
• Thisautomatically-markedquizisaccessedintheInteractiveTextbook.AprintablePDFWorksheetversionis alsoavailablethere.
ChapterReviewExercise
1 Find:
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2 Find:
3 Usethegivensubstitutiontofind:
4 Evaluate:
5 Evaluate:
6 Usethegivensubstitutiontofind:
7a Giventhat In = ∫ 1 0 xnex dx,provethat In = e nIn 1
b Henceshowthat I5 = 120 44e.
8a Deriveareductionformulafor In intermsof In 1 given In = ∫ x3(ln x)n dx
b Henceshowthat ∫ x3(ln x)3 dx = 1 128 x4 32(ln x)3 24(ln x)2 + 12ln x 3 + C
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9a If I2n = ∫ π 2 0 sin2n xdx,provethat I2n = 2
b Hencefind ∫ π 2 0 sin6 xdx
10a If In = ∫ 1 0 (1 + x2)n dx,provethat(2n + 1)In = 2n + 2nIn 1.
b If Jn = ∫ π 4 0 sec2n θ dθ,showthat Jn = In 1 andhencefindareductionformulafor Jn.
c Evaluate ∫ π 4 0 sec6 θ dθ.
11a Usethesubstitution u = a x toprovethat ∫ a 0 f (x) dx = ∫ a 0 f (a x) dx
b Henceshowthat ∫ π 0 x sin x 3 + sin2 x dx = π ln3 4 .
12a Giventhat In = ∫ sin2nx sin x dx,provethat I
= 2 2
sin(2n 1)x + In 1 [Hint:Write2nx as(2n 2)x + 2x.]
b Hencefind ∫ π 2 0 sin6x sin x dx
Appendix:AShortTableofIntegrals
Hereisashorttableofintegrals.Thegroupof CommonIntegrals listssimplifiedversionsofintegralsfoundin theexaminationReferenceSheet.Ineachcaseitisassumedthat a 0.Thistablemaybehelpfulwhilststudying thischapter,butthereadershouldaimtobecomeproficientatusingtheexaminationReferenceSheetassoonas possible.
Thegroupof OtherIntegrals mayoccasionallybeneeded,butdonotappearintheexaminationReferenceSheet. Studentsareexpectedtobeabletoapplyunfamiliarintegralformulaelikethesewhentheyaregiven.
COMMONINTEGRALS
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OTHERINTEGRALS
7 Mechanics
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Chapterintroduction
Somepracticalapplicationsofcalculusinmechanicsareconsideredinthischapter.Simplemathematical modelsenablesolutionstobefoundintermsofthefamiliarfunctionsofthiscourse.
ProblemslaterinthechapterrequiresomeknowledgeofNewton’slawsofmotion.Preparationforthatisdone inSection 7A,withpracticalapplicationsinvolvingslopesandpulleyspresentedinSection 7B.Acceleration andvelocityasfunctionsoftimeanddisplacementareinvestigatedinSection 7C.Thisknowledgeisusedto integrateNewton’ssecondlaw, F = ma,inordertodeterminedisplacementasafunctionoftime.Next,simple harmonicmotionispresented,firstlyintermsoftimeinSection 7D,andthenintermsofdisplacementin Section 7E.
TheworkinSections 7F and 7G takesthefirststepinmakingthetheoryofmotionmorerealisticby introducingaresistiveforce.Horizontalmotionwithfriction,andverticalmotioninaresistingmediumwith constantgravityareconsidered.Inthefollowingtwosections,projectilemotionisfirstreviewedandthen extendedtoincludearesistiveforceproportionaltothevelocity,basedontheworkdoneinSection 7G.The chapterconcludeswithacollectionofvariousproblemswhicheitherextendthetheorylearnttonewsituations orareharderexamplesofapplicationsofthattheory.Onespecificexampleinvestigatedisthesimpleharmonic motionapproximationforapendulum.
7A
ResolvingForces
Intheproblemsencounteredinthischapter,theequationsofmotionareeitherspecifiedexplicitlyorgiven indirectly,suchasbyabalanceofforces.Whilstthisisnotacourseinphysics,abasicunderstandingofthelaws ofmotionisrequired.
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SomeAssumptions
Asignificantsimplificationistheassumptionthatanobjectmayberepresentedbyapointmass,oftencalleda particle.Ifthescaleofthemotionislargecomparedwiththeobject,suchasinthecaseofaballbearingthrown 5m,thenthisassumptionisreasonable.Asecondsignificantassumptionisthatairisanidealfluidandisnot particulateinnature.Atlowtomediumspeedsthisisasatisfactoryassumption.
AthirdassumptionisthattheforcesduetotheorbitoftheEartharoundthesunandduetotherotationofthe Earthonitsaxisarenegligibleintheproblemsbeingconsidered.Bywayofexample,intheproblemofprojectile motionwithoutairresistance,theaccelerationduetogravityatthesurfaceoftheEarthisabout9.8m/s2.The accelerationduetotheorbitoftheEarthisabout6 × 10 4 m/s2,andattheequatortheaccelerationduetothe rotationoftheEarthisabout0.034m/s2.Thusthisthirdassumptionseemsreasonable.
Newton’sLawsofMotion
TheequationsofmotionencounteredinthiscourseareallderivedfromSirIsaacNewton’slawsofmotion, containedinhisbook Principia,publishedin1687.ItiswritteninLatin,thescientificlanguageoftheday.An earlytranslationofthelawsisasfollows:
LawI Everybodycontinuesinitsstateofrestorofuniformmotioninastraightlineexceptinsofarasit becompelledbyimpressedforcetochangethatstate.
LawII Therateofchangeofmomentumisproportionaltotheimpressedforceandtakesplaceinthe directioninwhichtheforceacts.
LawIII Toeveryactionthereisanequalandoppositereaction.
Inthesecondlaw,momentumisdefinedtobe mv,theproductofmasswithvelocity.Thestandardunitsusedare kilograms,metresandseconds.Theunitofforceiscalledthe newton,with1N = 1kgms 2.IftheseSIunitsare usedthen,inthesecondlaw,theconstantofproportionalityis1.Thus
F = d dt (mv), andifthemassisconstantthen
F = ma where a = v
Ifotherunitsareusedthentheseequationsmustbeappropriatelymodified. Itmayseemstrangetostatethatthemassisconstant.Inmanycasesthemassiscertainlynotconstant,suchasa rocketasitsfuelisburnt.Inthiscourse,however,themassisalwaysassumedtobeconstant.
1Newton’ssecondlawofmotion
Whenaforce F newtonsisappliedtoaconstantmass m kg,whichisfreetomove,theresulting accelerationis a ms 2,where
F = ma
Asstatedabove,1N = 1kgms 2.Anobviousquestiontoaskis:justhowbigisthat?Inanswertothat,first notethattheaccelerationduetogravityatthesurfaceoftheEarthisabout10ms 2 .
Thusthemassrequiredtogenerate1NofforceatthesurfaceoftheEarthis m = F a ≑ 1 10 kg(or100grams).
Toputthisintocontext,acalculatorisabout100gso,1Nisroughlytheweightthatisfeltwhenholdinga calculator.
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ForcesasVectors
Inmostpracticalsituationstherearenumerousforcesactingonanobject.Agoodwaytosimplifythesituation istoresolvetheseforcesintoasingletotalforce.Thiscanbeachievedbyusingavector F torepresentan individualforce.Themagnitudeof F isthemagnitudeoftheforceandthedirectionof F isthedirectioninwhich theforceacts.
Example1
Aforceof26Nisappliedinthedirectionof u ∼ = 12 i ∼ 5 j.Writetheforceinvectorformusing: amultipleofaunitvector, a arowvectorwithcorrectmagnitude. b
(24, 10) b
Noticethattheunitshavenotbeenincludedinthevectornotation.Unitsaresometimesincludedinavectorto emphasisethesituation,particularlywhenaprobleminvolvesmultipledifferentunits.Unitsarealwaysincluded, however,whenwritingdownacomponent.Thus,inthepreviousexample,thecomponentof F ∼ actingalongthe x-axisis24N.
2Forcesasvectors
Avector F maybeusedtorepresentaforce.Themagnitudeof F isthemagnitudeoftheforceandthe directionof F isthedirectioninwhichtheforceacts.
ComponentsofaForce
Strictlyspeaking,thefinalstatementbeforeBox 2 aboveshouldbewrittenas forcealong x-axis = F ∼ · i ∼ = 24N.
Thatis,tofindthecomponentofaforceactinginagivendirection,takethedotproductoftheforcewiththeunit vectorinthatdirection.
Example2
Findthecomponentoftheforce F = (24, 10)thatactsinthedirection u = (1,3).
Solution
Theunitvectorinthegivendirectionisˆu = 1 √10 (1,3).Thusthecomponentofforceinthisdirectionis F ˆ u = (24, 10) 1 √10 (1,3)
= 1 √10 (24 30) = 3 5 √10N(about 1.90N).
Thenegativesignindicatesthatthecomponentofforceisactingintheoppositedirectiontothegivenvector.
Insomesituationsitismoreconvenienttouseaprojectiontogettheanswer.Thisisbecauseacomponentalways hasadirectionassociatedwithit.Thatis,acomponentcanbewrittenasavector.Bywayofcomparison,hereis thesameproblemdonebyprojection.
Example3
Findthecomponentoftheforce F ∼ = (24, 10)thatactsinthedirection u ∼ = (1,3)byusingaprojection.
Solution
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(1,3)
Thusthecomponentoftheforceis 3 5 √10N,asbefore.
3Componentsofaforce
Thecomponentofaforce F inthedirection u iseitherascalarorvector,dependingoncontext.
• Thescalarcomponentis F ∼
Thevectorcomponentisproj
FreeBodyDiagrams
Whenanalysingseveralforcesactingonanobjectitisoftenhelpfultousea freebody diagram .Theobjectisdrawnasadotwithvectorsradiatingfromitrepresentingthe variousforcesandtheirdirections.Bywayofexample,thediagramontherightshows amass m acteduponbytwoforces F1 and F2.
TheResultantForce
Whenseveralforcesactonanobjecttheeffectisobservedtobecumulative.Thatis,theforcesareaddedand hencethetotalforcecanbecalculatedbyvectoraddition.Theoutcomeisoftencalledthe resultantforce.If theresultantforceisthezerovectorthen,byNewton’sfirstlaw,thereisnoacceleration(nordeceleration).The forcesinthissituationaresaidtobein equilibrium,whichisderivedfromtheLatin libra meaning balanced.If theresultantforceisnon-zerothen,byNewton’ssecondlaw,theobjectacceleratesinthedirectionofthatforce.
Example4
Threeforces F1 = 2 i j, F2 = i 3 j and F3 = 3 i + 6 j actonaboxsittingattheorigin.Allvaluesarein newtons.
a Whatistheresultantforce F r onthebox,andwhatisitsmagnitude?
b Explainwhytheboxbeginstomove.
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Solution
Thesituationisshowninthediagram.
a F r = F1 + F2 + F3 = 2 i ∼ j i ∼ 3 j + 3 i ∼ + 6 j
4 i + 2 j ∼ and |F ∼ r | = 2√5N.
b Theresultantforceisnon-zerosoitacceleratesinthedirectionof F ∼ r
4Theresultantforce
Whenthereismorethanoneforce,addthembyvectoradditiontoobtaintheresultantforce
• If F ∼ r = 0 thenthereisnoacceleration.Theforcesareinequilibrium.
• If F r 0 ∼ thentheobjectacceleratesinthedirectionof F r .
SomeCommentsabouttheExercise
Sofar,ithasnotbeennecessarytouseNewton’sthirdlaw.Itisneeded,however,inatleastonequestionofthe exercise.Thereaderisexpectedtoknowallthreelawswellenoughtoapplythemwithconfidencewhenrequired. Finally,aforcemaybetransmittedtoanobjectviaawireorsimilardeviceusedtopulltheobject.Insuchcases, theforceisoftenreferredtoasthe tension inthewire.Thewiremayalsobedescribedaslightandinextensible. Intheproblemsencounteredhere,thisisasuitableapproximation,butthereadershouldbeawarethatthephysics ofthesituationbecomesvastlymorecomplicatedifaheavyorstretchyropeisused.
Exercise7A
Note: Inthisexerciseuse g ≑ 9.8m/s2 fortheaccelerationduetogravity.
1 Inthediagram,a4-wheeldriveisboggedonamuddyroad.Awinchis pullingthevehiclewithaforceof1200N.Thechainconnectingthewinch tothevehiclemakesanangleof15◦ withthedirectionofmotion.Calculate, correcttothenearestnewton,themagnitudeofthecomponentoftheforce: a inthedirectionofmotion, b perpendiculartothedirectionofmotion.
FOUNDATION
4WD 15° direction of motion winch 1200N
2 Twoforcesofmagnitude30Nand16Nactawayfromapoint P andareperpendicular.Findthemagnitude anddirectionoftheresultantforce(measuredfromthe30Nforcecorrecttothenearestdegree).
3 Threeforcesactonanobjectofmass2kg.Theseforcesarerepresentedbythevectors12 i + 23 j ∼ ,9 i 7 j ∼ and 5 i + 14 j ∼ .Calculatethemagnitudeanddirection(asanangleofinclinationtothenearestdegree)of theaccelerationoftheobject.
4 Inthediagram,thevectors −−→ OP and −−→ OQ representforcesofmagnitude20Nand 16Nrespectively.
a Express −−→ OP and −−→ OQ incomponentform.
b Calculate,correctto2significantfigures,themagnitudeanddirection(asan angleofinclination)oftheresultantofthetwoforces.
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5 Useaprojectiontofindthecomponentoftheforce(5 i + 6 j + 8k)Ninthedirectionofthevector2 i +
6 Aboxofmass25kgisbeingraisedverticallybyaconnectingcablewithanaccelerationof0.2m/s2.Find theforce(thatis,thetension)inthecable.
DEVELOPMENT
.
7 Thediagramontherightshowstwoforcesofmagnitude5Nand7Nactingat35◦ toeachother.Find:
a themagnitudeoftheresultantforce(to4significantfigures), b theangle(indegreesto4significantfigures)thattheresultantforcemakeswith thelargerofthetwogivenforces.
8 Thediagramshowsanobjectofmass5kgwhichissuspendedfromahorizontal ceilingbytwostringsofequallength.Eachstringmakesanangleof60◦ withthe ceiling.Calculate,correctto3significantfigures,theequaltensionsinthetwo strings.
9 Twodogs,BrutusandNitro,aresimultaneouslytuggingonabone.Brutusispullingwithaforceof12N inadirection45◦ westofnorth,whileNitroispullingwithaforceof16Ninadirection30◦ southofeast. Calculate,correcttotwosignificantfigures,themagnitudeanddirectionoftheresultantforce.
10 Twoforces F1 and F2 areactingonaparticle P ofmass2kg. F1 hasmagnitude40Nandactsinthenortheasterlydirection,while F2 hascomponents40Nduenorthand20Ndueeast.Calculatethemagnitude (correctto3significantfigures)anddirection(asatruebearingcorrecttothenearestdegree)oftheresultant accelerationof P
11 Inthediagram,aflowerpotofmass m kgishungfromaceilingbytwochains.Let thetensionsinthechains AP and BP be T1 and T2 newtonsrespectively.Thethird forceactingat P istheweightoftheflowerpot.
a Byfindingthehorizontalcomponentoftheresultantofthethreeforcesacting at P,showthat T1 = √3 T2
b Byfindingtheverticalcomponentoftheresultantofthethreeforcesacting at P,showthat √3 T1 + T2 ≑ 19.6m
c Findthemassoftheflowerpot,giventhat T2 = 98N.
ceiling 60° 30° P flower pot (kg) m T1 T2
12 Aliftofmass600kgisraisedbyacableattachedtoitstop.Whentravellingupwardstheliftaccelerates fromrestto2m/sin5seconds.
a Showthatthemagnitudeoftheaccelerationis0.4m/s2
b Ifthecombinedmassofthepassengersis240kg,findthetensioninthecable.
13 Thesystemofforcesinthediagramisinequilibrium.Find:
a theexactmagnitudeof F,
b theangle θ,tothenearesttenthofadegree.
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14 Thediagramshowsanobjectofmass5kgbeingraisedbyforcesofmagnitude 75Nand50N.
a Findtheweightoftheobject.
b Find,correcttothenearestnewton,themagnitudeoftheresultantofthethree forcesactingontheobject.
c Find,correcttothenearestdegree,theanglethisresultantmakeswiththe upwardverticaldirection.
15 Atrailerisbeingtowedupahillbyapullingforceof(500 i + 300 j ∼ )N.Thegradientofthehillis0.2.Find, correcttothenearestnewton:
a themagnitudeofthepullingforce, b themagnitudeofthepullingforceinthedirectionofmotion.
16 Twoforcesofmagnitude p newtonsand q newtonshavearesultantof2√7Nwhentheyactat90◦ toeach other.Whentheyactat30◦ toeachotherthemagnitudeoftheresultantis2√13N.Findthevaluesof p and q.
17 Aloadof400kgisliftedbyacablethroughaverticaldistanceof48m.Theloadmovesupwardsfromrest withaconstantaccelerationof0.5m/s2 overthefirst36m,andthendeceleratesuniformlytorest.Taking g = 10m/s2,find:
a thetensioninthecablewhiletheloadisaccelerating,
b themaximumspeedreachedbytheload,
c thetensioninthecablewhiletheloadisdecelerating.
18 Alightinextensiblestringoflength50cmhasitsupperendfixedat P,whilea6kgmassisattachedtoits lowerend.Ahorizontalforce F appliedtothemasskeepsitinequilibrium30cmfromtheverticalline through P.Find: thetensioninthestring, a themagnitudeof F. b
CHALLENGE
19 Inthediagram ABCD isasquarewithside4cm.Thepoints E, F, G, H and J lie onthesidesofthesquaresothat
AE = BF = CG = HD = DJ = 1cm.
Forcesofmagnitude10N,3√10N,2√5N,10N, √10Nand5Nactat E inthe respectivedirections EB, EF, EG, EH, EJ and EA.
a Calculatethemagnitudeoftheresultantoftheseforces.
b Showthattheresultantisdirectedatanangleofarctan 4 3 to EB
20 TheverticesofasquareintheCartesianplaneare A(a, a), B( a, a), C( a, a)and D(a, a).Aparticleat P(x, y)isacteduponbyforcesalongthelinesjoining P tothefourverticesofthesquare.Thepoint O isthe originand k isarealscalar.
a Showthat P isinequilibriumwhentheforcesare k
b Showthattheresultantforceis4k
PO whentheforcesare k
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and k
7B
PlanesandPulleys
Theforcesinsimplesystemswithaflatplaneorpulleyareinvestigatedinthissection.Thereisalargevarietyof questionsthatmaybeasked,soonlyafewofthecommonideasarepresentedintheexamples.Thereisnonew theory.
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Thedownwardsforceorweightofamassduetogravityis mg.Whilstthereareafewproblemswheregravity isnotrelevant,itisafactorinmostquestions.Itshouldberoutinelyincludedinanyfreebodydiagram.Inthe exercisequestionstheaccelerationduetogravityisapproximatedwith g ≑ 9.8ms 2 .
SomeAssumptions
Theproblemsencounteredmayinvolveaplane,apulleyoraconnectingwireorstringbetweentwoobjects. Someassumptionsaremadeabouttheapparatusinordertosimplifythemathematics.Theseassumptionsare sometimesstatedinaquestionasareminder.
Wheneveraprobleminvolvesamassmovingonaplane,itisassumedthattheplaneissmooth.Thatis,thereis nofrictionbetweentheplaneandobject,whichisthusfreetomoveacrosstheplane.
Twoobjectsmaybeconnected,suchasbyawire.Itisassumedthatthewireislightandinextensible.Thatis,the massofthewireissmallenoughtobeignoredanditdoesnotstretch.Thewireisnormallyundertension,aforce whichpullsateachmasswithequalmagnitudebutinoppositedirections.
Thewirebetweentwomassesmaypassoverapulley,awheelthatislightandfreetospin.Thatis,themassof thepulleyissmallenoughtobeignoredandthereisnofrictionasthewheelturns.
SomeSimpleCalculus
Afewproblemspresentedinthetextandexerciserequirethevelocityordisplacementwhenaconstantforce F is appliedtoamass m overaperiodoftime.Integrationisallthatisrequired,startingwiththeacceleration:
dv dt = F m (fromNewton’ssecondlaw, F = ma)
so v = F m t + V0
where V0 istheinitialvelocityofthemass.Integratingagain:
x = F 2m t2 + V0 t + X0 where X0 istheinitialdisplacementofthemass.
Notethatthesesimpleresultsarenotpartofthecoursesomustbederivedeachtimetheyareneeded.
Inmanycasestheoriginsofdisplacementandtimemaybechosensothatboth V0 = 0and X0 = 0.Clearly,in thoseinstances, v = F
and
ASimplePulley
Inasimplepulleyproblem,twomasses m1 and m2 areconnectedbyawireovera pulley,asshowninthediagramontheright.Itshouldbeclearthattheonlyforces actingoneacharetensionandweight,whichbothactvertically.Theseareshowninthe accompanyingfreebodydiagrams.
Asusual,treatupwardsaspositiveandsupposethat m1 movesupwardswith acceleration a.Thenthesumoftheforcesgives
T m1g = m1a
Itfollowsthat m2 movesdownwardssothat
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Thesetwoequationshavefivevariables: a, g, m1, m2 and T .Ifthreeoftheseareknownthentheresultisapair ofsimultaneousequationswithtwounknownsthatcaneasilybesolved.Thisisthebasisofmostsimplepulley problems.
Example5
A3kgmassisattachedbyalightinextensiblewireviaafrictionlesspulleytoanunknownmass m kg.
The3kgmassisacceleratingdownwardsat2ms 2.Usetheapproximation g ≑ 10ms 2 fortheacceleration duetogravityinordertofind m andthetension T inthewire.
Solution
Thefreebodydiagramisshownontheright.
Forthe3kgmassthetotalforceis:
T 30 = 6
Fortheunknownmassthetotalforceis:
T 10m = 2m
Clearlyfromthefirstequation
T = 24N.
Thus,fromthesecondequation
12m = 24 so m = 2kg.
TheForceNormaltoaPlane
Thediagramshowsaboxofmass m kgsittingmotionlessonasmoothhorizontalplane. Thereisaforce mg actingdownwardsontheboxduetogravity.Iftherewerenoother forcethentheboxwouldacceleratedownwards.Hence,theremustbeanequaland oppositeforcekeepingtheboxinplace.ThisisanexampleofNewton’sthirdlaw. mg
Sincethisupwardsforceisperpendiculartotheplaneitiscalleda normal forceandsothepronumeral N isoften used.Itmayalsobecalleda normalreaction forceduetotheconnectionwithNewton’sthirdlaw.Wheneveran objectofinterest,likethebox,isincontactwithasurfacethereisalwaysanormalforceinvolved.Itshouldbe includedaspartofafreebodydiagram.
Example6
Aboxofmass2kgisstationaryonahorizontalsmoothfloor.Aconstantforceof6Nisappliedtotheboxat anangleof60◦ tothefloor.Consequently,theboxacceleratestotherightwithacceleration a butremainsin contactwiththefloor.Let N bethenormalreactionforceofthefloorontheboxandusetheapproximation g ≑ 10ms 2
a Find a,andhencefinditsdisplacementafter2seconds.
b Determine N,correcttoonedecimalplace.
Solution
Thefreebodydiagramisontheright.
a Thehorizontalforceis
6cos60◦ = 2a, hence a = 3cos60◦ = 3 2 ms 2
Integrating, v = 3 2 t + V0 butitisinitiallystationaryso V0 = 0.
Thus x = 3 4 t2 + X0.
Theinitialdisplacementiszeroso X0 = 0.
Hence,at t = 2thedisplacementis x = 3m.
b Thereisnoverticalaccelerationsothetotalverticalforceiszero.
Thus, N + 6cos30◦ 20 = 0 so N = 20 3√3 ≑ 14.8newtons
AMassonanInclinedPlane
Thekeytosolvingthepreviousexamplewastotakethecomponentsofforceshorizontallyandvertically.That is,inthedirectionsof i ∼ and j.Aforceinoneofthesedirectionshasnoeffectintheotherdirection.Thisis becausethecomponentofforceisfoundbyadotproductand i ∼ j = 0.Forexample,thecomponentof3 j inthe direction i is3 j ∼ i = 0.
Thislastobservationcanbeusedtogreatlysimplifythesituationforproblemsoninclinedplanes.Further,any perpendiculardirectionsmaybeused.Forthesakeoffamiliarity,thenextexampleusesthestandardvectors i ∼ and j ∼ ,andsomebasictrigonometry.
Example7
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Aparcelwithmass8kgisonaplankangledat30◦ tothehorizontal.Asusual, theforceduetogravityandthenormalforce N actontheparcel.Additionally,a force F duetofrictioniskeepingtheparcelfromslidingdowntheplankandis directeduptheslope.Findboth N and F.Usetheapproximation g = 10ms 2 asthe accelerationduetogravity.
Solution
Thesituationisshowninthefreebodydiagrambelowontheright.Sincethereisnomotion,thehorizontal andverticalcomponentsofforceaddtozero.Thehorizontalcomponentsofforcegive N cos60◦ F cos30◦ = 0
andfromtheverticalcomponents
sin60◦ + F sin30◦ 80 = 0.
Nowfrom(1), N = √3F so(2)becomes
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UsingDifferentBasisVectors
AlthoughthesolutiontoExample 7 isrelativelyeasy,thereisanothermethodthatavoidstheneedfor simultaneousequations.Thetrickistoalign i ∼ and j perpendiculartoandparallelwiththeslopingsurface.Here isthesameproblemdoneusingthisveryclevertrick.
Example8
SolveExample 7 byaligning i ∼ withthenormaland j uptheslopingplane.
Solution
Thefreebodydiagramisbelowontheright.Noticetheinclusionoftherevisedbasisvectors.Thefirsttaskis tofindthecomponentsoftheweightvector W ∼ .Itsmagnitudeis80Nso:
W = 80cos30◦ i 80sin30◦ j ∼ = 40(√3 i + j).
Thereisnomotioninthedirectionof i so
N 40√3 = 0 or N = 40√3N(exactlyasbefore).
Thereisnomotioninthedirectionof j so
F 40 = 0 or F = 40N(exactlyasbefore).
Noticehowthecarefulchoiceofdirectionsfor i and j ∼ hasgreatlysimplifiedtheanswer.Althoughworking isshownabove,thevaluesof N and F aresimplytheoppositeofthecomponentsof W ∼ .Thistrickwillnot alwaysbepractical,butshouldbeusedwheneverpossible.
AnInclinedPlaneandPulley
Thefinalexampleishardasitinvolvesbothaninclinedplaneandsimplepulley.Onceagain,thetrickof re-aligning i and j ∼ willbeusedontheslope.
Example9
Amass m kgat P isonasmoothslopeinclinedat45◦.Itisconnectedbyathin wireoverapulleyatthetopoftheslopetoanothermass m whichishangingfreely at Q.Thetopoftheslopeisatheight2h mabovetheground.Bothmassesare initiallyatrestatheight h.Thesizesofthemassesandpulleyarenegligiblewhen comparedwith h
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a Whichmassmovesdownwards?
b Howlongdoesittaketoreachtheground?
Solution
Supposethatthemassat P movesuptheslopewithacceleration a,whichmaybenegative.Clearlythemass at Q movesdownwardswiththesameacceleration.Letthetensioninthewirebe T .
a Align i withtheslopeat P and j withthenormal.Thefreebodydiagramis shown.The i componentofweightis
W = mg cos45◦ = mg √2
Theaccelerationinthedirectionof i is a so
Next,forthemassat Q,align i withthehorizontaland j ∼ withthevertical. Thefreebodydiagramisshown.
Theaccelerationinthedirectionof j is a so
mg + T = ma.
Subtractthisfromthepreviousequationtoget
mg(1 1 √2 ) = 2ma or a = g 2 (1 1 √2 ). i j mg T
Sincethisispositive,themassat Q fallsandhitstheground.
b Forthemassinitiallyat Q theaccelerationis
d2y dt2 = g 2 (1 1 √2 ).
Integrating, dy
dt = g 2 (1 1 √2 )t + V0.
Themassisinitiallyatrestso V0 = 0and
dy
dt = g 2 (1 1 √2 )t
Thus y = g 4 (1 1 √2 )t2 + Y0.
At t = 0themassisat Q where y = h so
Y0 = h and y = g 4 (1 1 √2 )t2 + h
Continuedonthenextpage
Themassreachesthegroundwhen y = 0.Thus
0 = g 4 (1 1 √2 )t2 + h t2 = 4h
g(1 1 √2 ) = 4h √2
g(√2 1) or t = 2 h √2
g(√2 1) .
Keenreadersmayliketocontinuethisexamplebyinvestigatingwhathappenstothemassat P oncethemass at Q stops.Althoughthealgebraiscomplicated,itcanbeshownthatthemassinitiallyat P travelsuptheslope atotaldistanceof 1 2 h(1 + √2)m.Thismeansitdoesnotreachthetopoftheslope,asthedistancefrom P to thetopis h√2m.
Exercise7B
Note: Inthisexerciseuse g ≑ 9.8m/s2 fortheaccelerationduetogravity.
1 Thediagramshowsanobjectofweight24Natrestat P onaninclinedplane.Find thecomponentoftheweight: a downtheplane, b perpendiculartotheplane.
FOUNDATION
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2 Acarofmass880kgisparkedonadrivewaythatisinclinedat10◦ tothehorizontal.Find,tothenearest newton,themagnitudeofthenormalreactionforceofthedrivewayonthecar.
3 Inthediagram,anobjectofmass10kgiskeptatrestonasmoothplaneinclined at30◦ tothehorizontalbyaforceof F newtonsactingparalleltotheplane.Find thevalueof F.
4 Anobjectofmass m ispulledalongasmoothhorizontalsurfacebyaforceofmagnitude F inclinedatan angleof θ abovethehorizontal.Iftheobjectstartsfromrest,showthatthedistanceithastravelledattime T isgivenby FT 2 cos θ 2m .
5 Amanofmass80kgisstandinginaliftthatisacceleratingdownwardsat3m/s2.Findthenormalreaction forceoftheflooroftheliftontheman.
6 Anobjectofmass1.2kgisonasmoothhorizontaltable.Itisconnectedtoafreelyhangingobjectof mass0.8kgbyalightinextensiblestringpassingoverasmoothpulleyattheedgeofthetable.Find: a theaccelerationofthetwoobjects, b thetensioninthestring.
DEVELOPMENT
7 Objectsofmass5kgand3kgareconnectedbyalightinextensiblestringthatpassesoverasmoothpulley. Withthemasseshangingvertically,thesystemisreleasedfromrest.Find:
a themagnitudeoftheaccelerationofthetwoobjects,
b thedistancetravelledbyeachobjectinthefirst2seconds.
8 Thediagramshowsobjectsofmass3kgand m kgattachedtotheendsofa lightinextensiblestringthatpassesoverasmoothpulley.The3kgobjectis acceleratingat4.9m/s2 upwards.Letthetensioninthestringbe T newtons.Find:
a thevalueof T ,
b thevalueof m 3kg m kg
9 Anobject A ofmass3kgisonthesurfaceofasmoothplaneinclinedat45◦ tothehorizontal.Itis connectedbyastringpassingoverasmoothpulleytoanobject B ofmass1kg.Object A isslidingdownthe planewhileobject B ismovingupwardstowardsthepulley.
a Byresolvingforcesat A downtheplaneandforcesat B vertically,findthetensioninthestring(to 4significantfigures),
b Findthemagnitudeoftheaccelerationoftheobjects(to3significantfigures).
10 Twoparticles A and B withmasses1.1kgand0.9kgrespectivelyareconnectedbyalightinextensible stringpassingoverasmoothpulley.With A and B hangingvertically,thesystemisreleasedfromrest withparticle A adistanceof d metresabovethefloor.Ifparticle A hitsthefloorafter T seconds,showthat 20d = gT 2
11 Twoobjectsofmass30gand20gareattachedtotheendsofalong,lightinextensiblestringthatpasses overasmoothpulley.Thesystemisreleasedfromrestwiththeobjectshangingvertically.Whenthesystem hasbeeninmotionfor5seconds,thestringsuddenlybreaks.Whenthishappens,the20gparticleisat point A.
a Showthattheinitialaccelerationofthetwoobjectsis1.96m/s2 .
b Findthespeedofthetwoobjectsattheinstantthestringbreaks.
c Findtheheightabovepoint A towhichthe20gobjectrises.
d Howlongdoesittakeforthe20gobjecttopassthroughpoint A againafterthestringbreaks?
12 Anobjectofmass m1 liesonasmoothhorizontaltableandisconnectedbyalightinextensiblestringtoa freelyhangingobjectofmass m2.Initiallytheobjectsareatrest,withthefirstobjectatadistance d from theedgeofthetable.Showthat:
a theaccelerationofthesystemis
m2g m1 + m2 ,
b thetimetakenforthefirstmasstoreachtheedgeofthetableis
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2d(m1 + m2)
m2g
13 Anobjectofmass2m isonasmoothplaneinclinedat θ abovethehorizontal.Itisconnectedbyalight inextensiblestringpassingoverasmoothpulleytoasecondobjectofmass m thatisfreelysuspended.Show thatthefirstobjectwillaccelerate up theinclinedplaneif θ < 30◦
14 Anobjectofmass m1 isonasmoothplaneinclinedat30◦ tothehorizontal.Itisconnectedbyalight inextensiblestringpassingoverasmoothpulleytoasecondobjectofmass m2 thatishangingfreely.When thesystemisreleasedfromrest,thefirstobjectmovesuptheplanetowardsthepulley.Find,intermsof m1, m2 and g:
a theaccelerationofthetwoobjects,
b thetensioninthestring.
15 Thediagramshowsobjectsofmass3kgand2kgonconnectedsmoothplanes inclinedatanglesof θ and2θ tothehorizontal.Theobjectsareattachedto theendsofalightinextensiblestringthatpassesoverasmoothpulley.Let T newtonsbethetensioninthestring,andsupposethatthe3kgobjectis acceleratingat a m/s2 upitsplane.
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3kg
2kg
a Find,intermsof a, T , g and θ,anequationforthemotionofthe3kgobjectupitsplane.
b Writedownasimilarequationforthemotionofthe2kgobjectdowntheotherplane.
c Showthatthesystemisinequilibriumwhencos θ = 3 4 .
16 Inthediagram,objectsofmass m1 and m2 areheldatrestonadjoiningsmooth inclinedplanes.Theyareconnectedbyalightinextensiblestringthatpasses overasmoothpulley.
a Showthatwhentheobjectsarereleased,theobjectofmass m1 will accelerate towardsthepulley if m1 < √3 m2
CHALLENGE
b Assumingthattheconditioninpart a issatisfied,showthattheobjectof mass m1 willhitthepulleywithspeed dg √3 m2 m1 m1 + m2 30°60° d md 1 m2
7C ForcesandAcceleration
Inmanysituationsitissimplyamatterofintegratingtheforceequationinordertofindtheotherdetailsofthe motion.AfewexampleswereencounteredinSection 7B.Thissectioncontainsmoreexamplesandintroduces someusefulalternativeexpressionsfortheacceleration¨ x.
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Example10
Abodyofmass4kgisacteduponbyavariableforce F = 48(5 t)newtonsfor5seconds.Ifthebodystarts fromrestat x = 0thenwhatisitsfinalvelocityandhowfarhasittravelled?
Solution
FromNewton’ssecondlaw,afterdividingthroughbythemass, dv dt = 60 12t
Integrating, v = 60t 6t2 + C
At t = 0thevelocityis v = 0,so C = 0and v = 60t 6t2
Henceat t = 5, v = 150m/s.
Integratingagaingivesthedisplacement:
x = 30t2 2t3 + D
At t = 0thebodyisat x = 0,so D = 0and
x = 30t2 2t3 .
Henceat t = 5, x = 500m.
OtherFormsofAcceleration
Therearesomeproblemswheretheequationofmotionismorenaturallyexpressed usingdisplacement.Forexample,itiseasytomeasurethedisplacement x ofoneendof aspringwhenamass m ishungfromit.Itisfoundthat x isproportionaltotheforceso theresultingequationis
v = kx (forsomeconstant k.) x m
Inordertodoanythingmeaningfulwiththisequationitisnecessarytorewritetheacceleration˙v asaderivative in x insteadof t.Thisisdoneasfollows.
dv dt = dv dx × dx dt (bythechainrule) = dv dx × v (bythedefinitionofvelocity) thus dv dt = v dv dx
Puttingthisintotheequationofmotionforthespringgives v dv dx = kx whichisavariableseparabledifferentialequation.
Example11
Aspringishungfromaceilingandextendedadistance a metresthenreleasedfromrest.Itisfoundthatthe equationofmotionforthespringis¨ x = x,where x metresisthedisplacementoftheendofthespringattime t seconds.Showthat a ≤ x ≤ a
Solution
Replacetheacceleration¨ x with v dv dx anddoubletoget
2v dv dx = 2x whichisvariableseparable.Nextintegratebothsidestoget
v2 = C x2
forsomeconstant C.Butat t = 0, x = a and v = 0so
0 = C a2 thus C = a2,andrearrangingtheequationgives
x2 + v2 = a2 .
Graphing v against x,thisistheequationofacircle,andhence a ≤ x ≤ a.Notethatthisdoesnotprovethat theendofthespringeverreaches x = a.Thatwillbedonelaterinthischapter. Noticethattheequationwasdoubledinthefirstlineofthesolutionabove.Thismadetheintegrationeasier. Thereisanotherapproachwhichisoftenused.
d2 x dt2 = v dv dx = 1 2 v dv dx + dv
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Example12
Aparticlestartsattheoriginwithvelocity v = 1m/s.Itisfoundthatitsdisplacement x metresattime t secondssatisfies¨ x = e 2x .
a Findanexpressionfor v2 intermsof x.
b Explainwhythevelocitymustalwaysbepositive.
c Henceshowthat v → √2as t →∞.
Solution
a Fromabove
forsomeconstant C.Nowapplytheconditions v(0) = 1and x(0) = 0.
1 = C 1
so C = 2andhence v2 = 2 e 2x
b Nowat t = 0, x = 0and v = 1,sotheparticlebeginsmovingtotheright.Thatis, x becomespositiveand v ispositive.Inorderfor v tochangesign,itmustfirstbezero,whichcannothappenfor x ≥ 0.Hence v remainspositive.
c Since v2 = 2 e 2x , v ≥ 1foralltimeandhenceas t →∞ sotoo x →∞
Thuslimt→∞ v2 = limx→∞ v2 = limx→∞ 2 e 2x = 2, andsince v > 0,takethepositivesquareroottogetlimt→∞ v = √2.
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5Otherformsofacceleration
Thefourcommonformsofaccelerationare d2 x dt2 = dv dt = v dv dx = d dx 1 2 v 2
Ineachproblem,choosetheformmostsuitableforintegration.
IntegratingTwice
Whenaccelerationiswrittenas d dx 1 2 v2 ,integrationyields v2 asanimplicitfunctionof x,suchas v2 = a2 x2 in Example 11.Rewritingthisequation:
dx dt 2 = a 2 x 2 ,
whichisanon-linearfirstorderdifferentialequation.Suchequationsareusuallytoodifficulttosolveinthis course,andnofurtherprogressispossible.
If,however,thesignof v canbeestablishedthentheappropriatesquarerootcanbetakenandasecond integrationperformed.Thesignof v maybedeterminedeithermathematically(asinExample 12)or,more typically,fromthephysicsofthesituation.
Example13
Experimentssuggestthataccelerationduetogravityisinverselyproportionaltothesquareofthedistanceto thecentreoftheplanet.Thus,if x isaltitudeand R istheradiusoftheEarththen
d2 x dt2 = k (x + R)2 where k isapositiveconstant.
Thenegativesignindicatestheaccelerationisdownwards.Airresistancehasbeenignored.Letthe accelerationduetogravityatthesurfaceoftheEarthbe g.
Anobjectisdroppedfrom x = R
a Showthatitlandswithspeed √Rg
b Findthetime t takentoreachthesurfaceoftheEarth.
Solution
Giventhat¨ x = g at x = 0,itfollowsthat g = kR 2.Hence k = R2g
a Rewritethegivendifferentialequationas
d dx
v2 = 2k(x + R) 2
so v2 = 2k(x + R) 1 + C
Usingtheinitialcondition v = 0at x = R,
C = kR 1
Hence
v2 = 2k x + R k R = k R × 2R (x + R) x + R = Rg R x R + x (fromthevalueof k above.)
Clearlyat x = 0, v2 = Rg andhencethespeedis √Rg
b Fromthephysicalsituation,thevelocityisalwaysnegative.Thus
v = √Rg × R x R + x sotakingreciprocalsandrearrangingyields
Integrating,
But x = R at t = 0so
and
Hence
Finally,put x = 0toget
Exercise7C
1 Ineachpart,thevelocity v isgivenasafunctionof x.Itisknownthat x = 1when t = 0.
Express: t intermsof x, i x intermsof t ii v =
2 Ineachmotionofthepreviousquestion,find¨ x usingtheresult¨ x = d dx ( 1 2 v2).
FOUNDATION
3 Ineachpart,theacceleration¨ x isgivenasafunctionof x.Byreplacing¨ x with d dx ( 1 2 v2),express v2 interms of x giventhat v = 0when x = 0.
x = 6x2 a x = 1 ex b x = 1 2x + 1 c x = 1 4 + x2 d
4 Ineachpart,theacceleration¨ x isgivenasafunctionof v.Byreplacing¨ x with dv dt ,express t intermsof v.
x = 2 v2 andwhen t = 0, v = 0 a
¨ x = 2 + v andwhen t = 0, v = 1 c
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x = v2 andwhen t = 0, v = 1 2 b
5 Ineachpart,theacceleration¨ x isgivenasafunctionof v.Byreplacing¨ x with v dv dx ,express x intermsof v
x = v2 4 andwhen x = 0, v = 1 a x = 3 v andwhen x = 0, v = 6 b
x = 2 + v andwhen x = 0, v = 0 c
6 Aparticleofmass m movesinastraightlinesubjecttoaforce F.Attime t,thedisplacementoftheparticle is x andthevelocityis v.Theparticlewasinitiallyatrestattheorigin.
If F = 6t 4and m = 2,find x when t = 4. a
If F = 2x + 1and m = 1.5,findthepositivevalueof v when x = 3. b
If F = 1 v + 2 and m = 0.25,find t when v = 4. c
If F = 1 v + 2 and m = 0.5,find x when v = 3. d
DEVELOPMENT
7 [Aformulafromphysics]Aparticlemoveswithconstantacceleration a,sothatitsequationofmotionis
x = a.Itsinitialvelocityis u.After t seconds,itsvelocityis v anditsdisplacementis s.Use d dx ( 1 2 v2)for accelerationtoshowthat v2 = u2 + 2as
8 Aballisthrownverticallyupwardsat20m/s.Taking g = 10m/s2,upwardsaspositive,andthegroundasthe originofdisplacement,theequationofmotionisthen¨ x = 10.
a Showthat v2 = 400 20x,andfindthegreatestheight.
b Explainwhy v = √400 20x whiletheballisrising.
c Integratetofindthedisplacement–timefunction,andfindhowlongittakestheballtoreachitsgreatest height.
9 Supposethatabulletmovesapproximatelyhorizontallythroughtheairwithdecelerationproportionaltothe squareofitsvelocity.Thatis,¨ x = kv2 forsomepositiveconstant k
a Abulletisfiredat500m/s.At100mitsspeedis300m/s.Use¨ x = v dv dx tofind x asafunctionof v.Then findthedistancetravelled,tothenearestmetre,whenitsspeedis100m/s.
b A differentbulletisfiredat640m/s.After1seconditsvelocityis200m/s.Use¨ x = dv dt tofindwhenits speedis100m/s,correcttothenearesttenthofasecond.
10 Aparticlehasacceleration¨ x = e x,andinitially v = 2and x = 0.Find v2 asafunctionof x,andexplainwhy v isalwayspositiveandatleast2.Thenbrieflyexplainwhathappensastimegoeson.
11
Aparticlehasvelocity v = 6 2x,andinitiallytheparticlewasattheorigin. Findtheaccelerationattheorigin.
a Showthat t = 1 2 ln 1 1 3 x ,andhencefind x asafunctionof t.
c
b Describethebehaviouroftheparticleas t →∞
12 Anobjectisinitiallyatrestattheorigin.Itmovesinastraightlineawayfromtheoriginwithacceleration 2(1 + v)m/s2 Find,correctto3significantfigures: howlongittakesforthevelocitytoreach20m/s, a thedistancetravelledwhenthevelocityreaches20m/s.
13
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b
Aparticle P ofmass m startsfromtheorigin O withvelocity u andmovesinastraightline.When OP = x, where x ≥ 0,thevelocity v of P isgivenby v = u + x k ,where k isapositiveconstant.
a Provethattheforceactingon P isatalltimesproportionalto v,andstatetheconstantofproportionality.
b Giventhatthevelocityis3u atthepoint A,find: thedistance OA intermsof k and u. i thetime,intermsof k,takenby P tomovefrom O to A ii
14 Aparticleofmass0.5kgisacteduponbyaforce F = x 1 2 newtons.Initiallytheparticleisatrest 5metresonthepositivesideoftheorigin.
a Find v2 intermsof x,andhenceexplainwhytheparticlecanneverbeattheorigin.
b Findwherethespeedoftheparticleis2√5m/s,justifyingyouranswer,anddescribethesubsequent motion.
15 Aparticleofmass2kgissubjecttoaforceof6x2 newtons.Initiallytheparticleisat x = 1withvelocity √2m/s.
Find v2 asafunctionof x a Thenfindthedisplacement-timefunction,andbrieflydescribethemotion. b
16
Aparticlestartsfromrestat x = 1onthe x-axis.Itsaccelerationisgivenby d2 x dt2 = 5 2x x3 .
a Explainwhytheparticlestartsmovinginthepositivedirection.
b Let v bethevelocityoftheparticle.Showthat v = √x2 + 4x 5 x ,explainingcarefullywhy v cannotever benegative.
c Describethebehaviourofthevelocityaftertheparticlepassesthrough x = 5 2
17 Theaccelerationofaparticlemovinginastraightlineisgivenby¨ x = 3(1 x2),where x metresisthe displacementoftheparticle.Initiallytheparticlewasattheoriginwithvelocity4m/s.
Find v2 asafunctionof x a Doestheparticleeverchangedirection?Justifyyouranswerwithclearreasoning. b
CHALLENGE
18 Newton’slawofgravitationsaysthatanobjectfallingtowardsaplanethasacceleration¨ x = kx 2,forsome positiveconstant k,where x isthedistancefromthecentreoftheplanet.Showthatifthebodystartsfrom restatadistance D fromthecentre,thenitsspeedatadistance x fromthecentreis 2k(D x) Dx
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19 Aprojectileisfiredverticallyupwardswithspeed V fromthesurfaceoftheEarth.
a Assumingthesameequationofmotionasinthepreviousquestion,andignoringairresistance,showthat k = gR2,where R istheradiusoftheEarth.
b Find v2 intermsof x andhencefindthemaximumheightoftheprojectileabovethecentreoftheEarth.
c [TheescapevelocityfromtheEarth]Giventhat R = 6400kmand g = 9.8m/s2,findtheleastvalueof V sothattheprojectilewillneverreturn.
20 Thevelocityofaparticlemovingonthepositive x-axisisgivenby v = 8 3e 2t m/s.
a Showthat ∫ v 8 v dv = ∫ 2 dx.
b Find,correctto3significantfigures,thedistancetravelledbytheparticleasitsspeedincreasesfrom 5m/sto7m/s.
7D SimpleHarmonicMotionandTime
Manythingsnaturallyexhibitrepeatedpatternsoroscillations.Thehumanbodyprovidesmanyexamplessuch astheheartbeating,thelungsbreathingorthevocalcordsvibrating.Elsewhereinnaturetherearethewaveson abeach,thetidesoftheoceanorthephasesofthemoon.Man-madephenomenaincludetheregularmotionof apistoninasteamengineorcombustionengine,ortheregularebbandflowoftrafficcausedbythephasesof trafficlights.Themathematicsbehindeachoftheseexamplesisextremelycomplicated,buteachcanbebasedon theoscillationsobservedinthegraphsofthesineandcosinefunctions.Thisiswhatwillbestudiedhereandin thenextsection.
SimpleHarmonicMotion
Theshapeofthegraphof x = sin t iscalled sinusoidal.Atranslation,dilationoranycombinationofthese twodoesnotchangethebasicshape.Thusthegraphsoffunctionslike x = 3cos(2t + π 2 )arealsosinusoidal. Wheneverthegraphofdisplacement x versustime t foraparticularmotionissinusoidal,thatmotioniscalled simpleharmonicmotion,or SHM forshort.
Accommodatingthecombinationsoftranslationsanddilations,bothverticallyandhorizontally,theequationof simpleharmonicmotioniscommonlywritteninoneoftwoways:
x = a cos(nt + α) + c,
x = a sin(nt + β) + c.
Sometimesathirdequationisusedandthatwillbediscussedlater.
Thephysicalmeaningsofthevalues a, n and c arethesameinboththese equations.Theconstant c iscalledthe centre ofmotion,asthewaveoscillates symmetricallyeithersideofthehorizontalline x = c,asseeninthegraphon theright.Thevalue a isthe amplitude ofthemotionbecausethewavemovesat most a awayfromthecentre.Thus x alwaysliesintherange c a ≤ x ≤ c + a Theconstant n isusedtocalculatethe period = 2π n ,whichisthetimebetweentwo peaksortwotroughs.
Atanytime t thequantities(nt + α)and(nt + β)areeachcalledthephaseofthemotion,astheirvaluesdetermine thelocation,ratherlikethewaythattimedictatesthephasesofthemoon.Theconstantangles α and β areeach calledthe initialphase ofthemotion,astheydeterminetheinitialdisplacement.When t = 0thetwoformsofthe equationgive
a cos α + c = a sin β + c andsocos α = sin β. Hencetheanglesarecomplementaryand β = π 2 α,thoughitisnotnecessarytomemorisethisformula.
Example14
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Aparticleismovinginsimpleharmonicmotionwithequation x = 2sin( π 2 t + π 6 ) + 3.
a Writedownitscentre,amplitudeandextremesofdisplacement.
b Determinetheperiod,initialphaseandlocationat t = 0.
c Determinewhentheparticleisnextatthesamelocation.
Solution
a Thecentreis x = 3,theamplitudeis2sotheextremesare x = 1and x = 5.
b Theperiodis2π ÷ π 2 = 4,withinitialphase π 6 .At t = 0, x = 2sin π 6 + 3 = 4.
c Solving x = 4gives
2sin( π 2 t + π 6 ) + 3 = 4
thussin( π 2 t + π 6 ) = 1 2 so
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3 ,2
, ... thatis t = 0, 4 3 ,4, Hencetheparticleisnextat x = 4when t = 4 3 .
Thefunctioninthislastexampleistheonegraphedaboveinthediscussion.Bychance,boththevaluesof a and c areintegerswhilstboth n and β areirrational.Ingeneral,therearenorestrictionsexceptthateachisareal number.
6Simpleharmonicmotion
Aparticlewithdisplacement x attime t isinsimpleharmonicmotionifeither
Thecentreofmotionis c,theamplitudeis a andtheperiodis 2π n .Thequantities(nt + α)and(nt + β)are eachcalledthephaseofthemotion.Eachangle α or β iscalledtheinitialphase.
SHMandTransformations
Asstatedearlier,motionwhichissimpleharmonicmaybetheresultofshiftsordilations.Thetransformations appliedto x = cos t inordertoobtain x = a cos(nt + α) + c areinvestigatedinExtension1.Herethehorizontal transformationswillbedonefirstandtheverticalsecond.Ofcourse,thatordercanbereversed.
Firstshift x = cos t leftby α toget x = cos(t + α).Bothofthesefunctionshaveperiod2π.Nowstretch horizontallybyfactor 1 n toget x = cos(nt + α)whichhasperiod 2π n .Nextstretchthewaveverticallybyfactor a,giving x = a cos(nt + α).Finallyshiftthegraphupby c sothatthecentre,alsocalledthemeanposition,is at x = c.Asimilarsequenceoftransformationsisappliedto x = sin t.Thefivegraphsthatfollowshowsucha sequencefrom x = sin
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= 2sin( π 2 t + π 6 ) + 3
Inafewcasesitmaybeconvenienttoapplythehorizontaltransformationsintheoppositeorderbywriting x = a cos n(t + α n ) + c.Inthiscaseahorizontalstretchbyfactor 1 n isfollowedbyashiftleftby α n .
ChoosingtheOrigin
Inmanypracticalproblemsthelocationoftheoriginisnotspecifiedormaybechangedwithoutsignificantly alteringtheproblem.Insuchproblems,theoriginshouldbechosensothattheequationofsimpleharmonic motionisoneofthefollowingfourcases.
x = a cos nt —themotionstartsatthetopandmovesdownwards,
x = a cos nt —themotionstartsatthebottomandmovesupwards,
x = a sin nt —themotionstartsatthecentreandmovesupwards,
x = a sin nt —themotionstartsatthecentreandmovesdownwards.
Noticethatineachcase c = 0.Inthefirstcase α = 0,andinthesecondcase α = π whichaccountsforthe negativesign.Likewise,inthethirdcase β = 0andinthefourthcase β = π
Example15
Aweightishungfromastandonatablebyaspringandsetinverticalmotion.Theweightoscillatesbetween 15cmand35cmabovethetable,andittakes2secondstocompleteonecycle.Itisfoundthatthemotionof theweightmaybemodelledbysimpleharmonicmotion.
a Whatisitsheightabovethetable 3 4 safteritpassesthroughthelowestpoint?
b Forhowlongineachcycleistheweightatorbelow18cm?Approximatetheanswercorrecttotwodecimal places.
Solution
Letthecentreofmotionbe x = 0,25cmabovethetable.Theprobleminvolvestimeafterthebottomofthe wave,solet t = 0there.Theamplitudeofthemotionis a = 10andtheperiodis2 = 2π n so n = π.Henceput
x = 10cos πt
a At t = 3 4 thedisplacementis
x = 10cos 3π 4 = 5√2
hencetheheightis25 + 5√2 ≑ 32.07cm.
35cm x
15cm
b Thequestionisequivalenttosolving 10cos πt ≤−7 orcos πt ≥ 0.7
Forequality,thefirstpositiveandnegativesolutionsare t ≑ ±0.253,thusthe weightisatorbelow18cmforabout0.51secondseachcycle. t x
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7Choosingtheorigin
Wheneverpossible,choosetheoriginofdisplacementandtimesothattheequationofmotionisoneof thefollowingcases:
x = a cos nt —themotionstartsatthetopandmovesdownwards,
x = a cos nt —themotionstartsatthebottomandmovesupwards,
x = a sin nt —themotionstartsatthecentreandmovesupwards,
x = a sin nt —themotionstartsatthecentreandmovesdownwards.
SpeedandAcceleration
Insimpleharmonicmotionthedisplacement x isgivenasafunctionoftime t.Henceitiseasytodifferentiateto findthevelocityandacceleration.Forsimplicityhere,assumethatthecentreofmotionistheorigin.
Let x = a cos(nt + α) then˙ x = na sin(nt + α), and¨ x = n2a cos(nt + α).
Hence¨ x = n2 x
Let x = a sin(nt + β) then˙ x = na cos(nt + β), and¨ x = n2a sin(nt + β).
Hence¨ x = n2 x
Severalimportantobservationscanbemadefromthisalgebra.Themaximumspeedis |x| = na,whichoccurs when | sin(nt + α)| = 1,or(nt + α) = π 2 + kπ.Atthesevaluescos(nt + α) = 0andso x = 0.Thatis,themaximum speedofanobjectinSHMoccursasitpassesthroughthecentreofmotion.Further,theminimumspeediszero, whichoccurswhen(nt + α) = kπ.Atthesevalues |x| = a.Thatis,theminimumspeedcorrespondswiththe turningpointsattheextremesofthemotion.
Themaximumaccelerationis |x| = n2a,whichoccurswhen | cos(nt + α)| = 1,or(nt + α) = kπ.Atthesevalues |x| = a.Thatis,themaximumaccelerationofanobjectinSHMcorrespondswiththeextremesofthemotion,asit ischangingdirection.Theminimumaccelerationiszerowhen(nt + α) = π 2 + kπ.Atthesevalues |x| = 0.Thatis, theminimumaccelerationoccursastheobjectpassesthroughthecentreofmotion.
Example16
Anobjectisinsimpleharmonicmotionwithaperiodof π 3 seconds.Atacertainmomentitspositionis3cm abovethecentreofmotion,anditismovingtowardsthecentreofmotionwithspeed24cm/s.
a Determineitsequationofmotionandhencefinditsmaximumspeed.
b Findwhenitnexthasthesamedisplacement,correcttotwodecimalplaces.
Solution
Letthecentreofmotionbe x = 0.Letthemomentitisobservedcorrespondwith t = 0.Fromtheperiod, n = 2π ÷ π 3 .Thatis n = 6.Sinceitisinitiallymovingdownwards,thevelocityisinitiallynegative.
Let x = a cos(6t + α) then˙ x = 6a sin(6t + α).
At t = 0,fromthegiveninformation,
a cos α = 3
and 6a sin α = 24 or a sin α = 4. (2)
Squaringandadding,itisclearthat
a2 = 25 thus a = 5(a > 0.)
Fromequations(1)and(2)itfollowsthat α isacute,so
α = cos 1 3 5 ≑ 0.9273
a Themaximumspeedis |x| = 6a = 30cm/s.
b Theobjectwillreturntoitsoriginalpositionwhen 5cos(6t + α) = 5cos α socos(6t + α) = cos α
Thefirsttimethishappensis,bythesymmetriesofcosine,when 6t + α = 2π α thus t = 1 3 (π α) ≑ 0.74s.
8Speedandacceleration
• Themaximumspeedisatthecentreofmotionwheretheaccelerationiszero.
• Themaximumaccelerationoccursattheextremesofthemotionwherethespeediszero.Thatis,atthe stationarypoints.
TheDifferentialEquationforSimpleHarmonicMotion
Aboveitwasshown:
for x = a cos(nt + α)
¨ x = n2a cos(nt + α),
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andfor x = a sin(nt + β)
¨ x = n2a sin(nt + β).
Lookcarefullyattheexpressionsfordisplacementandacceleration.Itshouldbeclearthatinbothcasesthe accelerationisrelatedtothedisplacementbytheautonomoussecondorderlineardifferentialequation
x = n 2 x
ApplyingNewton’sformulaforforcegives F = mn2 x.Thismeansthattheforcerequiredtokeepanobjectin SHMisproportionaltoitsdisplacement.Thenegativesignindicatesthattheforceisdirectedtowardsthecentre ofmotion.Theequationmaybeusedtotestwhetherornotagivenmotionissimpleharmonicwithcentrethe origin.Ifthecentreis x = c insteadthenthisequationbecomes
x = n 2(x c), andtheproofofthisisleftasanexercise.
(1)
9Thedifferentialequationforsimpleharmonicmotion
Ifanobjectismovinginsimpleharmonicmotionwithcentre x = c andperiod 2π n then x = n 2(x c).
Consequentlytheforceassociatedwiththismotionisdirectedtowardsthecentre.Thisequationmaybe usedasatestforsimpleharmonicmotion.
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AnotherFormofSimpleHarmonicMotion
Insomeproblemswheretheinitialdisplacementandvelocityareknown,itmaybebesttousethefollowing form:
x = A cos nt + B sin nt
ItiseasytoshowbydifferentiationthatthisisanequationforSHM.
x = nA sin nt + nB cos nt and¨ x = n2 A cos nt n2 B sin nt
= n2(A cos nt + B sin nt) thus¨ x = n2 x.
BythetestinBox 9,thisdifferentialequationconfirmsthatthemotionissimpleharmonicwithcentre x = 0and period 2π n
Example17
Onceagain,anobjectisinsimpleharmonicmotionwithaperiodof π 3 seconds.Atacertainmomentits positionis3cmabovethecentreofmotion,anditismovingtowardsthecentreofmotionwithspeed24cm/s. Let theequationofmotionbe x = A cos nt + B sin nt
a Determinethevaluesof n, A and B.
b Usethe t-formulaetofindwhentheobjectisnextatitsoriginalposition.
Solution
a Asinthepreviousexample, n = 6,and
At t = 0thedisplacementis x = 3so
andthevelocityis˙ x = 24so
thus B = 4.Hence x = 3cos6t 4sin6t
b Theobjectwillreturnatthefirstpositivesolutionof 3cos6t 4sin6t = 3.
Put τ = tan3t,thenthisbecomes 3 × 1 τ2 1 + τ2 4 × 2τ 1 + τ2 = 3 thus 3 3τ2 8τ = 3 + 3τ2 or 6τ2 + 8τ = 0.
Continuedonthenextpage
Henceforthefirstpositivesolution
τ = 4 3 , viz t = 1 3 (π tan 1 4 3 ) ≑ 0.74s(asbefore.)
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10Anotherformofsimpleharmonicmotion
Insomeproblemsforwhichtheinitialdisplacementandvelocityareknown,itmaybebesttouse
x = A cos nt + B sin nt
Thiscorrespondstosimpleharmonicmotionwithcentre x = 0andperiod 2π n
Ifitisnotpossibletoplacetheoriginatthecentreofmotionthensimplyshifttheabovefunctiontohavecentre at x = c bywriting
x = A cos nt + B sin nt + c.
Finally,itshouldbeobviousfromtheworkontrigonometryinYear11thattheamplitude a whenSHMiswritten inthiswayis
a = √A2 + B2
Theproofisleftasanexercise.
Exercise7D FOUNDATION
1 Aparticleismovinginsimpleharmonicmotionabouttheorigin.Itsdisplacement x cmafter t secondsis givenby x = 12cos π 2 t
a Whataretheamplitudeandperiodofthemotion?
b Differentiatetofind v and¨ x asfunctionsof t,andthenshowthat¨ x =
c Whataretheinitialdisplacementandvelocityoftheparticle?
d Whenistheparticlefirstattheorigin?
e Howlongisitbetweenvisitstotheorigin?
2 Aparticlemovessothatitsdisplacement x metresafter t secondsisgivenby x = 2sin4πt
a Writedowntheamplitudeandperiodofthemotion.
b Sketchthedisplacementfunctionfor0 ≤ t ≤ 1.
c Findthevelocityandaccelerationasfunctionsoftime.
d Findtheaccelerationasafunctionofdisplacementandhenceshowthattheparticleismovinginsimple harmonicmotion.
e Findthefirsttwotimestheparticleisatrest,anditsaccelerationatthosetimes.
f Whatisthegreatestspeedoftheparticle?
3 Aparticleismovinginsimpleharmonicmotionwithdisplacement x = 4 π sin πt,inunitsofmetresand seconds.
a Showthattheparticleisinitiallyattheorigin.
b Differentiatetofind v and¨ x asfunctionsoftime,andshowthat¨ x = π2 x.
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c Whataretheamplitudeandperiodofthemotion?
d Whatisthemaximumdistancefromtheoriginandwhatisthemaximumspeed?
e Sketchoneperiodofthegraphsof x, v and¨ x againsttime.
f Findthenexttwotimestheparticleisattheorigin,andthevelocitiesthen.
g Findthefirsttwotimestheparticleisstationary,andtheaccelerationsthen.
4 Ineachofthefollowingpartsfind a and n,andhencewritethedisplacementfunctionintheform x = a sin nt.
a AparticlemovinginSHMwithcentretheoriginandperiod π secondsstartsfromtheoriginwith velocity4m/s.
b AparticlemovinginSHMwithcentretheoriginandamplitude6metresstartsfromtheoriginwith velocity4m/s.
5a Aparticle’sdisplacementisgivenby x = b sin nt + c cos nt,where n > 0.Find v and¨ x asfunctionsof t Thenshowthat¨ x = n2 x,andhencethatthemotionissimpleharmonic.
b Bysubstitutingintothefunctionsfor x and v:
i find b and c ifinitiallytheparticleisatrestat x = 3, ii find b, c and n,andthefirsttimetheparticlereachestheorigin,iftheparticleisinitiallyatrestat x = 5,andtheperiodis1second.
6 Aparticlemovinginastraightlinestartedfromtheoriginwithvelocity4π cm/s.Itsdisplacementafter t secondsisgivenby x = a sin πt.
a ProvethatthemotionsatisfiesthedifferentialequationforSHM.
b Findthevalueof a
c Findthefirsttwotimesthatthespeedoftheparticleis2π cm/s.
7 Aparticle’sdisplacementis x = 12 2cos3t,inunitsofcentimetresandseconds.
a Differentiatetofind v and¨ x asfunctionsof t,showthattheparticleisinitiallystationaryat x = 10,and sketchthedisplacement–timegraph.
b Whataretheamplitude,periodandcentreofthemotion?
c Inwhatintervalistheparticlemoving,andhowlongdoesittaketogofromoneendtotheother?
d Findthefirsttwotimesaftertimezerowhentheparticleisclosesttotheorigin,andthespeedand accelerationthen.
e Findthefirsttwotimeswhentheparticleisatthecentre,andthespeedandaccelerationthen.
8 AparticleismovinginSHMaccordingtotheequation x = 6sin(2t + π 2 ).
a Whataretheamplitude,periodandinitialphase?
b Find˙ x and¨ x,andshowthat¨ x = n2 x,forsome n > 0.
c Findthefirsttwotimeswhentheparticleisattheorigin,andthevelocitythen.
d Findthefirsttwotimeswhenthevelocityismaximum,andthepositionthen.
e Findthefirsttwotimestheparticlereturnstoitsinitialposition,anditsvelocityandaccelerationthen.
DEVELOPMENT
9 Aparticleisoscillatinginsimpleharmonicmotionabouttheoriginwithperiod24secondsandamplitude 120metres.Initiallyitisattheoriginwithpositivevelocity.
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a Writedownfunctionsfor x and v,andstatethemaximumspeed.
b Whatisthefirsttimewhenitis30metres:(Answercorrecttofoursignificantfigures.) totherightoftheorigin, i totheleftoftheorigin? ii
c Findthefirsttwotimesitsspeedishalfitsmaximumspeed.
10 Aparticlemovesinsimpleharmonicmotionabouttheoriginwithperiod π 2 seconds.Initiallytheparticleis atrest4cmtotherightof O
a Writedowndisplacement–timeandvelocity–timefunctions.
b Findhowlongtheparticletakestomovefromitsinitialpositionto: apoint2cmtotherightof O, i apoint2cmontheleftof O. ii
c Findthefirsttwotimeswhenthespeedishalfthemaximumspeed.
11 Theequationofmotionofaparticleis x = sin2 t.Useadouble-angleidentitytoputtheequationintheform x = x0 a cos nt,andstatethecentre,amplitude,rangeandperiodofthemotion.
12 Aparticlemovesaccordingto x = 3 2cos2 2t,inunitsofcentimetresandseconds.
a Useadouble-angleidentitytoputtheequationintheform x = x0 a cos nt
b Findthecentreofmotion,theamplitude,therangeofthemotionandtheperiod.
c Whatisthemaximumspeedoftheparticle,andwhendoesitfirstoccur?
13 Thedisplacement x cmofaparticleafter t secondsisgivenby x = 2 + 3cos t + 3√3sin t
a Find¨ x andhenceprovethatthemotionissimpleharmonic.
b Whereisthecentreofmotion?
c Whatistheperiod?
d Express3cos t + 3√3sin t intheform a cos(t θ),where a > 0and0 < θ < π 2 .
e Hencestatetheamplitudeandtheinitialphase.
f Withinwhatintervaldoestheparticleoscillate?
14 Aparticle’sdisplacementisgivenby x = b sin nt + c cos nt,where n > 0.Find v asafunctionof t.Thenfind n, c and b,andthefirsttwotimestheparticleisattheorigin,if:
a theperiodis4π,theinitialdisplacementis6andtheinitialvelocityis3, b theperiodis6andwhen t = 0, x = 2and˙ x = 3.(Inthispartwritethetimescorrectto3decimal places.)
15 Giventhat x = a sin(nt + α)(inunitsofmetresandseconds),write v asafunctionoftime.Find a, n and α if a > 0, n > 0,0 ≤ α < 2π and: theperiodis6seconds,andinitially x = 0and v = 5, a theperiodis3π seconds,andinitially x = 5and v = 0, b theperiodis2π secondsandinitially x = 1and v = 1. c
16 Giventhat x = a cos(2t + α),find a and α if a > 0, π < α ≤ π and: initially x = 0and v = 6, a initially x = 1and v = 2√3. b
17
18
19
Aparticleismovinginsimpleharmonicmotionaccordingto x = a cos( π 8 t + α),where a > 0and 0 ≤ α < 2π.When t = 2itpassesthroughtheorigin,andwhen t = 4itsvelocityis4cm/sinthenegative direction.Findtheamplitude a andtheinitialphase α.
Aparticleismovinginsimpleharmonicmotionwithperiod8π secondsaccordingto x = a sin(nt + α), where x isthedisplacementinmetres,and a > 0and0 ≤ α < 2π.When t = 1, x = 3and v = 1.Find a and α correcttofoursignificantfigures.
Aparticlemovinginsimpleharmonicmotionhasperiod π 2 seconds.Initiallytheparticleisat x = 3with velocity v = 16m/s.
a Find x asafunctionof t intheform x = b sin nt + c cos nt.
b Find x asafunctionof t intheform x = a cos(nt ε),where a > 0and0 ≤ ε < 2π
c Findtheamplitudeandthemaximumspeedoftheparticle.
d Findthefirsttimetheparticleisattheorigin,usingeachoftheabovedisplacementfunctionsinturn. Provethatthetwoanswersobtainedarethesame.
20 Thetemperatureateachinstantofadaycanbemodelledbyasimpleharmonicfunctionoscillating between9◦ at4:00amand19◦ at4:00pm.Find,correcttothenearestminute,thetimesbetween4:00am and4:00pmwhenthetemperatureis:
21 Theriseandfallinsealevelduetotidescanbemodelledbysimpleharmonicmotion.Onacertainday,a channelis10metresdeepat9:00amwhenitislowtide,and16metresdeepat4:00pmwhenitishightide. Ifashipneeds12metresofwatertosaildownachannelsafely,atwhattimes(correcttothenearestminute) between9:00amand9:00pmcantheshippassthrough?
22 Showthat,foranyparticlemovinginsimpleharmonicmotion,theratiooftheaveragespeedoverone oscillationtothemaximumspeedis2: π.
CHALLENGE
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23 Themotionofaparticleinastraightlineisgovernedbythedisplacementfunction x = 4sin(3t + π 6 ) + 2sin3t.
a Provethatthemotionissimpleharmonic.
b Findtheamplitude.
24 AparticlemovinginSHMabouttheoriginstartsat x = 1withpositivevelocity.Thefirsttwotimesthe particleisat x = 5areattheendofeachofthefirsttwoseconds.
Provethattheperiodofthemotionis 2π cos 1 3 5 .
[Hint:Letthedisplacementfunctionbe x = a cos(nt + α).]
7E SimpleHarmonicMotionandDisplacement
Thefocusoftheprevioussectionwasonthedisplacement-timefunctionforsimpleharmonicmotion.Inthis sectionitisthedifferentialequation
¨ x = n 2(x c)
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thattakescentrestage.Firstlyitwillbeusedasatestforsimpleharmonicmotion.Thentheidentity d2 x dt2 = d dx 1 2 v2 willbeusedwithintegrationtofindthevelocityasafunctionofdisplacement.
SimpleHarmonicMotionandtheDifferentialEquation
Inthelastsectionitwasshownthatifaparticleisinsimpleharmonicmotionwithcentre x = 0andperiod 2π n then
x = n2 x
Thereisacleverwaytoprovethecorrespondingresultwhenthecentreisnotzero.Let y = x + c,thentheparticle isinsimpleharmonicmotionwithcentre y = c.Now
x = y c
so˙ x = y and¨ x = y.
Hencebysubstitution,theabovedifferentialequationbecomes
y = n2(y c).
Finally,sincethechoiceofpronumeralwasarbitrary,ifaparticleisinsimpleharmonicmotionwithcentre x = c andperiod 2π n then
¨ x = n2(x c).
Thisequationmaybetakenasatestforsimpleharmonicmotion.Thatis,ifthedisplacementofaparticle satisfiesthisequationthentheparticleisinsimpleharmonicmotion.Theproofthatthemotionissimple harmonicisgivenlater.
Example18
Thedisplacement x cmofaparticleattime t secondssatisfies¨ x = 4(x 3).Theparticleisinitiallyat x = 3 withvelocity6cm/s.Determinethedisplacement-timefunctionandfindwhentheparticleisfirstattheorigin.
Solution
Since¨ x = 22(x 3),theparticleisinsimpleharmonicmotionwithcentre x = 3andperiod π.Theparticle startsatitscentrewithpositivevelocity,soput
x = 3 + a sin2t
Thus˙ x = 2a cos2t soat t = 0
6 = 2a andhence x = 3 + 3sin2t
Continuedonthenextpage
Theparticlewillbeattheoriginwhen
sin2t = 1
or2t = , π 2 , 3π 2 , 7π 2 , sothefirstpositivesolutionis
t = 3π 4
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11Simpleharmonicmotionandthedifferentialequation
Ifthedisplacementofaparticlesatisfiestheequation
¨ x = n 2(x c)
thenitisinsimpleharmonicmotionwithcentre x = c andperiod 2π n .
Theproofthatthemotionmustbesimpleharmonicisrelativelystraightforwardwhenthecentreis x = 0.The substitution y = x + c canthenbeusedtoextendtheprooftomotionwithothercentres.
Proof Supposethat¨ x = n2 x.Regardlessofthevaluesof x(0)and˙ x(0),itisalwayspossibletofind a and α suchthat x(0) = a sin α and˙ x(0) = na cos α.(Theproofofthisisleftasanexercise.)Nowconsiderthe function
u = x a sin(nt + α)
Firstly, u(0) = x(0) a sin α = 0.
Next, ˙ u = ˙ x na cos(nt + α)
so u(0) = x(0) na cos α = 0.
Further,¨u = x + n2a sin(nt + α) = n2 x + n2a sin(nt + α)
so u = n2u
Thus d du 1 2 ˙ u2 = n2u or d du ˙ u2 = 2n2u
hence˙u2 = C n2u2 forsomeconstant C
Nowapplytheinitialconditions u(0) = u(0) = 0toget C = 0.
Thus˙u2 = n2u2
But u isarealfunctionandsquaresofrealsareeitherpositiveorzero.Thustheonlysolutiontothis equationisthat u = 0forall t.Hence
x = a sin(nt + α) andthemotionissimpleharmonic. ItisnowpossibletorevisitExample 11,concerningthemotionofaspring.Inthatproblem,
x = x
andthusthemotionissimpleharmonicwithcentre x = 0.Theinitialconditionsare x(0) = a and ˙ x(0) = 0,soitfollowsthatthedisplacement-timefunctionis
x = a cos t
Itisnowclearthat x takesallvaluesintherange a ≤ x ≤ a as t varies.
VelocityasaFunctionofDisplacement
Startingwiththedifferentialequationforacceleration
d dx 1 2 v2 = n2(x c) or d dx v2 = 2n2(x c)
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thus v2 = D n2(x c)2 .
Nowateachextremeofmotion,thevelocityiszero,thus
0 = D n2a2 .
Thatis D = n2a2 and v2 = n2a2 n2(x c)2 or v2 = n2 (x c)2 a2
Noticethattheterminsidetheouterbracketsisamonicquadraticwithitssquarecompleted.Studentsshould ensuretheyareabletoderivethisformulawhenrequired.Itcanbeusedtodeterminethecentre,periodand amplitudeofthemotion.Inpractice,however,itmaybebettertofirstcheckthatthemotionissimpleharmonic bydifferentiation,thusfindingthecentreand n.Thenset v = 0tofindtherangeandamplitude.
Example19
Themotionofaparticlesatisfies v2 = 2x2 + 8x + 10.Showthatthemotionissimpleharmonicthendetermine theperiod,centreofmotion,rangeandamplitude.
Solution
Firstly,differentiatetofindtheacceleration.
x = d dx x 2 + 4x + 5
= 2x + 4 = 2(x 2).
Hencethemotionissimpleharmonicwith n = √2.Theperiodis π√2andcentreis x = 2.Nowput v = 0to findtherange.
x2 4x 5 = 0
or(x + 1)(x 5) = 0, sotheextremesofthemotionare x = 1and x = 5.Takingtheaverageconfirmsthecentreis x = 2.Taking thedifferenceandhalvinggivestheamplitude a = 3.
12Velocityasafunctionofdisplacement
Firstcheckthatthemotionissimpleharmonicbydifferentiation,andsofindthecentreand n.Thenset v = 0tofindtherangeandamplitude.
FindingtheEquationofMotionfromtheGraph
Itiseasytodeterminetheequationofsimpleharmonicmotionfromitsgraph.Itissimplyamatterof determiningtheshiftsandstretchesappliedtosineorcosine.Aquickexampleisincludedhereasareminderof theprocess.
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Example20
Themotionofabodyisplottedinthegraphontheright.Assumingthemotiontobe simpleharmonic,findthedisplacement-timefunction.
Solution
Theperiodis4 = 2π n hence n = π 2
Theextremesofmotionare x = 1and x = 5.
Thecentreisthemean c = 5+ 1 2 = 2.
Theamplitudeishalftherange = 5+1 2 = 3.
Hence x = 2 + 3sin( π 2 t + α)with α acutesothattheinitialvelocityispositive.
Now˙ x = 3π 2 cos( π 2 t + α) andthevelocityiszeroattheextremewhen t = 1 3 .Thus 0 = cos( π 6 + α) whereby π 6 +
TheSlopeFieldforSHM
Thedifferentialequationforsimpleharmonicmotioncannotbesolvedusinghighschoolmathematics. Nevertheless,theslopefieldsuggeststhesolution x = R sin nt forSHMabouttheorigin.
Rewritingtheaccelerationintermsofdisplacementandvelocitygives
v dv dx = n 2 x.
Nowput v = n × y toget y dy dx = x x y
Theslopefieldforthisdifferentialequationisshownabove.Thepatternofslopessuggeststhatthesolution curvesareconcentriccircleswithcentreattheorigin.
Itislefttothereadertoshowthatacirclewithcentretheoriginandradius R canbeparameterisedby
x = R sin θ and y = R cos θ wheretheangle θ ismeasuredclockwisefrom(0, R).Thus, dx dt = dx dθ × dθ dt
= R cos θ × dθ dt and dx dt = v = ny = nR cos θ.
Hence dθ dt = n and θ = nt + C, where C issomeconstantinthedomain π < C ≤ π
Butat t = 0, x = 0so C = 0andfinally, x = R sin nt.
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Exercise7E
1 Thedisplacement x metresofaparticleafter t secondsisgivenby x = 3cos2t
a Findexpressionsfor v and¨ x asfunctionsof t,andfor¨ x and v2 intermsof x
b Findthevelocityandaccelerationoftheparticleat x = 2.
FOUNDATION
2 Themotionofaparticleisgovernedbytheequation¨ x = 9x (inunitsofmetresandseconds).Theparticle isstationarywhen x = 5.
a
Integratetofindanequationfor v2
b Whatisthespeedattheorigin?
Findthevelocityandaccelerationwhen x = 3.
c
3 Aparticleisoscillatingaccordingtotheequation¨ x = 16x (inunitsofcentimetresandseconds),andits speedattheoriginis24cm/s.
Integratetofindanequationfor v2 a
Whataretheamplitudeandtheperiod? b
c
Findthespeedandaccelerationwhen x = 2.
4 AparticlemovesinSHMaccordingtotheequation¨ x = 4x (inunitsofmetresandseconds).The amplitudeis6metres.
a Findthevelocity–displacementequation,theperiodandthemaximumspeed.
b Findthesimplestformofthedisplacement–timeequationifinitiallytheparticleis: stationaryat x = 6, i stationaryat x = 6, ii attheoriginwithpositivevelocity, iii attheoriginwithnegativevelocity. iv
5a Themotionofaballontheendofaspringismodelledbytheequation¨ x = 256x (inunitsof centimetresandseconds).Theballispulleddown2cmfromtheoriginandreleased.Findthespeedat thecentreofmotion.
b Themotionofanotherballontheendofaspringismodelledby¨ x + 1 4 x = 0(inunitsofcentimetresand seconds),anditsspeedattheequilibriumpositionis4cm/s.Howfarwasitpulleddownfromtheorigin before itwasreleased?
6 [Inthesequestionsusetheformula v2 = n2(a2 x2).]
a Aparticlemovinginsimpleharmonicmotionhasperiod π 2 minutes,anditstartsfromthemeanposition withvelocity4m/min.Findtheamplitude.
b Themotionofabuoyfloatingontopofthewavescanbemodelledassimpleharmonicmotionwith period3seconds.Ifthewavesriseandfall2metresabouttheirmeanposition,findthebuoy’sgreatest speed.
7 AparticleoscillatesinSHMbetweentwopoints A and B thatare20cmapart.Theperiodis8seconds. Let O bethemidpointof AB.
a Findthemaximumspeedandmaximumacceleration.
b Findthevelocityandaccelerationwhentheparticleis6cmfrom O
8 Theamplitudeofaparticlemovinginsimpleharmonicmotionis5metres,anditsaccelerationwhenitis 2metresfromitsmeanpositionis4m/s2.Findthespeedoftheparticlewhenitisatitsmeanposition,and alsowhenitis4metresfromitsmeanposition.
9 AparticleismovinginSHMwithperiod π secondsandmaximumspeed8m/s.Findtheamplitude,and findthespeedwhentheparticleis3metresfromitsmeanposition.
10 Aparticlemovesinastraightlinesothatitsaccelerationisproportionaltoitsdisplacement x fromthe origin O.When4cmonthepositivesideof O,itsvelocityis20cm/sanditsaccelerationis 6 2 3 cm/s2.Find theamplitudeofthemotion.
11 Aparticlemovesinsimpleharmonicmotionwithcentre O,andpassesthrough O withspeed10√3cm/s. Determinethespeedoftheparticlewhenitishalfwaybetweenitsmeanpositionandanendpoint.
12 Aparticlemovinginsimpleharmonicmotionabouttheoriginstartsattheoriginwithvelocity V.Provethat theparticlefirstcomestorestaftertravellingadistanceof V/n.
13 [Thegeneralcase]Supposethataparticleismovinginsimpleharmonicmotionwithamplitude a and equationofmotion¨ x = n2 x,where n > 0.
a Useintegrationtoprovethat
2).
b Findexpressionsfor: thespeedattheorigin, i thespeedandaccelerationhalfwaybetweentheoriginandthemaximumdisplacement. ii
14a Aparticlemovesinastraightlineaccordingtotheequation v2 = 9x2 + 18x + 27.Provethatthemotion issimpleharmonic,andfindthecentreofmotion,theperiodandtheamplitude.
b Repeatpart a for:
v2 = 80 + 64x 16x2 i
v2 = 2x2 8x 6 iii
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v2 = 9x2 + 108x 180 ii
v2 = 8 10x 3x2 iv
15a Showthatmotiondefinedby x = sin2 5t withunitsofmetresandminutessatisfies¨ x = n2(x c),for some c andsome n > 0,by: firstwritingthedisplacementfunctionas x = 1 2 1 2 cos10t, i differentiating x directlywithoutanyuseofdoubleangleidentities. ii
b Findthecentre,rangeandperiodofthemotion,andthenexttimeitvisitstheorigin.
16 Aparticlemovesinsimpleharmonicmotionaccordingtotheequation¨ x = 9(x 7),inunitsofcentimetres andseconds.Itsamplitudeis7cm.
a Findthecentreofmotion,andhenceexplainwhythevelocityattheoriginiszero.
b Integratetofind v2 asafunctionof x,completethesquareinthisexpression,andhencefindthe maximumspeed.
c Explainhow,althoughtheparticleisstationaryattheorigin,itneverthelessmovesawayfromtheorigin.
17 Aparticleismovingaccordingtotheequation x = 4cos3t 6sin3t
a Provethattheaccelerationisproportionaltothedisplacementbutoppositelydirected,andhencethatthe motionissimpleharmonic.
b Findtheperiod,amplitudeandmaximumspeedoftheparticle,andfindthemagnitudeoftheaccelerationwhentheparticleishalfwaybetweenitsmeanpositionandoneofitsextremepositions.
18 Themotionofaparticleisgovernedbytheequationby x = 3 + sin4t + √3cos4t Provethat¨ x = 16(x 3),andwritedownthecentreandperiodofthemotion. a Expressthemotionintheform x = x0 + a sin(4t + α),where a > 0and0 ≤ α < 2π. b Findthefirstthreetimesthattheparticleisatthecentre,anditsspeedthere.
c
19 Aparticlemovesaccordingtotheequation x = 10 + 8sin2t + 6cos2t.
a Provethatthemotionissimpleharmonic,andfindthecentreofmotion,theperiodandtheamplitude.
b Find,correcttofoursignificantfigures,whentheparticlefirstreachestheorigin.
20 [Simpleharmonicmotionistheprojectionofcircularmotionontoadiameter.] AFerriswheelofradius8metresmountedinthenorth-southplaneisturning anticlockwiseat1revolutionperminute.Attimezero,Zorbaislevelwiththe centreofthewheelandnorthofit.
a Let x and y beZorba’shorizontaldistancenorthofthecentreandheightabove thecentrerespectively.
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CHALLENGE
Showthat x = 8cos2πt and y = 8sin2πt. x y Zorba
b Findexpressionsfor˙ x,˙ y,¨ x and¨ y,andshowthat¨ x = 4π2 x and¨ y = 4π2y.
c Findhowfar(inradians)thewheelhasturnedduringthefirstrevolutionwhen: x : y = √3:1 i x :˙ y = √3:1 ii x = y iii
21 Aparticlemovesinsimpleharmonicmotionaccordingto¨ x = n2 x
a Provethat v2 = n2(a2 x2),where a istheamplitudeofthemotion.
b Theparticlehasspeeds v1 and v2 whenthedisplacementsare x1 and x2 respectively.Showthatthe period T isgivenby T = 2π x1 2 x2 2 v2 2 v1 2 , andfindasimilarexpressionfortheamplitude.
c Theparticlehasspeedsof8cm/sand6cm/swhenitis3cmand4cmrespectivelyfrom O.Findthe amplitude,theperiodandthemaximumspeedoftheparticle.
22 Aparticlemovinginsimpleharmonicmotionhasamplitude a andmaximumspeed V.Finditsvelocity when x = 1 2 a,anditsdisplacementwhen v = 1 2 V.Provealsothemoregeneralresults
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23 Twoballsonelasticstringsaremovingverticallyinsimpleharmonicmotionwiththesameperiod2π and withcentreslevelwitheachother.Thesecondballwassetinmotion α secondslater,where0 ≤ α < 2π, withtwicetheamplitude,sotheirequationsare
x1 = sin t and x2 = 2sin(t α).
Let x = sin t 2sin(t α)betheheightofthefirstballabovethesecond.
a Showthat¨ x = x,andhencethat x isalsosimpleharmonicwithperiod2π
b Showthatthegreatestverticaldifference A betweentheballsis A = √5 4cos α.Whatarethe maximumandminimumvaluesof A,andwhatformdoes x thenhave?
c Showthattheballsarelevelwhentan t = 4T 1 3T 2 ,where T = tan 1 2 α.Howmanytimesaretheylevelin thetimeinterval0 ≤ t < 2π?
d Ifthedistancebetweentheballsisknowntobegreatestwhen t = 0,whatvaluescould α have,andwhat formsdoes x have?
7F HorizontalResistedMotion
Ifanobjectmoveshorizontallythengravitymaybeeffectivelyignored.Typicallytheobjectisnotentirelyfreeto moveasthereisusuallyaresistiveforce,duetofrictionforexample,whichactsintheoppositedirectiontothe velocity.Intheproblemsofthissection,theequationsofmotionwilleitherbespecifiedinthequestionorwill needtobedeterminedbybalancingforces.
TwoCommonIntegrals
Alargenumberofproblemsencounteredinthissectionandthenextresultinintegralsoftheform
Noticethatinbothcasesthenumeratorisamultipleofthederivativeofthedenominator.Hencetheresultsare logarithmicfunctions.Thus
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forsomeconstant C.Andinthesecondinstance
Noticethatinbothcasesnoabsolutevaluesareused.Inthepracticalproblemsthatrequiretheseintegrals,the quantitieswillgenerallybepositive,andsonoabsolutevalueisneeded.Everyquestionshouldberoutinely checked,however,andtheabsolutevaluesre-insertediftheyarerequired.Theseformulaewillbeusedinthe exampleswithoutfurtherexplanation.
Example21
Arowingeightcrossesthefinishlinewithspeed5.5m/s,andstopsrowing.Inagreatlysimplifiedmathematicalmodel,theboatisslowedbytwodragforces.Theskindragisduetothesurfaceareaincontactwiththe waterandisequalto 1 10 mv2.Theformdragisduetotheshapeoftheboatpushingthewaterasideandisequal to 1 100 mv.Thustheforceequationis
mx = m(v + 10v2) 100 where x isthedistancepastthefinishline,and v isthespeedoftheboat.
a Find v asafunctionof x
b Howfar,correcttothenearestmetre,doestheboateventuallytravelasitcomestoastop?
Solution
a Dividethroughbythemass m anduse¨ x = v′v toget
v′v = v + 10v2 100 , thenre-arrangeandintegratewithrespectto x:
∫ 10v′ 1 + 10v dx = ∫ 1 10 dx
Solog(1 + 10v) = x 10 + C (forsomeconstant C.)
Continuedonthenextpage
At x = 0theboatspeedis v = 5.5so
C = log(56)
hencelog 1 + 10v 56 = x 10 .
Thus 1 + 10v 56 = e x 10
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or v = 1 10 56e x 10 1 .
b Let v → 0+ inthethirdlastlineabovetoget
x 10 = log 1 56
thus x = 10log(56) ≑ 40m.
Infact,accordingtothismathematicalmodel,theboatneverstopsmovingandonlyapproachesthisdistance inthelimitas t →∞.Asanexercise,find v then x asfunctionsof t andhenceexplainwhythisisthecase.
GeneralisedSolutions
Itisoftenefficienttosolveacertainproblemonce,usingsuitablepronumeralsinplaceofthevariousconstants. Oncethisgeneralisedsolutionisfound,itissimplyamatterofsubstitutingthevaluesoftheconstantstogetthe finalsolution.Inthenextexampleitisaparticularlyusefultechniqueasaratioofspeedsisspecified.
Example22
Anobjectofmass m moveshorizontally,startingattheoriginwithvelocity V0 > 0.Itexperiencesaresistance duetofrictionwhichisproportionaltothesquareofitsspeedandintheoppositedirection.
a Findanexpressionforthevelocity v intermsofthedisplacement x
b Foracertain5kgobjecttheresistanceisequalto 1 10 v2.Findhowfartheobjecthastravelledwhenithas slowedto 1 4 ofitsinitialspeed.
Solution
a Let k betheconstantofproportionalityforthefrictionalforce.Sinceitisoppositeindirectiontothe velocity,andtherearenootherforces: mx = kv2 or v dv dx = k m v2
so v′ v = k m .
Integratingwithrespectto x: ∫ v′ v dx = ∫ k m dx.
Thuslog v = k m x + C (forsomeconstant C)
or v = e k m x+C
so v = Ae kx/m (where A = eC .)
At x = 0 V0 = A
thus v = V0e kx/m
b Here k = 1 10 , m = 5and v = 1 4 V0.Thus
1 4 = e x/50 or ex/50 = 4.
Thus x = 50log(4) ≑ 69m.
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OtherMethods
Numerousmethodscanbeusedtosolvetheequationsencountered.Theaboveexampleisnowextendedto demonstratetwoofthosemethodsappropriatetothecourse.Inthefirstcase,thederivativeistreatedlikea fraction,andinthesecondcase,definiteintegralsareused.Teachersandstudentsareencouragedtoinvestigate othertechniques,suchasseparationofvariables,usinganintegratingfactor,andtreating(v2)likeavariable.
Example23
Startingwith v = V0e kx/m,findthedisplacementasafunctionoftime.
Solution
Firstreplace v withthederivative dx dt .
dx dt = V0e kx/m
Nowtreatthederivativelikeafractiontoget
dt
dx = ekx/m
V0
Thus t = mekx/m kV0 + C.
Recallthatat t = 0, x = 0,so
0 = m kV0 + C or C = m kV0 hence t = m kV0 ekx/m 1
Finally,re-arrangethisequationtoget,
ekx/m = kV0t + m m or x = m k log kV0t + m m .
Example24
Startingwith mv = kv2,findthetimetakenfortheobjecttoreduceinspeedfrom V0 to 1 2 V0
Solution
Onceagain,treatthederivativelikeafractionandre-arrange.
dv = m kv2
Thusthetimecanbewrittenasaprimitivefunctionof v asfollows:
) = ∫ m kv2 dv
Inthisquestion,thetimerequiredisthevalueof
But,bytheFundamentalTheoremofCalculus,thisisjustadefiniteintegral.
Thus
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Noticethatnoconstantofintegrationneededtobefoundbecauseadefiniteintegralwasused.Nowthatthe methodofdefiniteintegralshasbeenestablished,thetechniquecanbeusedinfutureproblemswithoutthe needforthedetailedexplanationgivenabove.
Exercise7F
FOUNDATION
1 Acertaindrag-racingcarofmass M kgiscapableofatopspeedof288km/h.Afteritreachesthistop speed,twodifferentretardingforcescombinetobringittorest.Firstthereisaconstantbrakingforceof magnitude 2 3 M newtons.Secondlythereisaresistiveforceofmagnitude 1 180 Mv2 newtons,where v m/sis thespeedofthecar,actingagainstaparachutereleasedfromtherear-endofthevehicle.Let x metresbethe distanceofthecarfromthepointatwhichthetworetardingforcesareactivated.
a Showthat x = 90ln 120 + 802 120 + v2 .
b Hencecalculate,tothenearestmetre,thedistancethatthedrag-racingcartravelsasitisbroughtfromits topspeedtorest.
2 Amonorailofmass10000kgispullingoutofastation S .Itsmotorprovidesapropellingforceof magnitude10000newtons,andasitmovesitexperiencesaresistiveforceofmagnitude100v2 newtons, where v metrespersecondisitsvelocity.
a Showthatthemaximumspeedthemonorailcanattainis36km/h.
b Showthat x = 50ln 100 100 v2 ,where x metresisthedistancethemonorailhastravelledfrom S
c Whatpercentage(tothenearestpercent)ofitsmaximumspeedhasthemonorailreachedwhenithas travelled50metres?
3
Aparticleofmass m kgexperiencesaresistanceof kv2 newtonswhenmovingalongthe x-axis,where k is apositiveconstantand v isthespeedoftheparticleinmetrespersecond.Themaximumspeedattainable bytheparticleis u metrespersecondunderavariablepropellingforceof P v newtons,where P isapositive constant.
Showthat k = P u3 a
Showthat dv dx = P m 1 v2 v u3 b
c Provethatthedistancetravelledasthespeedchangesfrom 1 3 u m/sto 2 3 u m/sis mu3 3P ln 26 19 metres.
d Whenthebrakesareapplied,thepropellingforceisnolongerinoperation.Ifthemaximumforceexerted bythebrakesis B newtons,provethattheminimumdistancetravelledincomingtorestfromaspeedof u m/sis mu3 2P ln 1 + P Bu metres.
DEVELOPMENT
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4 Asimplemodelofadoorclosingmechanismisaspringanddashpot.The springpullsthedoorclosedandthedashpot,agasoroilfilledpiston,resists themotion,whichensuresthatthedoordoesnotclosetoofastandslamshut. Aschematicdiagramisshownontheright.
Foraparticulardoorwithmass m,theforceexertedbythespringis 2my and theresistancefromthedashpotis 3my,where y isthedisplacementofthe doorfromthedoorjamb.
a Whatisthesignificanceoftheminussignineachforce?
b Writedowntheequationofmotionforthedoor.
doorjambdoor y
c Showthat,if y = f (t)and y = g(t)arebothsolutionstotheequationofmotion,andif A and B are constants,then y = Af (t) + Bg(t)isalsoasolution.
d Itisknownthatthefunction y = ekt isasolutionofthedifferentialequation. Showthattheonlypossiblevaluesoftheconstant k are k = 1and k = 2.
e Fromparts c and d,asolutionofthedifferentialequationis
y = Ae 2t + Be t .
When t = 0,itisknownthat y = 0and dy dt = 1.Findthevaluesof A and B.
5 Theenginesonasubmarineofmass m deliveramaximumdrivingforceof F newtons.Thewaterresists themotionwithaforceproportionaltothesquareofthespeed v.
a Explainwhy dv dt = 1 m F kv2 where k isapositiveconstant.
b Thesubmarineincreasesitsspeedfrom v1 to v2.Showthatthedistancetravelledduringthisperiodis m 2k × loge F kv1 2 F kv2 2 .
6 Asaparticleofunitmassmovesinastraightline,theonlyforceactingonitisaresistance,whichisinthe oppositedirectiontoitsvelocity, v.Thesizeofthisforceis v + v3.Initiallytheparticleisattheoriginand hasvelocity Q,where Q > 0.
a Usepartialfractionstoshowthatthetime t isgivenby t = 1 2 loge Q2(1 + v2) v2(1 + Q2) .
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b Hencefind v2 asafunctionof t.
c Determinethelimitof v as t →∞,andhenceexplainwhy v isalwayspositive.
d Showthatthevelocityisrelatedtothedisplacement x bytheformula x = tan 1 Q tan 1 v, andhencefindlimt→∞ x.
e Doesitfollowthat x = tan 1 Q v 1 + Qv ?Justifyyouranswer.
7a Findthevaluesof A and B suchthat
b Abodywithmass m = 4.5 × 106 kgisacteduponbyaforceof104(6 v)newtonswhere v isitsspeedin metrespersecond.Italsoexperiencesaresistanceproportionaltothesquareofitsspeed.Thatis:
dv dt = 104 m 6 v kv2
i Itsmaximumspeedis2m/s.Find k ii Showthatthebodyattainsaspeedof1.5m/s,startingfromrest,inalittleover2minutesand 41seconds.
8 Whenajetaircrafttouchesdowntwodifferentretardingforcescombinetoslowitdown.Iftheaircrafthas mass M kgandspeed v m/sthenthereisaconstantfrictionalforceof2M newtonsduetothebrakesanda forceof 1 10000 Mv2 newtonsduetothereversethrustoftheengines.Thereversethrustdoesnottakeeffect until3secondsaftertouchdown.
Let x bethedistanceinmetresofthejetfromitspointoftouchdownandlet t bethetimeinsecondsafter touchdown.
a Thejet’slandingspeedis72m/s.Showthat v = 66and x = 207attheinstantthereversethrustofthe enginestakeseffect.
b Showthatwhen t > 3, x = 207 + 5000ln 20000 + 662 20000 + v2
c Reversethrustisshutdownwhentheaircraftreachesaspeedof36m/s.Howfarfromthepointof touchdown,correcttothenearestmetre,doesthishappen?
d Thebrakesalonearethenusedtoreachthetaxispeedof7m/s.Howfarfromthepointoftouchdown, correcttothenearestmetre,doestheplanereachitstaxispeed?
9 Aboxofmass m ispushedacrossafloorwithaconstantforce mP.Astheboxmovesitisalsoretardedbya forceduetofrictionof mkv,where v isitsvelocityand k isapositiveconstant.Theboxisinitiallytravelling withvelocity Vi
a Writedownaforcebalanceequationandhenceshowthatthespeedwhichresultsinzeronetforceis V0 = P k
b Integrateoncetoshowthat v = V0 1 1 Vi V0 e kt .Hencefindlimt→∞ v.
c Drawagraphof v versus t inthecasewhen: Vi > V0 i
d Findthetimetakenfortheboxtoacceleratefrom
10
11
Anobjectismovingacrossaflatsurfaceforwhichthefrictionisproportionalto v3/2 andintheopposite direction.Thatis,theaccelerationisgivenbytheequation
v dv dx = kv3/2 , where v isthevelocitywhentheobjectisat x,and k isapositiveconstant.Initiallytheparticleisatthe originwithspeed V0 > 0.
ai Integratewithrespectto x andhenceshowthat v V
ii Whatvaluescan x take,andwhereistheparticlewhenitstopsmoving?
bi Find x asafunctionofthetime t.
ii Reviewyouranswertopart aii inlightofthisresult.
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CHALLENGE
Whenacertainobjectispushedacrossapolishedfloor,theresistanceduetofrictionisproportionalto v3/2 , where v isitsvelocity.Theresultingequationforaccelerationis dv dt = 1 v 3/2
Initiallytheobjectisatrest.
a Usethesubstitution u = √v toshowthat
b Usepartialfractionstoshowthat
c Henceshowthat
7G
VerticalResistedMotion
Inthissectionitisassumedthatwhenaparticleofmass m travelsverticallythroughafluidsuchasair,wateror oil,onlytwoforcesactontheparticle,namelyaforceduetogravityandaresistance R tothemotion.Typicallyit isfoundthattheresistanceisproportionaltoapowerofthespeed |v|.Thus
|R| = mk|v|n forsomeconstant k.
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Although n cantakevariousvalues,onlythecases n = 1and n = 2willbestudiedinanydetailinthiscourse. Itisalsoobservedthattheresistanceisoppositeindirectiontothevelocity v.Thusamoreconvenientformofthe resistanceis:
R = mk|v|n 1 v.
TheEquationsofMotionfor n = 1
Supposethataparticleofmass m isfiredupwardswithinitialvelocity V0 througha resistingmediumundertheinfluenceofgravity g.Let y betheheightoftheparticle abovethepointofprojectionattime t.Thatis,upwardsisthepositivedirectionas showninthediagramontheright.Let˙ y = v andnotethat y = 0when t = 0. y
Nowinthecase n = 1theresistancesimplifiesto
R = mkv
Theonlyotherforceactingontheparticleisitsweightduetogravity.Thusthecombinedforce F is F = mg mkv or my = mg mkv
so¨ y = (g + kv). mg v R
Thisdifferentialequationmustnowbesolvedalongwiththeinitialconditions
˙ y(0) = V0 and y(0) = 0.
Inanygivenproblem,ateachstep,thedifferentialequationisintegratedandthentheinitialconditionsareusedto determineanyunknownconstantofintegration.
13Theequationsfor n = 1
Summingtheforcesleadstotheequation
y = (g + kv).
Integratethisandfindtheconstantsusingtheinitialconditions: At t = 0,˙ y = V0 y = 0
Example25
Aparticleisprojectedupwardsinamediumforwhichtheresistancetothemotionisonefifthofthemass timesthevelocityandoppositeindirection.Theinitialvelocityis30m/s.Let y beitsheightinmetresabove thepointofprojectionafter t seconds.Use g ≑ 10m/s2,sothat
y = (10 + 1 5 v).
Find y asafunctionofthevelocity v andhencefinditsmaximumheight.
Solution
Notingthat¨ y = v dv dy ,itfollowsthat
v dv dy = 50 + v 5
so v 50 + v dv dy = 1 5 or 1
50
50 + v dv dy = 1 5
Thisisaseparableequation,soitcanbewrittenas:
∫ 1
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50
50 + v dv = ∫ 1 5 dy
Thus v 50log(50 + v) = 1 5 y + C
At t = 0,theparticleisattheoriginand v = 30,so C = 30 50log80.
Hence v 50log(50 + v) = 1 5 y + 30 50log80.
Rearrangethisequationtoget,afterafewlinesofalgebra,
y = 5 30 v + 50log 50 + v 80
Atthemaximumheight v = 0,thus ymax = 5 30 + 50log 5 8 (about32.5metres.)
TheDownwardJourneyfor n = 1
Afteraparticlereachesitsmaximumheightitwillfallundertheinfluenceofgravity.Itmay surprisesomethatthedifferentialequationremainsthesame.Theforcesinthefreebody diagramshowwhy.Theweightactsdownwardsandtheresistanceactsupwards,whichisin theoppositedirectionto v.Thus
F = mg + mk( v) or m ¨ y = mg mkv
so¨ y = g kv mg v R
Example26
Aparticlefallsfromrestthroughamediumforwhichtheresistanceisproportional tothevelocityandoppositeindirection.Let y beitsdisplacementinmetresfromits initialpositionafter t seconds.Let˙ y = v attime t andassumethat
a Find v asafunctionof t.
y = g kv y h
b Henceshowthattheterminalvelocityislimt→∞ v = g k
c Find y asafunctionof t.
d Showthatif h islargeenoughthentheapproximatetimetoreachthegroundat y = h isgivenby
t ≑ kh g + 1 k
Solution
a Since¨ y = v itfollowsthat
v = g kv
so kv g + kv = k.
Integratewithrespecttotimetoget: ∫ kv
g + kv dt = ∫ kdt
thuslog(g + kv) = kt + C1
At t = 0,thevelocityiszero,thus
C1 = log g
Hencelog(g + kv) = kt + log g.
Rearrangethisequationtoget
log g + kv g = kt
so g + kv g = e kt
or v = g k 1 e kt .
b limt→∞ v = limt→∞ g k 1 e kt = g k (1 0) = g k
c Frompart a,˙ y = g k 1 e kt .Integratewithrespecttotimetoget:
y = g k t + 1 k e kt + C2
At t = 00 = g k (0 + 1 k ) + C2
so C2 = g k2
Hence y = g k t + 1 k e kt 1
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d Theparticlereachesthegroundat y = h,sotheequationfor t is h = g k t + 1 k (e kt 1) or h = g k t + 1 k (e kt 1)
Thisiscalledatranscendentalequation,meaningthatitcannotbesolvedalgebraically.Thereasonforthis isthepresenceofboth t and e kt intheequation.If h islargeenough,however,thenthevalueof t willalso belargesothat e kt ≑ 0.Thus h ≑ g k t 1 k and t ≑ kh g + 1 k
TerminalVelocity
Theterminalvelocityofanobjectismostoftenassociatedwithfallingbodies,butissometimesreferredtoin otherproblems.Asdemonstratedintheaboveexample,onemethodoffindingtheterminalvelocityistofind v as afunctionof t andthendeterminethelimit,limt→∞ v(t).
Asimplerapproachistorecognisethattheterminalvelocitycorrespondstotheconstantsolutiontothe differentialequation.Thusintheaboveexample 0 = g kv so v = g k istheterminalvelocity.
Itiscommontowrite VT fortheterminalspeedsothat VT = g k inthiscase.
14Terminalvelocity
Thisistheconstantsolutionofthedifferentialequation.Thatis,solve˙v = 0.Itiscommontowrite VT for theterminalspeed.
AJourneyUpandDownfor n = 1
Sincetheforceequationisthesameforbothanupwardjourneyandadownwardjourney,itonlyneedstobe solvedoncewhenajourneyinvolvesbothdirections.
Example27
Aparticleofmass m isprojectedupwardwithaninitialvelocity V0.Theparticleisacteduponbygravity. Thereisalsoaresistiveforceequalto(mk)timesitsvelocitybutintheoppositedirection,where k isapositive constant.Let v beitsupwardvelocityattime t
a Useafreebodydiagramtohelpshowthat dv dt = (g + kv).
b Findthetime tm takenfortheparticletoreachitsmaximumheight.Givetheanswerintermsof g, k and V0
c Supposethat V0 ishalftheterminalspeed.Showthatwhen t = 2tm theparticleistravellingdownwardsat 2 3 ofitsinitialspeed.
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Solution
a Theforceduetogravityis mg andtheresistiveforceis mkv.Thus
m dv dt = mg mkv or dv dt = (g + kv). mg v R
b Re-arrangingandintegratingtheresultin a,
∫ k v g + kv dt = ∫ kdt
solog(g + kv) = kt + C.
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Theinitialvelocityis V0,so
log(g + kV0) = C
thuslog(g + kv) = kt + log(g + kV0).
Re-arrangethisandusethelogidentitiestoget kt = log g + kV0 g + kv .
Atthehighestpoint, v = 0sothat
tm = 1 k log g + kV0 g
c Fromthesecondlastequationinpart b:
g + kV0
g + kv = ekt
so g + kv
g + kV0 = e kt .
Nowput t = 2tm toget
g + kv
g + kV0 = e ktm 2 = g2 (g + kV0)2
thus g + kv = g2 g + kV0
and kv = g2 g2 gkV0 g + kV0 ,
so v = gV0
g + kV0 .
Thisisnegativesotheparticleistravellingdownwards.
Next,dividethroughby k,notingthat VT = g k ,toget
v = VT V0 VT + V0
Finallyusethefactthat VT = 2V0 toget
v = 2V0 2 2V0 + V0 = 2 3 V0.
AProblemwhen n = 2
Unlikethepreviouswork,when n = 2thejourneysupanddownmustbetreatedseparatelyasthegoverning equationisnotthesameforboth.Thisbecomesevidentwhentheresistanceisre-examined.
R = mk|v|n 1 v = mk|v|v when n = 2.
Theabsolutevalueissignificantbecause v changesfrompositivetonegativebetweentheupwardanddownward journeys.Asimplewaytoresolvethisissueistouseadifferentdefinitionforthedisplacementineachcase.
TheUpwardJourneyfor n = 2
Thesituationfortheupwardjourneyisdefinedinexactlythesamewayasfor n = 1. Thatis,supposethataparticleofmass m isfiredupwardswithinitialvelocity V0 througharesistingmediumundertheinfluenceofgravity g.Let y betheheightofthe particleabovethepointofprojectionattime t.Thatis,upwardsisthepositivedirection asshowninthediagramontheright.Let˙ y = v andnotethat y = 0when t = 0.It followsthat v ≥ 0so |v| = v y
Theresistanceisnow R = mkv2 andtheforceequationyields
y = (g + kv2) withtheusualinitialconditions
y(0) = V0 and y(0) = 0.
Thepresenceof v2 insteadof v ontherighthandsidegivesrisetodifferentintegralswhensolvingproblems.
Example28
Aparticleisfiredupwardswithinitialvelocity V0.Twoforcesactdownwardsontheparticle:itsweight mg andaresistiveforce mkv2,where v isthevelocityoftheparticleattime t andtheconstant k ispositive.Let y betheheightoftheparticleabovethestartingpoint.Itisgiventhat
¨ y = (g + kv2).
a Find Tup,thetimetakentoreachthemaximumheight,intermsoftheconstant k,theinitialvelocity V0 and theterminalvelocity VT = g k
b Supposethatthemaximumheightis H.Usepart a toshowthat
y = 1 k log cos kVT (Tup t) + H
where VT = g k istheterminalspeed.
Solution
a Firstnotethat
˙ v = k(VT 2 + v 2).
Somemayrecognisethisasatransformationofthederivativeofthetangentfunction.Nevertheless,thefull solutionisnowgiven.
˙ v VT 2 + v2 = k or VT × v VT 2 + v2 = kVT
thustan 1 v VT = kVT t + C1.
At t = 0, v = V0 so
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C1 = tan 1 V0 VT and kVT t = tan 1 V0 VT tan 1 v VT .
Thetime Tup tothemaximumheight,when v = 0,is
Tup = 1 kVT × tan 1 V0 VT .
b Theaboveequationforvelocityandtimecanbewrittenas
t = Tup 1 kVT × tan 1 v VT
or v = VT tan kVT (Tup t) .
Theprimitiveoftan u is log(cos u).Applythechainruletothenget
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y = 1 k log cos kVT (Tup t) + C2
Now y = H at t = Tup so H = 1 k log1 + C3 thatis, C2 = H
Thisgives y = 1 k log cos kVT (Tup t) + H.
TheDownwardJourneyfor n = 2
Thesituationdiffersfromtheupwardjourney.Supposethataparticleofmass m isfired downwardswithinitialvelocity V0 througharesistingmediumundertheinfluenceof gravity g.Let y bethedepthoftheparticlebelowthepointofprojectionattime t.That is,downwardsisthepositivedirectionasshownontheright.Let˙ y = v andnotethat y = 0when t = 0.Itfollowsthat v ≥ 0so |v| = v y
Theresistanceisupwardsso R = mkv2 andtheweightisdownwardsso w = mg.The forceequationyields
y = g kv2
withtheusualinitialconditions
y(0) = V0 and y(0) = 0. mg v R
15Displacementforthedownwardsjourneywhen n = 2
Inmostsituationsitisbesttoputtheoriginattheinitialpositionwithdownwardsaspositive.
AnObjectThrownDownfor n = 2
Therearethreecasesofnote:
0 ≤ V0 < VT , V0 = VT and V0 > VT .
Itcanbeshownthatwhen V0 = VT thevelocityneverchanges,sothatthedisplacementfunctionissimply y = VT t.
Studentsshouldensuretheyareabletoprovethisresultforthemselves.Theothertwocasesaredealtwithinthe exerciseandthefollowingexample.
Example29
Anobjectisthrowndownwardwithinitialspeedtwiceitsterminalvelocity.Itsmotionisaffectedbygravity andairresistance,whichisproportionaltothesquareofitsspeed.Puttheoriginatthepointofprojectionwith downwardaspositive.Thevelocity v satisfiestheequation
v = g kv2
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withtheinitialconditionthat v(0) = 2VT ,where VT = g k .
a Let w = v VT .Showthat w satisfiestheequation
w = (w 2 1) gk
andstatetheinitialconditionfor w
b Usepartialfractionstohelpfind t asafunctionof w
c Hencefindthetimetakenfortheobjecttoreduceinspeedfrom2VT to 11 10 VT .
Solution
a Itshouldbeclearthat v = VT w and˙v = VT w,thus
VT w = g kVT 2 w2 = g gw2
so w = (1 w2) g VT = (w2 1) gk
Theinitialconditionis v(0) = 2VT ,sofromthesubstitutionitfollowsthat
VT w(0) = 2VT , hence w(0) = 2.
b Treatingthederivativelikeafraction dt
dw = 1 (w2 1) gk or gk dt dw = 1 (w + 1)(w 1) Let 1 (w + 1)(w 1) = A w + 1 + B w 1 thenbythecover-uprule A = 1 2 and B = 1 2 .
Andso2 gk dt dw = 1 w + 1 1 w 1
Nowintegratetoget
2t gk = log(w + 1) log(w 1) + C,
andfromtheinitialcondition
0 = log3 + C
hence2t gk = log(w + 1) log(w 1) log3, or t = 1 2 gk log w + 1 3(w 1)
c When v = 11 10 VT itfollowsthat w = 11 10 so t = 1 2 gk log 11 + 10 3(11 10) = log7 2 gk .
Noticethattheequationinpart b canbesolvedfor w toget w = 3e2t√gk + 1 3e2t√gk 1
sothat v = VT × 3e2t√gk + 1 3e2t√gk 1
Itshouldbeclearthatthenumeratorofthefractionisalwaysgreaterthanthedenominator.Hencethevelocity isalwaysgreaterthantheterminalvelocity.Morespecifically,thevelocityisalwaysdecreasing,quicklyat firstandthenatadecreasingrateasitapproachesthelimitingvalue VT fromabove.Thusthegraphof v(t)is concaveupwithahorizontalasymptote.
Exercise7G
FOUNDATION
1 Anobjectofmass5kgisprojectedverticallyupwardswithvelocity40m/sandexperiencesaresistive forceinNewtonsofmagnitude0.2v2,where v isthevelocityoftheobjectattime t seconds.Assumethat g = 10m/s2
a Showthat dv dt = 250 v2 25 .
b Find,correcttothenearesttenthofasecond,thetimetoreachitsmaximumheight.
c Use dv dt = v dv dy tohelpfindthemaximumheight,correcttothenearestmetre.
2 Anobjectofmass0.5kgisprojectedupwardswithvelocity40m/sandexperiencesaresistiveforce innewtonsofmagnitude0.2v,where v isthevelocityoftheobjectattime t seconds.Assumethat g = 10m/s2
a Showthat¨ y = 50 2v 5 .
b Showthattheobjecttakes 5 2 ln 13 5 secondstoreachitsmaximumheight.
c Showthatthemaximumheightreachedis(100 + 125 2 ln 5 13 )metres.
3 Anobjectofmass100kgisfoundtoexperiencearesistiveforce,innewtons,ofone-tenththesquareofits velocity,inmetrespersecond,whenitmovesthroughtheair.Supposethattheobjectfallsfromrestunder gravity,andtake g = 9.8m/s2
a Showthatitsterminalvelocityisabout99m/s.
b Iftheobjectreaches80%ofitsterminalvelocitybeforestrikingtheground,showthatthepointfrom whichitwasdroppedwasabout511metresabovetheground.
DEVELOPMENT
4a Anobjectofmass1kgisprojectedverticallyupwardsfromthegroundat20m/s.Thebodyisunder theeffectofbothgravityandairresistancewhich,atanytime,hasamagnitudeof 1 40 v2,where v isthe velocityattime t.Put g = 10m/s2,andtakeupwardsasthepositivedirection.
Showthatthegreatestheightreachedbytheobjectis20ln2m. i
Showthatthetimetakentoreachthisgreatestheightis π 2 s. ii
b Havingreacheditsgreatestheighttheparticlefallsbacktoitsstartingpoint.Theparticleisstillunderthe effectofbothgravityandairresistance.Takedownwardsasthepositivedirection.
Writedowntheequationofmotionoftheobjectasitfalls. i
Findthespeedoftheobjectwhenitreturnstoitsstartingpoint. ii
5 Anobjectisprojecteddownwardswithinitialvelocity V0.Theairresistanceatspeed v hasmagnitude mkv, where k isapositiveconstant.Takedownwardsasthepositivedirection.
a Showthat t = 1 k loge g kV0 g kv
b Henceshowthat v = g k (1
c Integrateagaintoshowthat y
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,andthattheterminalvelocityis
d Supposethattheterminalvelocityofthisobjectis20m/sandthat g = 10m/s2.Theobjectisthrown verticallydownwardsfromalookoutatthetopofacliff atpreciselytheterminalvelocity.Atthesame instant,asimilarobjectisdroppedfromthesameheight.Showthatthedistancebetweenthetwofalling objectsafter t secondsis40(1 e 1 2 t )metres,andhencestatethelimitingdistancebetweenthem.
6 Aparticleofmass10kgisfoundtoexperiencearesistiveforce,innewtons,ofone-tenthofthesquareof itsvelocityinmetrespersecondwhenitmovesthroughtheair.Theparticleisprojectedverticallyupwards fromapoint O withavelocityof u metrespersecond.Thepoint A,verticallyabove O,isthehighestpoint reachedbytheparticlebeforeitstartstofalltothegroundagain.Assumethat g = 10m/s2
a Showthattheparticletakes √10tan 1 u 10√10 secondstoreach A from O
b Showthattheheight OA is50loge( 1000+u2 1000 )metres.
c Let w bethevelocityoftheparticlewhenitreturnsto O.Showthat w2 = 1000u2 1000 + u2 .
7a Aparticleofmass m fallsfromrest,fromapoint O,inamediumwhoseresistanceis mkv,where k isa positiveconstantand v isthevelocityattime t
Provethattheterminalvelocity V is V = g k i
Provethatthespeedattime t isgivenby V 1 e kt . ii
b Anidenticalparticleisprojectedupwardsfrom O withinitialvelocity U inthesamemedium.Suppose thatbothparticlesbegintheirmotionsimultaneously.
Provethatthesecondparticlereachesitsmaximumheightat t = 1 k ln g+kU g . i
Provethatthespeedofthefirstparticlewhenthesecondparticlereachesitsmaximumheightis UV U + V ii
8 Aparticle P1 ofmass m kgisdroppedfrompoint A andfallstowardspoint B,whichisdirectlyunderneath A.Attheinstantwhen P1 isdropped,asecondparticle P2,alsoofmass m kg,isprojectedupwardsfrom B towards A withaninitialvelocityequalinmagnitudetotwicetheterminalvelocityof P1.Eachparticle experiencesaresistanceofmagnitude mkv asitmoves,where v ms 1 isthevelocityand k isaconstant.
a Showthattheterminalvelocityof P1 is g k ,where g isaccelerationduetogravity.
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b Showthatthetimetakenforparticle P2 toreachvelocity v is t = 1 k ln 3g g + kv .
c Supposethattheparticlescollideattheinstantwhen P1 hasreached30%ofitsterminalvelocity.Show thatthevelocityof P2 whentheycollideis 11g 10k ms 1
9 Anobjectofmass1kgisdroppedfromalookoutontopofahighcliff.Lettheaccelerationduetogravity be10m/s2 .
a Atfirst,airresistancecausesadecelerationofmagnitude 1 10 v ,where v m/sisthespeedoftheobject t secondsafteritisdropped.
i Takingdownwardsaspositive,explainwhyitsequationofmotionis
¨ y = 10 1 10 v , where y isthedistancethattheobjecthasfalleninthefirst t seconds.
ii Showthat dv dy = 100 v 10v ,andhenceshowthatthespeed V oftheobjectwhenitis40metresbelow thelookoutsatisfiestheequation
V + 100loge 1 V 100 + 4 = 0.
b Aftertheobjecthasfallen40metresandreachedthisspeed V,averysmallparachuteopens,andair resistancenowcausesadecelerationtoitsmotionofmagnitude 1 10 v2
i Takingdownwardsaspositive,writeanexpressionforthenewacceleration¨ y oftheobject,where y nowisthedistancethattheobjecthasfalleninthefirst t secondsaftertheparachuteopens.
ii Showthat v2 = 100 (100 V 2)e 1 5 y,andhencefindtheterminalvelocityoftheobject.
iii Showthat t secondsaftertheparachuteopens,
t = 1 2 loge (v + 10)(V 10) (v 10)(V + 10)
iv Thesolutiontotheequationinpart aii is V ≑ 25.7m/s.Howlongaftertheparachuteopensdoesthe objectreach105%ofitsterminalvelocity?
10 Aparticleofmass m kgisprojectedverticallyupwards.Ataspeedof v m/s,itexperiencesairresistanceof magnitude mkv2.Itsinitialspeed U m/sisequaltotheterminalvelocityoftheparticlewhenitisfalling.It reachesitsmaximumheight H metresafter T seconds.
Showthat¨ y = g U2 U2 + v2 . a
b
c
d
e
Showthat v = U tan π 4
Showthat y =
Showthatwhen t
Whatpercentageofthemaximumheightisreachedduringthefirsthalfoftheascenttime?Giveyour answerasapercentagecorrecttoonedecimalplace.
Aparticleofmass m isprojectedupwardwithaninitialvelocity V0.Theparticleisacteduponbygravity andaresistiveforce( kmv),where v isitsvelocityattime t.Thus, dv dt = (g + kv).
Let VT = g k betheterminalspeedoftheparticle.
a Usethesubstitution v = VT × w tore-writethedifferentialequationintermsof w.
b Whatvalueof w isaconstantsolutionofthisdifferentialequation?Whatisthecorrespondingconstant valueof v?
c Usingappropriategraphingsoftwareorotherwise,drawtheslopefieldforthisequationwhen k = 1. Includepointsintheregion0 ≤ t ≤ 6and 4 ≤ w ≤ 2.Thenaddsolutioncurvesstartingateachof w(0) = 0, w(0) = 2and w(0) = 3.
d Whatvaluesof V0 dothesethreevaluesof w(0)correspondto?
e Howistheconstantsolutioninpart b evidentinthesolutioncurvesyouhavedrawn?
f Supposethat w(0) = 1 2 .Willthespeedoftheparticleincreaseordecrease?
12a Considerthefunction
Evaluate f (g). i
Showthat f ′(x) = 1 g x 2 . ii
Explainwhy f (x) > 0for x > g. iii
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CHALLENGE
.
b Abodyismovingverticallythrougharesistingmedium,withresistanceproportionaltoitsspeed.The bodyisinitiallyfiredupwardsfromtheoriginwithspeed V0.Let y metresbetheheightoftheobject abovetheoriginattime t seconds,andlet g betheconstantaccelerationduetogravity.Thus
d2y
dt2 = (g + kv),where k > 0.
Youmayassumethatthisequationisvalidforall t ≥ 0.
Find v asafunctionof t,andhenceshowthat k2 y = (g + kV0)(1 e kt ) gkt. i
Find T ,thetimetakentoreachthemaximumheight. ii
iii
Showthatwhen t = 2T ,
2 y = (g + kV0)
Usethisresultandpart a toshowthatthedownwardsjourneytakeslonger. iv
13 Supposeaparticleisprojectedupwardsinaresistingmediumundertheinfluenceofgravity.Itreachesa maximumheightthenfallsbacktoitsstartingposition.
a Theequationforverticalresistedmotionwhen n = 1is
˙ v = g kv.
Byconsideringtheupwardanddownwardjourneysseparately,andwithoutsolvingtheequation,explain whytheupwardjourneyisshorter.Thetriangleinequalitymaybehelpful.
b Dolikewiseforthecasewhen n = 2.
7H ProjectileMotion—FurtherExtension1
ThissectioncontainsquestionsontheExtension1topicProjectilemotion.Themotioniswithoutresistanceand thereisnonewtheory.Theexerciseprovidesfurtherpracticebeforetacklingresistedprojectilemotioninthenext section.
Example30
Abatterinawomen’scricketteamhitsaballfromneargroundlevel.Theballleavesthebatat20ms 1 andat anangleof30◦ tothehorizontal.Usetheapproximation g = 10ms 2 fortheaccelerationduetogravity.For simplicity,assumethattheoriginforthehorizontalandverticaldisplacements, x and y,isatthepointtheball leavesthebat.
a Startingwith¨ x = 0and¨ y = g,find x and y asfunctionsoftime.
b Hencefind y asafunctionof x infactoredform.
c Usetheanswertopart b tofindtherangeandmaximumheightoftheball.
d Showhowtheanswertopart c canbeobtaineddirectlyfrompart a
Solution
Theinitialconditionshorizontallyandverticallyare,at t = 0:
x(0) = y(0) = 0 with˙ x(0) = 20cos30◦ = 10√3 and˙ y(0) = 20sin30◦ = 10.
a Asgiven¨ x = 0 thus˙ x(t) = C1
For t = 0,10√3 = C1
sothat˙ x = 10√3.
Integrating: x = 10t√3 + D1.
For t = 0,0 = D1
so x = 10t√3.
b From x above, t = 1 10√3 x sothat
y = 1 √3 x 5 300 x 2
= √3 3 x 1 60 x 2
= 1 60 x(20√3 x).
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Likewise,¨ y = 10 thus˙ y = 10t + C2
For t = 0,10 = C2
sothat˙ y = 10 10t
Integrating: y = 10t 5t2 + D2.
For t = 0,0 = D2
so y = 10t 5t2 .
c Put y = 0toget,aftermultiplyingthroughby60,
0 = x(20√3 x).
Hencetherangeis xmax = 20√3.
Themaximumheightisatthevertexwhere x = 10√3and ymax = 5.
d Themaximumheightoccurswhen˙ y = 0,so t = 1.
Substitutethisvaluetoget x = 10√3and ymax = 5.
Solving y = 0(ordoublingthetimeabove)gives t = 2where xmax = 20√3.
Example31
Aprojectilehasrange R > 0anditreachesamaximumheight h > 0.Thusthetrajectoryhasequation
y = 4h R2 x(R x).
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Atacertainpoint P(a, b)onthetrajectorythetangentat P isperpendicularto OP.Itisknownthatthereis onlyonesuchvalueof a > 0.
Showthat a = 3 4 R andthat R = h√2.Hencefindtheangleofprojection.
Solution
Fromthegivenequation
y′ = 4h R2 (R 2x)
soat P thetangenthasgradient
y′(a) = 4h R2 (R 2a).
Thegradient m of OP isthenegativereciprocal:
m = R2
4h(2a R) , andfromthis,the y-coordinateat P is
b = aR2
4h(2a R) .
But P alsoliesontheparabolaso
b = 4h R2 a(R a).
Equatingthesevaluesof b gives
aR2
4h(2a R) = 4h R2 a(R a)
sofor a > 0
R2
4h(2a R) = 4h R2 (R a).
Re-arrangingthisequation: R4 = 16h2(R a)(2a R)
or16h2(a R)(2a R) + R4 = 0 16h2(2a2 3Ra + R2) + R4 = 0
so32h2a2 48h2Ra + R2(16h2 + R2) = 0.
Thisquadraticin a hasonlyonesolutionsoitis
a = 48h2R 2 × 32h2 = 3R 4
Further,thediscriminantiszerosothat
0 = 482h4R2 4 × 32h2 × R2(16h2 + R2)
Continuedonthenextpage
Dividingthroughby128h2R2 gives
0 = 18h2 16h2 R2
or0 = 2h2 R2 .
Further,sincebotharepositive,itfollowsthat R = h√2.
If α istheangleofprojectionthen
tan α = y′(0) = 4hR R2 = 4h2 √2 2h2 = 2√2.
Hence α = tan 1(2√2) ≑ 70.5◦ .
Althoughitisnotasked,tan α happenstobe4timesthegradientof OP.Keenreadersmayalsoliketoshow thatforaninitialspeed V therangeis
R = 4V 2 √2 9g , forwhich
a = V 2 √2 3g
Anenrichmentquestionintheexerciseconsidersarelatedproblemwherethereismorethanonevalueof a.
Enrichment—VectorsandCalculus
Ithasbeenconvenienttousevectorsforsomeprojectilemotionproblems.Ratesareinvolved,however,sothe derivativeisrequired,andthederivativeofavectorhasnotyetbeendefined.Hereisabriefoutlineofwhatis required.Itreliesonanintuitivenotionoflimitsandcontinuitywhenappliedtoavectorfunction.Thatis,
lim λ→c r ∼ (λ) = r ∼ (c)
providedeachcomponentof r ∼ (λ)iscontinuousat λ = c
Supposethataparticlehaspositionvector p ∼ (t) = x(t)
attime t.Thevelocityoftheparticleisthen definedtobe
v ∼ = limh→0 1 h (p(t + h)
(t)) , whichisanalogouswiththedefinitionforasimplefunctionoftime.Expandingthisdefinitiontocomponentform gives
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Likewisetheaccelerationvectorisdefinedtobe
thus
Ascanbeseenfromeachcaseabove,itissimplyamatterofdifferentiatingthecomponentstoobtainthe requiredresult.Thoughitwillnotbeneededhere,thesameistrueinthreedimensions,andtheproofisleftasan exercise.
Manyinterestingresultsarisefromthisdefinitionofdifferentiation.Themostobviousisthatthevelocityvector istangenttothepathoftheparticleateachpoint.Thisiseasilyshownintwodimensionsasfollows.The gradientatanypointonthepathis
dy dx = dy dt × dt dx (bythechainrule)
= y(t) ÷ x(t)(treatingthederivativelikeafraction.)
Buttherighthandsideisthegradientofthelinesegmentwhichrepresentsthevelocityvector.Thatis, v ∼ is parallelwiththetangentateachpointonthecurve.
Exercise7H
Note: ThesearefurtherquestionsontheExtension1topicProjectilemotion.
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FOUNDATION
1 Aprojectileisfiredwithvelocity v = 40m/sonahorizontalplaneatanangleofelevation α = 60◦.Take g = 10m/s2,andlettheoriginbethepointofprojection.
a Showthat˙ x = 20and˙ y = 10t + 20√3,andfind x and y
b Findtheflighttime,andthehorizontalrangeoftheprojectile.
c Findthemaximumheightreached,andthetimetakentoreachit.
d Anobserverclaimsthattheprojectilewouldhavehadagreaterhorizontalrangeifitsangleofprojection hadbeenhalved.Investigatethisclaimbyreworkingthequestionwith α = 30◦ .
2 Aprojectileisfiredataspeedof39m/sandatanangleofelevationoftan 1 12 5 .Itjustclearsatowerata horizontaldistanceof30mfromthepointofprojection.
a Obtain,byintegration,expressionsfor˙ x,˙ y, x and y,taking g = 10m/s2 .
b Findtheheightofthetower.
c Find,inm/scorrecttoonedecimalplace,thespeedoftheprojectileasitclearsthetower.
d Doestheprojectilereachitsgreatestheightbeforeorafteritclearsthetower?
3 Apebbleisthrownfromthetopofaverticalcliff withvelocity20m/satanangleofelevationof30◦.The cliff is75metreshighandoverlooksariver.
a Deriveexpressionsforthehorizontalandverticalcomponentsofthedisplacementofthepebblefromthe topofthecliff after t seconds.(Take g = 10m/s2.)
b Findthetimeittakesforthepebbletohitthewaterandthedistancefromthebaseofthecliff tothepoint ofimpact.
c Findthegreatestheightthatthepebblereachesabovetheriver.
d Findthevaluesof˙ x and˙ y attheinstantwhenthepebblehitsthewater.Hencefindthespeed(tothe nearestm/s)andtheacuteanglebelowthehorizontal(tothenearestdegree)atwhichthepebblehitsthe water.
e Thepathofthepebbleisaparabolicarc.Byeliminating t fromtheequationsfor x and y,findits equationinCartesianform.
4 Aplaneisflyinghorizontallyat363.6km/handitsaltitudeis600metres.Itistodropafoodparcelontoa largecrossmarkedonthegroundinaremotearea.
a Convertthespeedoftheplaneintometrespersecond.
b Deriveexpressionsforthehorizontalandverticalcomponentsofthefoodparcel’sdisplacementfromthe pointwhereitwasdropped.(Take g = 10m/s2.)
c Showthatthefoodparcelwillbeintheairfor2√30seconds.
d Findthespeedandangleatwhichthefoodparcelwillhittheground.
e Atwhathorizontaldistancefromthecross,correcttothenearestmetre,shouldtheplanedropthefood parcel?
5 Minghitagolfballfromlevelgroundwithinitialspeed50m/satanangleof45◦ abovethehorizontal.The ballhittheclubhouse75metresaway.Take g = 10m/s2 .
a Showthattheballhittheclubhouseafter 3 2 √2seconds,52.5metresabovetheground.
b Showthatthevelocityoftheballwhenitstrucktheclubhousewas5√58m/satanangleoftan 1 2 5 abovethehorizontal.
6 Jeffreyhitagolfballthatwaslyingonlevelground.Twosecondsintoitsflight,theballjustcleareda 28-metre-talltreewhichwas24√5metresfromwheretheballwashit.Let V m/sbetheinitialspeedofthe ball,andlet θ betheangleabovethehorizontalatwhichtheballwashit.Take g = 10m/s2 .
a Showthatthehorizontalandverticalcomponentsofthedisplacementoftheballfromitsinitialposition are x = Vt cos θ and y = 5t2 + Vt sin θ.
b Showthat V cos θ = 12√5and V sin θ = 24.
c Bysquaringandadding,find V.Thenfind θ,correcttothenearestminute.
d Find,correcttothenearestmetre,howfarJeffreyhittheball.
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DEVELOPMENT
7 Aballisthrownfromlevelgroundataninitialspeedof V m/sandatanangleofprojectionof α above thehorizontal.Assumethat, t secondsafterrelease,thehorizontalandverticaldisplacementsaregivenby x = Vt cos α and y = Vt sin α 1 2 gt2
a ShowthatthetrajectoryhasCartesianequation y =
b Henceshowthatthehorizontalrangeis V 2 sin2α g .
c When V = 30m/s,theballlands45metresaway.Take g = 10m/s2
i Findthetwopossiblevaluesof α
ii A2-metre-highfenceisplaced40metresfromthethrower.Examineeachtrajectorytoseewhether theballwillstilltravel45metres.
8 Aballisthrownwithinitialvelocity20m/satanangleofelevationof tan 1 4 3
a Showthattheparabolicpathoftheballhasparametricequations x = 12t and y = 16t 5t2 .
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b Hencefindtherangeoftheball,anditsgreatestheight.
c Supposethat,asshownopposite,theballisthrownuparoadinclinedat tan 1 1 5 tothehorizontal.Showthat:
i theballisabout9metresabovetheroadwhenitreachesitsgreatest height,
ii thetimeofflightis2.72seconds,andfind,correcttothenearesttenth ofametre,thedistancetheballhasbeenthrownuptheroad.
9 Sofiathrewaballwithvelocity20m/sfromapointexactlyonemetreabovethelevelgroundshewas standingon.Theballtravelledtowardsawallofatallbuilding16metresaway.Theplaneinwhichthe balltravelledwasperpendiculartothewall.Theballstruckthewall16metresabovetheground.Take g = 10m/s2
a Lettheoriginbethepointonthegrounddirectlybelowthepointfrom whichtheballwasreleased.Showthat, t secondsaftertheballwasthrown, x = 20t cos θ and y = 5t2 + 20t sin θ + 1,where θ istheangleabovethe horizontalatwhichtheballwasoriginallythrown.
b Theballhitthewallafter T seconds.Showthatboth4 = 5T cos θ and 3 = 4T sin θ T 2
c Henceshowthat16tan2 θ 80tan θ + 91 = 0.
d Hencefindthetwopossiblevaluesof θ,correcttothenearestminute.
10 Glennthefastbowlerrunsintobowlandreleasestheball2.4metresabovethe groundwithspeed144km/hatanangleof7◦ belowthehorizontal.Takethe origintobeonthegroundbeneaththepointofrelease,andtake g = 10m/s2
a Showthatthecoordinatesoftheball t secondsafteritsreleasearegivenby x = 40t cos7◦ , y = 2.4 40t sin7◦ 5
b Howlongwillitbe(tothenearest0.01seconds)beforetheballhitsthepitch?
c Calculatetheangle(tothenearestdegree)atwhichtheballwillhitthepitch.
d Thebatsmanisstanding19metresfromthepointofrelease.Iftheballlands morethan5metresinfrontofhim,itwillbeclassifiedasa short-pitched delivery.Isthisparticulardeliveryshort-pitched?
11 Twoparticles P1 and P2 areprojectedsimultaneouslyfromthepoints A and B, where AB ishorizontal.Themotiontakesplaceintheverticalplanethrough A and B.Theinitialvelocityof P1 is V1 atanangle θ1 tothehorizontal,andthe initialvelocityof P2 is V2 atanangle θ2 tothehorizontal.Youmayassume thattheequationsofmotionofaparticleprojectedwithvelocity V atan angle θ tothehorizontalare x = Vt cos θ and y = 1 2 gt2 + Vt sin θ
a Showthatiftheparticlescollidethenitisnecessarythat V1 sin θ1 = V2 sin θ2.
b Supposethat AB = 200metres, V1 = 30m/s, θ1 = sin 1 4 5 , θ2 = sin 1 3 5 , g = 10m/s2 andthatthe particlescollide.
i Showthat V2 = 40m/s,andthattheparticlescollideafter4seconds.
ii Findtheheightofthepointofcollisionabove AB.
iii Find,correcttothenearestdegree,theobtuseanglebetweenthedirectionsofmotionoftheparticles attheinstanttheycollide.
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12 Acricketerhitstheballfromgroundlevelwithavelocityofmagnitude20m/s atanangleofelevation α.Itfliestowardsahighwall20metresawayonlevel ground.Taketheoriginatthepointwheretheballwashit,andtake g = 10m/s2 .
a Showthattheballhitsthewallat h = 20tan α 5sec2 α.
b Showthat d dα (sec α) = sec α tan α
c Showthatthemaximumvalueof h occurswhentan α = 2.
d Findthemaximumheight.
e Findthespeedandangle(tothenearestminute)atwhichtheballhitsthe wall. x y 20m
13 Astoneispropelledatanangleof θ abovethehorizontalfromthetopofaverticalcliff 40metresabovea lake.Thespeedofpropulsionis20m/s.Take g = 10m/s2
a Showthat x(t)and y(t),thehorizontalandverticalcomponentsofthestone’sdisplacementfromthetop ofthecliff,aregivenby
x(t) = 20t cos θ, y(t) = 5t2 + 20t sin θ.
b Ifthestonehitsthelakeattime T seconds,showthat
x(T ) 2 = 400T 2 (5T 2 40)2 .
c Hencefind,bydifferentiation,thevalueof T that maximises x(T ) 2,andthenfindthevalueof θ that maximisesthedistancebetweenthefootofthecliff andthepointwherethestonehitsthelake.
14 Aparticle P1 isprojectedfromtheoriginwithspeed V atanangleofelevation θ.
a Assumingtheusualequationsofmotion,showthattheparticlereachesamaximumheightof
b Asecondparticle P2 isprojectedfromtheoriginwithvelocity 3 2 V atanangle 1 2 θ tothehorizontal.The twoparticlesreachthesamemaximumheight.
i Showthat θ = cos 1 1 8 .
ii Dothetwoparticlestakethesametimetoreachthismaximumheight?Justifyyouranswer.
15a Provethatthehorizontalrangeofaprojectileis V 2 sin2α g ,where V istheinitialspeed, α istheangleof projectionand g m/s2 istheaccelerationduetogravity.
b Agardensprinklersprayswatersymmetricallyaboutitsverticalaxisataconstantspeedof V m/s.The initialdirectionofthesprayvariescontinuouslybetween15◦ and45◦ abovethehorizontal.
i Explainwhy,fromafixedpoint O onlevelground,thesprinklerwillwetanannuluswithcentre O, innerradius V 2 2g metresandouterradius V 2 g metres.
ii Deducethatbyappropriatelylocatingthesprinklerrelativetoarectangulargarden6mby3m,the entiregardencanbewateredprovidedthat V 2 2g ≥ 1 + √7.
CHALLENGE
16 Aprojectileisfiredfromtheoriginwithvelocity V andangleofelevation α,where α isacute.Assumethe usualequationsofmotionandthatthetrajectoryisaparabola.
a Let k = V 2 2g .ShowthattheCartesianequationofthetrajectorycanbewrittenas x 2 tan2 α 4kx tan α + (4ky + x 2) = 0.
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b Showthattheprojectilecanpassthroughthepoint(X, Y)inthefirstquadrantbyfiringattwodifferent initialangles α1 and α2 onlyif X2 < 4k2 4kY.
c Supposethattan α1 andtan α2 arethetworealrootsofthequadraticequationintan α inpart a.Show that tan α1 tan α2 > 1,andhenceexplainwhyitisimpossiblefor α1 and α2 bothtobelessthan45◦ .
17 Thediagramshowstheparabolicpathofaparticlethatisprojectedfromthe origin O withspeed V atanangleof α abovethehorizontal.Itlandsat P,which liesonaplaneinclinedatanangleof β tothehorizontal.Thepathoftheparticle isperpendiculartotheplaneat P
Let OP = d,andassumethatthehorizontalandverticaldisplacementsofthe particlefrom O aregivenby x = Vt cos α and y = Vt sin α 1 2 gt2 x
a Writedownthecoordinatesof P intermsof d and β.
b Bysubstitutingthecoordinatesof P intothedisplacementequations,showthat d = 2V 2 cos2 α g cos2 β tan α cos β sin β .
c Byresolvingthehorizontalandverticalcomponentsofthevelocityat P,showthat cot β = gd cos β V 2 cos2 α tan α.
d Deducethattan α = cot β + 2tan β
18a Considerthefunction y = 2sin(x θ)cos x.
i Showthat dy dx = 2cos(2x θ).
ii Hence,orotherwise,showthat 2sin(x θ)cos x = sin(2x θ) sin θ
b Aprojectileisfiredfromtheoriginwithvelocity V atanangleof α to thehorizontalupaplaneinclinedat β tothehorizontal.Assumethatthe horizontalandverticalcomponentsoftheprojectile’sdisplacementare givenby x = Vt cos α and y = Vt sin α 1 2 gt2
i Theprojectilestrikestheplaneat(X, Y).Showthat X = 2V 2 cos2 α(tan α tan β) g .
ii Henceshowthattherange R oftheprojectileuptheplaneisgivenby R = 2V 2 cos α sin(α β) g cos2 β
iii Usepart aii toshowthatthemaximumpossiblevalueof R is V 2 g(1 + sin β)
iv Iftheangleofinclinationoftheplaneis14◦,atwhatangletothehorizontalshouldtheprojectilebe firedinordertoattainthemaximumpossiblerange?
19 Atallbuildingstandsonlevelground.Thenozzleofawatersprinklerispositionedatapoint P onthe groundatadistance d fromawallofthebuilding.Waterspraysfromthenozzlewithspeed V andthe nozzlecanbepointedinanydirectionfrom P.
a If V > gd,provethatthewatercanreachthewallabovegroundlevel.
b Supposethat V = 2 gd.Showthattheportionofthewallthatcanbesprayedwithwaterisaparabolic segmentofheight 15 8 d andarea 5√15 2 d2
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7I
ResistedProjectileMotion
Inwhatfollows,itisassumedthataparticleofmass m isprojectedfromtheoriginwithinitialvelocityvector u0 i + v0 j ∼ andisundertheinfluenceofgravityinamediumwithresistance.Likeverticalresistedmotion,the resistanceisproportionaltoapowerofthespeed, |v|n.Onlythecase n = 1isconsideredhereasthecomponent equationscannotbesolvedwhen n = 2.
Nevertheless,solutionsexistfor n = 2whentheresistanceisapproximatedwithanotherfunction.Somecases arepresentedasenrichmentattheendofthissection,andattheendoftheexercise.Themathematicsisquite complicatedbutmayprovidearewardinginvestigationforstudentsabletotakeonthechallenge.
ResistanceProportionaltoVelocity
Likeresistedverticalmotion,thegoverningequationsmaybederivedbysummingtheforces,thistimeinvector form.Thefreebodydiagramshowsaparticleat p withvelocity v ∼.Therearetwoforcesactingontheparticle:its weight mgj ∼ andaresistance R whichisproportionaltoitsspeedandintheoppositedirectiontoitsvelocity.
Thus R = mk|v ∼ | ˆ v ∼ = mkv
forsomeconstant k.Thetotalforceontheparticleisthus ma = mgj ∼ mkv or a ∼ = gj kv ∼ .
Thecomponentsofthisvectorequationprovidethegoverningequations:
d2 x dt
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and
Studentsareexpectedtobeabletoderivetheseequations,however,sometimestheymaybegiveninaquestion.
Example32
Aprojectileisfiredfromtheorigin.Airresistanceisproportionaltothevelocity.Thustheequationsofmotion are
d2 x dt2 = k dx dt and d2y dt2 = g k dy dt , where k = 1 5 and g ≑ 10.Theinitialvelocityis V ∼ = 15 i ∼ + 30 j
Find x(t)and y(t).HencedeterminetheCartesianequationofmotion.
Solution
Let˙ x = u and˙ y = v.Thedifferentialequationforthehorizontalcomponentisthen:
u = 1 5 u, whichistheequationforexponentialdecay,thus
u = Ae t/5
At t = 0,15 = A
hence˙ x = 15e t/5
Continuedonthenextpage
Integratingasecondtime
x = B 75e t/5
andat t = 0theprojectileisattheoriginso
B = 75
thus x = 75(1 e t/5).
Nowsolvethedifferentialequationfortheverticalcomponentofmotion.
v = 1 5 (50 + v) whichisshiftedexponentialdecay,thus,
v = Ce t/5 50.
At t = 0,30 = C 50,so C = 80andhence
y = 80e t/5 50.
Integratingasecondtime
y = 400e t/5 50t + D
andat t = 0theprojectileisattheoriginso
D = 400.
Thus y = 400(1 e t/5) 50t. (A)
Nowfromabove
x = 75(1 e t/5) (B) so75e t/5 = 75 x or et/5 = 75 75 x and t = 5log 75 75 x . (C)
Thuscombining(A),(B)and(C)gives:
y = 400x 75 250log 75 75 x
Thisisgraphedontherightalongwith y = 2x 1 45 x2,thetrajectorywhenthereisno resistance. x y
Enrichment—ResistanceProportionaltotheSquareofSpeed
Thefreebodydiagramisessentiallythesamewhentheresistanceisproportionaltothesquareofthevelocity, however,now
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andso
Writingoutthecomponentsofthisequationyields
u = k(u2 + v2) 1 2 u (1) and˙v = g k(u2 + v2) 1 2 v, (2) where u = x isthehorizontalcomponentofvelocityand v = y istheverticalcomponentofvelocity.Theseare coupledhighlynon-lineardifferentialequations,duetotheterm |v| = (u2 + v2) 1 2 appearingineach.
Thereisnoknownsolutiontotheseequations.Ifatrajectoryisrequired,thebestthatcanbedoneatschoollevel istoapproximatethecurvewithaseriesoflittlestraightlineelements.Thisisthemethodoutlinedinthelast questionoftheexercise.
Enrichment—Low-angleTrajectory
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Whentheangleofprojectionissmall,theinitialverticalcomponentofvelocitywillbesmall.Further,thevertical velocitywillapproachzeroneartheturningpoint.Thusitisclearthat u ≫ v (u ismuchgreaterthan v)forthe entirejourney.Consequently(u2 + v2) 1 2 ≑ |u| and |u| = u,sincethehorizontalcomponentofvelocityisalways positivebyobservation.Thusequations(1)and(2)abovemaybeapproximatedwith
and˙v = g kuv
Theseequationsarerelativelyeasytosolve.Also,becausetheapproximationisvalidateverypointonthecurve, thereshouldbegoodagreementbetweenthesolutionsof(3)and(4)andthesolutionsof(1)and(2).Solving(3) first,whichisvariableseparable,
u = u0 1 + ku0t and x = 1 k log(1 + ku0t).
Equation(4)thenbecomes v = g ku0v 1 + ku0
Thisequationcanbesolvedbyrearrangingitasfollows.
).
Noticethatthelefthandsideisthederivativeofaproduct,so
(1
which,afterintegration,applyingtheinitialconditionsandafewlinesofalgebra, gives v =
and
Enrichment—High-angleTrajectory
Whentheangleofprojectionisnearlyarightangle,thehorizontalcomponentofvelocityistinyandso v ≫ u forthemajorityofthejourney.Consequently(u2 + v2) 1 2 ≑ |v| andsoequations(1)and(2)abovemaybe approximatedwith u = ku|v| (5) and˙v = g kv|v| (6)
Aswithverticalprojections,theupwardanddownwardjourneysareconsideredseparately.Inbothcases,the algebraisdifficultandthesolutionsarelengthy.Itiscoveredinanenrichmentquestionoftheexercise.
Notethatnearthetopofthetrajectory v → 0andthus u ≫ v inthatvicinity.Consequentlytheapproximation (u2 + v2) 1 2 ≑ |v| isnotvalidneartheturningpoint.Thusitisexpectedthatthehigh-angletrajectoryapproximation willbegoodformostofthejourney,butpoorneartheturningpoint.
(3)
Enrichment—Mid-angleTrajectory
Whentheangleofprojectionisabout45◦,bothcomponentsofthevelocitywillberoughlyequalforasignificant portionofthetrajectory.Itseemsreasonablethentoapproximate(
|.Thatis, equations(1)and(2)willbeapproximatedwith
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Noticethatequation(7)maybeobtainedbyreplacing k with k√2inequation(3).Thusequation(7)hassolution
and
Likewise,equation(8)isobtainedbyreplacing k with k√2inequation(6).Thusthesolutionofequation(8)may beobtainedfromthesolutionofequation(6).
Likethehigh-angletrajectory,themid-angleapproximationisnotvalidneartheturningpoint.Thusitisexpected thatthemid-angletrajectoryapproximationwillbegoodformostofthejourney,butpoorneartheturningpoint.
Enrichment—GraphsoftheApproximations
Ineachcasetheinitialvelocityis |v ∼ 0| = 10,accelerationduetogravityis g = 10,thecoefficientofresistance is k = 0.1,theterminalvelocityis VT = 10andtheangleofprojectionis α.Thesolidlineistheappropriate approximation.Thedashedlineisthesolutiontothefullequations(1)and(2)obtainednumericallyusing avectorformofthefourthorderRunge-Kuttamethod,anapproximationmethodstudiedinsomeuniversity courses.Thetrajectorywithnoresistanceisshownwithadash-dotline.Aspredictedinthediscussionabove,the high-angleapproximationislessaccurateneartheturningpoint,butisclosetotheactualtrajectoryoverall.The mid-anglealsobreaksdownneartheturningpointasforetold,butagainisagoodapproximationoverall.
Aswasexpected,thelow-angleapproximationgivesexcellentresults.Noticethechangeinverticalscalein thethirdgraph.Evenwiththisverticalexaggeration,thedifferencebetweenthefullequationsolutionandthe approximationisbarelynoticeable.
Exercise7I
FOUNDATION
1 Anobjectofmass1kgisprojectedfromtheoriginwithinitialspeed30m/satanangleof60◦ above thehorizontal.Itissubjecttogravityaswellasairresistanceproportionalto 1 6 ofitsvelocityinboththe horizontalandverticaldirections,sothat
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a Showthat˙
b Showthat x = 90 1 e 1 6
c Showthattheparticlereachesitsgreatestheightwhen t = 6ln
d Find,correcttothenearestmetre,thehorizontaldistancetravelledwhentheparticlereachesitsgreatest height.
2 Anobjectofmass1kgisprojectedfromtheoriginwithinitialspeed20m/satanangleof45◦ abovethe horizontal.Itissubjecttogravityaswellasairresistanceproportionaltothevelocityinboththehorizontal andverticaldirectionswithconstantofproportionality2.Assumethat g = 10m/s2 .
a Showthatthehorizontalandverticalvelocitiesaregivenby
b Showthatthehorizontalandverticaldisplacementsaregivenby
c Showthattheobjectreachesamaximumheightof
DEVELOPMENT
3 Aparticleofmass1kgisprojectedfromtheoriginwithspeed V atanangleof α abovethehorizontal.It issubjecttobothgravityandanairresistanceproportionaltoitsvelocity,sothatitshorizontalandvertical componentsofaccelerationwhileitisrisingaregivenby¨ x = k ˙ x and¨ y = g k ˙ y.
a Showthat˙ x = V cos α e kt and˙
b Henceshowthat x =
c Whentheparticlereachesitsgreatestheight,showthatithastravelledahorizontaldistanceof V 2 sin2α 2 (g + Vk sin α) .
4 Anobjectofmass m kgisprojectedonlevelgroundfromtheoriginwithinitialvelocity v ∼ (0) = (10 i ∼ + 20 j ∼ )m/s.Theobjectexperiencesagravitationalforceof 10mj ∼ ,andtheairresistanceactingonthe projectileisgivenby 0.5mv.Sothetotalforce F actingontheprojectilecanbewrittenas F = ma = 10mj ∼ 0.5mv .
a Showthatthevelocityoftheprojectileis v(t) =
b Showthatthedisplacementoftheprojectileis r ∼ (t)
c Thegraphsof y = 1 3 x, y = 1 4 x,and y = 1 e 1 2 x areshownontheright.The x-coordinatesoftwointersectionpointsaregivencorrecttotwodecimal places.Usethediagramtoanswerthefollowingquestions.
Theprojectilejustclearsaverticalwallofheight d metreswhosebaseisa horizontaldistanceof d metresfromtheorigin.Considerthewalltohave negligiblethickness.
i Findthetimetheprojectilehasbeenintheairattheinstantitclearsthewall,andhencefindthevalue of d correcttoonedecimalplace.
ii Determinethedistancebetweenthebaseofthewallandthepointwheretheprojectilehitsthe ground.Giveyouranswerroundedtoonedecimalplace.
5 Aprojectileisfiredwithvelocity V = 30m/satanangleof45◦ abovethehorizontal.Airresistanceis proportionaltothevelocity.Thustheequationsofmotionare
and
whereitisknownthat k = 1 3 .Take
a Showthat,attime t,thehorizontaldisplacementis x =
b Findasimilarexpressionfortheverticaldisplacement y asafunctionof t.
c Henceshowthat y = 1 + √2 x 90ln
d Usegraphingsoftwaretoplotthefunctionin c andhenceestimatethehorizontalrange.
e Bywayofcomparison,showthetrajectoryfornoairresistanceonthesamegraph.
6 Aparticleofmass1kgisprojectedfromtheoriginwithspeed20m/satanangleofcos 1 4 5 below the horizontal.Theparticlemovesthroughamediumandexperiencesaresistance R ∼ thatisproportionaltothe velocityoftheparticle.So R ∼
a Showthat˙ x = 16e kt and˙ y = 1 k (10 12
.Usetheapproximation g = 10m/s2
10)
b ShowthattheCartesianequationofthepathoftheparticleis
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CHALLENGE
7 [Low-angleTrajectory]Aparticleisprojectedfromtheoriginwithspeed10m/satanangleofinclination of15◦.Resistanceisproportionaltothesquareofthespeedateachpointinthetrajectoryandtheconstant ofproportionalityis 1 10 m where m isthemassoftheparticle.Assumethattheaccelerationduetogravityis 10m/s2.Alsoassumethatthespeedmaybeapproximatedby(u2 + v2) 1 2 ≑ u.Thus,fromthetextabove: u = 1 10 u2 (3) and˙v = 10 1 10 uv (4)
a Showthatthehorizontalandverticalcomponentsoftheinitialvelocityare
b Considerthehorizontalcomponentofmotion.
i Equation(3)isvariableseparable.Determine u intermsof u0 and t
ii Hencedetermine x intermsof u0 and t.
c Considertheverticalcomponentofmotion.
i Usetheanswertopart bi andequation(4)toshowthat
(10 +
ii Showthat v = 10(10 +
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andhencefind y
d Set v = 0andhencefindthetimetakentoreachthemaximumheight,aswellasthecoordinatesofthat point.Givetheanswerscorrecttotwodecimalplaces.
8 [High-angleTrajectory]Aparticleisprojectedfromtheoriginwithspeed10m/satanangleofinclination of75◦.Resistanceisproportionaltothesquareofthespeedateachpointinthetrajectoryandtheconstant ofproportionalityis 1 10 m where m isthemassoftheparticle.Assumethattheaccelerationduetogravityis 10m/s2.Alsoassumethatthespeedmaybeapproximatedby(
and˙
a Showthatthehorizontalandverticalcomponentsoftheinitialvelocityare
b TheUpwardJourney:Inthiscase |v| = v.
i Showthat v =
ii Explainwhy0 ≤ t ≤ θ
iii Thenshowthat y = 10log cos(θ t) cos θ .
iv Henceshowthatthemaximumheightis
v Usepart i andequation(5)toshowthat d dt u cos(θ t) = 0.
vi Henceshowthat u = u0 cos θ sec(θ t).
vii Giventhat ∫ sec φ dφ = log | sec φ + tan φ| + C,usethesubstitution φ = θ t inpart vi tofindan expressionfor x intermsof u0, θ and φ
viii Henceshowthatthe x-coordinateoftheturningpointis
c TheDownwardJourney:Inthiscase |v| = v
i Put τ = t θ andhencefind du dτ and dv dτ intermsof u and v.Thenwritedownthecorresponding initialconditionsfor τ = 0intermsof u0 and θ
ii Usepartialfractionstoshowthat v = 10 (eτ e τ) eτ + e τ
iii Henceshowthat y = 10log 2sec θ eτ + e τ
iv Put y = 0intopart iii tofind τ andhencefindthetotaltimeofflight.
v Nowusepart ii andequation(5)toshowthat d dτ (eτ + e τ) u = 0.
vi Henceshowthat u = (2u0 cos θ)e τ 1 + e 2τ .
vii Showthat x = u0 cos θ log(sec θ + tan θ)when τ = 0.Thenusepart vi withthisinitialconditionto find x intermsof u0, θ and τ.
viii Usethevalueof τ inpart iv todeterminetherangeoftheprojectileintermsof u0 and θ,then approximatethistothreedecimalplaces.
9 [ANumericalApproximation]Theequationsofmotionforaprojectilewithairresistanceproportionalto thesquareofthespeedare
u = k(u 2 + v 2) 1 2 u and˙v = g k(u 2 + v 2) 1 2 v, where u = x isthehorizontalcomponentofvelocityand v = y istheverticalcomponentofvelocity.Thereis noknownsolutiontothispairofequations.Neverthelessthetrajectorycanbeapproximatedandplottedby followingthestepsbelow.
Inthiscase,theinitialspeedoftheprojectileis30m/s,theangleofprojectionis45◦,and k = 1 90 .Take g = 10m/s2.Giveyouranswerscorrectto2decimalplaces.
ai Recallthat,fromfirstprinciples,
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.Hence,for δt smallenough,
ii Findsimilarexpressionsfor x(t + δt), y(t + δt)and v(t + δt).
b Itshouldbeclearthat u(0) = v(0) = 15√2,andthat x(0) = y(0) = 0.
i Usepart ai with t = 0and δt = 0.1tofindtheapproximatevalueof u(0.1).
ii Similarlyfindtheapproximatevaluesof v(0.1), x(0.1)and y(0.1).
c Whataretheapproximatevaluesof u(0.2), v(0.2), x(0.2)and y(0.2)?
d Useaspreadsheetorappropriatemathematicalsoftwaretocontinuetofind x, y, u and v intimesteps of0.1.Henceplotthetrajectoryoftheprojectile.Whatistheapproximaterange,correcttothenearest metre?
7J
MiscellaneousProblems
Thereisnonewmathematicaltheorypresentedinthissection.Insteaditisacollectionofharderproblems deemedtobewithinthescopeofthesyllabus.Thelistofproblemspresentedhereisnotintendedtobe exhaustive.Teachersandstudentsareencouragedtofindotherpracticalapplicationsofthemathematicsthey haveencountered,andtosolvethoseproblems.Whenattemptingquestions,studentsareencouragedtodrawa diagram,whereappropriate,asoftenaproblemismoresimplysolvedfromthediagram.
Example33
Acarofmass m istravellingataconstantspeed V0 m/s.Thebrakesareapplied,resultinginaconstant decelerationuntilthecarcomestoastop.Giventhatthecarstopsin ℓ m,findtheforceexertedbythebrakes.
Solution
Let v m/sbethespeedofthecar x mbeyondwherethebrakesarefirstapplied.Sincedecelerationisinvolved, lettheaccelerationofthecarbe a,with a > 0,sothat
Integrating, 1 2 v2 = ax + C. At x = ℓ, v = 0so C = aℓ.Thusthevelocityisgivenby
At x = 0,
thus
Hencetheforcerequiredtostopthecaris:
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Noticethatforanygivenforce,thedistanceisproportionaltothesquareofthespeed.Thusacargoing just10%abovethespeedlimitwillrequireadistance21%greatertostop.Thisisobviouslyasignificantissue forroadsafety.
VariableGravity
Oneparticulartypeofproblemthatcanbesolvedinthiscourseismotionwithvariablegravity,asshownin previouslyExample 13 andhereinthefollowingexample.
Example34
Theaccelerationexperiencedbyanobjectduetogravityisinverselyproportionaltothesquareoftheradius r metresbetweenthecentreoftheEarthandthecentreoftheobject,andisdirectedtowardsthecentreofthe Earth.Thus,ignoringallotherforces,theequationofmotionis:
d2r dt2 = k r2 .
AssumethattheEarthisasphereofradius R andthattheaccelerationduetogravityatthesurfaceoftheEarth is g.Theobjectisprojectedverticallywithinitialvelocity V0
a ThesmallestinitialvelocityforwhichtheobjectneverreturnstotheEarthiscalledtheescapevelocity. Showthattheescapevelocityis V0 = 2gR.
b Giventhat V0 = 2gR,findthetimetakentoreachanaltitudeof R
Solution
a Beginbyfindingthevalueof k.AtthesurfaceoftheEarth r = R,so
g = k R2
hence k = gR2
Next,re-writetheaccelerationintermsofvelocity v andradius r
d dr v2 = 2gR2 r2
so v2 = 2gR2 r + C1 (forsomeconstant C1).
When t = 0, r = R and v = V0 so
V0 2 = 2gR + C1
thus C1 = V0 2 2gR and v2 = 2gR2 r + V0 2 2gR
Sincetheobjectneverreturnsand¨ r < 0, r mustalwaysincrease.
Sotakethelimitas r →∞ toget
limr→∞ v2 = V0 2 2gR whichmustbepositiveorzero.(Squarescannotbenegative.)Thus V0 2 2gR ≥ 0 hence V0 ≥ 2gR.
Thatis,theescapevelocityis V0 = 2gR
FortheEarth,thesequantitiesareapproximately:
g = 10m/s2 , R = 6400kmand V0 = 8√2 ≑ 11.3km/s.
b Inthecaseoftheescapevelocity, v2 = 2gR2 r ,whichiscontinuousfor r > 0,andisneverzero.Hence v can neverchangesign.Sinceitisinitiallypositive,itmustremainpositive.Thustakethepositivesquarerootto get dr dt = 2gR2 r or r 1 2 × dr dt = R 2g.
Integrating, 2 3 r 3 2 = Rt 2g + C2 (forsomeconstant C2.)
When t = 0, r = R so C
Thus Rt
and
Henceat r = 2R thetimeis
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√
) = √R 3√g (4 √2)(about690sforEarth.)
ProjectileMotion
Thereisanenormousnumberofdifferentquestionsthatcanbeaskedonthistopic.Someofthoseincludedinthe subsequentexercisemayalsobeappropriateforMathematicsExtension1candidates.Equally,manyworthwhile questionshavebeenomittedsimplyforthesakeofspace.
Example35
Anobjectisprojectedfromtheorigin O atanangle α withinitialspeed V.Ataparticularpoint P(x, y)onthe trajectory,thegradientof OP istan β andthegradientofthetangenttothetrajectoryat P is dy dx = tan β.You mayassumethat
y = Vt sin α 1 2 gt2 and x = Vt cos α
a Showthatthetimetakentoreach P
b Henceshowthattan α = 3tan β
Solution
a At P tan β =
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b Usingthevalueof t foundinpart a, tan β = dy dx
SimpleHarmonicMotion
Aharderapplicationofsimpleharmonicmotionisinthesmallangleapproximationofthemotionofapendulum. Itisalsoanexampleofhowaforcemayberesolvedintoitscomponentsindifferentdirections,ratherlikethe resolutionoftheinitialvelocityinprojectilemotion.
Example36
Amass m issuspendedfromafixedpointonalightinextensiblewireoflength L. Theballissetinmotion,swingingbackandforthalonganarcinaverticalplane.At time t theanglethatthewiremakeswiththeverticalis θ.Theonlyforcesactingon theballarethetension T actingalongthewirenormaltothearc,andgravityacting vertically.
a Resolvetheforceduetogravitynormallytofind T
b Next,resolvetheforcetangentiallytoshowthat
c Underwhatcircumstancescantheequationinpart b beapproximatedwith
d2θ dt2 = gθ L ?
d At t = 0,theballisreleasedfromrestatanangle θ = θ0,where θ0 issmallandpositive.Usetheresultof part c toshowthat θ = θ0 cos g L t
e Whatisthespeedoftheballatthebottomoftheswing?
Solution
a Sincethelengthofthewiredoesnotchangeitfollowsthatthesumofthenormal forcesiszero.Thus
T mg cos θ = 0 or T = mg cos θ
b Theonlyforceactingtangentiallyisthecomponentofweight,whichis mg sin θ Fromobservation,theforceactsintheoppositedirectiontothedisplacement, θL.
Hence
m d2 dt2 θL = mg sin θ
so d2θ dt2 = g sin θ L
c When θ issmall,itfollowsthatsin θ ≑ θ so
d2θ
dt2 = gθ L
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mg
d Thepreviousresultistheequationofsimpleharmonicmotionwith n2 = g L .Hencethegeneralsolutionis
θ = a cos g L t + φ where π 2 ≤ φ ≤ π 2 so ˙ θ = a g L sin g L t + φ
At t = 0,thevelocityiszero.Thatis θL = 0,so
0 = a g L sin φ hence φ = 0(for φ inthespecifieddomain) and θ = a cos g L t .
At t = 0,theinitialangleis θ0 = a cos0 hence a = θ0 and θ = θ0 cos g L t
Hereisasimplerargument.ThemotionisSHMandstartsfromresthenceacosinesolutionisappropriate, sincethecosinefunctionhasastationarypointat t = 0.Itfollowsthattheinitialdisplacementisthe amplitude.
e Atthebottomoftheswing θ = 0,so
cos g L t = 0
thatis g L t = π 2 + kπ.
Thespeedis v = ˙ θL ,so
L
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sin
(atthebottomoftheswing)
Exercise7J
FOUNDATION
1 Anobjectfallsfromthetopofabuilding.Anofficeworkerseestheobjectfallpastawindow2mhighin 1 5 s.Assumethatairresistanceisnegligibleanduse g = 10m/s2.Let y betheheightoftheobjectabovethe bottomofthewindowattime t
a Showthat y = H 5t2,where H isthedistancefromthebottomofthewindowtothetopofthebuilding.
b Usepart a andtheinformationgiveninthequestiontoformapairofequations.Solvethesesimultaneouslytofindthevalueof H.
2 Aprojectileisfiredverticallyundertheinfluenceofgravityalone.Lettheaccelerationduetogravitybe g ≑ 10m/s2.Thustheheight y1 attime t satisfies¨ y1 = 10.Theprojectileisfiredwithinitialspeedof 16m/s.Onesecondlateranotherprojectileisfiredfromthesamepoint.
a Findthetimeforthefirstprojectiletoreachitsmaximumheight.Hencefindthismaximumheight.
b Lettheinitialspeedofthesecondprojectilebe v. Showthatitsheightisgivenby y2 = 5(t 1)2 + v(t 1).
c Find v giventhatthetwocollidewhenthefirsthasreacheditsmaximumheight.
3 Ignoringairresistance,ahot-airballoonhastwoforcesactingonit:aconstantbuoyancyforce B and theforceduetogravity.Supposethatwhenthemassoftheballoonis M itdescendswithadownward acceleration d > 0.Ballastofmass m isthrownoutandtheresultisthattheballoonascendswithupward acceleration a > 0.
a Writedowntwoforcebalanceequationsfortheinformationgiven.
b Henceshowthat m = M a + d a + g
4 Theaccelerationduetogravityatdistance r fromthecentreofthemoonisgivenby d2r dt2 = gR2 r2 , where R ≑ 1737kmisitsradiusand g ≑ 1.625m/s2 istheaccelerationduetogravityatthesurfaceofthe moon.Aprojectileisfiredverticallyfromthesurfacewithinitialvelocity V0
a Deriveanexpressionfor v2,thesquareofthevelocityoftheprojectile,intermsof r
b Themaximumaltitudeoftheprojectileis R.Determine V0.
DEVELOPMENT
5 Thedeckofashipis1mbelowawharfatlowtideandis0.4mabovethewharfathightide.Onacertain daytheshiphasapreciouscargothatcanonlybesafelyoffloadedwhenthedeckisabovethelevelofthe wharf.Onthatday,lowtideisat1:10amandagainat1:35pm.Giventhatthetidecanbemodelledwith simpleharmonicmotion,betweenwhattimesinthemorningcanthecargobesafelyoffloaded?
6a
Showthattan( π 4 + α 2 ) = sin α + 1 cos α
b TheCartesianequationofaprojectilefiredatangle β andvelocity V is y = gx2 2V 2 sec2 β + x tan β
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Writethisequationasaquadraticintan β
c InagameofnetballtheGoalAttackisabouttoshootforgoal.Thelinejoiningthenetandthepointof projectionoftheballmakesanangle α withthehorizontal.Let h betheheightofthenetabovethepoint ofprojection.Let V betheminimumspeedrequiredtoscoreagoal.Thatis,thereisonlyoneangleof projection β atwhichtheshotmaybetakenwithspeed V
Letthecoordinatesofthenetbe(x, y).Showthat V 4 2gyV 2 g2 x2 = 0. i
Atthenet, y = h and x = h cot α.Showthat V 2 = gh(1 + cosec α). ii
Writedownanexpressionfortan β andhenceusepart a toshowthat β = π 4 + α 2 iii
7 TheCartesianequationofaprojectilefiredatangle θ andvelocity V maybewrittenas gx2 2V 2 tan2 θ x tan θ
a Letthisquadraticequationfortan θ havetwosolutions,tan θ1 andtan θ2. Showthattan(θ1 + θ2) = x y .
b Supposethatforacertainpoint P(a, b)thereisonlyonesolutionfor θ Provethatthisangleis
,wheretan
c Eachpointthatcanbehitbytheprojectileliesonorbelowaparabola.Byconsideringthediscriminant ofequation(∗),findanddescribethatparabola.
8 Anumberofballbearingsarefiredatthesamemoment,fromthesamepointandwiththesamespeed V but atdifferentanglesofprojectioninthesameverticalplane.Ignoringairresistance,thetimeequationsforthe positionofeachballbearingare
x = Vt cos α j and y = Vt sin α j 1 2 gt2 where α j istheangleofprojectionofthe jth ballbearing.Showthatatanytime,alltheballbearingslieona circle,andfindthecentreandradiusofthiscircle.
[Hint:Eliminate α j fromtheaboveequations.]
9 Athrill-seekergoesbungy-jumpingfromapoint O onabridgeaboveariver.Let x bethedistancebelow O andlet v bethevelocityofthethrill-seekertowardstheriver.
a Thepersonfreefallsforadistance ℓ.Thus d dx 1 2 v2 = g for0 ≤ x ≤ ℓ. Showthat v2 = 2gℓ when x = ℓ.
b For x >ℓ,thebungycordbeginstoslowthepersondownaccordingtotheequation d dx 1 2 v 2 = g gk(x ℓ) where k isapositiveconstant.
i Showthat v2 = 2gx gk(x ℓ)2 for x >ℓ
ii Showthatthefurthestthethrillseekerreachesis
xmax = 1 k 1 + kℓ + √1 + 2kℓ .
iii Find xmax intermsof ℓ when kℓ = 4.
10 Anobjectisfiredverticallyupwardswithinitialspeed V0 m/sfromgroundlevel.Ignoringairresistance, theaccelerationduetogravityvarieswithheight x abovegroundlevelaccordingtotheequation d dx 1 2 v 2 = 10R2 (R + x)2 , where R istheaverageradiusoftheEarth,and v isthevelocityoftheobject.
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a Whatisthesignificanceofthenegativesign?
b Findanexpressionfor h,themaximumheightreached,intermsof R and V0.
c Giventhat R = 6.4 × 106 mand V0 = 500m/s,evaluate h correcttothenearestmetre.
11 Thetemperature B ofabeakercontainingahotchemical,andthetemperature W ofacoolerwaterbathin whichthebeakerisplaced,bothsatisfyNewton’slawofcooling.Thusthetemperaturesarerelatedbythe equations
= k(B W)and
= 3 4 k(B W), where k isapositiveconstant.
a Bydifferentiating 3 4 B + W,showthat 3 4 B + W = C,where C isaconstant.
b Initiallythebeakeris120◦Candthewaterbathis22◦C.Find C andhenceshowthat dB dt = k( 7 4 B 112).
c Solvetheaboveequationtofind B asafunctionoftime.
d Aftertenminutesthetemperatureofthebeakeris92◦C.Findthetemperatureofthebeakeraftera furthertenminutes.
e Whatistheeventualtemperatureofthewaterbath?
12 Amodelforthepopulation P ofelephantsinSerengetiNationalParkis
P = 21000 7 + 3e t/3 where t isthetimeinyearsaftertheendof2008.
a ShowthatPsatisfiesthedifferentialequation
dP
dt = P 3 1 P 3000
b Whatwasthepopulationattheendof2008?
c Whatdoesthemodelpredictthattheeventualpopulationwillbe?
d Whatwastheannualpercentagerateofgrowthattheendof2008?
13 Aswateremptiesfromawatercoolerthroughasmallholeatthebottomitisfoundthatthedepthofthe watersatisfiestheequation
dy dt = k√y, where k isapositiveconstantand y isthedepthofwater.Initiallythedepthofthewateris y0 andittakes T secondstofullydrain.
a Showthat y y0 = 1 t T for0 ≤ t ≤ T .
b Supposethatittakes15secondsforhalfthewatertodrainout.Howlongdoesittaketoemptythefull cooler?
14 Thebobofapendulumisreleasedfromrest.Theinitialanglebetweenthewire andtheverticalis θ0 andthetensionalongthewireis T0.Attime t afteritis released,theanglebetweenthewireandtheverticalis θ andthetensionis T .You mayassumethattheonlyforcesactingonthebobareduetothetensionacting alongthewireandgravityactingvertically.Letthemassofthebobbe m andthe lengthofthewirebe L
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a Resolvetheforcesonthebobnormallyandtangentiallytoobtaintheequationsofmotion.
b Let ω = dθ dt .Showthat d2θ dt2 = d dθ 1 2 ω2 .
c Henceshowthat ω2 = 2 mL (T T0).
CHALLENGE
15 Abeadofmass m slidesalongawireintheshapeofthecurve y = 3 2 x2/3,where0 ≤ x ≤ 1.Attime t,letthe beadbeat P(x, y),where x and y arefunctionsof t.Thecoordinatesof P satisfytheequation 1 2 mx 2 + 1 2 my 2 + mgy = E, where E and g areconstants.
a When t = 0thebeadisreleasedfromrestatthepoint(1, 3 2 ).Find E
b Showthat˙ x = x1/3y,andhenceshowthat˙ y2 = 3g(3 2
) 3 + 2y
c Find˙ y and˙ x whenthebeadisattheorigin.
d Itisknownthat
Usethisresulttofindthetimeittakesforthebeadtotravelfrom(1, 3 2 )totheorigin.
e Asanexercise,usethemethodsofintegrationinChapter6toprovetheaboveresult.Whatispeculiar aboutthecase α = 1?
Chapter7Review
Reviewactivity
• Createyourownsummaryofthischapteronpaperorinadigitaldocument.
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Chapter7Multiple-choicequiz
• Thisautomatically-markedquizisaccessedintheInteractiveTextbook.AprintablePDFWorksheetversionis alsoavailablethere.
ChapterReviewExercise
Note: Inthisexerciseuse g ≑ 9.8m/s2 fortheaccelerationduetogravity.
1 Given a ∼ = 4 i ∼ + 3 j and b ∼ = 2 i ∼ + 5 j,findthemagnitudeandthedirection(asanangleofinclinationin degreescorrecttotwodecimalplaces)oftheresultantof a ∼ and b ∼
2 Threeforcesactonanobjectofmass5kg.Theseforcesarerepresentedbythevectors9 i ∼ 2 j, 3 i ∼ + 10 j and18 i ∼ j.Calculatethemagnitudeanddirection(asanangleofinclination)oftheaccelerationofthe object.
3 A2kgmassissuspendedfromaceilingbytwostrings,eachmakinganangleof30◦ withtheupward vertical.Taking g = 9.8m/s2,findthetensioninthestringscorrectto3significantfigures.
4 Thediagramontherightshowstwoforcesofmagnitude20Nand15N representedbythevectors −−→ OA and −−→ OB. a Express −−→ OA and −−→ OB ascomponentvectors.
b Calculatethemagnitudeoftheresultantofthetwoforces,correcttothenearest newton.
c Determinethedirectionoftheresultantasanangleofinclinationcorrecttothe nearestdegree.
5 Forcesofmagnitude5N,9Nand7Nareactingonanobject.Theforcesactparalleltothesides −−→ AB, −−→ BC and −−→ CA respectivelyofanequilateraltriangle ABC Findthemagnitudeoftheresultantofthethreeforces. a Findthedirectionoftheresultantrelativeto −−→ BA b
6 Anobjectofmass5kgisatrestonaplaneinclinedat35◦ tothehorizontal.Findthecomponentofthe weight: downtheplane, a perpendiculartotheplane. b
7 Objectsofmass6kgand4kgareconnectedbyalightinextensiblestringthatpassesoverasmoothpulley. Themasseshangverticallywhenthesystemisreleasedfromrest.Find: themagnitudeoftheaccelerationofthetwoobjects, a thetensioninthestring. b
8 Anobjectofmass10kgisonasmoothplaneinclinedat30◦ tothehorizontal.Itisconnectedtoafreely hangingobjectofmass8kgbyalightinextensiblestringpassingoverasmoothpulleyatthetopofthe slope.Find,correctto3significantfigures: theaccelerationofthe10kgobjectuptheplane, a thetensioninthestring. b
9
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10
11
Ineachpart v isgivenasafunctionof x,anditisknownthat x = 1when t = 0. Express:
t intermsof x, i x intermsof t ii
v = 2x 1 a v = 6x2 b
Ineachmotionofthepreviousquestion,find¨ x intermsof x
Ineachparttheacceleration¨ x isgivenasafunctionof x.Express v2 intermsof x giventhat v = 0when x = 0.
x = 6 a x = sin6x b
12
Giventhat¨ x = 4v,and v = 2when t = 0,find t asafunctionof v
13 Giventhat¨ x = 1 3v ,and v = 1when x = 0,find x asafunctionof v
14 Aparticleofmass m movesinastraightlinesubjecttoaforce F.Attime t,itsvelocityis v andits displacementis x.Theparticlewasinitiallyatrestattheorigin.
If F = 18t and m = 3,find x when t = 2. a
If F = 4x 3and m = 2,findthepositivevalueof v when x = 4. b
c
d
If F = 1 1 + v and m = 0.5,find t when v = 6.
If F = 2 + v2 and m = 2,find x when v = 2.
15 Aplanelandsonarunwayat100m/s.Itthenbrakeswithaconstantdecelerationuntilitstops2kmdown therunway.
a Explainwhytheequationofmotionis¨ x = k,forsomepositiveconstant k.Byintegratingwithrespect to x,find k,andfind v2 asafunctionof x
b Find: thevelocityafter1km, i whereitiswhenthevelocityis50m/s. ii
c Explainwhy,duringthebraking, v = √10000 5x ratherthan v = √10000 5x
d Integratetofindthedisplacement–timefunction,andfindhowlongittakestostop.
16 Whenfiredunderwater,atorpedomoveswithdecelerationproportionaltoitsvelocity,sothat¨ x = kv for somepositiveconstant k.Itsinitialspeedis 1 2 km/s,anditsspeedafterithasgone50metresis250m/s.
a Use¨ x = v dv dx tofind v asafunctionof x,thenfind x asafunctionof t
b Showthatittakes 1 5 ln2secondstogothefirst50metres,anddescribethesubsequentmotionofthe torpedo.
17 Aparticlemoveswithacceleration¨ x = 1 2 e x m/s2
a Initiallyitisattheoriginwithvelocity1m/s.Findanexpressionfor v2
b Explainwhy v isalwayspositive,andhencefindthedisplacementasafunctionoftime.
c Describewhathappenstotheparticleas t →∞
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18 Theaccelerationofaparticlemovinginastraightlineisgivenby¨ x = a + bv2,where a and b arepositive constants.Initiallytheparticlewasatrestattheorigin.Showthat: v2 = a b e2bx 1 a v = a b tan √abt b
19 Aparticleisoscillatinginsimpleharmonicmotionwithacceleration¨ x = 16x.Attime t = 0, x = 1and x = 4.
a Showbyintegratingthat v2 = 16(2 x2).
b Whatisthemaximumdisplacementoftheparticle?
c Find x asafunctionof t intheform x = a cos(nt + α).
20 Thedisplacement x attime t ofaparticlemovingonthe x-axisisgivenby x = 5 + √3sin3t cos3t
a Provethatthemotionoftheparticleissimpleharmonic.
b Express √3sin3t cos3t intheform R sin(3t α),where R > 0and0 < α < π 2 .
c Whataretheamplitudeandthecentreofthemotion?
d Whendoestheparticlefirstreachitsmaximumspeed?
21 Aparticleismovinginsimpleharmonicmotion.Itsmaximumspeedandaccelerationare2m/sand6m/s2 respectively.Findtheamplitudeandperiodofthemotion.
22 Apistoninacarengineismovingupanddowninsimpleharmonicmotionwithanamplitudeof5cm.The massofthepistonis0.5kganditismaking60oscillationspersecond.Find:
a themaximumspeedofthepiston,inm/scorrecttoonedecimalplace, b themaximumforceactingonthepiston,correcttothenearestnewton.
23 Aparticleisoscillatingonthe x-axisabouttheorigin.Itpassesthroughthepoint A,where x = 0.01metres, withspeed0.09m/sanditpassesthroughthepoint B,where x = 0.02metres,withspeed0.06m/s.
a Showthattheamplitudeofthemotionis √10 125 metres.
b Showthatthemotionhasperiod 2π √15 seconds.
c Find,correcttothenearesthundredthofasecond,thetimethattheparticletakestomovedirectlyfrom A to B
24 Anaircraftofmass M kgmovesalongahorizontalrunway,startingfromrest.Theaircraft’senginesexert aconstantthrustof T newtonsand,whenthespeedoftheaircraftis v m/s,itexperiencesaresistanceof magnitude kv2,where k > 0. Whenthespeedoftheaircraftis V m/s,showthatithastravelled M 2k ln T T kV 2 metres.
25 Anobjectofmass2kgisprojectedverticallyupwardswithinitialspeed20m/s.Itexperiencesair resistanceofmagnitude 1 10 v2 newtons,where v m/sisthespeedoftheobjectafter t seconds.Take g = 10m/s2 .
a Showthattheobjectreachesamaximumheightof10ln3metres.
b Find,correctto3significantfigures,thespeedoftheobjectwhenitreacheshalfitsmaximumheight.
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26 Achilddropsanair-filledballoonofmass30gramsfromabridge6metresaboveariver.Theballoon experiencesairresistanceofmagnitude0.6v newtons,where v m/sisthespeedoftheballoonafter t seconds.Take g = 10m/s2
a Writedownanequationofmotionfortheballoon’sdescent.
b Showthattheterminalspeedoftheballoonis0.5m/s.
c Showthat v = 1 2 1 e 20t
d Howlongdoesittake,to2significantfigures,fortheballoontoreachhalfitsterminalspeed?
27 Aparticleofmass m fallsfromrestinamediuminwhichtheresistancetoitsmotionis mkv,where k isa constantand v isthespeedoftheparticle.Astheparticlefalls, v approachesalimitingvalueof V.
a Showthat dv dt = g V (V v).
b Findthetimeittakesfortheparticletoreachaspeedof 1 2 V.
c Showthatthedistancefallenwhenaspeedof 1 2 V isreachedis V
28 Aparticle P1 isprojectedverticallyupwardsfromapoint A withinitialspeed U.Atthesameinstant,a secondparticle P2,alsoofmass m,isdroppedfromapoint B directlyabove A.Thedistance H between A and B isequaltothemaximumheightthat P1 wouldreachwereitnottocollidewith P2.Astheparticles move,theyeachexperienceairresistanceofmagnitude mkv2,where k > 0and v isthespeed.Attheinstant theparticlescollide, P2 hasreachedexactly50%ofitsterminalspeed.Let y1 bethedistanceof P1 above A, and y2 thedistanceof P2 below B.
a Showthat V = g k
b Showthat y1 = 1 2k ln g + kU2 g + kv1 2 ,where v1 isthespeedof P1.
c Henceshowthat H = 1 2k ln 1 + U2 V 2 .
d Giventhat y2 = 1 2k ln g g kv2 2 ,showthat,attheinstanttheparticlescollide, P2 hasfallenadistanceof 1 2k ln 4 3 .
e Deducethatthespeedof P1 attheinstanttheparticlescollideis 1 √3 V
29 Aprojectilehasinitialvelocityvector48 i + 36 j ∼ ,andairresistanceisnegligible.
a Obtain,byintegration,expressionsfor˙ x,˙ y, x and y,taking g = 10m/s2
b Findthemaximumheightreachedbytheprojectile.
c Findthehorizontalrangeoftheprojectile.
d Find,asacomponentvector,thevelocityoftheparticleafter1.6seconds.
30 Aparticleisprojectedfromtheoriginandfollowsaparabolicpathwithparametricequations x = 12t and y = 9t 5t2 (where x and y areinmetres).
a ShowthattheCartesianequationofthepathis y = 3 4 x 5 144 x2
b Findthehorizontalrange R andthegreatestheight H.
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c Findthegradientat x = 0,andhencefindtheangleofprojection.
d Find˙ x and˙ y when t = 0.Hence,byresolvingthesecomponents,findtheinitialvelocity,andconfirmthe angleofprojection.
e Findwhentheparticleis4metreshigh,andthehorizontaldisplacementthen.
31 EvetossedanappletoAdamwhowassittingnearher.AdamcaughtitatthesameheightthatEvereleased itfrom.Supposethattheinitialvelocityoftheapplewas v = 5m/s,atanangleoftan 1 2abovethe horizontal.
a Findtheinitialvaluesof˙ x and˙ y.
b Find˙ x, x,˙ y and y byintegrating¨ x = 0and¨ y = 10,takingtheorigintobethepointofreleaseofthe apple.
c Findthegreatestheightabovethepointofreleasereachedbytheapple.
d Showthattheapplewasintheairfor 2 5 √5seconds,andhencefindthehorizontaldistancetravelledby theapple.
e Find˙ x and˙ y atthetimeAdamcaughttheapple.Thenshowthatthefinalspeedwasthesameasthe initialspeed,andthefinaldirectionwastheoppositeoftheinitialdirection.
f Thepathoftheappleisaparabolicarc.Byeliminating t fromtheequationsfor x and y,finditsequation inCartesianform.
32 Abulletisfiredhorizontallyat200m/sfromawindow45metresabovelevelground.Iteventuallylands harmlesslyontheground.Assumethereisnoairresistance.
a Writedowntheinitialvaluesof˙ x and˙ y
b Taking g = 10m/s2 andtheoriginatthewindow,find˙ x, x,˙ y and y.HencefindtheCartesianequationof theparabolicpath.
c Findthehorizontaldistancethatthebullettravels.
d Find,correcttothenearestminute,theacuteangleatwhichthebulletlands.
33 Aballisthrownupahillwhichisinclinedat30◦ tothehorizontal.Theinitialvelocityoftheballis30m/s at anangleof45◦ abovethehorizontal.Supposethattheballisthrownfromtheorigin O atthefootofthe hill,anditlandsatapoint P onthesideofthehill.
a Startingwith¨ x = 0and¨ y = 10,showthattheparabolicpathoftheballhasparametricequations x = 15√2 t and y = 15√2 t 5t2
b FindtheCartesianequationoftheparabolicpath.
c Henceshowthat OP = 60(√3 1)metres.
34 Acannoncanfireashellwithaninitialspeed V andavariableangleofelevation α.Assumethat t seconds afterbeingfired,thehorizontalandverticaldisplacements x and y oftheshellaregivenby x = Vt cos α and y = Vt sin α 1 2 gt2 .
a ShowthattheCartesianequationoftheshell’spathmaybewrittenas gx 2 tan2 α 2xV 2 tan α + (2yV 2 + gx 2) = 0.
b Supposethat V = 200m/s, g = 10m/s2 andtheshellhitsatargetpositioned3kmhorizontally and0.5kmverticallyfromthecannon.Showthattan α = 4 ± √3 3 ,andhencefindthetwopossiblevalues of α,correcttothenearestminute.
35 Asmallpaintballisfiredfromtheoriginwithvelocity14m/stowardsan8mhighwall.Theoriginisat groundlevelandis10mfromthebaseofthewall.Supposethatthepaintballwasfiredatanangleof θ abovethehorizontal.
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a Startingwith¨ x = 0and¨ y = 9.8,showthat x = 14t cos θ and y = 14t sin θ 4.9t2
b ShowthatthetrajectoryofthepaintballhasCartesianequation y = mx m2 + 1 40 x 2,where m = tan θ.
c Ifthepaintballhitsthewallatheight h metres,showthat m = 2 ± √3 0.4h andhencedeterminethe maximumpossiblevalueof h.
d Supposethatthereisalargeholeinthewall,thebottomofwhichis3.9metresabovethegroundandthe topofwhichis5.9metresabovetheground.Determinethevaluesof m forwhichthepaintballwillpass throughthehole.
36 Aparticleofmass m isthrownfromthetop O ofatallbuildingwithinitialvelocity u atanangleof α abovethehorizontal.Theparticleexperiencestheeffectofgravityaswellasaresistanceproportionaltoits velocityinboththehorizontalandverticaldirections.Theequationsofmotionaregivenby¨ x = k ˙ x and ¨ y = k ˙ y g,where k isaconstantand g istheaccelerationduetogravity.
a Showthat˙ x = ue kt cos α.
b Show,bydifferentiating,that˙ y = 1 k (ku sin α + g) e kt g satisfiestheverticalequationofmotion.
c Findthevalueof t atwhichtheparticlereachesitsmaximumheight.
d Whatisthelimitingvalueofthehorizontaldisplacement?
≃ (equivalence)58–9
∼ , ¬ (negation)58,60–1
∧ (and)59–61
∨ (or)59–61
→, ⇒ (implication)61–3
↔, ⇔,iff (ifandonlyif)64,77
∃ (thereexists)61
∀ (forall)68
absolutevalue235,324 absolutevaluefunction231 acceleration280,282–5 forcesand288–91 formsof288–90 speedand298–9 algebra/algebraicapproach28,44, 118–19 ofdotproduct185 manipulation227–8 algebraicinequalities86–91 AM/GMinequality88,100 amplitude295,307 and(∧)59–61 negationof60–1 angles32–3,175,198 approximation343–4 arcs49
areaofatriangle189
Arganddiagram21–4,32 circlegeometry48–9 circles46–7,171 curvesin23–4,44–9 parabolas46–7 rays45–6 regions47–8 spirals132 straightlines44–5 arguments124,133–5 arithmeticmean88–9 assumptions280 azimuthangle175
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Béziercurve179–80 binomialexpansions128–9 bisection44–5,47–8,71 brackets60
calculus100–1,128–9,212–13,280, 316,334–5
Cartesianequations46–7,144, 173–4,178–9,181–2,341–2 Cauchy-Schwarzinequality187
centreofmotion299–300,306–7 chainrule,integralform230 circlegeometry48–9,100 circles46–7,132,171–5 arcs49
generalequations174–5 withothercentres172 parameters178 parametricequations144 intwodimensions171 inthreedimensions174–5 vectorequationsof171–5 collinearpoints154 collinearity39 commonfactors227–8 commutativelaw185 complements198
completingthesquare2,14,244–5 complexnumbers1 addition
Arganddiagram21–4 argument26–9 arithmetic2–9 conjugates5,8–9 divisionof5–6 equalityof7 geometryand22,36–40 i 2–4
imaginaryparts2–3,7,12,22, 27,29–30,125–6 mod-argform26–7,133–5 modulus26–7 polarform26,29 polynomials115–21 powers3–4,9,124–6 powersof i 3–4 productoftwo28–30 realparts4–5,7,12–13,22,27, 29–30,125–6 realisingthedenominator6 rootsof132–5 subtraction vectorsand32,36 z-plane21 complexplanes21,36–40,132–3 complexsquareroots15,17 generalcase16 Pythagoras15–16 compoundangleformula210 conclusion71 conditions70
conjugates/pairs5,8–9,13,30,124 constantofintegration248,251 constants236–7,295 continuity334 contradiction81–4 contraposition73,81 conventions6,261 conversestatements70–1 cosine190,210,253–4 cosinerule185 counterexamples57 cover-uprule238–9 curves44–9 diagrammatic23–4 parametric178–82 three3-D180–1
deMoivre’stheorem powersofcomplexnumbers 124–6 rootsofcomplexnumbers133–5 rootsofunity132–3 trigonometricidentities128–9 deMorgan’slaws60–1,63 definiteintegrals231–2 degenerateequations219 denominators227 withquadratics244–7 realising6 withrepeatedfactors240–1 derivatives313 diagrams freebody275,280–3,342 importanceof251 difference210–13 differenceoftwosquares12,244–5 differentialequations290–1, 299–300,305–7,321–2 differentiation298–9,307,334–5 dihedralangle199 dilations295 directedlinesegments36 direction36–7,39–40,147,274–5, 323–4 directioncosines190 directionvectors143,153,155,190 gradientin2D144–5 discriminants17–18 displacement288–9,295,299–300, 305–9,326 velocityasfunctionof307
distance26,32 frompointtoline166–7 distributivelaw60–1 distributivelawin3D185–6 divisibility77–9,95 divisiontheorem116–17 dotproduct154,187,198,282–3 algebraof185 areaofatriangle189 distributivelawin3D185–6 doubleangleidentities253 double-angleformula216–17 drag313–14
elimination45 enlargements31–2,36 equivalence(≃)58–9 equivalentstatements72–3 Euclideangeometry39 Eulerdiagram9 exactvalues211
factortheorem116–21 factorising/factoring115–16,250 factors235–6,239–40 falsestatements56–9,62,72 forall(∀)68 force accelerationand280,288–91 assumptions273 components274–5 deceleration349 drag313–14 equalandopposite281–2 normalforce281–2 resolving273–6 resultant275–6 astension276,280 asvectors274 fractiondecomposition235,240–1 freebodydiagrams275,280–3,342 friction280
FundamentalTheoremofAlgebra 118–20
FundamentalTheoremofArithmetic 83,111–13
geometry/geometricapproach22,28, 30–1,39,44,100 gradient–directionvectorin2D 144–5 graphs295,344 gravity280–3,320–1,323–4,341, 344,349–50
half-angleformula217 harmonicmotion295–301 helix180 horizontalresistedmotion313–16
identities253,258,261–3 if...then(—, →, ⇒)70–1 ifandonlyif(iff, ↔, ⇔)71–2,77 imaginaryaxis22 imaginarynumbers/values2–3,7,12, 22,27,29–30,125–6 implication61–3 →,
powersofcosandsin253–4 powersofsecandtan254–5 productstosums256–7 rationalisingthenumerator 246–7 reductionformulae261–3 repeatedapplication248–9, 290–1,342 standardintegrals225 substitution230–3 t-formulae233 trigonometricintegrals253–8 usingcommonfactors227–8 intersections47–8,153,156–8,207 intervals44–5,148–9,179–80 irrationalnumbers9,21 irreduciblequadratics120–1
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FundamentalTheoremof Calculus316
geometricmean88–9 geometryandcomplexnumbers22 multiplication30–2 multiplicationby i 22 opposite22 geometryandvectors
2Dproofs194–5
3Dproofs198–9,206–8
61 negationof63 proven63–5 inclinedplanes282–4 inconsistentequations158 induction94–7 inequality95–6 recursiveformulae96–7 inequality86–91 AM/GM88–9,100 calculus100–1 Cauchy-Schwarz187 combinations89 geometry100 induction95–6 known-result90 triangle188 initialconditions320–1,325–6 initialphase(motion)295 integercoefficients115–16 integerpowers9 integrals225,255–6 common271,313–14 definite231–2 integralformofchainrule230 leadingtologarithms235 miscellaneous266 recurrenceoforiginal251 integrands250 integration280 algebraicmanipulation227–8 thechainrule230 denominatorswithquadratics 244–7 limitsof231 partialfractions235–41 byparts248–51,262–3
letterpairing154 liketerms251 limits102,231,323,334 linesegments36 equationsof148 linearequations9 linearfactors235–6 linearnumerators245–6 linearquadratics120–1 linearterms244–5 lines arrangementsinaplane207 arrangementsinspace207–8 horizontal23–4 intersecting153,156–8,207 non-parallel155–6 parallel153–5,207 parametricequations143–5 perpendicular162–7,198,208 point-to-linedistance166–7 skew153,158,208 straight44,206 throughagivenpoint145–6 throughtheorigin143 throughtwogivenpoints 147,167 vectorequations143–9 vertical23–4 verticalin2D147 logarithmfunctions250 logarithmicfunctions313–14 logarithms,integralsleadingto235 logicalconnectors59–61 and(∧);or(∨)59–60 logicalvalues56–8 magnitude36–7,187,274,276 mass273–4,280,320–1 oninclinedplane282–3
middlepoint179–80 mod-argform26–7,133–5 product28–30 mod-augform125–6 motion273–6,288,295–301, 313–16,320–8,351 equationsof320–1 findingSHMfrom graphs308 naturalnumbers21 naturaloperations116 necessary70 negation58 ofor,and60–1 ofimplication63 ofquantifiers69 Newton’slawsofmotion273–6,288 non-parallellines155–6 normalforce281–2 notation121 numbers proofs77–9 structure9 numerators227 withhigherdegree237 linear245–6 rationalising246–7 opposites22–3,30 or(∨)59–60 negationof60–1 orderofprecedence60 origin22,30–1,143,298–9 choosing297–8
parabolas46–7 parallellines153–5,207 parameters/parametising145,148, 174–5,178–80,308–9 parametriccurves178–82 parametricequations143–5 curves144 parametricvectorequations143, 173–4 partialfractions235–41,257 cover-uprule238–9 findingconstants–substitution 236–7 linearfactors235–6 quadraticfactors239–40 repeatedfactors240–1 particles273,305–6,320–1 perfectsquares12 period295 perpendiculardistance167 perpendiculardotproduct189
perpendicularlines162–7,208 intwodimensions162–3 inthreedimensions164–6 planes45–6,186,280–5 arrangementsof207 defining206 inclined282–4 normalsto198,208,281–2 perpendicular208 seealso complexplanes points206,210–11 polarangle175 polarform26,29 polynomials237,250 complexnumbersand116 exceptionswith250 factors116–17,120–1 integercoefficients115–16 real coefficients117 reducing248–9 remainders116–17 positionvectors194–5 powers3–4,9,124–6,253–6 precedence60 premise71 problemsolving7,12–18,58,128–9, 218,349–53 generalisedsolutions314–15 methods315–16 productnotation121 products,sumsto212–13 productstosums210–13, 256–7 projectilemotion332–4,341–4,351 furtherExtension1332–4 projections165–6,275 anglesof343–4 pronumerals314–15 pronumeralsconvention6 proof logicandlanguageof56–65, 68–73 numberproofs77–9 proofbycontradiction81–3 proofbycontraposition81 proofbyinduction94–7,124 propositionalsymbol56 propositions59–61 provenimplications63–5 pulleys280–5 simple280–1,283–4 Pythagoras’theorem15–16, 166–7,198 Pythagoreanidentity216, 255–6 Pythagoreantriad218
quadraticequations9,12–18,28 complexcoefficients16–18 complexsquareroots15,17 realcoefficients9,12–13,28 quadraticfactors120–1,239–40 quadraticmethod14 quadratics denominatorswith244–7 irreducible120–1 withlinearterms244–5 quantifiers68–9 forall(∀)68 negationand69 thereexists(∃ )61 range307 ratiodivision148–9,179–80 rationalnumbers9,21 rationalisingthenumerator 246–7 rays45–6 realaxis22 realcoefficients117,120–1 complexzeroesand117–18 reallinearandquadraticfactors 120–1 realnumbers86,100 recursiveformulae–induction96–7 reductionformulae261–3 reflections22 regions44–8 remaindertheorem116–17 repeatedfactors240–1 resistance320 proportionaltosquareofspeed 342–3 proportionaltovelocity341–2 resistedmotion313–16 downwardfor n = 1321–4 downwardfor n = 2326 horizontal313–16 n = 1320–1 n = 2324,326–7 objectthrowndownwards 326–7 projectile332–4,341–4 upwardfor n = 1323–4 upwardfor n = 2325–6 vertical320–8 resultantforce275–6 results,startingwithknown90 rightangletriangle166–7 rootsofcomplexnumbers132–5 rootsofunity132–3 rotations22,30–1,36,39 intwodimensions181–2 Runge-Kuttaapproximation344
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sandwichtheorem102–3 secant254–6 sequences261 series261 shiftinginthecomplexplane32–3 SIunits273 side,alltermstoone86,95 simpleharmonicmotion(SHM) 295–301,351–2 acceleration280,288–91 differentialequation299–300 displacement305–9 findingfromgraphs308 formsof300–1 pendulum351–2 speed298–9 time295–301 transformations296–7 simultaneousequations7,15,156–7, 178–9,212,280–1 inconsistent158 sine210,253–4 sinusoidalgraphs295 skewlines153,158,208 slopefield308–9 solids198 some69 space arrangementsofplanesin207 arrangementsoftwolinesin 207–8 speed298–9,341–3 spheres172–3 vectorequationsof171–5 squareofspeed,resistance proportionalto342–3 squareroot–complex17–18 squares88
sumsanddifferencesof178–9 squeezetheorem102–3 standardintegrals225 StandardParametricVectorEquation 173–4 statements56,68–9 converse70–1 equivalent72–3 IfandOnlyIf71–2,77 negated81–3 proven57
stationarypoints210–11 substitution218,230–3,236–7, 257–8 sufficient70 sumoftwosquares4–5,12 sums,productsto210–13,256–7 sumstoproducts212–13 surds9 symbols144 tangents194,254–6 terminalvelocity323 terms alltooneside86 linear244–5 terms,alltooneside95 t-formulae216–17 trigonometricequations218 theorem83,102–3,111–13,116–17, 119–21,124–6,166–7,198,235 thereexists(∃ )68 3Dcurves180–1 time295–301,305–6,308 trajectories highangle343 lowangle343 midangle344 transformations30–1 translations295 triangle,area189 triangleinequality38–9 complexnumbers38 realnumbers38–9,86–7,188 vectorform188 triangles166–7 trigonometricequations, t-formulae 216–17 trigonometricfunctions211 trigonometricidentities deMoivre’stheorem128–9 productstosums210–13 sumstoproducts212–13 trigonometricintegrals253–8 powersofcosandsin253–4 powersofsecandtan254–6 productstosums256–7 t-substitution254–5,257–8 trigonometry178–9 truestatements56–9,62,72,77 truthtables58,62,83–4
union47–8 unitcircles132 unitvectors190,274–5 units273–4 unknowns241
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variablegravity349–50 vectorequations ofacircle/sphere171–5 findingviaparameter145 ofaline143–9 StandardParametric173–4 vectornotation274 vectors36–8,334–5 2Dgeometry194–5 3Dgeometry198–9 addition37–8 algebraicmanipulationof 185–90 differentbasis283 directioncosine190 directionvectors143–5 equal37 forcesas274,341–2 intersectinglines156–8,208 linesegment36,144,148 linethroughagivenpoint145–6 linethroughtheorigin143 linethroughtwogiven points147 parallellines153–5 parallelvectors153–4 parametricequations143–5 perpendicularlines162–7 perpendicularvectors162–3 ratiodivision148–9 skewlines158,208 subtraction37–8 unit190 verticallinein2D147 velocity308–9,320,323,343–4 resistanceproportionalto341–2 velocityasfunctionof displacement307 verticalresistedmotion320–8 zenithangle175,198 zeroes115–16 degreeandnumber119–20 realcoefficients117–18 z-planes45–6