📚 Year 10 Cambridge Additional Mathematics: Case Study Practice | Year 10 Cambridge 进阶数学：案例分析实战演练
Case studies bridge the gap between theoretical knowledge and real-world problem‑solving. In Cambridge IGCSE Additional Mathematics (0606), applying techniques to practical scenarios is essential for deep understanding. This article presents three detailed case studies that integrate quadratic functions, exponentials and logarithms, and calculus. Each case is broken down into clear steps, followed by reflective analysis. Work through them actively to sharpen your analytical skills.
案例分析可以弥合理论知识与实际问题解决之间的鸿沟。在剑桥 IGCSE 附加数学 (0606) 中，将技巧应用于实际情境对于深刻理解至关重要。本文将通过三个详细案例，综合运用二次函数、指数与对数以及微积分。每个案例均拆分为清晰的步骤，并附上反思分析。主动演练以增强你的分析能力。
1. The Case Study Mindset | 案例分析思维
Before diving in, adopt a structured approach: read the problem carefully, identify which mathematical topic applies, translate the words into equations, solve systematically and finally interpret the result in context.
在深入案例之前，请先建立结构化思维：仔细阅读问题，识别适用的数学主题，将文字转化为方程，系统求解，最后在情境中解读结果。
Always ask: What am I given? What am I trying to find? Which formula or technique links them? In Additional Mathematics, many problems combine multiple concepts, so breaking them into smaller, manageable parts is key.
始终问自己：已知什么？要求什么？哪个公式或技巧能联系两者？在附加数学中，许多问题综合了多个概念，因此将其拆分为小块、可管理的部分是关键。
2. Case 1: Projectile Motion & Quadratic Modelling | 案例一：抛体运动与二次模型
an initial speed of 20 m/s. The height h (in metres) after t seconds is modelled by h(t) = – want to find the maximum height reached and the time when the ball returns to the ground.
从地面以 20 m/s 的初速度向上抛出一个球。t 秒后的高度 h（米）由 h(t) = –5t² + 20t 建模。我们需要求出达到的最大高度以及球落回地面的时间。
ens downwards. The maximum height occurs at the vertex. The ground is reached when h(t)=0.
这是 t² 的系数为负的二次函数，因此抛物线开口向下。最大高度出现在顶点处。当 h(t)=0 时球回到地面。
The vertex’s t-coordinate is given by t = –b/(2a). Here a = –5, b = 20, so t = –20/(2 × –5) = 2 seconds. Substitute back to obtain height: h(2) = –5(2)² + 20(2) = –20 + 40 = 20 metres.
顶点的 t 坐标由 t = –b/(2a) 给出。这里 a = –5，b = 20，因此 t = –20/(2 × –5) = 2 秒。代回求得高度：h(2) = –5(2)² + 20(2) = –20 + 40 = 20 米。
tvertex = –b/(2a) ⟹ hmax = –5(2)² + 20(2) = 20 m
To find when the ball hits the ground, solve –5t² + 20t = 0. Factorising: –5t(t – 4) = 0, giving t = 0 (initial throw) or t = 4 seconds. Hence the total flight time is 4 s.
为求球击中地面的时间，解 –5t² + 20t = 0。因式分解：–5t(t – 4) = 0，得 t = 0（初始抛出）或 t = 4 秒。因此总飞行时间为 4 秒。
3. Interpreting Quadratic Parameters | 解读二次函数的参数
In the model h(t) = –5t² + 20t, the number –5 comes from ½(– velocity. Changing the initial velocity alters the vertex and the time to hit the ground.
在模型 h(t) = –5t² + 20t 中，–5 来自 ½(–g)，其中 g ≈ 10 m/s²。系数 20 是初速度。改变初速度会改变顶点和落地时间。
Suppose the initial speed were 25 m/s. Then h(t) = –5t² + 25t. The maximum height occurs at t = –25/(2 × –5) = 2.5 s, and h(2.5) = –5(2.5)² + 25(2.5) = 31.25 m. The ball lands when t = 5 s.
假设初速度为 25 m/s，则 h(t) = –5t² + 25t。最大高度在 t = –25/(2 × –5) = 2.5 秒，h(2.5) = –5(2.5)² + 25(2.5) = 31.25 米。球在 t = 5 秒时落地。
Real‑ elling often requires checking the domain: height cannot be negative, and time must be non‑negative. Always state the valid interval for t, e.g., 0 ≤ t ≤ 4 for the first case.
现实世界建模通常需要检查定义域：高度不能为负，时间必须非负。务必注明 t 的有效区间，例如在第一个案例中 0 ≤ t ≤ 4。
4. Case 2: Exponential Growth & Logarithms | 案例二：指数增长与对数
An investor places £1000 in an account offering 5% A after t years is A(t) = 1000 × 1.05ᵗ. How long will it take for the investment to double?
一位投资者将 1000 英镑存入年利率 5% 的复利账户。t 年后的金额 A 为 A(t) = 1000 × 1.05ᵗ。投资需要多久才能翻倍？
05ᵗ = 2000. Divide both sides by 1000 to get 1.05ᵗ = 2. Now use logarithms to solve for t.
建立方程 1000 × 1.05ᵗ = 2000。两边除以 1000 得 1.05ᵗ = 2。现在使用对数求解 t。
ides: ln(1.05ᵗ) = ln 2. Using the power rule: t · ln 1.05 = ln 2. Thus t = ln 2 / ln 1.05.
两边取自然对数（或常用对数）：ln(1.05ᵗ) = ln 2。使用幂法则：t · ln 1.05 = ln 2。因此 t = ln 2 / ln 1.05。
t = ln 2 ÷ ln 1.05 ≈ 0.6931 ÷ 0.04879 ≈ 14.2 years
Interpretation: it takes about 14.2 years for the investment to double. If the interest were compounded more frequently, the time would decrease slightly, but the logarithmic method remains the same.
解读：投资大约需要 14.2 年才能翻倍。如果复利更频繁，所需时间会略微缩短，但对数方法保持不变。
5. Using Logarithms to Analyse Change | 使用对数分析变化
alised. For any growth factor b (b > 1) and target multiple M, the time t satisfies bᵗ = M ⇒ t = ln M / ln b. For half‑life in decay, use b < 1.
翻倍时间公式可以推广。对于任意增长因子 b（b > 1）和目标倍数 M，时间 t 满足 bᵗ = M ⇒ t = ln M / ln b。对于衰变中的半衰期，使用 b < 1。
What if we wanted the investment to reach £1500? , t = ln 1.5 / ln 1.05 ≈ 0.4055 / 0.04879 ≈ 8.31 years. Notice that to grow by an extra 50% takes less than half the doubling time, highlighting the non‑linear nature of exponential growth.
如果我们希望投资达到 1500 英镑呢？那么 1.05ᵗ = 1.5，t = ln 1.5 / ln 1.05 ≈ 0.4055 / 0.04879 ≈ 8.31 年。注意到额外增长 50% 所需时间不到翻倍时间的一半，这体现了指数增长的非线性特性。
When solving logarithmic equations, always check the domain: the argument of a logarithm must be positive. The base b must be positive and not equal to 1. These constraints often filter out extraneous solutions.
解对数方程时，务必检查定义域：对数的真数必须为正。底数 b 必须为正且不等于 1。这些约束常常能过滤掉增根。
6. Recap of Logarithmic Laws for Case Studies | 案例必备的对数法则回顾
Before moving on, let’s consolidate the three fundamental log laws, as they are essential for manipulating exponential equations in context.
在继续之前，我们先巩固三条基本对数法则，它们在处理情境中的指数方程时至关重要。
| Law | Expression |
| Product rule | logₐ (xy) = logₐ x + logₐ y |
| Quotient rule | logₐ (x/y) = logₐ x – logₐ y |
| Power rule | logₐ (xⁿ) = n·logₐ x |
These laws also work for natural logs (ln) and common logs (lg). In our doubling problem, the power rule reduced t·ln 1.05 to handle the exponent.
这些法则同样适用于自然对数 (ln) 和常用对数 (lg)。在翻倍问题中，幂法则将 t·ln 1.05 拿下了指数。
Being fluent in these laws allows you to solve more complex scenarios, such as comparing two different saving plans or radioactive decay with background subtraction.
熟练运用这些法则使你能够解决更复杂的场景，例如比较两种不同的储蓄计划，或含有背景扣除的放射性衰变。
7. Case 3: Optimising Area with Calculus | 案例三：用微积分优化面积
A farmer has 20 metres of fencing and wants to enclose a rectangular area against a straight wall, using the wall as one side. What dimensions give the maximum enclosed area, and what is that maximum area?
一位农夫有 20 米围栏，想靠着一面笔直的墙围出一个矩形区域，用墙作为其中一边。什么样的尺寸能使围成的面积最大？最大面积是多少？
erpendicular sides each be x metres. Then the total fencing used is 2x + y = 20, so y = 20 – 2x. The area A = x · y = x(20 – 2x) = 20x – 2x².
设与墙平行的边长为 y 米，两个垂直的边各长 x 米。使用的围栏总长为 2x + y = 20，因此 y = 20 – 2x。面积 A = x · y = x(20 – 2x) = 20x – 2x²。
To maximise A, we differentiate: dA/dx = 20 – 4x. Set this to zero: 20 – 4x = 0 ⇒ x = 5 m. Then y = 20 – 2(5) = 10 m. The maximum area is A = 5 × 10 = 50 m².
为最大化 A，进行微分：dA/dx = 20 – 4x。令其为零：20 – 4x = 0 ⇒ x = 5 m。于是 y = 20 – 2(5) = 10 m。最大面积为 A = 5 × 10 = 50 m²。
A(x) = 20x – 2x² ⇒ A'(x) = 20 – 4x = 0 ⇒ x = 5, y = 10
8. Validating the Maximum with Second Derivative | 用二阶导数检验最大值
We should confirm that x = 5 gives a maximum, not a minimum. The second derivative is A''(x) = –4. Since A''(5) < 0, the function is concave down, confirming a local maximum.
我们应当确认 x = 5 给出的是最大值而非最小值。二阶导数为 A''(x) = –4。由于 A''(5) < 0，函数图形是凸向下的，确认了局部最大值。
lems, always check the endpoints of the feasible domain. Here x must be positive and y non‑negative: x > 0 and 20 – 2x ≥ 0 ⇒ → 0; at x = 10, y = 0, A = 0. So the stationary point indeed yields the absolute maximum.
在实际问题中，务必检查可行域的端点。此处 x 必须为正，y 非负：x > 0 且 20 – 2x ≥ 0 ⇒ 0 < x ≤ 10。当 x → 0⁺ 时 A → 0；当 x = 10 时 y = 0，A = 0。因此该驻点确实给出了绝对最大值。
t had been different, say a rope of length L, the analysis follows the same pattern: y = L – 2x, area = x(L – 2x), leading to optimal x = L/4, y = L/2.
如果周长约束不同，比如一根长度为 L 的绳子，分析模式相同：y = L – 2x，面积 = x(L – 2x)，最优解为 x = L/4，y = L/2。
9. Bridging Algebra and Calculus in Optimisation | 代数与微积分在优化中的衔接
restingly, this optimisation can also be tackled without calculus. The area expression 20x – 2x² is a quadratic. Its vertex occurs at x = –b/(2a) = –20/(2 × – giving the same optimal x. Thus quadratic theory and differentiation reinforce each other.
有趣的是，此优化问题也可不用微积分求解。面积表达式 20x – 2x² 是二次式。其顶点位于 x = –b/(2a) = –20/(2 × –2) = 5，给出相同的最优 x。因此二次函数理论与微分相互印证。
This duality is powerful: when you can model a scenario with a quadratic, vertex analysis instantly gives the optimum. When functions are more complex, differentiation becomes essential. Understanding both approaches builds mathematical flexibility.
这种双重性是强大的：当你能用二次式建模时，顶点分析立即给出最优解。当函数更复杂时，微分就变得必不可少。理解两种方法能构建数学灵活性。
Try modifying the case: suppose the farmer must leave a 1‑metre gap for a gate. Then the fencing equation becomes 2x + y + 1 = 20 ⇒ y = 19 – 2x. Area = x(19 – 2x). The new maximum shifts slightly—an excellent extension exercise.
尝试修改案例：假设农夫必须留出 1 米的缺口作为门。那么围栏方程变为 2x + y + 1 = 20 ⇒ y = 19 – 2x。面积 = x(19 – 2x)。新的最大值会略有偏移——这是一个极好的拓展练习。
10. Common Pitfalls and How to Avoid Them | 常见误区及规避方法
When working through case studies, students often rush to algebra without defining variables clearly. Always write down what each variable represents and its unit.
在进行案例分析时，学生常急于套用代数，而没有明确定义变量。务必写下每个变量代表什么及其单位。
Another mistake is forgetting to check constraints: time must be non‑negative, lengths positive, arguments of logs positive. In optimisation, verify that the stationary point is a maximum (or minimum) and check domain boundaries.
另一个错误是忘记检查约束：时间必须非负，长度为正，真数为正。在优化中，需验证驻点是最大值（或最小值）并检查定义域边界。
that ln 1 = 0 and logₐ 1 = 0, which can simplify expressions and reveal hidden solutions.
在对数问题中，避免除以可能为零的对数项。同时记住 ln 1 = 0 且 logₐ 1 = 0，这能简化表达式并揭示隐解。
11. Practice Extension – Mixed Skills | 拓展练习 – 混合技能
Try this: A toy rocket’s height is h(t) = – e rocket’s horizontal distance from launch is x = 3t, find the equation of the trajectory.
尝试此题：一个玩具火箭的高度为 h(t) = –4.9t² + 14.7t + 2。求 (a) 达到顶点的时间，(b) 最大高度，(c) 落地时间。然后，如果火箭离发射点的水平距离为 x = 3t，求轨迹方程。
is blends quadratic techniques with parametric thinking. Eliminating t from x = 3t and y = –4.9t² + 14.7t + 2 yields a quadratic trajectory. Such multi‑step problems are common in Additional Mathematics.
这融合了二次函数技巧与参数思维。从 x = 3t 和 y = –4.9t² + 14.7t + 2 中消去 t 可得一条二次轨迹。此类多步骤问题在附加数学中很常见。
consider an investment doubling problem with monthly compounding at a nominal rate of 4.8% per annum. Apply the formula A = P(1 + r/n)nt and solve using logs. This reinforces the power of logarithms beyond simple annual growth.
此外，考虑一个名义年利率 4.8%、按月复利的投资翻倍问题。应用公式 A = P(1 + r/n)nt 并用对数求解。这会强化对数在简单年增长之外的应用。
12. Key Takeaways & Exam Tips | 关键总结与考试建议
Case studies train you to ‘read’ mathematics. Always relate the mathematical result back to the context: state answers with units, interpret negative or extraneous solutions, and justify why a maximum is indeed a maximum.
案例分析训练你“阅读”数学。始终将数学结果与情境联系起来：回答时带上单位，解释负解或增根，并证明最大值确实是最大值。
In the exam, show all steps: define variables, write the equation, apply the technique (vertex formula, logs, differentiation), check conditions, and write a clear concluding sentence. Marks are awarded for method and interpretation, not just the final number.
考试中要展示所有步骤：定义变量，写出方程，应用技巧（顶点公式、对数、微分），检验条件，并写出清晰的结论句。评分不仅看最终数字，更重视方法与解读。
Finally, practise a variety of contexts – projectile motion, finance, geometry – so that you become confident in identifying the underlying mathematics quickly. The more case studies you explore, the more fluent your problem‑solving becomes.
最后，练习多种情境——抛体运动、金融、几何——以便你能自信地快速识别背后的数学。探索的案例越多，解题就会越流利。
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