Question

Which of the following population growth equation correctly represents the Verhulst-Pearl Logistic growth equation? [Where N =population density at time t; r = intrinsic rate of natural increase; K= carrying capacity]

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

Hint:

To remember the logistic growth term, think of it as "how much room is left for growth". \( (K-N) \) is the remaining capacity, and dividing by K, \( \frac{(K-N)}{K} \), turns it into a proportion. This proportion multiplies the exponential growth rate to slow it down.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The Logistic Growth Equation models population growth in an environment with limited resources. It starts exponentially but slows down as the population size (N) approaches the carrying capacity (K).

Step 2: Key Formula or Approach:
The equation starts with the exponential growth model, \( \frac{dN}{dt} = rN \). To account for environmental resistance, this is multiplied by a term that approaches 0 as N approaches K. This term is \( \frac{(K-N)}{K} \).

Step 3: Detailed Explanation:
The full equation is: \[ \frac{dN}{dt} = rN \left( \frac{K-N}{K} \right) \] Let's analyze the term \( \left( \frac{K-N}{K} \right) \): \[\begin{array}{rl} \bullet & \text{When N is very small compared to K, the term \( \frac{(K-N)}{K} \) is close to 1. The growth is nearly exponential (\( \frac{dN}{dt} \approx rN \)).} \\ \bullet & \text{As N increases and gets closer to K, the term \( \frac{(K-N)}{K} \) gets closer to 0, which slows down the population growth rate.} \\ \bullet & \text{When N = K, the term \( \frac{(K-N)}{K} \) becomes 0, and the population growth stops (\( \frac{dN}{dt} = 0 \)).} \\ \end{array}\] Option (B) correctly represents this relationship.

Step 4: Final Answer:
The correct Verhulst-Pearl Logistic growth equation is \( \frac{dN}{dt} = rN \frac{(K-N)}{K} \).