© Nuffield Foundation 2011

Nuffield Free-Standing Mathematics Activity

Gas guzzlers

Gas guzzlers

It is sometimes useful to find a function to model data, then use it to make predictions.

Think about ... How can you use data on the number of cars with large engines produced in previous years to predict the number of cars that will be produced in future years?

YearThousands of cars(engine size 2 litres or more)

1994 1558

1995 1600

1996 1715

1997 1844

1998 1980

1999 2145

2000 2262

2001 2451

2002 2647

2003 2869

2004 3118

2005 3314

2006 3512

2007 3687

Are there more cars with larger engines on the roads today?

Source: www.dft.gov.uk

Think about ... What type of function might provide a good model?

Finding an exponential model

N = N0ekt

Taking logs base e ln N = ln (N0ekt)

ln N = ln N0 + ln ektUsing the laws of logs

ln N = ln N0 + kt ln e

ln N = ln N0 + kt

y = c + mxCompare with:

Drawing a graph of ln N against t should give a straight line.

If so, its gradient will give k and its intercept will give ln N0.

Years after 1994 (t) Thousands of cars (N) ln N

0 1558

1 1600

2 1715

3 1844

4 1980

5 2145

6 2262

7 2451

8 2647

9 2869

10 3118

11 3314

12 3512

13 3687

Cars registered with engine size 2 litres or more

7.3777597.351158

7.4471687.5196927.5908527.670895

7.8042517.881182

7.9617198.0449478.1059118.163941

7.724005

8.212568

k = gradient = 0.0704 ln N0 = intercept = 7.318

ln N0 = 7.318

N0 = e7.318

= 1507

N = 1507e0.0704t

From the graph

Calculate N0

Exponential model:

Think about ... How good is this model?

e.g in 2000 i.e when t = 6 N = 1507e0.0704 × 6 = 2299

% error = predicted value – actual valueactual value

100

% error = = 1.6 %2299 – 2262

2262 100

Years after 1994 (t) Data (000s) Model (000s) % Error

0 1558 1507

1 1600 1617

2 1715 1735

3 1844 1861

4 1980 1997

5 2145 2143

6 2262 2299

7 2451 2467

8 2647 2647

9 2869 2840

10 3118 3047

11 3314 3269

12 3512 3508

13 3687 3763

0.9%

0.0%

– 0.1%

– 3.3%

1.1%

1.2%

0.7%

– 1.0%

– 2.3%

– 1.4%

– 0.1%– 2.1%

0.9%

1.6%

Comparison using percentage errors

Comparison using graph

N = 1507e0.0704tExponential model:

Prediction for 2008 t = 14

N = 1507e0.0704 × 14 = 1507e0.9856 = 4037

The actual values were 3731 and 3768 (thousand).

Compare the predicted and actual values.

Using the model to make predictions beyond 2007

Prediction for 2009 t = 15

N = 1507e0.0704 × 15 = 1507e1.056 = 4332

Think about ... Is there a better model?

YearThousands of cars

(engine size 2 litres or more)

2000 2262

2001 2451

2002 2647

2003 2869

2004 3118

2005 3314

2006 3512

2007 3687

2008 3731

2009 3768

Source: www.dft.gov.uk

Finding a new model

Draw a graph to show the data for 2000 to 2009.

Find a new model.

Reflect on your work

Why is the percentage error a better measure for the accuracy of a model than the difference between the actual value and the value predicted by the model?

What is indicated by a negative percentage error?

Gas guzzlers

Is your model valid for all values of t ?