Differentiation: EJC 2024 Promo Q4

Let the length of $BC$ be $x$ cm and angle $BAC$ be $\theta$.

By cosine rule $$\begin{aligned} x^2 &= 5^2 + 4^2 - 2(5)(4)\cos\theta\\ x^2 &= 41 - 40\cos\theta \quad\text{(1)} \end{aligned}$$ When $\theta = \dfrac{\pi}{3}$, $$\begin{aligned} x^2 &= 41 - 40\left(\tfrac{1}{2}\right) = 21, \qquad x = \sqrt{21} \quad\text{(since $x$ is positive)} \end{aligned}$$ Differentiating (1) with respect to time $t$, $$\begin{aligned} 2x\frac{\,\mathrm{d} x}{\,\mathrm{d} t} &= -40(-\sin\theta)\frac{\,\mathrm{d}\theta}{\,\mathrm{d} t}. \end{aligned}$$ Substituting in $\theta = \dfrac{\pi}{3}$, $x = \sqrt{21}$, $\dfrac{\,\mathrm{d}\theta}{\,\mathrm{d} t} = -0.2$, $$\begin{aligned} 2\sqrt{21}\,\frac{\,\mathrm{d} x}{\,\mathrm{d} t} &= -40\left(-\frac{\sqrt{3}}{2}\right)(-0.2) \end{aligned}$$ From GC, $\dfrac{\,\mathrm{d} x}{\,\mathrm{d} t}=-0.756$ (3 s.f.)

At that instant, $BC$ is decreasing at a rate of $0.756$ cm/s.