IAL 2023 Jan Q8 | 國際數學站

Figure 1

The curve $C$ shown in Figure 1 has polar equation

\begin{align*} r =\,& 1 - \sin \theta \qquad 0 \leqslant \theta < \frac{\pi}{2}\\[2mm] \end{align*}

The point $P$ lies on $C$ , such that the tangent to $C$ at $P$ is parallel to the initial line.

(a) Use calculus to determine the polar coordinates of $P$

(4)

The finite region $R$ , shown shaded in Figure 1, is bounded by

the line with equation $\theta = \frac{\pi}{2}$
the tangent to $C$ at $P$
part of the curve $C$
the initial line

(b) Use algebraic integration to show that the area of $R$ is

\begin{align*} \frac{1}{32}(a\pi + b\sqrt{3} + c)\\[2mm] \end{align*}

where $a$ , $b$ and $c$ are integers to be determined.

(6)

解答

(a)

The curve is $r = 1 - \sin\theta$ . The tangent at $P$ is parallel to the initial line, which means the $y$ -coordinate is at a maximum or minimum, so $\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0$ .

\begin{align*} y = &\,r\sin\theta\\[4mm] = &\,(1 - \sin\theta)\sin\theta\\[4mm] = &\,\sin\theta - \sin^2\theta \end{align*}

Differentiate with respect to $\theta$ :

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

Set this to zero to find the turning point:

\begin{align*} \cos\theta(1 - 2\sin\theta) = &\,0 \end{align*}

Since $0 \le \theta < \frac{\pi}{2}$ , $\cos\theta \neq 0$ , so we must have:

\begin{align*} 1 - 2\sin\theta = &\,0\\[4mm] \sin\theta = &\,\frac{1}{2}\\[4mm] \theta = &\,\frac{\pi}{6} \end{align*}

Substitute $\theta = \frac{\pi}{6}$ into the equation for $r$ :

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

The polar coordinates of $P$ are $\left( \frac{1}{2}, \frac{\pi}{6} \right)$ .

(b)

The region $R$ lies below the horizontal tangent at $P$ , between the vertical line $\theta = \frac{\pi}{2}$ (the $y$ -axis), the curve $C$ , and the initial line ( $x$ -axis). We can compute the area of $R$ by adding the area of the triangle formed by the origin, $P$ , and the $y$ -axis, to the polar area swept by the curve from $\theta = 0$ to $\theta = \frac{\pi}{6}$ .

First, find the Cartesian coordinates of $P$ :

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

The area of the triangle formed by $O(0,0)$ , $(0, y_P)$ , and $P(x_P, y_P)$ is:

\begin{align*} \text{Area of Triangle} = &\,\frac{1}{2} \cdot x_P \cdot y_P\\[4mm] = &\,\frac{1}{2} \left(\frac{\sqrt{3}}{4}\right) \left(\frac{1}{4}\right)\\[4mm] = &\,\frac{\sqrt{3}}{32} \end{align*}

Next, find the polar area of the sector from $\theta=0$ to $\theta=\frac{\pi}{6}$ :

\begin{align*} \text{Area of Sector} = &\,\frac{1}{2}\int_{0}^{\frac{\pi}{6}} r^2 \,\mathrm{d}\theta\\[4mm] = &\,\frac{1}{2}\int_{0}^{\frac{\pi}{6}} (1 - \sin\theta)^2 \,\mathrm{d}\theta\\[4mm] = &\,\frac{1}{2}\int_{0}^{\frac{\pi}{6}} (1 - 2\sin\theta + \sin^2\theta) \,\mathrm{d}\theta\\[4mm] = &\,\frac{1}{2}\int_{0}^{\frac{\pi}{6}} \left(1 - 2\sin\theta + \frac{1 - \cos 2\theta}{2}\right) \,\mathrm{d}\theta\\[4mm] = &\,\frac{1}{2}\int_{0}^{\frac{\pi}{6}} \left(\frac{3}{2} - 2\sin\theta - \frac{1}{2}\cos 2\theta\right) \,\mathrm{d}\theta\\[4mm] = &\,\frac{1}{2}\left[ \frac{3}{2}\theta + 2\cos\theta - \frac{1}{4}\sin 2\theta \right]_{0}^{\frac{\pi}{6}} \end{align*}

Evaluate the limits:

\begin{align*} \text{Upper limit: } & \frac{1}{2}\left( \frac{3}{2}\left(\frac{\pi}{6}\right) + 2\cos\left(\frac{\pi}{6}\right) - \frac{1}{4}\sin\left(\frac{\pi}{3}\right) \right)\\[4mm] = &\,\frac{1}{2}\left( \frac{\pi}{4} + \sqrt{3} - \frac{\sqrt{3}}{8} \right)\\[4mm] = &\,\frac{1}{2}\left( \frac{\pi}{4} + \frac{7\sqrt{3}}{8} \right) = \frac{\pi}{8} + \frac{7\sqrt{3}}{16}\\[4mm] \text{Lower limit: } & \frac{1}{2}\left( 0 + 2\cos(0) - 0 \right) = \frac{1}{2}(2) = 1 \end{align*}

The area of the sector is $\frac{\pi}{8} + \frac{7\sqrt{3}}{16} - 1$ .

Total area of $R$ is the sum of the triangle and the sector:

\begin{align*} \text{Total Area} = &\,\frac{\sqrt{3}}{32} + \left( \frac{\pi}{8} + \frac{7\sqrt{3}}{16} - 1 \right)\\[4mm] = &\,\frac{\sqrt{3}}{32} + \frac{4\pi}{32} + \frac{14\sqrt{3}}{32} - \frac{32}{32}\\[4mm] = &\,\frac{1}{32}\left( 4\pi + 15\sqrt{3} - 32 \right) \end{align*}

This is in the required form $\frac{1}{32}(a\pi + b\sqrt{3} + c)$ , where $a=4$ , $b=15$ , and $c=-32$ .