📚 SQA Advanced Higher Mathematics: Case Study Practical Drills | SQA 进阶数学：案例分析实战演练
Case studies in SQA Advanced Higher Mathematics challenge you to apply pure mathematical techniques to real‑world contexts. This article walks you through four complete practical drills, covering optimisation, differential equations, matrix diagonalisation and complex numbers, so you can sharpen your problem‑solving skills and exam technique.
SQA 进阶数学中的案例分析要求你把纯数学技术应用到真实情境中。本文将通过四个完整的实战演练，涵盖优化、微分方程、矩阵对角化和复数等主题，帮助你提升解题能力与考试技巧。
1. Overview of SQA AH Exam Case Studies | 考试案例分析概述
In the SQA Advanced Higher exam, extended‑response questions often come in the form of case studies. You are expected to translate a written scenario into mathematical language, select the appropriate model, perform calculations and interpret your results in context.
在 SQA 进阶数学考试中，扩展简答题常以案例分析的形式出现。你需要把文字情境转化为数学语言，选择合适的模型，进行计算并在具体情境中解释结果。
2. Case 1: Optimising a Cylindrical Can – Problem | 案例一：优化圆柱罐——问题陈述
A manufacturer wants to produce cylindrical tin cans with a fixed total surface area of 600 cm². The can has a top and a bottom. Find the radius r cm and height h cm that maximise the volume V cm³, and hence determine the maximum volume.
某制造商希望生产固定总表面积为 600 cm² 的圆柱形罐头，罐有顶盖和底盖。求使得体积 V cm³ 最大的半径 r cm 和高度 h cm，并计算最大体积。
3. Case 1: Solution Using Differentiation | 案例一：使用微分求解
The volume and surface area are given by V = πr²h and S = 2πr² + 2πrh. Since S = 600, we have 2πr² + 2πrh = 600, so h = (300 − πr²) / (πr).
体积和表面积公式为 V = πr²h 与 S = 2πr² + 2πrh。由 S = 600 得 2πr² + 2πrh = 600，因此 h = (300 − πr²) / (πr)。
Substitute h into V: V(r) = πr² × (300 − πr²)/(πr) = 300r − πr³. Differentiate with respect to r: dV/dr = 300 − 3πr². Set dV/dr = 0 to find the stationary point: 300 − 3πr² = 0 ⇒ r² = 100/π ⇒ r = 10/√π ≈ 5.64 cm.
将 h 代入 V：V(r) = πr² × (300 − πr²)/(πr) = 300r − πr³。对 r 求导：dV/dr = 300 − 3πr²。令 dV/dr = 0 求驻点：300 − 3πr² = 0 ⇒ r² = 100/π ⇒ r = 10/√π ≈ 5.64 cm。
Check the second derivative: d²V/dr² = −6πr < 0 for r > 0, confirming a maximum. Then h = (300 − π × 100/π) / (π × 10/√π) = 200 / (10√π) = 20/√π ≈ 11.28 cm. The maximum volume is V = π (100/π) (20/√π) = 2000/√π ≈ 1128.4 cm³.
检查二阶导数：d²V/dr² = −6πr < 0（r > 0），确认为极大值。可得 h = (300 − π × 100/π) / (π × 10/√π) = 200 / (10√π) = 20/√π ≈ 11.28 cm。最大体积为 V = π (100/π) (20/√π) = 2000/√π ≈ 1128.4 cm³。
4. Case 2: Newton’s Law of Cooling – Setting Up | 案例二：牛顿冷却定律——建立模型
A hot metal rod is removed from an oven at 180°C and placed in a room maintained at 20°C. According to Newton’s Law of Cooling, the rate of change of the rod’s temperature T(t) satisfies dT/dt = −k (T − 20), where k > 0 is a constant and t is measured in minutes.
一根热金属棒从 180°C 的烤箱中取出，放置在保持 20°C 的房间中。根据牛顿冷却定律，金属棒温度 T(t) 的变化率满足 dT/dt = −k (T − 20)，其中 k > 0 为常数，t 以分钟为单位。
After 4 minutes the rod’s temperature has fallen to 100°C. We need to find the value of k and predict when the rod will reach 30°C.
4 分钟后金属棒温度降至 100°C。我们需要求出 k 的值，并预测金属棒何时降至 30°C。
5. Case 2: Solving the Differential Equation | 案例二：求解微分方程
Separate variables: ∫ 1/(T − 20) dT = ∫ −k dt. This gives ln|T − 20| = −kt + C. Exponentiating gives T − 20 = Ae⁻ᵏᵗ, with A = e^C. Using the initial condition T(0) = 180: 180 − 20 = A ⇒ A = 160. So T(t) = 20 + 160e⁻ᵏᵗ.
分离变量：∫ 1/(T − 20) dT = ∫ −k dt。得到 ln|T − 20| = −kt + C。取指数得 T − 20 = Ae⁻ᵏᵗ，其中 A = e^C。利用初始条件 T(0) = 180：180 − 20 = A ⇒ A = 160。因此 T(t) = 20 + 160e⁻ᵏᵗ。
Using the data at t = 4, T(4) = 100: 100 = 20 + 160e⁻⁴ᵏ ⇒ 80 = 160e⁻⁴ᵏ ⇒ e⁻⁴ᵏ = 0.5. Taking natural logs: −4k = ln(0.5) = −ln 2 ⇒ k = (ln 2)/4 ≈ 0.1733 min⁻¹.
利用 t = 4 时 T(4) = 100：100 = 20 + 160e⁻⁴ᵏ ⇒ 80 = 160e⁻⁴ᵏ ⇒ e⁻⁴ᵏ = 0.5。取自然对数：−4k = ln(0.5) = −ln 2 ⇒ k = (ln 2)/4 ≈ 0.1733 min⁻¹。
6. Case 2: Interpreting Parameters and Prediction | 案例二：参数解释与预测
The value of k determines how quickly the rod cools. With k ≈ 0.1733, the model becomes T(t) = 20 + 160e⁻⁰·¹⁷³³ᵗ. To find the time t₃₀ when T = 30°C, set 30 = 20 + 160e⁻ᵏᵗ ⇒ 10 = 160e⁻ᵏᵗ ⇒ e⁻ᵏᵗ = 1/16.
k 的值决定了金属棒冷却的快慢。k ≈ 0.1733，模型变为 T(t) = 20 + 160e⁻⁰·¹⁷³³ᵗ。要求 T = 30°C 时的时间 t₃₀，令 30 = 20 + 160e⁻ᵏᵗ ⇒ 10 = 160e⁻ᵏᵗ ⇒ e⁻ᵏᵗ = 1/16。
Taking logs: −kt = ln(1/16) = −ln 16 = −4 ln 2. Thus t = (4 ln 2) / k. Since k = (ln 2)/4, we get t = (4 ln 2) / ((ln 2)/4) = 16 minutes. The rod reaches 30°C after 16 minutes.
取对数：−kt = ln(1/16) = −ln 16 = −4 ln 2。因此 t = (4 ln 2) / k。由于 k = (ln 2)/4，得 t = (4 ln 2) / ((ln 2)/4) = 16 分钟。金属棒在 16 分钟后达到 30°C。
7. Case 3: Frog Population Model – Matrix Formulation | 案例三：青蛙种群模型——矩阵建模
Two ponds, A and B, contain frog populations. Each day, 10% of the frogs in pond A move to pond B, while 20% of the frogs in pond B move to pond A. The state vector xₙ = [aₙ, bₙ]ᵀ gives the numbers in ponds A and B on day n. The transition matrix is M = [[0.9, 0.2], [0.1, 0.8]]. Initially, x₀ = [100, 100]ᵀ.
两个池塘 A 与 B 有青蛙种群。每天 A 中 10% 的青蛙移至 B，B 中 20% 的青蛙移至 A。状态向量 xₙ = [aₙ, bₙ]ᵀ 表示第 n 天池塘 A 和 B 中的数量。转移矩阵为 M = [[0.9, 0.2], [0.1, 0.8]]。初始时 x₀ = [100, 100]ᵀ。
The model is xₙ₊₁ = M xₙ, and the long‑term distribution can be found by analysing eigenvalues and eigenvectors.
模型为 xₙ₊₁ = M xₙ，可通过分析特征值与特征向量找出长期分布。
8. Case 3: Eigenvalues and Eigenvectors | 案例三：特征值与特征向量
Solve det(M − λI) = 0: (0.9 − λ)(0.8 − λ) − (0.2)(0.1) = 0 ⇒ λ² − 1.7λ + (0.72 − 0.02) = 0 ⇒ λ² − 1.7λ + 0.70 = 0. The eigenvalues are λ₁ = 1 and λ₂ = 0.7.
解 det(M − λI) = 0：(0.9 − λ)(0.8 − λ) − (0.2)(0.1) = 0 ⇒ λ² − 1.7λ + (0.72 − 0.02) = 0 ⇒ λ² − 1.7λ + 0.70 = 0。特征值为 λ₁ = 1 和 λ₂ = 0.7。
For λ₁=1, solve (M − I)v = 0: [[−0.1, 0.2], [0.1, −0.2]][x, y]ᵀ = [0, 0]. This gives −0.1x + 0.2y = 0 ⇒ x = 2y. An eigenvector is v₁ = [2, 1]ᵀ.
对 λ₁=1，解 (M − I)v = 0：[[−0.1, 0.2], [0.1, −0.2]][x, y]ᵀ = [0, 0]。得 −0.1x + 0.2y = 0 ⇒ x = 2y。一个特征向量为 v₁ = [2, 1]ᵀ。
9. Case 3: Long‑term Distribution and Diagonalisation | 案例三：长期分布与对角化
Since the dominant eigenvalue is 1, the population distribution tends to a steady state proportional to v₁. The total number of frogs is constant (200). Writing xₙ as a linear combination of eigenvectors, the component in the direction of v₁ dominates as n → ∞.
由于主导特征值为 1，种群分布趋向于与 v₁ 成比例的稳态。青蛙总数不变（200）。将 xₙ 表示为特征向量的线性组合，当 n → ∞ 时，沿 v₁ 方向的分量占主导。
The steady‑state vector is (200 / 3) × [2, 1]ᵀ = [400/3, 200/3]ᵀ ≈ [133.3, 66.7]. So in the long run, pond A has about 133 frogs and pond B about 67 frogs.
稳态向量为 (200 / 3) × [2, 1]ᵀ = [400/3, 200/3]ᵀ ≈ [133.3, 66.7]。因此长期来看，池塘 A 约有 133 只青蛙，池塘 B 约有 67 只。
10. Case 4: Complex Numbers in Trigonometric Integrals | 案例四：复数在三角积分中的应用
De Moivre’s theorem lets us express powers of cosine and sine in terms of multiple angles, which simplifies integration. For example, to evaluate ∫ cos⁴θ dθ, start with cosθ = (e^(iθ) + e^(−iθ)) / 2.
棣莫弗定理可以将余弦和正弦的幂表示为倍角形式，从而简化积分。例如，要计算 ∫ cos⁴θ dθ，可先令 cosθ = (e^(iθ) + e^(−iθ)) / 2。
Then cos⁴θ = (e^(iθ) + e^(−iθ))⁴ / 16. Expand using the binomial theorem: (e^(iθ) + e^(−iθ))⁴ = e^(4iθ) + 4e^(2iθ) + 6 + 4e^(−2iθ) + e^(−4iθ).
那么 cos⁴θ = (e^(iθ) + e^(−iθ))⁴ / 16。使用二项式定理展开：(e^(iθ) + e^(−iθ))⁴ = e^(4iθ) + 4e^(2iθ) + 6 + 4e^(−2iθ) + e^(−4iθ)。
11. Case 4: Evaluating a Definite Integral | 案例四：计算定积分
Group terms into cosines: e^(4iθ) + e^(−4iθ) = 2 cos 4θ, and e^(2iθ) + e^(−2iθ) = 2 cos 2θ. Thus cos⁴θ = (1/16)(2 cos 4θ + 8 cos 2θ + 6) = (1/8) cos 4θ + (1/2) cos 2θ + 3/8.
将项组合为余弦形式：e^(4iθ) + e^(−4iθ) = 2 cos 4θ，e^(2iθ) + e^(−2iθ) = 2 cos 2θ。因此 cos⁴θ = (1/16)(2 cos 4θ + 8 cos 2θ + 6) = (1/8) cos 4θ + (1/2) cos 2θ + 3/8。
Now integrate from 0 to π/2: ∫₀^(π/2) cos⁴θ dθ = [(1/32) sin 4θ + (1/4) sin 2θ + (3/8)θ]₀^(π/2). All sine terms vanish at both limits, leaving (3/8) × (π/2) = 3π/16.
现在从 0 到 π/2 积分：∫₀^(π/2) cos⁴θ dθ = [(1/32) sin 4θ + (1/4) sin 2θ + (3/8)θ]₀^(π/2)。正弦项在上下限处均为零，剩下 (3/8) × (π/2) = 3π/16。
This method avoids repeated use of double‑angle identities and is particularly powerful for higher powers of sine or cosine.
这种方法避免了反复使用倍角公式，对于正弦、余弦的高次幂尤为高效。
12. Exam Tips and Summary | 考试技巧与总结
Always identify the modelling assumptions (e.g. constant ambient temperature, unchanging transition probabilities) and state them clearly. When solving differential equations, don’t forget the constant of integration and check the domain of your solution.
一定要识别建模假设（如环境温度恒定、转移概率不变）并清晰陈述。在解微分方程时，不要忘记积分常数，并检查解的定义域。
For matrix models, use the dominant eigenvalue analysis to describe long‑term behaviour. With complex numbers, converting to exponential form can drastically reduce algebraic work. Practice these case‑study patterns regularly, and you’ll approach the exam with confidence.
对于矩阵模型，利用主导特征值分析来描述长期行为。涉及复数时，转化为指数形式可显著减少代数运算。经常练习这些案例模式，你就能自信地应对考试。
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