📚 WJEC Year 11 Further Mathematics: Interdisciplinary Problem-Solving Practice | WJEC 11年级进阶数学：跨学科综合题型训练
Interdisciplinary questions in WJEC Year 11 Further Mathematics require you to apply pure mathematical techniques (algebra, calculus, matrices, trigonometry, statistics) to real-world contexts such as physics, economics, biology, and computer science. This type of integrated problem-solving not only deepens your understanding but also prepares you for the kinds of challenges you will meet in A-levels and beyond.
WJEC 11年级进阶数学的跨学科综合题要求你将纯数学技巧（代数、微积分、矩阵、三角学、统计）应用于现实背景，如物理、经济、生物和计算机科学。这类综合解题训练不仅能加深理解，还能为你将来面对A-level及更高阶段的挑战做好准备。
1. Introduction to Interdisciplinary Questions | 跨学科综合题型介绍
In WJEC Further Mathematics, interdisciplinary questions often present a scenario from another subject and ask you to extract a mathematical model. You may need to identify variables, formulate equations, and choose the correct mathematical tool (e.g. differentiation for optimisation, integration for area, matrix multiplication for transformations).
在WJEC进阶数学中，跨学科题目通常会给出其他学科的情境，要求你提取数学模型。你可能需要识别变量、建立方程并选择正确的数学工具（例如用微分求最优化，用积分求面积，用矩阵乘法做变换）。
2. Algebra in Finance: Maximising Profit | 金融中的代数：利润最大化
Formulate the profit function P(x) and determine the number of units that maximise profit.
一家科技公司生产 x 件产品。成本函数为 C(x) = 2x² + 50x + 200（英镑），收入函数为 R(x) = 100x − x²。请建立利润函数 P(x)，并求出使利润最大化的产量。
6 ≈ 8.33. Since the number of units must be an integer, test x = 8 and 9. P(8) = −3(64) + 50(8) − 200 = −192 + 400 − 200 = 8; P(9) = −3(81) + 50(9) − 200 = −243 + 450 − 200 = 7. Hence, profit is maximised at 8 units.
利润 P(x) = R(x) − C(x) = (100x − x²) − (2x² + 50x + 200) = −3x² + 50x − 200。求导：dP/dx = −6x + 50。令导数为零：−6x + 50 = 0 → x = 50/6 ≈ 8.33。由于产量必须为整数，检验 x = 8 和 9。P(8) = −3(64) + 50(8) − 200 = 8；P(9) = −3(81) + 50(9) − 200 = 7。因此，生产 8 件时利润最大。
3. Functions and Graphs in Economics: Supply and Demand | 经济中的函数与图像：供给与需求
− 3p, where p is the price in pounds. Find the equilibrium price and quantity algebraically.
某产品的供给函数为 S(p) = 2p − 10，需求函数为 D(p) = 50 − 3p，其中 p 为价格（英镑）。用代数方法求均衡价格和数量。
p back into either function: S(12) = 2(12) − 10 = 14. Therefore, the equilibrium price is £ antity is 14 units. Graphically, the intersection of the two lines confirms this solution.
均衡点出现在供需相等时：S(p) = D(p) → 2p − 10 = 50 − 3p → 5p = 60 → p = 12。将 p 代回任一函数：S(12) = 2(12) − 10 = 14。因此，均衡价格为 £12，均衡数量为 14 件。从图形上看，两条直线的交点也证实了这一结果。
4. Vectors in Physics: Resultant Force | 物理中的向量：合力
sultant force vector, its magnitude, and the angle it makes with the positive i-direction.
两个力作用在一个质点上：F₁ = (3i + 4j) N，F₂ = (−2i + 6j) N。求合力向量、合力的大小以及与 i 正方向之间的夹角。
Resultant force R = F₁ + F₂ = (3 − 2)i + (4 + 6)j = i + 10j. Magnitude |R| = √1² + 10² 10) ≈ 84.3°. This kind of vector addition is fundamental in statics and dynamics problems.
合力 R = F₁ + F₂ = (3 − 2)i + (4 + 6)j = i + 10j。大小 |R| = √1² + 10² = √101 ≈ 10.05 N。与 i 正方向的夹角 θ 满足 tan θ = 10/1 = 10，因此 θ = tan⁻¹(10) ≈ 84.3°。这种向量相加是静力学和动力学问题的基础。
5. Matrices in Computer Graphics: Rotation | 计算机图形学中的矩阵：旋转
sing the transformation matrix M = [0 −1; 1 0]. Determine the coordinates of the image A′.
点 A 的坐标为 (2, 3)。使用变换矩阵 M = [0 −1; 1 0] 将其绕原点逆时针旋转 90°。求像点 A′ 的坐标。
s (−3, 2). This matrix technique is widely used in 2D computer graphics for rotating shapes and objects.
使用列向量方法计算：A′ = M × A = [0 −1; 1 0] × [2; 3] = [0×2 + (−1)×3; 1×2 + 0×3] = [−3; 2]。所以新点为 (−3, 2)。这种矩阵技术在二维计算机图形学中广泛用于旋转形状和物体。
6. Calculus in Kinematics: Velocity and Acceleration | 运动学中的微积分：速度与加速度
ity v(t) and acceleration a(t) functions. Find the time(s) when the particle is instantaneously at rest.
一个粒子在 t 秒后的位移 s（米）由 s(t) = t³ − 6t² + 9t + 4 给出。求速度 v(t) 和加速度 a(t) 的函数，并找出粒子瞬时静止的时刻。
v(t) = 0: 3t² − 12t + 9 = 0 → t² − 4t + 3 = 0 → (t − 1)(t − 3) = 0. Hence, the particle is momentarily at rest at t = 1 s and t = 3 s. Calculating position and acceleration at these times can provide further insight into the motion.
速度是一阶导数：v(t) = ds/dt = 3t² − 12t + 9。加速度是二阶导数：a(t) = dv/dt = 6t − 12。粒子瞬时静止时 v(t) = 0：3t² − 12t + 9 = 0 → t² − 4t + 3 = 0 → (t − 1)(t − 3) = 0。因此，粒子在 t = 1 秒和 t = 3 秒时瞬间静止。计算这些时刻的位置和加速度可以进一步了解运动状况。
7. Exponential Growth in Biology: Bacterial Colony | 生物中的指数增长：细菌群落
A bacterial colony grows according to N(t) = 200e0.4t, where N is the number of bacteria and t is time in hours. Calculate the doubling time, i.e. the time taken for the population to reach 400.
一个细菌群落按照 N(t) = 200e0.4t 增长，其中 N 是细菌数量，t 是时间（小时）。求细菌数量翻倍所需的时间（即达到 400 个所需的时间）。
Set N(t) = 400: 200e0.4t = 400 → e0.4t urs (about 1 hour 44 minutes). Exponential models like this are common in biology for population growth and in finance for compound interest.
令 N(t) = 400：200e0.4t = 400 → e0.4t = 2。取自然对数：0.4t = ln 2 → t = ln 2 / 0.4 ≈ 0.6931 / 0.4 ≈ 1.733 小时（约 1 小时 44 分钟）。这类指数模型在生物学种群增长和金融复利计算中都很常见。
8. Trigonometry in Engineering: Height of a Tower | 工程中的三角学：塔高测量
An engineer stands 50 metres from the base of a tower. The angle of elevation to the top of the tower is 30°. Calculate the height of the tower, correct to one decimal place.
一名工程师站在离塔基 50 米处，测得塔顶的仰角为 30°。计算塔的高度，精确到一位小数。
Using the tangent ratio: tan 30° = opposite / adjacent = height / 50. Since tan 30° = 1/√3, height = 50 × tan 30° = 50 / √3 ≈ 50 / 1.732 ≈ 28.9 m. The height of the tower is approximately 28.9 metres.
使用正切比值：tan 30° = 对边/邻边 = 高度 / 50。因为 tan 30° = 1/√3，高度 = 50 × tan 30° = 50 / √3 ≈ 50 / 1.732 ≈ 28.9 米。塔高约为 28.9 米。
9. Probability in Risk Assessment: Decision Tree | 风险评估中的概率：决策树
A business considers launching a new product. The probability of success is 0.6, yielding a profit of £50,000. The probability of failure is 0.4, resulting in a loss of £20,000. Calculate the expected value of the profit and advise whether the business should proceed.
一家企业考虑推出新产品。成功概率为 0.6，可获得利润 £50,000；失败概率为 0.4，会亏损 £20,000。计算利润的期望值，并建议企业是否应该进行。
Expected value = (0.6 × 50,000) + (0.4 × (−20,000)) = 30,000 − 8,000 = £ le. However, other factors such as cash flow and risk tolerance should also be considered.
期望
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