📚 Case Studies and Practical Exercises for Year 13 OCR Mathematics | Year 13 OCR 数学案例分析实战演练
This article provides a series of real-world case studies and worked examples aligned with the Year 13 OCR Mathematics specification. Each case reinforces core skills in pure mathematics, statistics, and mechanics through step-by-step bilingual exposition, helping students build confidence for the final examinations.
本文提供一系列与 Year 13 OCR 数学大纲紧密切合的真实案例与演练题。每个案例通过逐步的双语讲解巩固纯数、统计与力学中的核心技能，帮助学生建立考试信心。
1. Optimisation in Real-World Contexts | 真实情境中的优化问题
A drinks company wishes to manufacture a cylindrical aluminium can that holds a fixed volume V = 330 ml. The goal is to minimise the surface area A to save material cost. The surface area of a closed cylinder is A = 2πr² + 2πrh, and the volume is V = πr²h.
某饮料公司要生产容纳固定体积 V = 330 ml 的圆柱形铝罐，旨在最小化表面积 A 以节省材料成本。封闭圆柱的表面积 A = 2πr² + 2πrh，体积 V = πr²h。
First, express the height h in terms of the radius r: h = V / (πr²). Substituting into the area formula yields a function of one variable: A(r) = 2πr² + 2V / r.
首先，将高度 h 用半径 r 表示：h = V / (πr²)。代入面积公式得到关于单变量的函数：A(r) = 2πr² + 2V / r。
Differentiate A with respect to r: dA/dr = 4πr – 2V / r². Setting this derivative to zero gives the stationary condition: 4πr – 2V / r² = 0, which simplifies to r³ = V / (2π).
对 r 求导：dA/dr = 4πr – 2V / r²。令导数为零得到驻点条件：4πr – 2V / r² = 0，化简得 r³ = V / (2π)。
The second derivative is d²A/dr² = 4π + 4V / r³, which is strictly positive for all r > 0, confirming a minimum. Hence the optimal radius is r = ∛(V/(2π)), and the corresponding height is h = 2r, giving the familiar ‘diameter equals height’ economised can shape.
二阶导数为 d²A/dr² = 4π + 4V / r³，对一切 r > 0 恒正，确认该驻点为极小值。因此最优半径 r = ∛(V/(2π))，对应的 h = 2r，即“直径等于高度”的省材罐形。
2. Modelling with Differential Equations | 微分方程建模
Newton’s law of cooling states that the rate of change of temperature θ of an object is proportional to the difference between its temperature and the ambient temperature θₐ. A hot cup of coffee at 85°C is placed in a room at 20°C. The differential equation is dθ/dt = –k(θ – 20), with k > 0.
牛顿冷却定律指出，物体温度 θ 的变化率正比于其温度与环境温度 θₐ 的差值。一杯 85°C 的热咖啡放在 20°C 的室内。微分方程为 dθ/dt = –k(θ – 20)，k > 0。
Separating variables gives ∫ 1/(θ – 20) dθ = –k ∫ dt. Integrating yields ln|θ – 20| = –kt + C. After exponentiation, θ – 20 = Ae⁻ᵏᵗ, where A = eᶜ. Using the initial condition θ(0) = 85 gives A = 65, so θ(t) = 20 + 65 e⁻ᵏᵗ.
分离变量得 ∫ 1/(θ – 20) dθ = –k ∫ dt。积分后得 ln|θ – 20| = –kt + C。两边取指数后 θ – 20 = Ae⁻ᵏᵗ，其中 A = eᶜ。利用初始条件 θ(0) = 85 得 A = 65，故 θ(t) = 20 + 65 e⁻ᵏᵗ。
If the temperature drops to 50°C after 10 minutes, we find k: 50 = 20 + 65 e⁻¹⁰ᵏ → 30/65 = e⁻¹⁰ᵏ → –10k = ln(30/65) → k ≈ 0.0774. Then the time to reach 30°C is found by solving 30 = 20 + 65 e⁻ᵏᵗ → t ≈ 24.6 minutes.
若 10 分钟后温度降至 50°C，可求 k：50 = 20 + 65 e⁻¹⁰ᵏ → 30/65 = e⁻¹⁰ᵏ → –10k = ln(30/65) → k ≈ 0.0774。然后解 30 = 20 + 65 e⁻ᵏᵗ 得到降至 30°C 所需时间 t ≈ 24.6 分钟。
3. Complex Numbers in AC Circuits | 交流电路中的复数
In an RLC series circuit driven by an alternating voltage, the complex impedance Z = R + j(ωL – 1/(ωC)), where j is the imaginary unit (i in pure mathematics), ω is angular frequency, R is resistance, L is inductance, and C is capacitance. Resonance occurs when the reactive part vanishes, making the circuit purely resistive.
在交流驱动下的 RLC 串联电路中，复阻抗 Z = R + j(ωL – 1/(ωC))，其中 j 为虚数单位（纯数中用 i），ω 为角频率，R 为电阻，L 为电感，C 为电容。当虚部为零时发生谐振，电路呈纯电阻性。
Setting the imaginary part to zero gives ωL – 1/(ωC) = 0 → ω² = 1/(LC). Thus the resonant angular frequency is ω₀ = 1/√(LC). At this frequency, Z = R, and the current and voltage are in phase.
令虚部为零：ωL – 1/(ωC) = 0 → ω² = 1/(LC)。因此谐振角频率 ω₀ = 1/√(LC)。在此频率下 Z = R，电流与电压同相。
For instance, if R = 50 Ω, L = 0.2 H, C = 10 μF, then ω₀ = 1/√(0.2 × 10⁻⁵) ≈ 707.1 rad/s, and the phase angle φ = arctan((ωL – 1/(ωC))/R) is zero at resonance. Engineers use this to tune radio receivers to a desired station.
例如 R = 50 Ω, L = 0.2 H, C = 10 μF，则 ω₀ = 1/√(0.2 × 10⁻⁵) ≈ 707.1 rad/s，相位角 φ = arctan((ωL – 1/(ωC))/R) 于谐振时为零。工程师利用此原理调谐无线电接收器。
4. Vector Applications in Geometry | 向量的几何应用
Find the shortest distance from the point P(4, 1, 5) to the line passing through A(1, 2, 3) with direction vector d = 2i – j + k. The distance is given by |AP × d| / |d|.
求点 P(4, 1, 5) 到过 A(1, 2, 3) 且方向向量 d = 2i – j + k 的直线的最短距离。距离公式为 |AP × d| / |d|。
First, compute AP = P – A = (3, –1, 2). Then the cross product AP × d = ( (–1)(1) – (2)(–1), (2)(2) – (3)(1), (3)(–1) – (–1)(2) ) = (1, 1, –1). Its magnitude is |AP × d| = √(1² + 1² + (–1)²) = √3.
首先计算 AP = P – A = (3, –1, 2)。再求叉积 AP × d = ( (–1)(1) – (2)(–1), (2)(2) – (3)(1), (3)(–1) – (–1)(2) ) = (1, 1, –1)。其模 |AP × d| = √(1² + 1² + (–1)²) = √3。
The magnitude of the direction vector is |d| = √(2² + (–1)² + 1²) = √6. Hence the shortest distance = √3 / √6 = 1/√2 ≈ 0.707 units. This method is efficient for 3D geometry problems in OCR exams.
方向向量的模 |d| = √(2² + (–1)² + 1²) = √6。于是最短距离 = √3 / √6 = 1/√2 ≈ 0.707 单位。此法对 OCR 考试中的三维几何问题十分高效。
5. Probability Distributions in Quality Control | 质量控制中的概率分布
A factory produces components with a historical defect rate of 5%. A quality inspector tests a random sample of 20 components. If 3 or more are defective, the batch is rejected. We model the number of defectives X ~ B(20, 0.05).
某工厂生产元件，历史上的不良率为 5%。质检员随机抽检 20 个元件。若不良品数达到 3 个或更多，则整批拒收。建立不良品数 X ~ B(20, 0.05) 的模型。
The probability of rejection under the null hypothesis H₀: p = 0.05 is P(X ≥ 3) = 1 – P(X = 0) – P(X = 1) – P(X = 2). Computing: P(X = 0) = 0.95²⁰ ≈ 0.3585, P(X = 1) = 20 × 0.05 × 0.95¹⁹ ≈ 0.3774, P(X = 2) = C(20,2) × 0.05² × 0.95¹⁸ ≈ 0.1887.
在零假设 H₀: p = 0.05 下，拒收概率为 P(X ≥ 3) = 1 – P(X = 0) – P(X = 1) – P(X = 2)。计算：P(X = 0) = 0.95²⁰ ≈ 0.3585，P(X = 1) = 20 × 0.05 × 0.95¹⁹ ≈ 0.3774，P(X = 2) = C(20,2) × 0.05² × 0.95¹⁸ ≈ 0.1887。
Thus P(X ≥ 3) ≈ 1 – 0.9246 = 0.0754. If the acceptable significance level is 5%, this test has type I error rate ≈ 7.5%, slightly above the threshold; therefore the critical region should be adjusted to X ≥ 4 to bring the error below 5%.
故 P(X ≥ 3) ≈ 1 – 0.9246 = 0.0754。若可接受的显著性水平为 5%，此检验的第一类错误率约 7.5%，略高于阈值；因此应调整临界域为 X ≥ 4 以使错误率低于 5%。
6. Parametric Equations and Trajectories | 参数方程与运动轨迹
A projectile is launched from ground level with initial speed 18 m s⁻¹ at an angle of 35° above the horizontal. Taking g = 9.8 m s⁻², the parametric equations are x = 18t cos35°, y = 18t sin35° – ½ × 9.8 t².
Published by TutorHao | Year 13 Mathematics Revision Series | aleveler.com
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