📚 Year 10 WJEC Further Mathematics: Case Study Practical Exercises | 十年级 WJEC 进阶数学：案例分析实战演练
Welcome to our case study practical exercises for WJEC Year 10 Further Mathematics. This article presents a series of real-world scenarios that require the application of advanced mathematical techniques, including quadratics, polynomials, matrices, trigonometry, sequences and coordinate geometry. Each case study is carefully designed to mirror the problem-solving style of the WJEC specification, helping you build confidence and deepen your understanding while bridging theory and practice.
欢迎来到WJEC十年级进阶数学的案例分析实战演练。本文呈现了一系列需要运用高级数学技巧的真实场景，涵盖二次函数、多项式、矩阵、三角学、数列以及坐标几何等内容。每个案例都精心设计，模拟WJEC考试中的问题解决风格，旨在帮助你建立信心，加深理解，同时连接理论与实践。
1. Parabolic Bridge Design | 抛物线桥梁设计
A civil engineer models a pedestrian bridge arch using the quadratic equation y = −0.02x² + 2x, where y represents the height in metres and x is the horizontal distance in metres from the left base of the arch. The task is to determine the maximum height of the arch and the total span (the distance between the two points where the arch meets the ground).
一位土木工程师用二次方程 y = −0.02x² + 2x 对人行桥拱进行建模，其中 y 表示高度（米），x 表示距离拱左侧基座的水平距离（米）。任务是求出拱的最大高度以及总跨度（拱与地面两交点之间的距离）。
To find the maximum height, we complete the square: y = −0.02(x² − 100x) = −0.02[(x − 50)² − 2500] = −0.02(x − 50)² + 50. The vertex is at (50, 50), so the arch reaches a maximum height of 50 metres. The span is found by setting y = 0: −0.02x² + 2x = 0 ⇒ x(−0.02x + 2) = 0, giving x = 0 or x = 100. Hence the total span is 100 metres.
为求最大高度，我们进行配方：y = −0.02(x² − 100x) = −0.02[(x − 50)² − 2500] = −0.02(x − 50)² + 50。顶点坐标为 (50, 50)，因此拱的最大高度为 50 米。求跨度时令 y = 0：−0.02x² + 2x = 0 ⇒ x(−0.02x + 2) = 0，解得 x = 0 或 x = 100。故总跨度为 100 米。
2. Polynomial Division in Volume Analysis | 多项式除法与体积分析
An engineer is designing a rectangular storage tank. The volume of the tank is given by V = x³ + 6x² + 11x + 6 cubic metres. The length is (x + 1) metres and the width is (x + 2) metres. The engineer needs to determine an expression for the height of the tank.
一位工程师正在设计一个矩形储水罐。罐体体积由 V = x³ + 6x² + 11x + 6 立方米给出。长度是 (x + 1) 米，宽度是 (x + 2) 米。工程师需要求出罐高的表达式。
Since Volume = length × width × height, the height is V ÷ (length × width). First multiply the two known dimensions: (x + 1)(x + 2) = x² + 3x + 2. Now perform polynomial division: (x³ + 6x² + 11x + 6) ÷ (x² + 3x + 2). Using long division or synthetic division, we obtain x + 3 exactly. Therefore the height is (x + 3) metres.
由于体积 = 长 × 宽 × 高，所以高 = V ÷ (长 × 宽)。先将两个已知尺寸相乘：(x + 1)(x + 2) = x² + 3x + 2。然后进行多项式除法：(x³ + 6x² + 11x + 6) ÷ (x² + 3x + 2)。使用长除法或综合除法，恰好得到 x + 3。因此高度为 (x + 3) 米。
3. Matrix Transformations in Computer Graphics | 计算机图形学中的矩阵变换
A graphic designer defines a triangle with vertices A(1, 2), B(3, 2), and C(2, 4). She needs to rotate the triangle 90° anticlockwise about the origin. The transformation matrix for a 90° anticlockwise rotation is [[0, −1], [1, 0]]. Determine the new coordinates of the rotated triangle.
某图形设计师定义了一个三角形，顶点为 A(1, 2)、B(3, 2) 和 C(2, 4)。她需要将该三角形绕原点逆时针旋转 90°。逆时针旋转 90° 的变换矩阵为 [[0, −1], [1, 0]]。求旋转后三角形的新坐标。
Apply the matrix to each vertex column vector. For A: [0, −1; 1, 0] × [1; 2] = [0·1 + (−1)·2; 1·1 + 0·2] = [−2; 1], so A′(−2, 1). For B: [0, −1; 1, 0] × [3; 2] = [−2; 3] ⇒ B′(−2, 3). For C: [0, −1; 1, 0] × [2; 4] = [−4; 2] ⇒ C′(−4, 2). The rotated triangle has vertices (−2, 1), (−2, 3), and (−4, 2).
将矩阵作用于每个顶点的列向量。对于 A：[0, −1; 1, 0] × [1; 2] = [0·1 + (−1)·2; 1·1 + 0·2] = [−2; 1]，故 A′(−2, 1)。对于 B：[0, −1; 1, 0] × [3; 2] = [−2; 3] ⇒ B′(−2, 3)。对于 C：[0, −1; 1, 0] × [2; 4] = [−4; 2] ⇒ C′(−4, 2)。旋转后的三角形顶点为 (−2, 1)、(−2, 3) 和 (−4, 2)。
4. Linear Programming for Optimal Production | 最优生产的线性规划
A small factory manufactures two types of gadgets, X and Y. Each unit of X requires 2 hours on machine A and 1 hour on machine B; each unit of Y requires 1 hour on A and 2 hours on B. Machine A is available for at most 10 hours per day, and machine B for at most 8 hours. The profit per unit of X is £3 and per unit of Y is £4. Formulate the constraints and find the production plan that maximises daily profit.
一家小工厂生产两种产品 X 和 Y。每件 X 需要机器 A 工作 2 小时、机器 B 工作 1 小时；每件 Y 需要机器 A 工作 1 小时、机器 B 工作 2 小时。机器 A 每天最多可用 10 小时，机器 B 最多 8 小时。每件 X 的利润为 3 英镑，每件 Y 的利润为 4 英镑。列出约束条件并求使日利润最大化的生产计划。
Let x = number of X, y = number of Y. Constraints: 2x + y ≤ 10 (A), x + 2y ≤ 8 (B), x ≥ 0, y ≥ 0. Profit P = 3x + 4y. Determine the feasible region vertices: (0,0), (0,4) from B, (5,0) from A, and the intersection of 2x + y = 10 and x + 2y = 8. Solve simultaneously: from first, y = 10 − 2x; substitute into second: x + 2(10 − 2x) = 8 ⇒ x + 20 − 4x = 8 ⇒ −3x = −12 ⇒ x = 4, then y = 2. Vertex (4,2). Evaluate P: (0,0) → 0; (0,4) → 16; (5,0) → 15; (4,2) → 3·4 + 4·2 = 20. Maximum profit is £20 when 4 units of X and 2 units of Y are produced.
设 x = X 的数量，y = Y 的数量。约束条件：2x + y ≤ 10 (A)，x + 2y ≤ 8 (B)，x ≥ 0，y ≥ 0。利润 P = 3x + 4y。确定可行域顶点：(0,0)，由 B 得 (0,4)，由 A 得 (5,0)，以及 2x + y = 10 与 x + 2y = 8 的交点。联立求解：由第一式 y = 10 − 2x，代入第二式：x + 2(10 − 2x) = 8 ⇒ x + 20 − 4x = 8 ⇒ −3x = −12 ⇒ x = 4，则 y = 2。顶点 (4,2)。计算 P：(0,0) → 0；(0,4) → 16；(5,0) → 15；(4,2) → 3·4 + 4·2 = 20。最大利润为 20 英镑，此时生产 4 件 X 和 2 件 Y。
5. Trigonometric Height Measurement | 三角学测量高度
A surveyor stands 50 metres away from the base of a communication tower on level ground. Using a theodolite, she measures the angle of elevation to the top of the tower as 30°. Calculate the height of the tower, assuming the instrument height is negligible.
一名测量员站在平坦地面上，距离通信塔基座 50 米。她使用经纬仪测得塔顶的仰角为 30°。假设仪器高度可忽略不计，计算塔的高度。
Let h be the tower height. In the right-angled triangle formed, tan 30° = opposite / adjacent = h / 50. Since tan 30° = 1/√3, we have h = 50 × (1/√3) = 50/√3. Rationalising the denominator gives h ≈ 50 / 1.732 ≈ 28.87 metres. The tower is approximately 28.9 m tall (to 1 decimal place).
设塔高为 h。在构成的直角三角形中，tan 30° = 对边 / 邻边 = h / 50。因为 tan 30° = 1/√3，故 h = 50 × (1/√3) = 50/√3。分母有理化后计算得 h ≈ 50 / 1.732 ≈ 28.87 米。该塔高约 28.9 米（保留一位小数）。
6. Arithmetic Sequences in Loan Repayment | 等差数列与贷款偿还
A graduate borrows £5000 and plans to repay it in monthly instalments that form an arithmetic sequence. The first instalment is £200, and each subsequent instalment increases by £10. The lender agrees that the total sum of the instalments after n months can pay off the loan exactly. Find the number of months required.
某毕业生借款 5000 英镑，计划以等差数列的形式按月分期偿还。首月还款 200 英镑，此后每月还款额递增 10 英镑。贷方同意 n 个月后还款总额恰好还清贷款。求所需的月数。
The monthly payments form an arithmetic sequence with first term a = 200 and common difference d = 10. The total after n months is Sₙ = n/2 [2a + (n − 1)d] = n/2 [400 + 10(n − 1)] = n/2 (390 + 10n) = 5n² + 195n. Set Sₙ = 5000: 5n² + 195n = 5000 ⇒ n² + 39n − 1000 = 0. Solve using the quadratic formula: n = [−39 ± √(39² + 4000)] / 2 = [−39 ± √(1521 + 4000)] / 2 = [−39 ± √5521] / 2. √5521 ≈ 74.3, so n ≈ (−39 + 74.3)/2 ≈ 17.65. Since n must be an integer, we check n = 17: S₁₇ = 5(17)² + 195(17) = 1445 + 3315 = 4760, which is less than 5000. For n = 18: S₁₈ = 5(18)² + 195(18) = 1620 + 3510 = 5130, which exceeds 5000. The loan is fully settled in the 18th month, with a slightly larger final payment possible. The required number of months is 18.
每月还款构成首项 a = 200、公差 d = 10 的等差数列。n 个月后总还款额 Sₙ = n/2 [2a + (n − 1)d] = n/2 [400 + 10(n − 1)] = n/2 (390 + 10n) = 5n² + 195n。令 Sₙ = 5000：5n² + 195n = 5000 ⇒ n² + 39n − 1000 = 0。用二次公式求解：n = [−39 ± √(39² + 4000)] / 2 = [−39 ± √(1521 + 4000)] / 2 = [−39 ± √5521] / 2。√5521 ≈ 74.3，故 n ≈ (−39 + 74.3)/2 ≈ 17.65。由于 n 必须为整数，检验 n = 17：S₁₇ = 5(17)² + 195(17) = 1445 + 3315 = 4760，小于 5000。n = 18 时：S₁₈ = 5(18)² + 195(18) = 1620 + 3510 = 5130，超过 5000。贷款在第 18 个月完全清偿，最后一次还款额可能略高。所需月数为 18。
7. Algebraic Fractions in Parallel Circuits | 并联电路中的代数分式
Two resistors are connected in parallel. Their resistances are R₁ = x/(x+1) ohms and R₂ = (x+1)/(x+2) ohms. The total resistance R is given by 1/R = 1/R₁ + 1/R₂. Express R in terms of x in its simplest algebraic form.
两个电阻并联连接。它们的阻值分别为 R₁ = x/(x+1) 欧姆和 R₂ = (x+1)/(x+2) 欧姆。总电阻 R 满足 1/R = 1/R₁ + 1/R₂。将 R 用 x 的最简代数形式表示出来。
First compute 1/R₁ = (x+1)/x and 1/R₂ = (x+2)/(x+1). So 1/R = (x+1)/x + (x+2)/(x+1). Find a common denominator: x(x+1). Then 1/R = [(x+1)(x+1) + x(x+2)] / [x(x+1)] = [(x² + 2x + 1) + (x² + 2x)] / [x(x+1)] = (2x² + 4x + 1) / [x(x+1)]. Therefore R = x(x+1) / (2x² + 4x + 1). This is the simplified expression for the total resistance.
首先计算 1/R₁ = (x+1)/x，1/R₂ = (x+2)/(x+1)。因此 1/R = (x+1)/x + (x+2)/(x+1)。通分，公分母为 x(x+1)。则 1/R = [(x+1)(x+1) + x(x+2)] / [x(x+1)] = [(x² + 2x + 1) + (x² + 2x)] / [x(x+1)] = (2x² + 4x + 1) / [x(x+1)]。故 R = x(x+1) / (2x² + 4x + 1)。此即总电阻的简化表达式。
8. Exponential Growth of Bacteria | 细菌的指数增长
A biologist observes a colony of bacteria that doubles in size every 3 hours. Initially, there are 200 bacteria. Predict the number of bacteria after 15 hours. The growth model is N = N₀ · 2^(t / 3), where N₀ is the initial count and t is time in hours.
一位生物学家观察到某种细菌菌落每 3 小时数量翻倍。起初有 200 个细菌。预测 15 小时后的细菌数量。增长模型为 N = N₀ · 2^(t / 3)，其中 N₀ 为初始数量，t 为时间（小时）。
Substitute N₀ = 200, t = 15. The number of doubling periods is 15 / 3 = 5. Thus N = 200 · 2⁵ = 200 · 32 = 6400. After 15 hours, the colony will contain 6400 bacteria. This exponential model assumes unlimited resources and no deaths.
代入 N₀ = 200，t = 15。翻倍周期数为 15 / 3 = 5。因此 N = 200 · 2⁵ = 200 · 32 = 6400。15 小时后细菌数量将达到 6400。该指数模型假设资源无限且无死亡。
9. Circle Geometry and Tangents | 圆的几何与切线
A reflective dish is modelled by the circle x² + y² = 25. A signal receiver is placed at the point (3, 4) on the circle. Find the equation of the tangent line to the circle at that point, which represents the direction of the reflected signal.
一面反射碟形天线由圆 x² + y² = 25 建模。信号接收器位于圆上的点 (3, 4)。求该点处的切线方程，该切线代表反射信号的方向。
For the circle x² + y² = 25, the radius to point (3,4) has gradient 4/3. The tangent is perpendicular to the radius, so its gradient m = −3/4. Using the point-slope form: y − 4 = −3/4 (x − 3). Multiply through by 4: 4y − 16 = −3x + 9 ⇒ 3x + 4y = 25. This is the tangent line equation. A quick check: 3·3 + 4·4 = 9 + 16 = 25, confirming it passes through the given point.
对于圆 x² + y² = 25，到点 (3,4) 的半径斜率为 4/3。切线与半径垂直，故其斜率 m = −3/4。使用点斜式：y − 4 = −3/4 (x − 3)。两边乘以 4：4y − 16 = −3x + 9 ⇒ 3x + 4y = 25。此即切线方程。快速验算：3·3 + 4·4 = 9 + 16 = 25，证实它经过给定点。
10. Coordinate Geometry: Shortest Distance | 坐标几何：最短距离
A drone operator wants to fly the shortest possible route from the charging station at (5, 7) to a designated straight flight corridor modelled by the line 3x − 4y + 8 = 0. Calculate that minimum distance.
一名无人机操作员希望从位于 (5, 7) 的充电站飞往由直线 3x − 4y + 8 = 0 表示的一条指定直飞走廊，求最短飞行距离。
The perpendicular distance from a point (x₁, y₁) to a line Ax +
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