In the diagram, point Q is 400 m vertically above point A. Initially, a balloon is at point Q and moves horizontally in a straight line at a constant speed of 10 m per second away from an observer stationed at point A. At time t seconds, after the start, the balloon is at point P where QP is x m and the angle of elevation of the balloon from the observer at A is $\beta$ radians. Find exactly the rate of change of $\beta$ at the instant when $x$ is 100 m. A gardener designs a flower bed $ABC$ in the shape of an isosceles triangle inscribed in a circular plot of land with radius 6 meters, as shown in the diagram. The vertices of the triangle lie on the circumference of the circle and $AB = AC$. The angle $BAC$ is $\theta$, where $\theta$ is acute. Express the area of the flower bed $ABC$ in terms of $\theta$. As $\theta$ varies, use differentiation to find the exact maximum possible area of the flower bed, and show that it is a maximum.

1. Set up an equation relating the angle of elevation $\beta$ to the horizontal distance $x$.

From the right-angled triangle formed by the observer A, the point vertically below P, and P:
$\tan \beta = \frac{400}{x}$

2. Differentiate implicitly with respect to time $t$ to relate the rates of change.

Differentiating both sides with respect to $t$:
$\sec^2 \beta \cdot \frac{d\beta}{dt} = -400x^{-2} \cdot \frac{dx}{dt}$
$\sec^2 \beta \cdot \frac{d\beta}{dt} = -\frac{400}{x^2} \cdot \frac{dx}{dt}$

3. Evaluate $\sec^2 \beta$ at the instant when $x = 100$.

When $x = 100$:
$\tan \beta = \frac{400}{100} = 4$
Using the trigonometric identity $\sec^2 \beta = 1 + \tan^2 \beta$:
$\sec^2 \beta = 1 + 4^2 = 17$

4. Substitute the known values to find the exact rate of change of $\beta$.

Given the balloon's horizontal speed, $\frac{dx}{dt} = 10$.
Substitute $x = 100$, $\frac{dx}{dt} = 10$, and $\sec^2 \beta = 17$ into the differentiated equation:
$17 \frac{d\beta}{dt} = -\frac{400}{100^2} (10)$
$17 \frac{d\beta}{dt} = -\frac{4000}{10000}$
$17 \frac{d\beta}{dt} = -\frac{2}{5}$
$\frac{d\beta}{dt} = -\frac{2}{85}$

1. Relate the dimensions of the triangle to the centre of the circle using a perpendicular bisector.

Let $O$ be the centre of the circle and $M$ be the midpoint of $BC$.
Since $\triangle ABC$ is isosceles with $AB = AC$, the line of symmetry $AM$ passes through $O$ and is perpendicular to $BC$.
By the angle at centre theorem, $\angle BOC = 2\angle BAC = 2\theta$.
Since $\triangle OBC$ is isosceles ($OB = OC$), $OM$ bisects $\angle BOC$, so $\angle MOC = \theta$.

2. Express the base, height, and area of the flower bed in terms of $\theta$.

In the right-angled $\triangle OMC$, the radius $OC = 6$.
$MC = 6 \sin \theta \implies \text{Base } BC = 2 \times 6 \sin \theta = 12 \sin \theta$
$OM = 6 \cos \theta$
Since $\theta$ is acute, all angles in $\triangle ABC$ are acute, meaning $O$ lies inside the triangle. Thus, the height is:
$AM = AO + OM = 6 + 6 \cos \theta$

$\text{Area} = \frac{1}{2} \times BC \times AM$
$\text{Area} = \frac{1}{2}(12 \sin \theta)(6 + 6 \cos \theta)$
$\text{Area} = 36 \sin \theta(1 + \cos \theta)$

1. Derive the expression for the area of the flower bed $A$ in terms of $\theta$ (from part (b)(i)).

Let $O$ be the centre of the circle with radius $r = 6$. By the angle at centre theorem, $\angle BOC = 2\theta$.
Since $AB = AC$, the triangle is symmetric and $\angle AOB = \angle AOC = \frac{2\pi - 2\theta}{2} = \pi - \theta$.
The area of the flower bed is the sum of the areas of $\triangle AOB$, $\triangle AOC$, and $\triangle BOC$:
$A = \frac{1}{2}(6^2)\sin(\pi - \theta) + \frac{1}{2}(6^2)\sin(\pi - \theta) + \frac{1}{2}(6^2)\sin(2\theta)$
$A = 18\sin\theta + 18\sin\theta + 18\sin 2\theta$
$A = 36\sin\theta + 18\sin 2\theta$

2. Differentiate the area with respect to $\theta$ and set it to zero to find stationary points.

$\frac{\mathrm{d}A}{\mathrm{d}\theta} = 36\cos\theta + 36\cos 2\theta$
For a maximum area, we require $\frac{\mathrm{d}A}{\mathrm{d}\theta} = 0$:
$36\cos\theta + 36\cos 2\theta = 0$
$\cos 2\theta + \cos\theta = 0$

3. Solve the trigonometric equation for $\theta$.

Using the double angle identity $\cos 2\theta = 2\cos^2\theta - 1$:
$2\cos^2\theta - 1 + \cos\theta = 0$
$2\cos^2\theta + \cos\theta - 1 = 0$
$(2\cos\theta - 1)(\cos\theta + 1) = 0$
Since $\theta$ is an acute angle, $0 < \theta < \frac{\pi}{2}$, which implies $\cos\theta > 0$.
Therefore, $\cos\theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}$.

4. Calculate the exact maximum area.

Substitute $\theta = \frac{\pi}{3}$ into the area expression:
$A = 36\sin\left(\frac{\pi}{3}\right) + 18\sin\left(\frac{2\pi}{3}\right)$
$A = 36\left(\frac{\sqrt{3}}{2}\right) + 18\left(\frac{\sqrt{3}}{2}\right)$
$A = 18\sqrt{3} + 9\sqrt{3} = 27\sqrt{3} \text{ m}^2$

5. Perform the second derivative test to verify that the area is a maximum.

$\frac{\mathrm{d}^2A}{\mathrm{d}\theta^2} = -36\sin\theta - 72\sin 2\theta$
When $\theta = \frac{\pi}{3}$:
$\frac{\mathrm{d}^2A}{\mathrm{d}\theta^2} = -36\sin\left(\frac{\pi}{3}\right) - 72\sin\left(\frac{2\pi}{3}\right)$
$\frac{\mathrm{d}^2A}{\mathrm{d}\theta^2} = -36\left(\frac{\sqrt{3}}{2}\right) - 72\left(\frac{\sqrt{3}}{2}\right) = -54\sqrt{3}$
Since $\frac{\mathrm{d}^2A}{\mathrm{d}\theta^2} < 0$, the exact area $27\sqrt{3}$ is indeed a maximum.