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.

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]

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.