2023HSCShort answer

Q14 (c) (ii)

4 marks

Question

The point $P$ lies on the circle centred at $I(r,0)$ with radius $r>0$ , such that $\overrightarrow{IP}$ makes an angle of $t$ to the horizontal.

The point $Q$ lies on the circle centred at $J(-R,0)$ with radius $R>0$ , such that $\overrightarrow{JQ}$ makes an angle of $2t$ to the horizontal.

Note that $\overrightarrow{OP}=\overrightarrow{OI}+\overrightarrow{IP}$ and $\overrightarrow{OQ}=\overrightarrow{OJ}+\overrightarrow{JQ}$ .

Using part (i), or otherwise, find the values of $t$ , where $-\pi\leq t\leq\pi$ , that maximise the area of triangle $OPQ$ .

Uncropped source page 15 for question 14.c.ii; requires human review and cropping — Reviewed visual · source page 15

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Solution

Criteria	Marks
Provides correct solution	4
Finds the values of $t$ at the stationary points of the area, or equivalent merit	3
Obtains the area of triangle $OPQ$ in terms of $t$ , or equivalent merit	2
Obtains the components of $\overrightarrow{OP}$ or $\overrightarrow{OQ}$ in terms of $t$ , or equivalent merit	1

Sample answer:

\overrightarrow{OP}=\overrightarrow{OI}+\overrightarrow{IP}=\begin{pmatrix}r+r\cos t\\r\sin t\end{pmatrix}

\overrightarrow{OQ}=\overrightarrow{OJ}+\overrightarrow{JQ}=\begin{pmatrix}-R+R\cos2t\\R\sin2t\end{pmatrix}

Using part (i), we get that the area of triangle $OPQ$ is

\begin{aligned} \text{Area}&=\frac12\left|(r+r\cos t)(R\sin2t)-(r\sin t)(-R+R\cos2t)\right|\\ &=\frac12Rr\left|(1+\cos t)\sin2t-\sin t(-1+\cos2t)\right|\\ &=\frac12Rr\left|(\sin2t\cos t-\sin t\cos2t)+\sin2t+\sin t\right|\\ &=\frac12Rr\left|\sin t+\sin2t+\sin t\right|\\ &=\frac12Rr|\sin2t+2\sin t| \end{aligned}

Let $f(t)=\sin2t+2\sin t$ , so the area of triangle $OPQ$ is $\frac12Rr|f(t)|$ .

\begin{aligned} f'(t)&=2\cos2t+2\cos t\\ &=2(2\cos^2t+\cos t-1)\\ &=2(2\cos t-1)(\cos t+1) \end{aligned}

For all $t$ , $\cos t+1\ge0$ ; therefore $f'(t)$ has the same sign as $2\cos t-1$ .

$f(t)$ is odd, so $|f(t)|$ is even and we only need to study $|f(t)|$ on $[0,\pi]$ , where area of triangle $=\frac12Rr|f(t)|$ , so $f(t)\ge0$ on $[0,\pi]$ .

f'(t)\ge0\iff2\cos t-1\ge0\iff\cos t\ge\frac12

which will happen when

-\frac\pi3\le t\le\frac\pi3

By symmetry, since $|f(t)|$ is even, we see that the area of the triangle is maximum when

t=-\frac\pi3\quad\text{or}\quad t=\frac\pi3

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Syllabus alignment

Vectors

Introduction to VectorsRepresent and operate with two-dimensional vectors; use magnitude, direction, scalar products and vectors in geometric proofs and projectile motion.

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