Chapter 4.8, Problem 59E

To determine

To graph: The data mentioned below and the model $H\left(t\right)=12.13+2.77\sin \left(\frac{\pi t}{6}-1.60\right)$ .

  $\begin{array}{ccccccccccccc}\text{Month}& 1& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12\\ \text{Hours}, H& 9.67& 10.72& 11.92& 13.25& 14.37& 14.97& 14.72& 13.77& 12.48& 11.18& 10.00& 9.38\end{array}$

Explanation of Solution

Given information:

The data is mentioned below and the model is $H\left(t\right)=12.13+2.77\sin \left(\frac{\pi t}{6}-1.60\right)$ .

  $\begin{array}{ccccccccccccc}\text{Month}& 1& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12\\ \text{Hours}, H& 9.67& 10.72& 11.92& 13.25& 14.37& 14.97& 14.72& 13.77& 12.48& 11.18& 10.00& 9.38\end{array}$

Graph:

The data is mentioned below in which x axis represent the time of the months and y axis represent the number of hours of daylight also the model $H\left(t\right)=12.13+2.77\sin \left(\frac{\pi t}{6}-1.60\right)$ is represented in the same graph.

The graph is plotted in the real number line.

  

Interpretation:

Since the graph is made as according of the data mentioned above.

The model $H\left(t\right)=12.13+2.77\sin \left(\frac{\pi t}{6}-1.60\right)$ has the period of $\frac{2\pi *6}{\pi }=12$ .

Therefore the graph of the model is as on the points of the data.