PART I · CORE TECHNIQUES第一部分 · 核心技巧Computational fluency · 28 marks计算熟练度 · 28分
Parametric and Polar Computation参数与极坐标计算
Show all working. When eliminating the parameter, state the domain restrictions on $x$ that arise. When computing $dy/dx$ parametrically, write the formula $dy/dx = (dy/dt)/(dx/dt)$ explicitly before substituting.写出完整解题过程。消去参数时，需说明由此产生的 $x$ 的定义域限制。用参数法计算 $dy/dx$ 时，须先显式写出公式 $dy/dx = (dy/dt)/(dx/dt)$，再代入数值。
Q1MEDIUMCOREparametric curves and eliminating the parameter参数曲线与消参[6 marks]
For each parametric curve, eliminate the parameter to find a Cartesian equation, state any restrictions on $x$ or $y$, and identify the curve.对每条参数曲线，消去参数以求得直角坐标方程，说明 $x$ 或 $y$ 的限制条件，并判断曲线类型。
Q2MEDIUMCOREparametric first derivative, tangent lines, horizontal and vertical tangents参数一阶导数、切线、水平切线与竖直切线[8 marks]
The curve $C$ is given parametrically by $x = t^{3} - 3t$, $y = t^{2} - 1$, $t \in \mathbb{R}$.曲线 $C$ 由参数方程 $x = t^{3} - 3t$，$y = t^{2} - 1$，$t \in \mathbb{R}$ 给出。
(a)Find $\dfrac{dy}{dx}$ in terms of $t$.用 $t$ 表示 $\dfrac{dy}{dx}$。[2]
(b)Find the equation of the tangent line to $C$ at the point where $t = 2$. Give your answer in the form $y = mx + c$.求曲线 $C$ 在 $t = 2$ 处的切线方程，以 $y = mx + c$ 的形式给出。[3]
(c)Find all values of $t$ at which $C$ has a horizontal tangent, and all values at which it has a vertical tangent. For each, state the corresponding Cartesian point.求曲线 $C$ 具有水平切线的所有 $t$ 值，以及具有竖直切线的所有 $t$ 值，并分别写出对应的直角坐标点。[3]
Q3HARDCOREparametric second derivative and concavity参数二阶导数与凹凸性[6 marks]
(a)Compute $\dfrac{dy}{dx}$ and $\dfrac{d^{2}y}{dx^{2}}$ in terms of $t$. Recall that用 $t$ 计算 $\dfrac{dy}{dx}$ 和 $\dfrac{d^{2}y}{dx^{2}}$。注意
$$ \frac{d^{2}y}{dx^{2}} = \frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}. $$
[4]
(b)Determine whether the curve is concave up or concave down at the point where $t = 1$, and give the Cartesian coordinates of that point.判断曲线在 $t = 1$ 处是上凸还是下凸，并给出该点的直角坐标。[2]
Q4MEDIUMCOREpolar coordinates: conversion, curve identification, and conics极坐标：转换、曲线判定与圆锥曲线[8 marks]
Work with the standard conversions $x = r\cos\theta$, $y = r\sin\theta$, $r^{2} = x^{2}+y^{2}$. Recall that the polar equation of a conic with focus at the origin and directrix to the right is $r = \dfrac{ed}{1 + e\cos\theta}$, where $e$ is the eccentricity.使用标准转换公式 $x = r\cos\theta$，$y = r\sin\theta$，$r^{2} = x^{2}+y^{2}$。圆锥曲线以原点为焦点、准线在右侧时，极坐标方程为 $r = \dfrac{ed}{1 + e\cos\theta}$，其中 $e$ 为离心率。
(a)Convert the Cartesian point $(-\sqrt{3},\,1)$ to polar form $(r,\theta)$ with $r > 0$ and $\theta \in [0, 2\pi)$.将直角坐标点 $(-\sqrt{3},\,1)$ 转换为极坐标形式 $(r,\theta)$，要求 $r > 0$，$\theta \in [0, 2\pi)$。[2]
(b)Convert the polar equation $r = 4\sin\theta$ to Cartesian form and identify the curve. State its centre and radius.将极坐标方程 $r = 4\sin\theta$ 转换为直角坐标形式，判断曲线类型，并写出圆心与半径。[2]
(c)Identify the curve type and give one distinguishing feature for each polar equation below.判断下列各极坐标方程的曲线类型，并各给出一个判别特征。
(i) $r = 2 + 2\cos\theta$ [1]
(ii) $r = \sin 3\theta$ [1]
[2]
(d)Write $r = \dfrac{6}{2 + \cos\theta}$ in the standard polar conic form $r = \dfrac{ed}{1 + e\cos\theta}$ by dividing numerator and denominator by $2$. State the eccentricity $e$, the directrix distance $d$, and the type of conic.将 $r = \dfrac{6}{2 + \cos\theta}$ 的分子分母同除以 $2$，化为标准极坐标圆锥形式 $r = \dfrac{ed}{1 + e\cos\theta}$，写出离心率 $e$、准线距离 $d$ 及圆锥曲线的类型。[2]
PART II · DEFINITIONS AND PROOF第二部分 · 定义与证明Rigorous arguments · 26 marks严格论证 · 26分
Derivations and Structural Arguments推导与结构性论证
In each derivation, identify the geometric or analytic object you are partitioning, write down the Riemann-sum approximation, and pass to the limit. Cite results derived in earlier parts when you use them.在每道推导题中，说明所划分的几何或解析对象，写出黎曼和近似，然后取极限。引用前面各部分已推导的结论时须注明来源。
Q5HARDPROOFderiving the polar area element from a circular sector从扇形面积推导极坐标面积微元[8 marks]
Let $r = f(\theta)$ be a continuous non-negative function. The area swept by the radius vector from $\theta = \alpha$ to $\theta = \beta$ is claimed to be $A = \dfrac{1}{2}\displaystyle\int_{\alpha}^{\beta} r^{2}\,d\theta$.设 $r = f(\theta)$ 为连续非负函数。径向量从 $\theta = \alpha$ 扫到 $\theta = \beta$ 所围面积据称为 $A = \dfrac{1}{2}\displaystyle\int_{\alpha}^{\beta} r^{2}\,d\theta$。
(a)State the exact formula for the area of a circular sector of radius $\rho$ subtending angle $\Delta\theta$ at the centre.写出半径为 $\rho$、圆心角为 $\Delta\theta$ 的扇形面积的精确公式。[1]
(b)Partition $[\alpha,\beta]$ into $n$ subintervals of equal width $\Delta\theta = (\beta-\alpha)/n$ with sample points $\theta_i^{*}$. Write down a Riemann sum that approximates the swept area by $n$ circular sectors. Explain why the approximation improves as $n \to \infty$.将 $[\alpha,\beta]$ 等分为 $n$ 个宽度为 $\Delta\theta = (\beta-\alpha)/n$ 的子区间，取样本点 $\theta_i^{*}$。写出用 $n$ 个扇形近似扫掠面积的黎曼和，并解释为何当 $n \to \infty$ 时近似精度提高。[3]
(c)Pass to the limit $n\to\infty$ to obtain the integral formula $A = \dfrac{1}{2}\displaystyle\int_{\alpha}^{\beta} [f(\theta)]^{2}\,d\theta$, identifying the resulting expression as a definite integral by definition.令 $n\to\infty$ 取极限，得到积分公式 $A = \dfrac{1}{2}\displaystyle\int_{\alpha}^{\beta} [f(\theta)]^{2}\,d\theta$，并依定积分的定义说明所得表达式即为定积分。[2]
(d)State one condition on $f$ that the formula requires in order to avoid double-counting the swept area, and give an example of a polar curve where this condition can fail.给出该公式为避免重复计算扫掠面积所需的一个关于 $f$ 的条件，并举一个该条件可能不满足的极坐标曲线的例子。[2]
Q6HARDPROOFderiving the parametric arc length formula推导参数弧长公式[8 marks]
Let $C$ be a smooth parametric curve $x = f(t)$, $y = g(t)$, $a \le t \le b$, where $f'$ and $g'$ are continuous and the curve is traced exactly once.设 $C$ 为光滑参数曲线 $x = f(t)$，$y = g(t)$，$a \le t \le b$，其中 $f'$ 和 $g'$ 连续，且曲线恰好被描绘一次。
(a)Partition $[a,b]$ into $n$ equal subintervals. Write down the length of the straight-line chord connecting the points $(f(t_{i-1}),g(t_{i-1}))$ and $(f(t_i),g(t_i))$.将 $[a,b]$ 等分为 $n$ 个子区间，写出连接点 $(f(t_{i-1}),g(t_{i-1}))$ 与 $(f(t_i),g(t_i))$ 的弦长公式。[2]
(b)Apply the Mean Value Theorem to $f$ and $g$ on $[t_{i-1},t_i]$ to rewrite each chord length in terms of $f'(t_i^{*})$, $g'(t_i^{**})$, and $\Delta t$.对 $f$ 和 $g$ 在 $[t_{i-1},t_i]$ 上应用中值定理，将每段弦长用 $f'(t_i^{*})$、$g'(t_i^{**})$ 和 $\Delta t$ 表示。[2]
(c)Explain, with reference to continuity of $f'$ and $g'$, why the Riemann sum formed from these chord lengths converges to $\displaystyle\int_{a}^{b}\sqrt{[f'(t)]^{2}+[g'(t)]^{2}}\,dt$ as $n\to\infty$.结合 $f'$ 和 $g'$ 的连续性，解释由这些弦长构成的黎曼和在 $n\to\infty$ 时为何收敛到 $\displaystyle\int_{a}^{b}\sqrt{[f'(t)]^{2}+[g'(t)]^{2}}\,dt$。[2]
(d)Show that if $y=F(x)$ is an ordinary function parametrised as $x=t$, $y=F(t)$, the parametric arc length formula reduces to the familiar $\displaystyle\int_{a}^{b}\sqrt{1+[F'(t)]^{2}}\,dt$.证明：若将普通函数 $y=F(x)$ 参数化为 $x=t$，$y=F(t)$，则参数弧长公式化简为熟知的 $\displaystyle\int_{a}^{b}\sqrt{1+[F'(t)]^{2}}\,dt$。[2]
Q7HARDPROOFslopes of polar curves: derivation and application极坐标曲线的斜率：推导与应用[10 marks]
For a polar curve $r = f(\theta)$, write $x = r\cos\theta$ and $y = r\sin\theta$ as functions of $\theta$.对极坐标曲线 $r = f(\theta)$，将 $x = r\cos\theta$ 和 $y = r\sin\theta$ 视为 $\theta$ 的函数。
(a)Using the product rule, compute $\dfrac{dx}{d\theta}$ and $\dfrac{dy}{d\theta}$ in terms of $r$, $\dfrac{dr}{d\theta}$, $\cos\theta$, and $\sin\theta$.利用乘积法则，用 $r$、$\dfrac{dr}{d\theta}$、$\cos\theta$ 和 $\sin\theta$ 计算 $\dfrac{dx}{d\theta}$ 和 $\dfrac{dy}{d\theta}$。[2]
(b)Hence write down the slope formula由此写出斜率公式
$$\frac{dy}{dx} = \frac{\dfrac{dr}{d\theta}\sin\theta + r\cos\theta}{\dfrac{dr}{d\theta}\cos\theta - r\sin\theta}.$$
State the condition under which a polar curve has a horizontal tangent, and the condition for a vertical tangent.写出极坐标曲线具有水平切线的条件，以及具有竖直切线的条件。[2]
(c)Consider the cardioid $r = 1 + \cos\theta$. Find all values of $\theta \in [0, 2\pi)$ at which the cardioid has a horizontal tangent (exclude the cusp at the pole). You may use $\cos(2\theta) = 2\cos^{2}\theta - 1$ to factor the numerator of $dy/dx$.对心形线 $r = 1 + \cos\theta$，求 $\theta \in [0, 2\pi)$ 中所有使心形线具有水平切线的 $\theta$ 值（排除极点处的尖点）。可用 $\cos(2\theta) = 2\cos^{2}\theta - 1$ 对 $dy/dx$ 的分子进行因式分解。[4]
(d)Find the slope of the cardioid $r = 1 + \cos\theta$ at $\theta = \pi/3$.求心形线 $r = 1 + \cos\theta$ 在 $\theta = \pi/3$ 处的斜率。[2]
PART III · APPLICATIONS AND SYNTHESIS第三部分 · 应用与综合Extended problems · 28 marks综合题 · 28分
Arc Length, Surface Area, and Polar Area弧长、旋转体表面积与极坐标面积
Set up every integral with limits and integrand written out fully before evaluating. Use exact values (surds, $\pi$) throughout. Where a trigonometric identity is needed to simplify the integrand, state the identity before applying it.在求值之前，须完整写出每个积分的上下限和被积函数。全程使用精确值（根式、$\pi$）。若需三角恒等式化简被积函数，须在应用前先写出该恒等式。
Q8HARDAPPLIEDparametric arc length requiring a trigonometric identity需要三角恒等式的参数弧长[8 marks]
The parametric curve $x = \cos t + t\sin t$, $y = \sin t - t\cos t$, $0 \le t \le \pi$, traces the involute of a unit circle.参数曲线 $x = \cos t + t\sin t$，$y = \sin t - t\cos t$，$0 \le t \le \pi$，描绘的是单位圆的渐开线。
(a)Show that $\dfrac{dx}{dt} = t\cos t$ and $\dfrac{dy}{dt} = t\sin t$.证明 $\dfrac{dx}{dt} = t\cos t$ 且 $\dfrac{dy}{dt} = t\sin t$。[2]
(b)Hence show that由此证明
$$\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2} = t^{2},$$
identifying the trigonometric identity used.并说明所用的三角恒等式。[2]
(c)Compute the arc length of the curve for $0 \le t \le \pi$.计算曲线在 $0 \le t \le \pi$ 上的弧长。[2]
(d)The surface area generated by revolving this curve about the $x$-axis for $0 \le t \le \pi/2$ is $S = 2\pi\displaystyle\int_{0}^{\pi/2} y\,\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}\,dt$. Set up this integral, simplify the integrand as far as possible using parts (a) and (b), then evaluate it using integration by parts to obtain an exact answer.将该曲线在 $0 \le t \le \pi/2$ 上绕 $x$ 轴旋转所得旋转体的表面积为 $S = 2\pi\displaystyle\int_{0}^{\pi/2} y\,\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}\,dt$。写出该积分，利用第(a)、(b)部分的结果尽量化简被积函数，再用分部积分法求出精确值。[2]
Q9HARDAPPLIEDpolar area: inside one curve and outside another极坐标面积：在一条曲线内且在另一条曲线外[10 marks]
Consider the curves $r = 3\cos\theta$ (a circle) and $r = 1 + \cos\theta$ (a cardioid).考虑曲线 $r = 3\cos\theta$（圆）和 $r = 1 + \cos\theta$（心形线）。
(a)Find all intersection points of the two curves by solving $3\cos\theta = 1 + \cos\theta$ for $\theta \in [0, 2\pi)$. Then verify that the two curves also meet at the pole, and explain why the pole does not appear as a solution to the simultaneous equation.通过在 $\theta \in [0, 2\pi)$ 上求解 $3\cos\theta = 1 + \cos\theta$，找出两曲线的所有交点。然后验证两曲线也在极点相交，并解释为何极点不出现在联立方程的解中。[3]
(b)Determine which curve is the outer boundary and which is the inner boundary on the interval $\theta \in [-\pi/3, \pi/3]$, justifying your answer by evaluating both $r$-values at $\theta = 0$.通过在 $\theta = 0$ 处分别计算两曲线的 $r$ 值，判断在 $\theta \in [-\pi/3, \pi/3]$ 上哪条曲线为外边界，哪条为内边界。[2]
(c)Write down a definite integral, or sum of integrals, for the area of the region that lies inside $r = 3\cos\theta$ and outside $r = 1 + \cos\theta$. Evaluate the integral to obtain an exact answer.写出位于 $r = 3\cos\theta$ 内且位于 $r = 1 + \cos\theta$ 外的区域面积的定积分（或积分之和），并求出精确值。[5]
Q10HARDAPPLIEDpolar arc length and area synthesis极坐标弧长与面积综合[10 marks]
Consider the rose curve $r = 2\cos(2\theta)$.考虑玫瑰线 $r = 2\cos(2\theta)$。
(a)State the total number of petals and give the $\theta$-interval $[\alpha, \beta]$ that traces one petal in the first quadrant.写出花瓣的总数，并给出在第一象限描绘一片花瓣的 $\theta$ 区间 $[\alpha, \beta]$。[2]
(b)Compute the area enclosed by one petal. (Use the identity $\cos^{2}u = \tfrac{1}{2}(1+\cos 2u)$.)计算一片花瓣所围面积。（利用恒等式 $\cos^{2}u = \tfrac{1}{2}(1+\cos 2u)$。）[4]
(c)Set up, but do not evaluate, the integral giving the arc length of one petal. Show that the integrand simplifies to $\sqrt{4\cos^{2}(2\theta)+16\sin^{2}(2\theta)}$, and simplify further to $2\sqrt{1 + 3\sin^{2}(2\theta)}$.写出一片花瓣弧长的积分式（不需求值）。证明被积函数可化简为 $\sqrt{4\cos^{2}(2\theta)+16\sin^{2}(2\theta)}$，并进一步化简为 $2\sqrt{1 + 3\sin^{2}(2\theta)}$。[4]