Year 13 AQA Maths: Case Study Practical Exercises | Year 13 AQA 数学：案例分析实战演练

📚 Year 13 AQA Maths: Case Study Practical Exercises | Year 13 AQA 数学：案例分析实战演练

Year 13 AQA Mathematics often assesses your ability to apply pure, statistical and mechanical concepts to unfamiliar real‑world scenarios. Working through full case studies – from modelling a cooling corpse to optimising a box volume or testing a new drug – builds the fluency and confidence needed to tackle high‑mark exam questions. In this article, we walk through seven complete worked examples with detailed reasoning and exam‑style commentary, followed by targeted tips for the AQA papers.

Year 13 AQA 数学常常考查你将纯数、统计和力学概念应用到陌生实际情境中的能力。通过完整的案例――从模拟尸体的降温过程，到优化盒子体积，再到检验新药的有效性――你可以建立起应对高分考题所需的熟练度与信心。本文带你逐一拆解七个完整的实战案例，配有细节推理和考试风格的点评，最后还会给出针对 AQA 试卷的备考建议。

1. Introduction: Why Case Studies Matter | 引言：为什么案例研究重要

Case‑study questions on AQA papers blend multiple topics into a single narrative. You might need to formulate a differential equation, solve it analytically, then interpret the result in context, or combine a binomial hypothesis test with a loss function. The ability to translate a worded problem into clean mathematics is what examiners reward most. Practising whole‑case narratives helps you spot the underlying structure and avoid common pitfalls, such as incorrect units, forgotten domain restrictions, or misapplied significance levels.

AQA 试卷中的案例分析题会将多个知识点编织在一个情景中。你可能需要先建立微分方程、再解析求解，然后结合背景解释结果，或者将二项假设检验与损失函数结合起来。将文字问题转化为清晰的数学表达正是阅卷老师最看重的能力。演练完整的案例叙述能帮助你看出问题的底层结构，躲避常见陷阱，如单位错误、遗漏定义域限制或误判显著性水平。

2. Case 1: Maximising the Volume of an Open Box | 案例 1：最大化无盖盒子的体积

A rectangular sheet of cardboard 30 cm long and 20 cm wide has a square of side x cm cut from each corner. The flaps are folded up to form an open box. We need to find the value of x that gives the greatest possible volume and confirm it is a maximum.

一块长 30 cm、宽 20 cm 的矩形纸板，从每个角切去边长为 x cm 的正方形，再将四边折起做成一个无盖盒子。我们需要求出使盒子体积最大的 x 值，并确认这是一个极大值。

The volume V in cm³ is given by: V = x(30 − 2x)(20 − 2x). Expanding gives V = 4x³ − 100x² + 600x. Differentiating: dV/dx = 12x² − 200x + 600. For stationary points, set dV/dx = 0 and divide by 4: 3x² − 50x + 150 = 0. Solving this quadratic yields x = (50 ± √(2500 − 1800)) / 6 = (50 ± √700) / 6. Since √700 ≈ 26.46, the two roots are approximately 12.74 and 3.92. The physical constraint 20 − 2x > 0 means x < 10, so only x ≈ 3.92 cm is feasible. The second derivative d²V/dx² = 24x − 200 evaluates to −105.9 at x ≈ 3.92, confirming a maximum. The maximum volume is V ≈ 4(3.92)³ − 100(3.92)² + 600(3.92) ≈ 1056 cm³.

盒子体积 V（单位 cm³）为：V = x(30 − 2x)(20 − 2x)。展开得 V = 4x³ − 100x² + 600x。求导：dV/dx = 12x² − 200x + 600。令导数为零并除以 4：3x² − 50x + 150 = 0。解二次方程得 x = (50 ± √(2500 − 1800)) / 6 = (50 ± √700) / 6。√700 ≈ 26.46，两根约为 12.74 和 3.92。但物理限制 20 − 2x > 0，即 x < 10，故只有 x ≈ 3.92 cm 成立。二阶导数 d²V/dx² = 24x − 200，在 x ≈ 3.92 时值为 −105.9，确认为极大值。最大体积 V ≈ 4(3.92)³ − 100(3.92)² + 600(3.92) ≈ 1056 cm³。

3. Case 2: Newton’s Law of Cooling and Time of Death | 案例 2：牛顿冷却定律与死亡时间

When a body cools in a room with constant ambient temperature Tₐ, the rate of change of the body’s temperature T is proportional to (T − Tₐ). This gives the differential equation dT/dt = −k(T − Tₐ). Solving by separation of variables yields T(t) = Tₐ + (T₀ − Tₐ)e⁻ᵏᵗ, where T₀ is the temperature at t = 0. Forensic investigators use this model to estimate the time since death.

当尸体在恒温 Tₐ 的房间内冷却时，其温度 T 的变化率与温差 (T − Tₐ) 成正比。这给出微分方程 dT/dt = −k(T − Tₐ)。分离变量求解后得 T(t) = Tₐ + (T₀ − Tₐ)e⁻ᵏᵗ，其中 T₀ 为 t = 0 时的温度。法医利用该模型推断死亡时间。

Example: A body is discovered in a room held at 20 °C. Normal body temperature is 37 °C. Thirty minutes after discovery, the body’s temperature is measured at 35 °C. At discovery, its temperature was 37 °C? Wait – the typical scenario: at the moment of death, T₀ = 37 °C. The clock starts then. Suppose the body is found at some unknown time after death, and its temperature at discovery is, say, 30 °C (to illustrate). Let us construct a clean statement: The body is found at time t = t₁ with temperature 25 °C. But to find k we need a second reading. Often the problem gives the temperature at a known interval after discovery. Let us use: A body is found at midnight; its temperature is 30 °C. One hour later, the temperature has dropped to 28 °C. The ambient temperature is constant at 20 °C. Estimate how long before midnight death occurred, assuming normal body temperature at death was 37 °C. Then the measurements are taken relative to the unknown death time.

例题：在一间保持 20 °C 的房间内发现一具遗体。假设死亡时正常体温为 37 °C。午夜发现遗体时，其温度为 30 °C；一小时后降至 28 °C。试估算死亡时间（午夜前多少小时）。

Let t = 0 be the moment of death. Then T(0) = 37. Let t = t₁ be midnight, with T(t₁) = 30. The model is T(t) = 20 + 17e⁻ᵏᵗ. Also, at t = t₁ + 1, T = 28, so 28 = 20 + 17e⁻ᵏ⁽ᵗ₁⁺¹⁾ → 8 = 17e⁻ᵏᵗ₁ e⁻ᵏ. We also have 30 = 20 + 17e⁻ᵏᵗ₁ → 10 = 17e⁻ᵏᵗ₁. Dividing the two equations: 8/10 = e⁻ᵏ → e⁻ᵏ = 0.8 → k = −ln 0.8. Substituting into 10 = 17e⁻ᵏᵗ₁ gives e⁻ᵏᵗ₁ = 10/17. Hence t₁ = −(ln(10/17))/k. With k ≈ 0.2231, t₁ ≈ (ln 1.7)/0.2231 ≈ 2.38 hours. Death therefore occurred roughly 2 hours 23 minutes before midnight.

设 t = 0 为死亡时刻，则 T(0) = 37。设 t = t₁ 为午夜，此时 T(t₁) = 30。模型为 T(t) = 20 + 17e⁻ᵏᵗ。t = t₁ + 1 时 T = 28，故 28 = 20 + 17e⁻ᵏ⁽ᵗ₁⁺¹⁾ → 8 = 17e⁻ᵏᵗ₁ e⁻ᵏ。同时由 30 = 20 + 17e⁻ᵏᵗ₁ 得 10 = 17e⁻ᵏᵗ₁。两式相除：8/10 = e⁻ᵏ → e⁻ᵏ = 0.8 → k = −ln 0.8。代入 10 = 17e⁻ᵏᵗ₁ 有 e⁻ᵏᵗ₁ = 10/17，故 t₁ = −(ln(10/17))/k。k ≈ 0.2231，t₁ ≈ (ln 1.7)/0.2231 ≈ 2.38 小时。因此死亡时间大约在午夜前 2 小时 23 分。

4. Case 3: Logistic Growth Model for a Fish Population | 案例 3：鱼类种群的逻辑斯谛增长模型

The rate of change of a fish population P (in thousands) in a lake can be modelled by the logistic differential equation dP/dt = r P (1 − P/K), where r is the intrinsic growth rate and K is the carrying capacity. If r = 0.8 yr⁻¹, K = 1000 (meaning 1 000 000 fish) and P(0) = 100, we can solve for P(t) and predict the time needed to reach a sustainable harvest size.

一个湖泊中某种鱼类的种群数量 P（以千条为单位）的变化率可用逻辑斯谛微分方程 dP/dt = r P (1 − P/K) 建模，其中 r 为内禀增长率，K 为环境容纳量。若 r = 0.8 yr⁻¹，K = 1000（即 1,000,000 条鱼），且 P(0) = 100，我们可以解出 P(t) 并预测达到可持续捕捞规模所需的时间。

Separating variables: ∫ 1/[P(1 − P/K)] dP = ∫ r dt. Partial fractions give 1/[P(1 − P/K)] = 1/P + (1/K)/(1 − P/K). Integrating, we obtain ln|P| − ln|1 − P/K| = r t + C. Using P(0) = P₀ gives C = ln(P₀/(1 − P₀/K)). Hence ln[P/(1 − P/K)] = r t + ln(P₀/(1 − P₀/K)). Exponentiating and rearranging yields the explicit solution: P(t) = K / [1 + ((K − P₀)/P₀) e⁻ʳᵗ

Published by TutorHao | Year 13 Mathematics Revision Series | aleveler.com

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