Case Study Practical Exercises | 案例分析实战演练

📚 Case Study Practical Exercises | 案例分析实战演练

Case studies bridge the gap between abstract mathematical concepts and the messy, unpredictable problems we encounter in engineering, economics, and the natural sciences. In Year 12 CAIE Mathematics, you are expected not only to reproduce standard techniques but also to interpret a scenario, identify the relevant mathematical structure, and construct a solution from first principles. This article walks you through six carefully chosen practical exercises that mirror the style of extended questions in P1, M1, and S1 — each designed to strengthen your analytical thinking and build the confidence to tackle unfamiliar contexts under exam conditions.

案例分析在抽象的数学概念与我们在工程、经济以及自然科学中遇到的杂乱而不可预测的问题之间架起了桥梁。在 Year 12 CAIE 数学课程中，你不仅需要掌握标准解题技巧，还要能够解读实际情境、识别相关的数学结构并从基本原理出发构造解答。本文将通过六个精心挑选的实战练习，模拟 P1、M1 和 S1 中拓展题的风格——每一个练习都旨在强化你的分析思维，帮助你在考试条件下自信地应对陌生场景。

1. Optimising a Product’s Packaging | 产品包装的优化设计

A company manufactures cylindrical tins of volume 500 cm³. The material for the top and base costs 0.08 £/cm², while the curved surface costs 0.05 £/cm². Let the base radius be r cm and the height h cm. The total cost C in pounds is given by C = 0.16πr² + 0.10πrh. Using the fixed volume πr²h = 500, express C solely in terms of r and find the radius that minimises cost. Verify that your answer indeed yields a minimum by using the second derivative test. This classic constrained optimisation illustrates how calculus can drive real-world cost reduction — a favourite theme in P1 differentiial applications.

某公司生产容积为 500 cm³ 的圆柱形罐子。顶面和底面材料成本为 0.08 英镑/每平方厘米，侧面材料成本为 0.05 英镑/每平方厘米。设底面半径为 r cm，高度为 h cm。总成本 C（英镑）为 C = 0.16πr² + 0.10πrh。利用固定容积 πr²h = 500，将 C 表示为仅含 r 的表达式，并求出使成本最小的半径。利用二阶导数检验确认该解确实给出最小值。这个经典的有约束优化展示了微积分如何驱动现实世界的成本降低——这是 P1 微分应用中常见的主题。

Substituting h = 500 / (πr²) gives C(r) = 0.16πr² + 50r⁻¹. Differentiating, C'(r) = 0.32πr − 50r⁻². Setting C'(r) = 0 leads to r³ = 50/(0.32π) ≈ 49.7359, so r ≈ 3.68 cm. The second derivative C''(r) = 0.32π + 100r⁻³ is positive for all r > 0, confirming a minimum. The corresponding height is about 11.8 cm, making the tin roughly three times as tall as its radius. Such insights often surprise product designers who assume a taller, thinner can is cheaper.

代入 h = 500 / (πr²) 得到 C(r) = 0.16πr² + 50r⁻¹。求导得 C'(r) = 0.32πr − 50r⁻²。令 C'(r) = 0 解得 r³ = 50/(0.32π) ≈ 49.7359，因此 r ≈ 3.68 cm。二阶导数 C''(r) = 0.32π + 100r⁻³ 对所有 r > 0 恒为正，确认了极小值。对应高度约为 11.8 cm，罐子高度大约是其半径的三倍。这一结果常令产品设计师感到意外，他们往往认为更细高的罐子成本更低。

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2. Modelling Population Growth | 人口增长的数学建模

Ecologists studying a rare bird population on an island record the following data: at the start of 2021, there were 240 birds; by the start of 2022, 312 birds; and by the start of 2023, 405 birds. A scientist proposes an exponential model P(t) = Ae^kt, where t is the number of years since the start of 2021. Determine the constants A and k from the given data and predict the population at the start of 2030. Discuss one reason why this model might become unreliable over a longer period.

生态学家研究一个海岛上的珍稀鸟类种群，记录到以下数据：2021 年初有 240 只；2022 年初有 312 只；2023 年初有 405 只。一位科学家提出指数模型 P(t) = Ae^kt，其中 t 为自 2021 年初起经过的年数。利用给定数据确定常数 A 和 k，并预测 2030 年初的种群数量。讨论一个可能导致该模型在较长时间后不可靠的原因。

At t = 0, P(0) = 240, so A = 240. Using the point (1, 312): 240e^k = 312 → e^k = 1.3 → k = ln 1.3 ≈ 0.26236. For 2030, t = 9: P(9) = 240e^9k ds. However, unlimited exponential growth is unrealistic; factors like food scarcity, disease, or habitat limits would eventually slow the growth, calling for a logistic model. This exercise reinforces the importance of interpreting a model's domain of validity — a skill tested in P1 and S1.

当 t = 0 时，P(0) = 240，故 A = 240。利用点 (1, 312)：240e^k = 312 → e^k = 1.3 → k = ln 1.3 ≈ 0.26236。对于 2030 年，t = 9：P(9) = 240e^9k = 240 × (1.3)⁹ ≈ 240 × 10.6045 ≈ 2545 只。然而，不受限制的指数增长并不现实；食物短缺、疾病或栖息地限制等因素最终会减缓增长速度，需要采用逻辑斯谛模型。该练习强调了理解模型有效域的重要性——这是 P1 和 S1 中常考的能力。

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3. Projectile Motion with a Slope | 斜坡上的抛体运动

A golfer strikes a ball from a point on a hill that slopes downward at an angle of 20° to the horizontal. The initial speed is 25 m s⁻¹ at an angle of 30° above the horizontal. Take g ong and perpendicular to the slope, find the time of flight until the ball lands on the slope and the distance it travels along the slope. This M1 problem demands a slick coordinate transformation: rotating axes aligns the landing condition with a single coordinate, greatly simplifying the algebra.

一名高尔夫球手从一座与水平面成 20° 向下倾斜的山坡上击球。球的初速度为 25 m s⁻¹，方向与水平面成 30° 仰角。取 g = 9.8 m s⁻²。通过沿斜坡和垂直于斜坡方向分解运动，求球在空中飞行直到落在斜坡上的时间，以及球沿斜坡飞行的距离。这道 M1 题目需要巧妙的坐标变换：旋转坐标轴使落地条件与单一坐标对齐，从而大大简化代数运算。

Define axes: x down the slope, y perpendicular to the slope (upwards). The initial velocity components: u_x = 25 cos(30°+20°) = 25 cos 50°, u_y = 25 sin 50°. Gravity contributes components: a_x = g sin 20° down the slope, a_y = −g cos 20°. Landing occurs when y = 0. Equation for y: y = u_y t − ½ g cos 20° t² = 0 → t = 2u_y / (g os 20°) ≈ 2×25 sin 50° / (9.8 cos 20°) ≈ 38.30 / 9.208 ≈ 4.16 s. Then distance along slope x = u_x t + ½ g sin 20° t ≈ 66.8 + 29.0 ≈ 95.8 m. The rotated frame elegantly avoids solving simultaneous equations.

定义坐标轴：x 轴沿斜坡向下，y 轴垂直于斜坡向上。初速度分量：u_x = 25 cos(30°+20°) = 25 cos 50°，u_y = 25 sin 50°。重力加速度分量：a_x = g sin 20° 沿斜坡向下，a_y = −g cos 20°。落地条件为 y = 0。y 方向的方程：y = u_y t − ½ g cos 20° t² = 0 → t = 2u_y / (g cos 20°) ≈ 2×25 sin 50° / (9.8 cos 20°) ≈ 38.30 / 9.208 ≈ 4.16 s。沿斜坡的距离 x = u_x t + ½ g sin 20° t² ≈ 25 cos 50° × 4.16 + 0.5 × 9.8 × sin 20° × (4.16)² ≈ 66.8 + 29.0 ≈ 95.8 m。旋转参考系巧妙地避免了解联立方程组。

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4. Probability Distribution of a Discrete Random Variable | 离散型随机变量的概率分布

A fair six-sided die is rolled twice. Let the random variable X be the absolute difference between the two scores. Construct the probability distribution table for X and calculate E(X) and Var(X). This S1 case study trains you to define a discrete random variable from first principles, carefully counting equally likely outcomes. Often students rush and miscount the pairs that give a particular difference, so systematic listing is essential.

一个均匀的六面骰子投掷两次。设随机变量 X 为两次点数之差的绝对值。构建 X 的概率分布表，并计算 E(X) 和 Var(X)。这个 S1 案例训练你从基本原理出发定义离散随机变量，并仔细计数等可能结果。学生常因匆忙而数错产生某个差值的点数对，因此系统列举至关重要。

There are 36 equally likely ordered pairs. Possible values of X 3),(3,1) through (4,6),(6,4) → 4×2=8. Diff 3: 3×2=6. Diff 4: 2×2=4. Diff 5: 1×2=2. Thus P(X=x) = counts/36. E(X) = Σ x P(X=x) = (0×6 + 1×10 + 2×8 + 3×6 + 4×4 + 5×2)/36 = 70/36 ≈ 1.944. Var(X) = E(X²) − [E(X)]². E(X + 1²×10 + 2²×8 + 3²×6 + 4²×4 + 5²×2)/36 = (0+10+32+54+64+50)/36 = 210/36 ≈ 5.833. So Var(X) = 5.833 − (1.944)² ≈ 2.054. Systematic table construction is the bedrock of S1.

共有 36 种等可能的有序点数对。X 的可能取值为 0,1,2,3,4,5。计数：差为 0：(1,1)…(6,6) → 6 种。差为 1：(1,2),(2,1),(2,3),(3,2) … 直到 (5,6),(6,5) → 5×2 = 10。差为 2：4×2=8。差为 3：3×2=6。差为 4：2×2=4。差为 5：1×2=2。因此 P(X=x) = 频数/36。E(X) = Σ x P(X=x) = (0×6 + 1×10 + 2×8 + 3×6 + 4×4 + 5×2)/36 = 70/36 ≈ 1.944。Var(X) = E(X²) − [E(X)]²。E(X²) = (0²×6 + 1²×10 + 2²×8 + 3²×6 + 4²×4 + 5²×2)/36 = 210/36 ≈ 5.833。故 Var(X) = 5.833 − (1.944)² ≈ 2.054。系统的表格构建是 S1 的基础。

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5. Connected Particles on an Inclined Plane | 斜面上的连接体问题

Two particles of masses 3 kg and m kg are connected by a light inextensible string passing over a smooth pulley at the top of a plane inclined at 30° to the horizontal. The 3 kg mass rests on the plane, and m hangs freely. Given that the system is just on the point of moving up the plane (limiting equilibrium), and the coefficient of friction between the 3 kg mass and the plane is 0.4, determine the value of m. Take g ². This M1 scenario elegantly combines Newton’s laws with friction, demanding a clear free‑body diagram and resolution of forces.

质量分别为 3 kg 和 m kg 的两个质点用一根轻质不可伸长的绳子相连，绳子跨过位于斜面顶部的光滑滑轮。斜面与水平面成 30° 角。3 kg 的物体静止在斜面上，m 竖直悬挂。已知系统恰好处于即将沿斜面向上运动的临界状态（极限平衡），且 3 kg 物体与斜面之间的摩擦系数为 0.4，求 m 的值。取 g = 10 m s⁻²。这个 M1 情景将牛顿定律与摩擦力巧妙结合，需要清晰的受力图和力的分解。

For the 3 kg mass on the plane: forces parallel to slope — tension T up the slope, component of weight down slope = 3g sin 30° = 15 N, friction F down slope (since motion impends up, friction opposes). Limiting friction F = μR = 0.4 × 3g cos 30° = 0.4 × 30 × √3/2 = 6√3 ≈ 10.39 N. Equilibrium: T = 15 + 10.39 = 25.39 N. For the hanging mass m: T = mg = 10m. Equate: 10m = 25.39 → m 2.54 kg. The problem reinforces that impending motion determines the direction of friction — a common pitfall.

对斜面上的 3 kg 物体：沿斜面方向的力——张力 T 沿斜面向上，重力的分力沿斜面向下 = 3g sin 30° = 15 N，摩擦力 F 沿斜面向下（因为即将向上运动，摩擦力阻碍相对运动）。极限摩擦力 F = μR = 0.4 × 3g cos 30° = 0.4 × 30 × √3/2 = 6√3 ≈ 10.39 N。平衡方程：T = 15 + 10.39 = 25.39 N。对悬挂质量 m：T = mg = 10m。联立：10m = 25.39 → m ≈ 2.54 kg。本题强调了临界运动趋势决定摩擦力的方向——这是一个常见易错点。

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6. Binomial Distribution in Quality Control | 二项分布在质量控制中的应用

A factory produces light bulbs, and historically 5% are defective. A quality inspector randomly selects 20 bulbs. Find the probability that at most 2 are defective. If the factory implements improvements reducing the defective rate to 2%, what is the probability that a sample of 20 contains no defectives? Use a binomial model and discuss the assumption of independence. This S1 example links probability theory to everyday industrial sampling plans.

某工厂生产灯泡，历史上 5% 的产品有缺陷。质检员随机抽取 20 只灯泡。求至多 2 只有缺陷的概率。若工厂实施改进措施将缺陷率降至 2%，则 20 个样本中无缺陷品的概率是多少？使用二项模型并讨论独立性的假设。这个 S1 实例将概率理论与日常工业抽样方案联系起来。

Let X ~ B(20, 0.05). P(X P(2)=190 × 0.0025 × 0.95¹⁸ ≈ 190×0.0025×0.3972 ≈ 0.1887. Sum ≈ 0.9246. With improved rate p 6676. The binomial model assumes each bulb’s condition is independent, which is reasonable if production is random and defects occur independently. If bulbs came from the same batch with a common flaw, independence fails. Such critical evaluation is often worth a mark in CAIE S1.

设 X ~ B(20, 0.05)。P(X ≤ 2) = P(0) + P(1) + P(2)。P(0)=0.95²⁰ ≈ 0.3585；P(1)=20C1 × 0.05 × 0.95¹⁹ ≈ 20×0.05×0.3774 = 0.3774；P(2)=190 × 0.0025 × 0.95¹⁸ ≈ 190×0.0025×0.3972 ≈ 0.1887。总和 ≈ 0.9246。改进后缺陷率 p=0.02，无缺陷的概率 = 0.98²⁰ ≈ 0.6676。二项模型假设每只灯泡的状况相互独立，若生产随机且缺陷独立出现，则假设合理。若灯泡来自同一批次且存在共同缺陷，独立性将不成立。这类批判性评估在 CAIE S1 中通常值得一分。

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7. Integration as Accumulation — Water Flow | 积分作为累积量——水流问题

Water flows into a tank at a variable rate modelled by r(t) = 4 + 2 sin(πt/6) litres per minute, where t is the time in minutes after 8:00 a.m. Find the total volume of water that enters the tank between 8:00 a.m. and 8:30 a.m. Also determine the time at which the instantaneous flow rate equals the average flow rate over this interval. This P1 exercise showcases the Fundamental Theorem of Calculus in a fluid context, emphasising that definite integration gives net accumulation.

水以可变速率流入水箱，流速模型为 r(t) = 4 + 2 sin(πt/6) 升/分钟，其中 t 为上午 8:00 后的时间（分钟）。求从 8:00 到 8:30 期间流入水箱的总水量。此外，求瞬时流速等于该时间段平均流速的时刻。这个 P1 练习展示了微积分基本定理在流体情景中的应用，强调定积分给出净累积量。

Total volume = ∫₀³⁰ (4 + 2 sin(πt/6)) dt = [4t − (12/π) cos(πt/6)]₀³⁰. At t=30: 120 − (12/π) cos 5π = 120 − (12/π)×(−1) = 120 + 12/π. At t .64 litres. Average rate = total volume / 30 = 4 + 0.8/π ≈ 4.2546 L/min. Solve 4 + 2 sin(πt/6) = 4 + 0.8/π → sin(πt/6) = 0.4/π ≈ 0.1273. t = (6/π) arcsin(0.1273) ≈ 0.244 × 1.91? Actually arcsin(0.1273) ≈ 0.1276 rad, so t ≈ 6×0.1276/π ≈ 0.2438 — min, too small. Let’s recalc: total vol 127.64 L over 30 min, average = 4.2547. So 2 sin(πt/6) = 0.2547 → sin(πt/6) = 0.12735, arcsin ≈ 0.1277 rad, t π ≈ 0.244 min. But there’s another solution in the interval: sin symmetric about π/2, so πt/6 = π − 0.1277 → t . So two times: about 0.24 min and 5.76 min after 8:00 a.m. (The function is periodic, but within 30 min there may be more: solving πt/6 = 2π + 0.1277? That’s >30. Actually period = 12 min, so solutions repeat every 12 min. So also t min, and 5.76 + 12 = 17.76 min, and 24.24 min, 29.76 min. So six times. The question asked “the time” first occurrence.) This blend of integration and equation solving is typical P1 material.

总水量 = ∫₀³⁰ (4 + 2 sin(πt/6)) dt = [4t − (12/π) cos(πt/6)]₀³⁰。在 t=30 时：120 − (12/π) cos 5π = 120 − (12/π)×(−1) = 120 + 12/π。在 t=0 时：0 − (12/π) cos 0 = −12/π。差值为 120 + 24/π ≈ 127.64 升。平均流速 = 总水量 / 30 = 4 + 0.8/π ≈ 4.2546 升/分钟。解方程 4 + 2 sin(πt/6) = 4 + 0.8/π → sin(πt/6) = 0.4/π ≈ 0.1273。在 0 到 30 分钟内，解为 πt/6 = arcsin(0.1273) ≈ 0.1277 弧度，以及 πt/6 = π − 0.1277，因而 t ≈ 0.244 分钟和 5.756 分钟；此外周期为 12 分钟，故还有 12.24、17.76、24.24、29.76 分钟。积分与方程求解的结合是典型的 P1 内容。

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8. Normal Approximation to a Discrete Distribution | 离散分布的正态近似

In a large school, 60% of students travel by bus. On a day when 200 students are sampled, estimate the probability that between 110 and 130 (inclusive) travel by bus, using a normal approximation with continuity correction. Comment on why the approximation is justified. This S1 topic bridges discrete binomial probabilities and the continuous normal curve, reinforcing the Central Limit Theorem.

在一所大学校里，60% 的学生乘公交车上学。某天抽取 200 名学生，用正态近似（带连续性校正）估计乘公交车的人数在 110 至 130（含）之间的概率。评论为何该近似是合理的。这个 S1 主题将离散的二项概率与连续的正态曲线联系起来，强化了中心极限定理。

If X ~ B(200, 0.6), then μ = 200×0.6 = 120, σ² = 200×0.6×0.4 = 48, σ = √48 ≈ 6.928. Use Y ~ N(120, 48). With continuity correction: P(110 ≤ X ≤ 130) ≈ P(109.5 < Y < 130.5). Standardise: z₁ = (109.5 − 120)/6.928 ≈ −1.515, z₂ Φ(−1.515) = 1 − 0.935 = 0.065. Probability ≈ 0.935 − 0.065 = 0.870. The sample is large (n=200) and p is not extreme, so both np and nq exceed 5, satisfying the rule of thumb for normal approximation.

若 X ~ B(200, 0.6)，则 μ = 200×0.6 = 120，σ² = 200×0.6×0.4 = 48，σ = √48 ≈ 6.928。用 Y ~ N(120, 48)。带连续性校正：P(110 ≤ X ≤ 130) ≈ P(109.5 < Y < 130.5)。标准化：z₁ = (109.5 − 120)/6.928 ≈ −1.515，z₂ = (130.5 − 120)/6.928 ≈ 1.515。查正态分布表，Φ(1.515) ≈ 0.935，Φ(−1.515) = 1 − 0.935 = 0.065。概率 ≈ 0.935 − 0.065 = 0.870。样本容量大（n=200）且 p 不极端，因此 np 和 nq 均大于 5，满足正态近似的经验法则。

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9. Kinematics with Variable Acceleration | 变加速度运动学

A particle moves along a straight line so that its velocity v m s⁻¹ at time t seconds is given by v = 3t² − 12t + 9 for t d in the first 5 seconds. Distinguish carefully between displacement and distance when the particle changes direction. This P1/M1 crossover question sharpens the concept of integrating the absolute value of velocity.

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