A-LevelMathematicsNumerical methods9709/22 · May/June 2019 · Q610 Marks

The diagram shows the curve with parametric equations x = 3t - 6e⁻²ᵗ, y = 4t eᵗ, for 0 ≤ t ≤ 2. At the point P on the curve, the y-coordinate is 1. [Figure 6.1]

(i)Show that the value of t at the point P satisfies the equation t = ½e⁻ᵗ.[2]
(ii)Use the iterative formula tn+1 = ½e^(tn/2) with t1 = 0.7 to find the value of t at P correct to 3 significant figures. Give the result of each iteration to 5 significant figures.[3]
(iii)Find the gradient of the curve at P, giving the answer correct to 2 significant figures.[5]

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Mark scheme

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

[M1] Equate 4t²e⁻ᵗ to 1, rearrange to t² = ... and hence t = ...
[A1] Confirm t = ½eᵗ/² with necessary detail needed as answer is given
[M1] Use iterative process correctly at least once
[A1] Obtain final answer t = 0.715
[A1] Show sufficient iterations to 5 sf to justify answer or show a sign change in the interval [0.7145, 0.7155]
[B1] Obtain dx/dt = 3 + 12e⁻²ᵗ
[M1] Use product rule to find dy/dt
[A1] Obtain 8te⁻ᵗ - 4t²e⁻ᵗ
[M1] Divide correctly to obtain dy/dx
[A1] Substitute value from part (ii) to obtain 0.31

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