Differentiation: ACJC 2025 Promo Q9

(a)

With $\tan\beta = \dfrac{400}{x}$, so $\beta = \tan^{-1}\!\left(\dfrac{400}{x}\right)$, and $\dfrac{\,\mathrm{d} x}{\,\mathrm{d} t} = 10$: $$\begin{aligned} \frac{\,\mathrm{d}\beta}{\,\mathrm{d} x} &= \frac{1}{1 + \left(\tfrac{400}{x}\right)^2}\left(-\frac{400}{x^2}\right). \end{aligned}$$ When $x = 100$: $$\begin{aligned} \frac{\,\mathrm{d}\beta}{\,\mathrm{d} x} &= \frac{1}{1 + 16}\left(-\frac{400}{10000}\right) = -\frac{1}{425}\\ \frac{\,\mathrm{d}\beta}{\,\mathrm{d} t} &= \frac{\,\mathrm{d}\beta}{\,\mathrm{d} x}\frac{\,\mathrm{d} x}{\,\mathrm{d} t} = \left(-\frac{1}{425}\right)(10) = -\frac{2}{85} \text{ rad\,s}^{-1}. \end{aligned}$$

The rate of change of $\beta$ is negative, since $\beta$ is decreasing as the balloon moves away.

(b)

With circumradius $6$ and $AB = AC$, $AC = 12\cos\tfrac{\theta}{2}$: $$\begin{aligned} A &= \frac{1}{2}(AB)(AC)\sin\theta = \frac{1}{2}\left(12\cos\tfrac{\theta}{2}\right)^{\!2}\sin\theta\\ &= 72\cos^2\tfrac{\theta}{2}\sin\theta = 36\sin\theta(1 + \cos\theta). \end{aligned}$$ Differentiating: $$\begin{aligned} \frac{\,\mathrm{d} A}{\,\mathrm{d}\theta} &= 36\big[\cos\theta(1+\cos\theta) - \sin^2\theta\big]\\ &= 36\big[2\cos^2\theta + \cos\theta - 1\big]\\ &= 36(2\cos\theta - 1)(\cos\theta + 1) = 0. \end{aligned}$$ Since $\theta$ is acute, $\cos\theta = \tfrac{1}{2}$, so $\theta = \tfrac{\pi}{3}$. Then $$\begin{aligned} \frac{\,\mathrm{d}^2 A}{\,\mathrm{d}\theta^2}\bigg|_{\theta = \pi/3} &= -54\sqrt{3} < 0, \end{aligned}$$ so $A$ is a maximum. The maximum area is $$\begin{aligned} A &= 36\sin\frac{\pi}{3}\left(1 + \cos\frac{\pi}{3}\right) = 36\left(\frac{\sqrt{3}}{2}\right)\left(\frac{3}{2}\right) = 27\sqrt{3} \text{ m}^2. \end{aligned}$$