Chapter 5.5, Problem 65E

Breathing is cyclic and a full respiratory cycle from the beginning of inhalation to the end of exhalation takes about 5 s. The maximum rate of air flow into the lungs is about 0.5 L/s. This explains, in part, why the function $f(t)=\frac{1}{2}\sin (2\text{π}t/5)$ has often been used to model the rate of air flow into the lungs. Use this model to find the volume of inhaled air in the lungs at time t.

To determine

To find: The volume of inhaled air in the lungs at time t.

Answer to Problem 65E

The volume of inhaled air in the lungs is $\underset{\_}{\frac{5}{4\pi }\left[1-\cos \left(\frac{2\pi t}{5}\right)\right]\text{liters}}$.

Explanation of Solution

Given:

The function is $f\left(t\right)=\frac{1}{2}\sin \left(\frac{2\pi t}{5}\right)$.

Calculation:

The volume is given in terms of integral function as,

$V\left(t\right)={\displaystyle {\int }_0^{t}f\left(t\right)dt}$

Substitute $\left[\frac{1}{2}\sin \left(\frac{2\pi t}{5}\right)\right]$ for $f\left(t\right)$.

$\begin{array}{c}V\left(t\right)={\displaystyle {\int }_0^{t}f\left(t\right)dt}\\ ={\displaystyle {\int }_0^t\left[\frac{1}{2}\sin \left(\frac{2\pi t}{5}\right)\right]}dt\end{array}$ (1)

The region lies between $t=0$ and $t=t$.

Consider $u=\frac{2\pi t}{5}$ (2)

Differentiate both sides of the equation (2).

$\begin{array}{l}du=\frac{2\pi }{5}dt\\ dt=\frac{5}{2\pi }du\end{array}$

Calculate the lower limit value of u using equation (2).

Substitute 0 for t in equation (2).

$\begin{array}{c}u=\frac{2\pi \left(0\right)}{5}\\ =0\end{array}$

Calculate the upper limit value of u using equation (2).

Substitute t for t in equation (2).

$\begin{array}{c}u=\frac{2\pi \left(t\right)}{5}\\ =\frac{2\pi t}{5}\end{array}$

Apply lower and upper limits for u in equation (1).

Substitute u for $\left(\frac{2\pi t}{5}\right)$ and $\left(\frac{5}{2\pi }du\right)$ for dt in equation (1).

$\begin{array}{c}{\displaystyle {\int }_0^t\left[\frac{1}{2}\sin \left(\frac{2\pi t}{5}\right)\right]}dt={\displaystyle {\int }_0^{\frac{2\pi t}{5}}\frac{1}{2}\sin u\left(\frac{5}{2\pi }du\right)}\\ =\frac{1}{2}\times \frac{5}{2\pi }{\displaystyle {\int }_0^{\frac{2\pi t}{5}}\sin udu}\\ =\frac{5}{4\pi }{\displaystyle {\int }_0^{\frac{2\pi t}{5}}\sin udu}\end{array}$ (3)

Integrate equation (3).

$\begin{array}{c}\frac{5}{4\pi }{\displaystyle {\int }_0^{\frac{2\pi t}{5}}\sin udu}=\frac{5}{4\pi }{\left[-\cos u\right]}_0^{\frac{2\pi t}{5}}\\ =-\frac{5}{4\pi }\left[\cos \left(\frac{2\pi t}{5}\right)-\cos 0\right]\\ =-\frac{5}{4\pi }\left[\cos \left(\frac{2\pi t}{5}\right)-1\right]\\ =\frac{5}{4\pi }\left[1-\cos \left(\frac{2\pi t}{5}\right)\right]\end{array}$

Hence, the volume of inhaled air in the lungs is $\underset{\_}{\frac{5}{4\pi }\left[1-\cos \left(\frac{2\pi t}{5}\right)\right]\text{liters}}$.