Date | November 2011 | Marks available | 3 | Reference code | 11N.1.hl.TZ0.8 |
Level | HL only | Paper | 1 | Time zone | TZ0 |
Command term | Show that | Question number | 8 | Adapted from | N/A |
Question
The diagram below shows a circular lake with centre O, diameter AB and radius 2 km.
Jorg needs to get from A to B as quickly as possible. He considers rowing to point P and then walking to point B. He can row at \(3{\text{ km}}\,{{\text{h}}^{ - 1}}\) and walk at \(6{\text{ km}}\,{{\text{h}}^{ - 1}}\). Let \({\rm{P\hat AB}} = \theta \) radians, and t be the time in hours taken by Jorg to travel from A to B.
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.
Markscheme
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]
Examiners report
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.