📚 Case Study in Action: Practical Exercises for Year 13 CAIE Mathematics | Year 13 CAIE 数学：案例分析实战演练
In Year 13 CAIE Mathematics, case studies require you to translate real-world scenarios into mathematical models, solve them using techniques from Pure Mathematics, Mechanics, or Statistics, and critically interpret the outcomes. This article presents three detailed case studies that mirror the style of examination questions, guiding you step-by-step through optimisation, projectile motion, and normal approximation hypothesis testing. Each section pairs an English explanation with its Chinese equivalent, ensuring you master both the language of international assessment and deep conceptual understanding.
在 Year 13 CAIE 数学中，案例分析要求你把现实情境转化为数学模型，运用纯数学、力学或统计的技术求解，并批判性地解读结果。本文呈现三个详细的案例，模拟考试题型，一步步引导你经历优化、抛体运动和正态近似假设检验。每个部分都配对英文解释和中文翻译，确保你掌握国际测评的语言和深层的概念理解。
1. Introduction to Case Study Approach | 案例分析方法介绍
A case study in CAIE Mathematics is not just a word problem; it is an extended scenario that often spans multiple parts, testing your ability to model, compute, and evaluate. You will need to define variables, formulate equations, apply differentiation, integration, equations of motion, or statistical tests, and finally discuss the validity of your solution. Examiners look for clear logical steps, correct notation, and justified conclusions.
CAIE 数学中的案例研究不仅仅是一道文字题；它是一种扩展情境，常常跨越多个小问，考察你建模、计算和评价的能力。你需要定义变量、构建方程、运用微分、积分、运动方程或统计检验，最后讨论解的有效性。考官看重清晰的逻辑步骤、正确的符号和有依据的结论。
Before diving into specific cases, it is essential to adopt a structured method: (1) read and understand the real-world context; (2) identify the mathematical topic involved; (3) abstract the relevant quantities into symbols; (4) derive the governing equations; (5) solve using appropriate techniques; (6) interpret the answer back in context, checking for reasonableness.
在进入具体案例之前，必须采取结构化的方法：(1) 阅读并理解现实背景；(2) 确定涉及的数学主题；(3) 将相关量抽象为符号；(4) 推导控制方程；(5) 用恰当的技术求解；(6) 把答案放回情境中解释，检查合理性。
2. Case 1: Optimisation in Pure Mathematics | 案例1：纯数学中的优化问题
A manufacturer wants to produce cylindrical tin cans of a fixed volume V = 500 cm³. The cost of the metal is proportional to the total surface area of the can. Determine the radius r and height h that minimise the surface area, and thus the cost, assuming the can is closed at both ends. This is a classic constrained optimisation problem that requires expressing one variable in terms of the other and then using calculus.
一位制造商想要生产固定容积 V = 500 cm³ 的圆柱形罐头。金属成本与罐体的总表面积成正比。试确定能使表面积（从而成本）最小化的半径 r 和高 h，假设罐头两端封闭。这是一个经典的约束优化问题，需要把一个变量用另一个表示，再使用微积分求解。
The relevant volume and surface area formulas are: V = πr²h and A = 2πr² + 2πrh (two circular ends plus the curved side). With V fixed, we can write h = V/(πr²) and substitute into A, giving the area as a function of r alone.
相关的体积和表面积公式为：V = πr²h 和 A = 2πr² + 2πrh（两个圆底加曲面侧）。由于 V 固定，我们可以写出 h = V/(πr²)，并代入 A，便得到仅关于 r 的面积函数。
3. Setting Up the Model | 建立模型
Let V = 500 cm³. Then h = 500/(πr²). The surface area function becomes: A(r) = 2πr² + 2πr × (500/(πr²)) = 2πr² + 1000/r. The domain of r is r > 0. Our goal is to find r that minimises A(r), and then compute the corresponding h.
令 V = 500 cm³，则 h = 500/(πr²)。表面积函数变为：A(r) = 2πr² + 2πr × (500/(πr²)) = 2πr² + 1000/r。r 的定义域为 r > 0。我们的目标是求出使 A(r) 最小的 r，然后计算相应的 h。
At this stage, we check that the model makes sense: as r → 0⁺, A → ∞ (very tall, thin can has huge curved area); as r → ∞, A → ∞ (wide flat can has huge circular areas). Hence a minimum should exist. This preliminary reasoning is good practice before diving into differentiation.
在这个阶段，我们检验模型的合理性：当 r → 0⁺ 时，A → ∞（极细高的罐子曲面面积巨大）；当 r → ∞ 时，A → ∞（宽扁的罐子圆面积巨大）。因此最小值应当存在。在求导之前进行这种预先推理是好的习惯。
4. Solving the Optimisation Problem | 求解优化问题
Differentiate A(r) with respect to r: dA/dr = 4πr − 1000/r². Set the derivative equal to zero to locate stationary points: 4πr = 1000/r² → r³ = 1000/(4π) = 250/π → r = ³√(250/π). Using π ≈ 3.1416, 250/π ≈ 79.577, so r ≈ ³√79.577 ≈ 4.30 cm (to 3 s.f.).
对 A(r) 关于 r 求导：dA/dr = 4πr − 1000/r²。令导数为零以求驻点：4πr = 1000/r² → r³ = 1000/(4π) = 250/π → r = ³√(250/π)。取 π ≈ 3.1416，250/π ≈ 79.577，故 r ≈ ³√79.577 ≈ 4.30 cm（保留三位有效数字）。
To confirm that this stationary point is a minimum, compute the second derivative: d²A/dr² = 4π + 2000/r³. Since r > 0, both terms are positive, so d²A/dr² > 0, indicating a local minimum. There is only one stationary point and the domain boundaries give infinite area, so this is the global minimum.
为确认该驻点为最小值，计算二阶导数：d²A/dr² = 4π + 2000/r³。由于 r > 0，两项均为正，故 d²A/dr² > 0，表明是局部极小值。由于只有一个驻点且定义域边界处面积趋于无穷，这就是全局最小值。
The corresponding height is h = 500/(π × (4.30)²) ≈ 500/(π × 18.49) ≈ 500/58.09 ≈ 8.61 cm. Notice that h ≈ 2r; the minimum surface area occurs when the diameter equals the height. This is a well-known result for closed cylindrical containers.
相应的高为 h = 500/(π × (4.30)²) ≈ 500/(π × 18.49) ≈ 500/58.09 ≈ 8.61 cm。注意 h ≈ 2r；当直径等于高时表面积最小。这是封闭圆筒的一个众所周知的结果。
5. Interpreting the Solution | 解释解
The minimal surface area is A_min = 2π(4.30)² + 1000/4.30 ≈ 2π(18.49) + 232.6 ≈ 116.2 + 232.6 = 348.8 cm². The dimensions r ≈ 4.3 cm, h ≈ 8.6 cm are practical. In an exam, you might be asked to comment on the feasibility: a very wide or very narrow can would indeed use more material. The model assumes uniform metal thickness and ignores overlaps for seams, so it is an idealised case. Nonetheless, the result gives a starting point for design.
最小表面积为 A_min = 2π(4.30)² + 1000/4.30 ≈ 2π(18.49) + 232.6 ≈ 116.2 + 232.6 = 348.8 cm²。尺寸 r ≈ 4.3 cm, h ≈ 8.6 cm 是可行的。在考试中，可能会让你评价其可行性：极宽或极窄的罐子确实会使用更多材料。该模型假设金属厚度均匀且忽略搭接缝隙，所以是理想化情形。尽管如此，结果给出了设计的起点。
6. Case 2: Projectile Motion in Mechanics | 案例2：力学中的抛体运动
A ball is thrown from ground level with an initial speed of 20 m s⁻¹ at an angle of 30° to the horizontal. Assuming negligible air resistance and taking g = 9.8 m s⁻², find (a) the time of flight, (b) the maximum height reached, and (c) the horizontal range. This case tests your ability to resolve velocity into components and apply the constant acceleration equations separately in the horizontal and vertical directions.
一个小球从地面以初速 20 m s⁻¹、与水平面成 30° 角抛出。假设空气阻力可忽略，取 g = 9.8 m s⁻²，求 (a) 飞行时间，(b) 达到的最大高度，(c) 水平射程。这个案例考察你将速度分解为各分量、并分别在水平和竖直方向应用匀加速运动方程的能力。
In CAIE mechanics, the standard sign convention is to take upward as positive. Thus the initial vertical velocity u_y = u sinθ = 20 sin30° = 10 m s⁻¹, and the horizontal velocity u_x = u cosθ = 20 cos30° = 20 × (√3/2) ≈ 17.32 m s⁻¹. The acceleration in the vertical direction is a_y = −g = −9.8 m s⁻², while horizontal acceleration a_x = 0.
在 CAIE 力学中，标准符号惯例取向上为正。因此初始竖直速度 u_y = u sinθ = 20 sin30° = 10 m s⁻¹，水平速度 u_x = u cosθ = 20 cos30° = 20 × (√3/2) ≈ 17.32 m s⁻¹。竖直方向的加速度 a_y = −g = −9.8 m s⁻²，而水平加速度 a_x = 0。
7. Applying Equations of Motion | 应用运动方程
To find the time of flight T, consider the vertical motion from launch to landing. The ball returns to ground level, so vertical displacement s_y = 0. Using s_y = u_y t + ½ a_y t², we have 0 = 10 T + ½(−9.8) T² = 10 T − 4.9 T². Factorising gives T(10 − 4.9 T) = 0, so T = 0 (at launch) or T = 10/4.9 ≈ 2.04 s (2.0408 s). The time of flight is 2.04 s (3 s.f.).
为求飞行时间 T，考虑从抛出到落地的竖直运动。小球返回地面，故竖直位移 s_y = 0。利用 s_y = u_y t + ½ a_y t²，得 0 = 10 T + ½(−9.8) T² = 10 T − 4.9 T²。因式分解得 T(10 − 4.9 T) = 0，因此 T = 0（抛出时刻）或 T = 10/4.9 ≈ 2.04 s（2.0408 s）。飞行时间取 2.04 s（三位有效数字）。
Maximum height H occurs when the vertical velocity becomes zero. Using v_y = u_y + a_y t, set v_y = 0: 0 = 10 − 9.8 t → t = 10/9.8 ≈ 1.02 s (half the total flight time, as expected for symmetric motion). Then H = u_y t + ½ a_y t² = 10(1.02) + ½(−9.8)(1.02)² ≈ 10.2 − 5.10 = 5.10 m (to 3 s.f.). Alternatively, use v_y² = u_y² + 2a_y s_y: 0 = 100 − 19.6 H → H = 100/19.6 = 5.102 m ≈ 5.10 m.
最大高度 H 发生在竖直速度为零时。用 v_y = u_y + a_y t，令 v_y = 0：0 = 10 − 9.8 t → t = 10/9.8 ≈ 1.02 s（恰好是总飞行时间的一半，因运动具有对称性）。则 H = u_y t + ½ a_y t² = 10(1.02) + ½(−9.8)(1.02)² ≈ 10.2 − 5.10 = 5.10 m（三位有效数字）。另外可用 v_y² = u_y² + 2a_y s_y：0 = 100 − 19.6 H → H = 100/19.6 = 5.102 m ≈ 5.10 m。
The horizontal range R is found by multiplying the constant horizontal velocity by the total time of flight: R = u_x × T = 17.32 × 2.0408 ≈ 35.3 m (3 s.f.). Note that the range could also be expressed using the formula R = (u² sin 2θ)/g, which for 20 m s⁻¹ and 30° yields 400 × sin60° / 9.8 = 400 × 0.8660 / 9.8 ≈ 346.4/9.8 = 35.35 m, confirming the result.
水平射程 R 用恒定的水平速度乘总飞行时间求得：R = u_x × T = 17.32 × 2.0408 ≈ 35.3 m（三位有效数字）。注意射程也可用公式 R = (u² sin 2θ)/g 表示，对于 20 m s⁻¹ 和 30°，得 400 × sin60° / 9.8 = 400 × 0.8660 / 9.8 ≈ 346.4/9.8 = 35.35 m，从而验证结果。
8. Case 3: Statistical Modelling – Normal Approximation | 案例3：统计建模——正态近似
A factory produces bolts whose lengths are normally distributed with mean μ = 50.0 mm and standard deviation σ = 0.20 mm. A quality control inspector suspects that a new batch of bolts may have a different mean length. A random sample of 30 bolts from this batch yields a sample mean x̄ = 50.12 mm. Test at the 2% significance level whether there is evidence that the mean length of this batch differs from 50.0 mm. This case illustrates hypothesis testing for a population mean with known variance using the normal distribution.
一家工厂生产的螺栓长度服从正态分布，均值 μ = 50.0 mm，标准差 σ = 0.20 mm。质检员怀疑新批次的螺栓平均长度可能不同。从该批次随机抽取 30 个螺栓，样本均值 x̄ = 50.12 mm。在 2% 显著性水平下检验是否有证据表明该批次平均长度不同于 50.0 mm。这个案例演示在方差已知时用正态分布进行总体均值的假设检验。
We must carefully set up hypotheses: H₀: μ = 50.0; H₁: μ ≠ 50.0. This is a two-tailed test. The test statistic is Z = (x̄ − μ) / (σ/√n). Under H₀, Z ~ N(0,1). The significance level α = 0.02, so for a two-tailed test each tail has probability 0.01. The critical values are z = ±2.326 (or ±2.33, depending on tables; using 2.326 is more precise). We can use a table to find that P(Z > 2.326) = 0.01.
我们必须仔细设立假设：H₀: μ = 50.0；H₁: μ ≠ 50.0。这是一个双尾检验。检验统计量为 Z = (x̄ − μ) / (σ/√n)。在 H₀ 下，Z ~ N(0,1)。显著性水平 α = 0.02，因此对于双尾检验，每个尾部的概率为 0.01。临界值为 z = ±2.326（或 ±2.33，取决于表格；使用 2.326 更精确）。查表可知 P(Z > 2.326) = 0.01。
9. Conducting the Hypothesis Test | 进行假设检验
Calculate the observed test statistic: σ/√n = 0.20/√30 ≈ 0.20/5.477 = 0.03651 (to 4 d.p.). Then Z = (50.12 − 50.0) / 0.03651 ≈ 0.12 / 0.03651 ≈ 3.286. Compare with critical values: |3.286| > 2.326, so the test statistic falls in the critical region. We reject H₀ at the 2% significance level.
计算观测检验统计量：σ/√n = 0.20/√30 ≈ 0.20/5.477 = 0.03651（保留四位小数）。于是 Z = (50.12 − 50.0) / 0.03651 ≈ 0.12 / 0.03651 ≈ 3.286。与临界值比较：|3.286| > 2.326，因此检验统计量落入拒绝域。我们在 2% 显著性水平下拒绝 H₀。
Using the p-value approach: p-value = 2 × P(Z > 3.286) = 2 × (1 − Φ(3.286)). From normal tables, Φ(3.29) ≈ 0.9995, so p-value ≈ 2 × 0.0005 = 0.001. Since 0.001 < 0.02, we again reject H₀. There is strong evidence (significant at 2% level) that the mean bolt length has changed from 50.0 mm. The sample suggests the mean is slightly higher.
使用 p-值方法：p-值 = 2 × P(Z > 3.286) = 2 × (1 − Φ(3.286))。查正态分布表，Φ(3.29) ≈ 0.9995，故 p-值 ≈ 2 × 0.0005 = 0.001。由于 0.001 < 0.02，我们再次拒绝 H₀。有强证据（在 2% 水平显著）表明螺栓平均长度已从 50.0 mm 改变。样本显示均值略微偏高。
It is essential to interpret the conclusion in context: the test does not prove the new batch definitely has a different mean; it indicates that if H₀ were true, a sample mean as extreme as 50.12 would be very unlikely. Therefore, the quality controller should investigate the production process.
务必将结论放在情境中解释：检验并没有证明新批次的均值绝对不同；它表明如果 H₀ 成立，得到像 50.12 这样极端的样本均值的概率非常小。因此质检员应该调查生产过程。
10. Reflection and Common Pitfalls | 反思与常见陷阱
Case studies demand more than algorithmic application; they require thoughtful interpretation. Common pitfalls include forgetting to check stationary point type in optimisation, using the wrong sign for acceleration when upward is positive in projectiles, misinterpreting the alternative hypothesis as one-tailed when the question implies a two-tailed test, and failing to link the p-value to the significance level correctly. Always verify your final answer’s units and whether it makes physical or practical sense.
案例分析不仅要求机械式的应用，还要求深思熟虑的解读。常见陷阱包括：在优化中忘记检查驻点类型；当抛体取向上为正时错误使用加速度符号；当问题暗示双尾检验时却把备择假设错当成单尾；以及未能正确把 p-值与显著性水平相联系。始终验证最终答案的单位以及它是否有物理或实际意义。
Another frequent mistake is incomplete communication: in CAIE, you are expected to state your assumptions (e.g., smooth surface, particle model, normal distribution). In mechanics, always state positive direction. In statistics, clearly define the random variable and distribution. Practising case studies with full written solutions sharpens your ability to score high marks on structured questions and the longer, open-ended investigative parts of the exam.
另一个常见错误是表达不完整：在 CAIE 中，你要清楚声明假设（如光滑面、质点模型、正态分布）。在力学中，始终说明正方向。在统计中，明确定义随机变量和分布。通过完整的书面解答来练习案例研究，能够提升你在结构化问题以及考试中较长、开放式探究部分取得高分的能力。
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