Exercise 1.3.30

Question

In a normal respiratory cycle, the volume of air that moves into and out of the lungs is about 500 mL. The reserve and residue 
 volumes of air that remain in the lungs occupy about 2000 mL and a single respiratory cycle for an average human takes about 4 seconds. Find a model for the total volume of air $V(t)$ in the lungs as a function of time.

Khagan Gulusoy · Accepted Answer

From the exercise text, we can conclude that:
\begin{enumerate}
    \item The value of $V$ is minimal at $t=0$, the beginning of the respiratory cycle, and every $4$ units of time thereafter. At these points, $V=2000$ mL. 
    \item The value of $V$ is maximal when $t$ can be represented as $2+2x$ for some natural number $x$ divisible by two (at that $t$ the value of $V$ is equal to 2000).
\end{enumerate}
Consider the initial function to be the function $f_1$, such that $\forall x\in \mathbb{R}: f_1(x)=-\cos x$. \begin{itemize}\item Since the period of the function $f_1$ is equal to $2\pi$ and the period of the desired function is equal to $4$, we have: $$p=\frac{2\pi}{4}=\frac{\pi}{2},$$ where $p$ is the period of the desired function.
\item Our next step is finding the amplitude $a$ of the desired function:
$$a=\frac{1}{2}(2500-2000)=250.$$
\item Since the minimum of the graph of the equation $y=-250\cos \left(\dfrac{\pi}{2}x\right)$ is $-250$ and the minimum of the desired function is $2000,$ we must shift the graph $2250$ units up.
\end{itemize}
Thus, we define the function $V$ of time $t$ as follows:
$$\forall t\in [0,+\infty):\quad V(t)=-250\cos \left(\dfrac{\pi}{2}x\right)+2250.$$

Exercise 1.3.30

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Exercise 1.3.30

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