Mathematics: In 1859, 24 rabbits were released into the wild in Australia, where they had no natural predators. Their population grew exponentially, doubling every 6 months. a. Determine P(t), the function that gives the population at time t, and the differential equation describing the population growth. b. After how many years, rounded to one digit after the decimal point, did the rabbit population reach 1,000,000? c. Determine the rate of population change, in rabbits/year, midway through the third year.

You must subtract 30 into 4 to get 34...

In 1859, 24 rabbits were released into the wild

twiddleturd

20.01.2022

a) P' = P

$P(t) = 24e^{0.693t}$ 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

$\frac{dP}{dt} = P$

Where t is the range of 6 months, or half of a year.

P(t) would have the form of

$P(t) = P_0e^{kt}$

where P_0 = 24 is the initial population

After 6 month (t = 1), the population is doubled to 48

P(1) = 24e^k = 48

e^k = 2

k = ln(2) = 0.693

Therefore $P(t) = 24e^{0.693t}$

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:

$24e^{0.693t} = 1000000$

$e^{0.693t} = 1000000 / 24 = 41667$

0.693t = ln(41667) = 10.64

t = 10.64 / 0.693 = 15.35

So it would take 15.35 * 0.5 = 7.7 years to reach 1000000

c. $P' = P_0ke^{kt}$

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

$24e^{0.5k) = 48$

0.5k = ln2 = 0.693

k = 1.386

Therefore, $P' = 1.386*24e^{1.386t}$

At the mid of the 3rd year, where t = 2.5, we can calculate P'

$P' = 1.386*24e^{1.386*2.5} = 1064.67$ rabbits/year

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