A-LevelMathematicsDifferentiation9709/32 · May/June 2017 · Q46 Marks
The parametric equations of a curve are x = t² + 1, y = 4t + ln(2t − 1).
- (i)Express dy/dx in terms of t.[3]
- (ii)Find the equation of the normal to the curve at the point where t = 1. Give your answer in the form ax + by + c = 0.[3]
Mark scheme
Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.
- [B1] State dy/dt = 4 + 2/(2t-1)
- [M1] Use dy/dx = (dy/dt) / (dx/dt)
- [A1] Obtain answer dy/dx = (8t - 2) / (2t(2t - 1)) , or equivalent e.g. 2/t + 2/(4t²-2t)
- [M1] Use correct method to find the gradient of the normal at t = 1
- [M1] Use a correct method to form an equation for the normal at t = 1
- [A1] Obtain final answer x + 3y – 14 = 0, or horizontal equivalent
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