Eastern Standard Time for Johnson City, TN
Michelle Harvey
Ree’l Street
Gabriel Zimmer
Once again we are going to use the equation of oscillations:
y = M + a cos[(2p /365) + b sin[(2p /365)t]
Note that by using 365 in the denominator we will be able to use t in days instead of months.
Here is a look at our raw data for this problem:
|
Date |
t = # of days |
y = Sunset time |
|
January 15 |
15 |
5.6333 |
|
February 15 |
46 |
6.1667 |
|
March 1 |
60 |
6.400 |
|
March 15 |
74 |
6.6167 |
|
April 15 |
105 |
7.0500 |
With the average sunset time M = 6.5417 which is about 6:32.
Plotting the raw data gives the following graph:

We then use our transformation functions:
Y = (y - 6.5417) sec[(2p /365)t]
and
T = tan[(2p /365)t]
|
Y |
T |
|
.26411 |
-.93955 |
|
1.01299 |
-.53379 |
|
1.67606 |
-.27656 |
|
3.26806 |
.25632 |
|
-4.14565 |
-2.16767 |
Now that we have transformed our raw data, we can then plug it into Matlab and obtain the following values for our linear regression formula y = a + bt.
a = -.8678 b = .3265
which gives the regression line:
Y = -.8678 + .3265 T
This equation gives the following graph

Going back to our original equation we can model the sunset time with the approximating function:
y = 6.5417 - .8678 cos[(2p /365)t] + .3265 sin[(2p /365)t]
As an additional bonus, our problem told us that on June 15 (t = 166 days) the sunset time was 7:49. To test our approximating function we can plug in t = 166 and see how exact our function is.
y = 6.5417 - .8678 cos[(2p /365)(166)] + .3265 sin[(2p /365)(166)]
Solving for y, the sunset time, gives:
y = 6.74208
» 6:44As we can see our approximating function is right on target.
