曲线 $C$ 的极坐标方程为 $r^2=\mathrm{e}^{\sin\theta}\cos\theta$，其中 $-\frac12\pi\leqslant\theta\leqslant\frac12\pi$。

(a) 求 $C$ 上距离极点最远的点的极坐标，答案精确到小数点后 $3$ 位。

(b) 求 $C$ 上距离射线 $\theta=\frac12\pi$ 最远的点的极坐标，答案精确到小数点后 $3$ 位。

(c) 画出 $C$ 的草图。

(d) 求 $C$ 所围区域的面积，答案用精确形式表示。

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。

最大半径也对应最大半径平方。对r平方求导，化成sinθ的二次方程，舍掉范围外的根。

Let $R=r^2=\mathrm e^{\sin\theta}\cos\theta$. Then

At a maximum,

Thus $\sin\theta=(-1\pm\sqrt5)/2$. Rejecting the value less than $-1$,

The derivative is positive before this angle and negative afterwards, while $r=0$ at both endpoints. Substitution gives $r=1.207695548\ldots$.

Hence the polar coordinates, to $3$ decimal places, are

到竖直方向的垂直距离是横坐标。对x平方求导可避免根号；最后仍要回求极坐标中的r，而不是把x当作r。

The horizontal coordinate is $x=r\cos\theta\ge0$. Its square is

Differentiating,

At the interior maximum,

Solving gives

The other quadratic root is outside $[-1,1]$. Hence $\theta=0.3076036495\ldots$, and substitution into the polar equation gives $r=1.135814582\ldots$.

The derivative changes from positive to negative at this point. For $\theta<0$, distance from the upward half-line is at most $r\le1$, whereas this point in the upper half-plane has $x=1.082501560\ldots>1$. Thus it also gives the greatest distance from the stated half-line.

The polar coordinates to $3$ decimal places are

曲线在右半平面，从极点出发再回到极点，上半部比下半部大。标初始线交点及极点处竖直切向。

At $\theta=\pm\pi/2$, $r=0$, while at $\theta=0$ the point is $(1,0)$. Since $r\ge0$ and $\cos\theta\ge0$, the curve lies entirely in the right half-plane.

Sketch a single closed loop through the pole and $(1,0)$, with more of the loop above the initial line than below it. Both approaches to the pole have vertical tangent direction. At $(1,0)$ the tangent has positive gradient; in fact $\mathrm{d}y/\mathrm{d}x=2$ there. The rightmost point corresponds to (b), and the point with greatest radius to (a).

极坐标面积积分恰好是e的sinθ次方乘cosθ，直接识别复合函数的导数。

The enclosed area is