Specialist2026 NHT Exam 2 Section A · Q12CAS active1 mark

Time for logistic population growth as an integral

The growth of a population of emus in a certain region is given by the logistic differential equation $\frac{dP}{dt} = 0.02P\left(1 - \frac{P}{1000}\right)$ , where $P$ models the number of emus after $t$ years.

The initial population of emus is 50.

The time taken, in years, for the number of emus to grow to 500 is given by

1 mark

\int_{50}^{500} 0.02P\left(1 - \frac{P}{1000}\right) dP

\int_{0}^{500} \frac{50}{P} + \frac{50}{1000 - P} \, dP

\int_{0}^{500} 0.02P\left(1 - \frac{P}{1000}\right) dP

\int_{50}^{500} \frac{50}{P} + \frac{50}{1000 - P} \, dP

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