Chapter 5.4, Problem 59E

a.

To determine

To find the maximum velocity.

Answer to Problem 59E

The maximum velocity is $r=\frac{2r_0}{3}$ .

Explanation of Solution

Given velocity is $v=c\left(r_0-cr\right)r^2,\frac{r_0}{2}\le r\le r_0.$

Calculation: given in the question that average of velocity is

  $\begin{array}{l}v=c\left(r_0-cr\right)r^2,\frac{r_0}{2}\le r\le r_0\\ =c{r}_0{r}^2-c{r}^3\end{array}$

For any function $f\left(x\right)$ the condition of minimum and maximum is obtained by find $f^{\prime }\left(x\right)=0$ ,

  $\begin{array}{l}v=c{r}_0{r}^2-c{r}^3\\ \frac{dv}{dr}=2c{r}_{0}r-3c{r}^2\end{array}$

Critical points occur at

  $\begin{array}{l}{v}^{\prime }=0\\ 2c{r}_{0}r-3c{r}^2=0\\ r=\frac{2r_0}{3}\\ v^{\prime }=2c{r}_{0}r-3c{r}^2\\ {v^{\prime }}^{\prime }=2c{r}_0-6c<0\\ =2c\left(r_0-3\right)<0\end{array}$

It gives the condition of maximum.

Thus, the maximum velocity is $r=\frac{2r_0}{3}$ .

b.

To determine

To find the maximum velocity.

Answer to Problem 59E

The maximum value of $v$ is $\frac{1}{3}.$

Explanation of Solution

Given velocity is $v=c\left(r_0-cr\right)r^2,\frac{r_0}{2}\le r\le r_0.$

Calculation:

If plot the graph for $v=c{r}_0{r}^2-c{r}^3=c{r}^2\left(0.5-r\right)$ where $r_0=0.5$ it get $v$ is maximum at

  $r=\frac{2r_0}{3}=\frac{1}{3}.$

Graphical support:

  

Thus, the maximum value of $v$ is $\frac{1}{3}.$