Question

The population of a community is known to increase at a rate proportional to the number of people present at time t. The initial population P0 has doubled in 5 years. Suppose it is known that the population is 9,000 after 3 years.

Required:
a, What was the initial population P0?
b. How fast is the population growing at t = 10?

Answer 1

Explanation:

Let the initial population of a community be P0 and the population after time t is P(t).

If the population of a community is known to increase at a rate proportional to the number of people present at time t, this is expressed as:

`P(t) = P_0e^(kt)`

at t = 5 years, P(t) = 2P0

substitute:

`2P_0 = P_0e^(5k)\\2 = e^(5k)\\ln2 = lne^(5k)\\ln2 = 5k\\k = (ln2)/(5)\\k = 0.1386`

If the population is 9,000 after 3 years

at t = 3, P(t) = 9000

a) Substitute into the formula to get P0

`9000 = P_0e^(0.1386* 3)\\9000 = P_0e^(0.4158)\\9000 = 1.5156P_0\\P_0 = (9000)/(1.5156)\\ P_0 = 5938.24`

Hence the initial population is approximately 5938.

b) In order to know how fast the population growing at t = 10, we will substitute t = 10 into the formula as shown:

`P(10) = 5938.24e^(0.1386(10))\\P(10) = 5938.24e^(1.386)\\P(10) = 5938.24(3.9988)\\P(10) = 23,745.97`

Hence the population of the community after 10 years is approximately 23,746