Cross-disciplinary Integrated Problem-Solving for OCR Year 10 Further Maths | OCR Year 10 进阶数学：跨学科综合题型训练

📚 Cross-disciplinary Integrated Problem-Solving for OCR Year 10 Further Maths | OCR Year 10 进阶数学：跨学科综合题型训练

The OCR Year 10 Further Mathematics course equips you with powerful tools such as differentiation, exponential functions, vectors and sequences. These skills are not just for abstract exercises – they form the mathematical backbone of physics, economics, biology, computer science, engineering and more. This revision article presents cross-disciplinary problems that train you to recognise mathematical patterns in real-world situations and apply your knowledge confidently in an integrated context.

OCR Year 10 进阶数学课程赋予学生微分、指数函数、向量和数列等强有力的工具。这些技能不仅用于抽象练习，更构成了物理、经济、生物、计算机科学和工程等学科的数学支柱。本文提供跨学科综合题型训练，帮助你识别真实情境中的数学模式，并在整合性背景下自信地运用所学知识。

1. Physics – Kinematics with Quadratic Functions | 物理学——二次函数与运动学

an initial speed of 20 m/s. The height h (in metres) after t seconds is modelled by h(t) = – fferentiation to find the maximum height and then calculate when the ball hits the ground.

一个球从2米高处以20 m/s的初速度竖直向上抛出。t 秒后的高度 h（米）由函数 h(t) = –5t² + 20t + 2 模拟。我们将用微分求最大高度，并计算球何时落地。

Differentiating with respect to t gives the instantaneous velocity.

对 t 求导得到瞬时速度。

h'(t) = –10t + 20

The maximum height occurs when the velocity is zero, so we set h'(t) = 0.

最大高度出现在速度为零时，因此令 h'(t) = 0。

–10t + 20 = 0 ⇒ t = 2 s

Substituting t = 2 into h(t) gives the maximum height.

将 t = 2 代入 h(t) 得到最大高度。

h(2) = –5(2)² + 20(2) + 2 = –20 + 40 + 2 = 22 m

To find the time taken to reach the ground, we solve h(t) = 0 using the quadratic formula.

为求落地时间，我们用求根公式解 h(t) = 0。

–5t² + 20t + 2 = 0 ⇒ t = [ –20 ± √(20² – 4(–5)(2)) ] / (2 × –5)

This simplifies to t = [ –20 ± √400 + 40 ] / (–10). Only the positive root makes physical sense.

化简得 t = [ –20 ± √440 ] / (–10)。只有正根才有物理意义。

t ≈ [ –20 + 20.976 ] / (–10) ≈ –0.0976 (reject) and t ≈ [ –20 – 20.976 ] / (–10) ≈ 4.10 s

Therefore, the ball reaches a peak of 22 m after 2 seconds and lands after approximately 4.10 seconds. This example shows how quadratic functions and calculus describe vertical motion under gravity.

因此，球在2秒后达到最高点 22 m，约 4.10 秒后落地。本例展示了二次函数和微积分如何描述重力作用下的竖直运动。

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2. Economics – Profit Maximisation using Differentiation | 经济学——利用微分求利润最大化

A company's profit P (in £) depends on the number of units x (in hundreds) produced. The profit function is P(x) = –2x² + 120x – 500. We will find the production level that maximises profit.

某公司的利润 P（英镑）与产量 x（百件）的关系为 P(x) = –2x² + 120x – 500。我们要求出使利润最大的产量。

We differentiate the profit function with respect to x and set the derivative to zero.

对利润函数关于 x 求导并令其为零。

dP/dx = –4x + 120 = 0 ⇒ x = 30

This critical point corresponds to 3000 units (since x is in hundreds).

该临界点对应 3000 件产品（因为 x 以百为单位）。

We confirm it is a maximum by checking the second derivative: d²P/dx² = –4, which is negative, indicating a concave down curve.

我们通过二阶导数确认它是极大值：d²P/dx² = –4 为负，表明曲线开口向下。

Substituting x = 30 into P(x) gives the maximum profit.

将 x = 30 代入 P(x) 可求出最大利润。

P(30) = –2(30)² + 120(30) – 500 = –1800 + 3600 – 500 = £1300

Thus the firm should produce 3000 units to achieve a maximum profit of £1300. Marginal analysis using calculus is fundamental in economic optimisation.

因此，公司应生产 3000 件产品以获得最大利润 £1300。基于微积分的边际分析是经济优化中的基本方法。

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3. Biology – Exponential Growth and Logarithmic Equations | 生物学——指数增长与对数方程

A bacterial culture begins with 200 cells and grows according to N(t) = 200 e^0.4t, where t is time in hours. Determine how long it takes for the population to reach 1000 cells.

某细菌培养初始有200个细胞，并按 N(t) = 200 e^0.4t 增长，其中 t 为小时数。求种群达到 1000 个细胞所需的时间。

Set N(t) = 1000 and isolate the exponential term.

令 N(t) = 1000，将指数项分离出来。

1000 = 200 e^0.4t ⇒ e^0.4t = 5

Taking the natural logarithm of both sides allows us to solve for t.

对两边取自然对数即可解出 t。

ln( e^0.4t ) = ln 5 ⇒ 0.4t = ln 5

Using a calculator, ln 5 ≈ 1.6094, so t = 1.6094 / 0.4 ≈ 4.02 hours.

用计算器求得 ln 5 ≈ 1.6094，因此 t = 1.6094 / 0.4 ≈ 4.02 小时。

Exponential models and logarithms are essential for describing population dynamics, radioactive decay and many natural processes.

指数模型和对数是描述种群动态、放射性衰变以及许多自然过程的关键工具。

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4. Computer Science – Algorithm Complexity and Sequence Analysis | 计算机科学——算法复杂度与数列分析

2n + 1. We will examine how this sequence grows and identify the dominant term for large n.

某排序算法在处理数据量 n 时的基本操作次数为 T(n) = 3n² + 2n + 1。我们将分析这个数列的增长情况，并找出 n 很大时的主导项。

First, calculate the number of operations when n = 1000.

首先计算 n = 1000 时的操作次数。

T(1000) = 3(1000)² + 2(1000) + 1 = 3,000,000 + 2000 + 1 = 3,002,001

constant terms. To see this formally, consider the limit of the ratio T(n) / n² as n → ∞.

当 n 很大时，二次项 3n² 支配着一次项和常数项。为严格说明这一点，我们考虑 n → ∞ 时比值 T(n)/n² 的极限。

T(n)/n² = 3 + 2/n + 1/n² → 3 as n → ∞

This tells us that the algorithm's running time grows roughly in proportion to n² when the input is large. In computer science, we say the algorithm has O(n²) time complexity. Sequence analysis and limits help predict performance without running the code.

这告诉我们，当输入量很大时，算法的运行时间大致与 n² 成正比。在计算机科学中，我们说该算法具有 O(n²) 的时间复杂度。数列分析和极限能帮助我们在不运行代码的情况下预测性能。

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5. Chemistry – Reaction Rates and Linear Graphs | 化学——反应速率与线性图像

ts, the initial rate r is directly proportional to the concentration [A]. A student records the following data:

对于一级反应 A → 产物，初始速率 r 与浓度 [A] 成正比。某学生记录的数据如下：

[A] / mol L⁻¹	0.10	0.20	0.30
Rate / mol L⁻¹ s⁻¹	0.020	0.040	0.060

elled by the equation r = k[A], where k is the rate constant. Use the data to determine k.

这一关系可用方程 r = k[A] 模拟，其中 k 是速率常数。请根据数据求出 k。

From the first pair, substitute [A] = 0.10 and r = 0.020 into the equation.

代入第一对数据 [A] = 0.10 和 r = 0.020。

0.020 = k × 0.10 ⇒ k = 0.020 / 0.10 = 0.20 s⁻¹

40, which matches. The third point also agrees. Plotting r against [A] would give a straight line through the origin with gradient k. Linear equations are fundamental in analysing experimental data and deducing chemical kinetics.

检查其余点的一致性：当 [A] = 0.20 时，r = 0.20 × 0.20 = 0.040，吻合；第三点也一致。若绘制 r 对 [A] 的图像，将得到一条过原点且斜率为 k 的直线。线性方程在分析实验数据和推导化学动力学中十分重要。

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6. Geography – Population Modelling with Simultaneous Equations | 地理——人口模型与联立方程

Town A has a population of 50,000 in 2025 and grows by 2,000 people per year. Town B has 30,000 residents and grows by 3,500 per year. Assuming linear growth, in which year will the two towns have equal populations?

2025年 A 镇人口为 50000，每年增长 2000 人；B 镇人口为 30000，每年增长 3500 人。假设线性增长，哪一年两镇人口数相同？

Let t be the number of years after 2025. The populations are given by linear equations.

设 t 为 2025 年后的年数，人口数由线性方程给出。

P_A = 50000 + 2000t

P_B = 30000 + 3500t

Set them equal to find t.

使两者相等以求解 t。

50000 + 2000t = 30000 + 3500t

Subtract 30000 and 2000t from both sides.

两边同时减去 30000 和 2000t。

20000 = 1500t ⇒ t = 20000 / 1500 = 40/3 ≈ 13.33

The populations will be equal after about 13.33 years, i.e. during 2038. The ability to model and solve simultaneous linear equations helps geographers forecast demographic changes and plan resources.

大约 13.33 年后，即 2038 年，两镇人口相等。建立并求解联立线性方程的能力能帮助地理学家预测人口变化并规划资源。

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7. Finance – Compound Interest and Geometric Series | 金融——复利与等比数列

An investor deposits £1,000 at the end of each year into an account that pays 5% interest compounded annually. Calculate the total value of the investment after 10 years, just after the tenth deposit is made.

某投资者每年末存入 £1000，账户年利率为 5%，每年复利。计算第 10 年末（刚存完第 10 笔款后）的总投资价值。

, with the final deposit earning no interest: 1000 × 1.05⁰. This forms a geometric series.

第一笔存款增长 9 年：1000 × 1.05⁹；第二笔增长 8 年：1000 × 1.05⁸，依此类推，最后一笔不产生利息：1000 × 1.05⁰。这构成了一个等比数列。

S = 1000(1.05⁰ + 1.05¹ + 1.05² + ... + 1.05⁹)

Use the formula for the sum of a geometric series: Sₙ = a(rⁿ – 1)/(r – 1). Here a = 1000, r = 1.05, n = 10.

使用等比数列求和公式：Sₙ = a(rⁿ – 1)/(r – 1)，其中 a = 1000，r = 1.05，n = 10。

S = 1000 × (1.05¹⁰ – 1) / (1.05 – 1) = 1000 × (1.05¹⁰ – 1) / 0.05

Calculate 1.05¹⁰ ≈ 1.62889, so the numerator becomes 0.62889.

计算得 1.05¹⁰ ≈ 1.62889，因此分子为 0.62889。

S ≈ 1000 × 0.62889 / 0.05 = 1000 × 12.5778 = £12,577.80

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