Problem 129 - Precalculus Essentials

Question

Use this information to solve. Our cycle of normal breathing takes place every 5 seconds. Velocity of air flow, \(y,\) measured in liters per second, after \(x\) seconds is modeled by $$y=0.6 \sin \frac{2 \pi}{5} x$$ Velocity of air flow is positive when we inhale and negative when we exhale. Within each breathing cycle, when are we inhaling at a rate of 0.3 liter per second? Round to the nearest tenth of a second.

Step-by-Step Solution

Answer

The inhalation rate is 0.3 liters per second at 0.83 and 4.17 seconds into each breathing cycle.

Step 1: Understand the problem.

We need to find the time within each breathing cycle when the velocity of air flow is 0.3 liters per second. From the given equation, \(y=0.6 \sin \frac{2 \pi}{5} x\), we can see that the velocity of inhaling is modeled by a sinusoidal function.

Step 2: Set up the equation.

We are given that the velocity of inhaling, \(y\), is 0.3 liters per second. Replace \(y\) with 0.3 in the equation as follows: 0.3 = 0.6 \sin \frac{2 \pi}{5} x.

Step 3: Solve for \(x\)

To solve for \(x\), first divide both sides of the equation by 0.6. This simplifies to: \(\sin \frac{2 \pi}{5} x = 0.5\). Then, use the arcsine function to solve for the angle inside the sine function. Remember that the solutions to sine equations are of the form \(\frac{\pi}{2} + 2n\pi\) and \(\frac{3\pi}{2} + 2n\pi\), where \(n\) is an integer. This yields \(\frac{2 \pi}{5} x = \frac{\pi}{6} + 2n\pi\) or \(\frac{2 \pi}{5} x = \frac{5\pi}{6} + 2n\pi\). Now we can solve for \(x\) by multiplying through by \(\frac{5}{2 \pi}\).

Step 4: Find the times within the breathing cycle

Solving for \(x\) gives \(x = \frac{5}{6} + 5n\) and \(x = \frac{25}{6} + 5n\). To find the times that occur within a 5-second breathing cycle (0 < \(x\) < 5), we plug integers \(n\) into the two above equations one by one until we exceed the upper bound of 5 seconds. For \(n = 0\), we get \(x = 0.83\) and \(x = 4.17\). Both values are within the range (0, 5) and therefore represent the times within one breathing cycle when the inhalation rate is 0.3 liters per second.

Key Concepts

Sinusoidal Functions

Sinusoidal functions are periodic functions that exhibit a wave-like pattern. They are among the most essential functions in trigonometry and are primarily represented by sine, cosine, and tangent functions.
In the context of the breathing cycle, we use a sinusoidal function to model the velocity of air flow, which fluctuates as we inhale and exhale. The base function for a sine wave is expressed as:
\[ y = A imes ext{sin}(kx + heta) + D \]

• \(A\) is the amplitude, determining the height of the peaks from the centerline.
• \(k\) affects the period, which is how long it takes for the function to complete one cycle.
• \(\theta\) is the phase shift, adjusting the wave left or right.
• \(D\) is the vertical shift, moving the entire function up or down.

Our problem utilizes the equation \( y=0.6 \sin \frac{2 \pi}{5} x \), where the amplitude (0.6) indicates the maximum rate of inhalation or exhalation, and the period is derived from \( \frac{2\pi}{5} \), reflecting a cycle duration of 5 seconds.
Understanding these components allows us to predict changes in the wave, such as peak inhalation and exhalation rates.

Solving Equations

To solve the trigonometric equation in the breathing model, we identify when the inhalation rate reaches a specific value, namely 0.3 liters per second.
Start by equating the sinusoidal function to 0.3:\[ 0.3 = 0.6 \sin \frac{2 \pi}{5} x \]
Here's a step-by-step approach:

• Divide both sides by 0.6 to isolate the sine function: \( \sin \frac{2 \pi}{5} x = 0.5 \).
• Solve for \(x\) using the inverse sine, giving us solutions at \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \), along with their periodic equivalents.
• These correspond to the general solution forms \( \frac{\pi}{6} + 2n\pi \) and \( \frac{5\pi}{6} + 2n\pi \), where \( n \) is any integer.
• Solve for \( x \) by multiplying through by the reciprocal of \( \frac{2 \pi}{5} \), resulting in periods of \( \frac{5}{6} + 5n \) and \( \frac{25}{6} + 5n \).

By determining \( x \) within a single breathing cycle (0 to 5 seconds), we can find specific times at which the inhalation rate hits 0.3 liters per second.

Breathing Cycle Modeling

Modeling the breathing cycle with trigonometric equations offers invaluable insight into how our lungs handle air intake and output.
In our exercise, the sinusoidal function represents a continuous cycle of inhalation and exhalation over a fixed period, simulating a typical breathing pattern.
A complete cycle mimics the rise and fall of air flow velocity as follows:

• Positive velocity represents the inhalation phase, where air is drawn into the lungs.
• Negative velocity indicates exhalation, where air is expelled.

The periodic function is crucial since human breathing typically follows this structured rhythm. By understanding where the inhaling rate is 0.3 liters per second, we can pinpoint exact times of specific airflow during the cycle.
This information can be useful in respiratory studies, enhancing the predictability and monitoring of breathing patterns, especially for healthcare applications.