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
[tex]P = P_0e^{kt}[/tex]
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
[tex]P = 2P_0[/tex]
[tex]2P_0 = P_0e^{k5}[/tex]
[tex]e^{5k} = 2[/tex]
[tex]k = ln2/5 = 0.1386[/tex]
So after t = 3 years, population is P = 9000:
[tex]9000 = P_0e^{0.1386*3}[/tex]
[tex]P_0 = \frac{9000}{e^{0.1386*3}} = 5937.8[/tex]
After 10 years, population would be quadtripled (10 years is 2 times of 5 years):
[tex]P(10) = 4P_0 = 5937.8*4 = 23751[/tex]
The rate of change in population is the derivative of the population function with respect to t
[tex]P'(t) = P_0ke^{kt} = 5937.8*0.1386e^{0.1386t} = 823.15 e^{0.1386t}[/tex]
So after t = 10 years the rate of change in population would be
[tex]P'(10) = 823.15 e^{0.1386*10} = 3293 persons/years[/tex]
