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

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

Year 10 AQA Further Mathematics tests your ability to apply advanced algebra, trigonometry, calculus, and reasoning to unfamiliar contexts. This article presents a collection of cross-curricular problem-solving drills that connect concepts from physics, economics, biology, geography, and computer graphics to the core topics of your syllabus. Each section provides a realistic problem, followed by a step-by-step solution, so you can practise transferring your mathematical skills across disciplines.

Year 10 AQA 进阶数学考验你在陌生情境中应用高级代数、三角学、微积分和推理的能力。本文提供了一系列跨学科问题解决训练，将物理、经济学、生物学、地理和计算机图形学中的概念与课程的核心主题联系起来。每个部分给出一个真实问题以及逐步解答，让你能够练习将数学技能迁移到不同学科中。

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1. Quadratic Models in Physics: Projectile Motion | 物理中的二次模型：抛体运动

ebra to find the maximum height reached and the time taken for the ball to hit the ground.

一个球从平台上垂直向上抛出。t 秒后它离地面的高度 h（米）由 h(t) = -5t² + 20t + 2 给出。用代数方法求出达到的最大高度以及球落地所需的时间。

20, so t = -20/(2 × -5) = 2 seconds. Substitute t = 2 into h(t): h(2) = -5(4) + 40 + 2 = 22 metres. Therefore the maximum height is 22 m.

步骤1：最大高度出现在抛物线顶点。对于二次函数 ax² + bx + c，顶点的 t 坐标是 t = -b/(2a)。这里 a = -5, b = 20，所以 t = -20/(2 × -5) = 2 秒。将 t = 2 代入 h(t)：h(2) = -5(4) + 40 + 2 = 22 米。因此最大高度为 22 米。

t² + 20t + 2 = 0. Multiply by -1: 5t² - 20t - 2 = 0. Apply the quadratic formula t = [20 ± √400 + 40 ive root gives t ≈ 40.976/10 ≈ 4.10 s. So the ball lands after approximately 4.10 seconds.

步骤2：球在 h(t) = 0 时落地。解 -5t² + 20t + 2 = 0。乘以 -1：5t² - 20t - 2 = 0。应用二次公式 t = [20 ± √400 + 40] / 10 = [20 ± √440] / 10 ≈ [20 ± 20.976] / 10。正根得到 t ≈ 40.976/10 ≈ 4.10 秒。因此球大约在 4.10 秒后落地。

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2. Trigonometric Applications in Geography: Height of a Cliff | 地理学中的三角应用：悬崖高度

A geographer measures the angle of elevation to the top of a cliff from two points on level ground. Point A is 120 metres from the base of the cliff, and the angle is 34°. Point B is 40 metres closer to the cliff directly along the same line, and the angle is 48°. Determine the height of the cliff using trigonometric equations.

一位地理学家从水平地面上的两个点测量到悬崖顶部的仰角。A 点距离悬崖底部 120 米，仰角为 34°。B 点在同一直线上更靠近悬崖 40 米，仰角为 48°。使用三角方程求出悬崖的高度。

qual the same h. Calculate values: 120 tan 34° ≈ 120 × 0.6745 = 80.94 m, and 80 tan 48° ≈ 80 × 1.1106 = 88.85 m. The slight discrepancy is due to rounding of angles, but the model suggests a height of about 85–89 m. In practice, measurements can be averaged or checked with a third point.

设 h 为悬崖高度。从 A 点，tan 34° = h / 120，因此 h = 120 tan 34°。从 B 点，tan 48° = h / 80，得到 h = 80 tan 48°。两个表达式必须等于相同的 h。计算数值：120 tan 34° ≈ 120 × 0.6745 = 80.94 米，80 tan 48° ≈ 80 × 1.1106 = 88.85 米。微小差异是由角度四舍五入造成的，但模型表明高度约为 85–89 米。实际操作中，可以通过求平均值或增加第三个测量点来验证。

This problem reinforces the use of tangent ratios in right‑angled triangles and the importance of accurate trigonometric values. You may also rearrange to solve for the unknown distance if the height is given.

这个问题强化了在直角三角形中使用正切比以及精确三角值的重要性。你也可以在已知高度的情况下通过整理方程来求解未知距离。

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3. Exponential Growth in Biology: Bacterial Population | 生物学中的指数增长：细菌种群

hat a bacterial colony starts with 500 cells and grows according to the model N(t) = 500 e^0.4t, where t is time in hours. Calculate: (a) the population after 5 hours, (b) the time needed for the population to reach 20 000 cells. Use logarithms for part (b).

一位生物学家记录一个细菌菌落开始时为 500 个细胞，并按模型 N(t) = 500 e^0.4t 增长，其中 t 为时间（小时）。计算：(a) 5 小时后的种群数量，(b) 种群达到 20 000 个细胞所需的时间。第 (b) 部分使用对数。

(a) Substitute t = 5: N(5) = 500 e^0.4×5 e² ≈ 7.389, N(5) ≈ 500 × 7.389 = 3694.5, so about 3695 cells. (b) Set N(t) = 20 000: 500 e^0.4t = 20 000. Divide by 500: e^0.4t = 40. Take natural logarithms: ln(e^0.4t efore t ≈ 3.6889 / 0.4 = 9.22 hours. The population reaches 20 000 after about 9.22 hours.

(a) 代入 t = 5：N(5) = 500 e^0.4×5 = 500 e²。由于 e² ≈ 7.389，N(5) ≈ 500 × 7.389 = 3694.5，因此大约 3695 个细胞。(b) 令 N(t) = 20 000：500 e^0.4t = 20 000。除以 500：e^0.4t = 40。取自然对数：ln(e^0.4t) = ln 40，得到 0.4t = ln 40。所以 t = (ln 40) / 0.4。ln 40 ≈ 3.6889，因此 t ≈ 3.6889 / 0.4 = 9.22 小时。种群大约在 9.22 小时后达到 20 000。

Working with exponential models and natural logarithms is a key skill in Further Maths. You can also find the doubling time by solving e^0.4t = 2, yielding t = ln 2 / 0.4 ≈ 1.73 hours.

使用指数模型和自然对数是进阶数学的一项关键技能。你也可以通过解 e^0.4t = 2 来求倍增时间，得到 t = ln 2 / 0.4 ≈ 1.73 小时。

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4. Differentiation in Economics: Marginal Cost and Profit | 经济学中的微分：边际成本与利润

cost, in pounds, for producing x units is C(x) = 200 + 15x + 0.3x². The price per unit is £ C(x). (b) Determine the production level x that maximises profit, where profit π(x) = R(x) – C(x).

一家公司生产 x 件产品的总成本（英镑）为 C(x) = 200 + 15x + 0.3x²。单价为 40 英镑，因此总收入为 R(x) = 40x。(a) 通过对 C(x) 求导找到边际成本函数。(b) 确定使利润最大化的产量 x，其中利润 π(x) = R(x) – C(x)。

(a) The marginal cost is the derivative C' derivative of 200 is 0, derivative of 15x is 15, derivative of 0.3x² is 0.6x. Therefore C' This tells us the cost of producing one additional unit when x units are already produced.

(a) 边际成本是导数 C'(x)。逐项求导：200 的导数为 0，15x 的导数为 15，0.3x² 的导数为 0.6x。因此 C'(x) = 15 + 0.6x。它告诉我们已经生产了 x 件产品时，再多生产一件所需的成本。

(b) Profit π(x) = 40x – (200 + 15x + 0.3x²) = 25x – 200 – 0.3x². To maximise, find π'(x) = 25 – 0.6x and set it to zero: 25 – o x = 25 / 0.6 ≈ 41.67. Since x must be an integer, check x = 41 and x = 42. π(41) = 25×41 – 200 – 0.3×1681 = 1025 – 200 – 504.3 = 320.7; π(42) = 1050 – 200 – 0.3×1764 = 850 – = 320.8. Profit is slightly higher at x = 42, so the optimal production level is 42 units.

(b) 利润 π(x) = 40x – (200 + 15x + 0.3x²) = 25x – 200 – 0.3x²。为求最大化，求 π'(x) = 25 – 0.6x 并令其等于零：25 – 0.6x = 0，因此 x = 25 / 0.6 ≈ 41.67。由于 x 必须是整数，检验 x = 41 和 x = 42。π(41) = 25×41 – 200 – 0.3×1681 = 1025 – 200 – 504.3 = 320.7；π(42) = 1050 – 200 – 0.3×1764 = 850 – 529.2 = 320.8。x = 42 时利润略高，因此最优产量为 42 件。

ere marginal revenue equals marginal cost; here MR = 40 and MC = 15 + 0.6x, giving identical result.

微分使企业能够就生产水平做出明智决策。你使用了最大利润出现在边际收入等于边际成本的条件：这里 MR = 40，MC = 15 + 0.6x，结果相同。

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5. Integration in Physics: Total Displacement from Variable Velocity | 物理中的积分：变速运动的总位移

A particle moves along a straight line with velocity v(t) = 4t³ – of the particle between t = 1 second and t = 4 seconds by evaluating a definite integral.

一个粒子沿直线运动，速度 v(t) = 4t³ – 6t² + 2t（单位 m/s）。通过计算定积分求出粒子在 t = 1 秒到 t = 4 秒之间走过的总位移。

Displacement s is the integral of velocity with respect to time: s = ∫₁⁴ (4t³ – 6t² + 2t) dt. Find the antiderivative: t⁴ – 2t³ + t². No constant is needed for a definite integral. Evaluate from 1 to 4: [4⁴ – 2×4³ + 4²] – [1⁴ – 2×1³ + 1²] = [256 – 128 + 16] – [1 – 2 + 1] = 144 – 0 = 144. The total displacement is 144 metres.

位移 s 是速度对时间的积分：s = ∫₁⁴ (4t³ – 6t² + 2t) dt。求原函数：t⁴ – 2t³ + t²。定积分不需要常数。从 1 到 4 计算：[4⁴ – 2×4³ + 4²] – [1⁴ – 2×1³ + 1²] = [256 – 128 + 16] – [1 – 2 + 1] = 144 – 0 = 144。总位移为 144 米。

If the velocity changed sign during the interval, we would need to find total distance travelled by integrating the absolute value of velocity. Here v(t) stays positive for t>1, so displacement equals total distance.

如果速度在区间内变号，我们需要通过对速度的绝对值积分来求总路程。这里当 t>1 时 v(t) 保持正号，因此位移等于总路程。

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6. Matrices in Computer Graphics: Rotating a Point | 计算机图形学中的矩阵：旋转一个点

2D shape. The rotation through an angle θ anticlockwise is given by the matrix R = [cos θ –sin θ; sin θ ly a rotation of 60° to the point P(2, 3) using exact matrix multiplication and surd form.

一位图形设计师使用旋转矩阵来旋转一个二维图形。逆时针旋转角度 θ 的矩阵为 R = [cos θ –sin θ; sin θ cos θ]。使用精确的矩阵乘法和根式形式，对点 P(2, 3) 进行 60° 旋转。

For θ = 60°, cos 60° = 1/2, sin 60° = √3/2. Write P as a column vector [2; 3]. Multiply: R × P = [1/2 –√3/2; √3/2 1/2] [2; 3] = [ (1/2)×2 + (–√3/2)×3; (√3/2)×2 + (1/2)×3 ] = [1 – (3√3)/2; √3 + 3/2]. So the rotated point has coordinates (1 – (3√3)/2, √3 + 3/2). You can leave it in exact form.

对于 θ = 60°，cos 60° = 1/2，sin 60° = √3/2。将 P 写成列向量 [2; 3]。相乘：R × P = [1/2 –√3/2; √3/2 1/2] [2; 3] = [ (1/2)×2 + (–√3/2)×3; (√3/2)×2 + (1/2)×3 ] = [1 – (3√3)/2; √3 + 3/2]。因此旋转后的点坐标为 (1 – (3√3)/2, √3 + 3/2)。你可以保持精确形式。

Matrix transformations are fundamental in computer graphics. Further Maths also covers reflections, enlargements, and combined transformations, often requiring the multiplication of several matrices.

矩阵变换是计算机图形学的基础。进阶数学还涉及反射、缩放以及组合变换，通常需要进行多个矩阵相乘。

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7. Vectors in Mechanics: Resultant Force | 力学中的向量：合力

Two forces act on an object: F₁ = 3i + 4j N and F₂ = – the magnitude of the resultant force and the angle it makes with the positive i direction.

两个力作用在一个物体上：F₁ = 3i + 4j N，F₂ = –i + 2j N，其中 i 和 j 是垂直的单位向量。求合力的大小以及它与正 i 方向的夹角。

Resultant force R = F₁ + F₂ = (3 – 1)i + (4 + 2)j = 2i + 6j N. Magnitude |R| = √2² + 6² = √4 + 36 = √40 = 2√10 = (j-component)/(i-component) = 6/2 = 3. So θ = tan⁻¹(3) ≈ 71.6°. The resultant force is 2√10 N at an angle of approximately 71.6° from the i direction.

合力 R = F₁ + F₂ = (3 – 1)i + (4 + 2)j = 2i + 6j N。大小 |R| = √2² + 6² = √4 + 36 = √40 = 2√10 ≈ 6.32 N。与正 i 轴的夹角 θ 满足 tan θ = (j 分量)/(i 分量) = 6/2 = 3。所以 θ = tan⁻¹(3) ≈ 71.6°。合力大小为 2√10 N，与 i 方向夹角约为 71.6°。

Vector addition and resolution are essential for solving equilibrium problems in mechanics. You might also be asked to find the force needed to keep the system in equilibrium, which would be –R.

向量加法和分解对于解决力学中的平衡问题至关重要。你还可能被要求求出使系统保持平衡所需的力，它将是 –R。

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8. Sequences in Finance: Compound Interest and Savings | 金融中的数列：复利与储蓄

A person invests £2000 in a savings account paying 5% te the amount after 10 years. (b) Use logarithms to determine after how many years the investment will exceed £5000.

某人将 2000 英镑存入一个年利率 5% 的储蓄账户，复利按年计算。n 年后的金额构成等比数列：Aₙ = 2000 × (1.05)ⁿ。(a) 写出 10 年后的金额。(b) 使用对数确定投资何时超过 5000 英镑。

₁₀ = 2000 × (1.05)¹⁰. Using binomial expansion or calculator, (1.05)¹⁰ ≈ 1.62889, so A₁₀ ≈ £ n > log 2.5 / log 1.05 ≈ 0.39794 / 0.021189 ≈ 18.78. So after 19 years the balance exceeds £5000 for the first time.

(a) A₁₀ = 2000 × (1.05)¹⁰。使用二项式展开或计算器，(1.05)¹⁰ ≈ 1.62889，因此 A₁₀ ≈ 3257.78 英镑。(b) 令 2000 × (1.05)ⁿ > 5000。除以 2000：(1.05)ⁿ > 2.5。取对数：n log(1.05) > log 2.5。因此 n > log 2.5 / log 1.05 ≈ 0.39794 / 0.021189 ≈ 18.78。所以 19 年后余额首次超过 5000 英镑。

Geometric sequences are widely used in finance to model compound interest, depreciation, and loan repayments. Further Maths extends these ideas to sums of geometric series.

等比数列在金融中被广泛用于模拟复利、折旧和贷款偿还。进阶数学将这些概念延伸到等比级数的求和。

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9. Probability in Biology: Genetic Inheritance | 生物学中的概率：遗传继承

Tt) are crossed. Using a Punnett square, calculate the probability that a randomly selected offspring will be tall. Then, if four offspring are produced, find the probability that exactly three are tall, modelling with binomial distribution.

在豌豆植株中，高茎等位基因 (T) 对矮茎等位基因 (t) 为显性。将两株杂合高茎植株 (Tt × Tt) 杂交。使用庞纳特方格，计算随机选取的后代为高茎的概率。然后，如果产生四个后代，假设二项分布，求恰好有三个是高茎的概率。

tT, the probability of tall is 3/4. Let X be the number of tall offspring among 4. Then X ~ ⁴C₃ (0.75)³ (0.25)¹ = 4 × 0.421875 × 0.25 = 4 × 0.10546875 = 0.421875. So there is a 42.2% chance of exactly three tall offspring.

庞纳特方格给出基因型 TT、Tt、tT、tt，每种概率均为 1/4。由于高茎可以是 TT、Tt 或 tT，高茎的概率为 3/4。设 X 为四个后代中高茎的数量。则 X ~ B(4,

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