Differentiation: DHS 2024 Promo Q13

(a)

Let OB be x m.

From diagram, $$\cos\theta=\frac3{AB-x}\quad\text{and}\quad\sin\theta=\frac2x.$$ So we have $AB=(AB-x)+x=\dfrac3{\cos\theta}+\dfrac2{\sin\theta}$, where $\alpha=3,\beta=2$.

(b)

Differentiating the result in part (a) with respect to t, $$\frac{\,\mathrm{d}(AB)}{\,\mathrm{d} t}=(3\sec\theta\tan\theta-2\operatorname{cosec}\theta\cot\theta) \left(\frac{\,\mathrm{d}\theta}{\,\mathrm{d} t}\right)\ \ldots(*)$$ When $OB=\dfrac2{\sin\theta}=4$, $\theta=\dfrac16\pi$.

Subst $\theta=\dfrac16\pi$ and $\dfrac{\,\mathrm{d}\theta}{\,\mathrm{d} t}=-0.1$ into (*):

Using GC, $\dfrac{\,\mathrm{d}(AB)}{\,\mathrm{d} t}=0.493$ m/s (to 3 s.f.)

$\therefore$ Rate of increase in AB when OB = 4 is 0.493 m/s

(c)

$p\leq AB$ for all possible lengths p and angles $\theta$,

i.e. $\max p=\min AB$. $$\begin{aligned} AB&=\frac3{\cos\theta}+\frac2{\sin\theta}.\\ \frac{\,\mathrm{d}(AB)}{\,\mathrm{d}\theta} &=\frac{3\sin\theta}{\cos^2\theta}-\frac{2\cos\theta}{\sin^2\theta} =\frac{3\sin^3\theta-2\cos^3\theta}{\cos^2\theta\sin^2\theta} \end{aligned}$$ At $\min AB$, $\dfrac{\,\mathrm{d}(AB)}{\,\mathrm{d}\theta}=0$ $$\begin{aligned} \Rightarrow3\sin^3\theta&=2\cos^3\theta\\ \tan^3\theta&=\frac23\Rightarrow\tan\theta=\left(\frac23\right)^{1/3}\\ \therefore\theta&=\tan^{-1}\left(\frac23\right)^{1/3}. \end{aligned}$$

$\therefore AB$ is a minimum when $\theta=\tan^{-1}\left(\dfrac23\right)^{1/3}$.

Alternative $$\frac{\,\mathrm{d}(AB)}{\,\mathrm{d}\theta} =\frac{3\sin^3\theta-2\cos^3\theta}{\cos^2\theta\sin^2\theta} =\frac{3\cos^3\theta\left(\tan^3\theta-\frac23\right)}{\cos^2\theta\sin^2\theta}$$

Thus we have $$\begin{aligned} AB&=\frac{3\left[(2)^{2/3}+(3)^{2/3}\right]^{1/2}}{(3)^{1/3}} +\frac{2\left[(2)^{2/3}+(3)^{2/3}\right]^{1/2}}{(2)^{1/3}}\\ &=\left[(2)^{2/3}+(3)^{2/3}\right]^{1/2} \left[(3)^{2/3}+(2)^{2/3}\right]\\ &=\left[(2)^{2/3}+(3)^{2/3}\right]^{3/2} \end{aligned}$$ $$\max p=\left[(2)^{2/3}+(3)^{2/3}\right]^{3/2},\ \text{where }k=\frac23.$$