IAL 2023 June Q8 | 國際數學站

Figure 1

Figure 1 shows a sketch of the curve $C$ with equation

\begin{align*} r =\,& 6(1 + \cos \theta) \qquad 0 \leqslant \theta \leqslant \pi\\[2mm] \end{align*}

Given that $C$ meets the initial line at the point $A$ , as shown in Figure 1,

(a) write down the polar coordinates of $A$ .

(1)

The line $l_1$ also shown in Figure 1, is the tangent to $C$ at the point $B$ and is parallel to the initial line.

(b) Use calculus to determine the polar coordinates of $B$ .

(4)

The line $l_2$ also shown in Figure 1, is the tangent to $C$ at $A$ and is perpendicular to the initial line.

The region $R$ , shown shaded in Figure 1, is bounded by $C$ , $l_1$ and $l_2$ .

(c) Use algebraic integration to find the exact area of $R$ , giving your answer in the form $p\sqrt{3} + q\pi$ where $p$ and $q$ are constants to be determined.

(8)

解答

(a)

Curve $C$ has equation $r = 6(1+\cos\theta)$ . It meets the initial line at point $A$ where $\theta = 0$ .

\begin{align*} r = &\,6(1+\cos 0)\\[4mm] = &\,6(1+1)\\[4mm] = &\,12 \end{align*}

The polar coordinates of $A$ are $(12, 0)$ .

(b)

The tangent to $C$ at point $B$ is parallel to the initial line, which means $\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0$ . The $y$ -coordinate is given by:

\begin{align*} y = &\,r\sin\theta\\[4mm] = &\,6\sin\theta(1+\cos\theta)\\[4mm] = &\,6\sin\theta + 6\sin\theta\cos\theta\\[4mm] = &\,6\sin\theta + 3\sin 2\theta \end{align*}

Differentiate with respect to $\theta$ :

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}\theta} = &\,6\cos\theta + 6\cos 2\theta \end{align*}

Set this to zero:

\begin{align*} 6\cos\theta + 6\cos 2\theta = &\,0\\[4mm] \cos\theta + (2\cos^2\theta - 1) = &\,0\\[4mm] 2\cos^2\theta + \cos\theta - 1 = &\,0\\[4mm] (2\cos\theta - 1)(\cos\theta + 1) = &\,0 \end{align*}

This gives $\cos\theta = \frac{1}{2}$ or $\cos\theta = -1$ . Since $0 \le \theta \le \pi$ , we find $\theta = \frac{\pi}{3}$ or $\theta = \pi$ . The point $B$ is above the initial line, so $\theta = \frac{\pi}{3}$ . Substitute $\theta = \frac{\pi}{3}$ back to find $r$ :

\begin{align*} r = &\,6\left(1+\cos\frac{\pi}{3}\right)\\[4mm] = &\,6\left(1+\frac{1}{2}\right)\\[4mm] = &\,9 \end{align*}

The polar coordinates of $B$ are $\left(9, \frac{\pi}{3}\right)$ .

(c)

Region $R$ is bounded by the curve $C$ , the horizontal line $l_1$ (tangent at $B$ ), and the vertical line $l_2$ (tangent at $A$ ). Let $O$ be the pole. First, consider the trapezium formed by points $O(0,0)$ , $A(12,0)$ , $P$ (the intersection of $l_1$ and $l_2$ ), and $B$ . The coordinates of $B$ in Cartesian plane:

\begin{align*} x_B = &\,r\cos\left(\frac{\pi}{3}\right) = 9 \cdot \frac{1}{2} = \frac{9}{2}\\[4mm] y_B = &\,r\sin\left(\frac{\pi}{3}\right) = 9 \cdot \frac{\sqrt{3}}{2} = \frac{9\sqrt{3}}{2} \end{align*}

Line $l_1$ is $y = \frac{9\sqrt{3}}{2}$ . Line $l_2$ is $x = 12$ . So, $P$ is at $\left(12, \frac{9\sqrt{3}}{2}\right)$ . Let $Q$ be the projection of $B$ onto $l_1$ , thus the horizontal top length is $BP = 12 - \frac{9}{2} = \frac{15}{2}$ . The area of the trapezium $OAPB$ is:

\begin{align*} \text{Area}_{OAPB} = &\,\frac{1}{2} (OA + BP) \cdot y_B\\[4mm] = &\,\frac{1}{2} \left(12 + \frac{15}{2}\right) \left(\frac{9\sqrt{3}}{2}\right)\\[4mm] = &\,\frac{1}{2} \left(\frac{39}{2}\right) \left(\frac{9\sqrt{3}}{2}\right)\\[4mm] = &\,\frac{351\sqrt{3}}{8} \end{align*}

Now calculate the area of the sector of curve $C$ between $\theta=0$ and $\theta=\frac{\pi}{3}$ :

\begin{align*} \text{Area}_{sector} = &\,\frac{1}{2} \int_{0}^{\frac{\pi}{3}} r^2 \,\mathrm{d}\theta\\[4mm] = &\,\frac{1}{2} \int_{0}^{\frac{\pi}{3}} 36(1+\cos\theta)^2 \,\mathrm{d}\theta\\[4mm] = &\,18 \int_{0}^{\frac{\pi}{3}} (1 + 2\cos\theta + \cos^2\theta) \,\mathrm{d}\theta\\[4mm] = &\,18 \int_{0}^{\frac{\pi}{3}} \left( 1 + 2\cos\theta + \frac{1+\cos 2\theta}{2} \right) \,\mathrm{d}\theta\\[4mm] = &\,18 \int_{0}^{\frac{\pi}{3}} \left( \frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta \right) \,\mathrm{d}\theta\\[4mm] = &\,18 \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]_{0}^{\frac{\pi}{3}} \end{align*}

Evaluate at the upper limit $\frac{\pi}{3}$ :

\begin{align*} 18 \left( \frac{3}{2}\left(\frac{\pi}{3}\right) + 2\sin\left(\frac{\pi}{3}\right) + \frac{1}{4}\sin\left(\frac{2\pi}{3}\right) \right) = &\,18 \left( \frac{\pi}{2} + \sqrt{3} + \frac{1}{4}\left(\frac{\sqrt{3}}{2}\right) \right)\\[4mm] = &\,18 \left( \frac{\pi}{2} + \sqrt{3} + \frac{\sqrt{3}}{8} \right)\\[4mm] = &\,18 \left( \frac{\pi}{2} + \frac{9\sqrt{3}}{8} \right)\\[4mm] = &\,9\pi + \frac{81\sqrt{3}}{4} \end{align*}

The exact area of $R$ is the difference between the area of the trapezium $OAPB$ and the area of the sector:

\begin{align*} \text{Area}_R = &\,\frac{351\sqrt{3}}{8} - \left( 9\pi + \frac{81\sqrt{3}}{4} \right)\\[4mm] = &\,\frac{351\sqrt{3}}{8} - \frac{162\sqrt{3}}{8} - 9\pi\\[4mm] = &\,\frac{189\sqrt{3}}{8} - 9\pi \end{align*}

This is in the form $p\sqrt{3} + q\pi$ , with $p = \frac{189}{8}$ and $q = -9$ .