Chapter 4.4, Problem 59E

(a.)

To determine

To Show: $v$ is greatest when $r=\frac{2}{3}{r}_0$ .

Answer to Problem 59E

It has been shown that $v$ is greatest when $r=\frac{2}{3}{r}_0$ .

Explanation of Solution

Given:

The average flow velocity $v$ can be modelled by the equation, $v=c\left(r_0-r\right)r^2$ for $\frac{r_0}{2}\le r\le r_0$ , where $r_0$ is the rest radius of the trachea in cm and $c$ is a positive constant whose value depends in part on the length of the trachea.

Concept used:

A function attains maximum value at a point where its first derivative is zero and its second derivative is negative.

Calculation:

It is given that $v=c\left(r_0-r\right)r^2$ for $\frac{r_0}{2}\le r\le r_0$ .

Simplifying,

  $v=c\left(r_0{r}^2-r^3\right)$

Differentiating with respect to $r$ ,

  $v^{\prime }=c\left(2r_{0}r-3r^2\right)$

Simplifying,

  $v^{\prime }=cr\left(2r_0-3r\right)$

Differentiating again with respect to $r$ ,

  $v^{″}=c\left(2r_0-6r\right)$

Equating the first derivative to zero,

  $cr\left(2r_0-3r\right)=0$

This implies that either $r=0$ or $r=\frac{2}{3}{r}_0$ .

Now, it is given that $\frac{r_0}{2}\le r\le r_0$ .

So, $r=0$ is not a feasible solution.

Then, the only feasible solution is $r=\frac{2}{3}{r}_0$ .

Put $r=\frac{2}{3}{r}_0$ in $v^{″}=c\left(2r_0-6r\right)$ to get,

  $v^{″}=c\left(2r_0-6\left(\frac{2}{3}{r}_0\right)\right)$

Simplifying,

  $v^{″}=c\left(2r_0-4r_0\right)$

On further simplification,

  $v^{″}=-2c{r}_0<0$

Note that the above quantity is negative because $c$ is a positive constant and $r_0$ is the rest radius of the trachea, which must be positive.

This shows that $v$ is greatest when $r=\frac{2}{3}{r}_0$ .

Conclusion:

It has been shown that $v$ is greatest when $r=\frac{2}{3}{r}_0$ .

(b.)

To determine

The graph of $v$ over the interval $0\le r\le 0.5$ , when $r_0=0.5$ and $c=1$ .

Answer to Problem 59E

The graph of $v$ over the interval $0\le r\le 0.5$ , when $r_0=0.5$ and $c=1$ , has been provided.

Explanation of Solution

Given:

The average flow velocity $v$ can be modelled by the equation, $v=c\left(r_0-r\right)r^2$ for $\frac{r_0}{2}\le r\le r_0$ , where $r_0$ is the rest radius of the trachea in cm and $c$ is a positive constant whose value depends in part on the length of the trachea.

Concept used:

A function attains maximum value at a point where its first derivative is zero and its second derivative is negative.

Calculation:

It is given that $v=c\left(r_0-r\right)r^2$ for $\frac{r_0}{2}\le r\le r_0$ .

Put $r_0=0.5$ and $c=1$ in the above expression to get,

  $v=\left(0.5-r\right)r^2$

The graph of this function over the interval $0\le r\le 0.5$ is given as:

  

As shown previously, $v$ is greatest when $r=\frac{2}{3}{r}_0$ .

Put $r_0=0.5$ in $r=\frac{2}{3}{r}_0$ to get,

  $r=\frac{1}{3}\approx 0.3333$

It can be seen from the graph that in the given case, $v$ is greatest when $r\approx 0.3333$ , which supports the conclusion drawn in part (a).

Conclusion:

The graph of $v$ over the interval $0\le r\le 0.5$ , when $r_0=0.5$ and $c=1$ , has been provided.