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. What was the initial population P0? (Round your answer to one decimal place.) P0 = 5937.8 What will be the population in 10 years? (Round your answer to the nearest person.) 23751 persons How fast is the population growing at t = 10? (Round your answer to the nearest person.) 3293 persons/year

Answer 1

Answer:

P0 = 5937.8 people

P(t = 10) = 23751 people

P'(t = 10) = 3293 persons/year

Step-by-step explanation:

Let the population has the formula of

$P = P_0e^{kt}$

Where P0 is the initial population at t = 0 and k is the constant that we are looking fore.

Since the population doubled after t = 5 years

$P = 2P_0$

$2P_0 = P_0e^{k5}$

$e^{5k} = 2$

$k = ln2/5 = 0.1386$

So after t = 3 years, population is P = 9000:

$9000 = P_0e^{0.1386*3}$

$P_0 = \frac{9000}{e^{0.1386*3}} = 5937.8$

After 10 years, population would be quadtripled (10 years is 2 times of 5 years):

$P(10) = 4P_0 = 5937.8*4 = 23751$

The rate of change in population is the derivative of the population function with respect to t

$P'(t) = P_0ke^{kt} = 5937.8*0.1386e^{0.1386t} = 823.15 e^{0.1386t}$

So after t = 10 years the rate of change in population would be

$P'(10) = 823.15 e^{0.1386*10} = 3293 persons/years$