Maths Specialist CAS

MATHS QUEST 11 TI-NSPIRE CAS CALCULATOR COMPANION

Advanced General Mathematics

N O I T I D E D N 2

V C E M AT H E M AT I C S U N I T S 1 & 2

MATHS QUEST 11 TI-NSPIRE CAS CALCULATOR COMPANION

Advanced General Mathematics PAT R IC IC K S C O B L E MAR K DUNCA N

M A R K B A R NE NE S

T R A C EY H E R F T

R UT UT H B A K O G IA IA N I S

R OBYN WILLIAMS

KYLIE BOUCHER

J E N N I F E R N O LA N

G E O F F P H I LL I P S

N O I T I D E D N 2

First published 2013 by John Wiley & Sons Australia, Ltd 42 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 31771 6 978 1 118 31768 6 (flexisaver) Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL). Reproduction and communication for other purposes Except as permitted under the Act (for example, a fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher. Cover and internal design images: © vic&dd/Shutterstock.com

Typeset in India by Aptara Printed in Singapore by Craft Print International Ltd 10 9 8 7 6 5 4 3 2 1 Acknowledgements The authors and publisher would like to thank the following copyright holders, organisations and individuals for their permission to reproduce copyright material in this book. Images: Texas Te xas Instruments: Screenshots from TI-Nspire reproduced with permission of Te Texas xas Instruments

Every effort has been made to trace the ownership of copyright material. Information that will enable the publisher to rectify any error or omission in subsequent editions will be welcome. In such cases, please contact the Permissions Section of John Wiley & Sons Australia, Ltd.

Contents Introduction

vi

CHAPTER 10

Linear and non-linear graphs

CHAPTER 1

Number systems: real and complex

1

CHAPTER 11

Linear programming

CHAPTER 2

Transformations

7

Coordinate geometry 9

Vectors 13

67

CHAPTER 14

Statics of a particle

CHAPTER 5

Trigonometric ratios and their applications

23

Sequences and series

27

69

CHAPTER 15

Kinematics

CHAPTER 6

71

CHAPTER 16

Geometry in two and three dimensions

CHAPTER 7

Variation

65

CHAPTER 13

CHAPTER 4

Algebra

61

CHAPTER 12

CHAPTER 3

Relations and functions

51

37

CHAPTER 8

Further algebra

45

75

Introduction This booklet is designed as a companion to Maths Quest 11 Advanced General Mathematics Second Edition. It contains worked examples from the student text that have been re-worked using the TI-Nspire CX CAS 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 11 Advanced General Mathematics Second Edition.

vi

Introduction

CHAPTER 1

Number systems: real and complex WORKED EXAMPLE 4

Express each of the following in the form

 a 0.6

 b 0.23

 c 0.41

a b

, where a ∈ Z and b ∈ Z \{0}.

  d 2.1234

TH IN K

a

WR IT E

1

Write 0.6 in expanded form.

2

Multiply [1] by 10.

3

Subtract [1] from [2].

 0.6

a

b

c

d

  in .23

Write

2

Multiply [1] by 100.

3


4

State the simplest answer.

 in .41

Write

2

Multiply [1] by 10.

3


4


2 3


4


9 2

b

 0.23

=

3

0.232323 …

[1]

  = 23.232323 … 100 × 0.23

[2]

  = 23 99 × 0.23 = 0.23

c

the expanded form.

  in the expanded form. Write 2.1234 Multiply [1] by 1000.

1

6 =

=

the expanded form.

1

[2]

9 × 0.6 = 6


1

[1]

10 × 0.6 = 6.66666 …

 0.6 4

0.666666 …

=

 0.41

=

23 99

0.41111 …

[1]

10 × 0.41 = 4.11111 …

[2]

9 × 0.41 = 3.7

= 0.41

d

3.7

37 =

9

  2.1234

90

2.1234234 …

[1]

=

  = 2123.423423 … 1000 × 2.1234

[2]

  = 2121.3 999 × 2.1234

  2.1234

2121.3 =

999 21213

=

9999 2357

=

1111

CHAPTER 1

•


1

Note: The

CAS calculator can perform some of these calculations for you. On a Calculator page, press: MENU b 2: Number 2 2: Approximate to fraction 2 Complete the entry line as: 0.6666666666666666666666 ¢ approxFraction(5E-14) Then press ENTER ·. • • •

WORKED EXAMPLE 13

Expand and simplify the following where possible. a

7 ( 18

−

b

3)

−

2 3 ( 10

−

5 3)

c

TH IN K

a, 1 b

( 5

+

3 6 )(2 3

WR IT E

On a Calculator page, complete the entry lines as:

&

7 ( 18 − 3)

c

2 3( 10

−

−

5 3)

( 5 + 3 6) (2 3 − 2)

Press ENTER · after each entry. Note: The calculator must be set in either Real or Auto modes.

2

2

Write the answers.

Maths Quest 11 Advanced General Mathematics

a

3( 2 − 1) 7

b

30 − 2 30

c

(2 3 − 2) 5 − 6 3 + 18 2

−

2)

WORKED EXAMPLE 16

If z1 = 2 − 3i and z2 = −3 + 4i, find a z1

+ z2

b z1

− z2

c

3 z1 − 4 z2.

TH IN K

a,

1

b

WR IT E

On a Calculator page, press: MENU b 1: Actions 1 1: Define 1 Complete the entry lines as: Define z1 = 2 − 3i Define z2 = 3 + 4i z1 + z2 z1 − z2 3 z1 − 4 z2 Press ENTER · after each entry. •

&

•

c

•

2

Write the answers.

a

z1

+ z2 =

b

z1

− z2 =

c

−

1+i 5 − 7i 3 z1 − 4 z2 = 18 − 25i

WORKED EXAMPLE 17

Simplify a 2i(2 − 3i)

b

(2 − 3i)(−3 + 4i).

TH IN K

WR IT E

a

Expand the brackets.

a

2i(2 − 3i) = 4i − 6i2 = 6 + 4i

b

Expand the brackets as for binomial expansion and simplify.

b

(2 − 3i)(−3 + 4i) = −6 + 8i + 9i − 12i2 = 6 + 17i

Alternatively, on a Calculator page, complete the entry lines as: 2i × (2 − 3i) (2 − 3i) × (−3 + 4i) Press ENTER · after each entry.

CHAPTER 1

•


3

WORKED EXAMPLE 18

If z1 = 2 + 3i and z2 = −4 − 5i, find a z1

+

b z1

z2

+

z2

d z1 z 2 .

c z1 z 2

TH IN K

WR IT E

On a Calculator page, complete the entry lines as: Define z1 = 2 + 3i Define z2 = −4 − 5i conj( z1) + conj( z2) conj( z1 + z2) conj( z1) × conj( z2) conj( z1 × z2) Press ENTER · after each entry. Note: ‘conj’ can be typed directly onto the screen or can be found by pressing: MENU b 2: Number 2 9: Complex Number Tools 9 1: Complex Conjugate 1 • • • •

WORKED EXAMPLE 19

Express each of the following in a + bi form. a

4−i

b

2

3 −4i 3i

c

(3 − 2i)−1

TH IN K

WR IT E

b,


c

4−i

&

2

a,

d

3 − 4i 3i

(3 − 2i)

−

1

2 − 3i 2+i

Press ENTER · after each entry.

4

d


2 − 3i 2+i

WORKED EXAMPLE 21

Factorise each of the following quadratic expressions over C . 2 2 2 a 2 z + 6 z b 2 z − 6 c 2 z + 3 TH IN K

WR IT E

Ensure the calculator is set in Rectangular mode. On a Calculator & page, press: MENU b c 3: Algebra 3 C: Complex C 2: Factor 2 Complete the entry lines as: a, b

• • • •

cFactor(2 z2 + 6 z, z) cFactor(2 z2 − 6, z) cFactor(2 z2 + 3, z) Press ENTER · after each entry.

WORKED EXAMPLE 23

Solve the following using the formula for the solution of a quadratic equation. 2 2 a 2 z + 4 z + 5 = 0 b 2iz + 4 z − 5i = 0 TH IN K

WR IT E

Ensure the calculator is set in & Rectangular mode. On a Calculator b page, press: MENU b 3: Algebra 3 C: Complex C 1: Solve 1 Complete the entry lines as: a

• • • •

cSolve(2 z2 + 4 z + 5 = 0, z) cSolve(2iz2 + 4 z − 5i = 0, z) Press ENTER · after each entry.

CHAPTER 1

•


5

CHAPTER 2

Transformations WORKED EXAMPLE 4

Find the image rule for each of the following, given the original rule and translation. 2 a y = x, T −2, −3 b y = 2 x , T − 4, 5 c y = f ( x), T h, k TH IN K

a

b

WR IT E

1

State the image equations.

2

Find x and y in terms of x  and y.

3

Substitute into y = x .

y = x ⇒ y + 3 = x  + 2 ⇒ y = x  − 1

4

Express the answer without using the primes.

Given y = x under translation T −2, −3, the equation of the image (or image rule) is y = x − 1.

5

Note: The effect of the transformation can be illustrated on a CAS calculator. To do this, open a Graphs page. Complete the entry lines as: f 1( x ) = x f 2( x ) = x − 1 Press ENTER · after each entry.

a

x  = x − 2 y = y − 3 x = x  + 2 y = y + 3

1


2


x = x  + 4 y = y − 5

3

Substitute into y = 2 x 2.

y = 2 x 2 ⇒ y − 5 = 2( x  + 4)2 ⇒ y = 2( x  + 4)2 + 5

Express the answer without using the primes. Note: In first form of the answer, the turning point is ( −4, 5), which was the answer expected as (0, 0)  (−4, 5).

Given y = 2 x 2 under translation T −4, 5, the equation of the image (or image rule) is y = 2( x + 4)2 + 5 or y = 2 x 2 + 16 x + 37

4

b

x  = x − 4 y = y + 5

CHAPTER 2

•

Transformations

7

5

c

8

Note: The effect of the transformation can be illustrated on a CAS calculator. To do this, open a Graphs page. Complete the entry lines as: f 1( x ) = 2 x 2 f 2( x ) = 2( x + 4)2 + 5 Press ENTER · after each entry.

1


2


3

Substitute into y = f( x ).

y − k = f( x  − h) ⇒ y = f ( x  − h) + k

4

Express the answer without using the primes.

Given y = f ( x ) under translation T h, k , the equation of the image (or image rule) is y = f ( x − h) + k .


c

x  = x + h y = y + k x = x  − h y = y − k

CHAPTER 3

Relations and functions WORKED EXAMPLE 3

Find the range for the following functions. + a f : R → R, f ( x) = 4 x − 1 − 2 b f : R → R, f ( x) = x − 4 x + 5 x c f : R → R, f ( x) = 2 − 1 TH IN K

a

WR IT E

1

f ( x ) = 4 x − 1 is linear. The domain is x ∈ R+ or x ∈ (0, ∞).

2

f (0) = −1, but (0, −1) is not included and therefore this lower end of the range must be represented using a round bracket. State the range.

a

When x = 0, f (0) = 4(0) − 1 = −1 The range: y ∈ (−1, ∞).

−

b

1

− x 2

f ( x ) = − 4 x + 5 is an inverted parabola over the set of real −

numbers. Use x

=

b

b

x

−

= −

=

to

( 4)

−

2

2

2a determine the x -value of the turning point, as this can be used to indicate the maximum y-value of the graph.

c

2

Substitute this x -value into f ( x ) to determine the maximum y-value.

f (−2) = −4 + 8 + 5 = 9

3

State the range.

The range: y ∈ (−∞, 9].

1

f ( x ) = 2 x − 1 is an exponential graph over the set of real numbers. As x → −∞, 2 x → −1, y = −1 is an asymptote. Use a CAS calculator to draw this graph.

2

Use the graph and the information described above to state the range.

c

The range is: y ∈ (−1, ∞).

CHAPTER 3

•


9

WORKED EXAMPLE 4

If f : R → R, f ( x) = 2 x2 − 4 x + 1, find a f ( x2) b f (2 x + 1) TH IN K

WR IT E

To find f ( x 2), substitute x 2 for x and simplify.

a

b

To find f (2 x + 1), substitute 2 x + 1 for x and simplify.

a

f ( x ) = 2 x 2 − 4 x + 1 f ( x 2) = 2( x 2)2 − 4( x 2) + 1 = 2 x 4 − 4 x 2 + 1

b

f ( x ) = 2 x 2 − 4 x + 1 f (2 x + 1) = 2(2 x + 1)2 − 4(2 x + 1) + 1 = 2[4 x 2 + 4 x + 1] − 8 x − 4 + 1 = 8 x 2 + 8 x + 2 − 8 x − 4 + 1 = 8 x 2 − 1

Alternatively, on a Calculator page, complete the entry lines as: Define f ( x ) = 2 x 2 − 4 x + 1 f ( x 2) f (2 x + 1) Press ENTER · after each entry.

WORKED EXAMPLE 6

−

If f : (−∞, −1] → R, f ( x) = x2 + 2 x + 2, find the domain, range and rule of f 1( x), and sketch the − graphs of f and f 1 on the same set of axes. TH IN K

WR IT E −

1

First, determine if the inverse function, − f 1( x ), exists. Since f ( x ) = x 2 + 2 x + 2 is an upright parabola, it is necessary to −

locate the turning point using x 2

3

10

=

b

2a

.

Determine the range of f ( x ). The domain is x ∈ (−∞, −1], so substitute the end value of x to determine the range (this value is at the turning point). −

The domain of f ( x ) = the range of f 1( x ). − The range of f ( x ) = the domain of f 1( x ). − State the domain and range of f 1( x ).


x

=

2 2

=

−

1

Since the turning point occurs at x = −1, and the domain is x ∈ (−∞, −1], f ( x ) is a 1–1 function − and f 1( x ) exists. x = −1 ⇒ = 1 − 2 + 2 = 1 The point ( −1, 1) is the minimum point on the graph. f (−1)

Domain f ( x ): x ∈ (−∞, −1] − ⇒ range of f 1( x ) is y ∈ (−∞, −1] Range of f ( x ): y ∈ [1, ∞) − ⇒ Domain of f 1( x ) is [1, ∞)

4

−

To determine the rule of f 1( x ), let f ( x ) = y and interchange x and y. Then make y the subject.

Let y = x 2 + 2 x + 2 ⇒ y = ( x + 1)2 + 1 Interchange x and y ⇒ x = ( y + 1)2 + 1 ⇒ x − 1 = ( y + 1)2 x − 1 − 1

⇒ y = ± 5

Since f ( x ) = x 2 + 2 x + 2, and x ∈ (−∞, −1] − (left side of the parabola), then f 1 should x 1 1. be y Fully define the rule for the inverse function.

−

f 1: [1,

)

∞ →

R, f

−

1

( x) =

−

x − 1 − 1

−

=

6

−

−

Sketch the graphs over the required domains, showing the line y = x .

f ( x ) = x 2 + 2 x + 2

y 5 4 3 2 1

−5 −4 −3 −2 −− 1 10 −2 −3 −4 −5

7

y = x

1 2 3 4 5 x −

f 1( x ) = −1− √ x − 1

To view the graph and its inverse on a CAS calculator, open a new Graphs page. Complete the function entry lines as: f 1( x ) = x 2 + 2 x + 2 | x ≤ −1 f 2( x ) x 1 | x ≥ −1 1 f 3( x ) = x Press ENTER · after each entry. =

−

−

−

CHAPTER 3

•


11

CHAPTER 4

Algebra WORKED EXAMPLE 9

Transpose each of the following formulas to make the pronumerals indicated in brackets the subject. a

A =

4 3

r2 ( r)

π

b P

ab =

−

ac

d

( a)

c m

TH IN K

a

1

Write the equation.

2

Multiply both sides of the equation by 3.

3

Divide both sides by 4 π .

pq

=

a

A =

r2

π

3

2

3 A

4π r =

=

4π

4π r

2

r

3 A

Take the square root of both sides. Note: From an algebraic point of view we should write ± in front of the root. However, since r represents a physical quantity (radius of a sphere in this case), it can take only positive values. On a Calculator page, press: MENU b 3: Algebra 3 1: Solve 1 Complete the entry line as: solve( p = (a × b − a × c) ÷ d , a) Then press ENTER ·.

4

4

4π

1

( s)

3 × A = 3 π r2 × 3 3 A = 4π r2

3 A

b

rs

WR IT E

4π

4

−

=

2

r

3A =

4π

b

• • •

2

Write the answer.

a

dP =

b

−

c

Note: Capital P should be used in the answer.

CHAPTER 4

•

Algebra

13

c

1

Write the equation.

2

The inverse of x is x 2 so square both sides. Subtract pq from both sides.

3

4

Divide both sides by −r .

c

m

rs

m2 − pq = pq − rs − pq m2 − pq = −rs m

2

−

−

pq

rs

= −

r s

Multiply the numerator and denominator by −1 (optional).

−

m2 = pq − rs

−

5

pq

=

s

r

m

2

−

pq

= −

pq

r

−

m

2

=

r

WORKED EXAMPLE 12

Solve for x in 2( x + 5) = 3(2 x − 6). TH IN K 1

WR IT E

On a Calculator page, press: MENU b 3:Algebra 3 1:Solve 1 Complete the entry line as: solve(2( x + 5) = 3(2 x − 6), x ) Then press ENTER ·. • • •

2

14

Write the answer.


Solving 2( x + 5) = 3(2 x − 6) for x , gives x = 7.

WORKED EXAMPLE 13

Find the value of x that will make the following a true statement: TH IN K 1

x + 2 3

=

5−

x 2

.

WR IT E

On a Calculator page, press: MENU b 3:Algebra 3 1:Solve 1 Complete the entry line as: • • •

 x + 2 = 5 − x , x    3 2 

solve

Then press ENTER ·.

2

Write the answer.

Solving

x + 2 3

=

5−

x for x , gives x =

2

26 5

1

or 5 5.

WORKED EXAMPLE 14

Solve the following equation for x: TH IN K 1

2

x

3 +

2x

1 =

x−1

. WR IT E

On a Calculator page, press: MENU b 3:Algebra 3 1:Solve 1 Complete the entry line as: • • •

 2 + 3 = 1 , x   x 2 x x − 1  

solve


2

Write the answer.

Solving

2

x

3 +

2x

1 =

7

2

for x , gives x = or 1 5 .

x − 1

5

CHAPTER 4

•

Algebra

15

WORKED EXAMPLE 15

Solve the following pair of simultaneous equations graphically: a x + 2 y = 4 b y + 3 x = 17 2 x − 3 y = 4 x − y = 1 TH IN K

a

1

Rule up a set of axes. Label the origin and the x and y axes.

WR IT E

a

(See graph at step 7 on page 107.)

2

Find the x -intercept for the equation x + 2 y = 4, by making y = 0.

x -intercept: when y = 0, x + 2 y = 4 x + 2 × 0 = 4 x = 4 The x -intercept is at (4, 0).

3

Find the y-intercept for the equation x + 2 y = 4, by making x = 0. Divide both sides of the equation by 2.

y-intercept: when x = 0, x + 2 y = 4 0 + 2 y = 4 2 y = 4 y = 2 The y-intercept is at (0, 2).

Plot the points on graph paper and join them with the straight line. Label the graph.

(Refer to the graph at step 7.)

5

Find the x -intercept for the equation x – y = 1, by making y = 0.

6

Find the y-intercept for the equation x − y = 1, by making x = 0. Multiply both sides of the equation by −1.

x -intercept: when y = 0, x − y = 1 x − 0 = 1 x = 1 The x -intercept is at (1, 0). y-intercept: when x = 0, x − y = 1 0 − y = 1 − y = 1 − − y × 1 = 1 × −1 y = −1 The y-intercept is at (0, −1).

4

7

Plot the points on graph paper and join them with the straight line. Label the graph.

y x – y = 1

2 1 0 −1

16

x + 2 y = 4

(2, 1) 1

2

4

x

8

From the graph, read the coordinates of the point of intersection.

The point of intersection between the two graphs is (2, 1).

9

Verify the answer by substituting the point of intersection into the original equations.

Substitute x = 2 and y = 1 into x + 2 y = 4. LHS RHS = 4 = 2 + 2 × 1 = 2 + 2 = 4 LHS = RHS Substitute x = 2 and y = 1 into x − y = 1 LHS = 2 − 1 RHS = 1 = 1 LHS = RHS In both cases LHS = RHS; therefore, the solution set (2, 1) is correct.


b

1

Rearrange both equations to make y the subject. To do this, on a Calculator page, complete the entry lines as: solve( y + 3 x = 17, y) solve(2 x − 3 y = 4, y) Press ENTER · after each entry.

2

On a Graphs page, complete the function entry lines as: f 1( x )

b

2( x 2) −

=

3 f 2( x ) = 17 − 3 x Press ENTER · after each entry.

3

To find the point of intersection, complete the following steps. Press: MENU b 8:Geometry 8 1:Points & Lines 1 3:Intersection Point(s) 3. Click on each line. The point of intersection will appear. ENTER ·. • • • •

4

Write the answer.

The point of intersection between the two graphs is (5, 2).

CHAPTER 4

•

Algebra

17

WORKED EXAMPLE 18

Solve the following simultaneous equations. 2 x + 3 y = 4 3 x + 2 y = 10 TH IN K 1

WR IT E

On a Calculator page, press: MENU b 3:Algebra 3 1:Solve 1 Then complete the entry line as: solve(2 x + 3 y = 4 and 3 x + 2 y = 10, x ). Then press ENTER · Note: The term ‘and’ can be found in the catalogue k or can be typed. • • •

2

18

Answer the question.


Solution: x =

22 5

, y =

− 8

5

 22 − 8   . 5 

or  5 ,

WORKED EXAMPLE 21

A train (denoted as train 1) leaves station A and moves in the direction of station B with an average speed of 60 km/h. Half an hour later another train (denoted as train 2) leaves station A and moves in the direction of the first train with an average speed of 70 km/h. Find: a the time needed for the second train to catch up with the first train b the distance of both trains from station A at that time. TH IN K

WR IT E

Define the variables. Note: Since the first train left half an hour earlier, the time taken for it to reach the meeting point will be x + 0.5.

Let x = the time taken for train 2 to reach train 1. Therefore, the travelling time, t , for each train is: Train 1: t 1 = x + 0.5 Train 2: t 2 = x

2

Write the speed of each train.

Train 1: Train 2:

v1 = 60 v2 = 70

3

Write the distance travelled by each of the trains from station A to the point of the meeting. (Distance = speed × time.)

Train 1: Train 2:

d 1 = 60( x + 0.5) d 2 = 70 x

4

Equate the two expressions for distance. Note: When the second train catches up with the first train, they are the same distance from station A — that is, d 1 = d 2.

When the second train catches up w ith the first train, d 1 = d 2.

5

Solve the equation. On a Calculator page, press: MENU b 3:Algebra 3 1:Solve 1 Then complete the entry line as: solve(60( x + 0.5) = 70 x ,x ). Then press ENTER ·.

1

• • •

6

Substitute 3 in place of x into either of the two expressions for distance, say into d 2.

7

Evaluate.

8

Answer the questions.

Substitute x = 3 into d 2 = 70 x d 2 = 70 × 3 = 210

a b

The second train will catch up with the first train 3 hours after leaving station A. Both trains will be 210 km from station A.

CHAPTER 4

•

Algebra

19

WORKED EXAMPLE 23

Two hamburgers and a packet of chips cost $8.20, while one hamburger and two packets of chips cost $5.90. Find the cost of a packet of chips and a hamburger. TH IN K 1

Define the two variables.

WR IT E

Let x = the cost of one hamburger. Let y = the cost of a packet of chips.

2

Formulate an equation from the first sentence and call it [1]. Note: One hamburger costs $ x , two hamburgers cost $2 x . Thus, the total cost of cost of two hamburgers and one packet of chips is 2 x + y and it is equal to $8.20.

2 x + y = 8.20

[1]

3

Formulate an equation from the second sentence and call it [2]. Note: One packet of chips costs $ y, two packets cost $2 y. Thus, the total cost of two packets of chips and one hamburger is x + 2 y and it is equal to $5.90.

x + 2 y = 5.90

[2]

4

Solve for the simultaneous equations. On a Calculator page, press: MENU b 3:Algebra 3 1:Solve 1 Then complete the entry line as: solve( x + 2 y = 5.90 and 2 x + y = 8.20, x ). Then press ENTER ·. • • •

5

20

Answer the question and include appropriate units.


A hamburger costs $3.50 and a packet of chips costs $1.20.

WORKED EXAMPLE 25

Simplify 3

a

2

x

−

b

x−1

2 a x +

2

3

− x−

3

.

TH IN K 1

a

& b

WR IT E

On a Calculator page, press: MENU b 2:Number 2 7:Fraction Tools 7 4:Common Denominator 4 Complete the entry lines as:

a

•

&

•

b

• •

 3 − 2   x x − 1 

comDenom 

 2 − 2   + 3 − 3  

comDenom 

a

x

x


2

Write the answer.

a

3

x b

x

2 −

x

−

−

=

1

2a x +

x

2

−

2

3

− x −

3

x

2ax 3

=

−

2x 2

x

− 6a − 6

−

9

WORKED EXAMPLE 26

Simplify a

3 x 2 4

20 ×

b

9 x

x

2

+

6 y

2

4 ×

2 y 5 x

2

+

20

.

TH IN K

a

1

& b

WR IT E


a

3 x 2

b

4

20 ×

9 x

2

x + 4

6 y

2

&

×

2 y 2

5 x

+

20


2

Write the answers.

a

3 x 2 4

20 ×

9 x

=

2

b

x + 4

6 y

2

×

5 x 3 2 y

5 x

2

+

1 20

=

15 y

CHAPTER 4

•

Algebra

21

CHAPTER 5

Trigonometric ratios and their applications WORKED EXAMPLE 1

Determine the value of the pronumerals, correct to 2 decimal places. a

b 4

7

x

24° 25

50°

h

TH IN K

a

1

Label the sides, relative to the marked angles.

WR IT E

a 4

x

H

O

2

Write what is given.

Have: angle and hypotenuse ( H )

3

Write what is needed.

Need: opposite ( O) side

4

Determine which of the trigonometric ratios is required, using SOH–CAH–TOA.

5

b

50°

sin (θ) =

sin (50°) =

Substitute the given values into the appropriate ratio.

6

Transpose the equation and solve for x .

7

Round the answer to 2 decimal places.

1

Label the sides, relative to the marked angle.

O H

x

4

4 × sin (50°) = x x = 4 × sin (50°)

= 3.06 b

A 7 24° 25 h H

2


Have: angle and adjacent ( A) side

3


Need: hypotenuse ( H )

4

Determine which of the trigonometric ratios is required, using SOH–CAH–TOA.

5


cos (θ) =

cos (24°25′ ) =

A H

7 h

CHAPTER 5

•


23

6

Solve for h. On a Calculator page, complete the entry line as: 7   solve cos (24 25′ ) = ,h





h




7

Round the answer to 2 decimal places.

h ≈ 7.69

WORKED EXAMPLE 2

Find the angle θ , giving the answer in degrees and minutes. θ 12 18

TH IN K 1

WR IT E

Label the sides, relative to the marked angles.

θ

A 12

O

18

2


3


4

Determine which of the trigonometric ratios is required, using SOH –CAH–TOA.

tan (θ) =


tan (θ° ) =

5

6

Have: opposite ( O) and adjacent ( A) sides Need: angle O A

18 12

Transpose the equation and solve for θ , using the inverse tan function. − To calculate tan 1, on a Calculator page, complete the entry line as:

 tan−  18   ¢   12    DMS 1

Then press ENTER ·. Note: ¢ DMS is located in the catalogue.

7

Write the answer to the nearest minute.

θ°

= tan

−1

 18    12 

= 56° 19′

24


WORKED EXAMPLE 8

In the triangle ABC, a = 10 m, c = 6 m and C = 30°. a Show that the ambiguous case exists. b Find two possible values of A, and hence two possible values of B and b. TH IN K

a

1

&

WR IT E

Draw a labelled diagram of the triangle ABC and fill in the given information.

b

B c = 6

A 2

a = 10

30°

C

In part a it was shown that the ambiguous case of the sine rule exists. Therefore, on a Calculator page, complete the entry line as:

 10 = 6 , a | 0 ≤ a ≤ 180  sin (a) sin (30)  

solve 


3

Convert the angles to degrees and minutes.

4

Calculate the size of the angle B given each angle A.

5

To find the side length b, on a Calculator page, complete the entry lines as: solve  

A

= 56°27′ or A = 123°33′

If A = 56°27′,

= 180 − (30 + 56°27′) = 93°33′ If A = 123°33′, B = 180 − (30 + 123°33′) = 26°27′ B

  sin (93 33′) sin (30)  b 6   = ,b solve   sin (26 27′ ) sin (30)   b



=

6

, b 




6

Write the answers.

If B = 93°33′, b = 11.98 m If B = 26°27′, b = 5.35 m

CHAPTER 5

•


25

WORKED EXAMPLE 9

Find the third side of triangle ABC given a = 6, c = 10 and B = 76°, correct to 2 decimal places. TH IN K 1

WR IT E


B c

=

10

76°

A

2

Write the appropriate cosine rule to find side b.

3

On a Calculator page, complete the entry line as: solve(b2 = 62 + 102 − 2 × 6 × 10 × cos (76),b) Then press ENTER ·.

4

Since b represents the side length of a triangle, then b > 0.

2

b

a

=

6

C

b

= a2 + c2 − 2ac cos ( B)

b = 10.34, correct to 2 decimal places.

WORKED EXAMPLE 10

Find the smallest angle in the triangle with sides 4 cm, 7 cm and 9 cm. TH IN K 1

WR IT E B


c

A

26

2

Write the appropriate cosine rule to find the angle A.

3

On a Calculator page, complete the entry line as: solve(4 2 = 92 + 72 − 4 × 9 × 7 × cos (a),a) |0 ≤ a ≤ 180 Then press ENTER ·.

4

Round the answer to degrees and minutes.


2

a

A

=

7

a

b

=

9

=

4

C

= b2 + c2 − 2bc cos ( A)

= 25.2088 ° = 25°13′

CHAPTER 6

Sequences and series WORKED EXAMPLE 1

a b c

7

Find the next three terms in the sequence, b: {14, 7, 2 , . . .}. Find the 4th, 8th and 12th terms in the following sequence: e n = n2 − 3 n, n ∈ {1, 2, 3, . . .}. Find the 2nd, 3rd and 5th terms for the following sequence: k n + 1 = 2 k n + 1, k1 = −0.50.

TH IN K

a

1

WR IT E

In this example the sequence is listed and a simple pattern is evident. From inspection, the next term is half the previous term and so the sequence 7

7

7

7

7

7

a

The next three terms are 4 , 8 , 16 .

b

en = n − 3n

7

would be 14, 7, 2, 4 , 8 , 16.

b

2

On a Calculator page, complete the entry lines as: 14 Ans × 0.5 Press ENTER · repeatedly to generate the sequence.

1

This is an example of a functional definition. The nth term of the sequence is found simply by substitution into the expression 2 en = n − 3n.

2

Find the 4th term by substituting n = 4.

e4 = 4 − 3 × 4


e8 = 8 − 3 × 8


e12 = 12 − 3 × 12

3

4

2

2

= 4 2

= 40 2

= 108

CHAPTER 6

•


27

5

On a Calculator page, press MENU b 6: Statistics 6 4: List Operations 4 5: Sequence 5 Complete the entry line as: seq(e(n) = n2 − 3n,n,4,12,4) Then press ENTER ·. Note: The first 4 and the 12 are the highest and lowest values required, and the last 4 is the step value. • • • •

An alternative method to the one above for generating a sequence is shown below. Insert a new Lists and Spreadsheets page. Go to the title cell of column A and press: MENU b 3: Data 3 1: Generate Sequence 1 • • •

Complete the sequence fields as shown. Then press OK.

c

6

Scroll down to find the required terms.

1

This is an example of an iterative definition. We can find the 2nd, 3rd and 5th terms for the sequence k n + 1 = 2k n +1, k 1 = −0.50 by iteration.

2

3

Substitute to find k 2.

k 1 =

−

0.50 into the formula

Continue the process until the value of k 5 is found.

c

k n + 1 = 2k n + 1, k 1 =

−

0.50

k 2 = 2 × −0.50 + 1 = 0

k 3 = 2 × 0 + 1 = 1

k 4 = 2 × 1 + 1 = 3

k 5 = 2 × 3 + 1 = 7

28


4

Write the answer.

5

Follow the same steps as for the alternative method in part b; however, because the formula is defined as u(n), the rule needs to be entered as: u(n) = 2 × u(n − 1) + 1 with the initial term as: − 0.5 Press OK to view the sequence.

Thus k 2 = 0, k 3 = 1 and k 5 = 7.

CHAPTER 6

•


29

WORKED EXAMPLE 2

Given that a = 2 and t0 = 0.7, use the logistic equation to generate a sequence of 6 terms, and state whether the sequence is convergent, divergent or oscillating. If the sequence is convergent, state its limit. TH IN K 1

Insert a new Lists & Spreadsheet page. Go to the title cell of column A and press: MENU b 3: Data 3 1: Generate Sequence 1 Complete the sequence fields as shown. Then press OK. • • •

2

30

Scroll down to find the required terms. Note: Here the initial term corresponds to cell A1; however, we need to remember that this is actually t 0. Hence it is the sixth term when the sequence first converges to 0.5, not the fifth term.


WR IT E

WORKED EXAMPLE 6

Find the 16th and nth terms in an arithmetic sequence with the 4th term 15 and 8th term 37. TH IN K 1

Write the two equations that represent t 4 and t 8.

2

To solve equations [1] and [2] simultaneously, open a new Calculator page and press: MENU b 3: Algebra 3 1: Solve 1 Complete the entry line as: solve(a + 3d = 15 and a + 7d = 37,a) Then press ENTER ·.

WR IT E

t 4: a + 3d =

15 t 8: a + 7d = 37

[1] [2]

• • •

−

3

Write the answer.

If a t n

=

3 2

and

11n =

−

d

11 =

2

,

14

2

t 16 = 81

CHAPTER 6

•


31

WORKED EXAMPLE 7

Find the sum of the first 20 terms in the sequence t n: {12, 25, 38, . . .}. TH IN K 1

WR IT E

On a Lists & Spreadsheet page, place the cursor in the grey header cell of column A. Then press: MENU b 3: Data 3 1: Generate sequence 1 Complete the sequence fields as shown. • • •

32

2

Scroll down to the 20th term.

3

Write the answer.


S 20 = 2710

WORKED EXAMPLE 10

The fifth term in a geometric sequence is 14 and the seventh term is 0.56. Find the common ratio, r, the first term, a, and the nth term for the sequence. TH IN K

WR IT E 1

1

Write the general rule for the nth term of the geometric sequence.

t n = ar n −

2

Use the information about the 5th term to form an equation. Label it [1].

When n = 5, t n = 14 14 = a × r 5 − 1 14 = a × r 4

[1]

When n = 7, t n = 0.56 0.56 = a × r 7 − 1 0.56 = a × r 6

[2]

3

4

5

Similarly, use information about the 7th term to form an equation. Label it [2].

Solve equations simultaneously: Divide equation [2] by equation [1] to eliminate a. Solve for r .

[2] [1]] [1

6

gives

0.56

ar

4

=

ar

14

2

r = 0.04 r = ±

0.04

= ± 0.2 6

As there are two solutions, we must perform two sets of computations. Consider the positive value of r first. first. Substitute the value of r into into either of the two equations, e.g. [1], and solve for a.

If r = 0.2 Substitute r into into [1]: 4 a × (0.2) = 14 0.0016 a = 14 a = 8750

7

Substitute the values of r and and a into the general equation to find the expression for the nth term.

The nth term is: 1 t n = 8750 × (0.2)n −

8

Now consider the negative value of r .

If r = −0.2

9

Substitute the value of r into into either of the two equations, e.g. [1], and solve for a. (Note that the value of a is the same for both values of r .) .)

Substitute r into into [1] 4 − a = ( 0.2) = 14 0.0016 a = 14 a = 8750

10

Substitute the values of r and and a into the general formula to find the second expression for the nth term.

The nth term is: 1 t n = 8750 × (−0.2)n −

11

Write the two equations that represent t 5 and t 7.

t 5:

14 = a × r 4 6 t 7: 0.56 = a × r

[1] [2]

CHAPTER 6

•


33

12

To solve equations [1] and [2] simultaneously, open a new Calculator page and press: MENU b 3: Algebra 3 1: Solve 1 Complete the entry line as: solve(14 = a × r 4 and 0.56 = a × r 6,r ) Then press ENTER ·. • • •

13

Write the answer.

When r = −0.2 and a = 8750, 1 t n = 8750 × (−0.2)n − When r = 0.2 and a = 8750, 1 t n = 8750 × (0.2) n −

34


WORKED EXAMPLE 11

Find the sum of the first 5 terms ( S5) of these geometric sequences. a t n:

{1, 4, 16, . . .}

1 b t n = 2(2) n − , n ∈ {1,

TH IN K

a

1

On a Lists & Spreadsheet page, place the cursor in the grey header cell of column A. Then press: MENU b 3: Data 3 1: Generate sequence 1 Complete the sequence fields as shown.

2, 3, . . .}

c t n + 1 =

1

, t1 =

t 4 n

−

1 2

WR IT E

a

• • •

2

To sum the terms of any sequence, place the cursor in cell B1 and type =

3

Then press Catalogue and scroll down to sum(. Then press ENTER ·.

CHAPTER 6

•


35

b

4

To sum the first 5 terms, complete the entry line as: sum(a1:a5) Then press ENTER ·.

5

The answer will appear in cell B1.

6

Write the answer.

If t n: {1, 4, 16, . . .} then,

1

Write the sequence.

2

Compare the given rule with the general formula for the nth term of the geometric sequence t n = ar n − 1 and identify values of a and r ; the value of n is known from the question.

a = 2; r = 2; n = 5

3

Substitute values of a, r and n into the general formula for the sum and evaluate.

S 5

b

t n = 2(2)

2 (25 =

1 2

Write the sequence. This is an iterative formula, so the coefficient of t n is our r ; a = t 1; n is known from the question.

−

−

c

1 62

t n + 1 = r =

1 4

1 4

t n, t 1 =

;a=

−

1 2

− 1  1    2  4 

5

3

Substitute values of a, r and n into the general formula for the sum and evaluate.

S 5

=

−1

=


1 4

=

36

1)

2 1 2(32 1)

=

2

−

1 2

; n = 5

− 1 

−1

1 ×   1024 − 1  

−3

− 341 512

S 5 = 341.

, n ∈ {1, 2, 3, . . .}

−

=

c

n − 1

4

CHAPTER 7

Variation WORKED EXAMPLE 1

For the given data, establish whether direct variation exists between x and y using: a a numerical approach (clearly specify k, the constant of variation, if applicable) and b a graphical approach. c Confirm your result using a CAS calculator. x

4

7

8

10

y

5.2

9.1

10.4

13

y x TH IN K

a

1

Find the ratio

WR IT E/ DR AW y x

for each of the

a

Ratio =

y x

5.2

four pairs of values. One variable varies directly as the other if the ratio between any two corresponding values is constant.

First pair:

4

7 10.4 8 13

Fourth pair: Compare each of the four ratios and answer the question.

3

Copy and complete the table.

2

Plot the information from the table onto a set of axes. Join the given points and see if a straight line is obtained. Note: For a direct variation to exist between two variables x and y, a straight line passing through the origin (0, 0) must be obtained.

1.3

=

1.3

10

=

1.3

x

4

7

8

10

y

5.2

9.1

10.4

13

1.3

1.3

1.3

1.3

x 1

=

Since all four ratios are the same (that is, 1.3), y varies directly as x .

y

b

1.3

9.1

Second pair: Third pair:

2

=

b

y

13 10.4 9.1 5.2

0

4

7 8 10

x

The given points fit perfectly on a straight line. If the straight line is extended it will pass through the origin. Therefore y varies directly as x .

CHAPTER 7

•

Variation

37

3

Calculate the gradient using any two points on the straight line. Answer the question. Note: The gradient of the straight line will equal k , the constant of variation, if direct variation exists between the variables x and y.

Let ( x 1, y1) = (4, 5.2) and let ( x 2, y2) = (10, 13) m

=

y2

−

y1

x 2

−

x 1

13 =

−

10

5.2

−

4

7.8 =

6

1.3 The gradient of the straight line is equal to k , the constant of variation. =

c

1

On a Lists & Spreadsheet page enter the data into the spreadsheet. Label the columns x and y.

2

To draw the scatterplot of the data, on a Data & Statistics page: Tab to each axis to select ‘Click to add variable’. Place x on the horizontal axis and y on the vertical axis.

3

To check there is a linear relationship, press: • MENU b • 4: Analyze 4 • 6: Regression 6 • 1: Show linear ( mx + b) 1 In X List select x ; in Y List select y.

c

Interpret the graph; if all points are on the line, then it confirms that the relationship is one of direct variation.

The relationship is one of direct variation.

38


WORKED EXAMPLE 5

For the given data, establish the rule relating the variables x and y then graph the relationship using a CAS calculator. x

3

6

7

10

y

28.8

115.2

156.8

320

TH IN K 1

On a Calculator page, complete the entry lines as: {3,6,7,10} → x {28.8,115.2,156.8,320} → y Press ENTER · after each entry.

2

Find the rule that fits the data. To do this press: MENU b • 6: Statistics 6 • 1: Stat Calculations 1 • 9: Power Regression 9 Enter x in X List and y in Y List Then press ENTER ·.

3

Write down the rule for x and y.

4

WR IT E/ DI SP LAY

y = 3.2 x 2

Graph the rule. On a Graphs page, use the NavPad and the up arrow ` to display f 1( x ), which has already been defined in this document. Then press ENTER · to display the graph.

CHAPTER 7

•

Variation

39

WORKED EXAMPLE 8

For the data represented in the table below, establish whether an inverse variation exists between x and y using: a a numerical approach (clearly specify k, the constant of variation, if applicable) b a graphical approach. x

1

2

3

y

20

10

63

2

4

5

5

4

xy TH IN K

a

b

1

WR IT E/ DI SP LAY

Find the product of xy for each of the 5 pairs of values. Note: One variable varies inversely as the other if the product between any 2 corresponding values is constant.

2

Compare each of the five products and answer the question.

3

Copy and complete the table.

1

Calculate the values of

1

.

x

Place these values into a table.

a

Product = xy First pair: Second pair: Third pair: Fourth pair: Fifth pair:

1

1

1

1

{1, 2 , 3 , 4 , 5 } → x 2

{20, 10, 6 3 , 5, 4} → y Press ENTER · after each entry.

40


=

b

x

1

2

y

20

10

6

xy

20

20

x

1 1

1 y

1 On a Calculator page, enter the and y values by completing the x entry lines as:

20 20 = 20 = 20 = 20 =

Since the product of the corresponding values is the same in each case (that is 20), y varies inversely as x .

x

2

1 × 20 2 × 10 2 3 ×63 4 × 5 5 × 4

20

4

5

5

4

20

20

20

2

3

4

5

1

1

1

1

2

3

4

5

5

4

10

3

6

2 3

2 3

3

To find the rule that fits the data, press: • MENU b • 6: Statistics 6 • 1: Stat Calculations 1 • 3: Linear Regression (mx + b) 3 Enter x in X List and y in Y List Then press ENTER ·.

4

To graph the rule, open a Graphs page. Complete the entry line as f 2( x ) = f 1( x ) Then press ENTER · to view the graph. Adjust the window if necessary.

5

Answer the question.

The graph of y against

1

is a straight line

x

directed from, but not passing through, the origin, hence an open circle at the point (0, 0). 1 Therefore y ∝ . x

CHAPTER 7

•

Variation

41

WORKED EXAMPLE 16

My telephone bill consists of 2 parts: a fixed charge of $32 (paid whether any calls are made or not) and a charge proportional to the number of calls made. Last quarter I made 296 calls and my bill was $106. a Find the equation of variation. b Find the amount to be paid when 300 calls are made. TH IN K

a

1

Define each variable to be used.

2

Write the equation of variation.

3

Substitute the values for A and n into the equation. To solve the equation for k, on a Calculator page, press: • MENU b • 3: Algebra 3 • 1: Solve 1 Complete the entry line as: solve(106 = 296 × k + 32,k ) Then press ENTER ·.

4

b

WR IT E/ DI SP LAY

a

Let A = the total amount to be paid, in dollars Let n = the number of calls A = k n + 32

When n = 296 and A = 106, 106 = 296k + 32

1

5

Rewrite the equation substituting 4 in place of k .

1

Substitute n = 300 into the given equation.

2

Evaluate.

So b

1

A = 4 n + 32

When n = 300, 1 A = 4 (300) + 32 =

1 4

300 + 32

×

75 + 32 = 107 =

3

42

Answer the question and include the appropriate unit.


The amount to be paid when 300 calls are made is $107.

WORKED EXAMPLE 18

The following table shows the values of the total surface area, TSA, of spheres and their corresponding radii, r. Radius ( r) (cm) TSA (cm2)

1

2

3

4

5

12.57

50.27

113.1

201.06

314.16

Graph the values given in the table and comment on the shape of the graph. Using the graph, or otherwise, find the equation which relates total surface area of the sphere, TSA, and its radius r. TH IN K 1

On a Lists & Spreadsheet page, enter values and label in column A: radius: 1, 2, 3, 4, 5 and in column B: tsa 12.57, 50.27, 113.1, 201.06, 314.16

2

Draw the scatterplot of the data on a Data & Statistics page.

WR IT E/ DI SP LAY

Tab to each axis to select ‘Click to add variable’. Place radius on the horizontal axis and tsa on the vertical axis. To check the power relationship, press: • MENU b • 4: Analyse 4 • 6: Regression 6 • 7: Show power 7 In X List select x ; in Y List select y 3

Comment on the graph obtained.

The graph is not a straight line, passing through the origin, so direct variation does not exist between the two variables. Hence, there is no direct variation between the radius and the total surface area of the sphere.

4

Make assumptions about the graph obtained.

The graph resembles a parabola, so it is reasonable to assume that area is directly proportional to the square of the radius.

5

Write the variation statement for the assumption made.

TSA ∝ r 2

6

Write the variation equation.

TSA = kr 2

7

Transpose the equation to make k the subject.

TSA

k =

r 2

CHAPTER 7

•

Variation

43

8

Test the assumption by finding the values of r 2 and check that the ratio

TSA

r 2

is

constant. On the Lists & Spreadsheet page, label column C: ratio In the header (grey) cell, complete the entry line as: =tsa/(radius)^2 Then press ENTER ·.

9

10

Comment on the result obtained.

The ratio is constant for each corresponding pair (when rounded to 2 decimal places). Hence, TSA ∝ r 2 TSA = kr 2 TSA = 12.57r 2

Alternatively, once the values of r 2 have been calculated, rule up a table of values and plot TSA versus r 2.

r2

TSA

1

4

9

16

25

12.57

50.27

113.1

201.06

314.16

TSA 300 200 100 0 1 4 9 16 25

44

2

r

11

Comment on the graph obtained.

The graph is a straight line, passing through the origin.

12

Establish the value of k by substituting any pair of values from the table into the equation of variation and write the equation relating the two variables.

TSA = kr 2 When r = 1, TSA = 12.57, k = ? 12.57 = k × 1 12.57 = k k = 12.57 TSA = 12.57r 2


CHAPTER 8

Further algebra WORKED EXAMPLE 2

If 5 x3 + 2 x2 − 7 x + 1 = (2 a + b) x3 − ax2 − ( b − c) x + 1, then find the values of a, b and c. TH IN K 1

WR IT E

On a Calculator page, press: • MENU b • 3: Algebra 3 • 1: Solve 1 Complete the entry line as:

 2a + b = 5  − a = 2, a  solve       b − c = 7   Then press ENTER ·. Note: The template used here can be found in the Maths expression template.

2

Write the answer.

a = −2, b = 9 and c = 2

WORKED EXAMPLE 4

If x − 4 is a factor of x3 − 6 x2 + 2 x + 24, find the other factor. TH IN K 1

On a Calculator page, press: • MENU b • 3: Algebra 3 • 2: Factor 2 Complete the entry line as: factor( x 3 − 6 x 2 + 2 x + 24) Then press ENTER ·.

2

Write the answer.

WR IT E

The quadratic factor of x 3 − 6 x 2 + 2 x + 24 is x 2 − 2 x − 6.

CHAPTER 8

•

Further algebra

45

WORKED EXAMPLE 5

Express

4 x + 5

x − 3

in the form A +

b x − 3

.

TH IN K 1

WR IT E

On a Calculator page, press: • MENU b • 3: Algebra 3 • 9: Fraction Tools 9 • 1: Proper Fraction 1 Complete the entry line as:

 4 x + 5   x − 3  

propFrac 


2

Write the answer in the form A +

b x − 3

4 x + 5

x − 3

=

4+

17

x − 3

.

WORKED EXAMPLE 6

Express

x + 3 x 2

−

3 x − 40

in partial fraction form.

TH IN K 1

WR IT E

On a Calculator page, press: • MENU b • 3: Algebra 3 • 3: Expand 3 Complete the entry line as: x + 3  expand 2  x − 3 x − 40  Then press ENTER ·.

2

Write the answer in the form A

( x − 8)

46

B +

( x + 5)

.


x + 3 x

2

−

3 x − 40

x ∈ R \{−5, 8}

=

11 13( x − 8)

+

2 13( x + 5)

,

WORKED EXAMPLE 7

Express

2 x − 1 ( x − 2)( x + 1) 2

in partial fractions.

TH IN K 1

WR IT E

On a Calculator page, press: • MENU b • 3: Algebra 3 • 3: Expand 3 Complete the entry line as:

 2 x − 1   ( x − 2)( x + 1) 2  

expand 


2

Write the answer in the form A

B

( x − 2)

+

C

( x + 1)

+

( x + 1) 2

.

2 x − 1 ( x − 2)( x + 1) 2

=

1 3( x − 2)

−

1 3( x + 1)

+

1 ( x + 1) 2

,

x ∈ R \{−1, 2}

WORKED EXAMPLE 8

Express

5 x 2

+

x

3

9 x + 10 −

8

in partial fractions.

TH IN K 1

WR IT E


 5 x 2 + 9 x + 10   x 3 − 8  

expand 


2

Write the answer in the form A x − 2

+

Bx + C x

2

+

5 x 2

+

x

3

9 x + 10 −

8

=

4 x −

2

+

x + 3 x

2

+

2 x + 4

, x ∈ R \{2}

.

2 x + 4

CHAPTER 8

•

Further algebra

47

WORKED EXAMPLE 9

Express

x 2

+

5x − 2

x − 1

as a partial fraction.

TH IN K 1

WR IT E


 x 2 + 5x − 2   x − 1  

expand 


2

Write the answer.

x 2

+

5 x − 2

x − 1

=

4

x − 1

x + 6, x ∈ R\ {1}

+

WORKED EXAMPLE 10

Solve simultaneously: y = x and y = x2 + 3 x + 1. TH IN K 1

WR IT E

On a Calculator page, press: • MENU b • 3: Algebra 3 • 1: Solve 1 Complete the entry line as: solve( y = x and y = x 2 + 3 x + 1, x ) Then press ENTER ·.

2

48

Write the answer.


Solving y = x and y = x 2 + 3 x + 1 for x and y gives x = −1 and y = −1. That is, (−1, −1).

WORKED EXAMPLE 11

Solve simultaneously: y = x + 1 and x2 + y2 = 4. TH IN K 1

WR IT E

On a Calculator page, press: • MENU b • 3: Algebra 3 • 1: Solve 1 Complete the entry line as: solve( y = x + 1 and x 2 + y2 = 4, x ) Then press ENTER ·.

2

Write the answer.

Solving y = x + 1 and x 2 + y2 = 4 for x and y gives −

x = x

( 7 + 1) 2 7

=

−

2

1

−

and y

( 7

=

and y =

−

1)

2 7

or

+1

2

That is,

 − ( 

7 + 1) − ( 7 − 1) , or  2 2 

 

7

−1

2

,

+ 1 2   .

7

CHAPTER 8

•

Further algebra

49

WORKED EXAMPLE 12

Solve simultaneously: y = 2 x − 1 and

2 y

=

x TH IN K 1

−

. 3 WR IT E

On a Calculator page, press: • MENU b • 3: Algebra 3 • 1: Solve 1 Complete the entry line as:

 

solve y = 2 x − 1 and y =

2 x

−3

 

, x 


2

Write the answer.

Solving y

=

2x

−

1 and y

2 =

x −

x =

or x

=

( 41 − 7) and y = 4 41 + 7 4

and y =

−

−

3

for x and y gives

( 41 − 5) 4

41 + 5 4

That is,

 − ( 

50


41 − 7) − ( 41 − 5)  , or  4 4 

 

41 + 7 4

,

41 + 5  .  4 

CHAPTER 10

Linear and non-linear graphs WORKED EXAMPLE 11

Convert [2,

2π 3

] to Cartesian coordinates.

TH IN K 1

Find the x -coordinate.

WR IT E

x = r cos (θ )

 2π    3 

= 2 cos

=2×− x = −1 2

Find the y-coordinate.

1 2

y = r sin (θ ) = 2 sin 

2π

 3

= 2 ×

2π [2, — ] y

  

3

3 2

2π — 3

3 x

2

y = 3 3

State the Cartesian coordinates.

4

Alternatively, open a Calculator page. Make sure the setting is on AUTO. Complete the entry lines as:

Hence, the Cartesian coordinates are (−1, 3).

 2π   3   2π  P ¢ Ry   2, 3   P ¢ Rx  2,

Press ENTER · after each entry. Note: P ¢ Rx and P ¢ Ry can be found in the catalogue.

CHAPTER 10

•


51

WORKED EXAMPLE 16

Sketch the graph of r = θ for 0 < θ < 4π using a CAS calculator. TH IN K

WR IT E

Open a new Graphs page. Ensure the angle setting is on radians. Then press • MENU b • 3: Graph Entry/Edit 3 • 4: Polar 4 Complete the entry line as:

 r 1(θ ) = θ  ≤ ≤  0 θ 4π θ step = 0.13 Then press ENTER ·. The increment step is

π

(0.13) because this was 6 used in the introductory table. Compare the graphs of worked examples 15 and 16.

52


WORKED EXAMPLE 17

Sketch the graph of r = 8 for 0 ≤ θ ≤ 2π . TH IN K 1

2

WR IT E/ DI SP LAY

Construct a table of values for 0 ≤ θ ≤ 2π and find the corresponding r values to 2 decimal places.

θ

0

r

8

θ

π

r

8

Sketch the graph using a protractor and ruler to plot each of the points from the table. Remember r is the distance from the centre (the origin).

π

π

π

2π

5π

6 8

3

2

3

6

8

8

8

8

7π

4π

3π

5π

11π

6

3

2

3

6

8

8

8

8

8

2π 8

π

— 2

8

π

−8

8

−8 3π — 2

3

On a Graphs page, press • MENU b • 3: Graph Entry/Edit 3 • 4: Polar 4 Complete the entry line as:

 r 1(θ ) = 8   0 ≤ θ ≤ 2π θ step = 0.13 Then press ENTER ·. Zoom out to see the whole circle.

CHAPTER 10

•


53

WORKED EXAMPLE 18

Sketch the graph of r = 2 sin (θ ) for 0 ≤ θ ≤ 2π using a calculator. TH IN K

WR IT E

On a Graphs page, press • MENU b • 3: Graph Entry/Edit 3 • 4: Polar 4 Complete the entry line as:

 r 1(θ) = 2 sin (θ )   0 ≤ θ ≤ 2π θ step = π  12  Then press ENTER ·. Zoom out to see the whole circle.

WORKED EXAMPLE 21

Express the following complex numbers in polar form. a

z = 1 + i

b z

=

−

3+i

TH IN K

a

1

& b

WR IT E

Open a Calculator page. Enter the real and imaginary parts using the [ ] parentheses. Complete the entry lines as: [1,1] ¢ Polar  3,1  ¢ Polar



a

& b



Press ENTER · after each entry. Note: ¢ Polar can be found in the catalogue.

2

Write the answers.

a

z = 1 + i in polar form is z

b

z

−

=

3 + i in polar form is

 5π  .  6  

z = 2cis 

54


=

2cis

 π   4  .

WORKED EXAMPLE 22

Express

3 cis

 5π  in Cartesian form.  6  

TH IN K 1

WR IT E


 

5π 

 

5π 

P ¢ Rx  3, P ¢ Ry  3,

6

6

 

 


2

Write the answer.

3cis

 5π  =  6  

−3 2

+

3 2

i

CHAPTER 10

•


55

WORKED EXAMPLE 24

Sketch the graph that results from the addition 1 of ordinates of the functions y = x and y = , x whose graphs are shown at right.

y

y

=

1 –

x

3

y

x

=

2 1

3

2

−

0

1

−

1

−

2

x

3

1

−

2

−

3

−

TH IN K 1

WR IT E

Add the ordinates at the LHS end points of the graph: A small negative value and a large negative value should give a slightly larger negative value. Mark this point on the set of axes.

y

y

=

1 –

x

3

y

x

=

2 1

3

−

2

−

0

1

1

−

2

3

x

1

−

2

−

3

−

2

Add the ordinates at the first point of intersection. Because they are the same we simply double the y-value of this point. Mark the point on the set of axes.

y

y

=

1 –

x

y

3

x

=

2 1

3

−

2

−

0

1

−

1

2

3

x

1

−

2

−

3

−

3

4

At its x -intercept, the straight line graph y = x has an ordinate of zero while the non-linear graph has an undefined ordinate. This indicates that our graph will also be undefined at the point where x = 0, giving a vertical asymptote.

Vertical asymptote at x = 0

Add the ordinates at the second point of intersection. Mark the point on the set of axes.

y y

=

1 –

x

y

3 2 1

3

−

2

−

0

1

−

1

−

2

−

3

−

56


1

2

3

x

=

x

5

Add the ordinates at the RHS end points of the graph: A small positive value and a large positive value should give a slightly larger positive value. Mark this point on the set of axes.

y

y

=

1 –

x

3

y

x

=

2 1

3

−

2

−

0

1

−

1

2

3

y

1 – = x + x

x

1

−

2

−

3

−

6

Join the points with a smooth curve noting that the graph will asymptote toward the y-axis as x approaches zero, and toward the line y = x as x approaches positive or negative infinity.

y

3 2 1

−3

−2

−1

0

1

2

3

x

−1

−2 −3

7

To perform the addition of ordinates on the CAS calculator, open a new Graphs page. Complete the function entry lines as: f 1( x ) = x f 2( x ) =

1 x

f 3( x ) = f 1( x) + f 2( x )


CHAPTER 10

•


57

WORKED EXAMPLE 25

Sketch the reciprocal of the graph of the function y = x2 shown at right.

y

y

2

x

=

3 2 1

2

0

1

−

−

1

1

−

TH IN K

WR IT E

1

Sketch the given function and draw the horizontal line y = 1 on the same axes.

2

Mark any points where the ordinates of the graph are equal to 1.

3

Select several points on the original graph with ordinates greater than 1 (above the line). Estimate their value and mark their reciprocals below the line.

y

3 2

2

2 1

1 – 1 2 – 3

−2

3

3

1 –

1 2 – 3

0

−1

1

2

x

−1 −2

4

Select several points on the original graph with ordinates between 0 and 1 (between the line and the x -axis). Estimate their value and mark their reciprocals above the line.

y

3

3 3

2 2

1 – 2

1 1 – – 2 3

−2

3

2

2 2

1 1 – 3

1 – 3

1 – 2

0

−1

3

1 – 1 2 – 3

1

2

x

−1 5

Join the points with a smooth curve noting that the graph will asymptote toward the y-axis as x approaches zero, and toward the x -axis as x approaches positive or negative infinity.

y

3

3 3

2 2

1 – 2

1 1 – – 2 3

2

−

1 – 3

0

1

−

1

Alternatively, on a Calculator page, complete the entry line as: Define f ( x ) = x 2 Then press ENTER ·. On a Graphs page, complete the function entry lines as: f 1( x )

=

f 2( x )

=

f ( x ) 1

f ( x )


58


1 –

x 2

3

2 2

1 1 – 3

=

3

2

−

6

y

1 – 2

1

1 – 1 2 – 3

2

x

2

x

WORKED EXAMPLE 26

Sketch the square of the graph of the function y = y

2 shown below. x y

=

2 –

x

3 2 1 3

−

2

−

1

−

0 1

1

2

3

x

−

2

−

3

−

TH IN K 1

2

WR IT E

Sketch the given function and draw the horizontal lines y = 1 and y = −1 on the same axes. Mark any points where the ordinates of the graph are equal to 1 and change any ordinates of −1 to 1. (Squaring 1 results in 1.)

y

3 2 1

1

1

−3 −2 −1 −1

1

2

3

x

−2 −3 3

Select, estimate, and square several ordinates between the horizontal lines y = −1 and y = 1 marking these positive values closer to the x -axis.

y

4 3 2 1

1

1

1 – 4

1 – 1 2 – 4

−4 −3 −2 −1 −1 1

1

2

3

4

x

–—

2

−2 −3 −4

4

Select, estimate, and square several ordinates outside the horizontal lines y = −1 and y = 1 marking these positive values further away from the x -axis.

y

9 9

9

8 7 6 5 4

3 2 1

−4 −3 −2 2

3 2

1

1

1 – 4

1 –—

4

4

0 −1

–1

1

2

1 – 1 2 – 4

3

4

x

−2 −2 −3 −3 −4

CHAPTER 10

•


59

5

Join the points with a smooth curve noting that the graph will asymptote toward the y-axis as x approaches zero.

y

9 9

9 4

y = — 2

8

x

7 6 5 4

3 2 1 –

1

4

4 3 2

1

1 1 – 4

4

–4 1 –—

2

–3 –2

–1 0 –1 –2 –2 –3 –3 –4

6

Alternatively, on a Calculator page, complete the entry line as: Define f ( x )

2 =

x

Then press ENTER ·. On a Graphs page, complete the function entry lines as: f 1( x ) = f ( x ) f 2( x ) = ( f ( x ))2 Press ENTER · after each entry.

60


1

2

3

1 – 2

4

x

CHAPTER 11

Linear programming WORKED EXAMPLE 3

Sketch the graph of the inequation y − 4 x ≤ 8 and indicate the required region. TH IN K 1

Make y the subject of the inequation by adding 4 x to both sides.

2

On a Graphs page, delete = and complete the entry line as: y ≤ 8 + 4 x Then press ENTER ·.

3

Indicate the required region. Note: The calculator shades the required region.

WR IT E/ DR AW

y − 4 x ≤ 8 y ≤ 8 + 4 x

The required region.

CHAPTER 11

•

Linear programming

61

WORKED EXAMPLE 5

Sketch the following pair of simultaneous linear inequations, determine the point of intersection and indicate the required region. 2 x + 3 y ≤ 6 x − y ≥ 3 TH IN K 1

WR IT E/ DR AW

Make y the subject of the inequations.

2 x + 3 y ≤ 6 3 y ≤ 6 − 2 x y ≤

6 − 2 x

[1]

3

x − y ≥ 3 − y ≥ 3 − x y ≤ x − 3 2

[2]

On a Graphs page, delete = and complete the entry lines as: y ≤

6 − 2 x 3

y ≤ x − 3 Press ENTER · after each entry. To change the colour of the inequations, highlight the line and press: Ctrl / MENU b 4: Colour 4 2: Fill colour 2 Select a colour and press ENTER ·. • • • •

3

Indicate the required region.

4

To determine the point of intersection between the lines, press: MENU b 8: Geometry 8 1: Points & Lines 1 3: Intersection Point(s) 3 Move the flashing cursor to each line, press ENTER · to select each line and press ENTER · again to lock in the point of intersection.

The required region.

• • • •

State the point of intersection.

62


The point of intersection between y ≤ x − 3 is (3, 0).

y ≤

6 − 2 x 3

and

WORKED EXAMPLE 8

a

Sketch the following system of linear inequations and indicate the required region. x

b c

+ y ≤ 10, y ≥ x − 4, y ≤ 2 x + 1, x ≥ 0, y ≥ 0

Determine the coordinates of the vertices of the feasible region. Determine the maximum and minimum values of z = 3 x − y subject to the above constraints, using the corner-point method.

TH IN K

a

1

2

b

Make y the subject of the first inequation and write the others.

WR IT E

a

x + y ≤ 10 y ≤ 10 − x y ≥ x − 4 y ≤ 2 x + 1

[1] [2] [3]

x ≥ 0 y ≥ 0

[4] [5]

On a Graphs page, complete the entry lines as: y ≤ 10 − x y ≥ x − 4 y ≤ 2 x + 1 y ≥ 0 Press ENTER · after each entry. Note: x ≥ 0 cannot be drawn on the calculator. Change the colours as required.

3

Indicate the required region.

1

To determine the points of intersection between the lines, press: MENU b 8: Geometry 8 1: Points & Lines 1 3: Intersection Point(s) 3 Move the flashing cursor to each line, press ENTER · to select each line and press ENTER · again to lock in the point of intersection. Repeat for each pair of lines and state the points of intersection. Zoom out to see all the coordinates.

The required region. b

• • • •

2

To determine the points of intersection between the line x = 0 and y = 2 x + 1, substitute the value x = 0 into the equation y = 2 x + 1.

The points of intersection are: (3, 7), (7, 3), (4, 0). y = 2 x + 1 x = 0 ⇒ y = 2(0) + 1 ⇒ y = 1 That is the point (0, 1). From the graph, (0, 0) is also a corner point.

CHAPTER 11

•

Linear programming

63

c

64

1

To determine the maximum and minimum values, on a Calculator page, complete the entry line as: Define z = 3 x − y z | x = 0 and y = 0 z | x = 0 and y = 1 z | x = 3 and y = 7 z | x = 7 and y = 3 z | x = 4 and y = 0 Press ENTER · after each entry.

2

Write the maximum and minimum values of z.


c

zmin = −1 zmax = 18

at (0, 1) at (7, 3)

CHAPTER 12

Coordinate geometry WORKED EXAMPLE 20

ABCD is a parallelogram. The coordinates of A, B and C are (1, 5), (4, 2) and (2, −2) respectively. Find: a the equation of AD b the equation of DC c the coordinates of D. TH IN K

a

1

Draw the parallelogram ABCD. Note: The order of the lettering of the geometric shape determines the links in the diagram. For example: ABCD means join A to B to C to D to A. This avoids any ambiguity.

WR IT E/ DR AW

a

y

5

D

A

2

B

−41 1 −42

2 C

2

−

2

b

Find the gradient of BC.

mBC

=

x

4

2

−

−

2 4

−

4

−

2

=

3

State the gradient of AD.

Since mBC = 2 and AD||BC then mAD = 2

4

Using the given coordinates of A and the gradient of AD find the equation of AD.

y = 2 x + c Let ( x , y) = (1, 5):

1

Find the gradient of AB.

= 2

5 = 2(1) + c c = 3 Hence, the equation of AD is y = 2 x + 3. b

mAB

2 =

4

−

−

5 1

−

=

3

3

= −1

2

State the gradient of DC.

Since mAB = −1 and DC||AB then mDC = −1

3

Using the given coordinates of C and the gradient of DC find the equation of DC.

y = − x + c Let ( x , y) = (2, −2): −2 −(2) = +c c = 0

Hence, the equation of DC is y = − x . c

1

Solve simultaneously to find D, the point of intersection of the equations AD and DC.

c

Equation of AD: y = 2 x + 3 Equation of DC: y = − x

[1] [2]

CHAPTER 12

•

Coordinate geometry

65

66

2

On a Calculator page, complete the entry line as: solve( y = 2 x + 3 and y = − x , x ) Then press ENTER ·.

3

Write the solution.


Hence, the coordinates of D are ( −1, 1).

CHAPTER 13

Vectors WORKED EXAMPLE 14

a b

Draw a vector to represent a



=

3i

−



j.



Find the magnitude and direction of the vector a.



TH IN K

a

1

DR AW/ WR IT E

Draw axes with i and j as unit vectors in the x- and y-directions respectively.

a

y 2

j ~

1

i ~

−2 −1 O −1 2

Represent 3i − j as a vector from  in the positive 0 that is 3 units x-direction and 1 unit in the negative y-direction, and mark the angle between a and the x -axis as θ .

1

2

The magnitude of a (that is, a ) may   be found using Pythagoras’ theorem.

2

3

x

y 2

−2 −1 O −1 b

a



j ~

θ

1



b

1

i ~ 3

x

a ~

=

32 + ( −1) 2

=

10

On a Calculator page, complete the entry line as: 32 + (− 1) 2 Then press ENTER ·.

3

Find the value of angle θ using the tangent ratio.

tan (θ ) =

1 3

CHAPTER 13

•

Vectors

67

4

On a Calculator page, complete the entry line as:

tan

− 1  1 

 3  

Then press ENTER ·. Note: Ensure your calculator is in Degree mode.

5

68

Give the direction of vector a relative  to the positive x -axis.


Vector a makes an angle of −18.4° from the positive x -axis.

CHAPTER 14

Statics of a particle WORKED EXAMPLE 8

The forces acting on two bobs suspended from light inextensible strings are shown in the diagram at right. If the connecting string is at an angle of 10° to the horizontal find T and m.

T 2 N

T 1 N

60°

T N

10°

mg N TH IN K 1

30°

T N

5g N

WR IT E/ DR AW

Separate the two bobs. Draw a diagram representing the forces applied to the right bob.

T 1 sin (30 ) °

T cos (10 )

T 1 cos (30 )

°

°

T sin (10 ) 5g °

2

Calculate the force, T. Set up two equations. To solve these equations simultaneously, using a CAS calculator, let T 1 = x and

T 1 sin (30 °) = 5g + T sin (10°) T cos (10°) = T 1 cos (30°)

[1] [2]

T = y.

On a Calculator page, complete the entry line as: solve( x × sin(30) = 5g + y × sin(10) and y × cos(10) = x × cos(30), x ) | g = 9.8 Then press ENTER ·.

T = 124.1 N

CHAPTER 14

•

Statics of a particle

69

3

Draw a diagram representing the forces applied to the left bob.

T 2 sin (60 ) °

T sin (10 ) °

T cos (10 ) °

T 2 cos (60 ) °

mg 4

Calculate the value of m given T = 124.1 N.

T cos (10°) = T 2 cos (60°) Given T = 124.1 N ⇒

T 2

=

124.1cos (10) cos (60)

= 244.4 N T 2 sin (60 °) + T sin (10°) = mg

⇒ T 2

⇒

m=

⇒ m

5

70

State the answers.


244.4sin (60) + 124.1sin (10)

= 23.80 kg

T = 124.1 N m = 23.80 kg

9.8

CHAPTER 15

Kinematics WORKED EXAMPLE 11

A particle is travelling in a straight line with its velocity, v cm/s, at any time, t seconds, given as 8 v( t ) = , t ≥ 0. t + 1 Find the acceleration of the particle after 1 second. TH IN K 1

2

WR IT E/ DI SP LAY

Given the expression, v(t ) = want a(1).

8 t + 1

we

v (t ) =

8 t + 1

Find the acceleration equation by differentiating velocity with respect to time (a(t ) = v′(t )). To do this, on a Calculator page, complete the entry lines as: Define v(t ) = d dt

8 t + 1

( v( t )) −

Define a(t ) =

8

(t + 1)2

Press ENTER · after each entry. 3

Substitute t = 1 seconds into the formula for a(t ). To do this, complete the entry line as: a (1) Then press ENTER ·.

4

State the solution.

The acceleration of the particle at t = 1 seconds is −2 cm/s2.

CHAPTER 15

•

Kinematics

71

WORKED EXAMPLE 13

A particle is travelling in a straight line with its velocity, v (in m/s), at any time, t seconds, given as: 2 v( t ) = t + t , t ≥ 0 Calculate the exact distance travelled during the first 4 seconds of its motion. TH IN K 1

WR IT E/ DI SP LAY

On a Calculator page, press: MENU b 4: Calculus 4 3: Integral 3 Complete the entry line as: • • •

4

∫ (t 0

2

+ t ) dt


2

State the exact distance travelled.

The exact distance travelled during the first 1

4 seconds of its motion, is 29 3 metres.

72


WORKED EXAMPLE 14

A car accelerates from rest at 2 m/s 2 for 5 seconds. a Write an equation for the acceleration. b Write the equation for the velocity. c Calculate the distance covered in the first 5 seconds. TH IN K

WR IT E

a

1

The acceleration is 2 m/s 2.

a

b

1

To determine the equation for velocity, given a(t ) = 2, on a Calculator page, complete the entry line as:

b

a(t )

= 2

∫ (2)dt + c Then press ENTER ·.

c

2

Write the equation for velocity.

3

It is given that v = 0 when t = 0. Calculate the constant, c, using this information.

1

To calculate the distance, d (t ), covered in the first 5 seconds, on a Calculator page, complete the entry line as:

= 2t + c 0 = 2(0) + c c = 0 ∴ v(t ) = 2t , 0 ≤ t ≤ 5 v(t )

c

∫ (2t) dt 0


2

State the distance covered in the first 5 seconds.

The distance travelled in the first 5 seconds is 25 metres.

CHAPTER 15

•

Kinematics

73

CHAPTER 16

Geometry in two and three dimensions WORKED EXAMPLE 4

a b

Use a ruler and a pair of compasses to bisect a line, AB. Use a CAS calculator to bisect a line segment.

TH IN K

a

1

2

3

b

1

(a) Draw a line AB. (b) Place the compass point at A, with any radius (more than half the length of the line). (c) Draw a circle. With the same radius as in step 1, repeat for point B. The two circles will intersect at two points. Join these points with a straight line. Open a new Graphs page. To access the Geometry view, press: MENU b 2: View 2 2: Plane Geometry 2

DR AW/ DI SP LAY

a

A

B

b

• • •

2

To draw a line segment, complete the following steps. Press: MENU b 4: Points and Lines 4 5: Segment 5 Move the pencil to the position for the end of the line segment and press CLICK a. Repeat the process for the other end. • • •

CHAPTER 16

•


75

3

To bisect the line, press: MENU b 7: Constructions 7 3: Perpendicular bisector 3 Place the cursor over each of the end points of the line segment and press CLICK a to view the bisected line. • • •

WORKED EXAMPLE 5

a b

Use a ruler and compasses to construct a line parallel to a given line. Use a CAS calculator to construct a line parallel to a given line.

TH IN K

a

1

2

3

4

b

1

Pick any two points, A and B, on the given line. From point A, draw a circle of any radius (more than half the distance from A to B). With the same radius repeat step 1 at point B. Join the highest points (or lowest points) of the circles with a straight line. This will be parallel to AB.

a

Construct a straight line segment in the same manner as for worked example 4. To construct a parallel line, press: MENU b 7: Constructions 7 2: Parallel Press e until the word ‘segment’ appears.

b

• • •

2

76

DR AW/ DI SP LAY

Then press CLICK a and use the NavPad to place the parallel line in the required position.


A

B

WORKED EXAMPLE 6

a

TH IN K

a

b

Use a ruler and compasses to bisect any angle.

1

2

DR AW/ DI SP LAY

With any radius, and the point of the compass at the vertex V, draw an arc of a circle which crosses both arms of the angle. The crossings are labelled A and B.

a

B V

With any radius and the point of the compass at A, draw an arc inside the angle. This arc should be long enough so that the line representing half the angle would cross it.

3

With the same radius, repeat step 2, putting the point of the compass at B. The two arcs will cross at point C.

4

Join the vertex V to C. This line bisects the angle, namely: ∠AVC =

Use a CAS calculator to bisect any angle.

A

B V

A

∠AVB

2

C B V

b

1

Open a new Geometry page in a Plane Geometry view. Press: MENU b 7: Construction 7 4: Angle bisector 4 Move the cursor to the end point of the first arm of the angle and press CLICK a. Using the navigation pad move the cursor to where the vertex of the angle will be and press CLICK a. Move the cursor to the end of the other arm and again press CLICK a. The bisector of the angle will automatically appear.

A

b

• • •

2

Pressing ENTER · will show the three points and the bisecting line.

CHAPTER 16

•


77

WORKED EXAMPLE 8

Construct the perpendicular bisectors and median bisectors of each side of any triangle and investigate their properties. TH IN K 1

Draw any triangle and add circles centred at each vertex. The radius should be large enough so that perpendicular bisectors can be drawn.

2

Use the construction circles to draw perpendicular bisectors. The black lines join pairs of intersecting arcs. It should be clear that the bisectors all meet at a point. This point is called the circumcentre .

3

Use this point as a centre and draw a circle which just touches each vertex. To do this, put the compass point at the point where the bisectors met and the pencil point at any vertex. You should observe that the resultant circumcircle (or outcircle) just touches each vertex.

4

Alternatively, open a new Geometry page. Construct a triangle by pressing: MENU b 5: Shapes 5 2: Triangle 2 Press CLICK a and use the NavPad to generate the 3 vertices of the triangle. • • •

5

To construct the perpendicular bisectors, press: MENU b 7: Construction 7 3: Perpendicular Bisector 3 Move the cursor to the end points of each side of the triangle (vertices) and press CLICK a. • • •

78


DR AW/ DI SP LAY

6

To draw the circumcircle, press: MENU b 5: Shapes 5 1: Circle 1 Move the cursor over the intersection point (circumcentre) and press CLICK a. Move the cursor until it is over one of the vertices of the triangle and press CLICK a. • • •

7

8

Furthermore, from step 2, we can determine the midpoint of each side from the perpendicular bisectors. Note the use of short bars to indicate the bisection of the sides, at points P, Q and R.

Join each midpoint to its opposite vertex. These lines also meet at a single point, called the centroid , which has applications in physics and engineering, as it is effectively the point of symmetry of the triangle.

Q

R P

Q

R

P

CHAPTER 16

•


79

WORKED EXAMPLE 9 B

Construct the incentre of the triangle shown at right: a by hand b using a CAS calculator. TH IN K

a

1

2

(a) Construct the angle bisectors by drawing arcs centred at each vertex (A, B and C). (b) From the intersection of these arcs and the sides of the triangles, draw intersecting arcs between pairs of sides. In the figure at right this has been done to vertex A only, to keep the drawing uncluttered.

DR AW/ DI SP LAY

a

B

A

C

Complete the construction of the angle bisectors and observe that they, too, meet at a point — the incentre .

B

A

C

3

By placing the compass point at the incentre and carefully drawing a circle, it is possible to construct the incircle , which just touches each side of the triangle.

B

C

b

1

Construct a triangle using the method set out in worked example 8. Then press: MENU b 7: Construction 7 4: Angle Bisector 4 Move the cursor over each vertex and press CLICK a. This will bisect the angle at the second vertex in the order in which they were clicked. • • •

2

80

Repeat this process for each of the other angles of the triangle.


b

A

C

A

3

To draw the incircle, press: MENU b 5: Shapes 5 1: Circle 1 Move the cursor over the intersection point (incentre) and press CLICK a. Move the cursor until the circle fits inside the triangle, touching each side and press CLICK a. • • •

CHAPTER 16

•


81

Maths Specialist CAS

Recommend Documents