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Exponential Modelling and Curve Fitting. Mathematical Curves. Sometime it is useful to take data from a real life situation and plot the points on a graph. We then can find a mathematical equation for the curve formed by the points. - PowerPoint PPT Presentation
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Exponential Modelling and Curve Fitting
Mathematical CurvesSometime it is useful to take
data from a real life situation and plot the points on a graph.
We then can find a mathematical equation for the curve formed by the points.
The most usual curves that real life situations can be modelled by are:
Linear
Exponential
Power Functions
y
x
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
y
x
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
y
x
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
– 1
– 1
y mx c
x kxy Ar or y Ae
ky Ax
2 4 – 2
y
x
2
2
4
4
6
6
8
8
10
10
2
2
4
4
– 2
– 2
Revision on linear graph and log
y mx c
Gradient Y-intercept
lnA·B=lnA+lnBlnA·ekx =lnA+lnekx elny =y lnex=x
lnekx =kx
Exponential Graphs
The exponential model applies in these situations
• Investment• Economic Growth or Decline• Population Growth or Decline• Radio Active Decay• Cooling• etc
Exponential Graphs have an x in the power
ExampleIt is known that the data form an exponential graphFind out the equation forthe model
So we use the model
y= Aekx
x y
0 0.5
1 2.24
2 10.04
3 45.00
4 201.71
5 904.02
1 2 3 4 5 10 20 30 40 50
y
x
1
1
2
2
3
3
4
4
5
5
10
10
20
20
30
30
40
40
50
50
Now let’s look at the log version
x y lny
0 0.5 -0.693
1 2.24 0.81
2 10.04 2.31
3 45.00 3.81
4 201.71 5.31
5 904.02 6.81
1 2 3 4 5 6 7 8 9 10 – 1 – 2
y
x
1
1
2
2
3
3
4
4
5
5
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
– 1
– 1
– 2
– 2
We have now converted the exponential relationship to a linear one
ln y
x
How do we know if we are dealing with an Exponential
Function?
• If we have a situation where the graph of lny vs x is a straight line we know we are dealing with an exponential function
Exponential modelling
The linear lny and x graph is in the form of
kxy Ae
kxy Ae
ln lny A kx elny= ekx+lnA
y=ekxelnA
y=ekxA
Practical Applications
Time
(weeks)
No of fruit flies
ln (No of flies)
0 50 3.91
2 80 4.38
4 140 4.94
6 230 5.44
8 360 5.89
10 600 6.40
12 1000 6.91
Population of fruit flies
What’s the weekly increase
Rate?
No of Flies Vs Time in weeks
0
200
400
600
800
1000
1200
0 2 4 6 8 10 12 14
Weeks
No
of
Flie
s
No of Flies Expon. (No of Flies)
We can see this is an
exponential graph
Ln( No of Flies) Vs Time in Weeks
0.00
1.00
2.00
3.00
4.00
5.00
6.00
7.00
8.00
0 2 4 6 8 10 12 14
Weeks
LN
(N
o o
f fl
ies)
Ln(No of Flies) Linear (Ln(No of Flies))
This is the log version of the
graph
Time
(weeks)
No of fruit flies
ln (No of flies)
0 50 3.91
2 80 4.38
4 140 4.94
6 230 5.44
8 360 5.89
10 600 6.40
12 1000 6.91
Y intercept is 3.91
Gradient
1
1
y ym
x x
6.91 3.91
12 03
120.25
ln 0.25 3.91y x 0.25 3.91xy e 0.25 3.91.xy e e
0.25 49.9xy e 0.2550 xy e
ln y mx c
y=50×1.284x
y=50×(1+0.284)x
The rate is 28.4%
Using the calculator to find the equation
In Stats mode EXE
Enter data EXE after
each. F1 Grph
F6 Set
Graph type scatter
X List =List 1
Y List= List 2
Exit
F6 to scroll over
F2 Exp
50
0.25
as before
a
b
0.2550 xy e
X 1 2 3
Y 300 150 75
The data can be modelled by exponential equations
•Plot the raw data
•What is the exponential equation?
Power FunctionPower Function
Power Curve ModellingSometimes we have a power function rather than an exponential function.
Eg
ny ax
x y
1 3
2 4.24
3 5.20
4 6
5 6.71
1 2 3 4 5 1 2 3 4 5 6 7 8
y
x
1
1
2
2
3
3
4
4
5
5
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
0.53y x
Using the calculator to find power functions
In Stats mode EXE
Enter data F1 Grph
F6
F6 Set
Set as before
F6
F3 Pwr
0.5
3
0.5
3
a
b
y x
Power Curve Modelling
In this case if we plot lnx vs lny we will get a straight line
Eg
lnx x y lny
0 1 3 1.10
0.693 2 4.24 0.693
1.10 3 5.20 1.44
1.39 4 6 1.79
1.61 5 6.71 1.9
1.79 6 7.35 1.99
X Vs Lny
0.00
0.50
1.00
1.50
2.00
2.50
0 1 2 3 4 5 6 7 8
x values
ln y
val
ues
Not a linear relationship
Lnx vs Lny
0.00
0.50
1.00
1.50
2.00
2.50
0.00 0.50 1.00 1.50 2.00 2.50
Lnx
Ln
y
This relationship is now a linear
one
How to we know if we are dealing with an Power
Function?• If we have a situation where the graph of
lny vs lnx is a straight line we know we are dealing with an power function
ny ax
Tabulate Values
Plot the data with the independent variable on the x axis and the dependant on the y axis
If the data does not indicate a linear relation, decide whether it is
a. An exponential function
b. A power function
For an exponential function graph x against ln(y)
For a power function graph ln(x) against ln(y)
Draw the line of best fit
Mathematical Modelling
Sometimes either a power function or an exponential function appear to fit. We can use log graphs to choose between them but we need to plot the graph first
20.03#1
In October 1987 there was a major price crash in the NZ sharemarket
This table shows the average weekly share price of Ariadne shares over
a 5 week period following the crash.
EgEx
Week(x) 1 2 3 4 5Price(y) cents
120 44.1 16.2 6.0 2.2
) Draw a graph of lny against x
) Draw a graph of lny against lnx
a
b
Graphed Data
Price vs weeks
0
20
40
60
80
100
120
140
0 1 2 3 4 5 6
weeks
Pri
ce
(c
en
ts)
Price vs weeks
y = 325.6e-0.9993x
0
20
40
60
80
100
120
140
0 1 2 3 4 5 6
weeks
Pri
ce
(c
en
ts)
Exponential trend line
Good fit
Spread sheet will give us the equation
r=?
Price vs weeks
y = 164.86x-2.4201
0
20
40
60
80
100
120
140
160
180
0 1 2 3 4 5 6
weeks
Pri
ce
(c
en
ts)
Power trend line
Not so good
Spread sheet will give us the equation
r=?
lny vs x
0
1
2
3
4
5
6
0 1 2 3 4 5 6
weeks
ln s
ha
re p
ric
eWeek(x) 1 2 3 4 5Price(y) cents 120 44.1 16.2 6.0 2.2
) Draw a graph of lny against xa
2
ln 0.9993 5.7857
1
y x
R
=- +
=
Week(x) 1 2 3 4 5Price(y) cents 120 44.1 16.2 6.0 2.2
lny vs lnx
0
1
2
3
4
5
6
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8
lnx
lny
) Draw a graph of lny against lnxb
2
ln 2.4201ln 5.1051
0.9475
y x
R
=- +
=
In the exam you are required to test both y=axn and y=aekx as models for your data.
You need to record your working and the evidence you used in helping you to decide which model is most
appropriate for your data. Then find the model that best fits your original data.
Week(x) 1 2 3 4 5Price(y) cents
120 44.1 16.2 6.0 2.2
The data is for the average weeklyShare price of Air New Zealand shares over a 5 week period.
1) You need to plot the raw data
a) correct x and y scale
b) correct label for x and y
c) correct title
d) correct unit
Week(x) 1 2 3 4 5
Price(y) cents 120 44.1 16.2 6.0 2.2
Price vs weeks
y = 325.6e-0.9993x
0
20
40
60
80
100
120
140
0 1 2 3 4 5 6
weeks
Pri
ce
(c
en
ts)
Exponential trend line
Good fit
Price vs weeks
y = 164.86x-2.4201
0
20
40
60
80
100
120
140
160
180
0 1 2 3 4 5 6
weeks
Pri
ce
(c
en
ts)
Power trend line
Not so good
2) Test for the model Using graphics calculator:Exponential: A= 325.6 k=-0.9993 r2 =Power: a=164.86 n= -2.4201 r2= • The exponential model has the higher r2 value • and the two graphs of the given equation shows the
exponential to more closely follow the gathered data.
The best fit equation is y=325.6e-0.9993x
You are also required to make predictions for x and y.
You need to choose an x-value for which you don’t have raw data and use your model to predict the corresponding y-value.
You also need to choose a y-value for which you don’t have raw data and use your model to predict the corresponding x-value.
Week(x) 1 2 3 4 5Price(y) cents 120 44.1 16.2 6.0 2.2•y=325.6e-0.9993x
When x=2.5 y=325.6×e(-0.9993×2.5) =26.774
When y=10 go to equation Solver 10=325.6×e(-0.9993x)Press exe twice to get the answer!! x=3.5
•y=164.86x2.4201
When x=6 y=164.86×6 -2.4201=2.15
When y=10 go to equation Solver 10=164.86x -2.4201 Press exe twice to get the answer!! x=3.18
Analysing data
• Checking your model is appropriate by selecting either an x-value and calculate its y-value or a y-value and calculating its x-value.
Week(x) 1 2 3 4 5
Price(y) cents 120 44.1 16.2 6.0 2.2
Check closeness of fit by •using x=4 y=325.6×e(-0.9993×4) =5.98This value is only very slightly lower than the observed value of 6.0.
•Using x=1 y=325.6×e(-0.9993×1) =119.8 This value is also only very slightly lower than the observed value of 120.
•The model y=325.6e-0.9993x
Gives similar results to the observed values.
1. Graph your raw data
2. Test both y=axn and y=aekx as models for your data.(Record working and evidence)
Decide which model is the best fit.
3. Make predictions: a. Choose an x-value for which you do not have raw data
and use your model to predict the corresponding y-value. b. Choose a y-value for which you do not have raw data
and use your model to predict the corresponding x-value.
blocks 1 2 3 4 5 6 7 8 9 10 11 12
Distance 30.7 47.5 67.5 92.1 121.1 149 164.6 186.6 205.1 227.9 243.1 265.5
1. Graph your raw data
2. Test both y=axn and y=aekx as models for your data.(Record working and evidence)
Decide which model is the best fit.
3. Make predictions: a. Choose an x-value for which you do not have raw data
and use your model to predict the corresponding y-value. b. Choose a y-value for which you do not have raw data
and use your model to predict the corresponding x-value.
Excellence Questionsfor model y=27.4x0.912
1) Checking that your model is appropriate by selecting either an x-value and calculating its y-value and calculating its x-value
Answer:
When use 5 blocks, x=5,
y=27.450.912 = 118.7cm
This is slightly lower than the observed value of 121.1cm.
2) Explain how well your model fits the raw data, referring to at least 2 pieces of specific evidence from your graphs, or your calculations.
Answer: • The observed value and theoretical values are
very close, therefore a power model seem to fit well.
• The line of best fit drawn on the calculator passes through almost every point, 8 out of the 12 observed points. It again shows the power model fits the experiment well.
• Plotted on the log-log paper, the line of best fit is straighter than the semi-log paper. Hence, the power model is a better model than exponential.
3) Explain the theory behind the model Power model plotted on the log-log paper was a straight line.
The equation can be derived from
ln lny A kx elny= elnA+kx
y=elnAekx
y=Aekx
Semi-log for exponential
lny=nlnx+lna lny=lnxn + lna lny=lnxn · a y= axn
Log-log for power
4) Identify limitations of experiment and state how you could improve it.
Limitation• The carpet was uneven which would affect the distance
the golf ball would roll, thus affecting the accuracy of the results.
• The bricks used to increase the height of the ramp may not have been equal height.
• The ball might have lost some momentum when it dropped from the ramp to the carpet.
• The higher the blocks are, the more block displacement there is each time the ball was rolled.
Improvement:• Get a ramp that’s got a ground-piece• Place the bricks against a wall to eliminate the
displacement.
ny axln ln ny axln ln ln ny a x ln ln lny a n x
The linear graph for lnx and lny is in the form of
Lnx vs Lny
0.00
0.50
1.00
1.50
2.00
2.50
0.00 0.50 1.00 1.50 2.00 2.50
Lnx
Ln
y
Power function
The linear lny and x graph is in the form of
kxy Ae
ln lny A kx elny= ekx+lnA
y=ekxelnA
y=ekxA
Ln( No of Flies) Vs Time in Weeks
0.00
1.00
2.00
3.00
4.00
5.00
6.00
7.00
8.00
0 2 4 6 8 10 12 14
Weeks
LN
(N
o o
f fl
ies)
Ln(No of Flies) Linear (Ln(No of Flies))
Exponential function
1) You need to plot the raw data
a) correct x and y scale
b) correct label for x and y
c) correct title
d) correct unit
2) Test both y=axn and y=aekx as models for your data.(Record working and evidence)
Decide which model is the best fit.Exponential: a= 49.8491 b=0.1397 r2 =0.9077Power: a=38.6093 b= 0.6987 r2= 0.9973• The power model has the higher r2 value • The power graph follows more closely to the gathered
data.
The best fit equation is y=38.6093e0.6987x
3. Make predictions: a. Choose an x-value for which you do not have raw data
and use your model to predict the corresponding y-value.
b. Choose a y-value for which you do not have raw data
and use your model to predict the corresponding x-value.
Excellence Questionsfor model y=27.4x0.912
1) Checking that your model is appropriate by selecting either an x-value and calculating its y-value and calculating its x-value
Answer:
When use 5 blocks, x=5,
y=27.450.912 = 118.7cm
This is slightly lower than the observed value of 121.1cm.
2) Explain how well your model fits the raw data, referring to at least 2 pieces of specific evidence from your graphs, or your calculations.
Answer: • The observed value and theoretical values are
very close, therefore a power model seem to fit well.
• The line of best fit drawn on the calculator passes through almost every point, 8 out of the 12 observed points. It again shows the power model fits the experiment well.
• Plotted on the log-log paper, the line of best fit is straighter than the semi-log paper. Hence, the power model is a better model than exponential.
3) Explain the theory behind the model Power model plotted on the log-log paper was a straight line.
The equation can be derived from
ln lny A kx elny= elnA+kx
y=elnAekx
y=Aekx
Semi-log for exponential
lny=nlnx+lna lny=lnxn + lna lny=lnxn · a y= axn
Log-log for power
4) Identify limitations of experiment and state how you could improve it.
Limitation• The carpet was uneven which would affect the distance
the golf ball would roll, thus affecting the accuracy of the results.
• The bricks used to increase the height of the ramp may not have been equal height.
• The ball might have lost some momentum when it dropped from the ramp to the carpet.
• The higher the blocks are, the more block displacement there is each time the ball was rolled.
Improvement:• Get a ramp that’s got a ground-piece• Place the bricks against a wall to eliminate the
displacement.