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

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

In Year 12 Edexcel Mathematics, the ability to apply theoretical knowledge to real-world scenarios is essential for success, especially in the applied modules of Mechanics and Statistics. Case study drills provide an opportunity to blend pure mathematical techniques with contextual problem‑solving, building the skills needed for exam‑style modelling questions. This article presents a series of practical case studies, each targeting a key topic from the syllabus, and demonstrates how to break down a problem, formulate equations, solve them accurately, and interpret results in context.

在12年级Edexcel数学中，将理论知识应用于现实场景的能力至关重要，尤其在力学和统计学的应用模块中。案例分析练习提供了一个将纯数学技巧与情境问题解决相结合的机会，培养学生应对考试中建模题型所需的技能。本文通过一系列实战案例，每个案例针对教学大纲中的一个关键主题，展示如何分解问题、建立方程、精确求解并在情境中解释结果。

1. Mechanics: Projectile Motion Analysis | 力学：抛体运动分析

A particle is projected from a point on horizontal ground with speed 20 m s⁻¹ at an angle of 30° to the horizontal. Neglecting air resistance, we determine the time of flight, the horizontal range, and the maximum height reached. The initial velocity components are resolved: horizontally uₓ = 20 cos 30° = 17.32 m s⁻¹, and vertically uᵧ = 20 sin 30° = 10 m s⁻¹. Using the constant acceleration (suvat) equations, the time to reach the highest point is found from vᵧ = uᵧ + a t, where vᵧ = 0 at the apex and a = –9.8 m s⁻², giving t = 10 / 9.8 ≈ 1.02 s. The total time of flight is twice this, 2.04 s. The horizontal range is then R = uₓ × total time ≈ 17.32 × 2.04 ≈ 35.3 m. The maximum height H is calculated using s = uᵧ t + ½ a t² with t = 1.02 s, giving H ≈ 10 × 1.02 – 4.9 × 1.04 ≈ 10.2 – 5.1 = 5.1 m. These results must be stated with appropriate units and interpreted.

一个质点从水平地面上的一点以20 m s⁻¹的初速度与水平成30°角抛出。忽略空气阻力，我们求飞行时间、水平射程和最大高度。将初速度分解：水平分量 uₓ = 20 cos 30° = 17.32 m s⁻¹，竖直分量 uᵧ = 20 sin 30° = 10 m s⁻¹。利用匀加速运动（suvat）方程，到达最高点的时间由 vᵧ = uᵧ + a t 求出，最高点处 vᵧ = 0，a = –9.8 m s⁻²，得 t = 10 / 9.8 ≈ 1.02 s。总飞行时间为两倍，即2.04 s。水平射程 R = uₓ × 总时间 ≈ 17.32 × 2.04 ≈ 35.3 m。最大高度 H 用 s = uᵧ t + ½ a t² 计算，t = 1.02 s，得 H ≈ 10 × 1.02 – 4.9 × 1.04 ≈ 10.2 – 5.1 = 5.1 m。结果需标明适当单位并加以解释。

2. Statistics: Regression and Data Fitting | 统计学：回归与数据拟合

A biologist records the masses (kg) and lengths (cm) of 5 animals. The data are: (2.1, 30), (3.0, 42), (4.2, 55), (5.1, 68), (6.3, 80). We model the relationship using linear regression. First, calculate the means: x̄ = 4.14 kg, ȳ = 55 cm. The sums required are Sₓₓ = Σ(xᵢ – x̄)² = (2.1–4.14)² + … ≈ 10.632, and Sₓᵧ = Σ(xᵢ – x̄)(yᵢ – ȳ) ≈ 107.7. The slope b = Sₓᵧ / Sₓₓ ≈ 10.13, and the intercept a = ȳ – b x̄ ≈ 55 – 10.13 × 4.14 ≈ 13.0. The equation is y = 13.0 + 10.13x. To predict the length for a mass of 4.0 kg, substitute x = 4.0: y = 13.0 + 10.13 × 4.0 = 53.5 cm. The model’s validity should be discussed by examining the residual for this point or checking if extrapolation occurs.

一位生物学家记录了5只动物的质量（kg）和体长（cm）。数据为：(2.1, 30), (3.0, 42), (4.2, 55), (5.1, 68), (6.3, 80)。我们使用线性回归建模。首先计算均值：x̄ = 4.14 kg，ȳ = 55 cm。所需求和项：Sₓₓ = Σ(xᵢ – x̄)² = (2.1–4.14)² + … ≈ 10.632，Sₓᵧ = Σ(xᵢ – x̄)(yᵢ – ȳ) ≈ 107.7。斜率 b = Sₓᵧ / Sₓₓ ≈ 10.13，截距 a = ȳ – b x̄ ≈ 55 – 10.13 × 4.14 ≈ 13.0。回归方程为 y = 13.0 + 10.13x。要预测质量为4.0 kg时的体长，代入 x = 4.0：y = 13.0 + 10.13 × 4.0 = 53.5 cm。需检验模型的合理性，例如检查该点的残差或判断是否涉及外推。

3. Pure Mathematics: Optimisation Problem | 纯数学：优化问题

A rectangular enclosure is to be built against an existing wall, using 60 m of fencing for the other three sides. We find the dimensions that maximise the enclosed area. Let x be the side perpendicular to the wall, and y the side parallel to the wall. The constraint is 2x + y = 60, so y = 60 – 2x. The area A = x y = x(60 – 2x) = 60x – 2x². Differentiate: dA/dx = 60 – 4x. Set to zero: 60 – 4x = 0 ⇒ x = 15. Second derivative d²A/dx² = –4 < 0, confirming a maximum. Then y = 60 – 2(15) = 30. The maximum area is 15 × 30 = 450 m². This is a typical pure‑math modelling question where calculus is applied to a geometric constraint.

一个矩形围栏靠着一面已有的墙修建，另外三边使用60 m的围栏。求使围栏面积最大的尺寸。设 x 为垂直于墙的边，y 为平行于墙的边。约束条件为 2x + y = 60，所以 y = 60 – 2x。面积 A = x y = x(60 – 2x) = 60x – 2x²。求导：dA/dx = 60 – 4x。令导数为零：60 – 4x = 0 ⇒ x = 15。二阶导数 d²A/dx² = –4 < 0，确认是最大值。于是 y = 60 – 2(15) = 30。最大面积为 15 × 30 = 450 m²。这是一个典型的纯数学建模问题，将微积分应用于几何约束。

4. Mechanics: Friction on an Inclined Plane | 力学：斜面上的摩擦力

A block of mass 5 kg rests on a rough plane inclined at 20° to the horizontal. The coefficient of static friction is 0.4. We determine whether the block is in equilibrium or slides down. Resolving forces parallel to the plane: the component of weight down the slope is mg sin 20° = 5 × 9.8 × sin 20° ≈ 16.76 N. The normal reaction is R = mg cos 20° ≈ 46.0 N. The maximum static friction is F_max = μ R = 0.4 × 46.0 = 18.4 N. Since 16.76 < 18.4, the block remains stationary. In a second part, if a horizontal force P is applied to keep it from sliding when the plane angle increases, we can set up equilibrium equations to find P. Such multi‑step analysis tests fluency in resolving forces and applying friction laws.

一质量为5 kg的物块静止在粗糙斜面上，斜面与水平成20°角，静摩擦系数为0.4。判断物块是否平衡还是会下滑。分解平行于斜面的力：重力沿斜面的分量为 mg sin 20° = 5 × 9.8 × sin 20° ≈ 16.76 N。法向反作用力 R = mg cos 20° ≈ 46.0 N。最大静摩擦力 F_max = μ R = 0.4 × 46.0 = 18.4 N。由于 16.76 < 18.4，物块保持静止。第二部分，如果斜面角度增大，需施加水平力 P 防止其下滑，可建立平衡方程求解 P。这种多步骤分析考察力的分解和摩擦定律的熟练应用。

5. Statistics: Normal Distribution Application | 统计学：正态分布应用

The mass of a certain fruit is normally distributed with mean μ = 200 g and standard deviation σ = 15 g. We find the probability that a randomly selected fruit weighs between 185 g and 220 g. Standardise using Z = (X – μ)/σ. For X = 185, Z = (185 – 200)/15 = –1.00; for X = 220, Z = (220 – 200)/15 ≈ 1.33. Using standard normal tables: P(Z < 1.33) = 0.9082, and P(Z < –1.00) = 0.1587. The required probability is 0.9082 – 0.1587 = 0.7495. In a sample of 10 fruits, the probability that at least 8 weigh more than 185 g can be modelled using the binomial distribution with p ≈ 0.8413 (the probability X > 185). This crossover between normal and binomial distributions is common in exam case studies.

某种水果的质量服从正态分布，均值 μ = 200 g，标准差 σ = 15 g。求随机选取一个水果质量介于185 g与220 g之间的概率。标准化使用 Z = (X – μ)/σ。当 X = 185，Z = (185 – 200)/15 = –1.00；当 X = 220，Z = (220 – 200)/15 ≈ 1.33。查标准正态分布表：P(Z < 1.33) = 0.9082，P(Z < –1.00) = 0.1587。所求概率为 0.9082 – 0.1587 = 0.7495。在10个水果的样本中，至少8个质量大于185 g的概率可用二项分布建模，其中单次成功概率 p ≈ 0.8413（即 X > 185 的概率）。这种正态与二项分布的交叉在考试案例分析中很常见。

6. Pure Mathematics: Exponential Growth and Decay | 纯数学：指数增长与衰减

A bacterial culture grows according to the model N = N₀ e^{kt}, where N is the number of bacteria at time t hours. Initially there are 500 bacteria, and after 3 hours there are 2000. We find the growth rate k and the time taken to reach 10 000. Using N₀ = 500, after 3 hours: 2000 = 500 e^{3k} ⇒ e^{3k} = 4 ⇒ 3k = ln 4 ⇒ k = (ln 4)/3 ≈ 0.4621. The model becomes N = 500 e^{0.4621 t}. Set N = 10 000: 10 000 = 500 e^{0.4621 t} ⇒ e^{0.4621 t} = 20 ⇒ 0.4621 t = ln 20 ⇒ t = (ln 20)/0.4621 ≈ 6.48 hours. Contextual interpretation: the population doubles approximately every 1.5 hours, and the model assumes unlimited growth, which may not be realistic in the long term. This integrates logs and exponentials with modelling assumptions.

某细菌培养按模型 N = N₀ e^{kt} 增长，其中 N 是在 t 小时时的细菌数量。初始有500个细菌，3小时后有2000个。求增长率 k 和达到10 000所需的时间。使用 N₀ = 500，3小时后：2000 = 500 e^{3k} ⇒ e^{3k} = 4 ⇒ 3k = ln 4 ⇒ k = (ln 4)/3 ≈ 0.4621。模型变为 N = 500 e^{0.4621 t}。令 N = 10 000：10 000 = 500 e^{0.4621 t} ⇒ e^{0.4621 t} = 20 ⇒ 0.4621 t = ln 20 ⇒ t = (ln 20)/0.4621 ≈ 6.48 小时。情境解释：数量约每1.5小时翻倍，模型假设无限增长，长期看可能不现实。这结合了对数、指数与建模假设。

7. Mechanics: Moments and Equilibrium | 力学：力矩与平衡

A uniform beam AB of length 6 m and mass 20 kg rests horizontally on two supports at C and D, where AC = 1 m and DB = 2 m. A mass of 30 kg is placed at A. Find the reactions at the supports. Taking moments about C eliminates the reaction at C. Weight of beam acts at its centre 3 m from A, so distance from C is 2 m. The 30 kg weight is 1 m from C (at A, which is 1 m left of C? Wait, AC = 1 m, so A is 1 m to the left of C. Taking clockwise as positive, moment about C: 30g × 1 (anticlockwise, so negative) + 20g × 2 (clockwise) – R_D × 3 = 0, because CD = length – (AC + DB) = 6 – 1 – 2 = 3 m. So –30g + 40g – 3R_D = 0 ⇒ 10g = 3R_D ⇒ R_D = (10 × 9.8)/3 ≈ 32.67 N. Then vertically: R_C + R_D = 50g = 490 N, so R_C = 490 – 32.67 = 457.33 N. This problem tests the principle of moments and the ability to label distances accurately.

一根长6 m、质量20 kg的均质梁AB水平放置在C和D两个支点上，其中AC = 1 m，DB = 2 m。在A点放置一个30 kg的重物。求两支点的反力。对C点取矩可消去C处反力。梁的重力作用在中心，距A端3 m，因此距C点2 m。30 kg重物在A点，在C点左侧1 m。取顺时针为正，对C点的力矩：30g × 1（逆时针，负值）+ 20g × 2（顺时针）– R_D × 3 = 0，因为CD = 长度 – (AC + DB) = 6 – 1 – 2 = 3 m。所以 –30g + 40g – 3R_D = 0 ⇒ 10g = 3R_D ⇒ R_D = (10 × 9.8)/3 ≈ 32.67 N。竖直平衡：R_C + R_D = 50g = 490 N，因此 R_C = 490 – 32.67 = 457.33 N。此题考察力矩原理及准确标注距离的能力。

8. Statistics: Hypothesis Testing | 统计学：假设检验

A manufacturer claims that at most 5% of its light bulbs are defective. A customer suspects the proportion is higher and tests a random sample of 20 bulbs, finding 3 defectives. We conduct a hypothesis test at the 5% significance level. Let p be the true proportion of defectives. H₀: p = 0.05, H₁: p > 0.05. Under H₀, X ~ B(20, 0.05). We require P(X ≥ 3) = 1 – P(X ≤ 2). Using tables or calculator: P(X ≤ 2) = 0.9245 (approximately, using binomial cumulative distribution). Thus P(X ≥ 3) = 1 – 0.9245 = 0.0755. Since 0.0755 > 0.05, we do not have sufficient evidence to reject H₀ at the 5% level. The conclusion is that the sample does not provide strong enough evidence to dispute the manufacturer’s claim. This drill reinforces the structure of hypothesis testing: hypotheses, test statistic, critical region or p‑value, and conclusion in context.

某制造商声称其灯泡的次品率不超过5%。一位顾客怀疑比例更高，随机抽检了20个灯泡，发现3个次品。我们在5%显著性水平下进行假设检验。设 p 为真实次品率。H₀: p = 0.05，H₁: p > 0.05。在H₀下，X ~ B(20, 0.05)。需求 P(X ≥ 3) = 1 – P(X ≤ 2)。查表或用计算器：P(X ≤ 2) ≈ 0.9245（使用二项累积分布）。因此 P(X ≥ 3) = 1 – 0.9245 = 0.0755。由于 0.0755 > 0.05，在5%水平下我们没有充分证据拒绝H₀。结论是样本未提供足够强的证据质疑制造商的声称。此练习强化假设检验的结构：假设、检验统计量、临界区域或p值，以及情境下的结论。

9. Pure Mathematics: Modelling with Trigonometric Functions | 纯数学：三角函数建模

The depth of water in a harbour is modelled by D(t) = 5 + 3 sin(πt/6), where D is in metres and t is the time in hours after midnight. Determine the times when the depth is exactly 6.5 m between 0 and 24. Set 5 + 3 sin(πt/6) = 6.5 ⇒ 3 sin(πt/6) = 1.5 ⇒ sin(πt/6) = 0.5. Principal solution: πt/6 = π/6 ⇒ t = 1. Also, sin(πt/6) = 0.5 has a second solution in 0 ≤ t < 24: πt/6 = π – π/6 = 5π/6 ⇒ t = 5. The next cycle gives πt/6 = 2π + π/6 = 13π/6 ⇒ t = 13, and 2π + 5π/6 = 17π/6 ⇒ t = 17. Thus the depth is 6.5 m at 1 am, 5 am, 1 pm, and 5 pm. This case study links pure trigonometric equations to a periodic context, requiring careful handling of the general solution and the domain.

某海港的水深由模型 D(t) = 5 + 3 sin(πt/6) 给出，其中 D 以米为单位，t 是午夜后的小时数。求在0到24小时之间水深恰好为6.5 m的时刻。令 5 + 3 sin(πt/6) = 6.5 ⇒ 3 sin(πt/6) = 1.5 ⇒ sin(πt/6) = 0.5。主解：πt/6 = π/6 ⇒ t = 1。同时，在 0 ≤ t < 24 内 sin(πt/6) = 0.5 的第二个解为 πt/6 = π – π/6 = 5π/6 ⇒ t = 5。下一周期给出 πt/6 = 2π + π/6 = 13π/6 ⇒ t = 13，以及 2π + 5π/6 = 17π/6 ⇒ t = 17。因此水深为6.5 m的时刻为凌晨1点、5点、下午1点和5点。此案例将纯三角方程与周期情境相联系，需要仔细处理通解和定义域。

10. Mechanics: Connected Particles and Pulleys | 力学：连接体与滑轮

Two particles of masses 3 kg and 5 kg are connected by a light inextensible string passing over a smooth pulley. The 5 kg mass hangs vertically while the 3 kg mass rests on a rough horizontal table with coefficient of friction 0.2. The system is released from rest. We find the acceleration and the tension in the string. For the 5 kg mass: 5g – T = 5a. For the 3 kg mass: T – F = 3a, where friction F = μR = 0.2 × 3g = 0.6g. Adding the equations: 5g – 0.6g = 8a ⇒ 4.4g = 8a ⇒ a = (4.4 × 9.8)/8 = 5.39 m s⁻². Then T = 5g – 5a = 49 – 26.95 = 22.05 N. The motion must be interpreted: both particles accelerate in the direction of the heavier mass. This drill exemplifies the systematic approach of drawing free‑body diagrams, writing equations of motion, and solving simultaneously.

两个质量分别为3 kg和5 kg的质点通过一轻质不可伸长的绳子相连，绳子跨过一个光滑滑轮。5 kg的质量竖直悬挂，3 kg的质量静止在粗糙水平桌面上，摩擦系数为0.2。系统由静止释放。我们求加速度和绳中的张力。对5 kg质量：5g – T = 5a。对3 kg质量：T – F = 3a，其中摩擦力 F = μR = 0.2 × 3g = 0.6g。两式相加：5g – 0.6g = 8a ⇒ 4.4g = 8a ⇒ a = (4.4 × 9.8)/8 = 5.39 m s⁻²。然后 T = 5g – 5a = 49 – 26.95 = 22.05 N。需解释运动：两质点沿较重的质量方向加速。此练习示范了画受力图、列出运动方程并联立求解的系统性方法。

11. Statistics: Bivariate Data and Correlation | 统计学：双变量数据与相关性

An economist collects data on advertising spend (x, in £1000s) and sales revenue (y, in £1000s) for 8 months. The summary statistics are: Σx = 320, Σy = 480, Σx² = 14 200, Σy² = 30 400, Σxy = 21 100. We calculate the product moment correlation coefficient r to assess linear association. r = (S_{xy}) / √(S_{xx} S_{yy}), where S_{xy} = Σxy – (Σx Σy)/n. n = 8. S_{xy} = 21 100 – (320 × 480)/8 = 21 100 – 19 200 = 1900. S_{xx} = 14 200 – (320²)/8 = 14 200 – 12 800 = 1400. S_{yy} = 30 400 – (480²)/8 = 30 400 – 28 800 = 1600. Hence r = 1900 / √(1400 × 1600) = 1900 / √(2 240 000) = 1900 / 1496.66 ≈ 1.27? Wait, 1496.66 gives r > 1, which is impossible. Let’s recalculate: 1400 × 1600 = 2,240,000, square root is about 1496.66. Then 1900 / 1496.66 = 1.269, which is not possible for r. So there must be an arithmetic slip. Suppose we revise Σxy to a more reasonable figure, say 20 500, then S_{xy} = 20 500 – 19 200 = 1300, r = 1300 / 1496.66 = 0.868, which is plausible. Indeed, the calculation serves to remind us to check the feasibility of r (between –1 and 1). A high positive r would indicate a strong positive linear correlation between advertising and sales. This case study practises the formula and interpretation of correlation.

一位经济学家收集了8个月的广告支出（x，千英镑）和销售收入（y，千英镑）数据。汇总统计量为：Σx = 320，Σy = 480，Σx² = 14 200，Σy² = 30 400，Σxy = 21 100。我们计算积矩相关系数 r 来评估线性关联。r = (S_{xy}) / √(S_{xx} S_{yy})，其中 S_{xy} = Σxy – (Σx Σy)/n。n = 8。S_{xy} = 21 100 – (320 × 480)/8 = 21 100 – 19 200 = 1900。S_{xx} = 14 200 – (320²)/8 = 14 200 – 12 800 = 1400。S_{yy} = 30 400 – (480²)/8 = 30 400 – 28 800 = 1600。于是 r = 1900 / √(1400 × 1600) = 1900 / √(2 240 000) = 1900 / 1496.66 ≈ 1.27？1496.66使得 r > 1，这不可能。我们修正一下，假设 Σxy 为更合理值，例如 20 500，则 S_{xy} = 20 500 – 19 200 = 1300，r = 1300 / 1496.66 = 0.868，这符合常理。此计算提醒我们要检查 r 的可行性（介于 –1 与 1 之间）。较高的正 r 表明广告与销售之间存在强正线性相关。本案例练习了相关公式和解释。

12. Integrative Modelling and Exam Strategies | 综合建模与应试策略

Real Edexcel exam questions often combine multiple topics into a single case study. For instance, a problem might start with collecting data, fitting a regression line, then using a mechanics model to predict an outcome. To succeed, students should practise identifying the relevant pure technique in a wordy context, extracting key numbers, drawing diagrams, and clearly stating any assumptions (e.g., no air resistance, constant acceleration, or normal distribution). Checking the reasonableness of answers (like a correlation coefficient within [–1, 1] or a probability between 0 and 1) is vital. Time management during exams can be improved by regular timed drills on case studies.

真正的Edexcel考试题常将多个主题综合到单个案例中。例如，一道题可能从收集数据、拟合回归线开始，然后使用力学模型预测结果。要取得成功，学生应练习在文字冗长的情境中辨认相关的纯数学技巧，提取关键数字，画出示意图，并清晰陈述任何假设（如无空气阻力、匀加速、正态分布）。检查答案的合理性（如相关系数在[–1, 1]内，或概率介于0和1之间）至关重要。考试中的时间管理可通过定期限时案例分析训练来提高。

Published by TutorHao | Maths Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

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