Download Maths Primer - School of Earth Sciences

Year 1 Maths Primer
School of Earth Sciences
What is this Booklet?

Many people think of mathematics and want to walk away. However, all natural
sciences are quantitative and you will have to use numerical techniques to
understand what scientific data is telling you. If you are a maths-phobe, don’t
panic! The aim of this booklet is to revise basic concepts and put them to use.

You can’t avoid maths if you mean to study for a science degree in a leading
university. Remember that employers value a good Bristol science degree very
highly, and that is because Bristol science graduates are analytical, numerate and
imaginative.

This booklet has been prepared in response to student feedback from previous first
year classes. Many students felt they arrived in Bristol unprepared for the numerate
aspects of the first year course, as some time had elapsed since GCSE or ‘A’ level
mathematics courses.

This booklet outlines the quantitative methods you will use during the first year of
your geoscience degree. These methods will be used to explain geological
processes in lecture courses and you will use them to analyse scientific data in
practical classes. Understanding your course and doing assessed practical work will
be much easier if you are familiar with these methods and can apply them.

Spending some time going through this booklet now will mean that you will feel
more confident during your first year, you will get better marks, and you will have
more time for socialising and the other aspects of student life.

What is in this Booklet?

This booklet contains a series of short sections each describing some maths
methods or ideas. For each method or topic there is an explanation in non-technical
terms. The methods are stated simply with the key points highlighted, and followed
with worked examples. Common pitfalls and difficulties are pointed out. Each
section is independent and cross-referenced.

If you are good at maths and want to find out more about the geoscience behind the
examples, there are links to suggested further reading.

This is not meant to be a textbook. There are some very clear, well-written texts for
GCSE level mathematics, including revision textbooks (Letts and CSG guides) and
textbooks written for people older than 16 who have never studied mathematics to
GCSE level (Foundation Maths by Anthony Croft and Robert Harrison, published
by Prentice Hall is particularly good).

During your time at Bristol you will need to have a mathematics reference book on
your shelves. A good general reference book for the later years of your course is
Basic Mathematics for the Physical Sciences written by Robert Lambourne and
Michael Tinker and published in 2000 by John Wiley and Sons (ISBN
0471852074). Schaum’s Mathematical Handbook of Formulas and Tables edited
by Murray Spiegel and John Liu and published by Schaum in 1998 (ISBN
0070382034) is also very useful. In the autumn of 2004 Schaum published
Beginning Finite Math written by Seymour Lipschutz (ISBN 0071388974) which
includes similar material to that in this primer and also basic statistics.
1
What should I do now?

In the month before you start your degree, work through each section, making sure
you understand the methods, and testing this by trying the examples. After
committing the methods to memory using the guide below, attempt the examples
and make sure you know how to apply these methods in a geoscience context.
Learning Guide

There are three steps to committing these methods to your long-term memory,
and making sure they ‘sink in’.
Step 1 - Understand


Make sure you understand the basic concepts and the reasoning behind the
method.
Mark up the text if necessary – underline, highlight and make extra notes.
Go to step 2
Step 2 - Summarise


Make your own summary, perhaps working through an example again.
What is the main idea? What are the main points? How does the logic develop?
This is now in
SHORT TERM
MEMORY.
You will forget
80% of it if you
do not go onto
step 3.
2
Step 3 - Memorise



Memorise the methods in 25 minute blocks with 5 minute breaks.
After each break, test yourself by trying one of the examples without looking at
the text.
Write down the main points of the method without looking at the text.
The material
is on its way to
LONG TERM
MEMORY
List of Sections
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
Page
Dimensions and Units
Powers
Making and Using Equations
Solving Equations
Proportionality and Inequality
Straight lines, Curves and Gradients
The Exponential Function and Logarithms
Triangles
Trigonometric Ratios
Calculating Systematic Errors
Answers to Exercises and Geoscience Examples
4
13
19
26
34
37
41
49
53
61
69
Credits

This booklet was written by Jeremy Phillips with assistance from Mary Benton,
Mike Benton and Fiona Whitaker.

Geoscience questions were provided by Jeremy Phillips, Jon Blundy, Derek
Vance, David Speedyman, Pam Gill and Mike Benton.
3
1. Dimensions and Units
Learning
Summary:
After studying this section, you will be able to:
 work out the dimensions and units of quantities
 use dimensions to check that equations you are using or have rearranged are correct.
 put numbers into standard index form and evaluate expressions
using numbers in standard index form
 use consistent sets of units
Dimensions
When we use equations, we are usually dealing with quantities such as depth of
sediment, amount of molecules, mass of magma, rather than pure numbers.

These quantities are made up from a combination of independent dimensions,
usually Length, Mass, Time and Temperature. The dimensions of a particular
quantity can be identified from its definition.
Eg
Example 1: What are the dimensions of velocity, and what are its S.I. units?
The definition of a velocity is that it is a length per
L
unit time, so the dimensions of velocity are
,
T
which is usually written as L T-1, using the index laws
(see the section on Powers in this booklet).
The units of velocity follow directly from the
dimensions.
In the S.I. system, the unit of length is the metre (m),
and the unit of time is the second (s), so the S.I. units
for velocity are m s-1.
Key Point:
The dimensions of a particular quantity can be identified
from its definition, based on the independent dimensions
length, mass, time, temperature.
Key Point:
Units are written after the quantity to which they refer,
with a single space in between. A single space is left
between each unit, to avoid confusion. For example, a
velocity is written as 10 m s-1, not 10m s-1, nor 10 ms-1.
4
Eg
Example 2: What are the dimensions of volume flowrate?
A volume flowrate, Q, is a volume, V, per unit time.
A volume is defined as a length x length x length, so the
dimensions of volume are L3, and the dimensions of
volume flowrate are L3 T-1 .
Square brackets are often used to denote dimensions,
so we could write [V] = L3, and [Q] = L3 T-1 .

Some quantities are dimensionless by virtue of their definition. For example, the
strain of a stretching rubber band is the extension per unit length, so [strain] =
L/L = 1, so dimensionless.
Key Point:

If an equation correctly expresses a relationship between
variables in a physical process, each of its terms will have
the same dimensions.
Dimensions can be useful to assist a poor memory, for example in an exam.
Eg
Example 3: Which of these expressions is correct for the displacement
due to gravitational acceleration
z
1 2
gt
2
or
z
1 2
g t ?
2
(z is displacement, g is the gravitational acceleration and t is time).
Each term will have the same dimensions if the
expression is correct, so the dimensions on each side
of the equals sign will be the same.
For the first expression, does [z] = L = [gt2]? The
dimensions of gt2 are L T-2  T2 = L.
For the second expression, does [z] = L = [g2t]? The
dimensions of g2t are (L T-2)2  T = L2T-3.
The first expression is correct as the dimensions on each
side of the equals sign are the same.
Link:
Dimensions and units are combined using the algebra and index
laws. See section 2, p.13.
5
Key Point:
Pure constants have no dimensions.
They arise from mathematical operations.
Standard Index Form for Numbers

In Geoscience, we often encounter very large and very small numbers, such as the
diameter of the Earth, or the length of individual crystals in a magma.

We can use the index laws (see the section on Powers in this booklet) to help us do
calculations using large and small numbers. In your courses on Environmental
Systems and Cycles and The Dynamic Interior, you will meet standard index
notation. This is a conventional way to write large and small numbers.
For example, 14000000 = 1.4 x 10 x 10 x 10 x 10 x 10 x 10 x 10
= 1.4 x 107
Key Point:

This is called standard index form. The number is
expressed as a number between 1 and 10 multiplied by
10 to the appropriate index.
To convert large numbers to standard index form, count the zeros in the number.
This gives the value of the index of 10. Then make any adjustment required so that
the number in front lies between 1 and 10.
Eg
Example 4: Write 14000000 in standard index form.
14000000 is 14 followed by six zeros, so we can write
14000000 = 14 x 106.
This is not standard index form, because 14 does not lie
between 1 and 10. We can write 14 as 1.4 x 10, so in
standard index form
14000000 is 1.4 x 10 x 106 = 1.4 x 107.

To convert small numbers to standard index form, count the zeros in the number,
including the zero before the decimal point. This gives the value of the index of 10,
which will be negative if the number is less than one. The digits in front of the
index should be written to lie between 1 and 10.
6
Eg
Example 5: Write 0.00024 in standard index form.
0.00024 is four zeros followed by 24.
The value of the index is –4, and the 24 must be
written as 2.4 to lie between 1 and 10, so we write
0.00024 = 2.4 x 10-4 in standard index form.

We can use the index laws to help us do calculations without using a calculator.
Eg
Example 6: What is 100000/2000?
Write the numbers in standard index form.
100000 1  10 5

.
2000
2  10 3
In this form, we can treat the parts on either side of
the ‘x’ sign separately. On the left side of the ‘x’
1
sign, we have
and on the right side we have
2
10 5
 10 53  10 2 using the index law for division.
10 3
So
100000 1  10 5 1
1

  10 2   100  50
3
2000
2
2
2  10
As an alternative to standard index form, in the S.I. system of units different names are
introduced for each thousandfold increase or decrease in size. For example, the basic
unit of length is the metre. The next unit up, one thousand times larger, is called the
kilometre, and the next unit down, one thousand times smaller, is called the millimetre.
The prefixes used in S.I. units are listed in the table below.
multiple
10-9
10-6
10-3
1
103
106
109
1012
prefix
nano
micro
milli
no prefix
kilo
mega
giga
tera
symbol
n
µ
m
k
M
G
T
7
example
nanometre (nm)
micrometre (mm)
millimetre (mm)
metre (m)
kilometre (km)
megametre (Mm)
gigametre (Gm)
terametre (Tm)
Eg
Example 7: Write 1600 g in standard index form and using S.I. units.
1600 g is 16 followed by two zeros, so we can write
1600 g = 16 x 102 g.
To convert this to standard index form, we can write
16 = 1.6 x 10, so 1600 g = 1.6 x 10 x 102 g = 1.6 x 103 g
in standard index form.
In S.I. units, a thousandfold increase is denoted by the
prefix ‘kilo’, so we can write
1600 g = 1.6 kg.
Note that a lower case ‘k’ is used for this prefix – an
upper case ‘K’ is used for the Kelvin unit of temperature.
Consistent Units
Each term in an equation that correctly expresses a relationship between variables in a
physical process will have the same dimensions. For this reason, you must use
consistent sets of units when performing calculations using equations. One such system
is S.I. units, where the basic dimensions length, mass, time and temperature have units
of metres, kilogrammes, seconds and Kelvin respectively. Other quantities have units
that are combinations of these according to their definition.
Eg
Example 8: Momentum is defined as the product of the mass and
velocity of a body. What is the S.I. unit of momentum?
The S.I. unit of mass is the kilogramme. Velocity is defined
as length per unit time, so it has dimensions of L T-1, and the
S.I. unit of velocity is m s-1, following from its dimensions.
Momentum is the product of mass and velocity, so it has
dimension M x L T-1 = M L T-1. Following the dimensions,
the S.I. unit of momentum is kg m s-1.
Key Point:

Dimensions and units are combined using the algebra
and index laws.
Another system of units that can be more useful when measuring smaller objects is
the CGS system, where the basic dimensions length, mass, time and temperature
have units of centimetres, grammes, seconds and Kelvin respectively.
8
Key Point:
Eg
You need to use a consistent system of units when making
calculations. Always convert the units of quantities to a
consistent set before attempting any calculations.
Example 9: What is the volume of one kilogramme of basalt? The
density of basalt is about 2.6 g cm-3.
Density is defined as mass m per volume V, so we can
m
write the equation for density as   .
V
m
Re-arranging this equation, we find volume V  .

If we tried to substitute the values we are given into this
1 kg
equation, we would find V 
, which does not
2.6 g cm -3
make any sense as we have two different units for mass, kg
and g.
We can choose to do the calculation in either S.I. units or
CGS units by converting the units of all the quantities to
either consistent system first.
If we choose CGS, we need to convert the mass in
kilogrammes to mass in grammes. 1 kg = 1000 g, so the
1000
volume of 1000 g of basalt is equal to
 385 cm 3 .
2.6
Key Point:
You must use consistent sets of units when performing
calculations using equations.
Summary in Words

Physical quantities are made up from a combination of independent dimensions,
usually Length, Mass, Time and Temperature. The dimensions of a particular
quantity can be identified from its definition. Some quantities are dimensionless by
virtue of their definition. Pure constants have no dimensions. They arise from
mathematical operations.
9

If an equation correctly expresses a relationship between variables in a physical
process, each of its terms will have the same dimensions. For this reason, you must
use consistent sets of units when performing calculations using equations. One
such system is S.I. units, where the basic dimensions length, mass, time and
temperature have units of metres, kilogrammes, seconds and Kelvin respectively.
Other quantities have units that are combinations of these according to their
definition.

Units are written after the quantity to which they refer, with a single space in
between. A single space is left between each unit, to avoid confusion. For
example, a velocity is written as 10 m s-1, not 10m s-1, or 10 ms-1.

Standard index notation is a conventional way to write large and small numbers.
The number is expressed as a number between 1 and 10 multiplied by 10 to the
appropriate index.

You must use consistent sets of units when performing calculations using equations.
What should I do now?

Make sure you understand what dimensions are, how to work out the dimensions of
physical quantities, and how to use dimensions to check that equations are correct.
Make sure you know how to write units correctly, and you understand why you
need to always use consistent sets of units. Make sure you know how to write large
and small numbers in standard index form, and the shortcuts for doing so. Look at
the section on Index Laws in this booklet if you are unclear about how to
manipulate expressions using indices.

Try the examples below that test your application of the methods.

Try the Geoscience examples so you can see how to apply these methods in your
subject.
Exercises - Methods
1. Write down the four basic dimensions
2. The definition of density is mass per unit volume. What are the dimensions of
density?
3. The force associated with the motion of a mass can be calculated from
Newton’s Second Law,
Force = Mass x Acceleration.
What are the dimensions of Force?
4. The quantity c can be calculated from the formula E =mc2. If the dimensions of
E are ML2T-2, and m denotes a mass, what are the dimensions of c?
5. If e denotes energy and V denotes volume, write down [e] and [V]
6. The distance fallen by an object under the action of gravity, s, can be written in
terms of the elapsed time t, the velocity of the object u, and the acceleration due
to gravity g. Which of these formulae correctly expresses the relationship
between these quantities (a) s = gt + ½ ut2, or (b) s = ut + ½ gt2 ? Write down
all working and the rule that you used to work out the answer.
10
7. Write these numbers in standard index form: 101, 23570, 99x108, 187x10-3, 
8. Write these quantities in S.I. units: 16000g, 247 mm, 25000 MW, 100 mm2, 205
cm s-1, 0.002 km3
9. Without using a calculator, simplify the following expressions:
1
10 4 3 5 4 1
2
3
4
4
3
,
8

8
,
5
,
6
,
237

237
,
18
.
2

18
.
2
,
27
10 5
10. What is the volume of one kilogramme of basalt. The density of basalt is about
2.6 g cm-3. Express your answer in S.I. units.
11. Calculate the pressure due to a force of 105 kg m s-2 acting on an area of 3.7
cm2, and express your answer in standard index form.
Geoscience Examples
1.
Vapour transport
0.036
Evaporation
0.071
Precipitation
0.107
Ice (43.4)
Runoff 0.036
Lakes and rivers
(0.13)
Groundwater
(15.3)
Atmosphere (0.0155)
Precipitation
0.398
Evaporation
0.434
Ocean
(1400)
The above diagram shows a "box model" of the water cycle on Earth. It is
specifically designed to ignore all the complexities of, for example, every rain
shower or river that empties water into the oceans and to give a simple overall
quantitative picture of the storage of water in various reservoirs and its
movement around the planet by natural processes such as evaporation and
precipitation. There are no units on the diagram but the volumes of the various
reservoirs (in brackets) are in millions of km3 - 106km3 - and the fluxes in
millions of km3 per year - 106km3yr-1.
The oceans are the biggest reservoir of water on Earth, at a volume of
1400x106km3. What is the volume of the oceans expressed in units of m3?
The density of water is about 1000 kg m-3. What is the total mass of the
oceans? (The definition of density is given in example 9, p.9)
What is the second biggest reservoir of water on Earth? What is the mass of
water contained in it? Use the method in Example 6 to work out how many
times bigger the oceans are than this second-biggest reservoir.
11
2. The motion of fluids control many environmental and geological processes.
Fluid motion can take place in two forms: as ordered, uniform motion called
laminar flow, or as highly disordered motion called turbulent flow. The form of
motion depends on the properties of the fluids and the flow, and can be
measured by a parameter called the Reynolds number, defined as
Re 
Lu

where L is a length controlling the flow motion such as the width of a channel, u
is the velocity of the flow, is the density of the fluid and  is its viscosity (the
symbol is the Greek letter ‘rho’ and the symbol  is the Greek letter ‘mu’).
The viscosity (or stickiness) of a simple fluid can be measured as the ratio of the
shear stress to the velocity gradient in the fluid. If the dimensions of shear
stress and velocity gradient are M L-1 T-2 and T-1 respectively, what are
dimensions of viscosity and the Reynolds number?
(more information on how to write down equations from descriptions in words
can be found in section 3 of this booklet)
3. The Earth is 4.55 billion years old (4.55 Ga) and the dinosaurs died out 65
million years ago (65 Ma), yet our species is only 60000 years old (60 ka). If
the entire history of the Earth were to be condensed into one calendar year, on
what date did the dinosaurs die out, when did the first homo sapiens roam the
Earth, and when was Jesus born?
12
2. Powers
Learning
Summary:
After studying this section, you will be able to:
 find square roots and cube roots
 use index laws to evaluate expressions that contain powers
GCSE
Background: Use the terms square, positive square root, cube; use index notation for
squares, cubes and powers of 10
Reminders:

A number multiplied by itself is called a square and is written in the form
numberindex
The square of 3 is written 32 and has the value 3 x 3 = 9. In words, the value of
3 squared is 9.
In this example, the value of the index is 2. If the value of the index was 3, we
would be finding the cube of a number.
The cube of 4 is written as 43 and has the value 4 x 4 x 4 = 64. In words, the
value of 4 cubed is 64. When we cube a number, we are raising it to the power
of 3.

The square root of a number must be squared (that is multiplied by itself or raised
to the power 2) to give the number.

Any positive number has two square roots, one positive and one negative.
The square roots of 49 are 7 and –7. We can write this as
49  7, 49  7
Let’s check this. If the square roots of 49 are 7 and –7, then if we square these
numbers we should get 49.
72 = 7 x 7 = 49. –72 = -7 x –7 = 49. Remember that multiplying 2 negative
numbers together gives a positive number.

The cube root of a number must be cubed (that is raised to the power 3) to give the
number. So the cube root of 64 is 4. We can write this 3 64  4 .

This notation using an index is commonly used in science when using large
numbers. For example, the radius of the Earth is about 6 million metres or
6 x 10 x10 x10 x 10 x10 x10 m or 6 x 106 m.
13
Link
This notation is called standard index form and is described in section 1, p.6
How we deal with powers – the Index Laws
Using an index is a shorthand way of showing multiplication and division. In this
section we will illustrate this using examples and find some general rules for using
indices (this is the plural of index), also called powers, or exponents.
Eg
Example 1: Using powers of 2, find 16 x 32.
16 = 2 x 2 x 2 x 2 = 24
Do not confuse
24 (=16) with
2 x 4 (=8)!
(because 2 x 2 is 2 to the power 2,
2 x 2 x 2 is 2 to the power 3,
2 x 2 x 2 x 2 is 2 to the power 4.)
32 = 2 x 2 x 2 x 2 x 2 = 25
so, 16 x 32 = (2 x 2 x 2 x 2) x (2 x 2 x 2 x 2 x 2) = 29
Key Point:
This is an example of the index law for multiplication
ap x aq = ap+q

We say that a number is written in index form when it is written as a base, a, to
some power, p.
Key Point:

All numbers can be written in index form, because any
number raised to the power one is itself, so we can
write a1 = a.
If you have two numbers in index form which have the same base, then to multiply
them together you add their powers.
104 x 105 = 104+5 = 109

This does not work if the bases of each number are different.
14
Eg
Example 2: Using powers of 3, find 81/9.
81 = 3 x 3 x 3 x 3 = 34
9 = 3 x 3 = 32
Key Point:
so, 81/9 =
3 3 3 3
=3x3
3 3
This time,
34
= 34-2 = 32
32
This is an example of the index law for division
ap/aq = ap-q
Eg
Example 3: Find (23)4.
This is
(23) x (23) x (23) x (23) =
(2 x 2 x 2) x (2 x 2 x 2) x (2 x 2 x 2) x (2 x 2 x 2) =
212 = 23 x 4
Key Point:
This shows another index law
(ap)q = ap x q
Eg
Example 4: Use the index law for division to find 23/23.
Using the index law for division
23/23 = 23 – 3 = 20.
But any number divided by itself equals one,
so 20 = 1.
Key Point:
This shows another index law
a0 = 1
15
Eg
Example 5: Use the index law for division to find 1/32.
Using the index law above, any number raised to
the power zero equals 1.
So 1/32 can be written as 30/32.
Using the index law for division
30/32 = 30-2 = 3-2.
Key Point:
This demonstrates another index law
1/ap = a-p
Eg
Example 6: If
2 = 2y, find the value of y.
Remembering that the square root of a number can be
squared to give the number itself, and that we must do the
same thing to both sides of an equation, if we square both
sides of the equation we find
 2  = 2 = (2 ) .
2
y 2
Using the index law for multiplication on the right hand
side of the equals sign, we find
2 = 22y.
Because any number raised to the power one is itself, we
can write
If you are not
sure how to do
this, see the section
on Solving
Equations, p.26
21 = 22y.
We can solve this equation by equating the powers of 2,
1 = 2y, so y = 1/2.
Key Point:
This gives us a general law
n
a = a1/n
16
Summary in Words

When you have two numbers that can be written as the same base raised to different
powers, to multiply the numbers, add the powers.

When you have two numbers that can be written as the same base raised to different
powers, to divide the numbers, subtract the powers.

When you have a number that is written as a base raised to a power and you want to
raise it to another power, multiply the powers together.

Any number raised to the power zero has the value one.

The reciprocal of (that is, one over) a number that is written as a base raised to a
power can be re-written as the base raised to the same negative power.
What should I do now?

Learn the index laws and make sure you know how to write numbers in standard
index form. Make sure you understand the terminology of bases and indices, and
the operations using powers that are summarised by the index laws.

Try the examples below that test your application of the methods.

Try the Geoscience examples so you can see how to apply these methods in your
subject.
Exercises - Methods
1. Without using a calculator, work out the following, writing your answer in its
simplest form:
1
1
43 x 43 (52)2 (53)1/5
 2
2
3
3
2. Simplify:
1 1

b5 b3
 3.7
a 4.1 
 a  

3


b 17
b 14
 c 12 
 15 
c 
3
 b11 
 4.7 
b 
2.1
0
3. If y3 = 125, find the value of y.
4. Write out the answers to the following expressions:
1016
, (10 3 ) 3 / 2 , (53 ) 2 / 3 .
1012
17
1
c
1
3

1
c3
b0 + b3
Geoscience Examples
1. The mass of the atmosphere is approximately 1018.7 kg. The mass of CO2 in the
pre-Industrial atmosphere is 1015.15 kg. Use the index laws to work out the
concentration of CO2 in the pre-Industrial atmosphere in units of kg/kg.
By the late 1990s the amount of CO2 in the atmosphere, due to fossil fuel
burning by humans, had risen to 1015.26 kg. What is the modern concentration of
CO2 in the atmosphere and what has been the percentage rise since the
Industrial Revolution?
2. Oceanic crust can be formed at mid-ocean ridges and recycled at ocean margins.
The rate of formation of the Atlantic Ocean floor at the Mid-Atlantic Ridge is
estimated to be 1.3 x 102 km of new crust in 65 million years. Without using a
calculator, determine the spreading rate of the ocean floor away from the MidAtlantic Ridge in m/yr. How many years would it take to generate 1 m of new
crust?
The Earth’s magnetic field undergoes periodic reversals which are preserved in
the oceanic crust spreading away from mid-ocean ridges, as bands of alternate
positive and negative magnetic anomalies that are symmetric around the midocean ridge. In the Atlantic crust, the Gilbert reversal epoch is preserved in a
band of crust that has a width of 34 km perpendicular to the Mid-Atlantic Ridge.
Without using a calculator, determine the duration of this epoch.
18
3. Making and Using Equations
Learning
Summary:
After studying this section, you will be able to:
 use letters as symbols
 understand the terms equation, identity and function
 write down equations from a description in words
GCSE
Background: You should understand and be able to use the rules of general arithmetic,
especially multiplying out brackets, taking out common factors and
collecting like terms. You should be able to factorise simple quadratic
expressions. You should know the difference between equations,
formulae and identities. You should be able to solve simple linear
equations.
Reminders:

Letters are used as symbols to represent:
unknown quantities in an equation, which can be found, for example
x in x2+5x+6=0
variables in a formula, which can take many values, for example
v = u + at
numbers in an identity, which can take any values, for example
5( x  2)  5x  10

Manipulating and solving equations is included in a part of mathematics called
algebra. Some definitions used in algebra are
expression – this is any arrangement of letter symbols and numbers (a general
term), for example
y-x+3 is an expression.
equation - this connects two expressions which involve unknown quantities.
Equations have an equals sign in them, for example
y–x+3=2y-4 and y–x+3=0 are equations.
formula – this connects two expressions containing variables, the value of one
variable depending on the value of the others, for example
v2-u2=2as is a formula.
identity – this connects expressions which involve unspecified numbers. An
identity remains true whatever numerical values replace the letter symbols. It
has a  sign, for example
5( x  2)  5x  10 is an identity.
19
function – this is a relationship between two sets of values such that a value
from one set corresponds to a unique value from the second set, for example
y = ex where e is the exponential function.
Link: The rules for manipulating algebra are like those used in arithmetic and the
index laws (see p.14)
For example
3b+5b = 8b
a x a = a2
x4
 x2
x2
2c(3c-7) = 6c2-14c
Take particular care when multiplying out (or expanding) brackets
every term in
one bracket must
be multiplied by
every term in
the other
(u+v)(w+x) =
u(w+x) + v(w+x) =
uw+ux+vw+vx
(u-6)(u-2) =
u(u-2)-6(u-2) =
You can only
collect the terms
with the same
power of u
u2-2u-6u+12 =
u2-8u+12
The inverse process to expanding is called factorising.
For example
12b2-6b = 6b(2b-1)
Both terms can be divided by 6b, so 6b is a factor.
That example was done just by looking, but how can we systematically factorise
x2+5x+6 ?
20
Key Point:
To factorise an equation of the form x2+px+q
find two numbers that will add up to p and multiply
together to give q.
This method works if p and q are positive or negative.
Eg
Example 1: Factorise y2-6y+9.
Numbers that multiply together to give +9 are
1 x 9, -1 x -9, 3 x 3, -3 x -3.
The only pair of numbers that add together to give -6
are -3 x -3, so the answer must be
y2-6y+9 = (y-3)(y-3) = (y-3)2 .
Eg
Example 2: Factorise y2-3y-10.
Numbers that multiply together to give -10 are
1 x -10, -1 x 10, 2 x -5, -2 x -5.
The only pair of numbers that add together to give -3
are 2 x -5, so the answer must be
y2-3y-10 = (y+2)(y-5) .
Making equations from words
Often in Geoscience, quantitative relationships and instructions are expressed in words.
In order to use this information, you will need to convert it into a mathematical
expression e.g. an equation. This is a skill that you will develop with practice, and is
best developed in a classroom situation. However, here are a few definitions and
examples to help you understand what you might be being asked to do.
First, here are some examples of written terms that have a precise mathematical
meaning

Relationships:
sum – e.g. ‘the suspended solids concentration is the sum of the concentration of
sand and the concentration of mud’. To find the sum of two quantities, add
them together. To write this statement as an equation, we can write cT = cs + cm
where cT is the suspended solids concentration, cs is the sand concentration and
cm is the mud concentration.
21
difference – e.g. ‘the concentration of mud can be found from the difference
between the suspended solids concentration and the concentration of sand’. To
find the difference between two quantities, subtract one from the other. To
write this statement as an equation, we can write cm = cT - cs , where cT is the
suspended solids concentration, cs is the sand concentration and cm is the mud
concentration.
product – e.g. ‘the volume of suspended solids can be estimated as the product
of the suspended solids concentration and the total liquid volume’. To find the
product of two quantities, multiply them together. To write this statement as an
equation, we can write VT = cT VL , where VT is the volume of suspended solids,
cT is the suspended solids concentration and VL is the total liquid volume.
ratio – e.g. ‘the ratio of the sand concentration to the mud concentration has a
value of 1.7’. A ratio is the same as fraction, expressing the amount of, in this
case, sand concentration compared to mud concentration. To write this
c
statement as an equation, we can write s  1.7 , where cs is the sand
cm
concentration and cm is the mud concentration.
function – e.g. ‘the volume of suspended solids deposited is an exponential
function of the distance from the coast’. A function is is a relationship between
two sets of values such that a value from one set corresponds to a unique value
from the second set, for example y = ex where e is the exponential function. To
write the statement as an equation, we can write VT =k1ek2x , where VT is the
volume of suspended solids, k1 and k2 are unknown constants and x is distance
from the coast.

In the above examples, we have also seen the meanings of some written
instructions e.g.
determine, find, calculate, estimate – instructions to calculate the value of a
quantity from the relationship given.
express, re-arrange, define, solve, evaluate – instructions to re-arrange a
relationship so that a particular quantity is expressed in terms of the other
quantities.
Link:
Other terms you will come across include proportionality and inequality.
These have precise mathematical meaning and are described in detail on p.34
22
Eg
Example 1: Write the following statement in mathematical terms. The
speed of a pyroclastic flow depends on the product of the volume of the
current and the square root of the gravitational acceleration only.
If we denote the speed of the pyroclastic flow as u,
the volume of the pyroclastic flow as V and the
gravitational acceleration as g, we are told that
the speed depends on the product of the volume and
the square root of the gravitational acceleration.
If we were told the speed depends on the volume and
the square root of the gravitational acceleration, we
could write this as a functional relationship
u = f (V,g).
However, we are told that the speed depends on the
product of the volume and the square root of the
gravitational acceleration, so we write
u = f (Vg1/2).
Eg
Example 2: The deposition of sediment was constant at a rate of 2.7
m/My over the Carboniferous period from 345 million years ago to 280
million years ago. Calculate the total depth sedimented during the
Carboniferous period.
The deposition of sediment was constant at
2.7 m/My over the whole of the Carboniferous,
which lasted 345 – 280 = 65 My.
To calculate the total sediment depth, we multiply the
deposition rate by the time duration to find that the
total sediment depth
65 My  2.7 m/My = 175.5 m.
(Hint: multiply the units of each quantity together to
check that your equation gives you the desired quantity.
mm
 mm using the usual rules for
In this case My
My
algebra).
Summary in Words

Manipulating and solving equations is included in a part of mathematics called
algebra. Some definitions used in algebra are: expression (any arrangement of
letter symbols and numbers); equation (a connection between two expressions
which involve unknown quantities); formula (a connection between two expressions
23
containing variables, the value of one variable depending on the value of the
others); identity (a connection between two expressions which involve unspecified
numbers); function (a relationship between two sets of values such that a value from
one set corresponds to a unique value from the second set).

The rules for manipulating algebra are like those used in arithmetic and the index
laws.

To factorise an equation of the form x2+px+q find two numbers that will add up to p
and multiply together to give q. This method works if p and q are positive or
negative.
What should I do now?

Make sure you understand the differences between the algebraic relationships, and
that you know how to manipulate algebraic expressions, including those with
brackets. Learn the rule for factorizing quadratic expressions.

Try the examples below that test your application of the methods.

Try the Geoscience examples so you can see how to apply these methods in your
subject.
Exercises - Methods
1. Expand the brackets in the following expressions: x(x2+1), (a+b)(a-b), (a+b)(3x), b/2(b/3+1), (z+1/2)(z-3/5).
2. Factorise these expressions: 2x+6, 44y2-22y-22, 4a3+6a,
3
1 1
1
c  a, 2  3 .
8
8 x
x
3. Factorise these expressions: z2-z-56, q2-4q-45, x2-6x+9.
4. Solve these quadratic equations by factorising: x2+5x+6=0, a2-3a+2=0,
b2-3b-40=0.
5. If x=4 and y=7, find the sum, difference and product of 2+x and 12-y.
1
6. If x=25 and y=7, determine the ratio of 1  x 2 to 2y+2.
7. The speed, u, of a lava flow depends on the ratio of the square of its depth, h, to
its viscosity, . Write this statement in mathematical terms (the symbol . is the
Greek letter ‘mu’).
1
x 5
y   to make x the subject of the equation. If
2
3 3
y=18, estimate the value of x.
8. Re-arrange the equation
Geoscience Examples
1. The Shields parameter is the ratio of the product of the shear stress acting on an
object and its cross-sectional area to the weight of that object. If a stream flow
exerts a shear stress of 0.004 kg m-1 s-2 on a pebble of mass 3.7 g and cross-
24
sectional area of 0.0003 m2, estimate the value of the Shields parameter (the
acceleration due to gravity has the value 9.8 m s-2).
2. Although all groundwaters flow slowly through aquifers, some flow more
slowly than others. While studying the water supply to the town of Dijon in the
mid 18th century, town engineer Henry Darcy measured the elevation of water
in various wells and mapped the varying heights of the water table in the
district. He calculated the distances that water travelled from well to well. He
found that the volume of water flowing in a certain time divided by the area of
the aquifer through which it flows is equal to a constant multiplied by the ratio
of the vertical drop divided by the length of the flow through the aquifer. Using
suitable symbols, write down a formula that expresses this relationship, which is
known as Darcy’s Law. This formula defines the constant as the hydraulic
conductivity which is a measure of the permeability of the rock and also
depends on the properties of the fluid such as density and viscosity.
3. Oxygen, silicon and iron make up respectively 30%, 15% and 35% of the mass
of the whole Earth, and respectively 46%, 28% and 6% of the mass of the
Earth’s crust. Determine the ratios of the masses of these elements in the crust
to their masses in the whole Earth. If the mass of the Earth is 5.83 x 1024 kg,
what is the mass of each element in the whole Earth?
25
4. Solving Equations
Learning
Summary:
After studying this section, you will be able to:
 manipulate and solve linear equations
 solve simultaneous linear equations
 solve quadratic equations by factorising or by using the formula
Solving linear equations
In order to solve an equation, you need to manipulate it so that the unknown terms
are on one side of the equals sign and the numbers are on the other side. You
manipulate the equation by performing mathematic operations such as addition,
subtraction, division and multiplication to both sides of the equation.
Key Point:
Eg
An equation has two sides separated by an equals sign.
You MUST always do the same things to each side of
the equation.
Example 1: Solve the equation 5x=100.
We need to get x on its own on one side of the
equation. We can do this by dividing each side of
the equation by 5
Always do the same
thing to both sides
of the equation.
Eg
5 x 100

5
5
so x = 20
Example 2: Solve the equation x+4=7.
Always do the same
thing to both sides
of the equation.
We need to get x on its own on one side of the
equation. We can do this by subtracting 4 from each
side of the equation
x+4-4=7-4
x = 3.
26

It is easy to make mistakes when solving equations if you do not write out what
you do to each side of the equation. We suggest that you always write out what
you are doing in full. For the example above, you can use the following
shorthand,
x+4=7
x+4-4=7-4
x=3
-4
where the operations you carry out are written on the right hand side of the
vertical line.

More complicated examples involve brackets and fractions.
Eg
Example 3: Solve the equation 3(x-6)=8.
Check first to see if
each side has a
common factor. If it
does, divide each
side by the factor

3(x-6)=8
The first thing to do is to multiply out the bracket
3x-18 = 8.
Now we can add 18 to each side to get the x term on
its own
3x=8+18=26.
Now divide each side by 3 to get x on its own
26
1
x
8 .
3
3
Written out in full, this would look like:
3(x-6)=8
3x-18=8
3x-18+18=8+18
3x=26
26
1
x=
8
3
3

expand bracket
+18
3
Writing the answer out like this helps spot mistakes as you go.
27
Eg
Example 4: Solve the equation
Remove the
fractions first
1
1
( x  3)  ( 2 x  1)
2
3
1
1
( x  3)  ( 2 x  1)
2
3
The first thing to do is to remove the fractions. Multiply
both sides by the lowest common multiple of the
denominators of the fractions. In this case it is 2 x 3 = 6.
Multiply each side by 6
3(x-3)=2(2x+1) .
Now multiply out the brackets
3x-9=4x+2 .
Now subtract 3x from each side, to leave only one side
with an x term
-9=x+2 .
Now subtract 2 from each side to leave x on its own
-11 = x .
So the solution is x = -11.
Solving simultaneous equations
Sometimes we do not have enough information to solve an equation. Consider the
equation y = x +2. We can re-arrange it to make x the subject of the equation by
subtracting 2 from each side, so x = y –2, but we cannot work out the values of x and y.
If we knew the value of one of them we could work out the value of the other. To find
x and y you need another equation that relates them e.g. 2x = 5y +3. When dealing with
more than one unknown quantity and a set of equations you need at least as many
equations as unknown quantities to find a unique solution.
Key Point:

To solve a system of multiple equations and unknowns
you need at least as many equations as unknowns.
If you have a system of two equations and two unknowns, you can find a unique
solution. The equations are called simultaneous equations. We can find a solution
for the simultaneous equations above.
28
Eg
Example 5: Solve the simultaneous equations
You must do the same
thing to each term on
both sides of the
equation
x=y-2
2x = 5y – 3.
The object is to make either the x terms or the y terms the
same. We can make the x terms the same if we divide
the second equation by 2, so we have
x  y2
2x 5 y 3


2
2 2
The second equation becomes
5
3
x  y
2
2
and using the first equation, we know that x = y – 2, so we
can write
5
3
y  y2.
2
2
It is easiest to get rid of the fractions first, so if we multiply
both sides of the equation by 2, we find
5y – 3 = 2y – 4.
If we add 4 to each side of this equation, we find
5y + 1 = 2y.
If we subtract 2y from each side of this equation, we find
3y + 1 = 0 .
If we subtract 1 from each side of this equation, we find
3y = -1, so y = -1/3.
We can now put this value for y back into the first equation,
so we find
1
x    2 , so x = -7/3.
3
Solving quadratic equations
A quadratic equation has the form y = ax2+bx+c. It is an equation that contains a
term in which the variable is squared, and so it has two solutions.
You can solve them by methods including factorising (see the section on Making
and Using Equations on p.19) or by using a formula.
29
Eg
Example 6: Solve the equation x2-3x+2=0
If two numbers are
multiplied and the
result is zero, either
one or the other is
zero
Key Point:
To factorise the left hand side of this equation, we need to
find two numbers that add up to –3 and multiply together to
give 2. The only numbers that give 2 when multiplied
together are –1 x –2 or 1 x 2. The first combination
adds up to –3, so we find
(x-1)(x-2) = 0 .
Either (x-1) = 0 or (x-2) = 0, so the solution is x = 1 or
x = 2.
The solution of the quadratic equation ax2+bx+c=0
is given by
 b  b 2  4ac
x
.
2a
The sign ‘  ’ means ‘plus or minus’, so there are two solutions for x,
x
Eg
 b  b 2  4ac
 b  b 2  4ac
and x 
.
2a
2a
Example 7: Use the formula to solve the quadratic equation x2-3x+2=0
Make sure you get
the signs right and
operations in the
right order
In this equation, a=1, b= -3, c=2. Put these values into
the formula.
3  9  ( 4  1 2)
x
, so
2 1
3 1
, so the solution is
x
2
x = 1 or x = 2.
Summary in Words

In order to solve an equation, you need to manipulate it so that the unknown terms
are on one side of the equals sign and the numbers are on the other side. You
manipulate the equation by performing mathematic operations such as addition,
subtraction, division and multiplication to both sides of the equation.
30

An equation has two sides separated by an equals sign. You MUST always do the
same things to each side of the equation.

When dealing with more than one unknown quantity and a set of equations you
need at least as many equations as unknown quantities to find a unique solution.

A quadratic equation has the form y = ax2+bx+c. It is an equation that contains a
term in which the variable is squared, and so it has two solutions. You can solve
quadratic equations by methods including factorising or by using a formula.

To factorise an equation of the form x2+px+q find two numbers that will add up to p
and multiply together to give q. This method works if p and q are positive or
negative.
What should I do now?

Make sure you understand how to manipulate equations, factorise quadratic
expressions and solve simultaneous equations. Learn the formula for solving
quadratic equations and the rule for factorizing quadratic expressions.

Try the examples below that test your application of the methods.

Try the Geoscience examples so you can see how to apply these methods in your
subject.
Exercises - Methods
1. Solve the following equations: 2x+5=8, 3t-17=12, 5t+1=7-4t,
12d
14 1
 50,
 .
5
2d 2
2. Solve the following equations:
1
 x 1 3 1
4( y  7)  9,5( z  8)  2,4( x  6)  12,3    , ( x  4)  (2 x  5).
3
 8 8 4 2
3. Solve the sets of simultaneous equations: 2x=2y-3, 2x=5y-3; a-b=8, 2a+2b=32;
a+2b-3c=0, 2a+b+c=8, 5a-2b-c=7.
4. Solve the following quadratic equations using the formula: 2x2-6x+4=0, c2x2
 3x  2  0.
2c+1=0, z2-6z+7=0,
3
Geoscience Examples
1. The sinking velocity v of a sphere of radius r and density s in a liquid of
density l and viscosity  is given by Stokes’ Law
v
2 2 (  s   l)
.
gr
9

g is the acceleration due to gravity, g = 9.8 m s-2.
31
Sinking sphere experiments can be used to determine the density or viscosity of
a silicate liquid. Re-arrange Stokes’ Law in terms of (a) l and (b) .
In practice we would measure the distance fallen in one second for two noble
metal spheres of different density, but the same radius. Let the densities of the
noble metal spheres be s1 and s2 and the distance fallen in one second in each
case be L1 and L2.
L
(c) Express the ratio 1 in terms of l, s1 and s2.
L2
Here are the results of one such experiment carried out to determine the density
of a basaltic melt at 1500oC. Use them to calculate (d) l and (e) , using your
answers to (c) and (b). Be very careful to use SI units throughout.
Metal
Platinum
Palladium
Radius
1 mm
1 mm
Density at 1500oC
2.1225 x 104 kg m-3
1.1795 x 104 kg m-3
Distance fallen in 1 s
0.807 mm
0.396 mm
(f) At what velocity would a 6 mm diameter olivine crystal (density 3.028 x 103
kg m-3) sink through the melt?
2. The motion of fluids control many environmental and geological processes.
Fluid motion can take place in two forms: as ordered, uniform motion called
laminar flow, or as highly disordered motion called turbulent flow. The form of
motion depends on the properties of the fluids and the flow, and can be
measured by a parameter called the Reynolds number, defined as
Re 
Lu

where L is a length controlling the flow motion such as the width of a channel, u
is the velocity of the flow, is the density of the fluid and  is its viscosity.
Flows are generally laminar when their Reynolds number is less than about
1000 and are generally turbulent when their Reynolds number is greater than
about 4000. For a basaltic lava flow of width 3 m, at what velocity does the
flow become turbulent? If the flow speed was 0.25 m s-1, at what viscosity
would the flow become laminar? (At eruption temperatures, the density of
basalt is 2600 kg m-3 and its viscosity is 100 kg m-1 s-1).
3. The pH of rainwater can be calculated from the equation
[H+]2 + 10-6.4 [H+] – 4.32 x 10-12 = 0.
where [H+] is the concentration of hydrogen ions.
Solve this equation for [H+] using the quadratic formula, and calculate the pH of
rainwater given the definition of pH
pH = -log10 [H+].
4. Volcanic eruptions under glaciers can generate high-speed mudflows called
lahars as a consequence of the rapid melting of ice and mobilisation of
32
sediments by the meltwater. Lahars follow the local topography and flow down
river valleys. Field estimates suggest that the typical rate of propagation of
lahars generated during the 1985 eruption of Nevada del Ruiz, Colombia, can be
described by a relationship of the form
5.17u2 – 7.3u – 220 = 0
where u is the average speed of lahars observed 5 km from its source. Use the
quadratic formula to determine the average speed of the lahars. (Quadratic
expressions have two solutions, but in this case only one solution makes sense,
given that for these observations, the positive direction of motion is measured
away from the volcano).
5. The binary eutectic diagram explains the chemical behaviour of two immiscible
(unmixable) crystals of minerals A and B from a completely miscible (mixable)
melt. The liquidus line for mineral A separates the pure melt region of the
diagram from the region where melt and crystals of mineral A can co-exist. The
liquidus line for mineral B separates the pure melt region of the diagram from
the region where melt and crystals of mineral B can co-exist. The solidus line
separates the pure crystal region of the diagram from the region where melt and
crystals can co-exist. The eutectic is the point at which all phases can co-exist,
and is found at the intersection of the two liquidus lines with the solidus.
If the equation of the liquidus of mineral A is
T = 880xA+800,
and the equation of the liquidus of mineral B is
T = -564xA+1540
near the liquidus, where T is the temperature in oC and xA is the proportion of
mineral A (ranging from 0 to 1). Solve these equations simultaneously to find
the co-ordinates of the eutectic.
33
5. Proportionality and Inequality
Learning
Summary:
After studying this section, you will be able to:
 understand the meaning of proportionality
 manipulate and solve inequalities
Proportionality
When two quantities are proportional to each other, they are related in a precise
mathematical way.
Key Point:
The symbol used to represent proportion is . If y is
directly proportional to x , we write y x.
This is exactly the same as y being equal to x multiplied
by a constant k, so we could also write y = kx.
Here are some other forms of proportionality:
y is inversely proportional to x
y x
y = k/x
y is inversely proportional to the
square of x
y  1/x2
y = k/x2
y is proportional to the square of x
y x2
y = kx2
y is proportional to the square root
of x
y x1/2
y = kx1/2
To solve a problem involving proportionality, turn the proportionality relationship
into an equation using the rule above.
The constant k is called the ‘constant of proportionality’ for these relationships.
Inequalities
An equation describes an equality between its terms, where one side is exactly
equal to the other side. In some cases we need to deal with expressions where we
only know that one side is larger or smaller than the other. These expressions are
called inequalities, and instead of an equals sign the symbols ‘>’ (meaning greater
than) and ‘<’ (meaning less than) are used. Where one quantity is ‘greater than or
equal to’ another, the symbol ‘  ’ is used.
34
Eg
Example 1: Solve the inequality 3x-5>2(x-2).
We can use similar rules for inequalities as we do for
equations. First expand the bracket,
3x-5>2x-4 .
Now subtract 2x from each side,
x-5>-4 ,
and add 5 to each side
x>1 .
Eg
Example 2: Solve the inequality 5  x  3x  2
The first stage is to subtract 3x from each side
5  4x  2
Then subtract 5 from each side
 4 x  3
If this were an equation, the next step would be to divide
each side by –4. This will only work for an inequality if
you also change the direction of the inequality sign. Check
this with numbers. 3 < 5 but –3>-5.
So if we divide by –4 we must change the direction of the
inequality sign, so
3
x .
4
Key Point:
You can manipulate inequalities like equations except that
when you multiply or divide each side by a negative number,
you must change the direction of the inequality sign.
Summary in Words

When two quantities are proportional to each other, they are related in a precise
mathematical way. If y is directly proportional to x, this is exactly the same as y
being equal to a constant multiplied by x. To solve a problem involving
proportionality, turn the proportionality relationship into an equation using the rule
above.

Inequalities are expressions where we only know that one side is larger or smaller
than the other. Instead of the equals sign used in equations, the symbols ‘>’
(meaning greater than) and ‘<’ (meaning less than) are used. Where one quantity is
‘greater than or equal to’ another, the symbol ‘  ’ is used.

You can manipulate inequalities like equations except that when you multiply or
divide each side by a negative number, you must change the direction of the
inequality sign.
35
What should I do now?

Make sure you understand the mathematical meaning of proportionality, and how to
manipulate and solve inequalities.

Try the examples below that test your application of the methods.

Try the Geoscience examples so you can see how to apply these methods in your
subject.
Exercises - Methods
1. Write equivalent forms for the following statements: y is inversely proportional
to x; y is proportional to x3; z 3c; d z4.
2. Solve the following inequalities: 2z-4>3(z-3); 3x-5>1/3(x-2);
6  y  4 y  3; t 2  5t  6.
Geoscience Examples
1. The population of Beluga Whales in the Arctic Ocean is found to be
(i) proportional to the population of Arctic Cod (on which they feed) and
(ii) inversely proportional to the square root of the population of Bearded Seals
(which also feed on Arctic Cod).
There are an estimated 6 million Arctic Cod in the Arctic Ocean. If the constant
of proportionality for the relationship between the population of Arctic Cod and
Beluga Whales has the value 0.0017, what is the estimated number of Beluga
Whales in the Arctic Ocean?
If there are an estimated 40000 Bearded Seals in the Arctic Ocean, what is the
constant of proportionality between the populations of Bearded Seals and
Beluga Whales?
2. On the binary eutectic diagram for minerals A and B, the pure melt region is
defined as the region where
T > -564xA+1540
where T is the temperature in oC and xA is the proportion of mineral A (ranging
from 0 to 1).
If the temperature of a sample of pure melt is 1300oC, solve the inequality to
identify what proportion of mineral A the pure melt region corresponds to (hint:
re-arrange the inequality to make xA the subject and then substitute in the
temperature of the pure melt sample).
(For more details about the binary eutectic diagram, see section 4, Geoscience
example 5).
36
6. Gradients of Straight Lines and Curves
Learning
Summary:
After studying this section, you will be able to:
 recognise and write down the equation of a straight line
 find and interpret the gradients of straight lines
 approximate the gradients of curves using tangents
 recognise mathematical notation for gradients
Straight line graphs

The graph below shows how the depth of sediment at the bottom of a lake varies
with time (inferred from the age of the sediment). This is an example of a constant
rate process. The time taken to deposit 1 m of sediment is the same if you are at the
bottom or top of the sediment pile.
Key Point:
Graphs of constant rate processes show straight lines
when the process is plotted against time.
GCSE
Background
Recognise that equations of the form y = mx + c correspond to straight
line graphs
37
Eg
Example 1: Draw the graph of y – x –1 =0. This can be written more
compactly as y = x + 1

Look at the graph above – it goes up to the right. For every unit it moves across, it
moves one unit up. The gradient of the line is one.
Key Point:

the gradient is found by dividing the increase in y values by the increase
in x values
It crosses the y-axis at the point (0,1). This point is called the intercept.
Eg
Example 2: Draw the graph of y +(3/2)x – 4 = 0. This can be written as
y = -(3/2) x + 4.

This graph goes down to the right. For every two units it moves across it moves
three units down. The gradient of the line is –3/2.
38
Key Point:

The gradient is negative since the y value decreases as the
x value increases
The line crosses the y-axis at (0,4).
Key Point:
In the equation of the straight line y = m x + c, m is the gradient
of the line and c is the intercept of the line with the y-axis.
Summary in Words

On a graph, a straight line has a positive gradient if its y value increases as the x
value increases.

The gradient is found by dividing the increase in y values by the increase in x
values

The point at which the line crosses the y axis is called the intercept.

The equation of a straight line is written y = m x + c, where m is the gradient of the
line and c is the intercept.
What should I do now?

Learn the form of the equation of a straight line and make sure you understand the
terms in the equation. Make sure you know how to determine the gradient and
intercept of a straight line plotted on a graph.

Try the examples below that test your application of the methods.
Exercises - Methods
1. Find the gradients and intercepts of the following straight lines: y=2x+3,
5
7
y   x  4, y  x  2, y=5x104x-1.8x10-5.
2
3
2. Sketch the graphs of the straight lines in question 1.
Geoscience Examples
1. Sediment is deposited into a lake at a constant rate corresponding to an increase
of sediment depth of 20 m in 10000 years. If the age of the sediment at 100 m
depth in the lake is 1.05 million years (Ma), sketch a graph of the sediment age
from a depth of 100 m to the surface. What is the gradient of this graph?
39
2. The number of crinoid fragments found in a fossil bed varies from 7 per cm3 at
the top of the bed, to 23 per cm3 at the base of the bed, 5.6 m below. If the
number of fragments varies linearly through the bed, what is the gradient of the
straight line that describes the rate of decrease of the number of crinoid
fragments with height up the bed? What is the number of crinoid fragments per
cm3 2.4 m below the top of the bed?
40
7. The Exponential Function and Logarithms
Learning
Summary:
After studying this section, you will be able to:
 understand what the exponential function is and how to recognise
it
 understand how to express exponential growth and decay in
terms of natural logarithms
 be able to use and solve simple logarithmic expressions
The Exponential Function
Some processes do not occur at constant rate. The figure below shows graphs of the
equations y = 2t and y = 3t. These graphs increase at an increasing rate with time. If
you estimate the gradient at the points t = 1, 1.5 and 2 by placing a ruler so it just
touches the curve at that point and drawing the tangent (the straight line that just
touches the curve and which has the same gradient as the curve), you will see that the
gradient of the tangents increases as the value of t increases. If t denoted time, we
would say that the rate of this process increases with time.

If we plot the graph y = 2.71828x we find that at each value of x, the gradient of the
curve has the value 2.71828x.
41
At x = 1, the gradient of the line has the value 2.718281 = 2.71828.
At x = 2, the gradient of the line has the value 2.718282 = 7.38905.
Clearly, the number 2.71828 is a very special number that is called ‘e’ by
mathematicians. In fact, e is a number that has an inifinite number of decimal places,
like , and 2.71828 is an approximation to its value in the same way that 3.14 is an
approximation to the value of .

The function y = ex is called the exponential function. It can also be written
y = exp (x)

Many natural processes such as bacterial growth and radioactive decay have graphs
of this form.

The inverse function of the exponential function is the natural logarithm.
Logarithms

A logarithm is another name for an index whose base is a number. So if we write
102 = 100, we can also say ‘the logarithm of 100 to base 10 is 2’. We write this as
log10 (100) = 2.

Because we count in base 10, log10 is usually shortened to just ‘log’ (this is what
appears on your calculator button) and is known as a ‘common logarithm’.
42

We can write logarithms in bases other than 10. For example if we wanted to write
100 as a logarithm to base 5, we would write log5(100) = 2.637.

The logarithm of a number to base e is called a ‘natural logarithm’. The natural
logarithm of 100 would be written ‘ln(100)’.
How we deal with logarithms
Logarithms are similar to indices, and we can use similar rules to the index laws to
manipulate expressions that contain logarithms. The following rules apply to all
logarithms whatever their base.
Key Point:
The logarithm of the product ab is equal to the
logarithm of a + the logarithm of b
log(ab) = log(a) + log(b)
Eg
Example 1: What is log10(100) + log10(10)?
Write down the expression in terms of the base,
10 in this case,
log10(102) + log10(101).
The logarithm is the value of the index, so we have
log10(102) + log10(101) = 2 + 1 = 3.
By definition, log10(103) = 3, so
log10(102) + log10(101) = log10(103).
Key Point:
The logarithm of a/b is equal to the logarithm
of a - the logarithm of b
ln(a/b) = ln(a) - ln(b)
Using logarithms to solve exponential equations
Key Point:
Natural logarithms can be used to solve equations that
contain an exponential term, because the natural logarithm
is the inverse of the exponential function.
43
GCSE
Background:
The operation that reverses what has been done is called an
inverse operation.
For example,


the inverse of multiplying by 10 is dividing by 10,
the inverse of a squaring a number is taking the square root.
When you perform an operation and its inverse operation on a number, the result is the
number
e.g. 10 2  10 .
The same is true for unknown quantities e.g. (x5)1/5 = x.
Eg
Example 2: Solve the equation y = e2t to find t.
The inverse of the exponential function is the
natural logarithm. Our equation is
y = e2t .
remember when solving
equations to do the same
thing to both sides of the
equation
If we take the natural logarithm of each side of this
equation, we find
ln y = ln (e2t) = 2t,
because when we perform an operation and its inverse
on a quantity, the result is just the quantity, in this case
2t,
so our solution is
t
Eg
ln y
.
2
Example 3: Radioactive dating of rocks is based on the fact that the
amount of radioactive material decreases with time. The amount of
material present is given by
Q = Q0 exp(-t/)
where Q is the amount of material present, Q0 is the amount of material
originally present, t is the time since formation and  is a constant
known as the decay constant.
44
Solve this equation for .
Q = Q0 exp(-t/)
Take the logarithm of each side,
ln Q = ln (Q0 (exp (-t/)).
Using the rules for the logarithm of a product,
ln Q = ln Q0 + ln (exp (-t/))
remember logarithms
and exponentials are
inverse functions
ln Q = ln Q0 + (-t/)
so
ln Q = ln Q0 – t/
Re-arranging
t/ = ln Q0 – ln Q
which we can write as
t/ = ln (Q0/Q)
using the rules for logarithms. Multiplying both sides
by  and then dividing both sides by ln (Q0/Q), we find
 = t/ln (Q0/Q).
Summary in Words

The gradient of a curve can be estimated at a given point by drawing a tangent to
the curve at that point, which is a straight line which just touches the curve at that
point only. The gradient of the tangent is equal to the gradient of the curve at the
point at which they touch.

The exponential function y = ex has the property that the gradient of the function at
any point is equal to the value of the function. e is a number like  that has an
infinite number of decimal places and an approximate value of 2.71828.

The inverse of the exponential function is the natural logarithm. A logarithm is
another name for an index whose base is a number. So if we write 102 = 100, we
can also say ‘the logarithm of 100 to base 10 is 2’. The logarithm of a number to
base e is called a ‘natural logarithm’. The natural logarithm of 100 would be
written ‘ln(100)’.

Logarithms are similar to indices, and we can use similar rules to the index laws to
manipulate expressions that contain logarithms. The logarithm of the product ab is
equal to the logarithm of a plus the logarithm of b. The logarithm of a/b is equal to
the logarithm of a minus the logarithm of b.
45

An inverse operation is one that reverses what has been done. To solve an equation
that contains an exponential function, take the natural logarithm of each side of the
equation.

The exponential function and logarithms are much more important in mathematics
than these simple applications suggest. You will meet these functions again in the
second year where you will spend more time understanding how they are derived
and why they are important.
What should I do now?

Make sure you understand the definitions of the exponential function, logarithms
and inverse functions. Revise the index laws and make sure you know how to solve
equations that contain exponential terms.

Try the examples below that test your application of the methods.

Try the Geoscience examples so you can see how to apply these methods in your
subject.
Exercises – Methods
1. Write out the following terms in words: y=ex; r=et; log10(1000)=3; log7(7)=1;
ln(285)=5.652; ln(35)=3.555, log(98)=1.991.
2. Use the logarithm rules to simplify the following expressions: log10(100b),
log8(8c), log(10/z).
3. Solve the following equations: log(1000/a)=21; log5(25z2)=2.
4. Solve the following equations, showing all working: y=e3t, y=e3x-1.
5. Evaluate the equation y=2e-k/x when k=5 and x=3.
Geoscience Examples
1. Predicting future human population is important for estimating resource needs
or possible pollution impacts. A population where fertility and mortality are
constant is characterized by exponential growth.
Pt = Po e r t
where Pt is the population at time t, PO is the original population (when t = 0)
and r is the rate of change.
United Nations statistics indicate that the world population has been increasing
at the rate of 1.9% per year, since 1945. The world population in 1975 was
approximately 4 billion. Assuming an exponential growth model, (a) what will
the population of the world be in the year 2010? (b) When will the world
population be 9 billion?
2. The exponential and logarithmic functions are very useful when dealing
quantitatively with radioactivity - a key phenomenon for geoscience because it
can give us the information on the age of rocks and even the rate of circulation
in the oceans. One of the isotopic clocks we use is based on the radioactive
isotope of carbon, 14C. 14C is produced, among other places, in the atmosphere
46
by the collision of highly energetic subatomic particles (ultimately derived from
space) with an isotope of nitrogen, 14N. 14C is also radioactive - it is constantly
decaying back to 14N. The result of these two processes is that the ratio of 14C
to total carbon (14C/C) in the atmosphere is maintained in a dynamic
equilibrium at a value of 1.162x10-12. Every part of the surface Earth
containing carbon is more or less equilibrated with the atmosphere. The
consequence is that, for example, carbon contained in plants or the surface
oceans also has this ratio of 14C to total carbon. Once plants die, however, or if
surface ocean water sinks away from the surface to the deep ocean exchange
with the atmosphere stops. Decay of 14C, however, continues. The
consequence is that the longer plants are dead the further the 14C/C falls, away
from the atmospheric value of 1.162x10-12 until eventually, there is no 14C left at
all. If we know the rate at which decay occurs, and if we can measure the 14C/C
ratio of dead plant material, we can say how long it has been since the plant died
and stopped exchanging with the atmosphere. These principles are the basis for
the dating of such famous artefacts as the Turin shroud - in this case one is
dating the time at which the flax died that went to make the linen that eventually
became the Turin shroud!
One of the key applications of 14C in geosciences is for measuring the rate at
which the oceans circulate and mix. The oceans are constantly being mixed by
currents. Some of these you will probably have heard about - like the Gulf
Stream or North Atlantic Drift. These currents move water and heat around the
oceans and serve to partially equalise imbalances in the way the Sun heats the
Earth - more at the Equator than the Poles. What may be less familiar to you is
that the deep oceans have currents as well. In some areas of the world oceans,
surface water sinks from the surface to great depths. One key area is the North
Atlantic, where water that has traveled northwards in the Gulf Stream sinks and
travels southward in a water body known as North Atlantic Deep Water
(NADW). The only other part of the world's oceans where this process occurs
is the Southern Ocean (surrounding Antarctica) where NADW is joined by more
sinking water. The two water bodies mix and make their way around the world
ocean into the Pacific. In the Pacific this water upwells and travels back to the
Atlantic at the surface - the final leg of this journey being the Gulf Stream. So,
in essence, the circulation of the oceans can be described in terms of a big loop deep water flowing out from the Atlantic into the Pacific and surface water
returning the other way. Of course, real life is more complicated than this but
this picture captures the essence of the process.
How long does this loop take? This is a question that can be answered
approximately and partially using 14C. When the surface waters sink in the
North Atlantic and the Southern Ocean, they lose contact with the atmosphere
and the 14C/C ratio of the carbon dissolved in the water decreases - 14C is
decaying radioactively and is not being replaced by the action of cosmic rays on
14
N. In fact, the 14C/C ratio of deep Pacific water is not 1.162x10-12 as it is in
the atmosphere and most surface ocean waters but 1.029x10-12. If we knew the
rate of decay of 14C we could work out how long deep Pacific water had been
isolated from the atmosphere and thus travelling around the big circulation loop.
It turns out we do know the rate of decay and the expression giving the amount
of radioactive isotope left over (N) from an initial amount N0 is related to time
(t) by the following expression:
N0 = N et
 here is a constant. In a population of radioactive isotopes, eg. 14C, all the
atoms have a constant probability of decaying. For 14C, this probability is
47
0.0001209 per year and it is a constant through time and space. It is given the
symbol  and is called the decay constant.
Use this expression and the information given above on the 14C/C ratio of the
atmosphere and deep Pacific water to calculate the latter's "age" - the time since
it last equilibrated with the atmosphere.
48
8. Triangles
Learning
Summary:
After studying this section, you will be able to:
 relate the angles and length of sides of triangles
 find the length of the third side in a right-angled triangle
 find the length of a line joining two points on a graph
GCSE
Background:
You should understand and be able to use the rules of parallel lines to
prove that the sum of the internal angles of a triangle is 1800, recall the
definition of different types of quadrilaterals, understand and be able to
use Pythagoras’ theorem and trignometrical relationships in rightangled triangles.
Sum of angles of a triangle

The sum of the angles inside a triangle is 180o.

This is easy to see if you remember that if you rotate around a point on a
straight line, the angle you pass through is 180o to get from one part of the line
to the other. Now look at the triangle ABC in the figure below
B
X
Y
+
*
C
A

A straight line has been drawn which is parallel to one side of the triangle and
touches it at the point B. Using the properties of parallel lines, the angle A is
equal to the angle XBA (marked with a star) and the angle C is equal to the
angle CBY (marked with a cross). The sum of the angles of the triangle is angle
A + angle ABC + angle C which is equal to angle XBA + angle ABC + angle
CBY, which are the angles around the point B on the straight line. So the sum
of the angles of a triangle is 180o, same as the angles on a straight line.

Because any quadrilateral (four sided shape) can be divided into two triangles,
the angle sum of any quadrilateral is 360o.
49
Properties of right-angled triangles
Right-angled triangles have one angle of 90o, plus two other angles whose sum
is 90o. The right angle is opposite the longest side of the triangle, which is
called the hypotenuse, marked as h in the figure below
h
o
a

Pythagoras’ theorem relates the length of the sides of a right-angled triangle.
The theorem states that the area of the square on the hypotenuse of the triangle
is equal to the sum of the areas of the squares of the other two sides.
h2
h
o
o2
a
a2
Key Point:
Eg
Pythagoras’ theorem is usually abbreviated to
‘the square on the hypotenuse = the sum of the squares
on the other two sides’
h2 = a2 + o2
Example 1: The two shorter sides of a right-angled triangle have lengths
of 9 cm and 12 cm. Find the length of the hypotenuse.
Remember this
only applies to
right-angled
triangles!
Using Pythagoras’ theorem
h2 = a2 + o2, subtitute in the known lengths of a and o,
h2 = 92 + 122 = 81 + 144 = 225.
So h = 225  15.
50

Note that in the example, the length of each side is divisible by 3, to make a
smaller right-angled triangle with sides of length 3 cm, 4 cm and 5 cm. This is a
special right-angled triangle called (simply) ‘the 345 triangle’. If you have a
right-angled triangle whose hypotenuse has a length of 5 m and one side has a
length of 3 m, the other side has a length of 4 m. This also applies to multiples
of these lengths, so if you have a right-angled triangle whose hypotenuse has a
length of 10 miles and one side has a length of 8 miles, the other side has a
length of 6 miles.
Key Point:
Eg
For a right-angled triangle that has a hypotenuse of
length 5 units and one side of length 4 units long,
the remaining side will have a length of 3 units. This
triangle is called the 345 triangle.
Example 2: Calculate the length of the line joining the two points A and
B on the graph below.
y
25
B
20
15
10
5
A
C
0
0
5
10
15
20
x
On the graph, A is at (5,5) and B is at (18,20). To find
the length AB, construct the point C by drawing lines
parallel to each axis. Find the length AC and BC from
the coordinates.
AC is the difference in the x-coordinates = 18-5 =13.
BC is the difference in the y-coordinates = 20-5 =15.
By Pythagoras’ theorem, we have AB2 = AC2 + BC2, so
AB2 = 132+152 = 169 + 225 = 394, so
AB =
394  19.8
51
Summary in Words

The sum of the angles inside a triangle is 180o. Because any quadrilateral (four
sided shape) can be divided into two triangles, the angle sum of any quadrilateral is
360o.

Pythagoras’ theorem states that the square of the length of the longest side of a
right-angled triangle (called the hypotenuse) is equal to the sum of the squares of
the lengths of the other two sides.

For a right-angled triangle that has a hypotenuse of length 5 units and one side of
length 4 units long, the remaining side will have a length of 3 units. This triangle is
called the 345 triangle.
What should I do now?

Learn the angle sum for a triangle and Pythagoras’ Theorem. Make sure you
understand how to apply it and which triangles it can be used for.

Try the examples below that test your application of the methods.

Try the Geoscience examples so you can see how to apply these methods in your
subject.
Exercises - Methods
1. If two of the angles of a triangle are 37o and 45o, find the other angle.
2. If two of the angles of a triangle are 68o and 38o, find the other angle.
3. If the lengths of the two shorter sides of a right-angled triangle are 13 cm and
0.23 m, find the length of the longest side. Write out all working.
4. If the lengths of the two shorter sides of a right-angled triangle are 1012 km and
2304 km, find the length of the longest side. Write out all working.
5. If the length of the hypotenuse and one other side of a right-angled triangle are
200 m and 37 m respectively, find the length of the other side.
6. If the length of the hypotenuse and one other side of a right-angled triangle are
40 km and 32 km respectively, find the length of the other side.
7. A line joints the following pairs of points on a graph. In each case, calculate the
length of the line: (7,8) and (9,15); (100,-20) and (-500, 50).
Geoscience Examples
1. The angle between a sedimentary layer and the horizontal is termed the ‘dip’ of
that layer. A sandstone layer dips at 12o. What angle does it make to the
vertical?
2. The upper surface of an outcrop of volcanic rock appears as an approximately
straight line when mapped. If one end of the outcrop is 120 m due north of the
volcanic vent and the other end is 570 m due west of the volcanic vent, how
long is the outcrop?
52
9. Trigonometric ratios
Learning
Summary:
After studying this section, you will be able to:
 use the sine, cosine and tangent ratios to find angles and sides
in right-angled triangles
Trigonometric ratios

In a right-angled triangle the sides and angles are related by three trigonometric
ratios: the sine (abbreviated to sin), the cosine (abbreviated to cos) and the
tangent (abbreviated to tan).

To use these ratios, first name the sides of the triangle. Look at the triangle
below
h
o

a

We already know that the longest side is called the hypotenuse

Relative to the angle , (the Greek letter ‘theta’) one side is opposite it (called
‘opposite’ and labeled ‘o’ above), and one side is next to it (called ‘adjacent’
and labeled ‘a’ above).
Key Point:
sin  
opposite
hypotenuse
adjacent
hypotenuse
opposite
tan  
adjacent
cos  
53
Eg
Example 1: Find the length, to 2 decimal places, of the side labeled x in
this triangle.
14 cm
x
37
Make sure you
know which side
is which!
You have been given the length of the hypotenuse,
and you need to find the length of the opposite side,
so you need to use the sine,
sin 37 = x/14 .
Re-arrange this by multiplying each side by 14, so
14 x sin 37 = x, so x = 14 x 0.6018 = 8.4252 = 8.43 cm
to 2 decimal places.
Sine and Cosine Rules

The sine and cosine rules can be used to find the lengths and angles in a
triangle that is not right-angled. They apply to any triangles.

The sine rule is used when two of the angles and the length of one side is
known, or the lengths of two sides and the angle opposite one of these sides
is known.
A
b
c
C
a
B
Key Point:
The sine rule can be written
a
b
c
.


sin A sin B sin C
54
Eg
Example 2: In the triangle ABC, find the missing sides and angle.
A
c
B
55
9.5 cm
b
a
65
C
Remember that
the angles of a
triangle add up
to 180o.
Eg
We can use the sign rule to find the length of the side
c because we know the length a and angles A and C
9.5 cm
c

sin 55 sin 65
Multiplying each side by sin 65, we find
9.5 cm  sin 65
= 10.5 cm
c
sin 55
Angle B = 180o-55o – 65o = 60o.
Using the sine rule again,
9.5 cm
b
.

sin 55 sin 60
Multiplying each side by sin 60, we find
9.5 cm  sin 60
= 10.0 cm
b
sin 55
Example 3: Find the angle A in this triangle.
A
12 cm
b
c
a
7 cm
B
55
40 C
Using the sine rule
a
c
, we can write

sin A sin C
7
12
.

sin A sin 40
Multiplying both sides by sin A, we find,
12  sin A
7.
sin 40
Multiplying both sides by sin 40, we find,
12  sin A  7  sin 40 .
Dividing both sides by 12, we find,
7  sin 40
sin A 
 0.375 , so
12
A = sin-1(0.375) = 22o.

The cosine rule is used to find a side of a triangle when you are given two sides
and the angle between them, or to find an angle when you know the length of all
three sides.
A
b
c
C
a
B
Key Point:
The cosine rule can be written
a2 = b2 + c2 –2bc cos A .
56
Eg
Example 4: In the triangle ABC, c = 10 cm, b = 12 cm, angle A = 20o.
Find the length of side b to 2 decimal places.
C
12 cm
B
20
10 cm
A
Using the cosine rule a 2  b 2  c 2  2bc cos A ,
we can write
a 2  12 2  10 2  (2  12  10  cos 20 o ) ,
a  144  100  (240  0.94) ,
a = 4.30 to 2 d.p.
Summary in Words

In a right-angled triangle the sides and angles are related by three trigonometric
ratios. The sine of an angle is defined as the ratio of the length of the side opposite
to the angle to the length of the hypotenuse, the cosine of an angle is defined as the
ratio of the length of the side adjacent to the angle to the length of the hypotenuse
and the tangent of an angle is defined as the ratio of the length of the side opposite
to the angle to the length of the side adjacent to the angle. These ratios can be used
to work out the length of the sides and the angles in a right-angled triangle.

The sine and cosine rule relate the lengths of the sides and the angles in any
triangle. They can be used to find unknown lengths of sides and angles in nonright-angled triangles. The sine rule is used when two of the angles and the length
of one side is known, or the lengths of two sides and the angle opposite one of these
sides is known. The cosine rule is used to find a side of a triangle when you are
given two sides and the angle between them, or to find an angle when you know the
length of all three sides.
What should I do now?

Learn the definitions of the trigonometric ratios and the sine and cosine rule
formulae. Make sure you understand how to apply them and which triangles they
57
can be used for. Make sure you understand the convention for naming the different
sides of a right-angled triangle so that your trigonometric ratios are correct.

Try the examples below that test your application of the methods.

Try the Geoscience examples so you can see how to apply these methods in your
subject.
Exercises - Methods
1. Write out the formulae for the trigonometric ratios sine, cosine and tangent of a
right-angled triangle.
2. Given the angle and length of one side of a right-angled triangle, calculate the
length of the required side, showing all working:
(a) 37o, hypotenuse=14 cm, find the length of the opposite side;
(b) 45o, hypotenuse=10 km, find the length of the adjacent side;
(c) 26o, opposite=105 m, find the length of the hypotenuse;
(d) 18o, adjacent=12 mm, find the length of the opposite side.
3. Under what circumstances can the sine rule be used to relate the angles and
lengths of a triangle?
4. Find the missing sides and angles in the triangles below.
C
b
a
112
33
B
18
A
C
15
7
27
A
B
c
5. Under what circumstances can the cosine rule be used to relate the angles and
lengths of a triangle?
6. Find the missing sides and angles in the triangles below.
58
C
b
6
98
B
5
A
C
11
8
A
B
7
Geoscience Examples
1. The angle of repose of a pile of unconsolidated sediment is a measure of the
friction between individual sediment particles. The angle of repose is the angle
between the surface of the sediment and the horizontal. If the angle of repose of
a quartz sand is 31o and the diameter of the pile is 17 m, what is the height of
the centre of the pile?
2. The slope of layers, or beds, of rock can be stated as the dip of the rocks, in
which the dip is defined as the angle between the slope and the horizontal plane.
On a geological map, which shows the distribution of different rock units at the
surface of the Earth, the gradient of the slope of a rock unit can be determined
and from this the dip of the bed can be calculated.
If the gradient of dipping beds is 1 in 3, what is the dip?
3. If a borehole is drilled down though a sequence of rocks, the vertical thickness
of each unit, or bed, can be measured in the core that is extracted. This is not
the actual true thickness of the unit (which is measured perpendicular to the top
and bottom surfaces of the unit), unless the beds are exactly horizontal (i.e. with
zero dip). The true thickness of each unit can be calculated from its vertical
thickness if the dip of the beds is known by measuring in the field with an
instrument called a compass-clinometer.
If the vertical thickness of a unit, which dips at 47o, as measured from a
borehole core is 23m, what is its true thickness?
59
4. A carbonate building block is a flat, planar structure, consisting of a carbon
atom surrounded in a triangle by three oxygen atoms. Part of this structure is
shown below.
C
5.4x10-10 m
O
S
O
If the angle between between adjacent carbon oxygen bonds is 120o and the
length of these bonds is 5.4 x 10-10 m, what is the separation of each oxygen
atom (the length S on the figure)?
60
10. Calculating Systematic Errors
Learning
Summary:
After studying this section, you will be able to:
 understand the terms random error, systematic error, absolute
error and fractional error
 determine the systematic error in a quantity which is made up
from the sum or difference of two quantitiesthat have an error
associated with each of them.
 determine the systematic error in a quantity which is made up
from the product or quotient of two quantities that have an error
associated with each of them.
Error as uncertainty

In science, the term error means uncertainty. For example, the experimental
error in making a measurement is the uncertainty that exists in that
measurement which you cannot avoid even by being very careful. This
uncertainty may arise because of the limit of the sensitivity or the setup of the
instrument used to make the measurement, the capability of the scientist, the
environment in which the measurement is made, and other similar factors.

A random error can be caused by:
lack of perfection in the scientist;
the sensitivity of the measuring instrument;
changes in the environment in which the measurement is made.
Random errors are revealed by repeated measurement of a particular quantity.

A systematic error can result from:
an instrument having a zero error, that is, it is not set up correctly prior to use;
an instrument being incorrectly calibrated;
the scientist persistently making the measurement at the incorrect time or
position.
Systematic errors
We write quantities as the combination of a ‘best measurement’ and an absolute error.
The height of an object would be written h  h  h , where h is the best measurement
and h is the absolute error in h. The sign ‘  ’ is read as ‘plus or minus’ and so when
we write h  h  h , this is equivalent to writing h  h  h  h  h , so the value of
h lies between h  h  h and h  h .
The fractional error in h is the absolute error in h divided by the best measurement, so
h
we write f h 
.
h
The percentage error in h is the fractional error multiplied by 100%, so we write
h
ph 
 100 .
h
61
Combining systematic errors
Quantities are related to other quantities by formulas. Here are some examples.
V  V1  V2
p  p2  p2
u
1 2
gt
2
v
Q
A
In each example, the value of the quantity on the left hand side of the equals sign
depends on the values of the quantities on the right hand side of the equals sign.
Because these have errors associated with them, we need to know how to combine
these errors to estimate the error that these introduce into the value of the quantity on
the left hand side of these equations.
How do we combine the errors?
In the first equation, the volume V is related to the volumes V1 and V2 by
V  V1  V2 .
If V1 and V2 were measured subject to some error such that
V1  V1  V1 and V2  V2  V2 , where the overbar denotes the best measurement, and
 denotes the absolute error, then we want to find V, the absolute error in V. If we
write the formula for V out in full,
V  V  V1  V1  V2  V2 .
The largest possible value of V will include the largest value of the systematic error V.
The largest value of V is obtained by adding the quantities on the right hand side of the
equation above, so
V  V  V1  V1  V2  V2 .
If we group the best measurements and the absolute errors together, we find
V  V  (V1  V2 )  (V1  V2 ) .
If we equate the value of the absolute error of the right and left hand sides, we find
V  V1  V2 .
Key Point:
When you add two quantities that each have an absolute
error, the resulting error is equal to the sum of the
absolute errors, so you add the absolute errors.
62
Eg
Example 1: Find the systematic error in p = p1 - p2 if p1 = 10000  1000
Pa and p2 = 9000  500 Pa.
Two minuses
make a plus !
Key Point:
p  p1  p2 .
Writing this with overbars denoting the best measurement
and delta denoting the absolute error,
p  p  ( p1  p1 )  ( p2  p2 ).
The largest possible of p occurs when the left hand side
of this equation has its largest value, which is when
the right hand side of this equation has its largest value.
The largest value of the right hand side occurs when
p  p  ( p1  p1 )  ( p2  p2 ),
so when we expand the brackets
p  p  p1  p1  p2  p2 .
If we equate the absolute errors on each side, we find
p  p1  p2 .
So if we substitute in the values from the example, the
systematic error in p is the sum of the absolute errors
p = 1000 + 500 = 1500 Pa.
When you subtract two quantities that each have an absolute
error, the resulting error is equal to the sum of the
absolute errors, so you add the absolute errors.
63
Eg
Example 2: Calculate the systematic error in F = ma.
Do the same
thing to both
Writing the formula with overbars denoting the best
measurement and delta denoting the absolute error,
F  F  (m  m)(a  a) .
The largest possible value of F is given by
F  F  (m  m)(a  a) .
Multiplying out the brackets,
F  F  m a  m a  a m  ma .
In scientific measurements, the absolute errors should
be small compared to the ‘best measurements’. This
means that ma will be two small quantities
multiplied together, which gives a very small product,
which can be neglected compared to the other terms on
the right hand side,
F  F  m a  m a  a m .
Dividing both sides by m a ,
F
F m a m a a m




.
ma ma ma ma
ma
sides
Canceling like terms in each fraction,
F
F
a m

 1

.
ma ma
a
m
If we equate the ‘best measurements’ in the first equation
in this box, we can see that F  m a , so
F F
a m

 1

, or
F
F
a
m
F
a m
.
1
 1

F
a
m
Subtracting 1 from each side, we find
F a m
.


F
a
m
Key Point:
When you multiply two quantities that each have an absolute
error, the resulting fractional error is equal to the sum of the
fractional errors, so you add the fractional errors.
64
Eg
1 2
gt if the fractional
2
error in g is 0.01 and the fractional error in t is 0.05, and the best
measurement of u is 10 m s-1.
Example 3: Calculate the systematic error in u 
In the formula, u is found from the product of g and t2.
The fractional error in u is equal to the sum of the
fractional errors in g and t2.
t2 = t  t, so the fractional error in t2 is equal to the sum of
the fractional errors in t and t.
The fractional error in t is 0.05, so the fractional error in
t2 is 0.05 + 0.05 = 0.1.
The fractional error in g is 0.01, so the fractional error in
g t2 is 0.01+0.1 = 0.11.
The fractional error in u is defined as
u
,
fu 
u
so we can calculate the absolute error in u. Multiplying
each side of the equation by u , we find
u f u  u ,
and if we substitute for the ‘best measurement’ and the
fractional error, we find
u = 10 m s-1  0.11 = 1.1 m s-1.
65
Eg
Example 4: Calculate the systematic error in u 
Do the same
thing to both
Q
.
A
Writing the formula with overbars denoting the best
measurement and delta denoting the absolute error,
(Q  Q)
u  u 
.
( A  A)
The largest possible value of u is given by
(Q  Q)
u  u 
.
( A  A)
Multiplying each side by ( A  A) ,
(u  u)( A  A)  Q  Q , and multiplying out the
brackets,
u A  u A  A u  uA  Q  Q .
In scientific measurements, the absolute errors should
be small compared to the ‘best measurements’. This
means that ma will be two small quantities
multiplied together, which gives a very small product,
that can be neglected compared to the other terms on the
right hand side,
u A  u A  A u  Q  Q .
Dividing both sides by u A ,
u A u A A u Q Q




.
uA uA
uA
uA uA
sides
Canceling like terms in each fraction,
A u Q Q
1



.
A
u
uA uA
If we equate the ‘best measurements’ in the first equation
Q
in this box, we can see that u  , so Q  u A , so we
A
can replace u A by Q in the equation,
A u Q Q

 
, or
A
u
Q Q
A u
Q
1

 1
.
A
u
Q
Subtracting 1 from each side, we find
A u Q


.
A
u
Q
1
Key Point:
When you divide two quantities that each have an absolute
error, the resulting fractional error is equal to the sum of the
fractional errors, so you add the fractional errors.
66
Summary in Words

The fractional error is equal to the absolute error divided by the ‘best
measurement’.

If you are adding two quantities, the absolute error in the result is equal to the sum
of the absolute error in each quantity.

If you are subtracting two quantities, the absolute error in the result is equal to the
sum of the absolute error in each quantity.

If you are multiplying two quantities, the fractional error in the result is equal to the
sum of the fractional error in each quantity.

If you are dividing two quantities, the fractional error in the result is equal to the
sum of the fractional error in each quantity.
What should I do now?

Learn the rules for calculating the error in a quantity that is made up from a
combination of other quantities. Make sure you understand the terminology of
systematic error including definitions of absolute and fractional error.
Exercises - Methods
1. Convert these descriptions of measurements into mathematical statements:
(a)
(b)
(c)
(d)
A pressure of 10 GPa with an absolute error of 0.01 Gpa;
A length of 10 nm with a fractional error of 10%;
A volume of 1.47 m3 with a fractional error of 5%;
A mass of 6 x 1023 kg with an absolute error of 1 x 1020 kg.
2. Convert these absolute errors to fractional errors:
(a)
(b)
(c)
(d)
3  0.04 mm
101325  200 Pa
2 x 105  1000 km.
18.7  0.00001 g.
3. Convert these fractional errors to absolute errors:
(a)
(b)
(c)
(d)
3 mm  0.07%
101325 Pa  18%
2 x 105 km  1.5%
18.7 g  0.0003%
4. If a mass of 5 kg can be measured to an absolute error of  0.001 kg, determine
the absolute error involved in the following calculations:
(a) The mass is added to another mass of 5 kg measured with the same
systematic error;
(b) The mass is multiplied by a velocity of 10 m s-1  0.1 m s-1;
(c) The mass is subtracted from a mass of 7.2 kg  0.01 kg;
67
(d) The mass is divided by a volume of 0.002 m3  0.00004 m3;
(e) This mass is used in the formula d = Km2, where K has the value of 2.3.
What is the absolute error in d?
Geoscience Examples
1. The World Health Organisation estimated that in 1999 there were 4039000
deaths from acute respiratory infections (ARI), 1086000 deaths from malaria
and 2673000 deaths from HIV/AIDS, worldwide. If the absolute error in these
figures were  500 in each case, determine the absolute error in the total of the
deaths due to malaria and HIV/AIDS. Determine the fractional error in the ratio
of deaths due to malaria to deaths due to ARI, and in the ratio of deaths due to
HIV/AIDS to deaths due to ARI.
(More information on how to write down equations from descriptions in words
can be found in section 3, p 21 of this booklet)
68
11. Answers to Exercises and Geoscience Examples
Section 1
Dimensions and Units
Exercises
1.
2.
3.
4.
5.
6.
Length, mass, time and temperature
M L-3
M L T-2
L T-1
[e] = M L2 T-2, [V] = L3
If the formula is correct, the dimensions of each side and each term will be the
same.
The dimensions of each term of equation (a) can be written
L = L T-2 x L + L T-1 x T2 = L2 T-2 + L T.
The dimension of each term of equation (b) can be written
L = L T-1 x T + L T-2 x T2 = L + L . Equation (b) is correct.
7. 1.01 x 10-2, 2.357 x 104, 9.9 x 109, 1.87 x 10-1, 3.1415…
8. 16 kg, 0.247 m, 2.5 x 1010 W, 1 x 10-7 m2, 2.05 m s-1, 2 x 106 m3.
9. 10-1, 8-2, 625, 1/6, 237, 1, 3.
10. 3.85 x 10-4 m3
11. 2.70 x 108 kg m-1 s-2
Geoscience Examples
1.
2.
3.
1.4 x 1018 m3, 1.4 x 1021 kg, ice, 4.34 x 1019 kg, 32.3
M L-1 T-1, dimensionless
Boxing Day, 7 minutes to midnight on New Year’s Eve, 14 seconds to midnight
on New Year’s Eve.
Section 2
Powers
Exercises
1.
2.
3.
4.
16384, 625, 53/5, 1
b-8, b3, c-9, b13.23, c-10/3 , 1+b3, 1
5
104, 109/2, 52 = 25.
Geoscience Examples
1.
2.
10-3.55, kg/kg 10-3.44kg/kg, 2.11%
2 x 10-2 m/yr, 50 yrs, 1.7 x 106 yrs
69
Section 3
Making and Using Equations
Exercises
3.
4.
5.
6.
b2 b 2 z
3
 ,z  
6 2
10 10
1
1
1
1
1
2(x+3),
( y  1)( y  ),2a(2a 2  3), (3c  a), 2 (1  )
44
2
8
x
x
2
( z  8)( z  7), (q  9)(q  5), ( x  3)
x = -2 or x = -3, a = 1 or a = 2, b = 8 or b = -5
11, 1, 30
3/8
7.
uk
8.
x
1.
2.
x3 + x, a2-b2, 3a+3b-xa-xb,
h2

3
y  5 , x = 22
2
Geoscience Examples
1.
2.
3.
3.31 x 10-5
V
h
 k 
A
L
Ratios – O: 1.53, Si: 1.87, Fe: 0.17
Masses – O: 1.75 x 1024 kg, Si: 8.75 x 1023 kg, Fe: 2.04 x 1024 kg
Section 4
Solving Equations
Exercises
1.
2.
3.
4.
3
29
2
250
,t 
,t  , d 
, d  14
2
3
3
12
37  42
,
,3,1,2
4
5
3
40
16
33
3
25
x   , y  0; a 
,b  ;a  ,b  ,c 
2
3
3
14
2
14
x  2, x  1; c  1; z  3  2 , z  3  2; x  0.623, x  9.623
x
Geoscience Examples
1.
(a)  L   s 
9v
2 gr 2
2 gr 2
( s   L )
9v
(   L )
L
(c) 1  s1
L2 (  s 2   L )
(b)  
70
(d) Re-arrange the answer to (c) to give  L 
find L = 2709 kg m-3
 L1
 L2
 s1   s 2 
L
1   1
 L2





 and substitute to
(e) 50 kg m-1 s-1
(f) 5.00 x 10-4 m s-1
2.
51.3 m s-1, 1.95 kg m-1 s-1
3.
Only the positive root of this equation has any meaning, so [H+] = 4.148 x 10-7,
pH = 6.38
4.
7.27 m s-1
5.
xA = 0.51, T = 1251oC
Section 5
Proportionality and Inequality
Exercises
1.
2.
k
, y  kx3 , z  3kx, d  kz 4
x
3
3
z < 5, x  , y  , (t  2)(t  3)  0
8
5
y
Geoscience Examples
1.
2.
10000, 2 x 106
xA > 0.43
Section 6
Gradients of Straight Lines and Curves
Exercises
1.
The gradients and intercepts for the straight lines are, respectively: 2, 3; -5/2, 4;
7/3, -2; 5 x 10-4, -1.8 x 10-5
Geoscience Examples
1.
2.
The graph for the first part of this question is shown on p.37. The gradient of the
graph is 5 x 10-4 Ma m-1
The equation of the straight line is y = -2.86x +23, so the gradient = -2.86. The
number of fragments per cm3 2.4 m below the top of the bed is 16.
71
Section 7
The Exponential Function and Logarithms
Exercises
1.
y is equal to the exponential function of x; r is equal to the exponential function of
t; the logarithm of 1000 to the base 10 is 3; the logarithm of 7 to the base 7 is 1;
the natural logarithm of 285 is 5.652; the natural logarithm of 35 is 3.555; the
logarithm of 98 to the base 10 is 1.991.
2.
2 + log10 b, 1 + log8 c, 1 – log z
3.
log a = 18; log5 z = 0
4.
y = e3t
y = e3x-1
ln y = 3t ln y =3x-1
t = (ln y)/3
x = (ln y +1)/3
5.
y = 0.378
Geoscience Examples
1.
2.
7.78 billion, 2018
1005 years
Section 8
Triangles
Exercises
1.
2.
98o
74o
3.
h
4.
h
5.
6.
7.
197 m
24 km
7.28, 604
0.23  0.13  0.26 m
2304  1012  2516 km
2
2
2
2
Geoscience Examples
1.
2.
78o
582 m
Section 9
Trigonometric Ratios
Exercises
1.
2.
3.
4.
The formulae are given on p.54
8.43 cm, 7.07 km, 240 m, 3.9 mm
The sine rule is used when two of the angles and the length of one side is known,
or the lengths of two sides and the angle opposite one of these sides is known.
a = 17.1, b = 29.1, C = 35o; B = 76.6o, C = 76.4o, c = 15
72
5.
6.
The cosine rule is used to find a side of a triangle when you are given two sides
and the angle between them, or to find an angle when you know the length of all
three sides.
A = 45.7o, b = 8.31, C = 36.3o; A = 46.5o, B = 94.1o, C = 39.4o
Geoscience Examples
1.
2.
3.
4.
10.2 m
18.26o
15.7 m
9.35 x 10-10 m
Section 10
Calculating Systematic Errors
Exercises
1.
(a) 10  0.01 GPa
(b) 10  1 nm
(c) 1.47  0.0735 m3
(d) (6  0.001) x 1023 kg
2.
(a) 3 mm  1.3%
(b) 101325 Pa  0.2%
(c) 2 x 105 km  0.5%
(d) 18.7 g  5.3 x 10-5%
3.
(a) 3  0.0021 mm
(b) 101325  18239 Pa
(c) 2 x 105  3000 km
(d) 18.7  0.000056 g
4.
(a)
(b)
(c)
(d)
(e)
 0.002 kg
 0.51 kg m s-1
 0.009 kg
 50.5 kg m-3
 0.023 kg
Geoscience Examples
1.
 1000,  0.065%,  0.031%
73