Chapter 6.5, Problem 19E

To determine

To find the average volume inhaled velocity between $t=0\text{ and }t=T$

Answer to Problem 19E

The average volume inhaled velocity between $t=0\text{ and }t=T$ is $\frac{5}{4\pi }$

Explanation of Solution

Given information: $t=0\text{ and }t=T$

Calculation:

The result of problem 65 in section 5.5 is

  ${\displaystyle \underset{0}{\overset{t}{\int }}\frac{1}{2}}\sin \frac{2\pi t}{5}dt=\frac{5}{4\pi }\left(1-\cos \frac{2\pi t}{5}\right)$

Problem 65 also states that a full respiration cycle takes 5 seconds.

However the inhaling process takes only half the time the full cycle takes so

  $\begin{array}{l}{t}_0=0\\ T=\frac{5}{2}\end{array}$

Now to compute the average volume inhaled we use the formula showed in page 461

  $f_{ave}=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$

Use this we find the volume inhaled velocity between $t=0\text{ and }t=T$ as shown below

  $\begin{array}{l}{V}_{ave}=\frac{1}{T-t_0}{\displaystyle \underset{t_0}{\overset{T}{\int }}\frac{1}{2}\sin \frac{2\pi t}{5}}dt\\ =\frac{1}{\frac{5}{2}}\frac{5}{4\pi }\left(1-\cos \frac{2\pi }{2}\right)\\ =\frac{5}{8\pi }\left(1-\cos \pi \right)\\ =\frac{5}{8\pi }\left(1-\left(-1\right)\right)\\ =\frac{5}{4\pi }\end{array}$

Thus the average volume inhaled velocity between $t=0\text{ and }t=T$ is $\frac{5}{4\pi }$