Answer to Question #274305 in Calculus for Phenominal1

Question

a. When we cough, the trachea (windpipe) contracts to increase the velocity

of the air going out. This raises the questions 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

y can be modeled by the equation 𝑦 = 𝑐( 𝑟 , ,

0 − 𝑟)𝑟

2

𝑐𝑚/𝑠𝑒𝑐

𝑟0

2 < 𝑟 < 𝑟0

where 𝑟 is the rest radius of the trachea in centimeters and is a positive

𝑐

constant whose value depends in part on the length of the trachea. Show that

𝑦 is greatest when 𝑟 = (2/3)𝑟 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 to be and to be and graph over the interval .

0. 5 𝑐 1 𝑦 0 < 𝑟 < 0. 5

Compare what you see with the claim that 𝑦 is at a maximum when

𝑟 = (2/3)𝑟 .

Answer 1

a. 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 cm/s) can be modeled by the equation

$$v=c(r_0-r)r^2 , r_0/2\le r\le r_0$$

We are finding the absolute maximum of $v$ when $r$ takes values in a closed and bounded interval $[r_0/2, r_0]$

$$v'(r)=c(2r_0r-3r^2)$$

Find the critical number(s)

$$v'(r)=0=>c(2r_0r-3r^2)=0$$

$$r_1=0, r_2=\dfrac{2}{3}r_0$$

Critical numbers: $0, \dfrac{2}{3}r.$

Since $r\in [r_0/2, r_0]$ we take $r=\dfrac{2}{3}r_0$

$$v(\dfrac{1}{2}r_0)=c(r_0-\dfrac{1}{2}r_0)(\dfrac{1}{2}r_0)^2 =\dfrac{c}{8}r_0^3$$

$$v(\dfrac{2}{3}r_0)=c(r_0-\dfrac{2}{3}r_0)(\dfrac{2}{3}r_0)^2 =\dfrac{4c}{27}r_0^3$$

$$v(r_0)=c(r_0-r_0)(r_0)^2 =0$$

$$\dfrac{c}{8}r_0^3=\dfrac{4c}{32}r_0^3<\dfrac{4c}{27}r_0^3, c>0$$

Therefore  $v$ is greatest when $r=\dfrac{2}{3}r_0.$

b. Let $r_0=0.5, c=1$

For $0\le r\le0.5$

$$v=(0.5-r)r^2 , 0\le r\le 0.5$$

 $v$ is the absolute maximum with value of 1/54 when $r=\dfrac{2}{3}r_0=\dfrac{1}{3}.$