📚 AS CIE Further Mathematics: Cross-disciplinary Integrated Problem Training | AS CIE 进阶数学：跨学科综合题型训练
The AS Level CIE Further Mathematics syllabus is built around deliberate pairings: Further Pure Mathematics on one side, and either Further Mechanics or Further Probability & Statistics on the other. Examiners rarely test these in isolation. Questions routinely ask you to use complex numbers on a force diagram, differential equations on a damped spring, matrices on a system of equilibrium equations, or integration on a centre of mass. This article provides a curated set of cross-disciplinary problems of exactly that kind, each one worked through step by step.
AS Level CIE 进阶数学考纲的设计本身就是"成对"的：一边是进阶纯数学，另一边是进阶力学或进阶概率统计。考官极少把它们分开考，题目常常要求你在受力图上用复数、在阻尼弹簧上用微分方程、在平衡方程组上用矩阵、在质心问题上用积分。本文精选的正是这一类跨学科题目，逐题给出完整解法。
1. Complex Numbers and Force Resultants | 复数与合力
Complex numbers are the fastest route to a resultant when two-dimensional forces are given as vectors. Take two forces acting on a small particle: F₁ = (3 + 4i) N and F x-component and the imaginary part is the y-component. Adding real parts and imaginary parts separately gives the resultant immediately.
当二维力以复数形式给出时，复数是求合力最快的途径。设作用在小质点上的两个力：F₁ = (3 + 4i) N、F₂ = (5 − 2i) N，其中实部对应 x 分量、虚部对应 y 分量。实部与虚部分别相加即可直接得到合力。
F = F₁ + F₂ = (3 + 5) + (4 − 2)i = 8 + 2i N
The magnitude is |F| = √8² + 2² = √68 = 2√17 on is θ = arctan(2/8) ≈ 14.04° measured anticlockwise from the positive x-axis. In modulus–argument form, F = 2√17(cos 14.04° + i sin 14.04°).
合力大小为 |F| = √8² + 2² = √68 = 2√17 ≈ 8.25 N，方向为从正 x 轴逆时针量得的 θ = arctan(2/8) ≈ 14.04°。写成模—辐角形式即 F = 2√17(cos 14.04° + i sin 14.04°)。
Two further results follow at once. First, equilibrium: if the particle is to remain at rest, a third force F ector 90° anticlockwise without changing its modulus, so i(8 + 2i) = −2 + 8i has modulus 2√17 but argument 104.04°. More generally, multiplying by eiφ = cos φ + i sin φ rotates any force vector through an angle φ.
由此可立刻得到两个结论。第一是平衡：若质点要保持静止，必须施加第三个力 F₃ = −(8 + 2i) = −8 − 2i N，大小同为 8.25 N，方向为 194.04°。第二是旋转：乘以 i 相当于把向量逆时针旋转 90° 且模不变，故 i(8 + 2i) = −2 + 8i 的模仍为 2√17，辐角变为 104.04°。更一般地，乘以 eiφ = cos φ + i sin φ 可将任意力向量旋转 φ 角。
This technique links complex multiplication directly to rigid-body rotation problems in mechanics. Whenever a question describes a couple, a turning moment, or the rotation of a lamina by a given angle, the algebra is exactly the multiplication of complex numbers you already use in Further Pure Mathematics.
这一技巧把复数乘法与力学中的刚体旋转问题直接联系起来。凡是题目涉及力偶、转动力矩或薄板转过给定角度，其代数运算本质上就是进阶纯数学里已经熟悉的复数乘法。
2. Vector Kinematics and Differential Calculus | 向量运动学与微分
A particle moves so that its position vector in metres is r is the first derivative with respect to time and acceleration is the second, both differentiating the i and j components separately.
一质点的位置向量（单位：米）为 r(t) = (3t² − 2t)i + (4t³ − t)j。速度是位置对时间的一阶导数，加速度是二阶导数，对 i、j 分量分别求导即可。
v = dr/dt = (6t − 2)i + (12t² − 1)j m s⁻¹
