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 air-stream is related to the radius r of the trachea by the equation

\(v\left( r \right) = k\left( {{r_{\bf{0}}} - r} \right){r^{\bf{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{\bf{1}}}{{\bf{2}}}{r_{\bf{0}}} \le r \le {r_{\bf{0}}}\)

where k is a constant and \({r_{\bf{0}}}\) 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 \(\frac{{\bf{1}}}{{\bf{2}}}{r_{\bf{0}}}\) is prevented (otherwise the person would suffocate).

(a) Determine the value of r in the interval \(\left( {\frac{{\bf{1}}}{{\bf{2}}}{r_{\bf{0}}},{r_{\bf{0}}}} \right)\) 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 \(\left( {{\bf{0}},{r_{\bf{0}}}} \right)\).

Short answer

(a) The maximum value of v will occur at \(r = \frac{2}{3}{r_0}\).

(b) The maximum value of \(v\left( r \right)\) is \(\frac{4}{{27}}kr_0^3\).

(c) The graph is shown below:

Step-by-step solution

Differentiate the equation of \(v\left( r \right)\)

Differentiate the equation \(v\left( r \right) = k\left( {\frac{1}{2}{r_0} - r} \right){r^2}\) with respect to \(r\).

\(\begin{aligned}{c}v'\left( r \right) &= \frac{{\rm{d}}}{{{\rm{d}}r}}\left( {k\left( {\frac{1}{2}{r_0}{r^2} - {r^3}} \right)} \right)\\ &= k\left( {\frac{1}{2}{r_0}\left( {2r} \right) - 3{r^2}} \right)\\ &= kr\left( {2{r_0} - 3r} \right)\end{aligned}\)

Find the root of the equation \(v'\left( r \right) = {\bf{0}}\)

Solve the equation \(v'\left( r \right) = 0\).

\(\begin{aligned}{c}kr\left( {2{r_0} - 3r} \right) &= 0\\r &= 0,{\rm{ }}\frac{{2{r_0}}}{3}\end{aligned}\)

Find answer for part (a)

As \(\frac{1}{2}{r_0} \le r \le {r_0}\), the value of \(L\left( t \right)\) can be calculated as,

For \(t = \frac{1}{2}{r_0}\),

\(v\left( {\frac{1}{2}{r_0}} \right) = \frac{1}{8}kr_0^3\)

For \(r = \frac{2}{3}{r_0}\),

\(v\left( {\frac{2}{3}{r_0}} \right) = \frac{4}{{27}}kr_0^3\)

And,

\(v\left( {{r_0}} \right) = 0\)

Thus, the maximum value of v will occur at \(r = \frac{2}{3}{r_0}\).

Find answer for part (b)

The maximum value of \(v\left( r \right)\) is \(\frac{4}{{27}}kr_0^3\) at \(r = \frac{2}{3}{r_0}\).

Find answer for part (c)

Draw the figure by using the above values, so the graph below represents the curve of the function \(v\left( r \right) = k\left( {{r_0} - r} \right){r^2}\) as: