Chapter 13.5, Problem 41PPSE

Solution

i.

To graph: The given functions.

Given:

The functions $y=-30\cos \left(\frac{6\pi }{37}t\right)$ .

Calculation:

The graph of $y=-30\cos \left(\frac{6\pi }{37}t\right)$ is given below:

  

Conclusion:

Hence, the graph of functions is given above.

ii.

To find: The four times at which the level of water is highest.

  $4:40\text{ PM}$ , $5:00\text{ AM}$ , $5:20\text{ PM}$ and $5:40\text{ AM}$

Given:

The functions $y=-30\cos \left(\frac{6\pi }{37}t\right)$ .

Concept Used:

Calculation:

According to the obtained graph, the period of the function is,

  $18.5-6.3=12.20$

The water rises the highest at $4:40\text{ PM}$ first time. So,

The next time highest time is $4:40\text{ PM}+12:20=5:00\text{ AM}$ .

Similarly, the other two times are,

  $5:00\text{ PM}+12:20=5:20\text{ PM}$

  $5:20\text{ PM}+12:20=5:40\text{ AM}$

Conclusion:

Hence, the four times are $4:40\text{ PM}$ , $5:00\text{ AM}$ , $5:20\text{ PM}$ and $5:40\text{ AM}$ .

iii.

To find: The number of hours for which the water level is above $y=0$ during each cycle.

  $6\text{ hours }10\text{ min}$

Given:

The functions $y=-30\cos \left(\frac{6\pi }{37}t\right)$ .

Calculation:

The number of hours for which water level is above $y=0$ is,

  $\begin{array}{c}\text{hours}=\frac{\text{Period}}{2}\\ =\frac{12\text{ hours }20\text{ min}}{2}\\ =6\text{ hours }10\text{ min}\end{array}$

Conclusion:

The number of hours is $6\text{ hours }10\text{ min}$ .