📚 Case Study Practical Exercises in Pre-U OCR Further Mathematics | Pre-U OCR 进阶数学：案例分析实战演练
Case studies in the OCR Pre-U Further Mathematics specification challenge students to apply pure mathematics, mechanics and statistics in realistic, often open-ended situations. Unlike routine exercises, a case study requires you to formulate a model, choose appropriate mathematical techniques, carry out calculations and, crucially, interpret your results in the original context. This article presents three fully worked case studies – from mechanics, statistics and optimisation – alongside advice on common pitfalls and report writing, so you can approach your own case study with confidence.
OCR Pre-U 进阶数学中的案例分析题要求学生将纯数学、力学和统计知识应用于实际的、通常是开放式的情境。与常规习题不同，案例分析需要你建立模型、选择合适的数学方法、进行计算，并且最重要的是在原始情境中解读结果。本文通过三个完整的案例——分别来自力学、统计和优化——并附上常见错误分析与报告写作建议，帮助你自信地应对自己的案例分析任务。
1. Introduction to the Case Study Approach | 案例分析方法导论
A Pre-U case study typically provides a real-world scenario and a set of data or key assumptions. Your job is to develop a mathematical model, use the tools from Further Mathematics to solve it, and critically evaluate your findings. The examiners expect a logical structure, clear communication, and demonstration of both technical competence and thoughtful reflection. Remember that marks are awarded not only for correct mathematics but also for the quality of your reasoning and presentation.
Pre-U 案例分析通常给出一个真实场景以及一组数据或关键假设。你的任务是建立一个数学模型，运用进阶数学工具求解，并对结果进行批判性评估。考官期望看到清晰的逻辑结构、良好的表达，以及既展现技术能力又体现深入思考的过程。请记住，得分点不仅在于数学本身的正确性，还在于推理和展示的质量。
2. Case Study 1: Projectile Motion with Air Resistance – Building the Model | 案例一：含空气阻力的抛体运动——建立模型
A particle of mass 0.5 kg is projected vertically upwards from ground level with speed 20 m s⁻¹. Air resistance is modelled as a force of magnitude 0.2v N, where v is the speed of the particle at time t. The direction of the resistance always opposes the motion. By taking upwards as the positive direction, the net force on the particle while ascending is -mg – 0.2v. Newton’s second law gives m dv/dt = -mg – 0.2v, which simplifies to dv/dt = -9.8 – 0.4v.
一个质量为 0.5 kg 的质点从地面以 20 m s⁻¹ 的速度竖直向上发射。空气阻力被模拟为大小为 0.2v N 的力，其中 v 是质点在时刻 t 的速度。阻力的方向始终与运动方向相反。取向上为正方向，质点上升过程中所受合外力为 -mg – 0.2v。由牛顿第二定律得 m dv/dt = -mg – 0.2v，简化后为 dv/dt = -9.8 – 0.4v。
The initial condition is v(0) = 20. This first-order linear differential equation can be written in standard form as dv/dt + 0.4v = -9.8. The integrating factor is e^(∫ 0.4 dt) = e^(0.4t). Multiplying both sides and integrating yields v(t) = (20 + 24.5) e^(-0.4t) – 24.5, i.e. v(t) = 44.5 e^(-0.4t) – 24.5. The constant 24.5 arises from mg/k, where k = 0.2 N·m⁻¹·s.
初始条件为 v(0) = 20。这个一阶线性微分方程可写成标准形式 dv/dt + 0.4v = -9.8。积分因子为 e^(∫ 0.4 dt) = e^(0.4t)。两边同乘并积分后得到 v(t) = (20 + 24.5) e^(-0.4t) – 24.5，即 v(t) = 44.5 e^(-0.4t) – 24.5。常数 24.5 来自 mg/k，其中 k = 0.2 N·m⁻¹·s。
3. Case Study 1: Solving and Interpreting the Model | 案例一：求解模型并解读结果
To find the time at which the particle reaches its maximum height, set v(t) = 0: 44.5 e^(-0.4t) – 24.5 = 0 → e^(-0.4t) = 24.5/44.5 ≈ 0.55056. Hence t_max = -ln(0.55056)/0.4 ≈ 1.494 s. The height function can be obtained by integrating v(t) with initial height s(0) = 0: s(t) = ∫ (44.5 e^(-0.4t) – 24.5) dt = -111.25 e^(-0.4t) – 24.5t + C. Using s(0)=0 gives C = 111.25, so s(t) = 111.25(1 – e^(-0.4t)) – 24.5t. Substituting t_max yields s_max ≈ 13.5 m.
为求质点到达最高点的时间，令 v(t) = 0：44.5 e^(-0.4t) – 24.5 = 0 → e^(-0.4t) = 24.5/44.5 ≈ 0.55056。因此 t_max = -ln(0.55056)/0.4 ≈ 1.494 s。高度函数可由 v(t) 积分求得，初始高度 s(0) = 0：s(t) = ∫ (44.5 e^(-0.4t) – 24.5) dt = -111.25 e^(-0.4t) – 24.5t + C。由 s(0)=0 得 C = 111.25，于是 s(t) = 111.25(1 – e^(-0.4t)) – 24.5t。代入 t_max 可得 s_max ≈ 13.5 m。
Compare this with the case of no air resistance, where v = 20 – 9.8t, t_max = 20/9.8 ≈ 2.04 s and s_max ≈ 20.4 m. The drag significantly reduces both the time of ascent and the maximum height. Moreover, as t → ∞, v(t) → -24.5 m s⁻¹, which is the terminal velocity on descent; the negative sign indicates a downward direction. This analysis illustrates how a simple linear drag model can capture realistic behaviour while remaining analytically tractable.
与无空气阻力时比较：v = 20 – 9.8t，t_max = 20/9.8 ≈ 2.04 s，s_max ≈ 20.4 m。阻力显著缩短了上升时间并降低了最大高度。另外，当 t → ∞ 时，v(t) → -24.5 m s⁻¹，即下降过程中的终端速度；负号表示方向向下。这一分析展示了简单的线性阻力模型如何抓住真实行为的本质，同时仍能通过解析方法求解。
4. Case Study 2: Analysing Bivariate Data – Regression | 案例二：双变量数据分析——回归
An experiment measures two variables X and Y. The data are given in the table below. We suspect a linear relationship of the form Y = α + βX + ε.
某实验测量了两个变量 X 和 Y。数据如下表所示。我们猜测它们之间存在 Y = α + βX + ε 的线性关系。
| X | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Y | 2.1 | 3.9 | 6.2 | 7.8 | 10.0 |
First compute the summary statistics: n = 5, Σx = 15, Σy = 30.0, so x̄ = 3, ȳ = 6. Then Sxx = Σx² – n x̄² = 55 – 45 = 10. Sxy = Σxy – n x̄ ȳ = 109.7 – 90 = 19.7. Syy = Σy² – n ȳ² = 218.9 – 180 = 38.9. The least squares regression line of Y on X is given by ŷ = a + bx, where b = Sxy/Sxx = 19.7/10 = 1.97 and a = ȳ – b x̄ = 6 – 1.97×3 = 0.09. Thus ŷ = 0.09 + 1.97x.
先计算汇总统计量：n = 5，Σx = 15，Σy = 30.0，则 x̄ = 3，ȳ = 6。接着 Sxx = Σx² – n x̄² = 55 – 45 = 10；Sxy = Σxy – n x̄ ȳ = 109.7 – 90 = 19.7；Syy = Σy² – n ȳ² = 218.9 – 180 = 38.9。Y 对 X 的最小二乘回归直线为 ŷ = a + bx，其中 b = Sxy/Sxx = 19.7/10 = 1.97，a = ȳ – b x̄ = 6 – 1.97×3 = 0.09。因此回归方程为 ŷ = 0.09 + 1.97x。
The slope 1.97 suggests that for every unit increase in X, Y increases by about 2 units. The intercept 0.09 is very close to zero, indicating that the line passes near the origin. To assess the quality of the fit, we compute the coefficient of determination R² = (Sxy)²/(Sxx·Syy) = 19.7²/(10×38.9) ≈ 0.997, meaning the model explains 99.7% of the variation in Y.
斜率 1.97 表明 X 每增加一个单位，Y 大约增加 2 个单位。截距 0.09 非常接近零，意味着直线几乎过原点。为评估拟合优度，我们计算决定系数 R² = (Sxy)²/(Sxx·Syy) = 19.7²/(10×38.9) ≈ 0.997，即模型解释了 Y 中 99.7% 的变异。
5. Case Study 2: Hypothesis Testing and Conclusion | 案例二：假设检验与结论
We now test whether there is a significant linear relationship between X and Y. Set H₀: β = 0 against H₁: β ≠ 0 at the 5% significance level. The residual sum of squares is RSS = Syy – b·Sxy = 38.9 – 1.97×19.7 = 0.091. The error variance estimate is s² = RSS/(n-2) = 0.091/3 ≈ 0.03033, so s ≈ 0.1742. The standard error of b is SE(b) = s/√Sxx = 0.1742/√10 ≈ 0.0551. The test statistic t = b/SE(b) = 1.97/0.0551 ≈ 35.75.
我们现在检验 X 与 Y 之间是否存在显著的线性关系。设 H₀: β = 0，H₁: β ≠ 0，显著性水平 5%。残差平方和 RSS = Syy – b·Sxy = 38.9 – 1.97×19.7 = 0.091。误差方差估计 s² = RSS/(n-2) = 0.091/3 ≈ 0.03033，从而 s ≈ 0.1742。b 的标准误为 SE(b) = s/√Sxx = 0.1742/√10 ≈ 0.0551。检验统计量 t = b/SE(b) = 1.97/0.0551 ≈ 35.75。
With 3 degrees of freedom, the critical value from t-tables for a two-tailed test at 5% is about 3.182. Since 35.75 >> 3.182, we reject H₀ and conclude that the slope is significantly different from zero. The data provide very strong evidence of a linear relationship. A good case study would also comment on possible limitations, such as the small sample size and the assumption of constant variance, and perhaps suggest further data collection.
自由度为 3 时，t 分布 5% 显著性水平下的双侧临界值约为 3.182。由于 35.75 远大于 3.182，我们拒绝 H₀，认为斜率显著不为零。数据提供了极强的线性关系证据。一份优秀的案例分析还应讨论可能的局限性，例如样本量较小以及方差齐性的假设，或者建议进一步收集数据。
6. Case Study 3: Constrained Optimisation – Utility Maximisation | 案例三：约束优化——效用最大化
A consumer’s utility depends on the quantities x and y of two goods according to the Cobb-Douglas function U(x, y) = x⁰·⁵ y⁰·⁵. The unit prices are £2 and £3 respectively, and the consumer has a budget of £120. The problem is to maximise U subject to the budget constraint 2x + 3y = 120. This is a classic constrained optimisation problem that can be tackled using the method of Lagrange multipliers.
某消费者的效用取决于两种商品的数量 x 和 y，效用函数为柯布-道格拉斯型 U(x, y) = x⁰·⁵ y⁰·⁵。两种商品的单价分别为 £2 和 £3，消费者有 £120 的预算。问题是在预算约束 2x + 3y = 120 下最大化效用。这是一个经典的约束优化问题，可用拉格朗日乘数法处理。
Define the Lagrangian: L(x, y, λ) = x⁰·⁵ y⁰·⁵ + λ(120 – 2x – 3y). The first-order conditions are ∂L/∂x = 0.5 x⁻⁰·⁵ y⁰·⁵ – 2λ = 0, ∂L/∂y = 0.5 x⁰·⁵ y⁻⁰·⁵ – 3λ = 0, and ∂L/∂λ = 120 – 2x – 3y = 0. From the first two equations, eliminating λ gives (0.5 y⁰·⁵ / x⁰·⁵) / 2 = (0.5 x⁰·⁵ / y⁰·⁵) / 3, which simplifies to y/x = 2/3, hence 3y = 2x.
定义拉格朗日函数：L(x, y, λ) = x⁰·⁵ y⁰·⁵ + λ(120 – 2x – 3y)。一阶条件为 ∂L/∂x = 0.5 x⁻⁰·⁵ y⁰·⁵ – 2λ = 0，∂L/∂y = 0.5 x⁰·⁵ y⁻⁰·⁵ – 3λ = 0，以及 ∂L/∂λ = 120 – 2x – 3y = 0。由前两个方程消去 λ 得 (0.5 y⁰·⁵ / x⁰·⁵) / 2 = (0.5 x⁰·⁵ / y⁰·⁵) / 3，化简得 y/x = 2/3，即 3y = 2x。
7. Case Study 3: Solution and Economic Insight | 案例三：求解及经济含义
Substituting 3y = 2x into the budget constraint 2x + 3y = 120 gives 2x + 2x = 120 → 4x = 120 → x = 30. Then y = (2/3)×30 = 20. The maximum utility is U(30,20) = 30⁰·⁵ × 20⁰·⁵ = √600 ≈ 24.5. The Lagrange multiplier λ can be found from the first condition: λ = 0.25 y⁰·⁵ / x⁰·⁵ = 0.25 × √(20/30) = 0.25 × √(2/3) ≈ 0.204. In economic terms, λ represents the marginal utility of income – the additional utility gained from an extra £1 of budget at the optimum.
将 3y = 2x 代入预算约束 2x + 3y = 120，得 2x + 2x = 120 → 4x = 120 → x = 30。进而 y = (2/3)×30 = 20。最大效用为 U(30,20) = 30⁰·⁵ × 20⁰·⁵ = √600 ≈ 24.5。拉格朗日乘数 λ 可由第一个条件求出：λ = 0.25 y⁰·⁵ / x⁰·⁵ = 0.25 × √(20/30) = 0.25 × √(2/3) ≈ 0.204。经济学上，λ 代表收入的边际效用——即在最优解处每增加 £1 预算所能带来的额外效用。
This solution demonstrates the equi-marginal principle: at the optimum, the ratio of marginal utilities equals the price ratio, i.e. (∂U/∂x) / (∂U/∂y) = p₁/p₂. Here MUₓ = 0.5√(y/x), MUᵧ = 0.5√(x/y), and indeed (√(y/x)) / (√(x/y)) = y/x = 2/3, matching the price ratio 2/3. A full case study would also verify that the second-order conditions hold and discuss the realism of the Cobb-Douglas utility function.
这一求解展示了等边际原理：在最优解处，边际效用之比等于价格之比，即 (∂U/∂x) / (∂U/∂y) = p₁/p₂。这里 MUₓ = 0.5√(y/x)，MUᵧ = 0.5√(x/y)，而 (√(y/x)) / (√(x/y)) = y/x = 2/3，恰好等于价格比 2/3。完整的案例分析还应验证二阶条件是否成立，并讨论柯布-道格拉斯效用函数的现实性。
8. Key Pitfalls to Avoid in Case Studies | 案例分析中应避免的关键错误
Many students lose marks by rushing into calculations before fully understanding the problem. Always state your assumptions explicitly and justify why you have chosen a particular model. In mechanics, check the direction of forces and the sign convention; in statistics, check whether you are using the correct regression (Y on X or X on Y); in optimisation, ensure the constraint is correctly written and the objective function is appropriate.
许多学生因急于计算
Published by TutorHao | Pre-U 进阶数学 Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply