Download Topic guide 3.2: Processing data using numerical analysis

Unit 3: Analysis of scientific data and information
.
32
Link
Learners unfamiliar with standard
algebraic methods may wish to
study M/502/5009 Unit 15: Using
Mathematical Tools for Science from
Edexcel BTEC Level 2 in Applied
Science and A/502/5546 Unit 7:
Mathematical Calculations for Science
from Edexcel BTEC Level 3 in Applied
Science. Alternatively, brief revision
notes can be found at the following
websites:
Processing data using
numerical analysis
Data can be analysed for two aspects: patterns and correlation. The former
covers maximum/minimum points, gradients and rates of change, whereas
the latter is statistical analysis.
On successful completion of this topic you will:
•• be able to process data using numerical analysis (LO2).
To achieve a Pass in this unit you need to show that you can:
•• perform numerical analysis on scientific data using an algebraic
method (2.1)
•• demonstrate numerical analysis using calculus on standard polynomial
equations (2.2)
•• evaluate absolute errors in scientific data (2.3).
http://www.bbc.co.uk/schools/
gcsebitesize/maths/algebra/
simultaneoushirev1.shtml
http://www.bbc.co.uk/schools/
gcsebitesize/maths/algebra/
quadequationshirev1.shtml
http://www.bbc.co.uk/schools/
gcsebitesize/maths/algebra/
graphsrev1.shtml.
1
Unit 3: Analysis of scientific data and information
1 Analysis using algebraic methods
Once the equation of a trend line has been ascertained, it may need to be
manipulated, for example to determine the values of the equation at specific
points or to calculate the exact intersection between two trend lines.
In Topic guide 3.1, for example, the trend line was determined for results from an
experiment measuring plant root mass under different levels of incident light.
The experiment could have been repeated for a different plant that experienced
reduced growth at higher light levels. Finding the intersection point between the
two lines would provide an optimal light level that would suit both plants.
Figure 3.2.1: The graph on the left shows
linear trend lines representing the change
in root mass over a certain time in plants
exposed to different levels of light (the
red line represents a plant that gains
mass at higher levels of light, as shown in
Topic 3.1, whereas the blue line represents
a plant that loses mass at higher levels
of light). The graph on the right shows
the intersections between linear and
quadratic (parabolic) trend lines.
Root
mass
Root
mass
Light intensity
Light intensity
Intersecting linear plots can be solved easily using simultaneous linear equations.
We are given the following details for the left-hand graph in Figure 3.2.1:
y = –3x + 32 y = 2.5x – 5
Simultaneous equations are generally solved by two methods (elimination or
subtraction) but here we have two equations that are both equal to the same
variable y. Therefore, to solve them, we can equate the two equations, as shown:
–3x + 32 = 2.5x – 5
Terms with variables are grouped on one side, constant terms on the other:
–3x – 2.5x = –5 – 32
–5.5x = –37
x = –37
= 6.727
–5.5
This value for x is substituted back into either one of the original equations to find
the value for y:
y = 2.5x – 5 = 2.5(6.727) – 5 = 11.8175
The solution, and therefore the intersection point, is at (6.73, 11.82) rounded to
two decimal places.
If one line is quadratic then the intersection points can still be found by equating
the two trend lines. We are given the following details for the graph on the right in
Figure 3.2.1:
y = 0.05x2 + 0.9x + 1 y = 1.5x + 1
0.05x2 + 0.9x + 1 = 1.5x + 1
0.05x2 – 0.6x = 0
3.2: Processing data using numerical analysis
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Unit 3: Analysis of scientific data and information
This equation can then be factorised to:
x(0.05x – 0.6) = 0
Thus the solutions are:
x = 0 0.05x – 0.6 = 0
x = 0.6
= 12
0.05
Therefore the two trend lines intersect at x = 0 and x = 12.
Take it further
The following website provides an interactive presentation on some applications of
simultaneous equations and algebraic methods in chemistry:
http://mathsforchemistry.info/wordpress/index.php/2010/01/simultaneous-equationsin-chemistry/.
2 Analysis using calculus
Key terms
Calculus: The mathematical analysis
of change.
Differential calculus: The
mathematical analysis of the rate of
change, by studying the slopes of
curves.
Figure 3.2.2: Growth of algae over time
as an example of a non-linear relationship
between variables (in this case algal cell
number (y-axis) and time (x-axis)). Circles
show zero growth (trough) and fastest
growth rate (steepest linear slope).
There are plenty of examples in science where the relationship between two
variables is non-linear: in chemistry, equilibrium points between solutions; in
biology, the size of algal colonies in variable temperatures. Collected data from
such investigations, when plotted, may resemble the graph in Figure 3.2.2.
The circled regions mark areas of specific interest common to many scientific
investigations such as a minimum point (e.g. zero growth rate) and the steepest
positive gradient (fastest growth rate). Although this could be determined
manually from the graph, it is potentially quicker and certainly far more accurate
to do this mathematically using calculus.
Algal cell number (log scale)
10 000
1000
100
10
1
0
Link
Learners unfamiliar with basic
calculus may wish to study
A/502/5546 Unit 7: Mathematical
Calculations for Science from Edexcel
BTEC Level 3 in Applied Science.
0
1
2
3
4
5
Time (days)
Trend line equations can be analysed for gradient and maximum/minimum points
using differential calculus. This is a branch of mathematics that examines how
variables in an equation change with respect to each other. The key to this process
is determining what are called the first and second derivatives of the equation.
In this topic guide, only simple polynomial, natural exponential and natural
logarithmic equations will be examined, as these are the most common equations
for a trend line (Table 3.2.1).
3.2: Processing data using numerical analysis
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Unit 3: Analysis of scientific data and information
Table 3.2.1: The most common equations
for trend lines describing linear and nonlinear relationships between variables,
including first and second derivatives.
Standard derivatives (a,b are numbers, x,y are variables)
Equation
Linear
First derivative
Second derivative
y = ax + b
dy
=a
dx
d2y
=0
dx2
y = axb
dy
= abxb–1
dx
Polynomial
d2y
= (ab)(b – 1)xb–2
dx2
y = axb + cx
dy
= abxb–1 + c
dx
Natural exponential
y = aebx
dy
= abebx
dx
d2y
= ab2ebx
dx2
Natural logarithmic
y = alogebx
dy a
=
dx x
d2y
a
=– 2
dx2
x
Note that logex is often written as lnx and appears as such on calculators. It is
an abbreviation of the Latin term logarithmus naturalis. Typically logx (i.e. no
indication of the base number) refers to log10x. If you wish to differentiate such a
log, it needs to have the base changed to e:
alog10bx = a
3 logebx
loge10
Thus the first and second derivatives are as shown in Table 3.2.2:
Table 3.2.2: First and second derivatives
of the logarithmic equation (base 10).
Equation
First derivative
Second derivative
Equation
logarithm (base 10)
y = alog10bx
dy a/loge10
a
=
=
dx
x
loge10 3 x
d2y
a
=–
dx2
loge10 3 x2
The first derivative of an equation is the gradient equation. By differentiating an
equation, you can calculate the gradient at any given point.
Example
The trend line equation for the results of an experiment has the form
y = 2x2 – 4x – 5. Determine the gradient of this line at the value x = 1.2.
y = 2x2 – 4x – 5
dy
= 2 3 2 3 x2–1 – 4 = 4x – 4
dx
dy
Gradient at x = 1.2 is = 4(1.2) – 4 = 0.8
dx
Maximum and minimum points are solutions to the first derivative of the equation
when it equals zero; the nature of the equation at this point can be found either
visually (i.e. looking at the trend line near the point) or by using the value of the
second derivative at this point – a negative answer indicates that it is a maximum,
a positive answer indicates that it is a minimum.
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Unit 3: Analysis of scientific data and information
Key term
Integral calculus: The mathematical
analysis of accumulation, by studying
the areas under curves.
Activity
Find the first and second derivatives
for the following equations:
Example
Find the maximum and minimum points in the equation from the previous
example.
y = 2x2 – 4x – 5
dy
= 4x – 4
dx
dy
dy
Set to zero,  = 0 = 4x – 4
dx
dx
0 = 4x – 4
1 y = 2x – 6x – 4.2
x = 1 is the solution
2 y = 1.2x3 + 1.7x2 – 31x – 104
d2y
= 4 – this is a positive value, so the equation is a minimum at x = 1
dx2
2
3 y = 4e1.5x
4 y = –3e6x
5 y = 0.3 loge4x
A related branch, called integral calculus, can be used to calculate the area under
a trend line between two points (Figure 3.2.3).
Figure 3.2.3: Integral calculus is
used to calculate the area under a
trend line (blue) between two points
(here defined by broken red lines).
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
The process (known as definite integration) essentially involves the reverse
steps of differential calculus, and is carried out using the equations shown in
Table 3.2.3.
Table 3.2.3: Standard integrals
for different trend line equations.
Standard integrals (a,b are numbers, x,y are variables)
Equation
Integral
eaxb =
a b+1
x
b+1
linear/polynomial
y = axb
natural exponential
y = aebx
a
eaebx = ebx
b
natural logarithmic
y = aloge bx
ealoge bx = ax loge bx – ax
The area between the x-axis and curve of an equation between two points is the
difference of the two definite integrals at each point.
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Unit 3: Analysis of scientific data and information
Example
Find the area under the curve y = 3loge1.2x between x = 0.2 and x = 1.5.
3loge1.2x = 3xloge1.2x – 3x
3(0.2)loge(1.2 3 0.2) – 3(0.2) = –1.456
3(1.5)loge(1.2 3 1.5) – 3(1.5) = –1.855
Area = –1.855 – (–1.456) = –0.399 units2
It should be noted that several mathematical conventions and notations have been
omitted for simplicity, the most notable of which is the constant of integration.
When integrating an equation you must always include an unknown constant
factor, unless you are integrating between known values as shown above.
Although mathematically complex, calculus provides a far more robust method for
calculating rates of reactions, total heat release, peak oxygen output, etc. than doing
so manually from a plot of the results. Technology can be used to perform calculus
using applications such as MathCAD®, Autograph or a graphical display calculator.
Checklist
At the end of this section you should be able to use differential and integral calculus to:
 determine the gradient of a trend line
 calculate the rate of a reaction at a given point
 calculate the maximum/minimum point in a trend line
 calculate the area underneath a curve between two points.
Take it further
A more considered approach to differential and integral calculus for beginners is provided by the
Open University at http://labspace.open.ac.uk/course/view.php?id=7112 and by Paul Dawkins
at http://tutorial.math.lamar.edu/Classes/CalcI/CalcI.aspx (to a higher level of mathematics).
Explore how you can apply the above techniques using online technology, such as WolframAlpha®
(http://www.wolframalpha.com), to an investigation you are currently undertaking.
The BBC’s Bitesize website also covers introductory calculus: http://www.bbc.co.uk/bitesize/
higher/maths/calculus/.
The following website provides an interactive presentation on integrated rate equations in
chemistry: http://mathsforchemistry.info/wordpress/index.php/2010/06/integrated-rateequations-in-kinetics/.
Portfolio activity (2.1, 2.2)
To meet learning outcomes 2.1 and 2.2 for this unit, you should take data from an experiment that
you conducted during your course, preferably one where both the independent and dependent
variables are continuous.
You should then analyse the data, such as determining the linear trend line and its gradient, via a
manual method and then repeat the task using the mathematical methods covered in this guide.
If the trend line is non-linear in form, use software to ascertain the trend line and then, using
calculus, determine the values of any maximum or minimum points, or the point of highest
positive/negative gradient.
Do not forget that gradients typically refer to a rate of some kind – therefore units should be
included with the value.
3.2: Processing data using numerical analysis
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Unit 3: Analysis of scientific data and information
3 Evaluating absolute errors in data
Link
Different types of error and ways of
minimising them are discussed in
Unit 4: Quality assurance and quality
control, Topic guide 4.2.
Key term
Error: Deviation between an actual
value and an observed value or
approximation.
Types of errors
When collecting data from an experiment or other investigation, the measured
values may or may not be the true values that should have been detected. Various
factors influence the readings in such a way that there is a quantifiable uncertainty
in the stated figures; these values are typically known as errors and they impact on
the accuracy and precision of collected data.
•• Systematic errors are due to identifiable sources, such as the accuracy of
a meter.
•• Unsystematic or random errors are caused by unidentifiable or
unpredictable sources, such as fluctuations in temperature or differences in
physical skills and techniques between experimenters.
•• Results that are known to be an error due to carelessness or an accident are
classed as mistakes.
•• Gross errors are unnoticed mistakes that produce results greatly different
from the mean value of the other results.
By testing and calibrating all equipment before (and after) use, as well as
evaluating experimental procedures and theoretical models for inconsistencies,
systematic errors can be minimised.
Unsystematic errors cannot be directly controlled but their influence can be
reduced by training, paying due care and attention, and performing repeated
tests and collating more data; they are ultimately accounted for through
statistical analysis.
Accuracy and precision
Accuracy
Accuracy is the difference between the measured or calculated size of a quantity
and its actual value. It is affected by systematic errors, for example, a poorly
calibrated pH meter consistently reading 0.1 above the true value, or a calculation
that assumes the speed of light in air is the same as that in a vacuum.
Precision
Precision is the size of the variation in repeated measured or calculated values
of a quantity. It is affected by unsystematic errors and mistakes, for example,
mechanical vibrations in a room cause a balance to give results that refuse
to steady; an experimenter takes visual readings from a burette using an
inconsistent method.
Handling errors
Errors in data produced by variables will always have a minimum value (due to
the accuracy of the apparatus) that may or may not be measured easily, but
they will always ‘remain’ during subsequent calculations and thus need to be
handled correctly.
An absolute error is the difference between the exact value and the observed
value. A relative error is the absolute error divided by the magnitude of the exact
3.2: Processing data using numerical analysis
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Unit 3: Analysis of scientific data and information
value. For example, if the exact value is 50 and the observed value is 49.9, the
absolute error is 0.1 and the relative error is 0.1/50 = 0.002 (Table 3.2.4). The
handling of these errors is summarised in Table 3.2.5.
Table 3.2.4: Definitions of errors.
Term
Mathematical relationship
Dx = x – x0
absolute error
dx =
relative error
Table 3.2.5: How to handle errors
in different types of data sets.
Situation
Δx is the size of error
x is the measured value
x0 is the true value
dx is the relative error
What to do...
adding or subtracting variables
Add the absolute error values
e.g. (2.1 ± 0.5) + (1.2 ± 0.1) = 3.3 ± 0.6
multiplying or dividing variables
Add the relative percentage value of the errors
e.g. (2.0 ± 0.5) × (2.0 ± 0.1) = (2.0 ± 25%) 3 (2.0 ± 5%)
Answer is 4.0 ± 30% or 4.0 ± 1.2
equations of the errors (e.g. loge, ex)
The relative value of the error remains the same
e.g. log10(100 ± 0.5) = 2 ± 0.01
Key term
Percentage value of the error:
The relative error expressed per 100,
i.e. as a percentage.
Dx
x
Meaning of the values
Example
The energy released in a chemical reaction is determined by measuring the
change in temperature in 100 ± 1 ml of water over a given period of time. If the
specific heat capacity of water is 4.18 ± 0.01 Jg−1K−1, determine the amount of
energy released from the following information:
Initial temperature of water = 18.1 ± 0.1 °C
Final temperature of water = 23.8 ± 0.1 °C
Change in temperature, DT = (23.8 ± 0.1) – (18.1 ± 0.1) = 5.7 ± 0.2 °C
Equation to be used is Q = mcDT
1
% error in m =
3 100 = 1.00%
100
0.01
% error in c =
3 100 = 0.24%
4.18
0.2
% error in DT =
3 100 = 3.51%
5.7
Compound error in Q = 1 + 0.24 + 3.51 = 4.75%
Energy released, Q = 2.38 ± 0.11 kJ
Activity
A cylindrical block of aluminium has a radius of 3.0 ± 0.1 cm, height of 6.0 ± 0.2 cm and a mass of
460 ± 5 g. Calculate the density of aluminium and the error in the value.
3.2: Processing data using numerical analysis
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Unit 3: Analysis of scientific data and information
Checklist
At the end of this topic guide you should be able to evaluate absolute errors in scientific
data by:
 identifying sources of error
 determining procedures to account for and/or quantify errors
 calculating relative and compound errors.
Further reading
Cockett, M. and Doggett, G. (2003) Maths for Chemists, RSC Publishing
Croft, A. and Davison, R. (2010) Foundation Maths, Prentice Hall
Olive, J. (2003) Maths: A Student’s Survival Guide, CUP
Acknowledgements
The publisher would like to thank the following for their kind permission to reproduce their
photographs:
Corbis: Radius Images
Every effort has been made to trace the copyright holders and we apologise in advance for any
unintentional omissions. We would be pleased to insert the appropriate acknowledgement in any
subsequent edition of this publication.
3.2: Processing data using numerical analysis
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