Chapter 4.8, Problem 56E

a.

To determine

Use a graphing utility to graph the data and the model in the same viewing window.

Answer to Problem 56E

  

Explanation of Solution

Given information:

The numbers of hours $H$ of daylight in Denver, Colorado, on the $15th$ of each month starting with January are: $9.68,10.72,11.92,13.25,14.35,14.97,14.72,13.73,12.47,11.18,10.00and\mathrm{9.37.}$ A model for the data is, $H\left(t\right)=12.13+2.77\sin \left(\frac{\pi t}{6}-1.60\right)$ .

Where $t$ represents the month, with $t=1$ corresponding to January.

Use a graphing utility to graph the data and the model in the same viewing window.

Calculation:

The model for the data $H\left(t\right)=12.13+2.77\sin \left(\frac{\pi t}{6}-1.60\right)$ is as,

  

Hence the result is shown in graph.

b.

To determine

What is the period of the model?

Answer to Problem 56E

  $\boxed{period=12}$

Explanation of Solution

Given information:

The numbers of hours $H$ of daylight in Denver, Colorado, on the $15th$ of each month starting with January are: $9.68,10.72,11.92,13.25,14.35,14.97,14.72,13.73,12.47,11.18,10.00and\mathrm{9.37.}$ A model for the data is, $H\left(t\right)=12.13+2.77\sin \left(\frac{\pi t}{6}-1.60\right)$ .

Where $t$ represents the month, with $t=1$ corresponding to January.

What is the period of the model? Is it what you expected? Explain.

Calculation:

Consider, $H\left(t\right)=12.13+2.77\sin \left(\frac{\pi t}{6}-1.60\right)$ ,

  $\begin{array}{l}\omega =\frac{\pi }{6}\\ period=\frac{2\pi }{\omega }\\ =\frac{2\pi }{\pi /6}\\ =12\end{array}$

Hence the $\boxed{period=12}$ .

c.

To determine

What is the amplitude of the model?

Answer to Problem 56E

  $\boxed{Amplitude=2.77}$

Explanation of Solution

Given information:

The numbers of hours $H$ of daylight in Denver, Colorado, on the $15th$ of each month starting with January are: $9.68,10.72,11.92,13.25,14.35,14.97,14.72,13.73,12.47,11.18,10.00and\mathrm{9.37.}$ A model for the data is, $H\left(t\right)=12.13+2.77\sin \left(\frac{\pi t}{6}-1.60\right)$ .

Where $t$ represents the month, with $t=1$ corresponding to January.

What is the amplitude of the model? What does it represent in the context of the problem?

Calculation:

Consider, $H\left(t\right)=12.13+2.77\sin \left(\frac{\pi t}{6}-1.60\right)$ ,

  $Amplitude=2.77$

Hence the $\boxed{Amplitude=2.77}$ .