题目

Problem

Figure 1

Figure 1 shows a sketch of the curve $C$ with polar equation

\begin{align*} r =&\, 1 + \cos \theta \qquad 0 \leqslant \theta \leqslant \pi \end{align*}

and the line $l$ with polar equation

\begin{align*} r =&\, k\sec \theta \qquad 0 \leqslant \theta < \frac{\pi}{2} \end{align*}

where $k$ is a positive constant.

Given that

$C$ and $l$ intersect at the point $P$
$OP = 1 + \frac{\sqrt{3}}{2}$

(a) determine the exact value of $k$ .

(2)

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

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

\begin{align*} p\pi + q\sqrt{3} + r \end{align*}

where $p$ , $q$ and $r$ are simplified rational numbers to be determined.

(7)

解答

(a)

解法一

思路

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在交点 $P$ 上，曲线给出 $r=1+\cos\theta$ 。题目给了 $OP=r=1+\frac{\sqrt3}{2}$ ，所以可以先求 $\theta$ 。直线 $r=k\sec\theta$ 等价于 $r\cos\theta=k$ 。

答题过程

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At $P$ ,

\begin{align*} r=1+\frac{\sqrt3}{2} \end{align*}

Using $r=1+\cos\theta$ ,

\begin{align*} 1+\cos\theta =&\, 1+\frac{\sqrt3}{2}\\[2mm] \cos\theta =&\, \frac{\sqrt3}{2} \end{align*}

Hence

\begin{align*} \theta=\frac{\pi}{6} \end{align*}

Since $P$ lies on $r=k\sec\theta$ ,

\begin{align*} k =&\, r\cos\theta\\[4mm] =&\, \left(1+\frac{\sqrt3}{2}\right)\frac{\sqrt3}{2}\\[4mm] =&\, \boxed{\frac{\sqrt3}{2}+\frac34} \end{align*}

(b)

解法一

思路

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区域可以拆成两部分：从 $\theta=\frac{\pi}{6}$ 到 $\pi$ 的曲线面积，加上从 initial line 到 $OP$ 之间由直线 $l$ 形成的小三角形。曲线面积用极坐标积分，小三角形用 $\frac12xy$ 或 $\frac12ab\sin C$ 。

答题过程

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First find the area under $C$ from $\theta=\frac{\pi}{6}$ to $\theta=\pi$ :

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

Therefore

\begin{align*} \text{Area}_C =&\, \frac12\left[ \frac32\theta+2\sin\theta+\frac14\sin2\theta \right]_{\frac{\pi}{6}}^{\pi}\\[4mm] =&\, \frac12\left[ \frac{3\pi}{2} -\left( \frac{\pi}{4}+1+\frac{\sqrt3}{8} \right) \right]\\[4mm] =&\, \frac{5\pi}{8}-\frac12-\frac{\sqrt3}{16} \end{align*}

Now find the triangular area between the initial line and $OP$ .

At $P$ ,

\begin{align*} x_P =&\, r\cos\frac{\pi}{6} =k =\frac{\sqrt3}{2}+\frac34\\[4mm] y_P =&\, r\sin\frac{\pi}{6} =\left(1+\frac{\sqrt3}{2}\right)\frac12\\[4mm] =&\, \frac12+\frac{\sqrt3}{4} \end{align*}

\begin{align*} \text{Area}_{\triangle} =&\, \frac12 x_Py_P\\[4mm] =&\, \frac12 \left(\frac{\sqrt3}{2}+\frac34\right) \left(\frac12+\frac{\sqrt3}{4}\right)\\[4mm] =&\, \frac{7\sqrt3}{32}+\frac38 \end{align*}

Hence

\begin{align*} \text{Area}_R =&\, \left(\frac{5\pi}{8}-\frac12-\frac{\sqrt3}{16}\right) +\left(\frac{7\sqrt3}{32}+\frac38\right)\\[4mm] =&\, \frac{5\pi}{8} +\frac{5\sqrt3}{32} -\frac18 \end{align*}

Thus

\begin{align*} \boxed{ p=\frac58,\qquad q=\frac{5}{32},\qquad r=-\frac18 } \end{align*}

IAL 2024 June Q10

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程