answer :
P(t) = 3000 / (1 + 29*e^(-0.637t))
steps
logistic function or logistic curve is a common S-shape curve (sigmoid curve) with equation:
P(t) = K / (1 + A*e^(-rt))
1. There is a type of animal or plant
species that started with 100
individuals.
2. The environment can support up to
3000 individuals, which is called the
"carrying capacity".
3. After 4 years, the population has
grown to 600 individuals.
4. We need to find a "logistic function"
that can show how the population
changes over time.
A species with an initial population of 100 is growing in an environment where the carrying capacity is 3000.
After 4 years, the population is up to 600.
We want to find the logistic function that models this population as a function of time.
The logistic function is a model of population growth that takes into account the carrying capacity of the environment. It is given by the formula:
P(t) = K / (1 + A e^(-r(t-t0)))
Where:
P(t) is the population size at time t
K is the carrying capacity of the environment
r is the growth rate of the population
t0 is the time at which the population starts to grow
A is a constant that determines the initial population size
To find the logistic function that models the population described in the problem, we need to determine the values of K, r, t0, and A.
We know that the initial population size is 100, so A = 100.
After 4 years, the population is up to 600, so P(4) = 600. We can use this information to solve for r:
600 = K / (1 + A e^(-r(4-t0)))
600 = K / (1 + 100 e^(-4r))
600(1 + 100 e^(-4r)) = K
K = 600 + 60000 e^(-4r)
We also know that the carrying capacity is 3000, so K = 3000.
3000 = 600 + 60000 e^(-4r)
2400 = 60000 e^(-4r)
0.04 = e^(-4r)
ln(0.04) = -4r
r = ln(0.04) / -4
r ≈ 0.693
Now we can use the value of r to solve for t0. We know that the initial population size is 100, so we can use this to find the value of A at t0:
100 = K / (1 + A)
100 = 3000 / (1 + A)
1 + A = 30
A = 29
We can now use this value of A to solve for t0:
100 = 3000 / (1 + 29 e^(-r(t0)))
1 + 29 e^(-r(t0)) = 30
e^(-r(t0)) = 1/29
ln(1/29) = -r(t0)
t0 = ln(1/29) / -r
t0 ≈ 2.48
Now we have all the values we need to write the logistic function:
P(t) = 3000 / (1 + 29 e^(-0.693(t-2.48)))
This is the logistic function that models the population as a function of time. It predicts that the population will grow exponentially at first, but then level off as it approaches the carrying capacity of the environment.
We can model the population growth of the species using the logistic equation:
dP/dt = rP(1 - P/K)
where P is the population size, t is time, r is the growth rate, and K is the carrying capacity.
To find the logistic function that models this population as a function of time, we need to determine the values of r and K. We can use the information given in the problem to solve for these values:
The initial population size is 100, so P(0) = 100.
The carrying capacity is 3000.
After 4 years, the population size is 600, so P(4) = 600.
Using these values, we can solve for r and K:
P(t) = K / (1 + A*e^(-rt))
where A is a constant determined by the initial population size, and e is the base of the natural logarithm.
From the initial population size, we know that:
A = (K - P(0)) / P(0) = (3000 - 100) / 100 = 29
We can use the population size after 4 years to solve for r and K:
600 = K / (1 + 29*e^(-4r))
Multiplying both sides by the denominator:
600 + 29600e^(-4r) = K
Substituting K = 3000:
3000 = 600 + 29600e^(-4r)
Dividing both sides by 600:
5 = 29*e^(-4r)
Taking the natural logarithm of both sides:
ln(5) = ln(29) - 4r
Solving for r:
r = (ln(29) - ln(5)) / 4
r ≈ 0.637
Now that we have r and K, we can plug them into the logistic equation to get the logistic function:
P(t) = 3000 / (1 + 29*e^(-0.637t))
This function models the population of the species as a function of time, where P(t) is the population size at time t.
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