Modeling the Weather

The table shows Melbourne’s mean average daily maximum temperature (℃) for two year period 1999-2000.

Year| Jan| Feb| Mar| Apr| May| Jun| Jul| Aug| Sep| Oct| Nov| Dec| 1999| 25.7| 26.9| 24.5| 21.4| 18.0| 14.0| 13.5| 13.9| 17.2| 19.4| 22.2| 24.6| 2000| 26.0| 25.4| 24.7| 20.7| 17.5| 14.6| 14.8| 14.4| 17.5| 20.6| 22.9| 26.1|

1. Define appropriate variables and parameters, and identify any constraints for the data. Independent variable: months

Dependent variable: temperature

Range: 13.5≤y≤26.9

2. Use technology to plot the data points on a graph. Comment on any apparent trends shown in the graph. By plotting the data given from the table into graphmatica, the graph below can be obtained. GRAPH 1

Temperature (℃)

Months

This graph shows that if we connect these points, we will see a regular wavy curve appropriately.

3. What type of function models the behavior of the graph? Explain why you chose this function. I noticed that the sine graph best models the behavior of the graph. The reason is that GRAPH 1 simply looks like a sine function and it seems contain amplitude and period.

Sine function graph : y=sin(x)

4. Use your knowledge of the graphs of such functions to create a suitable equation that models the behavior of the data. Explain all steps you took to arrive at your equation. First of all, the general sine function is defined as: y=AsinB(x-C)+D * A represent the Amplitude

* B represent the n in Period (T=2πn)

* C represent the Horizontal translation

* D represent the Vertical translation

* To find A, I know the ymin=13.5 and ymax=26.9 from the table. Amplitude = ymax-ymin2 = 26.9-13.52 = 6.7

A is therefore 6.7.

* To find B, I know that the period (T)=12 months.

T=2π BT=12

B = 2π12 = π6

B is therefore π 6.

* To find D, I know the ymin=13.5 and ymax=26.9 from the table. D= ymax+ymin2 = 26.9+13.52 = 20.2

D is therefore 20.2.

* To find C, I chose one point from the GRAPH 1:（10,19.4） y=AsinB(x-C)+D 19.4 = 6.7 sinπ 6(10-C)+20.2

6.7 sinπ 6(10-C) = -0.8

sinπ 6(10-C) = -867

π 6(10-C) = sin-1(-867)

π 6(10-C) = -6.86

10-C = -41.16π

C = 10 + 41.16π

C is therefore 10 + 41.16π.

So, the suitable equation is y = 6.7sinπ6(x-10-41.16π)+20.2

5. On a new set of axes, draw your model function and the original data points. Comment on any differences.

GRAPH 2

y = 6.7sinπ6(x-10-41.16π)+20.2

Temperature (℃)

Months

GRAPH 1&2 (x)| GRAPH 1 (y)| GRAPH 2 (y)| Absolute Error| 1| 25.7| 25.8| 0.1|

2| 26.9| 26.9| 0|

3| 24.5| 26.2| 1.7|

4| 21.4| 23.9| 2.5|

5| 18.0| 20.6| 2.6|

6| 14.0| 17.2| 3.2|

7| 13.5| 14.6| 1.1|

8| 13.9| 13.6| 0.3|

9| 17.2| 14.2| 3|

10| 19.4| 16.5| 2.9|

11| 22.2| 19.8| 2.4|

12| 24.6| 23.2| 1.4|

13| 26.0| 25.8| 0.2|

14| 25.4| 26.9| 1.5|

15| 24.7| 26.2| 1.5|

16| 20.7| 23.9| 3.2|

17| 17.5| 20.6| 3.1|

18| 14.6| 17.2| 2.6|

19| 14.8| 14.6| 0.2|

20| 14.4| 13.6| 0.8|

21| 17.5| 14.2| 3.3|

22| 20.6| 16.5| 4.1|

23| 22.9| 19.8| 3.1|

24| 26.1| 23.2| 2.9|

Average Error=1.9875|

We can find the sine function graph is not same as the actual data. The actual data is not even as the model. Compare it with the actual data, it will be a small shift. The average error is very small. It is 1.9875.

6. (a) Use your equation to predict values for maximum daily temperature for the following year in: (i) August

y = 6.7sinπ6(x-10-41.16π)+20.2

When x = 32, y=6.7sinπ6(32-10-41.16π)+20.2

The graphmatica shows the result: y=13.5.

(ii) December

y = 6.7sinπ6(x-10-41.16π)+20.2

When x = 36, y=6.7sinπ6(36-10-41.16π)+20.2

The graphmatica shows the result: y=23.2.

(b) Discuss the reliability of your predictions.

According to the data of year 1999 and 2000, this sine function fit the data. It is...