题目

Problem

Figure 1

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

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

where $a$ is a positive constant.

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

(a) Determine the polar coordinates of $A$ .

(6)

The point $B$ on the curve has polar coordinates

\begin{align*} \left( a(2+\sqrt{3}), \frac{\pi}{6} \right) \end{align*}

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

(b) Use calculus to determine the exact area of the shaded region $R$ .

Give your answer in the form

\begin{align*} \frac{a^2}{4}(d\pi - e + f\sqrt{3}) \end{align*}

where $d$ , $e$ and $f$ are integers.

(7)

解答

(a)

解法一

思路

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切线平行于 initial line 表示水平切线。极坐标下可以先写

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

然后令 $\dfrac{\mathrm{d}y}{\mathrm{d}\theta}=0$ ，求出 $\theta$ 后再代回 $r=2a(1+\cos\theta)$ 。

答题过程

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Since

\begin{align*} r=2a(1+\cos\theta), \end{align*}

we have

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

Differentiate:

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}\theta} =&\,2a\{-\sin^2\theta+(1+\cos\theta)\cos\theta\}\\[2mm] =&\,2a(\cos\theta+\cos^2\theta-\sin^2\theta)\\[2mm] =&\,2a(\cos\theta+\cos2\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*}

The point $A$ is on the upper part of the curve, so

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

Then

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

Therefore

\begin{align*} \boxed{A=\left(3a,\frac{\pi}{3}\right)}. \end{align*}

(b)

解法一

思路

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阴影面积等于曲线从 $\theta=\frac{\pi}{6}$ 到 $\theta=\frac{\pi}{3}$ 扫出的极坐标面积，减去三角形 $OAB$ 的面积。

极坐标曲线面积公式是

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

答题过程

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The area swept out by the curve from $B$ to $A$ is

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

Expand the integrand:

\begin{align*} 2a^2\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (1+\cos\theta)^2\,\mathrm{d}\theta =&\,2a^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 swept area is

\begin{align*} 2a^2 \left[ \frac{3\theta}{2}+2\sin\theta+\frac{1}{4}\sin2\theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} =&\,2a^2\left(\frac{\pi}{4}+\sqrt{3}-1\right). \end{align*}

Now find the area of triangle $OAB$ .

From part (a),

\begin{align*} OA=3a. \end{align*}

From the question,

\begin{align*} OB=a(2+\sqrt{3}) \end{align*}

and

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

Therefore

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

So the shaded area is

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

Hence

\begin{align*} \boxed{\frac{a^2}{4}(2\pi-14+5\sqrt{3})}. \end{align*}

IAL 2022 Oct Q7

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程