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 » 6:44

As we can see our approximating function is right on target.