- The logistic equation is given as:
\[ \dfrac{dP}{dt} = 0.05P + 0.0005(P)^2 \]
Where time $t$ is measured in the weeks.
- What is the carrying capacity?
- What is the value of k?
This question aims to explain the carrying capacity K and value of relative growth rate coefficient k of the logistic equation that is given as:
\[ \dfrac{dP}{dt} = 0.05P + 0.0005(P)^2 \]
Logistic Differential Equations are used for modeling the growth of populations and other systems that have an exponentially increasing or decreasing function. A logistic differential equation is an ordinary differential equation that generates a logistic function.
The logistic population growth model is given as:
\[ \dfrac{dP}{dt} = kP(1 – \dfrac{P}{k} ) \]
Where:
t is the time taken by the population to grow.
k is the relative growth rate coefficient.
K is the carrying capacity of the logistic equation.
P is the population after the time t.
The carrying capacity K is the limiting value of the given population as time approaches infinity. The population must always tend towards the carrying capacity K. The relative growth rate coefficient k determines the rate at which the population is growing.
Expert Answer:
The general logistic equation for a population is given as:
\[ \dfrac{dP}{dt} = kP(1 – \dfrac{P}{k} ) \]
The logistic differential equation for the said population is given as:
\[ \dfrac{dP}{dt} = 0.05P + 0.0005(P)^2 \]
In order to calculate the carrying capacity $K$ and relative growth rate coefficient $k$, let’s modify the given logistic equation.
\[ \dfrac{dP}{dt} = 0.05P(1 + 0.01P ) \]
\[ \dfrac{dP}{dt} = 0.05P(1 + \dfrac{P}{100} ) \]
Now, compare it with the general logistic equation.
The value of the carrying capacity $K$ is given as:
\[ K = 100 \]
The value of the relative growth coefficient $k$ is given as:
\[ k = 0.05 \]
Alternative Solution:
Comparing both values that the equation gives,
The value of carrying capacity $K$ is:
\[ K = 100 \]
The value of relative growth coefficient is:
\[ k = 0.05 \]
Example:
Suppose that a population develops according to the logistic equation given:
\[ \dfrac{dP}{dt} = 0.08P – 0.0008(P)^2 \] where t is measured in weeks.
(a) What is the carrying capacity?
(b) What is the value of k?
The logistic equation given for the population is:
\[ \dfrac{dP}{dt} = 0.08P – 0.0008(P)^2 \]
Where time is measured in weeks.
The logistic equation for any population is defined as:
\[ \dfrac{dP}{dt} = kP(1 – \dfrac{P}{k} ) \]
Where $k$ is the relative growth coefficient and $K$ is the carrying capacity of the population.
In order to calculate the values of the carrying capacity and relative growth coefficients, let’s modify the given logistic equation for the population.
\[ \dfrac{dP}{dt} = 0.08P – 0.0008(P)^2 ) \]
\[ \dfrac{dP}{dt} = 0.08P( 1 – 0.01P ) \]
\[ \dfrac{dP}{dt} = 0.08P( 1 – \dfrac{P}{100} ) \]
Comparing the equation gives us:
\[ K = 100 \]
\[ k = 0.08 \]
Therefore, the value of carrying capacity $K$ is $100$ and the value of relative growth coefficient $k$ is $0.08$.