© nuffield foundation 2011 nuffield free-standing mathematics activity gas guzzlers
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© 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 ?