Exponential Modelling and Curve Fitting

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 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. ExampleIt is known that the data form an exponential graphFind out the equation forthe modelSo we use the modely= Aekx

7. Now let’s look at the log version 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?

11. We can see this is an exponential graph

12. This is the log version of the graph

13. Y intercept is 3.91 Gradient y=50×1.284x y=50×(1+0.284)x The rate is 28.4%

14. Using the calculator to find the equation

15. The data can be modelled by exponential equations • Plot the raw data • What is the exponential equation?

16. Power Function

17. Power Curve Modelling Sometimes we have a power function rather than an exponential function. Eg

18. Using the calculator to find power functions

19. Power Curve Modelling In this case if we plot lnx vs lny we will get a straight line Eg

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

25. Graphed Data

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

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

30. In the exam you are required to test bothy=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.

31. 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=-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

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. 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

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 • This 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.

39. Graph your raw data 2. Test bothy=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. Graph your raw data 2. Test bothy=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 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=27.450.912 = 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. 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. 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.

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

46. 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. 2) Test bothy=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 =0.9077 Power: 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

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 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=27.450.912 = 118.7cm This is slightly lower than the observed value of 121.1cm.