Chapter 1.5, Problem 57E

(a)

To determine

To find the population of bacteria after $1$ , $2$ and $3$ hours using the given values

Answer to Problem 57E

Therefore, the population after $1$ hour is $b_1=5.0\times {10}^6$ .

The population after $2$ hour is $b_2=9.0\times {10}^6$

The population after $3$ hour is $b_3=17.0\times {10}^6$

Explanation of Solution

Given information: $b_0=3.0\times {10}^6$ bacteria

Calculation:

Suppose a population of bacteria with the initial condition $b_0=3.0\times {10}^6$ , doubles every hour.

However, $1.0\times {10}^6$ individuals are removed after reproduction to be converted into valuable biological by-products.

Now, the population after $1$ hour will be

$\begin{array}{l}{b}_1=2b_0-1.0\times {10}^6\\ =2.\left(3.0\times {10}^6\right)-1.0\times {10}^6\\ =6.0\times {10}^6--1.0\times {10}^6\\ =\left(6.0-1.0\right)\times {10}^6\\ =5.0\times {10}^6\end{array}$

Therefore, the population after $1$ hour is

$b_1=5.0\times {10}^6$ .

Now, the population after $2$ hour will be

$\begin{array}{l}{b}_2=2b_1-1.0\times {10}^6\\ =2.\left(5.0\times {10}^6\right)-1.0\times {10}^6\\ =10.0\times {10}^6--1.0\times {10}^6\\ =\left(10.0-1.0\right)\times {10}^6\\ =9.0\times {10}^6\end{array}$

The population after $2$ hour is

$b_2=9.0\times {10}^6$

Again, the population after $3$ hour will be

$\begin{array}{l}{b}_3=2b_2-1.0\times {10}^6\\ =2.\left(9.0\times {10}^6\right)-1.0\times {10}^6\\ =18.0\times {10}^6-1.0\times {10}^6\\ =\left(18.0-1.0\right)\times {10}^6\\ =17.0\times {10}^6\end{array}$

The population after $3$ hour is $b_3=17.0\times {10}^6$

(b)

To determine

To find how many bacteria were harvested based on the given data.

Answer to Problem 57E

$3.0\times {10}^6$

Explanation of Solution

Given information: $b_0=3.0\times {10}^6$ bacteria

Calculation:

Since there are three harvests of $1.0\times {10}^6$ bacteria, the total number of bacteria that wereharvested is $3\cdot 1.0\times {10}^6=3.0\times {10}^6$ bacteria

Hence, the total number of bacteria that were harvested is

  $3.0\times {10}^6$

(c)

To determine

To write the discrete-time dynamical system using the obtained values

Answer to Problem 57E

$b_{l+1}=2.0b_t-1.0\times {10}^6$

Explanation of Solution

Given information: $b_0=3.0\times {10}^6$ bacteria As the population of bacteria doubles every hour and $1.0\times {10}^6$ individuals are removed afterreproduction

Calculation:

As the population of bacteria doubles every hour and $1.0\times {10}^6$ individuals are removed after reproduction to be converted into valuable biological by-products, the discrete-time dynamical system for the bacterial population is given by

$b_{l+1}=2.0b_t-1.0\times {10}^6$

(d)

To determine

To find how many bacteria could be removedin the population $b_3$ found in part $a$ using the given information.

Answer to Problem 57E

There would be $8.0\cdot 3.0\times {10}^6=24.0\times {10}^6$ bacteria

Explanation of Solution

Given information: $b_0=3.0\times {10}^6$ bacteria

Calculation:

Now, ifwe wish to harvest bacteria until the end of $3$ hours, then after $3$ hours the population would have doubled three times, and is ${2.0}^3=8.0$ times the initial condition.

Hence, there would be $8.0\cdot 3.0\times {10}^6=24.0\times {10}^6$ bacteria

So, we can harvest $7.0\times {10}^6$ bacteria and still have a population of $17.0\times {10}^6$ bacteria, whichmeans harvesting $4.0\times {10}^6$ bacteria more than the harvested bacteria $3.0\times {10}^6$ .

Again, the bacteria removed early never had a chance to reproduce.