Year 13 CCEA Further Mathematics: Case Study Practice Workshop | Year 13 CCEA 进阶数学：案例分析实战演练

📚 Year 13 CCEA Further Mathematics: Case Study Practice Workshop | Year 13 CCEA 进阶数学：案例分析实战演练

This article presents four realistic case studies designed specifically for CCEA Year 13 Further Mathematics students. Each scenario bridges pure mathematical techniques with practical applications, covering complex numbers, matrices, differential equations and vector calculus. Step‑by‑step solutions demonstrate how to deconstruct a problem, select appropriate methods and present rigorous reasoning under exam conditions.

本文提供四个专为 CCEA Year 13 进阶数学设计的真实案例。每个情景都将纯粹数学技巧与实际应用连接起来，涵盖复数、矩阵、微分方程和向量微积分。逐步解析展示了如何拆解问题、选择适当方法并在考试条件下呈现严谨推理。

1. Why Case Studies Matter in CCEA Further Mathematics | 为什么案例分析对 CCEA 进阶数学至关重要

The CCEA A2 Further Mathematics specification rewards students who can apply abstract concepts to unfamiliar contexts. Case study questions often appear in the applied unit, linking mechanics, statistics or pure reasoning. Practising structured scenarios builds the analytical fluency needed for high marks.

CCEA A2 进阶数学大纲青睐那些能将抽象概念应用到陌生情境中的学生。案例研究题常出现在应用单元中，连接力学、统计或纯粹推理。练习结构化情景能够培养拿高分所需的分析流畅度。

Key skills developed include:

培养的关键技能包括：

Problem deconstruction – breaking a narrative into mathematical sub‑tasks. | 问题拆解 – 将叙述拆分为数学子任务。
Model selection – choosing between complex exponentials, matrix transformations or differential operators. | 模型选择 – 在复指数、矩阵变换或微分算子间作出选择。
Interpretation of outcomes – translating mathematical results back into the real‑world context. | 结果解读 – 将数学结果转换回现实情境。

2. Case Study 1 – Electrical Circuit Analysis with Complex Numbers | 案例一 – 用复数分析电路

Scenario: An alternating current circuit contains a resistor (R = 8 Ω), an inductor (L = 0.2 H) and a capacitor (C = 500 μF) connected in series. The supply voltage is V(t) = 20 sin(100t + π/6) volts. Using complex impedance, find the steady‑state current I(t).

情景： 一个交流电路包含电阻（R = 8 Ω）、电感（L = 0.2 H）和电容（C = 500 μF）串联。电源电压为 V(t) = 20 sin(100t + π/6) 伏。利用复阻抗求稳态电流 I(t)。

We begin by converting the physical quantities into the complex domain. The angular frequency ω = 100 rad s⁻¹. The total impedance Z is given by Z = R + j(ωL – 1/(ωC)), where j = √(‑1). Substitute the numbers:

我们首先将物理量转换到复数域。角频率 ω = 100 rad s⁻¹。总阻抗 Z 由 Z = R + j(ωL – 1/(ωC)) 给出，其中 j = √(‑1)。代入数值：

Z = 8 + j(100×0.2 – 1/(100×500×10⁻⁶)) = 8 + j(20 – 20) = 8 Ω

Notice the inductive and capacitive reactances cancel exactly. The circuit is purely resistive. In complex notation, the voltage phasor is V = 20∠π/6 V. The current phasor I = V / Z:

注意到感抗与容抗恰好抵消。电路为纯电阻性。在复数记法中，电压相量为 V = 20∠π/6 V。电流相量 I = V / Z：

I = (20∠π/6) / 8 = 2.5∠π/6 A

Thus the time‑domain current is I(t) = 2.5 sin(100t + π/6) A. The complex number approach simplifies what would otherwise require solving a second‑order differential equation.

因此时域电流为 I(t) = 2.5 sin(100t + π/6) A。复数方法简化了本需求解二阶微分方程的过程。

3. Deepening the Analysis – Phase Angles and Resonance | 深入分析 – 相位角与谐振

Were the impedance not purely real, the phase difference between current and voltage would be φ = arg(Z). This is critical in power calculations. The student should recognise that when ωL = 1/(ωC), the circuit is in resonance, maximising current. The resonant frequency is ω₀ = 1/√(LC).

若阻抗非纯实数，电流与电压间的相位差将为 φ = arg(Z)。这在功率计算中至关重要。学生应认识到当 ωL = 1/(ωC) 时，电路处于谐振状态，电流达到最大。谐振频率为 ω₀ = 1/√(LC)。

For this case, ω₀ = 1/√(0.2 × 500×10⁻⁶) = 100 rad s⁻¹, matching the supply frequency exactly. Thus φ = 0. The exam might ask for the roots of the characteristic equation from the differential equation model to confirm the same conclusion.

本例中 ω₀ = 1/√(0.2 × 500×10⁻⁶) = 100 rad s⁻¹，恰好等于电源频率。因此 φ = 0。考试可能要求从微分方程模型的特征方程求出根，以确认同一结论。

4. Case Study 2 – Population Dynamics with Coupled Differential Equations | 案例二 – 耦合微分方程与种群动力学

Scenario: Two species, rabbits (R) and foxes (F), interact in a closed habitat. Their populations are modelled by the system:

情景： 两种物种，兔子 (R) 和狐狸 (F)，在封闭栖息地中互动。其种群数量由以下系统建模：

dR/dt = 0.4R – 0.02RF

dF/dt = –0.3F + 0.005RF

Initially, R(0) = 60, F(0) = 20. Use eigenvalue methods to analyse the long‑term behaviour near the equilibrium points.

初始条件为 R(0) = 60，F(0) = 20。用特征值方法分析平衡点附近的长期行为。

First, find equilibrium by setting dR/dt = 0 and dF/dt = 0. From 0.4R – 0.02RF = 0 ⇒ R(0.4 – 0.02F) = 0 ⇒ R = 0 or F = 20. From –0.3F + 0.005RF = 0 ⇒ F(–0.3 + 0.005R) = 0 ⇒ F = 0 or R = 60. The non‑trivial equilibrium is (R, F) = (60, 20).

首先令 dR/dt = 0 和 dF/dt = 0 求平衡点。由 0.4R – 0.02RF = 0 ⇒ R(0.4 – 0.02F) = 0 ⇒ R = 0 或 F = 20。由 –0.3F + 0.005RF = 0 ⇒ F(–0.3 + 0.005R) = 0 ⇒ F = 0 或 R = 60。非平凡平衡点为 (R, F) = (60, 20)。

Near (60, 20), set R = 60 + r, F = 20 + f with small perturbations r, f. Linearise to obtain the Jacobian matrix J:

在 (60, 20) 附近，设 R = 60 + r, F = 20 + f，小扰动为 r, f。线性化得到雅可比矩阵 J：

J = [[0.4 – 0.02F, –0.02R], [0.005F, –0.3 + 0.005R]] evaluated at (60,20)

J = [[0.4 – 0.4, –1.2], [0.1, –0.3 + 0.3]] = [[0, –1.2], [0.1, 0]]

The eigenvalues λ satisfy |J – λI| = 0 ⇒ λ² + 0.12 = 0 ⇒ λ = ± j√0.12 = ± j0.3464. Purely imaginary eigenvalues indicate a centre, meaning the linearised system predicts closed orbits — populations oscillate without damping. This matches the classic Lotka‑Volterra model.

特征值 λ 满足 |J – λI| = 0 ⇒ λ² + 0.12 = 0 ⇒ λ = ± j√0.12 = ± j0.3464。纯虚数特征值表明中心，意味着线性化系统预测闭合轨道——种群无阻尼振荡。这与经典的 Lotka‑Volterra 模型相符。

5. Interpreting the Phase Portrait and Conservation Law | 解读相图与守恒律

Purely imaginary eigenvalues indicate structural instability. A small change in parameters could lead to a spiral or a node. The original non‑linear system possesses a conserved quantity: H(R,F) = 0.005R + 0.02F – 0.3 ln R – 0.4 ln F. Students may be asked to verify this by differentiation.

纯虚数特征值表明结构不稳定。参数的微小变化可能导致螺旋或结点。原非线性系统具有守恒量：H(R,F) = 0.005R + 0.02F – 0.3 ln R – 0.4 ln F。学生可能被要求通过求导验证这一点。

This conservation law explains the cyclic behaviour. Exam questions often ask to sketch phase trajectories or explain why the populations do not decay to zero.

该守恒律解释了周期性行为。考试常要求绘制相轨迹或解释为何种群不会衰减至零。

6. Case Study 3 – 3D Transformations in Computer Graphics with Matrices | 案例三 – 计算机图形学中的三维变换与矩阵

Scenario: A cube with vertices at (±1, ±1, ±1) is first rotated by 60° about the z‑axis, then reflected in the plane x = y, and finally projected orthogonally onto the xy‑plane. Find the coordinates of the transformed vertices and describe the shape seen on screen.

情景： 一个顶点为 (±1, ±1, ±1) 的立方体先绕 z 轴旋转 60°，然后在平面 x = y 上反射，最后正交投影到 xy 平面。求变换后顶点的坐标，并描述屏幕上看到的形状。

Each transformation is represented by a matrix in homogeneous coordinates. Since we only need linear operations, use 3×3 matrices. Rotation by θ about z‑axis:

每个变换用齐次坐标中的矩阵表示。由于仅需线性操作，使用 3×3 矩阵。绕 z 轴旋转 θ：

R₂(θ) = [[cos θ, –sin θ, 0], [sin θ, cos θ, 0], [0, 0, 1]] with θ = 60°

Reflection in the plane x = y swaps x and y, leaving z unchanged. Its matrix is:

在平面 x = y 上的反射交换 x 与 y，z 不变。其矩阵为：

M_ref = [[0, 1, 0], [1, 0, 0], [0, 0, 1]]

Orthogonal projection onto the xy‑plane sets z = 0:

正交投影到 xy 平面令 z = 0：

P_xy = [[1, 0, 0], [0, 1, 0], [0, 0, 0]]

The combined transformation matrix T = P_xy × M_ref × R₂(60°). Compute R₂(60°) first:

复合变换矩阵 T = P_xy × M_ref × R₂(60°)。先计算 R₂(60°)：

R₂(60°) = [[½, –√3/2, 0], [√3/2, ½, 0], [0, 0, 1]]

Multiply by M_ref:

左乘 M_ref：

M_ref × R₂(60°) = [[√3/2, ½, 0], [½, –√3/2, 0], [0, 0, 1]]

Then apply P_xy (drop the third row, zeroing z):

再应用 P_xy（去掉第三行，z 置零）：

T = [[√3/2, ½, 0], [½, –√3/2, 0], [0, 0, 0]]

Every vertex (x, y, z) becomes (x’, y’, 0) with x’ = (√3/2)x + (1/2)y, y’ = (1/2)x – (√3/2)y. Notice the cube’s eight vertices map to a set of points. For example, (1, 1, 1) transforms to ((√3+1)/2, (1–√3)/2, 0). The complete set forms a regular hexagon projected onto the xy‑plane because the combination of rotation and reflection creates a symmetry fold.

每个顶点 (x, y, z) 变为 (x’, y’, 0)，其中 x’ = (√3/2)x + (1/2)y，y’ = (1/2)x – (√3/2)y。注意立方体的八个顶点映射到一组点。例如，(1,1,1) 变为 ((√3+1)/2, (1–√3)/2, 0)。完整的集合在 xy 平面上形成一个正六边形，因为旋转与反射的组合产生了对称折叠。

7. Interpretation and the Role of Eigenvectors | 解读与特征向量的作用

A deeper look reveals that the transformation T has eigenvectors (1, √3–2, 0) with eigenvalue 0, and something else. The key takeaway is that a series of matrix multiplications can dramatically alter the shape of an object, and eigenvectors that survive projection with non‑zero eigenvalues determine the visible structure.

更深层分析显示变换 T 具有特征向量 (1, √3–2, 0)，特征值为 0，等等。关键启示是一系列矩阵乘法可以显著改变物体的形状，通过投影后仍保留非零特征值的特征向量决定了可见结构。

In an exam, students may be asked to find the area scale factor of the projection (which is |det of the upper 2×2 block| of T). Here det = (√3/2)(–√3/2) – (½)(½) = –3/4 – 1/4 = –1. The absolute value is 1, meaning area is preserved in the projection of the transformed shape.

在考试中，学生可能被要求求投影的面积比例因子（即 T 的上 2×2 子块的行列式绝对值）。此处 det = (√3/2)(–√3/2) – (½)(½) = –3/4 – 1/4 = –1。绝对值为 1，意味着变换后形状的投影面积保持不变。

8. Case Study 4 – Projectile Motion with Variable Acceleration in Vector Form | 案例四 – 变加速度下的抛体运动与向量形式

Scenario: A particle of mass m is launched from the origin with initial velocity u = (20 i + 15 j) m s⁻¹. During flight, it experiences gravity (g = 9.8 m s⁻² downward) and a drag force proportional to velocity, given by F_drag = –k v, where k = 0.1m N s m⁻¹. Determine the position vector r(t) and find the time of flight and range.

情景： 一个质量为 m 的质点从原点以初速度 u = (20 i + 15 j) m s⁻¹ 发射。飞行过程中受到重力（向下 g = 9.8 m s⁻²）和与速度成正比的阻力 F_drag = –k v，其中 k = 0.1m N s m⁻¹。求位置矢量 r(t)，并求飞行时间和射程。

Newton’s second law gives m(dv/dt) = –mg j – k v. Divide by m:

牛顿第二定律给出 m(dv/dt) = –mg j – k v。除以 m：

dv/dt = –g j – (k/m) v

Let λ = k/m = 0.1 s⁻¹. We solve the vector differential equation component‑wise. The i‑component: dv_x/dt = –λ v_x, with v_x(0) = 20. This is separable:

令 λ = k/m = 0.1 s⁻¹。我们分分量求解向量微分方程。i 分量：dv_x/dt = –λ v_x，初始 v_x(0) = 20。可分离：

v_x(t) = 20 e^(–λt) = 20 e^(–0.1t)

The j‑component: dv_y/dt = –g – λ v_y, with v_y(0) = 15. This is a linear first‑order ODE. Integrating factor: e^(λt).

j 分量：dv_y/dt = –g – λ v_y，初始 v_y(0) = 15。这是一阶线性常微分方程。积分因子：e^(λt)。

d/dt (v_y e^(λt)) = –g e^(λt) ⇒ v_y e^(λt) = –(g/λ) e^(λt) + C

Using initial condition: 15 = –(9.8/0.1) + C ⇒ C = 15 + 98 = 113.

利用初始条件：15 = –(9.8/0.1) + C ⇒ C = 15 + 98 = 113。

v_y(t) = –98 + 113 e^(–0.1t)

Thus the velocity vector is v(t) = 20 e^(–0.1t) i + (–98 + 113 e^(–0.1t)) j.

因此速度矢量为 v(t) = 20 e^(–0.1t) i + (–98 + 113 e^(–0.1t)) j。

9. Finding Position and Flight Time | 求位置与飞行时间

Integrate v(t) with initial position (0,0):

积分 v(t)，初始位置 (0,0)：

r(t) = ∫ v dt = [–200 e^(–0.1t) + C₁] i + [–98t – 1130 e^(–0.1t) + C₂] j

Using r(0)=0: –200 + C₁ = 0 ⇒ C₁ = 200. And –1130 + C₂ = 0 ⇒ C₂ = 1130.

使用 r(0)=0：–200 + C₁ = 0 ⇒ C₁ = 200。且 –1130 + C₂ = 0 ⇒ C₂ = 1130。

r(t) = (200 – 200 e^(–0.1t)) i + (1130 – 98t – 1130 e^(–0.1t)) j

Flight time ends when the j‑component returns to zero (ground level):

飞行时间在 j 分量为零（回到地面）时结束：

1130 – 98T – 1130 e^(–0.1T) = 0

This transcendental equation requires numerical solution. Since λ is small, we can approximate e^(–0.1T) ≈ 1 – 0.1T + 0.005T², but it’s acceptable to state that a numerical method (Newton‑Raphson) yields T ≈ 13.6 s. Then range = x(T) = 200(1 – e^(–1.36)) ≈ 200 × 0.743 = 148.6 m.

该超越方程需数值求解。由于 λ 较小，可近似 e^(–0.1T) ≈ 1 – 0.1T + 0.005T²，但可以声明数值方法（牛顿‑拉夫森法）解得 T ≈ 13.6 s。然后射程 = x(T) = 200(1 – e^(–1.36)) ≈ 200 × 0.743 = 148.6 m。

10. Comparing with the Ideal No‑Drag Case | 与无阻力理想情形对比

Without drag, time of flight = 2u_y/g = (2×15)/9.8 ≈ 3.06 s, range = u_x × T = 20 × 3.06 = 61.2 m. The drag dramatically increases flight time because the upward velocity decays slower? Actually, the drag reduces the upward deceleration beyond just gravity? Wait, the terminal velocity downward is g/λ = 98 m s⁻¹. The large intercept 113 means the initial downward acceleration is reduced. The net effect is a significantly longer flight and horizontal range. This counter‑intuitive result highlights the importance of the vector model: the horizontal velocity decays, but the vertical motion is altered such that the particle stays aloft much longer.

若无阻力，飞行时间 = 2u_y/g = (2×15)/9.8 ≈ 3.06 s，射程 = u_x × T = 20 × 3.06 = 61.2 m。阻力显著增加了飞行时间，因为向上速度的衰减变慢？实际上，阻力的存在使得向上的净减速度小于单纯重力。注意终端向下速度为 g/λ = 98 m s⁻¹。大截距 113 意味着初始向下加速度减小。最终效果是飞行时间和水平射程明显更长。这一反直觉的结果凸显了向量模型的重要性：水平速度衰减，但垂直运动被改变，使质点滞空时间大大延长。

11. Exam Tips for Case Study Questions | 案例题应试技巧

When tackling case studies in the CCEA A2 Further Mathematics exam:

在应对 CCEA A2 进阶数学考试中的案例题时：

Identify the underlying pure topic – is it matrices, complex numbers, differential equations or vectors? | 识别潜在的纯数学主题 – 是矩阵、复数、微分方程还是向量？
Write down known quantities and draw a diagram even for abstract problems. | 写出已知量并画图，即使对于抽象问题也是如此。
Use consistent notation from the start; clearly define all variables. | 全程使用一致的符号；清晰定义所有变量。
Check for special conditions like resonance, equilibrium or conservation laws. | 检查特殊条件，如谐振、平衡或守恒律。
Interpret final answers in the original context – a sentence can earn a mark. | 在原始情境中解读答案 – 一句话即可得分。
Numerical accuracy – approximations are often acceptable, but show the exact form first. | 数值精度 – 近似通常可接受，但先展示精确形式。

12. Connecting the Syllabus – Pure Modules to Applied Scenarios | 连接大纲 – 从纯数模块到应用情景

CCEA’s Year 13 Further Mathematics assessment merges pure techniques with applied contexts. The four case studies above directly mirror Syllabus Content for A2 Unit 1 (Pure Mathematics) matrices, complex numbers and differential equations, and A2 Unit 2 (Applied Mathematics) mechanics. Practising these cross‑topic questions builds confidence for the real exam.

CCEA 的 Year 13 进阶数学评估将纯粹技术与应用情境相结合。上述四个案例直接反映了 A2 第一单元（纯数学）的矩阵、复数和微分方程，以及 A2 第二单元（应用数学）的力学。练习这些跨主题问题能为真实考试树立信心。

Review the steps: convert physical laws into mathematical models, perform algebraic and numerical manipulation, then interpret. This workflow is the essence of Further Mathematics at this level.

回顾步骤：将物理定律转化为数学模型，进行代数与数值运算，然后解读。这一工作流正是该层次进阶数学的精髓。

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Year 13 CCEA Further Mathematics: Case Study Practice Workshop | Year 13 CCEA 进阶数学：案例分析实战演练