📚 Interdisciplinary Integrated Question Training for Year 12 WJEC Mathematics | WJEC 12年级数学跨学科综合题型训练
Year 12 WJEC Mathematics lays the groundwork for applying pure mathematics, mechanics and statistics across a wide range of subjects. This article presents a series of integrated question exercises that connect calculus, algebra, probability, sequences and vectors to real scenarios in physics, biology, economics, finance, engineering and the social sciences. By working through these cross-disciplinary problems you will strengthen your modelling skills, learn to move fluently between a physical situation and its mathematical description, and see how the techniques examined in WJEC units are actually used outside the classroom.
12 年级 WJEC 数学为纯数学、力学和统计学的跨学科应用打下基础。本文提供一系列综合题型训练，把微积分、代数、概率、数列和向量与物理、生物、经济、金融、工程和社会科学中的真实情境联系起来。通过解决这些跨学科问题，你将强化建模能力，学会在现实情境与数学描述之间自如转换，并体会 WJEC 各单元所考查的方法在课堂之外的真实用途。
1. Kinematics in Physics: SUVAT and Calculus | 物理学中的运动学：匀加速运动与微积分
In mechanics, motion with constant acceleration is a perfect arena to combine differentiation and integration. Displacement s, velocity v and acceleration a are linked by v = ds/dt and a = dv/dt. The reverse process uses the area under a curve: s = ∫ v dt and v = ∫ a dt.
在力学中，匀加速运动是结合微分与积分的绝佳场景。位移 s、速度 v 和加速度 a 通过 v = ds/dt 与 a = dv/dt 相互关联，反过程使用曲线下方面积：s = ∫ v dt，v = ∫ a dt。
Starting from constant acceleration a and integrating once gives v = u + at. A second integration yields s = ut + ½at². Combining these with s = ½(u + v)t and v² = u² + 2as forms the SUVAT suite, in which each equation is missing exactly one kinematic variable.
从恒定加速度 a 出发，积分一次得 v = u + at，再次积分得 s = ut + ½at²。将这两个方程与 s = ½(u + v)t、v² = u² + 2as 组合，即构成 SUVAT 方程组，每个方程恰好缺失一个运动学变量。
| Equation | Missing variable |
|---|---|
| v = u + at | s |
| s = ½(u + v)t | a |
| s = ut + ½at² | v |
| v² = u² + 2as | t |
Example: A particle starts from rest and accelerates at 2 m s⁻². Find its displacement after 5 s using SUVAT, then verify the result by integral calculus.
例题：一质点从静止开始以 2 m s⁻² 加速。先用 SUVAT 求 5 s 后的位移，再用积分法验证结果。
Solution with SUVAT: u = 0, a = 2, t = 5, and we need s. Substituting into s = ut + ½at² gives s = 0 × 5 + ½ × 2 × 5² = 25 m.
用 SUVAT 求解：已知 u = 0，a = 2，t = 5，求 s。代入 s = ut + ½at² 得 s = 0 × 5 + ½ × 2 × 5² = 25 m。
Verification by integration: Since a = 2, integrating gives v = 2t + c, and v(0) = 0 so c = 0. Then s = ∫₀⁵ 2t dt = [t²]₀⁵ = 25 m, which matches the SUVAT result exactly.
用积分验证：由 a = 2 积分得 v = 2t + c，又 v(0) = 0，故 c = 0。于是 s = ∫₀⁵ 2t dt = [t²]₀⁵ = 25 m，与 SUVAT 结果完全一致。
Three exam habits matter here: choose the equation that omits the variable you neither know nor need; keep a consistent sign convention (taking upward as positive makes a mber that distance travelled differs from displacement once a particle reverses direction.
这里有三个应试习惯：选择恰好不含"既不知道也不需要"的那个变量的方程；保持符号约定一致（取向上为正时，重力加速度 a = −9.8 m s⁻²）；并记住一旦质点反向运动，路程与位移就不再相同。
2. Vectors and Forces in Equilibrium | 向量与力的平衡
Vectors are the shared language of mechanics and geometry. A force of magnitude F acting at angle θ to the horizontal has components F cos θ horizontally and F sin θ vertically. Equilibrium requires the vector sum of all forces to vanish, which in component form means ΣFₓ = 0 and ΣFᵧ = 0.
向量是力学与几何的共同语言。大小为 F、与水平方向成 θ 角的力，其分量为水平方向 F cos θ、竖直方向 F sin θ。平衡要求所有力的向量和为零，用分量表示即 ΣFₓ = 0 且 ΣFᵧ = 0。
On a smooth slope inclined at θ, the weight mg resolves into mg sin θ down the slope and mg cos θ perpendicular to the slope. If the surface is smooth the acceleration down the slope is simply g sin θ, independent of mass — a result worth memorising because it links trigonometry directly to Newton's second law.
在倾角为 θ 的光滑斜面上，重力 mg 分解为沿斜面向下的 mg sin θ 和垂直斜面的 mg cos θ。若斜面光滑，沿斜面的加速度就是 g sin θ，与质量无关——这个结论值得记牢，它把三角学与牛顿第二定律直接联系起来。
Example: A 5 kg block slides down a smooth slope of 30°. Find its acceleration. Two forces of 6 N and 8 N act at 90° to each other; find the resultant.
例题：质量 5 kg 的物块沿 30° 光滑斜面下滑，求其加速度；另有两力 6 N 与 8 N 互成 90°，求合力。
Solution: a = g 0° = 9.8 × 0.5 = 4.9 m s⁻² down the slope. For the two forces, the resultant has magnitude √6² + 8² = 10 N and acts at angle θ with tan θ = 8/6, so θ ≈ 53.1° from the 6 N force.
解答：a = g sin 30° = 9.8 × 0.5 = 4.9 m s⁻²，方向沿斜面向下。对于两力，合力大小 √6² + 8² = 10 N，方向与 6 N 力成 θ 角，tan θ = 8/6，故 θ ≈ 53.1°。
3. Exponential Models in Biology: Growth, Decay and Drug Clearance | 生物学中的指数模型：增长、衰减与药物清除
Whenever a quantity changes at a rate proportional to its current size, the model is the differential equation dN/dt = kN, whose solution is N = N₀eᵏᵗ. Positive k describes population growth, compound interest or viral spread; negative k describes radioactive decay, cooling or drug elimination.
只要某个量的变化率与其当前大小成正比，模型就是微分方程 dN/dt = kN，其解为 N = N₀eᵏᵗ。正的 k 描述种群增长、复利或病毒传播；负的 k 描述放射衰变、冷却或药物消除。
Two derived quantities appear constantly in exam questions. The doubling time satisfies t₂ = ln 2 / k, and the half-life satisfies t½ = ln 2 / |k|. Notice that both are independent of the starting value N₀, which is exactly why exponential models are so useful in biology and pharmacology.
考试题中经常出现两个由模型导出的量。倍增时间满足 t₂ = ln 2 / k，半衰期满足 t½ = ln 2 / |k|。注意两者都与初始值 N₀ 无关，这正是指数模型在生物学与药理学中如此实用的原因。
Example: A bacterial culture doubles every 3 hours. Find k, then the population multiple after 12 hours. A drug has a half-life of 6 hours; what fraction remains after 24 hours?
例题：某细菌培养物每 3 小时翻倍，求 k，并求 12 小时后种群变为原来的多少倍。某药物半衰期为 6 小时，24 小时后剩余比例为多少？
Solution: 2N₀ = N₀e³ᵏ gives k = ln 2 / 3 ≈ 0.231 h⁻¹. After 12 hours the multiple is e0.231 × 12 = e2.772 ≈ 16. For the drug, k = ln 2 / 6 ≈ 0.1155 h⁻¹, so the fraction remaining is e−0.1155 × 24 = e−2.772 ≈ 0.0625, i.e. about 6.25%.
解答：由 2N₀ = N₀e³ᵏ 得 k = ln 2 / 3 ≈ 0.231 h⁻¹。12 小时后倍数为 e0.231 × 12 = e2.772 ≈ 16 倍。对药物而言，k = ln 2 / 6 ≈ 0.1155 h⁻¹，故剩余比例为 e−0.1155 × 24 = e−2.772 ≈ 0.0625，即约 6.25%。
4. Probability in Medical Testing: Conditional Probability and Bayes | 医学检测中的概率：条件概率与贝叶斯
Screening tests connect probability to biology and public health. Define P(A|B) as the probability of A given that B has occurred, with P(A|B) = P(A ∩ B) / P(B). A test is described by its sensitivity P(+|disease) and its specificity P(−|no disease).
筛查检测把概率与生物学、公共卫生联系起来。定义 P(A|B) 为在 B 已发生的条件下 A 发生的概率，即 P(A|B) = P(A ∩ B) / P(B)。检测的两个关键指标
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