1. Exponential Modelling and Curve Fitting

2. 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.

3. The most usual curves that real life situations can be modelled by are:
Linear
Exponential
Power Functions

4. Revision on linear graph and log
Y-intercept
Gradient
lnA·B=lnA+lnB
lnA·ekx =lnA+lnekx
lnex=x
lnekx =kx
elny =y

5. Exponential Graphs The exponential model applies in these situations
Exponential Graphs have an x in the power
The exponential model applies in these situations
Investment
Economic Growth or Decline
Population Growth or Decline
Radio Active Decay
Cooling
etc

6. x y
0.5 1
2.24 2
10.04 3
45.00 4
201.71 5
904.02
Example It is known that the data form an exponential graph Find out the equation for the model So we use the model y= Aekx

7. Now let’s look at the log versionx y
lny
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
ln y
We have now converted the exponential relationship to a linear one x

8. 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

9. Exponential modelling
The linear lny and x graph is in the form of
elny= ekx+lnA
y=ekxelnA
y=ekxA

10. Practical Applications
Population of fruit flies
What’s the weekly increase
Rate?
Time
(weeks)
No of fruit flies
ln (No of flies) 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

11. We can see this is an exponential graph

12. This is the log version of the graph

13. y=50×1.284x y=50×(1+0.284)x The rate is 28.4% Y intercept is 3.91
Gradient
Time
(weeks)
No of fruit flies
ln (No of flies) 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=50×1.284x
y=50×( )x
The rate is 28.4%

14. Using the calculator to find the equation

15. 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?

16. Power Function

17. Power Curve Modelling x y 1 3 2 4.24 5.20 4 6 5 6.71
Sometimes we have a power function rather than an exponential function. Eg x y 1 3 2
4.24
5.20 4 6 5
6.71

18. Using the calculator to find power functions

19. Power Curve Modelling lnx x y lny 1 3 1.10 0.693 2 4.24 5.20 1.44 1.39
In this case if we plot lnx vs lny we will get a straight line Eg
lnx x y
lny 1 3
1.10
0.693 2
4.24
5.20
1.44
1.39 4 6
1.79
1.61 5
6.71
1.9
7.35
1.99

20. Not a linear relationship

21. This relationship is now a linear one

22. 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

23. Mathematical Modelling
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

24. 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
Week(x) 1 2 3 4 5
Price(y) cents
120
44.1
16.2
6.0
2.2

25. Graphed Data

26. Exponential trend line
Spread sheet will give us the equation
Good fit
r=?

27. Power trend line r=? Spread sheet will give us the equation
Not so good
r=?

28. Week(x) 1 2 3 4 5
Price(y) cents
120
44.1
16.2
6.0
2.2

29. Week(x) 1 2 3 4 5
Price(y) cents
120
44.1
16.2
6.0
2.2

30. 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.
The data is for the average weekly
Share price of Air New Zealand shares over a
5 week period.
Week(x) 1 2 3 4 5
Price(y) cents
120
44.1
16.2
6.0
2.2

31. You need to plot the raw data a) correct x and y scale
Week(x) 1 2 3 4 5
Price(y) cents
120
44.1
16.2
6.0
2.2
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

32. Exponential trend line
Good fit

33. Power trend line
Not so good

34. 2) Test for the model
Using graphics calculator:
Exponential: A= 325.6 k=
r2 =
Power: a=164.86 n=
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 x

35. 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.

36. When y=10 go to equation Solver 10=325.6×e(-0.9993x)
Week(x) 1 2 3 4 5
Price(y) cents
120
44.1
16.2
6.0
2.2
y=325.6e x
When x=2.5 y=325.6×e( ×2.5) =26.774
When y=10 go to equation Solver
10=325.6×e( x)
Press exe twice to get the answer!!
x=3.5
y=164.86x2.4201
When x=6 y=164.86× =2.15
10=164.86x Press exe twice to get the answer!!
x=3.18

37. 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.

38. Check closeness of fit by using x=4 y=325.6×e(-0.9993×4) =5.98
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( ×4) =5.98
This value is only very slightly lower than the observed value of 6.0.
Using x=1
y=325.6×e( ×1) =119.8
This value is also only very slightly lower than the observed value of 120.
The model y=325.6e x
Gives similar results to the observed values.

39. 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.

40. Decide which model is the best fit. 3. Make predictions:
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
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.

41. Excellence Questions for 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=27.4 = 118.7cm
This is slightly lower than the observed value of 121.1cm.

42. 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.

43. lny=nlnx+lna lny=lnxn + lna elny= elnA+kx lny=lnxn · a y= axn
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
Log-log for power
Semi-log for exponential
lny=nlnx+lna
lny=lnxn + lna
lny=lnxn · a
y= axn
elny= elnA+kx
y=elnAekx
y=Aekx

44. 4) Identify limitations of experiment and state how you could improve it.
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.

45. The linear graph for lnx and lny is in the form of
Power function

46. Exponential function elny= ekx+lnA y=ekxelnA y=ekxA
The linear lny and x graph is in the form of
elny= ekx+lnA
y=ekxelnA
y=ekxA
Exponential function

47. 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

48. Decide which model is the best fit.
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=
b=0.1397
r2 =0.9077
Power: a= b=
r2=
The power model has the higher r2 value
The power graph follows more closely to the gathered data.
The best fit equation is y= e0.6987x

49. 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.

50. Excellence Questions for 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=27.4 = 118.7cm
This is slightly lower than the observed value of 121.1cm.

51. 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.

52. lny=nlnx+lna lny=lnxn + lna elny= elnA+kx lny=lnxn · a y= axn
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
Log-log for power
Semi-log for exponential
lny=nlnx+lna
lny=lnxn + lna
lny=lnxn · a
y= axn
elny= elnA+kx
y=elnAekx
y=Aekx

53. 4) Identify limitations of experiment and state how you could improve it.
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.