Chapter 6.6, Problem 35AYU

(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 35AYU

The sinusoidal function is $y=1.6\sin \left(\frac{2\pi x}{365}-1.39\right)+12.15$ .

Explanation of Solution

Given:

The number of hours of sunlight on summer solstice of 2010 was 13.75 and the number of hours of sunlight on winter solstice was 10.55.

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:

  $\begin{array}{c}A=\frac{13.75-10.55}{2}\\ =\frac{3.2}{2}\\ =1.6\end{array}$

  $\begin{array}{c}B=\frac{13.75+10.55}{2}\\ =\frac{24.30}{2}\\ =12.13\end{array}$

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,

  $\begin{array}{c}y=A\sin \left(\omega x-\varphi \right)+B\\ =1.6\sin \left[\frac{2\pi }{365}\left(x-80.75\right)\right]+12.15\\ =1.6\sin \left(\frac{2\pi x}{365}-1.39\right)+12.15\end{array}$

(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 35AYU

The number of hours of sunlight is $12\text{ hours }26\text{ minutes}$ .

Explanation of Solution

Given:

The sinusoidal function is $y=1.6\sin \left(\frac{2\pi x}{365}-1.39\right)+12.15$ .

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 $y=1.6\sin \left(\frac{2\pi x}{365}-1.39\right)+12.15$ and simplify.

  $\begin{array}{c}y=1.6\sin \left(\frac{2\pi \left(91\right)}{365}-1.39\right)+12.15\\ \approx 12.43\text{ hours}\\ =12\text{ hours }26\text{ minutes}\end{array}$

So, the number of hours of sunlight is $12\text{ hours }26\text{ minutes}$ .

(c)

To determine

To graph: The curve of the function $y=1.6\sin \left(\frac{2\pi x}{365}-1.39\right)+12.15$ .

Answer to Problem 35AYU

The graph is shown in Figure 1.

Explanation of Solution

Calculation:

The graph of $y=1.6\sin \left(\frac{2\pi x}{365}-1.39\right)+12.15$ 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 35AYU

The number of hours is same.

Explanation of Solution

Calculation:

As determined in part (c), the actual number of hours is $12\text{ hours }26\text{ minutes}$ .

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

  

Figure 2

From the source in Figure 2, it is verified that the number of hours determined in part (c) is the same the number in the Old Farmer’s Almanac.