Chapter 6.6, Problem 38AYU

(a)

To determine

To find: A sinusoidal function $y=A\sin \left(\omega x-\varphi \right)+B$ that models the given data.

Answer to Problem 38AYU

The sinusoidal function is $y=1.29\sin \left(\frac{2\pi x}{365}-1.39\right)+12.14$ .

Explanation of Solution

Given:

The number of hours of sunlight, in Honolulu, Hawaii, on summer solstice of 2010 was 13.43 and the number of hours of sunlight on winter solstice was 10.85.

Calculation:

For any sinusoidal function $y=A\sin \left(\omega x-\varphi \right)+B$ , A is the amplitude, B is the vertical shift, $\omega$ is the angular frequency and $\varphi$ is the phase or the horizontal shift.

The value of A and B is determined as shown below:

  $\begin{array}{c}A=\frac{13.43-10.85}{2}\\ =\frac{2.58}{2}\\ =1.29\end{array}$

  $\begin{array}{c}B=\frac{13.43+10.85}{2}\\ =\frac{24.28}{2}\\ =12.14\end{array}$

In order to find the value of the angular frequency, use the formula $T=\frac{2\pi }{\omega }$ where value of T is substituted as 365, number of days in the year 2010.

  $\begin{array}{c}T=\frac{2\pi }{\omega }\\ \omega =\frac{2\pi }{T}\\ =\frac{2\pi }{365}\end{array}$

To find the value of the horizontal shift, the interval $\left[0,365\right]$ is first split into 4 equal intervals, based on the four seasons in a year. The intervals are $\left[0,91.25\right]$ , $\left[91.25,182.5\right]$ , $\left[182.5,273.75\right]$ and $\left[273.75,365\right]$ . Next, since the maximum of the sine curve occurs at $x=91.25$ and the maximum sunlight appears in the summer solstice at $x=172$ , the value for $\varphi$ is the difference between 172 and 91.25, that is 80.75.

So, the sinusoidal function is,

  $\begin{array}{c}y=A\sin \left(\omega x-\varphi \right)+B\\ =1.29\sin \left[\frac{2\pi }{365}\left(x-80.75\right)\right]+12.14\\ =1.29\sin \left(\frac{2\pi x}{365}-1.39\right)+12.14\end{array}$

(b)

To determine

To find: A prediction of the number of hours of sunlight on April 1, the 91st day of the year.

Answer to Problem 38AYU

The number of hours of sunlight is $12\text{ hours 22 minutes}$ .

Explanation of Solution

Given:

The sinusoidal function is $y=1.29\sin \left(\frac{2\pi x}{365}-1.39\right)+12.14$ .

Calculation:

In order to predict the number of hours of sunlight on April 1, the 91st day of the year, substitute 91 for x in the equation $y=1.29\sin \left(\frac{2\pi x}{365}-1.39\right)+12.14$ and simplify.

  $\begin{array}{c}y=y=1.29\sin \left(\frac{2\pi \left(91\right)}{365}-1.39\right)+12.14\\ \approx 12.37\text{ hours}\\ =12\text{ hours 22 minutes}\end{array}$

So, the number of hours of sunlight is $12\text{ hours 22 minutes}$ .

(c)

To determine

To graph: The curve of the function $y=1.29\sin \left(\frac{2\pi x}{365}-1.39\right)+12.14$ .

Answer to Problem 38AYU

The graph is shown in Figure 1.

Explanation of Solution

Calculation:

The graph of $y=1.29\sin \left(\frac{2\pi x}{365}-1.39\right)+12.14$ is plotted using a graphing utility as shown below:

  

Figure 1

(d)

To determine

To compare: The numbers of hours of sunlight for April 1 in the Old Farmer’s Almanac and the actual hours of daylight calculated in part (c).

Answer to Problem 38AYU

The number of hours determined in part (c) approximate the number of hours shown in the Old Farmer’s Almanac.

Explanation of Solution

Calculation:

As determined in part (c), the actual number of hours is $12\text{ hours 22 minutes}$ .

Checking the number of hours in the Old Farmer’s Almanac, the following is observed:

  

Figure 2

From the source in Figure 2, it can be observed that the number of hours determined in part (c) approximate the number of hours shown in the Old Farmer’s Almanac.