A-LevelMathematicsDifferentiation9709/21 · May/June 2017 · Q78 Marks
The parametric equations of a curve are x = t³ + 6t + 1, y = t⁴ – 2t³ + 4t² – 12t + 5.
- (i)Find dy/dx and use division to show that dy/dx can be written in the form at + b, where a and b are constants to be found.[5]
- (ii)The straight line x – 2y + 9 = 0 is the normal to the curve at the point P. Find the coordinates of P.[3]
Mark scheme
Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.
- [M1] Differentiate x and y and form dy/dx
- [A1] Obtain (4t³-6t²+8t-12)/(3t²+6)
- [M1] Carry out division at least as far as kt or equivalent
- [A1] Obtain (4/3)t
- [A1] Obtain (4/3)t - 2 with complete division shown and no errors seen
- [B1] State or imply gradient of straight line is 1/2
- [M1] Attempt value of t from their dy/dx = their negative reciprocal of gradient of line
- [A1] Obtain t = 0 and hence (1,5)
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