A simple model for the flow of air in and out of the lungs of a certain mammal is given by the following equation, where $$ \mathrm{V}(\mathrm{t}) $$ (measured in liters) is the volume of air in the lungs at time $$ t \geq 0 $$, $$ t $$ is measured in seconds, and $$ \mathrm{t}=0 $$ corresponds to a time at which the lungs are full and exhalation begins. Only a fraction of the air in the lungs is exchanged with each breath. The amount that is exchanged is called the tidal volume. Complete parts a through $$ c $$ below. $$ V^{\prime}(t)=-\frac{\pi}{6} \sin \left(\frac{\pi t}{6}\right) $$ a. Find the volume function V , assuming that $$ \mathrm{V}(0)=6 \mathrm{~L} $$. Notice that V changes over time at a known rate, $$ \mathrm{V}^{\prime} $$. Which equation below correctly gives the volume function? A. $$ \mathrm{V}(0)=\mathrm{V}(\mathrm{t})+\int_{0}^{t} \mathrm{~V}^{\prime}(\mathrm{x}) $$ dx. C . $$ \mathrm{V}(\mathrm{t})=\mathrm{V}(0)+\int_{a} \mathrm{~V}^{\prime}(\mathrm{t}) \mathrm{dt} $$. Find the volume function V , assuming that $$ \mathrm{V}(0)=6 \mathrm{~L} $$. $$ \mathrm{V}(\mathrm{t})=5+\cos \left(\frac{\pi t}{6}\right) $$ (Type an exact answer.) b. What is the breathing rate in breaths/minute? The breathing rate is $$ \square $$ breaths/minute. (Type an integer or a simplified fraction.)

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To solve the problem, we will follow the steps outlined in the prompt.

### Part a: Find the Volume Function $$ V(t) $$

We are given the derivative of the volume function:

$$
V'(t) = -\frac{\pi}{6} \sin\left(\frac{\pi t}{6}\right)
$$

To find the volume function $$ V(t) $$, we need to integrate $$ V'(t) $$ and apply the initial condition $$ V(0) = 6 \, \text{L} $$.

1. **Integrate $$ V'(t) $$**:

$$
V(t) = \int V'(t) \, dt = \int -\frac{\pi}{6} \sin\left(\frac{\pi t}{6}\right) \, dt
$$

Using the integral of sine, we have:

$$
\int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) + C
$$

In our case, $$ k = \frac{\pi}{6} $$:

$$
V(t) = -\frac{\pi}{6} \left(-\frac{6}{\pi} \cos\left(\frac{\pi t}{6}\right)\right) + C
$$

This simplifies to:

$$
V(t) = \cos\left(\frac{\pi t}{6}\right) + C
$$

2. **Apply the initial condition $$ V(0) = 6 $$**:

$$
V(0) = \cos(0) + C = 1 + C = 6
$$

Thus, $$ C = 5 $$.

3. **Final Volume Function**:

$$
V(t) = 5 + \cos\left(\frac{\pi t}{6}\right)
$$

### Part b: Breathing Rate in Breaths/Minute

To find the breathing rate, we need to determine the period of the function $$ V(t) $$. The function $$ V(t) $$ oscillates due to the cosine term.

1. **Determine the period of $$ V(t) $$**:

The period $$ T $$ of the function $$ \cos\left(\frac{\pi t}{6}\right) $$ can be found using the formula for the period of cosine:

$$
T = \frac{2\pi}{k}
$$

where $$ k = \frac{\pi}{6} $$:

$$
T = \frac{2\pi}{\frac{\pi}{6}} = 12 \, \text{seconds}
$$

2. **Convert the period to breaths per minute**:

Since one complete cycle (breath) takes 12 seconds, the number of breaths per minute is:

$$
\text{Breaths per minute} = \frac{60 \, \text{seconds}}{12 \, \text{seconds}} = 5 \, \text{breaths/minute}
$$

### Final Answers

- **Volume Function**: $$ V(t) = 5 + \cos\left(\frac{\pi t}{6}\right) $$
- **Breathing Rate**: $$ 5 $$ breaths/minute
The volume function is $$ V(t) = 5 + \cos\left(\frac{\pi t}{6}\right) $$ and the breathing rate is 5 breaths per minute.

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