Question

Which of the following equations depicts Verhulst-Pearl logistic population growth ?

• A. $\frac{dN}{dt} = rN \left(\frac{K-N}{N}\right)$
• B. $\frac{dN}{dt} = rN \left(\frac{K-N}{K}\right)$
• C. $\frac{dN}{dt} = rN \left(\frac{K}{K-N}\right)$
• D. $\frac{dN}{dt} = rN \left(\frac{K+N}{K}\right)$

Hint:

Logic Tip: The term $\left(\frac{K-N}{K}\right)$ represents the "unutilized capacity" of the environment. When $N = K$ (population hits carrying capacity), the term becomes zero, meaning population growth ($\frac{dN}{dt}$) completely stops!

Answer 1

The Correct Option is B

Concept:
In nature, a given habitat has enough resources to support a maximum possible number of individuals, beyond which no further growth is possible. This limit is called nature's carrying capacity ($K$). A population growing in a habitat with limited resources shows a logistic growth pattern, often described by the Verhulst-Pearl Logistic Growth equation.
Step 1:
Let $N$ = Population density at time $t$.
Let $r$ = Intrinsic rate of natural increase.
Let $K$ = Carrying capacity of the environment.
Step 2:
If resources were unlimited, the population would grow exponentially, represented by the differential equation: $\frac{dN}{dt} = rN$.
Step 3:
Because resources are limited, as the population ($N$) approaches the carrying capacity ($K$), the growth rate must slow down. The fraction of resources still available for population growth is represented mathematically as $\frac{K-N}{K}$.
Step 4:
By multiplying the exponential growth factor ($rN$) by the environmental resistance factor ($\frac{K-N}{K}$), we get the logistic growth equation: $$\frac{dN}{dt} = rN \left(\frac{K-N}{K}\right)$$
Step 5:
Reviewing the provided choices, Option (2) accurately depicts the standard Verhulst-Pearl logistic growth equation.