Answer:
a) P' = P
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
![\frac{dP}{dt} = P](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7Bdt%7D%20%3D%20P)
Where t is the range of 6 months, or half of a year.
P(t) would have the form of
![P(t) = P_0e^{kt}](https://tex.z-dn.net/?f=P%28t%29%20%3D%20P_0e%5E%7Bkt%7D)
where
is the initial population
After 6 month (t = 1), the population is doubled to 48
![P(1) = 24e^k = 48](https://tex.z-dn.net/?f=P%281%29%20%3D%2024e%5Ek%20%3D%2048)
![e^k = 2](https://tex.z-dn.net/?f=e%5Ek%20%3D%202)
![k = ln(2) = 0.693](https://tex.z-dn.net/?f=k%20%3D%20ln%282%29%20%3D%200.693)
Therefore ![P(t) = 24e^{0.693t}](https://tex.z-dn.net/?f=P%28t%29%20%3D%2024e%5E%7B0.693t%7D)
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:
![24e^{0.693t} = 1000000](https://tex.z-dn.net/?f=24e%5E%7B0.693t%7D%20%3D%201000000)
![e^{0.693t} = 1000000 / 24 = 41667](https://tex.z-dn.net/?f=e%5E%7B0.693t%7D%20%3D%201000000%20%2F%2024%20%3D%2041667)
![0.693t = ln(41667) = 10.64](https://tex.z-dn.net/?f=0.693t%20%3D%20ln%2841667%29%20%3D%2010.64)
![t = 10.64 / 0.693 = 15.35](https://tex.z-dn.net/?f=t%20%3D%2010.64%20%2F%200.693%20%3D%2015.35)
So it would take 15.35 * 0.5 = 7.7 years to reach 1000000
c. ![P' = P_0ke^{kt}](https://tex.z-dn.net/?f=P%27%20%3D%20P_0ke%5E%7Bkt%7D)
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
![24e^{0.5k) = 48](https://tex.z-dn.net/?f=24e%5E%7B0.5k%29%20%3D%2048)
![0.5k = ln2 = 0.693](https://tex.z-dn.net/?f=0.5k%20%3D%20ln2%20%3D%200.693)
![k = 1.386](https://tex.z-dn.net/?f=k%20%3D%201.386)
Therefore, ![P' = 1.386*24e^{1.386t}](https://tex.z-dn.net/?f=P%27%20%3D%201.386%2A24e%5E%7B1.386t%7D)
At the mid of the 3rd year, where t = 2.5, we can calculate P'
rabbits/year