Answer:
![A(t)=0.5cos(12\pi\cdot t)+4](https://tex.z-dn.net/?f=A%28t%29%3D0.5cos%2812%5Cpi%5Ccdot%20t%29%2B4)
Step-by-step explanation:
<u>Function Modeling</u>
Since the air in the lungs goes periodically from a minimum value to a maximum value, we can simulate its behavior as a sinusoid. Selecting a sine or a cosine will depend on the initial condition we'll assume since the question doesn't provide one. We'll set the initial state when the lungs are at maximum air content. The sinusoid that starts from maximum is the cosine, so our model is
![A(t)=A_ocos(wt)+M](https://tex.z-dn.net/?f=A%28t%29%3DA_ocos%28wt%29%2BM)
Where Ao is the amplitude of the oscillation, w is the angular frequency and M is the midline or y-displacement of the wave.
The values of A run from a min of 3.5 to a max of 4.5. That gives us twice the amplitude, thus
![\displaystyle A_o=\frac{4.5-3.5}{2}=0.5](https://tex.z-dn.net/?f=%5Cdisplaystyle%20A_o%3D%5Cfrac%7B4.5-3.5%7D%7B2%7D%3D0.5)
The vertical displacement or midline can be found as the shift from the center value:
![\displaystyle M=\frac{4.5+3.5}{2}=4](https://tex.z-dn.net/?f=%5Cdisplaystyle%20M%3D%5Cfrac%7B4.5%2B3.5%7D%7B2%7D%3D4)
We also know the cycle repeats 6 times per minute. If the time is expressed in minutes, then the frequency is f=6
Knowing that
![w=2\pi f](https://tex.z-dn.net/?f=w%3D2%5Cpi%20f)
Then
![w=12\pi](https://tex.z-dn.net/?f=w%3D12%5Cpi)
The model is
![\boxed{A(t)=0.5cos(12\pi\cdot t)+4}](https://tex.z-dn.net/?f=%5Cboxed%7BA%28t%29%3D0.5cos%2812%5Cpi%5Ccdot%20t%29%2B4%7D)
where t is expressed in minutes