The population of a town is growing according to the differential equation The g

Question

The population of a town is growing according to the differential equation

The growth constant, k, is equal to 0.04 year -1.

The size of the population at the start of the year 2000 was 25 thousand.

Since the population is growing exponentially, the population in year t is given by

Here, y is measured in thousands and t is measured in years since 2000.

(Enter your answer correct to one decimal place.)

Population: thousand.

(Enter your answer correct to two decimal places.)

Doubling time : years.

(Enter your answer correct to one decimal place.)

Population after three doubling times : thousand.

Explanation / Answer

Let's write the expression as P(t) = 4000e^(0.13)(t).

1. So the first question is what is P(6).
P(6) = 4e^(0.13)(6) = 4000e^(.78) = 8725.9

2.This question is what is t when P(t) = 8000.
4000e^(0.13)(t) = 8000, ln 4000 + .13t = ln 8000, t = (ln 8000 - ln 4000)/.13 = 5.33 years.

3. This question means what is P(3x5.33) or P(16).
P(16) = 4000e^(0.13)(16) = 4000e^(2.08) = 31976.3

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