📚 Year 13 Cambridge Mathematics: Case Study Practical Workout | 剑桥数学13年级：案例分析实战演练
Case studies in A Level Mathematics sharpen your ability to translate real scenarios into mathematical models, then apply techniques from pure, mechanics and statistics to solve them. This article presents eight carefully designed practical exercises that mirror the Cambridge Year 13 style. Work through each case, study the paired English–Chinese explanations, and build confidence for your exams.
在A Level数学中，案例分析能锻炼你将实际情境转化为数学模型，并运用纯数、力学和统计技巧求解的能力。本文精选了八个与剑桥13年级风格一致的实战演练。请逐题练习，研读中英配对讲解，为考试积累信心。
1. Optimisation of a Box Volume | 盒子体积的最优化
A manufacturer has 1200 cm² of sheet metal to produce an open‑top rectangular box with a square base. Determine the side length of the base and the height that give the greatest possible volume.
一名制造商有1200 cm²的金属板材，用来制作一个正方形底、无盖的长方体盒子。求能使体积最大的底边边长与盒高。
Let the side of the square base be x cm and the height be h cm. The surface area constraint is x² + 4xh = 1200, giving h = (1200 − x²) / (4x).
设正方形底边长为 x cm，高为 h cm。表面积限制条件为 x² + 4xh = 1200，可得 h = (1200 − x²) / (4x)。
The volume to be maximised is V = x²h. Substituting for h yields
V = x² × (1200 − x²)/(4x) = 300x − (1/4)x³.
要最大化的体积为 V = x²h。代入 h 得
V = x² × (1200 − x²)/(4x) = 300x − (1/4)x³.
Differentiate: dV/dx = 300 − (3/4)x². Setting dV/dx = 0 gives x² = 400, so x = 20 (since length must be positive). The second derivative d²V/dx² = −(3/2)x is negative at x = 20, confirming a maximum. Then h = (1200 − 400)/(80) = 10 cm. The maximum volume is V = 20² × 10 = 4000 cm³.
求导：dV/dx = 300 − (3/4)x²。令 dV/dx = 0 得 x² = 400，故 x = 20（长度取正值）。二阶导数 d²V/dx² = −(3/2)x 在 x = 20 处为负，确认为极大值。进而 h = (1200 − 400)/(80) = 10 cm。最大体积为 V = 20² × 10 = 4000 cm³。
2. Projectile Motion | 抛体运动
A football is kicked from ground level with an initial speed of 22 m s⁻¹ at an angle of 35° to the horizontal. Model the ball as a particle moving freely under gravity (g = 9.8 m s⁻²). Find the time of flight, the horizontal range, and the maximum height reached.
一个足球从地面以22 m s⁻¹的初速度、与水平成35°角踢出。把球视为在重力（g = 9.8 m s⁻²）作用下自由运动的质点。求飞行时间、水平射程与达到的最大高度。
Resolve the initial velocity: uₓ = 22 cos35° ≈ 18.02 m s⁻¹, u_y = 22 sin35° ≈ 12.62 m s⁻¹.
分解初速度：uₓ = 22 cos35° ≈ 18.02 m s⁻¹，u_y = 22 sin35° ≈ 12.62 m s⁻¹。
Time of flight from vertical motion: s_y = u_y t − ½gt². Setting s_y = 0 (returns to ground) gives t( u_y − ½gt ) = 0, so t = 2u_y / g ≈ 2×12.62 / 9.8 ≈ 2.58 s.
由竖直运动求飞行时间：s_y = u_y t − ½gt²。令 s_y = 0（返回地面）得 t( u_y − ½gt ) = 0，故 t = 2u_y / g ≈ 2×12.62 / 9.8 ≈ 2.58 s。
Horizontal range: R = uₓ × t ≈ 18.02 × 2.58 ≈ 46.5 m.
水平射程：R = uₓ × t ≈ 18.02 × 2.58 ≈ 46.5 m。
Maximum height when vertical velocity becomes zero: v_y² = u_y² − 2gH. Set v_y = 0 ⇒ H = u_y²/(2g) ≈ (12.62)²/(19.6) ≈ 8.13 m.
最大高度出现在竖直速度为零时：v_y² = u_y² − 2gH。令 v_y = 0 ⇒ H = u_y²/(2g) ≈ (12.62)²/(19.6) ≈ 8.13 m。
3. Normal Distribution in Quality Control | 正态分布与质量控制
A machine produces bolts with diameters normally distributed: X ∼ N(10.0 mm, 0.20² mm²). A bolt is acceptable if its diameter lies between 9.8 mm and 10.3 mm. Calculate the proportion of bolts that meet the specification. If a sample of 5 bolts is taken, find the probability that at least 4 are acceptable.
一台机器生产的螺栓直径服从正态分布：X ∼ N(10.0 mm, 0.20² mm²)。若直径在9.8 mm至10.3 mm之间为合格，计算合格品的比例。若随机抽取5个螺栓，求至少4个合格的概率。
Standardise the limits: z₁ = (9.8 − 10)/0.2 = −1.0, z₂ = (10.3 − 10)/0.2 = 1.5.
标准化边界：z₁ = (9.8 − 10)/0.2 = −1.0，z₂ = (10.3 − 10)/0.2 = 1.5。
Using the standard normal table: Φ(1.5) ≈ 0.9332, Φ(−1.0) ≈ 0.1587. Probability acceptable p = 0.9332 − 0.1587 = 0.7745.
查标准正态分布表：Φ(1.5) ≈ 0.9332，Φ(−1.0) ≈ 0.1587。合格概率 p = 0.9332 − 0.1587 = 0.7745。
Let Y be the number of acceptable bolts in a sample of 5, Y ∼ B(5, 0.7745). We require P(Y ≥ 4) = P(Y = 4) + P(Y = 5).
设Y为5个样本中的合格螺栓数，Y ∼ B(5, 0.7745)。需要 P(Y ≥ 4) = P(Y = 4) + P(Y = 5)。
P(Y = 4) = ₅C₄ (0.7745)⁴ (0.2255)¹ ≈ 5 × 0.3600 × 0.2255 ≈ 0.406; P(Y = 5) = (0.7745)⁵ ≈ 0.278. Sum ≈ 0.684. So roughly 68.4% chance.
P(Y = 4) = ₅C₄ (0.7745)⁴ (0.2255)¹ ≈ 5 × 0.3600 × 0.2255 ≈ 0.406；P(Y = 5) = (0.7745)⁵ ≈ 0.278。合计约 0.684。因此大约68.4%的概率。
4. Newton’s Law of Cooling | 牛顿冷却定律
A cup of coffee at 85 °C is placed in a room at 20 °C. After 5 minutes its temperature drops to 60 °C. Assuming Newton’s Law of Cooling, find the temperature after a further 5 minutes and the time taken to reach 40 °C.
一杯85 °C的咖啡放在20 °C的房间里。5分钟后温度降至60 °C。假设遵循牛顿冷却定律，求再经过5分钟后的温度，以及降至40 °C所需的时间。
The differential equation: dT/dt = −k (T − 20), with T(0) = 85. Separation of variables gives T − 20 = (85 − 20) e⁻ᵏᵗ, i.e. T = 20 + 65 e⁻ᵏᵗ.
微分方程：dT/dt = −k (T − 20)，且 T(0) = 85。分离变量得 T − 20 = (85 − 20) e⁻ᵏᵗ，即 T = 20 + 65 e⁻ᵏᵗ。
At t = 5, T = 60: 60 = 20 + 65 e⁻⁵ᵏ ⇒ e⁻⁵ᵏ = 40/65 = 8/13. Take natural log: −5k = ln(8/13) ⇒ k = −(1/5) ln(8/13) ≈ 0.0970.
当 t = 5 时 T = 60：60 = 20 + 65 e⁻⁵ᵏ ⇒ e⁻⁵ᵏ = 40/65 = 8/13。取自然对数：−5k = ln(8/13) ⇒ k ≈ 0.0970。
After 10 minutes (further 5): T = 20 + 65 e⁻¹⁰ᵏ = 20 + 65 (8/13)² ≈ 20 + 65 × 0.3787 ≈ 44.6 °C.
10分钟后（再经5分）：T = 20 + 65 e⁻¹⁰ᵏ = 20 + 65 (8/13)² ≈ 20 + 65 × 0.3787 ≈ 44.6 °C。
For T = 40: 40 = 20 + 65 e⁻ᵏᵗ ⇒ e⁻ᵏᵗ = 20/65 = 4/13. Taking ln: −kt = ln(4/13) ⇒ t = −ln(4/13)/k ≈ 1.178 / 0.0970 ≈ 12.1 minutes.
当T = 40时：40 = 20 + 65 e⁻ᵏᵗ ⇒ e⁻ᵏᵗ = 20/65 = 4/13。取对数：−kt = ln(4/13) ⇒ t ≈ 12.1 分钟。
5. Hypothesis Testing with the Binomial Distribution | 二项分布假设检验
A die is rolled 60 times, and the number ‘1’ appears 15 times. Test at the 5% significance level whether the die is biased towards showing ‘1’. State your null and alternative hypotheses clearly.
一颗骰子投掷60次，点数“1”出现了15次。在5%的显著性水平下检验这颗骰子是否偏向出现“1”。请明确陈述原假设与备择假设。
Let p = probability of rolling a ‘1’. H₀: p = 1/6; H₁: p > 1/6 (one‑tailed test, since suspicion is bias towards ‘1’). Under H₀, X ∼ B(60, 1/6).
设p为掷出“1”的概率。H₀: p = 1/6；H₁: p > 1/6（单侧检验，因为怀疑偏向“1”）。在原假设下，X ∼ B(60, 1/6)。
We need the probability of getting 15 or more ‘1’s if H₀ is true: P(X ≥ 15). Use normal approximation: mean μ = np = 10, variance σ² = npq = 60 × (1/6) × (5/6) = 50/6 ≈ 8.333, so σ ≈ 2.887. Apply continuity correction: P(X ≥ 15) ≈ P(Z ≥ (14.5 − 10)/2.887) = P(Z ≥ 1.559). From tables, Φ(1.56) ≈ 0.9406, so p‑value ≈ 1 − 0.9406 = 0.0594.
需求原假设下出现15次或更多“1”的概率：P(X ≥ 15)。用正态近似：均值 μ = np = 10，方差 σ² = npq ≈ 8.333，σ ≈ 2.887。连续性校正：P(X ≥ 15) ≈ P(Z ≥ (14.5 − 10)/2.887) = P(Z ≥ 1.559)。查表得 Φ(1.56) ≈ 0.9406，因此 p 值 ≈ 1 − 0.9406 = 0.0594。
Since 0.0594 > 0.05, the result is not significant at the 5% level. We do not reject H₀; there is insufficient evidence to claim the die is biased towards ‘1’. (Using exact binomial gives p‑value ≈ 0.047 if we consider exact computation? With B(60,1/6) exact P(X≥15) is about 0.042, but for teaching we use normal approximation; we can note that exact test might give different conclusion at boundary.)
由于0.0594 > 0.05，在5%水平下不显著。我们不拒绝原假设；没有足够证据表明骰子偏向“1”。（精确二项分布计算P(X≥15)≈0.042，可能得出不同结论，但此练习主要演示方法。）
6. Volume of Revolution | 旋转体体积
The region bounded by the curve y = √x, the x‑axis, and the lines x = 0 and x = 4 is rotated through 360° about the x‑axis. Find the exact volume of the solid generated.
由曲线 y = √x、x轴以及直线 x = 0 和 x = 4 所围成的区域绕x轴旋转360°。求所形成的旋转体体积的精确值。
The volume of revolution about the x‑axis is given by
V = π ∫ₐᵇ y² dx.
绕x轴旋转的体积公式为
V = π ∫ₐᵇ y² dx。
Here y² = (√x)² = x. So V = π ∫₀⁴ x dx = π [½ x²]₀⁴ = π × (½ × 16) = 8π.
此处 y² = (√x)² = x。因此 V = π ∫₀⁴ x dx = π [½ x²]₀⁴ = π × (½ × 16) = 8π。
The solid has a volume of exactly 8π cubic units.
该旋转体的体积精确为8π立方单位。
7. Geometric Distribution in Context | 几何分布的实际应用
A basketball player has a constant probability of 0.3 of scoring from the free‑throw line on any attempt. She takes independent shots until she scores. Find the probability that she needs exactly 4 attempts, and the expected number of attempts to score.
一名篮球运动员每次罚球命中的概率恒为0.3，且各次投篮相互独立。她持续投篮直至命中。求她恰好需要4次投篮才命中的概率，以及命中所需投篮次数的期望。
Let X be the number of attempts up to and including the first success. X follows a geometric distribution with p = 0.3. P(X = r) = (1 − p)ʳ⁻¹ p, for r = 1, 2, …
设X为直到首次命中为止的投篮次数。X服从参数 p = 0.3 的几何分布。P(X = r) = (1 − p)ʳ⁻¹ p，r = 1, 2, …
For exactly 4 attempts (first success on 4th): P(X = 4) = (0.7)³ × 0.3 = 0.343 × 0.3 = 0.1029.
恰好4次投篮命中的概率：P(X = 4) = (0.7)³ × 0.3 = 0.343 × 0.3 = 0.1029。
Expected value E(X) = 1/p = 1/0.3 ≈ 3.33 attempts.
期望值 E(X) = 1/p = 1/0.3 ≈ 3.33 次。
The variance is (1−p)/p² = 0.7/0.09 ≈ 7.78, but the main interest is the expected wait. On average she will score on her 3rd or 4th attempt.
方差为 (1−p)/p² ≈ 7.78，但主要关注的是期望等待次数。平均而言，她将在第3或第4次投篮时命中。
8. Complex Roots of Polynomial Equations | 多项式方程的复数根
Given that 2 + i is a root of the cubic equation x³ − 5x² + 9x − 5 = 0, find the other two roots. Express your answer in exact Cartesian form.
已知 2 + i 是三次方程 x³ − 5x² + 9x − 5 = 0 的一个根，求其余两个根，并用精确的笛卡尔形式表示。
Since the polynomial has real coefficients, complex roots occur in conjugate pairs. Therefore 2 − i is also a root. The product of these two factors is (x − (2+i))(x − (2−i)) = x² − 4x + 5.
由于多项式系数为实数，复根共轭成对出现。因此 2 − i 也是一个根。这两个因式的乘积为 (x − (2+i))(x − (2−i)) = x² − 4x + 5。
Divide the cubic by the quadratic: (x³ − 5x² + 9x − 5) ÷ (x² − 4x + 5) gives a linear quotient.
用二次因式除三次多项式：(x³ − 5x² + 9x − 5) ÷ (x² − 4x + 5) 得到一个一次商式。
Perform long division:
x³ − 5x² + 9x − 5 = (x² − 4x + 5)(x − a).
By expanding the right side: (x² − 4x + 5)(x − a) = x³ − (a+4)x² + (4a+5)x − 5a. Compare coefficients: constant term −5a = −5 ⇒ a = 1. Check: x² coefficient −(a+4) = −5 gives a=1, consistent. So the third root is x = 1.
进行长除：
x³ − 5x² + 9x − 5 = (x² − 4x + 5)(x − a)。
展开右边：(x² − 4x + 5)(x − a) = x³ − (a+4)x² + (4a+5)x − 5a。比较系数：常数项 −5a = −5 ⇒ a = 1。验证二次项系数 −(a+4) = −5 得 a=1，一致。因此第三根为 x
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