When a foreign object lodged in the trachea (windpipe) forces a person to cough, the diaphragm...

Question:

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 then 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 {eq}v {/eq} of the airstream is related to the radius {eq}r {/eq} of the trachea by the equation: {eq}v(r) = k (r_0 - r) r^2 , \frac{r_0}2 \geq r \geq r_0 {/eq} where {eq}k {/eq} is the constant and {eq}r_0 {/eq} is the normal radius of the trachea. The restriction on {eq}r {/eq} is due to the fact that the tracheal wall stiffens under pressure and a contraction greater than {eq}\frac{r_0}2 {/eq} is prevented, otherwise the person would suffocate.

A) Determine the value of r in the interval {eq}[\frac{r_0}2 , r_0 ] {/eq} at which {eq}v {/eq} has an absolute maximum.

B) What is the absolute maximum value of {eq}v {/eq} on the interval?

Absolute Maximum:

The maximum values of a function over an interval are determined:

from the critical points of the interval

and the ends of the interval

Among them is the value where the absolute extremum of the function is reached.

Answer and Explanation:

To determine the absolute maximum of the function over the interval, first, we need to determine the critical points of the interval:

{eq}v(r) = k({r_0} - r){r^2},\quad \frac{{{r_0}}}{2} \le r \le {r_0}\\ \\ v'(r) = k{r^2} + k({r_0} - r)2r\\ v'(r) = kr\left( {r + 2({r_0} - r)} \right)\\ v'(r) = kr\left( {r + 2{r_0} - 2r} \right)\\ v'(r) = kr\left( {2{r_0} - r} \right) = 0\\ \\ \left\{ \begin{array}{l} r = 0\\ 2{r_0} - r = 0 \to r = 2{r_0} \end{array} \right. {/eq}

According to the results we don't have critical points over the interval. So, the absolute maximum will be reached at the ends of the interval:

{eq}\left\{ \begin{array}{l} v(\frac{{{r_0}}}{2}) = k({r_0} - \frac{{{r_0}}}{2}){\left( {\frac{{{r_0}}}{2}} \right)^2} = k{\left( {\frac{{{r_0}}}{2}} \right)^3}\\ v({r_0}) = k({r_0} - {r_0}){r^2} = 0 \end{array} \right. {/eq}

A) So, the value of {eq}r {/eq} in the interval at which the {eq}v {/eq} has absolute maximum is {eq}r = \frac{{{r_0}}}{2}. {/eq}

B) For this value of the variable {eq}r {/eq}, the maximum value of the function is {eq}v\left( {\frac{{{r_0}}}{2}} \right) = k{\left( {\frac{{{r_0}}}{2}} \right)^3} {/eq}


Learn more about this topic:

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Find the Maximum Value of a Function: Practice & Overview

from CAHSEE Math Exam: Tutoring Solution

Chapter 10 / Lesson 12
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