9 ictorian certi cate of education supervisor to ......section a – continued 20 8 63(&0at +...

SPECIALIST MATHEMATICSWritten examination 2

Wednesday 6 June 2018 Reading time: 10.00 am to 10.15 am (15 minutes) Writing time: 10.15 am to 12.15 pm (2 hours)

QUESTION AND ANSWER BOOK

Structure of bookSection Number of

questionsNumber of questions

to be answeredNumber of

marks

A 20 20 20B 7 7 60

Total 80

t t a tt to to t a at o oo l l t a a l a ot a to t a a o t o o o

a o t olo al lato o o t a a o t al lato al lato o T to l a o a o o t a A ll t o al t a

t t a T tt to to t a at o oo la t o a a o

o t o ta

Materials supplied t o a a oo o 2 a o la t A t o lt l o t o

Instructions t o student number t a o a o o t a t at o name a student number a t o o a t o lt l o

t o a o t and o a t a o to t l ot at t a a t oo a not a to al All tt o t l

At the end of the examination la t a t o lt l o t o t o t o o t oo o a t o la t

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

T A A A T A T T 20 8

SUPERVISOR TO ATTACH PROCESSING LABEL HEREictorian Certi cate of Education 2018

STUDENT NUMBER

Letter

SECTION A – continued

20 8 AT A 2 T 2

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Question 1 et f x co ec x T e a o f i t an o ed

a dilation a acto o o t e x a i ollo ed a t an lation o unit o i ontall to t e i t ollo ed

a dilation a acto o 12

o t e y a i

T e ule o t e t an o ed a iA. g x 2co ec x

B. g x co ec 2x –

C. g x x( ) ( )= −( )3 2 1cosec

D. g x x( ) = −⎛⎝⎜

⎞⎠⎟2

31cosec

E. g x x( ) = −⎛⎝⎜

⎞⎠⎟3 1

2cosec

Question 2

et f x xx

( ) = +1 and g x tan2 x e e 02

< <x π .

f g x i e ual toA. in x ec2 xB. ec x tan2 xC. co x cot2 xD. co x co ec2 xE. co ec x co 2 x

SECTION A – Multiple-choice questions

Instructions for Section AAn e all ue tion in encil on t e an e eet o ided o ulti le c oice ue tion

oo e t e e on e t at i correct o t e ue tionA co ect an e co e an inco ect an e co e 0

a ill not e deducted o inco ect an eo a ill e i en i o e t an one an e i co leted o an ue tionnle ot e i e indicated t e dia a in t i oo a e not d a n to cale

Ta e t e acceleration due to gravity to a e a nitude g –2 e e g 8

SECTION A – continuedTURN OVER

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Question 3

T e i lied do ain o t e unction it ule f x x

x( )

arccos( )=

− −

3

22π

iA. B. – C. 0 2D. – 0 0 E. 2 2

Question 4

0

Im(z)

Re(z)

Q P

R

S

T

4

3

2

1

–1

–2

–3

–4

–4 –3 –2 –1 1 2 3 4

n t e A and dia am o n a o e 4 23

cis −⎛⎝⎜

⎞⎠⎟

π i e e ented t e ointA. PB. QC. RD. SE. T


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Question 5ic one o t e ollo in a o t e et o oint in t e com le lane eci ed t e elation

z z z z C: ,+( ) +( ) ={ ∈ }2 2 4 ?

Im(z)

Re(z)

Im(z)

Re(z)

Im(z)

Re(z)

Im(z)

Re(z)

Im(z)

Re(z)

A. B.

C. D.

E.

–4 –3 –2 –1

4

3

2

1

–1

–2

–3

–4

0 1 2 3 4–4 –3 –2 –1

4

3

2

1

–1

–2

–3

–4

0 1 2 3 4

–4 –3 –2 –1

4

3

2

1

–1

–2

–3

–4

0 1 2 3 4 –4 –3 –2 –1

4

3

2

1

–1

–2

–3

–4

0 1 2 3 4

–4 –3 –2 –1

4

3

2

1

–1

–2

–3

–4

0 1 2 3 4


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Question 6i en t at (z – 3i) i a acto o P(z) z3 + 2z2 + z + 18 ic one o t e ollo in tatement i false?

A. P(3i) 0B. P(–3i) 0C. P(z) a t ee linea acto o e CD. P(z) a no eal ootE. P(z) a t o com le con u ate oot

Question 7T e adient o t e line t at i perpendicular to t e a o t e elation 3y2 – 5xy – x2 1 at t e oint (1 2) i

A. 112

B. 127

C. 21

D. 712

E. 713

Question 8

in a uita le u titution 32 4 1 21

2

+ +

⎛

⎝⎜⎜

⎞

⎠⎟⎟∫ ( )xdx can e e e ed a

A. 34

12 21

2

+⎛⎝⎜

⎞⎠⎟∫ udu

B. 34

12 25

9

+⎛⎝⎜

⎞⎠⎟∫ udu

C. 31

2 25

9

+⎛⎝⎜

⎞⎠⎟∫ udu

D. 3 12 21

2

+⎛⎝⎜

⎞⎠⎟∫ udu

E. −+

⎛⎝⎜

⎞⎠⎟∫12 1

2 29

5

udu


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Question 9

1 10− ( )( )∫ cos x dx i e ui alent to

A. sin ( )2 5x dx( )∫B.

12

202sin ( )x dx( )∫C. cos ( )2 5x dx( )∫D. 2 102cos ( )x dx( )∫E. 2 52sin ( )x dx( )∫


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Question 10T e a o an antiderivative o a unction g i o n elo

x

y

ic one o t e ollo in could e t e e ent t e a o g?

x

y

x

y

x

y

x

y

x

y

A. B.

C.

E.

D.


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Question 11et a i j k= − +2 2 and b i j k .= + +2 3 6

T e acute an le et een a and i clo e t toA. 11B. 75C. 7D. 86E. 88

Question 12

L M

N

O

In t e dia am a o e LOM i a diamete o t e ci cle it cent e ON i a oint on t e ci cum e ence o t e ci cle

I r ON and s = MN , t en LN is e ual toA. 2 2r s

B. r s2

C. r s2

D. 2r s

E. 2r s

Question 13et i and e unit ectors in t e east and nort directions res ecti el

At time t, t 0, t e osition o article A is i en r i jA t t t= − +( ) + −2 5 6 5 8( ) and t e osition o

article B is i en r i j.B t t t= − + −( )( )3 2

article A ill e directl east o article B en t e ualsA. 1B. 2C. 1 and 2D. 2 and 4E. 4


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Question 14

90°

O150°

Q

P

R

T e dia ram a o e s o s a article at O in e uili rium in a lane under t e action o t ree orces o ma nitudes P, Q and R

ic one o t e ollo in statements is false?A. R = Q sin(60°)B. Q = R sin(60°)C. P = R sin(30°)D. Q cos(60°) = P cos(30°)E. P cos(60°) + Q cos(30°) = R

Question 15An 80 erson stands in an ele ator t at is acceleratin do n ards at 1 2 ms–2 T e reaction orce o t e ele ator oor on t e erson, in ne tons, isA. 688B. 704C. 784D. 880E. 896

Question 16A od o mass 2 is mo in in a strai t line it constant elocit en an e ternal orce o 8 is a lied in t e direction o motion or t secondsI t e od e eriences a c an e in momentum o 40 ms–1, t en t isA. 3B. 4C. 5D. 6E. 7


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Question 17 An o ect tra els in a strai t line relati e to an ori in OAt time t seconds its elocit , v metres er second, is i en

v tt t

t t( )

( ) ,

( ) ,=

− − ≤ ≤

− − − < ≤

⎧⎨⎪

⎩⎪

4 2 0 4

9 7 4 10

2

2

T e ra o v(t) is s o n elo

–2 2 4 6 8 10 12

4

2

0

–2

–4

v

t

T e o ect ill e ac at its initial osition en t is closest toA. 4 0B. 6 5C. 6 7D. 6 9E. 7 0

Question 18T e ei ts o all si ear old c ildren in a i en o ulation are normall distri uted T e mean ei t o a random sam le o 144 si ear old c ildren rom t is o ulation is ound to e 115 cmI a 95 con dence inter al or t e mean ei t o all si ear old c ildren is calculated to e (113 8, 116 2) cm, t e standard de iation used in t is calculation is closest toA. 1 20B. 7 35C. 15 09D. 54 02E. 88 13

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END OF SECTION ATURN OVER

Question 19A local su ermar et sells a les in a s t at a e negligible mass T e stated mass o a a o a les is 1 T e mass o t is articular t e o a le is no n to e normall distri uted it a mean o 115 rams and a standard de iation o 7 rams A articular a contains nine randoml selected a lesT e ro a ilit t at t e nine a les in t is a a e a total mass o less t an 1 isA. 0 0478B. 0 1132C. 0 4265D. 0 5373E. 0 9522

Question 20A arm ro s oran es and lemons T e oran es a e a mean mass o 200 rams it a standard de iation o 5 rams and t e lemons a e a mean mass o 70 rams it a standard de iation o 3 ramsAssumin masses or eac t e o ruit are normall distri uted, at is t e ro a ilit , correct to our decimal laces, t at a randoml selected oran e ill a e at least t ree times t e mass o a randoml selected lemon?A. 0 0062B. 0 0828C. 0 1657D. 0 8343E. 0 9172

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SECTION B – Question 1 – continued

Question 1 (10 mar s)onsider t e unction f it rule f (x) = 10 arccos(2 – 2x)

a. etc t e ra o f o er its ma imal domain on t e set o a es elo a el t e end oints it t eir coordinates 3 mar s

y

x1 2

30

20

10

–2 –1 0

SECTION B

Instructions for Section BAns er all uestions in t e s aces ro ided

nless ot er ise s eci ed, an exact ans er is re uired to a uestionIn uestions ere more t an one mar is a aila le, a ro riate or in must e s o n

nless ot er ise indicated, t e dia rams in t is oo are not dra n to scaleTa e t e acceleration due to gravity to a e ma nitude g ms–2, ere g = 9 8

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SECTION B – Question 1 – continuedTURN OVER

A ase is to e modelled rotatin t e ra o f a out t e y a is to orm a solid o re olution, ere units o measurement are in centimetres

b. i. rite do n a de nite inte ral in terms o y t at i es t e olume o t e ase 2 mar s

ii. ind t e olume o t e ase in cu ic centimetres 1 mar

c. ater is oured into t e ase at a rate o 20 cm3 s–1

ind t e rate, in centimetres er second, at ic t e de t o t e ater is c an in en t e de t is 5 cm 3 mar s

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SECTION B – continued

d. T e ase is laced on a ta le A ee clim s rom t e ottom o t e outside o t e ase to t e to o t e ase

at is t e minimum distance t e ee ill need to tra el? i e our ans er in centimetres, correct to one decimal lace 1 mar

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Question 2 (11 mar s)

In t e com le lane, L is t e line i en z z i+ = + −1 12

32

.

a. o t at t e cartesian e uation o L is i en y x= −13. 2 mar s

b. ind t e oint(s) o intersection o L and t e ra o t e relation z z 4 in cartesian orm 2 mar s

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c. etc L and t e ra o t e relation z z 4 on t e Ar and dia ram elo 2 mar s

–5 –4 –3 –2 –1 1 2 3 4 5

–5

–4

–3

–2

–1

1

2

3

5

4

Im(z)

Re(z)O

T e art o t e line L in t e ourt uadrant can e e ressed in t e orm Ar (z) =

d. tate t e alue o 1 mar

e. ind t e area enclosed L and t e ra s o t e relations z z 4, Arg( )z =π3

and Re( )z 3. 2 mar s

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SECTION B – continuedTURN OVER

f. T e strai t line L can e ritten in t e orm z k z , ere k C

ind k in t e orm rcis( ), ere is t e rinci al ar ument o k 2 mar s

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Question 3 (10 mar s)A 200 crate rests on a smoot lane inclined at to t e ori ontal An e ternal orce o F ne tons acts u t e lane, arallel to t e lane, to ee t e crate in e uili rium

a. n t e dia ram elo , dra and la el all orces actin on t e crate 1 mar

b. ind F in terms o 1 mar

T e ma nitude o t e e ternal orce F is c an ed to 780 and t e lane is inclined at = 30°

c. i. Ta in t e direction do n t e lane to e ositi e, nd t e acceleration o t e crate 2 mar s

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ii. n t e a es elo , s etc t e elocit –time ra or t e crate in t e ositi e direction or t e rst our seconds o its motion 1 mar

0

v

t

1

2

3

4

5

–11 2 3 4 5–1

iii. alculate t e distance t e crate tra els, in metres, in its rst our seconds o motion 1 mar

tartin rom rest, t e crate slides do n a smoot lane inclined at de rees to t e ori ontal A orce o 295 cos( ) ne tons, u t e lane and arallel to t e lane, acts on t e crate

d. I t e momentum o t e crate is 800 ms–1 a ter a in tra elled 10 m, nd t e acceleration, in ms–2, o t e crate 2 mar s

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e. ind t e an le o inclination, , o t e lane i t e acceleration o t e crate do n t e lane is 0 75 ms–2 i e our ans er in de rees, correct to one decimal lace 2 mar s

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CONTINUES OVER PAGE

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Question 4 (11 mar s)A as et all la er aims to t ro a as et all t rou a rin , t e centre o ic is at a ori ontal distance o 4 5 m rom t e oint o release o t e all and 3 m a o e oor le el T e all is released at a ei t o 1 75 m a o e oor le el, at an an le o ro ection to t e ori ontal and at a s eed o V ms–1 Air resistance is assumed to e ne li i le

3 m

4.5 m

1.75 m

T e osition ector o t e centre o t e all at an time, t seconds, or t 0, relati e to t e oint o

release is i en r i j,( ) cos( ) sin( ) .t Vt Vt t= + −( )α α 4 9 2 ere i is a unit ector in t e ori ontal

direction o motion o t e all and is a unit ector erticall u . is lacement com onents are measured in metres.

a. or t e la er s rst s ot at oal, V = 7 ms–1 and = 45°.

i. ind t e time, in seconds, ta en or t e all to reac its ma imum ei t. i e our

ans er in t e orm a bc

, ere a, b and c are ositi e inte ers. 2 mar s

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ii. ind t e ma imum ei t, in metres, a o e oor le el, reac ed t e centre o t e all. 2 mar s

iii. ind t e distance o t e centre o t e all rom t e centre o t e rin one second a ter release. i e our ans er in metres, correct to t o decimal laces. 2 mar s

b. or t e la er s second s ot at oal, V = 10 ms–1.

ind t e ossi le an les o ro ection, , or t e centre o t e all to ass t rou t e centre o t e rin . i e our ans ers in de rees, correct to one decimal lace. 3 mar s

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c. or t e la er s t ird s ot at oal, t e an le o ro ection is = 60°.

ind t e s eed V re uired or t e centre o t e all to ass t rou t e centre o t e rin . i e our ans er in metres er second, correct to one decimal lace. 2 mar s

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CONTINUES OVER PAGE

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Question 5 (9 mar s)A ori ontal eam is su orted at its end oints, ic are 2 m a art. T e de ection y metres o t e

eam measured do n ards at a distance x metres rom t e su ort at t e ori in O is i en t e

di erential e uation 80 3 42

2d ydx

x= − .

O

y

x

a. i en t at ot t e inclination, dydx

, and t e de ection, y, o t e eam rom t e ori ontal at

x = 2 are ero, use t e di erential e uation a o e to s o t at 80 12

2 23 2y x x x= − + . 2 mar s

b. ind t e an le o inclination o t e eam to t e ori ontal at t e ori in O. i e our ans er as a ositi e acute an le in de rees, correct to one decimal lace. 2 mar s

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c. ind t e alue o x, in metres, ere t e ma imum de ection occurs, and nd t e ma imum de ection, in metres. 3 mar s

d. ind t e ma imum an le o inclination o t e eam to t e ori ontal in t e art o t e eam ere x 1. i e our ans er as a ositi e acute an le in de rees, correct to one decimal

lace. 2 mar s

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Question 6 (5 mar s)A co ee mac ine dis enses co ee concentrate and ot ater into a 200 m cu to roduce a lon lac co ee. T e olume o co ee concentrate dis ensed aries normall it a mean o 40 m

and a standard de iation o 1.6 m .Inde endent o t e olume o co ee concentrate, t e olume o ater dis ensed aries normall

it a mean o 150 m and a standard de iation o 6.3 m .

a. tate t e mean and t e standard de iation, in millilitres, o t e total olume o li uid dis ensed to ma e a lon lac co ee. 2 mar s

b. ind t e ro a ilit t at a lon lac co ee dis ensed t e mac ine o er o s a 200 m cu . i e our ans er correct to t ree decimal laces. 1 mar

c. u ose t at t e standard de iation o t e olume o ater dis ensed t e mac ine can e ad usted, ut t at t e mean olume o ater dis ensed and t e standard de iation o t e olume o co ee concentrate dis ensed cannot e ad usted.

ind t e standard de iation o t e olume o ater dis ensed t at is needed or t ere to e onl a 1 c ance o a lon lac co ee o er o in a 200 m cu . i e our ans er in millilitres, correct to t o decimal laces. 2 mar s

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Question 7 (4 mar s)Accordin to medical records, t e lood ressure o t e eneral o ulation o males a ed 35 to 45 ears is normall distri uted it a mean o 128 and a standard de iation o 14. Researc ers su ested t at male teac ers ad i er lood ressures t an t e eneral o ulation o males. To in esti ate t is, a random sam le o 49 male teac ers rom t is a e rou as o tained and ound to a e a mean lood ressure o 133.

a. tate two ot eses and er orm a statistical test at t e 5 le el to determine i male teac ers elon in to t e 35 to 45 ears a e rou a e i er lood ressures t an t e

eneral o ulation o males. learl state our conclusion it a reason. 3 mar s

b. ind a 90 con dence inter al or t e mean lood ressure o all male teac ers a ed 35 to 45 ears usin a standard de iation o 14. i e our ans ers correct to t e nearest inte er. 1 mar

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 Certi cate of Education 2018

© VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2018

SPECMATH EXAM 2

Specialist Mathematics formulas

Mensuration

area of a trapezium 12 a b h+( )

curved surface area of a cylinder 2 rh

volume of a cylinder r2h

volume of a cone 13

2h

volume of a pyramid 13 Ah

volume of a sphere 43

3

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 – < Arg(z)

z1z2 = r1r2 cis ( 1 + 2)zz

rr

1

2

1

21 2= −( )cis θ θ

zn = rn cis ( ) (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 )

appro imate con dence interval for 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 vm

equation of motion R am

9 ictorian certi cate of education supervisor to ......section a – continued 20 8 63(&0at +...

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