Breathing is cyclic and a full respiratory cycle from the beginning of inhalation to the end of exhalation takes about 5 seconds. The maximum rate of air flow into the lungs is about 0.5 L/s. This explains, in part, why the function f(t)=1/2sin(2pi t/5) has often been used to model the rate of air flow into the lungs. This can then be used to show that the volume of inhaled air in the lungs at time t is given by V(t)=5/4π[1−cos((2/5)πt)]liters. Use this to compute the average volume (in liters) of inhaled air in the lungs in one respiratory cycle.
the average value is the sum of all values divided by the time. So, that is
(∫[0,10] v(t) dt)/10
= 1/10 ∫[0,10] 5/4π[1−cos((2/5)πt)] dt = 5π/4
The bounds should be 0 to 5, I'm not sure where the 10 is coming from. Also, the answer of 5pi/4 is incorrect
you are correct. It should be 5, not 10. But that does not change the answer. Better double-check the answer key.
I think it would change the answer because doesn't the 1/10 become 1/5 if we correct the error? Sorry, I just really have no idea how this problem works or how to do it.
To compute the average volume of inhaled air in the lungs in one respiratory cycle using the given function V(t) = (5/4π)[1 − cos((2/5)πt)], we need to find the integral of this function over one complete respiratory cycle.
In the given function, t represents time during the respiratory cycle. The function V(t) represents the volume of inhaled air in the lungs at time t.
To find the average volume of inhaled air, we need to integrate V(t) over one complete respiratory cycle and divide it by the duration of a respiratory cycle.
Given that a respiratory cycle takes about 5 seconds, we need to integrate V(t) from t = 0 to t = 5.
The integral of V(t) with respect to t from 0 to 5 can be calculated as follows:
∫[0 to 5] (5/4π)[1 − cos((2/5)πt)] dt
To solve this integral, we'll first integrate each term separately:
∫[0 to 5] (5/4π) dt - ∫[0 to 5] (5/4π)cos((2/5)πt) dt
The first term integrates to:
(5/4π)[t] from 0 to 5
= (5/4π)(5 - 0)
= (25/4π) liters
The second term involves the cosine function and requires integration by parts. By applying the substitution u = (2/5)πt and du = (2/5)π dt, the integral becomes:
-(5/4π) ∫cos(u) du
Integrating cos(u) gives us sin(u):
-(5/4π)sin(u)
Therefore, the second term integrates to:
-(5/4π)sin((2/5)πt) from 0 to 5
= -(5/4π)(sin(2π) - sin(0))
= 0 liters
Therefore, the average volume of inhaled air in the lungs in one respiratory cycle is:
(25/4π) liters
Keep in mind that this is just the calculation based on the given model, and actual measured values may differ.