Question

The population of a town grows at a rate proportional to the population present at time t. The initial population of 500 increases by 20% in 10 years. What will be the population in 30 years? (Round your answer to the nearest person.)How fast is the population growing at t = 30? (Round your answer to two decimal places.)

Answer 1

The population P after t years obeys the differential equation

dP/dt = kP where k is a positive constant; the initial condition is P(0) = 500. To solve this, we use separation of variables:

∫ 1 P dP = ∫ k dt

ln |P| = kt + C

|P| = e^Ce^kt

P = Aekt .

Using P(0) = 500 gives 500 = Ae⁰ ⇒ A = 500. Thus, P = 500e^kt. Furthermore, P(10) = 500 × 120% = 600 so

600 = 500e^10k

e^10k = 1.2

10k = ln(1.2)

k = ln(1.2)/10 ≈ 0.018.

Thus, P = 500e^0.018t .

i) The population after 30 years is therefore:

P = 500e^0.018(30) = 858.

ii) At t=30,population has grown by:

(858-500)/500 = 71.60% from the initial population.