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Suppose that a population develops according to the logistic equation.

  • 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 $0.08$.

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