A-LevelMathematicsDifferential equations9709/31 · Oct/Nov 2019 · Q48 Marks

The number of insects in a population t weeks after the start of observations is denoted by N. The population is decreasing at a rate proportional to Ne-0.02t. The variables N and t are treated as continuous, and it is given that when t = 0, N = 1000 and dN = -10. dt (i) Show that N and t satisfy the differential equation dN = -0.01e-0.02t N. dt (ii) Solve the differential equation and find the value of t when N = 800. (iii) State what happens to the value of N as t becomes large.

(i)Show that N and t satisfy the differential equation dN = -0.01e-0.02t N. dt[1]
(ii)Solve the differential equation and find the value of t when N = 800.[6]
(iii)State what happens to the value of N as t becomes large.[1]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.

[B1] State dN/dt = ke^-0.02t N and show k = -0.01 OE (-10=k×1×1000)
[B1] Separate variables correctly and integrate at least one side ∫(1/N)dN = ∫-0.01e^-0.02t dt
[B1] Obtain term ln N OE
[B1] Obtain term 0.5e^-0.02t OE
[M1] Use N = 1000, t = 0 to evaluate a constant, or as limits, in a solution with terms a ln N and be^-0.02t, where ab ≠ 0
[A1] Obtain correct solution in any form e.g. ln N - ln 1000 = 0.5(e^-0.02t - 1) OR ln(N/1000) = 0.5e^-0.02t - 0.5
[A1] Substitute N = 800 and obtain t = 29.6
[B1] State that N approaches 1000/√e Accept 606 or 607 or 606.5

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