A population of a species of animal is being studied. The population P is modelled by the differential equation dP/dt = kP(1 - P/1000) t ≥ 0 where k is a constant and t is the time in years from the start of the study. Initially, there are 100 animals and the population is increasing at a rate of 9 animals per year.

Mark scheme

Marking points from the official mark scheme. The mark or mark type is shown in brackets.

1. Part (a): M1 for substituting P=100 and dP/dt=9 into the differential equation.
2. Part (a): A1 for 9 = k(100)(1 - 100/1000) => 9 = k(100)(0.9) => 9 = 90k => k = 0.1 (Shown)
3. Part (b): M1 for separating variables to get ∫ (1 / (P(1 - P/1000))) dP = ∫ k dt
4. Part (b): M1 for using partial fractions on 1 / (P(1 - P/1000)) or 1000 / (P(1000 - P)). Setting up partial fractions A/P + B/(1000-P)
5. Part (b): A1 for 1/P + 1/(1000-P) (or equivalent form for A and B)
6. Part (b): M1 for integrating LHS to ln|P| - ln|1000-P| (or ln|P/(1000-P)|) and RHS to kt + c
7. Part (b): A1 for ln|P/(1000-P)| = kt + c (o.e.)
8. Part (b): M1 for substituting t=0, P=100 to find c. e.g. ln(100/900) = c => c = ln(1/9) = -ln 9
9. Part (b): A1 for correct constant c and rearrangement to P = 1000 / (1 + 9e^(-kt)) or P = 1000e^(kt) / (9 + e^(kt)) (given k=0.1 already)
10. Part (c): M1 for substituting P=500 and k=0.1 into their P(t) equation.
11. Part (c): A1 for solving for t: 500 = 1000 / (1 + 9e^(-0.1t)) => 1 + 9e^(-0.1t) = 2 => 9e^(-0.1t) = 1 => e^(-0.1t) = 1/9 => -0.1t = ln(1/9) => 0.1t = ln 9 => t = 10 ln 9 = 21.97... => t = 22.0 years (1 d.p.)