Chapter 10.1, Problem 31PS

Solution

a.

To write an equation that gives the buoy’s vertical position $y$ at time $t$ if the buoy is at its highest point when $t=0.$

  $y=1.75\cos \frac{\pi }{3}t$

Given:

It is given that the buoy moves a total of $3.5\text{ feet}$ from its low point its high point and then returns to its high point every $6$ seconds.

Calculation:

The amplitude is given as:

  $A=\frac{1}{2}\left[\text{maximum value}-\text{minimum value}\right]$

Here, buoy moves a total of $3.5\text{ feet,}$ so:

  $\begin{array}{l}A=\frac{1}{2}\left(3.5\right)\\ A=1.75\end{array}$

Also, the buoy returns to its high point every $6$ seconds, hence the period is $6.$ The period in terms of frequency can be written as:

  $\begin{array}{l}\text{period =}\frac{2\pi }{b}\\ 6=\frac{2\pi }{b}\\ b=\frac{2\pi }{6}\\ b=\frac{\pi }{3}\end{array}$

Now, to model the position use the function $y=a\cos bt$ and substitute the parameter values as:

  $y=1.75\cos \frac{\pi }{3}t$

Hence, an equation which gives the buoy’s vertical position $y$ at time $t$ is $y=1.75\cos \frac{\pi }{3}t$

b.

To explain that why chose the function $y=a\cos bt$ for part(a).

When the highest point occurs at $t=0,$ then it can be modelled using $y=a\cos bt$ . It cannot modelled using $y=a\sin bt$ . Here, the buoy is at its highest point when $t=0,$ so the function $y=a\cos bt$ is used to model its position with respect to time.