(a) 证明：笛卡尔方程为

的曲线，其极坐标方程为 $r^2=3\sin2\theta$。

曲线 $C$ 的极坐标方程为 $r^2=3\sin2\theta$，其中 $0\leqslant\theta\leqslant\frac12\pi$。

(b) 画出 $C$ 的草图，并写出 $C$ 上的点到极点的最大距离。

(c) 求 $C$ 所围区域的面积。

(d) 求 $C$ 上的点到极轴的最大距离。

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。

代入直角坐标与极坐标的关系，并用正弦倍角公式。除以r平方时要另外说明极点没有丢失。

Using $x=r\cos\theta$, $y=r\sin\theta$ and $x^2+y^2=r^2$, we obtain

For $r\ne0$, division by $r^2$ gives

The pole $r=0$ belongs to the Cartesian curve and is also represented by this polar equation, for example at $\theta=0$. Thus it is retained.

给定角区间只画第一象限的一个闭环，不画整个双纽线。正弦取最大值时极径最大，环关于四十五度射线对称。

At $\theta=0$ and $\theta=\frac\pi2$, $r=0$. For angles between these values, $r>0$.

Sketch a single loop in the first quadrant, starting and ending at the pole, symmetric about $\theta=\frac\pi4$. The tangents at the pole are the two coordinate axes.

Since $\sin2\theta\leqslant1$, the maximum radius occurs at $\theta=\frac\pi4$ and is $\boxed{\sqrt3}$.

一整个闭环由0至二分之π扫描，直接用极坐标面积积分。

整个环都在第一象限，到极轴的距离就是纵坐标。对纵坐标求导后用三倍角恒等式整理，最后与端点的零距离比较。

The distance from the initial line is

For $0<\theta<\frac\pi2$,

At an interior stationary point, multiplication by the positive factor $\sqrt{\sin2\theta}/\sqrt3$ gives

so $\sin3\theta=0$. In $0<\theta<\frac\pi2$, the only solution is $\theta=\frac\pi3$.

The distance is zero at both endpoints and positive inside, so this unique interior stationary point gives the maximum. Its distance is