📚 Case Study Practice: IGCSE Edexcel Further Pure Mathematics | IGCSE Edexcel 进阶数学：案例分析实战演练
In IGCSE Edexcel Further Pure Mathematics, applying abstract concepts to real-world situations is often the key to mastering the subject. This article presents a series of carefully selected case studies that demonstrate how differentiation, integration, complex numbers, matrices, vectors, series and exponential models are used in business, physics, engineering and everyday financial planning. Each case study walks you through the problem, the mathematical model, the solution steps and a final interpretation, helping you build the analytical skills required for the exam.
在IGCSE Edexcel进阶数学中，将抽象概念应用到实际场景往往是掌握这门学科的关键。本文通过一系列精心挑选的案例研究，展示了微分、积分、复数、矩阵、向量、级数和指数模型在商业、物理、工程和日常理财中的运用。每个案例都会引导你分析问题、建立数学模型、完成求解步骤并给出最终解释，帮助你培养考试所需的逻辑分析能力。
1. Profit Maximisation Using Differentiation | 案例一：用微分求利润最大化
A manufacturing company produces and sells a gadget. The total cost, in pounds, of producing q units is modelled by C(q) = 500 + 20q + 0.1q2. The revenue from selling q units is R(q) = 100q − 0.2q2. The profit function P(q) is the difference between revenue and cost.
某制造企业生产并销售一种小装置。生产q件产品的总成本（英镑）为C(q) = 500 + 20q + 0.1q²，销售收入为R(q) = 100q − 0.2q²。利润函数P(q)是收入与成本之差。
First, we write the profit function: P(q) = (100q − 0.2q2) − (500 + 20q + 0.1q2) = 80q − 0.3q2 − 500. To find the output that maximises profit, we differentiate P with respect to q.
首先写出利润函数：P(q) = (100q − 0.2q²) − (500 + 20q + 0.1q²) = 80q − 0.3q² − 500。为找到使利润最大的产量，我们对P关于q求导。
P′(q) = 80 − 0.6q
Setting P′(q) = 0 gives 80 − 0.6q = 0, hence q = 133.33. Since partial units are not possible, we check q = 133 and q = 134. The second derivative P″(q) = −0.6 < 0 confirms a maximum at the stationary point.
令P′(q) = 0，得80 − 0.6q = 0，因此q = 133.33。由于不能生产非整数件，我们检查q = 133和q = 134。二阶导数P″(q) = −0.6 < 0，确认该驻点为极大值点。
- At q = 133, P ≈ 80(133) − 0.3(133)² − 500 = £5 333.30.
- At q = 134, P ≈ 80(134) − 0.3(134)² − 500 = £5 333.20.
- 当q = 133时，P ≈ 80 × 133 − 0.3 × 133² − 500 = £5 333.30。
- 当q = 134时，P ≈ 80 × 134 − 0.3 × 134² − 500 = £5 333.20。
The analysis reveals that producing 133 units yields the highest profit, roughly £5 333.30. This case illustrates how differentiation helps businesses identify optimal production levels without costly trial and error.
分析表明，生产133件产品可获得最高利润，约为5 333.30英镑。这个案例展示了微分如何帮助企业确定最优生产水平，避免代价高昂的试错过程。
2. Kinematics – Motion Under Gravity | 案例二：运动学 – 重力作用下的运动
A ball is projected vertically upwards from ground level with an initial speed of 20 m s⁻¹. Assuming the acceleration due to gravity is 10 m s⁻² and air resistance is negligible, the height h (in metres) after t seconds is given by h = 20t − 5t².
一个小球从地面以20 m s⁻¹的初速度竖直向上抛出。设重力加速度为10 m s⁻²，空气阻力不计，则t秒后的高度h（米）由 h = 20t − 5t² 给出。
To determine the maximum height, we find when the velocity is zero. Velocity is the derivative of displacement, so v = dh/dt = 20 − 10t. Setting v = 0 yields t = 2 s.
要确定最大高度，我们需要求出速度为零的时刻。速度是位移的导数，即 v = dh/dt = 20 − 10t。令v = 0，得 t = 2 s。
h(2) = 20(2) − 5(2)² = 40 − 20 = 20 m
The ball reaches its peak of 20 metres after 2 seconds. The total time of flight is found when h = 0 again (excluding t = 0). Solving 20t − 5t² = 0 → 5t(4 − t) = 0 gives t = 4 s.
小球在2秒后到达最高点20米。求全程飞行时间，令h = 0（除去t = 0）。解方程 20t − 5t² = 0 → 5t(4 − t) = 0，得 t = 4 s。
The impact speed is given by v(4) = 20 − 10(4) = −20 m s⁻¹ (downwards). This neat symmetry between ascent and descent is characteristic of constant acceleration problems and often appears in exam kinematics questions.
落地速度为 v(4) = 20 − 10 × 4 = −20 m s⁻¹（向下）。这种上升与下降的对称性是匀加速运动问题的典型特征，经常出现在运动学考题中。
3. Exponential Growth of a Bacterial Colony | 案例三：细菌菌落的指数增长
A laboratory experiment starts with 1000 bacteria. The population grows according to the differential equation dN/dt = kN, where N is the number of bacteria and t is measured in hours. After 2 hours, the population is counted at 2500.
某实验开始时培养1000个细菌。菌落数量遵循微分方程 dN/dt = kN，其中N为细菌数量，t以小时计。2小时后，菌落数量达到2500。
The general solution is N = N₀ ekt. Substituting N₀ = 1000, t = 2, N = 2500 we obtain 2500 = 1000 e2k → e2k = 2.5. Taking natural logarithms gives 2k = ln 2.5, so k = ½ ln 2.5 ≈ 0.4581.
通解为 N = N₀ ekt。代入 N₀ = 1000, t = 2, N = 2500 得 2500 = 1000 e2k → e2k = 2.5。取自然对数得 2k = ln 2.5，因而 k = ½ ln 2.5 ≈ 0.4581。
We can now predict the population at t = 4 hours: N(4) = 1000 e4k = 1000 (e2k)² = 1000 × (2.5)² = 6250. This shows the power of exponential models – the population more than doubled again in the next two hours.
现在可以预测t = 4小时的菌落数量：N(4) = 1000 e4k = 1000 (e2k)² = 1000 × (2.5)² = 6250。这显示出指数模型的威力——在接下来的两小时内，数量又翻了一倍多。
The following table illustrates the growth pattern:
下表展示了增长规律：
| Time t (hours) | Population N |
|---|---|
| 0 | 1000 |
| 2 | 2500 |
| 4 | 6250 |
| 6 | 15625 (est.) |
Exponential growth is widely used in biology, finance and nuclear physics. The IGCSE Further Pure syllabus expects you to link the differential equation directly to the exponential function and to interpret constants in context.
指数增长广泛应用于生物学、金融学和核物理领域。IGCSE进阶数学大纲要求你能够直接把微分方程与指数函数建立联系，并结合实际情境解释参数的含义。
4. Complex Impedance in an AC Circuit | 案例四：交流电路中的复阻抗
In electrical engineering, alternating current (AC) circuits can be analysed using complex numbers. A simple series circuit contains a resistor and an inductor. The impedance Z is given by Z = 3 + 4j ohms, where the real part represents resistance and the imaginary part represents inductive reactance.
在电气工程中，交流电路可以用复数进行分析。一个简单的串联电路包含一个电阻和一个电感。其阻抗Z为 Z = 3 + 4j 欧姆，其中实部代表电阻，虚部代表感抗。
If a voltage of V = 10 V (rms) is applied, the current I is found using Ohm’s law: I = V / Z = 10 / (3 + 4j). To express the current in rectangular form, multiply numerator and denominator by the complex conjugate (3 − 4j).
如果施加电压为 V = 10 V（有效值），根据欧姆定律可求电流 I = V / Z = 10 / (3 + 4j)。为将电流表示为直角坐标形式，我们将分子分母同乘以共轭复数 (3 − 4j)。
I = 10(3 − 4j) / (9 + 16) = (30 − 40j) / 25 = 1.2 − 1.6j A
The magnitude of the current is |I| = √(1.2² + 1.6²) = √(1.44 + 2.56) = √4 = 2 A. The phase angle is arg(I) = arctan(−1.6 / 1.2) ≈ −53.1°, meaning the current lags behind the voltage by approximately 53.1°.
电流的模为 |I| = √(1.2² + 1.6²) = √(1.44 + 2.56) = √4 = 2 A。辐角为 arg(I) = arctan(−1.6 / 1.2) ≈ −53.1°，这意味着电流滞后电压约53.1°。
This case study reinforces operations with complex numbers, such as division using conjugates and conversion between Cartesian and polar forms – all essential skills in the Further Pure syllabus.
这个案例巩固了复数的运算，比如利用共轭进行除法，以及直角坐标与极坐标形式的相互转换——这些都是进阶数学大纲中的核心技能。
5. Matrix Transformation for Image Rotation | 案例五：图像旋转的矩阵变换
Computer graphics and robotics rely on matrix transformations to manipulate shapes. Consider a point P with coordinates (2, 0). We want to rotate it counterclockwise by 60° about the origin. The rotation matrix is given by:
计算机图形学和机器人学依赖矩阵变换来操控图形。考虑坐标为 (2, 0) 的点P，我们想让它绕原点逆时针旋转60°。旋转矩阵如下：
R = [ [cos 60° , −sin 60°] , [sin 60° , cos 60°] ]
Using cos 60° = 0.5 and sin 60° = √3/2 ≈ 0.866, the transformation is applied by multiplying R with the column vector P.
因为 cos 60° = 0.5，sin 60° = √3/2 ≈ 0.866，我们把旋转矩阵R与列向量P相乘即可完成变换。
P’ = R × P = [ [0.5, −0.866] , [0.866, 0.5] ] × [2; 0] = [1; 1.732]
The new coordinates are (1, 1.732). The distance from the origin remains √(1² + 1.732²) = √(1+3) = 2, confirming that rotation preserves length. This property is a direct consequence of the determinant of R being +1.
新坐标为 (1, 1.732)。与原点的距离保持为 √(1² + 1.732²) = √(1+3) = 2，验证了旋转变换保长。这是旋转矩阵行列式为+1的直接结果。
Further Pure students must be able to apply 2×2 matrices to vectors and interpret the geometrical effect of the transformation. Such case studies help visualise the abstract algebra.
进阶数学的学生必须能够将2×2矩阵作用于向量，并解释变换的几何效应。这种案例研究有助于将抽象的代数可视化。
6. Vector Navigation – Ship Journey | 案例六：向量导航 – 船舶航行
A ship leaves port O and sails with a constant velocity vector v = (10i + 24j) km h⁻¹, where i points east and j points north. After 2 hours, the ship reaches point A. We need to find its distance from O and the bearing.
一艘船离开港口O，以恒定速度向量 v = (10i + 24j) km h⁻¹ 航行，其中i指向正东，j指向正北。2小时后，船到达A点。我们需要求出它与O点的距离及方位角。
The position vector after 2 hours is OA = 2v = 20i + 48j. The distance is the magnitude |OA| = √(20² + 48²) = √(400 + 2304) = √2704 = 52 km.
2小时后的位置向量为 OA = 2v = 20i + 48j。距离为模长 |OA| = √(20² + 48²) = √(400 + 2304) = √2704 = 52 km。
The bearing from O is the angle measured clockwise from north. The velocity vector makes an angle θ north of east where tan θ = 24/10 = 2.4, thus θ ≈ 67.4°. The bearing is therefore 090° − 67.4° = 022.6° (to the nearest 0.1°). More directly, bearing = arctan(10/24) measured from north, giving the same result.
从O点出发的方位角是从正北顺时针度量的角度。速度向量东偏北的角度θ满足 tan θ = 24/10 = 2.4，故θ ≈ 67.4°。因此方位角为 090° − 67.4° = 022.6°（精确到0.1°）。更直接的方法是，从正北起算的方位角为 arctan(10/24)，结果相同。
At point A, the ship changes course and heads directly towards point B located at (60i + 20j). The new displacement vector AB = (60−20)i + (20−48)j = 40i − 28j. The new distance to travel is |AB| = √(40² + 28²) = √(1600 + 784) = √2384 ≈ 48.8 km.
在A点，船改变航向，径直朝位于 (60i + 20j) 的点B驶去。新的位移向量 AB = (60−20)i + (20−48)j = 40i − 28j。新航程的距离为 |AB| = √(40² + 28²) = √(1600 + 784) = √2384 ≈ 48.8 km。
Vector methods provide a powerful tool for solving navigation problems without drawing scale diagrams, making them indispensable for accurate route planning.
向量方法为解决导航问题提供了强有力的工具，无需绘制比例图便可精确规划航线，在实际应用中不可或缺。
7. Geometric Progression and Loan Interest | 案例七：等比数列与贷款利息
Grace borrows £5 000 from a bank at an annual compound interest rate of 5%. She decides to repay £1 000 at the end of each year. We want to model the outstanding debt after each payment and determine how many years it takes to repay the loan fully.
Grace向银行贷款5 000英镑，年利率5%，复利计息。她决定每年底偿还1 000英镑。我们希望建立模型，求出每次还款后剩余的债务，并计算需要多少年才能还清贷款。
Let the amount owed after n years (and after the n-th payment) be denoted by Dn. The recurrence relation is Dn = 1.05 Dn−1 − 1000, with D0 = 5000. This is a linear recurrence that can be solved by iteration or by using the sum of a geometric series.
设n年后（且已支付第n期款项后）的欠款为Dₙ。递推关系为 Dₙ = 1.05 Dₙ₋₁ − 1000，初始 D₀ = 5000。这个线性递推可通过迭代或等比数列求和解决。
After expanding, the outstanding debt after k payments is: Dk = 5000 × 1.05k − 1000 × (1.05k − 1) / 0.05. We need the smallest integer k such that Dk ≤ 0.
展开后，k次还款后的剩余债务为：Dₖ = 5000 × 1.05ᵏ − 1000 × (1.05ᵏ − 1) / 0.05。我们需要找到使 Dₖ ≤ 0 的最小整数k。
Evaluating for successive years:
- k = 5: D₅ ≈ 5000×1.2763 − 1000(1.2763−1)/0.05 = 6381.5 − 5526 = £855.50
- k = 6: D₆ ≈ 1.05×855.5 − 1000 = 898.28 − 1000 = −£101.72 (overpaid)
依次计算各年：
- k = 5：D₅ ≈ 5000×1.2763 − 1000×(1.2763−1)/0.05 = 6381.5 − 5526 = £855.50
- k = 6：D₆ ≈ 1.05×855.5 − 1000 = 898.28 − 1000 = −£101.72（多还）
The loan is fully repaid during the 6th year. To avoid overpayment, Grace could adjust the final payment to about £898.28. This case demonstrates how geometric series model real instalment plans, a topic regularly examined in Further Pure.
贷款在第6年还清。为避免多付，Grace可将最后一笔还款调整为约898.28英镑。该案例展示了等比数列如何模拟现实分期付款计划，这是进阶数学中常考的题型。
8. Area Between Curves in Architecture | 案例八：建筑学中的曲线间面积
An architect designs a parabolic archway following the curve y = 4 − x². The arch is to be constructed above the x‑axis between x = −2 and x = 2. The client asks for the cross‑sectional area of the arch to estimate the amount of material needed for the façade.
一位建筑师设计了一个抛物线拱门，其曲线方程为 y = 4 − x²。拱门建在x轴上方，从 x = −2 到 x = 2。客户要求计算拱门的横截面积，以估算外立面所需材料量。
The area bounded by the curve and the x‑axis is given by the definite integral ∫₋₂² (4 − x²) dx. We integrate term by term: ∫ 4 dx = 4x, ∫ x² dx = x³/3.
曲线与x轴围成的面积由定积分 ∫₋₂² (4 − x²) dx 给出。我们逐项积分：∫ 4 dx = 4x，∫ x² dx = x³/3。
Area = [4x − x³/3]₋₂²
Substituting the upper limit: 4(2) − (8)/3 = 8 − 8/3 = 16/3. Substituting the lower limit: 4(−2) − (−8)/3 = −8 + 8/3 = −16/3. Subtracting lower from upper yields (16/3) − (−16/3) = 32/3 ≈ 10.67 square units.
代入上限：4×2 − 8/3 = 8 − 8/3 = 16/3。代入下限：4×(−2) − (−8)/3 = −8 + 8/3 = −16/3。上限减下限得 (16/3) − (−16/3) = 32/3 ≈ 10.67 平方单位。
If the scale is 1 unit = 1 metre, the arch’
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