A-LevelMathematicsDifferential equations9709/31 · Oct/Nov 2022 · Q88 Marks

In a certain chemical reaction the amount, x grams, of a substance is increasing. The differential equation satisfied by x and t, the time in seconds since the reaction began, is dx/dt = kxe^(-0.1t), where k is a positive constant. It is given that x = 20 at the start of the reaction.

(a)Solve the differential equation, obtaining a relation between x, t and k.[5]
(b)Given that x = 40 when t = 10, find the value of k and find the value approached by x as t becomes large.[3]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2022). The mark or mark type is shown in brackets.

[B1] (a) Separate variables correctly
[B1] (a) Obtain term ln x
[B1] (a) Obtain term -10ke^(-0.1t)
[M1] (a) Use x = 20, t = 0 to evaluate a constant or as limits in a solution containing terms alnx,be^(−0.1t) where ab≠0
[A1] (a) Obtain ln x = 10k(1 - e^(-0.1t)) + ln 20 or equivalent ISW
[M1] (b) Use x = 40, t = 10 to find k or 10k
[A1] (b) Obtain 10k = 1.09654 or equivalent e.g. 10k = ln 2 / (1 - e⁻¹)
[A1] (b) State that x tends to 59.9 (Need a number, not an expression for that value 3 sf or better 59.87595.....)

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About this question

Topic

Differential equations, A-Level Mathematics (syllabus 9709). Structured question, 8 marks.

Source

A-Level Mathematics, Oct/Nov 2022 Paper 3 Variant 1, question 8.

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