2016 Specialist Mathematics Written examination 2€¦ · bi=−4 jk+α2 , where α is a real ......

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SPECIALIST MATHEMATICS Written examination 2 Monday 7 November 2016 Reading time: 11.45 am to 12.00 noon (15 minutes) Writing time: 12.00 noon to 2.00 pm (2 hours) QUESTION AND ANSWER BOOK Structure of book Section Number of questions Number of questions to be answered Number of marks A 20 20 20 B 6 6 60 Total 80 Students are permitted to bring into the examination room: pens, pencils, highlighters, erasers, sharpeners, rulers, a protractor, set squares, aids for curve sketching, one bound reference, one approved technology (calculator or software) and, if desired, one scientific calculator. Calculator memory DOES NOT need to be cleared. For approved computer-based CAS, full functionality may be used. Students are NOT permitted to bring into the examination room: blank sheets of paper and/or correction fluid/tape. Materials supplied Question and answer book of 23 pages. Formula sheet. Answer sheet for multiple-choice questions. Instructions Write your student number in the space provided above on this page. Check that your name and student number as printed on your answer sheet for multiple-choice questions are correct, and sign your name in the space provided to verify this. Unless otherwise indicated, the diagrams in this book are not drawn to scale. All written responses must be in English. At the end of the examination Place the answer sheet for multiple-choice questions inside the front cover of this book. You may keep the formula sheet. Students are NOT permitted to bring mobile phones and/or any other unauthorised electronic devices into the examination room. © VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2016 SUPERVISOR TO ATTACH PROCESSING LABEL HERE Victorian Certificate of Education 2016 STUDENT NUMBER Letter

Transcript of 2016 Specialist Mathematics Written examination 2€¦ · bi=−4 jk+α2 , where α is a real ......

SPECIALIST MATHEMATICSWritten examination 2

Monday 7 November 2016 Reading time: 11.45 am to 12.00 noon (15 minutes) Writing time: 12.00 noon to 2.00 pm (2 hours)

QUESTION AND ANSWER BOOK

Structure of bookSection Number of

questionsNumber of questions

to be answeredNumber of

marks

A 20 20 20B 6 6 60

Total 80

• Studentsarepermittedtobringintotheexaminationroom:pens,pencils,highlighters,erasers,sharpeners,rulers,aprotractor,setsquares,aidsforcurvesketching,oneboundreference,oneapprovedtechnology(calculatororsoftware)and,ifdesired,onescientificcalculator.CalculatormemoryDOESNOTneedtobecleared.Forapprovedcomputer-basedCAS,fullfunctionalitymaybeused.

• StudentsareNOTpermittedtobringintotheexaminationroom:blanksheetsofpaperand/orcorrectionfluid/tape.

Materials supplied• Questionandanswerbookof23pages.• Formulasheet.• Answersheetformultiple-choicequestions.

Instructions• Writeyourstudent numberinthespaceprovidedaboveonthispage.• Checkthatyournameandstudent numberasprintedonyouranswersheetformultiple-choice

questionsarecorrect,andsignyournameinthespaceprovidedtoverifythis.• Unlessotherwiseindicated,thediagramsinthisbookarenotdrawntoscale.• AllwrittenresponsesmustbeinEnglish.

At the end of the examination• Placetheanswersheetformultiple-choicequestionsinsidethefrontcoverofthisbook.• Youmaykeeptheformulasheet.

Students are NOT permitted to bring mobile phones and/or any other unauthorised electronic devices into the examination room.

©VICTORIANCURRICULUMANDASSESSMENTAUTHORITY2016

SUPERVISOR TO ATTACH PROCESSING LABEL HEREVictorian Certificate of Education 2016

STUDENT NUMBER

Letter

SECTION A – continued

2016SPECMATHEXAM2 2

Question 1Thecartesianequationoftherelationgivenbyx =3cosec2 (t)andy =4cot(t)–1is

A. y x+( )

− =1

16 91

2 2

B. yx

+( ) =+( )1

16 33

2

C. x y2 2

91

161+

+( )=

D. 4x–3y=15

E. yx

+( ) =−( )1

16 33

2

Question 2Theimplieddomainof y x a

b=

arccos ,whereb>0is

A. [–1,1]B. [a–b,a + b]C. [a–1,a+1]D. [a,a + bπ]E. [–b,b]

Question 3Thestraight-lineasymptote(s)ofthegraphofthefunctionwithrule f x x ax

x( ) = −3

2 ,whereaisanon-zerorealconstant,isgivenbyA. x =0only.

B. x =0andy =0only.

C. x =0andy = xonly.

D. x =0, x a= and x a= − only.

E. x =0andy = aonly.

SECTION A – Multiple-choice questions

Instructions for Section AAnswerallquestionsinpencilontheanswersheetprovidedformultiple-choicequestions.Choosetheresponsethatiscorrectforthequestion.Acorrectanswerscores1;anincorrectanswerscores0.Markswillnotbedeductedforincorrectanswers.Nomarkswillbegivenifmorethanoneansweriscompletedforanyquestion.Unlessotherwiseindicated,thediagramsinthisbookarenot drawntoscale.Taketheacceleration due to gravitytohavemagnitudegms–2,whereg=9.8

SECTION A – continuedTURN OVER

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Question 4Oneoftherootsofz3 + bz2 + cz=0is3–2i,wherebandcarerealnumbers.ThevaluesofbandcrespectivelyareA. 6,13B. 3,–2C. –3,2D. 2,3E. –6,13

Question 5IfArg − +( ) = −1 2

3ai π ,thentherealnumberais

A. − 3

B. −32

C. −13

D. 13

E. 3

Question 6Thepointscorrespondingtothefourcomplexnumbersgivenby

z z z z1 2 3 423

34

2 23

=

=

= −

= −cis cis cis cisπ π π π, , ,

44

aretheverticesofaparallelograminthecomplexplane.Whichoneofthefollowingstatementsisnottrue?

A. Theacuteanglebetweenthediagonalsoftheparallelogramis512

B. Thediagonalsoftheparallelogramhavelengths2and4

C. 1 2 3 4 0z z z z =

D. 1 2 3 4 0z z z z+ + + =

E. 1 2≤ ≤z forallfourofz1,z2,z3,z4

SECTION A – continued

2016SPECMATHEXAM2 4

Question 7Giventhatx=sin(t)–cos(t)and y t= ( )1

22sin ,then dy

dxintermsoftis

A. cos(t)–sin(t)

B. cos(t)+sin(t)

C. sec(t)+cosec(t)

D. sec(t)–cosec(t)

E. cos

cos sin2t

t t( )

( ) − ( )

Question 8Usingasuitablesubstitution, x e dxx

a

b3 2 4( )∫ ,wherea andbarerealconstants,canbewrittenas

A. e duu

a

b2( )∫

B. e duu

a

b2

4

4

( )∫C.

18

e duu

a

b( )∫

D. 14

24

4

e duu

a

b( )∫

E. 18 8

8

3

3

e duu

a

b( )∫

Question 9

If f x dydx

x x( ) = = −2 2 ,wherey0 = 0 = y(2),theny3usingEuler’sformulawithstepsize0.1is

A. 0.1f (2)

B. 0.6+0.1f (2.1)

C. 1.272+0.1f (2.2)

D. 2.02+0.1f (2.3)

E. 2.02+0.1f (2.2)

SECTION A – continuedTURN OVER

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Question 10

–1

O

–2

2

1

–3 –2 –1 1 2 3

y

x

Thedirectionfieldforthedifferentialequation dydx

x y+ + = 0 isshownabove.

Asolutiontothisdifferentialequationthatincludes(0,–1)couldalsoincludeA. (3,–1)B. (3.5,–2.5)C. (–1.5,–2)D. (2.5,–1)E. (2.5,1)

SECTION A – continued

2016SPECMATHEXAM2 6

Question 11Let

a i j k= + +3 2 α and

b i j k= − +4 2α ,whereαisarealconstant.

Ifthescalarresoluteof

a inthedirectionof

b is 74273

,thenαequalsA. 1B. 2C. 3D. 4E. 5

Question 12If

a i j k= − − +2 3 and

b i j k ,= − + +m 2 wheremisarealconstant,thevector

a b− willbeperpendiculartovector

b wheremequalsA. 0onlyB. 2onlyC. 0or2D. 4.5E. 0or–2

Question 13Aparticleofmass5kgissubjecttoforces12i

newtonsand9j

newtons.Ifnootherforcesactontheparticle,themagnitudeoftheparticle’sacceleration,inms–2,isA. 3B. 2 4 1 8. .

i j+C. 4.2D. 9E. 60 45

i j+

SECTION A – continuedTURN OVER

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Question 14Twolightstringsoflength4mand3mconnectamasstoahorizontalbar,asshownbelow.Thestringsareattachedtothehorizontalbar5mapart.

4 m3 m

5 m

mass

horizontal bar

T2T1

GiventhetensioninthelongerstringisT1andthetensionintheshorterstringisT2,theratioofthe

tensions TT1

2is

A. 35

B. 34

C. 45

D. 54

E. 43

Question 15 AvariableforceofFnewtonsactsona3kgmasssothatitmovesinastraightline.Attimetseconds,t≥0,itsvelocityvmetrespersecondandpositionxmetresfromtheoriginaregivenbyv =3–x2.ItfollowsthatA. F=–2xB. F=–6xC. F = 2x3–6xD. F = 6x3–18xE. F = 9x–3x3

SECTION A – continued

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Question 16Acricketballishitfromthegroundatanangleof30°tothehorizontalwithavelocityof20ms–1.Theballissubjectonlytogravityandairresistanceisnegligible.Giventhatthefieldislevel,thehorizontaldistancetravelledbytheball,inmetres,tothepointofimpactis

A. 10 3

g

B. 20g

C. 100 3

g

D. 200 3

g

E. 400g

Question 17Abodyofmass3kgismovingtotheleftinastraightlineat2ms–1.Itexperiencesaforceforaperiodoftime,afterwhichitisthenmovingtotherightat2ms–1.Thechangeinmomentumoftheparticle,inkgms–1,inthedirectionofthefinalmotionisA. –6B. 0C. 4D. 6E. 12

Question 18Orangesgrownonacitrusfarmhaveameanmassof204gramswithastandarddeviationof9grams.Lemonsgrownonthesamefarmhaveameanmassof76gramswithastandarddeviationof3grams.Themassesofthelemonsareindependentofthemassesoftheoranges.Themeanmassandstandarddeviation,ingrams,respectivelyofasetofthreeoftheseorangesandtwooftheselemonsareA. 764,3 29

B. 636,12

C. 764, 33

D. 636,3 10

E. 636,33

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END OF SECTION ATURN OVER

Question 19Arandomsampleof100bananasfromagivenareahasameanmassof210gramsandastandarddeviation of16grams.Assumingthestandarddeviationobtainedfromthesampleisasufficientlyaccurateestimateofthepopulationstandarddeviation,anapproximate95%confidenceintervalforthemeanmassofbananasproducedinthislocalityisgivenbyA. (178.7,241.3)B. (206.9,213.1)C. (209.2,210.8)D. (205.2,214.8)E. (194,226)

Question 20Thelifetimeofacertainbrandofbatteriesisnormallydistributedwithameanlifetimeof20hoursandastandarddeviationoftwohours.Arandomsampleof25batteriesisselected.Theprobabilitythatthemeanlifetimeofthissampleof25batteriesexceeds19.3hoursisA. 0.0401B. 0.1368C. 0.6103D. 0.8632E. 0.9599

2016SPECMATHEXAM2 10

SECTION B – Question 1–continued

Question 1 (9marks)a. Findthestationarypointofthegraphof f x

x xx

x R( ) = + +∈

4 02 3

, \{ }. Expressyouranswer

incoordinateform,givingvaluescorrecttotwodecimalplaces. 1mark

b. Findthepointofinflectionofthegraphgiveninpart a.Expressyouranswerincoordinateform,givingvaluescorrecttotwodecimalplaces. 2marks

SECTION B

Instructions for Section BAnswerallquestionsinthespacesprovided.Unlessotherwisespecified,anexactanswerisrequiredtoaquestion.Inquestionswheremorethanonemarkisavailable,appropriateworkingmust beshown.Unlessotherwiseindicated,thediagramsinthisbookarenotdrawntoscale.Taketheacceleration due to gravitytohavemagnitudegms–2,whereg=9.8

11 2016SPECMATHEXAM2

SECTION B – continuedTURN OVER

c. Sketchthegraphof f x x xx

( ) = + +4 2 3for x∈ −[ ]3 3, ontheaxesbelow,labellingthe

turningpointandthepointofinflectionwiththeircoordinates,correcttotwodecimalplaces. 3marks

y

x

–12

–8

–4

O

4

8

12

–3 –2 –1 1 2 3

Aglassistobemodelledbyrotatingthecurvethatisthepartofthegraphwhere x∈ − −[ ]3 0 5, . aboutthey-axis,toformasolidofrevolution.

d. i. Writedownadefiniteintegral,intermsofx, whichgivesthelengthofthecurvetoberotated. 1mark

ii. Findthelengthofthiscurve,correcttotwodecimalplaces. 1mark

e. ThevolumeofthesolidformedisgivenbyV a x dyc

b= ∫ 2 .

Findthevaluesofa,bandc.Donotattempttoevaluatethisintegral. 1mark

2016SPECMATHEXAM2 12

SECTION B – Question 2–continued

Question 2 (11marks)Alineinthecomplexplaneisgivenby z z i z C− = + − ∈1 2 3 , .

a. Findtheequationofthislineintheformy = mx + c. 2marks

b. Findthepointsofintersectionoftheline z z i− = + −1 2 3 withthecircle z − =1 3. 2marks

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SECTION B – continuedTURN OVER

c. Sketchboththeline z z i− = + −1 2 3 andthecircle z − =1 3 ontheArganddiagrambelow. 2marks

Im(z)

Re(z)

–4

–3

–2

–1O–1–2–3–4 4321

1

2

3

4

d. Theline z z i− = + −1 2 3 cutsthecircle z − =1 3 intotwosegments.

Findtheareaofthemajorsegment. 2marks

e. Sketchtheraygivenby Arg z( ) = − 34π ontheArganddiagraminpart c. 1mark

f. Writedowntherangeofvaluesofα α, ,∈R forwhicharaywithequationArg(z)=απ intersectstheline z z i− = + −1 2 3 . 2marks

2016SPECMATHEXAM2 14

SECTION B – Question 3–continued

Question 3 (11marks)Atankinitiallyhas20kgofsaltdissolvedin100Lofwater.Purewaterflowsintothetankatarateof10L/min.Thesolutionofsaltandwater,whichiskeptuniformbystirring,flowsoutofthetankatarateof5L/min.Ifxkilogramsistheamountofsaltinthetankaftertminutes,itcanbeshownthatthe

differentialequationrelatingxandtis dxdt

xt

++

=20

0.

a. Solvethisdifferentialequationtofindxintermsoft. 3marks

Asecondtankinitiallyhas15kgofsaltdissolvedin100Lofwater.Asolutionof160

kgof

saltperlitreflowsintothetankatarateof20L/min.Thesolutionofsaltandwater,whichiskeptuniformbystirring,flowsoutofthetankatarateof10L/min.

b. Ifykilogramsistheamountofsaltinthetankaftertminutes,writedownanexpressionfortheconcentration,inkg/L, ofsaltinthesecondtankattimet. 1mark

15 2016SPECMATHEXAM2

SECTION B – Question 3–continuedTURN OVER

c. Showthatthedifferentialequationrelatingyandtisdydt

yt

++

=10

13. 2marks

d. Verifybydifferentiationandsubstitutionintotheleftsidethat yt t

t=

+ ++

2 20 9006 10( )

satisfies

thedifferential equationinpart c.Verifythatthegivensolutionforyalsosatisfiestheinitial condition. 3marks

2016SPECMATHEXAM2 16

SECTION B – continued

e. Findwhentheconcentrationofsaltinthesecondtankreaches0.095kg/L.Giveyouranswerinminutes,correcttotwodecimalplaces. 2marks

17 2016SPECMATHEXAM2

SECTION B – continuedTURN OVER

CONTINUES OVER PAGE

2016SPECMATHEXAM2 18

SECTION B – Question 4–continued

Question 4 (10marks)Twoships,AandB,areobservedfromalighthouseatoriginO.RelativetoO,theirpositionvectorsattimethoursaftermiddayaregivenby

r i j

r i jA

B

t t

t t

= −( ) + +( )= −( ) + −( )5 1 3 1

4 2 5 2

wheredisplacementsaremeasuredinkilometres.

a. Showthatthetwoshipswillnotcollide,clearlystatingyourreason. 2marks

b. Sketchandlabelthepathofeachshipontheaxesbelow.Showthedirectionofmotionofeachshipwithanarrow. 3marks

y

x

10

8

6

4

2

O–10 –8 –6 –4 –2 2 4 6 8 10–2

–4

–6

19 2016SPECMATHEXAM2

SECTION B – continuedTURN OVER

c. Findtheobtuseanglebetweenthepathsofthetwoships.Giveyouranswerindegrees,correcttoonedecimalplace. 2marks

d. i. Findthevalueoft,correcttothreedecimalplaces,whentheshipsareclosest. 2marks

ii. Findtheminimumdistancebetweenthetwoships,inkilometres,correcttotwodecimalplaces. 1mark

2016SPECMATHEXAM2 20

SECTION B – Question 5–continued

Question 5 (10marks)Amodelrocketofmass2kgislaunchedfromrestandtravelsverticallyup,withaverticalpropulsionforceof(50–10t)newtonsaftertsecondsofflight,wheret∈[ ]0 5, .Assumethattherocketissubjectonlytotheverticalpropulsionforceandgravity,andthatairresistanceisnegligible.

a. Letvms–1bethevelocityoftherockettsecondsafteritislaunched.

Writedownanequationofmotionfortherocketandshowthatdvdt

t= −765

5 . 1mark

b. Findthevelocity,inms–1,oftherocketafterfiveseconds. 2marks

c. Findtheheightoftherocketafterfiveseconds.Giveyouranswerinmetres,correcttotwodecimalplaces. 2marks

21 2016SPECMATHEXAM2

SECTION B – continuedTURN OVER

d. Afterfiveseconds,whentheverticalpropulsionforcehasstopped,therocketissubjectonlytogravity.

Findthemaximumheightreachedbytherocket.Giveyouranswerinmetres,correcttotwodecimalplaces. 2marks

e. Havingreacheditsmaximumheight,therocketfallsdirectlytotheground.

Assumingnegligibleairresistanceduringthisfinalstageofmotion,findthetimeforwhichtherocketwasinflight.Giveyouranswerinseconds,correcttoonedecimalplace. 3marks

2016SPECMATHEXAM2 22

SECTION B – Question 6–continued

Question 6 (9marks)Themeanlevelofpollutantinariverisknowntobe1.1mg/Lwithastandarddeviationof 0.16mg/L.

a. LettherandomvariableX representthemeanlevelofpollutantinthemeasurementsfromarandomsampleof25sitesalongtheriver.

WritedownthemeanandstandarddeviationofX . 2marks

Afterachemicalspill,themeanlevelofpollutantfromarandomsampleof25sitesisfoundtobe1.2mg/L.Todeterminewhetherthissampleprovidesevidencethatthemeanlevelofpollutanthasincreased, astatisticaltestiscarriedout.

b. WritedownsuitablehypothesesH0andH1totestwhetherthemeanlevelofpollutanthasincreased. 2marks

c. i. Findthepvalueforthistest,correcttofourdecimalplaces. 2marks

ii. Statewithareasonwhetherthesamplesupportsthecontentionthattherehasbeenanincreaseinthemeanlevelofpollutantafterthespill.Testatthe5%levelofsignificance. 1mark

23 2016SPECMATHEXAM2

d. Forthistest,whatisthesmallestvalueofthesamplemeanthatwouldprovideevidencethatthemeanlevelofpollutanthasincreased?Thatis,find xc suchthatPr | . .X xc> =( ) =µ 1 1 0 05.Giveyouranswercorrecttothreedecimalplaces. 1mark

e. Supposethatforalevelofsignificanceof2.5%,wefindthatxc =1 163. .Thatis, Pr . | . .X > =( ) =1 163 1 1 0 025µ

Ifthemeanlevelofpollutantintheriver,μ,isinfact1.2mg/Lafterthespill,find Pr . | . .X < =( )1 163 1 2µ Giveyouranswercorrecttothreedecimalplaces. 1mark

END OF QUESTION AND ANSWER BOOK

SPECIALIST MATHEMATICS

Written examination 2

FORMULA SHEET

Instructions

This formula sheet is provided for your reference.A question and answer book is provided with this formula sheet.

Students are NOT permitted to bring mobile phones and/or any other unauthorised electronic devices into the examination room.

Victorian Certificate of Education 2016

© VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2016

SPECMATH EXAM 2

Specialist Mathematics formulas

Mensuration

area of a trapezium 12 a b h+( )

curved surface area of a cylinder 2π rh

volume of a cylinder π r2h

volume of a cone 13π r2h

volume of a pyramid 13 Ah

volume of a sphere 43π r3

area of a triangle 12 bc Asin ( )

sine ruleaA

bB

cCsin ( ) sin ( ) sin ( )

= =

cosine rule c2 = a2 + b2 – 2ab cos (C )

Circular functions

cos2 (x) + sin2 (x) = 1

1 + tan2 (x) = sec2 (x) cot2 (x) + 1 = cosec2 (x)

sin (x + y) = sin (x) cos (y) + cos (x) sin (y) sin (x – y) = sin (x) cos (y) – cos (x) sin (y)

cos (x + y) = cos (x) cos (y) – sin (x) sin (y) cos (x – y) = cos (x) cos (y) + sin (x) sin (y)

tan ( ) tan ( ) tan ( )tan ( ) tan ( )

x y x yx y

+ =+

−1tan ( ) tan ( ) tan ( )

tan ( ) tan ( )x y x y

x y− =

−+1

cos (2x) = cos2 (x) – sin2 (x) = 2 cos2 (x) – 1 = 1 – 2 sin2 (x)

sin (2x) = 2 sin (x) cos (x) tan ( ) tan ( )tan ( )

2 21 2x x

x=

3 SPECMATH EXAM

TURN OVER

Circular functions – continued

Function sin–1 or arcsin cos–1 or arccos tan–1 or arctan

Domain [–1, 1] [–1, 1] R

Range −

π π2 2, [0, �] −

π π2 2,

Algebra (complex numbers)

z x iy r i r= + = +( ) =cos( ) sin ( ) ( )θ θ θcis

z x y r= + =2 2 –π < Arg(z) ≤ π

z1z2 = r1r2 cis (θ1 + θ2)zz

rr

1

2

1

21 2= −( )cis θ θ

zn = rn cis (nθ) (de Moivre’s theorem)

Probability and statistics

for random variables X and YE(aX + b) = aE(X) + bE(aX + bY ) = aE(X ) + bE(Y )var(aX + b) = a2var(X )

for independent random variables X and Y var(aX + bY ) = a2var(X ) + b2var(Y )

approximate confidence interval for μ x z snx z s

n− +

,

distribution of sample mean Xmean E X( ) = µvariance var X

n( ) = σ2

SPECMATH EXAM 4

END OF FORMULA SHEET

Calculus

ddx

x nxn n( ) = −1 x dxn

x c nn n=+

+ ≠ −+∫ 11

11 ,

ddxe aeax ax( ) = e dx

ae cax ax= +∫ 1

ddx

xxelog ( )( ) = 1 1

xdx x ce= +∫ log

ddx

ax a axsin ( ) cos( )( ) = sin ( ) cos( )ax dxa

ax c= − +∫ 1

ddx

ax a axcos( ) sin ( )( ) = − cos( ) sin ( )ax dxa

ax c= +∫ 1

ddx

ax a axtan ( ) sec ( )( ) = 2 sec ( ) tan ( )2 1ax dxa

ax c= +∫ddx

xx

sin−( ) =−

12

1

1( ) 1 0

2 21

a xdx x

a c a−

=

+ >−∫ sin ,

ddx

xx

cos−( ) = −

12

1

1( ) −

−=

+ >−∫ 1 0

2 21

a xdx x

a c acos ,

ddx

xx

tan−( ) =+

12

11

( ) aa x

dx xa c2 2

1

+=

+

−∫ tan

( )( )

( ) ,ax b dxa n

ax b c nn n+ =+

+ + ≠ −+∫ 11

11

( ) logax b dxa

ax b ce+ = + +−∫ 1 1

product rule ddxuv u dv

dxv dudx

( ) = +

quotient rule ddx

uv

v dudx

u dvdx

v

=

2

chain rule dydx

dydududx

=

Euler’s method If dydx

f x= ( ), x0 = a and y0 = b, then xn + 1 = xn + h and yn + 1 = yn + h f (xn)

acceleration a d xdt

dvdt

v dvdx

ddx

v= = = =

2

221

2

arc length 1 2 2 2

1

2

1

2

+ ′( ) ′( ) + ′( )∫ ∫f x dx x t y t dtx

x

t

t( ) ( ) ( )or

Vectors in two and three dimensions

r = i + j + kx y z

r = + + =x y z r2 2 2

� � � � �ir r i j k= = + +ddt

dxdt

dydt

dzdt

r r1 2. cos( )= = + +r r x x y y z z1 2 1 2 1 2 1 2θ

Mechanics

momentum

p v= m

equation of motion

R a= m