题目

Problem

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

Figure 3

The curve $C_1$ has equation

\begin{align*} r=\sqrt3+\tan\theta\qquad 0<\theta<\frac{\pi}{2} \end{align*}

The tangent to $C_1$ is perpendicular to the initial line at the point $P$

(a) Use calculus to determine, in simplest form, the exact polar coordinates of $P$

(4)

Figure 3 shows a sketch of part of the curve $C_1$ and part of the curve $C_2$

The curve $C_2$ is a circle with centre at the pole $O$ .

The curves $C_1$ and $C_2$ intersect at $P$ .

The region $R$ , shown shaded in Figure 3, is bounded by $C_1$ , $C_2$ and the initial line.

(b) Use algebraic integration to determine the area of $R$ , giving the answer in the form

\begin{align*} a\pi+\frac{\sqrt3}{2}(\ln b+c) \end{align*}

where $a$ , $b$ and $c$ are rational numbers.

(8)

解答

(a)

解法一

思路

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切线垂直于初始线，表示切线是竖直的，所以用 $x=r\cos\theta$ ，令 $\dfrac{\mathrm{d}x}{\mathrm{d}\theta}=0$ 。这里 $r$ 是 $\theta$ 的函数。

#### 答题过程

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Since

\begin{align*} x=&\,r\cos\theta\\[2mm] =&\,(\sqrt3+\tan\theta)\cos\theta\\[2mm] =&\,\sqrt3\cos\theta+\sin\theta \end{align*}

Differentiate with respect to $\theta$ :

\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}\theta} =&\,-\sqrt3\sin\theta+\cos\theta \end{align*}

For a vertical tangent,

\begin{align*} -\sqrt3\sin\theta+\cos\theta =&\,0\\[4mm] \cos\theta =&\,\sqrt3\sin\theta\\[4mm] \tan\theta =&\,\frac{1}{\sqrt3}\\[4mm] \theta =&\,\frac{\pi}{6} \end{align*}

Then

\begin{align*} r =&\,\sqrt3+\tan\frac{\pi}{6}\\[4mm] =&\,\sqrt3+\frac{1}{\sqrt3}\\[4mm] =&\,\frac{4}{\sqrt3}\\[4mm] =&\,\frac{4\sqrt3}{3} \end{align*}

Therefore

\begin{align*} P=\left(\frac{4\sqrt3}{3},\frac{\pi}{6}\right) \end{align*}

(b)

解法一

思路

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$C_2$ 是以原点为圆心、经过 $P$ 的圆，所以半径就是 $P$ 的 $r$ 。阴影面积等于圆扇形面积减去 $C_1$ 从 $0$ 到 $\pi/6$ 下方的极坐标面积。

答题过程

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The radius of $C_2$ is

\begin{align*} \frac{4\sqrt3}{3} \end{align*}

So the sector area is

\begin{align*} \frac12\left(\frac{4\sqrt3}{3}\right)^2\left(\frac{\pi}{6}\right) =&\,\frac{4\pi}{9} \end{align*}

For $C_1$ ,

\begin{align*} \frac12\int_0^{\frac{\pi}{6}}r^2\,\mathrm{d}\theta =&\,\frac12\int_0^{\frac{\pi}{6}} (\sqrt3+\tan\theta)^2\,\mathrm{d}\theta\\[4mm] =&\,\frac12\int_0^{\frac{\pi}{6}} (3+2\sqrt3\tan\theta+\tan^2\theta)\,\mathrm{d}\theta\\[4mm] =&\,\frac12\int_0^{\frac{\pi}{6}} (2+2\sqrt3\tan\theta+\sec^2\theta)\,\mathrm{d}\theta \end{align*}

Hence

\begin{align*} \frac12\int_0^{\frac{\pi}{6}}r^2\,\mathrm{d}\theta =&\,\left[ \theta-\sqrt3\ln(\cos\theta)+\frac12\tan\theta \right]_0^{\frac{\pi}{6}}\\[4mm] =&\,\frac{\pi}{6} -\sqrt3\ln\left(\frac{\sqrt3}{2}\right) +\frac{\sqrt3}{6} \end{align*}

Therefore

\begin{align*} \text{Area of }R =&\,\frac{4\pi}{9} -\left( \frac{\pi}{6} -\sqrt3\ln\left(\frac{\sqrt3}{2}\right) +\frac{\sqrt3}{6} \right)\\[4mm] =&\,\frac{5\pi}{18} +\sqrt3\ln\left(\frac{\sqrt3}{2}\right) -\frac{\sqrt3}{6}\\[4mm] =&\,\frac{5\pi}{18} +\frac{\sqrt3}{2}\ln\left(\frac34\right) -\frac{\sqrt3}{6}\\[4mm] =&\,\frac{5\pi}{18} +\frac{\sqrt3}{2} \left(\ln\left(\frac34\right)-\frac13\right) \end{align*}

\begin{align*} a=\frac{5}{18},\qquad b=\frac34,\qquad c=-\frac13 \end{align*}

IAL 2025 June A Q6

题目

解答

(a)

解法一

思路

(b)

解法一

思路

答题过程