Problem 60 - Thomas Calculus in SI Units

Question

a. When we cough, the trachea (windpipe) contracts to increase the velocity of the air going out. This raises the questions of how much it should contract to maximize the velocity and whether it really contracts that much when we cough. Under reasonable assumptions about the elasticity of the tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity \(v\) can be modeled by the equation $$v=c\left(r_{0}-r\right) r^{2} \mathrm{cm} / \mathrm{s}, \quad \frac{r_{0}}{2} \leq r \leq r_{0}$$ where \(r_{0}\) is the rest radius of the trachea in centimeters and \(c\) is a positive constant whose value depends in part on the length of the trachea. Show that \(v\) is greatest when \(r=(2 / 3) r_{0} ;\) that is, when the trachea is about \(33 \%\) contracted. The remarkable fact is that X-ray photographs confirm that the trachea contracts about this much during a cough. b. Take \(r_{0}\) to be 0.5 and \(c\) to be 1 and graph \(v\) over the interval \(0 \leq r \leq 0.5 .\) Compare what you see with the claim that \(v\) is at a maximum when \(r=(2 / 3) r_{0}\)

Step-by-Step Solution

Answer

Velocity is maximized at \( r = \frac{2}{3} r_0 \). For \( r_0 = 0.5 \) and \( c = 1 \), the maximum velocity occurs for \( r \approx 0.333 \).

Step 1: Understand the Velocity Equation

The velocity equation given is: \[ v = c(r_0 - r)r^2 \] This equation determines the air flow velocity in the trachea when the radius contracts during a cough.

Step 2: Differentiate the Velocity Function

To find where the velocity is maximized, we need to differentiate the function with respect to \(r\):\[ \frac{dv}{dr} = c \left(-r^2 + 2r(r_0 - r)\right) \] Simplifying gives us:\[ \frac{dv}{dr} = c(-r^2 + 2r_0r - 2r^2) = c(2r_0r - 3r^2) \] The derivative simplifies to:\[ \frac{dv}{dr} = c(2rr_0 - 3r^2) \].

Step 3: Set Derivative to Zero to Find Critical Points

Set \( \frac{dv}{dr} \) to zero to find critical points:\[ c(2r_0r - 3r^2) = 0 \] This gives:\[ 2r_0r - 3r^2 = 0 \] Factor it:\[ r(2r_0 - 3r) = 0 \] The solutions are \( r = 0 \) or \( 2r_0 = 3r \). Solving for \(r\), we get \( r = \frac{2}{3}r_0 \).

Step 4: Verify Critical Points and Check Intervals

Calculate \( v \) at \( r = 0 \) and \( r = \frac{2}{3}r_0 \). For \( r = 0 \), \[ v = c \cdot r_0 \cdot 0^2 = 0 \]For \( r = \frac{2}{3}r_0 \), substituting back into the velocity equation:\[ v = c \left(r_0 - \frac{2}{3}r_0 \right)\left(\frac{2}{3}r_0\right)^2 \]\[ v = c \left( \frac{1}{3}r_0 \right)\left( \frac{4}{9}r_0^2 \right) = \frac{4}{27}c r_0^3 \] This is a positive number and hence is the maximum value for positive \(c\).

Step 5: Plug Given Values and Graph the Function

Given \(r_0 = 0.5\) and \(c = 1\), the velocity function becomes: \[ v = (0.5 - r)r^2 \] Graph this function for \( 0 \leq r \leq 0.5 \). At \( r = \frac{2}{3} \times 0.5 = \frac{1}{3} \approx 0.333 \), the graph should show the maximum value of \(v\). This confirms the given claims about the maximum velocity.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus used to determine the rate at which a function changes. When tackling an optimization problem involving a function, the process begins with taking the derivative of the function. The derivative, denoted as \( \frac{dv}{dr} \), tells us how the velocity \( v \) changes as the radius \( r \) of the trachea varies. This step is crucial because it helps identify where the velocity function, \( v = c(r_0 - r)r^2 \), may reach a maximum or minimum value. Once we have the derivative, we simplify it to make the next steps more manageable. In this problem, the derivative simplifies to \( c(2rr_0 - 3r^2) \). Finding where the function's rate of change is zero or changes sign can lead us to the critical points. These critical points are essential in determining where a maximum or minimum might occur in the function's behavior.

Critical Points

Critical points are the values of \( r \) where the derivative \( \frac{dv}{dr} \) becomes zero or is undefined. These points are fundamental because they help indicate potential locations of maxima or minima in a function. For this problem, solving the equation \( c(2r_0r - 3r^2) = 0 \) simplifies to finding the roots: \( r = 0 \) and \( r = \frac{2}{3}r_0 \). Though we found two critical points, not both correspond to maxima. By calculating the function's value at these critical points, you can infer which point offers the desired solution. For instance, when \( r = 0 \), the velocity \( v \) is zero, clearly not a maximum; whereas, at \( r = \frac{2}{3}r_0 \), the provided calculations show a positive and much higher value, indicating a maximum.

Graphing Functions

Graphing is an intuitive way to visualize functions' behavior and can reveal critical features like turning points, slopes, and intercepts. By projecting the velocity function \( v = (0.5 - r)r^2 \) over the interval \( 0 \leq r \leq 0.5 \), we can observe the expected maxima visually. When we graph this equation, it should clearly show that the function peaks when \( r \approx 0.333 \). This graphical representation offers a visual confirmation that matches the analytical critic point approach, reinforcing that the maximum is consistent with \( r = \frac{2}{3}r_0 \). Additionally, graphing functions can help compare theoretical predictions with expected real-world outcomes, such as the trachea contraction during a cough.

Maximization Problems

Maximization problems are common in calculus, aiming to find the highest value of a given function within a specified range. They often involve real-world scenarios where resources or conditions need to optimize performance, like maximizing the airflow velocity in this exercise. To solve a maximization problem, we follow a structured approach:

• Determine the function to be maximized.
• Find the derivative to explore how the function changes.
• Identify and calculate the critical points where this change is zero.
• Verify the maximum by evaluating the function at these points.
• Use graphical aids when possible to validate the analytical findings visually.

In this exercise, through differentiation and critical point analysis, we verified that the maximum air velocity occurs at a specific tracheal contraction, corroborated by both theoretical calculations and real-world observations.