Date | May 2013 | Marks available | 6 | Reference code | 13M.2.hl.TZ2.13 |
Level | HL only | Paper | 2 | Time zone | TZ2 |
Command term | Show that | Question number | 13 | Adapted from | N/A |
Question
A straight street of width 20 metres is bounded on its parallel sides by two vertical walls, one of height 13 metres, the other of height 8 metres. The intensity of light at point P at ground level on the street is proportional to the angle \(\theta \) where \(\theta = {\rm{A\hat PB}}\), as shown in the diagram.
Find an expression for \(\theta \) in terms of x, where x is the distance of P from the base of the wall of height 8 m.
(i) Calculate the value of \(\theta \) when x = 0.
(ii) Calculate the value of \(\theta \) when x = 20.
Sketch the graph of \(\theta \), for \(0 \leqslant x \leqslant 20\).
Show that \(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = \frac{{5(744 - 64x - {x^2})}}{{({x^2} + 64)({x^2} - 40x + 569)}}\).
Using the result in part (d), or otherwise, determine the value of x corresponding to the maximum light intensity at P. Give your answer to four significant figures.
The point P moves across the street with speed \(0.5{\text{ m}}{{\text{s}}^{ - 1}}\). Determine the rate of change of \(\theta \) with respect to time when P is at the midpoint of the street.
Markscheme
EITHER
\(\theta = \pi - \arctan \left( {\frac{8}{x}} \right) - \arctan \left( {\frac{{13}}{{20 - x}}} \right)\) (or equivalent) M1A1
Note: Accept \(\theta = 180^\circ - \arctan \left( {\frac{8}{x}} \right) - \arctan \left( {\frac{{13}}{{20 - x}}} \right)\) (or equivalent).
OR
\(\theta = \arctan \left( {\frac{x}{8}} \right) + \arctan \left( {\frac{{20 - x}}{{13}}} \right)\) (or equivalent) M1A1
[2 marks]
(i) \(\theta = 0.994{\text{ }}\left( { = \arctan \frac{{20}}{{13}}} \right)\) A1
(ii) \(\theta = 1.19{\text{ }}\left( { = \arctan \frac{5}{2}} \right)\) A1
[2 marks]
correct shape. A1
correct domain indicated. A1
[2 marks]
attempting to differentiate one \(\arctan \left( {f(x)} \right)\) term M1
EITHER
\(\theta = \pi - \arctan \left( {\frac{8}{x}} \right) - \arctan \left( {\frac{{13}}{{20 - x}}} \right)\)
\(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = \frac{8}{{{x^2}}} \times \frac{1}{{1 + {{\left( {\frac{8}{x}} \right)}^2}}} - \frac{{13}}{{{{(20 - x)}^2}}} \times \frac{1}{{1 + {{\left( {\frac{{13}}{{20 - x}}} \right)}^2}}}\) A1A1
OR
\(\theta = \arctan \left( {\frac{x}{8}} \right) + \arctan \left( {\frac{{20 - x}}{{13}}} \right)\)
\(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = \frac{{\frac{1}{8}}}{{1 + {{\left( {\frac{x}{8}} \right)}^2}}} + \frac{{ - \frac{1}{{13}}}}{{1 + {{\left( {\frac{{20 - x}}{{13}}} \right)}^2}}}\) A1A1
THEN
\( = \frac{8}{{{x^2} + 64}} - \frac{{13}}{{569 - 40x + {x^2}}}\) A1
\( = \frac{{8(569 - 40x + {x^2}) - 13({x^{2}} + 64)}}{{({x^2} + 64)({x^2} - 40x + 569)}}\) M1A1
\( = \frac{{5(744 - 64x - {x^2})}}{{({x^2} + 64)({x^2} - 40x + 569)}}\) AG
[6 marks]
Maximum light intensity at P occurs when \(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = 0\). (M1)
either attempting to solve \(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = 0\) for x or using the graph of either \(\theta \) or \(\frac{{{\text{d}}\theta }}{{{\text{d}}x}}\) (M1)
x = 10.05 (m) A1
[3 marks]
\(\frac{{{\text{d}}x}}{{{\text{d}}t}} = 0.5\) (A1)
At x = 10, \(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = 0.000453{\text{ }}\left( { = \frac{5}{{11029}}} \right)\). (A1)
use of \(\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{{\text{d}}\theta }}{{{\text{d}}x}} \times \frac{{{\text{d}}x}}{{{\text{d}}t}}\) M1
\(\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = 0.000227{\text{ }}\left( { = \frac{5}{{22058}}} \right){\text{ (rad }}{{\text{s}}^{ - 1}})\) A1
Note: Award (A1) for \(\frac{{{\text{d}}x}}{{{\text{d}}t}} = - 0.5\) and A1 for \(\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = - 0.000227{\text{ }}\left( { = - \frac{5}{{22058}}} \right){\text{ }}\).
Note: Implicit differentiation can be used to find \(\frac{{{\text{d}}\theta }}{{{\text{d}}t}}\). Award as above.
[4 marks]
Examiners report
Part (a) was reasonably well done. While many candidates exhibited sound trigonometric knowledge to correctly express θ in terms of x, many other candidates were not able to use elementary trigonometry to formulate the required expression for θ.
In part (b), a large number of candidates did not realize that θ could only be acute and gave obtuse angle values for θ. Many candidates also demonstrated a lack of insight when substituting endpoint x-values into θ.
In part (c), many candidates sketched either inaccurate or implausible graphs.
In part (d), a large number of candidates started their differentiation incorrectly by failing to use the chain rule correctly.
For a question part situated at the end of the paper, part (e) was reasonably well done. A large number of candidates demonstrated a sound knowledge of finding where the maximum value of θ occurred and rejected solutions that were not physically feasible.
In part (f), many candidates were able to link the required rates, however only a few candidates were able to successfully apply the chain rule in a related rates context.