Find the coordinates of ${\text{A}}$ in terms of $r$ and $\theta$.

As the wheel rolls, the point A traces out a curve. Show that the gradient of this curve is $\cot \left( {\frac{1}{2}\theta } \right)$.

Find the equation of the tangent to the curve when $\theta  = \frac{\pi }{3}$.

${\text{OX}} = {\text{OB}} - {\text{XB}} = r\theta  - r\sin \theta  = x$     (M1)A1

${\text{OY}} = {\text{ZB}} - {\text{ZC}} = r - r\cos \theta  = y$     A1

$\frac{{{\text{d}}x}}{{{\text{d}}\theta }} = r - r\cos \theta$     A1

$\frac{{{\text{d}}y}}{{{\text{d}}\theta }} = r\sin \theta$     A1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{r\sin \theta }}{{r - r\cos \theta }}$     M1

$= \frac{{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}{{2{{\sin }^2}\frac{\theta }{2}}}$     M1A1A1

$= \frac{{\cos \frac{\theta }{2}}}{{\sin \frac{\theta }{2}}}$

$= \cot \frac{\theta }{2}$     AG

when $\theta  = \frac{\pi }{3}$, gradient $= \sqrt 3$     A1

$x = \frac{\pi }{3}r - r\frac{{\sqrt 3 }}{2},{\text{ }}y = r - \frac{r}{2} = \frac{r}{2}$     A1

$y - \frac{r}{2} = \sqrt 3 \left( {x - \frac{\pi }{3}r + r\frac{{\sqrt 3 }}{2}} \right)\;\;\;{\text{or}}\;\;\;y = \sqrt 3 x + 2r - \frac{{\pi r}}{{\sqrt 3 }}$     A1

Many candidates were unable to find the coordinates of the point A which made (b) inaccessible. Many candidates reached the halfway point in (b) but were then unable to use the half angle formulae to obtain the required result. Many of the candidates who failed to solve (b) picked up the A1 in (c) for finding the gradient.

Many candidates were unable to find the coordinates of the point A which made (b) inaccessible. Many candidates reached the halfway point in (b) but were then unable to use the half angle formulae to obtain the required result. Many of the candidates who failed to solve (b) picked up the A1 in (c) for finding the gradient.

Many candidates were unable to find the coordinates of the point A which made (b) inaccessible. Many candidates reached the halfway point in (b) but were then unable to use the half angle formulae to obtain the required result. Many of the candidates who failed to solve (b) picked up the A1 in (c) for finding the gradient.