A population with $N = 400$ individuals increases in numbers till it reaches an asymptote at $K = 500$ individuals, $K$ being the carrying capacity. Assuming the intrinsic rate of natural increase ($r$) to be 0.01, what would be the population growth rate ($dN/dt$)?
Show Hint
Always pay attention to the term "reaches an asymptote".
This is the key term indicating that you should use the logistic growth formula rather than the exponential growth formula ($\frac{dN}{dt} = rN$).
Double check your basic arithmetic to avoid simple calculation errors.
Step 1: Understanding the Question:
The question describes a population growing until it reaches an asymptote at a carrying capacity ($K$).
This signifies Verhulst-Pearl logistic growth.
We are asked to calculate the population growth rate ($\frac{dN}{dt}$) given the population size ($N = 400$), carrying capacity ($K = 500$), and intrinsic rate of natural increase ($r = 0.01$). Step 2: Key Formula or Approach:
The equation for logistic population growth is:
\[ \frac{dN}{dt} = r N \left( \frac{K - N}{K} \right) \]
Where:
- $\frac{dN}{dt}$ is the rate of population change.
- $r$ is the intrinsic rate of natural increase.
- $N$ is the population size.
- $K$ is the carrying capacity. Step 3: Detailed Explanation:
Let's substitute the given values into the formula:
- $N = 400$
- $K = 500$
- $r = 0.01$
Now, plug these into the logistic growth equation:
\[ \frac{dN}{dt} = 0.01 \times 400 \times \left( \frac{500 - 400}{500} \right) \]
- Simplify the terms:
\[ 0.01 \times 400 = 4 \]
\[ \frac{500 - 400}{500} = \frac{100}{500} = 0.2 \]
- Now multiply the simplified terms:
\[ \frac{dN}{dt} = 4 \times 0.2 = 0.8 \]
- Thus, the population growth rate is 0.8. Step 4: Final Answer:
The calculated population growth rate is 0.8, which corresponds to option (A).
