As part of a game, a person stands $x$ m away from a wall and tries to throw an object through a \

Question

As part of a game, a person stands $x$ m away from a wall and tries to throw an object through a $0.40$ m wide hole in the wall. The point $B$ denotes the bottom of the hole, which is located $3.05$ m above the ground. The object is thrown from point $P$ at a height of $1.75$ m above the ground. The angle $APB$ is denoted by $\theta $ in radians. Show that $\theta ={{\tan }^{-1}}\left( \frac{1.7}{x} \right)-{{\tan }^{-1}}\left( \frac{1.3}{x} \right)$.[2] Using differentiation, find the value of $x$ which gives the stationary value of $\theta $, giving your answer to four significant figures.[3] Show that the value of $x$ in (ii) corresponds to a maximum value of $\theta $, and suggest why knowing this value of $x$ may be useful in this situation.[2] The person walks away from the wall at a constant rate of $0.1$ m/s. Find the rate of change of $\theta $ at the instant when the person is $1$ m away from the wall.[2]

Accepted Answer

(i) $\theta=\tan^{-1}\!\left(\dfrac{1.7}{x}\right)-\tan^{-1}\!\left(\dfrac{1.3}{x}\right)$ (ii) $x=1.487\text{ m}$ ($4$ s.f.) (iii) When $x=1.487$, $\frac{{{\text{d}}^{2}}\theta }{\text{d}{{x}^{2}}}=-0.0597<0$By the Second Derivative Test, this gives a maximum point. It gives the distance which maximises the angle APB, so it makes for easier throwing / more chance to throw object through the hole / gives the widest leeway / better accuracy (iv) $\theta$ is increasing at $0.00463\text{ rad s}^{-1}$ (3 s.f.)

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