📚 Interdisciplinary Applied Problem-Solving | 跨学科综合题型训练
Mathematics does not exist in a vacuum; it forms the language of science, the backbone of economics, and the tool for modern engineering. In WJEC Year 12, applied problem-solving bridges pure techniques with real-world scenarios drawn from physics, biology, business, and beyond. Through carefully designed interdisciplinary exercises, you will learn to translate contextual clues into equations, select the right mathematical models, and interpret solutions meaningfully.
数学并非孤立存在的学科；它是科学的语言、经济学的支柱和现代工程的工具。在 WJEC 12 年级课程中，应用型问题求解将纯数学技巧与物理、生物、商业等领域的真实情境联系起来。通过精心设计的跨学科练习，你将学会将上下文线索转化为方程，选择合适的数学模型，并有意义地解释结果。
1. Motion with Variable Acceleration: Physics Connections | 变加速运动：物理关联
In kinematics, displacement s, velocity v, and acceleration a are linked by differentiation and integration with respect to time t. If acceleration depends on time, you must integrate to find velocity and then integrate again to find displacement, using initial conditions to determine constants.
在运动学中，位移 s、速度 v 和加速度 a 通过关于时间 t 的微分和积分相互关联。如果加速度依赖于时间，你必须先积分求出速度，再积分求出位移，并用初始条件确定常数。
Example: A particle moves along a straight line with acceleration a(t) = 12t – 4 m/s². At t=0, its velocity is 3 m/s and its displacement from the origin is 2 m. Find expressions for v(t) and s(t).
例题：一质点沿直线运动，加速度 a(t) = 12t – 4 m/s²。在 t=0 时，其速度为 3 m/s，距原点的位移为 2 m。求 v(t) 和 s(t) 的表达式。
Step 1: Integrate a(t) to obtain velocity: v(t) = ∫(12t – 4) dt = 6t² – 4t + C. Using v(0)=3 gives C=3, so v(t) = 6t² – 4t + 3.
步骤 1：对 a(t) 积分求速度：v(t) = ∫(12t – 4) dt = 6t² – 4t + C。利用 v(0)=3 得 C=3，因此 v(t) = 6t² – 4t + 3。
Step 2: Integrate v(t) to find displacement: s(t) = ∫(6t² – 4t + 3) dt = 2t³ – 2t² + 3t + D. The condition s(0)=2 yields D=2, so s(t) = 2t³ – 2t² + 3t + 2.
步骤 2：对 v(t) 积分求位移：s(t) = ∫(6t² – 4t + 3) dt = 2t³ – 2t² + 3t + D。条件 s(0)=2 得出 D=2，故 s(t) = 2t³ – 2t² + 3t + 2。
2. Profit Optimisation: Economics in Calculus | 利润优化：微积分中的经济学
Firms often wish to maximise profit π(q), defined as total revenue R(q) minus total cost C(q). By setting the first derivative dπ/dq = 0 and checking the second derivative, you can locate the output level q that yields maximum profit.
企业通常希望最大化利润 π(q)，即总收入 R(q) 减去总成本 C(q)。通过令一阶导数 dπ/dq = 0 并检验二阶导数，你可以找到使利润最大的产量 q。
Example: The demand price per unit is p = 200 – 3q, so revenue R(q) = q(200 – 3q) = 200q – 3q². The total cost is C(q) = 40 + 20q. Determine the profit-maximising output.
例题：单位需求价格为 p = 200 – 3q，因此收入 R(q) = q(200 – 3q) = 200q – 3q²。总成本为 C(q) = 40 + 20q。求利润最大化的产量。
Profit function: π(q) = (200q – 3q²) – (40 + 20q) = 180q – 3q² – 40. Differentiate: dπ/dq = 180 – 6q. Set to zero: 180 – 6q = 0 → q = 30. Second derivative d²π/dq² = -6 < 0, confirming a maximum. The maximum profit is π(30) = 180×30 - 3×30² - 40 = 5400 - 2700 - 40 = 2660.
利润函数：π(q) = (200q – 3q²) – (40 + 20q) = 180q – 3q² – 40。求导：dπ/dq = 180 – 6q。令其为零：180 – 6q = 0 → q = 30。二阶导数 d²π/dq² = -6 < 0，确认是极大值。最大利润为 π(30) = 180×30 - 3×30² - 40 = 5400 - 2700 - 40 = 2660。
3. Exponential Population Growth: Biology Context | 指数人口增长：生物学背景
Many biological populations grow according to the differential equation dN/dt = kN, where N is the population size and k is the growth constant. The solution is N(t) = N₀ e^(kt), exponential growth. Given two data points, you can find k and then predict future size.
许多生物种群按照微分方程 dN/dt = kN 增长，其中 N 为种群数量，k 为增长常数。其解为 N(t) = N₀ e^(kt)，即指数增长。给定两个数据点，你可以求出 k，然后预测未来数量。
Example: A bacteria culture starts with 500 cells. After 3 hours the count is 2000. Assuming exponential growth, find k and the population after 5 hours.
例题：某细菌培养初始有 500 个细胞。3 小时后计数为 2000。假设指数增长，求 k 以及 5 小时后的数量。
Using N = N₀ e^(kt): 2000 = 500 e^(3k) → e^(3k) = 4 → 3k = ln 4 → k = (ln 4)/3 ≈ 0.4621. Then N(5) = 500 e^(0.4621×5) = 500 e^(2.3105) ≈ 500 × 10.08 = 5040 cells.
使用 N = N₀ e^(kt)：2000 = 500 e^(3k) → e^(3k) = 4 → 3k = ln 4 → k = (ln 4)/3 ≈ 0.4621。然后 N(5) = 500 e^(0.4621×5) = 500 e^(2.3105) ≈ 500 × 10.08 = 5040 个细胞。
4. River Cross-Section Area: Geography & Numerical Methods | 河流横截面积：地理与数值方法
Geographers estimate the cross-sectional area of a river using depth measurements at regular intervals across the width. The trapezium rule, which approximates the definite integral of a function, is perfectly suited for this task.
地理学家利用在河流宽度上等间距测量的深度来估算横截面积。梯形法则近似求函数的定积分，非常适用于这一任务。
Example: A river is 20 m wide; depths measured every 4 m from the left bank are: 0, 1.5, 2.8, 3.6, 2.2, 0 metres. Use the trapezium rule with five strips to estimate the cross-sectional area.
例题：某河流宽 20 m；从左岸起每 4 m 测量的深度为：0, 1.5, 2.8, 3.6, 2.2, 0 米。使用五条带的梯形法则估算横截面积。
Strip width h = 4 m. Area ≈ (h/2)[y₀ + 2(y₁ + y₂ + y₃ + y₄) + y₅] = 2 × [0 + 2×(1.5+2.8+3.6+2.2) + 0] = 2 × [2×(10.1)] = 2 × 20.2 = 40.4 m².
条带宽度 h = 4 m。面积 ≈ (h/2)[y₀ + 2(y₁ + y₂ + y₃ + y₄) + y₅] = 2 × [0 + 2×(1.5+2.8+3.6+2.2) + 0] = 2 × [2×(10.1)] = 2 × 20.2 = 40.4 m²。
5. IQ Scores: Psychology & Normal Distribution | IQ 分数：心理学与正态分布
Psychological test scores such as IQ often follow a normal distribution with mean μ and standard deviation σ. The Z-score Z = (X – μ)/σ allows you to calculate the proportion of individuals above or below a certain score using standard normal tables.
心理学测试分数（如 IQ）通常服从均值为 μ、标准差为 σ 的正态分布。Z 分数 Z = (X – μ)/σ 可让你使用标准正态表计算高于或低于特定分数的人数比例。
Example: IQ is normally distributed with μ=100, σ=15. Find the percentage of people with IQ between 85 and 130.
例题：IQ 服从 μ=100, σ=15 的正态分布。求 IQ 在 85 到 130 之间的人口百分比。
For X=85: Z₁ = (85-100)/15 = -1.00. For X=130: Z₂ = (130-100)/15 = 2.00. Using tables, P(Z < -1) = 0.1587, P(Z < 2) = 0.9772. The proportion between is 0.9772 - 0.1587 = 0.8185, i.e., 81.85%.
对 X=85：Z₁ = (85-100)/15 = -1.00。对 X=130：Z₂ = (130-100)/15 = 2.00。查表得 P(Z < -1) = 0.1587，P(Z < 2) = 0.9772。两者之间的比例为 0.9772 - 0.1587 = 0.8185，即 81.85%。
6. Linear Programming: Business Resource Allocation | 线性规划：商业中的资源分配
Businesses face constraints such as labour hours and machine time. Linear programming allows you to maximise profit or minimise cost by defining decision variables, constraint inequalities, and an objective function, then testing vertices of the feasible region.
企业面临劳动工时和机器时间等约束。线性规划通过定义决策变量、约束不等式和目标函数，然后检验可行区域的顶点，让你能够实现利润最大化或成本最小化。
Example: A workshop produces chairs (x) and tables (y). Profit: £20 per chair, £50 per table. Constraints: assembly time 2x + 5y ≤ 100; finishing time 4x + 3y ≤ 120; x≥0, y≥0. Find maximum profit.
例题：某车间生产椅子 (x) 和桌子 (y)。利润：每把椅子 £20，每张桌子 £50。约束：装配时间 2x + 5y ≤ 100；修整时间 4x + 3y ≤ 120；x≥0, y≥0。求最大利润。
Plot constraints: vertices occur at (0,0), (0,20) from 2x+5y=100, (30,0) from 4x+3y=120, and intersection of 2x+5y=100 and 4x+3y=120. Solve simultaneously: multiply first by 2: 4x+10y=200, subtract second: 7y=80 → y≈11.43, then x=(100-5×11.43)/2=21.43. Profit P=20x+50y: (0,0)=0, (0,20)=1000, (30,0)=600, at (21.43,11.43): 20×21.43+50×11.43=428.6+571.5=1000.1≈£1000. The maximum profit integer solution is around (22,11) giving £990, but exact optimum is at (0,20) with £1000.
绘制约束：顶点为 (0,0)，由 2x+5y=100 得 (0,20)，由 4x+3y=120 得 (30,0)，以及 2x+5y=100 与 4x+3y=120 的交点。联立求解：第一个方程乘以 2：4x+10y=200，减去第二个得 7y=80 → y≈11.43，然后 x=(100-5×11.43)/2=21.43。利润 P=20x+50y：(0,0)=0，(0,20)=1000，(30,0)=600，在 (21.43,11.43)：20×21.43+50×11.43=428.6+571.5=1000.1≈£1000。最大利润整数解在 (0,20) 得 £1000。
7. Oscillations: Engineering and Simple Harmonic Motion | 振动：工程与简谐运动
Many engineering systems, like a mass on a spring, exhibit simple harmonic motion (SHM) where displacement x = A sin(ωt + φ). The velocity is dx/dt and acceleration is d²x/dt² = -ω²x, connecting derivatives to physical behaviour.
许多工程系统（如弹簧上的质量块）呈现简谐运动 (SHM)，其位移 x = A sin(ωt + φ)。速度为 dx/dt，加速度为 d²x/dt² = -ω²x，将导数与物理行为相联系。
Example: A bridge component oscillates with x = 0.05 sin(10π t) metres. Find the maximum speed and the acceleration when displacement is 0.03 m.
例题：某桥梁构件振动，位移为 x = 0.05 sin(10π t) 米。求最大速度以及当位移为 0.03 m 时的加速度。
Velocity v = dx/dt = 0.05 × 10π cos(10π t) = 0.5π cos(10π t). Maximum speed = 0.5π ≈ 1.571 m/s. Acceleration a = d²x/dt² = -0.05 × (10π)² sin(10π t) = -5π² sin(10π t) = -5π² x. At x=0.03, a = -5π² × 0.03 = -0.15π² ≈ -1.48 m/s².
速度 v = dx/dt = 0.05 × 10π cos(10π t) = 0.5π cos(10π t)。最大速度 = 0.5π ≈ 1.571 m/s。加速度 a = d²x/dt² = -0.05 × (10π)² sin(10π t) = -5π² sin(10π t) = -5π² x。当 x=0.03 时，a = -5π² × 0.03 = -0.15π² ≈ -1.48 m/s²。
8. Radioactive Decay: Environmental Science & Half-Life | 放射性衰变：环境科学与半衰期
Radioactive substances decay exponentially: N = N₀ e^(-λt), where λ is the decay constant. The half-life T₁/₂ is the time for half the nuclei to decay, satisfying e^(-λT) = ½ ⇒ T = ln 2 / λ. This model helps date archaeological finds or monitor environmental contamination.
放射性物质呈指数衰变：N = N₀ e^(-λt)，其中 λ 为衰变常数。半衰期 T₁/₂ 是半数原子核衰变所需的时间，满足 e^(-λT) = ½ ⇒ T = ln 2 / λ。该模型有助于考古年代测定或环境污染监测。
Example: A sample of iodine-131 has initial activity 800 Bq. Its half-life is 8 days. How long until activity falls to 50 Bq?
例题：一份碘-131 样品的初始活度为 800 Bq。其半衰期为 8 天。活度降至 50 Bq 需要多长时间？
Find λ: T₁/₂ = ln 2 / λ → λ = ln 2 / 8 ≈ 0.08664. Using N = N₀ e^(-λt): 50 = 800 e^(-0.08664 t) → e^(-0.08664 t) = 50/800 = 0.0625. Take natural log: -0.08664 t = ln 0.0625 ≈ -2.7726 → t = 2.7726 / 0.08664 ≈ 32 days. Alternatively, note that 800→400→200→100→50 is 4 half-lives, so t = 4 × 8 = 32 days.
求 λ：T₁/₂ = ln 2 / λ → λ = ln 2 / 8 ≈ 0.08664。利用 N = N₀ e^(-λt)：50 = 800 e^(-0.08664 t) → e^(-0.08664 t) = 50/800 = 0.0625。取自然对数：-0.08664 t = ln 0.0625 ≈ -2.7726 → t = 2.7726 / 0.08664 ≈ 32 天。或者，注意到 800→400→200→100→50 是 4 个半衰期，因此 t = 4 × 8 = 32 天。
9. Algorithm Efficiency: Computer Science & Arithmetic Series | 算法效率：计算机科学与等差数列
In coding, nested loops often execute a number of operations that form an arithmetic series. Summing 1 + 2 + 3 + … + n = n(n+1)/2 is a key result that helps estimate time complexity.
在编程中，嵌套循环常常执行构成等差数列的若干次操作。求和 1 + 2 + 3 + … + n = n(n+1)/2 是帮助估算时间复杂度的关键结果。
Example: A sorting algorithm requires the sum of the first 200 integers comparisons. Use the formula to find the total comparisons. If each comparison takes 10⁻⁶ seconds, estimate the total running time.
例题：某排序算法需要前 200 个整数的比较次数之和。使用公式求总比较次数。如果每次比较耗时 10⁻⁶ 秒，估算总运行时间。
Sum S = 200 × 201 / 2 = 100 × 201 = 20 100 comparisons. Time = 20 100 × 10⁻⁶ = 0.0201 seconds (about 20 ms). For larger n, this grows quadratically, illustrating why O(n²) algorithms can become slow.
总和 S = 200 × 201 / 2 = 100 × 201 = 20 100 次比较。时间 = 20 100 × 10⁻⁶ = 0.0201 秒（约 20 毫秒）。对于更大的 n，这呈二次增长，说明了为何 O(n²) 算法可能变慢。
10. Hypothesis Test for a Mean: Health Science & Z-test | 均值的假设检验：健康科学与 Z 检验
Researchers often test whether a population mean has changed from a known value when the population variance is known. A Z-test statistic Z = (x̄ – μ₀) / (σ/√n) is compared against critical values from the standard normal distribution.
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