Chapter 4.5, Problem 81E

a.

To determine

To find: the period of the function $P=100-20\cos \left(\frac{5\pi t}{3}\right)$ .

Answer to Problem 81E

  $\frac{6}{5}\text{ sec}$

Explanation of Solution

Given:

The function $P=100-20\cos \left(\frac{5\pi t}{3}\right)$ approximates the blood pressure P (in millimeters of mercury) at time t (in seconds) for a person at rest.

Concept used:

The period of the function $y=a\cos bx$ is given by:

  $\text{Period}=\frac{2\pi }{b}$

Comparing the given function $P=100-20\cos \left(\frac{5\pi t}{3}\right)$ with $y=a\cos bx$ , we have $b=\frac{5\pi }{3}$

So, the period of given function is:

  $\begin{array}{c}\text{Period}=\frac{2\pi }{b}\\ =\frac{2\pi }{\frac{5\pi }{3}}\\ =\frac{6}{5}\text{ sec}\end{array}$

b.

To determine

To find: the number of heart beats per minute

Answer to Problem 81E

  $50\text{ heartbeats/min}$

Explanation of Solution

Given:

The function $P=100-20\cos \left(\frac{5\pi t}{3}\right)$ approximates the blood pressure P (in millimeters of mercury) at time t (in seconds) for a person at rest.

From part (a), the period of the given function is

  $\text{Period}=\frac{6}{5}\text{ sec}$

Since each period represents a heartbeat.

So, there is $\frac{5}{6}$ heartbeat in one second.

And thus, the number of heartbeats in one minute is:

  $60\times \frac{5}{6}=50\text{ heartbeats/min}$