(i)[3]
Demonstrate that $\frac{dy}{dx} = \frac{1 - 2t}{t^2}$.
(ii)[3]
Determine the exact coordinates of the only point on the curve at which the gradient is $3$.
Mathematics 9709 · AS & A Level · Differentiation
Differentiation — practice question
The curve is defined parametrically by $x = \ln(1 - 2t)$ and $y = \frac{2}{t}$, where $t < 0$.
Worked solution & mark scheme
This 6-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Write $\frac{dx}{dt} = \frac{-2}{1-2t}$ or $\frac{dy}{dt} = -2t^{-2}$” …
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