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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
DO
NO
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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
20 8 AT A 2 TD
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A
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
SECTION A – continued
2018 AT A 2 ( T) 4
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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
SECTION A – continuedTURN OVER
2018 AT A 2 ( T)D
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A
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
SECTION A – continued
2018 AT A 2 ( T) 6
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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( )∫
SECTION A – continuedTURN OVER
7 2018 AT A 2 ( T)D
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A
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.
SECTION A – continued
2018 AT A 2 ( T) 8
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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
SECTION A – continuedTURN OVER
2018 AT A 2 ( T)D
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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
SECTION A – continued
2018 AT A 2 ( T) 10
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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
11 2018 AT A 2 ( T)D
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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
2018 AT A 2 ( T) 12
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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
13 2018 AT A 2 ( T)D
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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
2018 AT A 2 ( T) 14
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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
15 2018 AT A 2 ( T)D
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SECTION B – Question 2 – continuedTURN OVER
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
2018 AT A 2 ( T) 16
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SECTION B – Question 2 – continued
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
17 2018 AT A 2 ( T)D
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A
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
2018 AT A 2 ( T) 18
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SECTION B – Question 3 – continued
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
19 2018 AT A 2 ( T)D
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SECTION B – Question 3 – continuedTURN OVER
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
2018 AT A 2 ( T) 20
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SECTION B – continued
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
21 2018 AT A 2 ( T)D
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A
SECTION B – continuedTURN OVER
CONTINUES OVER PAGE
2018 AT A 2 ( T) 22
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SECTION B – Question 4 – continued
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
23 2018 AT A 2 ( T)D
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SECTION B – Question 4 – continuedTURN OVER
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
2018 AT A 2 ( T) 24
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SECTION B – continued
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
25 2018 AT A 2 ( T)D
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SECTION B – continuedTURN OVER
CONTINUES OVER PAGE
2018 AT A 2 ( T) 26
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SECTION B – Question 5 – continued
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
27 2018 AT A 2 ( T)D
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RE
A
SECTION B – continuedTURN OVER
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
2018 AT A 2 ( T) 28
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SECTION B – continued
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
29 2018 AT A 2 ( T)D
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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