In a population showing sigmoid growth form, the population has reached its carrying capacity. What will be the value of d N/dt in this population.

Options

A. infinity
B. Equal to carring capacity
C. Zero
D. None of the above

Answer

To solve the question regarding the value of dN/dt in a population showing sigmoid growth form that has reached its carrying capacity, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Sigmoid Growth**: The sigmoid growth curve represents how populations grow over time. Initially, the population grows slowly, then experiences rapid growth (exponential phase), and finally levels off as it reaches the carrying capacity.

2. **Identify Carrying Capacity**: The carrying capacity is the maximum population size that an environment can sustain. At this point, the population growth stabilizes.

3. **Graph Interpretation**: On a graph where the x-axis represents time and the y-axis represents population size, the sigmoid curve will show a plateau at the carrying capacity. This plateau indicates that the population size is no longer increasing.

4. **Equilibrium of Birth and Death Rates**: When the population reaches its carrying capacity, the birth rate (natality) equals the death rate (mortality). This means that the number of individuals being born is equal to the number of individuals dying.

5. **Mathematical Representation**: The rate of change of the population size over time is represented by dN/dt. In mathematical terms, dN/dt = natality - mortality. Since natality equals mortality at carrying capacity, we can express this as:
\[ dN/dt = n - m = 0 \]
where n is the birth rate and m is the death rate.

6. **Conclusion**: Therefore, when the population has reached its carrying capacity, the value of dN/dt is 0, indicating no net change in population size.

### Final Answer:
The value of dN/dt in a population that has reached its carrying capacity is **0**.