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Chapter 4: Problem 85

Respiratory Cycle Suppose that the rate at which air is inhaled by a person during respiration is $$ r(t)=\frac{3}{5} \sin \frac{\pi t}{2} $$ liters per second, at time \(t .\) Find \(V(t)\), the volume of inhaled air in the lungs at any time \(t .\) Assume that \(V(0)=0 .\)

Short Answer

Expert verified
The function for the volume of inhaled air in the lungs at any time \(t\) is \(V(t) =\frac{-6}{5\pi}\cos(\frac{\pi t}{2}) + \frac{6}{5\pi}\).

Step by step solution

01

Write down the given function and initial condition

We are given that the rate of air inhaled is: \(r(t) = \frac{3}{5} \sin(\frac{\pi t}{2})\) And the initial condition is: \(V(0) = 0\)
02

Calculate the integral of the function

Since volume is the integral of the rate, we can find the volume function by taking the integral of \(r(t)\): \(V(t) = \int r(t) dt = \int \frac{3}{5} \sin(\frac{\pi t}{2}) dt\)
03

Evaluate the integral

We can evaluate the integral by applying the rules of integration: \(V(t) = \int \frac{3}{5} \sin(\frac{\pi t}{2}) dt = \frac{3}{5} \int \sin(\frac{\pi t}{2}) dt\) Now, we make a substitution: Let \(u = \frac{\pi t}{2}\), then \(du = \frac{\pi}{2} dt\) Therefore, \(dt = \frac{2}{\pi} du\) Substituting the values in the integral equation: \(V(t) = \frac{3}{5} \int \sin(u) \cdot \frac{2}{\pi} du = \frac{6}{5\pi} \int \sin(u) du\) Now, integrating the sine function: \(V(t) = \frac{6}{5\pi} (-\cos(u)) + C = \frac{-6}{5\pi} \cos(u) + C\) To find the value of the constant C, we use the initial condition \(V(0) = 0\).
04

Apply the initial condition

Given that \(V(0) = 0\), and substituting the value of \(u\) back in terms of \(t\): \(0 = \frac{-6}{5\pi} \cos(\frac{\pi(0)}{2}) + C = \frac{-6}{5\pi}\cos(0)+C\) Since \(\cos(0) = 1\): \(0 = \frac{-6}{5\pi} + C\) Solving for C: \(C = \frac{6}{5\pi}\)
05

Write the final volume function

Now that we have the constant C, we can write down the final volume function by substituting the value of C and the original value of u: \(V(t) = \frac{-6}{5\pi} \cos(\frac{\pi t}{2}) + \frac{6}{5\pi}\) This is the function for the volume of inhaled air in the lungs at any time \(t\).

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Key Concepts

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

Rate of Change
The concept of "rate of change" refers to how fast a certain quantity changes over time. In the context of the given exercise, the rate of change is represented by the function \(r(t) = \frac{3}{5} \sin\left(\frac{\pi t}{2}\right)\), which describes how quickly air is inhaled by a person. Here, the function is expressed in terms of time \(t\), with units of liters per second.
  • In graphs, the rate of change can be viewed as the slope at any point on a curve.
  • This particular function is sinusoidal, meaning it represents periodic breathing cycles of inhalation and exhalation.
Understanding the rate of change helps us see the dynamics of the respiratory cycle and how quickly or slowly air fills the lungs at any given second.
Volume Calculation
Volume calculation in calculus involves determining how much of a substance accumulates over a certain period. In this case, the exercise requires finding the total volume of air inhaled, \(V(t)\), by integrating the rate of inhalation function \(r(t)\).
  • To find \(V(t)\), you calculate the definite integral of \(r(t)\).
  • The volume calculation not only accounts for the constant inflow of air but also the way breathing patterns oscillate over time.
Solving for \(V(t)\) involves integrating the function derived from the rate of change to accumulate the total volume.
Definite Integral
A definite integral is a powerful calculus tool used to find the total accumulation of a quantity over an interval. In this task, we evaluate the definite integral of \(r(t)\) to find the total volume of air inhaled over time \(t\).
  • A definite integral requires evaluating the integral function at both the upper and lower limits of the interval.
  • In our exercise, the initial condition \(V(0)=0\) provides the lower limit of integration, ensuring accurate volume calculation from a starting point.
By integrating \(r(t)\), we proceed through the process of finding an antiderivative, applying limits, and using initial conditions to determine volume, ultimately reflecting the accumulated change from a point of rest.
Trigonometric Integration
Trigonometric integration involves integrating functions that include trigonometric terms like sine, cosine, and tangent. In the given problem, the integrand \( \frac{3}{5} \sin\left(\frac{\pi t}{2}\right) \) includes a sine function which characterized the oscillatory nature of inhalation during respiratory cycles.
  • Trigonometric integrals can often be solved by using substitutions to simplify the process.
  • For our function, we substitute \(u=\frac{\pi t}{2}\) to ease integration.
This substitution allows for easier calculation, converting the integrand into a more manageable form \(\int \sin(u) du\). Post substitution, applying integration rules to solve for the volume function illustrates the importance of understanding both the mechanics of integration and the behavior of trigonometric functions in dynamic systems like human respiration.

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Most popular questions from this chapter

The voltage in an AC circuit is given by $$V=V_{0} \sin \omega t$$ a. Show that the average (mean) voltage from \(t=0\) to \(t=\pi / \omega\) (a half-cycle) is \(V_{\mathrm{av}}=(2 / \pi) V_{0}\), which is \(2 / \pi\) \(\left(\right.\) about \(\left.\frac{2}{3}\right)\) times the maximum voltage \(V_{0}\). b. Show that the average voltage over a complete cycle is \(0 .\) Explain.

If \(f(x)=\int_{2}^{x} \frac{d t}{\sqrt{1+t^{3}}}\), where \(x>-1\), what is \(\left(f^{-1}\right)^{\prime}(0)\) ?

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Air Pollution According to the South Coast Air Quality Management District, the level of nitrogen dioxide, a brown gas that impairs breathing, present in the atmosphere on a certain June day in downtown Los Angeles is approximated by $$A(t)=0.03 t^{3}(t-7)^{4}+62.7 \quad 0 \leq t \leq 7$$ where \(A(t)\) is measured in pollutant standard index and \(t\) is measured in hours with \(t=0\) corresponding to 7 A.M. What is the average level of nitrogen dioxide present in the atmosphere from 7 A.M. to 2 P.M. on that day?

Evaluate \(\lim _{h \rightarrow 0} \frac{1}{h} \int_{2}^{2+h} \sqrt{5+t^{2}} d t\).
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