The diagram depicts the curve with parametric equations $x = 2 - \cos 2t$, $y = 2\sin^3 t + 3\cos^3 t + 1$ for $0 \leq t \leq \frac{1}{2}\pi$. Its end-points are $(1,4)$ and $(3,3)$.
(i)[5]
Show that $\frac{dy}{dx} = \frac{3}{2} \sin t - \frac{9}{4} \cos t$.
(ii)[3]
Give the coordinates of the minimum point, with each coordinate correct to 3 significant figures.
(iii)[3]
Find the exact gradient of the normal to the curve at the point on the curve with $x = 2$.
Worked solution & mark scheme
This 11-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Obtain $\dfrac{dx}{dt} = 2\sin 2t$” …