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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
ch1.indd 2 6/29/09 7:32:01 PM
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.
ch1.indd 3 6/29/09 7:32:12 PM
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 .
ch1.indd 4 6/29/09 7:32:21 PM
5Chapter 1 Geometry 2
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/∆ ∆
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.
ch1.indd 5 6/29/09 7:32:23 PM
6 Maths In Focus Mathematics HSC Course
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: XYZ∆;ABC <∆ XYZ∆ is three times larger than ABC∆
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.
ch1.indd 6 6/29/09 7:32:24 PM
7Chapter 1 Geometry 2
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.
ch1.indd 7 6/29/09 7:32:26 PM
8 Maths In Focus Mathematics HSC Course
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
ch1.indd 8 6/29/09 7:32:27 PM
9Chapter 1 Geometry 2
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
ch1.indd 9 6/29/09 7:32:28 PM
10 Maths In Focus Mathematics HSC Course
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
ch1.indd 10 7/6/09 10:19:18 PM
11Chapter 1 Geometry 2
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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12 Maths In Focus Mathematics HSC Course
EXAMPLES
1. Prove A C+ += in parallelogram ABCD .
Solution
°
° °° ° °° ° °
°
Let
Then ( , cointerior angles, )( ) ( , cointerior angles, )
A x
B x A B AD BCC x B C AB DC
xx
A C
180180 180180 180
`
+
+ + +
+ + +
+ +
<
<
== −= − −= − +==
2. Triangle ABC below is isosceles with AB AC= . Prove that the base angles of ABC∆ are equal by showing that ABD∆ and ACD∆ are congruent.
Solution
° ( )
( )ADB ADC
AB AC90 given
given+ += =
=
AD is common \ ABD ACD/∆ ∆ (RHS)
So ABD ACD+ += (corresponding angles in congruent )s∆ \ base angles are equal
ch1.indd 12 6/29/09 7:32:33 PM
13Chapter 1 Geometry 2
3. Prove that opposite sides in a parallelogram are equal.
Solution
Let ABCD be any parallelogram and draw in diagonal AC .
( , )( , )
DAC ACB AD BCBAC ACD AB DC
alternate salternate s
+ + +
+ + +
<
<
==
AC is common. ABC ADC/∆ ∆ (AAS)
( )( )
AB DCAD BC
corresponding sides in congruent ssimilarly
` ∆==
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
ch1.indd 13 6/29/09 7:32:34 PM
14 Maths In Focus Mathematics HSC Course
5. BD bisects .ABC+ Prove that .ABD CBD/∆ ∆
6. (a) Show that .ABC AED/∆ ∆ Hence prove that (b) ACD∆ 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 OAC∆ and OBC∆ are congruent, and hence that OC bisects AB .
9. CE and BD are altitudes of ,ABC∆ and ABC∆ 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) MNO∆ is congruent to MPO∆ (b) PMQ NMQ+ += (c) PMQ∆ is congruent to NMQ∆ (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
ch1.indd 14 6/29/09 7:32:37 PM
15Chapter 1 Geometry 2
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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16 Maths In Focus Mathematics HSC Course
25. ABCD is a rhombus. Prove (a) ADB∆ and BCD∆ are
congruent. Hence show (b) .ABE CBE+ += Prove (c) ABE∆ and CBE∆ 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.
ch1.indd 16 6/29/09 7:32:43 PM
17Chapter 1 Geometry 2
( )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
ch1.indd 17 6/29/09 7:32:45 PM
18 Maths In Focus Mathematics HSC Course
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 rl r h
r rr r
r
l rr
2
24
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
ch1.indd 18 7/6/09 10:19:19 PM
19Chapter 1 Geometry 2
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 .
ch1.indd 19 6/29/09 7:32:47 PM
20 Maths In Focus Mathematics HSC Course
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.
ch1.indd 20 6/29/09 7:32:49 PM
21Chapter 1 Geometry 2
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.
ch1.indd 21 6/29/09 7:32:53 PM
22 Maths In Focus Mathematics HSC Course
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
4 516 2541
2 3 8 4
5 425 1641
1 2 1 8
1 91 8182
2 2
2 2
2 2
2 2
2 2
2 2
= − − + − −= − + −= +== − − + −= − += +== − − − + − −= + −= +=
ch1.indd 22 6/29/09 7:32:56 PM
23Chapter 1 Geometry 2
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.
ch1.indd 23 6/29/09 7:32:58 PM
24 Maths In Focus Mathematics HSC Course
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 ABC∆ with vertices ( , ), ( , )A B0 8 3 0 and ( , )C 3 0−
show that (a) ABC∆ 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) ?ABC∆
8. Show that OAB∆ and OCD∆ 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 OAB∆ and OCB∆ 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
ch1.indd 24 6/29/09 7:32:58 PM
25Chapter 1 Geometry 2
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) BEC∆ is congruent to BDC∆ .
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) ADE∆ 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 WXYand∆ ∆ are similar.
10. In quadrilateral ABCE , AD ED DC= = 90°ACBand+ = . Also, AC BADbisects+ .
ch1.indd 25 6/30/09 10:54:03 AM
26 Maths In Focus Mathematics HSC Course
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 .ABC∆
13. ABCD is a kite.
Prove (a) ABC ADCand∆ ∆ are congruent. Prove (b) ABE ADEand∆ ∆ 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 ABC∆ 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.
ch1.indd 26 6/29/09 7:33:04 PM
27Chapter 1 Geometry 2
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 AEF∆ is similar to .ADC∆
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/∆ ∆
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) DPS∆ is similar to .DAC∆ Show (b) .PS QR< Show that (c) PQRS is a parallelogram.
ch1.indd 27 6/29/09 7:33:08 PM
28 Maths In Focus Mathematics HSC Course
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.
ch1.indd 28 6/29/09 7:33:13 PM
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
32 Maths In Focus Mathematics HSC Course
EXAMPLES
1.
2.
3.
4.
ch2.indd 32 6/12/09 12:49:57 AM
33Chapter 2 Geometrical Applications of Calculus
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
34 Maths In Focus Mathematics HSC Course
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
35Chapter 2 Geometrical Applications of Calculus
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
36 Maths In Focus Mathematics HSC Course
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
37Chapter 2 Geometrical Applications of Calculus
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
38 Maths In Focus Mathematics HSC Course
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
For stationary points,
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
39Chapter 2 Geometrical Applications of Calculus
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
For stationary points,
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
40 Maths In Focus Mathematics HSC Course
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
41Chapter 2 Geometrical Applications of Calculus
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
$
$ $
= +
= +
= +
= +
]]]
]
ggg
g
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
42 Maths In Focus Mathematics HSC Course
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
43Chapter 2 Geometrical Applications of Calculus
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
44 Maths In Focus Mathematics HSC Course
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
45Chapter 2 Geometrical Applications of Calculus
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
46 Maths In Focus Mathematics HSC Course
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
47Chapter 2 Geometrical Applications of Calculus
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
48 Maths In Focus Mathematics HSC Course
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
49Chapter 2 Geometrical Applications of Calculus
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
50 Maths In Focus Mathematics HSC Course
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
For stationary points,
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
51Chapter 2 Geometrical Applications of Calculus
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
For stationary points,
i.e.
y
x
x
0
10 00
4
`
===
l
CONTINUED
ch2.indd 51 6/12/09 12:50:25 AM
52 Maths In Focus Mathematics HSC Course
hen ,W x y0 2 0 3
3
5= = −= −] g
So ,0 3−^ h is a stationary point.
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
53Chapter 2 Geometrical Applications of Calculus
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
54 Maths In Focus Mathematics HSC Course
( )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
So ,1 6−^ h is a stationary point.
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
55Chapter 2 Geometrical Applications of Calculus
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=
For stationary points,
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
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56 Maths In Focus Mathematics HSC Course
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 1
3
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
57Chapter 2 Geometrical Applications of Calculus
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
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58 Maths In Focus Mathematics HSC Course
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= − += −
So ,2 1−^ h is a stationary point.
( )
(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.= −
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59Chapter 2 Geometrical Applications of Calculus
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
For stationary points,
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
So ,1 0−^ h is a stationary point.
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
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60 Maths In Focus Mathematics HSC Course
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
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61Chapter 2 Geometrical Applications of Calculus
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
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62 Maths In Focus Mathematics HSC Course
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= −
For stationary points,
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
63Chapter 2 Geometrical Applications of Calculus
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
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64 Maths In Focus Mathematics HSC Course
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
$
= + + +
= + + +
= + +
= + +
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65Chapter 2 Geometrical Applications of Calculus
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
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66 Maths In Focus Mathematics HSC Course
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.
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67Chapter 2 Geometrical Applications of Calculus
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
For stationary points,
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
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68 Maths In Focus Mathematics HSC Course
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
69Chapter 2 Geometrical Applications of Calculus
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
70 Maths In Focus Mathematics HSC Course
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
71Chapter 2 Geometrical Applications of Calculus
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
72 Maths In Focus Mathematics HSC Course
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
73Chapter 2 Geometrical Applications of Calculus
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
74 Maths In Focus Mathematics HSC Course
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
75Chapter 2 Geometrical Applications of Calculus
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
76 Maths In Focus Mathematics HSC Course
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
77Chapter 2 Geometrical Applications of Calculus
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
78 Maths In Focus Mathematics HSC Course
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
79Chapter 2 Geometrical Applications of Calculus
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
80 Maths In Focus Mathematics HSC Course
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= .
Challenge Exercise 2
ch2.indd 80 6/12/09 12:50:56 AM
81Chapter 2 Geometrical Applications of Calculus
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
84 Maths In Focus Mathematics HSC Course
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 dxδx 0
Σ=
="
#
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
85Chapter 3 Integration
( )
.
´
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 1250 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
86 Maths In Focus Mathematics HSC Course
( ) [( ) ( )]. . .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
87Chapter 3 Integration
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
88 Maths In Focus Mathematics HSC Course
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 4y 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
89Chapter 3 Integration
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
4# using 3 subintervals.
9. ( )x x dx2
0
2−# using 4 trapezia.
10. x dx1
0# using 5 subintervals.
11. x
dx15
21# using 4 subintervals.
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 9f (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 4f (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
90 Maths In Focus Mathematics HSC Course
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 c4 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
<<
FF
#
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
91Chapter 3 Integration
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
92 Maths In Focus Mathematics HSC Course
( ) ( ) ( )
( ) ( )
( . )
.
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
<<
FF
#
#
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 FF 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
93Chapter 3 Integration
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 31 105
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
94 Maths In Focus Mathematics HSC Course
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 4f(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
95Chapter 3 Integration
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
1# using 2 subintervals.
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
0# using 5 function values.
11. dxx1
1
7
3# using 7 function values.
12. dxx 1
12
5
2 −# using 6 subintervals.
13. ( )f x dx0
4# where values of f ( x ) are
given in the table:
x 0 1 2 3 4f(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 4f(t) 3.7 1.2 9.8 4.1 2.7
15. ( )f x dx0
3# where values of f ( x ) are
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
96 Maths In Focus Mathematics HSC Course
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
97Chapter 3 Integration
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
98 Maths In Focus Mathematics HSC Course
Let area ABCD be A ( x ) Let area ABGE be ( )A x h+ Then area DCGE is ( ) ( )A x h A x+ −
`
. .Area area areai.e. ( ) ( ) ( ) ( )
( )( ) ( )
( )
( )( ) ( )
( )
( ) ( ) ( )( ) ( )
lim lim lim
DCFE DCGE DHGEf 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 xA 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 CF a C
F a CA x F x F a
x b b aA b F b F a
0
If >
= += +
− == −== −
Defi nite Integrals
By the fundamental theorem of calculus, ( ) ( ) ( )f x dx F b F a
a
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
99Chapter 3 Integration
EXAMPLES
Evaluate
1. ( )x dx2 14
3+#
Solution
+x x( )
( ) ( )
x dx2 1
4 4 3 320 128
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
100 Maths In Focus Mathematics HSC Course
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
101Chapter 3 Integration
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
102 Maths In Focus Mathematics HSC Course
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
103Chapter 3 Integration
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
Find each indefi nite integral (primitive function)
ch3.indd 103 6/11/09 11:39:10 PM
104 Maths In Focus Mathematics HSC Course
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
105Chapter 3 Integration
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
Find each indefi nite integral (primitive function)
ch3.indd 105 6/11/09 11:39:12 PM
106 Maths In Focus Mathematics HSC Course
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
107Chapter 3 Integration
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
108 Maths In Focus Mathematics HSC Course
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#
So the area is 3210 units 2 .
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
109Chapter 3 Integration
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
110 Maths In Focus Mathematics HSC Course
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#
So the area is 8 units 2 .
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
# #
So the area is 3242 units 2 .
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
111Chapter 3 Integration
8. Find the area enclosed between the curve ,y x4= the x -axis and the lines 1x = − and 1.x =
9. Find the area enclosed between the curve ,y x3= the x -axis and the lines 2x = − and 2.x =
10. Find the area enclosed between the curve ,y x3= the x -axis and the lines 3x = − and 2.x =
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 =
13. Find the area enclosed between the curve ,y x2= the x -axis and the lines 3x = − 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 =
16. Find the area enclosed between
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
112 Maths In Focus Mathematics HSC Course
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
y3
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
y23
32
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
113Chapter 3 Integration
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
114 Maths In Focus Mathematics HSC Course
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 =
10. Find the area enclosed between
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
115Chapter 3 Integration
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 xy x
x xx x
xxx
14 248 16
0 8 168 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 xy x
x xx x
x xx
12 22
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
116 Maths In Focus Mathematics HSC Course
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
117Chapter 3 Integration
Volumes about the x -axis
πV y dxa
b2= #
Proof
Take a disc of the solid with width δ x .
ππ δ
r hy 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
118 Maths In Focus Mathematics HSC Course
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 yy x
V y dx
x dx
x x
99
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
119Chapter 3 Integration
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
= += +
=
= +
= +
= + − +
= −
=
^
^^^ ^ed
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
120 Maths In Focus Mathematics HSC Course
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
yy x
V x dy
y dy
yy
11
1
2
29 3
21 1
8 units
a
2
2
2
1
2
1
3
3
`
= −+ =
=
= +
= +
= + − −
=
−
−
3
b
^
c c
h
m m=<
GF
##
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
121Chapter 3 Integration
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
122 Maths In Focus Mathematics HSC Course
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
123Chapter 3 Integration
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
124 Maths In Focus Mathematics HSC Course
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 6v 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
125Chapter 3 Integration
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#
Challenge Exercise 3
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
126 Maths In Focus Mathematics HSC Course
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
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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
134 Maths In Focus Mathematics HSC Course
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 . ´y 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
135Chapter 4 Exponential and Logarithmic Functions
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
136 Maths In Focus Mathematics HSC Course
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
137Chapter 4 Exponential and Logarithmic Functions
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
138 Maths In Focus Mathematics HSC Course
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
139Chapter 4 Exponential and Logarithmic Functions
Function of a function rule
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
140 Maths In Focus Mathematics HSC Course
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
141Chapter 4 Exponential and Logarithmic Functions
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
142 Maths In Focus Mathematics HSC Course
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
143Chapter 4 Exponential and Logarithmic Functions
( ) 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
144 Maths In Focus Mathematics HSC Course
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
145Chapter 4 Exponential and Logarithmic Functions
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
146 Maths In Focus Mathematics HSC Course
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
147Chapter 4 Exponential and Logarithmic Functions
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
148 Maths In Focus Mathematics HSC Course
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
149Chapter 4 Exponential and Logarithmic Functions
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
150 Maths In Focus Mathematics HSC Course
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
151Chapter 4 Exponential and Logarithmic Functions
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
152 Maths In Focus Mathematics HSC Course
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
153Chapter 4 Exponential and Logarithmic Functions
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
154 Maths In Focus Mathematics HSC Course
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
155Chapter 4 Exponential and Logarithmic Functions
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
156 Maths In Focus Mathematics HSC Course
Function of a function rule
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
157Chapter 4 Exponential and Logarithmic Functions
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
158 Maths In Focus Mathematics HSC Course
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
159Chapter 4 Exponential and Logarithmic Functions
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
Integration is the inverse of differentiation.
ch4.indd 159 6/12/09 12:52:26 AM
160 Maths In Focus Mathematics HSC Course
1. Find the indefi nite integral (primitive function) of
(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
161Chapter 4 Exponential and Logarithmic Functions
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
162 Maths In Focus Mathematics HSC Course
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 +
3. Find the indefi nite integral (primitive function) of
(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
163Chapter 4 Exponential and Logarithmic Functions
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
Challenge Exercise 4
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
164 Maths In Focus Mathematics HSC Course
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
168 Maths In Focus Mathematics HSC Course
Proof
Circumference of the circle with radius 1 unit is given by:
( )πππ
C r22 12
===
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 360180
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
°,( °)
°
ππ
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 63665
radians
radian
radians
`
=
=
=
== l
,, ,
Press 180 1.145SHIFT
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 .
. .
0.41cos 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
170 Maths In Focus Mathematics HSC Course
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 017180
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.
1803
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
171Chapter 5 Trigonometric Functions
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
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172 Maths In Focus Mathematics HSC Course
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
173Chapter 5 Trigonometric Functions
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 sincos costan tan
− =− = −− = −
Third quadrant: Angle :π θ+
( )( )( )
π θ θπ θ θπ θ θ
sin sincos costan tan
+ = −+ = −+ =
Fourth quadrant: Angle :π θ2 −
( )( )( )
π θ θπ θ θπ θ θ
sin sincos costan tan
222
− = −− =− = −
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174 Maths In Focus Mathematics HSC Course
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
175Chapter 5 Trigonometric Functions
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 5 061.224−=
=
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
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176 Maths In Focus Mathematics HSC Course
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
177Chapter 5 Trigonometric Functions
7. (a) Show that .π ππ6
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= −
Which quadrant is the angle(b)
π4
7 in?
Find the exact value of (c) .πtan4
7
9. (a) Show that .π ππ3
43
= +
Which quadrant is the angle(b)
π3
4 in?
Find the exact value of (c)
.πcos3
4
10. (a) Show that .π ππ3
53
2= −
Which quadrant is the angle(b)
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 tanx
43= − 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
178 Maths In Focus Mathematics HSC Course
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
circumferencearc length
whole revolutionangle
`
=
=
=
=
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
179Chapter 5 Trigonometric Functions
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
rr
2
2
.Now
. .
.
.
.
θθ
θ
θ
rl 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 9312 5612 56
=
=
=
===
l
l
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180 Maths In Focus Mathematics HSC Course
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.
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181Chapter 5 Trigonometric Functions
Area of sector
θA r21 2=
( θ is in radians)
Proof
area of circlearea of sector
whole revolutionangleA 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
is subtended at the centre of a circle of radius 5 m.
Solution
2
θ
π
π
A r21
21 5
4
825 m
2
2
=
=
=
^ ch m
CONTINUED
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182 Maths In Focus Mathematics HSC Course
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
183Chapter 5 Trigonometric Functions
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
184 Maths In Focus Mathematics HSC Course
EXAMPLES
1. Find the area of the minor segment formed if an angle of π4
is subtended at the centre of a circle of radius 5 m.
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
185Chapter 5 Trigonometric Functions
. .. .
.
.
. . .
.
θ
θ
θ θ
cos
cos
sin
sin
Cab
a b c
A r
2
2 12 7 12 712 7 12 7 10
0 690 807
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 .
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186 Maths In Focus Mathematics HSC Course
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) ,∆OAB 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
187Chapter 5 Trigonometric Functions
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
188 Maths In Focus Mathematics HSC Course
Class Investigation
Use the calculator to explore other small angles and their trigonometric ratios. What do you notice?
A
B C
O 1x
In , 1∆OAC 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
sintancos
sin tan
x xx xx
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
189Chapter 5 Trigonometric Functions
Proof
Area
Area sector
Area
∆
∆
´ ´ ´
´ ´
´ ´
´ ´
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 ∆OAB area sector< OAB area< ∆OAC
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
190 Maths In Focus Mathematics HSC Course
EXAMPLES
1. Evaluate .lim sinx
x7x 0"
Solution
( )
7
( )
lim sin limsin
lim sinx
xx
x
xx
77
7 7
77
7 17
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 005660 00566180
1 180
0 00566 180 0 00566
0 32450 64939
6
`
`
`
=
=
===
=
=
=== l
The radius of the moon is 72 500 km.
ch5.indd 190 6/12/09 3:54:16 AM
191Chapter 5 Trigonometric Functions
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
192 Maths In Focus Mathematics HSC Course
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 tancot xx 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
193Chapter 5 Trigonometric Functions
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
194 Maths In Focus Mathematics HSC Course
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
195Chapter 5 Trigonometric Functions
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
196 Maths In Focus Mathematics HSC Course
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
197Chapter 5 Trigonometric Functions
The graph (c) siny x5 4= has amplitude 5 and period .π2
2. Sketch ( ) πsinf x x2
= +b l for ≤ ≤ π.x0 2
Solution
12
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
198 Maths In Focus Mathematics HSC Course
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
199Chapter 5 Trigonometric Functions
Method 1: Use table (b)
x 0π4
π2
3π4 π 5π
43π2
7π4 2π
cos 2x 1 0 −1 0 1 0 −1 0 12 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 23
== +=
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 x
2=
(g) ( ) tanf x x2 3= (h) cosy x3
2=
(i) siny x22
=
(j) ( ) cosf x x44
=
ch5.indd 199 6/12/09 3:59:39 AM
200 Maths In Focus Mathematics HSC Course
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
201Chapter 5 Trigonometric Functions
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
202 Maths In Focus Mathematics HSC Course
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
Janua
ry
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
203Chapter 5 Trigonometric Functions
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
Janua
ryM
arch
May Ju
ly
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
204 Maths In Focus Mathematics HSC Course
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 6.20am
11.55am
6.15pm
11.48pm
6.20am
11.55am
6.15pm
11.48pm
6.20am
11.55am
6.15pm
11.48pm
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
205Chapter 5 Trigonometric Functions
( )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
206 Maths In Focus Mathematics HSC Course
[ ( )] ( ) ( )sin cosdxd f x x f x= fl
FUNCTION OF A FUNCTION RULE
Proof
Let ( )Then
( ) and
( )( ) ( )
sin
cos
coscos
u f xy u
dxdu f x
dudy
u
dxdy
dudy
dxdu
u f xf 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
207Chapter 5 Trigonometric Functions
FUNCTION OF A FUNCTION RULE
[ ( )] ( ) ( )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
208 Maths In Focus Mathematics HSC Course
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 sincos sin
dxdy
u v v u
x x x xx x x x
22
2
2
= +
= += +
l l
hen ,
( )
π
π π π π
π π
π π
π π
π π
´ ´
cos sin
x
dxdy
4
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
FUNCTION OF A FUNCTION RULE
This result can be proven the same way as for sin f(x).
ch5.indd 208 6/12/09 4:02:01 AM
209Chapter 5 Trigonometric Functions
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
210 Maths In Focus Mathematics HSC Course
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= − +#
Integration is the inverse of differentiation.
ch5.indd 210 6/12/09 4:02:20 AM
211Chapter 5 Trigonometric Functions
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 12
=
= −
= − −= −
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
212 Maths In Focus Mathematics HSC Course
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
1. Find the indefi nite integral (primitive function) of
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
213Chapter 5 Trigonometric Functions
(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
=
5. Find the area enclosed between
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
214 Maths In Focus Mathematics HSC Course
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
6. Find the indefi nite integral (primitive function) of
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) θθlim tan2
θ 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
215Chapter 5 Trigonometric Functions
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 ,xdxdy
sin= 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
Challenge Exercise 5
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
216 Maths In Focus Mathematics HSC Course
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
220 Maths In Focus Mathematics HSC Course
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
221Chapter 6 Applications of Calculus to the Physical World
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
222 Maths In Focus Mathematics HSC Course
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
223Chapter 6 Applications of Calculus to the Physical World
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
224 Maths In Focus Mathematics HSC Course
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
225Chapter 6 Applications of Calculus to the Physical World
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
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226 Maths In Focus Mathematics HSC Course
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
227Chapter 6 Applications of Calculus to the Physical World
(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
228 Maths In Focus Mathematics HSC Course
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
229Chapter 6 Applications of Calculus to the Physical World
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
230 Maths In Focus Mathematics HSC Course
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
231Chapter 6 Applications of Calculus to the Physical World
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
232 Maths In Focus Mathematics HSC Course
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
233Chapter 6 Applications of Calculus to the Physical World
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
234 Maths In Focus Mathematics HSC Course
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
235Chapter 6 Applications of Calculus to the Physical World
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
236 Maths In Focus Mathematics HSC Course
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
237Chapter 6 Applications of Calculus to the Physical World
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
238 Maths In Focus Mathematics HSC Course
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
239Chapter 6 Applications of Calculus to the Physical World
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
240 Maths In Focus Mathematics HSC Course
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
241Chapter 6 Applications of Calculus to the Physical World
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
242 Maths In Focus Mathematics HSC Course
(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
243Chapter 6 Applications of Calculus to the Physical World
(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
244 Maths In Focus Mathematics HSC Course
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
245Chapter 6 Applications of Calculus to the Physical World
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?
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246 Maths In Focus Mathematics HSC Course
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
247Chapter 6 Applications of Calculus to the Physical World
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.
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248 Maths In Focus Mathematics HSC Course
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
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249Chapter 6 Applications of Calculus to the Physical World
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.
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250 Maths In Focus Mathematics HSC Course
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
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251Chapter 6 Applications of Calculus to the Physical World
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 .
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252 Maths In Focus Mathematics HSC Course
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
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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
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260 Maths In Focus Mathematics HSC Course
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
= += + += +
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262 Maths In Focus Mathematics HSC Course
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
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264 Maths In Focus Mathematics HSC Course
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
= + + + +=
( ) ( ) ( ) ( ) ( ) ( )´ ´ ´ ´ ´n2 5 2 3 5 2 4 5 2 5 5 2 6 5 2 7 53
7
+ = + + + + + + + + +/
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266 Maths In Focus Mathematics HSC Course
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 .
´n
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− = − .
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268 Maths In Focus Mathematics HSC Course
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
270 Maths In Focus Mathematics HSC Course
( )
( )
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+ + +
13. Is 405 a term of the series ?…0 3 6+ + +
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
272 Maths In Focus Mathematics HSC Course
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
274 Maths In Focus Mathematics HSC Course
( ) ( )
.
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+ + +
2. Find the sum of 30 terms of the series.
(a) …1 7 13+ + + (b) …15 24 33+ + + (c) …95 89 83+ + +
3. Find the sum of 25 terms of the series.
(a) …2 5 12− + + + (b) …5 4 13− − −
4. Find the sum of 50 terms of the series.
(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?
9. How many terms of the series …80 73 66+ + + give a sum of 495?
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.
14. How many terms of the series …20 18 16+ + + give a sum of 104?
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
276 Maths In Focus Mathematics HSC Course
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
278 Maths In Focus Mathematics HSC Course
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
280 Maths In Focus Mathematics HSC Course
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
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282 Maths In Focus Mathematics HSC Course
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?
11. Is 1200 a term of the series ?…2 16 128+ + +
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
284 Maths In Focus Mathematics HSC Course
(
´
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+ + +
2. Find the sum of 8 terms of the series .
(a) …1 7 49− + − + (b) …8 24 72+ + +
3. Find the sum of 15 terms of the series .
(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
286 Maths In Focus Mathematics HSC Course
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
288 Maths In Focus Mathematics HSC Course
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
290 Maths In Focus Mathematics HSC Course
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
292 Maths In Focus Mathematics HSC Course
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
294 Maths In Focus Mathematics HSC Course
(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
296 Maths In Focus Mathematics HSC Course
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
298 Maths In Focus Mathematics HSC Course
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
300 Maths In Focus Mathematics HSC Course
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
( )
( . )
( . ).
´n
A P r
6 4
24
1
2000 1 03
2000 1 034065 59
0
n
24
24
==
= += +==
So the amount is $4065.59.
(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
( )
( . )
( . ).
´n
A P r
6 12
72
1
2000 1 0 01
2000 1 014094 20
monthsn
72
==
= += +==
So the amount is $4049.20.
Always round money off to 2 decimal places (to the
nearest cent).
CONTINUED
ch7.indd 301 6/29/09 10:02:28 PM
302 Maths In Focus Mathematics HSC Course
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
304 Maths In Focus Mathematics HSC Course
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
´n 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
306 Maths In Focus Mathematics HSC Course
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
308 Maths In Focus Mathematics HSC Course
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
310 Maths In Focus Mathematics HSC Course
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
312 Maths In Focus Mathematics HSC Course
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
314 Maths In Focus Mathematics HSC Course
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
316 Maths In Focus Mathematics HSC Course
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.
Challenge Exercise 7
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
358 Maths In Focus Mathematics HSC Course
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
corresponding sides in congruent s
`
`
`
+ +
/Δ Δ
Δ
= ==
=
9. ° ( )
( )
( )
CDB BECACB ABC
CBCDB BEC
CE BD
90
is commonby AAS,
altitudes given
base angles of isosceles
corresponding sides in congruent s
`
`
+ ++ +
/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
base angles of isosceles
`
+ +
+ ++ ++ +
<
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
base angles of isosceles
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,
(opposite sides of gram)
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
360 Maths In Focus Mathematics HSC Course
24.
`
`
(corresponding sides in congruent s)Δ
°( )—
AD BCD C
DE ECADE BCE
AE BE
E
90
by SAS,
(opposite sides of gram)
(given)
is the midpoint given
+ +
/
<
Δ Δ
== ==
=
25. (a)
`
(
AD BCAB DC
DBADB BCD
is commonby SSS,
(opposite sides of gram)
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) 141
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
362 Maths In Focus Mathematics HSC Course
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
=
= −
= −
= −
e
e
o
o
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
364 Maths In Focus Mathematics HSC Course
(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
366 Maths In Focus Mathematics HSC Course
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
368 Maths In Focus Mathematics HSC Course
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
370 Maths In Focus Mathematics HSC Course
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
372 Maths In Focus Mathematics HSC Course
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
( )ab
b
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
374 Maths In Focus Mathematics HSC Course
24. 179
25. y
2x
Challenge exercise 2
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
376 Maths In Focus Mathematics HSC Course
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 e
d
g
n o
n
<
>
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
+
Challenge exercise 3
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
378 Maths In Focus Mathematics HSC Course
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
380 Maths In Focus Mathematics HSC Course
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 e1c 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
382 Maths In Focus Mathematics HSC Course
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
Challenge exercise 4
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
384 Maths In Focus Mathematics HSC Course
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
386 Maths In Focus Mathematics HSC Course
(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
388 Maths In Focus Mathematics HSC Course
(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
390 Maths In Focus Mathematics HSC Course
(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
392 Maths In Focus Mathematics HSC Course
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 π
x6 6
5=
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=
Challenge exercise 5
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
24d 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
394 Maths In Focus Mathematics HSC Course
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
396 Maths In Focus Mathematics HSC Course
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
Challenge exercise 6
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
398 Maths In Focus Mathematics HSC Course
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−
Practice assessment task set 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 ee
x
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
400 Maths In Focus Mathematics HSC Course
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 20
d 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
^
^
h
h
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
402 Maths In Focus Mathematics HSC Course
(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=
Challenge exercise 7
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
404 Maths In Focus Mathematics HSC Course
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
Challenge exercise 8
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
Practice assessment task set 3
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
406 Maths In Focus Mathematics HSC Course
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
408 Maths In Focus Mathematics HSC Course
:
.
.
´ ´
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 g
g 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
410 Maths In Focus Mathematics HSC Course
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=+ +
= − + + −
= −
f
d
c
p
n
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
412 Maths In Focus Mathematics HSC Course
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
414 Maths In Focus Mathematics HSC Course
(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
c
c
m
m
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
416 Maths In Focus Mathematics HSC Course
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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