(i)[3]
Show that this simplifies to $\frac{dy}{dx} = \frac{(t^2 - 9)(t - 2)}{t^2}$.
(ii)[3]
Find the coordinates of the only point on the curve where the gradient is equal to $0$.
Mathematics 9709 · AS & A Level · Differentiation
Differentiation — practice question
The curve is described parametrically by $x = 1 + \ln(t - 2)$ and $y = t + \frac{9}{t}$, with $t > 2$.
Worked solution & mark scheme
This 6-mark question has a full step-by-step worked solution and mark scheme. One marking point: “State $\frac{dx}{dt} = \frac{1}{t-2}$ or $\frac{dy}{dt} = 1-9t^{-2}$” …
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