曲线 $C$ 的极坐标方程为 $r^2=\tan^{-1}\big(\frac12\theta\big)$，其中 $0\le\theta\le2$。

(a) 画出 $C$ 的草图，并用精确形式写出 $C$ 上的点到极点的最大距离。

(b) 求由 $C$ 与射线 $\theta=2$ 所围区域面积的精确值。

现在考虑 $C$ 上满足 $0\le\theta\le\frac12\pi$ 的部分。

(c) 证明在距离射线 $\theta=\frac12\pi$ 最远的点处，

并验证此方程在 $0.6$ 与 $0.7$ 之间有一个根。

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。

半径随角度增大而增大，但端点角度是 $2$ 弧度，曲线应伸入第二象限，不能画成第一象限的闭曲线。

As $\theta$ increases from $0$ to $2$, $\tan^{-1}(\theta/2)$ increases from $0$ to $\pi/4$. The curve starts at the pole on the positive initial line, curves anticlockwise through the first quadrant, and ends in the second quadrant on the half-line $\theta=2$.

Since $r\ge0$, its greatest distance from the pole is

Near the pole, $r^2\sim\theta/2$, so $x\sim\sqrt{\theta/2}$ and $y\sim\theta\sqrt{\theta/2}$, giving $y\sim2x^3$. Thus the initial tangent is horizontal. For $\theta>0$,

At $\theta=2$, this gradient is

The endpoint tangent therefore slopes upwards to the right. Label the maximum radius at this endpoint and mark the initial line and the half-line $\theta=2$.

极坐标面积要积 $r^2$。反正切的积分用分部积分，剩下的一项可直接识别成对数导数。

The required area is

Integrate by parts, taking $u=\tan^{-1}(\theta/2)$ and $\mathrm{d}v=\mathrm{d}\theta$. Then

Consequently,

到正 $y$ 轴射线的距离是横坐标 $r\cos\theta$；在指定第一象限内不需要再取绝对值。求导为零后代入极坐标方程。

For $0\le\theta\le\pi/2$, the perpendicular distance from the half-line $\theta=\pi/2$ is

It is zero at both endpoints and positive in between, so its maximum is at an interior stationary point. Differentiating $r^2=\tan^{-1}(\theta/2)$ gives

At the maximum,

Multiplying by $r(\theta^2+4)$, which is positive in the interior, gives

Substituting for $r^2$ yields

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

Since $F$ is continuous and changes sign, it has a root between $0.6$ and $0.7$.