Breathing is cyclic and a full respiratory cycle from the beginning of inhalation to the end of exhalation takes about 3 s. The maximum rate of air flow into the lungs is about 0.3 L/s. This explains, in part, why the function
f(t) = (3/10)sin(2πt/3) has often been used to model the rate of air flow into the lungs. Use this model to find the volume of inhaled air in the lungs at time t.
V(t)= ????? liters
Yeah.
dV/dt = f(t) = (3/10) sin (2 pi t/3)
V = integral of that dt
= -(3/10)(3/2pi) cos (2 pi t/3)
= -(9/20pi)cos (2 pi t /3)
When i put this answer in it is marked wrong. Are u sure this is the right answer for this question?
Thanks!
Hey, I just gave you the method, not the answer
You have to put in the given period and amplitude
Oh, I take that back, their function includes the proper period and amplitude.
What were your bounds?
To find the volume of inhaled air in the lungs at time t using the given model, we need to integrate the rate of airflow function with respect to time.
The given rate of airflow function is: f(t) = (3/10)sin(2πt/3)
The rate of airflow represents the rate at which air is being inhaled into the lungs over time. Integrating this function will give us the cumulative volume of air inhaled at a given time.
Let's integrate f(t) with respect to t:
∫(3/10)sin(2πt/3) dt
To integrate this function, we can use the substitution method. Let's let u = 2πt/3. Therefore, du/dt = 2π/3.
Rearranging the equation, we get dt = (3/2π) du.
Substituting these values into the integral, we have:
∫(3/10)sin(u) * (3/2π) du
= (9/20π) ∫sin(u) du
= (-9/20π)cos(u) + C
Now, we need to substitute the value of u back into the equation.
u = 2πt/3
= (2π/3)t
Substituting back, we have:
V(t) = (-9/20π)cos(2πt/3) + C
The constant C is the integration constant. To determine this constant, we need to know the initial condition, which is the volume of inhaled air in the lungs at the starting point.
Once you have the value of C, you can evaluate V(t) for any given time t to find the volume of inhaled air in the lungs.