Yr12 Maths in Focus 2U HSC

Congruent: Two fi gures are congruent if they have the same size and shape. They are identical in every way

Polygon: A polygon is a closed plane fi gure with straight sides

Similar: Two fi gures are similar if they have the same shape but a different size. Corresponding angles

are equal and corresponding sides are in the same ratio

Vertex: A vertex is a corner of a fi gure (vertices is plural, meaning more than one vertex)

TERMINOLOGY

Geometry 2

1

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3Chapter 1 Geometry 2

INTRODUCTION

YOU STUDIED GEOMETRY IN the Preliminary Course. In this chapter, you will revise this work and extend it to include some more general applications of geometrical properties involving polygons .

You will also use the Preliminary topic on straight-line graphs to explore coordinate methods in geometry .

Plane Figure Geometry

Here is a summary of the geometry you studied in the Preliminary Course.

Vertically opposite angles

AEC DEBand+ + are called vertically opposite angles . AED CEBand+ + are also vertically opposite angles.

Vertically opposite angles are equal.

Parallel lines

If the lines are parallel, then alternate angles are equal.

If the lines are parallel, then corresponding angles are equal.

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4 Maths In Focus Mathematics HSC Course

If the lines are parallel, cointerior angles are supplementary (i.e. their sum is 180°).

TESTS FOR PARALLEL LINES

If alternate angles are equal, then the lines are parallel.

If corresponding angles are equal, then the lines are parallel.

If cointerior angles are supplementary, then the lines are parallel.

Angle sum of a triangle

The sum of the interior angles in any triangle is 180°, that is, a b c 801+ + =

If ,AEF EFD+ += then AB || CD .

If ,BEF DFG+ += then AB || CD .

If ,BEF DFE 180c+ ++ = then AB || CD .

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Exterior angle of a triangle

The exterior angle in any triangle is equal to the sum of the two opposite interior angles. That is,

x y z+ =

Congruent triangles

Two triangles are congruent if they are the same shape and size. All pairs of corresponding sides and angles are equal.

For example:

We write .ABC XYZ/Δ Δ

TESTS

To prove that two triangles are congruent, we only need to prove that certain combinations of sides or angles are equal.

Two triangles are congruent if • SSS : all three pairs of corresponding sides are equal • SAS : two pairs of corresponding sides and their included angles are equal • AAS : two pairs of angles and one pair of corresponding sides are equal • RHS : both have a right angle, their hypotenuses are equal and one

other pair of corresponding sides are equal

Similar triangles

Triangles, for example ABC and XYZ , are similar if they are the same shape but different sizes .

As in the example, all three pairs of corresponding angles are equal. All three pairs of corresponding sides are in proportion (in the same ratio).

The included angle is the angle between the 2 sides.

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This shows that all 3 pairs of sides are in proportion.

Two triangles are similar if: three pairs of • corresponding angles are equal three pairs of • corresponding sides are in proportion two pairs of • sides are in proportion and their included angles are equal

Ratios of intercepts

When two (or more) transversals cut a series of parallel lines, the ratios of their intercepts are equal.

That is, : :AB BC DE EF=

or BCAB

EFDE=

Pythagoras’ theorem

The square on the hypotenuse in any right angled triangle is equal to the sum of the squares on the other two sides.

That is, c a b2 2 2= + or c a b2 2= +

We write: XYZΔ;ABC <Δ XYZΔ is three times larger than ABCΔ

ABXY

ACXZ

BCYZ

26 3

412 3

515 3

= =

= =

= =

ABXY

ACXZ

BCYZ

` = =

TESTS

There are three tests for similar triangles.

If 2 pairs of angles are equal then the third pair must also be equal.

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Quadrilaterals

A quadrilateral is any four-sided fi gure

In any quadrilateral the sum of the interior angles is 360°

PARALLELOGRAM

A parallelogram is a quadrilateral with opposite sides parallel

Properties of a parallelogram: • opposite sides of a parallelogram are equal • opposite angles of a parallelogram are equal • diagonals in a parallelogram bisect each other each diagonal bisects the parallelogram into two • congruent triangles

TESTS A quadrilateral is a parallelogram if:

both pairs of • opposite sides are equal both pairs of • opposite angles are equal • one pair of sides is both equal and parallel the • diagonals bisect each other

These properties can all be proved.

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RHOMBUS

A rectangle is a parallelogram with one angle a right angle

Properties of a rectangle: the same as for a parallelogram, and also • diagonals are equal •

TEST

A quadrilateral is a rectangle if its diagonals are equal

Application

Builders use the property of equal diagonals to check if a rectangle is accurate. For example, a timber frame may look rectangular, but may be slightly slanting. Checking the diagonals makes sure that a building does not end up like the Leaning Tower of Pisa!

A rhombus is a parallelogram with a pair of adjacent sides equal

Properties of a rhombus: the same as for parallelogram, and also • diagonals bisect at right angles • diagonals bisect the angles of the rhombus •

It can be proved that all sides are equal.

If one angle is a right angle, then you can prove all angles are right angles.

RECTANGLE

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TESTS

A quadrilateral is a rhombus if: all sides are equal • diagonals bisect each other at right angles •

SQUARE

A square is a rectangle with a pair of adjacent sides equal

Properties of a square: the same as for rectangle, and also • diagonals are perpendicular • diagonals make angles of 45° with the sides •

TRAPEZIUM

A trapezium is a quadrilateral with one pair of sides parallel

KITE

A kite is a quadrilateral with two pairs of adjacent sides equal

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SQUARE

The sum of the exterior angles of any polygon is 360°

Areas

Most areas of plane fi gures come from the area of a rectangle.

RECTANGLE

A lb=

TRIANGLE

A x2=

A bh21=

A square is a special rectangle.

The area of a triangle is half the area of a rectangle.

A polygon is a plane fi gure with straight sides

A regular polygon has all sides and all interior angles equal

The sum of the interior angles of an n -sided polygon is given by

°( ) ´S n 1802= −

Polygons

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PARALLELOGRAM

A bh=

RHOMBUS

A xy21=

( x and y are lengths of diagonals)

TRAPEZIUM

( )A h a b21= +

CIRCLE

πA r2=

The following examples and exercises use these results to prove properties of plane fi gures.

You will study the circle in more detail. See Chapter 5.

The area of a parallelogram is the same as the area of

two triangles.

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EXAMPLES

1. Prove A C+ += in parallelogram ABCD .

Solution

°

° °

° ° °° ° °

°

Let

Then ( , cointerior angles, )

( ) ( , cointerior angles, )

A x

B x A B AD BC

C x B C AB DCx

x

A C

180

180 180180 180

`

+

+ + +

+ + +

+ +

<

<

== −= − −= − +==

2. Triangle ABC below is isosceles with AB AC= . Prove that the base angles of ABCΔ are equal by showing that ABDΔ and ACDΔ are congruent.

Solution

° ( )

( )

ADB ADC

AB AC

90 given

given

+ += ==

AD is common \ ABD ACD/Δ Δ (RHS)

So ABD ACD+ += (corresponding angles in congruent )sΔ \ base angles are equal

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3. Prove that opposite sides in a parallelogram are equal.

Solution

Let ABCD be any parallelogram and draw in diagonal AC .

( , )

( , )

DAC ACB AD BC

BAC ACD AB DC

alternate s

alternate s

+ + +

+ + +

<

<

==

AC is common. ABC ADC/Δ Δ (AAS)

( )

( )

AB DC

AD BC

corresponding sides in congruent s

similarly

` Δ==

opposite sides in a parallelogram are equal

1.

DE bisects acute angle ABC+ so that .ABD CBD+ += Prove that DE also bisects refl ex angle .ABC+ That is, prove .ABE CBE+ +=

2.

Prove that CD bisects .AFE+

3. Prove .VW XY<

4.

Given °,x y 180+ = prove that ABCD is a parallelogram.

1.1 Exercises

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5. BD bisects .ABC+ Prove that .ABD CBD/Δ Δ

6. (a) Show that .ABC AED/Δ Δ Hence prove that (b) ACDΔ is

isosceles.

7. ABCD is a square. Lines are drawn from C to M and N , the midpoints of AD and AB respectively. Prove that .MC NC=

8. OC is drawn so that it is perpendicular to chord AB and O is the centre of the circle. Prove that OACΔ and OBCΔ are congruent, and hence that OC bisects AB .

9. CE and BD are altitudes of ,ABCΔ and ABCΔ is isosceles ( ).AB AC= Prove that .CE BD=

10. ABCD is a kite where AB AD= and .BC DC= Prove that diagonal AC bisects both DAB+ and .DCB+

11. MNOP is a rhombus with .MN NO= Show that

(a) MNOΔ is congruent to MPOΔ (b) PMQ NMQ+ += (c) PMQΔ is congruent to NMQΔ (d) °MQN 90+ =

12. Show that a diagonal cuts a parallelogram into two congruent triangles.

13. Prove that opposite angles are equal in any parallelogram.

The altitude is perpendicular to the other side of the triangle

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14. ABCD is a parallelogram with .BM DN= Prove that AMCN is also a parallelogram.

15. ABCD and BCEF are parallelograms. Show that AFED is a parallelogram.

16. ABCD is a parallelogram with .DE DC= Prove that CE bisects .BCD+

17. In quadrilateral ABCD , AB CD= and .BAC DCA+ += Prove ABCD is a parallelogram.

18. ABCD is a parallelogram with .°AEB 90+ = Prove

(a) AB BC= (b) ABE CBE+ +=

19. Prove that the diagonals in any rectangle are equal.

20. Prove that if one angle in a rectangle is 90° then all the angles are 90°.

21. ABCD is a rhombus with .AD CD= Prove that all sides of the rhombus are equal.

22. ABCD is an isosceles trapezium. Prove the base angles ADC+ and BCD+ are equal.

23. Prove that ADC ABC+ += in kite ABCD .

24. In rectangle ABCD , E is the midpoint of CD . Prove .AE BE=

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25. ABCD is a rhombus. Prove (a) ADBΔ and BCDΔ are

congruent. Hence show (b) .ABE CBE+ += Prove (c) ABEΔ and CBEΔ are

congruent. Prove (d) .°AEB 90+ =

Surface Areas and Volumes

Areas are used in fi nding the surface area and volume of solids. Here is a summary of some of the most common ones.

SURFACE AREA VOLUME

( )S lb bh lh2= + + V lbh=

S x6 2= V x3=

( )πS r r h2= + πV r h2=

πS r4 2= πV r34 3=

You will need some of these formulae when you study maxima and minima problems in Chapter 2.

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( )πS r r l= + πV r h31 2=

In general, the volume of any prism is given by V Ah=

where A is the area of the base and h is its height

In general, the volume of any pyramid is given by

V Ah31=

Where A is the area of the base and h is its height

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While surface area and volume is not a part of the geometry in the HSC syllabus, the topic in Chapter 2 uses calculus to fi nd maximum or minimum areas, perimeters, surface areas or volumes. So you will need to know these formulae in order to answer the questions in the next chapter. Here are some questions to get you started.

EXAMPLE

Find the surface area of a cone whose height is twice the radius, in terms of r .

Solution

`

5

( )

h r

l r h

r r

r r

r

l r

r

2

2

4

5

5

2 2 2

2 2

2 2

2

2

== += += +=

==

Surface area ( )πS r r l= + where l is slant height 5( )πr r r= +

1. A rectangular prism has dimensions 12.5 mm, 84 mm and 64 mm. Find its

surface area and (a) volume. (b)

2. A sphere has a volume of π120 m 3 . Find the exact value of r.

3. A rectangular prism has dimensions x , x + 2 and 2 x – 1. Find its volume in terms of x .

1.2 Exercises

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4. A cylinder has a volume of 250 cm 3 . If its base has radius r and its

height is h , show that πh

r 250= .

5. Find the volume of a cylinder in terms of r if its height is fi ve times the size of its radius .

6. The ratio of the length to the breadth of a certain rectangle is 3:2. If the breadth of this rectangle is b units, fi nd a formula for the area of the rectangle in terms of b .

7. Find the volume of a cube with sides ( ) .x 2 cm+

8. What would the surface area of a cylinder be in terms of h if its height is a third of its radius?

9. A square piece of metal with sides 3 m has a square of side x cut out of each corner. The metal is then folded up to form a rectangular prism. Find its volume in terms of x .

10. A cone-shaped vessel has a height of twice its radius. If I fi ll the vessel with water to a depth of 10 cm, fi nd the volume of water to the nearest cm 3 .

11. The area of the base of a prism is given by 3 h + 2, where h is the height of the prism. Find a formula for the volume of the prism .

12. The area of the base of a pyramid is 6 h + 15 where h is the height of the pyramid. Find the volume of the pyramid in terms of h .

13. A rectangular pyramid has base dimensions x – 3 and 3 x + 5, and a height of 2 x + 1. Write a formula for the volume of the pyramid in terms of x .

14. The height of a rectangular prism is twice the length of its base. If the width of the base is x and the length is 3 x – 1, fi nd an expression for the

volume and (a) surface area of the prism. (b)

15. Find a formula for the slant height of a cone in terms of its radius r and height h .

16. A page measuring x by y is curved around to make an open cylinder with height y . Find the volume of the cylinder in terms of x and y .

17. The volume of a cylinder is 400 cm 3 . Find the height of the cylinder in terms of its radius r .

18. A cylinder has a surface area of 1500 cm 2 . Find a formula for its height h in terms of r .

19. The surface area of a cone is given by ( )πS r r l= + where l is the slant height. Find a formula for the slant height of a cone with surface area 850 cm 2 in terms of r .

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20. A rectangular timber post is cut out of a log with diameter d as shown. If the post has length x and breadth y , write y in terms of x when d = 900 mm.

d

DID YOU KNOW?

REGULAR SOLIDS

There are only fi ve solids with each face the same size and shape. These are called platonic solids. Research these on the internet.

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The skeletons opposite are those of radiolarians. These are tiny sea animals, with their skeletons shaped like regular solids.

A salt crystal is a cube. A diamond crystal is an octahedron.

Diamond crystal

Coordinate Methods in Geometry

Problems in plane geometry can be solved by using the number plane. You studied straight-line graphs in the Preliminary Course. Some of the

main results that you learned will be used in this section. You may need to revise that work before studying this section.

Here is a summary of the main formulae.

Distance

The distance between two points ( , )x y1 1 and ( , )x y2 2 is given by

d x x y y2 12

2 12= − + −_ _i i

Midpoint

The midpoint of two points ( , )x y1 1 and ( , )x y2 2 is given by

,Px x y y

2 21 2 1 2=+ +e o

Gradient

The gradient of the line between ( , )x y1 1 and ( , )x y2 2 is given by

m x xy y

2 1

2 1= −−

The gradient of a straight line is given by θtanm =

where θ is the angle the line makes with the x -axis in the positive direction.

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The gradient of the line ax by c 0+ + = is given by

mba= −

Equation of a straight line

The equation of a straight line is given by ( )y y m x x1 1− = −

where ( , )x y1 1 lies on the line with gradient m .

Parallel lines

If two lines are parallel, then they have the same gradient. That is, m m1 2=

Perpendicular lines

If two lines with gradients m1 and m2 respectively are perpendicular, then

m m m m1 1or1 2 21

= − = −

Perpendicular distance

The perpendicular distance from ( , )x y1 1 to the line ax by c 0+ + = is given by

| |

da b

ax by c2 2

1 1=+

+ +

EXAMPLES

1. Show that triangle ABC is right angled, where ( , ), ( , )A B3 4 1 1= = − − and ( , )C 2 8= − .

Solution

Method 1:

( ) ( )d x x y y2 12

2 12= − + −

( ) ( )( ) ( )

( ) ( )( )

( ( )) ( )( )

AB

AC

BC

1 3 1 4

4 5

16 25

41

2 3 8 4

5 4

25 16

41

1 2 1 8

1 9

1 81

82

2 2

2 2

2 2

2 2

2 2

2 2

= − − + − −= − + −= +== − − + −= − += +== − − − + − −= + −= +=

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AB AC

BC

41 4182

2 2

2

+ = +==

Since Pythagoras’ theorem is true, the triangle ABC is right angled.

Method 2:

´

m x xy y

m

m

m m

1 31 4

45

45

2 38 4

54

54

45

54

1

AB

AC

AB AC

2 1

2 1= −−

=− −− −

=−−

=

=− −

−

=−

= −

= −

= −

So AB and AC are perpendicular So triangle ABC is right angled at A .

2. Prove that points , , ,A B1 1 2 1− −^ ^h h and ( , )C 4 3 are collinear.

Solution

Collinear points lie on the same straight line, so they will have the same gradient.

( )( )

m x xy y

m

m

m m

2 11 1

32

32

4 2

3 1

64

32

AB

BC

AB BC

2 1

2 1= −−

=

= − −− −

= −−

=

=− −− −

=

=

So the points are collinear.

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1. Show that points ( , ),A 1 0−( , ), ( , )B C0 4 7 0 and ( , )D 6 4− are

the vertices of a parallelogram.

2. Prove that ( , ), ( , )A B1 5 4 6− and ( , )C 3 2− − are vertices of a right angled triangle.

3. Given ABCΔ with vertices ( , ), ( , )A B0 8 3 0 and ( , )C 3 0−

show that (a) ABCΔ is isosceles fi nd the length of the altitude (b)

from A fi nd the area of the triangle. (c)

4. Show that the points , ,X 3 2^ h

( , )Y 2 1− and ( , )Z 8 3 are collinear.

5. (a) Show that the points , ,A 2 5^ h

( , ), ( , )B C1 0 7 4− − and ( , )D 3 4− are the vertices of a kite.

Prove that the diagonals of (b) the kite are perpendicular.

Show that (c) CE AE2= where E is the point of intersection of the diagonals.

6. Find the radius of the circle that has its centre at the origin and a tangent with equation given by

.x y4 3 5 0− − =

7. (a) Find the equation of the perpendicular bisector of the line joining ,( )A 3 2 and ,( ).B 1 8−

Show that the point (b) ,( )C 7 9 lies on the perpendicular bisector.

What type of triangle is (c) ?ABCΔ

8. Show that OABΔ and OCDΔ are similar where ( , ), ( , ), ( , )0 7 2 0 0 14− and ( , )4 0− are the points A , B , C and D respectively and O is the origin.

9. (a) Prove that OABΔ and OCBΔ are congruent given ( , ), ( , ), ( , )A B C3 4 5 0 2 4− and O the origin.

Show that (b) OABC is a parallelogram.

10. The points ( , ), ( , ), ( , )A B C0 0 2 0 2 2 and ( , )D 0 2 are the vertices of a square. Prove that its diagonals make angles of 45° with the sides of the square.

11. Prove that ( , ), ( , ),P Q2 0 0 5− ( , )R 10 1 and ( , )S 8 4− are the

vertices of a rectangle .

12. The points ( , ), ( , )A B5 0 1 4− and ( , )C 3 0 form the vertices of a triangle.

Find (a) X and Y , the midpoints of AB and AC respectively.

Show that (b) XY and BC are parallel.

Show that (c) BC = 2 XY.

13. Show that the diagonals of a square are perpendicular bisectors, given the vertices of square ABCD where , , , , ,A a B a a C a0 0= − = − =^ ^ ^h h h and ( , )D 0 0= .

14. (a) Show that points ,X 3 2^ h and ( , )Y 1 0− are equal distances from the line .x y4 3 1 0− − =

Find (b) Z , the x -intercept of the line.

What is the area of triangle (c) XYZ ?

15. ABCD is a quadrilateral with , , , , ,A B C3 1 1 4 5 2− − −^ ^ ^h h h and , .D 4 3−^ h Show that the midpoints of each side are the vertices of a parallelogram.

1.3 Exercises

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Test Yourself 1 1. Triangle ABC is isosceles, with .AB AC=

D is the midpoint of AB and E is the midpoint of AC .

Prove that (a) BECΔ is congruent to BDCΔ .

Prove (b) BE DC= .

2. If the diagonals of a rhombus are x and y , show that the length of its side is

.x y

2

2 2+

3. If ( , ), ( , )A B4 1 7 5= − = − and ( , ),C 1 3= prove that triangle ABC is isosceles.

4. The surface area of a closed cylinder is 100 m2 . Write the height h of the cylinder in terms of its radius r .

5. ABCD is a parallelogram with ,°C AE ED45 perpendicular to+ = and CD DE= .

Show that (a) ADEΔ is isosceles. If (b) AE y= , show that the area of ABCE

is .y

2

3 2

6. In the fi gure AEF

prove (a) BACB

DECD=

fi nd the length of (b) AE.

7. Given ( , ), ( , ),A B1 3 2 4= − = − − ( , )C 5 4= − ( , ),D 6 3and = prove ABCD is a

parallelogram.

8. A parallelogram has sides 5 cm and 12 cm, with diagonal 13 cm.

Show that the parallelogram is a rectangle.

9. Prove that PQR WXYandΔ Δ are similar.

10. In quadrilateral ABCE , AD ED DC= = 90°ACBand+ = . Also, AC BADbisects+ .

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Prove ABCE is a parallelogram.

11. If ( , ), ( , )A B1 5 4 2= = and ( , )C 2 3= − , fi nd the coordinates of D such that ABCD is a parallelogram.

12. (a) Find the equation of AB if ( , )A 2 3= − − and ( , )B 4 5= .

(b) Find the perpendicular distance from ( , )C 1 3− to line AB .

(c) Find the area of .ABCΔ

13. ABCD is a kite.

Prove (a) ABC ADCandΔ Δ are congruent. Prove (b) ABE ADEandΔ Δ are congruent. Prove (c) AC is the perpendicular

bisector of BD .

14. ( , ), ( , )A B1 2 3 3 and ( , )C 5 1− are points on a number plane.

Show that (a) AB is perpendicular to BC . Find the coordinates of (b) D such that

ABCD is a rectangle. Find the point where the diagonals of (c)

the rectangle intersect. Calculate the length of the diagonal. (d)

15. The surface area of a box is 500 cm 2 . Its length is twice its breadth x .

Show that the height (a) h of the box is

given by hx

x3

250 2 2

= − .

Show that the volume of the box is(b)

.V x x3

500 4 3

= −

Challenge Exercise 1

1. In the fi gure, BD is the perpendicular bisector of AC . Prove that ABCΔ is isosceles.

2. Given E and D are midpoints of AC and AB respectively, prove that

(a) DE is parallel to BC

(b) .DE BC21=

3. Prove that the diagonals in a rhombus bisect the angles they make with the sides.

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4. Paper comes in different sizes, called A0, A1, A2, A3, A4 and so on. The largest size is A0, which has an area of one square metre. If the ratio of its length to breadth is :2 1 , fi nd the dimensions of its sides in millimetres, to the nearest millimetre.

5. The volume of a prism with a square base of side x is 1000 cm3 . Find its surface area in terms of x .

6. Prove that in any regular n- sided polygon

the size of each angle is °.n180 360−b l

7. A parallelogram ABCD has AB produced to E and diagonal AC produced to F so that .EF BC< Prove that AEFΔ is similar to .ADCΔ

8. ABCD is a rhombus with ( , ), ( , ),A B a b0 0 ( , )C a2 0 and ( , ).D a b− Show

the diagonals bisect each other at (a) right angles

all sides are equal (b) (c) AC bisects .BCD+

9. In the parallelogram ABCD , AC is perpendicular to BD . Prove that .AB AD=

10. Triangle ABC has P, Q and R as midpoints of the sides, as shown in the diagram below. Prove that .PQR CPR/Δ Δ

11. A pair of earrings is made with a wire surround holding a circular stone, as shown. Find the total length of wire needed for the earrings.

12. The sides of a quadrilateral ABCD have midpoints P, Q, R and S , as shown below.

Show that (a) DPSΔ is similar to .DACΔ Show (b) .PS QR< Show that (c) PQRS is a parallelogram.

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13. A plastic frame for a pair of glasses is designed as below. Find the length of plastic needed for the frame, to the nearest centimetre.

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Anti-differentiation: The process of fi nding a primitive (original) function from the derivative. It is the inverse operation to differentiation

Concavity: The shape of a curve as it bends around (it can be concave up or concave down)

Differentiation: The process of fi nding the gradient of a tangent to a curve or the derivative

Gradient of a secant: The gradient (slope) of a line between two points that lie close together on a curve

Gradient of a tangent: The gradient (slope) of a line that is a tangent to a curve at a point on a function. It is the derivative of the function

Horizontal point of infl exion: A stationary point (where the fi rst derivative is zero) where the concavity of the curve changes

Instantaneous rate of change: The derivative of a function

Maximum turning point: A local stationary point (where the fi rst derivative is zero) and where the curve is concave down. The gradient of the tangent is zero

Minimum turning point: A local stationary point (where the fi rst derivative is zero) and where the curve is concave up. The gradient of the tangent is zero

Monotonic increasing or decreasing function: A function is always increasing or decreasing

Point of infl exion: A point at which the curve is neither concave upwards nor downwards, but where the concavity changes

Primitive function: The original function found by working backwards from the derivative. Found by anti-differentiation

Rate of change: The rate at which the dependent variable changes as the independent variable changes

Stationary (turning) point: A local point at which the gradient of the tangent is zero and the tangent is horizontal. The fi rst derivative is zero

TERMINOLOGY

Geometrical Applications of Calculus

2

ch2.indd 30 6/12/09 12:49:44 AM

31Chapter 2 Geometrical Applications of Calculus

DID YOU KNOW?

Although Newton and Leibniz are said to have discovered calculus, elements of calculus were around before then. It was Newton and Leibniz who perfected the processes and developed the notation associated with calculus.

Pierre de Fermat (1601–65) used coordinate geometry to fi nd maximum and minimum values of functions. His method is very close to calculus. He also worked out a way of fi nding the tangent to a curve.

The 17th-century mathematicians who developed calculus knew that it worked, but it was not fully understood. Limits were not introduced into calculus until the nineteenth century.

INTRODUCTION

YOU LEARNED ABOUT differentiation in the Preliminary Course . This is the process of fi nding the gradient of a tangent to a curve. This chapter looks at how the gradient of a tangent can be used to describe the shape of a curve. Knowing this will enable us to sketch various curves and fi nd their maximum and minimum values. The theory also allows us to solve various problems involving maximum and minimum values.

Gradient of a Curve

To learn about the shape of a curve, we fi rst need to revise what we know about the gradient of a tangent. The gradient (slope) of a straight line measures the rate of change of y with respect to the change in x .

Since the gradient of a curve varies, we fi nd the gradient of the tangent at each point along the curve.

ch2.indd 31 6/12/09 12:49:54 AM


EXAMPLES

1.

2.

3.

4.

ch2.indd 32 6/12/09 12:49:57 AM


In the examples on the previous page, where the gradient is positive, the curve is going up, or increasing (reading from left to right).

Where the gradient is negative, the curve is going downwards, or decreasing.

The gradient is zero at particular points of the curves. At these points the curve isn’t increasing or decreasing. We say the curve is stationary at these points.

If 0,f x >l] g the curve is increasing If 0,f x <l] g the curve is decreasing If 0,f x =l] g the curve is stationary

A curve is monotonic increasing or decreasing if it is always increasing or decreasing; that is,

if f x 0>l] g for all x (monotonic increasing) or f x 0<l] g for all x (monotonic decreasing)

EXAMPLES

1. Find all x values for which the curve f x x x4 12= − +] g is increasing.

Solution

x2 4= −f xl] g 0>f xl] g for increasing curve

. x

x

x

2 4 0

2 4

2

i.e >>>

−

So the curve is increasing for x 2> .

This function is a parabola.

CONTINUED

ch2.indd 33 6/12/09 12:49:59 AM


2. Find the stationary point on the parabola y x x6 32= − + .

Solution

dx

dyx2 6= −

For stationary points,dx

dy

x

x

x

0

2 6 0

2 6

3

i.e.

=

− ===

When ,x y3 3 6 3 3

6

2= = − += −

] g

So the stationary point is ,3 6−^ h .

3. Find any stationary points on the curve y x x48 73= − − .

Solution

y x3 482= −l

`

For stationary points,

±

y

x

x

x

x

0

3 48 0

3 48

164

i.e. 2

2

2

=− =

===

l

When , ( )x y4 4 48 4 7

135

3= = − −= −

When , ( )x y4 4 48 4 7121

3= − = − − − −=] g

So the stationary points are ,4 135−^ h and ,4 121−^ h . You will use stationary points to sketch curves later in this chapter.

ch2.indd 34 6/12/09 12:50:00 AM


PROBLEM

What is wrong with this working out ? Find the stationary point on the curve y x x2 12= + − .

Solution

y x4 1= +l

.

y

x

x

x

04 1 0

4 1

0 25

For stationary points,i.e.

=+ =

= −= −

l

When . , ( . )x y0 25 4 0 25 1

1 1

0

= − = − += − +=

So the stationary point is . ,0 25 0−^ h . Can you fi nd the correct answer?

1. Find the parts of each curve where the gradient of the tangent is positive, negative or zero. Label each curve with +, − or 0.

(a)

(b)

(c)

(d)

2. Find all values of x for which the curve y x x2 2= − is decreasing.

3. Find the domain over which the function f x x4 2= −] g is increasing.

4. Find values of x for which the curve y x x3 42= − − is

decreasing (a) increasing (b) stationary. (c)

5. Show that the function f x x2 7= − −] g is always (monotonic) decreasing.

2.1 Exercises

ch2.indd 35 6/12/09 12:50:02 AM


6. Prove that y x3= is monotonic increasing for all 0≠x .

7. Find the stationary point on the curve ( ) xf x 3= .

8. Find all x values for which the curve y x x x2 3 36 93 2= + − + is stationary.

9. Find all stationary points on the curve

(a) y x x2 32= − − (b) f x x9 2= −] g (c) y x x x2 9 12 43 2= − + − (d) .y x x2 14 2= − +

10. Find any stationary points on the curve y x 2 4= −] g .

11. Find all values of x for which the curve ( ) x xf x 3 43= − + is decreasing.

12. Find the domain over which the curve y x x x12 45 303 2= + + − is increasing.

13. Find any values of x for which the curve y x x x2 21 60 33 2= −− + is

stationary (a) decreasing (b) increasing. (c)

14. The function f x x px2 72= + +] g has a stationary point at x 3= . Evaluate p .

15. Evaluate a and b if 3y x ax bx3 2= − + − has stationary

points at x 1= − and x 2= .

16. (a) Find the derivative of .y x x x3 27 33 2= − + − (b) Show that the curve is monotonic increasing for all values of x .

17. Sketch a function with 0f x >l] g for x < 2, 0f 2 =l] g and 0f x <l] g when x 2> .

18. Draw a sketch showing a curve

with dx

dy0< for x 4< ,

dx

dy0= when

x 4= and dx

dy0> for x 4> .

19. Sketch a curve with dx

dy0> for all

1≠x and dx

dy0= when x 1= .

20. Draw a sketch of a function that has 0f x >l] g for x 2< − , x 5> , 0f x =l] g for ,x 2 5= − and 0<f xl] g for x2 5< <− .

21. A function has ( )f 23 = and 0f 3 <l] g . Show this information on a sketch.

22. The derivative is positive at the point (−2, −1). Show this information on a graph.

23. Find the stationary points on the curve y x x3 1 2 4= − −] ]g g .

24. Differentiate y x x 1= + . Hence fi nd the stationary point on the curve, giving the exact value.

25. The curve ( ) 2 7 5f x ax x x x4 3 2= − + − + has a stationary point at .x 1= Find the value of a .

26. Show that ( ) xf x = has no stationary points.

27. Show that x

f x 13

=] g has no

stationary points.

ch2.indd 36 6/12/09 12:50:04 AM


Types of Stationary Points

There are three types of stationary points.

Local minimum point

The curve is decreasing on the left and increasing on the right of the minimum turning point.

x LHS Minimum RHSf xl] g < 0 0 > 0

Local maximum point

The curve is increasing on the left and decreasing on the right of the maximum turning point.

x LHS Maximum RHSf xl] g > 0 0 < 0

Local maximum and minimum points are also called turning points , as the curve turns around at these points. They can also be called relative maxima or minima.

Point of horizontal infl exion

The curve is either increasing on both sides of the infl exion or it is decreasing on both sides. It is not called a turning point as the curve does not turn around at this point.

The derivative has the same sign on both sides

of the infl exion.

ch2.indd 37 6/12/09 12:50:05 AM


x LHS Infl exion RHSf xl] g > 0 0 > 0

or < 0 0 < 0

The stationary points are important to the shape of a curve. A reasonably accurate sketch of the curve can be made by fi nding these points, together with the intercepts on the axes if possible.

EXAMPLES

1. Find the stationary point on the curve y x3= and determine which type it is.

Solution

dx

dyx3 2=

dx

dy

x

x

0

3 00


i.e. 2

=

==

0, 0x y

0When 3= =

=

So the stationary point is ,0 0^ h . To determine its type, check the curve on the LHS and the RHS.

x −1 0 1

dx

dy3 0 3

Since the curve is increasing on both sides, (0, 0) is a point of infl exion.

Substitute x = ±1 into dx

dy.

ch2.indd 38 6/12/09 12:50:06 AM


2. Find any stationary points on the curve f x x x x2 15 24 73 2= − + −] g and distinguish between them.

Solution

6 30 24f x x x2= − +l] g

( )

( ) ( )

f x

x x

x x

x x

x

0

6 30 24 0

5 4 01 4 0

1 4


i.e.

or

2

2

`

=− + =

− + =− − =

=

l

( )f 1 2 1 15 1 24 1 7

4= − + −=

3 2] ] ]g g g

So ,1 4^ h is a stationary point.

x 0 1 2f xl] g 24 0 −12

,1 4` ^ h is a maximum stationary point

( ) ( )f 4 2 4 15 4 24 4 7

23= − + −= −

3 2] ]g g

So ,4 23−^ h is a stationary point.

x 2 4 5f xl] g −12 0 24

,4 23` −^ h is a minimum stationary point.

Substitute x 0= and x 2= into ´(x)f . Take care that

there is no other stationary point between the x values

you choose.

ch2.indd 39 6/12/09 12:50:07 AM


1. Find the stationary point on the curve y x 12= − and show that it is a minimum point by checking the derivative on both sides of it.

2. Find the stationary point on the curve y x4= and determine its type.

3. Find the stationary point on the curve y x 23= + and determine its nature.

4. The function f x 2x x7 4= − −] g has one stationary point. Find its coordinates and show that it is a maximum turning point.

5. Find the turning point on the curve y x x3 6 12= + + and determine its nature.

6. For the curve y x4 2= −] g fi nd the turning point and determine its nature.

7. The curve y x x6 53 2= − + has 2 turning points. Find them and use the derivative to determine their nature.

8. Show that the curve f x x 15= +] g has a point of infl exion at ,0 1^ h .

9. Find the turning points on the curve y x x3 53 2= − + and determine their nature.

10. Find any stationary points on the curve f x 4 2 .x x2 3= − −] g What type of stationary points are they?

11. The curve y x x3 23= − + has 2 stationary points. Find their coordinates and determine their type.

12. The curve y x mx x7 55 3= + − + has a stationary point at x 2= − . Find the value of m .

13. For a certain function, .f x x3= +l] g For what value of x does the function have a stationary point? What type of stationary point is it?

14. A curve has ( 1) .f x x x= +l] g For what x values does the curve have stationary points? What type are they?

15. For a certain curve,

( ) ( ) .dx

dyx x1 22= − − Find the x

values of its stationary points and determine their nature.

16. (a) Differentiate P x x2 50= + with respect to x.

Find any stationary points (b) on the curve and determine their nature.

17. For the function ,h hA82 52 − +=

fi nd any stationary points and determine their nature.

18. Find any stationary points for the function 40 πV r r3= − and determine their nature (correct to 2 decimal places).

19. Find any stationary points on the

curve π rS r 1202= + (correct to 2

decimal places) and determine their nature.

20. (a) Differentiate .A x x3600 2= − Find any stationary points for (b)

A x x3600 2= − (to 1 decimal place) and determine their nature.

2.2 Exercises

ch2.indd 40 6/12/09 12:50:09 AM


Higher Derivatives

A function can be differentiated several times: differentiating • f x] g gives f xl] g differentiating • f xl] g gives f xm] g differentiating • f xm] g gives ,f xn ] g and so on

the other notation is • ,dx

d y

dx

d y2

2

3

3

and so on

EXAMPLES

1. Find the fi rst 4 derivatives of ( )f x x x x4 3 23 2= − + − .

Solution

( )x x x3 8 3= − +( )

( )

x x

x

6 8

0

= −

=( )x 6=

f

f

f

2

f

l

m

n

mm

2. Find the second derivative of y x2 5 7= +] g .

Solution

dx

dyx

x

dx

d yx

x

7 2 5 2

14 2 5

14 6 2 5 2

168 2 5

6

6

2

25

5

$

$ $

= +

= +

= +

= +

]]

]]

gg

gg

3. Find f 1−m] g if f x x 14= −] g

Solution

( )

( )

( )

f x x

f x x

f

4

12

1 12 112

3

2

2`

==

− = −=

l

m

m ] g

Only the fi rst 2 derivatives are used in sketching

graphs of curves.

Differentiating further will only give 0.

ch2.indd 41 6/12/09 12:50:10 AM


2.3 Exercises

1. Find the fi rst 4 derivatives of x x x x2 37 5 4− + − − .

2. If f x x 59 −=] g , fi nd .f xm] g

3. Find f xl] g and f xm] g if f x x x2 15 3− +=] g .

4. Find f 1l] g and f 2−m] g , given f t t t t3 2 5 44 3= − + −] g .

5. Find the fi rst 3 derivatives of x x x2 4 77 6 4− + − .

6. Differentiate y x x2 3 32= − + twice.

7. If f x x x x x2 5 14 3 2= − + − −] g , fi nd f 1−l] g and f 2m] g .

8. Differentiate x 4− twice.

9. If ,g x x=] g fi nd .g 4m] g

10. Given h t t t5 2 53 2= − + + , fi nd

dtd h

2

2

when t 1= .

11. Find the value of x for which

dx

d y3

2

2

= given y x x x3 2 53 2= − + .

12. Find all values of x for which f x 0>m] g given that f x 3 2x x x 9= − + +] g .

13. Differentiate x4 3 5−] g twice.

14. Find f xl] g and f xm] g if .f x x2= −] g

15. Find the fi rst and second

derivatives of .f xx

x3 1

5=−

+] g

16. Find dtd v

2

2

if ( ) .v t t3 2 1 2= + −] g 17. Find the value of b in

y bx x x2 5 43 2= − + + if dx

d y2

2

2

= −

when .x21=

18. Find the exact value of f 2m] g if ( )f x x x3 4= − .

19. Find f 1m] g if ( )f t t t2 1 7= −] g .

20. Find the value of b if f x bx x5 42 3= −] g and .f 1 3− = −m] g

Sign of the Second Derivative

The second derivative gives extra information about a curve that helps us to fi nd its shape.

Since f xm] g is the derivative of f xl] g , then f xm] g and f xl] g have the same relationship as f xl] g and f x] g .

( ) ( )

( ) ( )

( ) ( )

f x f x

f x f x

f x f x

0

0

0

That is if then is increasing

if then is decreasing

if then is stationary

><=

m l

m l

m l

ch2.indd 42 6/12/09 12:50:11 AM


Concavity

If f x 0>m] g then f xl] g is increasing. This means that the gradient of the tangent is increasing, that is, the curve is becoming steeper.

Notice the upward shape of these curves. The curve lies above the tangents. We say that the curve is concave upwards .

If f x 0<m] g then f xl] g is decreasing. This means that the gradient of the tangent is decreasing. That is, the curve is becoming less steep.

Notice the downward shape of these curves. The curve lies below the tangents. We say that the curve is concave downwards.

If f x 0=m] g then f xl] g is stationary. That is, it is neither increasing nor decreasing. This happens when the curve goes from being concave upwards to concave downwards, or when the curve changes from concave downwards to concave upwards. We say that the curve is changing concavity.

The curve has a point of infl exion as long as concavity changes .

( )´ xf is decreasing since

−4 is less then 21

− .

These points of infl exion are called horizontal

infl exions , as the tangent is horizontal.

( )´ xf is increasing since a

gradient of 21

− is greater

than a gradient of −6.

The point where concavity changes is called an

infl exion.

ch2.indd 43 6/12/09 12:50:12 AM


Class Investigation

How would you check that concavity changes?

The sign must change for concavity to change.

What type of point is it?

If • ,f x 0>m] g the curve is concave upwards . If • ,f x 0<m] g the curve is concave downwards. If • f x 0=m] g and concavity changes, there is a point of infl exion.

EXAMPLES

1. Does the curve y x4= have a point of infl exion?

Solution

dx

dyx

dx

d yx

4

12

3

2

22

=

=

For inflexions,

i.e.dx

d y

x

x

0

12 00

2

2

2

=

==

,x y0 0

0When 4= =

=

So ,0 0^ h is a possible point of infl exion. Check that concavity changes:

x −1 0 1

dx

d y2

2

12 0 12

Since concavity doesn’t change, (0, 0) is not a point of infl exion.

ch2.indd 44 6/12/09 12:50:14 AM


2. Find all values of x for which the curve xf x x x2 7 5 43 2= − − +] g is concave downwards.

Solution

( )

( )

f x x x

f x x

6 14 5

12 14

2= − −= −

l

m

For concave downwards, ( )f x

x

x

012 14 0

12 14

<<<

−m

x 161<`

3. Find the point of infl exion on the curve y x x x6 5 93 2= − + + .

Solution

y x x

y x

3 12 5

6 12

2= −= −

+l

m

For inflexions, y

x

x

x

06 12 0

6 12

2`

=− =

==

m

2, 2 6 5(2) 9

x y 2

3henW 3 2= = − + +

=] g

,2 3` ^ h is a possible point of infl exion Check that concavity changes:

x 1 2 3

dx

d y2

2

−6 0 6

Since concavity changes, ,2 3^ h is a point of infl exion.

1. For what values of x is the curve y x x x2 13 2= + − − concave upwards?

2. Find all values of x for which the curve y x 3 3= −] g is concave downwards.

3. Prove that the curve y x x8 6 4 2= − − is always concave downwards.

4. Show that the curve y x2= is always concave upwards.

5. Find the domain over which the curve f x 3 2x x7 1= − +] g is concave downwards.

6. Find any points of infl exion on the curve g x x x x3 2 93 2= − + +] g .

7. Find the points of infl exion on the curve y x x x6 12 244 2= − + − .

2.4 Exercises

ch2.indd 45 6/12/09 12:50:15 AM


8. Find the stationary point on the curve y x 23= − . Show that it is an infl exion.

9. Find all values of x for which the function f x 4 3 2x x x x2 12 12 1= + − + −] g is concave downwards.

10. Determine whether there are any points of infl exion on the curve

(a) y x6= (b) y x7= (c) y x5= (d) y x9= (e) y x12=

11. Sketch a curve that is always concave up .

12. Sketch a curve where f x 0<m] g for x 1> and f x 0>m] g for x 1< .

13. Find any points of infl exion on the curve – y x x x x8 24 4 94 3 2= +− − .

14. Show that x

f x 22

=] g is concave

upwards for all 0≠x .

15. For the function f x x x3 10 75 3= − +] g

Find any points of infl exion. (a)

Find which of these points (b) are horizontal points of infl exion (stationary points) .

16. (a) Show that the curve y x x x12 20 34 2= + − + has no points of infl exion.

Describe the concavity of the (b) curve.

17. If y ax x x12 3 53 2= +− − has a point of infl exion at x 2= , evaluate a .

18. Evaluate p if f x 4 2x px x6 20 11= − − +] g has a point of infl exion at x 2= − .

19. The curve 2 4 72 4 3y ax bx x x4 3 2= + − + − has points of infl exion at x 2= and x 1= − . Find the values of a and b .

20. The curve y x x x3 21 86 5= − + − has two points of infl exion.

Find these points . (a) Show that these points of (b)

infl exion are not stationary points.

If we combine the information from the fi rst and second derivatives, this will tell us about the shape of the curve.

EXAMPLES

1. For a particular curve, f 2 1= −] g , f 2 0>l] g and .f 2 0<m] g Sketch the curve at this point, showing its shape.

Solution

f 2 1= −] g means that the point ,2 1−^ h lies on the curve. If 0,f 2 >l] g the curve is increasing at this point. If ,f 2 0<m ] g the curve is concave downwards at this point.

ch2.indd 46 6/12/09 12:50:16 AM


2. The curve below shows the number of unemployed people P over time t months.

Describe the sign of (a) dtdP and .

dtd P

2

2

How is the number of unemployed people changing over time? (b) Is the unemployment rate increasing or decreasing? (c)

Solution

The curve is decreasing, so (a) dtdP 0< and the curve is concave upwards,

so .dtd P 0>

2

2

As the curve is decreasing, the number of unemployed people is (b) decreasing.

Since the curve is concave upwards, the gradient is increasing. This (c) means that the unemployment rate is increasing.

Remember that the gradient, or fi rst derivative, measures

the rate of change.

ch2.indd 47 6/12/09 12:50:17 AM


1. For each curve, describe the sign

of dx

dy and

dx

d y2

2

.

(a)

(b)

(c)

(d)

(e)

2. The curve below shows the population of a colony of seals.

Describe the sign of the fi rst (a) and second derivatives.

How is the population rate (b) changing?

3. Infl ation is increasing, but the rate of increase is slowing. Draw a graph to show this trend.

4. Draw a sketch to show the shape of each curve:

(a) andf x f x0 0< <l m] ]g g (b) andf x f x0 0> <l m] ]g g (c) andf x f x0 0< >l m] ]g g (d) andf x f x0 0> >l m] ]g g

5. The size of classes at a local TAFE college is decreasing and the rate at which this is happening is decreasing. Draw a graph to show this.

6. As an iceblock melts, the rate at which it melts increases. Draw a graph to show this information.

7. The graph shows the decay of a radioactive substance.

Describe the sign of dtdM and .

dtd M

2

2

2.5 Exercises

ch2.indd 48 6/12/09 12:50:18 AM


8. The population P of fi sh in a certain lake was studied over time, and at the start the number of fi sh was 2500.

During the study, (a) .dtdP 0<

What does this say about the number of fi sh during the study?

If at the same time, (b) ,d

d P 0>2

2

what can you say about the population rate?

Sketch the graph of the (c) population P against t .

9. The graph shows the level of education of youths in a certain rural area over the past 100 years.

Describe how the level of education has changed over this period of time. Include mention of the rate of change.

10. The graph shows the number of students in a high school over several years.

Describe how the school population is changing over time, including the rate of change.

dx

dy0>

dx

dy0<

dx

dy0=

dx

d y0>

2

2

dx

d y0<

2

2

dx

d y0

2

2

=

Here is a summary of the shape of a curve given the fi rst and second derivatives.

ch2.indd 49 6/12/09 12:50:20 AM


Determining Types of Stationary Points

We can determine the type of a stationary point by looking at the fi rst and second derivatives together.

If • f x 0=l] g and ,f x 0>m] g there is a minimum turning point.

If • f x 0=l] g and ,f x 0<m] g there is a maximum turning point.

If • ,f x 0=l] g f x 0=m] g and concavity changes, then there is a horizontal point of infl exion.

Class Investigation

There are three mistakes in this argument. Can you fi nd all of them?

−

2

2

y x x

dx

dyx

x

2 2

221

21

= =

= =

1

1c m


i.e.

±

dxdy

xx

x

0

21 0

11`

=

=

==

So there are stationary points at x 1= and – .x 1=

Minimum• turning point: , .f x f x0 0>=l m] ]g g • Maximum turning point: , .f x f x0 0<=l m] ]g g • Horizontal infl exion: ,f x f x0 0==l m] ]g g and concavity changes.

The curve is concave upwards at this point.

The curve is concave downwards at this point.

ch2.indd 50 6/12/09 12:50:24 AM


EXAMPLES

1. Find the stationary points on the curve f x x x x2 3 12 73 2= − − +] g and distinguish between them.

Solution

f x x x6 6 122= − −l] g For stationary points, f x 0=l] g

( ) ( )

x x

x x

x x

6 6 12 0

2 02 1 0

i.e. 2

2

− − =− − =

− + =

( ) ( )

x

f

2 1

2 2 2 3 2 12 2 713

or2

` = −

= − − += −

3] ]g g

So ( , )2 13− is a stationary point.

( ) ( )f 1 2 1 3 1 12 1 7

14

3 2− = − − − − − +=

] ]g g

So – , 1 14^ h is a stationary point.

Now ( )

( ) ( )

(concave upwards)

f x x

f

12 6

2 12 2 618

0>

= −= −=

m

m

So ,2 13−^ h is a minimum turning point.

( ) ( )

(concave downwards)

f 1 12 1 618

0<

− = − −= −

m

So ,1 14−^ h is a maximum turning point.

2. Find the stationary point on the curve y x2 35= − and determine its nature.

Solution

y x10 4=l


i.e.

y

x

x

0

10 00

4

`

===

l

CONTINUED

ch2.indd 51 6/12/09 12:50:25 AM


hen ,W x y0 2 0 3

3

5= = −= −

] g


Now

When ,

y x

x y

40

0 40 0

3

3

== =

m

m ] g

0= So ,0 3−^ h is a possible point of infl exion.

Check concavity on the LHS and RHS:

x −1 0 1

dx

d y2

2

−40 0 40

Since concavity changes, ,0 3−^ h is a horizontal point of infl exion.

2.6 Exercises

1. Find the stationary point on the curve y x x2 12= − + and determine its nature.

2. Find the stationary point on the curve y x3 14= + and determine what type of point it is.

3. Find the stationary point on the curve y x x3 12 72= − + and show that it is a minimum point.

4. Determine the stationary point on the curve y x x2= − and show that it is a maximum turning point.

5. Show that the curve f x x2 53= −] g has an infl exion and fi nd its coordinates.

6. Does the function f x x2 35= +] g have a stationary point? If it does, determine its nature.

7. Find any stationary points on the curve f x x x x2 15 36 503 2= + + −] g and determine their nature.

8. Find the stationary points on the curve y x x x3 4 12 14 3 2= − − + and determine whether they are maximum or minimum turning points.

9. Find any stationary points on the curve y x4 12 4= −] g and determine their nature.

10. (a) Find any stationary points on the curve y x x x2 27 1203 2= − + and distinguish between them.

Find any points of infl exion (b) on the curve.

11. Find any stationary points on the curve y x x3 4= − −] g and determine their nature.

12. Find any stationary points on the curve y x x x8 16 14 3 2= + + − and determine their nature.

ch2.indd 52 6/12/09 12:50:26 AM


13. The curve y ax x4 12= − + has a stationary point where .x 3= −

Find the value of (a) a. Hence, or otherwise, (b)

determine the nature of the stationary point.

14. The curve y x mx x8 73 2= − + − has a stationary point where .x 1= − Find the value of m .

15. The curve y ax bx x 53 2= + − + has a point of infl exion at , .1 2−^ h Find the values of a and b .

Curve Sketching

We can sketch curves by fi nding all of their important features, such as stationary points, points of infl exion and intercepts. Here is a summary of strategies for sketching a curve.

Find 1. stationary points .y 0=l^ h Find 2. points of infl exion .y 0=m^ h Find intercepts on axes. 3.

For • x -intercept, y 0= For • y -intercept, x 0=

Find 4. domain and range. Find any 5. asymptotes or limits. Use 6. symmetry, odd or even functions. Draw up a 7. table of values.

EXAMPLES

1. Find any stationary points and points of infl exion on the curve f x 3 2x x x3 9 1= − − +] g and hence sketch the curve.

Solution

( )

and ( )

f x x x

f x x

3 6 9

6 6

2= − −= −

l

m

First, fi nd the stationary points.

For stationary points, ( )

i.e.

( ) ( )

or

f x

x x

x x

x x

x

0

3 6 9 0

2 3 03 1 0

1 3

2

2

`

=− − =− − =

− + == −

l

CONTINUED

We cannot always fi nd the x -intercept.

ch2.indd 53 6/12/09 12:50:26 AM


( )f 3 3 3

26== −

33 − 2 ( )9 3 1− +] ]g g

So ( , )3 26− is a stationary point.

( )f 1 1 1

6− = − −

=

3 23− ( )9 1 1− − +] ]g g


We use the second derivative to determine their type.

( ) ( )

(concave upwards)

f 3 6 3 612

0>

= −=

m

,3 26` −^ h is a minimum turning point

( ) ( )

0 (concave wards)

1 6 1 612

down<

− = − −= −

f m

,1 6` −^ h is a maximum turning point

Next, fi nd any points of infl exion.

For inflexions, ( )i.e.

f x

x

x

x

06 6 0

6 6

1

=− =

==

m

( )f 1 1 3 1 9 1 1

10

3 2= − − += −

] ]g g

x 0 1 2

f xm ] g −6 0 6

Since concavity changes, ,1 10−^ h is a point of infl exion. Next, try to fi nd intercepts on the axes.

For intercept, :

( ) ( )

For intercept, :

y x

f

x y

0

0 0 3 0 9 0 11

0

3 2

-

-

=

= − − +==

] g

i.e. x x x3 9 1 03 2− − + = This is too hard to solve.

ch2.indd 54 6/12/09 12:50:27 AM


Now sketch the graph using an appropriate scale so that all stationary points and points of infl exion fi t on the graph.

2. Sketch the curve ,y x2 13= + showing any important features.

Solution

dx

dyx6 2=


i.e.dx

dy

x

x

0

6 00

2

=

==

When ,x y0 2 0 13= = +] g 1=

So (0, 1) is a stationary point.

(0, 1) 12(0)

dx

d yx

dx

d y

12

0

At

2

2

2

2

=

=

=

So (0, 1) is a possible point of infl exion.

x −1 0 1

dx

d y2

2−12 0 12

Since concavity changes, (0, 1) is a horizontal infl exion.

For intercept,

i.e.

.

.

x y

x

x

x

x

0

2 1 0

2 1

0 50 8

3

3

3

-

`

=+ =

= −= −= −

For y -intercept, x = 0

CONTINUED

ch2.indd 55 6/12/09 12:50:28 AM


We already know this point. It is (0, 1), the infl exion. We can also fi nd a point on either side of the infl exion.

When ,x y1 2 1 11

3= − = − += −

] g

When ,x y1 2 1 13

3= = +=

] g

1. Find the stationary point on the curve f x 2x x3 4= − −] g and determine its type. Find the intercepts on the axes and sketch the curve.

2. Sketch ,y x x6 2 2= − − showing the stationary point.

3. Find the stationary point on the curve y x 1 3= −] g and determine its nature. Hence sketch the curve.

4. Sketch 3,y x4= + showing any stationary points.

5. Find the stationary point on the curve y x5= and show that it is a point of infl exion. Hence sketch the curve.

6. Sketch .f x x7=] g

7. Find any stationary points on the curve y x x x2 9 24 303 2= − − + and sketch its graph.

8. (a) Determine any stationary points on the curve .y x x6 73 2= + −

Find any points of infl exion (b) on the curve.

Sketch the curve. (c)

9. Find any stationary points and infl exions on the curve y x x6 33 2= − + and hence sketch the curve.

10. Find any stationary points and infl exions on the curve .y x x x2 9 3 2 3= + − − Hence sketch the curve.

11. Sketch the function ,f x x x x3 4 12 14 3 2= + − −] g showing all stationary points.

12. Find the stationary points on the curve ( )y x x4 2 2= − +] g and hence sketch the curve.

2.7 Exercises

ch2.indd 56 6/12/09 12:50:29 AM


13. Find all stationary points and infl exions on the curve .4( )y x x2 1 2= + −] g Sketch the curve.

14. Show that the curve x

y1

2+

=

has no stationary points. By considering the domain and range of the function, sketch the curve.

15. Find any stationary points

on the curve .x

xy2

2

−= By also

considering the domain of the curve, sketch its graph.

Maximum and Minimum Values

A curve may have maximum and minimum turning points, but they are not necessarily the maximum or minimum values of the function.

You sketched this curve on page 55.

EXAMPLE

This curve has a maximum turning point at , .1 6−^ h We call it a relative maximum point, since it does not give the maximum value of the graph. Similarly, the curve has a relative minimum turning point at , .3 26−^ h The curve does not have any absolute maximum or minimum value since it increases to infi nity and decreases to negative infi nity.

If we restrict the domain of the curve to, say, 4 4,≤ ≤x− then the curve will have an absolute maximum and minimum value. The absolute maximum is the greatest value of the curve in the domain. The absolute minimum is the least value of the curve in the domain.

In order to fi nd the maximum or minimum value of a curve, fi nd the values of the function at the endpoints of the domain as well as its turning points.

The relative maximum or minimum point is also called a local maximum or minimum point.

CONTINUED

ch2.indd 57 6/29/09 10:02:56 AM


When , ( )x y4 4 3 4 9 4 1

75

3= − = − − − − − += −

2] ]g g

When , ( )x y4 4 3 4 9 4 1

19

3 2= = − − += −

] g

By restricting the curve to ≤ ≤ ,x4 4− the maximum value is 6 and the minimum value is −75.

EXAMPLES

1. Find the maximum and minimum values of y for the function f x 2x x4 3= − +] g in the domain 4 3.≤ ≤x−

Solution

f x x2 4= −l] g

For stationary points, ( )i.e.

f x

x

x

x

02 4 0

2 4

2

=− =

==

l

( ) ( )f 2 2 4 2 3

1

2= − += −


( )

(concave upwards)f x 2

0>=m

,2 1` −^ h is a minimum turning point

The highest part of the curve is at y 6= and the lowest part is at y 75.= −

ch2.indd 58 6/12/09 12:50:31 AM


fEndpoints: ( )

( ) ( )f

4 4 4 4 335

3 3 4 3 30

2

− = − − − +== − +=

2] ]g g

The maximum value of y is 35 and the minimum value is −l in the domain 4 3≤ ≤x− .

2. Find the maximum and minimum values of the curve y x x2 14 2= − + for ≤ ≤ .x2 3−

Solution

y x x4 43= −l


i.e.

( )( ) ( )

or ±

y

x x

x x

x x x

x

0

4 4 0

4 1 04 1 1 0

0 1

3

2

`

=− =− =

+ − ==

l

2When , ( )x y1 1 2 1 1

0

4= = − +=

So (1, 0) is a stationary point.

When ,x y1 1 2 1 1

0= − = − − − +

=

4 2] ]g g


When ,x y0 0 2 0 1

1

4= = − +=

2] ]g g

So (0, 1) is a stationary point. Now y x12 42= −m

When ,

(concave upwards)

x y1 12 1 48

0>

2= = −=

m ] g

CONTINUED

ch2.indd 59 6/12/09 12:50:32 AM


So (1, 0) is a minimum turning point.

When ,

(concave upwards)

x y1 12 1 48

0>

= − = − −=

2m ] g

So (−1, 0) is a minimum turning point.

When ,

(concave downwards)

x y0 12 0 44

0<

2= = −= −

m ] g

So (0, 1) is a maximum turning point.

Endpoints: When ,x y2 2 2 2 1

9

4= − = − − − +=

2] ]g g

When ,x y3 3 2 3 1

64

4 2= = − +=

] g

The maximum value is 64 and the minimum value is 0 in the domain ≤ ≤x2 3− .

1. Sketch y x x 22= + − for 2 2≤ ≤x− and fi nd the maximum value of y .

2. Sketch f x x9 2= −] g over the domain 4 2.≤ ≤x− Hence fi nd the maximum and minimum values of the curve over this domain.

3. Find the maximum value of the curve y x x4 42= − + for 3 3.≤ ≤x−

4. Sketch f x x x x2 3 36 53 2= + − +] g for 3 3,≤ ≤x− showing any stationary points. Find the maximum and minimum values of the function for 3 3.≤ ≤x−

2.8 Exercises

ch2.indd 60 6/12/09 12:50:33 AM


5. Find the maximum value of y for the curve y x 35= − for 2 1.≤ ≤x−

6. Sketch the curve f x x x3 16 52= − +] g for 0 4≤ ≤x and fi nd the maximum and minimum values of the function over this domain.

7. Find the relative and absolute maximum and minimum values of the function f x x x x3 4 12 34 3 2= + − −] g for 2 2.≤ ≤x−

8. Sketch y x 23= + over the domain 3 3≤ ≤x− and fi nd its minimum and maximum values in that domain.

9. Sketch y x 5= + for 4 4≤ ≤x− and fi nd its maximum and minimum values.

10. Show that x

y2

1−

= has no

stationary points. Find the maximum and minimum values of the curve for 3 3.≤ ≤x−

Problems Involving Maxima and Minima

Often a problem involves fi nding maximum or minimum values. For example, a salesperson wants to maximise profi t; a warehouse manager wants to maximise storage; a driver wants to minimise petrol consumption; a farmer wants to maximise paddock size.

PROBLEM

One disc 20 cm in diameter and one 10 cm in diameter are cut from a disc of cardboard 30 cm in diameter. Can you fi nd the largest disc that can be cut from the remainder of the cardboard?

EXAMPLES

1. The equation for the expense per year (in units of ten thousand dollars) of running a certain business is given by ,E x x6 122= − + where x is the number (in hundreds) of items manufactured.

Find the expense of running the business if no items are (a) manufactured.

Find the number of items needed to minimise the expense of the (b) business.

Find the minimum expense of the business. (c)

CONTINUED

ch2.indd 61 6/12/09 12:50:34 AM


Solution

(a) When , ( )x E0 0 6 0 1212

2= = − +=

So the expense of running the business when no items are manufactured is 12 × $10 000, or $120 000 per year.

(b) dxdE x2 6= −


i.e.

(concave upwards)

dxdE

x

x

x

dxd E

0

2 6 0

2 6

3

2 0>2

2

=

− ===

=

x 3` = gives a minimum value So 300 items manufactured each year will give the minimum expenses.

(c) When ,x E3 3 6 3 123

2= = − +=

] g

So the minimum expense per year is $30 000.

2. The formula for the volume of a pond is given by ,V h h h2 12 18 503 2= − + + where h is the depth of the pond in metres. Find the maximum volume of the pond.

Solution

V h h6 24 182= − +l

For stationary points, 0

i.e. 6 24 18 0

( )( )

V

h h

h h

h h

4 3 01 3 0

2

2

=− + =

− + =− − =

l

So h = 1 or 3

V h12 24= −m

When , ( )h V1 12 1 24= = −m

(concave downwards)

12

0<= −

So h = 1 gives a maximum V.

When 3, 12(3) 24h V= = −m

(concave wards)up

12

0>=

E is the expense in units of ten thousand dollars.

This will give any maximum or minimum values of E .

x is the number of items in hundreds.

ch2.indd 62 6/12/09 12:50:35 AM


So h = 3 gives a minimum V.

When ,h V1 2 1 12 1 18 1 50

85= = − + +

=

3 2] ] ]g g g

So the maximum volume is 58 m 3 .

In the above examples, the equation is given as part of the question. However, we often need to work out an equation before we can start answering the question. Working out the equation is often the hardest part!

EXAMPLES

1. A rectangular prism has a base with length twice its breadth. The volume

is to be 300 cm 3 . Show that the surface area is given by .S x x4 9002= +

Solution

Volume:

2

( )

´ ´V lbh

x x h

x h

xh

300 2

2300 1

2

2`

===

=

Surface area:

( )

( )

( )

S bh lh

x xh xh

x xh

x xh

lb2

2 2 2

2 2 3

4 6

2

2

2

= + += + += += +

Now we substitute (1) into this equation.

S x x

x

x x

4 62300

4 900

22

2

$= +

= +

2. ABCD is a rectangle with AB = 10 cm and BC = 8 cm. Length AE = x cm and CF = y cm.

CONTINUED

ch2.indd 63 6/12/09 12:50:35 AM


Show that triangles (a) AEB and CBF are similar. Show that (b) xy = 80. Show that triangle (c) EDF has area given by .A x x80 5 320= + +

Solution

(a) ° (given)(corresponding , )(similarly, )

EAB BCFBFC EBABEA FBC

AB DCAD BC

90s

+ ++ ++ +

+ <<

= ===

Δ AEB and Δ CBF are similar (AAA)

(b) (similar triangles have sides in proportion)yx

xy

10880

` =

=

Side (c) FD = y + 10 and side ED = x + 8

( ) ( )

( )

A bh

y x

xy y x

21

21 10 8

21 8 10 80

=

= + +

= + + +

We need to eliminate y from this equation.

If ,

then

xy

y x

80

80

=

=

Substituting xy 80= and y x80= into the area equation gives:

( )

( )

( )

A xy y x

x x

x x

x x

21 8 10 80

21 80 8 80 10 80

21 160 640 10

80 320 5

$

= + + +

= + + +

= + +

= + +

ch2.indd 64 6/12/09 12:50:36 AM


1. The area of a rectangle is to be 50 m 2 .

Show that the equation of its

perimeter is given by .xP x 1002= +

2. A rectangular paddock on a farm is to have a fence with a 120 m perimeter. Show that the area of the paddock is given by .A x x60 2= −

3. The product of two numbers is 20. Show that the sum of the

numbers is .xS x 20= +

4. A closed cylinder is to have a volume of 400 cm 3 .

Show that its surface area is

.π rS r 8002 2= +

5. A 30 cm length of wire is cut into 2 pieces and each piece bent to form a square.

Show that (a) .y x30= − Show that the total area (b)

of the 2 squares is given by

.x xA8

30 4502 − +=

6. A timber post with a rectangular cross-sectional area is to be cut out of a log with a diameter of 280 mm as shown.

Show that (a) .y x78400 2= − Show that the cross-sectional (b)

area is given by A x x78400 2= − .

7. A 10 cm ´ 7 cm rectangular piece of cardboard has equal square corners with side x cut out and folded up to make an open box.

Show that the volume of the box is V x x x70 34 42 3= − + .

8. A travel agency calculates the expense E of organising a holiday per person in a group of x people as .E x200 400= + The cost C for each person taking a holiday is .C x900 100= − Show that the profi t to the travel agency on a holiday with a group of x people is given by .P x x700 500 2= −

2.9 Exercises

ch2.indd 65 6/12/09 12:50:37 AM


9. Joel is 700 km north of a town, travelling towards it at an average speed of 75 km/h. Nick is 680 kmeast of the town, travelling towards it at 80 km/h.

Show that after t hours, thedistance between Joel andNick is given by .d t t952400 213800 12025 2= − +

10. Sean wants to swim from point A to point B across a 500 m wide river, then walk along the river bank to point C . The distance along the river bank is 7 km.

If he swims at 5 km/h and walks at 4 km/h, show that the time taken to reach point C is given by

.

.x xt

50 25

472 + + −=

When you have found the equation, you can use calculus to fi nd the maximum or minimum value. The process is the same as for fi nding stationary points on curves.

Always check that an answer gives a maximum or minimum value. Use the second derivative to fi nd its concavity, or if the second derivative is too hard to fi nd, check the fi rst derivative either side of this value.

EXAMPLES

1. The council wanted to make a rectangular swimming area at the beach using a straight cliff on one side and a length of 300 m of sharkproof netting for the other three sides. What are the dimensions of the rectangle that encloses the greatest area?

Solution

Let the length of the rectangle be y and the width be x .

There are many differently shaped rectangles that could

have a perimeter of 300 m.

ch2.indd 66 6/12/09 12:50:38 AM


Perimeter:

( )

Area

( ) substituting

x y

y x

A xy

x x

x x

dxdA x

2 300

300 2 1

300 2 1

300 2

300 4

m

2

`

+ == −== −

= −

= −

] g7 A


i.e.dxdA

x

x

x

0

300 4 0

300 4

75

=

− ===

(concave downwards)dxd A 4 0<

2

2

= −

So x 75= gives maximum A.

When , ( )x y75 300 2 75

150

= = −=

So the dimensions that give the maximum area are 150 m × 75 m.

2. Trinh and Soi set out from two towns. They travel on roads that meet at right angles, and they walk towards the intersection. Trinh is initially 15 km from the intersection and walks at 3 km/h. Soi is initially 10 km from the intersection and walks at 4 km/h.

Show that their distance apart after (a) t hours is given by .D t t25 170 3252 2= − +

Hence fi nd how long it takes them to reach their minimum distance (b) apart.

Find their minimum distance apart. (c)

Solution

After (a) t hours, Trinh has walked 3 t km. She is now t15 3− km from the intersection. After t hours, Soi has walked 4 t km. He is now t10 4− km from the intersection.

CONTINUED

ch2.indd 67 6/12/09 12:50:40 AM


1. The height, in metres, of a ball is given by the equation ,h t t16 4 2= − where t is time in seconds. Find when the ball will reach its maximum height, and what the maximum height will be.

2. The cost per hour of a bike ride is given by the formula ,C x x15 702= − + where x is the distance travelled in km. Find the distance that gives the minimum cost.

3. The perimeter of a rectangle is 60 m and its length is x m. Show that the area of the rectangle is given by the equation .A x x30 2= − Hence fi nd the maximum area of the rectangle.

4. A farmer wants to make a rectangular paddock with an area of 4000 m 2 . However, fencing costs are high and she wants the paddock to have a minimum perimeter.

2.10 Exercises

By Pythagoras’ theorem:

D t t

t t t t

t t

15 3 10 4225 90 9 100 80 16

25 170 325

2 2

2 2

2

= − + −= − + + − += − +

2] ]g g

(b) DtdD t50 170

2

= −

For stationary points, 0

i.e. 50 170 0

.

dtdD

t

t

t

50 170

3 4

2

=

− ===

(concave upwards)

So . gives minimum .dt

d D

t D

50 0

3 4

>2

2 2

2

=

=

Now . minutes minutes´0 4 60 24= . hours hours3 4 3` = and 24 minutes So Trinh and Soi are a minimum distance apart after 3 hours and 24 minutes. (c) When . , .t D3 4 25 3 42= = ( . )170 3 4 325− +2] g

D

36

366

=

==

So the minimum distance apart is 6 km.

ch2.indd 68 6/12/09 12:50:40 AM


Show that the perimeter is (a) given by the equation

.xP x 80002= +

Find the dimensions of (b) the rectangle that will give the minimum perimeter, correct to 1 decimal place.

Calculate the cost of fencing (c) the paddock, at $48.75 per metre.

5. Bill wants to put a small rectangular vegetable garden in his backyard using two existing walls as part of its border. He has 8 m of garden edging for the border on the other two sides. Find the dimensions of the garden bed that will give the greatest area.

6. Find two numbers whose sum is 28 and whose product is a maximum.

7. The difference of two numbers is 5. Find these numbers if their product is to be minimum.

8. A piece of wire 10 m long is broken into two parts, which are bent into the shape of a rectangle and a square as shown. Find the dimensions x and y that make the total area a maximum.

9. A box is made from an 80 cm by 30 cm rectangle of cardboard by cutting out 4 equal squares of side x cm from each corner. The edges are turned up to make an open box.

Show that the volume of the (a) box is given by the equation .V x x x4 220 24003 2= − +

Find the value of (b) x that gives the box its greatest volume.

Find the maximum volume of (c) the box.

10. The formula for the surface area of a cylinder is given by πS r h2= + ,r ] g where r is the radius of its base and h is its height. Show that if the cylinder holds a volume of π54 3,m the surface area is given by the

equation π πS r1082= 2r .+

Hence fi nd the radius that gives the minimum surface area.

11. A silo in the shape of a cylinder is required to hold 8600 m 3 of wheat.

ch2.indd 69 6/29/09 10:02:58 AM


Find an equation for the (a) surface area of the silo in terms of the base radius.

Find the minimum surface (b) area required to hold this amount of wheat, to the nearest square metre.

12. A rectangle is cut from a circular disc of radius 6 cm. Find the area of the largest rectangle that can be produced.

13. A poster consists of a photograph bordered by a 5 cm margin. The area of the poster is to be 400 cm 2 .

Show that the area of the (a) photograph is given by the

equation 500 10A x x4000= − −

Find the maximum area (b) possible for the photograph.

14. The sum of the dimensions of a box with a square base is 60 cm. Find the dimensions that will give the box a maximum volume.

15. A surfboard is in the shape of a rectangle and semicircle, as shown. The perimeter is to be 4 m. Find the maximum area of the surfboard, correct to 2 decimal places.

16. A half-pipe is to be made from a rectangular piece of metal of length x m. The perimeter of the rectangle is 30 m.

Find the dimensions of (a) the rectangle that will give the maximum surface area.

Find the height from the (b) ground up to the top of the half-pipe with this maximum area, correct to 1 decimal place.

17. Find the least surface area, to the nearest cm 2 , of a closed cylinder that will hold a volume of 400 cm 3 .

18. The picture frame shown below has a border of 2 cm at the top and bottom and 3 cm at the sides. If the total area of the border is to be 100 cm 2 , fi nd the maximum area of the frame.

19. A 3 m piece of wire is cut into two pieces and bent around to form a square and a circle. Find the size of the two lengths, correct to 2 decimal places, that will make the total area of the square and circle a minimum.

ch2.indd 70 6/12/09 12:50:43 AM


20. Two cars are travelling along roads that intersect at right angles to one another. One starts 200 km away and travels towards the intersection at 80 kmh -1 , while the other starts at 120 km away and travels towards the intersection at 60 kmh -1 .

Show that their distance apart after t hours is given by d t t10 000 46 400 54 4002 2= − + and hence fi nd their minimum distance apart.

21. X is a point on the curve .y x x2 52= − + Point Y lies directly below X and is on the curve .y x x4 2= −

y

x

X

Y

Show that the distance (a) d between X and Y is given by .d x x2 6 52= − +

Find the minimum distance (b) between X and Y .

22. A park is to have a dog-walking enclosure in the shape of a rectangle with a semi-circle at the end as shown, and a perimeter of 1200 m.

x

y

Show that (a)

.πy

x4

2400 2− −=

y

Show that the area of the (b) enclosure is given by

.πy y

A8

4800 4 2 2− −=

y

Find the dimensions (c) x and y (to the nearest metre) that maximises the area.

23. Points , ,,A a bB0 0−^ ^h h and ,O 0 0^ h form a triangle as shown and AB always passes through the point , .1 2−^ h

y

x

B (0, b)

A (-a, 0)

Show that (a) .ba

a1

2=−

Find values of (b) a and b that give the minimum area of triangle OAB .

24. Grant is at point A on one side of a 20 m wide river and needs to get to point B on the other side 80 m along the bank as shown.

x

B

20 m

80 m

A

Grant swims to any point on the other bank and then runs along the side of the river to point B . If he can swim at 7 km/h and run at 11 km/h, fi nd the distance he

ch2.indd 71 6/12/09 12:50:44 AM


Class Challenge

Can you solve either of these problems?

Heron’s problem 1. One boundary of a farm is a straight river bank, and on the farm stands a house and some distance away, a shed; each is sited away from the river bank. Each morning the farmer takes a bucket from his house to the river, fi lls it with water, and carries the water to the shed. Find the position on the river bank that will allow him to walk the shortest distance from house to river to shed. Further, describe how the farmer could solve the problem on the ground with the aid of a few stakes for sighting.

Lewis Carroll’s problem 2. After a battle at least 95% of the combatants had lost a tooth, at least 90% had lost an eye, at least 80% had lost an arm, and at least 75% had lost a leg. At least how many had lost all four?

Primitive Functions

This chapter uses differentiation to fi nd the gradient of tangents and stationary points of functions.

swims ( x ) to minimise the time taken to reach point B . Answer to the nearest metre.

25. A truck travels 1500 km at an hourly cost given by s 90002 + cents where s is the average speed of the truck.

Show that the cost for(a) the trip is given by

.sC s 90001500= +b l

Find the speed that minimises (b) the cost of the trip.

Find the cost of the trip to the (c) nearest dollar.

ch2.indd 72 6/12/09 12:50:45 AM


EXAMPLE

Sketch the primitive function (the original function) given the derivative function below.

Solution

Reversing what you would do to sketch the derivative function, the parts at the x -axis have a zero gradient, so show stationary points on the original function. There are stationary points at x 1 and x 2 .

The parts of the graph above the x -axis show a positive gradient, so the original function is increasing. This happens to the left of x 1 and to the right of x 2 .

The parts of the graph below the x -axis show a negative gradient, so the original function is decreasing. This happens between x 1 and x 2 .

Sketching this information gives:

CONTINUED

Sometimes you may know f xl] g and need to fi nd the original function, .f x] g This process is called anti-differentiation , and the original function is called the primitive function.

ch2.indd 73 6/12/09 12:50:47 AM


The only problem is, we do not have enough information to know where the curve turns around. There are many choices of where we could put the graph:

The primitive function gives a family of curves.

The examples below use what we know about differentiation to fi nd out how to reverse this to fi nd the primitive function.

EXAMPLES

1. Differentiate x 2 . Hence fi nd a primitive function of 2 x .

Solution

The derivative of x 2 is 2 x ` a primitive function of 2 x is x 2

2. Differentiate .x 52 + Hence fi nd a primitive function of 2 x .

Solution

The derivative of x 52 + is 2 x a primitive function of 2 x is x 52 +

3. Differentiate .x 32 − Hence fi nd a primitive function of 2 x .

Solution

The derivative of x 32 − is 2 x a primitive function of 2 x is x 32 −

ch2.indd 74 6/12/09 12:50:48 AM


Thus 2 x has many different primitive functions. In general, the primitive of 2 x is ,x C2 + where C is a constant (real number).

EXAMPLES

1. Find the primitive of x .

Solution

The derivative of x 2 is 2 x .

The derivative of x2

2

is x . So the primitive of x is .x C2

2

+

2. Find the primitive of x 2 .

Solution

The derivative of x 3 is 3 x 2 .

The derivative of x3

3

is x 2 .

So the primitive of x 2 is .x C3

3

+

EXAMPLES

1. The gradient of a curve is given by .dx

dyx x6 82= + If the curve passes

through the point , ,1 3−^ h fi nd the equation of the curve.

Solution

dx

dyx x

y x x C

y x x C

6 8

63

82

2 4

2

3 2

3 2`

= +

= + +

= + +

c cm m

Continuing this pattern gives the general primitive function of .xn

If dx

dyxn= then

nxy C

1

n 1

+= +

+

where C is a constant

Proof

dxd

nx C

nn x

x1 1

1n n

n

1 1 1

++ =

++

=

+ + −c ]m g

CONTINUED

ch2.indd 75 6/12/09 12:50:49 AM


The curve passes through ,1 3−^ h

C

C

3 2 1 12 4

9

` − == + +

− =

43 2+ C+] ]g g

Equation is .y x x2 4 93 2= + −

2. If f x x6 2= +m] g and = f ,0=f 1 2−l] ]g g fi nd f 3 .] g

Solution

( )

( )

f x x

f x x x C

x x C

6 2

62

2

3 2

2

2

= +

= + +

= + +

m

l c m

( )1 0=So

( )

( )

C

x x x

f x x x x C

x x x C

0 3 15

3 2 5

33

22

5

5

2

3 2

3 2

=− =

= + −

= + − +

= + − +

( ) C2 1+ +2

Now f

f`

l

l

]

c c

g

m m

Now ( )

So

( )

( ) ( )

f

C

C

f x x x x

f

2 0

0 2 28 4 10

6

5 6

3 3 3 5 3 627 9 15 6

15

3

3 2

3 2

`

− =

= − −= − + + +

− == + − −= + − −= + − −=

2+ ( ) C5 2− − +] ]g g

2.11 Exercises

1. Find the primitive function of (a) x2 3− (b) x x8 12 + + (c) x x45 3− (d) x 1− 2] g 6 (e)

2. Find f ( x ) if (a) ( )f x x x6 2= −l (b) ( )f x x x3 74 2= − +l (c) ( )f x x 2= −l (d) ( ) ( )( )f x x x1 3= + −l

(e) 2( )f x x=1

l

ch2.indd 76 6/12/09 12:50:50 AM


3. Express y in terms of x if

(a) dx

dyx5 94= −

(b) dx

dyx x24 2= −− −

(c) dx

dy x x5

32= −

(d) dx

dy

x2

2=

(e) dx

dyx x

32 13= − +

4. Find the primitive function of (a) x (b) x 3−

(c) x1

8

(d) 2 3x x2+− −1 2

(e) x x27 2−− −

5. If dx

dyx x3 523= − + and y 4= when

,x 1= fi nd an equation for y in terms of x .

6. If ( )f x x4 7= −l and ( ) ,f 2 5= fi nd ( ) .f x

7. Given f x x x3 4 22= + −l] g and ( )f 3 4− = , fi nd f 1] g .

8. Given that the gradient of the tangent to a curve is given

by dx

x2 6= −dy

and the curve

passes through , ,2 3−^ h fi nd the equation of the curve.

9. If tdtdx 3−= 2] g and x 7= when

,t 0= fi nd x when .t 4=

10. Given dx

d y8

2

2

= and dx

dy0= and

y 3= when ,x 1= fi nd the equation of y in terms of x .

11. If dx

d yx12 6

2

2

= + and dx

dy1= at the

point , ,1 2− −^ h fi nd the equation of the curve.

12. If ( )f x x6 2= −m and ,f 7= =f 2 2l] ]g g fi nd the function .f x] g

13. Given ( ) , ( )f x x f5 0 34= =m l and f 1− ,1=] g fi nd f 2 .] g

14. If dx

d yx2 1

2

2

= + and there is a

stationary point at (3, 2), fi nd the equation of the curve.

15. A curve has dx

d yx8

2

2

= and the

tangent at (−2, 5) makes an angle of °45 with the x -axis. Find the equation of the curve.

16. The tangent to a curve with

dx

d yx2 4

2

2

= − makes an angle

of °135 with the x -axis in the positive direction at the point , .2 4−^ h Find its equation.

17. A function has a tangent parallel to the line x y4 2 0− − = at the point ,0 2−^ h and ( )f x x x12 6 42= − +m . Find the equation of the function.

18. A curve has dx

d y6

2

2

= and the

tangent at ,1 3−^ h is perpendicular to the line .x y2 4 3 0+ − = Find the equation of the curve.

19. A function has ( )f 1 3=l and ( ) .f 1 5= Evaluate f (−2) given f x x6 18= +m] g .

20. A curve has dx

d yx12 24

2

2

= − and

a stationary point at , .1 4^ h Evaluate y when x 2= .

ch2.indd 77 6/12/09 12:50:51 AM


1. Find the stationary points on the curve y x x x6 9 113 2= + + − and determine their nature.

2. Find all x -values for which the curve y x x x2 7 3 13 2= − − + is concave upwards.

3. A curve has .dx

dyx x6 12 52= + − If the

curve passes through the point (2, −3), fi nd the equation of the curve.

4. If ( ) ,f x x x x3 2 25 4 3= − + − fi nd (a) ( )f 1− (b) ( )f 1−l (c) f 1−m] g

5. The height in metres of an object thrown up into the air is given by h t t20 2 2= − where t is time in seconds. Find the maximum height that the object reaches.

6. Find the stationary point on the curve y x2 2= and determine its nature.

7. Find the domain over which the curve y x x5 6 3 2= − − is decreasing.

8. For the curve y x x x3 8 48 14 3 2= + − + fi nd any stationary points and (a)

determine their nature sketch the curve. (b)

9. Find the point of infl exion on the curve .y x x x2 3 3 23 2= − + −

10. If ( ) ,f x x15 12= +m and ( ) ,2 5=( )f f2 = l fi nd ( ) .f x

11. A soft drink manufacturer wants to minimise the amount of aluminium in its cans while still holding 375 mL of soft drink. Given that 375 mL has a volume of 375 cm 3 ,

show that the surface area of a can is (a)

given by πS r2 7502= r +

fi nd the radius of the can that gives (b) the minimum surface area.

12. For the curve ,y x x x3 8 64 3 2= + + fi nd any stationary points (a) determine their nature (b) sketch the curve for (c) 3 3≤ ≤x− .

13. A rectangular prism with a square base is to have a surface area of 250 cm 2 .

Show that the volume is given by(a)

.x xV2

125 3−=

Find the dimensions that will give (b) the maximum volume.

14. Find all x -values for which the curve y x x3 18 42= − + is decreasing.

15. If ,dx

d yx6 6

2

2

= + and there is a stationary

point at (0, 3), fi nd the equation of the curve.

16. The cost to a business of manufacturing x products a week is given by .C x x300 90002= − + Find the number of products that will give the minimum cost each week.

17. Show that y x3= − is monotonic decreasing for all .≠x 0

Test Yourself 2

ch2.indd 78 6/12/09 12:50:52 AM


18. Sketch a primitive function for each graph.

(a)

(b)

(c)

19. A 5 m length of timber is used to border a triangular garden bed, with the other sides of the garden against the house walls.

Show that the area of the garden is(a)

.xA x21 25 2−=

Find the greatest possible area of the (b) garden bed.

20. For the curve ,y x x x3 72 53 2= + − + fi nd any stationary points on the (a)

curve and determine their nature fi nd any points of infl exion (b) sketch the curve .(c)

21. Find the domain over which the curve y x x x3 7 53 2= − + − is concave downwards .

22. A function has ( )f 3 5=l and ( ) .f 3 2= If ( ) ,f x x12 6= −m fi nd the equation of the function.

23. Find any points of infl exion on the curve y x x x6 2 14 3= − + + .

24. Find the maximum value of the curve y x x x3 24 13 2= + − − in the domain ≤ ≤x5 6− .

25. A function has ( )f 2 0<l and ( ) .f 2 0<m Sketch the shape of the function near x 2= .

ch2.indd 79 6/12/09 12:50:54 AM


1. Find the fi rst and second derivatives of

.x

x4 1

52 3+−

] g

2. Sketch the curve ,y x x 2 3= −] g showing any stationary points and infl exions.

3. Find all values of x for which the curve y x x x4 21 24 53 2= − − + is increasing.

4. Find the maximum possible area if a straight 8 m length of fencing is placed across a corner to enclose a triangular space.

5. Find the greatest and least values of the function ( )f x x x x4 3 183 2= − − in the domain ≤ ≤ .x2 3−

6. Show that the function 3( ) ( )f x x2 5 3= − has a horizontal point of infl exion at . , .0 6 0^ h

7. Two circles have radii r and s such that .r s 25+ = Show that the sum of areas of the circles is least when .r s=

8. The rate of change of V with respect to t

is given by .dtdV t2 1 2= −] g If V 5= when

,t21= fi nd V when .t 3=

9. (a) Show that the curve y x 1= − has no stationary points.

Find the domain and range of the (b) curve.

Hence sketch the curve. (c)

10. Find the radius and height, correct to 2 decimal places, of a cylinder that holds 200 cm 3 , if its surface area is to be a minimum.

11. A curve passes through the point (0, −1) and the gradient at any point is given by x x3 5+ −] ]g g . Find the equation of the curve.

12. The cost of running a car at an average

speed of ν km/h is given by c v15080

2

= +

cents per hour. Find the average speed, to the nearest km/h, at which the cost of a 500 km trip is a minimum.

13. Find the equation of a curve that is always concave upwards with a stationary point at (−1, 2) and y -intercept 3.

14. Given ( ) ( )( )f x x x x3 1= − +l and ( )f 0 0= fi nd the equation for (a) ( )f x fi nd any stationary points on the (b)

function sketch the function .(c)

15. The volume of air in a cubic room on a submarine is given by the formula ,V x x x4 27 24 23 2= − + − + where x is the side of the room in metres. Find the dimensions of the room that will give the maximum volume of air.

16. If ( )f x x 9= −m and ( ) ( ) ,f f1 2 7− = − =l fi nd ( ) .f 3

17. Given 5dx

dy

x1

2=

−^ h and y 1= when ,x 4=

fi nd

(a) dx

d y2

2

when x 6=

(b) y when x 6= .


ch2.indd 80 6/12/09 12:50:56 AM


18. Show that y xn= has a stationary point at (0, 0) where n is a positive integer.

If (a) n is even, show that (0, 0) is a minimum turning point.

If (b) n is odd, show that (0, 0) is a point of infl exion.

19. Find the maximum possible volume if a rectangular prism with a length twice its breadth has a surface area of 48 cm 2 .

20. (a) Find the stationary point on the curve y x 1k= + where k is a positive integer.

What values of (b) k give a minimum turning point?

21. Find the minimum and maximum values

of yxx

93

2=

−+ in the domain ≤ ≤ .x2 2−

22. The cost of running a car at an average

speed of km/hV is given by Vc75

1002

= +

cents per hour. Find the average speed (to the nearest km/h) at which the cost of a 1000 km trip is a minimum.

ch2.indd 81 6/12/09 12:50:58 AM

TERMINOLOGYTERMINOLOGY

Defi nite integral: The integral or primitive function restricted to a lower and upper boundary. It has the

notation ( )f x dxa

b# and geometrically represents the area

between the curve ( )f xy = , the x-axis and the ordinates x = a and x = b

Even function: A function where ( ) ( )f x f x− = . It is symmetrical about the y-axis

Indefi nite integral: General primitive function represented

by ( )f x dx# Integration: The process of fi nding a primitive function

Odd function: A function where ( ) ( )f x f x− = − . An odd function has rotational symmetry about the origin

TERMINOLOGY

Integration

3

ch3.indd 82 6/11/09 11:38:14 PM

83Chapter 3 Integration

DID YOU KNOW?

Box text...

DID YOU KNOW?

Integration has been of interest to mathematicians since very early times. Archimedes (287–212 BC)found the area of enclosed curves by cutting them into very thin layers and fi nding their sum. He found the formula for the volume of a sphere this way. He also found an estimation of π , correct to 2 decimal places.

INTRODUCTION

INTEGRATION IS THE PROCESS of fi nding an area under a curve . It is an important process in many areas of knowledge , such as surveying , physics and the social sciences . In this chapter , you will look at approximation methods of integrating, as well as shorter methods that lead to fi nding areas and volumes . You will learn how integration and differentiation are related .

Area within curve Archimedes

Approximation Methods

Mathematicians used rectangles in order to fi nd the approximate area between a curve and the x -axis.

ch3.indd 83 6/29/09 10:30:17 AM


EXAMPLES

1. Find an approximation for xdx

1

4# using the trapezoidal rule.

Solution

dxxdx

x1

1

4

1

4=# #

( ) ( ) [ ( ) ( )]

( ) [ ( ) ( )]

f x dx b a f a f b

xdx f f4 1 1 4

21

214

a

b

1

Z

Z

− +

− +

#

#

NOTATION

The area of each rectangle is ( ) δf x x where ( )f x is the height and δx is the width of each rectangle. As ,δx 0" sum of rectangles " exact area.

Area ( )

( )

δlim f x x

f x dx

δx 0Σ=

="

#

Now there are other, more accurate ways to fi nd the area under a curve. However, the notation is still used.

The symbol # comes from S , the sum of rectangles.

Trapezoidal rule

The trapezoidal rule uses a trapezium for the approximate area under a curve. A trapezium generally gives a better approximation to the area than a rectangle.

( ) ( ) [ ( ) ( )]f x dx b a f a f b21

a

bZ − +#

Proof

( )

( ) [ ( ) ( )]

A h a b

b a f a f b

21

21

= +

= − +

ch3.indd 84 6/11/09 11:38:32 PM


( )

.

´

21 3

11

41

23

45

1 875

= +

=

=

c m

2. Find an approximation for x dx1

3

0# using the trapezoidal rule with

2 subintervals.

Solution

( ) ( ) [ ( ) ( )]f x dx b a f a f b

x dx x dx x dx

21

1 0.5

.

1

a

b

3

0

3

0

3

0 5

Z

Z

− +

+

#

# # #

( . ) [ ( ) ( . )] ( . ) [ ( . ) ( )]

( . ) ( . ) ( . ) ( . )

. ( . ) . ( . )

.

f f f f0 5 0 0 0 5 1 0 5 0 5 1

0 5 0 0 5 0 5 0 5 1

0 25 0 125 0 25 1 125

0 3125

3 3 3 3

21

21

21

21

= − + + − +

= + + +

= +=

The function f (x) is x1

.

Two subintervals mean 2 trapezia.

The function is x3.

There is a more general formula for n subintervals. Several trapezia give a more accurate area than one. However, you could use the fi rst formula several times if you prefer using it.

ch3.indd 85 6/11/09 11:38:33 PM


( ) [( ) ( )]. . .f x dx h y y y y y y2

2 –n na

b

0 1 2 3 1Z + + + + + +#

here width of each trapeziumw h nb a= − ^ h

Proof

( ) ( ) ( ) ( ) . . . ( )f x dx h y y h y y h y y h y y2 2 2 2a

b

n n0 1 1 2 2 3 1Z + + + + + + + +−#

( )

[( ) ( )]

. . .

. . .

h y y y y y y

h y y y y y y

22 2 2 2

22

n n

n n

0 1 2 3 1

0 1 2 3 1

= + + + + + +

= + + + + + +

−

−

Seven subintervals mean 7 trapezia.

EXAMPLES

1. Find an approximation for ( ) ,t dt314

2

0+# using the trapezoidal rule

with 7 subintervals.

Solution

ch3.indd 86 6/11/09 11:38:34 PM


There are 4 trapezia.

( ) [( ) ( )]

( ) [( ) ( )]

. . .f x dx h y y y y y y

t dt h y y y y y y y y

22

32

2

n na

b

0 1 2 3 1

20 7 1 2 3 4 5 60

14

Z

Z

+ + + + + +

+ + + + + + + +

−#

#

where

( ) [( ) ( )] [( ) ( )

( ) ( ) ( ) ( )]}

h

t dt

714 0 2

322 0 3 14 3 2 2 3 4 3

6 3 8 3 10 3 12 3966

142 2 2 2 2

0

2 2 2 2

Z

= − =

+ + + + + + + +

+ + + + + + + +=

"#

2. Find an approximation for ,dx

x 12

2

3

−# using the trapezoidal rule

with 4 subintervals, correct to 3 decimal places.

Solution

( ) [( ) ( )]f x dx h y y y y y2

24 3a

b

0 1 2+ + + +Z#

here

.

w h nb a

43 2

0 25

= −

= −

=

.. . .x

dx1

22

0 252 1

23 1

2 22 25 1

22 5 1

22 75 1

23

2 − −+

−+

−+

−+

−. [( ) ( . . . )]

.

0 125 2 1 2 1 6 1 3333 1 1429

1 394

+ + + +=

Z

Z

c cm m< F#

Application

When surveyors need to fi nd the area of an irregular piece of land, they measure regular strips and use an approximation method such as the trapezoidal rule.

CONTINUED

ch3.indd 87 6/11/09 11:38:37 PM


1. x dx2

1

2# using 1 subinterval.

2. ( )x dx13

0

2+# using 1 subinterval.

3. xdx

1

5# using 1 subinterval.

4. x

dx31

2

+# using 1 subinterval.

5. x dx3

1

3# using

1 subinterval (a) 2 subintervals. (b)

6. logx dx3

2# using 2 subintervals.

7. x

dx4

2

0 +# using 2 strips.

3.1 Exercises

EXAMPLE

The table below gives the measurements of a certain piece of land:

x m 0 1 2 3 4

y m 3.7 5.9 6.4 5.1 4.9 Using the trapezoidal rule to fi nd its area, correct to 2 decimal places:

( )2

[( ) 2( )]

Where

44 0

1

( ) [( ) 2 ( )]

[(3.7 4.9) 2(5.9 6.4 5.1)]

21.7

f x dxh

y y y y y

h nb a

f x dx y y y y y

0 4 1 2 3

0 4 1 2 30 21

21

aZ

Z

= −

= −

= +

=

+ + + +

=

+ + + +

+ + +

b

4

#

#

So the area of the land is approximately 21.7 m 2 .

Use the trapezoidal rule to fi nd an approximation for

ch3.indd 88 6/11/09 11:38:38 PM


Simpson’s rule

This is generally more accurate than the trapezoidal rule, since it makes use of parabolic arcs instead of straight lines.

A parabola is drawn through points A , B and C to give the formula

8. logx dx1


9. ( )x x dx2

0

2−# using 4 trapezia.

10. x dx1


11. x

dx15


12. x

dx1

16

3 −# using 6 trapezia.

13. ( )f x dx9

1# where values of f ( x ) are

given in the table:

x 1 3 5 7 9

f (x) 3.2 5.9 8.4 11.6 20.1

14. ( )f t dt1

4# where values of ( )f t are

given in the table:

t 1 2 3 4

f (t) 8.9 6.5 4.1 2.9

15. ( )f x dx2

14# where values of f ( x )are given in the table:

x 2 4 6 8 10 12 14

f (x) 25.1 37.8 52.3 89.3 67.8 45.4 39.9

( ) ( ) ( )f x dx b a f a f a b f b6

42a

bZ

− + + +c m< F#

Proof

This proof is diffi cult and is only included for completion of the topic. It uses results that will be studied later in the chapter.

ch3.indd 89 6/11/09 11:38:38 PM


Let ( ) , and ( )f a y f y f b ya b20 1 2= = =+c m

Let the parabola passing through points ( , ), ( , )A h y B y00 1− and ( , )C h y2 be given by y ax bx c2= + +

Then y a h b h c

y ah bh c0

2

02

= − + − += − +

] ]g g (1)

( )y a b c

y c0 01

2

1

= + +=

] g (2)

( )y a h b h c

y ah bh c2

2

22

= + += + +

] g (3)

y y y ah bh c c ah bh c

ah c

4 4

2 60 1 2

2 2

2

+ + = − + + + + += +

(4)

( )

( )

ax bx c dx ax bx cx

ah bh ch ah bh ch

ah ch

h ah c

h y y y

3 2

3 2 3 2

32 2

32 6

34

h

h

h

h2

3 2

3 2 3 2

3

2

0 1 2

+ + = + +

= + + − − + −

= +

= +

= + +

− −]

c cg

m m; E#

Now , ( ), and ( )h b a y f a y f a b y f b2 20 1 2= − = = + =c m

` ( ) ( ) ( )

( ) ( )

f x dx

b a

f a f a b f b

b a f a f a b f b

32 4

2

64

2

a

b=

−

+ + +

= − + + +

cc

mm

<

<

F

F

#

You will study this later in the chapter.

One application of Simpson’s rule uses 3 function values (ordinates). •

Ordinates are y-values. You fi rst met this name in Chapter 5 of the Preliminary Course book.

ch3.indd 90 6/11/09 11:38:39 PM


EXAMPLES

1. Use Simpson’s rule with 3 function values to fi nd an approximation for

.x dx2

3#

Solution

Two applications use 5 function values. •

Three applications use 7 function values. •

How many function values would you need

for 4 applications of Simpson’s rule?

CONTINUED

ch3.indd 91 6/11/09 11:38:40 PM


( ) ( ) ( )

( ) ( )

( . )

.

f x dx b a f a f a b f b

x dx f f f

64

2

63 2 2 4

22 3 3

2 4 2 5 3

1 58

a

b

2

3

61

Z

Z

Z

− + + +

− + + +

= + +

cc

mm

<

<

F

F

#

#

2. Use Simpson’s rule with 5 function values to fi nd an approximation

for .x

dx10

2

+#

Solution

Z

=

=

( ) ( ) ( )

( ) ( ) ( ) ( )

. ..

f x dx b a f a f a b f b

xdx

xdx

xdx

f f f f f f

64

2

1 1 1

61 0 0 4

20 1 1

62 1 1 4

21 2 2

61

0 11 4

0 5 11

1 11

61

1 11 4

1 5 11

2 11

1 1

a

b

0

2

0

1

1

2

− + + +

+ ++

+− + + + + − + + +

++

++

++

++

++

+

Z

=

c

c cc c

m

m mm m

<

< <

< <

F

F F

F F

#

# # #

Five function values mean 2 applications of Simpson’s rule.

There is a general formula for n equal subintervals. There are different versions of this formula. This one uses the function values y 0 to y n .

The n 1+ function values give n subintervals.

ch3.indd 92 6/11/09 11:38:43 PM


Proof

( ) ( ) ( ) ( )

. . . ( )

( . . . )

[( ) ( . . .) ( . . .)]

f x dx h y y y h y y y h y y y

h y y y

h y y y y y y y y y y y y

h y y y y y y y

62 4

62 4

62 4

62 4

34 4 4 4

34 2

a

b

n n n

n n n

n I

0 1 2 2 3 4 4 5 6

2 1

0 1 2 2 3 4 4 5 6 2 1

0 3 5 2 4

= + + + + + + + +

+ + + +

= + + + + + + + + + + + +

= + + + + + + + +

− −

− −

#

You could also remember this rule as

.( ) [( ) (odds) (evens)]f x dx h y y3

4 2a

b

n0Z + + +#

EXAMPLES

1. Use Simpson’s rule with 7 ordinates to fi nd an approximation for

.log x dx1

4#

Solution

where

.

h

0 5

=

=

( ) [( ) ( ) ( )]f x dx h y y y y y y y

nb a3

4 2

64 1

a

b

0 6 1 3 5 2 4Z + + + + + +

−

= −

#

. [ ( . . . )

( )]

.

log log log log log log

log log

x dx3

0 5 1 4 4 1 5 2 5 3 5

2 2 3

1 105

1

4Z

Z

+ + + +

+ +

#

( ) [( ) ( . . .) ( . . .)]f x dx h y y y y y y y3

4 2na

b

0 1 3 5 2 4Z + + + + + + + +#

where h nb a= − for n subintervals ( 1n + function values)

Seven ordinates or function values give 6 subintervals.

Use the log key on your calculator.

CONTINUED

ch3.indd 93 6/11/09 11:38:45 PM


2. Use Simpson’s rule with 9 function values to fi nd an approximation

for .xdx

21

5#

Solution

( ) [( ) ( ) ( )]

where

.

.. . . .

.

f x dx h y y y y y y y y y

h nb a

xdx

34 2

85 1

0 5

13

0 511

51 4

1 51

2 51

3 51

4 51 2

21

31

41

0 8

a 0 8 1 3 5 7 2 4 6

2 2 2 2 2 2 2 2 2 21

Z

Z

Z

+ + + + + + + +

= −

= −

=

+ + + + + + + +

b

5 c cm m< F

#

#

3. Use Simpson’s rule to fi nd an approximation for ( )f x dx4

1# using the

values in the table below:

x 1 1.5 2 2.5 3 3.5 4

f(x) 8.6 11.9 23.7 39.8 56.7 71.4 93.2

Solution

[( ) ( ) ( )]

[( . . ) ( . . . ) ( . . )]

y y y y y y y4 2

8 6 93 2 4 11 9 39 8 71 4 2 23 7 56 7Z

+ + + + + +

+ + + + + +

Z( )

where

.

( )

.

f x dx h

h

f x dx

3

64 1

0 5

125 83

.

0 6 1 3 5 2 41

4

1

4

30 5

Z

= −

=

#

#

ch3.indd 94 6/11/09 11:38:50 PM


Computer Application

There are computer application packages that can calculate the area under curves by using rectangles, the trapezoidal rule or Simpson’s rule.

A graphics calculator will also do this.

Integration and the Primitive Function

Mathematicians found a link between fi nding areas under a curve and the primitive function. This made possible a simple method for fi nding exact areas.

1. x dx4

1

3# using 3 function values.

2. ( )x dx15

2

3−# using 3 ordinates.

3. xdx3


4. x

dx2

1

0 +# using 3 function values.

5. x dx4

2# using

3 function values (a) 5 function values. (b)

6. log x dx7

3# using 5 function

values.

7. x

dx1

5

2 +# using 6 subintervals.

8. log x dx6

3# using 7 function

values.

9. ( )x x dx4

3

0+# using 9 ordinates.

10. x dx4


11. dxx1

1

7


12. dxx 1

12

5

2 −# using 6 subintervals.

13. ( )f x dx0


given in the table:

x 0 1 2 3 4

f(x) 0.7 1.3 5.4 – 0.5 – 3.8

14. ( )f t dt4

2# where values of f ( t ) are

given in the table:

t 2 2.5 3 3.5 4

f(t) 3.7 1.2 9.8 4.1 2.7

15. ( )f x dx0


given in the table:

x 0 0.5 1 1.5 2 2.5 3

f(x) 15.3 29.2 38.1 56.2 69.9 94.8 102.5

3.2 Exercises

Use Simpson’s rule to fi nd an approximation for

ch3.indd 95 6/11/09 11:38:52 PM


EXAMPLE

The graph below shows the speed at which an object travels over time.

Find the distance it travels in (a) 1 s (i) 2 s (ii) 3 s (iii)

Find the area under the line between (b) (i) 0t = and 1 (ii) 0t = and 2 (iii) 0t = and 3

Solution

(a) The speed is a constant 30 metres per second ( )ms 1− The object travels 30 metres in 1 s (i) The object travels 60 metres in 2 s (ii) The object travels 90 metres in 3 s (iii)

(i) The area is (b) ´30 1 30= units 2 (ii) The area is ´30 2 60= units 2 (iii) The area is ´30 3 90= units 2

In this example, the line graph gives a rate of change. The area under the curve gives the information about the original data for this rate of change.

In the same way, the area under a rate of change curve will give the original data. This original data is the primitive function of the curve.

Notice that the area gives the distances.

ch3.indd 96 6/11/09 11:38:54 PM


DID YOU KNOW?

Many mathematicians in the 17th century were interested in the problem of fi nding areas under a curve. Isacc Barrow (1630–77) is said to be the fi rst to discover that differentiation and integration are inverse operations. This discovery is called the fundamental theorem of calculus .

Barrow was an Englishman who was an outstanding Greek scholar as well as making contributions in the areas of mathematics, theology, astronomy and physics. However, when he was a schoolboy, he was so often in trouble that his father was overheard saying to God in his prayers that if He decided to take one of his children, he could best spare Isaac.

Sir Isaac Newton (1643–1727), another English mathematician, and scientist and astronomer, helped to discover calculus. He was not interested in his school work, but spent most of his time inventing things, such as a water clock and sundial.

He left school at 14 to manage the estate after his stepfather died. However, he spent so much time reading that he was sent back to school. He went on to university and developed his theories in mathematics and science which have made him famous today.

Fundamental theorem of calculus

The area enclosed by the curve ( ),y f x= the x -axis and the lines x a= and x b= is given by

( ) ( ) ( )f x dx F b F aa

b= −# where F ( x ) is the primitive of function f ( x )

Proof

Consider a continuous curve ( )y f x= for all values of x a>

ch3.indd 97 6/11/09 11:38:56 PM


Let area ABCD be A ( x ) Let area ABGE be ( )A x h+ Then area DCGE is ( ) ( )A x h A x+ −

`

. .Area area area

i.e. ( ) ( ) ( ) ( )

( )( ) ( )

( )

( )( ) ( )

( )

( ) ( ) ( )

( ) ( )

lim lim lim

DCFE DCGE DHGE

f x h A x h A x f x h h

f xh

A x h A xf x h

f xh

A x h A xf x h

f x A x f x

A x f x

< << <

< <

< <

< <h h h0 0 0

+ − ++ −

+

+ −+

=

" " "

l

l

i.e. A ( x ) is a primitive function of f ( x ) ( ) ( )A x F x C= + where F ( x ) is the primitive function of f ( x ) (1)

Now A ( x ) is the area between a and x ( )A a 0=

Substitute in (1):

`

( ) ( )

( )

( )

( ) ( ) ( )

where ,

( ) ( ) ( )

A a F a C

F a C

F a C

A x F x F a

x b b a

A b F b F a

0

If >

= += +

− == −== −

Defi nite Integrals

By the fundamental theorem of calculus,

( ) ( ) ( )f x dx F b F aa

b= −# where F ( x ) is the primitive function of f ( x ) .

The primitive function of is .x Cnx

1n

n 1

++

+

Putting these pieces of information together, we can fi nd areas under simple curves.

This is the formula for differentiation from fi rst principles. You fi rst saw this in Chapter 8 of The Preliminary Course book.

ch3.indd 98 6/11/09 11:39:00 PM


EXAMPLES

Evaluate

1. ( )x dx2 14

3+#

Solution

+x x( )

( ) ( )

x dx2 1

4 4 3 320 12

8

3

4 23

4

2 2

+ =

= + − += −=

8 B#

2. x dx3 2

0

5#

Solution

x

( ) ( )

x dx3

5 0125

2 30

5

0

5

3 3

=

= −=

8 B#

x

nx

nb

na

1

1 1

nn

a

b

a

b

n n

1

1 1

=+

=+

−+

+

+ +

< F#

CONTINUED

Constant C will cancel out. That is, ( ) ( )4 4 C 3 3 C22 + + − + +

20 12

8.

= −

=

The graph shows the area that the defi nite integral

calculates.

ch3.indd 99 6/11/09 11:39:01 PM


3. x dx32

2

0−#

Solution

x−x dx3 2 3

0

2

0

2

3 3

− =

( ) ( )2 08

= − − −= −

: D#

4. x dx3

1

1

−#

Solution

( )

x dx x4

41

41

41

41

0

34

1

1

1

1

4 4

=

= −−

= −

=

−−; E#

Can you see why there is a negative solution?

The defi nite integral gives an area of zero. Can you see why?

ch3.indd 100 6/11/09 11:39:03 PM


1. x dx40

2#

2. ( )x dx2 13

1+#

3. x dx35

2

0#

4. ( )t dt4 72

1−#

5. ( )y dy6 51

1+

−#

6. x dx63

2

0#

7. ( )x dx12

2

1+#

8. x dx42

3

0#

9. ( )x x dx3 24

2

1−

−#

10. ( )x x dx4 6 33

2

1+ −#

11. x dx1

2

1−#

12. ( )x dx13

3

2+

−#

13. x dx2

5

2−#

14. x dx4

1#

15. ( )x x x dx3 41

3 2

0− +#

16. ( )x dx2 12

2

1−#

17. ( )y y dy1

3

1+

−#

18. ( )x dx2 2

3

4−#

19. t dt42

3

2−#

20. x dx3

4 2

2#

21. xx dx53 4

0#

22. xx x dx31 4

2

−−#

23. xx x x dx4 52 3 2

0

+ +#

24. xx x x dx2 31 3 2

0

− +#

25. x

x x dx3

34

5

2

3

+ +#

3.3 Exercises

Evaluate

Class Investigation

Look at the results of defi nite integrals in the examples and exercises. Sketch the graphs where possible and shade in the areas found.

Can you see why the defi nite integral sometimes gives a • negative answer? Can you see why it will sometimes be zero? •

Indefi nite Integrals

Sometimes it is necessary to fi nd a general or indefi nite integral (primitive function).

( ) ( )f x dx F x C= +# where F ( x ) is a primitive function of f ( x )

In Chapter 2 you learned the result for the primitive function of .xn This result is the same as the integral of .xn

x dxnx C

1n

n 1

=+

++

#

Simplify fi rst, then fi nd the defi nite integral.

ch3.indd 101 6/11/09 11:39:07 PM


EXAMPLES

Find each indefi nite integral (primitive function)

1. ( )x x x dx3 4 74 2− + −#

Solution

( )x x x dx x x x x C

x x x x C

3 4 75

33

42

7

52 7

4 25 3 2

53 2

: :− + − = − + − +

= − + − +

#

2. x dx5 9#

Solution

x dx x C

x C

5 510

2

910

10

= +

= +

c m#

3. xx x x dx2 7 35 2+ −b l#

Solution

( )x

xxx

xx dx x x dx

x x x C

2 7 3 2 7 3

52

27 3

5 24

5 2

+ − = + −

= + − +

b l# #

4. x

x dx13

+c m#

Solution

2

2

( )x

x dx x x dx

x xC

xx

C

1

223

21

32

33

2

2

3

+ = +

=−

+ +

= − + +

−

−

1

3

c m# #

5x dx9# is the same as x dx.5 9#

ch3.indd 102 6/11/09 11:39:08 PM


DID YOU KNOW?

John Wallis (1616–1703) found that the area under the curve 1 . . .y x x x2 3= + + + + is given by

2 3 4

. . . .xx x x2 3 4

+ + + +

He found this result independently of the fundamental theorem of calculus.

1. x dx2#

2. x dx3 5#

3. x dx2 4#

4. ( )m dm1+#

5. ( )t dt72 −#

6. ( )h dh52 +#

7. ( )y dy3−#

8. ( )x dx2 4+#

9. ( )b b db2 +#

10. ( )a a da13 − −#

11. ( )x x dx2 52 + +#

12. ( )x x x dx4 3 8 13 2− + −#

13. ( )x x x dx6 25 4 3+ +#

14. ( )x x dx3 97 6− −#

15. ( )x x x dx2 23 2+ − −#

16. ( )x x dx45 3+ +#

17. ( )x x dx4 5 82 − −#

18. ( )x x x dx3 24 3− +#

19. ( )x x dx6 5 43 2+ −#

20. ( )x x x dx3 24 3 2+ +− − −#

21. xdx

8#

22. 3x dx1

#

23. x

x x x dx3 23

6 5 4− +#

24. ( )x dx1 2 2−#

25. ( ) ( )x x dx2 5− +#

26. x

dx32#

27. xdx

3#

28. x

x x x dx4 3 75

3 5 2− − +#

29. ( )y y dy52 7− +−#

30. ( ) ( )t t dt4 12 − −#

31. x dx#

32. t

dt25#

33. x dx3#

34. x x dx#

35. 1 1xx

dx+e o#

3.4 Exercises


ch3.indd 103 6/11/09 11:39:10 PM


Function of a function rule

EXAMPLE

Differentiate 5( ) .x2 3+

Hence fi nd ( ) .x dx10 2 3 4+#

Find ( ) .x dx2 3 4+#

Solution

dxd x x

x

x dx x C

x dx x dx

x C

2 3 5 2 3 2

10 2 3

10 2 3 2 3

2 3101 10 2 3

101 2 3

5 4

4

4 5

4 4

5

$+ = +

= +

+ = + +

+ = +

= + +

] ]]

] ]] ]

]

g gg

g gg g

g

## #

n 1+

( )( )

( )ax b dx

a nax b

C1

n+ =+

++#

Proof

`

dxd ax b n ax b a

a n ax b

a n ax b dx ax b C

ax b dxa n

a n ax b dx

a nax b C

a nax b

C

1

1

1

11 1

11

1

n n

n

n n

n n

n

n

1

1

1

1

$+ = + +

= + +

+ + = + +

+ =+

+ +

=+

+ +

=+

+ +

+

+

+

+

] ] ]] ]

] ] ]] ] ] ]

] ]

]]

g g gg g

g g gg g g g

g g

gg

## #

Remember: Integration is the reverse of differentiation.

ch3.indd 104 6/11/09 11:39:11 PM


EXAMPLES

Find 1. 3( )x dx5 9−#

Solution

4

3( )

( )

( )´

x dxx

C

xC

5 95 4

5 9

205 9

4

− =−

+

=−

+

#

2. 8( )x dx3 −#

Solution

8( )( )

( )´

x dxx

C

xC

31 9

3

93

9

9

− =−

−+

= −−

+

#

3. x dx4 3+#

Solution

22

( )( )

( )

´x dx

xC

xC

4 34

23

4 3

64 3 3

+ =+

+

=+

+

13

#

1. (a) 2( )x dx3 4−# by (i) expanding (ii) the function of a function rule

(b) 4( )x dx1+#

(c) ( )x dx5 1 9−#

(d) 7( )y dy3 2−#

(e) 4( )x dx4 3+#

(f) 12( )x dx7 8+#

(g) ( )x dx1 6−#

(h) x dx2 5−#

(i) 4−( )x dx2 3 1+#

(j) ( )x dx3 7 2+ −#

(k) ( )x

dx2 4 5

13−

#

(l) x dx4 33 +#

3.5 Exercises


ch3.indd 105 6/11/09 11:39:12 PM


Class Exercise

Differentiate ( 1) 2 3 .x x+ − Hence fi nd .x

x dx2 3

3 2−

−#

Areas Enclosed by the x -axis

The defi nite integral gives the signed area under a curve. Areas above the x -axis give a positive defi nite integral.

(m) −2( )x dx2 −1

#

(n) ( )t dt3 3+#

(o) ( )x dx5 2 5+#

2. Evaluate

(a) ( )x dx2 1 4

1

2+#

(b) ( )y dy3 2 3

0

1−#

(c) 7( )x dx11

2−#

(d) ( )x dx3 2 5

0

2−#

(e) 2( )x

dx6

3 10

1 −#

(f) 6( )x dx54

5−#

(g) x dx26

3−#

(h) 4( )x

dx3 20

1

−#

(i) ( )n

dn2 1

530

2

+#

(j) ( )x

dx5 4

231

4

−#

ch3.indd 106 6/11/09 11:39:13 PM


Areas below the x -axis give a negative defi nite integral.

We normally think of areas as positive. So to fi nd areas below the x -axis, take the absolute value of the defi nite integral. That is,

( )f x dxAreaa

b= #

Always sketch the area to be found .

EXAMPLES

1. Find the area enclosed by the curve y x x2 2= + − and the x -axis.

Solution

Sketch .y x x2 2= + −

( )

( ) ( )( ) ( )

x x dx

x x x

2

22 3

2 222

32 2 1

21

31

4 238 2

21

31

421

Area2

2

1

2 3

1

2

2 3 2 3

= + −

= + −

= + − − − +−

−−

= + − − − + +

=

−

−

c dc c

m nm m

< F

#

So the area is 214 units 2 .

The integral will be positive as the area is

above the x -axis.

CONTINUED

ch3.indd 107 6/11/09 11:39:15 PM


2. Find the area bounded by the curve 4y x2= − and the x -axis.

Solution

Sketch 4.y x2= −

3

( )

( )( )

( )

x dx x x43

4

32 4 2

32

4 2

38 8

38 8

1032

22

3

2

2

2

3

− = −

= − −−

− −

= − − − +

= −

−−

c dc c

m nm m

< F#


3. Find the area enclosed between the curve ,y x3= the x -axis and the lines 1x = − and 3.x =

Solution

Sketch .y x3=

The integral will be negative as the area is below the x -axis.

Some of the area is below the x-axis, so its integral will be negative.

ch3.indd 108 6/11/09 11:39:18 PM


We need to fi nd the sum of the areas between 1x = − and 0, and 0x = and 3.

( )

x dx x4

40

41

41

34

1

0

1

4 4

=

= −−

= −

−−

0; E#

So A41

1 = units 2 .

x dx x4

43

40

481

34

0

3

0

3

4 4

=

= −

=

; E#

.A

A A

481

41

481

2021

So units

units

22

1 2

2

=

+ = +

=

Even and odd functions

Some functions have special properties.

For odd functions,

( )f x dx 0a

a=

−#

For even functions,

( ) ( )f x dx f x dx2a

a a

0=

−# #

ch3.indd 109 6/11/09 11:39:21 PM


EXAMPLES

1. Find the area between the curve ,y x3= the x -axis and the lines 2x = − and 2.x =

Solution

x dx 03

2

2=

−#

2 x dx

x24

242

40

8

Area 3

0

2

4

0

2

4 4

=

=

= −

=

c m; E

#


2. Find the area between the curve ,y x2= the x -axis and the lines 4x = − and 4.x =

Solution

x dx x dx

x

2

23

234

30

4232

2

4

42

0

4

3

0

4

3 3

=

=

= −

=

−

c m< F

# #


y x3= is an odd function since ( ) ( )f x f x .=− −

1. Find the area enclosed between the curve 1y x2= − and the x -axis.

2. Find the area bounded by the curve 9y x2= − and the x -axis.

3. Find the area enclosed between the curve 5 4y x x2= + + and the x -axis.

4. Find the area enclosed between the curve 2 3y x x2= − − and the x -axis.

5. Find the area bounded by the curve 9 20y x x2= − + − and the x -axis.

6. Find the area enclosed between the curve 2 5 3y x x2= − − + and the x -axis.

7. Find the area enclosed between the curve ,y x3= the x -axis and the lines 0x = and 2.x =

3.6 Exercises

y x2= is an even function since ( ) ( )f x f x .− =

ch3.indd 110 6/11/09 11:39:22 PM





11. Find the area bounded between the curve 3 ,y x2= the x -axis and the lines 1x = − and 1.x =

12. Find the area enclosed between the curve 1,y x2= + the x -axis and the lines 2x = − and 2.x =


14. Find the area enclosed between the curve ,y x x2= + and the x -axis.

15. Find the area enclosed between

the curve ,x

y 12

= the x -axis and

the lines 1x = and 3.x =


the curve ( )

,x

y3

22−

= the x -axis

and the lines 0x = and 1.x =

17. Find the area bounded by the curve ,y x= the x -axis and the line 4.x =

18. Find the area bounded by the curve 2,y x= + the x -axis and the line 7.x =

19. Find the area bounded by the curve 4 ,y x2= − the x -axis and the y -axis in the fi rst quadrant.

20. Find the area bounded by the x -axis, the curve y x3= and the lines x a= − and .x a=

Areas Enclosed by the y -axis

To fi nd the area between a curve and the y -axis, we change the subject of the equation of the curve to x . That is,

( ) .x f y=

The defi nite integral is given by

( )f y dy x dyora

b

a

b# #

Since x is positive on the right-hand side of the y -axis, the defi nite integral is positive . Since x is negative on the left-hand side of the y -axis, the defi nite integral is negative .

ch3.indd 111 6/11/09 11:39:24 PM


EXAMPLES

1. Find the area enclosed by the curve x = y 2 , the y -axis and the lines 1y = and 3.y =

Solution

y dy

y

3

33

31

832

Area

units

2

1

3

3

1

3

3 3

2

=

=

= −

=

= G

#

2. Find the area enclosed by the curve y = x 2 , the y -axis and the lines y 0= and y 4= in the fi rst quadrant.

Solution

Change the subject of the equation to x .

` ±

y x

y x

2=

=

In the fi rst quadrant,

2

y x

y xi.e.

=

=1

2

2

y dy

y

y

23

3

2

32 4

32 0

531

Area

units

4

0

0

4

3

0

4

3 3

2

=

=

=

= −

=

1

3R

T

SSSS

>

V

X

WWWW

H

#

The integral is positive since the area is to the right of the y-axis.

ch3.indd 112 6/11/09 11:39:25 PM


3. Find the area enclosed between the curve 1,y x= + the y -axis and the lines 0y = and 3.y =

Solution

3

y xy x

y x

y dy y dy

yy

yy

11

1

1 1

3 3

31 1

30 0

33 3

31 1

32 6

32

32 6

32

731

Area

units

2

2

2

0

12

1

3

3

0

1 3

1

3

3 3 3

2

= += +

− =

= − + −

= − + −

= − − − + − − −

= − + +

= +

=

^ ^

c c c c

h h

m m m m= =G G

# #

Some of the area is on the left of the y-axis.

1. Find the area bounded by the y -axis, the curve x = y 2 and the lines 0y = and 4.y =

2. Find the area enclosed between the curve x = y 3 , the y -axis and the lines 1y = and 3.y =

3. Find the area enclosed between the curve y = x 2 , the y -axis and the lines 1y = and 4y = in the fi rst quadrant.

4. Find the area between the lines y = x − 1, y 0= and 1y = and the y -axis.

5. Find the area bounded by the line y = 2 x + 1, the y -axis and the lines 3y = and 4.y =

6. Find the area bounded by the curve ,y x= the y -axis and the lines 1y = and 2.y =

3.7 Exercises

ch3.indd 113 6/11/09 11:39:27 PM


7. Find the area bounded by the curve x = y 2 - 2 y - 3 and the y -axis.

8. Find the area bounded by the curve x = - y 2 - 5 y - 6 and the y -axis.

9. Find the area enclosed by the curve 3 5,y x= − the y -axis and the lines 2y = and 3.y =


the curve ,x

y 12

= the y -axis and

the lines 1y = and 4y = in the fi rst quadrant.

11. Find the area enclosed between the curve y = x 3 , the y -axis and the lines 1y = and 8.y =

12. Find the area enclosed between the curve y = x 3 - 2 and the y -axis between y = -1 and y = 25.

13. Find the area enclosed between the lines 4y = and 1y = - x in the second quadrant.

14. Find the area enclosed between the y -axis and the curve x = y ( y - 2).

15. Find the area bounded by the curve y = x 4 + 1, the y -axis and the lines 1y = and 3y = in the fi rst quadrant, correct to 2 signifi cant fi gures.

Sums and Differences of Areas

EXAMPLES

1. Find the area enclosed between the curve y = x2, the y -axis and the lines y 0= and 4y = in the fi rst quadrant.

Solution

( )´

A x dx

x4 23

838

30

531

Area of rectangle

units

2

0

2

3

0

2

2

= −

= −

= − −

=

c m< F

#

This example was done before by fi nding the area between the curve and the y -axis.

ch3.indd 114 6/11/09 11:39:29 PM


2. Find the area enclosed between the curves y = x 2 , y = ( x − 4) 2 and the x -axis.

Solution

Solve simultaneous equations to fi nd the intersection of the curves.

`

( )

( )

y x

y x

x x

x xx

x

x

1

4 2

4

8 160 8 16

8 16

2

2

2

2 2

2

== −= −= − += − +==

]]

gg

Area x dx x dx

x x

4

3 34

38

30

30

38

531 units

2 4

3

2

0

2

2

3

0

2

2

4

2

= + −

= + −

= − + − −

=

]]

c c

gg

m m< <F F

# #

3. Find the area enclosed between the curve y = x 2 and the line y = x + 2.

Solution

Solve simultaneous equations to fi nd the intersection of the curve and line.

`

( )( )

( ) ( )

y x

y x

x x

x x

x x

x

12 2

2

2 02 1 0

2 1or

2

2

2

== += +

− − =− + =

= −

x dx x dx

x x dx

x x x

2

2

22

3

Area2 2

2

2

1

2

1

2

1

2 3

1

= + −

= + −

= + −

− −

−

−

]]

gg

< F

# ##

2

( )( )

( )( )

2 438

21 2

31

22 2 2

32

21

2 131

421 units

2 3 3

2

= + − −−

+ − −−

= + − − − +

=

c dc c

m nm m

ch3.indd 115 6/11/09 11:39:30 PM


Volumes

The volume of a solid can be found by rotating an area under a curve about the x -axis or the y -axis.

1. Find the area bounded by the line 1y = and the curve y = x 2 .

2. Find the area enclosed between the line y = 2 and the curve y = x 2 + 1.

3. Find the area enclosed by the curve y = x 2 and the line y = x .

4. Find the area bounded by the curve y = 9 - x 2 and the line y = 5.

5. Find the area enclosed between the curve y = x2 and the line y = x + 6.

6. Find the area bounded by the curve y = x 3 and the line y = 4 x .

7. Find the area enclosed between the curves y = ( x - 1) 2 and y = ( x + 1) 2 .

8. Find the area enclosed between the curve y = x 2 and the line y = -6 x + 16.

9. Find the area enclosed between the curve y = x 3 , the x -axis and the line y = -3 x + 4.

10. Find the area enclosed by the curves y = ( x - 2) 2 and y = ( x - 4) 2 .

11. Find the area enclosed between the curves y = x 2 and y = x 3 .

12. Find the area enclosed by the curves y = x 2 and x = y 2 .

13. Find the area bounded by the curve y = x 2 + 2 x - 8 and the line y = 2 x + 1.

14. Find the area bounded by the curves 1y = - x 2 and y = x 2 - 1.

15. Find the exact area enclosed between the curve 4y x2= − and the line x - y + 2 = 0.

3.8 Exercises

ch3.indd 116 6/11/09 11:39:32 PM


Volumes about the x -axis

πV y dxa

b2= #

Proof

Take a disc of the solid with width δ x .

ππ δ

r h

y x

Volume 2

2

==

Taking the sum of an infi nite number of these discs,

π δ

π

π

lim y x

y dx

y dx

Volumeδx

a

b

a

b

0

2

2

2

=

=

=

"

##

The disc is approximately a cylinder with radius y and

height x.δ

ch3.indd 117 6/11/09 11:39:33 PM


EXAMPLES

1. Find the volume of the solid of revolution formed when the curve x 2 + y 2 = 9 is rotated about the x -axis between 1x = and x = 3.

Solution

`

( ) ( )

π

π

π

π

π

x y

y x

V y dx

x dx

x x

9

9

9

93

27 9 931

328 units

a

b

2 2

2 2

2

2

1

3

3

1

3

3

+ == −

=

= −

= −

= − − −

=

^

c

h

m< F

##

2. Find the volume of the solid formed when the line y = 3 x - 2 is rotated about the x -axis from 1x = to x = 2.

Solution

π

π

π

π

π

π

´

y x

y x

V y dx

x dx

x

3 2

3 2

3 2

3 33 2

94

91

963

7 units

a

b

2

2

2

1

2

3

1

2

3 3

3

= −

= −

=

= −

= −

= −

=

=

2]

]]

d

g

gg

n< F

##

Use the function of a function rule

ch3.indd 118 6/11/09 11:39:34 PM


Volumes about the y -axis

The formula for rotations about the y -axis is similar to the formula for the x -axis.

PROBLEM

What is wrong with the working out of this volume ? Find the volume of the solid of revolution formed when the curve y x 12= + is rotated about the x -axis between 0x = and x = 2.

Solution

`

units

π

π

π

π

π

π

y x

y x

V y dx

x dx

x

1

1

1

31

32 1

30 1

35

31

3124

a

b

2

2 2 2

2

2 2

0

2

2 3

0

2

2 3 2 3

3 3

3

= +

= +

=

= +

= +

= + − +

= −

=

^

^^^ ê

d

h

hhh h on

= G

##

What is the correct volume?

The proof of this result is similar to that for volumes

about the x -axis.

πV x dya

b 2= #

ch3.indd 119 6/11/09 11:39:37 PM


EXAMPLE

Find the volume of the solid formed when the curve y = x 2 - 1 is rotated about the y -axis from y = -1 to y = 3.

Solution

x

π

π

π

π

π

y

y x

V x dy

y dy

yy

1

1

1

2

29 3

21 1

8 units

a

2

2

2

1

2

1

3

3

`

= −+ =

=

= +

= +

= + − −

=

−

−

3

b

^

c c

h

m m=

<

G

F

##

1. Find the volume of the solid of revolution formed when the curve y = x 2 is rotated about the x -axis from 0x = to x = 3.

2. Find the volume of the solid formed when the line y = x + 1 is rotated about the x -axis between x = 2 and x = 7.

3. Find the volume of the solid of revolution that is formed when the curve y = x 2 + 2 is rotated about the x -axis from 0x = to x = 2.

4. Determine the volume of the solid formed when the curve y = x 3 is rotated about the x -axis from 0x = to 1.x =

3.9 Exercises

ch3.indd 120 6/11/09 11:39:38 PM


5. Find the volume of the solid formed when the curve x = y 2 - 5 is rotated about the x -axis between 0x = and x = 3.

6. Find the volume of the solid of revolution formed by rotating the line y = 4 x - 1 about the x -axis from x = 2 to x = 4.

7. Find the volume of the hemisphere formed when the curve x 2 + y 2 = 1 is rotated about the x -axis between 0x = and 1.x =

8. The parabola y = ( x + 2) 2 is rotated about the x -axis from 0x = to x = 2. Find the volume of the solid formed.

9. Find the volume of the spherical cap formed when the curve x 2 + y 2 = 4 is rotated about the y -axis from 1y = to y = 2.

10. Find the volume of the paraboloid formed when y = x 2 is rotated about the y -axis from 0y = to y = 3.

11. Find the volume of the solid formed when y = x 2 - 2 is rotated about the y -axis from 1y = to 4.y =

12. The line y = x + 2 is rotated about the y -axis from y = -2 to y = 2. Find the volume of the solid formed.

13. Determine the volume of the solid formed when the curve y = x 3 is rotated about the x -axis from x = -1 to x = 4.

14. Find the volume of the solid formed when the curve y x= is rotated about the x -axis between 0x = and 5.x =

15. Find the volume of the solid formed when the curve 3y x= + is rotated about the x -axis from 1x = to x = 6.

16. Find the volume of the solid formed when the curve y x= is rotated about the y -axis between 1y = and 4.y =

17. Find the volume of the solid formed when the curve 4y x2= − is rotated about the y -axis from 1y = to y = 2.

18. Find the volume of the solid formed when the line x + 3 y - 1 = 0 is rotated about the y -axis from 1y = to y = 2.

19. Find the volume of the solid formed when the line x + 3 y - 1 = 0 is rotated about the x -axis from 0x = to x = 8.

20. The curve y = x 3 is rotated about the y -axis from 0y = to .y 1= Find the volume of the solid formed.

21. Find the volume of the solid of revolution formed if the area enclosed between the curves y = x 2 and y = ( x - 2) 2 is rotated about the x -axis.

22. The area bounded by the curve y = x 2 and the line y = x + 2 is rotated about the x -axis. Find the exact volume of the solid formed.

23. Show that the volume of a sphere

is given by the formula

πV r34 3= by rotating the

semicircle y r x2 2= − about the x -axis.

ch3.indd 121 6/11/09 11:39:40 PM


Application

The volume of water in a small lake is to be measured by fi nding the depth of water at regular intervals across the lake.

What could the volume of this lake be?

The table below shows the results of the measurements, where x stands for the regular intervals chosen, in metres, and y is the depth in metres.

x (m) 0 50 100 150 200 250 300

y (m) 10.2 39.1 56.9 43.2 28.5 15.7 9.8

Using Simpson’s rule to fi nd an approximation for the volume of the lake, correct to 2 signifi cant fi gures:

34 2

where6

6300 0

50

V y dx

y dx

f x dxh

y y y y y y y

hb a

π

π

2

2

0

0 6 1 3 5 2 4

a

aZ

=

=

+

= −

= −

=

+ + + + +

b

b

300

] _ _ _g i i i8 B

##

#

ch3.indd 122 6/11/09 11:39:41 PM


350

10.2 9.8 4 39. 43.2 15.7 2 56.9 28.5y dx 12 2 2 2 2 2 2

0Z + + + + + +2

300 ] ] ]g g g6 @#

Z

`

381099.3

V

381099.3

1197 258.866

1200 000 correct to 2 significant figures

y dxπ

π ´0

=

=

=

=

2300#

So the volume of the lake is approximately 1 200 000 m3 of water.

ch3.indd 123 6/11/09 11:39:45 PM


Test Yourself 3 1. (a) Use the trapezoidal rule with 2

subintervals to fi nd an approximation

to .xdx

21

2#

(b) Use integration to fi nd the exact

value of .xdx

21

2#

2. Find the indefi nite integral (primitive function) of

(a) x3 1+

(b) xx x5 2 −

(c) x

(d) 2 5x 7+] g

3. In an experiment, the velocity of a particle over time was found and the results placed in the table below.

t 0 1 2 3 4 5 6

v 1.3 1.7 2.1 3.5 2.8 2.6 2.2

Use Simpson’s rule to fi nd the approximate value of f t dt

0

6 ] g#

4. Evaluate

(a) x dx13

0

2−] g#

(b) x dx1

5

1−#

(c) x dx3 11 4

0−] g#

5. Draw a primitive function of the following curves.

(a)

(b)

6. Find the area bounded by the curve ,y x2= the x -axis and the lines 1x = − and 2.x =

7. Use Simpson’s rule with 5 function values to fi nd the approximate area enclosed between the curve ,xy 1= the x -axis and the lines 1x = and 3.x =

8. Find the area enclosed between the curves y x2= and 2 .y x2= −

9. The line y x2 3= − is rotated about the x -axis from 0x = to 3.x = Find the volume of the solid of revolution.

10. Evaluate x

x x x3 2 12

4 3 2

1

2 − + − .dx#

11. Find the area bounded by the curve ,y x3= the y -axis and the lines 0y = and 1.y =

12. Find the indefi nite integral (primitive function) of 7 3 .x 11+] g

13. Find the area bounded by the curve 2,y x x2= − − the x -axis and the lines 1x = and 3.x =

14. Find the volume of the solid formed if the curve 1y x2= + is rotated about the

(a) x -axis from 0x = to 2x = (b) y -axis from 1y = to 2y =

ch3.indd 124 6/11/09 11:39:46 PM


15. (a) Change the subject of 3y x 2= +] g to x . (b) Find the area bounded by the curve

3 ,y x 2= +] g the y -axis and the lines 9y = and 16y = in the fi rst quadrant.

(c) Find the volume of the solid formed if this area is rotated about the y -axis.

16. Evaluate t t dt3 6 52

0

4− +] g# .

17. Find the area bounded by the curve y x x2 152= + − and the x -axis.

18. Find the volume of the solid formed if the curve y x3= is rotated about the y -axis from 0y = to 1.y =

19. Find

(a) x dx5 2 1 4−] g#

(b) x dx4

3 5

#


1. (a) Find the area enclosed between the curves y x3= and .y x2=

(b) Find the volume of the solid formed if this area is rotated about the x -axis.

2. (a) Show that xf x x3= +] g is an odd function.

(b) Hence fi nd the value of f x dx2

2

−] g# .

(c) Find the total area between ,f x] g the x -axis and the lines x 2= − and .x 2=

3. The curve y 2x= is rotated about the x -axis between 1x = and 2.x = Use Simpson’s rule with 3 function values to fi nd an approximation of the volume of the solid formed, correct to 3 signifi cant fi gures.

4. Find the area enclosed between the curves y x 1 2= −] g and 5 .y x2= −

5. (a) Differentiate 1 .x4 9−] g

(b) Hence fi nd .x x dx13 4 8−] g#

6. (a) Differentiate .xx

3 42 1

2

2

−+

(b) Hence evaluate .x

x dx3 42 20

1

−^ h#

7. Find the area enclosed between the

curve ,x

y2

3−

= the y -axis and the lines

1y = and 3,y = using the trapezoidal rule with 4 subintervals.

8. Find the exact volume of the solid

formed by rotating the curve xy 1= about

the x -axis between 1x = and 3.x =

9. Show that the area enclosed between the

curve ,xy 1= the x -axis and the lines 0x =

and 1x = is infi nite by using Simpson’s rule with 3 function values.

10. (a) Sketch the curve .y x x x1 2= − +] ]g g (b) Find the total area enclosed between

the curve and the x -axis.

11. Find the exact area bounded by the parabola y x2= and the line .y x4= −

12. Find the volume of the solid formed when the curve y x 5 2= +] g is rotated about the y -axis from 1y = to 4.y =

ch3.indd 125 6/29/09 10:30:26 AM


13. (a) Find the derivative of 3 .x x +

(b) Hence fi nd .x

x dx3

2++#

14. (a) Evaluate .x x dx12

1

3− +] g#

(b) Use Simpson’s rule with 3 function

values to evaluate x x dx12

1

3− +] g# and

show that this gives the exact value.

15. Find the area enclosed between the curves y x= and .y x3=

16. For the shaded region below, fi nd the area (a) the volume when this area is rotated (b)

about the y -axis.

ch3.indd 126 6/29/09 10:30:32 AM

4

Exponential equation: Equation where the pronumeral is the index or exponent such as 3 9x =

Exponential function: A function in the form y ax= where the variable x is a power or exponent

Logarithm: A logarithm is an index. The logarithm is the power or exponent of a number to a certain base i.e. 2 8x = is the same as log 8 x

2=

TERMINOLOGY

Exponential and Logarithmic Functions

ch4.indd 132 6/12/09 12:52:00 AM

133Chapter 4 Exponential and Logarithmic Functions

DID YOU KNOW?

John Napier (1550–1617), a Scottish theologian and an amateur mathematician, was the fi rst to invent logarithms. These ‘natural’, or ‘Naperian’, logarithms were based on ‘ e ’ . Napier was also one of the fi rst mathematicians to use decimals rather than fractions. He invented the notation of the decimal, using either a comma or a point. The point was used in England, but a few European countries still use a comma.

Henry Briggs (1561–1630), an Englishman who was a professor at Oxford, decided that logarithms would be more useful if they were based on 10 (our decimal system). These are called common logarithms . Briggs painstakingly produced a table of logarithms correct to 14 decimal places. He also produced sine tables—to 15 decimal places—and tangent tables—to 10 decimal places.

The work on logarithms was greatly appreciated by Kepler, Galileo and other astronomers at the time, since they allowed the computation of very large numbers.

INTRODUCTION

THIS CHAPTER INTRODUCES A new irrational number, ‘ e ’, that has special properties in calculus . You will learn how to differentiate and integrate the exponential function .( )f x ex=

The defi nition and laws of logarithms are also introduced in this chapter, as well as differentiation and integration involving logarithms .

Differentiation of Exponential Functions

When differentiating exponential functions ( )f x ax= from fi rst principles, an interesting result can be seen. The derivative of any exponential function gives a constant which is multiplied by the original function.

EXAMPLE

Sketch the derivative (gradient) function of y 10x= .

Solution

CONTINUED

ch4.indd 133 6/12/09 12:52:06 AM


The graph of y 10x= always has a positive gradient that is becoming steeper. So the derivative function will always be positive, becoming steeper.

The derivative function of an exponential function will always have a shape similar to the original function.

We can use differentiation from fi rst principles to fi nd how close this derivative function is to the original function.

EXAMPLE

Differentiate ( )f x 10x= from fi rst principles.

Solution

( )x

10

=

=

( ) ( )

( )

lim

lim

lim

lim

fh

f x h f x

h

h

h

10 10

10 10 1

10 1

h

h

x h x

h

x h

x

h

h

0

0

0

0

+ −

= −

=−

−

"

"

"

"

+

l

Using the 10x key on the calculator, and fi nding values of h

10 1h − when h is small, gives the result:

or . ´2 3026 10Z( )x

( ) . ´dxd 10 2 3026 10

x

x xZ

f l

Drawing the graphs of . ý 2 3026 10x= and y 10x= together shows how close the derivative function is to the original graph.

12

10

8

6

4

2

1 2−1−2

y

y = 2.3026 × 10x

x

y = 10x

You can explore limits using a graphics package on a computer or a graphical calculator.

ch4.indd 134 6/12/09 12:52:09 AM


Similar results occur for other exponential functions. In general,

( )dxd a kax x= where k is a constant.

Application

If y ax= then dx

dyka

ky

x=

=

This means that the rate of change of y is proportional to y itself. That is, if y is small, its rate of change is small, but if y is large, then it is changing rapidly.

This is called exponential growth (or decay , if k is negative) and has many applications in areas such as population growth, radioactive decay, the cooling of objects, the spread of infectious diseases and the growth of technology.

Different exponential functions have different values of k .

EXAMPLES

1. ( ) . ´dxd 2 0 6931 2x xZ .

12

10

8

6

4

2

1 2 3−3 −2 −1

y

y = 2x

y = 0.6931 × 2x

x

2. ( ) . ´dxd 3 1 0986 3x xZ .

12

10

8

6

4

2

−3

y

y = 3x

y = 1.0986 × 3x

x1 32−2−1

You will study exponential growth and decay in

Chapter 6.

ch4.indd 135 6/12/09 12:52:10 AM


Notice that the derivative function of y 3x= is very close to the original function.

We can fi nd a number close to 3 that gives exactly the same graph for the derivative function. This number is approximately 2.71828, and is called e .

ex KEY

Use this key to fi nd powers of e.For example, to fi nd e2:Press SHIFT e 2 7.389056099e2x =To fi nd e:Press SHIFT 1 2.718281828e e1x =

EXAMPLES

1. Sketch the curve .y ex=

Solution

Use e x on your calculator to draw up a table of values:

x −3 −2 −1 0 1 2 3

y 0.05 0.1 0.4 1 2.7 7.4 20.1

( )dxd e ex x=

DID YOU KNOW?

The number e was linked to logarithms before this useful result in calculus was known. It is a transcendental (irrational) number. This was proven by a French mathematician, Hermite , in 1873. Leonhard Euler (1707–83) gave e its symbol, and he gave an approximation of e to 23 decimal places. Currently, e is known to about 100 000 decimal places.

Euler studied mathematics, theology, medicine, astronomy, physics and oriental languages. He did extensive research into mathematics and wrote more than 500 books and papers.

Euler gave mathematics much of its important notation. He caused π to become standard notation and used i for the square root of –1. He fi rst used small letters to show the sides of triangles and the corresponding capital letters for their opposite angles. Also, he introduced the symbol S for sums and f ( x ) notation.

A transcendental number is a number beyond ordinary numbers. Another transcendental number is π.

e is an irrational number like π.

ch4.indd 136 6/12/09 12:52:11 AM


2. Differentiate .e5 x

Solution

( )

( ) ( )

dxd e e

dxd e

dxd e

e

5 5

5

x x

x x

x

=

=

=

3. Find the equation of the tangent to the curve y e3 x= at the point (0, 3).

Solution

dx

dye3 x=

`

At ( , ),dx

dye

m

0 3 3

3

3

0=

==

Equation ( )

( )

y y m x x

y x

x

y x

3 0

3

3 3

31 1− = −

− = −== +

dx

dy gives the gradient of

the tangent.

CONTINUED

ch4.indd 137 6/29/09 10:53:24 AM


4.1 Exercises

1. Find, correct to 2 decimal places, the value of

(a) e .1 5 (b) e 2− (c) e2 .0 3

(d) e1

3

(e) e3 .3 1− −

2. Sketch the curve (a) y e2 x= (b) y e x= − (c) y ex= −

3. Differentiate (a) e9 x (b) ex− (c) e xx 2+ (d) x x x e2 3 5 x3 2− + − (e) 3( )e 1x + (f) 7( )e 5x + (g) 2( )e2 3x − (h) xex

(i) xex

(j) x ex2 (k) ( )x e2 1 x+

(l) xe

7 3

x

−

(m) ex5x

4. If ( ) ,f x x x e3 x3= + − fi nd ( )1f l and ( )f 1m in terms of e .

5. Find the exact gradient of the tangent to the curve y ex= at the point (1, e ).

6. Find the exact gradient of the normal to the curve y ex= at the point where .x 5=

7. Find the gradient of the tangent to the curve y e4 x= at the point where . ,x 1 6= correct to 2 decimal places.

8. Find the equation of the tangent to the curve y ex= − at the point (1, – e ).

4. Differentiate .e

x2 3x+

Solution

v u

.

( )

( )

( )

( )

dx

dy

vu v

e

e e x

ee xe e

ee xe

e

e x

e

x

2 2 3

2 2 3

2

1 2

1 2

x

x x

x

x x x

x

x x

x

x

x

2

2

2

2

2

= −

=− +

= − −

= − −

=− +

=− +

l l

This is the quotient rule from Chapter 8 of the Preliminary Course book.

ch4.indd 138 6/12/09 12:52:13 AM



Remember that the function of a function rule uses the result

.´dx

dy

du

dy

dxdu=

EXAMPLE

Differentiate .ex x5 32 + −

Solution

Let u x x5 32= + − Then y eu=

dxdu x2 5= + and

du

dyeu=

( )

( )

( )

´dx

dy

du

dy

dxdu

e x

e x

x e

2 5

2 5

2 5

u

x x

x x

5 3

5 3

2

2

=

= += += +

+ −

+ −

You studied this in Chapter 8 of the Preliminary Course book.

Can you see a quick way to do this?

Proof

Let ( )u f x=

Then y eu=

du

dyeu= and ( )x

dxdu f= l

( )x

( )x e

´dx

dy

du

dy

dxdu

e f

f ( )

u

f x

=

==

l

l

If y e ( )f x= then ( )x edx

dy( )f x= f l

9. Find the equation of the normal to the curve y ex= at the point where ,x 3= in exact form.

10. Find the stationary point on the curve y xex= and determine its nature. Hence sketch the curve.

11. Find the fi rst and second derivatives of y e7 x= . Hence show

that .dx

d yy

2

2

=

12. If ,y e2 1x= + show that .dx

d yy 1

2

2

= −

ch4.indd 139 6/12/09 12:52:14 AM


4.2 Exercises

(m) xe x

2

3

(n) x e x3 5

(o) xe

2 5

x2 1

+

+

2. Find the second derivative of .7( )e 1x2 +

3. If ( ) ,f x e x3 2= − fi nd the exact value of ( )f 1l and ( ) .f 0m

4. Find the gradient of the tangent to the curve y e x5= at the point where .x 0=

EXAMPLES

1. Differentiate e x5 2−

Solution

( )x ey f

e5

( )f x

x5 2

== −

l l

2. Differentiate x e x2 3 .

Solution

u +dx

dyvv u= l l

. .2 3

( )

x e e x

xe x2 3

x x

x

3 3 2

3

= += +

3. Given ,y e2 1x3= + show that ( ) .dx

d yy9 1

2

2

= −

Solution

( )

( )( )

y e

dx

dye

dx

d ye

e

e

y

2 1

6

18

9 2

9 2 1 19 1

x

x

x

x

x

3

3

2

23

3

3

= +

=

=

== + −= −

This is the product rule from Chapter 8 of the Preliminary Course book.

1. Differentiate (a) e x7 (b) e x− (c) e x6 2− (d) ex 12 + (e) ex x5 73 + + (f) e x5 (g) e x2− (h) e x10 (i) e xx2 + (j) x x e2 x2 1+ + − (k) 5( )x e x4+ (l) xe x2

ch4.indd 140 6/12/09 12:52:14 AM


Integration of Exponential Functions

Since ( ) ,dxd e ex x= then the reverse must be true.

5. Find the equation of the tangent to the curve y e x3x2= − at the point (0, 1).

6. Find the exact gradient of the normal to the curve y e x3= at the point where .x 1=

7. Find the equation of the tangent to the curve y ex2= at the point (1, e ).

8. If ( ) ,f x x x e4 3 x3 2 2= + − − fi nd ( )f 1−m in terms of e .

9. Find any stationary points on the curve y x e x2 2= and sketch the curve.

10. If ,y e ex x4 4= + − show that

.dx

d yy16

2

2

=

11. Prove ,dx

d y

dx

dyy3 2 0

2

2

− + = given

.y e3 x2=

12. Show dx

d yb y

2

22= for .y aebx=

13. Find the value of n if y e x3= satisfi es the equation

y

.dx

d

dx

dyny2 0

2

2

+ + =

14. Sketch the curve ,y ex x 22= + − showing any stationary points and infl exions.

15. Sketch the curve e

xy 1x

2

= + ,

showing any stationary points and infl exions.

e dx e Cx x= +#

Integration is the inverse of differentiation.

To fi nd the indefi nite integral (primitive function) when the function of a function rule is involved, look at the derivative fi rst.

EXAMPLE

Differentiate .e x2 1+

Hence fi nd .e dx2 x2 1+#

Find .e dxx2 1+#

Solution

( )dxd e e

e dx e C

e dx e dx

e C

2

2

21 2

21

x x

x x

x x

x

2 1 2 1

2 1 2 1

2 1 2 1

2 1

`

=

= +

=

= +

+ +

+ +

+ +

+

###

ch4.indd 141 6/12/09 12:52:15 AM


EXAMPLES

1. Find ( ) .e e dxx x2 − −#

Solution

( )

( )e e dx e e C

e e C

21

11

21

x x x x

x x

2 2

2

− = −−

+

= + +

− −

−

#

2. Find the exact area enclosed between the curve ,y e x3= the x -axis and the lines x 0= and .x 2=

Solution

Area

( )

( ) units

e dx

e

e e

e e

e

31

31

31

31

31 1

x

x

3

0

3

0

2

6 0

6 0

6 2

=

=

= −

= −

= −

2

; E#

3. Find the volume of the solid of revolution formed when the curve y ex= is rotated about the x- axis from x 0= to .x 2=

Solution

( )

y e

y e

e

x

x

x

2 2

2

`

===

In general

Use index laws to simplify ( ) .e 2x

e dx a e C1ax b ax b= ++ +#

Proof

( )dxd e ae

ae dx e C

e dx a ae dx

a e C

1

1

ax b ax b

ax b ax b

ax b ax b

ax b

`

=

= +

=

= +

+ +

+ +

+ +

+

###

ch4.indd 142 6/12/09 12:52:16 AM


( ) units

π

π

π

π

π

π

V y dx

e dx

e

e e

e

e

21

21

21

21

21

21

a

x

x

2

2

0

2

0

2

4 0

4

4 3

=

=

=

= −

= −

= −

b

2

cc

mm

; E

##

4.3 Exercises

(c) ( )e x dx2 3x 5

5

6+ −+#

(d) ( )e t dtt3 4

0−+1#

(e) ( )e e dxx x4 2

1+

2#

4. Find the exact area enclosed by the curve ,y e2 x2= the x- axis and the lines x 1= and .x 2=

5. Find the exact area bounded by the curve ,y e x4 3= − the x -axis and the lines x 0= and .x 1=

6. Find the area enclosed by the curve ,y x e x= + − the x -axis and the lines x 0= and ,x 2= correct to 2 decimal places.

7. Find the area bounded by the curve ,y e x5= the x -axis and the lines x 0= and ,x 1= correct to 3 signifi cant fi gures.

8. Find the exact volume of the solid of revolution formed when the curve y ex= is rotated about the x- axis from x 0= to .x 3=

9. Find the volume of the solid formed when the curve y e 1x= +− is rotated about the x -axis from x 1= to ,x 2= correct to 1 decimal place.

1. Find these indefi nite integrals.

(a) e dxx2#

(b) e dxx4#

(c) e dxx−#

(d) e dxx5#

(e) e dxx2−#

(f) e dxx4 1+#

(g) e dx3 x5−#

(h) e dtt2#

(i) ( )e dx2x7 −#

(j) ( )e x dxx 3 +−#

2. Evaluate in exact form.

(a) e dxx5

0

1#

(b) e dxx

0− −

2#

(c) e dx2 x3 4

1

+4#

(d) ( )x e dx3 x2 2

2−

3#

(e) ( )e dx1x2

0+

2#

(f) ( )e x dxx

1−

2#

(g) ( )e e dxx x2

0− −3#

3. Evaluate correct to 2 decimal places.

(a) e dxx

1

−3#

(b) e dy2 y3

0

2#

ch4.indd 143 6/12/09 12:52:17 AM


10. Use Simpson’s rule with 3 function values to fi nd an

approximation to ,xe dxx

1

2#

correct to 1 decimal place.

11. (a) Differentiate .x ex2

(b) Hence fi nd ( ) .x x e dx2 x+#

12. The curve y e 1x= + is rotated about the x- axis from x 0= to .x 1= Find the exact volume of the solid formed.

13. Find the exact area enclosed between the curve y e x2= and the lines y 1= and .x 2=

Application

The exponential function occurs in many fi elds, such as science and economics. P P e

0

kt= is a general formula that describes exponential growth .

P P e0

kt= - is a general formula that describes exponential decay .

Logarithms

‘Logarithm’ is another name for the index or power of a number. Logarithms are related to exponential functions, and allow us to solve equations like .2 5x = They also allow us to change the subject of exponential equations such as y ex= to x .

Defi nition

If ,y ax= then x is called the logarithm of y to the base a .

If ,y ax= then logx ya=

Logarithm keys

log is used for log x10

In is used for log xe

You will study these formulae in Chapter 6.

ch4.indd 144 6/12/09 12:52:17 AM


EXAMPLES

1. Find .log 5 310 correct to 1 decimal place.

Solution

log 5.3 0.724275869

0.7 correct to 1 decimal place10 =

=

2. Evaluate log 80e correct to 3 signifi cant fi gures.

Solution

.

. correct to 3 significant figures

log 80 4 382026634

4 38e =

=

3. Evaluate .log 813

Solution

Let

Then (by definition)

i.e.

So .

log

log

x

x

81

3 81

3 34

81 4

x

x

3

4

3

`

=

====

4. Find the value of .log41

2

Solution

Let

Then

So .

log

log

x

x

41

241

21

22

41 2

x

2

2

2

2

`

=

=

=

== −

= −

−

Use the log key.

Use the In key.

ch4.indd 145 6/12/09 12:52:18 AM


1. Evaluate log (a) 2 16 log (b) 4 16 log (c) 5 125 log (d) 3 3 log (e) 7 49 log (f) 7 7 log (g) 5 1 log (h) 2 128

2. Evaluate 3 log (a) 2 8 log (b) 5 25 + 1 3 – log (c) 3 81 4 log (d) 3 27 2 log (e) 10 10 000 1 + log (f) 4 64 3 log (g) 4 64 + 5 2 + 4 log (h) 6 216

(i) log

2

93

(j) log

log

8

64 4

2

8 +

3. Evaluate

(a) log21

2

(b) log 33

log (c) 4 2

(d) log251

5

(e) log 774

(f) log3

13 3

(g) log21

4

log (h) 8 2

(i) log 6 66

(j) log42

2

4. Evaluate correct to 2 decimal places.

log (a) 10 1200 log (b) 10 875 log (c) e 25 ln 140 (d) 5 ln 8 (e) log (f) 10 350 + 4.5

(g) log

2

1510

ln 9.8 + log (h) 10 17

(i) log

log

30

30

e

10

4 ln 10 – 7 (j)

4.4 Exercises

Class Investigation

Sketch the graph of 1. logy x2= . There is no calculator key for logarithms to the base 2. Use the defi nition of a logarithm to change the equation into index form, and the table of values:

x

y –3 –2 –1 0 1 2 3

On the same set of axes, sketch the curve 2. 2y x= and the line y x= . What do you notice?

ch4.indd 146 6/12/09 12:52:19 AM


5. Write in logarithmic form. (a) y3x = (b) z5x = (c) x y2 = (d) a2b = (e) b d3 = (f) y 8x= (g) y 6x= (h) y ex= (i) y ax= (j) e=Q x

6. Write in index form. (a) log x53 = (b) log x7a = (c) log a b3 = (d) 9 log yx = (e) log b ya = (f) logy 62= (g) logy x3= (h) logy 910= (i) 4lny = (j) logy x7=

7. Solve for x , correct to 1 decimal place where necessary.

(a) log x 610 =

(b) log x 53 =

(c) log 343 3x =

(d) log 64 6x =

(e) log x51

5 =

(f) log 321

x =

(g) 3.8ln x =

(h) log x3 2 1010 − =

(i) log x23

4 =

(j) log 431

x =

8. Evaluate y given that .log 125 3y =

9. If . ,log x 1 6510 = evaluate x correct to 1 decimal place.

10. Evaluate b to 3 signifi cant fi gures if . .log b 0 894e =

11. Find the value of log 2 1. What is the value of log a 1?

12. Evaluate log 5 5. What is the value of log a a?

13. (a) Evaluate ln e without a calculator.

Using a calculator, evaluate(b) (i) log e e

3 (ii) log e e

2 (iii) log e e

5 (iv) log e e

(v) log e 1e

(vi) e ln 2 (vii) e ln 3 (viii) e ln 5 (ix) e ln 7 (x) e ln 1 (xi) e ln e

14. Sketch the graph of .logy xe= What is its domain and range?

15. Sketch , logy y x10x10= = and

y x= on the same number plane. What do you notice about the relationship of the curves to the line?

16. Change the subject of logy xe= to x .

ch4.indd 147 6/12/09 12:52:19 AM


Proof

Let x am= and y an= Then logm xa= and logn ya=

` ( ) (by definition)

´

log

log log

xy a a

axy m n

x y

m n

m n

a

a a

=== += +

+

Class Discussion

1. Investigate these questions on the calculator. Can you see some patterns?

log (a) e e log (b) e e

2 log (c) e e

3 log (d) e e

4 log (e) e e

5 Can you write a rule for log e e

x ?

Evaluate using a calculator. Can you write a rule to show this pattern?2. (a) e ln 1 (b) e ln 2 (c) e ln 3 (d) e ln 4 (e) e ln 5

Can you write a rule for e ln x ?

Do these rules work if 3. x is negative?

Logarithm laws

Because logarithms are closely related to indices there are logarithm laws that correspond to the index laws.

( )log log logxy x ya a a= +

This corresponds to the law a a a´m n m n= + from Chapter 1 of the Preliminary Course book.

log log logyx x ya a a= −b l

This corresponds to the law .a a a÷m n m n= −

ch4.indd 148 6/12/09 12:52:20 AM


This corresponds to the law ( )a a .m n mn=

,log 36 26

= since .6 362 =

Proof

Let x am= and y an= Then logm xa= and logn ya=

` (by definition)

÷

log

log log

yx a a

a

yx m n

x y

m n

m n

a

a a

=

=

= −

= −

−

b l

log logx n xan

a=

Proof

Let x am= Then logm xa=

( )

(by )loglog

x a

a

x mnn x

definition

n m n

mn

an

a

====

EXAMPLES

1. Evaluate .log log3 126 6+

Solution

( )´log log log

log

3 12 3 12

36

2

6 6 6

6

+ ===

2. Given .log 3 0 685 = and . ,log 4 0 865 = fi nd (a) log 125 (b) .log 0 755 (c) log 95 (d) log 205

CONTINUED

ch4.indd 149 6/12/09 12:52:20 AM


Solution

(a) ( )

. .

.

´log log

log log

12 3 4

3 4

0 68 0 86

1 54

5 5

5 5

== += +=

(b) .

. .

.

log log

log log

0 7543

3 4

0 68 0 86

0 18

5 5

5 5

=

= −= −= −

(c)

.

.

´

log loglog

9 32 3

2 0 68

1 36

5 52

5

====

(d) ( )

.

.

´log loglog log

20 5 45 4

1 0 86

1 86

5 5

5 5

== += +=

3. Solve log 2 12 = log 2 3 + log 2 x .

Solution

log log log

log

x

x

12 3

32 2 2

2

= +=

So 12 = 3 x 4 = x

,log 5 15

= since .5 51 =

1. Use the logarithm laws to simplify

log (a) a 4 + log a y log (b) a 4 + log a 5 log (c) a 12 – log a 3 log (d) a b – log a 5 3 log (e) x y + log x z 2 log (f) k 3 + 3 log k y 5 log (g) a x – 2 log a y log (h) a x + log a y – log a z log (i) 10 a + 4 log 10 b + 3 log 10 c 3 log (j) 3 p + log 3 q – 2 log 3 r

2. Given log 7 2 = 0.36 and log 7 5 = 0.83, fi nd

log (a) 7 10 log (b) 7 0.4 log (c) 7 20 log (d) 7 25 log (e) 7 8 log (f) 7 14 log (g) 7 50 log (h) 7 35 log (i) 7 98 log (j) 7 70

4.5 Exercises

ch4.indd 150 6/12/09 12:52:21 AM


3. Use the logarithm laws to evaluate

log (a) 5 50 – log 5 2 log (b) 2 16 + log 2 4 log (c) 4 2 + log 4 8 log (d) 5 500 – log 5 4 log (e) 9 117 – log 9 13 log (f) 8 32 + log 8 16 3 log (g) 2 2 + 2 log 2 4 2 log (h) 4 6 – (2 log 4 3 + log 4 2) log (i) 6 4 – 2 log 6 12 2 log (j) 3 6 + log 3 18 – 3 log 3 2

4. If log a 3 = x and log a 5 = y , fi nd an expression in terms of x and y for

log (a) a 15 log (b) a 0.6 log (c) a 27 log (d) a 25 log (e) a 9 log (f) a 75 log (g) a 3 a

log (h) a 5a

log (i) a 9 a

log (j) a a125

5. If log a x = p and log a y = q , fi nd, in terms of p and q .

log (a) a xy log (b) a y 3

log (c) a xy

log (d) a x 2 log (e) a xy 5

log (f) a yx2

log (g) a ax

log (h) a ya

2

log (i) a a 3 y

log (j) a ayx

6. If log a b = 3.4 and log a c = 4.7, evaluate

log (a) a bc

log (b) a bc 2 log (c) a ( bc ) 2 log (d) a abc log (e) a a 2 c log (f) a b 7 log (g) a c

a log (h) a a 3 log (i) a bc 4 log (j) a b 4 c 2

7. Solve log (a) 4 12 = log 4 x + log 4 3 log (b) 3 4 = log 3 y – log 3 7 log (c) a 6 = log a x – 3 log a 2 log (d) 2 81 = 4 log 2 x log (e) x 54 = log x k + 2 log x 3

Change of base

Sometimes we need to evaluate logarithms such as log 2 7. We use a change of base formula.

loglog

logx

a

xa

b

b=

ch4.indd 151 6/12/09 12:52:22 AM


EXAMPLE

Find the value of ,log 25 correct to 2 decimal places.

Solution

.0 430676558Z

(by change of base)

.

loglog

log2

5

2

0 43

5 =

=

Proof

Let logy xa= Then x ay= Take logarithms to the base b of both sides of the equation:

`

log loglog

log

log

log

x ay a

a

xy

x

b by

b

b

b

a

==

=

=

You can use the change of base formula to fi nd the logarithm of any number, such as .log 25 You change it to either log x10 or ,log xe and use a calculator.

Exponential equations

You can also use the change of base formula to solve exponential equations such as .5 7x =

You studied exponential equations such as 2 8x = in the Preliminary Course. Exponential equations such as 2 9x = can be solved by taking logarithms of both sides, or by using the defi nition of a logarithm and the change of base formula.

You can use either log or In

ch4.indd 152 6/12/09 12:52:22 AM


EXAMPLES

1. Solve 5 7x = correct to 1 decimal place.

Solution

5 7x = Using the defi nition of a logarithm, this means:

(using change of base formula)

.

log

log

log

x

x

x

7

5

7

1 2

5 =

=

=

If you do not like to solve the equation this way, you can use the logarithm laws instead.

Taking logs of both sides:

.

log loglog log

log

log

x

x

5 75 7

5

7

1 2 1correct to decimal place

x

`

==

=

=

2. Solve 4 9y 3 =− correct to 2 decimal places.

Solution

y 3−4 9= Using the logarithm defi nition and change of base:

.

log

log

log

log

log

y

y

y

y

9 3

4

93

4

93

4 58

4 = −

= −

+ =

=

Using the logarithm laws:

Taking logs of both sides:

( )

.

log loglog log

log

log

log

log

y

y

y

4 93 4 9

34

9

4

93

4 58 2correct to decimal place

y 3 =− =

− =

= +

=

−

You can use either log or ln.

Use log x n log xa

n

a=

ch4.indd 153 6/12/09 12:52:23 AM


1. Use the change of base formula to evaluate to 2 decimal places.

log (a) 4 9 log (b) 6 25 log (c) 9 200 log (d) 2 12 log (e) 3 23 log (f) 8 250 log (g) 5 9.5 2 log (h) 4 23.4 7 – log (i) 7 108 3 log (j) 11 340

2. By writing each equation as a logarithm and changing the base, solve the equation correct to 2 signifi cant fi gures.

(a) 4 9x = (b) 3 5x = (c) 7 14x = (d) 2 15x = (e) 5 34x = (f) 6 60x = (g) 2 76x = (h) 4 50x = (i) 3 23x = (j) 9 210x =

3. Solve, correct to 2 decimal places. (a) 2 6x = (b) 5 15y = (c) 3 20x =

(d) 7 32m = (e) 4 50k = (f) 3 4t = (g) 8 11x = (h) 2 57p = (i) .4 81 3x = (j) .6 102 6n =

4. Solve, to 1 decimal place. (a) 3 8x 1 =+ (b) n35 71= (c) 2 12x 3 =− (d) 4 7n2 1 =− (e) 7 11x5 2 =+ (f) .8 5 7n3 =− (g) .2 18 3x 2 =+ (h) k7 3− .3 32 9= (i) 29 50=

x

(j) .6 61 3y2 1 =+

5. Solve each equation correct to 3 signifi cant fi gures.

(a) e 200x = (b) e 5t3 = (c) e2 75t = (d) e45 x= (e) e3000 100 n= (f) e100 20 t3= (g) e2000 50 . t0 15= (h) e15 000 2000 . k0 03= (i) Q Qe3 . t0 02= (j) . M Me0 5 . k0 016=

4.6 Exercises

ch4.indd 154 6/12/09 12:52:23 AM


Derivative of the Logarithmic Function

Drawing the derivative (gradient) function of a logarithm function gives a hyperbola.

EXAMPLE

Sketch the derivative function of .logy x2=

Solution

The gradient is always positive but is decreasing.

If logy xe= then 1dx

dyx=

Proof

dx

dy

dydx1=

Given logy xe= Then yx e=

y

dydx e

dx

dy

e

x

1

1

y

=

=

=

dx

dy

dy

dx

1= is a special result that

can be proved by differentiating from fi rst principles.

ch4.indd 155 6/12/09 12:52:24 AM



If .( ), then ( )( ) ( )

( )logy f x

dx

dyf x

f x f x

f x1e= = =l

l

Proof

Let ( )u f x= Then logy ue=

`

Also

.

.

.

( )

( )

( )( )

du

dyu

dxdu f x

dx

dy

du

dy

dxdu

u f x

f xf x

1

1

1

=

=

=

=

=

l

l

l

EXAMPLES

1. Differentiate ( 3 1).log x x2e − +

Solution

[ ( )]logdxd x x

x xx3 13 1

2 3e

22

− + =− +

−

2. Differentiate 3 4

1 .logxx

e −+

Solution

Let

( ) ( )

( ) ( )( ) ( )

( ) ( )

( ) ( )

log

log log

yxx

x x

dx

dy

x x

x xx x

x xx x

x x

3 41

1 3 4

11

3 43

1 3 41 3 4 3 1

1 3 43 4 3 3

1 3 47

e

e e

=−

+

= + − −

=+

−−

=+ −− − +

=+ −− − −

=+ −

−

ch4.indd 156 6/12/09 12:52:24 AM


3. Find the gradient of the normal to the curve ( )logy x 5e3= − at the

point where 2.x =

Solution

dx

dy

xx

533

2

=−

When 2,x =

( )dx

dy

m2 5

3 2

4

3

2

1

=−

=

The normal is perpendicular to the tangent

m m

m

m

1

4 1

41

i.e. 1 2

2

2`

= −= −

= −

4. Differentiate .logy x2=

Solution

´

´

log

log

log

loglog

log

log

y x

x

x

dx

dyx

x

2

21

21 1

21

e

e

ee

e

e

2=

=

=

=

=

5. Find the derivative of 2 x .

Solution

( )

´

ln

ln

ln

e

e

e

dx

dye

2

2

2

2 2

2 2

2

ln

ln

ln

ln

x x

x

x

x

x

2

2

2

`

===

=

==

This result comes from the Preliminary Course.

ch4.indd 157 6/12/09 12:52:25 AM


1. Differentiate

(a) logx xe+

(b) log x1 3e−

(c) ( )ln x3 1+

(d) ( )log x 4e2 −

(e) ( )ln x x5 3 93 + −

(f) ( )log x x5 1e2+ +

(g) lnx x x3 5 5 42 + − +

(h) ( )log x8 9 2e − +

(i) (2 4)(3 1)log x xe + −

(j) 2 74 1log

xx

e −+

(k) ( )log x1 e5+

(l) ( )ln x x 9−

(m) ( )log xe4

(n) ( )logx xe2 6+

(o) logx xe

(p) log

xxe

(q) ( ) logx x2 1 e+

(r) ( )logx x 1e3 +

(s) ( )log log xe e

(t) lnx

x2−

(u) log x

e

e

x2

(v) lne xx

(w) ( )log x5 e2

2. If ( ) 2 ,logf x xe= − fi nd ( )f 1l .

3. Find the derivative of .log x10

4. Find the equation of the tangent to the curve logy xe= at the point ( , )log2 2e .

5. Find the equation of the tangent to the curve ( 1)logy xe= − at the point where 2x = .

6. Find the gradient of the normal to the curve ( )logy x xe

4= + at the point ( , )log1 2e .

7. Find the exact equation of the normal to the curve logy xe= at the point where 5x = .

8. Find the equation of the tangent to the curve ( )logy x5 4e= + at the point where 3x = .

9. Find the point of infl exion on the curve .logy x x xe

2= −

10. Find the stationary point on the

curve lnx

xy = and determine its

nature.

11. Sketch, showing any stationary points and infl exions.

(a) logy x xe= − (b) ( )logy x 1e

3= − (c) lny x x=

12. Find the derivative of ( )log x2 53 + .

13. Differentiate (a) 3x (b) 10x (c) 2 x3 4−

14. Find the equation of the tangent to the curve y 4x 1= + at the point (0, 4) .

15. Find the equation of the normal to the curve logy x3= at the point where 3x = .

4.7 Exercises

ch4.indd 158 6/12/09 12:52:25 AM


logxdx

x dx x C1e= = +##

Integration and the Logarithmic Function

EXAMPLES

1. Find the area enclosed between the hyperbola ,y x1= the x -axis and

the lines 1x = and 2x = , giving the exact value.

Solution

log

log log

log

A x dx

x

1

2 1

2

e

e e

e

1

12

=

== −=

2

7 A#

So area is log 2e units 2 .

2. Find .x

x dx73

2

+#

Solution

( )log

xx dx

xx dx

x C

7 31

73

31 7e

3

2

3

2

3

+=

+

= + +

##

( )

( )( )log

f x

f xdx f x Ce= +

l#

CONTINUED


ch4.indd 159 6/12/09 12:52:26 AM



(a) 2 5

2x +

(b) x

x2 1

42 +

(c) 2

5x

x5

4

−

(d) 21x

(e) 2x

(f) 35x

(g) 3

2 3x x

x2 −

−

(h) x

x22 +

(i) x

x7

32 +

(j) x x

x2 5

12 + −

+

2. Find

(a) x

dx4 1

4−#

(b) x

dx3+#

(c) x

x dx2 73

2

−#

(d) x

x dx2 56

5

+#

(e) x x

x dx6 2

32 + +

+#

3. Evaluate correct to 1 decimal place.

(a) x

dx2 5

21

3

+#

(b) x

dx12 +

5#

(c) x

x dx23

2

1

7

+#

(d) x x

x dx2 1

4 120

3

+ ++#

(e) x xx dx

21

23

4

−−#

4. Find the exact area between the

curve ,y x1= the x- axis and the

lines 2x = and 3.x =

5. Find the exact area bounded by

the curve ,yx 1

1=−

the x- axis and

the lines 4x = and 7.x =

6. Find the exact area between the

curve ,y x1= the x- axis and the

lines y x= and 2x = in the fi rst quadrant.

7. Find the area bounded by the

curve ,yx

x12

=+

the x- axis and

the lines 2x = and 4x = , correct to 2 decimal places.

4.8 Exercises

3. Find .x x

x dx4

12 + +

+#

Solution

( )

( )log

x xx dx

x x

xdx

x xx dx

x x C

2 41

21

2 4

2 1

21

2 42 2

21 2 4e

2 2

2

2

+ ++ =

+ ++

=+ +

+

= + + +

##

#

ch4.indd 160 6/12/09 12:52:27 AM


8. Find the exact volume of the solid formed when the curve

yx

1= is rotated about the x- axis

from 1x = to 3x = .

9. Find the volume of the solid formed when the curve

yx2 12=

− is rotated about the

x -axis from 1x = to 5x = , giving an exact answer.

10. Find the area between the curve lny x= , the y- axis and the lines 2y = and 4y = , correct to 3 signifi cant fi gures.

11. Find the exact volume of the solid formed when the curve logy xe= is rotated about the y- axis from 1y = to 3y = .

12. (a) Show that

9

3 33

13

2xx

x x2 −+ =

++

− .

(b) Hence fi nd xx dx

93 3

2 −+# .

13. (a) Show that xx

x16

151

−− −

−= .

(b) Hence fi nd xx dx

16

−−# .

14. Find the indefi nite integral (primitive function) of 3 x2 1− .

15. Find, correct to 2 decimal places, the area enclosed by the curve logy x2= , the x- axis and the lines 1x = and 3x = by using Simpson’s rule with 3 function values.

ch4.indd 161 6/12/09 12:52:28 AM


Test Yourself 4 1. Evaluate to 3 signifi cant fi gures.

(a) e 12 − (b) log 9510 (c) log 26e (d) 7log4 (e) 3log4 (f) ln 50 (g) 3e +

(h) ln

e4

5 3

(i) eln 6 (j) eln 2

2. Differentiate (a) e x5 (b) e2 x1 − (c) 4log xe (d) ( )ln x4 5+ (e) xex

(f) lnx

x

(g) 10( )e 1x +


(a) e x4

(b) x

x92 −

(c) e x−

(d) 4

1x +

4. Find the equation of the tangent to the curve y e2 x3= + at the point where 0x = .

5. Find the exact gradient of the normal to the curve y x e x= − − at the point where 2x = .

6. Find the exact area bounded by the curve y e x2= , the x -axis and the lines 2x = and 5x = .

7. Find the volume of the solid formed if the area bounded by y e x3= , the x -axis and the lines 1x = and 2x = is rotated about the x -axis.

8. If .log 2 0 367 = and .log 3 0 567 = , fi nd the value of

(a) log 67 (b) log 87 (c) .log 1 57 (d) log 147 (e) .log 3 57

9. Find the area enclosed between the curve lny x= , the y -axis and the lines 1y = and 3y = .

10. (a) Use Simpson’s rule with 3 function values to fi nd the area bounded by the curve lny x= , the x -axis and the lines 2x = and 4x = . (b) Change the subject of lny x= to x . (c) Hence fi nd the exact area in part (a).

11. Solve (a) 3 8x = (b) 2 3x3 4 =− (c) log 81 4x = (d) log x 26 = (e) e12 10 . t0 01=

12. Evaluate

(a) e dx3 x2

0

1#

(b) xdx

3 21 −4#

(c) xx x x dx2 5 33 2

1

2 − + +#

13. Find the equation of the tangent to the curve y ex= at the point ( , )e4 4 .

14. Evaluate log 89 to 1 decimal place.

ch4.indd 162 6/12/09 12:52:28 AM


15. (a) Find the area bounded by the curve ,y ex= the x -axis and the lines 1x = and 2x = .

This area is rotated about the (b) x -axis. Find the volume of the solid of revolution formed.

16. Simplify (a) log logx y5 3a a+ (b) log log logk p2 3x x x− +

17. Find the equation of the normal to the curve lny x= at the point ( , )ln2 2 .

18. Find the stationary points on the curve y x ex3= and determine their nature.

19. Use the trapezoidal rule with 4 strips to fi nd the area bounded by the curve ( )lny x 12= − , the x -axis and the lines 3x = and 5x = .

20. Evaluate to 2 signifi cant fi gures (a) .log 4 510 (b) .ln 3 7


1. Differentiate log

e x

xx

e

2 + .

2. Find the exact gradient of the tangent to the curve y e logx xe= + at the point where 1x = .

3. If .log 2 0 6b = and .log 3 1 2b = , fi nd (a) log b6b (b) log 8b (c) .log b1 5b

2

4. Differentiate ( )loge xxe

4 9+ .

5. Find the shaded area, correct to 2 decimal places.

6. Find the derivative of .logx2 31

e −

7. Use Simpson’s rule with 5 function values to fi nd the volume of the solid formed when the curve y ex= is rotated about the y -axis from 3y = to 5y = , correct to 2 signifi cant fi gures.

8. Differentiate 5x .

9. Show that ( ) (1 2 )log logx x x xdxd

e e2 = + .

Hence evaluate ( )logx x dx2 1 2 e1+

3# ,

giving an exact answer.

10. Find dx3x# .

11. (a) Find the point of intersection of the curves logy xe= and logy x10= .

Find the exact equations of the (b) tangents to the two curves at this point of intersection.

Find the exact length of the interval (c) XY where X and Y are the y -intercepts of the tangents.

ch4.indd 163 6/12/09 12:52:32 AM


12. Use Simpson’s rule with 3 function values to fi nd the area enclosed by the curve ,y e x2= the y -axis and the line 3y = , correct to 3 signifi cant fi gures.

13. Find the derivative of .log

e

x xx

e

14. If y e ex x= + − , show dx

d yy

2

2

= .

15. Prove dx

d y

dx

dyy4 5 10 0

2

2

− − − = , given

y e3 2x5= − .

16. Find the equation of the curve that has ( )f x e12 x2=m and a stationary point at ( , )0 3 .

17. Sketch ( )logy x xe2= − .

ch4.indd 164 6/12/09 12:52:37 AM

TERMINOLOGY

Amplitude: Maximum displacement from the mean or centre position in vibrating or oscillating motion

Circular measure: The measurement of an angle as the length of an arc cut off by an angle at the centre of a unit circle. The units are called radians

Period: One complete cycle of a wave or other recurring function

Radian: A unit of angular measurement defi ned as the length of arc of one unit that an angle subtended at the centre of a unit circle cuts off

Sector: Part of a circle bounded by the centre and arc of a circle cut off by two radii

Segment: Part of a circle bounded by a chord and the circumference of a circle

T ERMINOLOGY

Trigonometric Functions

5

ch5.indd 166 6/12/09 2:44:17 AM

167

INTRODUCTION

IN THIS CHAPTER YOU will learn about circular measure, which uses radians rather than degrees. Circular measure is very useful in solving problems involving properties of the circle, such as arc length and areas of sectors and segments, which you will study in this chapter.

You will also study trigonometric graphs in greater detail than in the Preliminary Course. These graphs have many applications in the movement of such things as sound, light and waves.

Circular measure is also useful in calculus, and you will learn about differentiation and integration of trigonometric functions in this chapter.

Circular Measure

Radians

We are used to degrees being used in geometry and trigonometry. However, they are limited. For example, when we try to draw the graph of a trigonometric function, it is diffi cult to mark degrees along the x - or y -axis as there is no concept of how large a degree is along a straight line. This is one of the reasons we use radians.

Other reasons for using radians are that they make solving circle problems simple and they are used in calculus.

A radian is based on the length of an arc in a unit circle (a circle with radius 1 unit), so we can visualise a radian as a number that can be measured along a line.

11 unit

1c

1 unitt

The symbol for 1 radian is 1 c but we usually don’t use the symbol as radians are simply

distances, or arc lengths.

Chapter 5 Trigonometric Functions

ch5.indd 167 6/12/09 2:44:28 AM


Proof

Circumference of the circle with radius 1 unit is given by:

( )

πππ

C r2

2 1

2

===

The arc length of the whole circle is π.2 there are π2 radians in a whole circle. But there are 360° in a whole circle (angle of revolution).

°

°

ππ

2 360

180

So ==

One radian is the angle that an arc of 1 unit subtends at the centre of a circle of radius 1 unit.

°π 180radians =

1 unit

2π

EXAMPLES

1. Convert π2

3 into degrees.

Solution

°,

( °)

°

π

π

180

23

23 180

270

Since =

=

=

2. Change 60° to radians, leaving your answer in terms in π .

Solution

°

°

60° 60

ππ

π

π

π

´

180

1180

180

18060

3

radians

So radians

=

=

=

=

=

ch5.indd 168 6/12/09 2:44:36 AM

169Chapter 5 Trigonometric Functions

3. Convert 50° into radians, correct to 2 decimal places.

Solution

°

°

°

.

ππ

π

π

´

180

1180

50180

50

18050

0 87

radians

So radians

Z

=

=

=

=

4. Change 1.145 radians into degrees, to the nearest minute.

Solution

°

°

. ° .

. °

°

π

π

π ´

180

1 180

1 145 180 1 145

65 6

3665

radians

radian

radians

`

=

=

=

== l

,, ,

Press 180 1.145

SHIFT

π ´ ='

°

5. Convert ° 4138 l into radians, correct to 3 decimal places.

Solution

°

°°

°°

°

.

ππ

π ´

180

1180

38 41180

41

0 675

38

radians=

=

=

=

l l

6. Evaluate cos 1.145 correct to 2 decimal places.

Solution

For cos 1.145, use radian mode on your calculator .

. .

0.41

cos 1 145 0 413046135

correct to 2 ecimal placesd

==

The calculator may give the answer in degrees and

minutes. Simply change this into a decimal.

ch5.indd 169 6/12/09 2:44:36 AM


Notice from the examples that ° π1180

= and 1 radian .π180= You can use

these as conversions rather than starting each time from °.π 180=

To change from radians to degrees: Multiply by .π180

To change from degrees to radians: Multiply by .π180

1°

.

°

π

π

1800 017

180

1857

Notice that

radians

Also 1 radian

=

=

=

= l

While we can convert between degrees and radians for any angle, there are some special angles that we use regularly in this course. It is easier if you know these without having to convert each time.

180

3

2 360

°

°

°

°

°

°

°

π

ππ

ππ

π

π

290

2270

445

360

630

=

=

=

=

=

=

=

1. Change into degrees.

(a) π5

(b) 2π3

(c) 5π4

(d) 7π6

(e) π3

(f) 7π9

(g) 4π3

(h) 7π3

(i) π9

(j) 5π18

5.1 Exercises

ch5.indd 170 6/12/09 2:44:37 AM


2. Convert into radians in terms of π .

(a) °135 (b) °30 (c) °150 (d) °240 (e) °300 (f) °63 (g) °15 (h) °450 (i) °225 (j) °120

3. Change into radians, correct to 2 decimal places.

(a) °56 (b) °68 (c) °127 (d) °289 (e) °312

4. Change into radians, to 2 decimal places.

(a) ° 3418 l (b) ° 1235 l (c) ° 56101 l (d) ° 2988 l (e) ° 3950 l

5. Change these radians into degrees and minutes, to the nearest minute.

1.09 (a) 0.768 (b) 1.16 (c) 0.99 (d) 0.32 (e) 3.2 (f) 2.7 (g) 4.31 (h) 5.6 (i) 0.11 (j)

6. Find correct to 2 decimal places. sin 0.342 (a) cos 1.5 (b) tan 0.056 (c) cos 0.589 (d) tan 2.29 (e) sin 2.8 (f) tan 5.3 (g) cos 4.77 (h) cos 3.9 (i) sin 2.98 (j)

The calculator may give the answer in degrees and minutes.

Simply change this into a decimal.

Trigonometric Results

All the trigonometry that you studied in the Preliminary Course can be done using radians instead of degrees.

Special triangles

The two triangles that give exact trigonometric ratios can be drawn using radians rather than degrees.

1

12

π4

π4

2

1

3

π6

π3

ch5.indd 171 6/12/09 2:44:37 AM


EXAMPLE

Find the exact value of .π πcos cosec36 4

−

Solution

3 π π

ππ

cos cosec

cossin

6 4

36

4

1

323

12

23 3

2

23 3

22 2

23 3 2 2

−

= −

= −

= −

= −

= −

d n

π

π

π

sin

cos

tan

4 21

4 21

41

=

=

=

π

π

π

sin

cos

tan

3 23

3 21

33

=

=

=

π

π

π

sin

cos

tan

6 21

6 23

6 31

=

=

=

Angles of any magnitude

The results for angles of any magnitude can also be looked at using radians. If we change degrees for radians, the ASTC rule looks like this:

Using trigonometric ratios and these special triangles gives the results:

ch5.indd 172 6/12/09 2:44:38 AM


x

y

4th quadrant

0

1st quadrant

3rd quadrant

2nd quadrant

AS

CT

π2

3π2

π-θ

2π-θ

2π

π+θ

π

θ

We can summarise the trigonometric ratios for angles of any magnitude in radians as follows:

First quadrant: Angle θ : θsin is positive θcos is positive θtan is positive

Second quadrant: Angle π θ− :

( )

( )

( )

π θ θπ θ θπ θ θ

sin sin

cos cos

tan tan

− =− = −− = −

Third quadrant: Angle :π θ+

( )

( )

( )


sin sin

cos cos

tan tan

+ = −+ = −+ =

Fourth quadrant: Angle :π θ2 −

( )

( )

( )


sin sin

cos cos

tan tan

2

2

2

− = −− =− = −

ch5.indd 173 6/12/09 2:44:39 AM


EXAMPLES

1. Find the exact value of sin .π4

5

Solution

π π

π π π

π π

44

45

44

4

4

So

=

= +

= +

In the 3 rd quadrant, angles are in the form ,π θ+ so the angle is in the 3 rd quadrant, and θsin is negative in the 3 rd quadrant.

π π π

π

sin sin

sin

45

4

4

21

= +

= −

= −

c cm m

2. Find the exact value of cos .π6

11

Solution

π π

π π π

π π

26

12

611

612

6

26

So

=

= −

= −

In the 4 th quadrant, angles are in the form 2 ,π θ− so the angle is in the 4 th quadrant, and cos θ is positive.

π π π

π

cos cos

cos

611 2

6

6

23

= −

=

=

c cm m

1

12

π4

π4

2

1

3

π6

π3

ch5.indd 174 6/12/09 2:44:40 AM


3. Find the exact value of tan .π3

4−c m

Solution

π π

π π π

π π

33

34

33

3

3

So

=

− = − +

= − +

cc

mm

In the 2 nd quadrant, angles are in the form ( ),π θ− + so the angle is in the 2 nd quadrant, and θtan is negative in the 2 nd quadrant.

π π π

π

tan tan

tan

34

3

33

− = − − +

= −

= −

c cm m; E

You could change the radians into degrees before

fi nding these ratios.

We can use the ASTC rule to solve trigonometric equations. You studied these in the Preliminary Course using degrees.

You could solve all equations in degrees then

convert into radians where necessary.

EXAMPLES

1. Solve .cos x 0 34= for 0 2 .≤ ≤ πx

Solution

cos is positive in the 1 st and 4 th quadrants. Using a calculator (with radian mode) gives . .x 1 224= i.e. ..cos 0 341 224 =

This is an angle in the 1 st quadrant since .. π2

1 224 <

In the 4 th quadrant, angles are in the form of .π θ2 − So the angle in the 4 th quadrant will be . .π2 1 224− So . ,

. , . .

πx 1 224 2

1 224 5 06

1.224−==

2. Solve αsin2

1= − in the domain ≤ ≤α π0 2 .

Solution

Here the sin of the angle is negative. Since sin is positive in the 1 st and 2 nd quadrants, it is negative in the 3 rd and 4 th quadrants.

CONTINUED

2

1

3

π6

π3

ch5.indd 175 6/12/09 3:43:54 AM


πsin4 2

1=

In the 3 rd quadrant, angles are π θ+ and in the 4 th quadrant, π θ.2 −

,

,

,

α π π π π

π π π π

π π

42

4

44

4 48

4

45

47

So = + −

= + −

=

5.2 Exercises

1. Copy and complete the table, giving exact values.

π3

π4

π6

sin

cos

tan

cosec

sec

cot

2. Find the exact value, with rational denominator where relevant .

(a) πtan6

2

(b) πsin4

2c m

(c) πcos6

3c m

(d) π πtan tan3 6

+

(e) π πsin cos4 4

−

(f) π πtan cos3 3

+

(g) π πsec tan4 3

−

(h) π πcos cot4 4

+

(i) π πcos tan4 4

2

−c m

(j) π πsin cos6 6

2

+c m

3. Find the exact value, with rational denominator where relevant.

(a) π πcos sin6 4

2 2+

(b) π π π πsin cos cos sin3 4 3 4

−

(c) π π π πcos cos sin sin6 3 6 3

+

(d) π πsin cos4 4

2 2+

(e) π πsec sin3 6

2 2+

4. Find the exact value with rational denominator of

(a) π π π πsin cos sin cos3 4 4 3

+

(b) π π

π π

tan tan

tan tan

14 3

4 3

+

−

5. Show that

π π π πcos cos sin sin3 4 3 4

=+

.π π π πsin cos cos sin4 6 4 6

+

6. (a) Show that .π ππ4

34

= −

Which quadrant is the angle (b)

π4

3 in?

Find the exact value of cos (c) .π4

3

Remember that 2( )θ θtan tan2 =

1

12

π4

π4

ch5.indd 176 7/6/09 10:21:04 PM



56

= −

Which quadrant is the angle(b)

π6

5 in?

Find the exact value of (c) .πsin6

5 8. (a) Show that .π π π

47

42= −


π4

7 in?

Find the exact value of (c) .πtan4

7


43

= +


π3

4 in?

Find the exact value of (c)

.πcos3

4


53

2= −


π3

5 in?

Find the exact value of (c) πsin3

5 .

11. Find the exact value of each ratio. (a) πtan

43

(b) πcos6

11

(c) πtan3

2

(d) πsin4

5

(e) πtan6

7

12. (a) (i) Show that .π ππ6

136

2= +

(ii) Which quadrant is the

angle π6

13 in?

(iii) Find the exact value of

.πcos6

13

Find the exact value of (b) (i) πsin

49

(ii) πtan3

7

(iii) πcos4

11

(iv) πtan6

19

(v) πsin3

10

13. Solve for 0 2≤ ≤ πx

(a) cos x21=

(b) sin x2

1= −

(c) tan x 1= (d) tan x 3=

(e) cos x23= −

14. Simplify (a) π θsin −] g (b) ( )πtan x2 − (c) π αcos +] g (d) πcos x

2−c m

(e) 2π θtan −c m

15. Simplify π θsin12

2 −− c m

16. If tanx43= − and ,π πx

2< < fi nd

the value of cos x and sin x .

17. Solve –tan x 1 02 = for 0 2≤ ≤ πx

ch5.indd 177 6/29/09 11:53:52 AM


Circle Results

The area and circumference of a circle are useful in this section.

Area of a circle

πA r 2= where r is the radius of the circle

Circumference of a circle

πC 2= =π dr

We use equal ratios to fi nd the other circle results. The working out is simpler when using radians instead of degrees.

Length of arc

θl r= ( θ is in radians)

Proof

2

θ

πθ

πθ

θ

l

l

l

r

2 2

2

circumference

arc length

whole revolution

angle

`

=

=

=

=

ππ

rr

Using 360° instead of 2π gives a different formula. Can you fi nd it?

ch5.indd 178 6/12/09 3:48:18 AM


EXAMPLES

1. Find the length of the arc formed if an angle of π4

is subtended at the centre of a circle of radius 5 m.

Solution

θπ

π

l r

54

45 m

=

=

=

c m

2. The area of a circle is .450 cm2 Find, in degrees and minutes, the angle subtended at the centre of the circle by a 2.7 cm arc.

Solution

π

π

A

r

r

450450

450

2

==

=

=

ππ

r

r

2

2

.

Now

. .

.

.

.

θθ

θ

θ

r

l r

11 97

2 7 11 97

11 972 7

0 226

===

=

=°

°

°

°

°

°

radians

radian

. radians .

.

So .

π

π

π

θ

´

180

1 180

0 226 180 0 226

12 93

12 56

12 56

=

=

=

===

l

l

ch5.indd 179 6/12/09 3:49:27 AM


5.3 Exercises

1. Find the exact arc length of a circle if

radius is 4 cm and angle (a) subtended is π

radius is 3 m and angle (b)

subtended is π3

radius is 10 cm and angle (c)

subtended is π6

5

radius is 3 cm and angle (d) subtended is 30°

radius is 7 mm and angle (e) subtended is 45° .

2. Find the arc length, correct to 2 decimal places, given

radius is 1.5 m and angle (a) subtended is 0.43

radius is 3.21 cm and angle (b) subtended is 1.22

radius is 7.2 mm and angle (c) subtended is 55°

radius is 5.9 cm and angle (d) subtended is ° 1223 l

radius is 2.1 m and angle (e) subtended is ° 3582 l .

3. The angle subtended at the centre of a circle of radius 3.4 m is ° .29 51l Find the length of the arc cut off by this angle, correct to 1 decimal place.

4. The arc length when a sector of a circle is subtended by an angle of

π5

at the centre is π2

3 m. Find the

radius of the circle.

5. The radius of a circle is 3 cm and

an arc is π7

2 cm long. Find the

angle subtended at the centre of the circle by the arc.

6. The circumference of a circle is 300 mm. Find the length of the arc that is formed by an angle of

π6

subtended at the centre of the

circle.

7. A circle with area 60 cm2 has an arc 8 cm long. Find the angle that is subtended at the centre of the circle by the arc.

8. A circle with circumference 124 mm has a chord cut off it that subtends an angle of 40° at the centre. Find the length of the arc cut off by the chord.

9. A circle has a chord of 25 mm

with an angle of π6

subtended

at the centre. Find, to 1 decimal place, the length of the arc cut off by the chord.

10. A sector of a circle with radius

5 cm and an angle of π3

subtended at the centre is cut out of cardboard. It is then curved around to form a cone. Find its exact surface area and volume.

ch5.indd 180 6/12/09 3:50:08 AM


Area of sector

θA r21 2=

( θ is in radians)

Proof

area of circlearea of sector

whole revolution

angleA θ

πθ

πθ

θ

A

A

r

2

2

21 2

`

=

=

=

=

ππ

rr

2

2

EXAMPLES

1. Find the area of the sector formed if an angle of π4


Solution

2

θ

π

π

A r21

21 5

4

825 m

2

2

=

=

=

^ ch m

CONTINUED

ch5.indd 181 6/12/09 3:50:32 AM


2. The area of the sector of a circle with radius 4 cm is .π5

6 cm2 Find the

angle, in degrees, that is subtended at the centre of the circle.

Solution

θ=

°

°

θ

π θ

θπ

π θ

θ

θ

A r21

56

21 4

8

406

203

203 180

27

2

2

=

=

=

=

=

=

]

]

g

g

5.4 Exercise s

1. Find the exact area of the sector of a circle if

radius is 4 cm and angle (a) subtended is π

radius is 3 m and angle (b)

subtended is π3

radius is 10 cm and angle (c)

subtended is π6

5

radius is 3 cm and angle (d) subtended is 30°

radius is 7 mm and angle (e) subtended is 45°.

2. Find the area of the sector, correct to 2 decimal places, given

radius is 1.5 m and angle (a) subtended is 0.43

radius is 3.21 cm and angle (b) subtended is 1.22

radius is 7.2 mm and angle (c) subtended is 55°

radius is 5.9 cm and angle (d) subtended is ° 1223 l

radius is 2.1 m and angle (e) subtended is ° .3582 l

3. Find the area, correct to 3 signifi cant fi gures, of the sector of a circle with radius 4.3 m and an angle of 1.8 subtended at the centre.

4. The area of a sector of a circle is 20 cm2 . If the radius of the circle is 3 cm, fi nd the angle subtended at the centre of the circle by the sector.

5. The area of the sector of a circle that is subtended by an angle of

π3

at the centre is π6 m2 . Find the

radius of the circle.

6. Find the arc length (a) area of the sector of a circle (b)

with radius 7 cm if the sector is cut off by an angle of °30 subtended at the centre of the circle.

Remember that 180 .π = °

ch5.indd 182 6/12/09 3:50:41 AM


7. A circle has a circumference of 185 mm. Find the area of the

sector cut off by an angle of π5

subtended at the centre.

8. If the area of a circle is 200 cm2 and a sector is cut off by an angle

of π4

3 at the centre, fi nd the area

of the sector.

9. Find the area of the sector of a circle with radius 5.7 cm if the length of the arc formed by this sector is 4.2 cm.

10. The area of a sector is π103 cm2

and the arc length cut off by the

sector is π5

cm. Find the angle

subtended at the centre of the circle and fi nd the radius of the circle.

Area of minor segment

( )θ θsinA r21 2= −

( θ is in radians)

Proof

Area of minor segment area of sector area of triangle

( )

θ θ

θ θ

sin

sin

r r

r

21

21

21

2 2

2

−=

= −

= −

The area of the triangle is

given by 2A ab sin C.= 1

ch5.indd 183 6/12/09 6:09:54 AM


EXAMPLES

1. Find the area of the minor segment formed if an angle of π4


Solution

2

θ θ

π π

π

π

π

π

sin

sin

A r21

21

4 4

225

4 21

825

2 225

825

425 2

825 50 2

m

2

2

= −

= −

= −

= −

= −

= −

5

]^ ce

gh m

o

2. An 80 cm piece of wire is bent into a circle shape, and a 10 cm piece of wire is joined to it to form a chord. Find the area of the minor segment cut off by the chord, correct to 2 decimal places.

Solution

.

π

π

C

r

2 80

280

40

12 7

`

= =

=

=

Z

πr

Don’t round off the radius. Use the full value in your calculator’s display to fi nd θ.

ch5.indd 184 6/12/09 3:51:00 AM


. .. .

.

.

. . .

.

θ

θ

θ θ

cos

cos

sin

sin

Cab

a b c

A r

2

2 12 7 12 712 7 12 7 10

0 69

0 807

21

21 12 7 0 807 0 807

6 88 cm

2 2 2

2 2 2

2

2

`

Z

= + −

= + −

==

= −

= −2

] ]

]] ]

g g

gg g

5.5 Exercises

1. Find the exact area of the minor segment of a circle if

radius is 4 cm and the angle (a) subtended is π

radius is 3 m and the angle (b)

subtended is π3

radius is 10 cm and the angle (c)

subtended is π6

5

radius is 3 cm and the angle (d) subtended is 30°

radius is 7 mm and the angle (e) subtended is 45° .

2. Find the area of the minor segment correct to 2 decimal places, given

radius is 1.5 m and the angle (a) subtended is 0.43

radius is 3.21 cm and the (b) angle subtended is 1.22

radius is 7.2 mm and the (c) angle subtended is 55°

radius is 5.9 cm and the angle (d) subtended is ° 1223 l

radius is 2.1 m and the angle (e) subtended is ° 3582 l .

3. Find the area of the minor segment formed by an angle of 40° subtended at the centre of a circle with radius 2.82 cm, correct to 2 signifi cant fi gures.

4. Find the exact arc length (a) exact area of the sector (b) area of the minor segment, (c)

to 2 decimal places

if an angle of π7

is subtended at

the centre of a circle with radius 3 cm.

5. The area of the minor segment

cut off by an angle of π9

2 is

500 cm2 . Find the radius of the circle, correct to 1 decimal place.

6. Find the length of the chord, (a)

to 1 decimal place length of the arc (b) area of the minor segment, (c)

to 2 decimal places

cut off by an angle of π6

subtended at the centre of a circle with radius 5 cm.

7. A chord 8 mm long is formed by an angle of 45° subtended at the centre of a circle. Find

the radius of the circle (a) the area of the minor segment (b)

cut off by the angle, correct to 1 decimal place .

ch5.indd 185 6/12/09 3:51:10 AM


8. An angle of π8

is subtended at the

centre of a circle. If this angle

cuts off an arc of π4

5 cm, fi nd

the exact area of the sector (a) the area of the minor segment (b)

formed, correct to 1 decimal place.

9. A triangle OAB is formed where O is the centre of a circle of radius 12 cm, and A and B are endpoints of a 15 cm chord.

Find the angle ( (a) θ ) subtended at the centre of the circle, in degrees and minutes.

Find the area of (b) ,ΔOAB correct to 1 decimal place.

Find the area of the minor (c) segment cut off by the chord, correct to 2 decimal places.

Find the area of the major (d) segment cut off by the chord, correct to 2 decimal places.

10. The length of an arc is 8.9 cm and the area of the sector is .24 3 cm2 when an angle of θ is subtended at the centre of a circle. Find the area of the minor segment cut off by θ, correct to 1 decimal place.

11.

An arc, centre C , cuts side AC in D .

Find the exact length of arc (a) BD area of (b) ABD perimeter of (c) BDC

12.

The dartboard above has a radius of 23 cm. The shaded area has a height of 7 mm. If a player must hit the shaded area to win, fi nd

the area of the shaded part, to (a) the nearest square centimetre

the percentage of the (b) shaded part in relation to the whole area of the dartboard, to 1 decimal place

the perimeter of the (c) shaded area.

ch5.indd 186 6/12/09 3:51:24 AM


13. The hour hand of a clock is 12 cm long. Find the exact

length of the arc through (a) which the hand would turn in 5 hours

area through which the hand (b) would pass in 2 hours.

14.

Arc BC subtends an angle of 100° at the centre A of a circle with radius 4 cm. Find

the exact perimeter of (a) sector ABC

the approximate ratio of the (b) area of the minor segment to the area of the sector.

15. A wedge is cut so that its cross-sectional area is a sector of a circle, radius 15 cm and

subtending an angle of π6

at the centre. Find

the volume of the wedge (a) the surface area of the wedge. (b)

EXAMPLES

1. Find sin 0.0023.

Solution

Make sure your calculator is in radian mode . .sin 0 0023 0 002299997972=

2. Find tan 0.0023.

Solution

. .tan 0 0023 0 002300004056=

3. Find cos 0.0023.

Solution

. .cos 0 0023 0 999997355=

Small Angles

When we use radians of small angles, there are some interesting results.

ch5.indd 187 6/12/09 3:51:51 AM


Class Investigation

Use the calculator to explore other small angles and their trigonometric ratios. What do you notice?

A

B C

O 1x

In , 1ΔOAC OCZ when x is small. Arc (definition of a radian)AB x= AC x` Z when x is small

sin

cos

tan

xOCAC

x

x

xOCOA

xOAAC

x

x

1

11

1

1

Z

Z

Z

=

=

=

=

=

=

If x (radians) is a small angle then

1

sin

tan

cos

sin tan

x x

x x

x

x x xand < <

Z

Z

Z

Can you see why?

Use radians for the small angles. You could use a computer spreadsheet to fi nd these trigonometric ratios.

ch5.indd 188 6/12/09 3:53:57 AM


Proof

Area

Area sector

Area

Δ

Δ

´ ´ ´

´ ´

´ ´

´ ´

sin

sin

sin

tan tan

tan

OAB ab C

x

x

OAB r

x

x

OAC bh

AC

x x AC

x

21

21 1 1

21

21

21 1

21

21

21 1

21 1

1

21

2

2

=

=

=

=

=

=

=

=

= =

=

c m

Area ΔOAB area sector< OAB area< ΔOAC

sin tan

sin tan

x x x

x x x21

21

21< <

< <

`

`

Class Investigation

Check that sin tanx x x< < for πx2

0 < < by using your calculator.

Remember that x is in radians. Does this work for πx2

> ?

These results give the following result :

lim sinx

x 1x 0

="

Proof As 0,x" sinx x"

sinx

x 1"`

Also limtan x

1.x0x

="

Can you see why?

ch5.indd 189 7/6/09 10:21:08 PM


EXAMPLES

1. Evaluate .lim sinx

x7x 0"

Solution

( )

7

( )

lim sin limsin

lim sinx

xx

x

xx

77

7 7

77

7 1

7

x x

x

0 0

0

=

=

==

" "

"

2. The planet of Alpha Zeta in a faraway galaxy has a moon with a diameter of 145 000 km. If the distance between the centre of Alpha Zeta and the centre of its moon is . ´12 8 106 km, fi nd the angle subtended by the moon at the centre of the planet, in seconds.

Solution

´r

21 145 000

72 500

=

=

..

.

. .

.

.

°

°

°

°

°

α θ

α

απ

π

π

θ

´

´

tan

21

12 8 1072 500

0 00566

0 00566

180

1 180

0 00566 180 0 00566

0 3245

0 649

39

6

`

`

`

=

=

===

=

=

=== l

The radius of the moon is 72 500 km.

ch5.indd 190 6/12/09 3:54:16 AM


1. Evaluate, correct to 3 decimal places .

sin 0.045 (a) tan 0.003 (b) cos 0.042 (c) sin 0.065 (d) tan 0.005 (e)

2. Evaluate .lim sinxx

4x 0"

3. Find .θ

θ

limtan

3θ 0"

4. Find the diameter of the sun to the nearest kilometre if its distance from the Earth is 149000000 km and it subtends an angle of 31l at the Earth.

5. Given that the wingspan of an aeroplane is 30 m, fi nd the plane’s altitude to the nearest metre if the wingspan subtends an angle of 14l when it is directly overhead.

5.6 Exercises

Trigonometric Graphs

You drew the graphs of trigonometric functions in the Preliminary Course, using degrees.

You can also draw these graphs using radians. y xsin=

ch5.indd 191 6/12/09 3:54:35 AM


y xcos=

y xtan=

tany x= has asymptotes at πx2

= and π2

3 since tan x is undefi ned at those

points. By fi nding values for x on each side of the asymptotes, we can see where the curve goes.

We can also sketch the graphs of the reciprocal trigonometric functions, by fi nding the reciprocals of sin x , cos x and tan x at important points on the graphs.

Investigation

Use 1. , sin coscosec sec

x xx x1 1= = and tan

cotx

x 1= to complete the

table below.

x 0π2 π

3π2 2π

cosec x

sec x

cot x

ππcosec

sin22

1

11

1

e.g. =

=

=

ch5.indd 192 6/12/09 3:55:33 AM


Find any asymptotes 2.

(undefined)

ππsec

cos22

1

01

e.g. =

=

Discover what the values are on either side of the asymptotes.

Sketch each graph of the reciprocal trigonometric functions. 3.

Here are the graphs of the reciprocal ratios.

1

y

x2π

-1

y= cosec x

y= sin x

π 3π2

π2

x

y= sec x

1

y

-1

y= cos x

2π3π2

π2

π

ch5.indd 193 6/12/09 3:57:20 AM


x

y= tan x

y= cot x

y

3π2

π2

π 2π

We have sketched these functions in the domain 0 2≤≤ πx which gives all the angles in one revolution of a circle. However, we could turn around the circle again and have angles greater than 2 ,π as well as negative angles.

These graphs all repeat at regular intervals. They are called periodic functions.

Investigation

Here is a general sine function. Notice that the shape that occurs between 0 and 2π repeats as shown.

x

1

y

−1

−2π 2π 4π−4π

Draw a general cosine curve. How are the sine and cosine curves 1. related? Do these curves have symmetry? Draw a general tangent curve. Does it repeat at the same intervals as 2. the sine and cosine curves? Look at the reciprocal trigonometric curves and see if they repeat in a 3. similar way.

ch5.indd 194 6/12/09 3:58:38 AM


Use a graphics calculator or a graphing computer program to draw 4. the graphs of trigonometric functions. Choose different values of k to sketch these families of trig functions. (Don’t forget to look at values where k is a fraction or negative.)

(a) ( ) sinf x k x=

(b) ( ) cosf x k x=

(c) ( ) tanf x k x=

(d) ( ) sinf x kx=

(e) ( ) cosf x kx=

(f) ( ) tanf x kx=

(g) ( ) sinf x x k= +

(h) ( ) cosf x x k= +

(i) ( ) tanf x x k= +

(j) ( ) ( )sinf x x k= +

(k) ( ) ( )cosf x x k= +

(l) ( ) ( )tanf x x k= +

Can you see patterns in these families? Could you predict what the graph of ( ) sinf x a bx= looks like?

The trig functions have period and amplitude. The period is the distance over which the curve moves along the x -axis

before it repeats. The amplitude is the maximum distance that the graph stretches out

from the centre of the graph on the y -axis.

siny x= has amplitude 1 and period 2π

cosy x= has amplitude 1 and period 2π

tany x= has no amplitude and period π

siny a bx= has amplitude a and period 2πb

cosy a bx= has amplitude a and period 2πb

tany a bx= has no amplitude and period πb

ch5.indd 195 6/12/09 3:58:46 AM


EXAMPLES

1. Sketch in the domain ≤ ≤ π.x0 2 (a) siny x5= for (b) siny x4= for (c) siny x5 4= for

Solution

The graph of (a) siny x5= has amplitude 5 and period π.2

x

y = 5 sin x

y

5

−5

3π2

π2

π 2π

The graph (b) siny x4= has amplitude 1 and period .π π4

22

or

This means that the curve repeats every ,π2

so in the domain

0 2≤ ≤ πx there will be 4 repetitions.

x

y = sin 4x

y

1

−1

3π4

3π2

7π4

5π4

π4

π2

π 2π

ch5.indd 196 6/12/09 3:58:54 AM


The graph (c) siny x5 4= has amplitude 5 and period .π2

2. Sketch ( ) πsinf x x2

= +b l for ≤ ≤ π.x0 2

Solution

1

2

2

π

π1

Amplitude

Period

=

=

=

( ) ( )sinf x x k= + translates the curve k units to the left (see Investigation on page 194–5).

So ( ) πsinf x x2

= +c m will be moved π2

units to the left. The graph is the

same as ( ) sinf x x= but starts in a different position. If you are not sure where the curve goes, you can draw a table of values.

x 0 π2

π 3π2

2π

y 1 0 −1 0 1

This is the same as the graph y cos x.=

y = 5 sin 4x

y

5

−5

x3π4

3π2

7π4

5π4

π4

π2

π 2π

y

1

−1

x3π2

π2

π 2π

CONTINUED

ch5.indd 197 6/29/09 11:54:11 AM


3. Sketch tany x22

= for ≤ ≤ π.x0 2

Solution

There is no amplitude .

2

π

π21

Period =

=

y

x3π2

π2

π 2π

4. (a) Sketch cosy x2= and cosy x2= on the same set of axes for ≤ ≤ π.x0 2

(b) Hence or otherwise, sketch cos cosy x x2 2= + for ≤ ≤ π.x0 2

Solution

(a) cosy x2= has amplitude 2 and period 2π

cosy x2= has amplitude 1 and period 2π2

or π

y = 2 cos x

y = cos 2x22

y

1

2

−1

−2

x3π4

3π2

7π4

5π4

π4

π2

π 2π

ch5.indd 198 6/12/09 3:59:26 AM


Method 1: Use table (b)

x 0π4

π2

3π4 π

5π4

3π2

7π4 2π

cos 2x 1 0 −1 0 1 0 −1 0 1

2 cos x 2 2 0 2− −2 2− 0 2 2

cos 2x + 2 cos x 3 2 −1 2− −1 2− −1 2 3

Method 2: Add y values together on the graph itself

e.g. when 0:x

y 1 2

3

== +=

x

y = 2cos x

y = cos 2x 22 + 2 cos x

y = cos 2x22

y

1

2

3

−1

−2

−3

3π4

33ππ2

7π4

5π4

π4

π2

π 2π

Notice that the period of this graph is 2 .π

These graphs can be sketched more accurately

on a graphics calculator or in a computer program

such as Autograph.

5.7 Exercises

1. Sketch for 0 2≤ ≤ πx (a) cosy x= (b) ( ) sinf x x2= (c) siny x1= + (d) siny x2= − (e) ( ) cosf x x3= − (f) siny x4= (g) ( ) cosf x x 3= + (h) tany x5= (i) ( ) tanf x x 3= + (j) tany x1 2= −

2. Sketch for 0 2≤ ≤ πx (a) cosy x2= (b) tany x2= (c) siny x3= (d) ( ) cosf x x3 4= (e) cosy x6 3=

(f) tany x2

=

(g) ( ) tanf x x2 3=

(h) cosy x32

=

(i) siny x22

=

(j) ( ) cosf x x44

=

ch5.indd 199 6/12/09 3:59:39 AM


3. Sketch for ≤ ≤π πx− (a) siny x2= (b) cosy x7 4= (c) ( ) tanf x x4= (d) siny x5 4= (e) ( ) cosf x x2 2=

4. Sketch sin xy2

8= in the domain

≤ ≤ π.x0 4

5. Sketch for 0 2≤ ≤ πx (a) ( )πsiny x= +

(b) πtany x2

= +b l (c) ( ) ( )πcosf x x= −

(d) πsiny x32

= −b l (e) ( ) πcosf x x2

2= +b l

(f) πsiny x4 22

= +b l (g) πcosy x

4= −b l

(h) πtany x4

= +b l (i) ( ) πcosf x x2

21= + +b l

(j) – πsiny x2 32

= − b l

6. Sketch for 2 2≤ ≤x− (a) πsiny x= (b) 3 2πcosy x=

7. Sketch in the domain 0 2≤ ≤ πx (a) siny x= and siny x2= on the

same set of axes (b) sin siny x x2= +

8. Sketch for 0 2≤ ≤ πx (a) cosy x2= and siny x3= on

the same set of axes (b) cos siny x x2 3= +

9. By sketching cosy x= and cosy x2= on the same set of axes for 0 2≤ ≤ πx , sketch the graph of .cos cosy x x2= −

10. Sketch the graph of (a) cos siny x x= + (b) sin siny x x2= − (c) sin cosy x x2 2= + (d) cos cosy x x3 2= − (e) sin siny x x

2= −

Applications

The sine and cosine curves are used in many applications including the study of waves. There are many different types of waves, including water, light and sound waves. Oscilloscopes display patterns of electrical waves on the screen of a cathode-ray tube.

Search waves and the oscilloscope on the Internet for further information.

Simple harmonic motion (such as the movement of a pendulum) is a wave-like or oscillatory motion when graphed against time. In 1581, when he was 17 years old, Gallileo noticed a lamp swinging backwards and forwards in Pisa cathedral. He found that the lamp took the same time to swing to and fro, no matter how much weight it had on it. This led him to discover the pendulum.

Gallileo also invented the telescope. Find out more information about Gallileo’s life and his other discoveries.

ch5.indd 200 6/29/09 11:54:24 AM


Some trigonometric equations are diffi cult to solve algebraically, but can be solved graphically by fi nding the points of intersection between two graphs.

EXAMPLES

1. Find approximate solutions to the equation 1sin x x= − by sketching the graphs siny x= and 1y x= − on a Cartesian plane. Solution

When sketching the two graphs together, we use ≈ 3.14, 2 ≈ 6.28π π and so on to label the x -axis as shown . To sketch 1,y x= − fi nd the gradient and y -intercept or fi nd the x - and y- intercepts . x -intercept (where 0y = ) is 1 and y -intercept (where 0x = ) is –1.

The solution to sinx x 1= − is at the point of intersection of the two graphs. ≈ .x 2`

2. How many solutions are there for cos x x24

= in the domain 0 2≤ ≤ πx ?

Solution

Sketch cosy x2= and 4

y x= on the same set of axes.

2cosy x= has amplitude 1 and period π

4

y x= has x -intercept 0 and y -intercept 0. We can fi nd another point on the line

e.g. When 4x =

y

44

1

=

=

CONTINUED

x

y = x−1

y = sin x

y

1

1 2 3 4 5 6

−1

3π2

π2

π 2π

ch5.indd 201 6/30/09 10:20:41 AM


y= cos 2x22

y

y= x4

1

−1

1 2 3 4 5 63π2

π2

π 2π

There are 3 points of intersection of the graphs, so the equation

cos xx4

2 = has 3 solutions.

We can also look at applications of trigonometric graphs in real life situations.

EXAMPLE

The table shows the highest average monthly temperatures in Sydney.

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

°C 26.1 26.1 25.1 22.8 19.8 17.4 16.8 18.0 20.1 22.2 23.9 25.6

Draw a graph of this data, by hand or on a graphics calculator (a) or computer.

Is it periodic? Why would you expect it to be periodic? (b) What is the period and amplitude? (c)

Solution

Temperature

0

5

10

15

20

25

30

Janu

ary

Febru

ary

Marc

hApr

ilM

ayJu

ne July

Augus

t

Septem

ber

Octobe

r

Novem

ber

Decem

ber

Months

(a)

ch5.indd 202 7/6/09 10:21:09 PM


The graph looks like it is periodic, and we would expect it to be, (b) since the temperature varies with the seasons. It goes up and down, and reaches a maximum in summer and a minimum in winter.

This curve is approximately a cosine curve with one full period, (c) so the period is 12 months.

The maximum temperature is around 26° and the minimum is around 18°, so the centre of the graph is 22° with 4° either side. So the amplitude is 4.

1. Show graphically that

sin xx2

= has

2 solutions for (a) 0 2≤ ≤ πx 3 solutions for (b) .≤ ≤π πx−

2. Solve sin x x= for 0 2≤ ≤ πx graphically by sketching siny x= and y x= on the same number plane .

3. Solve 2 3cos x x= − for 0 2≤ ≤ πx by fi nding the points of intersection of the graphs cosy x= and 2 3y x= − .

4. Solve tan x x= graphically in the domain 0 2≤ ≤ πx .

5. Solve by graphical means 2sin x x= for 0 2≤ ≤ πx .

6. Draw the graphs of siny x= and cosy x= for 0 2≤ ≤ πx on the same set of axes. Use your graphs to solve the equation sin cosx x= for 0 2≤ ≤ πx .

7. The graph below show the times of sunsets in a city over a period of 2 years.

Sunsets

0123456789

10

Janu

ary

Marc

hM

ay July

Septem

ber

Novem

ber

Janu

ary

Marc

hM

ay July

Septem

ber

Novem

ber

Tim

e (p

m)

Find the period and (a) amplitude of the graph .

At approximately what time (b) would you expect the sun to set in July?

5.8 Exercises

To fi nd the centre, fi nd the average of 18 and 26 which

is 2

18 26+.

ch5.indd 203 6/12/09 4:01:04 AM


8. The graph shows the incidence of crimes committed over 24 years in Gotham City.

9. Below is a table showing the average daylight hours over several months.

Month Jan Feb Mar Apr May June July Aug

Daylight hours 15.3 14.7 13.2 13.1 12.7 12.2 12.5 13.8

Draw a graph to show this data. (a) Is it periodic? If so, what is the period? (b) Find the amplitude .(c)

10. The table below shows the high and low tides over a three-day period.

Draw a graph showing the tides . (a) Find the period and amplitude .(b) Estimate the height of the tide (c)

at around 8 am on Friday .

Day Friday Saturday Sunday

Time 6.20

am

11.55

am

6.15

pm

11.48

pm

6.20

am

11.55

am

6.15

pm

11.48

pm

6.20

am

11.55

am

6.15

pm

11.48

pm

Tide (m) 3.2 1.1 3.4 1.3 3.2 1.2 3.5 1.1 3.4 1.2 3.5 1.3

0

200

400

600

800

1000

1200

1400

1600

1800

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Approximately how many crimes were committed in the 10 th year? (a) What was the (b) (i) highest and (ii) lowest number of crimes? Find the amplitude and the period of the graph .(c)

ch5.indd 204 6/29/09 11:55:21 AM


( )sin cosdxd x x=

Proof

LetThen ( ) ( )

( )( ) ( )

( )

( )

( )

( ) ( ) since as and

.

sinsin

lim

limsin sin

limsin cos cos sin sin

limsin cos cos sin

lim sin cos cos sin

lim sin cos cos lim

cos

sin

f x xf x h x h

xh

f x h f x

hx h x

hx h x h x

hx h x h

xhh x

hh

x x h h

xh

h

1

1

0 1 1 0 1

h

h

h

h

h

h h

0

0

0

0

0

0 0" "

=+ = +

=+ −

=+ −

=+ −

=− +

= − +

= + =

=

"

"

"

"

"

" "

fl

]

c cc

g

m mm

Notice that the result lim sinh

h 1h 0

="

only works for radians. We always use

radians when using calculus.

Differentiation of Trigonometric Functions

D erivative of sin x

We can sketch the gradient (tangent) function of .siny x=

The gradient function is .cosy x=

This result is a formula in the Extension 1 Course

and you do not need to know it.

ch5.indd 205 6/29/09 11:55:32 AM


[ ( )] ( ) ( )sin cosdxd f x x f x= fl

FUNCTION OF A FUNCTION RULE

Proof

Let ( )

Then

( ) and

( )

( ) ( )

sin

cos

cos

cos

u f x

y u

dxdu f x

du

dyu

dx

dy

du

dy

dxdu

u f x

f x f x

$

$

==

= =

=

==

l

l

l

Derivative of cos x

We can sketch the gradient function of .cosy x=

The gradient function is .siny x= −

( )cos sindxd x x= −

Proof

Since

( )

π

π

π´

cos sin

cos sin

cos

sin

x x

dxd x

dxd x

x

x

2

2

12

= −

= −

= − −

= −

ccc

mmm

< F

ch5.indd 206 6/12/09 4:01:40 AM



[ ( )] ( ) ( )cos sindxd f x f x f x= − l

D erivative of tan x

The gradient function is harder to sketch for .tany x=

The gradient function is .secy x2=

It is easier to see the rule for differentiating tany x= by using cossintan x

xx = and the quotient rule.

Proof

( )

( ) ( )

tan cossin

tan cossin

cos

cos cos sin sin

coscos sin

cossec

x xx

dxd x

dxd

xx

vu v v u

x

x x x x

xx x

xx

1

2

2

2

2 2

2

2

=

=

= −

=− −

= +

=

=

l l

b l

( )tan secdxd x x2=

This result can be proven the same way as for sin f(x).

ch5.indd 207 6/12/09 4:01:51 AM


EXAMPLES

1. Differentiate ( ) .sin x5

Solution

[ ( )] ( ) ( )

[ ( )] ( )

sin cos

sin cos

dxd f x f x f x

dxd x x5 5 5`

=

=

l

2. Differentiate sin x ° .

Solution

( )°

°

π

π π

π

sin sin

cos

cos

dxd x

dxd x

x

x

180

180 180

180

=

=

=

c m

3. Find the exact value of the gradient of the tangent to the curve

y x2= sin x at the point where .πx4

=

Solution

( ) ( )cos sin

cos sin

dx

dyu v v u

x x x x

x x x x

2

2

2

2

= +

= += +

l l

hen ,

( )

π

π π π π

π π

π π

π π

π π

´ ´

cos sin

x

dx

dy4

4 42

4 4

16 21

2 21

16 2 2 2

16 28

322 8

W

2

2

2

2

=

= +

= +

= +

= +

=+

c cm m

[ ( )] ( ) ( ) .tan secdxd f x x f x2= fl


This result can be proven the same way as for sin f(x).

ch5.indd 208 6/12/09 4:02:01 AM


1. Differentiate sin 4 (a) x cos 3 (b) x tan 5 (c) x tan (d) (3 1)x + (e) ( )cos x− 3 sin (f) x (g) ( )cos x4 5 3− (h) ( )cos x2 3 (i) ( )tan x7 52 + (j) sin cosx x3 8+ (k) ( )πtan x x2+ + (l) x tan x sin 2 (m) x tan 3 x

(n) 2

sinxx

(o) 5

3 4sin xx +

(p) 9( )tanx x2 7+ (q) sin x2 (r) cos x3 53 (s) cose x2x − (t) ( )sin log x1 e− (u) ( )sin e xx + (v) ( )log sin xe (w) cose x2x3

(x) tan x

e7

x2

2. Find the derivative of .cos sinx x4

3. Find the gradient of the tangent to the curve 3tany x= at the

point where .πx9

=

4. Find the equation of the tangent to the curve ( )πsiny x= − at the

point , ,π6 2

1c m in exact form.

5. Differentiate ( ) .log cos xe

6. Find the exact gradient of the normal to the curve 3siny x= at

the point where .πx18

=

7. Differentiate etan x .

8. Find the equation of the normal to the curve 3 2siny x= at the

point where ,πx8

= in exact form.

9. Show that dx

d yy25

2

2

= − if

.cosy x2 5=

10. Given ( ) ,sinf x x2= − show that ( ) ( ) .f x f x= −m

11. Show that

[ ( )] .log tan tan cotdxd x x xe = +

12. Find the coordinates of the stationary points on the curve 2 siny x x= − for 0 2 .≤ ≤ πx

13. Differentiate (a) °tanx (b) °cos x3

(c) °sinx5

14. If ,sin cosy x x2 3 5 3= − show

that .dx

d yy9

2

2

= −

15. Find values of a and b if

,cos sindx

d yae x be x4 4x x

2

23 3= +

given .cosy e x4x3=

5.9 Exercises

cos sinx dx x C= +#

Integration of Trigonometric Functions

The integrals of the trigonometric functions are their primitive functions .

ch5.indd 209 6/12/09 4:02:11 AM


sin cosx dx x C= − +#

sin cosx dx x C− = +#

sin cosx dx x C` = − +#

sec tanx dx x C2 = +#

( ) ( )cos sinax b dx a ax b C1+ = + +#

( ) ( )sin cosax b dx a ax b C1+ = − + +#

( ) ( )sec tanax b dx a ax b C12 + = + +#

F unction of a function rule

Proof

=

[ ( )] ( )

( ) ( )

( ) ( )

( )

sin cos

cos sin

cos cos

sin

dxd ax b a ax b

a ax b dx ax b C

ax b dx a a ax b dx

a ax b C

1

1

`

+ = +

+ = + +

+ +

= + +

###

Similarly,

EXAMPLES

1. Find ( ) .sin x dx3#

Solution

( ) ( )sin cosx dx x C331 3= − +#


ch5.indd 210 6/12/09 4:02:20 AM


2. Evaluate 2 .sin x dx0

π

#

Solution π

1

=

=

2

( )

( )

π

sin cos

cos cos

x dx x

20

0 1

0 0−

= − − −

= − −

π2

; E#

3. Find ° .cos x dx#

Solution

°

°

π

ππ

π

cos cos

sin

sin

x dx x dx

x C

x C

180

180

1180

180

=

= +

= +

# #

4. Find the area enclosed between the curve ,cosy x= the x -axis and the

lines πx2

= and .πx2

3=

Solution

3

2

2

2

2

π π

cos sin

sin sin

x dx x

23

21 1

2

=

= −

= − −= −

π

π

π

π 3

; E#

area is 2 units 2 .

You studied areas in Chapter 3. The defi nite

integral is negative as the area is below the x-axis.

You could use

CONTINUED

2

2

Area cos x dx

2

2.

=

= −

=

3

π

π

#

ch5.indd 211 6/12/09 6:10:31 AM


5. Find the volume of the solid formed if the curve secy x= is rotated

about the x -axis from 0x = to .πx4

=

Solution

sec

sec

y x

y x2 2`

=

=

π

4

4

( )

π

π

π

π π

ππ

sec

tan

tan tan

V y dx

x dx

x

40

1 0

a

2

2

0

0

=

=

=

= −

= −=

π

b

c m; E

#

#

So the volume is units .π 3

The volume formula comes from Chapter 3.

5.10 Exercises


cos (a) x sin (b) x (c) sec x2

(d) °sinx4

sin 3 (e) x (f) sin x7− (g) sec x52 (h) ( )cos x 1+ (i) ( )sin x2 3− (j) ( )cos x2 1− (k) ( )πsin x− (l) ( )πcos x + (m) sec x2 72

(n) sin x42

(o) sec x33

2

(p) (3 )sin x− −

2. Evaluate, giving exact answers where appropriate .

(a) 2 cos x dx0

π

#

(b) 6

3 sec x dx2

π

π

#

(c) 2

sin x dx2

π

π#

(d) 2 cos x dx30

π

#

ch5.indd 212 6/12/09 4:02:47 AM


(e) 2 πsin x dx0

1

#

(f) 8 sec x dx22

0

π

#

(g) 12 cos x dx3 20

π

#

(h) 10 sin x dx50

−π

#

3. Find the area enclosed between the curve siny x= and the x -axis in the domain 0 2 .≤ ≤ πx

4. Find the exact area bounded by the curve 3 ,cosy x= the x -axis

and the lines 0x = and .πx12

=


the curve ,sec xy4

2= the x -axis

and the lines πx4

= and ,πx2

=

correct to 2 decimal places.

6. Find the volume, correct to 2 decimal places, of the solid formed when the curve πsecy x= is rotated about the x -axis from 0x = to 0.15.x =

7. Find, in exact form, the volume of the solid of revolution formed

by rotating the curve 2siny x= about the x -axis from 0x = to

.πx6

=

8. Find the exact area enclosed by the curve siny x= and the line

y21= for 0 2 .≤ ≤ πx

9. Find the exact area bounded by the curves siny x= and cosy x= in the domain 0 2 .≤ ≤ πx

10. (a) Show that the volume of the solid formed by rotating the curve cosy x= about the

x axis between 0x = and πx2

= is

.unitsπ 3

(b) Use the trapezoidal rule with 4 subintervals to fi nd an approximation to the volume of the solid formed by rotating the curve cosy x= about the x -axis

from 0x = to .πx2

=

11. A curve has y

dx

d18

2

2

= sin 3 x and a

stationary point at , .π6

2−c m Find

the equation of the curve.

ch5.indd 213 6/12/09 4:02:59 AM


Test Yourself 4 1. A circle with radius 5 cm has an angle of

π6

subtended at the centre. Find

the exact arc length (a) the exact area of the sector (b) the area of the minor segment to (c)

3 signifi cant fi gures.

2. Find the exact value of

(a) πtan3

(b) πcos6

(c) πsin3

2

(d) πcos4

3

3. Solve for 0 2≤ ≤ πx (a) tan x 1= − (b) sin x2 1=

4. Sketch for 0 2≤ ≤ πx (a) cosy x3 2=

(b) siny x72

=

5. Differentiate cos (a) x 2 sin (b) x (c) tan x 1+ (d) x sin x

(e) tanx

x

cos 3 (f) x tan 5 (g) x


sin 2 (a) x 3 cos (b) x (c) sec x52 (d) sin x1 +

7. Evaluate

(a) 4 cosx dx0

π

#

(b) 6

3 sec x dx2

π

π

#

8. Find the equation of the tangent to the

curve 3siny x= at the point , .π4 2

1e o

9. If ,cosx t2= show that .dtd x x4

2

2

= −

10. Find the exact area bounded by the curve

,siny x= the x -axis and the lines πx4

=

and .πx2

=

11. Find the volume of the solid formed if the curve secy x= is rotated about the

x -axis from 0x = to .πx6

=

12. Simplify

(a) lim sinx

x5x 0"

(b) θ

θlim tan2θ 0"

13. Find the gradient of the tangent to the curve cosy x3 2= at the point where

.πx6

=

14. A circle has a circumference of 8π cm. If

an angle of π7

is subtended at the centre

of the circle, fi nd the exact area of the sector (a) the area of the minor segment, to (b)

2 decimal places.

15. (a) Sketch 2cosy x= and y x32= on the

same set of axes for 0 2 .≤ ≤ πx

(b) Solve 2cos x x32= for 0 2 .≤ ≤ πx

Test Yourself 5

ch5.indd 214 6/12/09 4:03:12 AM


16. Find the exact area with rational denominator bounded by the curve

,siny x= the x -axis and the lines πx6

=

and .πx4

=

17. Find the area bounded by the curve 2 ,cosy x= the x -axis and the lines 0x = to .πx =

18. Find the equation of the normal to the

curve tany x= at the point , .π4

1c m

19. A curve has 6 2 ,xdx

dysin= and passes

through the point , .π2

3c m Find the

equation of the curve.

20. (a) Sketch 7y = sin 3 x and 2 1y x= − on the same number plane for 0 2 .≤ ≤ πx

(b) Solve 7 sin 3 2 1x x= − for 0 2 .≤ ≤ πx


1. Use Simpson’s rule with 5 function values to fi nd an approximation to

8

4 ,tanx dxπ

π

# correct to 2 decimal places.

2. Evaluate 12

8 ,sec x dx22

π

π

# in exact form.

3. The area of the sector of a circle is unitsπ4 2 and the length of the arc

bounded by this sector is π8

units. Find

the radius of the circle and the angle that is subtended at the centre.

4. If ( ) πcosf x x3= fi nd the period and amplitude of the (a)

function sketch (b) ( )f x for 0 4≤ ≤x .

5. Given y

sindx

dx9 3

2

2

=

fi nd (a) y if there is a stationary point at

,π2

1c m Show that (b)

y.

dx

dy9 0

2

2

+ =

6. Sketch 5y = sin πx +] g for 0 2 .≤ ≤ πx

7. Find the derivative of .°tanx

8. (a) Show that sec x cosec .tansec

xxx

2

=

(b) Hence, or otherwise, fi nd the exact

value of 4

3 .cosec secx x dxπ

π

#

9. Differentiate e sinx x2 .

10. (a) Find the stationary points on the curve siny x2 3= + over the domain 0 .≤ ≤ πx

(b) What is the maximum value of the curve?

(c) What is the amplitude?

11. The area of a sector in a circle of radius

4 cm is π3

8 cm2 . Find the area of the

minor segment, in exact form.

12. Find ° .sinx dx#

13. Find the gradient of the normal to the curve πcosy x x2= + at the point where 1.x =

ch5.indd 215 6/29/09 11:55:45 AM


14. Use the trapezoidal rule with 4 subintervals to fi nd, correct to 3 decimal places, an approximation to the volume of the solid formed by rotating the curve siny x= about the x -axis from 0.2x = to 0.6.x =

15. Differentiate ( ) .log sin cosx xe +

16. Find the exact area of the minor segment cut off by a quarter of a circle with radius 3 cm.

17. Sketch for .≤ ≤π πtany x x4

0 2= −c m

18. Find all the points of infl exion on the

curve for .≤ ≤π πcosy x x3 24

0 2= +c m

19. Find the exact area bounded by the curve ,cosy x= the x -axis and the lines

and .x xπ π6 4

= =

20. If ( ) ,cosf x x2 3= show that ( ) ( ) .f x f x9= −m

ch5.indd 216 6/29/09 11:55:58 AM

ch5.indd 217 6/12/09 2:46:04 AM

Acceleration: The rate of change of velocity with respect to time

At rest: Stationary (zero velocity)

Displacement: The movement of an object in relation to its original position

Exponential growth: Growth, or increase in a quantity, where the rate of change of a quantity is in direct proportional to the quantity itself. The growth becomes more rapid over time

Exponential decay: Decay, or decrease in a quantity, where the rate of change of a quantity is in direct proportion to the quantity itself. The decay becomes less rapid over time

Rate of change: The change of one variable with respect to another variable over time

Velocity: The rate of change of displacement of an object with respect to time involving speed and direction

TERMINOLOGY

Applications of Calculus to the Physical World

6

ch6.indd 218 6/30/09 2:06:09 PM

219Chapter 6 Applications of Calculus to the Physical World

DID YOU KNOW?

Galileo (1564–1642) was very interested in the behaviour of bodies in motion. He dropped stones from the Leaning Tower of Pisa to try to prove that they would fall with equal speed. He rolled balls down slopes to prove that they move with uniform speed until friction slows them down. He showed that a body moving through the air follows a curved path at a fairly constant speed.

John Wallis (1616–1703) continued this study with his publication Mechanica , sive Tractatus de Motu Geometricus . He applied mathematical principles to the laws of motion and stimulated interest in the subject of mechanics.

Soon after Wallis’ publication, Christiaan Huygens (1629−1695) wrote Horologium Oscillatorium sive de Motu Pendulorum , in which he described various mechanical principles. He invented the pendulum clock, improved the telescope and investigated circular motion and the descent of heavy bodies.

These three mathematicians provided the foundations of mechanics. Sir Isaac Newton (1642–1727) used calculus to increase the understanding of the laws of motion. He also used these concepts as a basis for his theories on gravity and inertia.

INTRODUCTION

CALCULUS IS USED IN many situations involving rates of change, such as physics or economics. This chapter looks at rates of change of a particle in motion , and exponential growth and decay . Both these types of rates of change involve calculus.

Galileo

Rates of Change

Simple rates of change

The gradient of a straight line measures the rate of change of y with respect to the change in x .

With a curve, the rate of change of y with respect to x is measured by the gradient of the tangent to the curve. That is, the derivative measures the rate of change.

ch6.indd 219 6/29/09 9:38:03 PM


EXAMPLES

1. The number of bacteria in a culture increases according to the formula ,B t t2 20004 2= − + where t is time in hours. Find

the number of bacteria initially (a) the number of bacteria after 5 hours (b) the rate at which the number of bacteria will be increasing after (c)

5 hours.

Solution

(a)

`

Initially,

( )

B t t

t

B

2 20000

2 0 0 20002000

4 2

4 2

= − +== − +=

So there are 2000 bacteria initially. (b)

4 2

When ,

( )

t

B

5

2 5 5 20003225

== − +=

So there will be 3225 bacteria after 5 hours. The rate of change is given by the derivative (c)

3

When ,

( ) ( )

dtdB t t

t

dtdB

8 2

5

8 5 2 5

990

3= −

=

= −

=

So the rate of increase after 5 hours will be 990 bacteria per hour.

2. The volume fl ow rate of water into a pond is given by R t4 3 2= + litres per hour. If there is initially no water in the pond, fi nd the volume of water after 12 hours.

The rate of change of a quantity Q with respect to time t is measured by .dtdQ

ch6.indd 220 6/29/09 9:38:06 PM


Volume fl ow rate is the rate of change of volume over time.

6.1 Exercises

1. Find a formula for the rate of change in each question.

(a) h t t20 4 2= − (b) D t t5 2 13 2= + + (c) A x x16 2 2= − (d) x t x x3 2 35 4= − + − (e) V e 4t= +

(f) θcosS 3 5=

(g) πS rr

2 502

= +

(h) 4D x2= −

(i) S r r800 400= +

(j) πV r34 3=

2. Find an original formula for each question given its rate of change.

(a) R t t4 12 2= − is the rate at which the height h of an object changes over time t

(b) R x8 13= + is the rate at which the area A of a fi gure changes with side x

(c) πR r4 2= is the rate at which the volume V changes with radius r

(d) sinR t7= is the rate at which the distance d of an object changes over time t

(e) R e8 3t2= − is the rate at which the speed s changes over time t .

3. If 7 5,h t t3= − + fi nd the rate of change of h when 3t = .

4. Given ( ) ,sinf t t2= fi nd the rate of

change when πt6

= .

5. A particle moves so that its distance x is given by 2 .x e3t= Find the exact rate of change when 4t = .

6. The volume of water fl owing through a pipe is given by .V t t32= + Find the rate at which the water is fl owing when 5t = .

Solution

`

`

`

i.e.

( )

When ,

When

( ) ( )L

R t

dtdV t

V t dtt t C

t V

C

C

V t t

t

V

4 3

4 3

4 340 0

0 0 0

412

4 12 121776

2

2

2

3

3

3

= +

= +

= += + += == + +== +== +=

#

So there will be 1776 L of water in the pond after 12 hours.

ch6.indd 221 6/29/09 9:38:06 PM


7. The rate of change of the angle sum S of a polygon with n sides is a constant 180. If S is 360 when 4,n = fi nd S when 7n = .

8. A particle moves so that the rate of change of distance D over time t is given by 2 1.R et= − If 10D = when 0,t = fi nd D when 3t = .

9. For a certain graph, the rate of change of y values with respect to its x values is given by .R x x3 2 12= − + If the graph passes through the point ( ),, 31− fi nd its equation.

10. The mass in grams of a melting iceblock is given by the formula ,M t t2 1002= − + where t is time in minutes. Show that the rate at which the iceblock is melting is given by R t1 4= − grams per minute and fi nd the rate at which it will be melting after 5 minutes.

11. The rate of change in velocity over time is given by

.dt

dyt t t4 2 3= + − If the initial

velocity is cms ,2 1− fi nd the velocity after 15 s.

12. The rate of fl ow of water into a dam is given by .LhR t500 20 1= + − If there is 15 000 L of water initially in the dam, how much water will there be in the dam after 10 hours?

13. The surface area in cm 2 of a balloon being infl ated is given by ,S t t t2 5 23 2= − + + where t is time in seconds. Find the rate of increase in the balloon’s surface area after 8 s.

14. According to Boyle’s Law, the pressure of a gas is given by

the formula ,PVk= where k is a

constant and V is the volume of the gas. If 100k = for a certain gas, fi nd the rate of change in the pressure when 20.V =

15. A circular disc expands as it is heated. The area, in cm 2 , of the disc increases according to the formula 4 ,A t t2= + where t is time in minutes. Find the rate of increase in the area after 5 minutes.

16. A balloon is infl ated so that its increase in volume is at a constant rate of 10 cm s .3 1− If its volume is initially 1 cm 3 , fi nd its volume after 3 s.

17. The number of people in a certain city is given by ,P e200 000 . t0 2= where t is time in years. Find the rate at which the population of the city will be growing after 5 years.

18. A radioactive substance has a mass that decreases over time t according to the formula 200 .M e 0.1t= − Find

the mass of the substance (a) after 20 years, to the nearest gram

the rate of its decrease after (b) 20 years.

19. If ,y e4x= show that the rate is given by 4R y= .

20. Given ,S e2 3t2= + show that the rate of change is 2( 3)R S= − .

ch6.indd 222 6/29/09 9:38:06 PM


Exponential Growth and Decay

Exponential growth or decay are terms that describe a special rate of change that occurs in many situations. Population growth and growth of bacteria in a culture are examples of exponential growth. The decay of radioactive substances, the cooling of a substance and change in pressure are examples of exponential decay.

When a substance grows or decays exponentially, its rate of change is directly proportional to the amount of the substance itself. That is, the more of the substance there is, the faster it grows. For example, in a colony of rabbits, there is fairly slow growth in the numbers at fi rst, but the more rabbits there are, the more rabbits will be born. In other species, such as koalas, the clearing of habitats and other factors may cause the population to decrease, or decay, exponentially.

The exponential function is a function where the rate of change (gradient) is increasing as the function increases.

In mathematical symbols, exponential growth or decay can be written as

,kQdtdQ

= where k is the growth or decay constant .

This is why it is called exponential growth.

ch6.indd 223 6/29/09 9:38:07 PM


kQdtdQ

= is solved by the equation Q Aekt=

Proof

`

`

´

log

log

log

dtdQ

kQ

dQdt

kQ

tkQ

dQ

kQ k

kt Q k

kt k Q

Q e

e e

Ae

1

1

1e

e

ekt k

kt k

kt

1

2

2

2

2

=

=

=

= +

= +− =

===

−

−

#

A is always the initial quantity

Proof

Given ,Q Aekt= suppose the initial quantity is Q0 i.e. Q Q0= when 0t = Substituting into the equation gives

Q Ae

Ae

A

0´k0

0

===

∴ A is the initial quantity. Sometimes the equation is written as Q Q ekt

0= .

Malthus was an economist.

DID YOU KNOW?

Thomas Malthus (1766–1834), at the beginning of the Industrial Revolution, developed a theory about population growth that we still use today. His theory states that under ideal conditions,

the birth rate is proportional to the size of the population. That is, dtdN

kN= (Malthusian Law of Population Growth).

Malthus was concerned that the growth rate of populations would be higher than the increase in food supplies, and that people would starve.

Was he right? Is this happening? How could we prove this?

When k is positive there is exponential growth.When k is negative there is exponential decay.

Initial means ‘at fi rst’ when t 0.=

ch6.indd 224 6/29/09 9:38:09 PM


EXAMPLES

1. The population of a colony of rabbits over time t months is given by .P e1500 . t0 046= Find

the initial population (a) the population after 2 years (b) when the population reaches 5000. (c)

Solution

Initially, (a) 0t = Substituting:

( )

P e

e

e

e

1500

1500

1500

1500 1

.

0.046 0´

t0 046

0

0

==== =

So the initial population is 1500 rabbits. (b) years months2 24= So 24t = Substituting:

.

P e

e

e

1500

1500

15004524 31

.

0.046 24

.

´

t0 046

1 104

====

So the population after 2 years is 4524 rabbits . (c) 5000P = Substituting:

P e

e

e

1500

5000 1500

15005000

.

.

.

t

t

t

0 046

0 046

0 046

==

=

.

. (since )

..

ln ln

ln

ln

ln

e

t e

t e

t

t

15005000

0 046

0 046 1

0 04615005000

26 2

. t0 046=

== =

=

=

So the population reaches 5000 after 26.2 months .

CONTINUED

ch6.indd 225 6/29/09 9:38:09 PM


2. The number of bacteria in a culture is given by .N Aekt= If 6000 bacteria increase to 9000 after 8 hours, fi nd

(a) k correct to 3 signifi cant fi gures the number of bacteria after 2 days (b) the rate at which the bacteria will be increasing after 2 days (c) when the number of bacteria will reach 1 000 000 (d) the growth rate per hour as a percentage. (e)

Solution

(a)

`

So

So

When ,

When ,

.

.

.

.

.

log loglog

log

N Ae

t N

Ae

A

N e

t N

e

e

e

e

k e

k

k

k

N e

0 6000

6000

60008 9000

9000 6000

60009000

1 5

1 58

8

8

1 5

0 0507

6000 .

kt

kt

k

k

k

e ek

e

e

t

0

8

8

8

8

0 0507

Z

== ===== ==

=

====

=

=

(b) When , ( hours in 2 days)t

N e

48 48

600068 344

.0 0507 48

===

#

So there will be 68 344 bacteria after 2 days.

(c) Rate: ( . )

.dtdN e

e

6000 0 0507

304 1

.

.

t

t

0 0507

0 0507

=

=

When ,

.

t

dtdN e

48

304 1

3464

.0 0507 48

=

=

=

#

So after 2 days the rate of growth will be 3464 bacteria per hour.

Another method:

.

,

. ´

dtdN kN

N

t N

dtdN

0 0507

48 68 344

0 0507 68 344 3464

When

=

== =

= =

A gives the initial number of bacteria.

Don't round k off, but put it in the calculator’s memory to use again.

The rate of growth is the derivative.

ch6.indd 226 6/29/09 9:38:09 PM


(d) When

.

..

.

.

.

.

log loglog

log

N

e

e

e

e

t e

t

t

t

1 000 000

1 000 000 6000

60001 000 000

166 7

166 70 0507

0 0507

0 0507

166 7

100 9

.

.

.

.

t

t

t

e et

e

e

0 0507

0 0507

0 0507

0 0507

Z

==

=

====

=

So the number of bacteria will be 1 000 000 after 100.9 hours.

(e) where growth constant

.

. %

dtdN kN k

k 0 0507

5 07

= =

==

So the growth rate is 5.07% per hour.

3. A 50 g mass of uranium decays to 35 g after 2 years. If the rate of decay of its mass is proportional to the mass itself, fi nd the amount of uranium left after 25 years.

Solution

When ,

SoWhen ,

.

.

.

SoWhen

.

log loglog

log

Q Ae

t Q

Ae

A

Q e

t Q

e

e

e

e

k e

k

k

Q e

t

Q e

0 50

50

502 35

35 50

5035

0 7

0 72

2

2

0 7

5025

500 579

.

.

kt

kt

k

k

k

e ek

e

e

t

0

2

2

2

2

0 1783

0 1783 25

`

Z

== ===== ==

=

====

=

=== #

−

−

. k0 1783 Z−

So there will be 0.579 grams, or 579 mg, left after 25 years.

Notice that k is negative for decay. You could use

the formula Q Ae kt= − for decay. What would

k be?

ch6.indd 227 6/29/09 9:38:10 PM


1. The number of birds in a colony is given by N e80 . t0 02= where t is in days.

How many birds are there in (a) the colony initially?

How many birds will there be (b) after 30 days?

After how many days will (c) there be 500 birds?

Sketch the curve of the (d) population over time.

2. The number of bacteria in a culture is given by ,N N e0

0.32t= where t is time in hours.

If there are initially 20 000 (a) bacteria, how many will there be after 5 hours?

How many hours, to the (b) nearest hour, would it take for the number of bacteria to reach 200 000?

3. The decay of radium is proportional to its mass. If 100 kg of radium takes 5 years to decay to 95 kg

show that the mass of radium (a) is given by 100M e . t0 01= −

fi nd its mass after 10 years (b) fi nd its half-life (the time (c)

taken for the radium to halve its mass).

4. A chemical reaction causes the amount of chlorine to be reduced at a rate proportional to the amount of chlorine present at any one time. If the amount of chlorine is given by the formula ,A A e0

kt= − and 100 L reduces to 65 L after 5 minutes, fi nd

the amount of chlorine after (a) 12 minutes

how long it will take for the (b) chlorine to reduce to 10 L.

5. The production output in a factory increases according to the equation ,P P e0

kt= where t is in years.

Find (a) P0 if the initial output is 5000 units.

The factory produces 8000 (b) units after 3 years. Find the value of k , to 3 decimal places.

How many units will the (c) factory produce after 6 years?

The factory needs to produce (d) 20 000 units to make a maximum profi t. After how many years, correct to 1 decimal place, will this happen?

6. The rate of depletion of rainforests can be estimated as proportional to the amount of rainforests. If 3 million m 2 of rainforest is reduced to 2.7 million m 2 after 20 years, fi nd how much rainforest there will be after 50 years.

6.2 Exercises

ch6.indd 228 6/29/09 9:38:10 PM


7. The annual population of a country is increasing at a rate of

6.9%; that is, . .dtdP P0 069= If the

population is 50 000 in 2015, fi nd a formula for the population (a)

growth the population in the year (b)

2020 the rate at which the (c)

population will be growing in the year 2020

in which year the population (d) will reach 300 000.

8. An object is cooling down according to the formula ,T T e0

kt= − where T is temperature in degrees Celsius and t is time in minutes. If the temperature is initially 90°C and the object cools down to °81 C after 10 minutes, fi nd

its temperature after half (a) an hour

how long (in hours and (b) minutes) it will take to cool down to 30°C .

9. In the process of the inversion of sugar, the amount of sugar present is given by the formula .S Aekt= If 150 kg of sugar is reduced to 125 kg after 3 hours, fi nd

the amount of sugar after (a) 8 hours, to the nearest kilogram

the rate at which the sugar (b) will be reducing after 8 hours

how long it will take to (c) reduce to 50 kg .

10. The mass, in grams, of a radioactive substance is given by ,M M e0

kt= − where t is time in years. Find

(a) M0 and k if a mass of 200 kg decays to 195 kg after 10 years

mass after 15 years (b) the rate of decay after 15 years (c) the half-life of the substance (d)

(time taken to decay to half its mass).

11. The number of bacteria in a culture increases from 15 000 to 25 000 in 7 hours. If the rate of bacterial growth is proportional to the number of bacteria at that time, fi nd

a formula for the number of (a) bacteria

the number of bacteria after (b) 12 hours

how long it will take for the (c) culture to produce 5 000 000 bacteria.

12. A population in a certain city is growing at a rate proportional to the population itself. After 3 years the population increases by 20%. How long will it take for the population to double?

ch6.indd 229 6/29/09 9:38:11 PM


13. The half-life of radium is 1600 years.

Find the percentage of (a) radium that will be decayed after 500 years.

Find the number of years that (b) it will take for 75% of the radium to decay.

14. The population of a city is P ( t ) at any one time. The rate of decline in population is proportional to the population P ( t ), that is,

( ) .( )

kP tdt

dP t= −

Show that (a) ( ) ( )P t P t e kt0= − is a

solution of the differential

equation ( ) .( )

kP tdt

dP t= −

What percentage decline in (b) population will there be after 10 years, given a 10% decline in 4 years?

What will the percentage rate (c) of decline in population be after 10 years?

When will the population fall (d) by 20%?

15. The rate of leakage of water out of a container is proportional to the amount of water in the container at any one time. If the container is 60% empty after 5 minutes, fi nd how long it will take for the container to be 90% empty.

16. Numbers of sheep in a certain district are dropping exponentially due to drought. A survey found that numbers had declined by 15% after 3 years. If the drought continues, how long would it take to halve the number of sheep in that district?

17. Anthony has a blood alcohol level of 150 mg/dL. The amount of alcohol in the bloodstream decays exponentially. If it decreases by 20% in the fi rst hour, fi nd

the amount of alcohol in (a) Anthony’s blood after 3 hours

when the blood alcohol level (b) reaches 20 mg/dL.

18. The current C fl owing in a conductor dissipates according

to the formula kCdtdC = − . If it

dissipates by 40% in 5 seconds, how long will it take to dissipate to 20% of the original current?

19. Pollution levels in a city have been rising exponentially with a 10% increase in pollution levels in the past two years. At this rate, how long will it take for pollution levels to increase by 50%?

20. If dtdQ

kQ= , prove that Q Aekt=

satisfi es this equation by differentiating (a) Q Aekt=

integrating (b) dtdQ

kQ= .

ch6.indd 230 6/29/09 9:38:15 PM


Class Investigation

Research some environmental issues. For example:

It is estimated that some 1. animals, such as pandas and koalas, will be extinct soon. How soon will pandas be extinct? Can we do anything to stop this extinction? How do the greenhouse effect 2. and global warming affect the earth? How soon will they have a noticeable effect on us? How long do radioactive substances such as radium and plutonium 3. take to decay? What are some of the issues concerning the storage of radioactive waste? The erosion and salination of Australian soil are problems that may 4. affect our farming in the future. Find out about this issue, and some possible solutions. The effect of blue-green algae in some of our rivers is becoming a 5. major problem. What steps have been taken to remedy this situation?

Look at the mathematical aspects of these issues. For example, what formulae are used to make predictions? What sorts of time scales are involved in these issues?

Research other issues, relating to growth and decay, that affect our environment.

Motion of a Particle in a Straight Line

In this study of the motion of a particle moving along a straight line, we ignore friction, gravity and other infl uences on the motion. We can fi nd the particle’s displacement, velocity and acceleration at any one time. The term ‘particle’, or ‘body’, can refer to something quite large (e.g. a car). It describes any moving object.

DID YOU KNOW?

Calculus was developed in the 17th century as a solution to problems about the study of motion . Some problems of the time included fi nding the speed and acceleration of planets, calculating the lengths of their orbits, fi nding maximum and minimum values of functions, fi nding the direction in which an object is moving at any one time and calculating the areas and volumes of certain fi gures.

ch6.indd 231 6/29/09 9:38:17 PM


Displacement

Displacement ( )x measures the distance of a particle from a fi xed point (origin). Unlike distance, displacement can be positive or negative, according to which side of the origin it is on. Usually, on the right-hand side it will be positive and on the left it will be negative .

When the particle is at the origin , its displacement is zero . That is, .0x =

Velocity

Velocity ( v ) is a measure of the rate of change of displacement with respect to time. Since the rate of change is the gradient of the graph of displacement against time,

.

.

.

vdtdx

xVelocity can also be written as

=

Velocity (unlike speed) can be positive or negative, according to which direction the object is travelling in. If the particle is moving to the right , velocity is positive . If it is moving to the left , velocity is negative .

When the object is not moving, we say that it is at rest . That is, .0v =

Acceleration

Acceleration is the measure of the rate of change of velocity with respect to time . This gives the gradient of the graph of velocity against time. That is,

.adtdv

dtd

dtdx

dtd x

2

2

= = =c m

Acceleration can also be written as ...x Acceleration can be positive or negative, according to which direction it is

in. If the acceleration is to the right , it is positive . If the acceleration is to the left , it is negative .

If the acceleration is in the same direction as the velocity, the particle is speeding up (accelerating). If the acceleration is in the opposite direction from the velocity, the particle is slowing down (decelerating).

If the object is travelling at a constant velocity , there is no acceleration. That is, .0a =

Motion graphs

You can describe the velocity of a particle by looking at the gradient function of a displacement graph. The acceleration is the gradient function of a velocity graph.

ch6.indd 232 6/29/09 9:38:19 PM


EXAMPLES

1. The graph below shows the displacement of a particle from the origin as it moves in a straight line.

When is the particle (a) at rest? (i) at the origin? (ii)

When is the particle moving at its greatest velocity? (b) Solution

(i) When the particle is at rest, velocity is zero. i.e. (a) .dtdx 0=

So the particle is at rest at the stationary points t3 and .t5 The particle is at the origin when (ii) 0,x = i.e. on the t -axis. So the

particle is at the origin at ,t t2 4 and .t6 The greatest velocity is at (b) t 4 (the curve is at its steepest). Notice that

this is where there is a point of infl exion.

2. The graph below shows the displacement x of a particle over time t .

x

tt1 t2

CONTINUED

ch6.indd 233 6/29/09 9:38:19 PM


Draw a sketch of its velocity (a) Sketch its acceleration (b) Find values of (c) t for which the particle is

at the origin and(i) at rest. (ii)

Solution

The velocity is the rate of change of displacement, or (a) dtdx . By noting

where the gradient of the tangent is positive, negative and zero, we draw the velocity graph.

-

-

0

+

+

++

+

+

+

x

tt1 t2

v

tt1 t2

Acceleration is the rate of change of velocity, or (b) dtd x

2

2

. By noting where

the gradient of the tangent is positive, negative and zero, we draw the acceleration graph.

ch6.indd 234 6/29/09 9:38:21 PM


v

t

+

+

+

+t1 t2

tt1 t2

a

(i) The particle is at the origin when (c) 0.x = This is at 0 and t 2 .

The particle is at rest when (ii) .v 0= This is at t 1 (on the t -axis on the velocity graph or the stationary

point on the displacement graph).

3. The graph below is the velocity of a particle. v

tt1 t3t2 t4 t5

CONTINUED

The gradient is a constant, so acceleration is constant.

ch6.indd 235 6/29/09 9:38:22 PM


Draw a sketch showing the displacement of the particle. (a) Find the times when the particle is (b)

at rest(i) at maximum velocity. (ii)

Explain why you can’t state accurately when the particle is at the (c) origin.

Solution For (a) 0 t1− and between t 3 and t 5 , the velocity is negative (below the

t -axis), so the displacement has dtdx 0< .

Between t 1 and t 3 , and all points > t 5 , the velocity is positive (above

the t -axis), so the displacement has dtdx 0> .

At t 1 , t 3 and t 5 , the velocity is zero, so dtdx 0= and there is a

stationary point.

v

tt1 t2 t3 t4 t5

-- -

-

-

-

-

0 0 0

+

+

+

+ +

+

Notice that at t 1 and t 5 , LHS < 0 and RHS > 0 so they are minimum turning points. At t 3 , LHS > 0 and RHS < 0 so it is a maximum turning point. Here is one graph that could describe the shape of the displacement .

x

tt1 t2 t3 t4 t5

(i) At rest, (b) 0v = . This occurs at t 1 , t 3 and t 5 .

ch6.indd 236 6/29/09 9:38:22 PM


Maximum velocity is at (ii) t 4 since it is furthest from the t -axis. Notice also that there is a point of infl exion at t 4 on the displacement graph.

You can’t tell when the particle is at the origin as the graph is not (c) accurate. For example, the graph in (a) is at the origin at t 4 and another point to the right of t 5 . However, when sketching displacement from the velocity (or the original function given the gradient or derivative function) there are many possible graphs, forming a family of graphs.

x

tt1 t2 t3 t4 t5

Notice that these cross the t -axis at various places, so we can’t tell where the displacement graph shows the particle at the origin.

1. The graphs at right and overleaf show the displacement of an object. Sketch the graphs for velocity and acceleration.

x

t

(a)

6.3 Exercises

ch6.indd 237 6/29/09 9:38:23 PM


x

t

(b)

x

t

(c)

(d)

(e)

2. The graph below shows the displacement of a particle as it moves along a straight line.

When is the particle at the origin? (a) at rest? (b) travelling with constant (c)

acceleration? furthest from the origin? (d)

3. The graph below shows a particle travelling at a constant acceleration.

Sketch the graphs that could represent

its velocity and (a) its displacement. (b)

4. The graph below shows the velocity of a particle.

x

t1 t2 t3 t4 t5 t6t

When is the particle at rest? (a) When is the acceleration (b)

zero? When is the velocity the (c)

greatest? Describe the motion of the (d)

particle at (i) t 2 and (ii) t 3 .

ch6.indd 238 6/29/09 9:38:23 PM


5. The graph below shows the displacement of a pendulum.

When is the pendulum at (a) rest?

When is the pendulum in (b) its equilibrium position (at the origin)?

6. Describe the motion of the particle at t1 for each displacement graph.

(a)

(b)

(c)

(d)

(e)

Motion and Differentiation

If displacement is given, then the velocity is the fi rst derivative and the acceleration is the second derivative.

Displacement x

.x

dtdxVelocity =

Acceleration xdtdv

dtd x

2

2

= =p

ch6.indd 239 6/29/09 9:38:24 PM


EXAMPLES

1. The displacement x cm of a particle over time t seconds is given by .x t t4 2= −

Find any times when the particle is at rest .(a) How far does the particle move in the fi rst 3 seconds? (b)

Solution

At rest means that the velocity (a) dtdxc m is zero.

0

–

dtdx t

dtdx

t

t

t

4 2

4 2 0

4 2

2

When

= −

=

===

So the particle is at rest after 2 seconds.

Notice that there is a stationary point at ,dtdx 0= so the particle turns

around at this time.

Initially, when (b) 0:t =

( )x 4 0 0

0

2= −=

So the particle is at the origin .

After 3 seconds, 3:t =

( )x 4 3 3

3

2= −=

So the particle is 3 cm from the origin.

However, after 2 seconds, the particle turns around, so it hasn’t simply moved from the origin to 3 cm away. We need to fi nd where it is when it turns around.

2:

( )

t

x 4 2 24

When2

== −=

So the particle is 4 cm from the origin.

The particle moves from the origin ( 0)x = to 4 cm away in the fi rst 2 seconds, so it has travelled 4 cm. It then turns and goes back to 3 cm from the origin. So in the 3 rd second it travels 1 cm back.

ch6.indd 240 6/29/09 9:38:25 PM


0 1 2 3 4 5

Total distance travelled in the fi rst 3 seconds is 4 1+ or 5 cm.

2. The displacement of a particle is given by 2 3 ,x t t cm2= − + + where t is in seconds.

Find the initial velocity of the particle. (a) Show that the particle has constant acceleration. (b) Find when the particle will be at the origin. (c) Find the particle’s maximum displacement from the origin. (d) Sketch the graph showing the particle’s motion. (e)

Solution

(a)

( )

vdtdx

t

t

v

2 2

0

2 0 2

2

Initially,

`

=

= − +== − +=

So the initial velocity is .2 cms 1−

(b) adtd x

2

2

2

=

= −

∴ the particle has a constant acceleration of .2 cms 2− −

(c)

( )( )

x

t t

t t

t

0

2 3 01 3 0

1 3

At the origin

i.e.

or

2

`

=− + + =

− + − == −

So the particle will be at the origin after 3 s.

For maximum displacement, (d)

dtdx

t

t

t

0

2 2 0

2 2

1

=

− + ===

(Maximum displacement occurs when .t 1= ) ,

( ) ( )

t

x

1

1 2 1 34

When2

== − + +=

So maximum displacement is 4 cm.

A time of –1 does not exist.

CONTINUED

ch6.indd 241 6/29/09 9:38:25 PM


(e)

3. The displacement of a particle is given by the formula ,sinx t1 3 2= + where x is in metres and t is in seconds.

Sketch the graph of (a) x a s a function of t . Find the times when the particle will be at rest. (b) Find the position of the particle at those times. (c)

Solution

Graph has period (a) π and amplitude 3.

(b)

cos

vdtds

t6 2

=

=

When the particle is at rest

`

, , , . . .

, , , . . .

π π π

π π π

cos

cos

v

t

t

t

t

0

6 2 0

2 0

22 2

32

5

4 43

45

i.e.

===

=

=

This is the graph of a parabola. Notice that x can be positive or negative, but t is never negative.

ch6.indd 242 6/29/09 9:38:25 PM


(c) sinx t1 3 2= +

( )

( )

π

π

π

π

sin

sin

t

x

t

x

4

1 32

1 3 1

4

43

1 32

3

1 3 1

2

When

When

=

= +

= +=

=

= +

= + −= −

Similarly, when , , ,…π πt x4

54

7= moves between 4 m and –2 m from the origin.

Investigation

In physics, equations for displacement and velocity are given by

21s ut at2+= and ,v u at= + where u is the initial velocity and a is the

acceleration. Differentiate the displacement formula to show the formula for velocity. What happens when you differentiate again?

1. The displacement of a particle is given by 9x t t cm,3= − where t is time in seconds.

Find the velocity of the (a) particle after 3 s.

Find the acceleration after 2 s. (b) Show that the particle is (c)

initially at the origin, and fi nd any other times that the particle will be at the origin.

Find after what time the (d) acceleration will be 30 cms .2−

2. A particle is moving such that its displacement is given by ,s t t2 8 32= − + where s is in metres and t is in seconds.

Find the initial velocity. (a) Show that acceleration is (b)

constant and fi nd its value. Find the displacement after 5 s. (c) Find when the particle will be (d)

at rest. What will the particle’s (e)

displacement be at that time? Sketch the graph of the (f)

displacement against time.

6.4 Exercises

ch6.indd 243 6/29/09 9:38:26 PM


3. A projectile is fi red into the air and its height in metres is given by ,h t t40 5 42= − + where t is in seconds.

Find the initial height. (a) Find the initial velocity. (b) Find the height after 1 s. (c) What is the maximum height (d)

of the projectile? Sketch the graph of the (e)

height against time.

4. The displacement in cm after time t s of a particle moving in a straight line is given by .x t t2 2= − −

Find the initial displacement. (a) Find when the particle will be (b)

at the origin. Find the displacement (c)

after 2 s. How far will the particle (d)

move in the fi rst 2 seconds? Find its velocity after 3 s. (e)

5. The equation for displacement of a particle is given by 1x e mt2= + after time t seconds.

Find the initial velocity. (a) Find the exact acceleration (b)

after 1 s. Show that the acceleration is (c)

always double the velocity. Sketch the graph of velocity (d)

over time t .

6. The displacement of a pendulum is given by cosx t2 cm= after time t seconds.

Find the equation for the (a) velocity of the pendulum.

Find the equation for its (b) acceleration.

Find the initial displacement. (c)

Find the times when the (d) pendulum will be at rest.

What will the displacement (e) be at these times?

When will there be zero (f) displacement?

Show that acceleration (g) .a x4= −

7. An object is travelling along a straight line over time t seconds, with displacement according to the formula .x t t t6 2 1 m3 2= + − +

Find the equations of its (a) velocity and acceleration.

What will its displacement be (b) after 5 s?

What will its velocity be (c) after 5 s?

Find its acceleration after 5 s. (d)

8. The displacement, in centimetres, of a body is given by ( ) ,x t4 3 5= − where t is time in seconds.

Find equations for velocity (a) .x and acceleration .. .x

Find the values of (b) .,x x and ..x after 1 s.

Describe the motion of the (c) body after 1 s.

9. The displacement of a particle, in metres, over time t seconds

is ,s ut gt21 2= + where u 5= and

.g 10= − Find the equation of the (a)

velocity of the particle. Find the velocity after 10 s. (b) Show that the acceleration is (c)

equal to g .

10. The displacement in metres after

t seconds is given by .stt

3 12 5=

+−

Find the equations for velocity and acceleration.

ch6.indd 244 6/29/09 9:38:26 PM


11. The displacement of a particle is given by ( ),logx t 1e= + where x is in centimetres and t is in seconds.

Find the initial position of the (a) particle.

Find the velocity after 5 s. (b) Find the acceleration after 5 s. (c) Describe the motion of the (d)

particle at 5 s. Find the exact time when the (e)

particle will be 3 cm to the right of the origin.

12. Displacement of a particle is given by ,x t t t4 33 2= − + where s is in metres and t is in seconds.

Find the initial velocity. (a) Find the times when the (b)

particle will be at the origin. Find the acceleration after 3 s. (c)

13. The displacement of a particle moving in simple harmonic motion is given by ,sinx t3 2= where x is in centimetres and t is in seconds.

Sketch the graph of the (a) displacement of the particle from t 0= to 2 .πt =

Find equations for the (b) velocity .x and acceleration .. .x

Find the exact acceleration(c)

after .π s3

Show that (d) . .x4= −.x

14. The height of a projectile is given by ,h t t7 6 2= + − where height is in metres and time is in seconds.

Find the initial height. (a) Find the maximum height (b)

reached.

When will the projectile (c) reach the ground?

Sketch the graph showing (d) the height of the projectile over time t .

How far will the projectile (e) travel in the fi rst 4 s?

15. A ball is rolled up a slope at a distance from the base of the slope, after time t seconds, given by x t t15 3 2= − metres.

How far up the slope will the (a) ball roll before it starts to roll back down?

What will its velocity be (b) when it reaches the base of the slope?

How long will the motion of (c) the ball take altogether?

16. The displacement of a particle is given by .x t t t2 3 423 2= − +

Show that the particle is (a) initially at the origin but never returns to the origin.

Show that the particle is (b) never at rest.

17. A weight on the end of a spring has a height given by sinh t3 2 cm= where t is time in seconds.

Find its initial height . (a) Find when the spring is at (b)

its maximum distance from the origin .

What is the acceleration at (c) these times?

ch6.indd 245 6/29/09 9:38:27 PM


18. A particle moves so that its displacement x cm is given by x et2= at t seconds.

Find the exact velocity of the (a) particle after 4 seconds .

When is the particle at rest? (b) Where is the particle at that (c)

time?

19. A particle is moving in a straight line so that its displacement x cm over time t seconds is given by .x t t49 2= −

For how many seconds does (a) the particle travel?

Find the exact time at which (b) the particle comes to rest .

How far does the particle (c) move altogether?

Motion and Integration

If the fi rst derivative of the displacement of a particle gives its velocity , then the reverse is true. That is, the primitive function (indefi nite integral) of the velocity gives the displacement . Similarly, the integral of the acceleration gives the velocity .

x v dt

v a dt

=

=

##

Displacement is the area under the velocity–time graph.

EXAMPLES

1. The velocity of a particle is given by .v t t3 2 12= + + If initially the particle is 2 cm to the left of the origin, fi nd the displacement after 5 s. Solution

( )

x v dt

t t dtt t t C

3 2 12

3 2

=

= + += + + +

##

When ,t x

C

C

x t t t

0 2

2 0 0 0

2

3 2

3 2

`

`

= = −− = + + +

== + + −

When ,t

x

5

5 5 5 2125 25 5 2

153

3 2

== + + −= + + −=

So after 5 s the particle will be 153 cm to the right of the origin.

Initially means t 0=

ch6.indd 246 6/29/09 9:38:27 PM


2. The acceleration of a particle is given by ( )

.at

61

2 ms 22

= −+

− If the

particle is initially at rest 1 m to the right of the origin, fi nd its exact displacement after 9 s. Solution

1−

( )

[ ( ) ]( )

v a dt

tdt

t dt

tt

C

tt

C

61

2

6 2 1

6 211

61

2

2

2

=

= −+

= − +

= −−+

+

= ++

+

−

e o##

#

When ,

( )

6 2

t v

C

C

C

v tt

0 0

0 6 00 1

2

2

2

12So

`

`

= =

= ++

+

= += −

= ++

−

x v dt

tt

dt61

2 2

e2

=

= ++

−

( )logt t t C3 2 1 2= + + − +

c m##

When ,

( ) ( ) ( )

( )

( ) ( ) ( )

log

log

loglog

log

t x

C

C

x t t t

x

0 1

1 3 0 2 0 1 2 0

3 2 1 2 1

3 9 2 9 1 2 9 1243 2 10 18 1

226 2 10

e

e

e

e

e

e

2

2

2

`

`

= == + + − +== + + − +

= + + − += + − += +

When

( )log

t 9

2 113 10

=

= +

So after 9 s the displacement will be ( ) .log2 113 10 me+

‘At rest’ means v 0.=

You must fi nd C before going on to fi nd the

displacement.

ch6.indd 247 6/29/09 9:38:27 PM


1. The velocity of a particle is given by .v t3 5 cms2 1= − − If the particle is at the origin initially, fi nd its displacement after 3 s.

2. A particle has velocity given by

2 3 .tdtdx ms 1= − − After 3 s the

particle is at the origin. Find its displacement after 7 s.

3. The velocity of a particle is given by .v t4 3 cms 1= − − If the particle is 5 cm from the origin after 2 s, fi nd the displacement after 7 s.

4. The velocity of a particle is given by .v t t8 3 cms3 2 1= − − If the particle is initially at the origin, fi nd

its acceleration after 5 s (a) its displacement after 3 s (b) when it will be at the origin (c)

again.

5. The velocity of a particle is given by .x e3 cmst3 1= −: Given that the particle is initially 2 cm to the right of the origin, fi nd its exact displacement after 1 s.

6. A particle has a constant acceleration of .12 ms 2− If the particle has a velocity of 2 ms 1− and is 3 m from the origin after 5 s, fi nd its displacement after 10 s.

7. The acceleration of an object is given by .t6 4 cms 2+ − The particle is initially at rest at the origin. Find

its velocity after 5 s (a) its displacement after 5 s. (b)

8. A projectile is accelerating at a constant rate of . .9 8 ms 2− − If it is initially 2 m high and has

a velocity of 4 ,ms 1− fi nd the equation of the height of the projectile.

9. A particle is accelerating according to the equation ( ) .a t3 1 ms2 2= + − If the particle is initially at rest 2 m to the left of 0, fi nd its displacement after 4 s.

10. The velocity of a particle is given by .v e 1 mst 1= − − If the particle is initially 3 m to the right of the origin, fi nd its exact position after 5 s.

11. A particle accelerates according to the equation .a e t2= − If the particle has initial velocity of zero cms 1− and initial displacement of 1 ,cm− fi nd its displacement after 4 s, to the nearest centimetre.

12. The acceleration of a particle is given by .sina t9 3 cms 2= − − If the initial velocity is 5 cms 1− and the particle is 3 cm to the left of the origin, fi nd the exact displacement after .π s

13. The velocity of a particle is given

by .xt

t3

ms 12

=+

−: If the particle

is initially at the origin, fi nd its displacement after 10 s, correct to 2 decimal places.

14. The acceleration of an object is given by .a e mst3 2= − If initially the object is at the origin with velocity 2 ,ms 1− − fi nd its displacement after 3 seconds, correct to 3 signifi cant fi gures.

6.5 Exercises

ch6.indd 248 6/29/09 9:38:27 PM


15. The velocity of a particle is given by .cosv t4 2 ms 1= − If the particle is 3 m to the right of the origin after ,π s fi nd the exact

displacement after (a) π6

s

acceleration after (b) π6

s .

16. The acceleration of a particle is

given by ( )

at 3

22=

+ ms −2 and the

particle is initially at rest 2 ln 3 m to the left of the origin.

Evaluate its velocity after (a) 2 seconds.

Find its displacement after (b) n seconds.

Show that (c) ≤ v0 1< for all t .

17. The acceleration of a particle is 25x e t5=..

ms −2 and its velocity is 5 ms −1 initially. Its displacement is initially 1 m.

Find its velocity after (a) 9 seconds.

Find its displacement after (b) 6 seconds.

Show that the acceleration (c) .x x25=..

Find the acceleration when (d)

the particle is 2 m to the right of the origin.

ch6.indd 249 6/29/09 9:38:28 PM


1. A particle moves so that its displacement after t seconds is x t t4 52 3= − metres. Find

its initial displacement, velocity and (a) acceleration

when the particle is at the origin (b) its maximum displacement. (c)

2. The velocity of a particle is given by 6 12 .V t t ms2 1= − − Find its displacement and acceleration after 3 seconds if it is initially 5 m to the right of the origin.

3. A city doubles its population in 25 years. If it is growing exponentially, when will it triple its population?

4. The displacement of a particle after t seconds is 2 .x e cmt3=

Find its initial velocity. (a) Find its acceleration after 3 seconds, (b)

to 1 decimal place. Show that (c) 9 .x x=..

5. A particle accelerates at 8 2 .cosx t ms 2= −..

The particle is initially at rest at the origin. Find its displacement after

π6

seconds.

6. If a particle has displacement ,sinx t2 3= show that its acceleration is given by –9 .x x=..

7. A particle has displacement 12 36 9x t t t cm3 2= − + − at time t seconds.

When is the particle at rest? (a) What is the(b) (i) displacement(ii) velocity(iii) acceleration after 1 s? Describe the motion of the particle (c)

after 1 s.

8. The acceleration of a particle is .6 12x t ms 2= − −..

The particle is initially at rest 3 metres to the left of the origin.

Find the (a) (i) acceleration(ii) velocity(iii) displacement after 5 seconds. Describe the motion of the particle (b)

after 5 seconds.

9. The graph below shows the displacement of a particle.

When is the particle at the origin? (a) When is it at rest? (b) When is it travelling at its greatest (c)

speed? Sketch the graph of its (d) (i) velocity(ii) acceleration.

10. A radioactive substance decays by 10% after 80 years.

By how much will it decay after (a) 500 years?

When will it decay to a quarter of (b) its mass?

11. The equation for displacement of a particle over time t seconds is sinx t= metres.

When is the displacement 0.5 m? (a) When is the velocity (b) 0.5 ?ms 1−

Find the acceleration after (c) .π4

s

Test Yourself 6

ch6.indd 250 6/29/09 9:38:28 PM


12. The height of a ball is h t t20 5 2= − metres after t seconds.

Find the height after 1 s. (a) What is the maximum height of (b)

the ball? What is the time of fl ight of the ball? (c)

13. A bird population of 8500 increases to 12 000 after 5 years. Find

the population after 10 years (a) the rate at which the population is (b)

increasing after 10 years when the population reaches 30 000. (c)

14. The graph below shows the velocity of a particle.

Sketch a graph that shows (a) (i) displacement (ii) acceleration When is the particle at rest? (b)

15. A particle is moving with acceleration 10 .a e mst2 2= − If it is initially 4 m to the right of the origin and has a velocity of 2 ,ms 1− − fi nd its displacement after 5 seconds, to the nearest metre.

Challenge Exercise 6 1. (a) Find an equation for the

displacement of a particle from the origin if its acceleration is given by

cosdtdv t6 3 cms 2= − and initially the

particle is at rest 2 cm to the right of the origin .

(b) Write the acceleration of the particle in terms of x .

2. The displacement of a particle is given by ( ) ,x t 13 6= + where x is in metres and t is in seconds .

Find its initial displacement and (a) velocity .

Find its acceleration after 2 s in (b) scientifi c notation, correct to 3 signifi cant fi gures .

Show that the particle is never at (c) the origin.

3. (a) Find an equation for the displacement of a particle from the origin if its acceleration is given by

16 4cos tdtdv cms 2= − − and initially the

particle is at rest 1 cm to the right of the origin .

(b) Find the exact velocity of the particle when it is 0.5 cm from the origin.

4. The velocity, in ,ms 1− of an object is given by .cosv t5 5=

Show that if the object is initially at (a) the origin, its acceleration will always be –25 times its displacement .

Find the maximum acceleration of (b) the object .

Find the acceleration when the object (c) is 0.3 m to the right of the origin .

ch6.indd 251 6/29/09 9:38:29 PM


5. The population of a fl ock of birds over t years is given by the formula .P P e0= . t0 0151

How long will it take, correct (a) to 1 decimal place, to increase the population by 35%?

What will be the percentage increase (b) in population after 10 years, to the nearest per cent?

6. The Logistic Law of Population Growth, fi rst proposed by Verhulst in 1837, is

given by ,kN bNdtdN 2−= where k and b

are constants. Show that the equation

( )

NbN k bN e

kNt

0 0

0=+ − k−

is a solution of this

differential equation ( 0N is a constant).

7. A particle moves so that its velocity is given by .x te cmst 12= −: Find its exact acceleration after 1 s .

ch6.indd 252 6/29/09 9:38:31 PM

Terminology Bold Terminology text...

TERMINOLOGY

Terminology Bold Terminology text...

TERMINOLOGY

Annuity: An annuity is a fi xed sum of money invested every year that accumulates interest over a number of years

Arithmetic sequence: A set of numbers that form a pattern where each successive term is a constant amount (positive or negative) more than the previous one

Common difference: The constant amount in an arithmetic sequence that is added to each term to get the next term

Common ratio: The constant amount in a geometric sequence that is multiplied to each term to get the next term

Compound interest: Interest is added to the balance of a bank account so that the interest and balance both earn interest

Geometric sequence: A set of numbers that form a pattern where each successive term is multiplied by a constant amount to get the next term

Limiting sum: (or sum to infi nity) The sum of infi nite terms of a geometric series where the common ratio r obeys the condition −1 < r < 1

Partial sum: The sum of a certain fi nite number of terms of a sequence

Sigma notation: The Greek letter stands for the sum of a sequence of a certain number of terms, called a partial sum

Sequence: A set of numbers that form a pattern or obey a fi xed rule

Series: The sum of a sequence of n terms

Superannuation: A sum of money that is invested every year (or more frequently) as part of a salary to provide a large amount of money when a person retires from the paid workforce

Term: A term refers to the position of the number in a sequence. For example the fi rst term is the fi rst number in the sequence

TERMINOLOGY

Series

7

ch7.indd 258 6/29/09 10:01:23 PM

259Chapter 7 Series

DID YOU KNOW?

Box text...

INTRODUCTION

THE INFINITE SUM OF a sequence of numbers (or terms) is called a series . Many series occur in real life—think of the way plants grow, or the way money accumulates as a certain amount earns interest in a bank.

You will look at series in general, arithmetic, geometric series and their applications.

DID YOU KNOW?

Number patterns and series of numbers have been known since the very beginning of civilisation. The Rhind Papyrus , an Egyptian book written about 1650 BC, is one of the oldest books

on mathematics. It was discovered and deciphered by Eisenlohr in 1877, and it showed that the ancient Egyptians had a vast knowledge of mathematics. The Rhind Papyrus describes the problem of dividing 100 loaves among 5 people in a way that shows the Egyptians were exploring arithmetic series.

Around 500 BC the Pythagoreans explored different polygonal numbers:

• triangular numbers: 1 2 3 4 …+ + + +

1 1 2+ 1 2 3+ + 1 2 3 4+ + +

• square numbers: 1 3 5 7 …+ + + +

1 1 3+ 1 3 5+ + 1 3 5 7+ + +Could you fi nd a series for pentagonal or hexagonal

numbers?

A sequence forms a pattern. Some patterns are easy to see and some are diffi cult. People are sometimes asked to identify sequences and their patterns in tests of intelligence.

General Series

ch7.indd 259 6/29/09 10:01:41 PM


1. 5, 8, 11, ...

2. 8, 13, 18, ...

3. …11 22 33+ + +

4. 100, 95, 90, ...

5. …7 5 3+ + +

6. 99, 95, 91, ...

7. , , ,…21 1 1

21

8. . . .1 3 1 9 2 5 f+ + +

9. 2, 4, 8, ...

10. …4 12 36+ + +

11. , , , ,…1 2 4 8− −

12. , , , ,…3 6 12 24− −

13. , , ,…21

41

81

14. , , ,…52

154

458

15. …1 4 9 16 25+ + + + +

16. 1, 8, 27, 64, ...

17. 0, 3, 8, 15, 24, ...

18. …3 6 11 18 27+ + + + +

19. …2 9 28 65+ + + +

20. 1, 1, 2, 3, 5, 8, 13, ...

DID YOU KNOW?

The numbers 1, 1, 2, 3, 5, 8, … are called Fibonacci numbers after Leonardo Fibonacci (1170–1250). The Fibonacci numbers occur in natural situations.

For example, when new leaves grow on a plant’s stem, they spiral around the stem. The ratio of the number of turns to the number of spaces between successive leaves gives the

series of fractions 11

,21

,32

,53

,85

,138

,2113

,3421

,5534

,8955

,14489

, …

Find the next 3 terms in each sequence or series of numbers.

7.1 Exercises

ch7.indd 260 6/29/09 10:01:44 PM

261Chapter 7 Series

Research Fibonacci numbers and fi nd out where else they appear in nature .

The Fibonacci ratio is the number of turns divided by the number of spaces.

Formula for the nth term of a series

When a series follows a mathematical pattern, its terms can be described by a formula. We call any general or n th term of a series T n where n stands for the number of the term and must be a positive integer.

CONTINUED

EXAMPLE

For the series …6 13 20+ + + fi nd (a) T 1 (b) T 2 (c) T 3 (d) T n

Solution

The fi rst term is 6, so when (a) 1:n = 6T1 = The 2 nd term is 13 so (b) 13T2 = The 3 rd term is 20 so (c) 20T3 = Each term is 7 more than the previous term. (d)

The 1 st term is 6

The term is

´

13 6 7

3 20 6 7 7

6 2 7

The 2nd term is

rd

= += + += +

ch7.indd 261 6/29/09 10:01:45 PM


Following this pattern:

The 4th term

´6 7 7 7

6 3 7

= + + += +

and so on 6 ( 1) 7´T nn = + −

T n

n

6 7 7

7 1n = + −

= −

If we are given a formula for the n th term, we can fi nd out much more about the series.

EXAMPLES

1. The n th term of a series is given by the formula 3 4.T nn = + Find the fi rst 3 terms and hence write down the series.

Solution

( )

( )

( )

T

T

T

3 1 4

7

3 2 4

10

3 3 4

13

1

2

3

= +== +== +=

So the series is …7 10 13+ + +

2. The n th term of a series is given by .t n5 1n = − Which term of the series is equal to 104?

Solution

t n

n

n

5 1 104

5 105

21

n = − ===

So 104 is the 21st term of the series.

3. The n th term of a series is given by 103 3 .u nn = − Find the fi rst 3 terms. (a) Find the value of (b) n for the fi rst negative term in the series.

Notice that the product with 7 in each term involves a number one less than the number of the term.

ch7.indd 262 6/29/09 10:01:46 PM

263Chapter 7 Series

Solution

(a) ( )

( )

( )

u

u

u

103 3 1

100

103 3 2

97

103 3 3

94

1

2

3

= −== −== −=

So the series is . . . .100 97 94+ + + For the fi rst negative term, we want (b)

i.e.

u

n

n

n

0

103 3 0

3 103

3431

<<<

>

n

−− −

Since n is an integer, 35n = ∴ the 35th term is the fi rst negative term .

4. The n th term of a series is given by the formula .t 2 1nn= − Which

term of the series is equal to 4095?

Solution

t 2 1nn= −

We are given that the n th term is 4095 and need to fi nd n .

log log

log

log

log

n

n

n

4095 2 1

4096 2

4096 22

2

4096

12

n

n

n10 10

10

10

10

= −===

=

=

So 4096 is the 12 th term of the sequence .

You could use base e instead of base 10.

You could use trial and error to guess what

power of 2 is equal to 4096 if you don’t want to

use logarithms.

1. Find the fi rst 3 terms of the series with n th term as follows

(a) T n8 5n = − (b) T n2 3n = + (c) u n6 1n = − (d) T n8 5n = − (e) t n20n = −

(f) u 3nn=

(g) Q 2 7nn= +

(h) t n4 2nn= −

(i) T n n8 1n2= − +

(j) T n nn3= +

7.2 Exercises

ch7.indd 263 6/29/09 10:01:46 PM


2. Find the fi rst 3 terms of each series. (a) T n3 2n = − (b) t 4n

n= (c) T n nn

2= +

3. Find the 50 th term of each series. (a) T n7 1n = − (b) t n2 5n = + (c) T n5 2n = − (d) T n40 3n = − (e) U n8 7n = −

4. Find the 10 th term of each series. (a) T 2 5n

n= + (b) t 10 3n

n= − (c) T n2n

n= − (d) u n n5 3n

2= − + (e) T n 2n

3= +

5. Which of these are terms of the series ?T n3 5n = +

68 (a) 158 (b) 205 (c) 266 (d) 300 (e)

6. Which of these are terms of the series ?T 5 1n

n= − 3126 (a) 124 (b) 15 634 (c) 78 124 (d) 0 (e)

7. Which term of the series with n th term t n9 15n = − is equal to 129?

8. Is 255 a term of the series with n th term ?T 2 1n

n= −

9. Which term of the series with n th term u n 5n

3= + is 348?

10. Which term of T n 3n2= − is equal

to 526?

11. For the series ,T nn3= fi nd

the 12 th term (a) (b) n if the n th term is 15 625 .

12. A series is given by the formula T n n3 2n

2= + − . Find the 25 th term. (a) Find (b) n if the n th term is .252−

13. Find the value of n that gives the fi rst term of the series T n3 2n = + greater than 100 .

14. Find the value of n that gives the fi rst term of the series T 2n

n= larger than 500 .

15. Find the values of n where the series T n4 1n = − is greater than 350 .

16. Find the values of n for which the series T n400 5n = − is less than 200 .

17. Find the value of n that gives the fi rst negative term of the series T n1000 2n = − .

18. Find the value of n that gives the fi rst positive term of the series T n2 300n = − .

19. Find the value of (a) n that gives the

fi rst negative term of the series u n80 6n = −

the fi rst negative term. (b)

20. Find the fi rst negative term of the series T n50 7n = − .

ch7.indd 264 6/29/09 10:01:47 PM

265Chapter 7 Series

Sigma Notation

Series are often written in sigma notation . The Greek letter sigma (Σ) is like an English ‘S’. Here it stands for the sum of a series.

Application

The mean of a set of scores is given by

fΣx = fxΣ

where fΣ is the sum of frequencies and fxΣ is the sum of the scores × frequencies.

Notice that there are 5 terms, since they include

both the 3 and the 7. We work out the number of

terms by 7 3 1− + .

CONTINUED

EXAMPLES

1. Evaluate rr

2

1

5

=/

Solution

rr

2

1

5

=/ means the sum of terms where the formula is r 2 with r starting at 1

and ending at 5 .

1 4 9 16 25

55

= + + + +=

So r 1 2 3 4 5r

2

1

52 2 2 2 2= + + + +

=/

2. Evaluate ( )n2 5

3

7

+/

Solution

( )n2 53

7

+/ means the sum of terms where the formula is 2 5n + with n

starting at 3 and ending at 7. Can you work out how many terms there are in this series?

11 13 15 17 19

75

= + + + +=

( ) ( ) ( ) ( ) ( ) ( )´ ´ ´ ´ ń2 5 2 3 5 2 4 5 2 5 5 2 6 5 2 7 53

7

+ = + + + + + + + + +/

ch7.indd 265 6/29/09 10:01:47 PM


3. Write ( )… k7 11 15 4 3+ + + + + in sigma notation.

Solution

In this question, the formula is 4 3n + where the last term is at .n k= We need to fi nd the number of terms. The fi rst term is 7 so we guess that 1n = for this term .

ń

n

1

4 3 4 1 3

7

When =+ = +

=

So the fi rst term is at 1n = and the last term is at n k= .

We can write the series as ( )n4 3k

1+/ .

4. How many terms are there in the series

(a) ( )n3 71

100

−/ ?

(b) 2 ?5

50n/

Solution

The fi rst term is when (a) 1n = and the last term is when 100n = . There are 100 terms .

The fi rst term is when (b) 5n = and the last term is when 50n = . This is harder to work out. Look at example 2 where there are 5 terms. We can work out the number of terms by 50 5 1− + . There are 46 terms .

If n is not 1, we could try 2 then 3 and so on until we fi nd the fi rst term.

The number of terms in the series ( )f nq

p

/ is 1p q− +

1. Evaluate

(a) ( )3 2n

n 1

4

+=/

(b) nn

2

2

5

=/

(c) (5 6)n2

6

−/

(d) ( )r3 1r 1

10

+=/

(e) ( )k 1k

3

2

5

−=/

(f) 1n3

5

/

(g) ( )n nn

2

1

5

−=/

(h) | |p3 22

4

−/

(i) ( )2n

n 1

6

=/

(j) ( )n3 2 5n

3

6

− −/

7.3 Exercises

There are 10 1 10 − + terms.

ch7.indd 266 6/29/09 10:01:48 PM

267Chapter 7 Series

2. How many terms are there in each series?

(a) ( )7n

n 1

65

=/

(b) ( )n 1n

2

2

100

+=/

(c) ( )n3 45

80

+/

(d) ( )r3r

2

1

200

=/

(e) ( )n7 110

20

+/

(f) ( )n n32

7

45

+/

(g) nn

3

12

108

=/

(h) 4n

n 9

74

=/

(i) ( )n2 33

77

−/

(j) ( )n7n

11

55

+/

3. Write these series in sigma notation.

(a) …1 3 5 7 11+ + + + + (b) …7 14 21 70+ + + + (c) 1 8 27 64 125+ + + + (d) … ( )n2 8 14 6 4+ + + + − (e) … n9 16 25 36 2+ + + + + (f) …1 2 3 50− − − − − (g) … ´3 6 12 3 2n+ + + +

(h) …121

41

5121+ + + +

(i) ( ) ( ) …a a d a d2+ + + + + + ( )a n d1+ −5 ?

(j) …a ar ar ar ar2 3+ + + + + n 1−

Arithmetic Series

In an arithmetic series each term is a constant amount more than the previous term. The constant is called the common difference .

EXAMPLES

1. Find the common difference of the series …5 9 13 17 21+ + + + +

Solution

We can see that the common difference between the terms is 4. To check this, we notice that , ,9 5 4 13 9 4 17 13 4− = − = − = and 21 17 4− = .

2. Find the common difference of the series …85 80 75 70 65+ + + + +

Solution

We can see that the common difference between the terms is 5− . To check this, we notice that , ,80 85 5 75 80 5 70 75 5− = − − = − − = − and 65 70 5− = − .

ch7.indd 267 6/29/09 10:01:49 PM


Generally, if , , , ,…T T T T Tn n1 2 3 1− are consecutive terms of an arithmetic series then .…d T T T T T Tn n2 1 3 2 1= − = − = = − −

If T 1 , T 2 and T 3 are consecutive terms of an arithmetic series then d T T T T2 1 3 2= − = −

EXAMPLES

1. If …x5 31+ + + is an arithmetic series, fi nd x .

Solution

For an arithmetic series,

i.e. 5 31

18

T T T T

x x

x

x

x

2 5 31

2 36

2 1 3 2

`

− = −− = −− =

==

2.

Evaluate (a) k if ( ) ( ) ( ) …k k k2 3 2 6 1+ + + + − + is an arithmetic series. Write down the fi rst 3 terms of the series. (b) Find the common difference (c) d .

Solution

For an arithmetic series, (a) T T T T2 1 3 2− = −

So ( ) ( ) ( ) ( )k k k k

k k k k

k k

k

k

3 2 2 6 1 3 2

3 2 2 6 1 3 2

2 3 3

0 3

3

+ − + = − − ++ − − = − − −

= −= −=

The series is (b) ( ) ( ) ( ) …k k k2 3 2 6 1+ + + + − + Substituting 3:k =

´

´

T k

T k

T k

2

3 2

5

3 2

3 3 2

11

6 1

6 3 1

17

1

2

3

= += +== += +== −= −=

The sequence is (c) 5 11 17 . . .+ + + 11 5 17 11 6− = − = So common difference 6d = .

Notice that 2

5 31x

+=

which is the average of T1 and T2. We call x the arithmetic mean.

ch7.indd 268 6/29/09 10:01:49 PM

269Chapter 7 Series

Terms of an arithmetic series

( ) ( ) ( ) [ ( ) ]… …a a d a d a d a n d2 3 1+ + + + + + + + + − + is an arithmetic series with fi rst term a , common difference d and n th term given by

( 1)T a n dn = + −

Proof

Let the fi rst term of an arithmetic series be a and the common difference d . Then fi rst term is a

second term is a d+ third term is 2a d+ fourth term is 3 ,a d+ and so on. The n th term is ( 1) .a n d+ −

EXAMPLES

1. Find the 20th term of the series … .3 10 17+ + +

Solution

, ,a d n3 7 20= = =

( )

( )

´

T a n d

T

1

3 20 1 7

3 19 7

136

n

20

= + −= + −= +=

2. Find an expression for the n th term of the series 2 8 14 . . . .+ + +

Solution

,a d2 6= =

( )

( )

T a n d

n

n

n

1

2 1 6

2 6 6

6 4

n = + −= + −= + −= −

3. Find the fi rst positive term of the series … .50 47 44− − − −

Solution

,a d50 3= − = For the fi rst positive term,

0T >n

CONTINUED

ch7.indd 269 6/29/09 10:01:50 PM


( )

( )

a n d

n

n

n

1 0

50 1 3 0

50 3 3 0

3 53 0

i.e. >>>>

+ −− + −

− + −−

n

n

3 53

1732

>

>

18n` = gives the fi rst positive term

( )

50 17 3´T 50 18 1 3

1

18 = − + −= − +=

So the fi rst positive term is 1.

4. The 5th term of an arithmetic series is 37 and the 8th term is 55. Find the common difference and the fi rst term of the series.

Solution

( )

( )

T a n d

T a d

1

5 1 37n

5

= + −= + − =

( )a d4 37 1i.e. + = (8 1) 55T a d8 = + − =

7 55 (2)a di.e. + =( ) ( ): a d2 1 4 37− + =

d

d

3 18

6

==

Put in ( ):d 6 1=

( )a

a

a

4 6 37

24 37

13

+ =+ =

= So 6 13.d aand= =

n must be a positive integer.

1. The following series are arithmetic. Evaluate all pronumerals .

(a) …y5 9+ + + (b) …x8 2+ + + (c) …x45 99+ + + (d) …b16 6+ + + (e) …x 14 21+ + +

(f) ( ) …x32 1 51+ − + + (g) ( ) …k3 2 3 21+ + + + (h) ( ) ( ) …x x x3 2 5+ + + + + (i) ( ) ( ) …t t t5 3 3 1− + + + + (j) ( ) ( ) ( ) …t t t2 3 3 1 5 2− + + + + +

7.4 Exercises

ch7.indd 270 6/29/09 10:01:50 PM

271Chapter 7 Series

2. Find the 15 th term of each series. (a) …4 7 10+ + + (b) …8 13 18+ + + (c) …10 16 22+ + + (d) …120 111 102+ + + (e) …3 2 7− + + +

3. Find the 100 th term of each series. (a) …4 2 8− + + + (b) …41 32 23+ + + (c) …18 22 26+ + + (d) …125 140 155+ + + (e) …1 5 9−− − −

4. What is the 25 th term of each series?

(a) …14 18 22− − − − (b) . . . …0 4 0 9 1 4+ + + (c) . . . …1 3 0 9 0 5+ + +

(d) …1 221 4+ + +

(e) …152 2 2

53+ + +

5. For the series ,…3 5 7+ + + write an expression for the n th term.

6. Write an expression for the n th term of the following series .

(a) …9 17 25+ + + (b) …100 102 104+ + + (c) …6 9 12+ + + (d) …80 86 92+ + + (e) …21 17 13− − − − (f) …15 10 5+ + +

(g) …87 1 1

81+ + +

(h) …30 32 34− − − − (i) . . . …3 2 4 4 5 6+ + +

(j) …21 1

41 2+ + +

7. Find which term of …3 7 11+ + + is equal to 111.

8. Which term of the series …1 5 9+ + + is 213?

9. Which term of the series …15 24 33+ + + is 276?

10. Which term of the series …25 18 11+ + + is equal to 73?−

11. Is zero a term of the series ?…48 45 42+ + +

12. Is 270 a term of the series ?…3 11 19+ + +


14. Find the fi rst value of n for which the terms of the series …100 93 86+ + + become less than 20.

15. Find the values of n for which the terms of the series …86 83 80− − − − are positive.

16. Find the fi rst negative term of …54 50 46+ + + .

17. Find the fi rst term that is greater than 100 in the series …3 7 11+ + + .

18. The fi rst term of an arithmetic series is -7 and the common difference is 8. Find the 100 th term of the series.

19. The fi rst term of an arithmetic series is 15 and the 3 rd term is 31.

Find the common difference (a) Find the 10 th term of the (b)

series.

20. The fi rst term of an arithmetic series is 3 and the 5 th term is 39. Find its common difference.

21. The 2 nd term of an arithmetic series is 19 and the 7 th term is 54. Find its fi rst term and common difference.

ch7.indd 271 6/29/09 10:01:51 PM


22. Find the 20 th term in an arithmetic series with 4 th term 29 and 10 th term 83.

23. The common difference of an arithmetic series is 6 and the 5 th term is 29. Find the fi rst term of the series.

24. If the 3 rd term of an arithmetic series is 45 and the 9 th term is 75, fi nd the 50 th term of the series.

25. The 7 th term of an arithmetic series is 17 and the 10 th term is 53. Find the 100 th term of the series.

26. (a) Show that …log log logx x x5 5

25

3+ + + is an arithmetic series.

(b) Find the 80 th term. (c) If 4,x = evaluate the 10 th term

correct to 1 decimal place.

27. (a) Show that . . .3 12 27+ + + is an arithmetic series.

(b) Find the 50 th term in simplest form.

28. Find the 25 th term of …log log log4 8 162 2 2+ + +

29. Find the 40 th term of …b b b5 8 11+ + +

30. Which term is 213 y of the series ?…y y y28 33 38+ + +

Partial sum of an arithmetic series

The sum of the fi rst n terms of an arithmetic series ( n th partial sum) is given by the formula:

( )S a ln2n = + where l nlast or th term=

Proof

Let the last or n th term be l .

`

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( )

( )

…

…

…

S a a d a d l

S l l d l d a

S a l a l a l a l

n a l

S n a l

2 1

2 2

2 1 2

2

n

n

n

n

= + + + + + += + − + − + += + + + + + + + + += +

= +

In general,

[ ( ) ]S n a n d2

2 1n = + −

ch7.indd 272 6/29/09 10:01:51 PM

273Chapter 7 Series

Proof

( )

( )

[ ( ) ]

[ ( ) ]

l n

l a n d

S n a l

n a a n d

n a n d

1

2

21

22 1

Since th term,

n`

== + −

= +

= + + −

= + − We use this formula when the n th term is unknown.

EXAMPLES

1. Evaluate .…9 14 19 224+ + + +

Solution

,a d9 5= = First we fi nd n .

( )

( )

( )

( )

´

T

T a n d

n

n

n

n

n

S n a l

S

224

1

224 9 1 5

9 5 5

5 4

220 5

44

2

244 9 224

22 233

5126

n

n

n

44

`

`

== + −= + −= + −= +==

= +

= +

==

2. For what value of n is the sum of n terms of …2 11 20+ + + equal to 618?

Solution

, ,

[ ( ) ]

[ ( ) ]

( )

( )

´

a d S

S n a n d

n n

n n

n n

n n

2 9 618

22 1

6182

2 2 1 9

1236 4 9 9

9 5

9 5

n

n

2

= = =

= + −

= + −

= + −= −= −

CONTINUED

ch7.indd 273 6/29/09 10:01:52 PM


( ) ( )

.

n n

n n

n

0 9 5 123612 9 103

12 11 4or

2

`

= − −= − += −

But n cannot be negative, so 12.n =

3. The 6th term of an arithmetic series is 23 and the sum of the fi rst 10 terms is 210. Find the sum of 20 terms.

Solution

( )

( )

[ ( ) ]

[ ( ) ]

T a n d

T a d

a d

S n a n d

S a d

1

6 1 23

5 23

22 1

210 2 10 2101

n

n

6

10

= + −= + − =

+ =

= + −

= + − =

(1)

( )

( )

a d

a d

5 2 9 210

2 9 42 2

+ =+ =

( ) ( ): ( )´ a d1 2 2 10 46 3+ =

( ) ( ):

( ):

( )

d

d

d

a

a

a

2 3 4

4

4 1

5 4 23

20 23

3

Substitute in

− − = −=

=+ =

+ ==

[ ( ) ( ) ]

[ ( )]

´

S2

20 2 3 20 1 4

10 6 19 4

10 82

820

20 = + −

= +==

4. Evaluate .r3 2

r 1

50

+=/

Solution

( ) ( ) ( ) ( )…

…

´ 1 ´ 2 ´ 3 ´ 50r3 2 3 2 3 2 3 2 3 2

5 8 11 152r 1

50

+ = + + + + + + + +

= + + + +=/ Arithmetic series with , , ,a d l n5 3 152 50= = = =

( )

( )

´

S n a l

S

2

250 5 152

25 157

3925

n

50

= +

= +

==

n must be a positive integer.

It would take a long time to add these up!

ch7.indd 274 6/29/09 10:01:52 PM

275Chapter 7 Series

1. Find the sum of 15 terms of the series.

(a) …4 7 10+ + + (b) …2 7 12+ + + (c) …60 56 52+ + +


(a) …1 7 13+ + + (b) …15 24 33+ + + (c) …95 89 83+ + +


(a) …2 5 12− + + + (b) …5 4 13− − −


(a) …50 44 38+ + + (b) …11 14 17+ + +

5. Evaluate (a) …15 20 25 535+ + + + (b) …9 17 25 225+ + + + (c) …5 2 1 91+ − − − (d) …81 92 103 378+ + + + (e) …229 225 221 25+ + + + (f) …2 6 14 94− + + + + (g) …0 9 18 216− − − − (h) …79 81 83 229+ + + + (i) …14 11 8 43+ + + −

(j) …121 1

43 2 25

41+ + + +

6. Evaluate

(a) n4 7n 1

20

−=/

(b) r5 3r 1

15

−=/

(c) r4 6r 3

20

−=/

(d) n5 3n 1

50

+=/

(e) n4 35

40

−/

7. How many terms of the series …45 47 49+ + + give a sum of 1365?

8. For what value of n is the sum of the arithmetic series …5 9 13+ + + equal to 152?


10. The sum of the fi rst 5 terms of an arithmetic series is 110 and the sum of the fi rst 10 terms is 320. Find the fi rst term and the common difference.

11. The sum of the fi rst 5 terms of an arithmetic series is 35 and the sum of the next 5 terms is 160. Find the fi rst term and the common difference.

12. Find S 25 , given an arithmetic series whose 8th term is 16 and whose 13th term is 81.

13. The sum of 12 terms of an arithmetic series is 186 and the 20th term is 83. Find the sum of 40 terms.


15. The sum of the fi rst 4 terms of an arithmetic series is 42 and the sum of the 3rd and 7th term is 46. Find the sum of the fi rst 20 terms.

16. (a) Show that ( ) ( ) ( ) …x x x1 2 4 3 7+ + + + + + is an arithmetic series.

(b) Find the sum of the fi rst 50 terms of the series.

7.5 Exercises

ch7.indd 275 6/29/09 10:01:53 PM


EXAMPLE

Find the common ratio of the series …3 6 12+ + +

Solution

By looking at this sequence, each term is multiplied by 2 to get the next term. If you can’t see this, we divide the terms as follows:

36

612 2= =

17. The 20th term of an arithmetic series is 131 and the sum of the 6th to 10th terms inclusive is 235. Find the sum of the fi rst 20 terms.

18. The sum of 50 terms of an arithmetic series is 249 and the sum of 49 terms of the series is 233. Find the 50th term of the series.

19. Prove that T S Sn n n 1= − − for any series.

20. Find the sum of all integers between 1 and 100 that are not multiples of 6.

Class Investigation

Look at the working out of question 14 in the previous set of exercises. Why are there two values for n ?

DID YOU KNOW?

Here is an interesting series:

4 1

131

51

71

91 … .π

= −− + − +

Gottfried Wilhelm Leibniz (1646–1716) discovered this result. It is interesting that while π is an irrational number, it can be written as the sum of rational numbers.

Geometric Series

In a geometric series each term is formed by multiplying the preceding term by a constant. The constant is called the common ratio.

ch7.indd 276 6/29/09 10:01:54 PM

277Chapter 7 Series

If …T T T1 2 3+ + + is a geometric series then

rT

T

T

T

1

2

2

3= =

In general, if …T T T T T Tn n1 2 3 4 1+ + + + + +− is a geometric series then

… .rT

T

T

T

T

T

T

T

n

n

1

2

2

3

3

4

1

= = = = =−

When r is negative, the signs alternate.

x is called the geometric mean.

EXAMPLES

1. Find x if …x5 45+ + + is a geometric series.

Solution

For a geometric series T

T

T

T

1

2

2

3=

±±

xx

x

x

545

225

22515

2

=

=

==

…( ) .

, …( ).

x r

x r

15 5 15 45 3

15 5 15 45 3

If the series is

If the series is

= + + + == − − + − = −

2. Is …41

61

181+ + + a geometric series?

Solution

÷

´

T

T

61

41

61

14

32

1

2 =

=

=

÷

≠

´

T

T

T

T

181

61

181

16

31

2

3

1

2

=

=

=

∴ the series is not geometric.

ch7.indd 277 6/29/09 10:01:54 PM


Terms of a geometric series

… …a ar ar ar arn2 3 1+ + + + + +− is a geometric series with fi rst term a , common ratio r and n th term given by T arn

n 1= −

Proof

Let the fi rst term of a geometric series be a and the common ratio be r . Then fi rst term is a second term is ar third term is ar2 fourth term is ,ar3 and so on n th term is ar n 1−

EXAMPLES

1.

Find the 10 th term of the series (a) …3 6 12+ + + Write an expression for the (b) n th term of the series .

Solution

This is a geometric series with (a) 3a = and 2r = . We want the 10 th term, so 10n = .

( )

´

T ar

T 3 2

3 21536

nn 1

1010 1

9

====

−

−

(b) n 1−( )

T ar

3 2n

n 1==

−

2. Find the common ratio of …32

154

758+ + + and hence fi nd the 8th

term in index form.

Solution

÷ ÷

´

r154

32

758

154

154

23

52

= =

=

=

c m

ch7.indd 278 6/29/09 10:01:55 PM

279Chapter 7 Series

´

T ar

T32

52

32

52

3 52

n

8

8 1

7

7

8

=

=

=

=

−

n 1−

ccmm

3. Find the 10th term of the series …5 10 20− + − +

Solution

, ,

( )

( )( )

a r n

T ar

T

5 2 10

5 2

5 25 512

2560

nn 1

1010 1

9

= − = − =

== − −= − −= − −=

−

−

4. Which term of the series …4 12 36+ + + is equal to 78 732?

Solution

This is a geometric series with 4a = and 3r = . The n th term is 78 732 .

n 1−( )

( )log log

log

log

log

T ar

n

n

n

n

78732 4 3

19 683 3

19 683 31 3

3

19 6831

9 1

10

nn

n

n

1

1

10 101

10

10

===== −

= −

= −=

−

−

−

So the 10 th term is 78 732 .

Terms can become very large in geometric series.

You could use logarithms to the base e or ln instead of logarithms to the base 10.

CONTINUED

ch7.indd 279 6/29/09 10:01:55 PM


5. The third term of a geometric series is 18 and the 7th term is 1458. Find the fi rst term and the common ratio.

Solution

T ar

T ar

ar

18

18

n

3

== =

=

3 1−

2

n 1−

(1)

T ar

ar

1458

14587

7 1

6

= ==

−

(2)

( ) ( ):

( ):

( )

±±±

±

¸ r

r

r

aa

a

2 1 81

813

3 1

3 189 18

2

Substitute into

4

4

2

====

===

6. Find the fi rst value of n for which the terms of the series …51 1 5+ + +

exceed 3000.

Solution

,a r51 5= =

For terms to exceed 3000,

( )

( )

.

log loglog log

log

log

log

log

T

ar

n

n

n

n

n

3000

3000

51 5 3000

5 15 000

5 15 0001 5 15 000

15

15 000

5

15 0001

6 97

7

>

>

>

>>>

>

>

>

n

n

n

n

1

1

1

`

−

−

+

=

−

−

−

n 1−

So the 7th term is the fi rst term to exceed 3000.

Always divide the equations in geometric series to eliminate a.

n must be an integer.

ch7.indd 280 7/6/09 10:22:04 PM

281Chapter 7 Series

1. Are the following series geometric? If they are, fi nd the common ratio.

(a) …5 20 60+ + +

(b) …4 3 241− + − +

(c) …43

143

493+ + +

(d) …7 565 3

31+ + +

(e) …14 42 168− + − +

(f) …131

98

278+ + +

(g) . . . …5 7 1 71 0 513+ + +

(h) …241 1

207

10081− + +

(i) …63 9 187+ + +

(j) …187 15 120− + − +

2. Evaluate all pronumerals in these geometric series .

(a) …x4 28+ + + (b) …y3 12− + + + (c) …a2 72+ + + (d) …y 2 6+ + + (e) …x 8 32+ + + (f) …p5 20+ + + (g) …y7 63+ + + (h) …m3 12− + − + (i) ( ) …x3 4 15+ − + + (j) ( ) …k3 1 21+ − + +

(k) ,…t41

91+ + +

(l) ,…t31

34+ + +

3. Write an expression for the n th term of the following series .

(a) …1 5 25+ + + (b) . . …1 1 02 1 0404+ + + (c) …1 9 81+ + + (d) …2 10 50+ + + (e) …6 18 54+ + + (f) …8 16 32+ + +

(g) …41 1 4+ + +

(h) …1000 100 10− + − (i) …3 9 27− + − +

(j) …31

152

754+ + +

4. Find the 6 th term of each series. (a) …8 24 72+ + + (b) …9 36 144+ + + (c) …8 32 128− + + (d) …1 5 25− + − +

(e) …32

94

278+ + +

5. What is the 9 th term of each series?

(a) …1 2 4+ + + (b) …4 12 36+ + + (c) . . …1 1 04 1 0816+ + + (d) …3 6 12− + − +

(e) …43

83

163− + −

6. Find the 8 th term of each series. (a) …3 15 75+ + + (b) . . . …2 1 4 2 8 4+ + + (c) …5 20 80− + −

(d) …21

103

509− + − +

(e) …18147 2

2710 3

95+ + +

7. Find the 20th term of each series, leaving the answer in index form.

(a) …3 6 12+ + + (b) …1 7 14+ + + (c) . . . …1 04 1 04 1 042 3+ + +

(d) …41

81

161+ + +

(e) …43

169

6427+ + +

7.6 Exercises

ch7.indd 281 6/29/09 10:01:56 PM


8. Find the 50 th term of …1 11 121+ + + in index form.

9. Which term of the series …4 20 100+ + + is equal to 12 500?

10. Which term of …6 36 216+ + + is equal to 7776?


12. Which term of …3 21 147+ + + is equal to 352 947?

13. Which term of the series

…8 4 2− + − is ?128

1

14. Which term of …54 18 6+ + +

is ?243

2

15. Find the value of n if the n th term of the series

…2 121 1

81− + − + is

12881− .

16. The fi rst term of a geometric series is 7 and the 6 th term is 1701. Find the common ratio.

17. The 4 th term of a geometric series is −648 and the 5 th term is 3888.

Find the common ratio .(a) Find the 2 nd term. (b)

18. The 3 rd term of a geometric series

is 52 and the 5 th term is 1

53 . Find

its fi rst term and common ratio.

19. Find the value of n for the fi rst term of the series …5000 1000 200+ + + that is less than 1.

20. Find the fi rst term of the series

…72

76 2

74+ + + that is greater

than 100.

Partial sum of a geometric series

The sum of the fi rst n terms of a geometric series ( n th partial sum) is given by the formula:

( )| |

( )| |

Sr

a rr

Sr

a rr

11

1

11

1

for

for

>

<

n

n

=−

−

=−−

n

n

Proof

The sum of a geometric series can be written . . .S a ar ar arn = + + + +2 n 1− (1) . . .rS ar ar ar arn = + + + +3 n2 (2)

(1) − (2):

( )

(

(

S r a ar

a r

Sr

a r

1

1

11

n

n

− = −= −

=−−

n

n)n)

(2) − (1) gives the formula

( )

Sr

a r11

n =−

−n

You studied absolute values in the Preliminary Course | r | < 1 means 1 r 1.< <− What does | r | 1> mean?

The two formulae are the same, just rearranged.

ch7.indd 282 6/29/09 10:01:57 PM

283Chapter 7 Series

EXAMPLES

1. Find the sum of the fi rst 10 terms of the series …3 12 48+ + +

Solution

This is a geometric series with ,a r3 4= = and we want 10n = . Since 1r > we use the fi rst formula.

( )

( )

( )

Sr

a r11

4 13 4 1

33 4 1

4 11 048 575

n

10

10

10

=−

−

=−

−

=−

= −=

n

2. Evaluate .2n

3

11

/

Solution

2 2 2 . . . 2

8 16 32 . . . 2048

2n

3

113 4 5 11= + + + +

= + + + +

/

Geometric series with , ,a r n8 2 9= = =

( )

( )

( )

´

Sr

a r

S

11

2 18 2 1

18 512 1

8 511

4088

n

n

9

9

=−

−

=−

−

=−

==

3. Evaluate .…60 20 632

8120+ + + +

Solution

, ,a r T

T ar

6031

8120

8120

n

n

= = =

= =n 1−

CONTINUED

Can you see why n 9?= Count the terms carefully.

First fi nd n.

ch7.indd 283 6/29/09 10:01:58 PM


(

´

n

n

Sr

a r

S

6031

8120

31

2431

31

2431

31

31

1 5

6

11

131

60 131

32

60 1729

1

60729728

23

898171

n

n

n

n

n

1

1

1

1 5

6

6

`

=

=

=

=

− ==

=−−

=−

−

=−

=

=

−

−

−

−

n)

cc

c

c

c

mm

m

m

m

= G

4. The sum of n terms of …1 4 16+ + + is 21 845. Find the value of n .

Solution

, ,

( )

( )

a r S

Sr

a r

n

1 4 21845

11

218454 1

1 4 1

34 1

65 535 4 1

65 536 4

4 48

n

n

n

n

n

n

n8

`

= = =

=−

−

=−

−

= −

= −===

n

So 8 terms gives a sum of 21 845 .

ch7.indd 284 6/29/09 10:01:58 PM

285Chapter 7 Series

1. Find the sum of 10 terms of the series .

(a) …6 24 96+ + + (b) …3 15 75+ + +


(a) …1 7 49− + − + (b) …8 24 72+ + +


(a) …4 8 16+ + +

(b) …43

83

163− + −

4. Evaluate (a) …2 10 50 6250+ + + +

(b) …18 9 421

649+ + + +

(c) …3 21 147 7203+ + + +

(d) …43 2

41 6

43 182

41+ + + +

(e) …3 6 12 384− + − + +

5. Evaluate

(a) 2n

n

1

81

=

−/

(b) 31

nn

1

6

=/

(c) 5n

n

2

103

=

−/

(d) 21

rr

1

8

1=

−/

(e) 4k

k

1

71

=

+/

6. Find the 9th term (a) sum of 9 terms of the (b)

series …7 14 28+ + +

7. Find the sum of 30 terms of the series . . . ,1 09 1 09 1 092 3 f+ + + correct to 2 decimal places.

8. Find the sum of 25 terms of the series . . ,1 1 12 1 122 f+ + + correct to 2 decimal places.

9. Find the value of n if the sum of n terms of the series …11 33 99+ + + is equal to 108 251.

10. How many terms of the series

…21

41

81+ + + give a sum of

?10241023

11. The common ratio of a geometric series is 4 and the sum of the fi rst 5 terms is 3069. Find the fi rst term.

12. Find the number of terms needed to be added for the sum to exceed 1 000 000 in the series …4 16 64+ + +

13. Lucia currently earns $25 000. Her wage increases by 5% each year. Find

her wage after 6 years (a) her total earnings (before tax) (b) in 6 years.

14. Write down an expression for the series … ( )2 10 50 2 5− + − + − k 1−

in sigma notation (a) as a sum of (b) n terms.

15. Find the sum of the fi rst 10 terms of the series

[ ( )]… …n3 7 13 2 2 1n+ + + + + − +

7.7 Exercises

ch7.indd 285 6/29/09 10:01:58 PM


Limiting sum (sum to infi nity)

In some geometric series the sum becomes very large as n increases. For example,

…2 4 8 16 32 64 128+ + + + + + +

We say that this series diverges, or it has an infi nite sum. In some geometric series, the sum does not increase greatly after a few

terms. We say this series converges and it has a limiting sum (the sum is limited to a fi nite number).

EXAMPLE

For the series …21

41

81

1612 1+ + + + + + , notice that after a while the

terms are becoming closer and closer to zero and so will not add much to the sum of the whole series.

Evaluate, correct to 4 decimal places, the sum of (a) (i) 10 terms (ii) 20 terms. Estimate its limiting sum. (b)

PUZZLES

A poor girl saved a rich king from drowning one day. The king offered the 1. girl a reward of sums of money in 30 daily payments. He gave the girl a choice of payments:Choice 1: $1 the fi rst day, $2 the second day, $3 the third day and so on.Choice 2: 1 cent the fi rst day, 2 cents the second day, 4 cents the third day and so on, the payment doubling each day. How much money would the girl receive for each choice? Which plan would give the girl more money?Can you solve 2. Fibonacci’s Problem?A man entered an orchard through 7 guarded gates and gathered a certain number of apples. As he left the orchard he gave the guard at the fi rst gate half the apples he had and 1 apple more. He repeated this process for each of the remaining 6 guards and eventually left the orchard with 1 apple. How many apples did he gather? (He did not give away any half apples.)

This is a hard one!

Can you estimate the sum of this series?

ch7.indd 286 6/29/09 10:01:59 PM

287Chapter 7 Series

Solution

(a) ,a r221= =

(i) 10n =

(

.

Sr

a r11

121

2 121

21

2 121

3 9961

n

n

10

10

=−−

=−

−

=−

=

)

ce

c

m o

m

(ii) 20n =

(

.

Sr

a r11

121

2 121

21

2 121

4 0000

n

20

20

=−−

=−

−

=−

=

)n

ce

c

m o

m

The limiting sum is 4. (b)

Can you see why the series …2 4 8 16+ + + + diverges and the series

…21

412 1+ + + + converges?

The difference is in the common ratio. Only geometric series with common ratios | |r 1< will converge and have a limiting sum.

Sr

a1∞ =

−

| | 1r < is the necessary condition for the limiting sum to exist.

ch7.indd 287 6/29/09 10:01:59 PM


Proof

(

Sr

a r11

n =−− )n

For | |r 1< , as n increases, r n decreases and approaches zero

e.g. When :r21=

21

1024110

=c m and 21

1 048 576120

=c m

`

( )S

ra

ra11 0

1

∞ =−−

=−

EXAMPLES

1. Find the limiting sum of …2 121

41+ + + +

Solution

In the previous example, we guessed that the limiting sum was 4. Here we will use the formula to fi nd the limiting sum.

,a r221= =

´

Sr

a1

121

2

212

212

4

∞ =−

=−

=

=

=

2. Find the sum of the series …6 232+ + +

Solution

,a r662

232

31= = = =

ch7.indd 288 6/29/09 10:02:00 PM

289Chapter 7 Series

´

Sr

a1

131

6

326

623

9

∞ =−

=−

=

=

=

3. Which of the following series have a limiting sum?

(a) …43

1615 1

6411+ + +

(b) …100 50 25+ + +

(c) …3 241 1

1611+ + +

Solution

(a)

r

43

1615

1615

16411

141

1>

= =

=

So this series does not have a limiting sum.

(b)

r10050

5025

21

1<

= =

=

So this series does have a limiting sum.

(c) r3

241

241

11611

43

1<

= =

=

So this series does have a limiting sum.

CONTINUED

ch7.indd 289 6/29/09 10:02:00 PM


4. Evaluate 232∞ n

n 2=c m/

Solution

…

…

…

232 2

32 2

32 2

32

294 2

278 2

8116

98

2716

8132

∞ n

n 2

2 3 4

= + + +

= + + +

= + + +

=c c c c

c c cm m m m

m m m/

,a r98

32= =

´

Sr

a1

132

98

3198

98

13

232

∞ =−

=−

=

=

=

1. Which of the following series have a limiting sum? Find its limiting sum where it exists.

(a) …9 3 1+ + +

(b) …41

21 1+ + +

(c) …16 4 1− + −

(d) …32

97

5449+ + +

(e) …132

94+ + +

(f) …85

81

401+ + +

(g) …6 36 216− + − +

(h) …241 1

87 1

4827− + − +

(i) …91

61

41+ + +

(j) …254

258− + −

2. Find the limiting sum of each series.

(a) …40 20 10+ + + (b) …320 80 20+ + + (c) …100 50 25− + −

(d) …6 3 121+ + +

(e) …52

356

24518+ + +

(f) …72 24 8− + −

(g) …12 231− + − +

(h) …43

21

31− + −

(i) …12 9 643+ + +

(j) …32

125

9625− + − +

7.8 Exercise s

ch7.indd 290 6/29/09 10:02:01 PM

291Chapter 7 Series

3. Find the difference between the limiting sum and the sum of 6 terms of

(a) …56 28 14− + − (b) …72 24 8+ + +

(c) …151

251+ + +

(d) …21

41

81+ + +

(e) …141

1615

6445+ + +

4. Evaluate

(a) 51∞ r

r

1

1

−

=c m/

(b) 72∞ n

n 1=c m/

(c) 41∞ k

3c m/

(d) 53∞ n

n

1

1

−

=c m/

(e) 32∞ p

p

2

2

−

=c m/

(f) 215

∞ n

n 1=c m/

(g) 612

∞ n

1c m/

(h) 324

∞ n

n

1

3−

−

=c m/

(i) 912

∞ n 2

2−

−

c m/

(j) 52

21∞ n

n 0=c m/

5. A geometric series has limiting

sum 6 and common ratio .31

Evaluate the fi rst term of the series.

6. A geometric series has a limiting sum of 5 and fi rst term 3. Find the common ratio.

7. The limiting sum of a geometric

series is 9 31 and the common

ratio is 52 . Find the fi rst term of

the series.

8. A geometric series has limiting sum 40 and its fi rst term is 5. Find the common ratio of the series.

9. A geometric series has limiting

sum 652− and a fi rst term of .8−

Find its common ratio.

10. The limiting sum of a geometric

series is 103− and its fi rst term

is 21− . Find the common ratio of

the series.

11. The second term of a geometric series is 2 and its limiting sum is 9. Find the values of fi rst term a and common ratio r .

12. A geometric series has 3 rd term 12 and 4 th term–3. Find a , r and the limiting sum.

13. A geometric series has 2 nd term

32 and 4 th term

278 . Find a , r and

its limiting sum.

14. The 3 rd term of a geometric series

is 54 and the 6 th term is .1112583

Evaluate a , r and the limiting sum.

15. The 2 nd term of a geometric

series is 154 and the 5 th term is

40532 . Find the values of a and r

and its limiting sum.

16. The limiting sum of a geometric series is 5 and the 2 nd term is

151 . Find the fi rst term and the

common ratio.

17. The series …x xx4 16

+ + + has

a limiting sum of 87 . Evaluate x.

ch7.indd 291 6/29/09 10:02:01 PM


18. (a) For what values of k does the limiting sum exist for the series …?k k k2 3+ + +

Find the limiting sum of the (b)

series when k32= − .

Evaluate (c) k if the limiting sum of the series is 3.

19. (a) For what values of p will the limiting sum exist for the series ?…p p1 2 4 2− + −

Find the limiting sum when (b)

p51= .

Evaluate (c) p if the limiting sum

of the series is 87.

20. Show that in any geometric series the difference between the limiting sum and the sum of

n terms is .r

ar1 −

n

Applications of Series

General applications

Arithmetic and geometric series are useful in solving real life problems involving patterns. It is important to choose the correct type of series and know whether you are asked to fi nd a term or a sum of terms.

EXAMPLES

1. A stack of cans on a display at a supermarket has 5 cans on the top row. The next row down has 2 more cans and the next one has 2 more cans and so on.

Calculate the number of cans in the 11 th row down. (a) If there are 320 cans in the display altogether, how many rows are (b)

there?

Solution

The fi rst row has 5 cans, the 2 nd row has 7 cans, the 3 rd row 9 cans and so on. This forms an arithmetic series with 5a = and .d 2=

For the 11 th row, we want (a) .n 11=

( )

( ) ´´

T a n d

T

1

5 11 1 2

5 10 2

25

n

11

= + −= + −= +=

So there are 25 cans in the 11 th row.

ch7.indd 292 6/29/09 10:02:02 PM

293Chapter 7 Series

If there are 320 cans altogether, this is the sum of cans in all rows. (b)

[ ( ) ]

[ ( ) ]

[ ]

[ ]

( )( )

´ ´

S

S n a n d

n n

n n

n n

n n

n nn n

320

22 1

3202

2 5 1 2

210 2 2

22 8

4

0 4 32016 20

So n

n

2

2

=

= + −

= + −

= + −

= +

= += + −= − +

,

,

n n

n n

16 0 20 0

16 20

− = + == = −

Since n must be a positive integer, then 16n = .

There are 16 rows of cans.

2. A layer of tinting for a car window lets in 95% of light. What percentage of light is let in by (a) (i) 2 layers (ii) 3 layers (iii) 10 layers

of tinting? How many layers will let in 40% of light? (b)

Solution

(i) 1 layer lets in 95% of light .(a) So 2 layers let in 95% × 95% of light .

% % . .

.

. %

´ 5 ´95 95 0 9 0 95

0 9025

90 25

===

So 2 layers let in 90.25% of light . (ii) 1 layer lets in 95% or 0.95 of light . 2 layers let in 0.95 × 0.95 or 0.95 2 of light . 3 layers let in 0.95 2 × 0.95 or 0.95 3 of light .

. .

. %0 95 0 857

85 7

3 ==

So 3 layers let in 85.7% of light .

To convert a percentage to a decimal, divide by 100 and to convert a decimal to a percentage, multiply

by 100.

CONTINUED

ch7.indd 293 6/29/09 10:02:03 PM


(iii) The number of layers forms the geometric series . . . …0 95 0 95 0 952 3+ + + with . , .a r0 95 0 95= = . For 10 layers, 10n =

. ( . )

. ( . )

.

.

. %

T ar

T 0 95 0 95

0 95 0 95

0 950 5987

59 87

–

–n

n 1

1010 1

9

10

======

So 10 layers let in 59.87% of light.

We want to fi nd (b) n when the n th term is 40% or 0.4 .

. . ( . )

.

. ..

.

.

.

log loglog

log

log

T ar

n

n

n

0 4 0 95 0 95

0 95

0 4 0 950 95

0 95

0 4

17 9

nn

n

n

1

=====

=

=

−

n 1−

So around 18 layers of tinting will let in 40% of light.

The limiting sum can also be used to solve various problems.

EXAMPLES

1. Write 0.5• as a fraction .

Solution

. . …

…

0 5 0 5555555

105

1005

10005

•=

= + + +

This is a geometric series with a105

21= = and r

101= .

Sr

a1

1101

21

10921

∞ =−

=−

=

0.5• is called a recurring

decimal.

ch7.indd 294 6/29/09 10:02:04 PM

295Chapter 7 Series

´21

910

95

=

=

2. A ball is dropped from a height of 1 metre and bounces up to 31 of

its height. It continues bouncing, rising 31 of its height on each bounce

until it eventually reaches the ground. What is the total distance through which it travels?

Solution

1 m

13

m 13

m19

m 19

m

Notice that there is a series for the ball coming downwards and another series upwards. There is more than one way of calculating the total distance. Here is one way of solving it.

Total distance 1

31

31

91

91

271

271

1 231

91

271

f

f

= + + + + + + +

= + + + +c m

…31

91

271+ + + is a geometric series with a

31= and

31r =

´

Sr

a1

131

31

3231

31

23

21

∞ =−

=−

=

=

=

Total distance 1 221

1 1

2

= +

= +=

c m

So the ball travels 2 metres altogether.

Can you fi nd the total distance a different way?

ch7.indd 295 6/29/09 10:02:05 PM


Investigation

1. In the second example above, in theory will the ball ever stop?

2. Kim owes $1000 on her credit card. If she pays back 10% of the amount owing each month, she will never fi nish paying it off. Is this true or false?

1. A market gardener plants daffodil bulbs in rows, starting with a row of 45 bulbs. Each successive row has 5 more bulbs.

Calculate the number of bulbs (a) in the 34th row.

Which row would be the (b) fi rst to have more than 100 bulbs in it?

The market gardener plants (c) 10 545 bulbs. How many rows are there?

2. A stack of logs has 1 on the top, then 3 on the next row down, and each successive row has 2 more logs than the one on top of it.

How many logs are in the (a) 20th row?

Which row has 57 logs? (b) If there are 1024 logs (c)

altogether, how many rows are in the stack?

3. A set of books is stacked in layers, where each layer contains three books fewer than the layer below. There are 6 books in the top layer, 9 in the next layer and so on. There are n layers altogether.

Write down the number of (a) books in the bottom layer.

Show that there are (b) ( 3)n n23 +

books in the stack altogether.

4. A painting appreciates (increases its value) by 16% p.a. It is currently worth $20 000.

How much will it be worth in (a) (i) 1 year? (ii) 2 years? (iii) 3 years? How much will it be worth in (b)

11 years? How long will it take for it to (c)

be worth $50 000?

5. Water evaporates from a pond at an average rate of 7% each week.

What percentage of water is (a) left in the pond after

(i) 1 week? (ii) 2 weeks? (iii) 3 weeks? What percentage is left after (b)

15 weeks? If there was no rain, (c)

approximately how long would it take for the pond to only have 25% of its water left?

7.9 Exercise s

ch7.indd 296 6/29/09 10:02:06 PM

297Chapter 7 Series

6. A timber fence is to be built on sloping land, with the shortest piece of timber 1.2 m and the longest 1.8 m. There are 61 pieces of timber in the fence.

What is the difference in (a) height between each piece of timber?

What length of timber is (b) needed for the fence altogether?

7. A logo is made with vertical lines equally spaced as shown. The shortest line is 25 mm, the longest is 217 mm and the sum of the lengths of all the lines is 5929 mm.

25 mm

217 mm

How many lines are in the (a) logo?

Find the difference in length (b) between adjacent lines.

8. In a game, a child starts at point P and runs and picks up an apple 3 m away. She then runs back to P and puts the apple in a bucket. The child then runs to get the next apple 6 m away, and runs back to P to place it in the bucket. This continues until she has all the apples in the bucket.

How far does the child run to (a) get to the k th apple?

How far does the child run (b) to fetch all k apples, including return trips to P ?

The child runs 270 m to fetch (c) all apples and return them to the bucket. How many apples are there?

9. The price of shares in a particular company is falling by an average of 2% each day.

What percentage of their (a) value are they after 2 days?

How many days will it take (b) for the shares to halve in value?

After how many days will (c) the shares be worth 25% of their value?

10. A southern brown bandicoot population in WA is decreasing by 5% each year.

What percentage of the (a) population is left after 5 years?

After how many years will the (b) population be only 50%?

How many years will it take (c) for the population to decrease by 80%?

ch7.indd 297 6/29/09 10:02:09 PM


11. Write each recurring decimal as a fraction.

(a) 0.4•

(b) 0.7•

(c) 1.2•

(d) 0.25• •

(e) 2.81

• •

(f) 0.23•

(g) 1.47•

(h) 1.015•

(i) 0.132• •

(j) 2.361

• • •

12. A frog jumps 0.5 metres. It then jumps 0.1 m and on each subsequent jump, travels 0.2 m of the previous distance. Find the total distance through which the frog jumps.

13. A tree 3 m high grows by another

252 m after 5 years, then by

another 12523 m after 5 more years

and so on, each year growing 54

more than the previous year’s growth. Find the ultimate height of the tree.

14. An 8 cm seedling grows by 454

cm in the fi rst week, and then

keeps growing by 53 of its previous

week’s growth. How tall will it grow?

15. An object rolls 0.5 m in the fi rst second. Then each second

after, it rolls by 65 of its previous

roll. Find how far it will roll altogether.

16. A lamb grows by 52 of its previous

growth each month. If a lamb is 45 cm tall,

how tall will it be after (a) 6 months?

what will its fi nal height be? (b)

17. A 100 m cliff erodes by 72 of its

height each year. What will the height of the (a)

cliff be after 10 years? After how many years will the (b)

cliff be less than 50 m high?

18. An elastic string drops down 60 cm and then bounces back

to 32 of its initial height. It keeps

bouncing, each time rising back

to 32 of its previous height. What

is the total distance through which the string travels?

19. Mary bounces a ball, dropping it from 1.5 m on its fi rst bounce.

It then rises up to 52 of its

height on each bounce. Find the distance through which the ball travels.

ch7.indd 298 6/29/09 10:02:17 PM

299Chapter 7 Series

21. Kate receives an email chain letter that she is asked to send to 8 friends. If Kate forwards this email on to 8 friends and each of them sends it on to 8 friends, and so on,

describe the number of (a) people receiving the email as a sequence (including Kate’s email)

how many people would (b) receive the email when it is sent for the 9 th time?

how many people would (c) have received the email altogether if it is sent 9 times?

20. A roadside wall has a zig zag pattern on it as shown.

The two longest lines are each 2 m long, then the next two lines are

143 m long and each subsequent pair of lines are

87 of the length of the

previous lines. Find the total distance of the lines.

Compound interest

Compound interest refers to when the interest earned on an investment is then added to the amount of the investment. This means that each amount of interest earned is calculated on the previous interest as well as the investment.

Money that earns compound interest grows according to a geometric series.

EXAMPLE

An amount of $2000 is invested at the rate of 6% p.a. Find the amount in the bank at the end of n years.

Solution

After 1 year: The amount in the bank is the original amount plus the interest of 6% earned on the amount.

Amount 1 $2000 6% $2000

$ ( %)

$ ( . )

$ ( . )

2000 1 6

2000 1 0 06

2000 1 06

of= += += +=

CONTINUED

You may remember that p.a. means per annum or

annually.

ch7.indd 299 6/29/09 10:02:26 PM


After 2 years: The amount in the bank is the previous amount plus the interest of 6% earned on this amount.

Amount Amount % of amount

$ ( . ) % of $ ( . )

2 1 6 1

2000 1 06 6 2000 1 06

= += +

2

$2000(1.06) (1 6%)

$2000(1.06)(1 0.06)

$ ( . )( . )

$ ( . )

2000 1 06 1 06

2000 1 06

= += +=

=

Similarly after 3 years the amount is $2000(1.06) 3 and so on. The amount after n years will be n$2000( ) ..1 06

Factorise using a common factor of $2000(1.06).

Factorise using a common factor of P (1 r)+ .

In general the compound interest formula is

n(1 )P

r

n

A P r whereprincipal (initial amount)

interest rate as a decimal

number of time periods

= +===

Proof

Let $ P be invested at the rate of r p.a. compound interest for n years where r is a decimal.

After 1 year:

Amount of

( )

P r P

P r1

= += +

After 2 years: Amount ( ) of ( )P r r P r1 1= + + +

( ) ( )

( )

P r r

P r

1 1

1 2

= + += +

After 3 years:

2Amount ( ) of ( )

( ) ( )

( )

P r r P r

P r r

P r

1 1

1 1

1

2

2

3

= + + += + += +

Following this pattern, the amount after n years will be n( ) .P r1 +

ch7.indd 300 6/29/09 10:02:27 PM

301Chapter 7 Series

EXAMPLES

1. Find the amount that will be in the bank after 6 years if $2000 is invested at 12% p.a. with interest paid

yearly (a) quarterly (b) monthly .(c)

Solution

2000P = (a)

n

6

%

.

( )

( . )

( . ).

r

n

A P r

12

0 12

6

1

2000 1 0 12

2000 1 123947 65

6

===

= += +==

So the amount is $3947.65.

(b) For quarterly interest, it is divided into 4 amounts each year.

0.12

. quarterly

.

¸r

0 12 4

0 03

p.a.===

Also the interest is paid in 4 times a year.

quarters

( )

( . )

( . ).

ń

A P r

6 4

24

1

2000 1 03

2000 1 034065 59

0

n

24

24

==

= += +==


(c) For monthly interest, it is divided into 12 amounts each year.

0.12

. monthly

.

¸r

0 12 12

0 01

p.a.===

Also the interest is paid in 12 times a year.

72

( )

( . )

( . ).

ń

A P r

6 12

72

1

2000 1 0 01

2000 1 014094 20

monthsn

72

==

= += +==


Always round money off to 2 decimal places (to the

nearest cent).

CONTINUED

ch7.indd 301 6/29/09 10:02:28 PM


2. Geoff wants to invest enough money to pay for a $10 000 holiday in 5 years’ time. If interest is 8% p.a., how much does Geoff need to invest now?

Solution

, % .A r10 000 8 0 08= = = and 5n = We want to fi nd the principal P .

( )

( . )

( . )

..

A P r

P

P

P

P

1

10 000 1 0 08

1 08

1 0810 000

6805 83

n

5

5

5

= += +=

=

=

So Geoff will need to invest $6805.83 now.

3. An amount of $1800 was invested at 6% p.a. and is now worth $2722.66. For how many years was the money invested if interest is paid twice a year?

Solution

, .P A1800 2722 66= = Interest is paid twice a year .

% .

.

.

¸r 6 0 06

0 06 2

0 03

p.a.

twice a year

= ===

We want to fi nd n

( )

. ( . )

( . ). .

. .

. ..

.

.

log loglog

log

log

A P r

n

n

n

1

2722 66 1800 1 0 03

1800 1 03

18002722 66 1 03

1 5126 1 03

1 5126 1 031 03

1 03

1 5126

14

n

n

n

n

n

n

= += +=

=

===

=

=

Since interest is paid in twice a year, the number of years will be 14 2¸ . So the money was invested for 7 years.

ch7.indd 302 6/29/09 10:02:29 PM

303Chapter 7 Series

1. Find the amount of money in the bank after 10 years if

$500 is invested at 4% p.a. (a) $7500 is invested at 7% p.a. (b) $8000 is invested at 8% p.a. (c) $5000 is invested at 6.5% p.a. (d) $2500 is invested at 7.8% p.a. (e)

2. Sam banks $1500 where it earns interest at the rate of 6% p.a. Find the amount after 5 years if interest is paid

annually (a) twice a year (b) quarterly. (c)

3. Chad banks $3000 in an account that earns 5% p.a. Find the amount in the bank after 10 years if interest is paid

quarterly (a) monthly. (b)

4. I put $350 in the bank where it earns interest of 8% p.a. with interest paid annually. Find the amount there will be in the account after 2 years.

5. How much money will there be in an investment account after 3 years if interest of 4.5% p.a. is paid twice a year on $850?

6. Find the amount of money there will be in a bank after 8 years if $1000 earns interest of 7% p.a. with interest paid twice a year.

7. Abdul left $2500 in a building society account for 4 years, with interest of 5.5% p.a. paid yearly. How much money did he have in the account at the end of that time?

8. Find the amount of money there will be after 15 years if $6000 earns 9% p.a. interest, paid quarterly.

9. How much money will be in a bank account after 5 years if $500 earns 6.5% p.a. with interest paid monthly?

10. Find the amount of interest earned over 4 years if $1400 earns 6% p.a. paid quarterly .

11. How much money will be in a credit union account after 8 years if $8000 earns 7.5% p.a. interest paid monthly?

12. Elva wins a lottery and invests $500 000 in an account that earns 8% p.a. with interest paid monthly. How much will be in the account after 12 years?

13. Calculate the amount deposited 4 years ago at 5% p.a. if the current amount in the account is

$5000 (a) $675 (b) $12 000 (c) $289.50 (d) $12 800. (e)

14. How much was banked 3 years ago if the amount in the bank now is $5400 and interest is 5.8% p.a. paid quarterly?

15. How many years ago was an investment made if $5000 was invested at 6% p.a. paid monthly and is now worth $6352.45?

7.10 Exercises

ch7.indd 303 6/29/09 10:02:30 PM


16. Find the number of years that $10 000 was invested at 8% p.a. with interest paid twice a year if there is now $18 729.81 in the bank.

17. Jude invested $4500 fi ve years ago at x % p.a. Evaluate x if Jude now has an amount of

$6311.48 (a) $5743.27 (b) $6611.98 (c) $6165.39 (d) $6766.46 in the bank. (e)

18. Hamish is given the choice of a bank account in which interest is paid annually or quarterly. If he deposits $1200, fi nd the difference in the amount of interest paid over 3 years if interest is 7% p.a.

19. Kate has $4000 in a bank account that pays 5% p.a. with interest paid annually and Rachel has $4000 in a different account paying 4% quarterly. Which person will receive the most interest over 5 years and by how much?

20. A bank offers investment account A at 8% p.a. with interest paid twice a year and account B with interest paid at 6% p.a. at monthly intervals. If Georgie invests $5000 over 6 years, which account pays the most interest? How much more does it pay?

Annuities

An annuity is a fund where a certain amount of money is invested regularly (often annually, which is where the name comes from) for a number of years. Scholarship and trust funds, certain types of insurance and superannuation are examples of annuities.

While superannuation and other investments are affected by changes in the economy, due to their long-term nature, the amount of interest they earn balances out over the years. In this course, we average out these changes, and use a constant amount of interest annually.

EXAMPLES

1. A sum of $1500 is invested at the beginning of each year in a superannuation fund. If interest is paid at 6% p.a., how much money will be available at the end of 25 years?

Solution

It is easier to keep track of each annual amount separately. The fi rst amount earns interest for 25 years, the 2 nd amount earns interest for 24 years, the 3 rd amount for 23 years and so on.

ch7.indd 304 6/29/09 10:02:31 PM

305Chapter 7 Series

The 25 th amount earns interest for 1 year. 1500P = and 6% 0.06r = =

n

n

25

24

23

1

( )

( . )

( . )

( . )

( . )

( . ).

.

.

( . )

A P r

A

A

A

A

1

1500 1 0 06

1500 1 06

1500 1 06

1500 1 06

1500 1 06

1500 1 06

n

1

2

3

25

= += +====

=

Total superannuation amount

24

…

( . ) ( . ) ( . ) … ( . )

( . . . … . )

( . . . … . )

T A A A A

1500 1 06 1500 1 06 1500 1 06 1500 1 06

1500 1 06 1 06 1 06 1 06

1500 1 06 1 06 1 06 1 06

1 2 3 2525 23 1

25 24 23 1

2 3 25

= + + + +

= + + + += + + + += + + + +

. . . . . . .1 06 1 06 1 06 1 062 3 25+ + + + is a geometric series with . , .a r1 06 1 06= = and 25n =

( )

.. ( . )

.

.

.

´

Sr

a r

S

T

11

1 06 11 06 1 06 1

58 16

1500 58 16

87 234 57

n

25

25

=−

−

=−

−

===

n

So the total amount of superannuation after 25 years is $87 234.57.

2. An amount of $50 is put into an investment account at the end of each month. If interest is paid at 12% p.a. paid monthly, how much is in the account at the end of 10 years?

Solution

%

.

P

r

50

12

0 12

===

The monthly interest rate is . . .¸0 12 12 0 01= Monthly interest over 10 years gives

ń 10 12

120

==

Since the last amount is deposited at the beginning of the year, it earns interest

for 1 year.

CONTINUED

ch7.indd 305 6/29/09 10:02:32 PM


The fi rst amount earns interest for 119 months, since it is deposited at the end of the month. The 2 nd amount earns interest for 118 months, the 3 rd amount for 117 months and so on. The 120 th amount earns no interest as it goes in at the end of the last month.

( )

( . )

( . )

( . )

( . )

( . ).

.

.

( . )

A P r

A

A

A

A

1

50 1 0 01

50 1 01

50 1 01

50 1 01

50 1 01

50 1 01

n

n

n

1119

2118

3117

1200

= += +====

=

Total amount

119 118

119 118

119

0 1

…

( . ) ( . ) ( . ) … ( . )

( . . . … . )

( . . . … . )

( . . … . )

T A A A A

50 1 01 50 1 01 50 1 01 50 1 01

50 1 01 1 01 1 01 1 01

50 1 01 1 01 1 01 1 01

50 1 1 01 1 01 1 01

1 2 3 120117 0

117 0

2 119

1 2

= + + + +

= + + + += + + + += + + + += + + + +

. . . . . .1 1 01 1 01 1 011 2 119+ + + + is a geometric series with , .a r1 1 01= = and 120n =

120

( )

.( . )

.

.

.

´

Sr

a r

S

T

11

1 01 11 1 01 1

230 04

50 230 04

11501 93

n

120

=−

−

=−

−

===

n

So the total amount after 10 years is $11 501.93.

1. Find the amount of superannuation available at the end of 20 years if $500 is invested at the beginning of each year and earns 9% p.a.

2. Michael works for a company that puts aside $1200 at the beginning of each year for his superannuation. If the money earns interest at the rate of 5% p.a., fi nd the amount of superannuation that Michael will have at the end of 30 years.

7.11 Exercises

ch7.indd 306 6/29/09 10:02:34 PM

307Chapter 7 Series

3. How much superannuation will there be at the end of 20 years if $800 is invested at 10% p.a. at the beginning of each year?

4. Rachel starts working for a business at the beginning of 2005. If she retires at the end of 2034, how much superannuation will she have if $1000 is invested at the beginning of each year at 9.5% p.a.?

5. A sum of $1500 is invested at the end of each year for 15 years at 8% p.a. Find the amount of superannuation available at the end of the 15 years.

6. How much superannuation will there be at the end of 18 years if $690 is invested at 8.5% p.a. at the beginning of each year?

7. If Phan pays $750 into a superannuation fund at the beginning of each year, how much will she have at the end of 29 years if the interest is 6.8% p.a.?

8. A sum of $1000 is invested at the end of each year for 22 years, at 9% p.a. Find the amount of superannuation available at the end of the 22 years.

9. Matthew starts work at the beginning of 2010. If he retires at the end of 2037, how much superannuation will he have if he invests $700 at the beginning of each year at 12.5% p.a.?

10. An apprentice starts work for a small business. If she invests $400 at the beginning of each year, how much superannuation will she have at the end of 25 years if the money earns 15.5% p.a.?

11. Lliam wants to save up $15 000 for a car in 5 years’ time. He invests $2000 at the end of each year in an account that pays 7.5% p.a. interest. How much more will Lliam have to pay at the end of 5 years to make up the $15 000?

12. A school invests $5000 at the end of each year at 6% p.a. towards a new library. How much will the school have after 10 years?

13. Jacques puts aside $500 at the end of each year for 5 years. If the money is invested at 6.5% p.a., how much will Jacques have at the end of the 5 years?

14. An employee contributes $200 into a superannuation fund at the end of each year. If the interest rate on this fund is 11.5% p.a., how much will the employee have at the end of 20 years?

15. Mohammed’s mother invests $200 for him each birthday up to and including his 18th birthday. The money earns 6% p.a. How much money will Mohammed have on his 18th birthday?

16. Xuan is saving up for a holiday. She invests $800 at the end of each year at 7.5% p.a. How much will she have for her holiday after 5 years’ time?

17. A couple saves $3000 at the end of each year towards a deposit on a house. If the interest rate is 5% p.a., how much will the couple have saved after 6 years?

ch7.indd 307 6/29/09 10:02:35 PM


18. Lucia saves up $2000 each year and at the end of the year she invests it at 6% p.a.

She does this for 10 years. (a) What is her investment worth?

Lucia continues investing (b) $2000 a year for 5 more years. What is the future value of her investment?

19. Jodie starts work in 2012 and puts $1000 in a superannuation fund at the end of the year. She keeps putting in this same amount at the end of every year until she retires at the end of 2029. If interest is paid at 10% p.a., calculate how much Jodie will have when she retires.

20. Michelle invests $1000 at the end of each year. The interest rate is 8% p.a.

How much will her (a) investment be worth after 6 years?

How much more would (b) Michelle’s investment be worth after 6 years if she had invested $1200 each year?

21. Jack cannot decide whether to invest $1000 at the end of each year for 15 years or $500 for 30 years in a superannuation fund. If the interest rate is 5% p.a., which would be the better investment for Jack?

22. Kate is saving up to go overseas in 8 years’ time. She invests $1000 at the end of each year at 7% p.a. and estimates that the trip will cost her around $10 000. Will she

have enough? If so, how much over will it be? If she doesn’t earn enough, how much will she need to add to this money to make it up to the $10 000?

23. Farmer Brown puts aside part of the farm’s earnings at the beginning of each month to buy a new truck in 10 years’ time when the old one wears out. He invests $400 each month at 9% p.a. and estimates the cost of a new truck at $80 000. Will the investment earn enough to buy the new truck? What is the difference?

24. Marcus and Rachel want to save up $25 000 for a deposit on an apartment in 6 years’ time. They aim to pay around half the deposit each. Marcus invests an inheritance of $9000 in a bank account where it earns 8% p.a. Rachel invests $100 at the beginning of each month where it earns interest of 9% p.a.

What is the future value (a) of Marcus’s investment after 6 years?

How much will Rachel’s (b) investment be worth after 6 years?

Will they be able to pay the (c) deposit in 6 years’ time?

25. Jenny puts aside $20 at the end of each month for 3 years. How much will she have then if the investment earns 8.2% p.a., paid monthly?

ch7.indd 308 6/29/09 10:02:36 PM

309Chapter 7 Series

Loan repayments

The formula for compound interest and the geometric series can help with working out regular loan repayments.

EXAMPLES

1. A sum of $20 000 is borrowed at 12% p.a. and paid back at regular monthly intervals over 4 years. Find the amount of each payment.

Solution

Let M stand for the monthly repayment. The number of payments will be 4 12´ or 48 . Monthly interest will be 12% 12¸ or 1% (0.01). Each month, interest for that month is added to the loan and the repayment amount is taken off. The interest added to the fi rst month will be ( . )20 000 1 0 01 1+

( . )20 000 1 01or 1. The interest added to other months will be ( . )A 1 0 01 1+ or ( . )A 1 01 1 . The amount owing for the fi rst month:

1

1( . )

( . )

A M

M

20 000 1 0 01

20 000 1 011 = + −

= −

The amount owing for the 2 nd month is what was owing from the 1 st month, together with that month’s interest, minus the repayment.

1

1

1

1

( . )

( . )

( . ) ( . ) ( )

( . ) ( . )

( . ) ( . )

A A M

A M

M M A

M M

M

1 0 01

1 01

20 000 1 01 1 01

20 000 1 01 1 01

20 000 1 01 1 01 1

substituting in

2 11

1

12

2 1

= + −= −= − −= − −= − +

7 A

Similarly,

1

2 1

1

1

1

( . )

( . ) ( . ) ( . )( . ) ( . ) ( . )

( . ) ( . . )

( . ) ( . . )

A A M

M MM M

M M

M

1 0 01

20 000 1 01 1 01 1 1 0120 000 1 01 1 01 1 1 01

20 000 1 01 1 01 1 01

20 000 1 01 1 01 1 01 1

3 21

3 1

3 2

3 2

= + −= − + −= − + −= − + −= − + +

7 A

The 20 000(1.01) 1 is the interest added for the

month and the M is the repayment subtracted from

the balance.

CONTINUED

ch7.indd 309 6/29/09 10:02:37 PM


Continuing this pattern, after 4 years (48 months) the amount owing will be 48 146 2( . ) ( . . . . )A M20 000 1 01 1 01 1 01 1 01 1 01 148

47 f= − + + + + + .

But the loan is paid out after 48 months. So 0A48 =

48 47 2( . ) ( . . . . )M0 20 000 1 01 1 01 1 01 1 01 1 01 146 1f= − + + + + +

( . . . . ) ( . )M 1 01 1 01 1 01 1 01 1 20 000 1 0147 46 2 1 48f+ + + + + =

. . . .

( . ) M

1 01 1 01 1 01 1 01 1

20 000 1 0147 46 2 1

48

f=

+ + + + +

. . . .1 01 1 01 1 01 1 01 147 46 2 1f+ + + + +. . . .1 1 01 1 01 1 01 1 011 2 46 47f= + + + + +

This is a geometric series with , .a r1 1 01= = and 48n = .

( )

.( . )

.

.( . )

.

Sr

a r

S

M

11

1 01 11 1 01 1

61 223

61 22320 000 1 01

526 68

n

48

48

48

`

=−

−

=−

−

=

=

=

n

So the monthly repayment is $526.68.

2. A store charges 10% p.a. for loans and repayments do not have to be made until the 4 th month. Ivan buys $8000 worth of furniture and pays it off over 3 years.

How much does Ivan owe after 3 months? (a) What are his monthly repayments? (b) How much does Ivan pay altogether? (c)

Solution

Let P stand for the payments each month.

(´3 12 3 3

33

Number of payments months of no repayments)

months

= −=

Monthly interest rate

10% 12

0.1 12

.

¸¸

r

0 0083•

==

=

After 3 months, the amount owing is (a)

( )

( . )

( . ).

A P r1

8000 1 0 0083

8000 1 00838201 67

•

•

n

3

3

= +

= +

==

So the amount owing after 3 months is $8201.67.

ch7.indd 310 6/29/09 10:02:39 PM

311Chapter 7 Series

The fi rst repayment is in the 4 th month .(b)

1

1

4

5

6

6

( . )

( . )

( . )

( . )

( . ) ( . )

( . ) ( . )

( . ) ( . )

( . ) ( . ) ( . )

( . ) ( . ) ( . )

( . ) ( . . )

( . ) ( . . )

A

A

A

A P

A P P

P P

P

A P P

P P

P P

P

8000 1 0083

8000 1 0083

8000 1 0083

8000 1 0083

8000 1 0083 1 0083

8000 1 0083 1 0083

8000 1 0083 1 0083 1

8000 1 0083 1 0083 1 1 0083

8000 1 0083 1 0083 1 1 0083

8000 1 0083 1 0083 1 0083

8000 1 0083 1 0083 1 0083 1

•

•

•

•

• •

• •

• •

• • •

• • •

• • •

• • •

11

22

33

44

51

5

5

61

1

2 1

6

=

=

=

= −

= − −

= − −

= − +

= − + −

= − + −

= − + −

= − + +

1

2 1

1

8

9

B

C

Continuing this pattern,

( . ) ( . . . . . . )A P8000 1 0083 1 0083 1 0083 1 0083 1• • • •

3632= − + + + +36 31 1

The loan is paid out after 36 months. So 0A36 =

( . ) ( . . . . . . )P0 8000 1 0083 1 0083 1 0083 1 0083 1• • • •36 32= − + + + +31 1

36(1.008 1.008 . . . 1.008 1) 8000(1.008 )P 3 3 3 3• • • •

+ + + + =32 31 1

. . . . . . .

( . )P

1 0083 1 0083 1 0083 1 0083 1

8000 1 0083• • • •

•

31 2 1

36

=+ + + + +32

1. . . . . .1 0083 1 0083 1 0083 1• • •

+ + + +32 31

. . . . . .1 1 0083 1 0083 1 0083• • •31= + + + +1 32

This is a geometric series with , .a r1 1 0083•

= = and 33n = .

( )

.

( . )

.

.( . )

.

Sr

a r

S

P

11

1 0083 1

1 1 0083 1

37 804

37 8048000 1 0083

285 30

•

•

•

n

n

33

33

36

`

=−

−

=−

−

=

=

=

So the monthly repayment is $285.30.

Ivan pays (c) $285.30 33´ or $9414.94 altogether.

If you round off the repayment, your answer will

be slightly different.

The fi rst repayment is in the 4th month.

ch7.indd 311 6/29/09 10:02:40 PM


1. An amount of $3000 is borrowed at 22% p.a. and repayments made yearly for 5 years. How much will each repayment be?

2. The sum of $20 000 is borrowed at 18% p.a. interest over 8 years. How much will the repayments be if they are made monthly?

3. David borrows $5000 from the bank and pays back the loan in monthly instalments over 4 years. If the loan incurs interest of 15% p.a., fi nd the amount of each instalment.

4. Mr and Mrs Nguyen mortgage their house for $150 000.

Find the amount of the (a) monthly repayments they will have to make if the mortgage is over 25 years with interest at 6% p.a.

If the Nguyens want to pay (b) their mortgage out after 15 years, what monthly repayments would they need to make?

5. A loan of $6000 is paid back in equal annual instalments over 3 years. If the interest is 12.5% p.a., fi nd the amount of each annual instalment.

6. The Smith family buys a car for $38 000, paying a 10% deposit and taking out a loan for the balance. If the loan is over 5 years with interest of 1.5% monthly, fi nd

the amount of each monthly (a) loan repayment

the total amount that the (b) Smith family paid for the car.

7. A $2000 loan is offered at 18% p.a. with interest charged monthly, over 3 years.

If no repayment need be paid (a) for the fi rst 2 months, fi nd the amount of each repayment.

How much will be paid back (b) altogether?

8. Bill thinks he can afford a mortgage payment of $800 each month. How much can he borrow, to the nearest $100, over 25 years at 11.5% p.a.?

9. Get Rich Bank offers a mortgage

at %217 p.a. over 10 years and

Capital Bank offers a mortgage at

%215 p.a. over 25 years.

Find the amount of the (a) monthly repayments for each bank on a loan of $80 000.

Find the difference in the (b) total amount paid on each mortgage.

10. Majed buys a $35 000 car. He puts down a 5% deposit and pays the balance back in monthly instalments over 4 years at 12% p.a. Find the total amount that Majed pays for the car.

11. Amy borrowed money over 7 years at 15.5% p.a. and she pays $1200 a month. How much did she borrow?

7.12 Exercises

ch7.indd 312 6/29/09 10:02:41 PM

313Chapter 7 Series

12. NSW Bank offers loans at 9% p.a. with an interest-free period of 3 months, while Sydney Bank offers loans at 7% p.a. Compare these loans on an amount of $5000 over 3 years and state which bank offers the best loan and why.

13. Danny buys a plasma TV for $10 000. He pays a $1500 deposit and borrows the balance at 18% p.a. over 4 years.

Find the amount of each (a) monthly repayment .

How much did Danny pay (b) altogether?

14. A store offers furniture on hire purchase at 20% p.a. over 5 years, with no repayments for 6 months. Ali buys furniture worth $12 000.

How much does Ali owe after (a) 6 months?

What are the monthly (b) repayments?

How much does Ali pay for (c) the furniture altogether?

15. A loan of $6000 over 5 years at 15% p.a. interest, charged monthly, is paid back in 5 annual instalments.

What is the amount of each (a) instalment?

How much is paid back (b) altogether?

ch7.indd 313 6/29/09 10:02:42 PM


1. Find a formula for the n th term of each sequence .

9, 13, 17, … (a) 7, 0, −7, … (b) 2, 6, 18, … (c) 200, 50, (d) 12 ,

21 …

(e) , , ,2 4 8− − …

2. For the series …156 145 134+ + + Find the 15 th term .(a) Find the sum of 15 terms. (b) Find the sum of 14 terms .(c) Write a relationship between (d)

T 15 , S 15 and S 14 . Find the value of (e) n for the fi rst

negative term.

3. Evaluate

(a) (9 7)r1

50

r−

=/

(b) 1n3

7 b l/ as a fraction

(c) ( )5 3n

2

12

/

(d) (7 6)n1

100

−/

(e) 251∞

n

n

3

2

=

−

c m/

4. A bamboo blind has 30 slats. When the blind is down, each gap between slats, and between the top and bottom of the window, is 3 mm. When the blind is up, the slats have no gaps between them.

Show that when the blind is up, the (a) bottom slat rises 90 mm.

How far does the next slat rise? (b) Explain briefl y why the distances the (c)

slats rise when the blind is up form an arithmetic series .

Find the distance the 17 th slat from (d) the bottom rises .

What is the sum of the distances that (e) all slats rise?

5. Find the amount invested in a bank account at 9.5% p.a. if the balance in the account is $5860.91 after 6 years .

6. State whether each series is (i) arithmetic (ii) geometric (iii) neither . (a) …97 93 89+ + +

(b) …32

21

83+ + +

(c) …5 20 45+ + + (d) . . . …1 6 0 4 0 6− − + − (e) . . . …3 4 7 5 11 6+ + + (f) …48 24 12+ + +

(g) …51 1 5− + − +

(h) …105 100 95+ + +

(i) 121 1

41 1 ...+ + +

(j) …log log logx x x10 102

103+ + +

7. The n th term of the series …8 13 18+ + + is 543. Evaluate n .

8. Amanda earned $20 000 in one year. At the beginning of the next year she received a salary increase of $450.

Test Yourself 7

3 mm

ch7.indd 314 6/29/09 10:02:43 PM

315Chapter 7 Series

She now receives the same increase each year.

What will her salary be after 10 years? (a) How much will Amanda earn (b)

altogether over the 10 years?

9. The 11 th term of an arithmetic series is 97 and the 6 th term is 32. Find the fi rst term and common difference.

10. A series has n th term given by 5.T n3n = −

Find the 4 th term (a) the sum of 4 terms (b) which term is 5827 .(c)

11. A series has terms …x5 45+ + + Evaluate x if the series is

arithmetic (a) geometric. (b)

12. Convert each recurring decimal to a fraction .

(a) 0.4•

(b) 0.72•

(c) 1.57• •

13. If ,x x2 3+ and 5 x are the fi rst 3 terms of an arithmetic series, calculate the value of x.

14. Find the 20 th term of 3, 10, 17, … (a) 101, 98, 95, … (b) 0.3, 0.6, 0.9, … (c)

15. Find the limiting sum (sum to infi nity) of …81 27 9+ + +

16. Karl puts $300 aside as an annuity for his son at the beginning of each year. If interest is 7% p.a., how much will his son receive at the end of 15 years?

17. For each series, write an expression for the sum of n terms .

(a) …5 9 13+ + + (b) . . …1 1 07 1 072+ + +

18. (a) For what values of x does the geometric series …x x1 2+ + + have a limiting sum?

(b) Find the limiting sum when x53= .

(c) Evaluate x when the limiting sum

is 1 21 .

19. The fi rst term of an arithmetic series is 4 and the sum of 10 terms is 265. Find the common difference.

20. Every week during a typing course, Tony improves his typing speed by 3 words per minute until he reaches 60 words per minute by the end of the course.

If he can type 18 words per minute in (a) the fi rst week of the course, how many words per minute can he type by week 8?

How many weeks does the course (b) run for?

21. A farmer borrows $50 000 for farm machinery at 18% p.a. over 5 years and makes equal yearly repayments on the loan at the end of each year.

How much does he owe at the end (a) of the fi rst year, just before he makes the fi rst repayment?

How much is each yearly repayment? (b)

22. A ball drops from a height of 1.2 metres

then bounces back to 53 of this height.

On the next bounce, it bounces up to 53

of this height and so on. Through what distance will the ball travel?

23. If ,x x2 7 2+ − and 15 6x + are consecutive terms in a geometric series, evaluate x.

24. (a) If $2000 is invested at 4.5% p.a., how much will it be worth after 4 years?

(b) If interest is paid quarterly, how much would the investment be worth after 4 years?

ch7.indd 315 6/29/09 10:02:50 PM


25. Evaluate …8 14 20 122+ + + + .

26. (a) Calculate the sum of all the multiples of 7 between 1 and 100.

(b) Calculate the sum of all numbers between 1 and 100 that are not multiples of 7 .

27. Scott borrows $200 000 to buy a house. If the interest is 6% p.a. and the loan is over 20 years,

how much is each monthly (a) repayment?

how much does Scott pay altogether? (b)

28. The sum of n terms of the series …214 206 198+ + + is 2760. Evaluate n.

29. Evaluate n if the n th term of the series …4 12 36+ + + is 354 292.


1. The n th term of a sequence is given by

.n

nT1n

2

+=

What is the 9 th term of the sequence? (a)

Which term is equal to (b) 18201 ?

2. For the sequence , , . . .π π π4

34

5 evaluate

the common difference (a) the 7 th term (b) the sum of 6 terms. (c)

3. Evaluate the sum of the fi rst 20 terms of the series

(a) …3 5 9 17 33 65+ + + + + + (b) …5 2 10 8 15 32− + − + − +

4. A factory sells shoes at $60 a pair. For 10 pairs of shoes there is a discount, whereby each pair costs $58. For 20 pairs, the cost is $56 a pair and so on. Find

the price of each pair of shoes on an (a) order of 100 pairs of shoes .

the total price on an order of 60 pairs (b) of shoes .

5. Which term of the sequence

, , , . . .97

4514

22528 is equal to ?

28125224

6. Find the sum of all integers between 1 and 200 that are not multiples of 9.

7. Find the least number of terms for which

the sum of the series …20 454+ + + is

greater than 24.99.

8. Find the values of n for which the sequence given by T n n4n

2= − is negative.

9. The sum of the fi rst 5 terms of a geometric sequence is 77 and the sum of the next 5 terms is 2464.− Find the 4 th term of the sequence.

10. Jane’s mother puts $300 into an account at the beginning of each year to pay for Jane’s education in 5 years’ time. If 6% p.a. interest is paid quarterly, how much money will Jane’s mother have at the end of the 5 years?

11. The geometric series …x x x2 4 82 3+ + + has a limiting sum of 3. Evaluate x.

12. Find the amount of money in a bank account if $5000 earns 8.5% p.a. for 4 years, then 6.5% p.a. for 3 years, with interest paid monthly for all 7 years.

Hint:

3 2 1, 5 4 1= + = + and so on.

ch7.indd 316 6/29/09 10:02:57 PM

317Chapter 7 Series

13. Find the value of k for which 81 920 is the 8 th term of the geometric sequence …k5 80+ + +

14. Kim borrows $10 000 over 3 years at a rate of 1% interest compounded each month. If she pays off the loan in three equal annual instalments, fi nd

the amount Kim owes after one (a) month

the amount she owes after the (b) fi rst year, just before she pays the fi rst instalment

the amount of each instalment (c) the total amount of interest Kim pays. (d)

15. (a) Find the limiting sum of the series …cos cosx x1 2 4+ + + in simplest form .

(b) Why does this series have a limiting sum?

16. A superannuation fund paid 6% p.a. for the fi rst 10 years and then 10% p.a. after that. If Thanh put $5000 into this fund at the end of each year, how much would she have at the end of 25 years?

ch7.indd 317 6/29/09 10:03:03 PM

TERMINOLOGYTERMINOLOGY

8

Complement: The complement of an event E is when the event E does not occur

Equally likely outcomes: Each outcome has the same chance of occurring

Independent events: Events are independent if the result of one event does not infl uence the outcome of another event. There is no overlap between the events

Multi-stage events: A series of successive independent events

Mutually exclusive results: Two events with the same sample space that cannot both occur at the same time

Non-mutually exclusive results: Two events with the same sample space that can occur at the same time i.e. there is some overlap

Probability tree: A diagram that uses branches to show multi-stage events and sets out the probability on each

branch with the sample space listed at the right of each branch

Random experiments: Experiments that are made with no pattern or order where each outcome is equally likely to occur

Sample space: The set of all possible outcomes in an event or series of events

Tree diagram: A diagram that uses branches to show multi-stage events where the probabilities on each branch are equal

Venn diagram: A special diagram to show the sample space for non-mutually exclusive events using circles for each event drawn inside a rectangle which represents the sample space

TERMINOLOGY

Probability

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319Chapter 8 Probability

DID YOU KNOW?

Girolamo Cardano (1501–76) was a doctor and mathematician who developed the fi rst theory of probability. He was a great gambler, and he wrote De Ludo Aleae (‘On Games of Chance’). This work was largely ignored, and it is said that the fi rst book on probability was written by Christiaan Huygens (1629–95).

The main study of probability was done by Blaise Pascal (1623–62), whom you have already heard about, and Pierre de Fermat (1601−65).

INTRODUCTION

PROBABILITY IS THE STUDY of how likely it is that something will happen. It is used to make predictions in different areas, ranging from gambling to determining the rate of insurance premiums. For example an actuary looking at death-rate statistics can estimate the probable age to which someone will live, and set life insurance premiums accordingly .

Another example of where probability is useful is in biology. The probability of certain diseases or genetic defects can be calculated in high-risk families.

Probability is also closely related to statistics and data analysis, as well as games of chance such as card games, tossing coins, backgammon and so on. It is also relevant to buying raffl e and lottery tickets.

In this chapter, you will revise simple probability that you have learnt in previous years and extend this to more complex probability involving multi-stage events.

Simple Probability

Mutually exclusive events

Mutually exclusive events means that if one event occurs, the other cannot. For example, when rolling a die, a 6 cannot occur at the same time as a 2.

We can measure probability in theory as long as the events are random. However, even then, probability only gives us an approximate idea of the likelihood of certain events happening.

For example, in Lotto draws, there is a machine that draws out the balls at random and a panel of supervisors checks that this happens properly. Each ball is independent of the others and is equally likely to be drawn out.

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The probability of an event E happening, P ( E ), is given by the number of ways the event can occur, n ( E ), compared with the total number of outcomes possible n ( S ) (the size of the sample space)

( )( )( )

P En Sn E

=

If P ( E ) = 0 the event is impossible

If P ( E ) = 1 the event is certain (it has to happen)

0 ≤ P ( E ) ≤ 1

The sum of all (mutually exclusive) probabilities is 1

EXAMPLES

1. A container holds 5 blue, 3 white and 7 yellow marbles. If one marble is selected at random, fi nd the probability of getting

a white marble (a) a white or blue marble (b) a yellow, white or blue marble (c) a red marble. (d)

In a horse race, it is diffi cult to measure probability as the horses are not all equally likely to win. Other factors such as ability, training, experience and weight of the jockey all affect it. The likelihood of any one horse winning is not random.

Class Discussion

Discuss these statements:

The probability of one particular football team winning a 1.

competition is 161 as there are 16 teams.

The probability of Tiger Woods winning the US Open golf 2.

tournament is 781 if there are 78 players in the tournament.

A coin came up tails 8 times in a row. So there is a greater chance 3. that the next time it will come up heads.

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Getting a red marble is impossible!

Solution

The size of the sample space, or total number of marbles is 5 + 3 + 7 or 15.

(a) ( )P W153

51

=

=

(b) ( )P W B15

3 5

158

or = +

=

(c) ( )P Y W B15

7 3 5

1515

1

or or = + +

=

=

(d) ( )P R150

0

=

=

2. The probability that a traffi c light will turn green as a car approaches

it is estimated to be 125 . A taxi goes through 192 intersections where there

are traffi c lights. How many of these would be expected to turn green as the taxi approached?

Solution

It is expected that 125 of the traffi c lights would turn green.

192 80´125 =

So it would be expected that 80 traffi c lights would turn green as the taxi approached.

1. Peter is in a class of 30 students. If one student is chosen at random to make a speech, fi nd the probability that the student chosen will be Peter.

2. A pack of cards contains 52 different cards, one of which is the ace of diamonds. If one card is chosen at random, fi nd the probability that it will be the ace of diamonds.

8.1 Exercise s

ch8.indd 321 7/6/09 11:42:11 PM


‘Die’ is singular for ‘dice’.

3. There are 6 different newspapers sold at the local newsagent each day. Kim sends her little brother to buy her a newspaper one morning but forgets to tell him which one. What is the probability that he will buy the correct newspaper?

4. A raffl e is held in which 200 tickets are sold. If I buy 5 tickets, what is the probability of my winning the prize in the raffl e?

5. In a lottery, 200 000 tickets are sold. If Lucia buys 10 tickets, what is the probability of her winning fi rst prize?

6. A bag contains 6 red balls and 8 white balls. If one ball is drawn out of the bag at random, fi nd the probability that it will be

white (a) red. (b)

7. A shoe shop orders in 20 pairs of black, 14 pairs of navy and 3 pairs of brown school shoes. If the boxes are all mixed up, fi nd the probability that one box selected at random will contain brown shoes.

8. Abdul has 12 CDs in his car glovebox. The labels have become mixed up. If he chooses one of the CDs at random, fi nd the probability that it is his favourite CD.

9. A bag contains 5 black marbles, 4 yellow marbles and 11 green marbles. Find the probability of drawing 1 marble out at random and getting

a green marble (a) a yellow or a green marble. (b)

10. A die is thrown. Calculate the probability of throwing

a 6 (a) an even number (b) a number less than 3. (c)

11. A book has 124 pages. If the book is opened at any page at random, fi nd the probability of the page number being

either 80 or 90 (a) a multiple of 10 (b) an odd number (c) less than 100. (d)

12. In the game of pool, there are 15 balls, each with the number 1 to 15 on it. In Kelly Pool, each person chooses a number at random to determine which ball to sink. If Tracey chooses a number, fi nd the probability that her ball will be

an odd number (a) a number less than 8 (b) the 8 ball. (c)

13. At a school dance, each girl is given a ticket with a boy’s name on it. The girl then must dance with that boy for the next dance. If the tickets are given out at random and there are 50 boys at the dance, what is the probability that Jill will get to dance with her boyfriend?

14. Find the probability of a coin coming up heads when tossed. If the coin is double-headed, fi nd the probability of tossing a head.

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15. In a bag of caramels, there are 21 with red wrappers and 23 with blue wrappers. If Leila chooses a caramel at random from the bag, fi nd the probability that she will choose one with a blue wrapper.

16. A student is chosen at random to write about his/her favourite sport. If 12 students like tennis best, 7 prefer soccer, 3 prefer squash, 5 prefer basketball and 4 prefer swimming, fi nd the probability that the student chosen will write about

soccer (a) squash or swimming (b) tennis. (c)

17. A school has 875 students. If 5 students are chosen at random to help show some visitors around, fi nd the probability that a particular student will be chosen.

18. A box containing a light globe

has a 201 probability of holding

a defective globe. If 160 boxes are checked, how many would be expected to be defective?

19. There are 29 red, 17 blue, 21 yellow and 19 green chocolate beans in a packet. If Kate chooses one at random, fi nd the probability that it will be red or yellow.

20. The probability of breeding a

white budgerigar is .92 If Mr Seed

breeds 153 budgerigars over the year, how many would be expected to be white?

21. A biased coin is weighted so that heads comes up twice as often as tails. Find the probability of tossing a tail.

22. A die has the centre dot painted white on the 5 so that it appears as a 4. Find the probability of throwing

a 2 (a) a 4 (b) a number less than 5. (c)

23. Discuss these statements. The probability of one (a)

particular horse winning the

Melbourne cup is 201 if there are

20 horses in the race. The probability of Greg (b)

Norman winning a masters golf

tournament is 151 if there are

15 players in the tournament. A coin came up tails 8 times (c)

in a row. So the next toss must be a head.

A family has three sons. (d) There is more chance of getting a daughter next time.

The probability of a Holden (e) winning the car race at Bathurst

this year is 476 as there are

6 Holdens in the race and 47 cars altogether.

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Complementary events

When we fi nd the probabilities of events, the total of all the possible events will always add up to 1.

EXAMPLE

A ball is chosen at random from a bag containing 5 blue, 3 red and 7 yellow balls. The probabilities are as follows:

(blue)

(red)

(yellow)

Total probability

P

P

P

155

153

157

155

153

157

1515

1

=

=

=

= + +

=

=

The complement of an event happening is the event not happening. That is, the complement of P ( E ) is P (not E ). We can write this as P E^ hL

EXAMPLE

A die is thrown. Find the probability of throwing a 6 (a) not throwing a 6. (b)

Solution

(a) P 661=] g

(b) (not 6) ( , , , , or )P P 1 2 3 4 5

65

=

=

In general,( ) ( ) 1

or ( ) 1 ( )

P E P E

P E P E

+ == −

LL

ch8.indd 324 6/29/09 10:17:35 PM


Proof

Let e be the number of ways E can happen out of a total of n events. Then the number of ways E will not happen is .n e−

Then ( )

( )

( )

P E ne

P E nn e

nn

ne

ne

P E

1

1

=

= −

= −

= −

= −

L

EXAMPLES

1. The probability of a win in a raffl e is .350

1 What is the probability of losing?

Solution

( )P P1

1350

1

350349

lose win= −

= −

=

] g

2. The probability of a tree surviving a fi re is 72%. Find the probability of the tree failing to survive a fi re.

Solution

% %

%

P P1100 72

28

failing to survive surviving= −= −=

^ ^h h

1. The probability of a bus arriving

on time is estimated at .3318 What

is the probability that the bus will not arrive on time?

2. The probability of a seed

producing a pink fl ower is .97

Find the probability of the fl ower producing a different colour.

3. If a baby has a 0.2% chance of being born with a disability, fi nd the probability of the baby being born without any disabilities.

4. The probability of selecting a card with the number 5 on it is 0.27. What is the probability of not selecting this card?

8.2 Exercise s

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5. There is a 62% chance of a student being chosen as a prefect. What is the probability of a student not being selected as a prefect?

6. A machine has a 1.5% chance of breaking down at any given time. What is the probability of the machine not breaking down?

7. The life of a certain type of computer is about 7 years. If the probability of its needing repairs

in that time is ,231 fi nd the

probability that it will not need repairs.

8. A certain traffi c light has a

probability of 1813 of being green.

Find the probability of the light not being green when a car comes to the traffi c light.

9. A certain organism in a river has a probability of 0.79 of surviving a fl ood. What is the probability of its not surviving?

10. A city has an 8.1% chance of being hit by an earthquake. What is its chance of not having an earthquake?

11. The probabilities when 3 coins are tossed are as follows:

( )

( )

( )

( )

P

P

P

P

381

2 183

1 283

381

heads

heads and tail

head and tails

tails

=

=

=

=

Find the probability of tossing at least one head.

12. The probabilities of a certain number of seeds germinating when 4 seeds are planted are as follows:

(0 )

P

P

P

P

P

4494

3498

24916

14918

493

seeds

seeds

seeds

seed

seeds

=

=

=

=

=

]]]]

gggg

Find the probability of at least 1 seed germinating.

13. The probabilities of 4 friends being chosen for a soccer team are as follows:

P

P

P

P

4151

3154

2156

1152

chosen

chosen

chosen

chosen

=

=

=

=

]]]]

gggg

Find the probability of none of the friends being (a)

chosen at least 1 of the friends being (b)

chosen.

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EXAMPLES

1. One card is drawn from a set of cards numbered 1 to 10. Find the probability of drawing out an odd number or a multiple of 3.

Solution

The odd cards are 1, 3, 5, 7 and 9. The multiples of 3 are 3, 6 and 9. The numbers 3 and 9 are both odd and multiples of 3. So there are 6 numbers that are odd or multiples of 3: 1, 3, 5, 6, 7 and 9

( )P 3

106

53

odd or multiple of =

=

The trick with this question is not to count the 3

and the 9 twice.

14. A dog breeder tries to produce a dog with a curly tail. If 2 puppies are born, the probabilities are as follows:

P

P

P

114

115

112

no curly tails

1 curly tail

2 curly tails

=

=

=

^^^

hhh

Find the probability that at least 1 puppy will have a curly tail.

15. The probabilities of 3 new cars passing a quality control check are as follows:

P

P

P

P

3161

2167

1163

0165

passing

passing

passing

passing

=

=

=

=

^^^^

hhhh

Find the probability that at least 1 car will fail the check.

Non-mutually exclusive events

All the examples of probability given so far in this chapter are mutually exclusive. This means that if one event occurs, then another one cannot. For example, if a die is thrown, a 6 cannot occur at the same time as a 2.

Sometimes, there is an overlap where more than one event can occur at the same time. We call these non-mutually exclusive events.

It is important to count the possible outcomes carefully when this happens. If there are not too many outcomes, we can simply list them, but if this is diffi cult, we can use a Venn diagram to help.

CONTINUED

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Using a Venn diagram:

Count all the numbers inside the two circles.

From the Venn diagram:

Probability is .106

53=

2. In year 7 at Mt Random High School, every student must do art or music. In a group of 100 students surveyed, 47 do music and 59 do art. If one student is chosen at random from year 7, fi nd the probability that this student does

both art and music (a) only art (b) only music. (c)

Solution

47 59 106+ = But there are only 100 students! This means 6 students have been counted twice. i.e. 6 students do both art and music.

Students doing music only: 47 6 41− = Students doing art only: 59 6 53− = A Venn diagram shows this information.

(a) bothP100

6503= =] g

(b) art onlyP10053=^ h

(c) music onlyP10041=^ h

There are 6 numbers inside the circles and 10 numbers altogether.

DID YOU KNOW?

Venn diagrams are named after John Venn (1834–1923), an English probabilist and logician.

ch8.indd 328 6/29/09 10:17:37 PM


EXAMPLE

From 100 cards, numbered from 1 to 100, one is selected at random. Find the probability that the card selected is even or less than 20.

Solution

Some cards are both even and less than 20 (i.e. 2, 4, 6, 8, 10, 12, 14, 16, 18).

( )

( )

( )

P

P

P

20100

9

10050

2010019

even and

even

<

<

=

=

=

( ) ( ) ( ) ( )P P P P20 20 20

10050

10019

1009

10060

53

even or even even and< < <= + −

= + −

=

=

There is a formula that can be used for non-mutually exclusive events.

or andP A B P A P B P A B= + −] ] ] ]g g g g

Notice that P A P B+] ]g g counts andP A B] g twice, since it occurs in both P ( A ) and P ( B ). This can be adjusted by subtracting P ( A and B ).

1. A number is chosen at random from the numbers 1 to 20. Find the probability that the number chosen will be

divisible by 3 (a) less than 10 or divisible by 3 (b) a composite number (c) a composite number or a (d)

number greater than 12.

2. A set of 50 cards is labelled from 1 to 50. One card is drawn out at random. Find the probability that the card will be

a multiple of 5 (a) an odd number (b) a multiple of 5 or an odd (c)

number a number greater than 40 or (d)

an even number less than 20. (e)

8.3 Exercise s

ch8.indd 329 6/29/09 10:17:37 PM


3. A set of 26 cards, each with a different letter of the alphabet on it, is placed in a box and one is drawn out at random. Find the probability that the letter on the card drawn will be

a vowel (a) a vowel or one of the letters (b)

in the word ‘random’ a consonant or one of the (c)

letters in the word ‘movies’.

4. A set of discs is numbered 1 to 100 and one is chosen at random. Find the probability that the number on the disc will be

less than 30 (a) an odd number or a number (b)

greater than 70 divisible by 5 or less than 20. (c)

5. In Lotto, a machine holds 45 balls, each with a number between 1 and 45 on it. The machine draws out one ball at a time at random. Find the probability that the fi rst ball drawn out will be

less than 10 or an even (a) number.

between 1 and 15 inclusive, (b) or divisible by 6

greater than 30 or an odd (c) number.

6. A class of 28 students puts on a concert with all class members performing. If 15 dance and 19 sing in the performance, fi nd the probability that any one student chosen at random from the class will

both sing and dance (a) only sing (b) only dance. (c)

7. A survey of 80 people with dark hair or brown eyes showed that 63 had dark hair and 59 had brown eyes. If one of the people surveyed is chosen at random, fi nd the probability that the person will

have dark hair but not brown (a) eyes

have brown eyes but not dark (b) hair

have both brown eyes and (c) dark hair.

8. A list is made up of people with experience of either computers or digital cameras. On the list of 20 people, 13 have computer experience while 9 have experience with a digital camera. If one name is chosen at random from the list, fi nd the probability that the person will have experience with

both computers and digital (a) cameras

computers only (b) digital cameras only. (c)

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9. In a group of 75 students, all do either History or Geography. Altogether 54 do History and 31 do Geography. If I select one student at random, fi nd the probability that he/she will do

only Geography (a) both History and Geography (b) History but not Geography. (c)

10. In a group of 20 dogs at obedience school, 14 dogs will walk to heel and 12 will stay when told. If one dog is chosen at random, fi nd the probability that the dog will

both walk to heel and stay (a) walk to heel but not stay (b) stay but not walk to heel. (c)

Multi-Stage Events

Product rule of probability

Class Discussion

Break up into pairs and try these experiments with one doing the activity and one recording the results.

Toss two coins as many times as you can in a 5 minute period and 1. record the results in the table:

Result Two heads One head and one tail Two tails

Tally

Compare your results with others in the class. What do you notice? Is this surprising?

Roll two dice as many times as you can in a 5 minute period, fi nd the 2. total of the two uppermost numbers on the dice and record the results in the table:

Total 2 3 4 5 6 7 8 9 10 11 12

Tally

Compare your results with others in the class. What do you notice? Is this surprising?

Why don’t these results appear to be equally likely?

The counting of all possible outcomes (the size of the sample space) is important. This is why we use tables and tree diagrams.

You learned about tree diagrams in your earlier

studies.

ch8.indd 331 7/6/09 11:42:12 PM


Compare these probabilities with your results in the

experiments.

EXAMPLES

Find the size of the sample space and the probability of each outcome for each question by using a table or tree diagram.

1. Tossing two coins

Solution

Using a table gives

H T

H HH HT

T TH TT

Using a tree diagram gives

HT

H

T

H

T

Since there are four separate outcomes (HH, HT, TH, TT) each outcome

has a probability of 41 .

Remember that each outcome when tossing 1 coin is 21 .

Notice that ´21

21

41= .

2. Rolling 2 dice and recording the sum of the uppermost numbers.

Solution

A tree diagram would be too diffi cult to draw for this question. Using a table:

1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

Since there are 36 outcomes, each has a probability of 361

Remember that each outcome when rolling 1 die is 61 .

Notice that ´61

61

361=

Can you see why a tree diagram is too diffi cult here?

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If A and B are independent events, then the probability of both occurring is the product of their probabilities.

P AB P A P B$=] ] ]g g g

EXAMPLES

1. The probability of getting a 6 when rolling a die is 61 . Find the

probability of getting a double 6 when rolling two dice.

Solution

double ´P 6

61

61

361

=

=

] g

2. The probability that a certain missile will hit a target is

87 . Find the

probability that the missile will hit two targets (a) miss two targets. (b)

Solution

(a)

´P 287

87

6449

hits =

=

] g

(b)

P 187

81

miss = −

=

] g

´P 2

81

8

641

1misses =

=

] g

Sometimes the outcomes change when looking at more than one event.

Using these answers, could you calculate the probability

that the missile hits one target and not the other?

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EXAMPLES

1. Maryam buys 5 tickets in a raffl e in which 95 tickets are sold altogether. There are two prizes in the raffl e. What is the probability of her

winning both fi rst and second prizes? (a) winning neither prize? (b) winning at least one of the prizes? (c)

Solution

Probability of winning fi rst prize (a) 955=

After winning fi rst prize, Maryam’s winning ticket is taken out of the draw. She then has 4 tickets left in the raffl e out of a total of 94 tickets left.

Probability of winning second prize 944=

WW ´P

955

944

8932

=

=

] g

Probability of not winning fi rst prize (b) 1955

9590

= −

=

After not winning fi rst prize, Maryam’s 5 tickets are all left in the draw, but the winning ticket is taken out, leaving 94 tickets in the raffl e.

Probability of winning second prize 945=

Probability of not winning second prize 1945

9489

= −

=

LL ´P

9590

9489

893801

=

=

] g

(c) W LLP P1

1 893801

89392

at least one = −

= −

=

] ]g g

2. I choose 3 balls at random from a bag containing 7 blue and 5 red balls.

Find the probability of getting 3 blue balls if (a) I replace each ball before choosing the next one (i) I don’t replace each ball before choosing the next one. (ii)

Find the probability of getting at least one red ball (without (b) replacement).

ch8.indd 334 6/29/09 10:17:40 PM


Solution

(i) (a) P B127=] g

So

´ ´P BBB

127

127

127

1728343

=

=

] g

(ii) P B127=] g

After the fi rst blue ball has been chosen, the bag now contains 6 blue and 5 red balls.

ndP B2116=] g

After the second blue ball has been chosen, the bag contains 5 blue and 5 red balls.

3rdP B105=] g

So ´ ´P BBB

127

116

105

447

=

=

] g

(b) P R R

P BBB

1

1

1447

4437

at least one no= −= −

= −

=

] ]]

g gg

1. If 2 dice are thrown, fi nd the probability of throwing two 6s.

2. Find the probability of getting 2 heads if a coin is tossed twice.

3. A coin is tossed 3 times. Find the probability of tossing 3 tails.

4. A card has a picture on one side and is blank on the other. If the card is thrown into the air twice, fi nd the probability that it will land with the picture side up both times.

5. A box contains 2 black balls, 5 red balls and 4 green balls. If I draw out 2 balls at random, replacing the fi rst before drawing out the second, fi nd the probability that they will both be red.

6. The probability of a conveyor belt in a factory breaking down at any one time is 0.21. If the factory has 2 conveyor belts, fi nd the probability that at any one time

both machines will break down (a) neither machine will break (b)

down.

8.4 Exercise s

ch8.indd 335 6/29/09 10:17:41 PM


7. The probability of a certain plant fl owering is 93%. If a nursery has 3 of these plants, fi nd the probability that they will all fl ower.

8. An archery student has a 69% chance of hitting a target. If she fi res 3 arrows at a target, fi nd the probability that she will hit the target each time.

9. A bag contains 8 yellow and 6 green lollies. If I choose 2 lollies at random, fi nd the probability that they will both be green

if I replace the fi rst lolly (a) before selecting the second

if I don’t replace the fi rst lolly. (b)

10. I buy 10 tickets in a raffl e in which 250 tickets are sold. Find the probability of winning both fi rst and second prizes.

11. Two cards are drawn from a deck of 20 red and 25 blue cards (without replacement). Find the probability that they will both be red.

12. Find the probability of winning the fi rst 3 prizes in a raffl e if Peter buys 5 tickets and 100 tickets are sold altogether.

13. The probability of a pair of small parrots breeding an albino bird

is 332 . If they lay three eggs, fi nd

the probability of the pair not breeding any albinos (a) having all three albinos (b) breeding at least one albino. (c)

14. A photocopier has a paper jam on average around once every 2400 sheets of paper.

What is the probability that a (a) particular sheet of paper will jam?

What is the probability that (b) two particular sheets of paper will jam?

What is the probability that (c) two particular sheets of paper will not jam?

15. In Yahtzee, 5 dice are rolled. Find the probability of rolling

fi ve 6’s (a) no 6’s (b) at least one 6. (c)

16. The probability of a faulty computer part being manufactured at Omega

Computer Factory is 5000

3 .

If 2 computer parts are examined, fi nd the probability that

both are faulty (a) neither are faulty (b) at least one is faulty. (c)

17. A set of 10 cards is numbered 1 to 10 and two are drawn out at random with replacement. Find the probability of drawing

two odd numbers (a) two numbers that are (b)

divisible by 3 two numbers less than 4. (c)

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18. A bag contains 5 white, 4 black and 3 red marbles. If 2 marbles are selected from the bag at random without replacement, fi nd the probability of selecting

two red marbles (a) two black marbles (b) no white marbles (c) at least one white marble. (d)

19. The probability of an arrow hitting a target is 85%. If 3 arrows are shot, fi nd the probability as a percentage, correct to 2 decimal places, of

all arrows hitting the target (a) no arrows hitting the target (b) at least one arrow hitting the (c)

target.

20. A coin is tossed n times. Find the probability in terms of n of tossing

all heads (a) no tails (b) at least one tail. (c)

Tree diagrams and probability trees

When using the product rule to fi nd the probability of successive events occurring, sometimes there is more than one possible result. For example, when tossing two coins, there are two ways of getting a head and a tail (HT and TH). We add these results together.

P A B P A P Bor = +] ] ]g g g

We use tree diagrams or probability trees to combine the product and addition rules.

We use the product rule by multiplying along the branches and the addition rule by adding up the probabilities from different branches.

This is called the addition rule of

probability.

This is a probability tree as it has the

probabilities listed.

EXAMPLES

1. If 2 coins are tossed, fi nd the probability of tossing a head and a tail.

Solution

CONTINUED

ch8.indd 337 7/6/09 11:42:14 PM


( ) ( )

´ ´

P P HT P TH

21

21

21

21

41

21

41

head and tail = +

= +

=

=

+

]c c

gm m

2. A person has probability of 0.2 of winning a prize in a competition. If he enters 3 competitions, fi nd the probability of his winning

2 competitions (a) at least 1 competition. (b)

Solution

Probability of losing is 0.8 (a)

( ) ( ) ( ) ( )

(0.2 0.2 0.8) (0.2 0.8 0.2) (0.8 0.2 0.2)

. . .

.

´ ´ ´ ´ ´ ´P W P WWL P WLW P LWW2

0 032 0 032 0 032

0 096

= + += + += + +=

(b)

( ) ( )

1 (0.8 0.8 0.8)

.

.

´ ´P W P LLL1

1 0 512

0 488

at least one = −= −= −=

Probability of losing is 1 – 0.2.

ch8.indd 338 6/29/09 10:17:44 PM


3. A bag contains 3 red, 4 white and 7 blue marbles. Two marbles are drawn at random from the bag

replacing the fi rst before the second is drawn (a) without replacement (b)

Find the probability of drawing out a red and a white marble in these cases.

Solution

(a)

( ) ( ) ( )

´ ´

P R W P RW P WR

143

144

144

143

19612

19612

496

and = +

= +

= +

=

c cm m

(b)

The probabilities in each set of branches must add

up to 1.

CONTINUED

ch8.indd 339 6/29/09 10:17:47 PM


( ) ( ) ( )

´ ´

P R W P RW P WR

143

134

144

133

18212

18212

9112

and = +

= +

= +

=

c cm m

1. Two coins are tossed. Find the probability of getting

2 heads (a) a head followed by a tail (b) a head and a tail. (c)

2. Three coins are tossed. Find the probability of getting

3 tails (a) 2 heads and 1 tail (b) at least 1 head. (c)

3. In a set of 30 cards, each one has a number on it from 1 to 30. If 1 card is drawn out, then replaced and another drawn out, fi nd the probability of getting

two 8s (a) a 3 on the fi rst card and an (b)

18 on the second card a 3 on one card and an 18 on (c)

the other card.

4. Five cards are labelled A , B , C , D and E . If 2 are selected at random, with replacement, fi nd the probability that they will be

both (a) A s an (b) A and a D.

5. A bag contains 5 red marbles and 8 blue marbles. If 2 marbles are chosen at random, with the fi rst

replaced before the second is drawn out, fi nd the probability of getting

2 red marbles (a) a red and a blue marble. (b)

6. A certain breed of cat has a 35% probability of producing a white kitten. If a cat has 3 kittens, fi nd the probability that she will produce

no white kittens (a) 2 white kittens (b) at least 1 white kitten. (c)

7. The probability of a certain type of photocopier in a school needing a service on any one day is 0.3. Find the probability that a school with 2 of these photocopiers will need to service, on a particular day,

1 machine (a) both machines (b) neither of them (c)

8. The probability of rain in May

each year is given by .103 A

school holds a fete in May for three years running. Find the probability that it will rain at

2 of the fetes (a) 1 fete (b) at least 1 fete (c)

8.5 Exercise s

ch8.indd 340 6/29/09 10:17:48 PM


9. A certain type of plant has a probability of 0.85 of producing a variegated leaf. If I grow 3 of these plants, fi nd the probability of getting a variegated leaf in

2 of the plants (a) none of the plants (b) at least 1 plant. (c)

10. A bag contains 6 white balls and 5 green balls. If 2 balls are chosen at random, fi nd the probability of getting a white and a green ball

with replacement (a) without replacement. (b)

11. A bag contains 3 yellow balls, 4 pink balls and 2 black balls. If 2 balls are chosen at random, fi nd the probability of getting a yellow and a black ball


12. Alan buys 4 tickets in a raffl e in which 100 tickets are sold altogether. There are two prizes in the raffl e. Find the probability that Alan will win

fi rst prize (a) both prizes (b) 1 prize (c) no prizes (d) at least 1 prize. (e)

13. Mary buys 20 tickets in a lottery that has 5000 tickets altogether. Find the probability that Mary will win

fi rst and second prize (a) second prize only (b) neither fi rst nor (c)

second prize.

14. Two musicians are selected at random to lead their band. One person is chosen from Band A , which has 8 females and 7 males, and the other is chosen from Band B , which has 6 females and 9 males. Find the probability of choosing

2 females (a) 1 female and 1 male. (b)

15. The two machines in a workshop

each have a probability of 451

of breaking down. Find the probability that at any one time

neither machine will be (a) broken down

1 machine will be broken (b) down.

16. Two tennis players are said to

have a probability of 52 and

43 respectively of winning a

tournament. Find the probability that

1 of them will win (a) neither one will win. (b)

17. If 4 dice are thrown, fi nd the probability that the dice will have

four 6’s (a) only one 6 (b) at least one 6. (c)

ch8.indd 341 6/29/09 10:17:48 PM


18. In a batch of 100 cars, past experience would suggest that 3 could be faulty. If 3 cars are selected at random, fi nd the probability that

1 is faulty (a) none are faulty (b) all 3 cars are faulty. (c)

19. In a certain poll, 46% of people surveyed liked the current government, 42% liked the opposition and 12% had no preference. If 2 people from the survey are selected at random, fi nd the probability that

both will prefer the (a) opposition

1 will prefer the government (b) and the other will have no preference

both will prefer the (c) government.

20. A manufacturer of X brand of soft drink surveyed a city and found that 31 people liked X drinks best, 19 liked another brand better and 5 did not drink soft drink. If any 2 people are selected at random from that city, fi nd the probability that

one person would like the (a) X brand of soft drink

both people would not drink (b) soft drink.

21. A bag contains 5 red, 6 blue, 2 white and 7 green balls. If 2 are selected at random (without replacement), fi nd the probability of getting

2 red balls (a) a blue and a white ball (b) 2 green balls (c) at least 1 red ball. (d)

22. In a group of people, 32 are Australian born, 12 were born in Asia and 7 were born in Europe. If 2 of the people are selected at random, fi nd the probability that

they were both born in Asia (a) at least 1 of them will be (b)

Australian born both were born in Europe. (c)

23. There are 34 men and 32 women at a party. Of these, 13 men and 19 women are married. If 2 people are chosen at random, fi nd the probability that

both will be men (a) 1 will be a married woman (b)

and the other an unmarried man both will be married. (c)

24. In a certain city, the probability that the pollution level will be high is 0.27. If the pollution is monitored for 4 successive days, find the probability that the pollution levels will be

high on 2 days (a) high on 1 day (b) low on at least 1 day. (c)

ch8.indd 342 7/6/09 11:42:19 PM


25. At City Heights School it was found that 75% of students in year 12 study 13 units, 21% study 12 units and 4% study 11 units. If 2 students are selected at random from year 12, fi nd the probability that

1 student will study 12 units (a) at least 1 student will study (b)

13 units.

26. Three dice are rolled. Find the probability of rolling

3 sixes (a) 2 sixes (b) at least 1 six. (c)

27. A set of 5 cards, each labelled with the letters A, B, C, D and E, is placed in a hat and two selected at random without replacement. Find the probability of getting

(a) D and E Neither (b) D nor E on either card At least one (c) D.

28. The ratio of girls to boys at a school is four to fi ve. Two students are surveyed at random from the school. Find the probability that the students are

both boys (a) a girl and a boy (b) at least one girl. (c)

29. The number of cats to dogs at a pet hotel is in the ratio of 4 to 7. If 3 pets are chosen at random, fi nd the probability that

they are all dogs (a) just one is a dog (b) at least one is a cat. (c)

30. A set of 20 cards is numbered 1 to 20 and three are selected at random with replacement. Find the probability of selecting

all three 10’s (a) no 10’s (b) at least one 10.(c)

ch8.indd 343 6/29/09 10:17:51 PM


Test Yourself 8 1. The probability that a certain type

of seed will germinate is 93%. If 3 of this type of seeds are planted, fi nd the probability that

all will germinate (a) just 1 will germinate (b) at least 1 will germinate. (c)

2. A game is played where the differences of the numbers on 2 dice are taken.

Draw a table showing the sample (a) space (all possibilities).

Find the probability of getting a (b) difference of

(i) 3 (ii) 0 (iii) 1 or 2.

3. Mark buys 5 tickets in a raffl e in which 200 are sold altogether.

What is the probability that he will(a) (i) win(ii) not win the raffl e? If the raffl e has 2 prizes, fi nd the (b)

probability that Mark will win just 1 prize.

4. In a class of 30 students, 17 study history, 11 study geography and 5 study neither. One of these students is chosen at random. Find the probability that this student will

study geography but not history (a) study both history and geography. (b)

5. ‘In the casino, when tossing 2 coins, 2 tails came up 10 times in a row. So there is less chance that 2 tails will come up next time.’ Is this statement true? Why?

6. A set of 100 cards numbered 1 to 100 is placed in a box and one drawn at random. Find the probability that the card chosen will be

odd (a) less than 30 (b) a multiple of 5 (c) less than 30 or a multiple of 5 (d) odd or less than 30. (e)

7. One game has a probability of 53 of

winning and a second game has a

probability of 32 of winning. If Jenny

plays one of each game, fi nd the probability that she wins

both games (a) one game (b) neither game. (c)

8. A bag contains 5 black and 7 white marbles. Two are chosen at random from the bag (a) with replacement (b) without replacement. Find the probability of getting a black and a white marble.

9. There are 7 different colours and 8 different sizes of leather jackets in a shop. If Jean selects a jacket at random, fi nd the probability that she will select one the same size and colour as her friend does.

10. Each of a certain type of machine in a factory has a probability of 4.5% of breaking down at any time. If the factory has 3 of these machines, fi nd the probability that at any one time

all will be broken down (a) at least one will be broken down. (b)

ch8.indd 344 6/29/09 10:17:51 PM


11. A bag contains 4 yellow, 3 red and 6 blue balls. Two are chosen at random. Find the probability of choosing

2 yellow balls (a) a red and a blue ball (b) 2 blue balls. (c)

12. In a group of 12 friends, 8 have seen the movie Star Wars 20 and 9 have seen the movie Mission Impossible 6 . Everyone in the group has seen at least one of these movies. If one of the friends is chosen at random, fi nd the probability that this person has seen

both movies (a) only (b) Mission Impossible 6.

13. A game of chance has a 52 probability of

a win or a 83 probability of a draw.

If I play one of these games, fi nd the (a) probability of losing.

If I play 2 of these games, fi nd the (b) probability of

(i) a win and a draw (ii) a loss and a draw (iii) 2 wins.

14. A card is chosen at random from a set of 10 cards numbered 1 to 10. A second card is chosen from a set of 20 cards numbered 1 to 20. Find the probability that the combination number these cards make is

911 (a) less than 100 (b) between 300 and 500. (c)

15. A loaded die has a 32 probability of

coming up 6. The other numbers have an equal probability of coming up. If the

die is rolled, fi nd the probability that it comes up

2 (a) even. (b)

16. Amie buys 3 raffl e tickets. If 150 tickets are sold altogether, fi nd the probability that Amie wins

1st prize (a) only 2nd prize (b) 1st and 2nd prizes (c) neither prize. (d)

17. A bag contains 6 white, 8 red and 5 blue balls. If two balls are selected at random, fi nd the probability of choosing a red and a blue ball


18. A group of 9 friends go to the movies. If 5 buy popcorn and 7 buy ice creams, fi nd the probability that one friend chosen at random will have

popcorn but not ice cream (a) both popcorn and ice cream. (b)

19. The probability that an arrow will hit a

target is 98 . If 3 arrows are fi red, fi nd the

probability that

2 hit the target (a) at least 1 hits the target. (b)

20. The probability of winning Game A is 53

and the probability of winning Game B is

107 . Find the probability of winning

both games (a) neither game (b) one game. (c)

ch8.indd 345 6/29/09 10:17:53 PM

Test Yourself 4



1. In a group of 35 students, 25 go to the movies and 15 go to the football. If all the students like at least one of these activities, fi nd the probability that a student chosen at random will

go to both the movies and the (a) football

only go to the movies. (b)

2. A certain soccer team has a probability of 0.5 of winning a match and a probability of 0.2 of drawing. If the team plays 2 matches, fi nd the probability that it will

draw both matches (a) win at least 1 match (b) not win either match. (c)

3. A game of poker uses a deck of 52 cards with 4 suits (hearts, diamonds, spades and clubs). Each suit has 13 cards, consisting of an ace, cards numbered from 2 to 10, a jack, queen and king. If a person is dealt 5 cards fi nd the probability of getting

four aces (a) a fl ush (all cards the same suit). (b)

4. If a card is drawn out at random from a set of playing cards fi nd the probability that it will be

an ace or a heart (a) a diamond or an odd number (b) a jack or a spade. (c)

5. Bill does not select the numbers 1, 2, 3, 4, 5 and 6 for Lotto as he says this combination would never win. Is he correct?

6. In a set of 5 cards, each has one of the letters A, B, C, D and E on it. If two cards are selected at random without replacement, fi nd the probability that

both cards are A’s (a) one card is an A and the other is a D (b) neither card is an A or D. (c)

7. In the game of Yahtzee, 5 dice are rolled. Find the probability of rolling

all 6’s (a) all the same number (b)

8. Out of a class of 30 students, 19 play a musical instrument and 7 play both a musical instrument and a sport. Two students play neither.

One student is selected from the class (a) at random. Find the probability that this person plays a sport but not a musical instrument.

Two people are selected at random (b) from the class. Find the probability that both these people only play a sport.

9. A game involves tossing 2 coins and rolling 2 dice. The scoring is shown in the table.

Result Score (points)

2 heads and double 6

5

2 heads and double (not 6)

3

2 tails and double 6 4

2 tails and double (not 6)

2

Find the probability of getting (a) 2 heads and a double 6.

Find the probability of getting 2 tails (b) and a double that is not 6.

What is the probability that Andre will (c) score 13 in three moves?

What is the probability that Justin (d) will beat Andre’s score in three moves?

ch8.indd 346 6/29/09 10:17:54 PM

127PRACTICE ASSESSMENT TASK 1

1. Prove that the opposite sides of a parallelogram are equal.

2. Find all values of x for which the curve y x2 1 2= −^ h is decreasing.

3. Find ( ) .x x dx3 2 12 − +#

4. Find the maximum value of the curve y x x3 42= + − in the domain .≤ ≤x1 4−

5. The area of a rectangle is 4 m 2 . Find its minimum perimeter.

6. Show that ABDE is a parallelogram, given that ACDF is a parallelogram and .BC FE=

7. Find the primitive function of .x x3 48 +

8. Sketch the curve ,y x x x3 9 23 2= − − + showing all stationary points and infl exions.

9. Find the volume of the solid of revolution formed when the curve y x 13= + is rotated about

the (a) x -axis from x 0= to x 2= the (b) y -axis from y 1= to .y 9=

10. Find the area enclosed between the curve y x 12= − and the x -axis.

11. If ( ) ,f x x x x2 5 93 2= − + − fi nd ( )f 3l and ( ) .f 2−m

12. Evaluate ( ) .x x dx6 42

1+

3#

13. ABC is an isosceles triangle with D the midpoint of BC . Show that AD is perpendicular to BC .

14. Find the volume of the solid of revolution formed, correct to 3 signifi cant fi gures, if y x 22= + is rotated about

the (a) x -axis from x 0= to x 2= the (b) y -axis from y 2= to .y 3=

15. Find the domain over which the curve y x x x3 7 3 13 2= + − − is concave upwards.

16. If ( ) ,f x x x x2 7 94 3= − − + fi nd 1) 1) .,(1), ( (f f fl m What is the geometrical signifi cance of these results? Illustrate by a sketch of ( )y f x= at .x 1=

17. A piece of wire of length 4 m is cut into 2 parts. One part is bent to form a rectangle with sides x and 3 x , and the other part is bent to form a square with sides y . Prove that the total area of the rectangle and square is given by ,A x x7 4 12= − + and fi nd the dimensions of the rectangle and square when the area has the least value.

Practice Assessment Task SET 1

pat_1.indd 127 6/12/09 12:39:18 AM


18. The gradient function of a curve is given by ( ) .f x x4 3= −l If ( ) ,f 2 3= − fi nd ( ) .f 1−

19. Evaluate ( ) .x dx2 10

3+#

20. The following table gives values for

( ) .f xx1

2=

x 1 2 3 4 5

f(x) 141

91

161

251

Use the table together with Simpson’s

rule to evaluate xdx

1 2

5# correct to 3

signifi cant fi gures.

21. Show that Δ ABC is right angled at .B+

22. Find ( ) .x dx3 5 7+#

23. Evaluate ( ) .x x dx52

0

2+ −#

24. Find the point of infl exion on the curve ( ) .f x x x x3 4 13 2= − + −

25. Find an approximation to xdx

1

3# by using

Simpson’s rule with 3 function values (a) the trapezoidal rule with (b)

2 subintervals.

26. Find the stationary points on the curve ( )f x x x2 34 2= − + and distinguish between them.

27. Evaluate x dx5 1−1

2

# as a fraction.

28. (a) Find the area enclosed between the curve y x 12= − and the y -axis between y 1= and y 2= in the fi rst quadrant, to 3 signifi cant fi gures.

(b) This area is rotated about the y -axis. Find the exact volume of the solid formed.

29. Evaluate .y dy4

9

#

30. Find the area enclosed between the curve ( )y x 1 2= − and the line .y 4=

31. Prove Δ ABC and Δ CBD are similar.

32. Show that the curve y x2 4= has a minimum turning point at ( , ) .0 0

33. Find the maximum volume of a rectangular prism with dimensions x , 2 x and y and whose surface area is 12 m 2 .

34. If a function has a stationary point at ( , )1 2− and ( )xf ,x2 4= -m fi nd ( ) .f 2

35. Find the area enclosed between the curves y x2= and .y x x2 42= − + +

36. Find the minimum surface area of a closed cylinder with volume 100 m 3 , correct to 1 decimal place.

37. A curve has a stationary point at (3, 2) . If ( )xf ,x6 8= −m fi nd the equation of the function.

38 . Sketch the function ( ) ,f x x x x3 9 53 2= − − + showing any stationary points and infl exions.

39. Prove that the area of a rhombus is

A xy21= where x and y are the lengths of

the diagonals.

pat_1.indd 128 6/12/09 12:39:22 AM


40.

Given BG CD< and ,GF DE< prove

.ACAB

AEAF=

41. Given ( )f x x x2 12= − −l , evaluate (2)f if ( ) .f 0 5=

42. (a) Complete the statement: The primitive function of . . .xn =

(b) Explain why there is a constant C in the primitive function.

43. The graph below is the derivative of a function.

y

x

Which graph could represent the function? There may be more than one answer.

(a)

(b) y

x

(c) y

x

y

x

pat_1.indd 129 6/12/09 12:39:25 AM


(d) y

x

44. A function has dx

dy > 0. To which of the

following graphs does this apply?

(a) y

x

(b) y

x

(c) y

x

(d) y

x

45. The area of a rectangle with sides x and y is 45. The perimeter is given by:

(a) P x x45 2= +

(b) P x x45= +

(c) 2P x x90= +

(d) 2P x x45= +

46. The area enclosed between the curve – ,y x 13= the y -axis and the lines y 1= and 2y = is given by

(a) ( )x dy11

2 3 −#

(b) ( )y dy11

2+#

(c) ( )y dy11

23 +#

(d) ( )y dy11

23 +#

pat_1.indd 130 6/12/09 12:39:26 AM


47.

D

A

B

O

C

Given O is the centre of the circle, triangles OAB and OCD can be proven to be congruent using the test

(a) SSS (b) SAS (c) AAS (d) RHS.

48. The area bounded by the curve ,y x3= the x -axis and the lines 2x = − and 2x = is

8 units (a) 2 2 units (b) 2 0 units (c) 2 4 units (d) 2 .

49. y

x

The curve has

(a) ,dx

dy

dx

d y0 0> >

2

2

(b) ,dx

dy

dx

d y0 0> <

2

2

(c) ,dx

dy

dx

d y0 0< >

2

2

(d) ,dx

dy

dx

d y0 0< <

2

2

50. A cone with base radius r and height h has a volume of 300 cm 3 . Its slant height is given by

(a) π

πlh

h 9003

= +

(b) π

lh

h 9002

= +

(c) π

lh

h 810 0002

= +

(d) π

πlh

h 9003

= +

pat_1.indd 131 6/12/09 12:39:28 AM


1. The population of a city over t years is given by the formula .P e100 000 . t0 71= After how many years, to 1 decimal place, will the population become 1 million?

2. Evaluate xdx

1

3# correct to 3 decimal places.

3. The acceleration of a particle is given by msa t6 2= − and the particle is initially at rest at the origin. Find where it will be after 3 seconds.

4. Find .log251

5

5. Write log xe as an equation with x in terms of y . Hence fi nd the value of x, to 3 signifi cant fi gures, when . .y 1 23=

6. Differentiate .x e x3 2+

7. Solve .log161 4x =

8. If 100 g of a substance decays to 80 g after 3 years, fi nd, to 1 decimal place,

its mass after 10 years (a) the rate of decay after 10 years (b) when it will decay to 50 g. (c)

9. A particle has displacement x in centimetres over time t in seconds according to the formula .sinx t2 3=

Find its velocity after (a) s.π2

Show that the acceleration is given (b) by .x x9= −p

10. Water is fl owing out of a pool at the rate given by R t20= litres per minute. If the volume of water in the pool is initially 8000 L, fi nd

the volume after 5 minutes (a) how long it will take to empty the (b)

pool.

11. Find the volume, to 1 decimal place, of the solid formed by rotating y ex= about the x -axis from x 1= to .x 3=

12. Find the derivative of ( ) .log x4 3e3+

13. Find .x x

x dx3 3 2

2 12 + −

+#

14. The volume in litres of a rectangular container that is leaking over time t minutes is given by .V t t4 1002= − + + Find

the initial volume (a) the volume after 10 minutes (b) the rate of change in volume after (c)

10 minutes how long it will take, to 1 decimal (d)

place, until the container is empty.

15. Find ( ) .x x x dx6 2 42 1− + −#

16. Find the exact value of .x

dx5

31

7

+#

17. The velocity v cms -1 of a particle over

time t is given by .t

v1

2+

= If the particle

is initially 3 cm from the origin, fi nd its displacement after 10 s, (a)

to 2 signifi cant fi gures its acceleration after 5 s, (b)

to 1 signifi cant fi gure .

18. Find ( ) .e dx1x4 +#

19. Differentiate .ex

x2

20. The rate at which an epidemic of measles in a certain city is spreading is proportional to the number of people

with measles. That is, .dtdP kP= If 40

people initially have measles and after 10 days, 110 have measles, fi nd


pat_2.indd 253 6/12/09 3:14:34 AM


the value of (a) k to 3 signifi cant fi gures how many people will have measles (b)

after 6 weeks after how long, to the nearest day, (c)

300 people will have measles the rate at which the disease will be (d)

spreading after (i) 10 days (ii) 6 weeks.

21. (a) Find the value of log 73 by changing the base to e .

Differentiate (b) log x3 by changing the base to e .

22. Find the exact volume of the solid of revolution formed when the curve y ex= is rotated about the x -axis from x 1= to .x 3=

23. A particle is moving such that its displacement after t seconds is given by sinx t3 2= metres.

Find the initial velocity and (a) acceleration.

Find the maximum displacement. (b) Find the times when the particle will (c)

be at rest. Prove that the acceleration is given (d)

by .a x4= −

24. Find the area enclosed between the curve ,logy xe= the y -axis and the lines y 1= and ,y 2= correct to 3 signifi cant fi gures.

25. The acceleration of a particle is given by .mst6 12 2− − If the particle is initially at rest 2 m to the left of the origin, fi nd its displacement after 5 seconds.

26. Solve .7 3x2 =

27. Find the equation of the tangent to y ex 1= + at the point where .x 1= −

28. Find the stationary point on the curve y xe x2= and determine its nature .

29. The length of an arc in a circle of radius 2 cm is 1.6 cm. Find the area, correct to 2 decimal places, of the

sector (a) minor segment cut off by this arc .(b)

30. A certain chemical treatment of blue-green algae in a river causes it to decrease at a rate proportional to the amount of blue-green algae in the river. If 250 kg of blue-green algae reduces to 150 kg after 3 months, fi nd how long it will take, to the nearest month, to reduce the blue-green algae to 20 kg.

31. An angle of °30 is subtended at the centre of a circle with radius 5 cm. Find the exact

arc length (a) area of the sector. (b)

32. Sketch cos xy2

2= for ≤ ≤ .πx0 2

33. Find 2 ( ) .sin x dx20

π

#

34. Differentiate ( )log sinxe .

35. Find the derivative of ( )tan e 1x5 + .

36. Evaluate 3 ,sec x dx2

0

π

# giving the exact

value.

37. (a) Sketch siny x2= for ≤ ≤ πx0 2 . (b) Find the area bounded by the curve

,siny x2= the x -axis and the lines x 0= and .πx =

38. Evaluate 4

0tanx dx

π

# correct to 3 decimal

places by using Simpson’s rule with three ordinates.

39. Differentiate (a) sine xx (b) tan x3 (c) πcos x

22 3 −b l

pat_2.indd 254 6/12/09 3:14:35 AM


40. A particle moves so that its displacement after time t seconds is given by .x e5 mt2=

Find its velocity after 2 s, to (a) 3 signifi cant fi gures.

Show that its acceleration is given by (b) .a x4=

Find the initial acceleration. (c)

41. Evaluate ,x

dx30

5

+# giving the exact value.

42. A gardener wishes to make a rectangular garden bed using a wall as one of the sides. She has 24 m of edging strip to place around the garden. What dimensions will give the maximum garden area?

43. Find the volume of the solid formed when the curve cosy x= is rotated about the x -axis from x 0= to .πx

3=

44. Sketch the curve ,y x x x3 4 12 14 3 2= − − + showing any stationary points.

45. Find the equation of the tangent to the curve tany x3= at the point where .πx

4=

46. Find the exact area enclosed between the

curve siny x= and the line y21= for the

domain ≤ ≤ .πx0 2

47. Differentiate ( ) .sin ex3

48. Find the exact area bounded by the curve ( ),logy x 4e= + the y -axis and the lines y 0= and .y 1=

49. Find the exact value of

(a) πcos4

7

(b) .π πsin 23

−b l

50. The length of an arc in a circle of radius 6 cm is π cm.7 Find the area of the

sector (a) minor segment cut off by this arc .(b)

51. Differentiate ( )tan log x 1e + .

52. Find .x

x x dx3 2 52

2 − +#

53. Find the volume of the solid formed, correct to 2 decimal places, when the curve logy xe= is rotated about the x -axis from x 1= to ,x 4= by using the trapezoidal rule with 3 subintervals.

54. Find ( ) .πsine x dxx5 −#

55. Solve graphically sin x x 1= − for ≤ ≤ πx0 2 .

56. Find the exact area enclosed between the curve ,y ex= the x -axis, the y -axis and the line .x 2=

57. Evaluate 3

.πcos x dx2

+π

π

b l#

58.

xπ2

π 3π2

2π

y

-1

1

2

3

4

5

-2

The graph has the equation (a) – cosy x4 2= (b) cosy x4 2= + (c) x – cosy 2 2= (d) – .cosy x2 4=

59. A radioactive substance decays by 60% after 200 years. This information can be shown by the equation of its mass

(a) . MeM0 6 k200= (b) . MeM0 6 k200= − (c) . MeM0 4 k200= (d) . MeM0 4 k200= −

pat_2.indd 255 6/12/09 3:14:35 AM


60. The value of sin π3

−b l is (a)

23−

(b) 23

(c) 21−

(d) 21 .

61. A particle has a displacement of .cosx t4 7= Its acceleration can be written as

(a) x x49=p (b) x x196= −p (c) x x196=p (d) x x49= −p .

62. Evaluate x

x dx2 5

32 −

#

(a) ( )ln x C3 2 52 − +

(b) ( )ln x

C4

3 2 52 −+

(c) ( )ln x

C12

2 52 −+

(d) ( )ln x

C3

4 2 52 −+

63. The rate at which a waterfall is fl owing over a cliff is m s .R t t4 3 2 3 1= + − Find the amount of water fl owing after a minute if the amount of water is 10 970 m 3 after 20 seconds.

223 220 m (a) 3 8800 m (b) 3 225 370 m (c) 3 226 250 m (d) 3

pat_2.indd 256 6/12/09 3:14:36 AM


1. Find the 5th term of . . .4 252 1

2511+ + +

2. The probability that a seed will produce a

plant with a white fl ower is 71 . If 4 seeds

are planted, fi nd the probability that no plants will have white fl owers (a) exactly 1 plant will have white fl owers (b) at least 1 plant will have white fl owers. (c)

3. By writing the recurring decimal . . . .0 444 as an infi nite geometric series or otherwise, express it as a rational number.

4. Find in index form the 10th term of

. . .343

163+ + +

5. George deposits $400 at the beginning of each year in an investment account that pays 13% interest per annum. How much will there be in the account after 5 years?

6. The second term of a geometric series is 52 and the 4th term is 13. Find two series that satisfy these requirements.

7. Find which term -370 is in the series . . .1 8 17 + − −

8. A man promises his son a sum of money on his 18 th birthday, made up of $10 for his fi rst year of life, $15 for his second year, $20 for his third year and so on, up to 18 years. How much will his son receive?

9. Two dice are thrown and the total of the dice is noted. Find the probability of throwing

double l’s (a) any double (b) at least one 3 (c) a total of 6 (d) a total of at least 8. (e)

10. Find the sum of the fi rst 9 terms of . . .a a a

3 9+ + +

11. A plant grows so that it increases its height each month by 0.2 of its previous month’s height. If it grows to 3 m, fi nd its height in the fi rst month.

12. The population of an insect colony is given by the formula ,P t t 3 000 0003= − + where t is time in days, Find

the initial population .(a) the population after a week .(b) the rate at which the population will (c)

be increasing after a week.

13. Find the stationary point on the curve 3( )y x 2= − and determine its nature. Hence sketch the curve.

14. Each card in a set of 100 cards has a different number on it, ranging from 1 to 100. If one card is chosen at random, fi nd the probability of getting

an even number less than 30 (a) an odd number or a number divisible (b)

by 9.


pat_3.indd 347pat_3.indd 347 7/16/09 5:28:04 PM7/16/09 5:28:04 PM


15. Two families travelling on holidays drive along roads that intersect at right angles. One family is initially 230 km from the intersection and drives towards the intersection at an average of 65 kmh −1 . The other family is initially 125 km from the intersection and travels towards it at an average of 80 kmh −1 .

Show that their distance (a) apart after t hours is given by .d t t10 625 49 000 68 5252 2= − +

Hence fi nd how long it will take them (b) to reach their minimum distance apart.

Find their minimum distance apart. (c)

16. Find the 50 th term of . . .3 7 11+ + + and calculate the sum of 50 terms .

17. If I buy 10 tickets, fi nd the probability of my winning both fi rst and second prizes in a raffl e in which 100 tickets are sold.

18. The n th term of a series is given by 7 n − 3

Find the fi rst 3 terms and the 12 th (a) term.

Evaluate (b) .( )n7 3n 1

20

−=

/

Which term is 200? (c)

19. Find the exact value of π πsin4

+b l

20 . Sketch the graph of (a) y x x5 62= − + (b) 2 4 for 0 ≤ ≤ πsiny x x= (c) y e x= −

21. The 4 th term of an arithmetic series is 18 and the 8 th term is 62. Find the series.

22. A bag contains 7 red and 5 white marbles. Two are drawn out at random. Find the probability of drawing out

2 white marbles (a) 1 white and 1 red marble.(b)

23. I put $2000 in the bank, where it earns interest at the rate of 12% p.a., paid quarterly. How much will there be in the account after 3 years?

24. The displacement of a pendulum is given by cosx t3 4= cm after time t seconds.

Find the equation for the velocity of (a) the pendulum.

Find the equation for its acceleration. (b) Find the initial displacement. (c) Find the times when the pendulum (d)

will be at rest. What is the displacement at these (e)

times? When will there be zero (f)

displacement? Show that acceleration (g) .a x16= −

25. Express 0.17o as a rational number.

26. A coin is tossed and a die thrown. Find the probability of getting

a head and a six (a) a tail and an odd number. (b)

27. I borrow $5000 at 18% interest p.a. and make equal monthly payments over 3 years, at the end of which the loan is fully paid out. Find the amount of each monthly payment.

28. Prove that ABC3 is right angled.



29. The geometric series . . .x x x2 3+ + + has a sum to infi nity of 5. Find the value of x .

30. Prove that if one pair of adjacent sides is equal in a rhombus, then all sides are equal.

31. Josie buys 5 lottery tickets in which 200 000 tickets are sold. What is the probability that she wins 1st prize?

32. The mass (in grams) of a radioactive substance is given by M M e kt

0= − , where t is time in years. Find

(a) k if its mass is halved after 28 years when the mass will be 35% of the (b)

original amount the time taken to reduce the mass (c)

by 40% .

33. Evaluate . . .7 12 17 872+ + + + .

34. A rectangle is cut from a circular disc of radius 15 cm. Find the area of the largest rectangle that can be produced.

35. The equation for displacement of a particle is given by x e3 2 metrest4= + after time t seconds.

Find the initial velocity .(a) Find the exact acceleration after 1 s. (b) Show that the acceleration is always (c)

4 times greater than the velocity. Sketch the graph of velocity over time (d) t.

36. Find the fi rst value of n for which the

sum of the series . . .20 454+ + + is

greater than 24.85.

37. If ,siny x7= show that dx

d yy49

2

2

= − .

38. Evaluate .52∞

rr 2=/

39. A farmer wants to fence a rectangular paddock and a square paddock with a combined area of 5000 m 2 . If the length of the rectangular paddock is to be 3 times the size of its breadth

fi nd the dimensions of the paddocks (a) that give the largest perimeter

calculate the cost of fencing the (b) paddocks, at $19.95 per metre .

40. Two dice are rolled, and the numbers of the dice are totalled. Find the probability of rolling

a total of 8 (a) a total less than 7 (b) a total greater than 9 (c) a total of 4 or 5 (d) an odd total .(e)

41. Sketch the function ( )f x x x x3 4 124 3 2= + − showing all stationary points.

42. The area of a circle is 5 π and an arc 3 cm long cuts off a sector with an angle of θ subtended at the centre. Find

(a) θ in degrees and minutes the area of the minor segment cut off (b)

by the arc, correct to 2 decimal places .

43. Ian pays $50 into a superannuation fund at the beginning of each month, where it earns interest of 4% p.a. How much will be in the fund at the end of 20 years?

44. (a) Show that . . .log log log3 9 27+ + + is an arithmetic series.

Find the exact sum of 20 terms of (b) the series.

45. Sketch the curve ,y x x x2 3 36 13 2= + − + showing any stationary points and points of infl exion.



46. A chemical reaction causes the amount of hydrogen to be reduced at a rate proportional to the amount of hydrogen present at any one time. If the amount of hydrogen is given by the formula A A e0= ,kt− and 800 mL reduces to 650 mL after 3 minutes, fi nd

the amount of hydrogen after 1 hour (a) how long it will take for the (b)

hydrogen to reduce to 100 mL .

47. Find the domain over which the curve 3 24 7y x x x3 2= + − + is concave

downwards. (a) x 1< − (b) x 1> −

x4 2< <−(c) ,x x4 2< >−(d)

48. I borrow $10 000 over 5 years at 1.85% monthly interest. How much do I need to pay each month?

49. Find the probability of drawing out a blue and a white ball from a bag containing 7 blue and 5 white balls if the fi rst ball is not replaced before taking out the second.

(a) 12170

(b) 14470

(c) 17 4241225

(d) 13270

50. The limiting sum of a geometric series exists when

(a) ||r 1> (b) || 1r < (c) || 1≥r (d) || ≤r 1

51. Find the limiting sum of the series

. . .32

21

83+ + +

2 (a)

(b) 131

(c) 232

(d) 98

52. Which statement is the same as ?3 7x = There may be more than one answer.

(a) logx37=

(b) log x 7 3 =

(c) log x73 =

(d) log

logx

3

7=

53. The n th term of the series …7 49 343+ + + is

7 (a) n (b) 7 n 1− (c) n7 1− (d) 7n

54 . An amount of $6000 is invested at 3.5% p.a. with interest paid quarterly. Find the balance after 10 years.

$8501.45 (a) $8463.59 (b) $6546.16 (c) $6899.96 (d)

55. The limiting sum of . . .53

52

154+ + + is

(a) 51

(b) 109

(c) 154

(d) 156



56. In a group of 25 students, 19 catch a train to school and 21 catch a bus. If one of these students is chosen at random, fi nd the probability that the student only catches a bus to school.

(a) 256

(b) 2521

(c) 53

(d) 203

57. The formula for the sum 1. . . . . .1 1 03 1 03 1 03 n2+ + + + − is

(a) .

. ( . )S

1 03 11 03 1 03 1n 1

=−

−−

(b) .

. ( . )S

1 03 11 03 1 03 1n

=−

−

(c) .

.S1 03 1

1 03 1n 1

=−

−−

(d) ..S

1 03 11 03 1n

=−−



Sample Examination Papers

This chapter contains two sample Mathematics HSC papers. They are designed to give you some practice in working through an examination. These papers contain work from the Year 11 Preliminary Course, as up to 20% of questions are allowed to contain this work. You may need to revise this work before you try these papers.

If you can, set yourself a time limit and work under examination conditions. Give yourself 3 hours to do each Mathematics paper. Try not to look up any notes while working through these papers.

MATHEMATICS—PAPER 1 Time allowed — Three hours

( Plus 5 minutes ’ reading time )

DIRECTIONS TO CANDIDATES

• All questions may be attempted. • All questions are of equal value. All necessary working should be shown • in every question, as marks are awarded for this. Badly arranged or careless work may not receive marks.

QUESTION 1 (a) Find, correct to 2 decimal places, the

value of . .

. .´15 7 2 4

28 3

(b) Factorise 3 x 2 - 11 x + 6.

(c) Solve the equation .x x2 3

1 5− − =

(d) The volume of a cone is given by

πV r31 2= . If a cone has volume 12 m 3 ,

fi nd its radius correct to 2 decimal places.

(e) Two cities are 3295 km apart. Write this number correct to 2 signifi cant fi gures.

(f) Solve | | .x 3 7<+

QUESTION 2 (a)

(i) Copy the diagram into your examination booklet. (ii) Show that AC = CE . (iii) Find the length of DE .

(b)

(i) Copy the diagram into your examination booklet. (ii) Use the sine rule to calculate MO correct to 1 decimal place. (iii) Find MP to the nearest metre.

QUESTION 3 (a) Differentiate

(i) x x5 13+ +

(ii) ln x x3 1+

(iii) (2 x + 3) 5

(b) Find (i) x e dx−( )x−#

(ii) ( )θ θsin d1π

0+#

samp.indd 352 7/9/09 8:41:40 PM

353SAMPLE EXAMINATION PAPERS

(c) (i) Rationalise the denominator of

.2 15−

(ii) Find integers a and b such that

.a b2 15−

= +

QUESTION 4 (a) (i) Plot points A (3, 2) and B (-1, 5)

on a number plane. (ii) Show that line AB has equation

3 x + 4 y - 17 = 0. (iii) Find the perpendicular distance from the origin to line AB . (iv) Find the area of triangle OAB where O is the origin.

(b)

Find the length of side BC using the cosine rule and give your answer correct to the nearest centimetre.

QUESTION 5 (a) Kate earns $24 600 p.a. In the

following year she receives a pay increase of $800. Each year after that, she receives a pay increase of $800.

(i) What percentage increase did Kate receive in the fi rst year? (ii) What will be Kate’s salary after 12 years? (iii) What will Kate’s total earnings be over the 12 years?

(b) A function is given by y = x 3 - 3 x 2 - 9 x + 2.

(i) Find the coordinates of the stationary points and determine their nature. (ii) Find the point of infl exion. (iii) Draw a sketch of the function.

QUESTION 6 (a) A plant has a probability of

72 of

producing white fl owers. If 3 plants are chosen at random, fi nd the probability that

(i) 2 plants will produce white fl owers (ii) at least 1 plant will produce white fl owers .

(b) Solve sinx23= for ° ≤ ≤ °.x0 360

A particle (c) P moves so that it is initially at rest at the origin, and its acceleration is given by a = 6 t + 4 cms -2 .

(i) Find the velocity of P when t = 5. (ii) Find the displacement of P when t = 2.

QUESTION 7 (a)

ABCD is a parallelogram with AD = 1, AB = 2 and °ADX 60+ = . AX is drawn so that it bisects DC .

(i) Copy the diagram into your examination booklet. (ii) Show that ADX is an equilateral triangle. (iii) Show that AXB is a right-angled triangle. (iv) Find the exact length of side BX .

(b) On a number plane, shade in the region given by the two conditions y ³ x 2 and x + y £ 3.

(c) The number of people P with chickenpox is increasing but the rate at which the disease is spreading is slowing down over t weeks.

(i) Sketch a graph showing this information.

(ii) Describe dtdP and

dtd P

2

2

for this

information.

samp.indd 353 6/30/09 5:25:53 AM


QUESTION 8 (a) Consider the function given by

logy x10= . (i) Copy and complete the table, to 3 decimal places, in your examination booklet.

x 1 2 3 4 5

y 0

(ii) Apply the trapezoidal rule with 4 subintervals to fi nd an approximation, to 2 decimal places, of .log x dx101

5#

(b) A population of mice at time t in weeks is given by P P ekt

0= , where k is a constant and P 0 is the population when t = 0.

(i) Given that 20 mice increase to 100 after 6 weeks, calculate the value of k , to 3 decimal places. (ii) How many mice will there be after 10 weeks? (iii) After how many weeks will there be 500 mice?

QUESTION 9 (a)

The surface area of a cylinder is given by the formula ( ) .S r r hπ2= +

A cylinder is to have a surface area of 160 cm 2 .

(i) Show that the volume is given by V r rπ80 3= − . (ii) Find the value of r , to 2 decimal places, that gives the maximum volume. (iii) Find the maximum volume, to 1 decimal place.

(b) Kim invests $500 at the beginning of each year in a superannuation fund. The money earns 12% interest per annum. If she starts the fund at the beginning of 1996, what will the fund be worth at the end of 2025?

QUESTION 10 (a)

(i) Find the exact point of intersection of the curve y = e x and the line y = 4. (ii) Find the exact shaded area enclosed between the curve and the line.

(b) (i) The quadratic equation x 2 + ( k - 1) x + k = 0 has real and equal roots. Find the exact values of k in simplest surd form.

(ii) If k = 5, show that x 2 + ( k - 1) x + k > 0 for all x .

(c) A parabola has equation y = x 2 + 2 px + q . (i) Show that the coordinates of its vertex are (- p , q - p 2 ). (ii) Find the coordinates of its focus. (iii) Find the distance between the point ( m , 3 m 2 + q ) and point P vertically below it on the parabola x 2 = 8 y when q > 0.

samp.indd 354 6/30/09 5:25:54 AM


y = x2 + 2px + q

x2 = 8y

P

(m, 3m2 + q)

y

x

(iv) Find the minimum distance between these two points when m + q = 5.

MATHEMATICS—PAPER 2 Time allowed — Three hours

( Plus 5 minutes ’ reading time )

QUESTION 1 (a) Solve | |x 3 5− = . (b) If f ( x ) = 5 - x 2 , fi nd x when f ( x ) = -4 .

(c) Find the exact value of πsin6

5 .

(d) Factorise fully a 3 - 2 a 2 - 4 a + 8 . (e) Simplify 2 24 150− . (f) Evaluate log a 50 if log a 5 = 1.3 and

log a 2 = 0.43 . (g) Find the midpoint of (-3, 4) and

(0, -2) .

QUESTION 2 Plot points ,,A 5 1

21−e o ,B 1 1

21e o and

C (-4, -1) on a number plane. Show that the equation of line (a) AB is

given by 3 x + 4 y - 9 = 0. Find the equation of the straight line (b)

l through C that is perpendicular to AB . Find the point of intersection (c) P of the

two lines, AB and l . Find the area of triangle (d) ABC. Find the coordinates of (e) D such that

ADCB is a rectangle.

QUESTION 3 (a) Differentiate

(i) x cos x (ii) e 5 x (iii) ( )log x2 1e

2 − (b) Find the indefi nite integral (primitive

function) of (i) (3 x - 2) 4 (ii) 3 sin 2 x

(c) Evaluate ( )e e dxx x

0

3− −# .

(d) For a certain curve, dx

d yx18 6

2

2

= − . If

there is a stationary point at (2, −1), fi nd the equation of the curve.

QUESTION 4 (a) Mark plays a game of chance that has

a probability of 52 of winning the game.

Hong plays a different game in which

there is a probability of 83 of winning.

Find the probability that (i) both Mark and Hong win their games (ii) one of them wins the game (iii) at least one of them wins the game .

(b) (i) On a number plane, shade the region where y ³ 0, x £ 3 and y £ x + 2 .

(ii) Find the area of this shaded region. (iii) This area is rotated about the x -axis. Find the volume of the solid formed.

(c) (i) Sketch the graph of y x2 1= − on a number plane. (ii) Hence solve x2 1 3<− .

QUESTION 5 (a) For the curve y = 2 x 3 - 9 x 2 + 12 x - 7

(i) fi nd any stationary points on the curve and determine their nature (ii) fi nd any points of infl exion (iii) sketch the curve in the domain -3 £ x £ 3 .

samp.indd 355 6/30/09 5:25:56 AM


(b) A plane leaves Bankstown airport and fl ies for 850 km on a bearing of 100°. It then turns and fl ies for 1200 km on a bearing of 040°.

(i) Draw a diagram showing this information. (ii) How far from the airport is the plane, to the nearest km?

(c) Simplify ( )( )θ θ θθcosec cot cosec cot+ − .

QUESTION 6 (a) (i) Find the equation of the normal

to the curve y = x 2 at point P (-2, 4). (ii) This normal cuts the parabola

again at point Q . Find the coordinates of Q . (iii) Find the shaded area enclosed between the parabola and the normal, to 3 signifi cant fi gures.

(b) The graph below shows the displacement of a particle over time t .

(i) When is the particle at the origin? (ii) When is the particle at rest?

(c) The temperature T of a metal is cooling exponentially over t minutes. It cools down from 97°C to 84°C after 5 minutes. Find

(i) the temperature after 15 minutes (ii) when it cools down to 20°C.

QUESTION 7 (a) (i) Find the area bounded by the

curve tany x= , the x -axis and the lines

x = 0 and πx4

= , by using Simpson’s rule

with 5 function values (to 2 decimal places).

(ii) Differentiate ln (cos x ).

(iii) Hence evaluate tan x dx

π

0

4# to 2 decimal places.

(b) A bridge is 40 metres long, held by wires at A and C with angles of elevation of 47° and 23° as shown.

(i) Find the length of AD to 3 signifi cant fi gures. (ii) Find the height of the bridge BD to 1 decimal place.

(c) Triangle BEC is isosceles with BC = CE . Also °BEC 50+ = , °ABE 130+ = , and °ADC 80+ = .

Prove that ABCD is a parallelogram.

QUESTION 8 (a) (i) Sketch the curve

≤ ≤ πsiny x x3 2 0 2for= . (ii) On the same set of axes, sketch

xy2

= .

(iii) How many roots does the equation

3 sin xx2

2 = have in this domain?

(b) Solve 2 sin x - 1 = 0 for 0° £ x £ 360°. (c) If log p3x = and log q2x = , write in

terms of p and q (i) log 12x (ii) log x2x

(d) A stack of oranges has 1 orange at the top, 3 in the next row down, then each row has 2 more oranges than the previous one.

samp.indd 356 6/30/09 5:25:57 AM


(i) How many oranges are in the 12th row? (ii) If there are 289 oranges stacked altogether, how many rows are there?

QUESTION 9 (a) The diagram below shows the graph

of a function y = f ( x ).

(i) Copy this diagram into your writing booklet. (ii) On the same set of axes, draw a sketch of the derivative ( )f xl of the function.

(b) ‘A bag contains yellow, blue and white marbles. Therefore if I choose one marble at random from the bag, the

probability that it is blue is 31 .’

Is this statement true or false? Explain why in no more than one sentence.

(c) Solve the equation ( )ln lnx x2 2 3= + .

(d)

The diagram shows the graph of y = e x and the tangent to the curve at x = k .

(i) Find the gradient of the tangent at x = k .(ii) Find the equation of the tangent at x = k .

(iii) Find the value of k if the tangent has an x -intercept of 2 .

(e) Find the area of the sector to 1 decimal place .

QUESTION 10 (a) The graph of y = ( x + 2) 2 is drawn

below. Copy the graph into your writing booklet.

(i) Shade the region bounded by the curve, the x -axis and the line x = 1. (ii) This area is rotated about the x -axis. Find the volume of the solid of revolution formed.

(b) The accountant at Acme Business Solutions calculated that the hourly cost of running a business car is s 2 + 7500 cents where s is the average speed of the car. The car travels on a 3000 km journey.

(i) Show that the cost of the journey

is given by .sC s 75003000= +b l

(ii) Find the speed that minimises the cost of the journey .(iii) Find the cost of the trip to the nearest dollar .

samp.indd 357 6/30/09 5:25:58 AM


Chapter 1: Geometry 2

Exercises 1.1

1. °°°°

( )

( )

( )

ABE ABDCBE CBD

ABDABE

ABD CBD

180180180

180straight angle

straight angle

given

+ ++ +

++

+ +

= −= −= −=

= —

2.

( )AFB is a straight angle+

° ( )°

° ( ° )

( )

( )

DFB xx

AFC xCFE x x

xAFC CFE

CD AFE

AFB180 180

180 180 2

bisects

is a straight angle

vertically opposite angles`

`

`

+

++

+ ++

+= − −=== − + −

==

3.

°WBC BCY x x2 115 65 2

180+ ++ = + + −

=

These are supplementary cointerior angles . VW XY` <

4. °°

( )x yA D

180180

given

`+ ++ =

+ =

These are supplementary cointerior angles.

( )AB DC

A BAlso similarly`

+ +<

=

These are supplementary cointerior angles.

.AD BCABCD is a parallelogram

`

`

<

5. °

,

( )

( )

ADB CDBABD CBD

BD

ABD CBD

BD ABC

110

is commonby AAS

given

bisects

`

+ ++ + +

/Δ Δ

= ==

6.

,

( )

( )

( )

AB AEB E

BC DEABC AED

(a)

by SAS

given

base angles of isosceles

given

`

+ +

/Δ Δ

Δ===

`

( )corresponding angles in congruent sΔ

°°

( )

BCA EDA

ACD BCAEDA

ADCACD

BCD180180

(b)

since base angles are equal, is isosceles

is a straight angle

+ +

+ ++

+Δ

=

= −= −=

7.

`

°

,

( )

( )

( )

( )

DC BCB D

DM AD

BN AB

DM BNMDC NBC

MC NCc

90

21

21

and

by SAS

given

given

given

orresponding sides in congruent s

`

`

+ +

/Δ Δ

Δ

== =

=

=

=

=

8. °

( )

( )

( )

OCA OCBOA OB

OCOAC OBC

AC BC

OC AB

90

is commonby RHS,

bisects

given

equal radii


`

`

`

+ +

/Δ Δ

Δ

= ==

=

9. ° ( )

( )

( )

CDB BECACB ABC

CBCDB BEC

CE BD

90

is commonby AAS,

altitudes given



`

`

+ ++ +

/Δ Δ

Δ

Δ

= ==

=

10.

`

( )

( )

( )

( )

AB ADBC DC

ACABC ADCDAC BAC

AC DABBCA DCA

AC DCB

is commonby SSS,

So bisectsAlso

bisects

given

given

corresponding angles in congruent s

corresponding angles in congruent s

`

`

+ +

++ +

+

/Δ Δ

Δ

Δ

==

=

=

11. ( )

( )

NMO MOPPMO MON

MO

MNO MPO

MN PO

PM ON

(a)

is commonby AAS,

alternate angles,

alternate angles,

`

+ ++ +

/

<

<

Δ Δ

==

( )

( )

( )

PMO MONMN NO

MON NMOPMO NMOPMQ NMQ

PM ON(b)

i.e.

alternate angles,

given


`

+ +

+ ++ ++ +

<

Δ

=====

)(corresponding sides in congruent sΔ

( )

( ( ))

MN NOPM NO

PM MNPMQ NMQ

MQPMQ NMQ

(c)

is commonby SAS,

given

from b

`

`

+ +

/Δ Δ

==

==

`

( )corresponding angles in congruent sΔ

°°

( )

MQN MQP

MQN MQPMQN MQP

PQN18090

(d)

But straight angle

+ +

+ ++ +

+

=

+ == =

12.

Answers

Ans_PART_1.indd 358 6/30/09 5:39:04 AM

359ANSWERS

Let ABCD be a parallelogram with diagonal AC .

,

( )

( )

DAC ACBBAC ACD

ACAAS ADC ABC

AD BC

AB DC

is commonby

alternate angles,

alternate angles,

`

+ ++ +

/

<

<

Δ Δ

==

13.

`

`

( )

( s)

BD

ADC ABCADC ABC

A C

12

Similarly, by using diagonal we canprove

opposite angles are equal

see question

corresponding angles in congruent+ +

+ +

/Δ ΔΔ=

=

14.

`

.

( )

( )

( )

AB DCBM DN

AB BM DC DNAM NC

AM NC

CN

ABCD

i.e.AlsoSince one pair of sides is both parallel andequal, is a parallelogram

opposite sides in gram

given

is a gram

AM

<

<

<

==

− = −=

15.

`

( )AD BCBC FEAD FE

opposite sides in gram

( )similarly

<===

`

( )

( )

AD BCBC FEAD FE

ABCD

BCEF

Alsoand

is a gram

is a gram

<

<

<

<

<

Since one pair of sides is both parallel and equal, AFED is a parallelogram.

16.

`

`

( )

( )

DEC DCEDEC ECBDCE ECB

CE BCD

AD BCAlso,

bisects


alternate angles,

+ ++ ++ +

+

<

Δ===

17.

`

`

(corresponding sides in congruent s)Δ

.

AB CDBAC DCA

ACABC ADC

AD BC

ABCD

is commonby SAS,

Since two pairs of opposite sides are equal,is a parallelogram

(given)

(given)+ +

/Δ Δ

==

=

18.

`

`

(a)( )diagonals bisect each other in gram<

(corresponding sides incongruent s)Δ

°

AE EC

AEB CEBEB

ABE CBEAB BC

90is commonby SAS,

(given)+ +

/Δ Δ

=

= =

=

(b) (corresponding angles in congruent s)Δ

ABE CBE+ +=

19.

Let ABCD be a rectangle

`

BCD/ Δ`

°

AD BCD C

DCADC

AC DB

90is common

by SAS,

(opposite sides in gram)

(corresponding sides in congruent s)

+ +<

Δ

Δ

== =

=

20.

Let ABCD be a rectangle with °D 90+ =

`

)

( )

D C AD BC

B C AB DC

( and cointerior angles,

and cointerior angles,

+ +

+ +

<

<

( )A B AD BCand cointerior angles,+ + <

° °

°° °

°° °

°

C

B

A

180 90

90180 90

90180 90

90all angles are right angles

` +

+

+

= −

== −

== −

=

21.

`

`

( )

(

AD CDAD BCAB CDAB AD BC CD

Also

all sides of the rhombus are equal

given

(opposite sides of gram)

similarly)

<

==== = =

22.

`

`

(

)

BE ADAD BEAD BCBE BC

BCD BECADC BECADC BCD

AD BE

ConstructThenButThen

Also,


given)

(base angles of isosceles )

(corresponding angles,

+ ++ ++ +

<

<

<

Δ

======

23.

`

`

(corresponding angles in congruent s)Δ

AD ABDC BC

ACADC CADC ABC

is commonby SSS,

(given)

(given)

+ +/Δ ΔΑΒ

==

=

Ans_PART_1.indd 359 6/30/09 5:39:06 AM


24.

`

`

(corresponding sides in congruent s)Δ

°( )—

AD BCD C

DE ECADE BCE

AE BE

E

90

by SAS,


(given)

is the midpoint given

+ +

/

<

Δ Δ

== ==

=

25. (a)

`

(

AD BCAB DC

DBADB BCD

is commonby SSS,


similarly)

/

<

Δ Δ

==

(b) (corresponding angles in congruent s)Δ

ABE CBE+ +=

`

(c)( )

AB BCABE CBE

BEABE CBE

is commonby SAS,

(adjacent sides in rhombus)

found+ +

/Δ Δ

==

`

(d)

But

(corresponding angles in congruent s)Δ

°°

(

AEB BEC

AEB BECAEB BEC

AEC18090

is a straight angle)

+ +

+ ++ +

=

+ == =

Exercises 1.2

1. (a) 14 452 mm2 (b) 67 200 mm3 2. 90 m3

3. V x x x2 3 23 2= + −

4. ππ

π

π

V r hr h

hr

hr

250250

250

2

2

2

==

=

=

5. πV 5 3= r 6. Ab2

3 2

= 7. V xx x x

26 12 8

3

3 2

= += + + +] g

8. πS h24 2=

9. V x xx x x3 2

4 12 9

2

3 2

= −= − +

] g

10. 262 cm 3 11. V h h3 22= +

12. V h h2 52= +

13. ( )V x x x31

6 5 34 153 2= − − −

14. (a) V x x x18 12 23 2= − + (b) S x x54 30 42= − +

15. l h r2 2= +

16. Vx y

π4

2

= 17. hrπ

4002

= 18. h rr

ππ750 2

=−

19. l rr

ππ850 2

=−

20. y x810 000 2= −

Exercises 1.3

1. Show m m 4AB CD= = and

m m74

AD BC= = −

2. Show ;´ ´m m47

74

1AC BC = − = −

∴ right-angled triangle with °C 90+ =

3. (a) ,AB AC BC73 6= = = (b) 8 units (c) 24 units 2

4. Show m m51

XY YZ= =

5. (a) Show ,AB AD 26= =

BC CD 80= = (b) Show ´m m 1AC BD = −

( , ), ,E CEAE

1 2 72 6 218 3 2

(c) = − = == =

6. r 1=

7. (a) x y2 3 13 0− + = (b) Substitute (7, 9) into the equation (c) Isosceles

8.

`

°AOB COD

OBOD

OAOC

OBOD

OAOC

90

24

2

714

2

+ += =

= =

= =

=

Since 2 pairs of sides are in proportion and their included angles are equal, .OCDΔ;OAB<Δ

9. (a) OB is common OA BC

AB OCOAB OCB

5

20by SSS` /Δ Δ

= == =

(b) m m m m131

2Show andOA BC AB OC= = = = −

10. °ABC 90+ = and AB BC 2= = So ABC is isosceles CAB ACB`+ += But °CAB ACB 90+ ++ = (angle sum of triangle) CAB ACB 45`+ += = Similarly, other angles are 45°.

11. PR QS 145= = Since diagonals are equal, PQRS is a rectangle.

12. (a) ( , ), ,X Y2 2 1 0= − = −^ h (b) m m 2XY BC= = − So XY BC< (c) ,XY BC5 20 2 5= = = So BC XY2=

Ans_PART_1.indd 360 6/30/09 5:39:08 AM

361ANSWERS

13. ´ ´m m 1 1 1AC BD = − = − So AC and BD are perpendicular.

,AC BDa a2 2

Midpoint midpoint= = −d n So AC and BD bisect each other. So AC and BD are perpendicular bisectors.

14. (a) X Y 1Distance from distance from unit= =

(b) ,Z41

0= d n (c) 1

41

units 2

15. Midpoint AB : ,W 2 121= −d n

Midpoint BC : ,X 2 3= − −^ h Midpoint CD : ,Y 4

21

21= −d n

Midpoint AD : ,Z21

2= −d n m m

83

WX ZY= =

So WX ZY<

m m57

XY WZ= = −

So XY WZ< WXYZ is a parallelogram.

Test yourself 1

1. (a) AB AC= (given) So BD EC= (midpoints)

)DBC ECB (base s in isoceles+ + + T= BC is common ( )BEC BDC SAS` /Δ Δ (b) s)TBE DC (corresponding sides in` /=

2.

12

y12

x

c a b

x y

x y

x y

cx y

x y

2 2

4 4

4

4

2

2 2 2

2 2

2 2

2 2

2 2

2 2

= +

= +

= +

=+

=+

=+

d en o

3. 2 2

2 2

2 2

( ) ( )

( ) ( )

( ) ( )

AB

BC

AC

7 4 5 15

7 1 5 310

4 1 1 35

= − + − − −== − + − −== − + − −=

Since ,≠AB AC BC= ABCΔ is isosceles.

4. ππ

h50 2

=−r

r

5. °° °°

( )

( )

ADEEAD

ADE

BC AD4590 4545

(a)

So is isosceles.

corresponding s

sumof

++

+

+

<

Δ

Δ== −=

(b) AE DE yCD yAB yAB CE

( , )

(

CD DE

(isoceles )

given

oppositesides in gram)

(given)<

= === <

Δ=

So ABCE is a trapezium.

( )

( )´ ´

´ ´

A h a b

y y y

y y

y

21

21

2

21

3

2

3 2

= +

= +

=

=

6. (a)

`

BACB

GFCG

GFCG

DECD

BACB

DECD

(equal ratio of intercepts)

(similarly)

=

=

=

(b) 20.4 cm

7.

,

m

m

m

m

BC AD CD ABABCD

2 54 4

0

5 64 3

7

1 63 3

0

2 14 3

7

So is a parallelogram.

BC

CD

AD

AB

< <

=− −

− − − =

=−

− − =

=− −

− =

=− − −− − =

8.

` °

DC BC

DBC

ABCD

12 5144 2516913

90So is a rectangle.

2 2 2 2

2

2

(Pythagoras)+

+ = += +====

9.

WXYΔ;`

..

..

.

39°Y PPQR

3 49 18

1 95 13

2 7

(two pairs of sides in proportion, with included s equal)

(given)+ +

+

<Δ

= =

= =

10.

`

—

( )

BAD

BAC xDAC xDCA xBAC DCA

A

DAC

LetThen (given C bisects )

base s of isosceles

++++ +

+

+ Δ

====

Ans_PART_1.indd 361 6/30/09 5:39:09 AM


These are equal alternate angles.

°

°°

°

( )

( )

AB ECEDA xDEA DAE

DAEx

xEAC x x

EAC ACB

AED

2

2180 2

909090

base s of isosceles

sum of isosceles

)DAC(exterior of`

`

++ +

+

+

+ +

+

+

<

Δ

Δ

==

= −

= −= − +==

+ Δ

These are equal alternate angles. AE CB` < So ABCE is a parallelogram.

11. ,1 0−^ h 12. (a) x y4 3 1 0− − =

(b) 2.4 units (c) 12 units 2

13.

`

AB ADBC DC

ACABC ADC

(a)

is common(SSS)

(given)

(given)

/Δ Δ

==

AB ADBAE DAE

AEABE ADE

(b)

is common(corresponding s in s)

(SAS)

(given)

`

+ + +

/

/

Δ Δ

Δ==

(c)`

`

°°

.

BE DEAC BDBEA DEA

BEA DEABEA DEA

AC BD

18090

bisects

But

So is perpendicular to

(straight )

(corresponding sides in s)

(coresponding s in s)+ ++ ++ +

+

=

=+ =

= =

+

/

/

Δ

Δ

14. : , :

´

AB m BC m

m m

AB BC

21

2

21

2 1

(a)

So and are perpendicular.

1 2

1 2

= = −

= − = −

(b) ( , )3 2−

(c) ( , )321

(d) 5 units

15. (a) x xhx xhx xh

xx

h

500 4 6500 4 6250 2 3

3250 2

2

2

2

2

= +− =− =− =

(b) V x h

xx

x

xx

x x

2

23

250 2

23

250 2

3500 4

2

22

2

3

=

= −

= −

= −

ee

oo

Challenge exercise 1

1. BD is common °

( )—

ADB CDBAD DC

ABD CBDAB BC

ABC

BD AC

90

by SAS,

So is isosceles.

(given)

bisects given

(corresponding sides in congruent s)

`

`

+ +

/Δ Δ

ΔΔ

= ==

=

2.

`

`

`

AD AB

ABAD

AE AC

ACAE

ABAD

ACAE

A

21

21

21

21

(a)

is common+

=

=

=

=

=

Since two pairs of sides are in proportion and their included angles are equal, ;

`

ABCΔADEADE ABC

DE BCThese are equal corresponding angles`

+ +<

<

Δ=

`

,ABC; Δ

`

ADE

ABAD

BCDE

BCDE

DE BC

21

21

21

(b) Since <Δ

= =

=

=

3.

Let ABCD be a rhombus with AD DC= To prove: ADE CDE+ +=

Proof `

`

`

`

( )

AD DCADC

DAE DCEAE EC

ADE CDEADE CDE

is isosceles

by SAS,

(given)

diagonals bisect each other

(corresponding angles in congruent s)

+ +

+ +/

Δ

Δ Δ

Δ

=

==

=

(Note: We can prove other pairs of angles equal similarly.)

4. ´1189 841mm mm

5. S x x240002= +

Ans_PART_1.indd 362 6/30/09 5:39:10 AM

363ANSWERS

6.

°

°

( )

( )

´g n

n

nn

n

0

180 360

180360

2 18Each an le =

−

=−

= −

°

c m

7.

EAF ACDADC ABCABC AEFADC AEF

(alternate angles, )

(opposite angles in gram)

(corresponding angles, )

AB DC

BC EF

`

+ ++ ++ ++ +

<

<

<

====

Since 2 pairs of angles equal, third is equal by angle sum of Δ ADC;ΔAEF` <Δ

8 .

(a) Midpoint of : , ( , )BDa a b b

a2 2

0+ + − =d n

Midpoint of : , ( , )ACa

a2

0 22

0 00

+ + =d n

diagonals BD and AC bisect each other at E (a, 0) BD is vertical and AC is horizontal the diagonals are perpendicular

(b) 2

2

2

( ) ( )

( ) ( )

( ) ( )

( ) ( )

AB a ba b

BC a a ba b

CD a a ba b

AD a ba b

0 0

2 0

2 0

0 0

2

2 2

2

2 2

2 2

2 2

2

2 2

= − + −= += − + −= += − + += += − + − −= +

all sides are equal

(c)

`

`

(from (a))(from (a))

is common

So bisects(

BC CDBE EDCESSS CBE

BCE DCEAC BCD

bycorresponding s in congruent+ ++

+

/Δ ΔΔ

==

= s)

CDE

9. °

DE BEAEB AED 90

(digonals bisect in gram)

(given)+ +<=

= =

AE is common by SAS, ADE ABE/Δ Δ AB AD (corresponding sides in congruent` Δ= s)

10. P RPACP

RBCR

11

( and are midpoints)= =

PR AB (equal ratios on lines)` < < Similarly PQ CB QR ACand< < PQ CB

AC RQ

QPR PRCCPR PRQ

(alternate s, )

(alternate s, )

+ ++ +

+

+

<

<

==

PR is common by AAS, PQR CPR/Δ Δ

11. 188 mm

12. P SDADP

DCDS

D21

(a)

is common

( , are midpoints)

+

= =

Since 2 pairs of sides are in proportion and the included angles are equal, .DAC;ΔDPS<Δ

(b) PADP

SCDS

11= =

(PS AC equal rations on lines)` < <

Similarly, QA

BQ

RCBR=

QR AC` <

PS QR` <

(c) ( ,PDAP

QB

AQP Q

11

are midpoints)= =

PQ DB (equal ratios on lines)` < <

Similarly SR DB< PQ SR` < Since , .PS QR PQ SR< < PQRS is a parallelogram.

13. 70 cm

Chapter 2: Geometrical applications

of calculus

Problem

. , .0 25 1 125− −^ h Exercises 2.1

1. (a)

Ans_PART_1.indd 363 6/30/09 5:39:11 AM


(b)

(c)

(d)

2. x41< 3. x 0<

4. (a) .x 1 5< (b) .x 1 5> (c) .x 1 5=

5. ( )x 2 0<= −fl for all x

6. y x3 0>2=l for all ≠x 0

7. ( , )0 0 8. ,x 3 2= − 9. (a) ( , )1 4− (b) ( , )0 9

(c) ( , )1 1 and ( , )2 0 (d) ( , ),0 1 ( , )1 0 and ( , )1 0−

10. ( , )2 0 11. x1 1< <− 12. ,x x5 3< >− −

13. (a) ,x 2 5= (b) x2 5< < (c) ,x x2 5< >

14. p 12= −

15. ,a b121

6= = −

16. (a) dx

dyx x3 6 272= − +

(b) The quadratic function has a 0>b ac4 288 0<2 − = −

x x x3 6 27 0So for all>2 − +The function is monotonic increasing for all x.

17.

x2

y

18.

x4

y

19.

1

y

x

20.

5-2

y

x

21.

x

y

3

2

Ans_PART_1.indd 364 6/30/09 2:56:14 PM

365ANSWERS

22.

x

y

-1-2

23. ( , )2 0 and ,32

38113c m

24. ; ,x

xx

x

x

2 11

2 1

3 232

92 3

++ + =

++ −

−f p

25. .a 1 75= −

26. x2

10! 27.

x3

04!

−

Exercises 2.2

1. ( , ); y y0 1 0 0on LHS, on RHS< >− l l

2. ( , )0 0 minimum 3. ( , )0 2 infl exion

4. ( , );2 11− show ( )f x 0>l on LHS and ( )f x 0<l on RHS.

5. ( , )1 2− − minimum 6. ( , )4 0 minimum

7. ( , )0 5 maximum, ( , )4 27− minimum

8. ( ) , ( )f f x0 0 0>=l l on LHS and RHS

9. ( , )0 5 maximum, ( , )2 1 minimum

10. ( , )0 3− maximum, ( , )1 4− minimum, ( , )1 4− − minimum

11. ( , )1 0 minimum, ( , )1 4− maximum

12. m 6121= −

13. x 3= − minimum 14. x 0= minimum, x 1= − maximum

15. x 1= infl exion, x 2= minimum

16. (a) dxdP

x2

502

= −

(b) (5, 20) minimum, ,5 20− −^ h maximum

17. ,121d n minimum

18. (2.06, 54.94) maximum, . , .20 6 54 94− −^ h minimum

19. (4.37, 54.92) minimum, . , .4 37 54 92− −^ h maximum

20. (a) dxdA

xx

x

x

x

36003600

3600

3600 2

2

2

2

2

2

= − −−

=−

−

(b) (42.4, 1800) maximum, . ,42 4 1800− −^ h minimum

Exercises 2.3

1. ; ;x x x x x x7 10 4 1 42 40 126 4 3 5 3 2− + − − + ;x x x x x210 120 24 840 240 244 2 3− + − +

2. ( )f x x72 7=m 3. ( ) , ( )f x x x f x x x10 3 40 64 2 3= − = −l m

4. ( ) , ( )f f1 11 2 168= − =l m

5. ; ;x x x x x x7 12 16 42 60 486 5 3 5 4 2− + − + x x x210 240 964 3− +

6. ,dx

dyx

dx

d y4 3 4

2

2

= − =

7. ( ) , ( )f f1 16 2 40− = − =l m 8. ,x x4 205 6− − −

9. ( )g 4321= −m 10.

dtd h

262

2

= when t 1=

11. x187= 12. x

31>

13. ( ) ; ( )x x20 4 3 320 4 34 3− −

14. ( ) ;f xx2 2

1= −−

l

( )( )

f xx4 2

13

= −−

m

15. ( )

; ( )( )

xx

xx3 1

163 1

962 3

= −−

=−

ffl m^ h

16. dtd v

t24 162

2

= + 17. b32= 18. ( )f 2

4 2

38

3 2= =m

19. ( )f 1 196=m 20. .b 2 7= −

Exercises 2.4

1. x31> − 2. x 3< 3. y 8 0<= −m 4. y 2 0>=m

5. x 231< 6. ( , )1 9 7. ( , )1 17− and ( , )1 41− −

8. ( , ); y y0 2 0 0on LHS, on RHS< >− m m 9. x2 1< <−

10. (a) No—minimum at (0, 0) (b) Yes—infl exion at (0, 0) (c) Yes—infl exion at (0, 0) (d) Yes—infl exion at (0, 0) (e) No—minimum at (0, 0)

11. y

x

Ans_PART_1.indd 365 6/30/09 5:39:16 AM


12.

x1

y

13. None: (2, 31) is not an infl exion since concavity does not change.

14. ( )f xx12

4=m

x 0>4 for all x 0!

So x12

0>4

for all x 0!

So the function is concave upward for all .x 0!

15. (a) (0, 7), (1, 0) and ,1 14−^ h (b) (0, 7)

16. (a) dx

d yx12 24

2

2

2= +

≥x 02 for all x So ≥x12 02 for all x ≥x12 24 242 + So ≠x12 24 02 + and there are no points of infl exion. (b) The curve is always concave upwards.

17. a 2= 18. p 4= 19. ,a b3 3= = −

20. (a) , , ,0 8 2 2−^ ^h h (b)

dx

dyx x6 15 215 4= − +

At

, :

, :

≠

≠

dx

dy

dx

dy

0 8 6 0 15 0 21

210

2 2 6 2 15 2 21

270

At 5 4

5 4

− = − +

=

= − +

= −

^ ] ]

^ ] ]

h g g

h g g

So these points are not horizontal points of infl exion.

Exercises 2.5

1. (a) ,dx

dy

dx

d y0 0> >

2

2

(b) ,dx

dy

dx

d y0 0< <

2

2

(c) ,dx

dy

dx

d y0 0> <

2

2

(d) ,dx

dy

dx

d y0 0< >

2

2

(e) ,dx

dy

dx

d y0 0> >

2

2

2. (a) ,dtdP

dtd P

0 0> <2

2

(b) The rate is decreasing.

3.

4. (a)

(b)

(c)

(d)

5.

Ans_PART_1.indd 366 6/30/09 5:39:17 AM

367ANSWERS

6.

7. ,dtdM

dtd M

0 0< >2

2

8. (a) The number of fi sh is decreasing. (b) The population rate is increasing. (c)

9. The level of education is increasing, but the rate is slowing down.

10. The population is decreasing, and the population rate is decreasing.

Exercises 2.6

1. (1, 0) minimum 2. (0,1) minimum

3. ( , ), y2 5 6 0>− =m 4. ( . , . ), y0 5 0 25 0 so maximum<m

5. ( , ); ( ) ( , ),f x0 5 0 0 5at− = −m ( ) , ( )f x f x0 0on LHS RHS< >m m

6. Yes—infl exion at (0, 3)

7. ,2 78− −^ h minimum, ,3 77− −^ h maximum

8. (0, 1) maximum, ,1 4− −^ h minimum, ,2 31−^ h minimum

9. (0, 1) maximum, (0.5, 0) minimum, . ,0 5 0−^ h minimum

10. (a) (4, 176) maximum, (5, 175) minimum (b) (4.5, 175.5)

11. (3.67, 0.38) maximum

12. ,0 1−^ h minimum, ,2 15−^ h maximum, ,4 1− −^ h minimum

13. (a) a32= − (b) maximum, as y 0<m

14. m 521= − 15. ,a b3 9= = −

Exercises 2.7

1.

2.

3.

4.

Ans_PART_1.indd 367 6/30/09 5:39:19 AM


5.

6.

7.

8. (a) ,0 7−^ h minimum, ,4 25−^ h maximum (b) ,2 9−^ h (c)

9.

10.

11.

12.

Ans_PART_1.indd 368 6/30/09 5:39:22 AM

369ANSWERS

13.

14. ( )

≠dx

dy

x12

02

=+−

15.

Exercises 2.8

1. Maximum value is 4.

2. Maximum value is 9, minimum value is −7.

3. Maximum value is 25.

4. Maximum value is 86, minimum value is −39.

5. Maximum value is −2.

6. Maximum value is 5, minimum value is .1631−

7. Absolute maximum 29, relative maximum −3, absolute minimum −35, relative minimum −35, −8

8. Minimum −25, maximum 29

Ans_PART_1.indd 369 6/30/09 5:39:25 AM


9. Maximum 3, minimum 1

10. Maximum ∞, minimum −∞

Problem

The disc has radius cm.7

30 (This result uses Stewart’s

theorem—check this by research.)

Exercises 2.9

1.

´

A xyxy

x y

P x y

x x

x x

5050

2 2

2 250

2100

`

==

=

= +

= +

= +

2.

( )

x yy xy xA xy

x xx x

2 2 1202 120 2

60

6060 2

+ == −= −== −= −

3. xy

y xS x y

x x

2020

20

=

=

= +

= +

4. 400

2 2

2 2

2

ππ

ππ π

π ππ

π

V r hr h

rh

S r rh

r rr

r r

400

400

800

2

2

2

2

22

2

==

=

= +

= +

= +

e o

5. x yy x

3030

(a)`

+ == −

(b) The perimeter of one square is x , so its side is

.x41

The other square has side .y41

2( )

( )

A x y

x y

x y

x x

x x x

x x

x x

x x

41

41

16 16

16

16

30

16900 60

162 60 900

16

2 30 450

830 450

2 2

2 2

2 2

2

2 2

2

2

2

= +

= +

=+

=+ −

= + − +

= − +

=− +

= − +

d dn n

6. (a) x y

y x

y x

28078 40078 400

78 400

2 2 2

2 2

2

+ === −= −

(b) A xy

x x78 400 2

== −

7.

( ) ( )( )( )

V x x xx x x xx x x

x x x

10 2 7 270 20 14 470 34 4

70 34 4

2

2

2 3

= − −= − − += − += − +

8. Profit per person Cost Expenses( ) ( )

For people, ( )

x xx xx

x P x xx x

900 100 200 400900 100 200 400700 500

700 500700 500 2

= −= − − += − − −= −= −= −

Ans_PART_1.indd 370 6/30/09 5:39:28 AM

371ANSWERS

9.

After t hours, Joel has travelled 75 t km. He is t700 75− km from the town. After t hours, Nick has travelled 80 t km. He is t680 80− km from the town.

2( ) ( )d t tt t

t tt t

700 75 680 80490 000 105 000 5625 462 400108 800 6400952 400 213 800 12 025

2

2

2

2

= − + −= − + +

− += − +

10. The river is 500 m, or 0.5 km, wide Distance AB :

..

Speedtime

distance

Timespeed

distance

.

d xx

tx

0 50 25

50 25

2 2

2

2

`

= += +

=

=

=+

Distance BC :

Timespeed

distanced x

tx

7

47

= −

=

= −

So total time taken is:

.

tx x

50 25

472

=+

+ −

Exercises 2.10

1. 2 s, 16 m 2. 7.5 km

3.

( )

( ) from ( )

x yy xy xA xy

x xx x

2 2 602 60 2

30 1

30 130 2

+ == −= −== −= −

Max. area 225 m2

4.

`

(a)

( )

from ( )

A xy

y xP x y

x x

x x

40004000

1

2 2

2 24000

1

28000

= =

=

= +

= +

= +

c m

(b) 63.2 m by 63.2 m(c) $12 332.89

5. 4 m by 4 m 6. 14 and 14 7. -2.5 and 2.5

8. . ,x 1 25 m= .y 1 25 m=

9. (a) ( ) ( )( )

V x x xx x x

x x x

30 2 80 22400 220 4

2400 220 4

2

2 3

= − −= − += − +

(b) cmx 632=

(c) 7407.4 cm 3

10.

( )

π π

ππ

π

π

ππ

V r h

hr

rS r r h

r rr

r r

5454

54

2

254

2108

2

2

2

2

2

= =

=

=

= +

= +

= +

d n

Radius is 3 m.

11. (a) 2πS r r17 2002= + (b) 2323.7 m 2

12. 72 cm 2

13. (a)

( ) ( )

xy

y xA y x

xy y x

x x x x

x x

x x

400400

10 1010 10 100

40010

40010 100

4004000

10 100

500 104000

`

=

=

= − −= − − +

= − − +

= − − +

= − −

c cm m

(b) 100 cm 2

14. 20 cm by 20 cm by 20 cm 15. 1.12 m 2

16. (a) 7.5 m by 7.5 m (b) 2.4 m

17. 301 cm 2 18. cm16061 2 19. 1.68 cm, 1.32 cm

20. ( ) ( )

km

d t tt t

t tt t

200 80 120 6040 000 32 000 6400

14 400 14 400 360010 000 46 400 54 40024

2 2 2

2

2

2

= − + −= − +

+ − += − +

21. (a) d x x x xx x x x

x x

2 5 42 5 4

2 6 5

2 2

2 2

2

= − + − −= − + − += − +

] ]g g (b) unit21

Ans_PART_1.indd 371 6/30/09 5:39:29 AM


22. (a) Perimeter ( ) where

2

π

π

π

π

π

π

´

x y r ry

x yy

x yy

yy

x

y yx

y yx

221

22

1200 221

2

22

12002

2

6002 4

4

2400 2

= + + =

= + +

= + +

− − =

− − =

− −=

e o

(b)

2

π

ππ

π π

π π

π

A xy r

y yy

y

y y y y

y y y y

y y y

21

4

2400 2

21

2

4

2400 2

8

8

4800 4 2

8

8

4800 4

2

2 2 2

2 2 2

2 2

= +

=− −

+

=− −

+

=− −

+

=− −

e eo o

(c) m, mx y168 336= =

23. (a) Equation AB : y mx b

ab

x b

= +

= +

Substitute ,1 2−^ h ( )a

bb

ab

b

a b abb ab a

aa

b

2 1

21

1

12

= − +

= − +

= − += − += −

−=

]]

gg

(b) ,a b2 4= =

24. 26 m

25. (a)

So

std

t sd

s1500

=

=

=

Cost of trip taking t hours:

´

C s t

s s

s s

s s

9000

90001500

15009000 1500

15009000

2

2

= +

= +

= +

= +

]]

c

gg

m

(b) 95 km/h (c) $2846

Exercises 2.11

1. (a) x x C32 − + (b) x

x x C3

43

2+ + +

(c) x

x C6

64− + (d)

xx x C

3

32− + +

(e) x C6 +

2. (a) ( )f x xx

C22

32

= − + (b) ( )f xx

x x C5

75

3= − + +

(c) ( )f xx

x C2

22

= − + (d) ( )f xx

x x C3

33

2= − − +

(e) 2

( )f xx

C3

2= +

3

3. (a) y x x C95= − + (b) yx

x C3

23

1= − + +−

−

(c) yx x

C20 3

4 3

= − + (d) y x C2= − +

(e) yx x

x C4 3

4 2

= − + +

4. (a) x

C3

2 3

+ (b) x

C2

2

− +−

(c) x

C71

7− + (d)

2 3x x C2 6+ +1 1

(e) x

x C6

26

1− + +−

−

5. yx

x x4

5414

3= − + − 6. ( )f x x x2 7 112= − +

7. ( )f 1 8= 8. y x x2 3 192= − + 9. x 1631=

10. y x x4 8 72= − + 11. y x x x2 3 23 2= + + −

12. ( )f x x x x 53 2= − − + 13. ( ) .f 2 20 5=

14. yx x

x3 2

12 24213 2

= + − + 15. yx

x3

415 14

313

= − −

16. yx

x x3

2 3 4323

2= − + − 17. ( ) 2 4 2f x x x x x4 3 2= − + + −

18. y x x3 8 82= + + 19. 77f 2− =] g 20. y 0=

Test yourself 2

1. ( , )3 11− − maximum, ( , )1 15− − minimum

2. x 161> 3. y x x x2 6 5 333 2= + − −

4. (a) -8 (b) 26 (c) -90 5. 50 m

6. (0, 0) minimum 7. x 1> −

Ans_PART_1.indd 372 6/30/09 5:39:30 AM

373ANSWERS

8. (a) (0, 1) maximum, ( , )4 511− − minimum, ( , )2 79− minimum

(b)

9. ,21

1−c m 10. ( )f xx

x x2

56 49 59

32= + − +

11. (a) ππ

ππ π

π ππ

π

V r hr h

rh

S r rh

r rr

r r

375375

2 2

2 2375

2750

2

2

2

2

22

2

==

=

= +

= +

= +

d n

(b) 3.9 cm

12. (a) (0, 0) and ( , )1 1− (b) (0, 0) minimum, ( , )1 1− point of infl exion (c)

13. (a)

( )

S x xhx xh

x xh

xx

h

x

xh

xx

h

2 4250 2 4

250 2 4

4250 2

4

2 125

2125

2

2

2

2

2

2

= += +

− =− =

−=

− =

h

( )

V x

xx

x

x x

x x

2125

2

125

2125

2

22

2

3

=

= −

=−

= −

d n

(b) . cm . cm . cm´ ´6 45 6 45 6 45

14. x 3< 15. y x x3 33 2= + + 16. 150 products

17. For decreasing curve, dx

dy0<

dx

dyx3 2= −

( )≠x x0 0 0since for all< >2 monotonic decreasing function

18. (a)

(b)

(c)

19. (a) x yy x

y x

A xy

x x

5 2525

25

21

21

25

2 2 2

2 2

2

2

+ = == −= −

=

= −

(b) 6.25 m 2

20. (a) ( , )4 171− minimum, ( , )6 329− maximum (b) ( , )1 79− (c)

21. x 1<

22. f x x x x2 3 31 683 2= − − +] g

23. (0, 1) and ,3 74−^ h

Ans_PART_1.indd 373 6/30/09 5:39:32 AM


24. 179

25. y

2x


1.

5

( );

( )

( )x

x x

x

x x x4 1

20 120 1

4 1

8 60 420 9 15

2 4

2

2

3 2

+− −

+− − − +

2.

3. ,x x21

4< >−

4. 16 m 2

5. 27; -20.25

6. ( . ) ( . )f f0 6 0 6 0= =l m and concavity changes

7. Show sum of areas is least when .r s 12 5= =

8. 2565

9. (a) ≠dx

dy

x2 1

10=

−

(b) Domain: ;≥x 1 range: ≥y 0 (c)

10. . cm,r 3 17= . cmh 6 34=

11. yx

x x3

15 13

2= − − −

12. 110 km/h

13. may be other solution.)y x x2 3 (There2= + +

14. (a) ( )f xx x x4 3

22

34 3 2

= − −

(b) ( , ), , , ,0 0 3 1141

1127− − −c cm m

(c)

15. m m m´ ´4 4 4

16. ( )f 3 2261= − 17. (a) -2 (b) -1

18. y 0=l at (0, 0); (a) y 0>m on LHS and RHS (b) y 0<m on LHS, y 0>m on RHS

19. cm2131 3 20. (a) (0, 1) (b) , , , , . . .k 2 4 6 8=

21. minimum −1; maximum 51−

22. 87 kmh −1

Chapter 3: Integration

Exercises 3.1

1. 2.5 2. 10 3. 2.4 4. 0.225 5. (a) 28 (b) 22

6. 0.39 7. 0.41 8. 1.08 9. 0.75 10. 0.65

11. 0.94 12. 0.92 13. 75.1 14. 16.5 15. 650.2

Exercises 3.2

1. 48.7 2. 30.7 3. 1.1 4. 0.41 5. (a) 3.4475 (b) 3.4477 6. 2.75 7. 0.693 8. 1.93 9. 72 10. 5.25 11. 0.558 12. 0.347 13. 3.63 14. 7.87 15. 175.8

Exercises 3.3

1. 8 2. 10 3. 125 4. -1 5. 10 6. 54 7. 331

8. 16 9. 50 10. 5232

11. 32

12. 2141

13. 0

14. 432

15. 141

16. 431

17. 0 18. 231

19. 0

20. 692

21. 10141

22. 1243− 23. 22

32

24. 231

25. 0.0126

Ans_PART_1.indd 374 6/30/09 5:39:34 AM

375ANSWERS

Exercises 3.4

1. x

C3

3

+ 2. x

C2

6

+ 3. x

C5

2 5

+ 4. m

m C2

2

+ +

5. t

t C3

73

− + 6. h

h C8

58

+ + 7. y

y C2

32

− +

8. x x C42 + + 9. b b

C3 2

3 2

+ + 10. a a

a C4 2

4 2

− − +

11. x

x x C3

53

2+ + + 12. x x x x C44 3 2− + − +

13. xx x

C5 2

65 4

+ + + 14. x x

x C8 7

39

8 7

− − +

15. x x x

x C2 3 2

24 3 2

+ − − + 16. x x

x C6 4

46 4

+ + +

17. x x

x C3

42

58

3 2

− − + 18. x x x

C5

32 2

5 4 2

− + +

19. x x

x C2

33

54

4 3

+ − + 20. xx

x C2

232

1− − − +−−

−

21. x

C71

7− + 22.

3xC

4

3+

4

23. x

x x C4

43 2− + +

24. x xx

C23

423

− + + 25. x x

x C3 2

310

3 2

+ − + 26. x C3− +

27. x

C21

2− + 28. x x

x xC

423

47

2 4− − + − +

29. y y

y C3 6

53 6

+ + +−

30. t t

t t C4 3

2 44 3

2− − + +

31. x

C3

2 3

+ 32. t

C21

4− + 33.

xC

43 43

+

34. x

C5

2 5

+ 35. x

x C3

2 3

+ +

Exercises 3.5

1. (a) (i) x x x C3 12 163 2− + + (ii) 3( )x

C9

3 4−+

(b) ( )x

C5

1 5++ (c)

( )xC

50

5 1 10−+

(d) ( )y

C24

3 2 8−+ (e)

( )xC

15

4 3 5++

(f) ( )x

C91

7 8 13++ (g)

( )xC

7

1 7

−−

+

(h) ( )x

C3

2 5 3−+ (i)

( )xC

9

2 3 1 3

−+

+−

(j) ( )x C3 7 1− + +− (k) ( )x

C16 4 5

12

−−

+

(l) ( )x

C16

3 4 33 4++ (m) 2( )x C2 2− − +

1

(n) 5( )t

C5

3++2 (o)

( )xC

35

5 2 7++2

2. (a) 288.2 (b) 141− (c)

81− (d) 60

32

(e) 61

(f) 71

(g) 432

(h) 81− (i) 1

51

(j) 53

Exercises 3.6

1. units131 2 2. 36 units 2 3. 4.5 units 2 4. units10

32 2

5. units61 2 6. 14.3 units 2 7. 4 units 2 8. 0.4 units 2

9. 8 units 2 10. 24.25 units 2 11. 2 units 2 12. units931 2

13. units1132 2 14. units

61 2 15. units

32 2 16. units

31 2

17. units531 2 18. 18 units 2 19. . unitsπ 3 14 2=

20. unitsa2

42

Exercises 3.7

1. units2131 2 2. 20 units 2 3. 4

32

units2

4. 1.5 units 2 5. 141

units2 6. 231

units2

7. 1032

units2 8. 61

units2 9. 397

units2

10. 2 units 2 11. 1141

units2 12. 60 units 2

13. 4.5 units 2 14. 131

units2 15. 1.9 units 2

Exercises 3.8

1. 131

units2 2. 131

units2 3. 61

units2

4. 1032

units2 5. 2065

units2 6. 8 units 2

7. 32

units2 8. 16632

units2 9. 0.42 units 2

10. 32

units2 11. 121

units2 12. 31

units2

13. 36 units2 14. 232

units2 15. ( )π 2 units2−

Problem

π15

206units3

Ans_PART_1.indd 375 6/30/09 5:39:36 AM


Exercises 3.9

1. π

5243

units3 2. π

3485

units3

3. π

15376

units3 4. π7

units3 5. 39π2

units3 6. 758π

3units3

7. 2π3

units3 8. 992π

5units3 9.

5π3

units3 10. 9π2

units3

11. 27π2

units3 12. 64π3

units3 13. 16 385π

7units3

14. 25π2

units3 15. 65π2

units3 16. 1023π

5units3

17. 5π3

units3 18. 13π units3 19. 344π27

units3

20. 3π5

units3 21. 2π5

units3 22. 72π5

units3

23. `

( )

π

π

π

π

π

π

y r xy r x

V y dx

r x dx

r xx

rr

rr

r r

r

3

3 3

32

32

34

units

a

b

r

r

r

r

2 2

2 2 2

2

2 2

23

33

33

3 3

33

= −= −

=

= −

= −

= − − − −−

= − −

=

−

−

]

d ed

g

n on

<

>

F

H

##

Test yourself 3

1. (a) 0.535 (b) 0.5

2. (a) x

x C2

3 2

+ + (b) x

x C2

5 2

− + (c) x

C3

2 3

+

(d) ( )x

C16

2 5 8++

3. 14.83 4. (a) 2 (b) 0 (c) 251

5. (a)

(b)

6. 3 units 2 7. 1.1 units 2

8. 232

units2 9. π9 units3 10. 421

11. 43

units2

12. 12( )x

C84

7 3++ 13. 3 units 2 14. (a)

π15

206units3

(b) π2

units3 15. (a) ±x y 3= − (b) 332

units2

(c) π

25

units3

16. 36 17. 8531

units2 18. π

53

units3

19. (a) ( )x

C2

2 1 5−+ (b)

xC

8

6

+


1. (a) 121

units2 (b) π

352

units3

2. (a) Show ( ) ( )f x f x− = − (b) 0 (c) 12 units 2

3. 27.2 units 3 4. 9 units 2 5. (a) 8( )x x36 13 4 −

(b) 9( )x

C36

14 −+

6. (a) ( )x

x3 4

222 2−

− (b)

81

7. 7.35 units 2 8. π

32

units3

9. ( ) ∞f 001= =

10. (a)

(b) 3.08 units 2

11. 6

17 17units2 12.

215π6

units3

13. (a) ( )

x

x

x

x

2 3

3 6

2 3

3 2

++ =

+

+ (b)

x xC

32 3+

+

14. (a) 632

(b) 632

15. 125

units2

16. (a) a3

8units

22

(b) πa2 units3 3

Ans_PART_1.indd 376 6/30/09 5:39:38 AM

377ANSWERS

Practice assessment task set 1

1.

Let ABCD be a parallelogram with diagonal AC . ACD BAC+ += (alternate , AB DCs+ < )

DAC BCA+ += (alternate , AD BCs+ < )

AC is common by AAS ACD ACB/Δ Δ AB DC` = and AD BC= )(corresponding sides in congruent

opposite sides are equal`

Δs

2. x21< 3. x x x C3 2− + + 4. 24 5. 8 m

6. AC FD= (opposite sides of < gram equal)

( )BC FEAB AC BC

FD FEED

given

`

== −= −=

Also AB ED< (since ACDF is < gram) since AB ED= and ,AB ED< ABDE is a parallelogram

7. x

x C3

29

2+ +

8.

9. (a) π

7198

units3 (b) π

596

units3

10. 131

units2 11. ( ) , ( )f f3 20 2 16= − = −l m 12. 68

13. AB ACBD CD

(given)

(given)

==

AD is common. ABD ACD

ADB ADCby SSS`

` + +/Δ Δ=

(corresponding s+ in congruent sΔ )

But °ADB ADC 180+ ++ = ( BDC+ is a straight + )

°ADB ADC 90`+ += = AD BC` =

14. (a) 78.7 units 3 (b) 1.57 units 3 15. x97> −

16. ( ) ;1 18=( ) , f1 2= −( ) ,f 1 3= fl m curve is decreasing and concave upwards at ( , )1 3

17. P x yy xy x

8 4 44 4 8

1 2

= + == −= −

A x y

x xx x xx x

3

3 1 23 1 4 47 4 1

2 2

2 2

2 2

2

= += + −= + − += − +

] g

Rectangle ,´72

76

m m square with sides 73

m

18. ( )f 1 0− = 19. 12 20. 0.837

21. AB

BC

AC

AB BC

AC

24576321024401600576 10241600

2 2

2 2

2 2

2 2

2

======

+ = +==

ABC` Δ is right angled at B+ (Pythagoras’ theorem)

22. 8( )x

C24

3 5++ 23. 5

31− 24. (1, 1)

25. (a) 1.11 (b) 1.17

26. ,0 3^ h maximum, ( , )1 2 minimum, ( , )1 2− minimum

27. 2158

28. (a) .1 58 units2 (b) π

25

units3

Ans_PART_1.indd 377 6/30/09 5:39:40 AM


29. 1232

30. 1032

units2

31. B+ is common °BDC ACB 90+ += = (given) CBD;Δ ( )ABC AAA` <Δ

32. Show ( ) ( ) ( )f x f x f x0 0and >= =l m m on both LHS and RHS of ( , )0 0

33. 232

m3 34. ( )f 2 16= − 35. 9 units2

36. .119 3 m2 37. ( )f x x x x4 3 203 2= − − +

38.

39.

Let ABCD be a rhombus with AC x= and .BD y= °AEB 90+ =

(diagonals perpendicular in rhombus)

DE BE y21= =

(diagonals bisect each other)

ACBΔ has area ´ ´x y xy21

21

41=

ADCΔ has area ´ ´x y xy21

21

41=

ABCD has area xy xy xy41

41

21+ =

40. ( )

( )

ACAB

ADAG

ADAG

AEAF

ACAB

AEAF

BG CD

GF DE

equal ratios of intercepts,

equal ratios of intercepts,

`

<

<

=

=

=

41. ( )f 2 132=

42. (a) nx

C1

n 1

++

+

(b) Since ( )dxd

C 0= , the primitive function

could include C.

43. (c), (d) 44. (a), (b) 45. (c)

46. (d) 47. (b) 48. (a) 49. (b) 50. (d)

Chapter 4: Exponential and

logarithmic functions

Exercises 4.1

1. (a) 4.48 (b) 0.14 (c) 2.70 (d) 0.05 (e) -0.14

2. (a)

(b)

(c)

3. (a) e9 x (b) ex− (c) e x2x + (d) x x e6 6 5 x2 − + −

(e) e e3 1x x + 2] g (f) e7 x x 6( )e 5+ (g) e e4 2 3x x −] g (h) e x 1x +] g (i)

( )

x

e x 1x

2

− (j) xe x 2x +] g

(k) ( ) ( )x e e e x2 1 2 2 3x x x+ + = + (l) ( )

( )

x

e x

7 3

7 10x

2−−

(m) ( )

ee xe

e

x5 5 5 1x

x x

x2

− =−

4. ( ) e1 6= −( ) ;e f1 6= −fl m 5. e 6. ee15

5− = −−

7. 19.81 8. ex y 0+ = 9. x e y e3 03 6+ − − =

Ans_PART_1.indd 378 6/30/09 5:39:41 AM

379ANSWERS

10. , mine11− −c m

11. ;dx

dye

dx

d ye y7 7x x

2

2

= = =

12.

`

`

;dx

dye

dx

d ye

y ey e

dx

d yy

2 2

2 11 2

1

x x

x

x

2

2

2

2

= =

= +− =

= −

Exercises 4.2

1. (a) e7 x7 (b) e x− − (c) e6 x6 2− (d) xe2 x 1+2

(e) ( )x e53 x x2 5 73+ + + (f) e5 x5 (g) e2 x2− − (h) e10 x10

(i) e2 1x2 + (j) x e2 2 x1+ − − (k) ( ) ( )e x e5 1 4 x x4 4 4+ +

(l) ( )e x2 1x2 + (m) ( )

x

e x3 2x

3

3 − (n) ( )x e x5 3x2 5 +

(o) ( )

( )

x

e x

2 5

4 2x

2

2 1

+++

2. ( ) ( )e e e28 1 7 1x x x2 2 5 2+ +

3. ( ) e0 9=( ) ;e f1 3=f 2−m 4. 5

5. x y 1 0+ − = 6. e31

3−

7. y ex e2= − 8. ( ) e1 18 4 2− = − −f m

9. 2−( , ) ; ( , )min maxe0 0 1−

10.

( )

dx

dye e

dx

d ye e

e ey

4 4

16 16

1616

x x

x x

x x

4 4

2

2

4 4

4 4

= −

= +

= +=

−

−

−

11.

( ) ( )

y e

dx

dye

dx

d ye

dx

d y

dx

dyy

e e ee e e

3

6

12

3 2

12 3 6 2 312 18 60

LHS

RHS

x

x

x

x x x

x x x

2

2

2

2

2

2

2

2 2 2

2 2 2

=

=

=

= − +

= − += − +==

dx

d y

dx

dyy3 2 0

2

2

` − + =

12. y ae

dx

dybae

dx

d yb ae

b y

bx

bx

bx2

2

2

2

=

=

=

=

13. n 15= −

14.

15.

Exercises 4.3

1. (a) e C21 x2 + (b) e C

41 x4 + (c) e Cx− +− (d) e C

51 x5 +

(e) e C21 x2− +− (f) e C

41 x4 1 ++ (g) e C

53 x5− +

(h) e C21 t2 + (i) e x C

71

2x7 − + (j) ex

C2

x 32

+ +−

2. (a) ( )e51

15 − (b) ee

11

122

− = −− (c) ( )e e32

17 9 −

(d) ( )e e1921

14 2− − (e) e21

1214 + (f) e e 1

212 − −

(g) e e21

1216 3+ −−

Ans_PART_1.indd 379 6/30/09 5:39:44 AM


3. (a) 0.32 (b) 268.29 (c) 37 855.68 (d) 346.85 (e) 755.19

4. ( )e e e e 1 units4 2 2 2 2− = − 5. ( )e e41

units3 2− −

6. 2.86 units 2 7. 29.5 units 2 8. ( )π

e2

1 units6 3−

9. 4.8 units 3 10. 7.4 11. (a) ( )x x e2 x+ (b) x e Cx2 +

12. πe units3

13. ( )e21

5 units4 2−

Exercises 4.4

1. (a) 4 (b) 2 (c) 3 (d) 1 (e) 2 (f ) 1 (g) 0 (h) 7

2. (a) 9 (b) 3 (c) −1 (d) 12 (e) 8 (f ) 4 (g) 14 (h) 14 (i) 1 (j) 2

3. (a) −1 (b) 21

(c) 21

(d) −2 (e) 41

(f ) 31− (g)

21−

(h) 31

(i) 121

(j) 121−

4. (a) 3.08 (b) 2.94 (c) 3.22 (d) 4.94 (e) 10.40 (f ) 7.04 (g) 0.59 (h) 3.51 (i) 0.43 (j) 2.21

5. (a) log y x3 = (b) log z x5 = (c) log y 2x = (d) log a b2 = (e) log d 3b = (f ) log y x8 = (g) log y x6 = (h) log y xe = (i) log y xa = (j) log Q xe =

6. (a) 3 5x = (b) a 7x = (c) a3b = (d) x y9 = (e) a by = (f ) 2 6y = (g) x3y = (h) 10 9y = (i) e 4y = (j) x7y =

7. (a) x 1 000 000= (b) x 243= (c) x 7= (d) x 2= (e) x 1= − (f ) x 3= (g) .x 44 7= (h) x 10 000= (i) x 8= (j) x 64=

8. y 5= 9. 44.7 10. 2.44 11. 0 12. 1

13. (a) 1 (b) (i) 3 (ii) 2 (iii) 5 (iv) 21

(v) -1 (vi) 2

(vii) 3 (viii) 5 (ix) 7 (x) 1 (xi) e

14. Domain: ;x 0> range: all real y

15. Curves are symmetrical about the line .y x=

16. x ey=

Exercises 4.5

1. (a) log y4a (b) log 20a (c) log 4a (d) logb5a

(e) zlog yx3 (f ) log y9k

3 (g) logyx

a 2

5

(h) log zxy

a

(i) log ab c104 3 (j) log

r

p q3 2

3

2. (a) 1.19 (b) -0.47 (c) 1.55 (d) 1.66 (e) 1.08 (f ) 1.36 (g) 2.02 (h) 1.83 (i) 2.36 (j) 2.19

3. (a) 2 (b) 6 (c) 2 (d) 3 (e) 1 (f ) 3 (g) 7

(h) 21

(i) -2 (j) 4

4. (a) x y+ (b) x y− (c) 3 x (d) 2 y (e) 2 x (f ) x y2+ (g) x 1+ (h) y1 − (i) x2 1+ (j) y3 1−

5. (a) p q+ (b) 3 q (c) q p− (d) 2 p (e) p q5+ (f ) p q2 − (g) p 1+ (h) q1 2− (i) q3 + (j) p q1− −

6. (a) 1.3 (b) 12.8 (c) 16.2 (d) 9.1 (e) 6.7 (f ) 23.8 (g) -3.7 (h) 3 (i) 22.2 (j) 23

7. (a) x 4= (b) y 28= (c) x 48= (d) x 3= (e) k 6=

Exercises 4.6

1. (a) 1.58 (b) 1.80 (c) 2.41 (d) 3.58 (e) 2.85 (f ) 2.66 (g) 1.40 (h) 4.55 (i) 4.59 (j) 7.29

2. (a) .x 1 6= (b) .x 1 5= (c) .x 1 4= (d) .x 3 9= (e) .x 2 2= (f ) .x 2 3= (g) .x 6 2= (h) .x 2 8= (i) .x 2 9= (j) .x 2 4=

3. (a) .x 2 58= (b) .y 1 68= (c) .x 2 73= (d) .m 1 78= (e) .k 2 82= (f ) .t 1 26= (g) .x 1 15= (h) .p 5 83= (i) .x 3 17= (j) .n 2 58=

4. (a) .x 0 9= (b) .n 0 9= (c) .x 6 6= (d) .n 1 2= (e) .x 0 2= − (f ) .n 2 2= (g) .x 2 2= (h) .k 0 9= (i) .x 3 6= (j) .y 0 6=

5. (a) .x 5 30= (b) .t 0 536= (c) .t 3 62= (d) .x 3 81= (e) .n 3 40= (f ) .t 0 536= (g) .t 24 6= (h) .k 67 2= (i) .t 54 9= (j) .k 43 3= −

Ans_PART_1.indd 380 6/30/09 5:39:46 AM

381ANSWERS

Exercises 4.7

1. (a) x11+ (b) x

1− (c) x3 13+

(d) x

x4

22 −

(e) x x

x5 3 9

15 33

2

+ −+

(f) x

xx

x x5 1

52

5 110 2 52

++ =

++ +

(g) x x6 51+ + (h)

x8 98−

(i) ( ) ( )x x

x2 3 16 5

+ −+

(j) ( ) ( )x x x x4 1

42 7

24 1 2 7

30+

−−

=+ −

−

(k) 4( )logx x5

1 e+ (l) ( )lnx x x91

1 8− −c m

(m) ( )logx x4

e3 (n) 5( )logx x x x6 2

1e

2+ +c m

(o) log x1 e+ (p) log

x

x1 e

2

−

(q) logxx

x2 1

2 e

+ + (r) ( )logx

xx x

13 1e

32

++ +

(s) logx x

1

e

(t) ( )

log

x x

x x x

2

2 e

2−

− −

(u) ( )

( )

log

log

x x

e x x2 1

e

xe

2

2 −

(v) loge x x1x

e+c m

(w) log

x

x10 e

2. ( )1 = −21

fl 3. logx 10

1

e

4. logx y2 2 2 2 0e− − + =

5. y x 2= − 6. 52− 7. logx y5 5 25 0e+ − − =

8. logx y5 19 19 19 15 0e− + − = 9. , log21

21

21

41

e −d n 10. ,e e

1c m maximum

11. (a)

(b)

(c)

12. ( ) logx2 5 3

2

e+ 13. (a) ln3 3x (b) ln10 10x

(c) ´ln3 2 2 x3 4−

14. ln x y4 4 4 0$ − + = 15. log logx y3 3 1 9 3 0e e$ + − − =

Exercises 4.8

1. (a) ( )log x C2 5e + + (b) ( )log x C2 1e2 + +

(c) ( )ln x C25 − + (d) log logx C x C21

21

2ore e+ +

(e) ln x C2 + (f) log x C35

e + (g) ( )log x x C3e2 − +

(h) ( )ln x C21

22 + + (i) ( )log x C23

7e2 + +

(j) ( )log x x C21

2 5e2 + − +

2. (a) ( )ln x C4 1− + (b) ( )log x C3e + +

(c) ( )ln x C61

2 73 − + (d) ( )log x C121

2 5e6 + +

(e) ( )log x x C21

6 2e2 + + +

3. (a) 0.5 (b) 0.7 (c) 1.6 (d) 3. 1 (e) 0.5

4. .log log log3 2 1 5 unitse e e2− = 5. log 2 unitse

2

6. ( . )log0 5 2 unitse2+ 7. 0.61 units 2

8. 3π log unitse3 9. 2 9π log unitse

3

10. 47.2 units 2 11. ( )π

e e2

1 units2 4 3−

12. (a)

( ) ( )

( )

( )( )

( )

x x

x x

x

x x

x

xx x

xx

31

32

3 3

1 3

3 3

2 3

93 2 6

93 3

RHS

LHS

2

2

=+

+−

=+ −

−

++ −

+

=−

− + +

=−+

=

xx

x x93 3

31

32

2`

−+ =

++

−

(b) ( ) ( )log logx x C3 2 3e e+ + − +

Ans_PART_1.indd 381 6/30/09 5:39:48 AM


13. (a) x

xx

x

xx

11

5

11

15

16

RHS

LHS

= −−

=−− −

−

=−−

=

xx

x16

11

5`

−− = −

−

(b) ( )logx x C5 1e− − +

14. log

C2 3

3

e

x2 1

+−

15. .1 86 units2

Test yourself 4

1. (a) 6.39 (b) 1.98 (c) 3.26 (d) 1.40 (e) 0.792 (f) 3.91 (g) 5.72 (h) 72.4 (i) 6 (j) 2

2. (a) e5 x5 (b) e2 x1− − (c) x1

(d) x4 54+

(e) ( )e x 1x +

(f) ln

xx1

2

− (g) 9( )e e10 1x x +

3. (a) e C41 x4 + (b) ( )ln x C

21

92 − + (c) e Cx− +−

(d) ( )ln x C4+ +

4. x y3 3 0− + = 5. e

e12

2

−+

6. ( )e e21

1 units4 6 2−

7. ( )π

e e6

1 units6 6 3−

8. (a) 0.92 (b) 1.08 (c) 0.2 (d) 1.36 (e) 0.64

9. ( )e e 1 units2 2−

10. (a) 2.16 units 2 (b) x ey= (c) .2 16 units2

11. (a) .x 1 9= (b) .x 1 9= (c) x 3= (d) x 36= (e) .t 18 2=

12. (a) ( )e23

12 −

(b) ln31

10

(c) ln861

3 2+

13. e x y e3 04 4− − =

14. 0.9

15. (a) ( )e e 1 units2−

(b) ( )π

e e2

1 units2 2 3−

16. (a) ylog xa5 3

(b) log3x

2pk

17. lnx y2 2 4 0+ − − =

18. (0, 0) point of infl exion, ( , )e3 27 3− − − minimum

19. 5.36 units 2

20. (a) 0.65 (b) 1.3


1. ( )

( ) ( ) log

e x

e x x e x1

2 1

x

x xe

2 2

2 2

+

+ − + 2. 2 e

3. (a) 2.8 (b) 1.8 (c) 2.6

4. ( )loge x e x9 41x x

e4 4 8+ +c m 5. .0 42 units2

6. x2 32−−

7. 12 units3 8. log5 5xe

9. ( ) ( );log log logdxd

x x x x1 2 18 3e e e2 = + 10.

logC

33

e

x

+

11. (a) (1, 0) (b) ; logx y x y1 0 10 1 0e $− − = − − =

(c) log

110

1units

e

−f p 12. 0.645 units 2

13. ( )log log

e

e x xe x1x

xe

xe

2

+ −

log log

e

x x x1x

e e=+ −

14.

( )

y e e

dx

dye e

dx

d ye e

e ey

x x

x x

x x

x x

2

2

= +

= −

= − −

= +=

−

−

−

−

15.

( ) ( )

y e

dx

dye

dx

d ye

dx

d y

dx

dyy

e e ee e e

3 2

15

75

4 5 10

75 4 15 5 3 2 1075 60 15 10 100

LHS

RHS

x

x

x

x x x

x x x

5

5

2

2

5

2

2

5 5 5

5 5 5

= −

=

=

= − − −

= − − − −= − − + −==

16. ( )f x e x3 6x2= −

17.

Ans_PART_1.indd 382 6/30/09 5:39:50 AM

383ANSWERS

Chapter 5: Trigonometric functions

Exercises 5.1

1. (a) 36° (b) 120° (c) 225° (d) 210° (e) 540° (f) 140°(g) 240° (h) 420° (i) 20° (j) 50°

2. (a) 43π

(b) π6

(c) 5π6

(d) 4π3

(e) 5π3

(f) 7π20

(g) π

12

(h) 5π2

(i) 5π4

(j) 2π3

3. (a) 0.98 (b) 1.19 (c) 1.78 (d) 1.54 (e) 0.88

4. (a) 0.32 (b) 0.61 (c) 1.78 (d) 1.54 (e) 0.88

5. (a) 62° 27’ (b) 44° 0’ (c) 66° 28’ (d) 56° 43’ (e) 18° 20’ (f) 183° 21’ (g) 154° 42’ (h) 246° 57’(i) 320° 51’ (j) 6° 18’

6. (a) 0.34 (b) 0.07 (c) 0.06 (d) 0.83 (e) −1.14(f) 0.33 (g) −1.50 (h) 0.06 (i) −0.73 (j) 0.16

Exercises 5.2

1.

π3

π4

π6

sin23

2

121

cos 21

2

1

23

tan 3 13

1

cosec3

2 2 2

sec 2 23

2

cot3

11 3

2. (a) 31

(b) 21

(c) 8

3 3 (d)

34 3

(e) 0 (f) 2

2 3 1+

(g) 2 3− (h) 2

2 2+ (i)

23 2 2−

(j) 2

2 3+

3. (a) 141

(b) 4

6 2− (c)

23

(d) 1 (e) 441

4. (a) 4

6 2+ (b) 3 2−

5. π π π π

π π π π

´ ´

´ ´

cos cos sin sin

sin cos cos sin

3 4 3 4

21

2

123

2

1

2 2

1

2 2

3

4 6 4 6

2

123

2

121

2 2

3

2 2

1

LHS

RHS

LHS RHS

+

= +

= +

= +

= +

=

= +

=

So π π π π π π

cos cos sin sin sin cos3 4 3 4 4 6

+ = +π π

cos sin4 6

6. (a) π π π

ππ

43

44

4

4

= −

= −

(b) 2nd (c) 2

1−

7. (a) π π π

ππ

65

66

6

6

= −

= −

(b) 2nd (c) 21

8. (a) π π π

ππ

47

48

4

24

= −

= −

(b) 4th (c) 1−

9. (a) π π π

ππ

34

33

3

3

= +

= +

(b) 3rd (c) 21−

10. (a) π π π

ππ

35

36

3

23

= −

= −

(b) 4th (c) 23

−

11. (a) 1− (b) 23

(c) 3− (d) 2

1− (e) 3

1

12. (a) (i) π π π

ππ

613

612

6

26

= +

= +

(ii) 1st (iii) 23

(b) (i) 2

1 (ii) 3 (iii)

2

1− (iv) 3

1 (v)

23

−

13. (a) ,π π3 3

5 (b) ,

π π4

54

7 (c) ,

π π4 4

5 (d) ,

π π3 3

4

(e) ,π π6

56

7

14. (a) sin θ (b) tan x− (c) − cos α (d) sin x (e) cot θ

Ans_PART_1.indd 383 6/30/09 5:39:52 AM


15. θsin2

16. ;cos sinx x54

53= − =

17. , , ,π π π π

x4 4

34

54

7=

Exercises 5.3

1. (a) π4 cm (b) π m (c) π

325

cm (d) π2

cm (e) π4

7mm

2. (a) 0.65 m (b) 3.92 cm (c) 6.91 mm (d) 2.39 cm (e) 3.03 m

3. 1.8 m 4. 7.5 m 5. π

212

6. 25 mm 7. 1.83

8. 1397

mm 9. 25.3 mm 10. π

SA36

175cm ,2=

π

V648

125 35cm3=

Exercises 5.4

1. (a) π8 cm2 (b) π2

3m2 (c)

π3

125cm2 (d)

π4

3cm2

(e) π

849

mm2

2. (a) .0 48 m2 (b) .6 29 cm2 (c) .24 88 mm2 (d) .7 05 cm2 (e) .3 18 m2

3. .16 6 m2 4. θ 494= 5. 6 m 6. (a)

π6

7 cm (b)

π12

49cm2

7. π8

6845mm2 8. 75 cm 2 9. 11.97 cm 2

10. , 3π

rθ15

cm= =

Exercises 5.5

1. (a) π8 cm2 (b) π

46 9 3

m2−

(c) π3

125 75cm2−

(d) ( )π

4

3 3cm2

− (e)

( )π8

49 2 2mm2

−

2. (a) .0 01 m2 (b) 1.45 cm 2 (c) 3.65 mm 2 (d) 0.19 cm 2 (e) 0.99 m 2

3. 0.22 cm 2 4. (a) π7

3 cm (b)

π149

cm2 (c) 0.07 cm 2

5. 134.4 cm 6. (a) 2.6 cm (b) π6

5 cm (c) 0.29 cm 2

7. (a) 10.5 mm (b) 4.3 mm 2

8. (a) π

425

cm2 (b) 0.5 cm 2

9. (a) °77 22l (b) 70.3 cm 2 (c) 26.96 cm 2 (d) 425.43 cm 2

10. 9.4 cm 2

11. (a) π

911

cm (b) π π

2218

12118

396 121cm2− =

−

(c) ( )π π

229

119

11 18cm+ =

+

12. (a) 5 cm 2 (b) 0.3% (c) 15.6 cm

13. (a) π10 cm (b) π24 cm2

14. (a) ( )π π

89

209

4 18 5cm+ =

+ (b) 3:7

15. (a) π

2225

cm3 (b) ( )π π

2105

1802

15 7 24cm2+ =

+

Exercises 5.6

1. (a) 0.045 (b) 0.003 (c) 0.999 (d) 0.065 (e) 0.005

2. 41

3. 31

4. 1 343 622 km 5. 7367 m

Exercises 5.7

1. (a) y

xπ2

3π2

π 2π

-1

1

(b) y

xπ2

3π2

π 2π

-2

2

Ans_PART_2.indd 384 6/30/09 5:59:13 AM

385ANSWERS

(c) y

xπ2

3π2

π 2π

1

-1

2

(d) y

xπ2

3π2

π 2π

3

2

1

(e) y

xπ2

π 2π

3

-3

3π2

(f) y

xπ 2π

4

-4

3π2

π2

(g)

xπ 2π

y

3

4

2

1

3π2

π2

(h)

x

y

5

2πππ2

3π2

(i)

3

x

y

π 2π3π2

π2

(j)

xπ 2π

y

1

3π2

π2

2. (a)

x

y

1

-1

2ππ 3π2

π4

7π4

5π4

3π4

π2

Ans_PART_2.indd 385 6/30/09 5:59:15 AM


(b) y

xπ 2π3π

45π4

3π2

7π4

π2

π4

(c) y

1

–1

x2ππ

32π3

4π3 3

5ππ

(d)

3

–3

x

y

π4

3π4

π2

5π4

3π2

7π 2ππ4

(e) y

-6

x

6

2π3

5π3

2π4π3

ππ3

(f) y

x2ππ

(g) y

x5π3

4π3

2π3

π3

2ππ

(h) y

-3

3

x2ππ 3π

2

π2

(i) y

−2

2

x2ππ 3π

2π2

(j) y

-4

4

x2ππ 3π

2π2

Ans_PART_2.indd 386 6/30/09 5:59:17 AM

387ANSWERS

3. (a) y

-1

1

xπ-π

- - - 3π4

3π4

π4

π2

π4

π2

(b) y

-7

7

xπ-π - - 3π

4π2

π4

π4

π2

3π4

-

(c)

xπ-π

y

- 3π4

π2

π4

π2

π4-3π

4-

(d) y

5

x

-5

-π - - - π3π4

π4

π4

π2

3π4

π2

(e) y

-2

xπ-π --

π2

π2

-3π4

2

π4

π4

3π4

4.

πx

y

8

-8

3π 4π2π

5. (a)

x

y

1

-1

π 2π3π2

π2

(b)

x

y

3π2

π2

π 2π

(c)

x

y

1

-1

π 2ππ2

3π2

(d)

x

y

3

-3

π 2π3π2

π2

Ans_PART_2.indd 387 6/30/09 5:59:19 AM


(e)

x

y

2

-2

π 2π3π2

π2

(f)

x

y

4

-4

π 2π5π4

3π4

π2

π4

3π2

7π4

(g)

x

y

1

-1

π 2π5π4

3π2

π4

π2

7π4

3π4

(h)

x

y

1

π 2π7π4

3π4

π4

π2

5π4

3π2

(i)

3π2

π2

x

y

3

2

1

-1π 2π

(j)

x

y

2

1

5

4

3

-1π 2π3π

2π2

6. (a)

x

y

1

-1

-2 -1 1 2

(b)

x

y

3

-3

-2 -1 1 2

7. (a)

y = sin x

y = sin 2x

x

y

1

-1

π 3π2

π2

2π

(b)

x

y

1

2

-1

-2

ππ2

3π2

2π

Ans_PART_2.indd 388 6/30/09 5:59:22 AM

389ANSWERS

8. (a)

x

y

1

2

3

4

5

−1

−2

−3

−4

−5

π

y = 2 cos x

y = 3 sin x

π2

3π2

2π

(b)

y = 2 cos x + 3 sin x

x

y

1

−1

−2

−3

−4

−5

2

3

4

5

π 2π3π2

π2

9.

x

y

-1

-2

1

2 y= cos 2x - cos x

π 2π3π2

π2

10. (a)

y = cos x + sin x

-1

-2

1

2

y

x2πππ

23π2

(b)

y = sin 2x – sin x

–1

–2

1

2

y

xπ 3π

2π2

2π

(c)

x

–1

–2

1

2

–3

y = sin x + 2 cos 2x

2ππ

y

3π2

π2

(d)

y=3 cos x − cos 2x

x

−1

−2

1

2

−3

−4

3

y

2πππ2

3π2

Ans_PART_2.indd 389 6/30/09 5:59:24 AM


(e)

y = sin x - sinx2

3π2

2πx

y

-1

-2

1

2

-3

-4

3

π2

π

Exercises 5.8

1. (a)

x

y

1

-1y = sin x

6

y = x2

541 2π2

2π3π2

π3

There are 2 points of intersection, so there are 2 solutions to the equation. (b)

x

y

1

-π2

π2

-1

y = sin x

y = x

2

1-1-2 32 π-π -3

There are 3 points of intersection, so there are 3 solutions to the equation.

2. x 0= 3. .x 1 5= 4. , .x 0 4 5= 5. ,x 0 1=

6. . ,x 0 8 4=

π2

3π2

π 2π

1

-1

y

x

y = cos x

y = sin x

7. (a) Period 12 months, amplitude 1.5 (b) 5.30 p.m.

8. (a) 1300 (b) (i) 1600 (ii) 1010 (c) Amplitude 300, period 10 years

9. (a)

0

2

4

6

8

10

12

14

16

18

January February March April May June July August

(b) It may be periodic - hard to tell from this data. Period would be about 10 months. (c) Amplitude is 1.5

10. (a)

(b) Period 24 hours, amplitude 1.25 (c) 2.5 m

Exercises 5.9

1. (a) 4 cos 4 x (b) sin x3 3− (c) sec x5 52 (d) ( )sec x3 3 12 + (e) ( )sin x− (f) 3 cos x

(g) ( )sin x20 5 3− − (h) ( )sinx x6 2 3−

(i) ( )secx x14 52 2 + (j) cos sinx x3 3 8 8−

0

0.5

1

1.5

2

2.5

3

3.5

4

6.20

am

11.55 am

6.15

pm

11.48 pm

6.20

am

11.55 am

6.15

pm

11.48 pm

6.20

am

11.55 am

6.15

pm

11.48 pm

Ans_PART_2.indd 390 6/30/09 5:59:25 AM

391ANSWERS

(k) ( )πsec x x22 + + (l) sec tanx x x2 +

(m) sin sec tan cosx x x x3 2 3 2 3 22 +

(n) cos sin cos sin

xx x x

xx x x

42 2

22 2

− = −

(o) ( )

sin

sin cos

x

x x x

5

3 5 5 3 4 52

− +

(p) ( ) ( )sec tanx x x9 2 7 7 2 72 8+ +

(q) ( )sin cos sin cosx x x x2 45 5 5r 2−

(s) ( ) ( )sin cos loge x x x2 21

1txe+ − −

(u) ( ) ( )cose e x1x x+ + (v) sincos

cotxx

x=

(w) ( )sin coscos sin

e x e xe x x2 2 3 2

3 2 2 2

x x

x

3 3

3

− += −

(x)

( )tan

tan sec

tan

tan secx

e x e x

x

e x x7

2 7 7 7

7

2 7 7 7

x x

x

2

2 2 2

2

2 2

−

−=

2. ( )

cos sin sinsin cos sin

x x xx x x

44

2 3 5

3 2 2

−= −

3. 12 4. πx y6 3 12 6 3 0− + − =

5. cossin

tanxx

x− = − 6.

3 3

29

2 3− = −

7. sec xetanx2 8. πx y8 2 48 72 2 2 0+ − − =

9.

( )

cos

sin

cos

y x

dx

dyx

dx

d y

x

2 5

10 5

25 2 5

2

2

=

= −

= −

cos x

y

50 5

25

= −

= −

10. ( )( )( )

( )

sincos

sin

f x xf x xf x x

f x

22

2

= −= −== −

l

m

11. [ ( )]log tan

tansec

tantan

tantan

tantan cot

dxd

x

xx

xx

xx

xx x

1

1

LHS

RHS

e

2

2

2

=

=

= +

= +

= +=

( )log tan tan cotdxd

x x xe` = +7 A

12. ,π π3

33

−d n maximum,

,π π3

53

35

− −d n minimum

13. (a) °π

sec x180

2 (b) °π

sinx60−

(c) °π

cosx900

14.

( )

sin cos

cos sin

sin cos

sin cos

y x x

dx

dyx x

dx

d yx x

x xy

2 3 5 3

6 3 15 3

18 3 45 3

9 2 3 5 39

2

2

= −

= +

= − +

= − −= −

15. ,a b7 24= − = −

Exercises 5.10

1. (a) sin x C+ (b) cos x C− + (c) tan x C+

(d) °π cos x C45− + (e) cos x C

31

3− + (f) cos x C71

7 +

(g) tan x C51

5 + (h) ( )sin x C1+ +

(i) ( )cos x C21

2 3− − + (j) ( )sin x C21

2 1− +

(k) ( )πcos x C− + (l) ( )πsin x C+ +

(m) tan x C72

7 + (n) cosx

C82

− +

(o) tanx

C93

+ (p) ( )cos x C3− − +

2. (a) 1 (b) 33

13

2 3− = (c)

2

22= (d)

31−

(e) π1

(f) 21

(g) 43

(h) 51−

3. 4 units 2 4. 3 2

162

= units 2 5. 0.86 units 2

6. 0.51 units 3 7. π4

units 3

8. π π

33 3

3 3− =

− units 2 9. 2 2 units 2

10. (a)

2

( )

( )

π

π

π

ππ

ππ

cos

sin

sin sin

V y dx

x dx

x

20

1 0units

πa

b2

0

2

0

3

=

=

=

= −

= −=

π

; E

#

#

(b) .3 1 units 3

11. siny x2 3= −

Ans_PART_2.indd 391 6/30/09 5:59:27 AM


Test yourself 5

1. (a) π6

5cm (b)

π12

25cm2 (c) 0.295 cm 2

2. (a) 3 (b) 23

(c) 23

(d) 2

1−

3. (a) ,π π

x4

34

7= (b) ,

π πx

6 65

=

4. (a)

(b)

5. (a) sin x− (b) cos x2 (c) sec x2 (d) cos sinx x x+

(e) sec tan

xx x x

2

2 − (f) sin x3 3− (g) sec x5 52

6. (a) cos x C21

2− + (b) sin x C3 + (c) tan x C51

5 +

(d) cosx x C− +

7. (a) 2

1 (b)

32 3

8. π

x y3 2 14

30+ − − =

9. cos

sin

x t

dtdx

t

dtd x

2

2 2

2

2

=

= −

cos t

x

4 2

4

= −

= −

10. 2

1 units 2 11.

π3

units 3 12. (a) 5 (b) 2

13. 3 3− 14. (a) π7

8cm2 (b) 0.12 cm 2

15. (a)

(b) .x 0 6=

16. 2

3 2− units 2

17. 2 units 2 18. πx y4 8 8 0+ − − = 19. cosy x3 2= −

20. (a)

(b) . , . ,x 0 9 2 3 3=


1. 0.27 2. 21

13

16

3 3− =

−e o 3. r 64= units, θπ

512=

4. (a) ,2 3Period amplitude= =

(b)

5. (a) siny x3= −

(b)

( )sin sinsin sin

dx

d yy

x xx x

9

9 3 9 39 3 9 30

LHS

RHS

2

2

= +

= + −= −==

Ans_PART_2.indd 392 6/30/09 5:59:29 AM

393ANSWERS

6.

7. °π

sec x180

2

8. (a)

´

´

tansec

cossincos

cos sincos

cos sinsec cosec

xx

xxx

x xx

x xx x

1

1

1 1

RHS

LHS

2

2

2

=

=

=

=

==

sec cosectansec

x xxx2

` =

(b) log log321

3e e=

9. ( )cos sinx x x e2 2 2 sinx x2+ 10. (a) ,π4

4d n and ,π3

24

d n (b) 4Maximum = (c) 1Amplitude =

11. ( )π π

83 2

33

4 2 3 3cm cm2 2− =

−e o

12. °π cosx C180− + 13.

21−

14. 0.204 units 3 15. sin coscos sin

x xx x

+−

16. ( )π π

29

21

4

9 2cm2− =

−d n

17.

18. , , , , , , ,π π π π8

085

089

08

130d d d dn n n n

19. 2

121

22 1

− =−

units 2

20. ( )( )( )

( )( )

cossincoscos

f x xf x xf x x

xf x

2 36 318 39 2 39

== −= −= −= −

l

m

Chapter 6: Applications of calculus to

the physical world

Exercises 6.1

1. (a) R t20 8= − (b) R t t15 42= + (c) R x16 4= −

(d) R t t15 4 24 3= − + (e) R et= (f) θsinR 15 5= −

(g) πRr

2100

3= − (h) R

x

x

42=

− (i) R

r800

4002

= −

(j) πR r4 2=

2. (a) h t t C2 42 3= − + (b) A x x C2 4= + +

(c) πV r C34 3= + (d) cosd t C7= − +

(e) s e t C4 3t2= − +

3. 20 4. 1 5. 6 e 12 6. 13 7. 900 8. e2 53 +

9. y x x x 63 2= − + + 10. ;RdtdM

t R1 4 19= = − = −

[i.e. melting at the rate of 19 g per minute (g min -1 )]

11. .11 079 25 cm− per second (cms -1 ) 12. 21 000 L

13. 165 cm 2 per second (cm 2 s -1 ) 14. 0.25−

15. 41 cm 2 per minute (cm 2 min -1 ) 16. 31 cm 3

17. 108 731 people per year 18. (a) 27 g (b) .2 7− (i.e. decaying at a rate of 2.7 g per year)

19. y e

dx

dye

y

4

4

x

x

4

4

=

=

=

20.

( )

( )( )

S e

dtdS

e

eS

2 3

2 2

2 2 3 32 3

t

t

t

2

2

2

= +

=

= + −= −

Ans_PART_2.indd 393 6/30/09 5:59:31 AM


Exercises 6.2

1. (a) 80 (b) 146 (c) 92 days (d)

2. (a) 99 061 (b) 7 hours

3. `

`

( ) ,

,

.

.

.ln

t MM et M

ee

kk

M e

0 1001005 95

95 1000 950 95 50 01

100

a When

When

So .

kt

k

k

t

5

5

0 01

= === ==== −==

−

−

−

−

(b) 90.25 kg (c) 67.6 years

4. (a) 35.6 L (b) 26.7 minutes

5. (a) P 50000 = (b) .k 0 157= (c) 12 800 units (d) 8.8 years

6. 2.3 million m 2 7. (a) P e50 000 . t0 069= (b) 70 599 (c) 4871 people per year (d) 2040

8. (a) . °65 61 C (b) 1 hour 44 minutes

9. (a) 92 kg (b) Reducing at the rate of 5.6 kg per hour (c) 18 hours

10. (a) ; .M k200 0 002530 = = (b) 192.5 g (c) Reducing by 0.49 g per year (d) 273.8 years

11. (a) B e15 000 . t0 073= (b) 36 008 (c) 79.6 hours

12. 11.4 years 13. (a) 19% (b) 3200 years

14. (a) ( ) ( )( )

( )

( )

P t P t e

dt

dP tkP t e

kP t

kt

kt

0

0

=

= −

= −

−

−

(b) 23% (c) 2% decline per year (d) 8.5 years

15. 12.6 minutes 16. 12.8 years

17. (a) 76.8 mg/dL (b) 9 hours 18. 15.8 s 19. 8.5 years

20. (a) Q Ae

dt

dQkAe

kQ

kt

kt

=

=

=

(b)

´

ln

lnln

dt

dQkQ

dQdt

kQ

tkQ

dQ

k QdQ

kQ C

kt Q C

kt C Q

e Qe e Q

Ae Q

1

1

1 1

1

So

kt C

kt C

kt

1

1

1

1

=

=

=

=

= +

= +− =

===

−

−

#

#

Exercises 6.3

1. (a) v

t

a

t

(b) v

t

a

t

(c)

t t

v a

Ans_PART_2.indd 394 6/30/09 5:59:33 AM

395ANSWERS

(d)

(e)

2. (a) , ,t t t2 4 6 (b) 0 to , ,t t t1 3 5 (c) 0 to t1 (d) t5

3. (a)

(b)

4. (a) , , ,O t t t2 4 6 (b) , ,t t t1 53 (c) t5 (d) (i) At rest, accelerating to the left . (ii) Moving to the left with zero acceleration.

5. (a) , , ,…π π π4 4

34

5 (b) , , , ,…

ππ

π0

2 23

6. (a) At the origin, with positive velocity and positive constant acceleration (moving to the right and speeding up). (b) To the right of the origin, at rest with negative constant acceleration. (c) To the left of the origin, with negative velocity and positive acceleration (moving to the left and slowing down). (d) To the right of the origin, with negative velocity and acceleration (moving to the left and speeding up) . (e) To the left of the origin, at rest with positive acceleration.

Exercises 6.4

1. (a) 18 cms -1 (b) 12 cms -2 (c) When , ;t x0 0= = after 3 s (d) After 5 s

2. (a) 8 ms 1− − (b) ;a 4= constant acceleration of 4 ms -2 (c) 13 m (d) after 2 s (e) 5 m−

(f)

3. (a) 4 m (b) 40 ms -1 (c) 39 m (d) 84 m

(e)

4. (a) 2 cm (b) After 1 s (c) -4 cm (d) 6 cm (e) -7 cms -1

5. (a) 2 ms -1 (b) e4 ms2 2− (c) ( )a e e v4 2 2 2t t2 2= = =

(d)

6. (a) sinv t2 2= − (b) cosa t4 2= − (c) 1 cm

(d) , , , , . . .π

ππ

02 2

3s (e) ±1 cm (f) , , . . .

π π π4 4

34

5s

(g) cosa t x4 2 4= − = −

7. (a) ;v t t a t3 12 2 6 122= + − = + (b) 266 m (c) 133 ms -1 (d) 42 ms -1

Ans_PART_2.indd 395 6/30/09 5:59:34 AM


8. (a) 4 3( ) , ( )x t x t20 4 3 320 4 3• •= − = −• (b) , , ,x x x1 20 320cm cms cms1 2= = =− −o p (c) The particle is on the RHS of the origin, travelling to the right and accelerating.

9. (a) v t5 10= − (b) 95 ms 1− − (c) a 10 g= − =

10. 32( )

,( )

vt

at3 1

173 1

102=+

=+

−

11. (a) At the origin (b) 61

cms 1− (c) 361

cms 2− −

(d) The particle is moving to the right but decelerating (e) ( )e 1 s3 −

12. (a) 3 ms -1 (b) When , ,t 0 1 3s s s= (c) 10 ms -2

13. (a)

(b) ,cos sinx t x t6 2 12 2= = −o p (c) 6 3− cms -2 (d) ( )sin sinx t t x12 2 4 3 2 4= − = − = −p

14. (a) 7 m (b) 16 m (c) After 7 s

(d)

(e) 10 m

15. (a) 18.75 m (b) -15 ms -1 (c) 5 s

16. (a) At the origin: x 0=

( )

,

t t tt t t

t t t

2 3 42 02 3 42 00 2 3 42 0

3 2

2

2

− + =− + =

= − + =

Since ,t 0= the particle is initially at the origin .

( ) ( )( )

t tb ac2 3 42 0

4 3 4 2 42327

0<

2

2 2

− + =− = − −

= −

So the quadratic equation has no real roots . So the particle is never again at the origin .

(b) dtdx

t t6 6 422= − +

At rest:

( ) ( )( )

dtdx

t tt tb ac

0

6 6 42 07 0

4 1 4 1 727

0<

2

2

2 2

=

− + =− + =− = − −

= −

So the quadratic equation has no real roots. So the particle is never at rest .

17. (a) 0 cm (at the origin) (b) , , , . . .π π π4 4

34

5s (c) ±12 cm

18. (a) 8 e 16 cms -1 (b) 0 s (initially) (c) 1 cm

19. (a) 7 s (b) 2

7 or

27 2

s (c) 49 cm

Exercises 6.5

1. 12 cm 2. 28 m 3. -42.5 cm

4. (a) 570 cms -2 (b) 135 cm (c) After 0.5 s

5. ( 1)e cm3 + 6. 163 m 7. (a) 95 cms -1 (b) 175 cm

8. .h t t4 9 4 22= − + + 9. 262 m 10. ( )e 3 m5 −

11. -744 cm 12. ( )π2 3 cm− 13. 1.77 m 14. 893 m

15. (a) ( )3 3 m+ (b) 4 3 ms 2− −

16. (a) 154

ms -1 (b) ( )lnxn

n3

22 3 m= − +

(c)

( )

( )

( )

≥

vt

t

t

tt

tt

t

32

32

3 3

2 3 6

3 32 6 6

3 92

0

= −+

=+

+ −

=+

+ −

=+

When :t v0 0= = When :t v0 0> > Also t2 0> and t3 9 0>+ when t 0>

≤

t t

tt

v

2 3 9

3 92

1

0 1

So <

<

<`

+

+

17. (a) 5 e 45 ms −1 (b) e 30 m (c) x ex

2525

t5==

p (d) 50 ms -2

Test yourself 6

1. (a) 0 m, 0 ms -1 , 8 ms -2 (b) 0, 0.8 s (c) 0.38 m

2. -76 m, -66 ms -2 3. 39.6 years

Ans_PART_2.indd 396 6/30/09 5:59:36 AM

397ANSWERS

4. (a) 6 cms 1− (b) 145 855.5 cms -2 (c)

( )

x ex ex e

ex

26189 29

t

t

t

t

3

3

3

3

=====

o

p

5. 1 m

6.

( )

sincos

sinsin

x tx tx t

tx

2 36 3

18 39 2 39

=== −= −= −

o

p

7. (a) 2, 6 s (b) (i) 16 cm (ii) 15 cms 1− (iii) 18 cms 2− − (c) Particle is 16 cm to the right of the origin, travelling at 16 cms 1− to the right. Acceleration is 18 cms 2− − (to the left), so the particle is slowing down.

8. (a) (i) 18 ms 2− (ii) 15 ms 1− (iii) -28 m (b) Particle is 28 m to the left of the origin, travelling at 15 ms 1− to the right, with 18 ms 2− acceleration (to the right), so the particle is speeding up.

9. (a) , ,t t t1 3 5 (b) ,t t2 4 (c) andt tafter3 5

(d) (i)

(ii)

10. (a) 48.2% (b) 1052.6 years

11. (a) , , , , . . .π π π π6 6

56

136

17s (b) , , , , . . .

π π π π3 3

53

73

11s

(c) 2

1ms 2− −

12. (a) 15 m (b) 20 m (c) 4 s

13. (a) 16 941 (b) 1168 birds/year (c) 18.3 years

14. (a) (i)

(ii)

(b) ,t t1 3

15. 55 033 m


1. (a) ( )

coscos

x tt

32

338

3

2 4 3= − + =

−

(b) x x938= − −p c m

2. (a) 1 m, 0 ms -1 (b) . ´3 26 107 ms -2 (c) Show ( )t 1 03 6+ = has no solution for ≥t 0

3. (a) cosx t4= (b) ±2 3 cms -1

4.

`

( )

( )

,

( )

cos

cos

sin

sin

sin

cos

sin

v t

x t dt

t Ct x

CC

x t

adtd

t

tx

5 5

5 5

50 0

0 0

5

5 5

25 525

a

When

=

== += == +==

=

= −= −

#

(b) 25 ms 2− (c) .7 5ms–2−

5. (a) 19.9 years (b) 16%

Ans_PART_2.indd 397 6/30/09 5:59:38 AM


6. ( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

NbN k bN e

kN

dtdN

bN k bN e

kN k k bN e

bN k bN e

k N k bN e

bN k bN e

k N bN k bN e bN

bN k bN e

k N bN k bN e

bN k bN e

bk N

bN k bN e

k N

bbN k bN e

kN

kN bN

kt

kt

kt

kt

kt

kt

kt

kt

kt

kt

kt

kt

0 0

0

0 02

0 0

0 02

20 0

0 02

20 0 0 0

0 02

20 0 0

0 02

202

0 0

20

0 0

02

2

=+ −

=+ −

− − −

=+ −

−

=+ −

+ − −

=+ −

+ −

−+ −

=+ −

−+ −

= −

−

−

−

−

−

−

−

−

−

−

−

−f p

7

7

7

7

7

7

7

7

A

A

A

A

A

A

A

A

7. e3 cms 2−


1. 3.2 years 2. 1.099 3. 27 m 4. 2−

5. , .x e x 3 42y= = 6. x e3 2 x2 2+ 7. ±x21=

8. (a) 47.5 g (b) 3.5 g/year (c) After 9.3 years

9. (a) 0 cms 1− (b) sina t x18 3 9= − = −

10. (a) 7750 L (b) 28 minutes

11. .622 1 units3 12. x4 312

+

13. ( )log x x C31

3 3 2e2 + − +

14. (a) 100 L (b) 40 L (c) 16 L− per minute, i.e. leaking at the rate of 16 L per minute (d) 12.2 minutes

15. logx x x C2 4 e3 2− + + 16. log3 2e

17. (a) 7.8 cm (b) .0 06 cms 2− −

18. e x C41 x4 + + 19.

ex1 2

x2

−

20. (a) .k 0 101= (b) 2801 (c) 20 days (d) (i) 11 people per day (ii) 283 people per day

21. (a) 1.77 (b) logx 3

1

e

22. ( )πe

e2

1 units2

4 3−

23. (a) ,v a6 0ms ms1 2= =− − (b) 3 m

(c) , , , . . .π π π4 4

34

5seconds

(d) ( )

sinsin

a tt

x

12 24 3 24

= −= −= −

24. .4 67 units2 25. 27 m− 26. .x 0 28=

27. x y 2 0− + = 28. ,e2

121− −c m minimum

29. (a) .1 60 cm2 (b) .0 17 cm2

30. 15 months

31. (a) π6

5cm (b)

π12

25cm2

32.

33. 1 34. cot x 35. ( )sece e5 1x x5 2 5 + 36. 3

37. (a)

(b) 2 units2

38. 0.348

39. (a) ( )sin cose x xx + (b) x xtan sec3 2 2

(c) π

sin x6 32

− −c m 40. (a) 546 ms 1− (b)

( )a e

ex

204 54

t

t

2

2

===

(c) 20 ms -2

41. ln ln ln8 338− = 42. ´6 12m m

43. π

23

units3

Ans_PART_2.indd 398 6/30/09 5:59:39 AM

399ANSWERS

44.

45. π

x y6 12

30− − − =

46. π π

33 3

3 3units2− =

−

47. ( ) ( )sin cose e e3 x x x2

48. ( )e5 units2−

49. (a) 2

1 (b)

23

−

50. (a) 21 π cm 2 (b) ( )π21 9 cm2+

51. ( )sec log

x

x 1e2 +

52. lnx x x C3 25− − +

53. .8 32 units3 54. π πcose x C51 1x5 + +

55. .x 1 8Z

56. ( )e 1 units2 2− 57. 1−

58. (c) 59. (d) 60. (a)

61. (d) 62. (b) 63. (c)

Chapter 7: Series

Exercises 7.1

1. 14, 17, 20 2. 23, 28, 33 3. 44, 55, 66

4. 85, 80, 75 5. , ,1 1 3− − 6. 87, 83, 79

7. , ,2 221

3 8. 3.1, 3.7, 4.3 9. 16, 32, 64

10. 108, 324, 972 11. 16, 32− , 64 12. 48, 96− , 192

13. , ,161

321

641

14. , ,13516

40532

121564

15. 36, 49, 64 16. 125, 216, 343 17. 35, 48, 63

18. 38, 51, 66 19. 126, 217, 344 20. 21, 34, 55

Exercises 7.2

1. (a) , ,T T T3 11 191 2 3= = = (b) , ,T T T5 7 91 2 3= = = (c) , ,u u u5 11 171 2 3= = = (d) , ,T T T3 2 71 2 3= = − = − (e) , ,t t T19 18 171 2 3= = = (f) , ,u u u3 9 271 2 3= = = (g) , ,Q Q Q9 11 151 2 3= = = (h) , ,t t t2 12 581 2 3= = = (i) , ,T T T8 31 701 2 3= = = (j) , ,T T T2 10 301 2 3= = =

2. (a) , ,T T T1 4 71 2 3= = = (b) , ,t t t4 16 641 2 3= = = (c) , ,T T T2 6 121 2 3= = =

3. (a) 349 (b) 105 (c) 248 (d) -110 (e) -342

4. (a) 1029 (b) -59 039 (c) 1014 (d) 53 (e) 1002

5. (a), (b), (d) 6. (b), (d) 7. 16 th term 8. Yes

9. 7 th term 10. 23 rd term 11. (a) 1728 (b) 25 th term

12. (a) -572 (b) 17 th term

13. n 33= 14. n 9= 15. , , , . . .n 88 89 90=

16. , ,n 41 42 43= 17. n 501= 18. n 151=

19. (a) n 14= (b) -4 20. -6

Exercises 7.3

1. (a) 128 (b) 54 (c) 70 (d) 175 (e) 220

(f) 6047

(g) 40 (h) 21 (i) 126 (j) 1024

2. (a) 65 (b) 99 (c) 76 (d) 200 (e) 11 (f) 39 (g) 97 (h) 66 (i) 75 (j) 45

3. (a) n2 1n 1

6

−=/ (b) n7

n 1

10

=/ (c) n

n

3

1

5

=/ (d) k6 4

k

n

1−

=/

(e) kk

n2

3=/ (f) ( )n

n 1

50

−=/ (g) .3 2k

k

n

0=/ (h)

21

nn 0

9

=/

(i) ( )a k d1k

n

1+ −

=/ (j) ark

k

n1

1

−

=/

Exercises 7.4

1. (a) y 13= (b) x 4= − (c) x 72= (d) b 11= (e) x 7=

(f ) x 4221= (g) k 4

21= (h) x 1= (i) t 2= − (j) t 3=

Ans_PART_2.indd 399 6/30/09 5:59:41 AM


2. (a) 46 (b) 78 (c) 94 (d) -6 (e) 67

3. (a) 590 (b) -850 (c) 414 (d) 1610 (e) -397

4. (a) -110 (b) 12.4 (c) -8.3 (d) 37 (e) 1554

5. T n2 1n = +

6. (a) T n8 1n = + (b) T n2 98n = + (c) T n3 3n = +

(d) T n6 74n = + (e) T n4 25n = − (f) T n20 5n = −

(g) Tn

86

n = + (h) T n2 28n = − − (i) .T n1 2 2n = +

(j) Tn4

3 1n = −

7. 28 th term 8. 54 th term 9. 30 th term

10. 15 th term 11. Yes 12. No

13. Yes 14. n 13= 15. , , . . .n 30 31 32=

16. -2 17. 103 18. 785

19. (a) d 8= (b) 87

20. d 9= 21. ,a d12 7= =

22. 173 23. a 5=

24. 280 25. 1133

26. (a) log loglog log

log

log loglog log

log

T T x x

x x

x

T T x x

x x

x

2

3 2

2 1 52

5

5 5

5

3 2 53

52

5 5

5

− = −= −=

− = −= −=

Since T T T T2 1 3 2− = − it is an arithmetic series with .logd x5= (b) log logx x80 or5 5

80 (c) 8.6

27. (a)

´

´ ´

T T

T T

12 3

4 3 32 3 3

327 12

9 3 4 33 3 2 3

3

2 1

3 2

− = −

= −= −=

− = −

= −= −=

Since T T T T2 1 3 2− = − it is an arithmetic series

with .d 3=

(b) 50 3

28. 26 29. 122 b 30. 38 th term

Exercises 7.5

1. (a) 375 (b) 555 (c) 480

2. (a) 2640 (b) 4365 (c) 240

3. (a) 2050 (b) 2575−

4. (a) 4850− (b) 4225

5. (a) 28 875 (b) 3276 (c) 1419− (d) 6426 (e) 6604 (f) 598 (g) 2700− (h) 11 704 (i) 290− (j) 1284

6. (a) 700 (b) 285− (c) 1170 18 terms− ] g (d) 6525 (e) 2286−

7. 21 8. 8 9. 11 10. ,a d14 4= =

11. ,a d3 5= − = 12. 2025 13. 3420

14. 8 and 13 terms 15. 1010

16. (a) ( ) ( )x x2 4 1+ − +

( ) ( )x xx3 7 2 4

3= + − += +

(b) ( )x25 51 149+

17. 1290 18. 16

19. S S T

S S Tn n n

n n n

1

1`

= +− =

−

−

20. 4234

Exercises 7.6

1. (a) No (b) Yes, r43= − (c) Yes, r

72=

(d) No (e) No (f ) No (g) Yes, .r 0 3=

(h) Yes, r53= − (i) No (j) Yes, r 8= −

2. (a) x 196= (b) y 48= − (c) ±a 12=

(d) y32= (e) x 2= (f ) ±p 10=

(g) ±y 21= (h) ±m 6= (i) x 4 3 5!=

(j) 3 7±k 1= (k) ±t61= (l) ±t

32=

3. (a) T 5nn 1= − (b) .T 1 02n

n 1= − (c) T 9 –n

n 1=

(d) .T 2 5nn 1= − (e) .6 3Tn

n 1= − (f) .T 8 2

2n

n

n

1

2

==

−

+

(g) ·T41

4

4

n

n 2

=

= −

n 1− (h) T 1000 10

10n

n

n

1

2

= −= − −

−

+

]]

gg

(i) T 3 3

3n

n

n

1= − −= −

−]]

gg

(j) T31

52

n

n 1

=−d n

4. (a) 1944 (b) 9216 (c) -8192

(d) 3125 (e) 72964

5. (a) 256 (b) 26 244 (c) 1.369

(d) -768 (e) 1024

3

6. (a) 234 375 (b) 268.8 (c) -81 920

(d) 156 250

2187 (e) 27

7. (a) ´3 219 (b) 7 19 (c) 1.04 20

Ans_PART_2.indd 400 6/30/09 5:59:44 AM

401ANSWERS

(d) 41

21

2119

21=d n (e)

43 20d n

8. 11 49 9. 6 th term

10. 5 th term 11. No

12. 7 th term 13. 11 th term

14. 9 th term 15. n 5= 16. r 3=

17. (a) r 6= − (b) −18

18. , ±a r1

210

= = 19. n 7= 20. 20872

Exercises 7.7

1. (a) 2 097 150 (b) 7 324 218

2. (a) 720 600 (b) 26 240

3. (a) 131 068 (b) 65 53632 769

4. (a) 7812 (b) 356455

(c) 8403 (d) 273 (e) 255

5. (a) 255 (b) 729364

(c) 97 656.2

(d) 1128127

(e) 87 376

6. (a) 1792 (b) 3577

7. 148.58 8. 133.33

9. n 9= 10. 10 terms

11. a 9= 12. 10 terms

13. (a) $33 502.39 (b) $178 550.21

14. (a) 1−( )2 5 k

k

n

1−

=/ (b)

( ) ( )S

3

5 1

3

1 5n

n n

= −− −

=− −

15. 2146

Puzzles

1. Choice 1 gives $465.00. Choice 2 gives $10 737 418.23!

2. 382 apples

Exercises 7.8

1. (a) Yes LS 1321= (b) No (c) Yes LS 12

54= (d) No

(e) Yes LS 3= (f) Yes LS 3225= (g) No

(h) Yes LS 1225= − (i) No (j) Yes LS 1

73=

2. (a) 80 (b) 42632

(c) 6632

(d) 12 (e) 107

(f) 54

(g) 1072− (h)

209

(i) 48 (j) 3916−

3. (a) 127

(b) 274

(c) 12500

1 (d)

641

(e) 40963645

4. (a) 141

(b) 52

(c) 481

(d) 221

(e) 3 (f) 5 (g) 52

(h) 531− (i) 1

54

(j) 56

5. a 4= 6. r52= 7. a 5

53= 8. r

87= 9. r

41= −

10. r32= − 11. 3, ,a r a r

32

631

and= = = =

12. 192, , 153a r41

53

LS= = − =

13. 1, , 3, 1, ,a r a r32

32

43

LS LS= = = = − = − = −

14. 150, , 375a r53

LS= = =

15. , , 1a r52

32

51

LS= = = 16. , ,a r a r352

253

and= = = =

17. x3221= 18. (a) k1 11 1− (b)

52− (c) k

43=

19. (a) p21

21

1 1− (b) 75

(c) p141=

20.

Sr

ar

a r

r

a a r

ra a ar

rar

1 1

1

1

1

1

1

LS n− =−

−−

−

=−

− −

=−

− +

=−

n

n

n

n

^^

hh

Exercises 7.9

1. (a) 210 (b) 13th (c) 57

2. (a) 39 (b) 29th (c) 32

3. (a) n3 3+

(b) [ ( ) ]

[ ( ) ]

( )

( )

( )

´ ´

Sn

a n d

n n

n n

n n

n n

22 1

21

2 6 1 3

21

12 3 3

21

3 9

23

3

n = + −

= + −

= + −

= +

= +

4. (a) (i) $23 200 (ii) $26 912 (iii) $31 217.92 (b) $102 345.29 (c) 6.2 years

5. (a) (i) 93% (ii) 86.49% (iii) 80.44% (b) 33.67% (c) 19 weeks

6. (a) 0.01 m (b) 91.5 m

7. (a) 49 (b) 4 mm

8. (a) 3 k m (b) ( )k k3 3 m+ (c) 9

9. (a) 96.04% (b) 34 (c) 68.6

10. (a) 77.4% (b) 13.5 (c) 31.4

11. (a) 94

(b) 97

(c) 192

(d) 9925

(e) 2119

(f) 307

Ans_PART_2.indd 401 6/30/09 5:59:45 AM


(g) 19043

(h) 1450

7 (i)

990131

(j) 2999361

12. 0.625 m 13. 15 m 14. 20 cm 15. 3 m

16. (a) 74.7 cm (b) 75 m

17. (a) 4.84 m (b) After 3 years

18. 300 cm 19. 3.5 m 20. 32 m

21. (a) 1, 8, 64, … (b) 16 777 216 people (c) 19 173 961 people

Exercises 7.10

1. (a) $740.12 (b) $14 753.64 (c) $17 271.40 (d) $9385.69 (e) $5298.19

2. (a) $2007.34 (b) $2015.87 (c) $2020.28

3. (a) $4930.86 (b) $4941.03

4. $408.24 5. $971.40

6. $1733.99 7. $3097.06

8. $22 800.81 9. $691.41

10. $1776.58 11. $14 549.76 12. $1 301 694.62

13. (a) $4113.51 (b) $555.32 (c) $9872.43 (d) $238.17 (e) $10 530.59

14. $4543.28 15. 4 years 16. 8 years

17. (a) x 7= (b) x 5= (c) x 8=

(d) .x 6 5= (e) .x 8 5=

18. $7.68 19. Kate $224.37

20. Account A $844.94

Exercises 7.11

1. $27 882.27 2. $83 712.95

3. $50 402.00 4. $163 907.81

5. $40 728.17 6. $29 439.16

7. $67 596.72 8. $62 873.34

9. $164 155.56 (28 years) 10. $106 379.70

11. $3383.22 12. $65 903.97

13. $2846.82 14. $13 601.02

15. $6181.13 16. $4646.71 17. $20 405.74

18. (a) $26 361.59 (b) $46 551.94

19. $45 599.17

20. (a) $7335.93 (b) $1467.18

21. $500 for 30 years 22. Yes, $259.80 over

23. No, shortfall of $2013.75

24. (a) $14 281.87 (b) $9571.96 (c) No, they will only have $23 853.83 .

25. $813.16

Exercises 7.12

1. $1047.62 2. $394.46 3. $139.15

4. (a) $966.45 (b) $1265.79

5. $2519.59

6. (a) $592.00 (b) $39 319.89

7. (a) $77.81 (b) $2645.42

8. $78 700

9. (a) Get Rich $949.61, Capital Bank $491.27 (b) $33 427.80 more through Capital Bank

10. $43 778.80 11. $61 292.20

12. NSW Bank $175.49 a month ($5791.25 altogether) Sydney Bank $154.39 a month ($5557.88 altogether) Sydney Bank is better

13. (a) $249.69 (b) $13 485.12

14. (a) $13 251.13 (b) $374.07 (c) $20 199.77

15. (a) $1835.68 (b) $9178.41

Test yourself 7

1. (a) T n4 5n = + (b) T n14 7n = −

(c) T 2 3nn 1

:= − (d) T 20041

n

n 1

=−d n

(e) T 2nn= −] g

2. (a) 2 (b) 1185 (c) 1183 (d) T S S15 15 14= −

S S T15 14 15= +

(e) n 16=

3. (a) 11 125 (b) 114013

(c) 3 985 785 (d) 34 750

(e) 21

4. (a) Each slat rises 3 mm so the bottom one rises up ´30 3 mm or 90 mm. (b) 87 mm (c) 90, 87, 84, . . . which is an arithmetic sequence with a 90= , d 3= − (d) 42 mm (e) 1395 mm

5. $3400.01

6. (a) (i) (b) (ii) (c) (i) (d) (iii) (e) (i) (f) (ii) (g) (ii) (h) (i) (i) (i) (j) (i)

Ans_PART_2.indd 402 6/30/09 5:59:47 AM

403ANSWERS

7. n 108=

8. (a) $24 050 (b) $220 250

9. ,a d33 13= − =

10. (a) 59 (b) 80 (c) 18 th term

11. (a) x 25= (b) ±x 15=

12. (a) 94

(b) 1813

(c) 13319

13. x 3=

14. (a) 136 (b) 44 (c) 6

15. 12121

16. $8066.42

17. (a) T n4 1n = + (b) .T 1 07nn 1= −

18. (a) x1 1< <− (b) 221

(c) x31=

19. d 5=

20. (a) 39 words/min (b) 15 weeks

21. (a) $59 000 (b) $15 988.89

22. 4.8 m 23. ,x172

2= −

24. (a) $2385.04 (b) $2392.03

25. 1300

26. (a) 735 (b) 4315

27. (a) $1432.86 (b) $343 886.91

28. n 20=

29. n 11=


1. (a) 8.1 (b) 19 th term

2. (a) π4

(b) π4

9 (c)

π4

33

3. (a) 2 097 170 (b) -698 775

4. (a) $40 (b) $2880

5. 6 th term 6. 17 823

7. 5 terms 8. , ,n 1 2 3=

9. 56− 10. $1799.79

11. x83= 12. $8522.53 13. k 20=

14. (a) $10 100 (b) $11 268.25 (c) $4212.41 (d) $2637.23

15. (a) cosec 2 x (b) ≤ ≤cos x1 1− So ≤ ≤cos x0 12 | | ≤cos x 12 So the limiting sum exists.

16. $240 652.62 .

Chapter 8 : Probability

Exercises 8.1

1. 301

2. 521

3. 61

4. 401

5. 20 000

1

6. (a) 74

(b) 73

7. 373

8. 121

9. (a) 2011

(b) 43

10. (a) 61

(b) 21

(c) 31

11. (a) 621

(b) 313

(c) 21

(d) 12499

12. (a) 158

(b) 157

(c) 151

13. 501

14. ,21

1 15. 4423

16. (a) 317

(b) 317

(c) 3112

17. 175

1 18. 8 19.

4325

20. 34 21. 31

22. (a) 61

(b) 31

(c) 65

23. (a) False: outcomes are not equally likely. Each horse and rider has different skills . (b) False: outcomes are not equally likely. Each golfer has different skills . (c) False: outcomes are not dependent on the one before. Each time the coin is tossed, the probability is the same . (d) False: outcomes are not dependent on the one before. Each birth has the same probability of producing a girl or boy. (e) False: outcomes are not equally likely. Each car and driver has different skills.

Exercises 8.2

1. 115

2. 92

3. 99.8% 4. 0.73 5. 38%

6. 98.5% 7. 2322

8. 185

9. 0.21 10. 91.9%

11. 87

12. 4946

13. (a) 152

(b) 1513

14. 117

15. 1615

Exercises 8.3

1. (a) 103

(b) 53

(c) 2011

(d) 107

2. (a) 51

(b) 21

(c) 53

(d) 53

(e) 5019

Ans_PART_2.indd 403 6/30/09 5:59:48 AM


3. (a) 265

(b) 269

(c) 1312

4. (a) 10029

(b) 2013

(c) 259

5. (a) 4527

(b) 94

(c) 32

6. (a) 143

(b) 2813

(c) 289

7. (a) 8021

(b) 8017

(c) 4021

8. (a) 101

(b) 2011

(c) 207

9. (a) 257

(b) 152

(c) 7544

10. (a) 103

(b) 52

(c) 103

Exercises 8.4

1. 361

2. 41

3. 81

4. 41

5. 12125

6. (a) 0.0441 (b) 0.6241 7. 80.4% 8. 32.9%

9. (a) 499

(b) 9115

10. 2075

3 11.

9919

12. 16 170

1

13. (a) 35 93729 791

(b) 35 937

8 (c)

35 93735 929

14. (a) 2400

1 (b)

5760 0001

(c) 5760 0005755 201

15. (a) 7776

1 (b)

77763125

(c) 77764651

16. (a) 25 000 000

9 (b)

25 000 00024 970 009

(c) 25 000 000

29 991

17. (a) 41

(b) 100

9 (c)

1009

18. (a) 221

(b) 111

(c) 227

(d) 2215

19. (a) 61.41% (b) 0.34% (c) 99.66%

20. (a) 21

n (b)

21

n (c) 1

21

22 1

n n

n

− = −

Exercises 8.5

1. (a) 41

(b) 41

(c) 21

2. (a) 81

(b) 83

(c) 87

3. (a) 900

1 (b)

9001

(c) 450

1 4. (a)

251

(b) 252

5. (a) 16925

(b) 16980

6. (a) 27.5% (b) 23.9% (c) 72.5%

7. (a) 0.42 (b) 0.09 (c) 0.49 8. (a) 1000189

(b) 1000441

(c) 1000657

9. (a) 0.325 (b) 0.0034 (c) 0.997

10. (a) 12160

(b) 116

11. (a) 274

(b) 61

12. (a) 251

(b) 825

1 (c)

82564

(d) 165152

(e) 16513

13. (a) 1 249750

19 (b)

124 975498

(c) 1 249 7501 239771

14. (a) 7516

(b) 7538

15. (a) 20251936

(b) 2025

88

16. (a) 2011

(b) 203

17. (a) 1296

1 (b)

324125

(c) 1296671

18. (a) 1 000 000

84 681 (b)

1 000 000912 673

(c) 1 000 000

27

19. (a) 17.6% (b) 11% (c) 21.2%

20. (a) 30251488

(b) 121

1 21. (a)

191

(b) 956

(c) 19021

(d) 3817

22. (a) 42522

(b) 425368

(c) 425

7 23. (a)

6517

(b) 715133

(c) 2145496

24. (a) 0.23 (b) 0.42 (c) 0.995 25. (a) 33% (b) 94%

26. (a) 216

1 (b)

725

(c) 21691

27. (a) 101

(b) 103

(c) 52

28. (a) 8125

(b) 8140

(c) 8156

29. (a) 1331343

(b) 1331336

(c) 1331988

30. (a) 8000

1 (b)

80006859

(c) 80001141

Test yourself 8

l . (a) 80.4% (b) 1.4% (c) 99.97%

2. (a) 1 2 3 4 5 6

1 0 1 2 3 4 5

2 1 0 1 2 3 4

3 2 1 0 1 2 3

4 3 2 1 0 1 2

5 4 3 2 1 0 1

6 5 4 3 2 1 0

(b) (i) 61

(ii) 61

(iii) 21

3. (a) (i) 401

(ii) 4039

(b) 79639

4. (a) 154

(b) 101

5. False: the events are independent and there is the

same chance next time 41d n

6. (a) 21

(b) 10029

(c) 51

(d) 2511

(e) 2516

7. (a) 52

(b) 157

(c) 152

8. (a) 7235

(b) 6635

9. 561

10. (a) 0.009% (b) 12.9% 11. (a) 131

(b) 133

(c) 265

Ans_PART_3.indd 404 6/30/09 3:49:21 PM

405ANSWERS

12. (a) 125

(b) 31

13. (a) 409

(b) (i) 103

(ii) 16027

(iii) 254

14. (a) 200

1 (b)

20081

(c) 10011

15. (a) 151

(b) 54

16. (a) 01

5 (b)

7450147

(c) 1

3725 (d)

33

725577

17. (a) 36180

(b) 17140

18. (a) 92

(b) 31

19. (a) 24364

(b) 729728

20. (a) 5021

(b) 253

(c) 5023


1. (a) 71

(b) 74

2. (a) 0.04 (b) 0.75 (c) 0.25

3. (a) 54 145

1 (b)

1 26433

73 4. (a)

134

(b) 5225

(c) 134

5. No—any combination of numbers is equally likely to win.

6. (a) 0 (b) 101

(c) 103

7. (a) 7776

1 (b)

12961

8. (a) 103

(b) 14512

9. (a) 144

1 (b)

1445

(c) 144

7 (d)

1443


1. 625324

2. (a) 24011296

(b) 2401864

(c) 24011105

3. 94

4. 43

9 5. $2929.08 6. …± ±104 52 26+ +

7. 44 th term 8. $945

9 . (a) 361

(b) 61

(c) 3611

(d) 365

(e) 125

10. a

65619841

11 . 2.4 m

12. (a) 3 000 000 (b) 3 000 336 (c) 146 insects per day

13. (2, 0), infl exion

14. (a) 507

(b) 2011

15. (a)

( ) ( ) ( )d t tt t t t

t t

230 65 125 8052 900 29 900 4225 15 625 20 000 640010 625 49 900 68 525

Pythagoras2 2 2

2 2

2

= − + −= − + + − += − +

(b) 2.3 h (c) 109.7 km

16. 199; 5050 17. 110

1

18. (a) , , ,T T T T4 11 18 811 2 3 12= = = =

(b) 1410 (c) 29 th term

19. 2

1−

20. (a)

(b)

(c)

21. – . . .15 4 7− + + 22. (a) 335

(b) 6635

23. $2851.52

24. (a) sinv t12 4= − (b) cosa t48 4= − (c) 3 cm

(d) , , , . . .π π

t 04 2

s= (e) ±3 cm (f) , , , . . .π π π

t8 8

385

s=

(g) ( )cos

cosa t

tx

48 416 3 416

= −= −= −

Ans_PART_3.indd 405 6/30/09 3:49:25 PM


25. 458

26. (a) 121

(b) 41

27. $180.76

28. . .

,.

ACAB BC

AC AB BCABC B

16 2569 6 12 8256

Sinceis right angled at

2 2

2 2 2 2

2 2 2

+Δ

= =+ = +

== +

29. x65=

30.

Let ABCD be a rhombus with .AB BC=

AB DC= and AD BC=

(opposite sides in a parallelogram)

AB BC DC ADall sides are equal

`

`

= = =

31. 40 000

1

32. (a) .k 0 025Z (b) after 42.4 years (c) 20.6 years

33. 76 473 34. 450 cm2

35. (a) 12 ms 1− (b) e48 ms4 2− (c)

( )

x e

x e

x ee

x

3 2

12

484 12

4

t

t

t

t

4

4

4

4

= +

=

==

=

..

.

.

(d)

36. n 4=

37.

( )

sin

cos

sin

sin

y x

dx

dyx

dx

d yx

xy

7

7 7

7 7 7

49 749

2

2

=

=

= −

= −= −

38. 101

39. (a) Square . .´46 3 46 3m m, rectangle . .´30 9 92 7m m (b) $8626.38

40. (a) 365

(b) 125

(c) 61

(d) 367

(e) 21

41.

42. (a) °θ 76 52= l (b) 0.92 cm2 43. $18 399.86

44. (a) . . .. . .

. . .

log log loglog log loglog log log

3 9 273 3 33 2 3 3 3

2 3

+ + += + + += + + +

Arithmetic series, since log log log log

log2 3 3 3 3 2 3

3− = −

=

(b) 210 log 3

45.

46. (a) 12.6 mL (b) 30 minutes 47. (a)

48. $277.33 49. (d) 50. (b) 51. (c) 52 . (c), (d)

53. (d) 54. (a) 55. (c) 56. (a) 57. (d)

Sample examination papers

Mathematics—Paper 1

1. (a) 0.75 (b) ( ) ( )x x3 2 3− −

(c) ( )

( )

x x

x xx x

xx

16

2 16

31

6 5

3 2 1 303 2 2 30

2 3028

− − =

− − =− + =

+ ==

d dn n

(d)

.

π

π

π

π

r

r

r

r

r

1231

3636

36

3 39

2

2

2

=

=

=

=

=

Ans_PART_3.indd 406 6/30/09 3:49:30 PM

407ANSWERS

(e) 3300 x

xxx

xx

3 74

3 73 7

1010 4

(f)

or

<<<>>< <`

+

− ++ −

−−

] g

2. (a) (ii) Since , : :AB BD AB AD 1 2= = ( )

( )

ABC ADEACB AEDA

BC DE

is common

corresponding s,

similarly

+ ++ ++

+ <==

ADEΔ;ABC

ADAB

AEAC

AE ACAC CE AC

CE AC

21

22

`

`

`

`

<Δ

= =

=+ =

=

(iii) ADAB

DEBC

21= =

.

..´

DEDE

3 421

2 3 46 8 cm

` =

==

(b) (ii) )

° °

( )

(

NOMNMO

10532

straight angle

angle sum of

++ T

==

° °

°°

sin sin

sin sin

sinsin

Aa

Bb

MO

MO

43 325

325 43

=

=

=

.6 4 mZ

(iii)

m

° °.

°. °

sin sin

sin sin

sinsin

Aa

Bb

MP

MP

75 536 4

536 4 75

8Z

=

=

=

3. (a) (i) 2

2

x x x x

dx

dyx x

xx

5 1 5 1

21

152

115

3 3

2 2

+ + = + +

= + = +

1

1−

(ii)

´

ln lnx x x x

dx

dyx x

x x

xx

31

3

31

3 1

3 1

1

2

2

2

+ = +

= −

= −

= −

−

−

(iii) ´dx

dyx

x

5 2 3 2

10 2 3

4

4

= +

= +

]]

gg

(b) (i) x

e Cx

e C2 1

12

x x2 2

−−

+ = + +− −

(ii) π

θ θ π π

ππ

cos cos cos 0 0

1 12

− + = − + − − +

= − − + += +

0

] ]]

g gg

; E

(c) (i) ( )

´2 1

5

2 1

2 1

2 1

5 2 5

15 2 5

5 2 5

2 2− +

+=

−

+

=+

= +

( )

,

´

a b

5 2 5 5 5 2

5 25 2

5 505 50

ii

`

+ = += += +

= =

4. (a) (i)

( ) ( , )( ) ( )

( )

A x y3 2 3 4 17 03 3 4 2 17 0

9 8 17 00 0

ii Substitute into

true

+ − =+ − =

+ − ==

A` lies on the line

( , )( ) ( )

( )

B x y1 5 3 4 17 03 1 4 5 17 0

3 20 17 00 0

Substitute into

true

− + − =− + − =

− + − ==

B` lies on the line Since both A and B lie on the line ,x y3 4 17 0+ − = this is the equation of AB

(iii) | |

| ( ) ( ) |

| |

.

da b

ax by c

3 4

3 0 4 0 17

9 16

17

25

17

517

3 4 units

2 2

1 1

2 2

=+

+ +

=+

+ −

=+

−

=

=

=

(iv) Length AB : 2 2

2

( ) ( )

( ) ( )

d x x y y

3 1 2 5

16 9

255

2 1 2 1

2

= − + −

= − − + −= +==

6 @

Ans_PART_4.indd 407 7/1/09 2:26:17 AM


:

.

.

´ ´

OAB A bh21

21

5 3 4

8 5

Area

units2

Δ =

=

=

(b) °

.

.

coscos

c a b CBC

BC

ab26 4 2 6 4 8749 49

49 497 cm

2 2 2

2 2 2

Z

Z

= + −= + −

=

] ]g g

5. (a) (i) % . %´24 600

800100 3 25=

(ii) Arithmetic series with ,a d24 600 800= = and n 12=

–

–

T a n d

T

1

24 600 12 1 80033 400

n

12

= += +=

]]g

g

So Kate earns $33 400 in the 12 th year.

(iii) [ ( ) ]

[ ( ) ]´

Sn

a n d2

2 1

212

2 24 600 12 1 800

348 000

n = + −

= + −

=

So Kate earns $348 000 over the 12 years. (b) (i)

yy

y x x xx xx

3 9 23 6 96 6

3 2

2

= − − += − −= −

l

m

For stationary points, 0=yl

( )( ),

x xx x

x

3 6 9 03 3 1 0

3 1

i.e. 2

`

− − =− + =

= −

,

,

x y

x y

3 3 3 3 9 3 225

1 1 3 1 9 1 27

When

When

3 2

3 2

= = − − += −

= − = − − − − − +=

] ]] ] ]

g gg g g

So ( , )3 25− and ( , )1 7− are stationary points. ( , ), ( )

( , ), ( )( )

( )

y

y

3 25 6 3 60

1 7 6 1 60

At

Atminimum point

maximum point

>

<

− = −

− = − −

m

m

( , )1 7` − is a maximum, ( , )3 25− minimum stationary point ( )

xxx

06 6 0

6 61

iii.e.

=− =

==

yFor inflexions, m

When ,x 1= 3 ( ) ( )y 1 3 1 9 1 29

2= − − += −

( , )1 9` − is a point of infl exion

(iii)

6. (a)

(i) ( , ) ( ) ( )( )

´ ´ ´ ´

´ ´

P P PP

2 1

72

72

75

72

75

72

75

72

72

34360

W N WWN WNWNWW

= ++

= +

+

=

(ii) ( ) ( )

´ ´

P P1

175

75

75

343218

at least one W NNN= −

= −

=

(b) °, ° °°, °

x 60 180 6060 120

(first, second quadrants)= −=

(c) (i) ( )v t dt

t t C

6 4

3 42

= += + +#

`

`

,

,

t v

CC

v t tt

v

0 0

0 3 0 4 0

3 45

3 5 4 595

When

When

cms

2

1

= == + +== +== += −

2

2

] ]

] ]

g g

g g

(ii) ( )x t t dt

t t C

3 4

2

2

2

= += + +3

#

,

,

t x

CCt t

t

0 0

0 2 0

22

2 2 216

When

When

cm

3 2

2

2

`

`

`

= == + +== +== +=

x

3

3

x

0 ] ]

] ]

g g

g g

7. (a)

c cc

( )

1

( )

(DC AB

DX AD

ADXDAX DXA

ADX

DX DC

2

180 60 260

21

ii

is isosceles

is equilateral

opposite sides of gram)

given`

`

'

`

+ +

<

Δ

= =

= =

= = −=

= −c m

Ans_PART_4.indd 408 6/30/09 4:12:16 PM

409ANSWERS

c c

c

( , )ADC XCB AD BCcointerior s,+ + + <

[ ,XC CB 1 similar to part (ii)]= =

( ) 180 60

120

XCB

CXB

iii

is isosceles

+

Δ

= −

=

c cc

c c c

c

(180 120 ) 230

180 (60 30 )

90(

¸CXB CBX

AXB

AXB

DXC

is right angled

straight angle)

`

`

+ +

++

Δ

= = −=

= − +

=

( ) (AXa b

BXBX

BX

BX

ADX1

2 11

3

iv equilateral)2 2 2

2 2 2

2

2

`

Δ== += += +==

c

43

(b) Sketch y x2= and x y 3+ = as unbroken lines.

2

( , ) ≥≥

y x1 00 1

Substitute into(false)

2

( , ) ££

x y1 0 31 0 3

Substitute into(true)

++

(c) (i)

(ii) The curve is increasing so .dtdP

0>

The curve is concave downwards so .dtd P

0<2

2

8. (a) (i) x 1 2 3 4 5

y 0 0.301 0.477 0.602 0.699

( ) ( )

.

log log log log1 2 2 3

1 73Z

= + + +

( ) ( ) [ ( ) ( )]

( ) ( ) ( )

( ) ( ) ( )

( ) [ ( ) ( )]

( ) [ ( ) ( )]

( ) ( )

log

log log log log

f x dx b a f a f b

x dx f f

f f

f f

f f

21

21

2 1 1 2

21

3 2 2 3

21

4 3 3 4

21

5 4 4 5

21

21

21

3 421

4 5

(ii)a

101

5

= − +

= − +

+ − +

+ − +

+ − +

+ + + +

b

6

6

@

@

##

(b) ( ) ,

,

.

ln lnln

ln

t PP e

P

P et P

e

e

ee

k ek

k

k

0 2020

206 100

100 20

20100

55

66

65

0 268

i When

When( )

kt

k

k

k

k

00

0

6

6

6

6

`

Z

= ===== ==

=

====

=

When( )

,P etP e

e

20102020292

ii

mice

.

. ( )

.

t0 268

0 268 10

2 68

=====

(iii) ,

.

.

.

ln lnln

ln

Pe

e

ee

t et

t

t

500500 20

20500

2525

0 2680 268

0 26825

12

When

(i.e. after 12 weeks)

.

.

.

.

t

t

t

t

0 268

0 268

0 268

0 268

==

=

====

=

=

9. (a) ( )( )π

π

π

Sr r h

r hr

hr

r

1602 160

2160

2160

i`

=+ =

+ =

= −

Ans_PART_4.indd 409 6/30/09 4:12:19 PM


ππ

π ππ

r r

V r h

r r r

r r

80

80

80

2

2

3

= −

=

= −

= −

c m

rZ

.

VV

±

±

π

ππ

π

π

r

rr

r

r

80 30

80 3 080 3

380

380

(ii)For max./min. volume,i.e

2

2

2

2

= −=

− ==

=

=

.2 91

l

l

. , ( . )

.

VV

ππ

rr

r

62 91 6 2 91

02 91

When

cm(maximum)<

`

= −= = −

=

m

m

(iii) . , ( . ) ..

πr V2 91 80 2 91 2 914206

Whencm

3

3

= = −=

] g

(b) 1996 to 2025 inclusive is 30 years. ( . ) ( . ) ( . ) . . .

( . )500(1.12 1.12 . . . 1.12 )500(1.12 1.12 . . . 1.12 )

A 500 1 12 500 1 12 500 1 12500 1 12

30 29 28

1

30 29 1

1 2 30

= + + ++

= + + += + + +

. . . . . .. , .( )

.

. ( . )

.( . )

$ .

a r

Sr

a r

S

A

1 12 1 12 1 121 12 1 12

1

1

1 12 1

1 12 1 12 1

270 29500 270 29

135146 30

is a geometric series

n

n

1 2 30

30

30

`

Z

+ + += =

=−

−

=−

−

==

10.

( , )( )lnln

log

y y ee

x

44

44 4

(a) (i) Substitute into

point of intersection isby definition of

x

x

`

`

= ==

=

(ii)

ln 4

ln4

ln

ln

A OABC e dx

e4 4

4 4 4 1

Area of rectangleln

x

x

0

4

0

2

= −

= −

= − +( )

( )

ln

ln

e e4 4

4 4 3 units

= − −

= −

06 @

#

(b) (i) For real, equal roots, 0Δ = i.e.

( )

( ) ( ) ( )( )

±

±

±

±

±

b ac

k kk k k

k k

ka

b b ac

4 0

1 4 1 02 1 4 0

6 1 0

24

2 1

6 6 4 1 1

26 32

26 4 2

3 2 2

2

2

2

2

2

2

− =− − =− + − =

− + =

=− −

=− − − −

=

=

=

] ] ]g g g

,( )

( ) ( )

,

kx k x k x x

ab ac

a x xx

51 4 5

04

4 4 1 54

00 0 4 5 0

(ii) When

Since andfor all

>

<> < >

2 2

2

2

2

`

Δ

ΔΔ

=+ − + = + +

= −= −= −

+ +

(c) (i)

( )

y x px qy q x px

y q p x px p

y q p x p

222

2

2

2 2 2

2 2

= + +− = +

− + = + +− − = +^ h

This is in the form ( ) ( ),x h a y k42− = − where a is the focal length and ( , )h k is the vertex. h p= − and k q p2= −

( , )p q pvertex is 2` − −

(ii) a

a

4 1

41

`

=

=

Count up 41

units for the focus

focus is ,p q p412− − +c m

(iii) For P : x m= since it is vertically below ,m m q3 2 +^ h

,

x mm ym

y

P mm

8

8

8

When

So

2

2

2

==

=

= d n

Ans_PART_4.indd 410 6/30/09 4:12:26 PM

411ANSWERS

≥

m qm

mq

mq m q

38

823

823

0 0

Distance

since and >

22

2

22

= + −

= +

= +

( ) m qq m

55

iv`

+ == −

Dm

q

mm

dmdD m

823

823

5

846

1

2

2

= +

= + −

= −

For stationary points dmdD

0=

m

m

m

m

846

1 0

846

1

46 8

468

234

− =

=

=

=

=

So there is a stationary point at .m234=

To determine its nature

dmd D

846

0>2

2

=

So concave upwards. minimum turning point

m234

When =

2

Dm

m8

235

8

23234

5234

42321

2

= + −

= + −

=

d n

So minimum distance is 42321

units.

Mathematics—Paper 2

1. (a) (i) xx3 5

8− =

=

(ii) xx3 5

2− = −

= −

(b)

±

xx

xx

5 49 0

93

2

2

2

− = −− =

==

(c)

( )

π

ππ

π

sin

sin

sin

65

6

6

21

2nd quadrant

= −

=

=

c m

(d) ( ) ( )( )( )( )( )( )

a a aa aa a a

a a

2 4 22 42 2 2

2 2

2

2

2

− − −= − −= − + −= + −] ]g g

(e) ´ ´ ´´ ´ ´

2 4 6 25 6

2 2 6 5 6

4 6 5 6

6

−= −= −= −

(f) ( )

. ..

´

´

log log

log loglog log

50 5 2

5 22 5 22 1 3 0 433 03

a a

a a

a a

2

2

== += += +=

(g) ,

,

,

Px x y y

2 2

23 0

24 2

121

1

1 2 1 2=+ +

= − + + −

= −

fdc

pn

m

2.

(a) Substitute ,A 5 121−c m into x y3 4 9 0+ − =

´ ´3 5 4 121

9

0

LHS

RHS

= + − −

==

A lies on the line

Substitute ,B 1 121c m into x y3 4 9 0+ − =

´ ´3 1 4 121

9

0

LHS

RHS

= + −

==

` B lies on the line AB has equation x y3 4 9 0+ − = ( ) x y

y x

y x

m

3 4 9 04 3 9

43

49

43

b

1`

+ − == − +

= − +

= −

l is perpendicular to AB , so m m 11 2 = −

Ans_PART_4.indd 411 6/30/09 4:12:28 PM


m

m

43

1

34

2

2`

− = −

=

Equation of l :

( )

( )

( )

y y m x x

y x

y xxx y

134

4

3 3 4 44 16

0 4 3 13

1 1− = −

− − = − −

+ = += += − +

(c)

( ) :´2 3

( )( )

( ) : ( )( )

( ) ( ):

´

x yx y

x yx y

xxx

4 3 13 0 13 4 9 0 2

1 4 16 12 52 0 39 12 27 0 4

3 4 25 25 025 25

1

− + =+ − =

− + =+ − =

+ + == −= −

Substitute x 1= − in (1): ´ y

yy

y

4 1 3 13 09 3 0

9 33

− − + =− =

==

So point of intersection is .( , )1 3−

(d) ( ) ( ):

( )

( )

:

( ) ( )

.

´ ´

d x x y y

AB

d

CP

d

A bh

5 1 121

121

4 3

16 9

255

1 4 3 1

3 4

9 16

255

21

21

5 5

12 5 units

2 12

2 12

22

2 2

2 2

2 2

2

= − + −

= − + − −

= + −= +==

= − − − + − −= += +==

=

=

=

c m

( ) ,

,

,

( , )

ACx x y y

D x y

xx x

x

xx

2 2

24 5

2

1 121

21

141

2

21

21

1 10

e Midpoint

where

1 2 1 2

1 2

=+ +

= − +− + −

= −

=

=+

= +

= +=

AC BDMidpoint midpoint=

f

fc

p

pm

yy y

y

y

2

141

2

121

221

121

1 2=+

− =+

− = +

y4− =

(f) ( , )D 0 4So = −

3. (a) (i)

( )cos sincos sin

dx

dyu v v u

x x xx x x

1$

= +

= + −= −

l l

(ii) dx

dye5 x5=

(iii) dx

dy

xx

2 142

=−

(b) (i) ( )

( )´

xC

xC

3 5

3 2

15

3 2

5

5

−+

=−

+

(ii) ´ cos

cos

x C

x C

321

2

23

2

− +

= − +

(c)

3

[ ] [ ]

e e

e e

e e e ee ee e

11

1 12

x x

x x

0

3

3 3 0 0

3 3

3 3

−−

= += + − += + − −= + −

−

−

− −

−

−

0

<

6

F

@

(d) ( )dx

dyx dx

x x C

18 6

9 62

= −

= − +

#

( , ),dx

dy

CC

C

2 1 0

0 9 2 6 224

24

At

2

− =

= − += +

− =

] ]g g

( )

dx

dyx x

y x x dx

x x x C

9 6 24

9 6 24

3 3 24

2

2

3 2

` = − −

= − −= − − +#

Substitute ( , )2 1− :

CC

Cy x x x

1 3 2 3 2 24 236

353 3 24 35

3 2

3 2`

− = − − += − +== − − +

] ] ]g g g

4. (a)

Ans_PART_4.indd 412 6/30/09 4:12:32 PM

413ANSWERS

(i) ( ) ´P52

83

203

WW =

=

(ii) ( ) ( ) ´ ´P P52

85

53

83

4019

WL LW+ = +

=

(iii) ( ) ( )

´

P P1 1

153

85

85

at least W LL= −

= −

=

(b) (i)

(ii)

.

( )

( )( )

( )

.

´ ´

A bh

A x dx

xx

21

21

5 5

12 5

2

22

23

2 32

22 2

12 5

units

or

units

2

2

2

3

2 2

2

2

=

=

=

= +

= +

= + −−

+ −

=

−

−

3

<

< =

F

F G

#

or

`

( )

( )

( )

( ) ( )

π

π

π

π

π

π

π

π

π

´

´

y x

y x

V y dx

x dx

x

V r h

2

2

2

1 3

2

3

3 2

3

2 2

3125

0

3125

31

31

5 5

3125

iii

units

units

a

2 2

2

2

2

3

2

3

3 3

2

2

3

3

= += +

=

= +

=+

=+

−− +

= −

=

=

=

=

−

−

b

3

]

]

g

g

=

=

<

G

G

F

##

(c) (i)

(ii) x1 2< <−

5. (a) y x x x

dx

dyx x

dx

d yx

2 9 12 7

6 18 12

12 18

3 2

2

2

2

= − + −

= − +

= −

,

( ) ( ),

dx

dy

x xx x

x xx

0

6 18 12 03 2 0

2 1 01 2

(i) For stationary points

2

2

=

− + =− + =

− − ==

, ( ), ( )

x yx y

1 2 1 9 1 12 1 7 22 2 2 9 2 12 2 7 3

WhenWhen

3 2

3 2

= = − + − = −= = − + − = −

] ]] ]g gg g

So ( , )1 2− and ( , )2 3− are stationary points.

At ( , ) ( )dx

d y1 2 12 1 18 6

2

2

− = − = −

( , )1 2− is a maximum turning point

At ( , ) ( )dx

d y2 3 12 2 18 6

2

2

− = − =

` ( , )2 3− is a minimum turning point

.

dx

d y

xxx

0

12 18 012 18

1 5

(ii) For points of inflexion2

2

=

− ===

. ,( . ) ( . ) ( . ).

xy

1 52 1 5 9 1 5 12 1 5 7

2 5

When3 2

== − + −= −

Check concavity:

x 1.25 1.5 1.75

dx

d y2

2

3− 0 3

Concavity changes, so . )2 5( . ,1 5 − is a point of infl exion. ,

3,

2 9 12 7

x

y

x

y

3

2 3 9 3 12 3 7178

3 3 32

(iii) When

When

3 2

3 2

= −= − − + − −= −== − + −=

] ] ]

] ] ]

g g g

g g g

Ans_PART_4.indd 413 6/30/09 4:12:34 PM


(b) (i)

(ii)

°´ ´

coscos

c a b ab C

c

2850 1200 2 850 1200 1203182 500

31825001784

2 2 2

2 2

= + −= + −===

So the plane is 1784 km from the airport. (c) ( ) ( )θ θ θ θ

θ θθ θ

cosec cot cosec cotcosec cot

cot cot11

2 2

2 2

+ −= −= + −=

6. (a)

`

At ( , ) ( )

dx

dyx

dx

dy

m

2

2 4 2 2

4

(i)

1

=

− = −

= −

Normal is perpendicular to tangent

( )

m m

m

m

y y m x x

y x

y xx y

14 1

41

441

2

4 16 20 4 18 1

1 2

2

2

1 1

` = −− = −

=

− = −

− = − −

− = += − +

_]

ig

(ii) y x2= (2) Substitute (2) in (1):

( ) ( ),

,

x xx x

x xx x

x x

x

0 4 184 18 0

2 4 9 02 0 4 9 0

2 4 9

241

2

2

= − +− − =

+ − =+ = − =

= − =

=

`

( ):

,

x

y

Q

241

2

241

5161

241

5161

Substitute in

2

=

=

=

=

d

c

n

m

(iii)

:PQ x yx y

xy

4 18 018 4

4 418

− + =+ =

+ =

4

4

( ) ( ) ( )

.

xx dx

x x x

4 418

8 418

3

8

241

4

18 241

3

241

8

2

4

18 2

3

2

12 8

Area

units

2

2

2 3

2

2

2 3

2 3

2

= + −

= + −

= + −

−−

+−

−−

=

−

−

1

1

2 c

c c c

m

m m mR

T

SSSS

<

=

V

X

WWWW

F

G

#

(b) (i) The particle is at the origin when ,x 0= i.e. at ,t t1 3 and t5

(ii) At rest, dtdx

0= (at the stationary points,

i.e. t2 and t4 )

,

,

.

ln ln

ln

ln

T T e

t TT

T et T

e

e

e

k ek

k

kT e

0 9797

975 84

84 97

9784

9784

55

59784

0 02997

(c)When

When

So .

´

kt

kt

k

k

k

t

0

0

5

5

5

0 029

`

== ==== ==

=

=

= −= −

−=

==

−

−

−

−

−

−

(i) tT e

159763

When. ´0 029 15

===

−

So the temperature is 63°C after 15 minutes.

Ans_PART_4.indd 414 6/30/09 4:12:38 PM

415ANSWERS

(ii)

.

.

..

ln ln

ln

ln

Te

e

e

t et

t

t

2020 97

9720

9720

0 0290 029

0 0299720

54 9

When.

.

.

t

t

t

0 029

0 029

0 029

==

=

=

= −= −

−=

=

−

−

−

So the temperature is 20°C after 54.9 minutes.

7. (a) (i)

4

( ) ( ) ( )

ππ

π π π

tan tan tan

tan tan tan

f x dxh

y y y y y

x dx

34 2

316

04

416 16

32

8

a0 4 1 3 2

0

2

Z

Z

+ + + +

+

+ + +

π

.0 35 unitsZ

b

cc

mm

7

;

A

G

#

#

(ii) cossin

tandx

dyxx

x

= −

= −

(iii)

) 44 (

[ ( )]

.

π

tan ln cos

ln cos ln cos

x dx x

40

0 35

00

= −

= − − −

=

ππ

c m

:

=

D

G

#

(b) (i)

° °

°°

.

sin sin

sin sin

sinsin

Aa

Bb

AD

AD

23 11040

11040 23

16 6 m

=

=

=

=

(ii) °.

. °.

sin

sin

BD

BDBD

4716 6

16 6 4712 2

=

==

So the height is 12.2 m

( )sum of quadrilateral+

°° °°° °

°° ( ° ° °)

°

( )

( )

( )

CBE

CBEDCB

ABC

DAB

DAB DCB ABC ADC

5050 50100130 5080360 100 80 80

100

(c)

andABCD is a parallelogram

base s of isosceles

exterior of

opposite s equal

`

`

++

+

+

+ + + +

+

+

+

ΔΔ

== +== −== − + +

== =

8. (a) (i) & (ii) ,π

π322

Amplitude period= = =

(iii) 4 points of intersection, so 4 roots

(b)

°, ° °°, °

( )

sinsin

sin

xx

x

x

2 1 02 1

21

30 180 3030 150

1st, 2nd quadrants

− ==

=

= −=

(c) (i) ´log log

log loglog logq p

12 2 3

2 32 2 32

x x

x x

x x

2

2

== += += +

] g

(ii) log log logx x

q

2 21

x x x= += +

(d) (i) . . .1 3 5+ + + is an arithmetic series with ,a d1 2= =

( )( ) ´

nT a n d

T

121

1 12 1 223

When

n

12

== + −= + −=

So there are 23 oranges in the 12th row. (ii) Total number of oranges is 289, so S 289n =

[ ( ) ]

[ ( ) ]

[ ]

´

´

Sn

a n d

nn

n nn nn

n

nn

22 1

2892

2 1 1 2

578 2 2 22

2289

28917

n

2

2

= + −

= + −

= + −=====

So there are 17 rows of oranges altogether.

Ans_PART_4.indd 415 6/30/09 4:12:41 PM


9. (a)

(b) The statement would only be true if there were equal numbers of each colour. It is probably false. ( )

( ) ( ),

( )≠

ln ln

ln

x xx x

x xx x

x

x

2 32 3

2 3 03 1 0

3 1

13

1

(c)

Butso the solution is

does not exist

x

2

2

2

`

= += +

− − =− + =

= −−

=−

(d) (i)

Whendx

dye

x k

dx

dye

x

k

=

=

=

So gradient m ek=

(ii) ,( )

( )

( )

x k y ey y m x x

y e e x ke x ke

y e x ke ee x k 1

When k

k k

k k

k k k

k

1 1

= =− = −− = −

= −= − += − +

(iii) Substitute ( , )2 0 into the equation ( )

( )e ke k

kk

0 2 13

3 03

k

k

= − += −

− ==

`

θ

°

°°

°°

°

.

ππ

π

π

π

´

´ ´

A r

180

1180

53180

53

18053

21

21

718053

22 7

(e) radians

cm

2

2

2

=

=

=

=

=

=

=

10. (a) (i)

(ii)`

( )

( )

( ) ( )

π

π

π

π

π

π

´

y xy x

V y dx

x dx

x

22

2

1 5

2

5

1 2

5

2 2

5243

0

5243

units

a

b

2

2 4

2

4

2

1

5

2

1

5 5

3

= += +

=

= +

=+

=+

−− +

= −

=

−

−

]]

gg

=

=

<

G

G

F

##

(b) (i) std

=

t sd

s3000

So =

=

Cost of trip taking t hours:

´

C s t

s s

s s

s s

7500

75003000

30007500 3000

30007500

2

2

= +

= +

= +

= +

]]

c

gg

m

( )

C s ss s

dsdC

s

s

30007500

3000 7500

3000 1 7500

3000 17500

(ii)

1

2

2

= +

= +

= −

= −

−

−

c]

d

mg

n

For minimum cost, dsdC

0=

.( )

s

s

ss

s

3000 17500

0

17500

0

17500

7500

750086 6 km/h

speed is positive

2

2

2

2

− =

− =

=

===

d n

Check:

( )

.

.

dsd C

s

ss

dsd C

3000 15000

300015000

86 6

300086 6

15000

0

When

>

2

23

3

2

2

3

=

=

=

=

−

d

d

n

n

Concave upwards So minimum when .s 86 6=

..

$ .

C 3000 86 686 67500

519 6155196 15

(iii)

cents

= +

==

d n

Ans_PART_4.indd 416 6/30/09 4:12:47 PM

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Yr12 Maths in Focus 2U HSC