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.

1. Define a horizontal reference line from $P$ and calculate the vertical distances to $A$ and $B$.

Let $M$ be the point on the wall that is horizontally level with $P$.
Height of $P$ above ground = $1.75\text{ m}$.
Height of $A$ above ground = $3.05 + 0.40 = 3.45\text{ m}$.
Height of $B$ above ground = $3.05\text{ m}$.

The vertical distance from $M$ to $A$ is:
$AM = 3.45 - 1.75 = 1.7\text{ m}$

The vertical distance from $M$ to $B$ is:
$BM = 3.05 - 1.75 = 1.3\text{ m}$

The horizontal distance $PM$ is $x\text{ m}$.

2. Use trigonometric ratios to find the angles and subtract them to show the result.

In the right-angled triangle $APM$:
$\tan(\angle APM) = \frac{AM}{PM} = \frac{1.7}{x}$
$\angle APM = \tan^{-1}\left(\frac{1.7}{x}\right)$

In the right-angled triangle $BPM$:
$\tan(\angle BPM) = \frac{BM}{PM} = \frac{1.3}{x}$
$\angle BPM = \tan^{-1}\left(\frac{1.3}{x}\right)$

Since $\theta = \angle APB = \angle APM - \angle BPM$:
$\theta = \tan^{-1}\left(\frac{1.7}{x}\right) - \tan^{-1}\left(\frac{1.3}{x}\right)$ (Shown)

1. Differentiate $\theta$ with respect to $x$ using the chain rule.

$\frac{d\theta}{dx} = \frac{1}{1 + \left(\frac{1.7}{x}\right)^2} \left(-\frac{1.7}{x^2}\right) - \frac{1}{1 + \left(\frac{1.3}{x}\right)^2} \left(-\frac{1.3}{x^2}\right)$
$\frac{d\theta}{dx} = -\frac{1.7}{x^2 + 1.7^2} + \frac{1.3}{x^2 + 1.3^2}$

2. Equate the derivative to zero and simplify to form an equation in $x$.

For a stationary value, $\frac{d\theta}{dx} = 0$:
$\frac{1.7}{x^2 + 1.7^2} = \frac{1.3}{x^2 + 1.3^2}$
$\frac{1.7}{x^2 + 2.89} = \frac{1.3}{x^2 + 1.69}$
$1.7(x^2 + 1.69) = 1.3(x^2 + 2.89)$
$1.7x^2 + 2.873 = 1.3x^2 + 3.757$

3. Solve for $x$ and round the answer to four significant figures.

$0.4x^2 = 0.884$
$x^2 = 2.21$
$x = \sqrt{2.21} \approx 1.487 \text{ (since } x > 0)$

1. Use the second derivative test to show that the stationary point is a maximum.

From (ii), $\frac{d\theta}{dx} = \frac{1.3}{x^2 + 1.69} - \frac{1.7}{x^2 + 2.89}$ and the stationary point occurs at $x = \sqrt{2.21}$.

Differentiating with respect to $x$:
$\frac{d^2\theta}{dx^2} = -1.3(x^2 + 1.69)^{-2}(2x) - (-1.7)(x^2 + 2.89)^{-2}(2x)$
$\frac{d^2\theta}{dx^2} = \frac{3.4x}{(x^2 + 2.89)^2} - \frac{2.6x}{(x^2 + 1.69)^2}$

Substitute $x = \sqrt{2.21}$ and $x^2 = 2.21$:
$\frac{d^2\theta}{dx^2} = \frac{3.4\sqrt{2.21}}{(2.21 + 2.89)^2} - \frac{2.6\sqrt{2.21}}{(2.21 + 1.69)^2}$
$\frac{d^2\theta}{dx^2} = \frac{3.4\sqrt{2.21}}{5.1^2} - \frac{2.6\sqrt{2.21}}{3.9^2}$
$\frac{d^2\theta}{dx^2} \approx 0.194 - 0.254 = -0.060$

Since $\frac{d^2\theta}{dx^2} < 0$, the value $x = \sqrt{2.21}$ corresponds to a maximum value of $\theta$.

2. Suggest why knowing this optimal value of x is useful in the context of the game.

Knowing this value of $x$ gives the optimal standing distance where the angle $\theta$ subtended by the hole is maximized. This provides the largest apparent target size for the thrower, giving them the maximum margin of error and thus the highest chance of successfully throwing the object through the hole.

1. Find the rate of change of $\theta$ with respect to $x$ when $x = 1$.

Differentiating the expression from (i) with respect to $x$ (as previously done in part ii):
$\frac{d\theta}{dx} = \frac{1}{1 + \left(\frac{1.7}{x}\right)^2} \left(-\frac{1.7}{x^2}\right) - \frac{1}{1 + \left(\frac{1.3}{x}\right)^2} \left(-\frac{1.3}{x^2}\right)$
$\frac{d\theta}{dx} = -\frac{1.7}{x^2 + 1.7^2} + \frac{1.3}{x^2 + 1.3^2}$
At $x = 1$:
$\frac{d\theta}{dx} = -\frac{1.7}{1^2 + 2.89} + \frac{1.3}{1^2 + 1.69}$
$= -\frac{1.7}{3.89} + \frac{1.3}{2.69} \approx 0.046253 \text{ rad/m}$

2. Apply the connected rates of change formula to calculate the rate of change of $\theta$ with respect to time.

Since the person walks away from the wall at $0.1\text{ m/s}$, the distance $x$ is increasing, so $\frac{dx}{dt} = 0.1$.
Using the chain rule for connected rates of change:
$\frac{d\theta}{dt} = \frac{d\theta}{dx} \times \frac{dx}{dt}$
$\frac{d\theta}{dt} \approx 0.046253 \times 0.1$
$= 0.0046253 \approx 0.00463 \text{ rad/s} \text{ (to 3 s.f.)}$