📚 Year 13 CCEA Mathematics: Cross-Disciplinary Integrated Problem-Solving Training | Year 13 CCEA 数学：跨学科综合题型训练
In Year 13 CCEA Mathematics, students are increasingly required to apply pure mathematical techniques to real-world problems from physics, economics, biology and engineering. This article provides a structured training in cross-disciplinary problem-solving, covering key topics such as differential equations, optimisation, exponential models, probability distributions and vectors, all contextualised within authentic scenarios. The aim is to build fluency in translating interdisciplinary contexts into mathematical models and back again, a skill essential for A2 assessments and beyond.
在 Year 13 CCEA 数学课程中，学生需要将纯数学技巧应用于物理、经济学、生物学和工程等现实情境。本文提供跨学科问题解决的结构化训练，涵盖微分方程、最优化、指数模型、概率分布和向量等核心主题，并全部置于真实场景中。目标是培养将跨学科情境转化为数学模型并还原解释的能力，这对 A2 考试和后续学习至关重要。
1. Modelling Population Growth: Exponential and Logistic Models | 种群增长模型：指数与逻辑斯谛模型
Ecologists often model population size P(t) over time t using differential equations. The simplest model is exponential growth: dP/dt = kP, where k is the growth rate. This yields P(t) = P₀eᵏᵗ. However, limited resources lead to the logistic model: dP/dt = rP(1 – P/K), where K is the carrying capacity. Solve this to obtain P(t) = K / (1 + ((K – P₀)/P₀)e⁻ʳᵗ). Interpret the behaviour as t → ∞.
生态学家常用微分方程模拟种群数量 P(t) 随时间 t 的变化。最简单的模型是指数增长：dP/dt = kP，其中 k 为增长率，解得 P(t) = P₀eᵏᵗ。但资源有限时会引入逻辑斯谛模型：dP/dt = rP(1 – P/K)，K 为环境容纳量，求解得到 P(t) = K / (1 + ((K – P₀)/P₀)e⁻ʳᵗ)。试解释 t → ∞ 时的行为。
A fish farm stocks 500 trout in a lake with carrying capacity 10 000. The intrinsic growth rate is r = 0.8 year⁻¹. Find the time needed for the population to reach 5000. Differentiate to find the population growth rate when P = 3000.
某养鱼场在容纳量为 10000 的湖中投放 500 尾鳟鱼，内禀增长率 r = 0.8 年⁻¹。求种群数量达到 5000 所需的时间。求导计算当 P = 3000 时的种群增长速率。
Using the logistic solution: 5000 = 10000 / (1 + (9500/500)e⁻⁰·⁸ᵗ). This simplifies to 1 + 19e⁻⁰·⁸ᵗ = 2, so 19e⁻⁰·⁸ᵗ = 1 ⇒ e⁻⁰·⁸ᵗ = 1/19 ⇒ t = –ln(1/19)/0.8 ≈ 3.68 years. Rate of change: dP/dt = 0.8 × 3000 × (1 – 3000/10000) = 0.8 × 3000 × 0.7 = 1680 fish/year.
代入逻辑斯谛解：5000 = 10000 / (1 + (9500/500)e⁻⁰·⁸ᵗ)，化简得 1 + 19e⁻⁰·⁸ᵗ = 2，所以 19e⁻⁰·⁸ᵗ = 1 ⇒ e⁻⁰·⁸ᵗ = 1/19 ⇒ t = –ln(1/19)/0.8 ≈ 3.68 年。增长速率：dP/dt = 0.8 × 3000 × (1 – 3000/10000) = 0.8 × 3000 × 0.7 = 1680 尾/年。
2. Optimisation in Economics: Revenue, Cost and Profit | 经济学中的最优化：收入、成本与利润
A firm’s total revenue R and total cost C are functions of output q. Profit Π(q) = R(q) – C(q). Marginal revenue and marginal cost are R'(q) and C'(q). Profit maximisation requires Π'(q) = 0 and Π”(q) < 0, i.e. R'(q) = C'(q). This is a classic application of differentiation often examined in CCEA A2 papers.
企业总收入 R 和总成本 C 是产量 q 的函数。利润 Π(q) = R(q) – C(q)。边际收入与边际成本分别为 R'(q) 和 C'(q)。利润最大化要求 Π'(q) = 0 且 Π”(q) < 0，即 R'(q) = C'(q)。这是 CCEA A2 考试中经典的微分应用题型。
Given a demand function p = 120 – 2q and total cost C = 2q² + 10q + 200, find the output that maximises profit and calculate the maximum profit. Verify the second-order condition.
已知需求函数 p = 120 – 2q，总成本 C = 2q² + 10q + 200，求使利润最大化的产量并计算最大利润，验证二阶条件。
Revenue R = pq = (120 – 2q)q = 120q – 2q². Profit Π = (120q – 2q²) – (2q² + 10q + 200) = 110q – 4q² – 200. Π’ = 110 – 8q = 0 ⇒ q = 13.75. Π” = –8 < 0, maximum. Max profit = 110(13.75) – 4(13.75)² – 200 = 1512.5 – 756.25 – 200 = 556.25. Answer: q = 13.75 units, profit £556.25.
收入 R = pq = (120 – 2q)q = 120q – 2q²。利润 Π = 110q – 4q² – 200。Π’ = 110 – 8q = 0 ⇒ q = 13.75。Π” = –8 < 0，为极大值。最大利润 = 110(13.75) – 4(13.75)² – 200 = 1512.5 – 756.25 – 200 = 556.25。答案：产量 13.75 单位，利润 556.25 英镑。
3. Kinematics with Calculus: Variable Acceleration | 微积分在运动学中的应用：变加速运动
In mechanics, displacement s, velocity v and acceleration a are related by differentiation with respect to time t: v = ds/dt, a = dv/dt = d²s/dt². When acceleration is not constant, integration is required to find velocity and displacement from an acceleration-time function. CCEA papers often feature problems where acceleration is given as a function of time or displacement.
在力学中，位移 s、速度 v 和加速度 a 通过对时间 t 的微商相联系：v = ds/dt，a = dv/dt = d²s/dt²。当加速度不恒定时，需要积分从加速度-时间函数求速度和位移。CCEA 试卷中常出现加速度作为时间或位移函数的问题。
A particle moves along a straight line with acceleration a = 6t – 4 ms⁻². Initially t = 0, s = 0, v = 2 ms⁻¹. Find expressions for v(t) and s(t). Determine the distance travelled in the first 3 seconds.
一质点做直线运动，加速度 a = 6t – 4 ms⁻²。初始条件 t = 0, s = 0, v = 2 ms⁻¹。求 v(t) 和 s(t) 的表达式，并求前 3 秒内经过的路程。
Integrate a: v = ∫(6t – 4) dt = 3t² – 4t + C. Using v(0)=2 ⇒ C = 2, so v = 3t² – 4t + 2. Then s = ∫(3t² – 4t + 2) dt = t³ – 2t² + 2t + D. s(0)=0 ⇒ D=0, so s = t³ – 2t² + 2t. At t=3, s = 27 – 18 + 6 = 15 m. Need to check if velocity changes sign: v = 3t² – 4t + 2 = 0 has discriminant 16 – 24 < 0, so v > 0 always. Distance = 15 m.
积分 a 得：v = ∫(6t – 4) dt = 3t² – 4t + C。由 v(0)=2 得 C = 2，故 v = 3t² – 4t + 2。再积分得 s = t³ – 2t² + 2t + D。由 s(0)=0 得 D=0，所以 s = t³ – 2t² + 2t。t=3 时，s=27-18+6=15 m。检查速度是否改变符号：v = 3t² – 4t + 2，判别式 16 – 24 < 0，因此 v 始终为正，路程 = 15 m。
4. Exponential Decay: Radioactivity and Medicine | 指数衰减：放射性衰变与医药
The rate of decay of a radioactive substance is proportional to the amount present: dN/dt = –λN, leading to N = N₀e⁻λᵗ. Half-life T₁/₂ is given by ln2/λ. Similar exponential models describe drug concentration in the bloodstream. Understanding constant of proportionality and solving equations involving e⁻ᵏᵗ is a core skill.
放射性物质的衰变速率与现有数量成正比：dN/dt = –λN，解得 N = N₀e⁻λᵗ。半衰期 T₁/₂ = ln2/λ。类似的指数模型也描述药物在血液中的浓度。理解比例常数并求解含 e⁻ᵏᵗ 的方程是核心技能。
A patient receives an injection that leads to an initial concentration of 8 mg/L. The concentration halves every 6 hours. Find the concentration after 24 hours and the time when it drops below 0.5 mg/L. Use C = 8 × (1/2)^(t/6) or C = 8e⁻ᵏᵗ with k = ln2/6.
某患者接受注射后初始血药浓度为 8 mg/L，浓度每 6 小时减半。求 24 小时后的浓度，以及浓度降至 0.5 mg/L 以下的时间。使用 C = 8 × (1/2)^(t/6) 或 C = 8e⁻ᵏᵗ，其中 k = ln2/6。
After 24 h: C = 8 × (1/2)^(24/6) = 8 × (1/2)⁴ = 8 × 1/16 = 0.5 mg/L. For C < 0.5, using the exponential form: 8e⁻ᵏᵗ < 0.5 ⇒ e⁻ᵏᵗ < 0.0625. Taking ln: –kt < ln(0.0625) = –2.7726. With k = 0.1155, t > 24 h. So just after 24 hours the concentration drops below 0.5 mg/L.
24 小时后：C = 8 × (1/2)⁴ = 0.5 mg/L。降至 0.5 以下：8e⁻ᵏᵗ < 0.5 ⇒ e⁻ᵏᵗ < 0.0625，取对数得 –kt < ln(0.0625) = –2.7726，k = 0.1155，所以 t > 24 小时。因此 24 小时刚过浓度就降到 0.5 以下。
5. Vectors in Mechanics: Forces and Equilibrium | 向量在力学中的应用：力与平衡
Forces are vector quantities with magnitude and direction. A particle in equilibrium has zero resultant force. Problems often involve resolving forces into i and j components, finding magnitude of a force as √(x² + y²), and using dot product to find angles between forces. CCEA expects fluency in vector notation and equilibrium conditions.
力是具有大小和方向的向量。处于平衡状态的质点合力为零。问题常涉及将力分解为 i 和 j 分量，求力的大小如 √(x² + y²)，以及用点积求力之间的夹角。CCEA 要求熟练掌握向量表示法和平衡条件。
Three forces F₁ = (2i + 3j) N, F₂ = (–i + 4j) N, and F₃ = (ai + bj) N act on a particle at rest. Find a and b. Then find the magnitude of F₃ and the angle it makes with the i direction.
三个力 F₁ = (2i + 3j) N，F₂ = (–i + 4j) N，F₃ = (ai + bj) N 作用于静止的质点。求 a 和 b，F₃ 的大小及其与 i 方向的夹角。
Equilibrium: F₁ + F₂ + F₃ = 0 ⇒ (2–1+a)i + (3+4+b)j = 0 ⇒ a + 1 = 0 and 7 + b = 0 ⇒ a = –1, b = –7. Magnitude of F₃ = √((-1)² + (–7)²) = √50 = 5√2 ≈ 7.07 N. Angle θ = tan⁻¹(7/1) with negative i component, so direction is third quadrant: angle 180 + tan⁻¹(7) ≈ 180 – 81.87 = 261.87° from positive i (or 81.87° below negative i axis).
平衡条件：F₁ + F₂ + F₃ = 0 ⇒ (2–1+a)i + (3+4+b)j = 0 ⇒ a + 1 = 0 且 7 + b = 0 ⇒ a = –1，b = –7。F₃ 的大小 = √(1+49) = √50 ≈ 7.07 N。与正 i 方向的夹角 θ = tan⁻¹(7)，因 a 负 b 负，方向为第三象限：θ = 180 + 81.87 = 261.87°（或说是与负 i 轴夹 81.87° 向下）。
6. Probability Distributions in Quality Control | 质量管理中的概率分布
Binomial and normal distributions are frequently used in industrial quality control. If the probability of a defective item is p, the number of defectives in a sample of n follows B(n, p). For large n and moderate p, normal approximation N(np, np(1–p)) can be applied with continuity correction. CCEA questions often link probability with hypothesis testing.
二项分布与正态分布常用于工业质量控制。若次品率为 p，则抽样 n 件中的次品数服从 B(n, p)。当 n 较大且 p 适中时，可用正态近似 N(np, np(1–p)) 并附连续性校正。CCEA 题目常将概率与假设检验相结合。
A factory produces 5% defective items. A batch of 200 items is inspected. Find the probability that there are exactly 15 defectives. Use a suitable approximation to find the probability that more than 12 are defective. What does this suggest if the batch actually contains 20 defectives?
某工厂次品率为 5%。检查 200 件产品。求恰好有 15 件次品的概率。使用适当的近似求次品超过 12 件的概率。若某批次实际有 20 件次品，这说明了什么？
Exact binomial: P(X=15) = ⁿC₁₅(0.05)¹⁵(0.95)¹⁸⁵. Approximate by Poisson with λ = np = 10, P(15) = e⁻¹⁰ × 10¹⁵ / 15! ≈ 0.0347. Normal approximation: np = 10, npq = 9.5, σ = √9.5 ≈ 3.082. P(X > 12) using continuity correction: P(X > 12.5). z = (12.5 – 10)/3.082 ≈ 0.811. P(Z > 0.811) = 1 – 0.7910 = 0.209. With 20 defectives observed, expected is only 10, so it strongly suggests the defective rate has increased above 5% — a hypothesis test would reject the null hypothesis at typical significance levels.
精确二项：P(X=15) = ²⁰⁰C₁₅(0.05)¹⁵(0.95)¹⁸⁵。泊松近似 λ=10，P(15) ≈ e⁻¹⁰×10¹⁵/15! ≈ 0.0347。正态近似：np=10, npq=9.5, σ≈3.082。P(X > 12) 用连续性校正：P(X > 12.5)，z = (12.5–10)/3.082≈0.811，P(Z>0.811)=0.209。实际观察到 20 件次品，而期望仅 10 件，强烈提示次品率已上升超过 5%——假设检验会在典型显著性水平下拒绝原假设。
7. Differential Equations in Engineering: Cooling and Heating | 工程中的微分方程：冷却与加热
Newton’s law of cooling states that the rate of change of temperature T of an object is proportional to the difference between its temperature and the ambient temperature Tₐ: dT/dt = –k(T – Tₐ). Solving this yields T = Tₐ + (T₀ – Tₐ)e⁻ᵏᵗ. This first-order linear DE appears in many engineering contexts, including electrical circuits and heat exchangers.
牛顿冷却定律指出：物体温度 T 的变化率与物体温度和环境温度 Tₐ 之差成正比：dT/dt = –k(T – Tₐ)。解得 T = Tₐ + (T₀ – Tₐ)e⁻ᵏᵗ。这一阶线性微分方程出现在许多工程情境中，包括电路和热交换器。
A metal bar heated to 200°C is placed in a room at 20°C. After 5 minutes its temperature is 100°C. Find the temperature after 10 minutes. How long until it reaches 30°C? Interpret the constant k.
一根金属棒加热至 200°C 后放入 20°C 的房间。5 分钟后温度为 100°C。求 10 分钟后的温度，以及降至 30°C 所需的时间。解释常数 k 的含义。
Using T = 20 + (200 – 20)e⁻ᵏᵗ = 20 + 180e⁻ᵏᵗ. At t=5, 100 = 20 + 180e⁻⁵ᵏ ⇒ 80/180 = e⁻⁵ᵏ ⇒ e⁻⁵ᵏ = 4/9. Hence k = –(1/5)ln(4/9) ≈ 0.162 min⁻¹. After 10 min: T = 20 + 180e⁻¹·⁶² ≈ 20 + 180 × 0.198 = 55.6°C. For T=30: 30 = 20 + 180e⁻ᵏᵗ ⇒ 10/180 = e⁻ᵏᵗ ⇒ e⁻ᵏᵗ = 1/18 ⇒ t = –ln(1/18)/0.162 ≈ 17.8 minutes. k depends on thermal properties and surface area.
使用 T=20+180e⁻ᵏᵗ。t=5 时，100=20+180e⁻⁵ᵏ ⇒ e⁻⁵ᵏ=4/9，得 k≈0.162 min⁻¹。10 分钟后：T=20+180e⁻¹·⁶²≈20+35.6=55.6°C。T=30：30=20+180e⁻ᵏᵗ⇒e⁻ᵏᵗ=1/18⇒ t≈17.8 分钟。k 取决于热属性和表面积。
8. Geometric Sequences in Finance: Compound Interest and Loans | 金融中的几何序列：复利与贷款
Compound interest leads to geometric progression in savings and loan repayments. The future value of an investment S = P(1 + r/n)^(nt) where r is annual rate, n compounding periods per year. Regular savings over time form a geometric series, and the sum formula is used to calculate annuity values. CCEA financial mathematics questions often require solving exponential equations using logs.
复利导致储蓄和贷款还款额呈几何级数增长。投资终值 S = P(1 + r/n)^(nt)，其中 r 为年利率，n 为每年复利次数。定期储蓄构成几何序列，求和公式用于计算年金价值。CCEA 金融数学题常需用对数求解指数方程。
A person invests £5000 at 4% per annum compounded quarterly. How long will it take to double? If they instead deposit £200 at the start of each quarter into the same account, find the total after 10 years using the sum of a geometric series.
某人以年利率 4%，每季度复利投资 £5000。需多久可翻倍？若改为每季度初存入 £200，利用几何级数求和计算 10 年后的总金额。
Doubling time: 10000 = 5000(1.01)^(4t) ⇒ 2 = (1.01)^(4t). Taking logs: 4t ln(1.01) = ln2 ⇒ t = ln2/(4 ln1.01) ≈ 0.6931/(4 × 0.00995) ≈ 17.4 years. For quarterly payments at start: each payment grows for different periods. Quarter 1 payment: 200(1.01)^(40); quarter 2: 200(1.01)^(39), …, last payment 200(1.01). Total = 200(1.01 + 1.01² + … + 1.01⁴⁰). This is a geometric series with a=200×1.01, r=1.01, n=40. Sum = a(rⁿ–1)/(r–1) = 202(1.01⁴⁰–1)/0.01 ≈ 202(1.4889 – 1)/0.01 = 202×48.89 = £9875.78.
翻倍时间：10000=5000(1.01)^(4t)⇒ 2=(1.01)^(4t)，取对数：4t ln1.01 = ln2⇒ t = ln2/(4 ln1.01)≈17.4 年。每季初付款：第一笔 200 增长 40 个季度：200(1.01)⁴⁰；第二笔(1.01)³⁹，…，最后一笔 200(1.01)。总和 = 200(1.01 + 1.01² + … + 1.01⁴⁰) = 几何级数。a=200×1.01=202, r=1.01, n=40。Sum = 202(1.01⁴⁰–1)/0.01 ≈ 202(1.4889–1)/0.01 = £9875.78。
9. Projectile Motion with Parametric Equations | 参数方程描述抛体运动
A projectile launched with initial speed u at angle θ above horizontal follows the path: x = u cos θ t, y = u sin θ t – ½ gt². Eliminating t gives the Cartesian equation. The range, maximum height, and time of flight are standard results derived from the quadratic. CCEA Mechanics often combines these with algebraic solving of trigonometric equations.
以初速 u、仰角 θ 发射的抛体遵循路径：x = u cos θ t，y = u sin θ t – ½ gt²。消去 t 可得直角坐标方程。射程、最大高度和飞行时间是从二次方程导出的标准结果。CCEA 力学常将此类问题与三角方程的代数求解结合。
A ball is kicked from ground level with speed 20 ms⁻¹ at 30°. Find the time of flight, maximum height, and horizontal range. Then determine the angle needed to achieve exactly the same range with the same speed. Use g = 9.8 ms⁻².
足球从地面以 20 ms⁻¹、仰角 30° 踢出。求飞行时间、最大高度和水平射程。再求用相同初速获得相同射程所需的另一个角度。取 g = 9.8 ms⁻²。
Time of flight: T = 2u sin θ / g = 2×20×sin30° / 9.8 = 40×0.5/9.8 ≈ 2.04 s. Max height H = (u sin θ)²/(2g) = (10)²/19.6 = 5.10 m. Range R = u² sin2θ / g = 400×sin60°/9.8 = 400×0.8660/9.8 ≈ 35.35 m. For same range, sin2θ = sin60° ⇒ 2θ = 60° or 120° ⇒ θ = 30° or 60°. Thus the complementary angle 60° gives the same range.
飞行时间：T=2u sinθ / g = 40×0.5/9.8≈2.04 s。最大高度 H=(u sinθ)²/(2g)=100/19.6≈5.10 m。射程 R=u² sin2θ / g =400×sin60°/9.8≈35.35 m。相同射程要求 sin2θ=sin60°，所以 2θ=60° 或 120°，得 θ=30° 或 60°。互余角 60° 给出相同射程。
10. Integration Applied to Areas and Volumes in Design | 积分在设计与体积计算中的应用
Definite integrals calculate areas under curves and volumes of revolution, applicable in industrial design, architecture and biomechanics. Volume V = π ∫ₐᵇ y² dx when rotating the curve y = f(x) around the x-axis. For a region between two curves, the area is ∫(top – bottom) dx. These concepts are assessed in technical contexts.
定积分可计算曲线下方面积和旋转体体积，应用于工业设计、建筑和生物力学。体积 V = π ∫ₐᵇ y² dx 表示曲线 y = f(x) 绕 x 轴旋转。两曲线间的面积为 ∫(上方 – 下方) dx。这些概念常在技术情境下考查。
A designer creates a vase shape by rotating the curve y = √(x) + 1 between x = 0 and x = 4 about the x-axis. Find the internal volume. Then find the area of the curved surface using A = 2π ∫ y √(1 + (dy/dx)²) dx, though this is beyond CCEA, discuss the approach conceptually.
设计师通过将曲线 y = √x + 1（x 从 0 到 4）绕 x 轴旋转创建一个花瓶形状。求内部体积。再讨论曲面面积的概念公式 A = 2π ∫ y √(1 + (dy/dx)²) dx，虽超出 CCEA 范围，但进行概念讨论。
Volume V = π ∫₀⁴ (√x + 1)² dx = π ∫₀⁴ (x + 2√x + 1) dx = π [½x² + (4/3)x^(3/2) + x]₀⁴ = π (8 + (4/3)×8 + 4) = π (12 + 32/3) = π (68/3) ≈ 71.2 cubic units. Surface area would involve dy/dx = 1/(2√x), then length element √(1 + 1/(4x)). The integral can be evaluated but is lengthy. The main point is that integration links curves to real objects.
体积 V = π ∫₀⁴ (√x + 1)² dx = π ∫₀⁴ (x+2√x+1) dx = π [½x² + (4/3)x^(3/2) + x]₀⁴ = π(8 + 32/3 + 4) = π(68/3) ≈ 71.2 立方单位。曲面面积涉及 dy/dx = 1/(2√x)，长度微元为 √(1 + 1/(4x))。积分虽可求但较长。重点是积分将曲线与真实物体联系起来。
11. Statistical Sampling and the Central Limit Theorem in Research | 研究中的统计抽样与中心极限定理
When conducting experiments, researchers often deal with sample means. The Central Limit Theorem states that for a sufficiently large sample size n, the sampling distribution of the mean is approximately normal with mean μ and standard error σ/√n, regardless of the population distribution shape. This underpins confidence intervals and hypothesis tests in psychology, medicine and social sciences.
实验研究常涉及样本均值。中心极限定理表明：对足够大的样本量 n，样本均值的抽样分布近似正态，均值为 μ，标准误为 σ/√n，无论总体分布形状如何。这是心理学、医学和社会科学中置信区间和假设检验的基础。
A psychologist measures reaction times with a known population standard deviation σ = 0.2 s. She takes a sample of 100 people and finds the mean reaction time is 0.85 s. Construct a 95% confidence interval for the true mean. If she wants the interval width to be at most 0.05 s, what sample size is needed?
某心理学家测量反应时间，已知总体标准差 σ = 0.2 s。她抽取 100 人的样本，得到平均反应时间 0.85 s。构建真实均值的 95% 置信区间。若希望区间宽度不超过 0.05 s，需要多大样本量？
95% CI: x̄ ± z × σ/√n. z = 1.96. ME = 1.96 × 0.2/√100 = 1.96 × 0.02 = 0.0392. Interval: (0.8108, 0.8892). For width ≤ 0.05, half-width ME ≤ 0.025. ME = 1.96 × 0.2/√n ≤ 0.025 ⇒ √n ≥ (1.96×0.2)/0.025 = 15.68 ⇒ n ≥ 245.86, so n = 246 needed.
95% 置信区间：x̄ ± z×σ/√n。z=1.96，误差限 = 1.96×0.2/10 = 0.0392。区间 (0.8108, 0.8892)。宽度≤0.05 即半宽≤0.025：1.96×0.2/√n ≤ 0.025 ⇒ √n ≥ 15.68 ⇒ n ≥ 245.86，需 n=246。
12. Numerical Methods: Iteration in Engineering Design | 数值方法：工程设计中的迭代
Many equations arising from science cannot be solved analytically; numerical methods like Newton-Raphson and fixed-point iteration are used. CCEA requires students to apply iterative schemes and discuss convergence. The Newton-Raphson formula: xₙ₊₁ = xₙ – f(xₙ)/f'(xₙ) is particularly efficient for finding roots.
许多科学中的方程无法解析求解，需用数值方法如牛顿-拉弗森法和不动点迭代。CCEA 要求学生应用迭代格式并讨论收敛性。牛顿-拉弗森公式：xₙ₊₁ = xₙ – f(xₙ)/f'(xₙ) 是求根的高效方法。
An engineer needs to solve e⁻ˣ = x. Starting with x₀ = 0.5, apply three iterations of Newton-Raphson. Show the improvement in precision. Discuss what happens if a poor starting value like x₀ = 2 is chosen.
工程师需解方程 e⁻ˣ = x。取 x₀ = 0.5，应用三次牛顿-拉弗森迭代，展示精度的提高。讨论若选择不良初值如 x₀ = 2 会怎样。
Let f(x) = e⁻ˣ – x = 0. Then f'(x) = –e⁻ˣ – 1. Newton-Raphson: xₙ₊₁ = xₙ – (e⁻ˣ – xₙ)/(–e⁻ˣ – 1) = xₙ + (e⁻ˣ – xₙ)/(e⁻ˣ + 1). Start x₀ = 0.5: f(0.5) = e⁻⁰·⁵ – 0.5 ≈ 0.6065 – 0.5 = 0.1065. f'(0.5) = –0.6065 – 1 = –1.6065. x₁ = 0.5 – 0.1065/(–1.6065) ≈ 0.5663. Next iteration: f(0.5663) = e⁻⁰·⁵⁶⁶³ – 0.5663 ≈ 0.5676 – 0.5663 = 0.0013. f’ = –0.5676 –1 = –1.5676. x₂ = 0.5663 – 0.0013/(–1.5676) ≈ 0.56714. Third iteration: f(0.
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