Download presentation

Presentation is loading. Please wait.

Published byMacy Keeton Modified over 4 years ago

1
© Nuffield Foundation 2011 Nuffield Free-Standing Mathematics Activity Gas guzzlers

2
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?

3
Year Thousands of cars (engine size 2 litres or more) 19941558 19951600 19961715 19971844 19981980 19992145 20002262 20012451 20022647 20032869 20043118 20053314 20063512 20073687 Are there more cars with larger engines on the roads today? Source: www.dft.gov.ukwww.dft.gov.uk

4
Think about... What type of function might provide a good model?

5
Finding an exponential model N = N 0 e kt Taking logs base e ln N = ln ( N 0 e kt ) ln N = ln N 0 + ln e kt Using the laws of logs ln N = ln N 0 + kt ln e ln N = ln N 0 + kt y = c + mx Compare 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 N 0.

6
Years after 1994 ( t )Thousands of cars ( N )ln N 01558 11600 21715 31844 41980 52145 62262 72451 82647 92869 103118 113314 123512 133687 Cars registered with engine size 2 litres or more 7.377759 7.351158 7.447168 7.519692 7.590852 7.670895 7.804251 7.881182 7.961719 8.044947 8.105911 8.163941 7.724005 8.212568

8
k = gradient = 0.0704ln N 0 = intercept = 7.318

9
ln N 0 = 7.318 N 0 = e 7.318 = 1507 N = 1507 e 0.0704 t From the graph Calculate N 0 Exponential model: Think about... How good is this model? e.g in 2000 i.e when t = 6 N = 1507 e 0.0704 × 6 = 2299 % error = predicted value – actual value actual value 100 % error = = 1.6 % 2299 – 2262 2262 100

10
Years after 1994 ( t )Data (000s)Model (000s)% Error 015581507 116001617 217151735 318441861 419801997 521452143 622622299 724512467 82647 928692840 1031183047 1133143269 1235123508 1336873763 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

11
Comparison using graph

12
N = 1507e 0.0704t Exponential model: Prediction for 2008 t = 14 N = 1507 e 0.0704 × 14 = 1507 e 0.9856 = 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 = 1507 e 0.0704 × 15 = 1507 e 1.056 = 4332 Think about... Is there a better model?

13
Year Thousands of cars (engine size 2 litres or more) 20002262 20012451 20022647 20032869 20043118 20053314 20063512 20073687 20083731 20093768 Source: www.dft.gov.uk www.dft.gov.uk Finding a new model Draw a graph to show the data for 2000 to 2009. Find a new model.

14
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 ?

Similar presentations

Presentation is loading. Please wait....

OK

STRAIGHT LINE GRAPHS y = mx + c.

STRAIGHT LINE GRAPHS y = mx + c.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To ensure the functioning of the site, we use **cookies**. We share information about your activities on the site with our partners and Google partners: social networks and companies engaged in advertising and web analytics. For more information, see the Privacy Policy and Google Privacy & Terms.
Your consent to our cookies if you continue to use this website.

Ads by Google

Ppt on hunter commission Ppt on forest conservation in india Ppt on management by objectives Ppt on power sharing in democracy power Ppt on 2nd world war File type ppt on cyber crime Ppt on trade fair 2016 Ppt on solar power satellites sps Ppt on math quiz for grade 3 Ppt online application form 2016/2017