Chapter 1.5, Problem 58E

a.

To determine

To find:The population of bacteria after 1, 2, and 3 hours.

Answer to Problem 58E

The population of bacteria after 1, 2, and 3 hours that go to next stage is $5.0\times {10}^6$ , $9.0\times {10}^6$ , and $17.0\times {10}^6$ .

Explanation of Solution

Given:

The population of bacteria doubles every hour but that $1.0\times {10}^6$ individual are removed before reproduction to be converted into valuable biological by-products. The initial population of the bacteria is $b_0=3.0\times {10}^6$ .

Formula used:

The next hour bacteria population is $b_{t+1}=2b_t-1.0\times {10}^6$ .

Calculation:

In the beginning the bacteria population is $b_0=3.0\times {10}^6$ .

Since, population of bacteria doubles every hour, and before the next stage culture $1.0\times {10}^6$ bacteria are removed, therefore,

The population of bacteria after 1hour $b_1=2b_0=2\times 3.0\times {10}^6=6.0\times {10}^6$ .

The population of bacteria after 1 hour before reproduction $6.0\times {10}^6-1.0\times {10}^6=5.0\times {10}^6$ .

The population of bacteria after 2 hour $b_2=2b_1=2\times 5.0\times {10}^6=10.0\times {10}^6$ .

The population of bacteria after 1 hour before reproduction $10.0\times {10}^6-1.0\times {10}^6=9.0\times {10}^6$ .

The population of bacteria after 1 hour $b_3=2b_2=2\times 9.0\times {10}^6=18.0\times {10}^6$ .

The population of bacteria after 1 hour before reproduction $18.0\times {10}^6-1.0\times {10}^6=17.0\times {10}^6$ .

Conclusion:

The populationof bacteria after 1, 2, and 3 hours that go to next stage is $5.0\times {10}^6$ , $9.0\times {10}^6$ , and $17.0\times {10}^6$ .

b.

To determine

To write:The discrete- time dynamical system for bacteria population growth.

Answer to Problem 58E

The discrete- time dynamical system for bacteria population growth is $b_{t+1}=2b_t-1.0\times {10}^6$

Explanation of Solution

Given:

The population of bacteria doubles every hour but that $1.0\times {10}^6$ individual are removed before reproduction to be converted into valuable biological by-products.

Concept used:

The population of bacteria doubles every hour but that $1.0\times {10}^6$ individual are removed.

Explanation: (From data in part (a))

Let at any time “ t ” the population of bacteria is $b_t$ , after next hour the population will become $2\times b_t$ . But from this population $1.0\times {10}^6$ bacteria are removed, therefore the population of bacteria at the end of t^th hour or the start of next hour is $b_{t+1}$ given by $b_{t+1}=2b_t-1.0\times {10}^6$ .

Conclusion:

The discrete- time dynamical system for bacteria population growth is $b_{t+1}=2b_t-1.0\times {10}^6$ .

c.

To determine

To compare:The population growth with the problem if individual bacteria are removed after each hour to the population growth with the problem if individual bacteria are removed after 3^rd hour.

Answer to Problem 58E

If bacteria are not removed then the population of bacteria after 1, 2, and 3 hours that go to next stage will be $6.0\times {10}^6$ , $12.0\times {10}^6$ , and $24.0\times {10}^6$ that high ( worst) population growth because even $1.0\times {10}^6$ bacteria are removed after 3^rd hour the population of bacteria is very high equal to $24.0\times {10}^6-1.0\times {10}^6=23.0\times {10}^6$ .

Explanation of Solution

Given:

Given and calculated data in part (a).

Concept used:

Given in part (a).

If the $1.0\times {10}^6$ individual bacteria are removed after each hour then the population of bacteria after 1, 2, and 3 hours that go to next stage is $5.0\times {10}^6$ , $9.0\times {10}^6$ , and $17.0\times {10}^6$ .

But if bacteria are not removed then the population of bacteria after 1, 2, and 3 hours that go to next stage will be $6.0\times {10}^6$ , $12.0\times {10}^6$ , and $24.0\times {10}^6$ that high ( worst) population growth because even $1.0\times {10}^6$ bacteria are removed after 3^rd hour the population of bacteria is very high equal to $24.0\times {10}^6-1.0\times {10}^6=23.0\times {10}^6$ .