Chapter 5, Problem 10PS

a.

To determine

Use a graphing utility to graph both models in the same viewing window. Use a viewing window of $0\le \text{t}\le \text{365}$

Answer to Problem 10PS

Explanation of Solution

Given information: The number of hours of daylight that occur at any location on Earth depends on the time of year and the latitude of the location. The equations below model the numbers of hours of daylight in Seward, Alaska ( $\text{6}0°$ latitude), and New Orleans, Louisiana ( $\text{3}0°$ latitude).

  $D=12.2-6.4\cos \left[\frac{\pi \left(t+0.2\right)}{182.6}\right]$ Seward

  $D=12.2-1.9\cos \left[\frac{\pi \left(t+0.2\right)}{182.6}\right]$ New Orleans

In these models, $\text{D}$ represents the number of hours of daylight and t represents the day, with $\text{t}=0$ corresponding to January $\text{1}$ .

Calculation:

Let us consider the numbers of hours of daylight in Seward, Alaska ( $\text{6}0°$ latitude), and New Orleans, Louisiana ( $\text{3}0°$ latitude) plotted against in a graph below,

  

Hence, the graph is shown above.

b.

To determine

Find the days of the year on which both cities receive the same amount of daylight.

Answer to Problem 10PS

April ${\text{1}}^{st}$ and October ${\text{2}}^{\text{nd}}$ of a leap year.

Explanation of Solution

Given information:

The number of hours of daylight that occur at any location on Earth depends on the time of year and the latitude of the location. The equations below model the numbers of hours of daylight in Seward, Alaska ( $\text{6}0°$ latitude), and New Orleans, Louisiana ( $\text{3}0°$ latitude).

  $D=12.2-6.4\cos \left[\frac{\pi \left(t+0.2\right)}{182.6}\right]$ Seward

  $D=12.2-1.9\cos \left[\frac{\pi \left(t+0.2\right)}{182.6}\right]$ New Orleans

In these models, $\text{D}$ represents the number of hours of daylight and t represents the day, with $\text{t}=0$ corresponding to January $\text{1}$ .

Calculation:

The two regions receive same amount of daylight on only two days of the year as can be seen from the above plot.

The first day on which the daylight received is same is the ${\text{91}}^{\text{st}}$ day of the year that is, on the April ${\text{1}}^{st}$ of a leap year, both the regions receive same daylight.

The second day on which the daylight received is same is the ${\text{74}}^{\text{th}}$ day of the year that is, on the October ${\text{2}}^{\text{nd}}$ of a leap year, both the regions receive same daylight.

Hence, on April ${\text{1}}^{st}$ and October ${\text{2}}^{\text{nd}}$ of a leap year, both the regions receive same daylight.

c.

To determine

Which city has the greater variation in the number of hours of daylight? Which constant in each model would you use to determine the difference between the greatest and least numbers of hours of daylight?

Answer to Problem 10PS

Seward.

  $\boxed{3.8}$

  $\boxed{12.8}$

Explanation of Solution

Given information

The number of hours of daylight that occur at any location on Earth depends on the time of year and the latitude of the location. The equations below model the numbers of hours of daylight in Seward, Alaska ( $\text{6}0°$ latitude), and New Orleans, Louisiana ( $\text{3}0°$ latitude).

  $D=12.2-6.4\cos \left[\frac{\pi \left(t+0.2\right)}{182.6}\right]$ Seward

  $D=12.2-1.9\cos \left[\frac{\pi \left(t+0.2\right)}{182.6}\right]$ New Orleans

In these models, $\text{D}$ represents the number of hours of daylight and t represents the day, with $\text{t}=0$ corresponding to January $\text{1}$ .

Calculation:

Seward experiences a greater variation in daylight hours tha New orleans .

The numbers of hours of daylight in Seward, Alaska is expressed as,

  $D=12.2-6.4\cos \left[\frac{\pi \left(t+0.2\right)}{182.6}\right]$

The numbers of hours of daylight in New Orleans, Louisiana is expressed as,

  $D=12.2-1.9\cos \left[\frac{\pi \left(t+0.2\right)}{182.6}\right]$

Since the varying part in the above expressions is trigonometric part, so,the contants that should be used to calculate the difference between the greatest and the least number of daylight hours is the coefficient of the trigonometric functions.

Since the value of a cosine function ranges from $-1$ to $1$ , so the difference between the greatest and the least number of daylight in Seward are given as below,

  $\begin{array}{l} \left(12.2+6.4\right)-\left(12.2-6.4\right)\\ =12.8\end{array}$

But the difference between the greatest and least number of daylight in New Orleans, are as given below

  $\begin{array}{l}\left(12.2+1.9\right)-\left(12.2-1.9\right)\\ =3.8\end{array}$

Hence, Seward has greater variation and the greatest difference is $\boxed{12.8}$ while the least difference is $\boxed{3.8}$ .

d.

To determine

Determine the period of each model.

Answer to Problem 10PS

  $\boxed{P_1=365.2}$

  $\boxed{P_2=365.2}$

Explanation of Solution

Given information

The number of hours of daylight that occur at any location on Earth depends on the time of year and the latitude of the location. The equations below model the numbers of hours of daylight in Seward, Alaska ( $\text{6}0°$ latitude), and New Orleans, Louisiana ( $\text{3}0°$ latitude).

  $D=12.2-6.4\cos \left[\frac{\pi \left(t+0.2\right)}{182.6}\right]$ Seward

  $D=12.2-1.9\cos \left[\frac{\pi \left(t+0.2\right)}{182.6}\right]$ New Orleans

In these models, $\text{D}$ represents the number of hours of daylight and t represents the day, with $\text{t}=0$ corresponding to January $\text{1}$ .

Calculation:

The period $P_1$ for the expression the numbers of hours of daylight in Seward can be calculated by determining the reciprocal of the coefficient of $t$ in the argument of the given cosine function of the expression and multiplying it by $2\pi$ as follows,

  $\begin{array}{l}{P}_1=\frac{1}{\frac{\pi }{182.6}}\times 2\pi \\ =\frac{2\pi \times 182.6}{\pi }\\ =365.2\end{array}$

The period $P_2$ for the expression the numbers of hours of daylight in New Orleans can be calculated by determining the reciprocal of the coefficient of $t$ in the argument of the given cosine function of the expression and multiplying it by $2\pi$ as follows,

  $\begin{array}{l}{P}_2=\frac{1}{\frac{\pi }{182.6}}\times 2\pi \\ =\frac{2\pi \times 182.6}{\pi }\\ =365.2\end{array}$

Hence $\boxed{P_1=365.2}$ , $\boxed{P_2=365.2}$