Answer:
a) P' = P
[tex]P(t) = 24e^{0.693t}[/tex] where t is step of 6 months
b) 7.7 years
c)1064.67 rabbits/year
Step-by-step explanation:
The differential equation describing the population growth is
[tex]\frac{dP}{dt} = P[/tex]
Where t is the range of 6 months, or half of a year.
P(t) would have the form of
[tex]P(t) = P_0e^{kt}[/tex]
where [tex]P_0 = 24[/tex] is the initial population
After 6 month (t = 1), the population is doubled to 48
[tex]P(1) = 24e^k = 48[/tex]
[tex]e^k = 2[/tex]
[tex]k = ln(2) = 0.693[/tex]
Therefore [tex]P(t) = 24e^{0.693t}[/tex]
where t is step of 6 months
b. We can solve for t to get how long it takes to get to a population of 1,000,000:
[tex]24e^{0.693t} = 1000000[/tex]
[tex]e^{0.693t} = 1000000 / 24 = 41667[/tex]
[tex]0.693t = ln(41667) = 10.64[/tex]
[tex]t = 10.64 / 0.693 = 15.35[/tex]
So it would take 15.35 * 0.5 = 7.7 years to reach 1000000
c. [tex]P' = P_0ke^{kt}[/tex]
We need to resolve for k if t is in the range of 1 year. In half of a year (t = 0.5), the population is 48
[tex]24e^{0.5k) = 48[/tex]
[tex]0.5k = ln2 = 0.693[/tex]
[tex]k = 1.386[/tex]
Therefore, [tex]P' = 1.386*24e^{1.386t}[/tex]
At the mid of the 3rd year, where t = 2.5, we can calculate P'
[tex]P' = 1.386*24e^{1.386*2.5} = 1064.67[/tex] rabbits/year
