题目

Problem

The curve shown in Figure 1 has polar equation

\begin{align*} r =&\, 4a(1 + \cos \theta), \qquad 0 \leqslant \theta < \pi \end{align*}

where $a$ is a positive constant.

The tangent to the curve at the point $A$ is parallel to the initial line.

(a) Show that the polar coordinates of $A$ are $\Big( 6a, \frac{\pi}{3} \Big)$

(6)

The point $B$ lies on the curve such that angle $AOB = \frac{\pi}{6}$

The finite region $R$ , shown shaded in Figure 1, is bounded by the line $AB$ and the curve.

(b) Use calculus to determine the area of the shaded region $R$ , giving your answer in the form $a^2 (n\pi + p\sqrt{3} + q)$ , where $n, p$ and $q$ are integers.

(7)

题目中文翻译

图 1 中曲线的极坐标方程为

\begin{align*} r=&\,4a(1+\cos\theta),\qquad0\leqslant\theta<\pi, \end{align*}

其中 $a$ 为正常数。

曲线在点 $A$ 处的切线平行于极轴。

(a) 证明 $A$ 的极坐标为 $\Big(6a,\dfrac{\pi}{3}\Big)$ 。

(6)

点 $B$ 在曲线上，且 $\angle AOB=\dfrac{\pi}{6}$ 。

图 1 中阴影所示的有限区域 $R$ 由直线 $AB$ 和曲线围成。

(b) 用微积分求阴影区域 $R$ 的面积，并将答案写成 $a^2(n\pi+p\sqrt3+q)$ 的形式，其中 $n$ 、 $p$ 和 $q$ 为整数。

(7)

解答

(a)

解法一

思路

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极坐标曲线要判断切线是否平行于 initial line，可以转成

\begin{align*} y=r\sin\theta. \end{align*}

切线平行于 initial line 表示水平切线，所以令 $\dfrac{\mathrm{d}y}{\mathrm{d}\theta}=0$ 。求出对应的 $\theta$ 后，再代回 $r=4a(1+\cos\theta)$ 。

答题过程

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For the polar curve,

\begin{align*} r=4a(1+\cos\theta). \end{align*}

Using $y=r\sin\theta$ ,

\begin{align*} y=&\,4a(1+\cos\theta)\sin\theta. \end{align*}

Differentiate with respect to $\theta$ :

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}\theta} =&\,4a\Big\{-\sin^2\theta+(1+\cos\theta)\cos\theta\Big\}\\[2mm] =&\,4a(\cos\theta+\cos^2\theta-\sin^2\theta)\\[2mm] =&\,4a(\cos\theta+\cos 2\theta). \end{align*}

At $A$ , the tangent is parallel to the initial line, so

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}\theta}=0. \end{align*}

Hence

\begin{align*} \cos\theta+\cos2\theta=&\,0\\[2mm] \cos\theta+2\cos^2\theta-1=&\,0\\[2mm] 2\cos^2\theta+\cos\theta-1=&\,0\\[2mm] (2\cos\theta-1)(\cos\theta+1)=&\,0. \end{align*}

Since $0\leqslant\theta<\pi$ ,

\begin{align*} \cos\theta=\frac{1}{2} \quad\Longrightarrow\quad \theta=\frac{\pi}{3}. \end{align*}

Then

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

Therefore the polar coordinates of $A$ are

\begin{align*} \boxed{\Big(6a,\frac{\pi}{3}\Big)}. \end{align*}

(b)

解法一

思路

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阴影区域是「曲线从 $B$ 到 $A$ 扫出的极坐标面积」减去三角形 $OAB$ 。

由 (a) 知 $A$ 的角是 $\frac{\pi}{3}$ ，又 $\angle AOB=\frac{\pi}{6}$ ，所以 $B$ 的角是 $\frac{\pi}{6}$ 。先用

\begin{align*} \frac{1}{2}\int_{\alpha}^{\beta}r^2\,\mathrm{d}\theta \end{align*}

求曲线扇形面积，再减去

\begin{align*} \frac{1}{2}(OA)(OB)\sin\angle AOB. \end{align*}

答题过程

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From part (a),

\begin{align*} A=\Big(6a,\frac{\pi}{3}\Big). \end{align*}

Since $\angle AOB=\dfrac{\pi}{6}$ , the polar angle of $B$ is

\begin{align*} \frac{\pi}{3}-\frac{\pi}{6} =\frac{\pi}{6}. \end{align*}

Let $S$ be the area swept out by the curve from $B$ to $A$ . Then

\begin{align*} S=&\,\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \{4a(1+\cos\theta)\}^2\,\mathrm{d}\theta\\[2mm] =&\,8a^2\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (1+\cos\theta)^2\,\mathrm{d}\theta. \end{align*}

Expanding the integrand,

\begin{align*} S=&\,8a^2\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (1+2\cos\theta+\cos^2\theta)\,\mathrm{d}\theta. \end{align*}

Using

\begin{align*} \cos^2\theta=\frac{1+\cos2\theta}{2}, \end{align*}

the area is

\begin{align*} S=&\,8a^2 \Big[ \frac{3\theta}{2}+2\sin\theta+\frac{1}{4}\sin2\theta \Big]_{\frac{\pi}{6}}^{\frac{\pi}{3}}. \end{align*}

At the upper and lower limits, respectively,

\begin{align*} \frac{3}{2}\cdot\frac{\pi}{3} +2\sin\frac{\pi}{3} +\frac14\sin\frac{2\pi}{3} =&\,\frac{\pi}{2}+\sqrt3+\frac{\sqrt3}{8},\\[3mm] \frac{3}{2}\cdot\frac{\pi}{6} +2\sin\frac{\pi}{6} +\frac14\sin\frac{\pi}{3} =&\,\frac{\pi}{4}+1+\frac{\sqrt3}{8}. \end{align*}

Subtracting,

\begin{align*} S=&\,8a^2\Big[ \Big(\frac{\pi}{2}+\sqrt3+\frac{\sqrt3}{8}\Big)\\[2mm] &\,\hspace{2pt}-\Big(\frac{\pi}{4}+1+\frac{\sqrt3}{8}\Big)\Big]\\[2mm] =&\,8a^2\Big(\frac{\pi}{4}+\sqrt3-1\Big). \end{align*}

Now find the area of triangle $OAB$ .

At $B$ ,

\begin{align*} OB=&\,4a\Big(1+\cos\frac{\pi}{6}\Big)\\[2mm] =&\,4a\Big(1+\frac{\sqrt{3}}{2}\Big)\\[2mm] =&\,4a+2a\sqrt{3}. \end{align*}

Also $OA=6a$ and $\angle AOB=\dfrac{\pi}{6}$ . Therefore

\begin{align*} \text{area of triangle }OAB =&\,\frac{1}{2}(6a)(4a+2a\sqrt{3}) \sin\frac{\pi}{6}\\[2mm] =&\,\frac{1}{2}(6a)(4a+2a\sqrt{3}) \cdot\frac{1}{2}\\[2mm] =&\,6a^2+3a^2\sqrt{3}. \end{align*}

So the shaded area is

\begin{align*} &\,8a^2\Big(\frac{\pi}{4}+\sqrt{3}-1\Big)\\[2mm] &\,\hspace{2pt}-(6a^2+3a^2\sqrt{3})\\[2mm] =&\,2\pi a^2+8a^2\sqrt{3}-8a^2\\[2mm] &\,\hspace{2pt}-6a^2-3a^2\sqrt{3}\\[2mm] =&\,a^2(2\pi+5\sqrt{3}-14). \end{align*}

Therefore

\begin{align*} \boxed{n=2,\qquad p=5,\qquad q=-14}. \end{align*}

FP2 2022 June Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程