题目

Problem

The curve $C$ has polar equation $r^2=\dfrac{1}{\theta^2+1}$ , for $0\le\theta\le\pi$ .

(a) Sketch $C$ and state the polar coordinates of the point of $C$ furthest from the pole.

(3)

(b) Find the area of the region enclosed by $C$ , the initial line, and the half-line $\theta=\pi$ .

(4)

\bigg(\theta+\frac1\theta\bigg)\cot\theta-1=0

and verify that this equation has a root between $1.1$ and $1.2$ .

(5)

题目中文翻译

曲线 $C$ 的极坐标方程为 $r^2=\dfrac{1}{\theta^2+1}$ ，其中 $0\le\theta\le\pi$ 。

(a) 画出 $C$ 的示意图，并写出 $C$ 上距极点最远的点的极坐标。

(3)

(b) 求由 $C$ 、极轴及射线 $\theta=\pi$ 围成的区域的面积。

(4)

\bigg(\theta+\frac1\theta\bigg)\cot\theta-1=0

并验证此方程在 $1.1$ 与 $1.2$ 之间有一个根。

(5)

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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取非负极径，随θ增加半径减小，但到π时并未到达极点。两端点的切线方向也要区分。

答题过程

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Here $r=(1+\theta^2)^{-1/2}$ , with

\frac{\mathrm{d}r}{\mathrm{d}\theta} =-\frac\theta{(1+\theta^2)^{3/2}}\leqslant0.

Sketch an upper-half-plane arc running anticlockwise from $(1,0)$ at $\theta=0$ to $(-1/\sqrt{1+\pi^2},0)$ at $\theta=\pi$ . The radius decreases strictly after the starting point and never reaches zero.

The tangent at $(1,0)$ is vertical. At the other endpoint its gradient is $-(1+\pi^2)/\pi$ , so it slopes downwards to the right.

The furthest point from the pole has polar coordinates $\boxed{(1,0)}$ .

(b)

解法一

思路

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两个边界射线给积分上下限0与π，代入r平方后是标准反正切积分。

答题过程

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\begin{align*} A=&\,\frac12\int_0^\pi\frac1{1+\theta^2}\,\mathrm{d}\theta\\[4mm] =&\,\frac12\big[\tan^{-1}\theta\big]_0^\pi\\[4mm] =&\,\boxed{\frac12\tan^{-1}\pi}. \end{align*}

(c)

解法一

思路

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最大垂距点在上半平面的内部，对纵坐标求导。整理时只能除内部非零的θ与正弦，最后检查连续函数变号。

答题过程

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The distance from the initial line is

y=r\sin\theta=\frac{\sin\theta}{\sqrt{1+\theta^2}}.

It is zero at both endpoints and positive inside. Thus its maximum occurs at an interior point, where

\begin{align*} 0=&\,\frac{\mathrm{d}y}{\mathrm{d}\theta}\\[4mm] =&\,\frac{\cos\theta}{\sqrt{1+\theta^2}}\\[4mm] &\,-\frac{\theta\sin\theta}{(1+\theta^2)^{3/2}}. \end{align*}

Multiplying by $(1+\theta^2)^{3/2}$ gives

(1+\theta^2)\cos\theta-\theta\sin\theta=0.

Since $0<\theta<\pi$ , divide by $\theta\sin\theta>0$ to obtain

\boxed{\bigg(\theta+\frac1\theta\bigg)\cot\theta-1=0}.

Let $F(\theta)$ be the left-hand side. In radians,

\begin{align*} F(1.1)=&\,0.0225632\ldots>0,\\[4mm] F(1.2)=&\,-0.2094815\ldots<0. \end{align*}

Since $F$ is continuous on $[1.1,1.2]$ , it has a root in $\boxed{(1.1,1.2)}$ .

CIE 9231 2023 June Paper 12 Q5

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