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

Step-by-step 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:

$2P_0 = P_0e^{5k}\\2 = e^{5k}\\ln2 = lne^{5k}\\ln2 = 5k\\k = \frac{ln2}{5}\\k = 0.1386$

If the population is 9,000 after 3 years

at t = 3, P(t) = 9000

a) Substitute into the formula to get P0

$9000 = P_0e^{0.1386\times 3}\\9000 = P_0e^{0.4158}\\9000 = 1.5156P_0\\P_0 = \frac{9000}{1.5156}\\ P_0 = 5938.24$

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