Chapter 4.5, Problem 83E

a.

To determine

Create a scatter plot of the data.

Answer to Problem 83E

  

Explanation of Solution

Given information:

The table shows the percent y (in decimal form) of the moon’s face illuminated on day x in the year 2018, where x = 1 corresponds to January 1. (Source: U.S. Naval Observatory)

Create a scatter plot of the data.

Calculation:

Below table shows the percent $y$ of moons face illuminated on day $x$ in year 2014.

  $\begin{array}{cc}x& y\\ 1& 0\\ 8& 0.5\\ 16& 1\\ 24& 0.5\\ 30& 0\\ 37& 0.5\end{array}$

Plot the points in coordinate plane.

Scatter plot is as shown below:

  

b.

To determine

Find a trigonometric model for the data.

Answer to Problem 83E

  $\boxed{y=0.5\sin \left(0.217x-1.736\right)+0.5}$

Explanation of Solution

Given information:

The table shows the percent y (in decimal form) of the moon’s face illuminated on day x in the year 2018, where x = 1 corresponds to January 1. (Source: U.S. Naval Observatory)Create a scatter plot of the data.

Find a trigonometric model for the data.

Calculation:

Below table shows the percent $y$ of moons face illuminated on day $x$ in year 2014.

  $\begin{array}{cc}x& y\\ 1& 0\\ 8& 0.5\\ 16& 1\\ 24& 0.5\\ 30& 0\\ 37& 0.5\end{array}$

Consider the sine model $y=a\sin \left(bs-c\right)+d$

Here maximum percent is $1$ and minimum percent is $0$ .

  $a=\frac{1}{2}$ [max percent-min percent is $0$ .

  $=\frac{1}{2}\left(1-0\right)$

  $=\frac{1}{2}$

Thus the amplitude is $=\frac{1}{2}$ .

Here one cycle is completed between two minimum percents.

  $p$ =difference of time of two consecutive percents.

  $=30-1$

  $=29$

  $b=\frac{2\pi }{p}$

  $=\frac{2\pi }{29}\approx 0.217$

Here starting point is at $x=8$

Let the left endpoint is $\frac{c}{b}=8\to c=8\times 0.217\to c=\mathrm{1.736.}$

Average of minimum and maximum values is

  $d=\frac{1}{2}$ (min+max)

  $=\frac{1}{2}\left(1+0\right)$

  $=0.5$

Hence, the value of $d$ is $0.5$ .

Required sine function is

  $\boxed{y=0.5\sin \left(0.217x-1.736\right)+0.5}$

c.

To determine

How well does the model fit the data?

Answer to Problem 83E

  $\boxed{y=0.5\sin (0.217x-1.736)+0.5}$

Explanation of Solution

Given information:

The table shows the percent y (in decimal form) of the moon’s face illuminated on day x in the year 2018, where x = 1 corresponds to January 1. (Source: U.S. Naval Observatory)Create a scatter plot of the data.

Add the graph of your model in part (b) to the scatter plot. How well does the model fit the data?

Calculation:

Below table shows the percent $y$ of moons face illuminated on day $x$ in year 2014.

  $\begin{array}{cc}x& y\\ 1& 0\\ 8& 0.5\\ 16& 1\\ 24& 0.5\\ 30& 0\\ 37& 0.5\end{array}$

Graph is as shown below:

  

Hence, the equation fit the data

  $\boxed{y=0.5\sin (0.217x-1.736)+0.5}$

d.

To determine

What is the period of the model?

Answer to Problem 83E

  $\boxed{29days}$

Explanation of Solution

Given information:

The table shows the percent y (in decimal form) of the moon’s face illuminated on day x in the year 2018, where x = 1 corresponds to January 1. (Source: U.S. Naval Observatory)Create a scatter plot of the data.

What is the period of the model?

Calculation:

Below table shows the percent $y$ of moons face illuminated on day $x$ in year 2014.

  $\begin{array}{cc}x& y\\ 1& 0\\ 8& 0.5\\ 16& 1\\ 24& 0.5\\ 30& 0\\ 37& 0.5\end{array}$

Period of the model is

  $\frac{2\pi }{b}=\frac{2\pi }{0.217}$

  $=29$

Hence, the period is $\boxed{29days}$

e.

To determine

Estimate the percent of the moon’s face illuminated on March 12, 2018.

Answer to Problem 83E

  $\boxed{0.95}$

Explanation of Solution

Given information:

The table shows the percent y (in decimal form) of the moon’s face illuminated on day x in the year 2018, where x = 1 corresponds to January 1. (Source: U.S. Naval Observatory)Create a scatter plot of the data.

Estimate the percent of the moon’s face illuminated on March 12, 2018.

Calculation:

Below table shows the percent $y$ of moons face illuminated on day $x$ in year 2014.

  $\begin{array}{cc}x& y\\ 1& 0\\ 8& 0.5\\ 16& 1\\ 24& 0.5\\ 30& 0\\ 37& 0.5\end{array}$

Here March 12 corresponds to the day $x=71$ .

Substitute $x=71$ in $y=0.5\sin \left(0.217x-1.736\right)+05.$

  $y=0.5\sin \left(0.217\times 71-1.736\right)+0.5$

  $=0.95$

Hence the percent of illumination on March 12, 2014 is $\boxed{0.95}$