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Problem 61

When a foreign object lodged in the trachea (windpipe)

forces a person to cough, the diaphragm thrusts upward

causing an increase in pressure in the lungs. This is accompanied by a contraction of the trachea, making a narrower

channel for the expelled air to flow through. For a given

amount of air to escape in a fixed time, it must move faster

through the narrower channel than the wider one. The

greater the velocity of the airstream, the greater the force on

the foreign object. $X$ -rays show that the radius of the circular tracheal tube contracts to about two-thirds of its normal

radius during a cough. According to a mathematical model

of coughing, the velocity $v$ of the airstream is related to the

radius $r$ of the trachea by the equation

$$v(r)=k\left(r_{0}-r\right) r^{2} \quad \frac{1}{2} r_{0} \leqslant r \leqslant r_{0}$$

where $k$ is a constant and $r_{0}$ is the normal radius of the trachea. The restriction on $r$ is due to the fact that the tracheal wall stiffens under pressure and a contraction greater than

$\frac{1}{2} r_{0}$ is prevented (otherwise the person would suffocate).

(a) Determine the value of $r$ in the interval $\left[\frac{1}{2} r_{0}, r_{0}\right]$ at

which $v$ has an absolute maximum. How does this

compare with experimental evidence?

(b) What is the absolute maximum value of $v$ on the interval?

(c) Sketch the graph of $v$ on the interval $\left[0, r_{0}\right] .$

Answer

a. $r=\frac{2}{3} r_{0}, \quad$ agrees with experimental evidence

b. $v\left(\frac{2}{3} r_{0}\right)=\frac{4}{27} r_{0}^{3}$

c. SEE GRAPH

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