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.

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

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

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

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

Now let’s look at the log version x 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

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 elny= ekx+lnA y=ekxelnA y=ekxA

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

We can see this is an exponential graph

This is the log version of the graph

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×(1+0.284)x The rate is 28.4%

Using the calculator to find the equation

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 Function

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

Using the calculator to find power functions

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

Not a linear relationship

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

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

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

Graphed Data

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

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

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

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

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

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

Exponential trend line Good fit

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.

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-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 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.

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(-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.

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.

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.

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50.912 = 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.

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

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.

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

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

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

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= 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

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 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50.912 = 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.

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

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.