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Chapter 6: Problem 19

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 0.5 \(\mathrm{L} / \mathrm{s} .\) This explains, in part, why the function \(f(t)=\frac{1}{2} \sin \frac{2 \pi t}{5}\) has often been used to model the rate of air flow into the lungs. If inhalation occurs during the interval 0\(\leqslant t \leqslant 2.5\) what is the average rate of air flow during inhalation?

Short Answer

Expert verified
The average rate of air flow during inhalation is \(\frac{1}{2\pi}\) L/s.

Step by step solution

01

Understanding the Function

The function given is \(f(t) = \frac{1}{2} \sin \frac{2 \pi t}{5}\), which models the rate of airflow in liters per second. The variable \(t\) represents time in seconds during the breathing cycle.
02

Define the Interval for Calculation

The interval for inhalation is \([0, 2.5]\) as it indicates the time from the start of inhalation to its completion. We'll calculate the average flow rate over this interval.
03

Set Up the Average Value Formula

The average value of a function \(f(t)\) over the interval \([a, b]\) is given by: \[\text{Average} = \frac{1}{b-a} \int_{a}^{b} f(t) \, dt\]Substitute \(a = 0\) and \(b = 2.5\), so the formula becomes: \[\text{Average} = \frac{1}{2.5-0} \int_{0}^{2.5} \left(\frac{1}{2} \sin \frac{2 \pi t}{5}\right) \, dt\]
04

Compute the Definite Integral

Calculate the integral: \[\int_{0}^{2.5} \frac{1}{2} \sin \frac{2 \pi t}{5} \, dt\]Use the substitution \(u = \frac{2\pi}{5}t\), so \(du = \frac{2\pi}{5} \, dt\) or \(dt = \frac{5}{2\pi} \, du\). Transform the limits: - when \(t = 0\), \(u = 0\) - when \(t = 2.5\), \(u = \pi\)The integral becomes:\[\frac{1}{2} \times \frac{5}{2\pi} \int_{0}^{\pi} \sin(u) \, du\]
05

Integrate and Simplify

Integral of \(\sin(u)\) is \(-\cos(u)+ C\). Evaluate from \(0\) to \(\pi\): \[\frac{1}{2} \times \frac{5}{2\pi} \left[-\cos(u) \right]_{0}^{\pi} = \frac{1}{2} \times \frac{5}{2\pi} \left[-\cos(\pi) + \cos(0)\right]\] Substituting the values, \(-\cos(\pi) = 1\) and \(\cos(0) = 1\), results in:\[\frac{1}{2} \times \frac{5}{2\pi} (1+1) = \frac{5}{2\pi} \times \frac{2}{2} = \frac{5}{2\pi}\]
06

Calculate the Average Function Value

Now, compute the average rate of air flow:\[\text{Average} = \frac{1}{2.5} \times \frac{5}{2\pi} = \frac{1}{2\pi}\]This represents the average rate of airflow during inhalation in liters per second.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Respiratory Cycle
In human physiology, the respiratory cycle encompasses two main phases: inhalation and exhalation. This cycle is crucial for gas exchange, allowing oxygen to enter the body and carbon dioxide to be expelled. Typically, a full respiratory cycle takes about 5 seconds.
During inhalation, the diaphragm contracts and moves downward, creating a vacuum that pulls air into the lungs. Exhalation follows as the diaphragm relaxes, pushing air out. Understanding this cyclical process is vital to grasp how the rate of air flow into the lungs changes over time.
By analyzing the inhalation phase, scientists can model breathing patterns using mathematical functions, helping to study and perhaps improve respiratory health.
Trigonometric Functions
Trigonometric functions like sine and cosine are mathematical tools derived from the unit circle. They help model periodic phenomena due to their wave-like oscillations, which are ideal for representing cycles like breathing.
In our problem, the function used is a modified sine function, specifically: \(f(t) = \frac{1}{2} \sin \frac{2 \pi t}{5}\).
  • The value of \(\frac{1}{2}\) scales the function, representing half of the maximum air flow rate observed in the respiratory cycle.
  • The term \(\frac{2\pi t}{5}\) adjusts the function's frequency to match a complete respiratory cycle over 5 seconds.
These adjustments indicate how air flow varies with time, capturing the natural rhythm of breathing.
Average Value Theorem
The Average Value Theorem is fundamental in calculating the mean value of a function over an interval. This theorem states that the average value of a continuous function \(f(t)\) over the interval \([a, b]\) can be found using the formula:\[\text{Average} = \frac{1}{b-a} \int_{a}^{b} f(t) \, dt.\]
This can be applied to find the average air flow during inhalation. By integrating the function \(f(t) = \frac{1}{2} \sin \frac{2 \pi t}{5}\) over the interval \([0, 2.5]\) and dividing by the interval's length, we find the average rate of air flow.
In this exercise, executing these calculations provides the average value of \(\frac{1}{2\pi}\), which offers an insight into the normal breathing pattern during inhalation.

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