Question

when a foreign object lodged in the trachea 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)=kleft(r_{0}-r
ight) r^{2}, quad \frac{1}{2} r_{0} leq r leq 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}
ight ) at which v has an absolute maximum.

( t= )

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

( v= )

(c) sketch the graph of v on the interval ( left0, r_{0}
ight ).

Explanation:

Step1: Find the derivative of \(v(r)\)

Given \(v(r)=k(r_{0}-r)r^{2}=kr_{0}r^{2}-kr^{3}\).
Using the power rule \((x^{n})^\prime = nx^{n - 1}\), the derivative \(v^\prime(r)=2kr_{0}r-3kr^{2}=kr(2r_{0}-3r)\).

Step2: Find the critical points

Set \(v^\prime(r) = 0\).
Since \(kr(2r_{0}-3r)=0\) and \(k
eq0\) (a non - zero constant), we have two cases:

• Case 1: \(r = 0\) (but \(r\in[\frac{1}{2}r_{0},r_{0}]\), so we discard \(r = 0\)).
• Case 2: \(2r_{0}-3r=0\), which gives \(r=\frac{2}{3}r_{0}\).

We also need to check the endpoints of the interval \(r=\frac{1}{2}r_{0}\) and \(r = r_{0}\).
When \(r=\frac{1}{2}r_{0}\), \(v(\frac{1}{2}r_{0})=k(r_{0}-\frac{1}{2}r_{0})(\frac{1}{2}r_{0})^{2}=k(\frac{1}{2}r_{0})(\frac{1}{4}r_{0}^{2})=\frac{1}{8}kr_{0}^{3}\).
When \(r=\frac{2}{3}r_{0}\), \(v(\frac{2}{3}r_{0})=k(r_{0}-\frac{2}{3}r_{0})(\frac{2}{3}r_{0})^{2}=k(\frac{1}{3}r_{0})(\frac{4}{9}r_{0}^{2})=\frac{4}{27}kr_{0}^{3}\).
When \(r = r_{0}\), \(v(r_{0})=k(r_{0}-r_{0})r_{0}^{2}=0\).

Step3: Compare the values

Since \(\frac{4}{27}kr_{0}^{3}>\frac{1}{8}kr_{0}^{3}>0\) (because \(\frac{4}{27}\approx0.148\) and \(\frac{1}{8}=0.125\)).

Answer:

(a) \(r=\frac{2}{3}r_{0}\)
(b) \(v=\frac{4}{27}kr_{0}^{3}\)