Subject: Mathematics — Sequences and Series
Question: A 2 metre long object is fixed vertically upwards from the mid-point of two locations $A$ and $B$, 8 metre apart. The speed of the object after $t$ seconds is given by
$$\frac{ds}{dt} = (2t + 1)\text{ metres per second}.$$
Let $\alpha$ and $\beta$ be the angles subtended by the object at $A$
and $B$, respectively after one and two seconds, find the value of
$\cos(\alpha - \beta)$.
Answer: $\frac{5}{\sqrt{26}}$
Solution: We have
$$\frac{ds}{dt} = 2t + 1$$
$$\Rightarrow s = t^{2} + t + k,\text{ where }k\text{ is a constant.}$$
When $t = 0$, the distance of the top of the object from the mid point
$M$ of $AB$ is 2 meters,
$$\therefore 2 = k.$$
Also $s = t^{2} + t + 2$
where $s$ is the distance of the top of the object from $M$ after $t$
seconds.
Let $PQ$ and $RS$ be the positions of the object after one and two
seconds respectively, then
$$QM = 1 + 1 + 2 = 4$$
$$PM = QM - PQ = 4 - 2 = 2$$
$$SM = 4 + 2 + 2 = 8$$
$$RM = SM - SR = 6$$
$$AM = MB = 4$$
[IMAGE] Now,
$$AP^{2} = AM^{2} + PM^{2}$$
$$AP^{2} = 16 + 4 = 20$$
$$AQ^{2} = 16 + 16 = 32$$
$$BR^{2} = 16 + 36 = 52$$
$$BS^{2} = 16 + 64 = 80$$
By the cosine formula,
$$\cos\alpha = \frac{AP^{2} + AQ^{2} - PQ^{2}}{2 \cdot AP \cdot AQ}$$
$$\cos\alpha = \frac{20 + 32 - 4}{2\sqrt{20}\sqrt{32}}$$
$$\cos\alpha = \frac{48}{2 \times 8 \times \sqrt{10}} = \frac{3}{\sqrt{10}}$$
and
$$\cos\beta = \frac{BS^{2} + BR^{2} - SR^{2}}{2 \cdot BS \cdot BR}$$
$$\cos\beta = \frac{80 + 52 - 4}{2\sqrt{80}\sqrt{52}}$$
$$\cos\beta = \frac{128}{2 \times 4\sqrt{5} \times 2\sqrt{13}} = \frac{8}{\sqrt{65}}.$$
Therefore,
$$\sin\alpha = \frac{1}{\sqrt{10}}\quad\text{and}\quad\sin\beta = \frac{1}{\sqrt{65}}.$$
$$\cos(\alpha - \beta) = cos\alpha \cdot cos\beta + sin\alpha \cdot sin\beta$$
$$\cos(\alpha - \beta) = \frac{3}{\sqrt{10}} \times \frac{8}{\sqrt{65}} + \frac{1}{\sqrt{10}} \times \frac{1}{\sqrt{65}}$$
$$\cos(\alpha - \beta) = \frac{25}{5\sqrt{26}} = \frac{5}{\sqrt{26}}.$$
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