📚 Pre-U CIE Further Mathematics: Case Study Practical Exercises | Pre-U CIE 进阶数学：案例分析实战演练
This article presents a series of engaging case studies designed to consolidate key concepts in the CIE Pre-U Further Mathematics syllabus. Each case study links abstract theory to real-world contexts, encouraging you to apply pure mathematics, mechanics, probability, and numerical methods in practical problem-solving scenarios.
本文通过一系列精心设计的案例分析，巩固 CIE Pre-U 进阶数学大纲中的关键概念。每个案例都将抽象理论与现实情境联系起来，引导你在纯数学、力学、概率和数值方法的实际解题中加以运用。
1. Matrix Transformations in Computer Graphics | 计算机图形中的矩阵变换
A graphic designer wants to rotate a 2D shape by 60° anticlockwise about the origin and then reflect it across the line y = x. The original shape is defined by the vertices (1,0), (0,3) and (-2,1). Determine the single transformation matrix that combines both operations and find the new coordinates of the vertices.
一名图形设计师希望将二维图形绕原点逆时针旋转 60°，再关于直线 y = x 进行反射。原图形的顶点为 (1,0)、(0,3) 和 (-2,1)。求合并这两个操作的单一变换矩阵，并计算顶点的新坐标。
The rotation matrix for 60° is R = [ [cos60°, -sin60°], [sin60°, cos60°] ] = [ [1/2, -√3/2], [√3/2, 1/2] ]. The reflection matrix across y = x is S = [ [0,1], [1,0] ]. The combined matrix is M = SR = [ [0,1], [1,0] ] × [ [1/2, -√3/2], [√3/2, 1/2] ] = [ [√3/2, 1/2], [1/2, -√3/2] ]. Apply M to each vertex: (1,0) → (√3/2, 1/2); (0,3) → (3/2, -3√3/2); (-2,1) → (-√3 + 1/2, -1 – √3/2).
60° 的旋转矩阵 R = [[½, -√3/2], [√3/2, ½]]，关于 y=x 的反射矩阵 S = [[0,1], [1,0]]。合成矩阵 M = SR = [[√3/2, ½], [½, -√3/2]]。将其作用于各顶点，得到新坐标如上。
2. Complex Impedance in AC Circuits | 交流电路中的复阻抗
An AC circuit consists of a resistor R = 10 Ω, an inductor with reactance XL = 5 Ω and a capacitor with reactance XC = 12 Ω connected in series. The supply voltage is V = 230∠30° V. Express the total impedance Z as a complex number in Cartesian form, and calculate the current phasor I.
一个交流电路由电阻 R = 10 Ω、感抗 XL = 5 Ω 和容抗 XC = 12 Ω 串联组成，电源电压为 V = 230∠30° V。用笛卡尔形式表示总阻抗 Z，并计算电流相量 I。
Impedance Z = R + j(XL – XC) = 10 + j(5 – 12) = 10 – j7 Ω. In polar form, |Z| = √(10² + (-7)²) = √149 ≈ 12.21 Ω, arg(Z) = arctan(-7/10) ≈ -35.0°. Current I = V / Z = (230∠30°) / (12.21∠-35.0°) ≈ 18.84∠65.0° A. Cartesian form: I ≈ 18.84(cos65° + j sin65°) ≈ 7.96 + j17.08 A.
阻抗 Z = 10 + j(5 – 12) = 10 – j7 Ω。模 |Z| ≈ 12.21 Ω，辐角 ≈ -35.0°。电流 I = 230∠30° / 12.21∠-35.0° ≈ 18.84∠65.0° A，笛卡尔形式约为 7.96 + j17.08 A。
3. Hyperbolic Functions and the Catenary | 双曲函数与悬链线
A uniform flexible cable hangs between two poles of equal height, separated by 20 m. The cable dips 4 m below the tops of the poles. Using the catenary equation y = c cosh(x/c), find the parameter c and determine the length of the cable.
一根均匀柔软的缆绳悬挂在两根等高的杆子之间，杆距 20 m。缆绳比杆顶低 4 m。使用悬链线方程 y = c cosh(x/c)，求参数 c 并计算缆绳长度。
Set coordinate origin at the lowest point, so the poles are at x = ±10 m. The sag gives y(10) – c = 4, so c cosh(10/c) – c = 4 ⇒ cosh(10/c) – 1 = 4/c. Solve numerically: try c = 10, LHS = cosh(1)-1 ≈ 0.543, RHS=0.4; c = 12, LHS=cosh(0.833)-1≈0.361, RHS=0.333; c ≈ 12.5 gives ≈0.337 vs 0.32; refined solution c ≈ 13.2. More precisely by iterative method, c ≈ 13.3. Length s = 2 ∫₀¹⁰ √(1+sinh²(x/c)) dx = 2c sinh(10/c) ≈ 2×13.3×sinh(0.752) ≈ 22.4 m.
取最低点为原点，杆位于 x = ±10 m。下垂量 4 m 给出方程 cosh(10/c) – 1 = 4/c。通过数值求解得到 c ≈ 13.3。缆绳长度为 s = 2c sinh(10/c) ≈ 22.4 m。
4. Polar Curve Length and Surface Area | 极坐标曲线长度与旋转表面积
A curve is defined in polar coordinates by r = 2 + sin θ, for 0 ≤ θ ≤ π. Calculate the exact length of the curve and the surface area generated when the curve is rotated about the initial line.
一条曲线在极坐标下定义为 r = 2 + sin θ，0 ≤ θ ≤ π。计算该曲线的精确长度，以及曲线绕极轴旋转所形成的表面积。
Arc length L = ∫₀π √(r² + (dr/dθ)²) dθ. Here dr/dθ = cos θ, so r²+(dr/dθ)² = (2+sin θ)²+cos²θ = 4 + 4 sin θ + sin²θ + cos²θ = 5 + 4 sin θ. Thus L = ∫₀π √(5 + 4 sin θ) dθ. This requires elliptic functions; no simple elementary form, so the exact length is expressed as an elliptic integral: L = 2√5 E(2/√5) – …, but a numerical approximation using Simpson’s rule gives L ≈ 7.64. Surface area about initial line (x-axis): S = 2π ∫₀π r sin θ √(r²+(dr/dθ)²) dθ = 2π ∫₀π (2+sin θ) sin θ √(5+4 sin θ) dθ, approximated as 40.3 square units.
弧长 L = ∫₀π √(5 + 4 sin θ) dθ，该积分没有初等形式，可用辛普森法则近似为 7.64。绕极轴旋转的表面积 S = 2π ∫₀π (2+sin θ) sin θ √(5+4 sin θ) dθ ≈ 40.3 平方单位。精确结果需借助椭圆积分，可证明其为 2√5 E(2/√5) 等。
5. Differential Equation: Logistic Population Growth | 微分方程：逻辑斯谛人口增长
The population of a small island follows the logistic model dP/dt = 0.04P(1 – P/500), where P is the number of individuals and t is in years. Initially P(0)=100. Find the explicit solution for P(t) and determine the time taken for the population to reach 400.
一个小岛的人口遵循逻辑斯谛模型 dP/dt = 0.04P(1 – P/500)，其中 P 为个体数量，t 以年为单位。初始条件 P(0)=100。求 P(t) 的显式解，并计算人口达到 400 所需的时间。
Separate variables: dP / [P(1 – P/500)] = 0.04 dt ⇒ ∫ [1/P + 1/(500-P)] dP = 0.04t + C. Integrating: ln|P| – ln|500-P| = 0.04t + C ⇒ ln(P/(500-P)) = 0.04t + C. P(0)=100 gives ln(100/400)=ln(1/4) = -ln 4, so C = -ln 4. Thus P/(500-P) = exp(0.04t – ln 4) = (1/4)e0.04t. Solving for P: P = 500 / (1 + 4e-0.04t). Set P=400: 400 = 500/(1+4e-0.04t) ⇒ 1+4e-0.04t = 1.25 ⇒ e-0.04t = 0.0625 ⇒ -0.04t = ln(0.0625) = -ln 16 ⇒ t = ln(16)/0.04 ≈ 69.3 years.
分离变量并积分，利用部分分式求得通解，代入初始条件得特解 P(t) = 500/(1+4e⁻⁰·⁰⁴ᵗ)。令 P=400，解得 t = ln(16)/0.04 ≈ 69.3 年。
6. Vector Forces and Equilibrium in 3D | 三维空间中的向量力与平衡
A particle is in equilibrium under three forces: F₁ = 3i – 2j + 4k N, F₂ = -i + 5j – 2k N, and an unknown force F₃. Find F₃ and determine the direction cosines of the resultant of F₁ and F₂.
一个质点在三个力作用下平衡：F₁ = 3i – 2j + 4k N，F₂ = -i + 5j – 2k N，以及未知力 F₃。求 F₃，并计算 F₁ 和 F₂ 合力的方向余弦。
Equilibrium ⇒ F₁ + F₂ + F₃ = 0 ⇒ F₃ = -(F₁ + F₂). Sum F₁+F₂ = (3-1)i + (-2+5)j + (4-2)k = 2i + 3j + 2k. Hence F₃ = -2i – 3j – 2k N. Resultant of F₁ and F₂ is R = 2i+3j+2k, magnitude |R|=√(4+9+4)=√17. Direction cosines: cos α = 2/√17, cos β = 3/√17, cos γ = 2/√17.
由平衡条件得 F₃ = -(F₁+F₂) = -2i – 3j – 2k N。F₁ 和 F₂ 的合力 R = 2i+3j+2k，模为 √17，方向余弦为 cos α=2/√17, cos β=3/√17, cos γ=2/√17。
7. Poisson Process: Call Centre Arrivals | 泊松过程：呼叫中心来电
Calls arrive at a helpline according to a Poisson process with an average rate of 3 calls per minute. Find the probability that in a randomly chosen 2-minute interval, there are exactly 5 calls. Also calculate the probability that more than 2 minutes pass with no calls.
某服务热线来电服从泊松过程，平均每分钟 3 通。求在随机选取的 2 分钟间隔内恰好接到 5 通电话的概率，并计算连续超过 2 分钟没有来电的概率。
For 2 minutes, mean λ = 3×2 = 6. P(X=5) = e⁻⁶·6⁵/5! ≈ 0.1606. Inter-arrival time T ~ Exp(3). P(T > 2) = e⁻³ˣ² = e⁻⁶ ≈ 0.00248. Alternatively, in a Poisson process, the number of calls in 2 minutes with no calls gives directly e⁻⁶. So probability of no calls in 2 minutes is e⁻⁶, and the event “more than 2 minutes with no calls” is equivalent to waiting time > 2 min, thus probability is e⁻⁶ ≈ 0.00248.
2 分钟内的平均来电数 λ = 6。P(X=5) = e⁻⁶·6⁵/5! ≈ 0.1606。相邻来电间隔时间 T 服从指数分布 Exp(3)，故 P(T>2) = e⁻⁶ ≈ 0.00248，即超过 2 分钟无来电的概率极小。
8. Hypothesis Testing: Binomial Proportion | 假设检验：二项比例
A manufacturer claims that at most 5% of its products are defective. A random sample of 200 items contains 16 defectives. Test the claim at the 5% significance level, using a suitable binomial model and a normal approximation where appropriate.
某制造商声称其产品次品率不超过 5%。随机抽取 200 件产品，发现 16 件次品。请用合适的二项模型并在适当时使用正态近似，以 5% 显著性水平检验该声明。
Let p be true proportion. H₀: p = 0.05, H₁: p > 0.05. Under H₀, X ~ B(200, 0.05) with mean μ = 10, variance σ² = 200×0.05×0.95 = 9.5. Using normal approximation with continuity correction: test statistic z = (15.5 – 10)/√9.5 ≈ 1.787. Critical value for one-tailed 5% is 1.645. Since 1.787 > 1.645, reject H₀. There is sufficient evidence that the defect rate exceeds 5%. Alternatively, exact binomial P(X≥16) = 1 – P(X≤15) ≈ 0.037 (using tables), which is <0.05, confirming rejection.
设真实次品率为 p。原假设 H₀: p = 0.05，备择假设 H₁: p > 0.05。在 H₀ 下，X~B(200,0.05)，均值 10，方差 9.5。用正态近似并连续性校正得 z ≈ (15.5-10)/√9.5 = 1.787 > 1.645，拒绝 H₀。精确二项 P(X≥16) ≈ 0.037 < 0.05，同样拒绝。有充分证据表明次品率超过 5%。
9. Numerical Methods: Newton-Raphson for Transcendental Equation | 数值方法：牛顿-拉夫逊法解超越方程
Use the Newton-Raphson method to find the real root of the equation x³ – cos x = 0, starting with x₀ = 1. Perform three iterations and give your final approximation to four decimal places.
用牛顿-拉夫逊法求方程 x³ – cos x = 0 的实根，初始值 x₀ = 1。进行三次迭代，并将最终近似值保留四位小数。
Define f(x) = x³ – cos x, then f'(x) = 3x² + sin x. Iteration formula: xn+1 = xn – (xn³ – cos xn) / (3xn² + sin xn). Start x₀=1: f(1)=1-0.5403=0.4597; f'(1)=3+0.8415=3.8415; x₁=1-0.4597/3.8415≈0.8803. x₁≈0.8803: f≈0.6825-0.6372=0.0453; f’≈3(0.7750)+0.7701=3.0951; x₂=0.8803-0.0453/3.0951≈0.8657. x₂≈0.8657: f≈0.6489-0.6474=0.0015; f’≈3(0.7494)+0.7619=3.0101; x₃=0.8657-0.0015/3.0101≈0.8652. The root is approximately 0.8652.
令 f(x)=x³-cos x，f'(x)=3x²+sin x。迭代三次得 x₁≈0.8803, x₂≈0.8657, x₃≈0.8652。方程的实根约为 0.8652。
10. Linear Programming: Resource Allocation | 线性规划：资源分配
A factory produces two products, A and B. Each unit of A requires 2 hours on machine I and 1 hour on machine II; each unit of B requires 1 hour on machine I and 3 hours on machine II. Machine I is available for 10 hours per day and machine II for 12 hours per day. Profit per unit is £40 for A and £50 for B. Formulate a linear programming model to maximise daily profit and find the optimal production mix graphically.
一家工厂生产两种产品 A 和 B。每件 A 需机器 I 2 小时、机器 II 1 小时；每件 B 需机器 I 1 小时、机器 II 3 小时。机器 I 每天可用 10 小时，机器 II 每天可用 12 小时。A 的利润为每件 £40，B 为 £50。建立线性规划模型以最大化日利润，并用图解法求最优生产组合。
Let x = number of A, y = number of B. Maximise P = 40x + 50y subject to: 2x + y ≤ 10 (Machine I), x + 3y ≤ 12 (Machine II), x, y ≥ 0. Plot constraints: line 2x+y=10 from (5,0) to (0,10); line x+3y=12 from (12,0) to (0,4). The feasible region is a quadrilateral with vertices (0,0), (5,0), (0,4) and intersection of the two lines. Solve 2x+y=10 and x+3y=12 simultaneously: multiply second by 2: 2x+6y=24, subtract first: 5y=14 ⇒ y=2.8, then x=12-3(2.8)=3.6. Intersection at (3.6, 2.8). Evaluate P at vertices: (0,0):0; (5,0):200; (0,4):200; (3.6,2.8):40(3.6)+50(2.8)=144+140=284. Maximum profit £284 at x=3.6, y=2.8. Since integer solutions are required, check integer points nearby: (3,3): 40*3+50*3=270; (4,2): 260; (3,2): 220. The integer optimum is (3,3) with profit £270 if fractional products not allowed.
设 x 为产品 A 的数量，y 为 B 的数量。目标函数 max P=40x+50y，约束：2x+y≤10，x+3y≤12，x,y≥0。图解得可行域顶点，计算交点 (3.6,2.8)，利润 £284。若要求整数解，附近点 (3,3) 利润 £270 为最优。
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