Case Study Practical Exercises in CAIE Further Mathematics | CAIE 进阶数学案例分析实战演练

📚 Case Study Practical Exercises in CAIE Further Mathematics | CAIE 进阶数学案例分析实战演练

In CAIE Further Mathematics, applying theoretical concepts to real-world scenarios is essential for deeper understanding. This article presents ten case studies that demonstrate how topics such as complex numbers, matrices, series, and differential equations are used to solve practical problems. Each case includes step-by-step reasoning and calculations to strengthen your problem-solving skills.

在 CAIE 进阶数学中，将理论概念应用于实际场景对于深入理解至关重要。本文通过十个案例展示复数、矩阵、级数和微分方程等主题如何解决实际问题。每个案例包含逐步推理和计算，以提升您的解题能力。

1. Complex Numbers Case Study: AC Circuit Impedance | 复数案例分析：交流电路阻抗计算

In an AC circuit, a resistor R = 10 Ω and an inductor L = 0.1 H are connected in series. The supply voltage is V = 230 V at angular frequency ω = 100 rad/s. The total impedance is Z = R + iωL = 10 + 10i Ω. Using Ohm’s law, the current I = V / Z.

在一个交流电路中，电阻 R = 10 Ω 和电感 L = 0.1 H 串联连接。电源电压为 V = 230 V，角频率 ω = 100 rad/s。总阻抗 Z = R + iωL = 10 + 10i Ω。根据欧姆定律，电流 I = V / Z。

We compute I = 230 / (10 + 10i) = 23 / (1 + i). Multiply numerator and denominator by the conjugate (1 – i): I = 23(1 – i) / (1² – i²) = 23(1 – i)/2 = 11.5 – 11.5i A.

计算 I = 230 / (10 + 10i) = 23 / (1 + i)。分子和分母同时乘以共轭 (1 – i)：I = 23(1 – i) / (1² – i²) = 23(1 – i)/2 = 11.5 – 11.5i A。

The magnitude |I| = √(11.5² + 11.5²) = 11.5√2 ≈ 16.26 A, and the phase angle φ = arctan(–11.5/11.5) = –45°. So current lags voltage by 45°. In polar form:

幅值 |I| = √(11.5² + 11.5²) = 11.5√2 ≈ 16.26 A，相位角 φ = arctan(–11.5/11.5) = –45°。因此电流滞后电压 45°。用极形式表示：

I = 11.5 – 11.5i = 16.26∠–45° A

This case shows how complex numbers simplify AC circuit analysis, turning differential equations into algebraic ones.

这个案例展示了如何用复数简化交流电路分析，将微分方程转化为代数方程求解。

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2. Matrix Transformation Case Study: Image Processing | 矩阵变换案例分析：图像处理

A unit square has vertices at coordinates (1,1), (2,1), (2,2), (1,2). These points can be represented as column vectors. A shear transformation parallel to the x‑axis is given by the matrix T = [[1, 0.5],[0, 1]]. Find the images of the vertices.

一个单位正方形的顶点坐标为 (1,1), (2,1), (2,2), (1,2)。这些点可用列向量表示。一个平行于 x 轴的剪切变换由矩阵 T = [[1, 0.5],[0, 1]] 给出。求各顶点的像。

Multiply each point by T: (1,1) → (1×1+0.5×1, 0×1+1×1) = (1.5, 1); (2,1) → (2.5, 1); (2,2) → (3, 2); (1,2) → (2, 2). The transformed shape is a parallelogram.

用 T 乘每个点：(1,1) → (1.5, 1); (2,1) → (2.5, 1); (2,2) → (3, 2); (1,2) → (2, 2)。变换后的形状是一个平行四边形。

In image processing, such matrices perform rotations, scaling, and shearing. Composing transformations multiplies their matrices, enabling efficient manipulation of digital graphics.

在图像处理中，这类矩阵可实现旋转、缩放和剪切。复合变换就是矩阵相乘，从而高效地操控数字图形。

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3. Roots of Polynomials Case Study: Electrical Resistance Network | 多项式根与系数案例分析：电阻网络

Three resistors have resistances α, β, γ ohms, which are the roots of the cubic equation x³ – 6x² + 11x – 6 = 0. Find the equivalent resistance R_eq when they are connected in parallel.

三个电阻的阻值 α, β, γ 欧姆是三次方程 x³ – 6x² + 11x – 6 = 0 的根。求它们并联时的等效电阻 R_eq。

For parallel resistors, 1/R_eq = 1/α + 1/β + 1/γ. From Vieta’s formulas: Σα = 6, Σαβ = 11, αβγ = 6. Hence 1/R_eq = (βγ+γα+αβ)/αβγ = 11/6, giving R_eq = 6/11 Ω.

并联电阻满足 1/R_eq = 1/α + 1/β + 1/γ。由韦达定理：Σα = 6，Σαβ = 11，αβγ = 6。因此 1/R_eq = (βγ+γα+αβ)/αβγ = 11/6，得 R_eq = 6/11 Ω。

This example illustrates how symmetric sums of roots are used without solving the cubic explicitly, a powerful technique in further mathematics.

该例题展示了如何不显式求解三次方程而利用根的对称和，这是进阶数学中的有力技巧。

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4. Series Case Study: Loan Repayment Schedule | 级数案例分析：贷款还款计划

A loan of £10,000 is taken at an annual compound interest rate of 5%. The borrower repays £R at the end of each year for 5 years. Find R so that the loan is fully paid off.

一笔 £10,000 贷款年复利率 5%。借款人在每年年末偿还固定金额 £R，共 5 年。求 R 使得贷款刚好还清。

The amount owed after 5 years if no payments were made would be 10000 × 1.05⁵. The payments form a geometric sequence: the first payment earns interest for 4 years, so its future value is R × 1.05⁴, the next R × 1.05³, …, the last payment R. Equating the final values gives 10000 × 1.05⁵ = R(1.05⁴ + 1.05³ + 1.05² + 1.05 + 1).

若未还款，5 年后的欠款为 10000 × 1.05⁵。还款构成几何序列：第一笔还款获得 4 年利息，终值为 R × 1.05⁴，下一笔 R × 1.05³，……，最后一笔为 R。令终值相等：10000 × 1.05⁵ = R(1.05⁴ + 1.05³ + 1.05² + 1.05 + 1)。

The sum of the geometric progression is S = (1.05⁵ – 1)/(1.05 – 1) = (1.05⁵ – 1)/0.05. Compute 1.05⁵ ≈ 1.27628, so S ≈ (0.27628)/0.05 = 5.5256. Therefore R = 10000 × 1.27628 / 5.5256 ≈ £2310.60.

几何级数和为 S = (1.05⁵ – 1)/(1.05 – 1) = (1.05⁵ – 1)/0.05。计算得 1.05⁵ ≈ 1.27628，故 S ≈ 0.27628/0.05 = 5.5256。因此 R = 10000 × 1.27628 / 5.5256 ≈ £2310.60。

Series are indispensable in financial mathematics; this method generalises to any number of periods or interest rates.

级数在金融数学中不可或缺；此方法可推广至任意期数和利率。

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5. Proof by Induction Case Study: De Moivre’s Theorem | 数学归纳法案例分析：棣莫弗定理

De Moivre’s theorem states that (cos θ + i sin θ)^n = cos nθ + i sin nθ for all integers n. Prove it for positive integers by induction.

棣莫弗定理指出对所有整数 n 有 (cos θ + i sin θ)^n = cos nθ + i sin nθ。用归纳法证明正整数情形。

Base case n = 1: LHS = cos θ + i sin θ, RHS = cos(1θ) + i sin(1θ). True.

归纳奠基 n = 1：左边 = cos θ + i sin θ，右边 = cos θ + i sin θ。成立。

Inductive step: assume true for n = k, i.e. (cos θ + i sin θ)^k = cos kθ + i sin kθ. For n = k+1: (cos θ + i sin θ)^{k+1} = (cos θ + i sin θ)^k (cos θ + i sin θ) = (cos kθ + i sin kθ)(cos θ + i sin θ). Multiply using complex multiplication: = cos kθ cos θ – sin kθ sin θ + i(sin kθ cos θ + cos kθ sin θ) = cos(kθ+θ) + i sin(kθ+θ) = cos((k+1)θ) + i sin((k+1)θ). Hence true for n = k+1.

归纳递推：假设 n = k 成立，即 (cos θ + i sin θ)^k = cos kθ + i sin kθ。对于 n = k+1，有 (cos θ + i sin θ)^{k+1} = (cos θ + i sin θ)^k (cos θ + i sin θ) = (cos kθ + i sin kθ)(cos θ + i sin θ)。复数乘法展开：= cos kθ cos θ – sin kθ sin θ + i(sin kθ cos θ + cos kθ sin θ) = cos(kθ+θ) + i sin(kθ+θ) = cos((k+1)θ) + i sin((k+1)θ)。所以 n = k+1 成立。

By mathematical induction, the theorem holds for all positive integers n. This proof underpins many results in further mathematics.

由数学归纳法，定理对所有正整数 n 成立。这个证明是进阶数学中许多结论的基础。

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6. Calculus Case Study: Draining Tank Problem | 微积分案例分析：水槽排水问题

A cylindrical water tank has constant cross-sectional area A = 2 m². Water drains through a hole at the bottom at a rate proportional to the square root of the water depth h: dV/dt = –k√h, with k = 0.1 m^{2.5}/s. Initially h₀ = 3 m. Find the time to empty the tank.

一个圆柱形水箱截面积 A = 2 m²。水从底部小孔流出，速率正比于水深 h 的平方根：dV/dt = –k√h，其中 k = 0.1 m^{2.5}/s。初始水深 h₀ = 3 m。求水排空所需时间。

Since volume V = A h, we have dV/dt = A dh/dt = –k√h. Thus dh/dt = –(k/A)√h. Separate variables: dh/√h = –(k/A) dt. Integrate both sides: ∫ h^{–½} dh = –(k/A) ∫ dt → 2√h = –(k/A)t + C.

由于体积 V = A h，有 dV/dt = A dh/dt = –k√h。因此 dh/dt = –(k/A)√h。分离变量：dh/√h = –(k/A) dt。积分两边：∫ h^{–½} dh = –(k/A) ∫ dt → 2√h = –(k/A)t + C。

Apply initial condition h = 3 at t = 0: 2√3 = C. So 2√h = –(k/A)t + 2√3. When the tank is empty, h = 0: 0 = –(k/A)t + 2√3 → t = (2√3 A)/k = (2√3 × 2)/0.1 = 40√3 ≈ 69.28 seconds.

应用初始条件 t = 0 时 h = 3：2√3 = C。所以 2√h = –(k/A)t + 2√3。水排空时 h = 0：0 = –(k/A)t + 2√3 → t = (2√3 A)/k = (2√3 × 2)/0.1 = 40√3 ≈ 69.28 秒。

This separable differential equation models many natural outflow processes and reinforces integration techniques.

这个可分离的微分方程模拟了许多自然流出过程，并强化了积分技巧。

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7. Vectors Case Study: Ship Interception | 向量案例分析：船舶相遇问题

At noon, ship A is at position vector r_A = (2i + 3j) km and moves with constant velocity v_A = (4i + j) km/h. Ship B is at r_B = (5i + j) km and moves with v_B = (–i + 2j) km/h. Determine whether the ships collide, and if so, when and where.

正午时，船 A 的位置向量为 r_A = (2i + 3j) km，以常速度 v_A = (4i + j) km/h 运动。船 B 的位置为 r_B = (5i + j) km，速度 v_B = (–i + 2j) km/h。判断两船是否会相撞，若是，求出碰撞的时间和位置。

Position after t hours: r_A(t) = (2+4t)i + (3+t)j, r_B(t) = (5 – t)i + (1+2t)j. For collision, the position vectors must be equal: 2+4t = 5 – t → 5t = 3 → t = 0.6 h; 3+t = 1+2t → t = 2 h. The times do not match, so no collision. Instead, find the time of minimum distance.

t 小时后的位置：r_A(t) = (2+4t)i + (3+t)j，r_B(t) = (5 – t)i + (1+2t)j。若要碰撞，位置向量必须相等：2+4t = 5 – t → 5t = 3 → t = 0.6 h；3+t = 1+2t → t = 2 h。时间不一致，所以不会相撞。改求最小距离的时刻。

Relative position vector s = r_B – r_A = (3 –5t)i + (–2 + t)j. Square of distance D² = (3–5t)² + (–2+t)². Expand: 25t² –30t +9 + t² –4t +4 = 26t² –34t +13. Minimise by d(D²)/dt = 0: 52t –34 = 0 → t = 34/52 = 17/26 ≈ 0.654 h

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