Nuffield Free-Standing Mathematics Activity
Gas guzzlers
© Nuffield Foundation 2011
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?
Are there more cars
with larger engines
on the roads today?
Source: www.dft.gov.uk
Year
Thousands 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
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)
Using the laws of logs ln N = ln N0 + ln ekt
ln N = ln N0 + kt ln e
ln N = ln N0 + kt
Compare with:
y = c + mx
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.
Cars registered with engine size 2 litres or more
Years after 1994 (t)
Thousands of cars (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
ln N
7.351158
7.377759
7.447168
7.519692
7.590852
7.670895
7.724005
7.804251
7.881182
7.961719
8.044947
8.105911
8.163941
8.212568
k = gradient = 0.0704
ln N0 = intercept = 7.318
From the graph
Calculate N0
ln N0 = 7.318
N0 = e7.318
= 1507
Exponential model:
N = 1507e0.0704t
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 value
 100
actual value
% error =
2299 – 2262
 100
2262
= 1.6 %
Comparison using percentage errors
Years after 1994 (t)
Data (000s)
Model (000s)
% Error
0
1558
1507
– 3.3%
1
1600
1617
1.1%
2
1715
1735
1.2%
3
1844
1861
4
1980
1997
0.9%
0.9%
5
2145
2143
– 0.1%
6
2262
2299
1.6%
7
2451
2467
0.7%
8
2647
2647
0.0%
9
2869
2840
– 1.0%
10
3118
3047
– 2.3%
11
3314
3269
– 1.4%
12
3512
3508
– 0.1%
13
3687
3763
– 2.1%
Comparison using graph
Using the model to make predictions beyond 2007
Exponential model: N = 1507e0.0704t
Prediction for 2008 t = 14
N = 1507e0.0704 × 14 = 1507e0.9856
= 4037
Prediction for 2009 t = 15
N = 1507e0.0704 × 15 = 1507e1.056 = 4332
The actual values were 3731 and 3768 (thousand).
Compare the predicted and actual values.
Think about ... Is there a better model?
Finding a new model
Year
Thousands 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
Draw a graph to
show the data for
2000 to 2009.
Find a new model.
Source:
www.dft.gov.uk
Gas guzzlers
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?
Is your model valid for all values of t ?