Question

Breathing is cyclic and a full respiratory cycle from the beginning of inhalation to the end of exhalation takes about 5 s. The maximum rate of air flow into the lungs is about 0.6 L/s. This explains, in part, why the function f(t) = 3/5 sin(2πt/5) 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)=___

Answer 1

To find the volume of inhaled air at time t, integrate the function f(t) = 3/5 sin(2πt/5). The resulting volume function after integration and setting the constant of integration to zero, due to the initial condition, is v(t) = -3/(10π) cos(2πt/5).

Step-by-step explanation:

The student asks how to find the volume of inhaled air in the lungs at any time t using the function f(t) = 3/5 sin(2πt/5), which models the rate of air flow into the lungs during breathing. To determine the volume v(t), we need to integrate the rate function f(t), since the rate of flow represents the derivative of the volume with respect to time.

Step-by-step explanation:

1. Write down the rate of airflow function f(t) = 3/5 sin(2πt/5).
2. Integrate the function to find the volume function v(t). The integration of the sine function is negative cosine.
3. Perform the integration: \( v(t) = \int f(t)\, dt = \int \frac{3}{5} \sin(\frac{2\pi t}{5}) \, dt = -\frac{3}{5} \cdot \frac{5}{2\pi} \cos(\frac{2\pi t}{5}) + C \)
4. As we are interested in the volume inhaled from the start of inhalation, we set the constant of integration C to 0 assuming that at t=0, the volume is zero.
5. After determining the constant, the equation for the volume of air inhaled at time t is v(t) = -\frac{3}{10\pi} \cos(\frac{2\pi t}{5}).

The volume function v(t) provides us the amount of air inhaled into the lungs up to time t. This is an example of applying calculus to a real-world problem, specifically in understanding physiological processes such as breathing.