Relate the solution to the problem.

1. Relate the solution to the problem. • The model y = 43.4368e-0.02896x is used to predict the difference above room temperature of a cup of coffee as it cools. By comparing the raw data to predictions based on the model, we can decide whether the model is appropriate.

2. Relate the solution to the problem.

3. Relate the solution to the problem. • The predicted values from my model are very close to the actual data obtained in the experiment. This means that my model will provide good estimates of the temperature of the cup of coffee as it cools over the range of times up to 20 minutes. Initially my model slightly underestimates the temperature, then it slightly overestimates the temperature and at 20 minutes it starts to underestimate the temperature again.

4. Explain how the theory of the situation relates to the model. • Linear regression is the most common tool for finding lines of best fit for data, but cannot be used directly for data which have a non-linear relationship. By transforming non-linear relationships into linear equations, we can then compare the graphs of the transformed data to see which one is a better model. The more appropriate model is the one for which the straight line is a better fit for the transformed data.

5. Explain how the theory of the situation relates to the model. • For an exponential model, the transformation is: y = aekx taking the natural log of both sides ln y = ln (aekx) ln y = ln a + ln (ekx) ln y = ln a + kx

6. Explain how the theory of the situation relates to the model. • If we then plot x against ln y we get a straight line graph with k as the gradient. ln a is the y-intercept so a = ey-intercept • In the graphs of my transformed data, the straight line fits the log-linear transformed graph better than the log-log transformed graph so the exponential model is more appropriate for my data.

7. Discuss any limitations of the model. • Although my model is good for predicting values within the time range of the experiment which was 20 minutes, I need to be careful about using my model for extrapolation. My model has an asymptote on the x-axis. This implies that the temperature will keep cooling if you let the time get longer and longer. However, in actuality, the temperature of the coffee would stop changing once it reached room temperature.

8. Consider the nature of the underlying variables. • The phenomenon of coffee cooling over time is a natural one so we would expect that an exponential model would fit the data best as the coffee should cool by the same proportion over equal time intervals. • We can check this by investigating the proportional change in the difference above room temperature and compare this to the proportional change indicated in our model.

9. Consider the nature of the underlying variables. • Proportional change in the data • The temperature is changing by a factor of about 0.9 every 4 minutes which suggests that an exponential model may be appropriate.

10. Consider the nature of the underlying variables. • The model y = 43.4368e-0.02896x indicates a proportional change of e-0.02896 = 0.97. This value is quite different from the 0.9 expected from the data. • This indicates that although an exponential model seems suitable as the difference above room temperature seems to be changing by a constant proportion, perhaps the equation of the model that we have obtained is not that appropriate for our data.

11. Consider the nature of the underlying variables. • Exponential models have a y-intercept. In the coffee cooling experiment, we recorded a temperature of 44 °C when the time was 0 seconds. The presence of a y-intercept supports the fact that an exponential model is more appropriate for this data as a power model does not have a y-intercept.

12. Consider the possibility of using an alternative model. • The original graph of the difference above room temperature, as well as the calculations done previously show that my model initially slightly underestimates the temperature of the coffee, then it slightly overestimates the temperature and at 20 minutes it starts to underestimate the temperature again.

13. Consider the possibility of using an alternative model. • The predictions from my model are very close to the raw data and since there is not a permanent shift of the curve from below the data points at one end to above at the other end (as it shifts back to being below the point again at t=20), it is not suitable to fit two different curves to this data.

14. Graph of raw data