题目

Problem

The curve $C$ has polar equation $r=\theta\mathrm{e}^{\frac18\theta}$ , for $0\leqslant\theta\leqslant2\pi$ .

(a) Sketch $C$ .

(2)

(b) Find the area of the region bounded by $C$ and the initial line, giving your answer in the form

(p\pi^2+q\pi+r)\mathrm{e}^{\frac12\pi}+s,

where $p$ , $q$ , $r$ and $s$ are integers to be determined.

(6)

\theta\cos\theta+\left(\tfrac18\theta+1\right)\sin\theta=0

and verify that this equation has a root between $5$ and $5.05$ .

(5)

题目中文翻译

曲线 $C$ 的极坐标方程为 $r=\theta\mathrm{e}^{\frac18\theta}$ ，其中 $0\leqslant\theta\leqslant2\pi$ 。

(a) 画出 $C$ 的草图。

(2)

(b) 求由 $C$ 与极轴围成的区域的面积，答案写成

(p\pi^2+q\pi+r)\mathrm{e}^{\frac12\pi}+s

的形式，其中 $p$ 、 $q$ 、 $r$ 和 $s$ 为待求的整数。

(6)

\theta\cos\theta+\left(\tfrac18\theta+1\right)\sin\theta=0,

并验证该方程在 $5$ 与 $5.05$ 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

展开

半径严格递增，角度从0走到2π，画一圈向外展开的螺线，并标清起终点的初始线交点。

答题过程

展开

Since

\frac{\mathrm{d}r}{\mathrm{d}\theta} =\mathrm e^{\theta/8}(1+\theta/8)>0,

the radius increases strictly as $\theta$ increases. Sketch one anticlockwise turn of an outward spiral, beginning at the pole when $\theta=0$ and ending on the positive initial line at $r=2\pi\mathrm e^{\pi/4}$ when $\theta=2\pi$ .

The curve is tangent to the initial line at the pole and crosses it obliquely at its final point. Mark the pole and initial line.

(b)

解法一

思路

展开

面积积分中要先平方r，所以指数变为θ/4。分部积分两次，并在最后代上下限。

答题过程

展开

The area is

A=\frac12\int_0^{2\pi}\theta^2\mathrm e^{\theta/4}\,\mathrm{d}\theta.

For the first integration by parts,

\begin{align*} &\,\int\theta^2\mathrm e^{\theta/4}\,\mathrm{d}\theta\\[4mm] =&\,4\theta^2\mathrm e^{\theta/4}-8\int\theta\mathrm e^{\theta/4}\,\mathrm{d}\theta. \end{align*}

For the remaining integral,

\begin{align*} \int\theta\mathrm e^{\theta/4}\,\mathrm{d}\theta =&\,4\theta\mathrm e^{\theta/4}-4\int\mathrm e^{\theta/4}\,\mathrm{d}\theta \\[4mm] =&\,4\theta\mathrm e^{\theta/4}-16\mathrm e^{\theta/4}+C. \end{align*}

Therefore

\begin{align*} A=&\,\big[(2\theta^2-16\theta+64)\mathrm e^{\theta/4}\big]_0^{2\pi} \\[4mm] =&\,\boxed{(8\pi^2-32\pi+64)\mathrm e^{\pi/2}-64}. \end{align*}

Thus $p=8$ , $q=-32$ , $r=64$ , $s=-64$ .

(c)

解法一

思路

展开

离初始线的垂直距离是纵坐标的绝对值。最大距离在非零纵坐标的内点出现，因此可令纵坐标导数为零。

答题过程

展开

The signed vertical coordinate is

y=r\sin\theta=\theta\mathrm e^{\theta/8}\sin\theta.

The distance from the initial line is $|y|$ . It is zero at the endpoints and at $\theta=\pi$ , so its positive global maximum occurs at an interior point with $y\ne0$ . At such a point $\mathrm{d}y/\mathrm{d}\theta=0$ .

Differentiating,

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}\theta} =&\,\theta\mathrm e^{\theta/8}\cos\theta\\[4mm] &\,+\mathrm e^{\theta/8}\bigg(1+\frac\theta8\bigg)\sin\theta. \end{align*}

Since the exponential factor is positive, the required condition is

\boxed{\theta\cos\theta+(1+\theta/8)\sin\theta=0},

as required.

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

\begin{align*} f(5)=&\,-0.1399410190\ldots,\\[4mm] f(5.05)=&\,0.1335675315\ldots. \end{align*}

The function is continuous, so the change of sign verifies a root between $5$ and $5.05$ .

CIE 9231 2025 June Paper 11 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