A-LevelMathematicsDifferentiation9709/32 · Oct/Nov 2023 · Q24 Marks
2 The parametric equations of a curve are x = (ln t)², y = e^(2-t²), for t > 0. Find the gradient of the curve at the point where t = e, simplifying your answer.
- Find the gradient of the curve at the point where t = e, simplifying your answer.[4]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.
- B1 Obtain dx/dt = 2ln t / t Any equivalent form.
- B1 Obtain dy/dt = -2te^(2-t²). Any equivalent form.
- M1 Correct use of chain rule for dy/dx = (dy/dt) / (dx/dt) and substitute t=e Condone an error between correct combination of the derivatives and attempt to substitute e.
- A1 Obtain -e^(4-e^2) ISW Accept -0.0337(405..) . Accept -e^(e^2)e^(-e^2) and -e^2e^(2-e^2) . Allow M1A1 for a correct decimal answer following B1B1 seen.
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