题目

Problem

The curve $C$ has Cartesian equation $x^2+xy+y^2=a$ , where $a$ is a positive constant.

(a) Show that the polar equation of $C$ is $r^2=\dfrac{2a}{2+\sin2\theta}$ .

(3)

(b) Sketch the part of $C$ for $0\le\theta\le\frac14\pi$ .

(2)

The region $R$ is enclosed by this part of $C$ , the initial line and the half-line $\theta=\frac14\pi$ .

(c) It is given that $\sin2\theta$ may be expressed as $\dfrac{2\tan\theta}{1+\tan^2\theta}$ . Use this result to show that the area of $R$ is

\frac12a\int_0^{\pi/4}\frac{1+\tan^2\theta}{1+\tan\theta+\tan^2\theta}\,\mathrm d\theta

and use the substitution $t=\tan\theta$ to find the exact value of this area.

(8)

题目中文翻译

曲线 $C$ 的直角坐标方程为 $x^2+xy+y^2=a$ ，其中 $a$ 是正常数。

(a) 证明 $C$ 的极坐标方程为 $r^2=\dfrac{2a}{2+\sin2\theta}$ 。

(3)

(b) 画出 $C$ 在 $0\le\theta\le\frac14\pi$ 范围内的部分的草图。

(2)

区域 $R$ 由 $C$ 的这一部分、极轴和射线 $\theta=\frac14\pi$ 围成。

\frac12a\int_0^{\pi/4}\frac{1+\tan^2\theta}{1+\tan\theta+\tan^2\theta}\,\mathrm d\theta

并利用代换 $t=\tan\theta$ 求此面积的精确值。

(8)

原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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代入 $x=r\cos\theta$ 、 $y=r\sin\theta$ ，用平方和与二倍角恒等式消去直角坐标。

答题过程

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Substituting $x=r\cos\theta$ , $y=r\sin\theta$ into $x^2+xy+y^2=a$ gives

\begin{align*} &\,r^2\cos^2\theta+r^2\sin\theta\cos\theta\\[4mm] &\,\hspace{2pt}+r^2\sin^2\theta=a,\\[4mm] r^2(1+\sin\theta\cos\theta)=&\,a,\\[4mm] r^2\bigg(1+\frac12\sin2\theta\bigg)=&\,a. \end{align*}

Therefore

\boxed{r^2=\frac{2a}{2+\sin2\theta}}.

(b)

解法一

思路

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曲线是第一象限内的一段椭圆弧，半径严格递减。端点切线既不能随意画成水平，也不能画成竖直。

答题过程

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As $\theta$ increases from $0$ to $\pi/4$ , $\sin2\theta$ increases from $0$ to $1$ , so $r$ decreases strictly from $\sqrt a$ to $\sqrt{2a/3}$ . The endpoints in Cartesian coordinates are

\begin{align*} &\,(\sqrt a,0),\\[4mm] &\,\bigg(\sqrt{\frac a3},\sqrt{\frac a3}\bigg). \end{align*}

Implicit differentiation of $x^2+xy+y^2=a$ gives

\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{2x+y}{x+2y},

so the endpoint gradients are $-2$ and $-1$ respectively. The arc slopes downwards to the right and is concave down; indeed $\mathrm{d}^2y/\mathrm{d}x^2=-6a/(x+2y)^3<0$ in this quadrant. Draw only the arc between the initial line and $\theta=\pi/4$ , marking the pole and these bounding rays.

(c)

解法一

思路

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先将极坐标面积式化成题目指定形式，再完整替换积分上下限。分母配方后使用反正切积分。

答题过程

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The polar area formula and part (a) give

\begin{align*} R=&\,\frac12\int_0^{\pi/4}r^2\,\mathrm{d}\theta\\[4mm] =&\,a\int_0^{\pi/4}\frac1{2+\sin2\theta}\,\mathrm{d}\theta\\[4mm] =&\,\frac a2\int_0^{\pi/4}\frac1{1+\frac{\tan\theta}{1+\tan^2\theta}}\,\mathrm{d}\theta\\[4mm] =&\,\frac a2\int_0^{\pi/4}\frac{1+\tan^2\theta}{1+\tan\theta+\tan^2\theta}\,\mathrm{d}\theta, \end{align*}

as required. Now let $t=\tan\theta$ . Then $\mathrm{d}t=(1+\tan^2\theta)\,\mathrm{d}\theta$ , and the limits are $t=0,1$ . Thus

\begin{align*} R=&\,\frac a2\int_0^1\frac1{t^2+t+1}\,\mathrm{d}t\\[4mm] =&\,\frac a2\int_0^1\frac1{(t+1/2)^2+3/4}\,\mathrm{d}t\\[4mm] =&\,\frac a{\sqrt3}\bigg[\tan^{-1}\bigg(\frac{2t+1}{\sqrt3}\bigg)\bigg]_0^1\\[4mm] =&\,\frac a{\sqrt3}\bigg(\frac\pi3-\frac\pi6\bigg)\\[4mm] =&\,\boxed{\frac{\pi a}{6\sqrt3}}. \end{align*}

CIE 9231 2022 June Paper 13 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

(c)

解法一

思路

答题过程