A-LevelMathematicsDifferentiation9709/23 · Oct/Nov 2017 · Q69 Marks

The parametric equations of a curve are x = 2e²ᵗ + 4eᵗ, y = 5te²ᵗ.

(i)Find dy/dx in terms of t and hence find the coordinates of the stationary point, giving each coordinate correct to 2 decimal places.[6]
(ii)Find the gradient of the normal to the curve at the point where the curve crosses the x-axis.[3]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.

B1 (i) Obtain dx/dt = 4e²ᵗ + 4eᵗ
M1 (i) Use product rule to find dy/dt
A1 (i) Obtain dy/dt = 5e²ᵗ + 10te²ᵗ
A1 (i) Obtain dy/dx = (5e²ᵗ + 10te²ᵗ) / (4e²ᵗ + 4eᵗ) or equivalent
M1 (i) Equate first derivative of the form (ae²ᵗ + bte²ᵗ) / (ce²ᵗ + deᵗ) to zero and solve to find t
A1 (i) Obtain t = -½ from completely correct work
A1 (i) Obtain (3.16, -0.92)
B1 (ii) Identify t = 0
M1 (ii) Substitute t = 0 in expression for first derivative and find negative reciprocal
A1 (ii) Obtain -5/8 or equivalent

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Topic: Differentiation, A-Level Mathematics (syllabus 9709). Structured question, 9 marks.
Source: A-Level Mathematics, Oct/Nov 2017 Paper 2 Variant 3, question 6.

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