Edexcel IALMathematicsGeneral Applications12 Marks

A population of insects, P, is modelled by the differential equation dP/dt = kP(1000-P), where t is the time in days and k is a positive constant. Given that when t=0, P=100, and when t=10, P=400.

(a)Show that ∫ 1/(P(1000-P)) dP = ∫ k dt can be written as 1/1000 ln|P/(1000-P)| = kt + C.[5]
(b)Find the value of k, correct to 3 significant figures.[4]
(c)Find the time taken for the population to reach 800.[3]

Mark scheme

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

M1: Separates variables correctly. ∫ 1/(P(1000-P)) dP = ∫ k dt.
M1: Uses partial fractions for 1/(P(1000-P)), setting A/P + B/(1000-P).
A1: Correct partial fractions: 1/1000P + 1/(1000(1000-P)).
M1: Integrates both sides. 1/1000 ln|P| - 1/1000 ln|1000-P| = kt + C.
A1: Shows the given form: 1/1000 ln|P/(1000-P)| = kt + C. (Must include constant of integration).
M1: Uses initial condition t=0, P=100 to find C.
A1: Correct C = 1/1000 ln(100/900) = 1/1000 ln(1/9).
M1: Uses second condition t=10, P=400 to find k.
A1: Correct k=0.000157 (3 s.f.).
M1: Substitutes P=800 and their C and k into the general solution.
A1: Sets up the equation correctly (e.g. 1/1000 ln(4) = kt + 1/1000 ln(1/9)).
A1: t ≈ 23.3 days (3 s.f.).

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