A-LevelMathematicsDifferentiation9709/31 · May/June 2017 · Q48 Marks
The parametric equations of a curve are x = ln cos θ, y = 3θ - tan θ, where 0 ≤ θ < ½π.
- (i)Express dy/dx in terms of tan θ.[5]
- (ii)Find the exact y-coordinate of the point on the curve at which the gradient of the normal is equal to 1.[3]
Mark scheme
Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.
- Use chain rule to differentiate x, dx/dθ = (-sinθ)/cosθ M1
- State dy/dθ = 3 - sec²θ B1
- Use dy/dx = (dy/dθ) / (dx/dθ) M1
- Obtain correct dy/dx in any form e.g. (3-sec²θ) / (-tanθ) A1
- Obtain dy/dx = (tan²θ-2) / tanθ, or equivalent A1
- Equate gradient to -1 and obtain an equation in tanθ M1
- Solve a 3 term quadratic (tan²θ + tanθ - 2 = 0) in tanθ M1
- Obtain θ = π/4 and y = 3π/4 - 1 only A1
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