Show that $t = \frac{2}{3}(2\cos \theta + \theta )$.

Find the value of $\theta$ for which $\frac{{{\text{d}}t}}{{{\text{d}}\theta }} = 0$.

What route should Jorg take to travel from A to B in the least amount of time?

Give reasons for your answer.

angle APB is a right angle

$\Rightarrow \cos \theta = \frac{{{\text{AP}}}}{4} \Rightarrow {\text{AP}} = 4\cos \theta$     A1

Note: Allow correct use of cosine rule.

${\text{arc PB}} = 2 \times 2\theta = 4\theta$     A1

$t = \frac{{{\text{AP}}}}{3} + \frac{{{\text{PB}}}}{6}$     M1

Note: Allow use of their AP and their PB for the M1.

$\Rightarrow t = \frac{{4\cos \theta }}{3} + \frac{{4\theta }}{6} = \frac{{4\cos \theta }}{3} + \frac{{2\theta }}{3} = \frac{2}{3}(2\cos \theta + \theta )$     AG

[3 marks]

$\frac{{{\text{d}}t}}{{{\text{d}}\theta }} = \frac{2}{3}( - 2\sin \theta + 1)$     A1

$\frac{2}{3}( - 2\sin \theta + 1) = 0 \Rightarrow \sin \theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi }{6}$ (or 30 degrees)     A1

[2 marks]

$\frac{{{{\text{d}}^2}t}}{{{\text{d}}{\theta ^2}}} = - \frac{4}{3}\cos \theta < 0\,\,\,\,\left( {{\text{at }}\theta = \frac{\pi }{6}} \right)$     M1

$\Rightarrow t$ is maximized at $\theta = \frac{\pi }{6}$     R1

time needed to walk along arc AB is $\frac{{2\pi }}{6}{\text{ (}} \approx {\text{1 hour)}}$

time needed to row from A to B is $\frac{4}{3}{\text{ (}} \approx {\text{1.33 hour)}}$

hence, time is minimized in walking from A to B     R1

[3 marks]

The fairly easy trigonometry challenged a large number of candidates.

Part (b) was very well done.

Satisfactory answers were very rarely seen for (c). Very few candidates realised that a minimum can occur at the beginning or end of an interval.