Q. 35 - Precalculus Enhanced with Graphing Utilities

Question

According to the Old Farmer’s Almanac,

in Miami, Florida, the number of hours of sunlight on the

summer solstice of 2011 was 13.75, and the number of hours.

(a) 

Find a sinusoidal function of the form

$y=A\sin (\omega x-\varphi )+B$ that models the data.

of sunlight on the winter solstice was 10.55.

(b) 

Use the function found in part (a) to predict the number

of hours of sunlight on April 1, the 91st day of the year.

(c) Draw a graph of the function found in part (a).

(d) 

Look up the number of hours of sunlight for April 1

in the Old Farmer’s Almanac, and compare the actual

hours of daylight to the results found in part (b).

Step-by-Step Solution

Answer

(a) $y=1.6\sin (0.01721x+1.3768)+12.15$

(b) 12.47 hours of sunlight

(c)

(d) 

Part (a) Step 1. Given

The number of hours of sunlight on the

summer solstice of 2011 was 15.30, and the number of hours of sunlight on the winter solstice was 9.10.

Part (a) Step 2. Calculation

To find sinusoidal function

We first find the amplitude using the formula:

$$\begin{gathered} A=\frac{high-low}{2} \\ A=\frac{13.75-10.55}{2} \\ A=\frac{3.2}{2} \\ A=1.6 \end{gathered}$$

Now we find the average of high and low point

$$\begin{gathered} B=\frac{high+low}{2} \\ B=\frac{13.75+10.55}{2} \\ B=\frac{24.3}{2} \\ B=12.15 \end{gathered}$$

Part (a) Step 3. Calculation of time period

Time period is given by:

$$\begin{gathered} t=\frac{2\mathrm{\pi }}{\omega } \\ 365=\frac{2\mathrm{\pi }}{\omega } \\ \omega =\frac{2\mathrm{\pi }}{365} \\ \omega =0.01721 \end{gathered}$$

Therefore, the final equation will be:

in order to find phase shift we consider equilibrium position during spring equinox i.e. on March ${21}^{st}$ in 2011.

therefore,

$$\begin{gathered} shift=\frac{\varphi }{\omega } \\ 80=\frac{\varphi }{0.01721} \\ \varphi =1.3768 \end{gathered}$$

Part (a) Step 4. Final equation

Substituting all the values we get:

$$\begin{gathered} y=A\sin (\omega x+\varphi )+B \\ y=1.6\sin (0.01721x+1.3768)+12.15 \end{gathered}$$

Part (b) Step 1. Given

The number of hours of sunlight on the

summer solstice of 2011 was 15.30, and the number of hours of sunlight on the winter solstice was 9.10.

Part (b) Step 2. Calculation

Substituting the value of 

$$\begin{gathered} x=91 as April 1, is the {91}^{st }day \\ therefore, \\ y=1.6\sin (0.01721(91)+1.3768)+12.15 \\ y=12.47 \end{gathered}$$

Part (c) Step 1. Given

The number of hours of sunlight on the

summer solstice of 2011 was 15.30, and the number of hours of sunlight on the winter solstice was 9.10.

Part (c) Step 2. Graphing the expression