Case Studies in Pre-U Edexcel Further Mathematics | Pre-U Edexcel 进阶数学：案例分析实战演练

📚 Case Studies in Pre-U Edexcel Further Mathematics | Pre-U Edexcel 进阶数学：案例分析实战演练

In Pre-U Further Mathematics, the ability to apply abstract concepts to concrete practical problems is not only examined but essential for deeper understanding. This article brings together a series of real-world case studies drawn from across the full Further Mathematics syllabus – including pure, mechanics, statistics and decision mathematics – to demonstrate how topics such as complex numbers, hyperbolic functions, differential equations, matrices, vectors, probability models and optimisation are used in authentic contexts. Each case study provides step‑by‑step reasoning, modelling assumptions and, where appropriate, interpretations of the mathematical results.

在 Pre‑U 进阶数学中，将抽象概念应用于具体实际问题的能力不仅是考试重点，更是深入理解学科的关键。本文汇集了一系列源自真实场景的案例，覆盖进阶数学的全部领域——纯数学、力学、统计与决策数学——展示复数、双曲函数、微分方程、矩阵、向量、概率模型以及最优化等主题如何在真实情境中发挥作用。每个案例都提供逐步推理、建模假设，并酌情解读数学结果。

1. Modelling a Hanging Chain with Hyperbolic Functions | 用双曲函数模拟悬链线

A flexible chain of uniform density hanging freely between two supports assumes a shape known as a catenary. By considering the equilibrium of a small segment, the governing differential equation reduces to d²y/dx² = (μg/T₀)√(1 + (dy/dx)²), where μ is the mass per unit length, g is gravitational acceleration and T₀ is the horizontal tension. Solving yields the hyperbolic cosine function.

一根均匀柔软的链条自由悬挂在两个支撑之间，其形状被称为悬链线。通过考虑微段的平衡，控制微分方程可简化为 d²y/dx² = (μg/T₀)√(1 + (dy/dx)²)，其中 μ 是单位长度质量，g 是重力加速度，T₀ 是水平张力。求解得到双曲余弦函数。

The general equation of the catenary is

y = a cosh(x/a) + c, where a = T₀/(μg)

悬链线的一般方程为 y = a cosh(x/a) + c，其中 a = T₀/(μg)。

Given two supports at the same height separated by a distance L and a chain of length S, the parameter a is found by solving S = 2a sinh(L/(2a)). This transcendental equation can be tackled using numerical methods such as the Newton‑Raphson technique, a common tool in the Pre‑U syllabus. Once a is known, the sag can be computed as a[cosh(L/(2a))−1].

给定两个等高支架相距 L，链条总长为 S，参数 a 可通过求解方程 S = 2a sinh(L/(2a)) 得到。这个超越方程可用 Pre‑U 大纲中的牛顿‑拉弗森法等数值方法求解。求得 a 后，垂度可计算为 a[cosh(L/(2a))−1]。

This case study reinforces hyperbolic identities, differentiation of hyperbolic functions and numerical solution of equations. The catenary also arises in architectural arches and overhead power cables, making the mathematics directly relevant to engineering.

本案例巩固了双曲恒等式、双曲函数求导以及方程数值解法。悬链线也出现在拱形建筑和架空电缆中，使得相关数学与工程直接关联。

---

2. Alternating Current Circuits Using Complex Numbers | 用复数分析交流电路

In AC circuit theory, sinusoidal voltages and currents are conveniently represented as phasors using complex numbers. A resistor R, inductor L and capacitor C are assigned impedances Z_R = R, Z_L = jωL, Z_C = 1/(jωC), where ω is the angular frequency. The total impedance of a series RLC circuit is

Z = R + j(ωL − 1/(ωC))

在交流电路理论中，正弦电压和电流用复数相量表示非常方便。电阻 R、电感 L 和电容 C 分别赋予阻抗 Z_R = R、Z_L = jωL、Z_C = 1/(jωC)，其中 ω 为角频率。串联 RLC 电路的总阻抗为 Z = R + j(ωL − 1/(ωC))。

The complex current phasor I is obtained from I = V_s / Z, where V_s is the supply voltage phasor. The magnitude |I| and argument arg(I) give the amplitude and phase shift of the physical current. Resonance occurs when the reactive part vanishes, i.e. ωL = 1/(ωC), yielding the resonant frequency ω₀ = 1/√(LC).

复数电流相量 I 由 I = V_s / Z 得出，其中 V_s 为电源电压相量。幅值 |I| 和辐角 arg(I) 分别给出物理电流的振幅和相位差。当电抗部分消失，即 ωL = 1/(ωC) 时发生谐振，谐振频率为 ω₀ = 1/√(LC)。

Students practise arithmetic with complex numbers in both Cartesian and polar forms, converting between the two. They also apply De Moivre’s theorem when analysing power and harmonics. A typical problem: given R = 50 Ω, L = 0.2 H, C = 50 μF and V_s = 230∠0° V at 50 Hz, find the magnitude of the current and the phase angle.

学生练习复数在笛卡尔形式和极坐标形式之间的运算与转换。在分析功率和谐波时还用到棣莫弗定理。典型题目：已知 R = 50 Ω，L = 0.2 H，C = 50 μF，电源电压 V_s = 230∠0° V，频率 50 Hz，求电流大小及相位角。

The calculation:

ω = 2π×50 = 100π rad s⁻¹, Z = 50 + j(100π×0.2 − 1/(100π×50×10⁻⁶)) ≈ 50 + j(62.832 − 63.662) = 50 − j0.83 Ω

计算过程：ω = 2π×50 = 100π rad s⁻¹，Z = 50 + j(100π×0.2 − 1/(100π×50×10⁻⁶)) ≈ 50 + j(62.832 − 63.662) = 50 − j0.83 Ω。

Hence |Z| ≈ √(50² + 0.83²) ≈ 50.007 Ω, arg(Z) ≈ −0.95°. The current magnitude is 230/50.007 ≈ 4.60 A with phase angle +0.95°, nearly in phase. This illustrates how complex impedance elegantly captures both magnitude and phase behaviour.

因此 |Z| ≈ √(50² + 0.83²) ≈ 50.007 Ω，arg(Z) ≈ −0.95°。电流大小为 230/50.007 ≈ 4.60 A，相位角 +0.95°，几乎同相。这体现了复数阻抗如何优雅地同时描述幅值和相位行为。

---

3. Matrix Encryption and Decryption | 矩阵加密与解密

Hill cipher is a classic application of matrices in cryptography. Letters are mapped to numbers (A=0, B=1, …, Z=25) and grouped into vectors of size n. An invertible n×n matrix K over ℤ₂₆ serves as the encryption key. Encryption: C = K·P (mod 26). Decryption: P = K⁻¹·C (mod 26), requiring K⁻¹ modulo 26 to exist, i.e. det(K) and 26 must be coprime.

希尔密码是矩阵在密码学中的经典应用。字母与数字对应（A=0, B=1, …, Z=25），分组成大小为 n 的向量。ℤ₂₆ 上的 n×n 可逆矩阵 K 作为加密密钥。加密：C = K·P (mod 26)。解密：P = K⁻¹·C (mod 26)，需要 K⁻¹ 模 26 存在，即 det(K) 与 26 互质。

A typical Pre‑U task: using the 2×2 key matrix

K = [ 3 3 ; 2 5 ]

det(K) = 3×5 − 3×2 = 9, which is coprime to 26. Its inverse modulo 26 can be found: 9×3 = 27 ≡ 1 mod 26, so det⁻¹ = 3. Then

K⁻¹ = 3 × [ 5 −3 ; −2 3 ] mod 26 = [ 15 −9 ; −6 9 ] mod 26 = [ 15 17 ; 20 9 ]

典型的 Pre‑U 任务：使用 2×2 密钥矩阵 K = [3 3; 2 5]，det(K)=3×5−3×2=9，与 26 互质。其模 26 逆元可求：9×3=27≡1 mod 26，故 det⁻¹=3。然后 K⁻¹ = 3 × [5 −3; −2 3] mod 26 = [15 −9; −6 9] mod 26 = [15 17; 20 9]。

Encrypting the word “HI” (7,8):

[ 3 3 ; 2 5 ] [7;8] = [21+24; 14+40] = [45;54] mod 26 = [19;2] → “TC”

加密单词 “HI” (7,8)：[3 3; 2 5] [7;8] = [21+24; 14+40] = [45;54] mod 26 = [19;2] → “TC”。

Decryption returns: [15 17; 20 9] [19;2] = [285+34; 380+18] = [319;398] mod 26 = [7;8] → “HI”. This hands‑on activity consolidates matrix multiplication, determinants, modular arithmetic and the concept of invertibility in a different number system.

解密返回：[15 17; 20 9] [19;2] = [285+34; 380+18] = [319;398] mod 26 = [7;8] → “HI”。这一实操活动巩固了矩阵乘法、行列式、模运算以及不同数系下的可逆性概念。

---

4. Relative Motion and Vector Algebra | 相对运动与向量代数

In mechanics, relative motion problems often involve an aeroplane navigating in wind or a boat crossing a river with a current. The velocity of the aircraft relative to the ground v_g is the vector sum of its airspeed v_a and wind speed v_w : v_g = v_a + v_w. Given v_w and desired ground track, vector resolution is used to determine the required heading and the groundspeed.

在力学中，相对运动问题常涉及飞机在风中航行或船只在河流中横渡。飞机相对于地面的速度 v_g 是空速 v_a 与风速 v_w 的矢量和：v_g = v_a + v_w。已知 v_w 和计划的地面航迹，可用向量分解确定所需的航向和地速。

Consider an aircraft that must fly due north at a ground speed of 250 km h⁻¹ when the wind blows from 060° at 50 km h⁻¹. Drawing a vector triangle, the airspeed vector closes the triangle. By the sine and cosine rules, the airspeed magnitude and the heading can be found.

考虑一架飞机必须以 250 km h⁻¹ 的地速向正北飞行，风从 060° 方向以 50 km h⁻¹ 吹来。绘制向量三角形，空速向量闭合三角形。运用正弦定理和余弦定理可求得空速大小和航向。

Southernly wind component: v_w = 50 sin60° towards north? Actually wind from 060° means blowing towards 240°. Setting cardinal directions as vectors, v_g = 250 j, v_w = 50(cos240° i + sin240° j) = 50(−1/2 i − √3/2 j) = −25 i − 43.3 j. Then v_a = v_g − v_w = 250 j − (−25 i − 43.3 j) = 25 i + 293.3 j. Thus |v_a| = √(25² + 293.3²) ≈ 294.4 km h⁻¹, heading θ = arctan(25/293.3) ≈ 4.9° east of north.

南风分量：v_w = 50 sin60° 向北？实际上风来自 060° 表示吹向 240°。设定方向向量，v_g = 250 j，v_w = 50(cos240° i + sin240° j) = 50(−1/2 i − √3/2 j) = −25 i − 43.3 j。则 v_a = v_g − v_w = 250 j − (−25 i − 43.3 j) = 25 i + 293.3 j。因此 |v_a| = √(25² + 293.3²) ≈ 294.4 km h⁻¹，航向 θ = arctan(25/293.3) ≈ 4.9° 东偏北。

Relative velocity vector problems also involve closest approach between two moving objects. By considering the velocity of one relative to the other and setting up position vectors, the minimum separation is found via perpendicular condition or calculus. These exercises sharpen vector manipulation skills and spatial reasoning.

相对速度向量问题还涉及两个运动物体之间的最近距离。通过考虑一个物体相对另一个的速度，建立位置向量，利用正交条件或微积分可求出最小间隔。这些练习能强化向量运算能力和空间推理能力。

---

5. Population Dynamics with Differential Equations | 微分方程与人口动态

The logistic differential equation models population growth limited by resources: dP/dt = rP(1 − P/K), where r is the intrinsic growth rate and K the carrying capacity. This separable first‑order differential equation is solved by partial fractions to give

P(t) = K / (1 + ((K−P₀)/P₀) e⁻ʳᵗ)

逻辑斯谛微分方程模拟受资源限制的人口增长：dP/dt = rP(1 − P/K)，其中 r 为内禀增长率，K 为环境容纳量。这个可分离的一阶微分方程通过部分分式法求解，得到 P(t) = K / (1 + ((K−P₀)/P₀) e⁻ʳᵗ)。

Given data for a yeast population: initial count P₀ = 10 at t=0, K = 600, and observed P = 100 at t = 5 hours. First, find r. Rearranging the solution: r = (1/t) ln( (P(K−P₀)) / (P₀(K−P)) ). Substituting: r = (1/5) ln( (100×590) / (10×500) ) = (1/5) ln(11.8) ≈ 0.494 h⁻¹. The model predicts P(10) = 600 / (1 + (590/10)e⁻⁴.⁹⁴) ≈ 600 / (1 + 59×0.0072) ≈ 600 / 1.4248 ≈ 421.

对于酵母种群的数据：初始数量 P₀ = 10，t=0；K=600；在 t=5 小时观察到 P=100。首先求 r。变换解式：r = (1/t) ln( (P(K−P₀)) / (P₀(K−P)) )。代入：r = (1/5) ln( (100×590) / (10×500) ) = (1/5) ln(11.8) ≈ 0.494 h⁻¹。模型预测 P(10) = 600 / (1 + (590/10)e⁻⁴.⁹⁴) ≈ 600 / (1 + 59×0.0072) ≈ 600 / 1.4248 ≈ 421。

This case study requires students to interpret logistic growth, perform algebraic manipulation, solve for parameters using logarithms, and make predictions. Extensions include analysing the inflection point at P=K/2 where the growth rate is maximal, and connecting to the differential form to sketch slope fields – a topic directly examined in Pre‑U Further Maths.

本案例要求学生理解逻辑斯谛增长，进行代数操作，利用对数求解参数并做出预测。延伸内容包括分析拐点 P=K/2 处增长率最大，并将微分形式与斜率场草图联系起来——这正是 Pre‑U 进阶数学的直接考查内容。

---

6. Planetary Orbits in Polar Coordinates | 极坐标与行星轨道

Kepler’s laws can be derived from Newtonian gravity, and the equation of a planetary orbit is elegantly expressed in polar coordinates (r, θ) with the origin at the sun. The general conic section equation is r = l / (1 + e cosθ), where e is the eccentricity and l the semi‑latus rectum. For planets, 0 < e < 1 yields an ellipse.

开普勒定律可从牛顿引力推导出，行星轨道方程用极坐标 (r, θ)（原点在太阳）优雅表达。一般圆锥曲线方程为 r = l / (1 + e cosθ)，其中 e 为离心率，l 为半正焦弦长。对于行星，0 < e < 1 得到椭圆。

Using the polar form, the area swept out per unit time is constant: dA/dt = (1/2) r² dθ/dt = constant. This is Kepler’s second law. Integration over one full orbit gives the period T. By combining with r = l/(1+ecosθ) and the relationship with the semi‑major axis a = l/(1−e²), one derives T² ∝ a³.

利用极坐标形式，单位时间扫过的面积恒定：dA/dt = (1/2) r² dθ/dt = 常数，即开普勒第二定律。对整个轨道积分可得周期 T。结合 r = l/(1+ecosθ) 以及半长轴 a = l/(1−e²) 的关系，可导出 T² ∝ a³。

A practical task: given the orbit of Earth has e ≈ 0.0167 and semi‑major axis a = 1 AU, compute the closest and farthest distances from the sun (perihelion and aphelion). Using a = l/(1−e²), l = a(1−e²). Perihelion r_min = l/(1+e) = a(1−e) = 1 − 0.0167 = 0.9833 AU. Aphelion r_max = l/(1−e) = a(1+e) = 1.0167 AU. This reinforces the interpretation of polar curves, parameters, and their geometric meaning.

实践任务：已知地球轨道 e ≈ 0.0167，半长轴 a = 1 AU，计算离太阳最近和最远的距离（近日点和远日点）。由 a = l/(1−e²) 得 l = a(1−e²)。近日点 r_min = l/(1+e) = a(1−e) = 1 − 0.0167 = 0.9833 AU。远日点 r_max = l/(1−e) = a(1+e) = 1.0167 AU。这巩固了对极坐标曲线、参数及其几何意义的理解。

---

7. Solving Non‑Linear Equations by Numerical Methods | 数值方法求解非线性方程

Transcendental equations appear frequently in modelling; exact solutions are rare. The Newton‑Raphson method, derived from the tangent line approximation, is xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ). The Pre‑U syllabus covers its derivation, condition for convergence, and application.

超越方程在建模中频繁出现，解析解罕见。牛顿‑拉弗森方法源于切线逼近，迭代公式为 xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ)。Pre‑U 大纲涵盖其推导、收敛条件及应用。

Consider finding the root of f(x) = x³ − 2x − 5 = 0 near x₀ = 2. Derivative f'(x) = 3x² − 2. Iteration:

n	xₙ	f(xₙ)	f'(xₙ)	xₙ₊₁
0	2	−1	10	2.1
1	2.1	0.061	11.23	2.0946
2	2.0946	0.0005	11.16	2.09455

The root converges rapidly to 2.09455, correct to 5 decimal places. Students also practise error estimation and awareness of cases where the method fails (e.g. stationary point near root).

根迅速收敛到 2.09455，精确至 5 位小数。学生也练习误差估计并注意方法失效的情形（例如根附近有驻点）。

Another numerical method is the trapezium rule for approximate integration, used when a function has no elementary antiderivative. For ∫₀¹ e^(−x²) dx with 4 strips, h = 0.25: approximates 0.7468. Comparisons with the actual value (0.7468…) show its accuracy. These techniques underpin many scientific computations.

另一种数值方法是用于近似积分的梯形法则，当函数没有初等原函数时使用。例如 ∫₀¹ e^(−x²) dx，取 4 个条带，h = 0.25，近似值为 0.7468，与实际值（0.7468…）的比较展示其精度。这些技术是众多科学计算的基础。

---

8. Weather Prediction via Markov Chains | 马尔可夫链与天气预报

A simple probabilistic model for weather states (Sunny, Cloudy, Rainy) can be described by a Markov chain with a transition matrix M. Each entry M_ij represents the probability of moving from state i to state j in one day. If the initial state vector is v₀, then after n days the state probability is vₙ = v₀ Mⁿ.

天气状态（晴、多云、雨）的简单概率模型可用马尔可夫链及转移矩阵 M 描述。每个元素 M_ij 表示一天内从状态 i 转移到状态 j 的概率。若初始状态向量为 v₀，则 n 天后的状态概率为 vₙ = v₀ Mⁿ。

Example transition matrix:

State	Sunny	Cloudy	Rainy
Sunny	0.7	0.2	0.1
Cloudy	0.3	0.4	0.3
Rainy	0.2	0.3	0.5

If today is Sunny, v₀ = [1, 0, 0]. The probability of weather in 3 days is v₃ = v₀ M³. Compute M², then M³, multiply by v₀. This yields a vector, e.g. [0.535, 0.282, 0.183], indicating a 53.5% chance of sun three days later. The long‑term stationary distribution can be found by solving π = π M with sum π_i = 1, which involves solving a system of linear equations – a direct application of matrix algebra.

若今天为晴天，v₀ = [1, 0, 0]。3 天后的天气概率为 v₃ = v₀ M³。计算 M²，再得 M³，与 v₀ 相乘得到向量，例如 [0.535, 0.282, 0.183]，表示 3 天后有 53.5% 的概率为晴天。长期稳态分布可通过求解 π = π M 且 ∑π_i = 1 得到，这涉及求解线性方程组——直接应用矩阵代数。

This case vividly demonstrates the power of matrices in modelling sequential random events, linking probability, vector‑matrix multiplication, and eigenvalues/ eigenvectors in the advanced study.

该案例生动展示了矩阵在建模顺序随机事件中的威力，将概率、向量‑矩阵乘法以及高级学习中的特征值/特征向量联系起来。

---

9. Optimisation Using Linear Programming | 线性规划优化

Linear programming (LP) is a decision mathematics technique for maximising or minimising a linear objective function subject to linear inequalities. The feasible region is a convex polygon (or polytope). The optimal solution occurs at a vertex. The simplex method or graphical method (for two variables) is used.

线性规划（LP）是决策数学中的一种技术，用于在满足线性不等式约束下最大化或最小化线性目标函数。可行域为凸多边形（或多胞体），最优解出现在顶点。二维情形下用图解法，高维则用单纯形法。

A manufacturer produces two types of souvenirs, A and B. Each A requires 2 hours of crafting and 1 hour of finishing; each B requires 1 hour crafting and 3 hours finishing. Weekly available hours: 100 for crafting, 90 for finishing. Profit: £40 per A, £50 per B. Maximise profit P = 40x + 50y. Constraints: 2x + y ≤ 100, x + 3y ≤ 90, x, y ≥ 0.

一家制造商生产两种纪念品 A 和 B。每个 A 需 2 小时制作和 1 小时修整；每个 B 需 1 小时制作和 3 小时修整。每周可用工时：制作 100 小时，修整 90 小时。利润：每个 A £40，每个 B £50。最大化利润 P = 40x + 50y。约束条件：2x + y ≤ 100，x + 3y ≤ 90，x, y ≥ 0。

Graphical method: plot constraints, identify vertices (0,0), (50,0), (0,30), and intersection of 2x+y=100 and x+3y=90. Solving simultaneously gives x=42, y=16. Evaluating P at each: (0,0):0; (50,0):2000; (0,30):1500; (42,16): 40×42+50×16 = 1680+800 = 2480. So optimal is 42 A and 16 B per week, profit £2480. This reinforces solving linear systems, graph plotting, and integer considerations.

图解法：绘制约束条件，确定顶点 (0,0)、(50,0)、(0,30)，以及 2x+y=100 与 x+3y=90 的交点。联立解得 x=42, y=16。计算各点 P 值：(0,0):0；(50,0):2000；(0,30):1500；(42,16): 40×42+50×16 = 1680+800 = 2480。因此每周生产 42 个 A 和 16 个 B 最优，利润 £2480。这巩固了线性方程组求解、图形绘制和整数考量。

---

10. Probability Generating Functions in Queueing | 概率生成函数在排队论中的应用

A probability generating function (PGF) for a discrete random variable X is G_X(t) = E(t^X) = Σ P(X=k) t^k. PGFs are especially useful for sums of independent random variables and analysing branching processes. In queueing theory, the number of arrivals in a fixed time may follow a Poisson distribution with PGF G(t) = e^(λ(t−1)).

离散随机变量 X 的概率生成函数（PGF）为 G_X(t) = E(t^X) = Σ P(X=k) t^k。PGF 对独立随机变量之和以及分析分支过程特别有用。在排队论中，固定时间内的到达数可能服从泊松分布，其 PGF 为 G(t) =

Published by TutorHao | Pre-U 进阶数学 Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）