Problem 64 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

Maximum velocity occurs when \( r = \frac{2}{3}r_0 \). This confirms the trachea contracts about 33% during a cough.

Step 1: Understanding the Formula

We are given the formula for the velocity of airflow as \( v = c(r_0-r)r^2 \) where \( r_0 \) is the rest radius of the trachea. This formula will allow us to determine the velocity based on changes in the radius \( r \).

Step 2: Objective of the Exercise

We need to find the value of \( r \) that maximizes the velocity \( v \). Furthermore, we want to verify whether \( v \) is maximized when \( r = \frac{2}{3}r_0 \).

Step 3: Calculating the Derivative of v

To maximize \( v \), first find the derivative \( \frac{dv}{dr} \). This becomes \( c(2r_0r - 3r^2) \) when we apply the product rule on the function \( v = c(r_0-r)r^2 \).

Step 4: Setting the Derivative to Zero

Set \( \frac{dv}{dr} = 0 \) to find the critical points, giving us the equation \( 2r_0r - 3r^2 = 0 \). Factoring out \( r \), we get \( r(2r_0 - 3r) = 0 \). This gives solutions \( r = 0 \) and \( r = \frac{2}{3}r_0 \).

Step 5: Analyzing the Critical Points

Since \( r = 0 \) is not practical (as it would mean no airflow), the feasible solution is \( r = \frac{2}{3}r_0 \), suggesting maximal velocity when the trachea contracts by \( 33\% \).

Step 6: Substitute and Graph with Given Values

Substitute \( r_0 = 0.5 \) and \( c = 1 \) into the equation \( v = (0.5 - r)r^2 \) and graph this for \( 0 \leq r \leq 0.5 \). Locate the peak of this graph.

Step 7: Conclusion from Graph

The graph demonstrates that the peak of \( v \) occurs very close to \( r = \frac{2}{3} \times 0.5 = 0.333 \), which confirms the theoretical prediction that the trachea contracts by about 33% during a cough.

Key Concepts

Derivatives

In calculus, derivatives are fundamental for understanding how functions change. A derivative gives the rate at which a function varies with respect to one of its variables. If you think of a graph, the derivative at a point is the slope of the tangent line to the curve at that point. This concept helps us determine points where functions reach their maximum or minimum values without needing to visually inspect the entire graph.
When dealing with a function like the velocity equation in our exercise, which is given by \( v = c(r_0-r)r^2 \), we take the derivative with respect to \( r \) to assess how small changes in \( r \) affect velocity \( v \). This involves applying the product rule, which is essential when dealing with products of functions.

Critical Points

Critical points in calculus are where the derivative of a function is zero or undefinable. Finding these points is crucial in solving optimization problems. They indicate where functions could potentially reach their maximum or minimum values.
For the velocity function \( v = c(r_0-r)r^2 \), we derived \( \frac{dv}{dr} = c(2r_0r - 3r^2) \) and set it to zero to find critical points. Solving \( 2r_0r - 3r^2 = 0 \) led us to two solutions: \( r = 0 \) and \( r = \frac{2}{3}r_0 \).
In practical terms, \( r = 0 \) is not feasible because it implies no airflow. The critical point \( r = \frac{2}{3}r_0 \) shows that velocity is maximized when the windpipe contracts to this proportion of its original radius.

Product Rule

The product rule is an essential differentiation technique in calculus. It is applied when you need to find the derivative of a product of two functions. Suppose you have a function \( u(x) \) multiplied by another function \( v(x) \). The product rule states that the derivative of the product is given by:

• \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \)

In our problem, the velocity function \( v = c(r_0-r)r^2 \) involves a product of two functions, \( (r_0-r) \) and \( r^2 \). By applying the product rule, we derived \( c(2r_0r - 3r^2) \).
This result is crucial to find the critical points, allowing us to further analyze where the function might reach maximum velocity.

Optimization in Calculus

Optimization in calculus deals with finding maximum or minimum values of functions given certain constraints. It's widely used in problems where you're trying to make something as large or as small as possible. This could involve maximizing efficiency, profit, speed, or in our case, the velocity of airflow during a cough based on the contraction of the trachea.
This exercise involves an optimization problem where, through derivatives and critical points, we determine that the velocity is maximized when the trachea contracts to two-thirds of its original size, or 33%. This conclusion came from setting the derivative of the velocity function to zero and solving for \( r \). The verification through actual X-ray observations of coughing confirms the powerful predictive ability of calculus in real-world applications.
The given example effectively illustrates how calculus tools can be applied to predict and verify biological phenomena, showing the practical side of mathematical optimization.