📚 Year 12 Cambridge Mathematics: Case Study Practical Drill | 剑桥AS数学：案例分析实战演练
In Cambridge AS Level Mathematics, applying theoretical knowledge to real-world problems is essential for mastering the syllabus and achieving high marks in the examination. This article presents a series of carefully selected case studies that cover key topics from Pure Mathematics 1, Mechanics 1, and Statistics 1. Each case study is dissected step by step to build problem-solving fluency and deepen conceptual understanding.
学习剑桥AS数学时，将理论知识应用于实际问题是掌握考纲并在考试中取得高分的关键。本文精选了一系列案例分析，覆盖了纯数学1、力学1和统计1的核心主题。每个案例都逐步解析，帮助您提升解题熟练度并加深概念理解。
1. Maximising Revenue with Quadratic Functions | 利用二次函数最大化收入
A company sells a product and finds that its total revenue R, in thousands of pounds, is modelled by R(x) = 50x − 0.5x², where x is the number of units sold (in hundreds). Determine the number of units that maximise revenue and calculate the maximum revenue.
一家公司在销售某产品时，发现其总收入R（以千英镑计）可由 R(x) = 50x − 0.5x² 建模，其中x为销售单位数（以百为单位）。试求使收入最大化的销售单位数，并计算最大收入。
We recognise this as a quadratic function with a negative coefficient of x², so the graph is a downward-facing parabola. The maximum occurs at the vertex. Differentiating: R'(x) = 50 − x. Setting the derivative equal to zero gives x = 50. The second derivative R”(x) = −1 is negative, confirming a maximum. Substituting back, maximum revenue = 50(50) − 0.5(50)² = 2500 − 1250 = 1250 (thousand pounds). Thus, selling 5000 units yields a maximum revenue of £1,250,000.
我们注意到这是一个x²项系数为负的二次函数，因此图像是开口向下的抛物线，最大值出现在顶点处。求导：R'(x) = 50 − x。令导数等于零，得x = 50。二阶导数R”(x) = −1为负，确认是最大值。回代得最大收入 = 50×50 − 0.5×2500 = 2500 − 1250 = 1250（千英镑）。因此，销售5000单位可获得最大收入1,250,000英镑。
2. Minimising Surface Area of a Cylinder | 最小化圆柱体表面积
A manufacturer needs to produce a cylindrical can with a volume of 500 cm³. The material for the curved surface costs the same per cm² as the material for the top and bottom. Find the radius and height that minimise the total surface area, and hence the material cost.
某制造商需生产一个容积为500 cm³的圆柱形罐子，侧面材料与顶盖、底面的单位面积成本相同。试求使总表面积（从而材料成本）最小的半径和高。
Let the radius be r cm and height h cm. Volume constraint: πr²h = 500 ⇒ h = 500/(πr²). Total surface area A = 2πr² + 2πrh = 2πr² + 2πr(500/(πr²)) = 2πr² + 1000/r. Differentiate with respect to r: dA/dr = 4πr − 1000/r². Set to zero: 4πr = 1000/r² ⇒ r³ = 1000/(4π) = 250/π ⇒ r = ∛(250/π). The second derivative is positive for all r > 0, confirming a minimum. Height h = 500/(πr²). Using r ≈ 4.30 cm gives h ≈ 8.60 cm. This is the design that uses the least material.
设半径为r cm，高为h cm。体积约束：πr²h = 500 ⇒ h = 500/(πr²)。总表面积 A = 2πr² + 2πrh = 2πr² + 2πr・(500/(πr²)) = 2πr² + 1000/r。对r求导：dA/dr = 4πr − 1000/r²。令其为零：4πr = 1000/r² ⇒ r³ = 1000/(4π) = 250/π ⇒ r = ∛(250/π)。二阶导数对所有r>0为正，确认是极小值。高 h = 500/(πr²)。代入r ≈ 4.30 cm，得h ≈ 8.60 cm。该设计用料最省。
3. Related Rates: Filling a Conical Tank | 相关变化率：圆锥水箱注水
Water is poured into an inverted conical tank of height 6 m and base radius 2 m at a constant rate of 0.5 m³/min. Find the rate at which the water level rises when the depth of water is 3 m.
水以恒定速率0.5 m³/min 注入一个倒置的圆锥水箱，水箱高6 m，底面半径2 m。求水深为3 m时，水位的上升速率。
The volume of water in the cone is V = (1/3)πr²h. By similar triangles, r/h = 2/6 ⇒ r = h/3. Substitute: V = (1/3)π(h/3)² h = (π/27) h³. Differentiate with respect to time t: dV/dt = (π/9) h² (dh/dt). Given dV/dt = 0.5 m³/min and h = 3 m, we have 0.5 = (π/9) × 9 × (dh/dt) = π (dh/dt). Hence dh/dt = 0.5/π ≈ 0.159 m/min. The water level is rising at about 15.9 cm per minute at that instant.
锥内水的体积 V = (1/3)πr²h。由相似三角形，r/h = 2/6 ⇒ r = h/3。代入得：V = (1/3)π(h/3)² h = (π/27) h³。对时间t求导：dV/dt = (π/9) h² (dh/dt)。已知 dV/dt = 0.5 m³/min，h = 3 m，代入得 0.5 = (π/9) × 9 × (dh/dt) = π (dh/dt)。因此 dh/dt = 0.5/π ≈ 0.159 m/min。此时水位以每分钟约15.9 cm的速度上升。
4. Motion of a Particle: Displacement to Acceleration | 质点运动：位移到加速度
A particle moves along a straight line such that its displacement s (in metres) from a fixed point after t seconds is given by s(t) = t³ − 6t² + 9t + 2. Determine the times when the particle is at rest, and find its acceleration at those times.
一质点沿直线运动，t秒后其离开固定点的位移s（米）由 s(t) = t³ − 6t² + 9t + 2 给出。求质点静止的时刻，并求这些时刻的加速度。
Velocity v = ds/dt = 3t² − 12t + 9. For at rest, v = 0 ⇒ 3t² − 12t + 9 = 0 ⇒ divide by 3: t² − 4t + 3 = 0 ⇒ (t−1)(t−3) = 0. So t = 1 s and t = 3 s. Acceleration a = dv/dt = 6t − 12. At t = 1, a = 6(1) − 12 = −6 m/s²; at t = 3, a = 6(3) − 12 = 6 m/s². The particle stops momentarily but is accelerating in opposite directions at the two instants.
速度 v = ds/dt = 3t² − 12t + 9。静止时v = 0 ⇒ 3t² − 12t + 9 = 0 ⇒ 除以3：t² − 4t + 3 = 0 ⇒ (t−1)(t−3) = 0。故 t = 1 s 和 t = 3 s。加速度 a = dv/dt = 6t − 12。t = 1时，a = 6×1 − 12 = −6 m/s²；t = 3时，a = 6×3 − 12 = 6 m/s²。质点在两个瞬间暂时停止，但加速度方向相反。
5. Equilibrium of Forces on an Inclined Plane | 斜面上力的平衡
A block of mass 10 kg rests on a rough plane inclined at 30° to the horizontal. The coefficient of friction between the block and the plane is 0.4. A horizontal force P is applied to the block. Find the minimum value of P required to just move the block up the plane.
一质量为10 kg的木块放在与水平面成30°角的粗糙斜面上，木块与斜面间的摩擦系数为0.4。施加一水平力P作用于木块。求使木块刚好向上滑动所需的最小P值。
Resolve forces parallel and perpendicular to the plane. Weight component down the plane: 10g sin 30°. Normal reaction R = 10g cos 30° + P sin 30° (since P has an upward component perpendicular to plane). Maximum friction F_max = μR = 0.4 R. For motion up the plane, P cos 30° must overcome friction plus weight component: P cos 30° = 10g sin 30° + 0.4 (10g cos 30° + P sin 30°). Taking g = 9.8, sin 30° = 0.5, cos 30° ≈ 0.866. Then P × 0.866 = 10×9.8×0.5 + 0.4 (10×9.8×0.866 + P×0.5). Compute: 49 + 0.4 (84.868 + 0.5P) = 49 + 33.9472 + 0.2P = 82.9472 + 0.2P. So 0.866P − 0.2P = 82.9472 ⇒ 0.666P ≈ 82.9472 ⇒ P ≈ 124.6 N. Thus a horizontal force of about 125 N is required.
分解平行和垂直于斜面的力。沿斜面向下的重力分力：10g sin 30°。法向反作用力 R = 10g cos 30° + P sin 30°（因P有垂直于斜面的向上分量）。最大摩擦力 F_max = μR = 0.4 R。向上运动时，P cos 30° 必须克服摩擦和重力分力：P cos 30° = 10g sin 30° + 0.4 (10g cos 30° + P sin 30°)。取g = 9.8，sin 30° = 0.5，cos 30° ≈ 0.866。代入：P × 0.866 = 10×9.8×0.5 + 0.4 (10×9.8×0.866 + P×0.5)。计算：49 + 0.4 (84.868 + 0.5P) = 49 + 33.9472 + 0.2P = 82.9472 + 0.2P。故 0.866P − 0.2P = 82.9472 ⇒ 0.666P ≈ 82.9472 ⇒ P ≈ 124.6 N。因此，约需125 N的水平力。
6. Normal Distribution: Quality Control of Light Bulbs | 正态分布：灯泡质量控制
The lifetimes of a certain brand of light bulb are normally distributed with mean 800 hours and standard deviation 40 hours. A batch is considered acceptable if at least 95% of bulbs last more than 740 hours. Determine whether a batch meets this quality standard, and find the probability that a randomly chosen bulb lasts between 760 and 840 hours.
某品牌灯泡的寿命服从正态分布，均值为800小时，标准差为40小时。若一批次中至少有95%的灯泡寿命超过740小时，则视为合格。试判断该批次是否符合此质量标准，并求随机抽取一只灯泡寿命在760至840小时之间的概率。
For X ~ N(800, 40²), standardise: Z = (X − 800)/40. Proportion lasting more than 740 hours: P(X > 740) = P(Z > (740 − 800)/40 = −1.5). Using tables, P(Z > −1.5) = 1 − Φ(−1.5) = Φ(1.5) ≈ 0.9332, or 93.32%. This is less than 95%, so the batch does not meet the standard. For P(760 < X < 840): z₁ = (760−800)/40 = −1, z₂ = (840−800)/40 = 1. Then P(−1 < Z < 1) = Φ(1) − Φ(−1) = 0.8413 − 0.1587 = 0.6826. Thus, about 68.3% of bulbs have lifetimes in that range.
设 X ~ N(800, 40²)，标准化：Z = (X − 800)/40。寿命超过740小时的比例：P(X > 740) = P(Z > (740−800)/40 = −1.5)。查表，P(Z > −1.5) = 1 − Φ(−1.5) = Φ(1.5) ≈ 0.9332，即93.32%。该值低于95%，故此批次不合格。对于P(760 < X < 840)：z₁ = (760−800)/40 = −1，z₂ = (840−800)/40 = 1。则 P(−1 < Z < 1) = Φ(1) − Φ(−1) = 0.8413 − 0.1587 = 0.6826。因此，约68.3%的灯泡寿命在此区间。
7. Probability: Arrangements with Restrictions | 概率：受限排列
Six books – three Mathematics (M1, M2, M3), two Physics (P1, P2), and one Chemistry (C) – are to be arranged on a shelf. Find the number of distinct arrangements if the three Mathematics books must be together, and the two Physics books must not be adjacent.
六本书——三本数学（M1, M2, M3）、两本物理（P1, P2）和一本化学（C）——要排列在书架上。若三本数学书必须放在一起，且两本物理书不得相邻，求不同的排列数。
Treat the block of three Maths books as a single item, giving 4 items: Maths block, P1, P2, C. The block can be internally arranged in 3! = 6 ways. Arranging the 4 items without restrictions gives 4! = 24 ways, so 6 × 24 = 144 arrangements without any Physics restriction. Now subtract arrangements where Physics books are together within this case. Treat Physics pair as one block alongside Maths block and C: 3 items (Maths block, Physics block, C). Physics block can be arranged internally in 2! = 2 ways. Then 3! × 2 = 6 × 2 = 12 ways, multiplied by 6 for the Maths block internal arrangements gives 72. So valid arrangements = 144 − 72 = 72. Alternatively, direct counting with Physics non-adjacent: arrange C and Maths block (2 items) in 2! ways, then choose 2 of the 3 gaps for Physics books in ³P₂ = 6 ways, all multiplied by 6 gives 2×6×6 = 72.
将三本数学书视为一个整体，得到4个物件：数学块、P1、P2、C。该块内部有3! = 6种排列。4个物件无限制排列有4! = 24种，故 6×24 = 144 种（未施加物理限制）。现减去物理书相邻的排列：将物理书视为一个整体，与数学块、C共3个物件，内部排列2! = 2种。则 3! × 2 = 6×2 = 12，乘以数学块内部6种得72。故符合要求的排列数 = 144 − 72 = 72。另一种直接计数法：物理书不相邻，先排列C和数学块（2个物件）有2!种，再从3个空隙中选2个放置物理书，有³P₂ = 6种，全部乘以6得 2×6×6 = 72。
8. Curve Sketching and Tangent/Normal Analysis | 曲线描绘与切线法线分析
Consider the curve y = x³ − 3x² + 4. Find the coordinates of any stationary points, determine their nature, and write the equations of the tangent and normal lines at the point where x = 1.
考虑曲线 y = x³ − 3x² + 4。求所有驻点的坐标并判断其性质，写出在 x = 1 处的切线和法线方程。
Differentiate: dy/dx = 3x² − 6x = 3x(x − 2). Stationary points when dy/dx = 0 ⇒ x = 0 or x = 2. At x = 0, y = 4; at x = 2, y = 8 − 12 + 4 = 0. Second derivative d²y/dx² = 6x − 6. At x = 0, d²y/dx² = −6 < 0, so (0,4) is a local maximum. At x = 2, d²y/dx² = 6 > 0, so (2,0) is a local minimum. At x = 1, y = 1 − 3 + 4 = 2. Gradient m_t = 3(1)² − 6(1) = −3. Tangent: y − 2 = −3(x − 1) or y = −3x + 5. Normal gradient m_n = 1/3. Normal: y − 2 = (1/3)(x − 1) or y = (1/3)x + 5/3.
求导：dy/dx = 3x² − 6x = 3x(x − 2)。驻点满足 dy/dx = 0 ⇒ x = 0 或 x = 2。x = 0时 y = 4；x = 2时 y = 8 − 12 + 4 = 0。二阶导数 d²y/dx² = 6x − 6。x = 0时 d²y/dx² = −6 < 0，故 (0,4) 为极大值点。x = 2时 d²y/dx² = 6 > 0，故 (2,0) 为极小值点。当 x = 1，y = 1 − 3 + 4 = 2。切线斜率 m_t = 3×1² − 6×1 = −3。切线方程：y − 2 = −3(x − 1) 即 y = −3x + 5。法线斜率 m_n = 1/3。法线方程：y − 2 = (1/3)(x − 1) 即 y = (1/3)x + 5/3。
9. Integration: Area Between Curves in Context | 积分：情境中曲线间面积
Two car prototypes are tested on a straight track. Their velocity functions (m/s) are modelled as v₁(t) = 2t² − 4t + 5 and v₂(t) = t + 3 for t between 0 and 4 seconds. Interpret the area between the two velocity curves over this interval, and calculate it.
两款原型车在直线赛道上测试。它们的速度函数（m/s）分别建模为 v₁(t) = 2t² − 4t + 5 和 v₂(t) = t + 3，t 在0到4秒之间。阐释此区间内两速度曲线间的面积含义，并计算其值。
The area between v₁(t) and v₂(t) from t=0 to t=4 represents the difference in distance travelled by the two cars, assuming the faster car is ahead. Determine which function is larger. At t=0, v₁=5, v₂=3; difference v₁ − v₂ = 2t² − 5t + 2. Solve 2t² − 5t + 2 = 0 → (2t − 1)(t − 2) = 0, so t = 0.5 and t = 2. Between 0 and 0.5, v₁ > v₂. Between 0.5 and 2, v₂ > v₁. After 2, v₁ > v₂. The total absolute area is ∫₀⁰·⁵ (v₁−v₂) dt + ∫⁰·⁵² (v₂−v₁) dt + ∫²⁴ (v₁−v₂) dt. Integrating: ∫(2t²−5t+2)dt = (2/3)t³ − (5/2)t² + 2t. Evaluate appropriately to get total area = 11/3 m (approx 3.67 m). This is the total difference in distances covered, accounting for overtaking.
v₁(t) 与 v₂(t) 在 t=0 到 t=4 之间的面积代表两车行驶距离之差，假设较快者在前。先判断哪个函数更大。在 t=0，v₁=5，v₂=3；差 v₁ − v₂ = 2t² − 5t + 2。解 2t² − 5t + 2 = 0 → (2t − 1)(t − 2)=0，得 t=0.5 和 t=2。在0到0.5之间 v₁ > v₂；0.5到2之间 v₂ > v₁；2以后 v₁ > v₂。总面积需绝对值：∫₀⁰·⁵ (v₁−v₂) dt + ∫⁰·⁵² (v₂−v₁) dt + ∫²⁴ (v₁−v₂) dt。积分：∫(2t²−5t+2)dt = (2/3)t³ − (5/2)t² + 2t。分别代入计算得总面积为 11/3 m（约3.67 m）。这便是考虑超车过程的总行程差。
10. Modelling with Differential Equations: Cooling | 微分方程建模：冷却
According to Newton’s law of cooling, the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature. A cup of coffee at 85 °C is placed in a room at 20 °C. After 5 minutes, the coffee has cooled to 60 °C. Form a differential equation, find the temperature function, and determine the time needed to reach 40 °C.
根据牛顿冷却定律，物体温度的变化率与其温度和环境温度之差成正比。一杯85 °C的咖啡放在20 °C的房间中，5分钟后冷却至60 °C。建立微分方程，求出温度函数，并确定达到40 °C所需的时间。
Let T(t) be temperature at time t minutes. Then dT/dt = −k(T − 20), with k > 0. Separate variables: dT/(T − 20) = −k dt. Integrate: ln|T − 20| = −kt + C. Initially t=0, T=85: ln(65) = C. So T − 20 = 65 e⁻ᵋᵗ. Using T(5)=60: 40 = 65 e⁻⁵ᵏ → e⁻⁵ᵏ = 40/65 = 8/13 → −5k = ln(8/13) → k = (1/5) ln(13/8). Approx k ≈ 0.0972. Temperature function: T(t) = 20 + 65 e⁻ᵋᵗ with k as above. Set T=40: 20 = 65 e⁻ᵋᵗ → e⁻ᵋᵗ = 20/65 = 4/13. Take ln: −kt = ln(4/13) → t = −(1/k) ln(4/13) = (5 ln(13/4))/ln(13/8). Evaluate: t ≈ (5×1.17865)/0.4855 ≈ 12.14 minutes. It takes about 12.1 minutes to cool to 40 °C.
设 T(t) 为t分钟时的温度。则 dT/dt = −k(T − 20)，k > 0。分离变量：dT/(T − 20) = −k dt。积分：ln|T − 20| = −kt + C。初始 t=0, T=85：ln(65) = C。故 T − 20 = 65 e⁻ᵏᵗ。利用 T(5)=60：40 = 65 e⁻⁵ᵏ → e⁻⁵ᵏ = 40/65 = 8/13 → −5k = ln(8/13) → k = (1/5) ln(13/8)。近似 k ≈ 0.0972。温度函数：T(t) = 20 + 65 e⁻ᵏᵗ（k同上）。设 T=40：20 = 65 e⁻ᵏᵗ → e⁻ᵏᵗ = 20/65 = 4/13。取对数：−kt = ln(4/13) → t = −(1/k) ln(4/13) = (5 ln(13/4))/ln(13/8)。计算：t ≈ (5×1.17865)/0.4855 ≈ 12.14 分钟。降至40 °C约需12.1分钟。
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