| Date | May 2009 | Marks available | 7 | Reference code | 09M.2.hl.TZ2.10 |
| Level | HL only | Paper | 2 | Time zone | TZ2 |
| Command term | Find and Show that | Question number | 10 | Adapted from | N/A |
Question
A helicopter H is moving vertically upwards with a speed of 10 ms−1 . The helicopter is \(h\) m directly above the point Q which is situated on level ground. The helicopter is observed from the point P which is also at ground level and PQ \( = 40\) m. This information is represented in the diagram below.
When \(h = 30\),
(a) show that the rate of change of \({\rm{H}}\hat {\text{P}}{\text{Q}}\) is \(0.16\) radians per second;
(b) find the rate of change of PH.
Markscheme
(a) let \({\text{H}}\hat {\text{P}}{\text{Q}} = \theta \)
\(\tan \theta = \frac{h}{{40}}\)
\({\sec ^2}\theta \frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{1}{{40}}\frac{{{\text{d}}h}}{{{\text{d}}t}}\) M1
\(\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{1}{{4{{\sec }^2}\theta }}\) (A1)
\( = \frac{{16}}{{4 \times 25}}\) \(\left( {\sec \theta = \frac{5}{4}{\text{ or }}\theta = 0.6435} \right)\) A1
\( = 0.16\) radians per second AG
(b) \({x^2} = {h^2} + 1600\), where PH \( = x\)
\(2x\frac{{{\text{d}}x}}{{{\text{d}}t}} = 2h\frac{{{\text{d}}h}}{{{\text{d}}t}}\) M1
\(\frac{{{\text{d}}x}}{{{\text{d}}t}} = \frac{h}{x} \times 10\) A1
\( = \frac{{10h}}{{\sqrt {{h^2} + 1600} }}\) (A1)
\(h = 30\) , \(\frac{{{\text{d}}x}}{{{\text{d}}t}} = 6\) ms–1 A1
Note: Accept solutions that begin\(x = 40\sec \theta \) or use \(h = 10t\) .
[7 marks]
Examiners report
For those candidates who realized this was an applied calculus problem involving related rates of change, the main source of error was in differentiating inverse tan in part (a). Some found part (b) easier than part (a), involving a changing length rather than an angle. A number of alternative approaches were reported by examiners.
