Year 13 OCR Further Mathematics: Case Study Practical Exercises

📚 Year 13 OCR Further Mathematics: Case Study Practical Exercises | 案例分析实战演练

Case studies offer a powerful way to connect pure mathematical techniques with real-world problem-solving. In this revision article, we walk through eight carefully chosen practical examples that mirror the style and depth expected in the Year 13 OCR Further Mathematics exam. Each case study introduces a scenario, identifies the relevant mathematical tools, works through a solution step by step, and highlights common pitfalls. By engaging with these examples, you will develop the confidence to tackle modelling, proof, and decision-making questions that require both fluency and insight.

案例学习是将纯粹的数学技巧与现实问题解决联系起来的有力方式。在这篇复习文章中，我们将走过八个精心挑选的实践案例，这些案例反映了 Year 13 OCR 进阶数学考试所期望的风格与深度。每个案例都会引入一个情境，确定相关的数学工具，逐步展示解决过程，并指出常见误区。通过接触这些例子，你将培养起应对既需要流利度又需要洞察力的建模、证明与决策问题所需的信心。

1. Complex Numbers in Drone Navigation | 无人机导航中的复数应用

A delivery drone starts at a warehouse represented by the complex number z₀ = 2 + 3i, where the real part gives east-west position (metres) and the imaginary part gives north-south position. It first flies on a bearing of 30° for 10 metres, then turns clockwise by 90° and travels a further 6 metres. Using complex multiplication to encode rotations and translations, we can find the drone’s final position efficiently.

一架送货无人机从仓库出发，仓库位置用复数 z₀ = 2 + 3i 表示，实部代表东西位置（米），虚部代表南北位置。它首先沿 30° 方位飞行 10 米，然后顺时针旋转 90°，再飞行 6 米。利用复数乘法来编码旋转与平移，我们可以高效地求出无人机的最终位置。

The first displacement vector can be written as 10(cos 30° + i sin 30°) = 10(√3/2 + i/2) = 5√3 + 5i. Adding this to z₀ gives an intermediate point z₁ = (2+5√3) + (3+5)i.

第一个位移向量可写成 10(cos 30° + i sin 30°) = 10(√3/2 + i/2) = 5√3 + 5i。将其加到 z₀ 上，得到中间位置 z₁ = (2+5√3) + (3+5)i。

A clockwise rotation of 90° corresponds to multiplication by –i. Therefore, the second displacement of 6 metres east initially (i.e. along the positive real axis) becomes a vector 6 × (–i) = –6i after the rotation. Adding this to z₁ yields the final complex coordinate z_final = (2+5√3) + (3+5–6)i = (2+5√3) + 2i.

顺时针旋转 90° 相当于乘以 –i。因此，最初向东的 6 米位移（沿正实轴）旋转后变为向量 6 × (–i) = –6i。将其加到 z₁ 上，得到最终复数坐标 z_final = (2+5√3) + (3+5–6)i = (2+5√3) + 2i。

Thus the drone ends at approximately (2+8.66) + 2i = 10.66 + 2i, i.e. 10.66 m east and 2 m north of the warehouse. This method elegantly avoids breaking vectors into components manually.

因此，无人机最终大约位于 (2+8.66) + 2i = 10.66 + 2i，即在仓库以东 10.66 m、以北 2 m 处。这种方法巧妙地避免了手动将向量分解为分量。

2. Matrix Transformations in Computer Graphics | 计算机图形中的矩阵变换

A graphics engine represents a triangular logo with vertices A(1, 2), B(3, 4) and C(2, 5). The designer wants to apply an anti-clockwise rotation of 90° about the origin, followed by a reflection in the y-axis. By finding the composite transformation matrix, we can compute the new coordinates directly.

图形引擎用顶点 A(1, 2)、B(3, 4) 和 C(2, 5) 表示一个三角形徽标。设计师希望先绕原点逆时针旋转 90°，再关于 y 轴反射。通过求出复合变换矩阵，我们可以直接计算新坐标。

The rotation matrix for 90° anti-clockwise is R =
| 0 –1 |
| 1 0 |.
The reflection in the y-axis is given by S =
| –1 0 |
| 0 1 |.
The combined transformation matrix is M = S × R (apply R first, then S).

逆时针旋转 90° 的矩阵为 R =
| 0 –1 |
| 1 0 |。
关于 y 轴的反射矩阵为 S =
| –1 0 |
| 0 1 |。
组合变换矩阵为 M = S × R（先应用 R，再应用 S）。

Perform the multiplication:
M =
| –1 0 | × | 0 –1 | = | (–1)×0+0×1 (–1)×(–1)+0×0 | = | 0 1 |
| 0 1 | | 1 0 | | 0×0+1×1 0×(–1)+1×0 | | 1 0 |.

计算乘法：
M =
| –1 0 | × | 0 –1 | = | 0 1 |
| 0 1 | | 1 0 | | 1 0 |。

Now transform each vertex by M:
A’: (0·1+1·2, 1·1+0·2) = (2, 1);
B’: (0·3+1·4, 1·3+0·4) = (4, 3);
C’: (0·2+1·5, 1·2+0·5) = (5, 2).
The new triangle has vertices (2,1), (4,3), (5,2). This compact matrix approach is central to real-time rendering.

现在用 M 变换每个顶点：
A’: (0·1+1·2, 1·1+0·2) = (2, 1)；
B’: (0·3+1·4, 1·3+0·4) = (4, 3)；
C’: (0·2+1·5, 1·2+0·5) = (5, 2)。
新三角形的顶点为 (2,1)、(4,3)、(5,2)。这种紧凑的矩阵方法对于实时渲染至关重要。

3. Maclaurin Series Approximation for Small Angles | 小角度的麦克劳林级数近似

In a physics experiment, a pendulum’s angle θ is measured as 0.1 rad, but the exact value of sin θ is needed for a period correction. Instead of relying on a calculator, we can use the Maclaurin series sin x ≈ x – x³/3! to obtain a high-quality approximation and estimate the error.

在物理实验中，一个单摆的角度 θ 测量值为 0.1 rad，但需要精确的 sin θ 值来进行周期修正。与其依赖计算器，我们可以使用麦克劳林级数 sin x ≈ x – x³/3! 来获得高质量的近似并估计误差。

The Maclaurin series for sin x is sin x = x – x³/3! + x⁵/5! – … . Truncating after the cubic term, the approximation error is bounded by the next term, |x⁵|/120. For x = 0.1, we compute:
sin(0.1) ≈ 0.1 – (0.1)³/6 = 0.1 – 0.001/6 = 0.1 – 0.0001667 = 0.0998333.

sin x 的麦克劳林级数为 sin x = x – x³/3! + x⁵/5! – … 。截断到三次项后，近似误差被下一项 |x⁵|/120 所界定。对于 x = 0.1，计算可得：
sin(0.1) ≈ 0.1 – (0.1)³/6 = 0.1 – 0.001/6 = 0.1 – 0.0001667 = 0.0998333。

The true value of sin(0.1) is approximately 0.0998334, so the error is only about 1 × 10⁻⁷. This demonstrates that the series gives an excellent approximation for small angles, and we can confidently use it in further calculations, such as estimating the pendulum length using T = 2π√(L/g).

sin(0.1) 的真实值约为 0.0998334，因此误差仅为大约 1 × 10⁻⁷。这表明该级数对小角度给出了极好的近似，我们可以放心地将其用于后续计算，例如用 T = 2π√(L/g) 估算摆长。

Always check the size of the next term to confirm the precision required. In exam problems, you may be asked to find the range of x for which the error is less than a given tolerance.

始终检查下一项的大小来确认所需的精度。在考试问题中，你可能需要找出使误差小于给定容许值的 x 的范围。

4. Modelling Cooling with First-Order Differential Equations | 用一阶微分方程建模冷却过程

Newton’s law of cooling states that the rate of change of an object’s temperature T is proportional to the difference between T and the ambient temperature Tₐ. A forensic scientist notes that a body initially at 80 °C in a room at 20 °C cools to 60 °C after 1 minute. The differential equation dT/dt = –k(T – 20) models the situation. By solving this equation, we can determine the cooling constant k and predict the time when the temperature reaches 30 °C.

牛顿冷却定律指出，物体温度 T 的变化率与 T 和环境温度 Tₐ 之差成正比。一位法医科学家注意到，在 20 °C 的房间中，一个初始温度为 80 °C 的物体在 1 分钟后冷却至 60 °C。微分方程 dT/dt = –k(T – 20) 对这个情况进行建模。通过求解该方程，我们可以确定冷却常数 k 并预测温度降至 30 °C 所需的时间。

Separate variables and integrate: ∫ 1/(T – 20) dT = ∫ –k dt ⇒ ln|T – 20| = –kt + C. Using T(0)=80, we find C = ln 60. Hence T – 20 = 60 e⁻ᵏᵗ.

分离变量并积分：∫ 1/(T – 20) dT = ∫ –k dt ⇒ ln|T – 20| = –kt + C。代入 T(0)=80，得 C = ln 60。因此 T – 20 = 60 e⁻ᵏᵗ。

After 1 minute, T = 60: 60 – 20 = 60 e⁻ᵏ ⇒ 40 = 60 e⁻ᵏ ⇒ e⁻ᵏ = 2/3 ⇒ –k = ln(2/3) ⇒ k = ln(3/2) ≈ 0.4055 min⁻¹.

1 分钟后，T = 60：60 – 20 = 60 e⁻ᵏ ⇒ 40 = 60 e⁻ᵏ ⇒ e⁻ᵏ = 2/3 ⇒ –k = ln(2/3) ⇒ k = ln(3/2) ≈ 0.4055 min⁻¹。

To find the time when T = 30: 30 – 20 = 60 e⁻ᵏᵗ ⇒ 10 = 60 e⁻ᵏᵗ ⇒ e⁻ᵏᵗ = 1/6 ⇒ –kt = ln(1/6) = –ln 6 ⇒ t = (ln 6)/k = ln 6 / ln(3/2) ≈ 1.7918 / 0.4055 ≈ 4.42 minutes. This case study highlights how differential equations provide crucial information in forensic investigations.

要求出 T = 30 时的时间：30 – 20 = 60 e⁻ᵏᵗ ⇒ 10 = 60 e⁻ᵏᵗ ⇒ e⁻ᵏᵗ = 1/6 ⇒ –kt = ln(1/6) = –ln 6 ⇒ t = (ln 6)/k = ln 6 / ln(3/2) ≈ 1.7918 / 0.4055 ≈ 4.42 分钟。这个案例研究凸显了微分方程如何在法医调查中提供关键信息。

5. Area Enclosed by a Cardioid in Polar Coordinates | 极坐标中心形线所围面积

A designer wants to create a heart-shaped logo using the cardioid r = 1 + cos θ. To estimate the amount of gold leaf required, the area enclosed by this polar curve must be calculated. The standard formula for polar area is A = ½ ∫ r² dθ, integrated over a full period 0 ≤ θ ≤ 2π.

一位设计师想要用心脏线 r = 1 + cos θ 创建一个心形徽标。为了估算所需金箔的量，必须计算这条极坐标曲线所围成的面积。极坐标面积的标准公式为 A = ½ ∫ r² dθ，积分区间为一个完整周期 0 ≤ θ ≤ 2π。

Compute r² = (1 + cos θ)² = 1 + 2cos θ + cos²θ. Use the identity cos²θ = ½(1 + cos 2θ) to rewrite: r² = 1 + 2cos θ + ½ + ½ cos 2θ = 3/2 + 2cos θ + ½ cos 2θ.

计算 r² = (1 + cos θ)² = 1 + 2cos θ + cos²θ。利用恒等式 cos²θ = ½(1 + cos 2θ) 改写：r² = 1 + 2cos θ + ½ + ½ cos 2θ = 3/2 + 2cos θ + ½ cos 2θ。

Now integrate from 0 to 2π: A = ½ ∫₀²π (3/2 + 2cos θ + ½ cos 2θ) dθ. The integral of cos θ over a full period is zero, as is that of cos 2θ. Therefore, A = ½ [ (3/2)θ ]₀²π = ½ × (3/2 × 2π) = ½ × 3π = 3π/2.

现在从 0 到 2π 积分：A = ½ ∫₀²π (3/2 + 2cos θ + ½ cos 2θ) dθ。cos θ 在一个完整周期上的积分为零，cos 2θ 也是如此。因此，A = ½ [ (3/2)θ ]₀²π = ½ × (3/2 × 2π) = ½ × 3π = 3π/2。

The enclosed area is therefore 3π/2 square units. This result appears frequently in OCR questions, and remembering symmetry properties can save time. For a general r = a(1 + cos θ), the area is 3πa²/2.

因此所围面积为 3π/2 平方单位。这个结果经常出现在 OCR 考题中，记住对称性可以节省时间。对于一般的 r = a(1 + cos θ)，面积为 3πa²/2。

6. Simple Harmonic Motion in a Mass-Spring System | 质量弹簧系统的简谐运动

A 0.5 kg mass attached to a spring oscillates horizontally. Its displacement x (in metres) from equilibrium is described by x(t) = A cos(ωt + φ). The motion is determined by ω = √(k/m), where k = 50 N/m is the spring constant. Initially, the mass is displaced 0.2 m to the right and released from rest. We need the amplitude, phase constant, period, and maximum speed.

一个 0.5 kg 的质量块连接在弹簧上水平振动。它相对平衡位置的位移 x（米）由 x(t) = A cos(ωt + φ) 描述。该运动由 ω = √(k/m) 决定，其中 k = 50 N/m 是弹簧的劲度系数。初始时，质量块被向右拉至 0.2 m 处并从静止释放。我们需要求出振幅、初相、周期和最大速率。

First, ω = √(50/0.5) = √100 = 10 rad/s. The initial conditions give x(0) = A cos φ = 0.2 and v(0) = –Aω sin φ = 0 (released from rest). Since ω ≠ 0, sin φ = 0 so φ = 0 or π. A must be positive, so φ = 0 and A = 0.2.

首先，ω = √(50/0.5) = √100 = 10 rad/s。初始条件给出 x(0) = A cos φ = 0.2 以及 v(0) = –Aω sin φ = 0（从静止释放）。由于 ω ≠ 0，sin φ = 0，因此 φ = 0 或 π。A 必须为正，故 φ = 0，且 A = 0.2。

The period T = 2π/ω = 2π/10 = π/5 ≈ 0.628 s. Velocity is v(t) = –Aω sin(ωt + φ) = –2 sin(10t). The maximum speed is |v_max| = Aω = 0.2 × 10 = 2 m/s, occurring when sin(10t) = ±1.

周期 T = 2π/ω = 2π/10 = π/5 ≈ 0.628 s。速度为 v(t) = –Aω sin(ωt + φ) = –2 sin(10t)。最大速率为 |v_max| = Aω = 0.2 × 10 = 2 m/s，出现在 sin(10t) = ±1 时。

This case shows how differential equation solutions (SHM) link directly to measurable physical quantities, a common OCR mechanics question.

这个案例展示了微分方程的解（简谐运动）如何直接与可测量的物理量联系起来，这是 OCR 力学中常见的问题。

7. Continuous Random Variables – Probability Density Function | 连续随机变量 – 概率密度函数

A manufacturer tests the lifespan X (in years) of a component. The probability density function is modelled by f(x) = kx(1 – x) for 0 ≤ x ≤ 1, and 0 otherwise. To fully characterise this distribution, we need to find the constant k, the expected value E(X), the variance Var(X), and the probability that a component fails before half a year, P(X < 0.5).

一家制造商对某部件的使用寿命 X（年）进行测试。概率密度函数模型为 f(x) = kx(1 – x)，0 ≤ x ≤ 1，其他处为 0。为了完整刻画这个分布，我们需要求出常数 k、期望值 E(X)、方差 Var(X) 以及部件在半年内失效的概率 P(X < 0.5)。

Since ∫₀¹ f(x) dx = 1, we set up: k ∫₀¹ x(1 – x) dx = 1. Compute ∫₀¹ (x – x²) dx = [½x² – ⅓x³]₀¹ = ½ – ⅓ = 1/6. Hence k × (1/6) = 1 ⇒ k = 6.

由于 ∫₀¹ f(x) dx = 1，我们建立方程：k ∫₀¹ x(1 – x) dx = 1。计算 ∫₀¹ (x – x²) dx = [½x² – ⅓x³]₀¹ = ½ – ⅓ = 1/6。因此 k × (1/6) = 1 ⇒ k = 6。

Now E(X) = ∫₀¹ x·6x(1 – x) dx = 6 ∫₀¹ (x² – x³) dx = 6 [⅓x³ – ¼x⁴]₀¹ = 6 (⅓ – ¼) = 6 × 1/12 = ½. Next, E(X²) = ∫₀¹ x²·6x(1 – x) dx = 6 ∫₀¹ (x³ – x⁴) dx = 6 [¼x⁴ – ⅕x⁵]₀¹ = 6 (¼ – ⅕) = 6 × 1/20 = 0.3. Thus Var(X) = E(X²) – [E(X)]² = 0.3 – 0.25 = 0.05.

现在 E(X) = ∫₀¹ x·6x(1 – x) dx = 6 ∫₀¹ (x² – x³) dx = 6 [⅓x³ – ¼x⁴]₀¹ = 6 (⅓ – ¼) = 6 × 1/12 = ½。接着，E(X²) = ∫₀¹

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Year 13 OCR Further Mathematics: Case Study Practical Exercises | 案例分析实战演练