A-LevelMathematicsDifferential equations9709/33 · May/June 2021 · Q79 Marks

For the curve shown in the diagram, the normal to the curve at the point P with coordinates (x, y) meets the x-axis at N. The point M is the foot of the perpendicular from P to the x-axis. The curve is such that for all values of x in the interval 0 < x < ¼π, the area of triangle PMN is equal to tan x. [Figure X.X]

(a)(i)Show that MN/y = dy/dx.[1]
(a)(ii)Hence show that x and y satisfy the differential equation ½y² dy/dx = tan x.[2]
(b)Given that y = 1 when x = 0, solve this differential equation to find the equation of the curve, expressing y in terms of x.[6]

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Mark scheme

Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.

[B1] Justify the given statement MN/y = dy/dx
[M1] Express the area of PMN in terms of y and dy/dx and equate to tan x
[A1] Obtain the given equation correctly
[M1] Separate variables and integrate at least one side
[A1] Obtain term 1/6 y³
[M1] Obtain term of the form ± ln cos x
[M1] Evaluate a constant or use x = 0 and y = 1 in a solution containing terms ay³ and ±lncosx, or equivalent
[A1] Obtain correct answer in any form, e.g. 1/3 y³ = -ln cos x + 1/6
[A1] Obtain final answer y = ³√(1-6ln cosx) OE

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