Chapter 8.1, Problem 23E

a.

To determine

To calculate city’s radius if the population density approaches zero at the edge of the city.

Answer to Problem 23E

The radius of the city is 2 miles.

Explanation of Solution

Given information:

The decrease in Washerton’s population density as you move away from the city can be approximated by the function $10,000\left(2-r\right)$

Calculation:

Calculation density at end of the city is zero.

  $\begin{array}{l}10000\left(2-r\right)=0\\ 2-r=0\left[\text{divide both sides by 10000}\right]\\ r=2\left[\text{subtract 2 from both sides}\right]\end{array}$

Therefore, The radius of the city is 2 miles.

b.

To determine

To state the approximate area.

Answer to Problem 23E

The area is $2\pi r\Delta \text{r}$

Explanation of Solution

Given information:

Radius of the ring is r and its thickness is $\Delta \text{r}$ .

If the ring is straighten out the circumference of the ring become the length of the rectangle and its thickness becomes breadth of the rectangle.

Therefore,

The area is $2\pi r\Delta \text{r}$

c.

To determine

To explain why the population of the ring in part (b) is approximately $10,000\left(2-r\right)\left(2\pi r\right)\Delta \text{r}$ .

Answer to Problem 23E

  $\text{population=population density }\times \text{ area}$

Explanation of Solution

Given information:

Radius of the ring is r and its thickness is $\Delta \text{r}$ .

The population of the ring is $10,000\left(2-r\right)\left(2\pi r\right)\Delta \text{r}$ because

  $\text{population=population density }\times \text{ area}$

And $10,000\left(2-r\right)$ represents function of population density at a distance r from the city centre, $\left(2\pi r\right)\Delta \text{r}$ represents area.

d.

To determine

To estimate the total population of Washerton by setting up and evaluating a definite integral.

Answer to Problem 23E

The population is approximately 83,776

Explanation of Solution

Given information:

The population density function is $10,000\left(2-r\right)$ and area is $\left(2\pi r\right)\Delta \text{r}$

Calculation:

Since the radius is evaluated as 2 miles.

Seek the cumulative effect of population density for $0\le r\le 2$

The population is

  $\begin{array}{l}={\displaystyle \sum 10000\left(2-r\right)2\pi r\Delta r}\\ ={\displaystyle \underset{0}{\overset{2}{\int }}10000\left(2-r\right)2\pi r\Delta r}\\ =10000{\left[2\pi r^2-\frac{2}{3}\pi r^3\right]}_0^2\\ =83,775.8\end{array}$

Therefore,

The population is approximately 83,776