a) A population of bacteria has an initial size 200. After one hour, the population has reached 220. The population grows in such a way that the rate of growth doubles each hour.(2 marks)i. What is the population after five hours? ii. Write a recurrence relation for growth of bacteria, be, after n hours. (3 marks)
Question
a) A population of bacteria has an initial size 200. After one hour, the population has reached 220. The population grows in such a way that the rate of growth doubles each hour.(2 marks)i. What is the population after five hours? ii. Write a recurrence relation for growth of bacteria, be, after n hours. (3 marks)
Solution
a) i. To find the population after five hours, we first need to determine the growth rate. The growth rate is the increase in population size per unit time. In this case, the population increased from 200 to 220 in one hour, so the growth rate is (220-200)/200 = 0.1 or 10%.
However, the problem states that the rate of growth doubles each hour. So, after the first hour, the growth rate is 10%, after the second hour it's 20%, after the third hour it's 40%, after the fourth hour it's 80%, and after the fifth hour it's 160%.
To find the population after five hours, we apply these growth rates consecutively to the initial population:
After 1 hour: 200 + 2000.1 = 220 After 2 hours: 220 + 2200.2 = 264 After 3 hours: 264 + 2640.4 = 369.6 After 4 hours: 369.6 + 369.60.8 = 665.28 After 5 hours: 665.28 + 665.28*1.6 = 1729.728
So, the population after five hours is approximately 1730 (if we round to the nearest whole number).
a) ii. A recurrence relation for the growth of the bacteria, be, after n hours can be written as:
be(n) = be(n-1) + 2^n * be(n-1)
This equation states that the population at any given hour is equal to the population from the previous hour plus the previous hour's population multiplied by two to the power of the current hour. This reflects the fact that the growth rate doubles each hour.
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