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 \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 $\left(a(2+\sqrt3),\dfrac{\pi}{6}\right)$ .

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\sqrt3) \end{align*}

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

(7)

解答

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切线平行于初始线，说明切线水平。极坐标中可以写 $y=r\sin\theta$ ，水平切线对应 $\dfrac{\mathrm{d}y}{\mathrm{d}\theta}=0$ 。

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Since

\begin{align*} y =&\,r\sin\theta\\[4mm] =&\,2a(1+\cos\theta)\sin\theta\\[4mm] =&\,2a\sin\theta+2a\sin\theta\cos\theta\\[4mm] =&\,2a\sin\theta+a\sin2\theta \end{align*}

Differentiate:

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}\theta} =&\,2a\cos\theta+2a\cos2\theta \end{align*}

For a horizontal tangent,

\begin{align*} 2a\cos\theta+2a\cos2\theta =&\,0\\[4mm] \cos\theta+\cos2\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*}

In the required part of the curve,

\begin{align*} \cos\theta=\frac12 \quad \Rightarrow \quad \theta=\frac{\pi}{3} \end{align*}

Then

\begin{align*} r =&\,2a\left(1+\cos\frac{\pi}{3}\right)\\[4mm] =&\,2a\left(1+\frac12\right)\\[4mm] =&\,3a \end{align*}

Therefore

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

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阴影面积等于曲线从 $\theta=\frac{\pi}{6}$ 到 $\theta=\frac{\pi}{3}$ 扫过的极坐标面积，减去三角形 $OAB$ 的面积。三角形两边分别是 $OA=3a$ 和 $OB=a(2+\sqrt3)$ ，夹角是 $\frac{\pi}{6}$ 。

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The polar area under the curve from $\theta=\frac{\pi}{6}$ to $\theta=\frac{\pi}{3}$ is

\begin{align*} \frac12\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}r^2\,\mathrm{d}\theta =&\,\frac12\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left(2a(1+\cos\theta)\right)^2\,\mathrm{d}\theta\\[4mm] =&\,2a^2\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (1+2\cos\theta+\cos^2\theta)\,\mathrm{d}\theta\\[4mm] =&\,2a^2\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left(\frac32+2\cos\theta+\frac12\cos2\theta\right) \,\mathrm{d}\theta \end{align*}

\begin{align*} \frac12\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}r^2\,\mathrm{d}\theta =&\,2a^2 \left[ \frac32\theta+2\sin\theta+\frac14\sin2\theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}\\[4mm] =&\,2a^2 \left( \frac{\pi}{4}+\sqrt3-1 \right) \end{align*}

The area of $\triangle OAB$ is

\begin{align*} \frac12\cdot 3a\cdot a(2+\sqrt3)\cdot \sin\frac{\pi}{6} =&\,\frac{3a^2}{4}(2+\sqrt3) \end{align*}

Therefore the shaded area is

\begin{align*} R =&\,2a^2\left(\frac{\pi}{4}+\sqrt3-1\right) -\frac{3a^2}{4}(2+\sqrt3)\\[4mm] =&\,\frac{a^2\pi}{2}+2a^2\sqrt3-2a^2 -\frac{3a^2}{2}-\frac{3a^2\sqrt3}{4}\\[4mm] =&\,\frac{a^2}{4}(2\pi-14+5\sqrt3) \end{align*}