Maths Quest Specialist 12 Textbook 4E TI-Nspire CAS Companion

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MATHS QUEST 12TI-NSPIRE C AS CALCULATOR COMPANION

SpecialistMathematics





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



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)

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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



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





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






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



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)


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.




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,




WORKED EXAMPLE 14

Sketch the graph of 5 x2 + 9( y − 2)2 = 45.


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.


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)


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= =

−


Graphs page press:

• MENUb


• 2: Equation2

• 6: Conic6

• 1:


x xy y x y5 0 9 0 36 9 02 2+ + + + + =

− −

then press ENTER·.





(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




WORKED EXAMPLE 15

Sketch the graph of the relation described by the rule: 25 x2 + 150 x + 4 y2 − 8 y + 129 = 0.


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.


Graphs page press:

• MENUb


• 2: Equation2

• 6: Conic6

• 1:


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=

+−


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




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.


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]




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

− + +

− −


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

+ −− × +

− + +− −


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

+ −

− × +

=

+

−

−

+

− + +

− −

=

+

+

−

+ −




WORKED EXAMPLE 27

Sketch the graph of the function y x x

x

5 6

4

2

==

− +

−

.


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


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




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·.





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= =


• 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°




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


• 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







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= −




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·.



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−





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


• MENUb

• 2: Number2

•

9: Complex Number Tools9• 2: Real Part2 or

• 3: Imaginary Part3


real(7 + 6i)

imag(10)

real(2 + i − 3i2)

i i iimag

1 3

2

2 3− − −


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− − −

=

−




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


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


• MENUb

• 2: Number2


• 1: Complex Conjugate1


conj (8 + 5i)

conj (−2 − 3i)


2 Write the answers. a 8 − 5i

b −2 + 3i




WORKED EXAMPLE 16

If z = a + bi, find a and b such that−

−

= −

z

zi

5 15

14 3 .

THINK WRIT E


• MENUb

•3: Algebra3

• C: ComplexC

• 1: Solve1


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


• MENUb

• 2: Number2


•5: Magnitude5


|8 − 6i|Then press ENTER·.

2 Write the answer. | z | = |8 − 6i| = 10




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


entry line as:

1 + iThen press:

• MENUb

• 2: Number2


• 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

=

−




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


• 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)


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 )



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


entry lines as:

Define z = x + yi

Then press ENTER·.To answer part a press:

• MENUb

• 2: Number2


• 2: Real Part2 or

• 3: Imaginary Part3 or

• 5: Magnitude5


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 .





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


• MENUb

• 4: Calculus4• 1: Derivative1


d

dx x (tan (6 ))

d

dx

x 2 tan

4

3


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

=




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


• MENUb

•

4: Calculus4• 9: Tangent Line9


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




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


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= − −


entry line as:

f x x

x ( ) :

sin ( )=


d

dx

f x ( ( ) )2

2





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


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

= − +




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


• MENUb• 4: Calculus4

• 3: Integral3


x dx

1

25

2∫ −

x dx

3

49

2∫ −

−

x dx

20

16

2∫ +


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

∫ + =

+

−




WORKED EXAMPLE 21

Differentiate the equation y2 + 3 x2 = 4 to find dy

dx in terms of x.

THINK WRIT E


• MENUb

•4: Calculus4

• E: Implicit DifferentiationE


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


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= ±

−

< <

−




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


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= −





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


• MENUb

• 4: Calculus 4

• 3: Integral 3


x dx

x x dx

( 3)

(4 (2 1) )

7

2 4

∫

∫

+

× +

x

x x dx

3 12

3∫ +

+


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= +




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


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·.


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)




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−

+

−




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


tCollect x x x

2sin

2cos

2−

Then press ENTER·.


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

+

+




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


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}




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}




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


∫ −+ +

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




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


x x dx cos ( ) 1 sin ( )0

2∫ +

π

Then press ENTER·.




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.





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


• 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


• MENUb

• 4: Calculus4

• D: Differential Equation SolverD.


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= + −




WORKED EXAMPLE 10

Find the general solution to f x e( ) 2.

x

2′′ = −

THINK WRIT E


• MENUb

•

4: Calculus4• D: Differential Equation SolverD


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


• 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.



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


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) |




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 ).






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


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.



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


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.




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




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


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


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




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


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)


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 )


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.


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≠ =

−


x (0)

x (1)

x (3)


b

Therefore the body travels 16 m in the first three

seconds of motion.




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.






3 Using the CAS calculator solve the

equation for t .


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.




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.


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.




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.


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= − +




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



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~






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


On a Calculator page, press:

• MENUb

• 1: Actions1

• 1: Define1


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


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= +




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


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

=

=

=

−

−




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=

−




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− = + −




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: 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





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


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


vector notation.

π ( ) ( )= − + −

v ie

j2 12 2 1

2




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




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


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


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


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


vector notation.

=

a t

v

dt ( )

= −−

a t i t j( ) 2






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.


( )+ =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.




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




4 Evaluate the vector constant of

integration using the fact that

= −

v i j(1) 2 3 .


= −

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·.




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






6 Use CAS to solve the vector

equation.

First define the vector R, by


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

=

=



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°

~

~




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




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

=

=




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




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




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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Maths Quest Specialist 12 Textbook 4E TI-Nspire CAS Companion