Case Study Practical Exercises for Year 9 WJEC Further Mathematics | 九年级 WJEC 进阶数学：案例分析实战演练

📚 Case Study Practical Exercises for Year 9 WJEC Further Mathematics | 九年级 WJEC 进阶数学：案例分析实战演练

In Further Mathematics, real-world problems often require a multi-step approach: identifying the underlying mathematical model, formulating equations, solving them, and interpreting results in context. This article presents a series of case studies designed to sharpen your applied skills, covering topics such as algebra, geometry, functions, trigonometry, sequences, calculus, vectors and probability. Each case study models a practical scenario you might encounter in engineering, finance, science or daily life, helping you to bridge the gap between theory and practice.

在进阶数学中，现实世界的问题通常需要多步骤方法：识别潜在的数学模型、建立方程、求解并在情境中解释结果。本文提供一系列案例研究，旨在提升你的应用技能，涵盖代数、几何、函数、三角学、数列、微积分、向量和概率等主题。每个案例模拟了工程、金融、科学或日常生活中可能遇到的实际情景，帮助你跨越理论与实践的鸿沟。

1. Investment Growth Using Exponential Functions | 使用指数函数的投资增长

A sum of £500 is invested in a savings account that offers 4% nnum. The amount after t years is modelled by the exponential function A(t) = 500 × (1.04)ᵀ. We want to find how many years it takes for the investment to double.

一笔500英镑存入年利率为4%的复利储蓄账户。t年后的金额由指数函数 A(t) = 500 × (1.04)ᵀ 建模。我们需要求出投资翻倍所需的年数。

Set up the equation 1000 = 500 × (1.04)ᵀ, which simplifies to 2 = (1.04)ᵀ g 1.04 ≈ 0.3010 / 0.0170 ≈ 17.7 years. Thus the investment doubles in just under 18 years.

建立方程 1000 = 500 × (1.04)ᵀ，简化为 2 = (1.04)ᵀ。取对数得 log 2 = t log 1.04，因此 t = log 2 / log 1.04 ≈ 0.3010 / 0.0170 ≈ 17.7 年。所以投资在不到18年时翻倍。

If the interest were compounded quarterly, the model would become A = 500 × (1 + 0.04/4)^4t = 500 × (1.01)^4t. Comparing the two scenarios shows how compounding frequency affects growth.

如果利息按季度复利，模型将变为 A = 500 × (1 + 0.04/4)^4t = 500 × (1.01)^4t。比较这两种情况可以看出复利频率对增长的影响。

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2. Maximising Area with Fixed Perimeter | 固定周长下的最大面积

A farmer has 80 metres of fencing and wants to enclose a rectangular field against an existing wall, using the wall as one side of the rectangle. The objective is to maximise the area that can be enclosed.

一位农民有80米长的围栏，想利用一面已有的墙作为矩形的一边，围出一个矩形场地。目标是最大化可围成的面积。

rea A = x × y = x(80 − 2x) = 80x − 2x². This is a quadratic function opening downwards, so the maximum occurs at the vertex. The x-coordinate of the vertex is x = −b/(2a) = −80/(2 × (−2)) = 20. Then y = 80 − 40 = 40, giving maximum area A = 20 × 40 = 800 m².

设与墙平行的一边为 y 米，两个垂直边各为 x 米。所用围栏满足 y + 2x = 80，因此 y = 80 − 2x。面积 A = x × y = x(80 − 2x) = 80x − 2x²。这是一个开口向下的二次函数，最大值出现在顶点处。顶点的 x 坐标为 x = −b/(2a) = −80/(2 × (−2)) = 20。于是 y = 80 − 40 = 40，最大面积 A = 20 × 40 = 800 平方米。

ndeed gives the highest area. This case illustrates how quadratic optimisation can solve practical planning problems.

我们可以用数值表验证：x = 15 时，A = 15 × 50 = 750；x = 25 时，A = 25 × 30 = 750。计算出的最优解确实给出了最大面积。该案例说明了二次优化如何解决实际规划问题。

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3. Modelling the Path of a Projectile | 射弹路径的建模

model h = 12t − 5t², where 5 is half the acceleration due to gravity (using g ≈ 10 m/s²).

一个足球从地面以12米/秒的初始垂直速度踢出。t秒后的高度 h（米）由二次模型 h = 12t − 5t² 给出，其中5是重力加速度的一半（使用 g ≈ 10 米/秒²）。

f the parabola. The time to reach maximum height is t = −12/(2×(−5)) = 1.2 s, giving h = 12×1.2 − 5×(1.2)² = 14.4 − 7.2 = 7.2 metres.

为求球落地的时间，解 h = 0：12t − 5t² = 0 → t(12 − 5t) = 0，得 t = 0 或 t = 12/5 = 2.4 秒。最大高度在抛物线的顶点处达到。到达最高点的时间为 t = −12/(2×(−5)) = 1.2 秒，高度 h = 12×1.2 − 5×(1.2)² = 14.4 − 7.2 = 7.2 米。

6 metres, by solving 12t − 5t² > 6 → −5t² + 12t − 6 > 0. The roots are t ≈ 0.633 and t ≈ 1.767, so the ball is above 6 m for about 1.134 seconds.

该模型假设空气阻力可忽略不计。通过分析二次函数，我们还可以求出高度超过6米的时间段，解 12t − 5t² > 6 → −5t² + 12t − 6 > 0。根为 t ≈ 0.633 和 t ≈ 1.767，因此球在6米以上的时间约为1.134秒。

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4. Solving Trigonometric Problems in Navigation | 导航中的三角问题

A ship sails from a port on a bearing of 060° for 20 km to point A, then changes course to a bearing of 150° and sails 30 km to point B. We need to find the distance and bearing from the starting port to point B.

一艘船从港口以060°的方位角航行20公里到达A点，然后改变航向至150°的方位角，航行30公里到达B点。我们需要计算从出发港口到B点的距离和方位角。

southeast. Calculate the resultant displacement using vector components or the cosine rule. The angle between the two legs is 150° − 60°? g as a vector. Leg 1: (20 cos 60°, 20 sin 60°) in an (east, north) system where 0° is east? °. 150° is 150° from north, giving components: north = 30 cos 150° = 30 cos 150° = 30 × (−√3/2)? Wait, cos 150° = −cos 30° = −√3 splacement = 20 sin 60° + 30 sin 150° = 20×0.8660 + 30×0.5 = 17.32 + 15 = 32.32 km. The distance from port is √( (−15.98)² + (32.32)² ) ≈ √255 + 1044 = √1299 ≈ 36.0 km. The bearing α satisfies tan α = east displacement / north displacement? t direction is in the southeast quadrant. The angle from south towards east: tan θ = east/|north| om north is 180° − 63.7° = 116.3°. Thus the port to B is about 36 km on a bearing of 116°.

绘制示意图。第一段航程与正北方向成60°角，第二段航程的方位角为150°。使用向量分量或余弦定理计算合位移。将每段航程表示为向量：第一段北向分量 = 20 cos 60° = 10 km，东向分量 = 20 sin 60° ≈ 17.32 km；第二段北向分量 = 30 cos 150° = −25.98 km，东向分量 = 30 sin 150° = 15 km。总北向位移 = 10 − 25.98 = −15.98 km，总东向位移 = 17.32 + 15 = 32.32 km。距离 = √( (−15.98)² + (32.32)² ) ≈ 36.0 km。方位角 α 满足 tan α = 东向位移 / 北向位移的绝对值（考虑象限）。由于北向为负，合向量指向东南，从正北顺时针度量的角度为 180° − arctan(32.32/15.98) ≈ 180° − 63.7° = 116.3°。因此从港口到B点的距离约36公里，方位角约116°。

This problem integrates non-right-angled triangle trigonometry and vector addition, key concepts in Further Mathematics that are essential for navigation and surveying.

该问题综合了非直角三角形三角学和向量加法，这是进阶数学中的关键概念，对导航和测量至关重要。

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5. Sequences and Loan Repayments | 数列与贷款还款

A student borrows £1000 and agrees to pay back £75 at the end of each month. Interest of 1.5% per month is charged on the outstanding balance. We need to model the balance over time and determine when the loan will be fully repaid.

一名学生借款1000英镑，同意在每月底偿还75英镑。未偿还余额每月收取1.5%的利息。我们需要模拟余额的变化，并确定贷款何时能完全还清。

5 − 75 = 1015 − 75 = 940; B₂ = 940×1.015 − 75 = 954.1 − 75 = 879.1; B₃ ≈ 879.1×1.015 − 75 = 892.29 − 75 = 817.29. Continuing this, we find the balance becomes zero after about 15 months. This is an example of an arithmetico-geometric sequence.

设 Bₙ 为 n 个月后的余额。递推关系为 Bₙ = Bₙ₋₁ × 1.015 − 75，且 B₀ = 1000。我们可以计算出前几项：B₁ = 1000×1.015 − 75 = 940；B₂ = 940×1.015 − 75 = 879.1；B₃ ≈ 879.1×1.015 − 75 = 817.29。继续计算，余额约在15个月后变为零。这是一个算术—几何数列的例子。

olve the closed form. The sequence can be shown to be Bₙ = (1000 − 5000) × (1.015)ⁿ + 5000? r a recurrence uₙ = ruₙ₋₁ − d, the general solution is uₙ = (u₀ − d/(1−r)) rⁿ + d/(1−r) if |r| 00/4000 = 1.25, so n = log 1.25 / log 1.015 ≈ 0.09691/0.006466 ≈ 14.99. So the loan is repaid in the 15th month.

为求得确切的月数，可以求解闭式。递推关系 uₙ = ruₙ₋₁ − d 的通解为 uₙ = (u₀ − d/(1−r)) rⁿ + d/(1−r)，其中 r=1.015，d=75，固定点 L = 75/0.015 = 5000。于是 Bₙ = (1000 − 5000)×1.015ⁿ + 5000 = −4000×1.015ⁿ + 5000。令 Bₙ = 0 得 1.015ⁿ = 1.25，n = log 1.25 / log 1.015 ≈ 14.99。因此贷款在第15个月还清。

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6. Rates of Change in Biology (Differentiation) | 生物学中的变化率（微分）

The population of a bacterial culture is modelled by P(t) = 200e^0.3t, where t is time in hours. Find the growth rate after 5 hours and the time at which the population reaches 1000.

一个细菌培养物的种群数量由模型 P(t) = 200e^0.3t 给出，其中 t 为时间（小时）。求5小时后的增长率，以及种群达到1000所需的时间。

First, the growth rate is the derivative P'(t) = 200 × 0.3 e^0.3t = 60 e^0.3t. At t = 5, P'(5) = 60 e^1.5 ≈ 60 × 4.4817 = 268.9 bacteria per hour. To find when P(t) = 1000, solve 200 e^0.3t = 1000 → e^0.3t (ln 5)/0.3 ≈ 1.6094/0.3 ≈ 5.36 hours. So the population reaches 1000 in about 5.36 hours.

增长率即导数 P'(t) = 200 × 0.3 e^0.3t = 60 e^0.3t。当 t = 5 时，P'(5) = 60 e^1.5 ≈ 268.9 个/小时。为求 P(t) = 1000 的时间，解 200 e^0.3t = 1000 → e^0.3t = 5 → t = (ln 5)/0.3 ≈ 5.36 小时。

This case study demonstrates how calculus is used in life sciences to measure instantaneous rates of change, and how exponential equations are solved with natural logarithms.

本案例展示了微积分如何在生命科学中用于测量瞬时变化率，以及如何利用自然对数求解指数方程。

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7. Using Vectors in Physics | 物理中的向量应用

e magnitude and direction of the resultant force. Then, if a third force F₃ is added to keep the particle in equilibrium, determine its magnitude and direction.

一个质点受到两个力的作用：F₁ = 5 N 向东，F₂ = 12 N 向北。求合力的大小和方向。接着，如果添加第三个力 F₃ 使质点保持平衡，确定其大小和方向。

an (east, north) coordinate system. The resultant R = F₁ + F₂ = (5, 12). Its magnitude is |R| = √5² + 12² = √25 + 144 ≈ 67.4° north of east. Thus the resultant is 13 N at an angle of 67.4° to the horizontal.

将力表示为向量：F₁ = (5, 0)，F₂ = (0, 12)，采用（东，北）坐标系。合力 R = (5, 12)，大小 |R| = √5² + 12² = 13 N。方向角 θ 满足 tan θ = 12/5，θ ≈ 67.4°，即东偏北67.4°。

west direction, specifically 180° + 67.4° = 247.4° from east (or bearing). This application shows how vector addition directly solves static equilibrium problems.

要达到平衡，第三个力必须与 R 大小相等、方向相反。因此 F₃ = −R = (−5, −12)，对应13 N朝向西南，具体为从正东方向转角 180° + 67.4° = 247.4°。该应用展示了向量加法如何直接解决静态平衡问题。

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8. Probability in Quality Control | 质量控制中的概率

A factory produces light bulbs in batches of 20. Historically, 5% of bulbs are defective. A quality control inspector randomly selects 3 bulbs from a batch without replacement. Calculate the probability that exactly one of the selected bulbs is defective.

某工厂以每批20只生产灯泡。历史数据显示5%的灯泡有缺陷。质检员从一批中无放回地随机抽取3只灯泡。计算恰好抽到一只缺陷灯泡的概率。

ealistically it varies). Using the hypergeometric distribution: number of ways to choose 1 defective out of 1 = ¹C₁, and 2 non-defective out of 19 = ¹⁹C₂. Total ways to choose 3 out of 20 = ²⁰C₃. So probability = (¹C₁ × ¹⁹C₂) / ²⁰C₃ = (1 × 171) / 1140 = 171/1140 = 0.15 exactly. (Check: ¹⁹C₂ = 19×18/2 = 171, ²⁰C₃ = 20×19×18/(3×2)=1140).

这批有20只灯泡，预期缺陷数为1（但实际概率模型中我们假设该批中恰有1只缺陷）。使用超几何分布：选1只缺陷的方式 = ¹C₁，选2只合格的方式 = ¹⁹C₂。总方式 = ²⁰C₃。概率 = (1 × 171) / 1140 = 0.15。

135375. Comparing the two highlights the effect of sampling without replacement, which is important when batch sizes are small.

如果是有放回抽样，概率将遵循二项分布：P(X=1) = ³C₁ × (0.05)¹ × (0.95)² = 0.135375。两者比较突出了无放回抽样的影响，在批量较小时这一点很重要。

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9. Simultaneous Equations in Business Break-Even Analysis | 商业盈亏平衡分析中的联立方程

A company produces handmade mugs. The total cost £ producing x mugs is modelled by C = 500 + 4x (fixed cost plus variable cost). The revenue £ elling x mugs is R = 10x. Find the break-even point and the profit when 120 mugs are sold.

一家公司生产手工马克杯。生产 x 个杯子的总成本 £C 为 C = 500 + 4x（固定成本加可变成本）。销售 x 个杯子的收入 £R 为 R = 10x。求盈亏平衡点，并计算售出120个杯子时的利润。

profit function is P = R − C = 10x − (500 + 4x) = 6x − 500. For x = 120, P = 6×120 − 500 = 720 − 500 = £220. Thus a profit of £220 is made.

盈亏平衡发生在收入等于成本时：10x = 500 + 4x → 6x = 500 → x ≈ 83.33，因此公司需售出至少84个杯子才能开始盈利。利润函数为 P = R − C = 6x − 500。当 x = 120 时，P = 720 − 500 = 220 英镑。

This linear model can be extended by introducing a more realistic demand curve, but even simple simultaneous equations provide powerful decision-making tools for small businesses.

该线性模型可通过引入更真实的需求曲线进行扩展，但即使简单的联立方程也能为小企业提供强大的决策工具。

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10. Functions and Graphs: Modelling Cooling | 函数与图像：冷却建模

ced in a room at 20 °C. Its temperature T (°C) after t minutes is given by T(t) = 20 + 70e^−kt, where k is a positive constant. If after 5 minutes the temperature is 60 °C, find k and determine the time when the tea reaches 30 °C.

一杯90°C的热茶放在20°C的房间中。t分钟后其温度 T (°C) 由 T(t) = 20 + 70e^−kt 给出，其中 k 是一个正常数。如果5分钟后温度为60°C，求 k，并确定茶冷却到30°C所需的时间。

Use the data: 60 = 20 + 70e^−5k → 40 = 70e^−5k → e^−5k 7) → k = −ln(4/7)/5 = ln(7/4)/5 ≈ (0.55962)/5 = 0.1119. To find when T = 30: 30 = 20 + 70e^−kt → 10 = 70e^−kt → e^−kt ln 7)/k ≈ 1.94591/0.1119 ≈ 17.39 minutes. So the tea cools to 30 °C in about 17.4 minutes.

代入数据：60 = 20 + 70e^−5k → 40 = 70e^−5k → e^−5k = 4/7。取自然对数：−5k = ln(4/7) → k = ln(7/4)/5 ≈ 0.1119。求 T = 30 的时间：30 = 20 + 70e^−kt → e^−kt = 1/7 → −kt = −ln 7 → t = (ln 7)/k ≈ 17.39 分钟。

This case study combines exponential decay with transformations of functions. The graph of T(t) has a horizontal asymptote at 20 °C, representing the room temperature limit. Understanding such behaviour is crucial in many scientific fields.

该案例结合了指数衰减与函数变换。T(t) 的图像有一条水平渐近线在20°C，代表室温极限。理解这种行为在许多科学领域至关重要。

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