Chapter 13.7, Problem 40PPSE

Solution

a.

The plot of the data shown in the table of water temperatures.

The data is plotted.

Given:

The below table shows the water temperature at a buoy in the Gulf of Mexico on several days of the year:

Day of the year	16	47	75	106	136	167	198	228	258	289	319	350
Temperature ( $°F$ )	71	69	70	73	77	82	85	86	84	82	78	74

Calculation:

The data shown in the table is plotted:

  

Conclusion:

The given data is plotted.

b.

The cosine model for the data.

The Cosine function is:

  $y=8.5\cos \frac{2\pi }{365}\left(x-228\right)+77.5$

Given:

The graphs of the functions $y=\tan x,y=4\tan x\left(4>0\right)\text{ and }y=-4\tan x\left(-4<0\right)$ are shown in part (a).

Concept used:

The cosine function is:

  $y=a\cos b\left(x-h\right)+k$

Where, amplitude a is the half of the difference between the maximum and minimum values.

  $\text{b=}\frac{\text{2π}}{\text{period}}$

h is the phase shift and k is the vertical shift.

Calculation:

The cosine function is:

  $y=a\cos b\left(x-h\right)+k$

Here, amplitude a will be the half of the difference between the minimum and maximum temperature values.

So, $\begin{array}{c}a=\frac{86-69}{2}\\ =\frac{17}{2}\\ a=8.5\end{array}$

Now, $\text{period=}\frac{\text{2π}}{\text{b}}$

Here, period is 365 days.

So, $\begin{array}{c}\text{period=}\frac{\text{2π}}{\text{b}}\\ \text{b=}\frac{\text{2π}}{\text{period}}\\ \text{b=}\frac{\text{2π}}{\text{365}}\end{array}$

Since the maximum temperature occurs at $x=228$ , then the function will shift to 228 units to the right side and phase shift $h=228$

Now, the vertical shift, being the midline of the graph, is the average of the minimum and maximum values.

So, $\begin{array}{c}k=\frac{86+69}{2}\\ =\frac{155}{2}\\ k=77.5\end{array}$

Now, substituting all these values of a , b , h and k in the cosine function,

  $y=a\cos b\left(x-h\right)+k$

  $y=8.5\cos \frac{2\pi }{365}\left(x-228\right)+77.5$

Conclusion:

The Cosine function is:

  $y=8.5\cos \frac{2\pi }{365}\left(x-228\right)+77.5$