In a population of organisms, the logistic growth equation is dN/dt=rN(1-N/K). If r=0.2 day^-1, K=500, and the initial population size N_0=50, what is the population size after one day?

In a population of organisms, the logistic growth equation is $\frac{d N}{d t} = r N (1 - \frac{N}{K})$ . If $r = 0.2 day^{- 1}$ , $K = 500$ , and the initial population size $N_{0} = 50$ , what is the population size after one day?

medium MCQ 3 marks Class 12 · Biology · Organisms and Populations

Answer

The correct answer is 60 — Calculate $N (1) = \frac{500}{8.3683} \approx 59.8$ . Round to the nearest whole number: $60$ .

Write the explicit solution of the logistic equation: $N (t) = \frac{K}{1 + ( \frac{K - N _{0}}{N _{0}} ) e ^{- r t}}$ .
Compute the ratio $(\frac{K - N _{0}}{N _{0}}) = (\frac{500 - 50}{50}) = 9$ .
Compute the exponential term $e^{- r t} = e^{- 0.2} \approx 0.8187$ .
Find the denominator: $1 + 9 \times 0.8187 = 1 + 7.3683 = 8.3683$ .
Calculate $N (1) = \frac{500}{8.3683} \approx 59.8$ . Round to the nearest whole number: $60$ .

Common mistake: Ignoring the

\frac{K - N _{0}}{N _{0}}

factor in the denominator

Concept tested: Logistic growth equation