A-LevelMathematicsDifferentiation9709/21 · May/June 2023 · Q58 Marks

The diagram shows the curve with parametric equations x = 4e^(2t), y = 5e^(-t) cos 2t, for -¼π ≤ t ≤ ¼π. The curve has a maximum point M.

(a)Find an expression for dy/dx in terms of t.[3]
(b)Find the coordinates of M, giving each coordinate correct to 3 significant figures.[5]

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Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

[M1] Use product rule to find dy/dt
[A1] Obtain dy/dt = -5e⁻ᵗ cos 2t - 10e⁻ᵗ sin 2t or (unsimplified) equivalent (do not condone poor use of brackets.
[A1] Obtain dy/dx = (-5e⁻ᵗ cos 2t – 10e⁻ᵗ sin 2t) / (8e^(2t)) OE following their expression for dy/dt
[M1*] Equate dy/dx to zero and simplify at least as far as tan 2t = ... now condoning any error with dx/dt
[A1] Obtain tan 2t = -½
[A1] Obtain t = -0.231... allow t = -0.232 .
[DM1] Substitute negative value of t in expressions for x and y
[A1] Obtain x = 2.52 and y = 5.64 or greater accuracy.

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