The curve is given parametrically by $x = \ln(\cos \theta)$ and $y = 3\theta - \tan \theta$, with $0 \leq \theta < \frac{1}{2}\pi$.
(i)[5]
Express $\frac{dy}{dx}$ using $\tan \theta$.
(ii)[3]
Find the exact $y$-coordinate of the point on the curve where the gradient of the normal is $1$.
(b(ii))[3]
Find the exact $y$-coordinate of the point on the curve for which the normal has gradient $1$.
Worked solution & mark scheme
This 11-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Differentiate $x$ by applying the chain rule, giving $\frac{dx}{d\theta}=-\frac{\sin\theta}{\cos\theta}$” …