Chapter 4.4, Problem 76E

Solution

a.

To find: The period of the given function.

The period of the given function is $1$ .

Given information:

The given function is, $P=120+30\sin 2\pi t$ .

Calculation:

The given function is,

  $P=120+30\sin 2\pi t$

The standard function equation is given by,

  $y=a\sin \left(bt\right)+k$

Comparing both the equation to get,

  $\begin{array}{c}a=20\\ b=2\pi \\ k=120\end{array}$

The period of the function is,

  $\begin{array}{c}\text{Period}=\frac{2\pi }{b}\\ =\frac{2\pi }{2\pi }\\ \text{Period}=1\end{array}$

Therefore, the period of the given function is $1$ .

b.

To find: The heartbeats for every minute.

There are sixty heartbeats in a minute.

Given information:

The given function is, $P=120+30\sin 2\pi t$ .

Calculation:

The given function is of the form,

  $y=a\sin \left(bt+c\right)+d$

From this the period is given by $\left|\frac{2\pi }{b}\right|$ .

Since, there are $60$ seconds in a minute and 1 heartbeat every second, there are $60$ heartbeats in a minute.

Therefore, there are $60$ heartbeats in a minute.

c.

To graph: The function to model a $10$ -sec time interval.

Given information:

The given function is, $P=120+30\sin 2\pi t$ .

Graph:

  

Interpretation:

The given function is of the form,

  $y=a\sin \left(bt+c\right)+d$

From this the period is given by $\left|\frac{2\pi }{b}\right|$ .

Therefore, the graph is drawn for the given function to model a $10$ -sec time interval.