Question

In a normal respiratory cycle the volume of air that moves into and out of the lungs is about 508 mL. The reserve and residual volumes of air that remain in the lungs occupy about 2020 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.

Answer 1

V(t)=2020+508sin((pi/2)t))

Step-by-step explanation:

breathing can be seen as an oscillatory function, so it would be perfect to use a sine or cosine function.

for this reason this is an equation of the form

V(t)=C+Asin(Wt)

as we know that there is always a fixed amount of air in the lungs (2020ml) which indicates that it is a constant amount.

on the other hand, the amount of air that moves in the lungs varies between 0l and 508ml for this reason this will be the amplitude

W=2pi/T

the period (T) is defined as the amount of time it takes for a function to repeat itself, in our case it is 4

W=2pi/4=pi/2

considering all of the above

V(t)=2020+508sin((pi/2)t))

Answer 2

`v(t) = 2274 + 254 sin((\pi)/(2)t)`

Step-by-step explanation:

`V_(min) = 2020 ml`

`V_(max) = 2020 + 508 =2528 mL`

Assuming the given function is cyclic function

`v_(avg) = (v_(max) - v_(min))/(2) = (2528 + 2020)/(2) = 2274 ml`

Amplitude
`= A = (V_(max) - V_(min))/(2)= (2528 - 2020)/(2)= 254`

time period is 4 sec

we know that
`\omega  =  (2\pi)/(t) = (2\pi)/(4) = (pi)/(2) rad/s`

so function is

`V(t) = V_(avg) + Asin (\omega t})`

`v(t) = 2274 + 254 sin((\pi)/(2)t)`