Question

How We Cough When we cough, the trachea (windpipe) contracts to increase the velocity of the air going out. This raises the question 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 \(v(\) in \(\mathrm{cm} / \mathrm{sec})\) can be modeled by the equation $$v=c\left(r_{0}-r\right) r^{2}, \quad \frac{r_{0}}{2} \leq r \leq r_{0}$$ where \(r_{0}\) is the rest radius of the trachea in \(\mathrm{cm}\) and \(c\) is a positive constant whose value depends in part on the length of the trachea. (a) 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 to the claim that \(v\) is a maximum when \(r=(2 / 3) r_{0}\) .

Short answer

Velocity of airflow during a cough is maximized when the radius of the trachea is two-thirds of its resting value (it is contracted by about 33%). This can also be shown empirically by plotting the function for given values.

Step-by-step solution

Understanding the function

The first step is to understand that the velocity \(v\) is a function of \(r\), the trachea's radius during a cough. The aim is to find the value of \(r\) which results in the maximum velocity.

Differentiate the velocity function

To find the maximum of a function, it's necessary to take the derivative and set it equal to zero, then solve for the variable. The derivative of \(v\) with respect to \(r\) is given by the equation \(v' = c\left(2r-r_{0}\right) r \).

Find the critical points

Setting the derivative \(v'\) equal to zero and solving for \(r\), we find that \(r = 0\) and \(r =(2 / 3) r_{0}\) are critical points.

Determine the maximum

We can determine that \(r =(2 / 3) r_{0}\) results in a maximum of \(v\) by either evaluating the second derivative or by checking the behavior of the function around this point.

Evaluate and plot the function

In the second part of the exercise, we are asked to evaluate and plot the function for \(r_{0}=0.5\) and \(c=1\). For this particular case, our function becomes \(v=(0.5-r) r^{2}\). Plotting this function over the interval \(0 \leq r \leq 0.5\) will further confirm that the maximum velocity is obtained for \(r=(2 / 3) r_{0}\).