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MATHS QUEST 12TI-NSPIRE C AS CALCULATOR COMPANION
SpecialistMathematics
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VCE MATHEMATICS UNITS 3 &
RAYMOND ROZEN | PAULINE HOLLAND | BRIAN HODGSON
HOWARD LISTON | JENNIFER NOLAN | GEOFF PHILLIPS
MATHS QUEST 12TI-NSPIRE C AS CALCULATOR COMPANION
SpecialistMathematics
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First published 2013 byJohn Wiley & Sons Australia, Ltd42 McDougall Street, Milton, Qld 4064
Typeset in 10/12 pt Times LT Std
© John Wiley & Sons Australia, Ltd 2013
The moral rights of the authors have been asserted.
ISBN: 978 1 118 31811 9978 1 118 31809 6 (flexisaver)
Reproduction and communication for educational purposesThe Australian Copyright Act 1968 (the Act) allows a maximum of onechapter or 10% of the pages of this work, whichever is the greater, to bereproduced and/or communicated by any educational institution for itseducational purposes provided that the educational institution (or the bodythat administers it) has given a remuneration notice to Copyright AgencyLimited (CAL).
Reproduction and communication for other purposesExcept as permitted under the Act (for example, a fair dealing for thepurposes of study, research, criticism or review), no part of this book maybe reproduced, stored in a retrieval system, communicated or transmitted
in any form or by any means without prior written permission. Allinquiries should be made to the publisher.
Cover and internal design images: © vic&dd/Shutterstock.com
Typeset in India by Aptara
Illustrated by Aptara and Wiley Composition Services
Printed in Singapore byCraft Print International Ltd
10 9 8 7 6 5 4 3 2 1
AcknowledgementsThe authors and publisher would like to thank the following copyright
holders, organisations and individuals for their permission to reproducecopyright material in this book.
ImagesTexas Instruments: Screenshots from TI-Nspire reproduced withpermission of Texas Instruments
Every effort has been made to trace the ownership of copyright material.Information that will enable the publisher to rectify any error or omissionin subsequent editions will be welcome. In such cases, please contact thePermissions Section of John Wiley & Sons Australia, Ltd.
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Contents
Introduction vi
CHAPTER 1
Coordinate geometry 1
CHAPTER 2
Circular functions 15
CHAPTER 3
Complex numbers 21
CHAPTER 4
Relations and regions of thecomplex plane 27
CHAPTER 5
Differential calculus 29
CHAPTER 6
Integral calculus 37
CHAPTER 7
Differential equations 47
CHAPTER 8
Kinematics 53
CHAPTER 9
Vectors 63
CHAPTER 10
Vector calculus 71
CHAPTER 11
Mechanics 79
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Introduction
This booklet is designed as a companion to Maths Quest 12 Specialist Mathematics Fourth Edition.
It contains worked examples from the student text that have been re-worked using the TI-Nspire CXCAS calculator with Operating System v3.
The content of this booklet will be updated online as new operating systems are released by Texas
Instruments.
The companion is designed to assist students and teachers in making decisions about the judicious use of
CAS technology in answering mathematical questions.
The calculator companion booklet is also available as a PDF file on the eBookPLUS under the
preliminary section of Maths Quest 12 Specialist Mathematics Fourth Edition.
vi Introduction
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2 Maths Quest 12 Specialist Mathematics
4 To determine the turning points, press:
• MENUb
• 3: Algebra3
• 1: Solve1
• MENUb
• 4: Calculus4
• 1: Derivative1
Complete the entry line as:
solved
dx f x x ( 1( )) 0,=
then press ENTER·.
5 To find the y-coordinates of the stationary
points by substitution, complete the entry
lines as:
f 1 2 )(
f 1 2 )( −
Press ENTER· after each entry.Describe the nature and coordinates of
the stationary points, as deduced from the
graph.
Solvingd
dx
x
x
2
30
2 +
= for x gives
x x 2 or 2.= =
−
The coordinates of the stationary points are:
Local minimum
2,
2 2
3
Local maximum 2,2 2
3
− −
6 Sketch the graph of y x
x
2
3
2
=+ .
y
x 0
−5
5
10 x = 0 (Asymptote)
−5 5 10−10
−10
y = x (Asymptote)1–3
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4 Maths Quest 12 Specialist Mathematics
7 Sketch the graph of the ellipse. y
x 0
−2
2
4
6
−4−6 −2 62 4
(6, 2)(1, 2)(−4, 2)
(1, 5)
(1, −1)
= 1+( x − 1)
2
––––––25
( y − 2)2
––––––9
10 + 6√6–––––––
50,
10 − 6√6–––––––
50,
3 − 5√5––––––
3, 0
3 + 5√5––––––
3, 0
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CHAPTER 1 • Coordinate geometry 5
WORKED EXAMPLE 13
Sketch the graph of x y( 2)
9
( 4)
161
2 2−
++
= .
THINK WRIT E/DISPLAY
1 Compare x y( 2)
9
( 4)
161
2 2−
++
= with
x h
a
y k
b
( ) ( )1
2
2
2
2
−+
−= .
h = 2, k = −4
So the centre is (2,
−
4).a2 = 9 b2 = 16
a = 3 b = 4
2 The major axis is parallel to the y-axis as
b > a.
3 The extreme points (vertices) parallel to
the x -axis for the ellipse are:
(−a + h, k ) (a + h, k )
Vertices are:
(−3 + 2, −4) (3 + 2, −4)
= (−1, −4) = (5, −4)
4 The extreme points (vertices) parallel to
the y-axis for the ellipse are:
(h, −b + k ) (h, b + k )
and
(2, −4 − 4) (2, 4 − 4)
= (2, −8) = (2, 0)
5 Find the x - and y-intercepts.On a Calculator page, complete the entry
lines as:
solve x y
x y( 2)
9
( 4)
161, 0
2 2−+
+=
=
solve x y
y x ( 2)
9
( 4)
161, 0
2 2−+
+=
=
Press ENTER· after each entry.
The x -intercept is x = 2.
The y-intercepts are:
y y12 4 5
3,
12 5
3=
−=
+− −
6 To sketch the graph of the ellipse, on a
Graphs page press:
• MENUb
• 3: Graph Entry/Edit3
• 2: Equation2
• 4: Ellipse4
• 1:Complete the entry line as:
x y2
3
4
41
2
2
2
2
( )( )−+
−
=
−
then press ENTER·.
Note that the viewing window has been
changed.
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6 Maths Quest 12 Specialist Mathematics
7 Sketch the graph of the ellipse.
x
0
−1 621 3 5
y
−4
−8
−10
2
(2, −4)
= 1+( x − 2)
2
––––––9
( y + 4)2
––––––16
(−1, −4)
(5, −4)
(2, −8)
4
(2, 0)
−2
−6
−12 + 4√5––––––––
30,
−12 − 4√5––––––––
30,
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CHAPTER 1 • Coordinate geometry 7
WORKED EXAMPLE 14
Sketch the graph of 5 x2 + 9( y − 2)2 = 45.
THINK WRIT E/DISPLAY
1 Rearrange and simplify by dividing both
sides by 45 to make the RHS = 1.
5 x 2 + 9( y − 2)2 = 45
2 Simplify by cancelling. x y5
45
9( 2)
45
45
45
2 2
+
−
=
x y
9
( 2)
51
2 2
+−
=
3 Compare x y
9
( 2)
51
2 2
+−
= with
x h
a
y k
b
( ) ( )1
2
2
2
2
−+
−= .
h = 0, k = 2 and so the centre is (0, 2).
a2 = 9 b2 = 5 as a, b > 0
a = 3 b = 5
4 Major axis is parallel to the x -axis as a > b.
5 The extreme points (vertices) parallel to
the x -axis for the ellipse are:(−a + h, k ) (a + h, k )
Vertices are:
(−
3+
0, 2) (3+
0, 2)= (−3, 2) = (3, 2)
6 The extreme points (vertices) parallel to
the y-axis for the ellipse are:
(h, −b + k ) (h, b + k )
and (0, 5−
+ 2) (0, 5 + 2)
or (0, 2 − 5) (0, 2 + 5)
≈ (0, −0.24) ≈ (0, 4.24)
7 Find the x -intercepts.
On a Calculator page, complete the
entry line as:
solve + − = = x y x y(5 9( 2) 45, ) | 02 2
then press ENTER·.
x x 3 5
5,
3 5
5= =
−
8 To sketch the graph of the ellipse, on a
Graphs page press:
• MENUb
• 3: Graph Entry/Edit3
• 2: Equation2
• 6: Conic6
• 1:
Complete the entry line as:
x xy y x y5 0 9 0 36 9 02 2+ + + + + =
− −
then press ENTER·.
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8 Maths Quest 12 Specialist Mathematics
9 Sketch the graph of the ellipse.
(0, 2 − )5
y
x
0
−2
2
4
6
−2−3−4 −1 3 41 2
(3, 2)(0, 2)(−3, 2)
5 x 2 + 9( y − 2)2 = 45(0, 2 + )5
−3√5––––
5, 0 3√5
––––5
, 0
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CHAPTER 1 • Coordinate geometry 9
WORKED EXAMPLE 15
Sketch the graph of the relation described by the rule: 25 x2 + 150 x + 4 y2 − 8 y + 129 = 0.
THINK WRIT E/DISPLAY
1 To locate the intercepts, on a Calculator
page, complete the entry lines as:
solve x x y y x
y
25 150 4 8 129 0,
| 0
2 2( + + − + =
=
solve x x y y y
x
25 150 4 8 129 0,
| 0
2 2( + + − + =
=
Make a record of the intercepts.
2 To sketch the graph of the ellipse, on a
Graphs page press:
• MENUb
• 3: Graph Entry/Edit3
• 2: Equation2
• 6: Conic6
• 1:
Complete the entry line as:
x xy y x y25 0 4 0 8 129 02 2+ + + + + =
−
then press ENTER·.
3 Write the x -intercepts. x 15 4 6
5=
−−
x 15 4 6
5=
+−
4 Sketch the graph of the ellipse.
0
6
−3 −1−5
(−5, 1)(−3, 1)
(−3, 6)
(−3, −4)
(−1, 1)
y
x
−4
−15 + 4√6––––––––
5, 0
−15 − 4√6––––––––
5, 0
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10 Maths Quest 12 Specialist Mathematics
WORKED EXAMPLE 16
Determine the Cartesian equation of the curve with parametric equations x = 2 + 3 sin ( t) and
y = 1 − 2 cos ( t ) where t ∈ R. Describe the graph and state its domain and range.
THINK WRIT E/DISPLAY
1 Use a CAS calculator to sketch the graph
in a Graphs page, in parametric mode, by
completing the entry line as:= +
= −
x t t
y t t
1( ) 2 3 sin ( )
1( ) 1 2 cos ( )
Then press ENTER·.
2
Rewrite the parameters by isolating cos (t )and sin (t ).
x 2
3
−
=
sin (t ) and
y 1
2
−
− =
cos (t )
3 Square both sides of each equation then
add.
x y( 2)
9
( 1)
4
2 2−
+−
= sin2 (t ) + cos2 (t )
= 1
4 Describe the relation. This represents an ellipse with centre (2, 1).
5 The domain is the range of the parametric
equation x = 2 + 3 sin (t ).Domain is [2 − 3, 2 + 3] = [−1, 5]
6 The range is the range of the parametric
equation y = 1 – 2 cos (t ).
Range is [1 − 2, 1 + 2] = [−1, 3]
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CHAPTER 1 • Coordinate geometry 11
WORKED EXAMPLE 25
Express each of the following as partial fractions.
a x x
x x
5 10 52
( 2)( 4)
2+ −
− +
b x x x
x x
2 5 3 7
2
3 2
2
− + +
− −
THINK WRIT E/DISPLAY
a
&
b
1 On a Calculator page, press:
• MENUb
• 3: Algebra3
• 3: Expand3
Complete the entry lines as:
x x
x x
x x x
x x
expand 5 10 52
( 2) ( 4)
expand 2 5 3 7
2
2
3 2
2
+ −− × +
− + +− −
Press ENTER· after each entry.
2 Write the answers. a
b
x x
x x x x
x x x
x x x x x
5 10 52
( 2) ( 4)
2
4
2
25
2 5 3 7
2
1
1
3
22 3
2
3 2
2
+ −
− × +
=
+
−
−
+
− + +
− −
=
+
+
−
+ −
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12 Maths Quest 12 Specialist Mathematics
WORKED EXAMPLE 27
Sketch the graph of the function y x x
x
5 6
4
2
==
− +
−
.
THINK WRIT E/DISPLAY
1 Use a CAS calculator to express the
rational function as partial fractions by
completing the following steps. Press:• MENUb
• 2: Number2
• 7: Fraction Tools7
• 1: Proper Fraction1
Complete the entry line as:
x x
x propFrac
5 6
4
2 − +−
then press ENTER·.
2
Express the function as partial fractions. y x x
2
41=
−
+ −
3 Sketch the graphs of y1 = x − 1
(asymptote) and y2 = x
2
4− on the same
axes.
y
x 0
3−1
x = 4
42
y1 = x− 1
y2 =
2 —— x − 4
1
4 Determine any x -intercepts. y = 0, x 2 − 5 x + 6 = 0
( x − 2)( x − 3) = 0
⇒ x = 2 and x = 3
5 Determine the y-intercept. x = 0, y 6
4=
−
y 3
2=
−
6 Add the two graphs by addition of ordinates
to obtain the graph of y x x
x
5 6
4
2
=− +
−
.
3–2
+
y1 = x− 1
y = x− 1
y2 =
2 —— x − 4
2 —— x − 4
x = 4
y
x
(2, 0)
1−1 2 3 4
(3, 0)
(0, − )
0
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CHAPTER 1 • Coordinate geometry 13
7 Open a Graphs page, and complete the
entry lines as;
f x x
f x x
f x f x f x
1 1
2 2
4
3 1 2
( )
( )
( ) ( ) ( )
= −
=
−
= +
then press ENTER·.
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CHAPTER 2 • Circular functions 15
CHAPTER 2
Circular functionsWORKED EXAMPLE 2
If cosec ( x) = 43 and, 0 ≤ x ≤ 90°, find x (to the nearest tenth of a degree).
THINK WRIT E
1 Express the equation cosec ( x ) = 4
3
in terms of sin ( x ).
x x
cosec ( ) 1
sin ( )
4
3= =
2 On a Calculator page, press:
• MENUb
• 3: Algebra3
• 1: Solve1
Complete the entry line as
x x x solve
1
sin( )
4
3, | 0 90=
≤ ≤
Then press ENTER·.
Alternatively, the three reciprocal
functions are built into the TI-Nspire.
They can be accessed by theµ key, or
through the catalogue, or you can simply
use the letter keys and enter csc, sec orcot as needed.
3 Write the solution. Solving cosec ( x ) = 4 for
x x [0, 90 ], 48.5904∈ ° = °
4 Round off the answer to 1 decimal place. x = 48.6°
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16 Maths Quest 12 Specialist Mathematics
WORKED EXAMPLE 12
Solve cosec ( x) = 1.8 over the interval 0 ≤ x ≤ 4π . Give your answer(s) correct to 2 decimal places.
THINK WRIT E
1 On a Calculator page, press:
• MENUb
• 3: Algebra3
•
1: Solve1Complete the entry line as
solve (csc( x ) = 1.8, x ) | 0 ≤ x ≤ 4π
Then press ENTER·.
2 Write the solution. Solving cosec ( x ) = 1.8 over the interval
0 ≤ x ≤ 4π gives
3 Round the answers to 2 decimal places. x = 0.59, 2.55, 6.87, 8.84
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CHAPTER 2 • Circular functions 19
WORKED EXAMPLE 16
Find the exact value ofπ
cot5
12.
THINK WRIT E
1 Express5
12
π as the sum of
4
π and
6
π . cot
5
12cot
4 6
π π π
= +
2 Express cot in terms of its reciprocal,1
tan.
1
tan4 6
π π
=+
3 Use the appropriate compound-angle
formula to expand the denominator.
1
tan4
tan6
1 tan4
tan
π π
π π
=
+
−
6
4 Express in simplest fraction form.
1 tan 4
tan6
tan 4
tan6
π π
π π =
−
+
5 Simplify.1 (1)
1
1
3
1
3
=−
+
1
1
1
3
1
3
=
−
+
3 1
3
3 1
3
=
−
+
3 1
3 1
=−
+
6 Rationalise the denominator. ( 3 1) ( 3 1)
( 3 1) ( 3 1)=
− −
+ −
7 Simplify. 3 2 3 1
3 1=
− +
−
4 2 3
2=
−
2 3= −
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20 Maths Quest 12 Specialist Mathematics
Note: It is possible to check the answer
using a calculator.
On a Calculator page, complete the entry
line as:
π
cot
5
12
then press ENTER·.
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CHAPTER 3 • Complex numbers 21
CHAPTER 3
Complex numbersWORKED EXAMPLE 1
Using the imaginary number i, write down an expression for:
a 16− b 5− .
THINK WRIT E
a
&
b
1 Change the document settings to
Rectangular mode. To do this, press:
•
HOMEc• 5: Settings5
• 2: Settings2
• 2: Document Settings2
Tab down to Real or Complex and
select Rectangular.
a
&
b
2 On a Calculator page, complete the
entry lines as:
16−
5−
Press ENTER· after each entry.
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22 Maths Quest 12 Specialist Mathematics
WORKED EXAMPLE 4
Simplify z = i4 − 2i2 + 1 and w = i6 − 3i4 + 3i2 − 1.
THINK WRIT E
1 On a Calculator page, complete the entry
lines as:
i4 − 2i2 + 1
i6 − 3i4 + 3i2 − 1Press ENTER· after each entry.
2 Write the answer. z = 4
w = −8
WORKED EXAMPLE 5
Evaluate each of the following.
a Re(7 + 6i) b Im(10) c Re(2 + i − 3i3) di i i
Im 1 3
2
2 3− − −
THINK WRIT E
a ,b ,
c &
d
1 On a Calculator page, press:
• MENUb
• 2: Number2
•
9: Complex Number Tools9• 2: Real Part2 or
• 3: Imaginary Part3
Complete the entry lines as:
real(7 + 6i)
imag(10)
real(2 + i − 3i2)
i i iimag
1 3
2
2 3− − −
Press ENTER· after each entry.
2 Write the answers. a Re(7 + 6i) = 7
b Im(10) = 0
c Re(2 + i − 3i3) = 2
d i i i
Im 1 3
21
2 3− − −
=
−
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CHAPTER 3 • Complex numbers 23
WORKED EXAMPLE 10
Determine Re( z2w) + Im(zw2) for z = 4 + i and w = 3 − i.
THINK WRIT E
1 On a Calculator page, complete the entry
lines as:
Define z = 4 + i
Define w = 3 – iPress ENTER· after each entry.
Then press:
• MENUb
• 2: Number2
• 9: Complex Number Tools9
• 2: Real Part2
Complete the entry line as:
real( z2 × w) + imag( z × w2)
Then press ENTER·.
Note: The imaginary part can be found in
the same menu as the real part.
2Write the answer. Re( z
2
w) + Im( zw2
) = 37
WORKED EXAMPLE 12
Write down the conjugate of each of the following complex numbers.
a 8 + 5i b −2 − 3i
THINK WRIT E
a
&
b
1 On a Calculator page, press:
• MENUb
• 2: Number2
• 9: Complex Number Tools9
• 1: Complex Conjugate1
Complete the entry lines as:
conj (8 + 5i)
conj (−2 − 3i)
Press ENTER· after each entry.
2 Write the answers. a 8 − 5i
b −2 + 3i
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24 Maths Quest 12 Specialist Mathematics
WORKED EXAMPLE 16
If z = a + bi, find a and b such that−
−
= −
z
zi
5 15
14 3 .
THINK WRIT E
1 On a Calculator page, press:
• MENUb
•3: Algebra3
• C: ComplexC
• 1: Solve1
Complete the entry line as:
cSolve z
zi z
5 15
14 3 ,
−−
= −
Then press ENTER·.
2 a is the real part of z, b is the
imaginary part.
a = 2, b = −3
WORKED EXAMPLE 17
Find the modulus of the complex number z = 8 − 6i.
THINK WRIT E
1 On a Calculator page, press:
• MENUb
• 2: Number2
• 9: Complex Number Tools9
•5: Magnitude5
Complete the entry line as:
|8 − 6i|Then press ENTER·.
2 Write the answer. | z | = |8 − 6i| = 10
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CHAPTER 3 • Complex numbers 25
WORKED EXAMPLE 23
Express each of the following in polar form, r cis (θ ), where θ = arg( z), −π < θ ≤ π .
a z = 1 + i b = − z i1 3
THINK WRIT E
a
&
b
1 On a Calculator page, complete the
entry line as:
1 + iThen press:
• MENUb
• 2: Number2
• 9: Complex Number Tools9
• 6: Convert to Polar6
Then press ENTER·.
a
&
b
2 Write the answer. For a, i1 2 cis 4
π
+ =
3 Use the relationship reiθ = r cos (θ ) + ir sin
(θ ) to express the answer in the required
form. The calculator always gives θ in
principle valued form.
Key in i1 3− and repeat the above
procedure.
4 Write the answer.For b, i e1 3 2
i3
− =
π −
2 cis3
π
=
−
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26 Maths Quest 12 Specialist Mathematics
WORKED EXAMPLE 36
a If f ( z) = z3 + 7 z2 + 16 z + 10, find all factors of f ( z) over C .
b Factorise P( z) = z3 − (3 −i) z2 + 2 z − 6 + 2i.
THINK WRIT E
a
&
b
1 On a Calculator page, press:
• MENUb
•
3: Algebra3• C: ComplexC
• 2: Factor2
Complete the entry lines as
cFactor ( z3 + 7 z2 + 16 z + 10, z)
cFactor ( z3 − (3 − i) z2 + 2 z − 6 + 2i, z)
Press ENTER· after each entry.
a
&
b
2 Write the answers in the required
form.
For a, the three factors of P( z) are
( z + 1), ( z + 3 − i) and ( z + 3 + i)
For b, = − + + −P z z i z i z i( ) ( 3 )( 2 )( 2 )
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CHAPTER 4 • Relations and regions of the complex plane 27
CHAPTER 4
Relations and regions
of the complex planeWORKED EXAMPLE 16
Express each of the following expressions in Cartesian form.
a Re( z + 5) b Im( z − 2 − 3i) c | z − 4 + 2i |
THINK WRIT E
a ,
b
&
c
1 On a Calculator page, complete the
entry lines as:
Define z = x + yi
Then press ENTER·.To answer part a press:
• MENUb
• 2: Number2
• 9: Complex Number Tools9
• 2: Real Part2 or
• 3: Imaginary Part3 or
• 5: Magnitude5
Complete the entry line as:
real( z + 5)
imag( z − 2 − 3i)
| z − 4 + 2i |Press ENTER· after each entry.
a ,
b
&
c
2 Write the answers. For a, Re( z + 5) = x + 5.
For b, Im( z − 2 − 3i) = y − 3.
For c , | − + | = − + + + z i x x y y4 2 8 4 202 2 .
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CHAPTER 5 • Differential calculus 29
CHAPTER 5
Differential calculusWORKED EXAMPLE 1
Differentiate the following expressions with respect to x.
a y = tan (6 x) b y x
2tan 4
3=
THINK WRIT E
a
&
b
1 On a Calculator page, press:
• MENUb
• 4: Calculus4• 1: Derivative1
Complete the entry lines as:
d
dx x (tan (6 ))
d
dx
x 2 tan
4
3
Press ENTER· after each entry.
a
&
b
2 Write the solutions. For a,d
dx x
x [tan (6 )]
6
[cos (6 )]2=
For b,d
dx
x
x 2 tan
4
3
8
3 cos 4
3
2
=
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30 Maths Quest 12 Specialist Mathematics
WORKED EXAMPLE 4
Find the equation of the tangent to the curve y = 3 x + cos (2 x) + tan ( x) where x4.
π
=
THINK WRIT E
1 On a Calculator page, press:
• MENUb
•
4: Calculus4• 9: Tangent Line9
Complete the entry lines as:
tangentLine x x x x 3 cos (2 ) tan ( ), ,4
π + +
Then press ENTER·.
2 Write the solution. Equation of the tangent is y = 3 x + 1
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CHAPTER 5 • Differential calculus 31
WORKED EXAMPLE 6
Find, using calculus, f″ ( x) if f ( x) is equal to:
a ecos (2 x) + loge ( x) b x
x
sin ( ).
THINK WRIT E
a
&
b
1 On a Calculator page, complete the
entry line as:
f ( x ) : = ecos (2 x ) + ln( x )
Then press ENTER·.
Note: The syntax used here is
another way of defining a function or
variable. You can use the Define or
Store methods if you prefer.
a
&
b
2 Complete the entry line as:
d
dx f x ( ( ) )
2
2
Then press ENTER·.
3 Write the solution. The second derivative,
f '' x x x e x
( ) [4 sin (2 ) 4 cos (2 )] 1 x 2 cos (2 )
2= − −
4 On a Calculator page, complete the
entry line as:
f x x
x ( ) :
sin ( )=
Complete the entry line as:
d
dx
f x ( ( ) )2
2
Press ENTER· after each entry.
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32 Maths Quest 12 Specialist Mathematics
5 Write the solution. The second derivative,
f x
x x
x x
x
( ) 3
4
1 sin ( )
cos ( )5
2
3
2
′′ = −
−
6 You may rearrange the answer to
a form similar to that given in the
solution obtained manually as follows.
Press:• MENUb
• 3: Algebra3
• 2: Factor2
Then select and paste the previous
answer to obtain the entry line:
factor
x x
x x
x
3
4
1 sin ( )
cos ( )5
2
3
2
−
−
Then press ENTER·.
7 Write the solution. The second derivative:
f x x x x x
x
( )(3 4 ) sin ( ) 4 cos ( )
4
2
5
2
′′ = − −
WORKED EXAMPLE 16
Find the equation of the normal to the curve with equation:
y x
2 cos2
1=
− at the point where x 3= .
THINK WRIT E
1 On a calculator page, press:
• MENUb
• 4: Calculus4
• A: NormalLineA
Complete the entry line as:
normalLine x
x 2 cos2
, , 31
−
Then press ENTER·.
2 Write your solution in an appropriate
form.
The equation of the normal is y x
2
3
2 3.
π
= − +
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CHAPTER 5 • Differential calculus 33
WORKED EXAMPLE 18
Find the antiderivative for each of the following expressions:
a x
1
25 2−
b x
3
49 2−
−
c x
20
16 2+
.
THINK WRIT E
a ,
b
&
c
1 On a Calculator page, press:
• MENUb• 4: Calculus4
• 3: Integral3
Complete the entry lines as:
x dx
1
25
2∫ −
x dx
3
49
2∫ −
−
x dx
20
16
2∫ +
Press ENTER· after each entry.
Note: The calculator finds the second
form of the antiderivative in part b . Also,
it does not include the constant. You will
have to do that yourself.
2 Write your solutions, remembering to
include the constant of integration. a
b
c
x dx
x c
1
25
sin52
1∫ − =
+
−
x dx
x c
3
49 3 sin
72
1∫ − =
+
−− −
x dx
x c
20
16 5 tan
42
1
∫ + =
+
−
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34 Maths Quest 12 Specialist Mathematics
WORKED EXAMPLE 21
Differentiate the equation y2 + 3 x2 = 4 to find dy
dx in terms of x.
THINK WRIT E
1 On a Calculator page, press:
• MENUb
•4: Calculus4
• E: Implicit DifferentiationE
Complete the entry line as:
impDif ( y2 + 3 x 2 = 4, x , y)
Then press ENTER·.
2 Substitute for y as in part a (which
is preferable in this straightforwardequation) or continue to use the calculator
to make y the subject in the equation.
Press:
• MENUb
• 3: Algebra3
• 1: Solve1
Complete the entry line as:
Solve ( y2 + 3 x 2 = 4, y)
Then press ENTER·.
3 Express the domain, 3 x 2 − 4 ≤ 0
shown in the screen in a more appropriate
form. Take care to change ≤ to < as y is in
the denominator in the solution.
x 2 3
3
2 3
3< <
−
4 Write your solution, remembering to
include the domain.
dy
dx
x
x x
3
4 3;
2 3
3
2 3
32= ±
−
< <
−
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CHAPTER 5 • Differential calculus 35
WORKED EXAMPLE 23
For x2 y2 + y = 2, find the gradient of the tangent, dy
dx, at the point (1, −2). Hence, determine the
equation of the tangent at this point.
THINK WRIT E
1 The gradient of the implicit function can
be found using a CAS calculator.To do this, on a Calculator page, press:
• MENUb
• 4: Calculus4
• E: Implicit DifferentiationE
Complete the entry line as:
impDif ( x 2 × y2 + y = 2, x,y) | x = 1 and
y = −2
Then press ENTER·.
2 The gradient of the tangent at (1, −2) is 8
3.
Use this information and the general
equation of a straight line to determine the
equation of the tangent.
y y m x x
y x
y x
y x
( )
2 8
3( 1)
2 8
3
8
3
8
3
8
32
1 1− = −
− = −
+ = −
= − −
−
y x 8
3
14
3= −
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CHAPTER 6 • Integral calculus 37
CHAPTER 6
Integral calculusWORKED EXAMPLE 1
Find the antiderivative of the following expressions.
a ( x + 3)7 b 4 x(2 x2 + 1)4 c x
x x
3 12
3
+
+
THINK WRIT E
a
&
b
1 On a Calculator page, press:
• MENUb
• 4: Calculus 4
• 3: Integral 3
Complete the entry lines as:
x dx
x x dx
( 3)
(4 (2 1) )
7
2 4
∫
∫
+
× +
x
x x dx
3 12
3∫ +
+
Press ENTER· after each entry.
Note: The calculator cannot
determine the solution for part c .
It simply returns the input youentered. You will have to do this
problem manually. Also, it does not
include the constant for any of the
antiderivatives. Ensure you include
the constant with your answers.
a
&
b
2 Write your solutions, remembering
to include the constant of integration. x dx x
c
x x dx x
c
( 3) ( 3)
8
4 (2 1) (2 1)
5
78
2 42 5
∫
∫
+ = +
+
+ = +
+
c 1 Recognise that 3 x 2 + 1 is the
derivative of x 3 + x . Let u = x 3 + x .
c Let u = x 3 + x.
2 Finddu
dx .
du
dx x 3 1
2= +
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38 Maths Quest 12 Specialist Mathematics
WORKED EXAMPLE 2
Antidifferentiate the following functions with respect to x.
a f x x
x x( )
( )
3
62 3=
+
+ b f x x x x( ) ( ) cos ( )1 32 3
= − −
THINK WRIT E
a 1 On a Calculator page, press:
• MENUb• 4: Calculus 4
• 3: Integral 3
Complete the entry line as:
x
x x dx
3
( 6 )2 3∫ +
+
Then press ENTER·.
The answer is in an equivalent but
more complex form than the solution
found manually.
a
2 Collect the terms in factorised
form over a common denominator as
follows:
Press:
• MENUb
• 3: Algebra 3
• 2: Factor 2
Complete the entry line as
factor x
x x x
( 9)
432( 6)
1
432
1
1442 2
++
+ −
−
Then press ENTER·.
3 Write your solution, remembering to
include the constant of integration.
x
x x dx
x x c
3
( 6 )
1
4 ( 6)2 3 2 2∫ +
+ = −
+ +
b 1 Express in integral notation. b x x x dx ( 1) cos (3 )2 3∫ − −
2 Recognise that x 2 − 1 is a multiple of
the derivative of 3 x − x 3.
3 Let u = 3 x − x 3. Let u = 3 x − x 3.
4 Finddu
dx .
du
dx x 3 3
2= −
= 3(1 − x 2)
= −3( x 2 − 1)
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CHAPTER 6 • Integral calculus 39
5 Substitute u for 3 x − x 3 and
x du
dx ( 1)
1
3
2− =
−
.
So x x x dx ( 1) cos (3 )2 2∫ − −
u u
dx dx
u du
dx dx
cos 1
3
cos
3
∫
∫
= ×
=
−
−
= udu
cos
3∫ −
= u du1
3cos∫
−
6 Antidifferentiate with respect to u. = u
c
sin
3+
−
7 Replace u with 3 x − x 3. = x x
c
sin (3 )
3
3−
+
−
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40 Maths Quest 12 Specialist Mathematics
WORKED EXAMPLE 7
Find the antiderivative of the following expressions.
a sin2 x
2
b 2 cos2 x
4
THINK WRIT E
a 1 Express in integral notation. a x
dx sin2
2∫
2 Use identity 1 to change sin2 x
2 . = x dx [1 cos ( )]
12∫ −
3 Take the factor of1
2 to the front of the
integral.
= x dx [1 cos ( )]12 ∫ −
4 Antidifferentiate by rule. = x x c[ sin ( )]12
− +
5 Simplify the answer. =
x
x c2
1
2 sin ( )− +
6 If you use a calculator for problems
such as these, you may find the
answer expressed in a form that is
different from the ones above.
The screen dump shows the result, in
the first line, of part a done using a
CAS calculator.
The compact form shown in the
second line can be obtained as
follows.
Press:
• MENUb
• 3: Algebra 3
• B: Trigonometry B
• 2: Collect2
Complete the entry line as:
tCollect x x x
2sin
2cos
2−
Then press ENTER·.
7 Write your solution, remembering to
include the constant of integration.
x dx
x x csin
2
sin ( )
2
2∫
=
−+
b 1 Express in integral notation. b x dx 2 cos4
2
∫
2 Use identity 2 to change cos2 x
4
. =
x dx 2 1 cos
2
1
2∫ ( ) +
3 Simplify the integral. = x
dx 1 cos2∫ +
4 Antidifferentiate by rule. = x x
c2 sin2
+
+
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CHAPTER 6 • Integral calculus 41
WORKED EXAMPLE 12
For each of the following rational expressions:
i express as partial fractions ii antidifferentiate the result.
a x
x x
7
2 3( )( )
+
+ − b
x
x x
2 3
3 42
−
− −
THINK WRIT E
a i 1 Express the rational expression
as two separate fractions with
denominators ( x + 2) and
( x − 3) respectively.
a i x
x x
a
x x
7( 2)( 3) ( 2) ( 3)
+
+ −
=
+
+
−
2 Express the partial fractions
with the original common
denominator.
= a x x
x x
( 3) ( 2)
( 2)( 3)
− + +
+ −
3 Equate the numerator on the
left-hand side with the right-
hand side.
so x + 7 = a( x − 3) + b( x + 2)
4 Let x = −2 so that a can beevaluated.
Let x = −2, and thus 5 = −5a
5 Solve for a. a = −1
6 Let x = 3 so that b can be
evaluated.
Let x = 3, and thus 10 = 5b
7 Solve for b. b = 2
8 Rewrite the rational expression
as partial fractions.
Therefore x
x x x x
7
( 2)( 3)
1
2
2
3
+
+ −
=
+
+
−
−
9 A CAS calculator can convert
expressions in partial fractionform as follows.
Press:
• Menub
• 3: Algebra 3
• 3: Expand 3
Complete the entry line as:
expand x
x x
7( 2)( 3)
++ −
Then press ENTER·.
10 Write the answer. x
x x x x
7( 2)( 3)
23
12
+
+ −
=
−
−
+
ii 1 Express the integral in partial
fraction form.
ii x
x x dx
7
( 2) ( 3)∫ +
+ −
= x x
dx 1
2
2
3∫ + +
−
−
2 Antidifferentiate by rule. = −loge (| x + 2 |) + 2 loge (| x − 3 |) + c;
x ∈ R \{−2, 3}
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42 Maths Quest 12 Specialist Mathematics
3 Simplify using log laws. = loge x
x
( 3)
2
2−
| + |
+ c; x ∈ R \{−2, 3}
b i 1 Factorise the denominator. b i x
x x
2 3
3 42
−
− − =
x
x x
2 3
( 4)( 1)
−
− +
2 Express the partial fractions
with denominators ( x − 4) and
( x + 1) respectively.
= a
x
b
x 4 1−
+
+
3 Express the right-hand side
with the original common
denominator.
= a x b x
x x
( 1) ( 4)
( 4)( 1)
+ + −
− +
4 Equate the numerators. So 2 x − 3 = a( x + 1) + b( x − 4)
5 Let x = 4 to evaluate a. Let x = 4, 5 = 5a
6 Solve for a. a = 1
7 Let x = −1 to evaluate b. Let x = −1, −5 = −5b
8Solve for b. b = 1
9 Rewrite the rational expression
as partial fractions.
Therefore x
x x
2 3
3 42
−
− − =
x x
1
4
1
1−
+
+
ii 1 Express the integral in its
partial fraction form.
ii x
x x dx
2 3
3 42∫ −
− −
= x x
dx 1
4
1
1∫ − +
+
2 Antidifferentiate by rule. = loge (| x − 4 |) + loge (| x + 1|) + c
3 Simplify using log laws. = loge (| x − 4 |)(| x + 1|) + c; x ∈ R \{−1, 4}
or loge (| x 2
− 3 x
− 4 |) + c; x
∈ R
\{−
1, 4}
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CHAPTER 6 • Integral calculus 43
WORKED EXAMPLE 16
Evaluate the following definite integrals.
a x
x x dx
2
5 42
0
2
∫ −
+ + b x x dxcos ( ) 1 sin ( )
0
2
∫ +
π
THINK WRIT E
a 1 Write the integral. a x x x
dx 25 4
20
2
∫ −+ +
2 Factorise the denominator of the
integrand.
Consider: x
x x
2
5 42
−
+ +
= x
x x
2
( 1)( 4)
−
+ +
3 Express in partial fraction form
with denominators x + 1 and x + 4.=
a
x
b
x 1 4+ +
+
4 Express the partial fractions with
the original common denominator.=
a x b x
x x
( 4) ( 1)
5 42
+ + +
+ +
5 Equate the numerators. x − 2 = a( x + 4) + b( x + 1)
6 Let x = −1 to find a. Let x = −1, −3 = 3aa = −1
7 Let x = −4 to find b. Let x = −4, −6 = −3b
b = 2
8 Rewrite the integral in partial
fraction form.
So x
x x dx
2
5 42
0∫
−
+ +
= x x
dx 1
1
2
40
2
∫ + +
+
−
9 Antidifferentiate the integrand. = x x [ log ( 1 ) 2 log ( 4 )]e e 0
2| + | + | + |
−
10 Evaluate the integral.=
[−loge (3)+
2 loge (6)]−
[−loge (1)+
2 loge (4)]
= −loge (3) + 2 loge (6) − 2 loge (4)
11 Simplify using log laws. = 2 loge (1.5) − loge (3)
= loge (2.25) − loge (3)
= loge (0.75)
(or approx. −2.88)
12 You may attempt this problem using
a CAS calculator.
On a Calculator page, press
• MENUb
• 4: Calculus 4
• 3: Integral 3
Complete the entry line as:
∫ −+ +
x
x x dx
2
( 5 4)20
2
Then press ENTER·.
Pressing Ctrl/ENTER·will
give an approximate answer.
13 Write the solution. ∫ −+ +
=
≈ −
x
x x dx
2
( 5 4)log
3
4
2.88
e20
2
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44 Maths Quest 12 Specialist Mathematics
b 1 Write the integral. b x x dx cos ( ) 1 sin ( )0
2∫ +
π
2 Let u = 1 + sin ( x ) to antidifferentiate. Let u = 1 + sin ( x )
3 Findu
dx .
du
dx x cos ( )=
4 x dx
u
dx dx cos ( ) = x dx
du
dx dx cos ( ) =
5 Change terminals by finding u
when x = 0 and x =2
π
.
When x = 0, u = 1 + sin (0)
= 1
When x 2
π
= , u = 1 + sin2
π
= 1 + 1
= 2
6 Simplify the integrand. So x x dx cos ( ) 1 sin ( )0
2∫ +
π
= u dudx
dx
1
2
1
2
∫ = u du
1
2
1
2
∫
7 Antidifferentiate the integrand. = u2
3
3
2
1
2
8 Evaluate the integral. = 2 12
3
3
2 2
3
3
2× − ×
= 4 2
3
2
3−
or
4 2 2
3
−
9 Using CAS, press
• MENUb
• 4: Calculus4
• 3: Integral3
Complete the entry line as:
x x dx cos ( ) 1 sin ( )0
2∫ +
π
Then press ENTER·.
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CHAPTER 6 • Integral calculus 45
WORKED EXAMPLE 19
Find the area bounded by the curves y = x2 − 2 and y = 2 x + 1.
THINK WRIT E
1 On a Calculator page,
complete the entry line as:
x x dx ( 2 (2 1))
2
1
3
∫ − − +
−
Then press ENTER·.
2 State the solution. The area bounded by the two curves is 102
3
square units.
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CHAPTER 7 • Differential equations 47
CHAPTER 7
Differential equationsWORKED EXAMPLE 7
Find the general solution to dy
dx x
x2
1
1 2
= +
−
.
THINK WRIT E
1 On a Calculator page, press:
• MENUb
• 4: Calculus4
• D: Differential Equation SolverDComplete the entry line as:
deSolve y x
x
x y2 1
1, ,
2′ = +
−
,
then press ENTER·.
2 Write the solution. The general solution is y = sin−1 ( x ) + x 2 + c.
WORKED EXAMPLE 8
Find the particular solution to h′( t) t
t 92
=
+
where h(4) = 1.
THINK WRIT E
1 On a Calculator page, press:
• MENUb
• 4: Calculus4
• D: Differential Equation SolverD.
Complete the entry line as:
deSolve h' t
t
h t h
9and (4) 1, , ,
2=
+ =
then press ENTER·.
2 Write the solution. The particular solution is h t t ( ) 9 4.2= + −
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48 Maths Quest 12 Specialist Mathematics
WORKED EXAMPLE 10
Find the general solution to f x e( ) 2.
x
2′′ = −
THINK WRIT E
1 On a Calculator page, press:
• MENUb
•
4: Calculus4• D: Differential Equation SolverD
Complete the entry line as:
deSolve y e x y2, , ,
x
2′′ = −
then press ENTER·.
2 Write the solution. The general solution is
y e x cx d 4 . x 2 2
= − + +
WORKED EXAMPLE 11
Find the particular solution to y″ ( x) = 4 sin (2 x + π ) given that y′(0) = 1 and y(0) = 4.
THINK WRIT E
1 On a Calculator page, press:
• MENUb
• 4: Calculus4
•
D: Differential Equation SolverDComplete the entry line as:
deSolve( y″ = 4 sin (2 x + π ) and y′(0) = 1
and y(0) = 4, x , y),
then press ENTER·.
2 Write the solution. Note that it is different
from but equivalent to the solution found
manually.
The particular solution is
y = sin (2 x ) − x + 4.
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CHAPTER 7 • Differential equations 49
WORKED EXAMPLE 13
Find general solutions for each of the following differential equations.
a dy
dx y=
b y′( x) = tan ( y)
THINK WRIT E
a 1 On a Calculator page, press:• MENUb
• 4: Calculus4
• D: Differential Equation SolverD
Complete the entry line as:
deSolve( y′ = y, x , y),
then press ENTER·.
a
2 Write the solution, making sure to
write the integration constant in a
conventional form.
The general solution is y = Ae x .
b 1 Express the derivative asdy
dx . b y′( x ) = tan ( y)
dx
dy ytan( )=
2 Invert both sides of the differential
equation.
dx
dy =
y
1
tan ( ), y ≠ 0
3 Simplify y
1tan ( )
. = y y
cos ( )sin ( )
4 Let u = sin ( y) to antidifferentiate
using the ‘derivative present’ method.
x = y
ydy
cos ( )
sin ( )∫ Let u = sin ( y)
5 Finddu
dy.
du
dy = cos ( y)
6 Substitute u and cos ( y) = du
dy into
the integral.
x = u
du
dydy
1
∫ ×
7 Simplify the integrand. = du
u∫ 8 Antidifferentiate by rule. = loge (|u |) + c
9 Replace u by sin ( y). x = loge (| sin ( y) |) + c
10 Subtract c from both sides. x − c = loge (| sin ( y) |)
11 Express the equation in exponential
form.
e x − c = | sin ( y) |
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50 Maths Quest 12 Specialist Mathematics
12 Write the constant in a more suitable
form.
| sin ( y) | = e−ce x
13 Remove the modulus sign and take
sin−1 of both sides to make y the
subject.
sin ( y) = ±e−ce x
y = sin−1 ( Ae x ), where A = ±e−c
14 State the solution. Therefore, the general solution is y = sin−1 ( Ae x ).
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52 Maths Quest 12 Specialist Mathematics
WORKED EXAMPLE 24
Use the numerical integration method to find the value of y when x = 15 if dy
dx x2= , given that y = 1
at x = 0.
THINK WRIT E
1
Substitute appropriate values andexpressions, a = 0, b = 1 and f ( x ) = 2 x , in
the formula. On a Calculator page, press:
• MENUb
• 4: Calculus4
y x dx (15) 2 10
15
∫ = +
• F: Numerical CalculationsF
• 3: Numerical Integral3
Complete the entry line as:
nInt(2 x , x ,0,15) + 1,
then press ENTER·.
2 Write the answer. The value of y when x = 15 is 226.
3 Find the antiderivative of 2 x . y x c2
= +
4 Evaluate the constant by applying the
initial condition y(0) = 1.
y x 12
= +
5
Substitute x = 15 and compare the newanswer to the previous one.
y = 225 + 1 = 226The answers are the same.
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CHAPTER 8 • Kinematics 53
CHAPTER 8
KinematicsWORKED EXAMPLE 3
The position of a particle moving in a straight line is given by:
x( t) = 2 t3 + t loge ( t) − 4, t > 0.
Find:
a the velocity at any time t b the acceleration at any time t.
THINK WRIT E
a 1 To determine the velocity at any
time, differentiate x with respectto t , using a CAS calculator by
completing the following steps.
Define the function x (t ). To do this,
on a Calculator page, press:
• MENUb
• 1: Actions1
• 1: Define1
Complete the entry line as:
Define x (t ) = 2t 3 + t ln(t ) − 4
then press ENTER·.
a
2 Differentiate x with respect to t .To do this press:
• MENUb
• 4: Calculus4
• 1: Derivative1
Complete the entry line as:d
dt x t ( ( ))
then press ENTER·.
3 Write the solution using the correct
notation and variables.
The velocity at any time is given as:
v(t ) = loge (t ) + 6t 2 + 1.
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54 Maths Quest 12 Specialist Mathematics
b 1 To determine the acceleration at any
time, differentiate the velocity with
respect to t . To do this, complete the
entry line as:
+ × +d
dt t t (ln( ) 6 1)2
then press ENTER·.
b See the screen above.
2
Write the solution using the correctnotation and variables.
The acceleration at any time is given as:a t t
t
( ) 12 1
= + .
WORKED EXAMPLE 4
Find the acceleration in terms of x if x( t) = sin (2 t) − cos (2 t).
THINK WRIT E
1 To find the acceleration given the
position, on a Calculator page, complete
the entry line as:
d
dt t t (sin (2 ) cos (2 ))
2
2 −
then press ENTER·.
Note: The second derivative template is
located in the Maths expression template.
2 Write the solution in terms of x . a(t ) = 4 cos (2t ) − 4 sin (2t )
a(t ) = −
4 (sin (2t ) − cos (2t ))But x = sin (2t ) − cos (2t )
a( x ) = −4 x
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CHAPTER 8 • Kinematics 55
WORKED EXAMPLE 6
The acceleration of a particle moving in straight line is given by: dv
dte t5 6 4
t= − + cm/s2, where v is the velocity at any time.
If the particle starts at the origin with a velocity of −1 cm/s, find:
a the velocity at any time t
b the displacement x ( t) from the origin at any time t
c the displacement from the origin 1 second. THINK WRIT E
a Solving the differential equation, to find
the velocity press:
• MENUb
• 4: Calculus4
• D: Differential Equation SolverD
Complete then entry line as:
v e t v v t deSolve ' 5 6 4 and 0 1, ,t ( )( )= − + = −
then press ENTER·.
a
b
&
c
Solving the differential equation, to find
the displacement press:
• MENUb
• 4: Calculus4
• D: Differential Equation SolverD
Complete then entry line as: x e t t
x x t
deSolve ( ' 5 3 4 5and 0 0, , )
t 2
( )= − + −
=
then press ENTER·.
To find the displacement after 1 second,
complete as shown.
b
&
c
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56 Maths Quest 12 Specialist Mathematics
WORKED EXAMPLE 7
The acceleration of a body travelling in a straight line is given by:
a( t) = 6 t − 2 m/s2; when t = 0, x = 0 and v = −1.
a Find the displacement at any time t.
b Find the distance travelled in the first 3 seconds.
THINK WRIT E
a1
To find the displacement at anytime t , first find the velocity by
antidifferentiating a with respect
to t , using a CAS calculator by
completing the following steps.
To do this, on a Calculator page,
complete the entry lines as:
Define a(t ) = 6t − 2
a t dt ( ( ))∫ Press ENTER· after each entry.
To find the constant of integration
complete the entry lines as:
Define v(t ) = 3t 2
− 2t + csolve(v(0) = −1,c)
Press ENTER· after each entry.
a
2 State the velocity at any time. v(t ) = 3t 2 − 2t − 1
3 Antidifferentiate v to find the
displacement x by completing the
entry line as:
t t dt (3 2 1)2∫ − −
then pressing ENTER·.
The constant of integration can be
found by completing the entry lines
as:Define x (t ) = t 3 − t 2 − t + d solve( x (0) = 0,d )
Press ENTER· after each entry.
4 State the displacement at any time t . x (t ) = t 3 − t 2 − t
b 1 To find when the object might have
changed direction, find t when v = 0.
Complete the entry lines as:
Define x (t ) = t 3 − t 2 − t
Define v(t ) = 3t 2 − 2t − 1
solve(v(t ) = 0,t )Press ENTER· after each entry.
Since t t 1
3, 1≠ =
−
Complete the entry lines as:
x (0)
x (1)
x (3)
Press ENTER· after each entry.
b
Therefore the body travels 16 m in the first three
seconds of motion.
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CHAPTER 8 • Kinematics 57
2 Write the solution given that
Distance = | x (3) − x (0) | + | x (1) − x (0) |.Solving v(t ) = 0, for t gives
t = 0 as t 1
3≠
−
x (0) = 0
x (1) = −1
x (3) = 15
Distance = | x (3) − x (0) | + | x (1) − x (0) | = 16
Therefore the body travels 16 m in the first3 seconds of motion.
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CHAPTER 8 • Kinematics 59
3 Using the CAS calculator solve the
equation for t .
Complete the entry line as:
solve(0 = 24.5 − 9.8 × t ,t )
then press ENTER·.
4 State the solution. Solving 0 = 24.5 − 9.8 × t for t gives t = 2.5.
Therefore, the ball takes 2.5 seconds to reach its
maximum height.
c 1 Sketch another diagram for the ball
falling from its maximum height.
c
Ground
Platform
Max. heightu = 0
16 m
30.625 m
a = −9.8 m/s2
2 List the given information and what
has to be found.
Both a and s are negative as their
direction is downwards.
Given: u = 0
a = −9.8
s = −(30.625 + 16)
= −46.625
Require: t = ?
3 Select an appropriate formula. s ut at 1
2
2= +
4 Using a CAS calculator to solve the
equation for t , complete the entry line
as:
solve(−46.625 = 0 × t + 1
2 × −9.8 × t 2,t )
Then press ENTER·.
5 Write the solution. Solving −46.625 = 0 × (t ) +1
2 × (−9.8) × t 2 for t
gives t = 3.085.
Therefore, it takes approximately 3.085 seconds
for the ball to fall to the ground from its maximum
height.
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60 Maths Quest 12 Specialist Mathematics
d 1 Add the time travelling up to the
time travelling down.
d Total time = 2.5 + 3.085
= 5.585
2 State the answer. Therefore, the ball is in the air for approximately
5.585 seconds.
e 1 List the given information and what
has to be found.
e u = 0
a = −9.8
s = −30.625
v = ?
2 Select an appropriate formula. v2 = u2 + 2as
3 Using the CAS calculator to solve
the equation for v, complete the entry
line as:
solve(v2 = 02 + 2 × −9.8 × −30.625, v)
Then press ENTER·.
4 Write the solution. Solving v2 = 02 + 2 × −9.8 × −30.625 for v gives
v = −24.5 or v = 24.5.
Since the object is travelling downward, v = −24.5.
Therefore, the speed of the ball when it returns to
the level of the platform is 24.5 m/s.
f 1 List the information and what has to
be found.
f u = 0
a = −9.8
s =
−
46.625v = ?
2 Select an appropriate formula. v2 = u2 + 2as
3 Use the CAS calculator to solve for v.
Complete the entry line as:
solve(v2 = 02 + 2 × −9.8 × −46.625, v)
then press ENTER·.
4 Write the solution. Solving v2 = 02 + 2 × −9.8 × −46.625 for v gives
v = −30.23 or v = 30.23.
Therefore, the ball hits the ground with an
approximate speed of 30.23 m/s.
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CHAPTER 8 • Kinematics 61
WORKED EXAMPLE 16
The acceleration of a particle travelling in a straight line is given by = + dv
dtv 1, where v is the
velocity at any time t. At time t = 0, the body is at rest at the origin.
Find, as a function of time:
a the velocity
b the acceleration
c the position.
THINK WRIT E
a 1 Write the acceleration as given. a dv
dt v 1= +
2 Invert both sides of the differential
equation.
ordt
dv v
1
1=
+
3 Express t in integral form. t v
dv1
1=
+
4 Antidifferentiate the integrand, using
the CAS calculator.
Complete the entry line as:
vdv
1
1∫ +
then press ENTER·.
Given that t = 0 when v = 0, to find
the constant of integration, complete
the entry line as:
solve(0 = ln(|0 + 1 |) + c,c)
then press ENTER·.
Rearrange the equation to make v
the subject by completing the entry
line as:
solve(t = ln(|v + 1 |),v)then press ENTER·.
5 Write the solution. t v
dv1
1∫ =+
= loge (|v + 1 |) + c
Solving 0 = loge (|0 + 1 |) + c for c
gives c = 0.
So, t = loge (|v + 1 |)Then, solving t = loge (|v + 1 |) for v
gives v = −et − 1 or v = et − 1.
Since v = 0 when t = 0
Then v = et − 1
b 1 Differentiate v with respect to t tofind the acceleration (or we could
substitute v into the original equation
for acceleration).
b a
v
dt =
a d e
dt
( 1)t
=
−
2 State the answer. a = et
c 1 Express x as the antiderivative of v. c x e dt ( 1)t ∫ = −
2 Antidifferentiate the integrand. x e t c= − +
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62 Maths Quest 12 Specialist Mathematics
3 Substitute t = 0 and x = 0, the initial
condition given.
When t = 0, x = 0
0 = e0 − 0 + c
4 Solve for c. c = −1
5 State the answer. x = et − t − 1
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CHAPTER 9 • Vectors 63
CHAPTER 9
VectorsWORKED EXAMPLE 9
Find the unit vector in the direction of u~ : a
a in the figure at right
b in u i j k2 3 2 3= − +
.
THINK WRIT E
a 1 Express the vector in component
form.
a u i j6 3= +
2 Compute the magnitude of the
vector, u.
u
6 3
45
3 5
2 2= +
=
=
3 Divide each component of the
original vector by the magnitude to
get u.
u i j
i j
ˆ 6
3 5
3
3 5
2 5
5
5
5
= +
= +
4 Comfirm that u has a magnitude of 1. u
x y
20
25
5
25
25
25
1
2 2= +
= +
=
=
C (6, 3)u~ 3 j
~
6i~
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CHAPTER 9 • Vectors 65
WORKED EXAMPLE 15
Let u i j5 2= +
. Find a vector parallel to u such that the dot product is 87.
THINK WRIT E
1 Let the required vector v k u=
. Let v = k i j(5 2 )+
= ki kj5 2+
2 Find the dot product of u v.
. .u v
= .i j ki k j(5 2 ) (5 2 )+ +
3 Simplify. = 25k + 4k
= 29k
4 Equate the result to the given dot product 87. 29k = 87
5 Solve for k . k = 3
6 Substitute k = 3 into vector v. v
= i j15 6+
7 An alternative method is to use a CAS
calculator. Let the required vector v k u=
.
u i j5 2= +
So v k i j(5 2 )= +
8 Using a CAS calculator, define u
and v
by
completing the following steps.
On a Calculator page, press:
• MENUb
• 1: Actions1
• 1: Define1
Complete the entry line as:
Define u [5,2]=then press ENTER·.
Repeat for v k [5,2]=
9 Using a CAS calculator solve .u v 87=
for
k , by completing the following steps. Press:
• MENUb
• 3: Algebra3
• 1: Solve1
After ‘solve’, then press:
• MENUb
• 7: Matrix and Vector7
• C: VectorC
• 3: Dot Product3
Complete the entry line as:
u v k solve(dotP( , ) 87, )=
then press ENTER·.
10 Write the solution. Solving .u v 87=
for k gives
k 3=
Therefore,
v i j15 6= +
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66 Maths Quest 12 Specialist Mathematics
WORKED EXAMPLE 19
Determine the value of p if the following vectors are linearly dependent.
a i pj k= + −
, b i j k3 8= + +
and c i j k2= − +
THINK WRIT E
1 Let a mb nc= +
. Let a mb nc= +
Solve a mb nc= +
for m, n and p.
2 To solve this equation for m, n and
p define vectors a b c, and
as shown
previously.
Then press:
• MENUb
• 3: Algebra3
• 1: Solve1
Complete the entry line as:
a m b n c m n psolve( ,{ , , })= × + ×
Then press ENTER·.
3 Write the solution. Solving a m b n c= × + × for m, n and p gives:
m
n
p
0.5
1.5
3
=
=
=
−
−
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CHAPTER 9 • Vectors 67
WORKED EXAMPLE 21
Let u i j k2 3= + +−
and v i j k3 2= + −
.
Find:
a the scalar resolute of v on u
b the vector resolute of v
parallel to u
, namely v
c the vector resolute of v perpendicular to u
, namely v⊥
.
THINK WRIT E
a 1 To find the scalar resolute of v
on u, find u
.
Using a CAS calculator, define v and
u as shown previously and find u
by
completing the entry line as:
unitV(u)
Then press ENTER·.
a
2 Write u. u i j k ˆ
14
7
3 14
14
14
14=
−
+ +
3 To find .u vˆ
(the scalar resolute of v
on u), first use CAS to
define w u=
.
Then, to find the dot product .u vˆ
,
complete the entry line as:
dotP(w,v)
Then press ENTER·.
4 Write the scalar resolute of
v
on u.
.u vˆ 14
14=
−
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68 Maths Quest 12 Specialist Mathematics
b 1 To find the vector resolute of v
parallel to u (that is, .u v u( ˆ ) ˆ
) using
CAS, complete the entry line as
follows:
dotP(w,v) × w
Then press ENTER·.
b
2 Write the vector resolute of v parallel
to u.
.u v u i j k ( ˆ ) ˆ 1
7
3
14
1
14= − −
c To find the vector resolute of v
perpendicular to u (that is, .v u v u( ˆ ) ˆ−
)
using CAS, complete the entry line as:
v 1
7
3
14
1
14−
− −
then press ENTER·.
Write the solution in an appropriate form.
c .v u v u i j k ( ˆ ) ˆ 20
7
31
14
13
14− = + −
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CHAPTER 9 • Vectors 69
WORKED EXAMPLE 25
Let a particle’s position as a function of time be given by u ti t j2 cos ( ) 3 sin ( ) .= +
Find the
equation of the path and sketch its graph.
THINK WRIT E
1 Express the i and j
components of u
in
terms of their functions.
x = 2 cos (t )
y = 3 sin (t )
2 In this case, first eliminate the constants in
front of the trigonometric functions.
x
2 = cos (t )
y
3 = sin (t )
3 Square both sides of the equation. x
4
2
= cos2 (t )
y
9
2
= sin2 (t )
4 Add the 2 equations. x
4
2
+ y
9
2
= cos2 (t ) + sin2 (t )
5 Use the trigonometric identity
cos2 (θ) + sin2 (θ) = 1.
x
4
2
+ y
9
2
= 1
6 Use CAS to sketch the graph using the
parametric equations.
On a Graphs page, press:
• MENUb
• 3: Graph Entry/Edit3
• 3: Parametric3
Complete the entry line as shown:
x t t
y t t
t tstep
1( ) 2cos ( )
1( ) 3sin ( )0 6.28 0.13
=
=
≤ ≤ =
then press ENTER·.
7 This is the equation of an ellipse of centre
(0, 0) where a = 2 and b = 3.
0
y
x 2
3
−2
−3
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CHAPTER 10 • Vector calculus 71
CHAPTER 10
Vector calculusWORKED EXAMPLE 3
An object has a position vector, in metres, rr t t t i e t j( ) (3 sin ( )) ( 2 ) t2π = − + +
−
; t ≥ 0 seconds.
a Find the velocity vector vv t ( )
.
b Find the velocity of the object at t = 2 s.
c Find the speed of the object at t = 2 s.
d Find the average velocity in the first 2 seconds of the body’s motion.
THINK WRIT E
a 1 Define the position vector,r t )( . To
do this press:
• MENUb
• 3: Actions3
• 1: Define1
Then complete as shown, then press
ENTER·.
To findv t )( . To do this press:
• MENUb
• 4: Calculus4
• 1: Derivative1
Complete the entry line as:
dt r t )( )(
a
2 Write the solution using correct
vector notation.
π π ( )( )( ) ( )= − + − +−
v t t t i e j6 cos 2t
b 1 Substitute t = 2 b
2 Write the solution using correct
vector notation.
π ( ) ( )= − + −
v ie
j2 12 2 1
2
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72 Maths Quest 12 Specialist Mathematics
c 1 To find the speed, find the norm of
the vector.
c
2 State the speed correct to 2 decimal
places.
The speed of the object is 9.05 m/s.
d 1 The average velocity in the first
2 seconds, is( ) ( )−
−
r r 2 0
2 0
d
2 State the average velocity = +v i j6 1.567av
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CHAPTER 10 • Vector calculus 73
WORKED EXAMPLE 5
A body moves in such a way that its position vector in metres at an instant t seconds is given by:
r t t t i t t j t( ) 15 1
21 12
1
3, 02 3= − +
+ −
≥
.
Find:
a the velocity vector
t ( )
b the acceleration vector
a t
( )c the angle between the velocity vector and the acceleration vector of the body at a time t = 1 s, to
the nearest degree
d the time when the body has an acceleration of magnitude 9.8 m/s2.
THINK WRIT E
a 1 Differentiater , the position vector, with
respect to t , using a CAS calculator by
completing the following steps.
Define the position functionr t ( ). To do
this press:
• MENUb
• 1: Actions1
•
1: Define1Complete the entry line as:
Define = − + −
r t t t t t ( ) 15
1
21,12
1
3
2 3
then press ENTER·.
To findv t ( ) differentiate
r with respect
to t , press:
• MENUb
• 4: Calculus4
• 1: Derivative1
Complete the entry line as:d
dt r t ( ( ))
then press ENTER·.
a
2 Write the solution using correct
vector notation.
=
v t
r
dt ( )
= − + −
v t t i t j( ) (15 ) (12 )2
b 1 To find the acceleration vectora t ( )
complete the following steps.
Define the velocity functionv t ( ), by
completing the entry line as:
Define = − −v t t t ( ) [15 , 12 ]2
Then press ENTER·.
Differentiatev t ( ) with respect to t , by
completing the entry line as:d
dt v t ( ( ))
Then press ENTER·.
b
2 Write the solution using correct
vector notation.
=
a t
v
dt ( )
= −−
a t i t j( ) 2
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CHAPTER 10 • Vector calculus 75
d State the accelerationa t ( ) at any time, t .
Find the magnitude of the acceleration at
any time, t.
Use the CAS calculator to solve for t
when the acceleration is equal to 9.8 m/s2.
Complete the entry line as:
( )+ =t t solve 1 4 9.8,2
Then press ENTER·. Note: Only positive solutions for t are
allowed.
d = −−
a t i tj( ) 2
= +
a t t ( ) 1 4 2
Solving + =t 1 4 9.82 for t gives ≈ ≥t t 4.87, 0.
When t = 4.87 s the magnitude of the acceleration
is 9.8 m/s2.
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76 Maths Quest 12 Specialist Mathematics
WORKED EXAMPLE 10
An object is thrown off a building ( t = 0 s) on a windy day. The acceleration of the object in m/s2
is given by a t i e j( ) 9.8
t
112
10= −
−
. At a time, t = 1, the object has a velocity in m/s of v i j(1) 2 3= −
.
The building is 50 m above the ground and hence the initial position of the object is taken to be
r i j0 50= +
.
a What is the initial acceleration of the object?
bDetermine the velocity vector
v t
( ) for all times t
≥
0.c Determine the position vector r t ( )
for all times t ≥ 0.
THINK WRIT E
a 1 Define the acceleration functiona t ( ),
by completing the entry line as:
Define =
−−
a t e( )
1
12, 9.8
t
10
Then press ENTER·.
Substitute t = 0 intoa t ( ) to find the
initial acceleration.
a
2 Write the solution = −
a i j(0) 1
29.8
b 1 The velocity vector is found by
integrating the acceleration
t ( ) with
respect to time.
b
2 The vector constant of integration
must be found. To do this using the
CAS calculator, define the velocity
vectorv t ( ) by completing the entry
line as:
Define = + +
−
v t
t m e n( )
12, 98
t
10
Then press ENTER·.
3 State the integrand and the velocity
vectorv t ( ) with the vector constant
of integration expressed as +
i nj
∫ = + v t a t dt c( ) ( )
= + + +
−
v t t
i e j mi nj( )12
98
t
10
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CHAPTER 10 • Vector calculus 77
4 Evaluate the vector constant of
integration using the fact that
= −
v i j(1) 2 3 .
Complete the entry line as:
= −
v m nsolve( (1) [2, 3],{ , })
Then press ENTER·.
5 State the values of m and n and the
velocity vector.
=m 23
12 and n = −91.674
= +
+ −
−
v t t
i e j( )12
23
1298 91.674
t
10
c 1
2
Redefine the velocity vectorv t ( ) with
the correct values of m and n.
The position vectorr t ( ) is found
by integrating the vectorv t ( ) with
respect to time.
c
3 State the integrand and the positionvector
r t ( ), with the vector constant
of integration expressed as +
pi qj
∫ = + r t v t dt d ( ) ( )
= + +
+ − +
−
−
r t t
t p i e t q j( )24
23
12980 91.674
t 210
4 The vector constant of integration
must be found. To do this using the
CAS calculator, define the position
vectorr t ( ) by completing the entry
line as:
Define
= + + − +
−−
r t
t
t p e t q( ) 24
23
12 , 980 91.674
t 2
10
Then press ENTER·.
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78 Maths Quest 12 Specialist Mathematics
5 Use = +
r i j(0) 0 50 to solve the vector
constants p and q. To do this, complete
the entry line as:
solve(r (0) = [0, 50],{ p, q})
then press ENTER·.
6 State the values of p and q and the
position vector.
p = 0 and =q 1030
= +
+ − +
−
−
r t
t
t i e t j( ) 24
23
12 980 91.674 1030
t 2
10
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80 Maths Quest 12 Specialist Mathematics
6 Use CAS to solve the vector
equation.
First define the vector R, by
completing the entry line as:
Define r = [a − b × cos (42),
b × sin (42) − 9.8]
Then press ENTER·.
7 To solve the equation complete the
entry line as:
solve(r = [0, 0], {a, b})
Then press ENTER·.
8 Write the solution correct to 1 decimal
place, showing the correct units. A
B
10.9 N
14.6 N
=
=
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CHAPTER 11 • Mechanics 81
WORKED EXAMPLE 6
A car of mass 800 kg is parked in a street which has an angle of elevation
of 15°. The i
direction is parallel down the street and the
direction is
perpendicular to the street.
The car is subject to three forces, namely its weight,W
, the normal
reaction force, N
, of the road acting on the car and the applied force of the
brake (this is actually a static friction force) F
.
a Draw a vector diagram indicating the three forces,W , N and F , acting on the car, taking the car
as a particle.
b What is the magnitude of the resultant force R?
c Resolve the weight,
, into its components and express it as a vector using i − j
notation.
d Calculate the magnitude of N
, the normal contact force, and the magnitude of the applied force
of the brake, F
.
15°i
j
15°
~
~
© J o h n W i l e y & S o n s A u s t r a l i a / J
e n n i f e r W r i g h t
THINK WRIT E
a 1 A stationary car parked on a street
will have a vertical weight force, a
normal reaction force and a static
frictional force resisting its sliding or
rolling down the street.
a j
N
F
W
i15°
~
~
~~
~
2 Draw the force vector diagram.
b 1 The car is in equilibrium since it is
stationary.
b
2 Apply Newton’s First Law of Motion:
the resultant force, R, must be zero.
R 0=
3 Therefore, the magnitude of the
resultant force, R, is zero.
R = 0 N
c 1 Draw a diagram showing theresolution of the weight, W
, into
components parallel to i and j
.
c
W x
W y
j
W
i15°
15°
~
~
~
2 The component of W
parallel to i, x ,
is W sin (15°).
F = 207g
j
i15°
~
~
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82 Maths Quest 12 Specialist Mathematics
3 Substitute W = 800g and evaluate. = 800g sin (15°)i
≈ 207gi
4 The component of W parallel to j
,W y
is W cos (15°).
W y = W cos (15°)
5 Substitute W = 800g and evaluate. =−800g cos (15°) j
≈ −773g j
6 Express W
in vector notation. W = i x
+ W j y
W = 207gi
− 773g j
d 1 The components of the net forces
parallel to the i and j
vectors are
both zero.
d g F i N g j i j(207 ) ( 773 ) 0 0− + − = +
2 Use CAS to solve for F and N by
completing the entry line:
solve([207 × g − f , n − 773 × g] = [0, 0],
{ f , n})
Then press ENTER·.
3 Write the solutions for F and N . F = 207g
N = 773g
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CHAPTER 11 • Mechanics 83
WORKED EXAMPLE 11
A skier of mass 60 kg begins a ski run from the top of a mountain down an 18° slope. A constant
friction force of 37 N acts on the skier as he descends.
a Draw a vector force diagram depicting the three forces which act on the skier.
b Write the equation of motion using Newton’s Second Law of Motion.
c Determine the magnitude of the normal contact force and the acceleration of the skier.
d If there were no frictional force present, find the acceleration of the skier.
THINK WRIT E
a Three significant forces act on the skier: the
weight force W
, the normal reaction force N
and the force of friction F .
a N
F
W
j
i
18°
Mass = 60 kg
~
~
~~
~
b 1 Use Newton’s second law. b R m a=
2 The vector W
is to be resolved into
components parallel to i
and j
.
R W F i N W j
R ma i ma j
( sin ( ) ) ( cos ( ))
x y
θ θ = − + −
= +
3 We know that m = 60 and a y = 0. R a i j60 0 x = +
c 1 But:
W = 60 × 9.8 = 584 N
F = 37 N
θ = 18°
c W = 584 N
F = 37 N
θ = 18°
2 Use CAS to solve the vector equation,
R ai j60 0= +
, by completing the entry
line as:
solve([60 × 9.8 × sin (18) − 37, n − 60 ×
9.8 × cos (18)] = [60 × a, 0], {n, a})
Then press ENTER·.
3 Write the solutions correct to
2 decimal places, with the correct
units.
N
a
559.22 N
2.41 m/s2
=
=
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84 Maths Quest 12 Specialist Mathematics
d 1 To find the acceleration when no
frictional force is present, edit the
resultant force equation by completing
the entry line:
solve([60 × 9.8 × sin (18), n − 60 × 9.8 ×
cos (18)] = [60 × a, 0], {n, a})
Then press ENTER·.
d
2 Write the solution, with the correct units. a = 3.03 m/s2
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CHAPTER 11 • Mechanics 85
WORKED EXAMPLE 15
Two carts connected by a light, inextensible rope are being accelerated
across a tarmac by a tractor. The leading card (CartL) has a mass of
35 kg; the trailing cart (CartT) has a mass of 25 kg. The coefficient
of friction on the tarmac is = 0.20. The tractor exerts a pulling force
of 190 N on the cart of mass 35 kg, as shown at right. The breaking
tension in the rope is 150 N.
a Draw a force vector diagram for each of the two masses.b Calculate the magnitude of the acceleration of the two carts.
c Calculate the tension in the rope which connects the two masses.
d If the two masses are to be accelerated at a higher rate, what is the greatest acceleration of the
carts before the rope breaks?
THINK WRIT E
a 1 Each cart can be represented as a
particle with its own set of forces
acting.
a
W 1
N 1
F 1 T ~~
~
~
Mass 1 = 25 kg
(CartT)
2 The trailing cart (CartT) will have
four forces acting on it. These forces
are the normal contact force, N 1 ; theweight force, W 1
; the force of friction,
F 1
; and the tensile force, T , acting to
the right.
3 The leading cart (CartL) will have
five forces, namely the normal
contact force, N 2
; the weight force,
W 2
; the friction and tensile forces,
F 2
and T , acting to the left; and the
force of the tractor, A, acting towards
the right.
AF 2
T
N 2
W 2
Mass 2 = 35 kg
~
~
~ ~
~
(CartL)
b 1 Use Newton’s second law to write
the equation of motion for CartT.
b CartT: R T F i N g j( ) ( 25 )1 1= − + −
ai j25 0= +
2 Equate the j
components of the
equation.
N 1 − 25g = 0
3 Solve for N 1. N 1 = 25g
4 Find the friction force acting on
CartT using F 1 = N 1.
F 1 = µ N 1= 0.2(25g)
= 5g
5 Equate the i components of the
equation of motion.
T − F 1 = 25a
6
Substitute F 1 = 5g into the equationand call it equation 1.
T − 5g = 25a [1]
7 Use Newton’s second law to write
the equation of motion for CartL.CartL: R F T i N g j(190 ) ( 35 )2 2= − − + −
ai j35 0= +
8 Equate the j
component of the equation. N 2 − 35g = 0
9 Solve for N 2. N 2 = 35g
10 Find the friction force F 2 acting on
CartL.F 2 = 0.2(35g)
= 7g
Tractor
25kg
CartT CartL
35kg
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86 Maths Quest 12 Specialist Mathematics
11 Equate the i components of the
equation of motion for CartL.
190 − F 2 − T = 35a
12 Substitute F 2 = 7g into the equations
and call it equation 2.
190 − 7g − T = 35a [2]
13 Use CAS to solve equations 1 and 2,
by completing the entry line as:solve(t − 5 × 9.8 = 25 × a and
190 − 7 × 9.8 − t = 35 × a, {t , a})then pressing ENTER·.
14 Write the solution. a = 1.2 m/s2, correct to 1 decimal place.
c Write the solution. c T = 79 N correct to the nearest whole number.
Therefore the tension in the rope is 79 N.
d 1 To find the maximum acceleration
substitute T = 150 into equation 1.Substitute T = 150 into equation 1:
150 − 5g = 25a
2 Solve for a. a = 4.04
3 State the solution. The maximum acceleration of the carts before the
rope breaks is 4.04 m/s2.
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