📚 IGCSE Cambridge Additional Mathematics: Case Study Practical Exercises | IGCSE 剑桥进阶数学：案例分析实战演练
Case studies bridge the gap between theoretical concepts and real-world applications. In IGCSE Cambridge Additional Mathematics, combining functions, calculus, and equations in a single scenario sharpens problem-solving skills. This article presents a detailed case study of a parabolic arch bridge, guiding you through modelling, analysis, optimisation, and even a trigonometric wind-load challenge. Every stage is written the way an examiner expects to see it on paper: clear modelling statements, defined variables, correct notation, and units throughout.
案例分析能够弥合理论概念和实际应用之间的差距。在 IGCSE 剑桥进阶数学中，将函数、微积分和方程融汇于一个场景，可以增强解决问题的能力。本文通过一个抛物线拱桥的详细案例，带你一步步经历建模、分析、优化，甚至三角函数风荷载挑战。每一个步骤都按考官期望的书写方式呈现：明确的建模说明、定义的变量、规范记号以及完整的单位。
1. Setting the Scene: The Parabolic Arch | 场景设定：抛物线拱桥
ace the origin at the left base of the arch, so the arch touches the ground at (0, 0) and (20, 0). The highest point is at (10, 5). Choosing the origin at the left base keeps every x-value positive, which makes the domain easy to state: 0 ≤ x ≤ 20. The y-axis measures vertical height above ground level in metres.
想象一座抛物线形的石拱桥。拱跨度为 20 米，中心最高处距地面 5 米。我们将原点设在拱的左基脚，因此拱与地面相交于 (0, 0) 和 (20, 0)。最高点在 (10, 5) 处。把原点设在左基脚可以让所有 x 值都为正，便于直接写出定义域：0 ≤ x ≤ 20。y 轴表示距地面的竖直高度，单位为米。
- Span (horizontal width): 20 m | 跨度（水平宽度）：20 米
- Maximum height (rise): 5 m at the centre | 最大高度（矢高）：中心处 5 米
- Known points: (0, 0), (10, 5), (20, 0) | 已知点：(0, 0)、(10, 5)、(20, 0)
- Line of symmetry: x = 10 | 对称轴：x = 10
2. Modelling the Arch Equation | 建立拱门方程
working is 5 = a × 10 × (10 - 20) = a × 10 × (-10) = -100a, so a = -0.05. Notice that a is negative, which is exactly what we expect because the parabola opens downwards, like a bridge arch.
因为抛物线的 x 截距为 0 和 20，其方程可写成因式分解形式 y = a x (x - 20)。代入顶点 (10, 5) 求 a：计算过程为 5 = a × 10 × (10 - 20) = a × 10 × (-10) = -100a，得 a = -0.05。注意 a 为负数，这正是我们预期的结果，因为抛物线开口向下，形状如同拱桥。
Therefore the arch is modelled by:
因此拱的数学模型为：
y = -0.05x² + x, 0 ≤ x ≤ 20
Expanding -0.05x(x - 20) gives -0.05x² + x, which is the standard quadratic form used later for differentiation and integration. Always state the domain; a bridge arch exists only between its two bases, and marks are often awarded for this statement alone.
将 -0.05x(x - 20) 展开得 -0.05x² + x，这就是后面用于求导和积分的一般二次形式。务必写出定义域；拱桥只存在于两个基脚之间，仅这一句说明往往就能得分。
3. Verifying Key Points | 验证关键点
symmetric about the vertical line x = 10, so the height at x = 10 + t always equals the height at x = 10 - t. Verification is not optional decoration: it proves your model matches the physical situation.
检验端点：当 x = 0 时 y = 0；当 x = 20 时 y = -0.05 × 400 + 20 = -20 + 20 = 0。最大值出现在 x = 10，得 y = -0.05 × 100 + 10 = -5 + 10 = 5。拱关于竖直线 x = 10 对称，因此 x = 10 + t 处的高度总等于 x = 10 - t 处的高度。验证并非可有可无的装饰：它证明你的模型与实际情况相符。
| x (m) | 0 | 5 | 10 | 15 | 20 |
| y (m) | 0 | 3.75 | 5 | 3.75 | 0 |
4. Completing the Square: Vertex Form | 配方：顶点式
20x). Completing the square inside the bracket gives x² - 20x = (x - 10)² - 100, so y = -0.05[(x - 10)² - 100] = -0.05(x - 10)² + 5.
同一个模型可以通过配方改写成顶点式，直接暴露顶点坐标。从 y = -0.05x² + x 出发，提取 -0.05：y = -0.05(x² - 20x)。括号内配方得 x² - 20x = (x - 10)² - 100，因此 y = -0.05[(x - 10)² - 100] = -0.05(x - 10)² + 5。
y = -0.05(x - 10)² + 5
r the maximum height or the position of the highest point, completing the square is often faster and less error-prone than differentiating, and it also gives you a free check on your derivative work later.
这立即确认了三个事实：顶点为 (10, 5)，最大值为 5，对称轴为 x = 10。在考试中，如果题目要求最大高度或最高点位置，配方往往比求导更快、更不易出错，而且还能为后续的求导计算提供免费的检验。
5. Interpreting the Gradient: Differentiation | 梯度解读：求导
tationary point. At the right base, x = 20, the gradient is -1. The gradient of the tangent equals the tangent of the angle the arch makes with the horizontal, so tan θ = 1 gives θ = 45° at each base.
对模型逐项求导：dy/dx = -0.1x + 1。在左基脚 x = 0 处，梯度为 1；在顶点 x = 10 处，梯度为 0，证实这是一个驻点；在右基脚 x = 20 处，梯度为 -1。切线的梯度等于拱与水平方向夹角的正切，由 tan θ = 1 可得两个基脚处的倾角均为 45°。
dy/dx = -0.1x + 1
be a maximum. Writing this justification earns the final method mark in most mark schemes.
二阶导数为 d²y/dx² = -0.1，对所有 x 都为负，说明曲线处处上凸（开口向下），因此 x = 10 处的驻点必为极大值点。写出这一论证可在多数评分标准中获得最后的方法分。
6. Optimisation: The Largest Rectangular Clearance | 优化：最大矩形净空
the largest rectangular opening that fits under the arch, centred on the axis of symmetry? 5 - 0.05t². Therefore the cross-sectional area is A = w h = 2t(5 - 0.05t²) = 10t - 0.1t³.
一个工程实际问题：能放进拱下的最大矩形开口是多少？设开口从 x = 10 - t 延伸到 x = 10 + t，则宽度 w = 2t。其高度等于 x = 10 + t 处的拱高，即 h = -0.05(10 + t)² + (10 + t) = 5 - 0.05t²。因此横截面积 A = w h = 2t(5 - 0.05t²) = 10t - 0.1t³。
A = 10t - 0.1t³, t > 0
Differentiate and set to zero: dA/dt = 10 - 0.3t² = 0, so t² = 100/3 and t = 10/√3 e d²A/dt² = -0.6t is negative for t > 0, confirming a maximum. The optimum width is 20/√3 ≈ 11.55 m, the optimum height is 10/3 ≈ 3.33 m, and the maximum area is 200√3/9 ≈ 38.49 m².
求导并令其为零：dA/dt = 10 - 0.3t² = 0，得 t² = 100/3，t = 10/√3 ≈ 5.77 米。二阶导数 d²A/dt² = -0.6t 在 t > 0 时为负，确认是极大值。最优宽度为 20/√3 ≈ 11.55 米，最优高度为 10/3 ≈ 3.33 米，最大面积为 200√3/9 ≈
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