Question

1. the volume of air in a persons lungs can be modeled with a periodic function. the graph below represents the volume of air, in ml, in a persons lungs over time t, measured in seconds. write an equation in terms of y, volume of air in a persons lungs in ml, and t, time in seconds, to represent the given context.

Explanation:

Step1: Determine the amplitude

The amplitude \(A\) is half the distance between the maximum and minimum values.
\(A=\frac{3000 - 1300}{2}=\frac{1700}{2} = 850\)

Step2: Determine the vertical shift

The vertical shift \(D\) is the mid - value between the maximum and minimum values.
\(D=\frac{3000 + 1300}{2}=\frac{4300}{2}=2150\)

Step3: Determine the period

The period \(P\) is the distance between two consecutive maxima. Here, \(P=8 - 2=6\). Then the frequency \(B=\frac{2\pi}{P}=\frac{2\pi}{6}=\frac{\pi}{3}\)

Step4: Determine the phase shift

We can use the form \(y = A\cos(B(t - C))+D\). Let's assume a cosine function (since at \(t = 2\) we have a maximum). If we take the general form \(y=A\cos(Bt)+D\) (since the phase shift \(C = 2\) can be incorporated into the standard form. Using the point \((2,3000)\) and \(y = A\cos(Bt)+D\), when \(t = 2\), \(y = 3000\), \(A = 850\), \(B=\frac{\pi}{3}\), \(D = 2150\)

The equation of a cosine function is \(y=A\cos(Bt)+D\)

Substituting \(A = 850\), \(B=\frac{\pi}{3}\), \(D = 2150\)

\(y = 850\cos(\frac{\pi}{3}t)+2150\)

Answer:

\(y = 850\cos(\frac{\pi}{3}t)+2150\)