Problem 83
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 0.5 L/s. This explains, in part, why the function \( f(t) = \frac{1}{2} \sin (2\pi t/5) \) 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 \( t \).
Step-by-Step Solution
Verified Answer
The inhaled volume at time \( t \) is \( V(t) = \frac{5}{4\pi}(1 - \cos(\frac{2\pi t}{5})) \).
1Step 1: Understanding the Problem
We are given that the rate of air flow into the lungs is modeled by the function \( f(t) = \frac{1}{2} \sin (\frac{2\pi t}{5}) \), where \( t \) is measured in seconds. We need to find the volume of air inhaled into the lungs at a time \( t \). This involves integrating the rate function over a specified time interval.
2Step 2: Setting Up the Integral
The volume of air inhaled at time \( t \) is the integral of \( f(x) \) from 0 to \( t \), which can be written as \( V(t) = \int_{0}^{t} \frac{1}{2} \sin \left( \frac{2\pi x}{5} \right) \, dx \).
3Step 3: Calculating the Integral
To integrate \( \frac{1}{2} \sin \left( \frac{2\pi x}{5} \right) \), we'll use the substitution method. Let \( u = \frac{2\pi x}{5} \), then \( du = \frac{2\pi}{5} \, dx \) or \( dx = \frac{5}{2\pi} \, du \).
4Step 4: Substituting and Simplifying
Substitute \( x \) with \( u \): \( V(t) = \int_{u=0}^{u=\frac{2\pi t}{5}} \frac{1}{2} \sin(u) \cdot \frac{5}{2\pi} \, du = \frac{5}{4\pi} \int_{0}^{\frac{2\pi t}{5}} \sin(u) \, du \).
5Step 5: Integrating the Function
The integral of \( \sin(u) \) is \( -\cos(u) \), thus the integral becomes: \( \frac{5}{4\pi} \left[ -\cos(u) \right]_{0}^{\frac{2\pi t}{5}} \).
6Step 6: Evaluating the Integral
Evaluate \( -\cos(u) \) from \( 0 \) to \( \frac{2\pi t}{5} \): \(-\cos\left(\frac{2\pi t}{5}\right) + \cos(0)\). Since \( \cos(0) = 1 \), this simplifies to \( 1 - \cos\left(\frac{2\pi t}{5}\right) \).
7Step 7: Calculating the Inhaled Volume
Substitute back into the integrated function: \( V(t) = \frac{5}{4\pi} (1 - \cos(\frac{2\pi t}{5})) \). This gives the volume of air inhaled up to time \( t \).
8Step 8: Final Answer
Thus, the volume of air inhaled at a time \( t \) using the model \( f(t) = \frac{1}{2} \sin \left(\frac{2\pi t}{5}\right) \) is \( V(t) = \frac{5}{4\pi} \left( 1 - \cos\left(\frac{2\pi t}{5}\right) \right) \).
Key Concepts
Respiratory Cycle ModelingTrigonometric FunctionsSubstitution MethodIntegral Evaluation
Respiratory Cycle Modeling
Respiratory cycle modeling is a fascinating aspect of how mathematics can simulate real-world biological processes. In this context, the respiratory cycle refers to the full process of breathing in and out, which typically lasts about 5 seconds in a healthy adult at rest. By modeling the breathing cycle, scientists can predict how much air flows into the lungs over time, allowing for insights into various health conditions and the effectiveness of different treatments.
In our example, the maximum airflow rate is about 0.5 liters per second, which can be mathematically represented by the function \( f(t) = \frac{1}{2} \sin \left(\frac{2\pi t}{5}\right) \). This function helps to understand the cyclic nature of how air enters and leaves the lungs. This model assumes a repeating cycle, with smooth transitions that are typical of sinusoidal functions, and aids in predicting inhaled air volume during different stages of the respiratory cycle.
In our example, the maximum airflow rate is about 0.5 liters per second, which can be mathematically represented by the function \( f(t) = \frac{1}{2} \sin \left(\frac{2\pi t}{5}\right) \). This function helps to understand the cyclic nature of how air enters and leaves the lungs. This model assumes a repeating cycle, with smooth transitions that are typical of sinusoidal functions, and aids in predicting inhaled air volume during different stages of the respiratory cycle.
Trigonometric Functions
Trigonometric functions, like sine and cosine, are integral in modeling periodic phenomena, such as oscillations or waves. In the respiratory cycle modeling, the function \( \sin \left(\frac{2\pi t}{5}\right) \) represents the sinusoidal nature of breathing. This makes sense, as breathing can be thought of as a repeating wave form.
The sine function oscillates between -1 and 1, and when scaled and shifted, can effectively represent the rate of airflow in the lungs over time. The factor \( \frac{1}{2} \) scales the amplitude, adjusting the maximum value of the function to fit the realistic flow rates. Similarly, \( \frac{2\pi}{5} \) scales the x-axis, harmonizing each cycle with the typical duration of a single respiratory cycle.
Understanding trigonometric functions is essential as they provide insight into the periodic nature of various physical processes and phenomenon, including everything from daily temperature changes to the natural rhythms of the human body.
The sine function oscillates between -1 and 1, and when scaled and shifted, can effectively represent the rate of airflow in the lungs over time. The factor \( \frac{1}{2} \) scales the amplitude, adjusting the maximum value of the function to fit the realistic flow rates. Similarly, \( \frac{2\pi}{5} \) scales the x-axis, harmonizing each cycle with the typical duration of a single respiratory cycle.
Understanding trigonometric functions is essential as they provide insight into the periodic nature of various physical processes and phenomenon, including everything from daily temperature changes to the natural rhythms of the human body.
Substitution Method
The substitution method is an effective technique for solving integrals, particularly when dealing with functions that are products or composites. In the exercise, we encounter the integral of the function \( \frac{1}{2} \sin \left(\frac{2\pi x}{5}\right) \).
The substitution method allows us to simplify the integration process. By setting \( u = \frac{2\pi x}{5} \), we transform both the variable and the differential. This substitution modifies the integral into a simpler form, making it easier to solve: \( dx = \frac{5}{2\pi} \, du \).
After substitution, the integral becomes \( \frac{5}{4\pi} \int \sin(u) \, du \), which is straightforward to evaluate. By substituting the variable and simplifying the integrand, we effectively reduce the problem to a basic trigonometric integral, demonstrating the power and utility of the substitution method.
The substitution method allows us to simplify the integration process. By setting \( u = \frac{2\pi x}{5} \), we transform both the variable and the differential. This substitution modifies the integral into a simpler form, making it easier to solve: \( dx = \frac{5}{2\pi} \, du \).
After substitution, the integral becomes \( \frac{5}{4\pi} \int \sin(u) \, du \), which is straightforward to evaluate. By substituting the variable and simplifying the integrand, we effectively reduce the problem to a basic trigonometric integral, demonstrating the power and utility of the substitution method.
Integral Evaluation
Evaluating integrals is a key skill in calculus, necessary for determining the area under curves, which corresponds to the cumulative effect over time in many disciplines. In our example, once the substitution simplifies the integral to \( \frac{5}{4\pi} \int \sin(u) \, du \), we can easily solve for the function's antiderivative.
The antiderivative of \( \sin(u) \) is \( -\cos(u) \). When evaluated from the limits of \( u = 0 \) to \( u = \frac{2\pi t}{5} \), we find that the result is \( -\cos\left(\frac{2\pi t}{5}\right) + \cos(0) \). Since \( \cos(0) = 1 \), this boils down to evaluating \( 1 - \cos\left(\frac{2\pi t}{5}\right) \).
This definite integration ultimately provides us with the volume \( V(t) = \frac{5}{4\pi} \left( 1 - \cos\left(\frac{2\pi t}{5}\right) \right) \), offering insight into the total amount of air inhaled over a specific time period. Integral evaluation thus bridges the gap between instantaneous rates and cumulative totals.
The antiderivative of \( \sin(u) \) is \( -\cos(u) \). When evaluated from the limits of \( u = 0 \) to \( u = \frac{2\pi t}{5} \), we find that the result is \( -\cos\left(\frac{2\pi t}{5}\right) + \cos(0) \). Since \( \cos(0) = 1 \), this boils down to evaluating \( 1 - \cos\left(\frac{2\pi t}{5}\right) \).
This definite integration ultimately provides us with the volume \( V(t) = \frac{5}{4\pi} \left( 1 - \cos\left(\frac{2\pi t}{5}\right) \right) \), offering insight into the total amount of air inhaled over a specific time period. Integral evaluation thus bridges the gap between instantaneous rates and cumulative totals.
Other exercises in this chapter
Problem 82
A bacteria population starts with 400 bacteria and grows at a rate of \( r(t) = (450.268)e^{1.12567t} \) bacteria per hour. How many bacteria will there be afte
View solution Problem 82
Let \( f(x) = \left\\{ \begin{array}{ll} 0 & \mbox{if \) x 2 \(} \end{array} \right.\) and $$ g(x) = \int^x_0 f(t) \, dt $$ (a) Find an expression for \( g(x) \
View solution Problem 83
Find a function \( f \) and a number \( a \) such that \( \displaystyle 6 + \int^x_a \frac{f(t)}{t^2} \, dt = 2 \sqrt{x} \) for all \( x > 0 \)
View solution Problem 85
Dialysis treatment removes urea and other waste products from a patient's blood by diverting some of the bloodflow externally through a machine called a dialyze
View solution