📚 Year 10 SQA Advanced Mathematics: Case Study Practical Exercises | Year 10 SQA 进阶数学：案例分析实战演练
Mastering advanced mathematics goes beyond memorising formulas; it requires the ability to apply concepts to unfamiliar, real-world scenarios. This article presents a series of case study practical exercises aligned with the Year 10 SQA Advanced Mathematics curriculum, designed to strengthen problem-solving skills and deepen conceptual understanding through contextual challenges.
掌握进阶数学不仅仅是记忆公式，更要求学生具备将概念应用于陌生且真实世界场景的能力。本文提供了一系列贴合 Year 10 SQA 进阶数学课程的案例分析实战演练，旨在通过情境化挑战来强化解决问题的技巧并加深对概念的理解。
1. Optimisation in Packaging Design | 包装设计中的最优化问题
cm. The material for the base costs twice as much per cm² as the material for the sides. The goal is to minimise the total material cost. First, express h in terms of x using the volume constraint: V = x²h = 500, so h = 500 / x². Then, formulate the cost function C(x). If the side material costs p per cm², the base costs 2p per cm². The area of the base is x², and the total area of the four sides is 4xh = 4x(500 / x²) = 2000 / x. Hence, C(x) = 2p · x² + p · (2000 / x). To minimise cost, differentiate C with respect to x and set C'(x) = 0: C'(x) = 4p x – inimum. This case study elegantly links calculus with practical manufacturing constraints.
一家公司希望设计一个体积为 500 cm³ 的无盖长方形盒子。盒子的底面必须是边长为 x cm 的正方形，高度为 h cm。底面材料每平方厘米的成本是侧面材料的两倍。目标是使总材料成本最小化。首先，利用体积约束将 h 表示为 x 的函数：V = x²h = 500，因此 h = 500 / x²。然后，建立成本函数 C(x)。如果侧面材料的成本为每平方厘米 p，则底面成本为 2p。底面积为 x²，四个侧面的总面积为 4xh = 4x(500 / x²) = 2000 / x。因此，C(x) = 2p · x² + p · (2000 / x)。为使成本最小化，对 C 关于 x 求导并令 C'(x)=0：C'(x) = 4p x – 2000p / x² = 0，得到 4x³ = 2000，故 x = ³√500 ≈ 7.94 cm。二阶导数验证了极小值。本案例巧妙地将微积分与实际制造约束联系起来。
2. Modelling the Spread of a Rumor | 谣言传播的建模
rumor and the number who have not. This is a classic logistic growth model. Let y be the number of students who know the rumor after t hours. The differential equation is dy/dt = k y (1200 – sing partial fractions, we find the solution: y = 1200 / (1 + A e⁻¹²⁰⁰ᵏᵗ), where A = (1200 – rumor will reach half the school population, which occurs when y = 600, leading to t ≈ 5.4 hours. The exercise demonstrates how differential equations model social dynamics and highlights the power of calculus in predicting trends.
在一所拥有 1200 名学生的学校里，一个谣言在 t = 0 时从 5 名学生开始传播。传播速率与已听闻谣言的学生数量和未听闻的数量均成正比。这是一个经典的逻辑斯谛增长模型。设 y 为 t 小时后得知谣言的学生人数。微分方程为 dy/dt = k y (1200 – y)，其中 k 为正的常数。在 t = 0 时，y=5。1 小时后，有 50 名学生得知谣言。通过分离变量并使用部分分式，我们得到解：y = 1200 / (1 + A e⁻¹²⁰⁰ᵏᵗ)，其中 A = (1200 – 5)/5 = 239。利用条件 y(1)=50，可确定 k：50 = 1200 / (1 + 239 e⁻¹²⁰⁰ᵏ)，解得 k ≈ 0.000071。该模型可预测谣言何时会传遍学校一半人口，即 y=600 时，得到 t ≈ 5.4 小时。这一练习展示了微分方程如何对社会动态进行建模，并突显了微积分在预测趋势中的强大功能。
3. Kinematics of a Decelerating Vehicle | 车辆减速的运动学分析
A car travels along a straight road with velocity v(t) = 20 – istance traveled from t = 0 to the point where the car comes to rest. For the first phase, the displacement s₁ is the integral of v(t) from 0 to 5: s₁ = ∫₀⁵ (20 – 0.8t²) dt = [20t – (0.8/3)t³]₀⁵ = 100 – (0.8/3)×125 = 100 – 100/3 ≈ 66.67 m. At t = 5, velocity is v(5) = 20 – 0.8×25 = 0 m/s. Wait, v(5) = 0, so the car stops exactly at t = 5? That would mean no second phase. Let's adjust: suppose v(t) = 24 – 0.8t², then v(5) = 24 – 20 = 4 m/s. Then braking from 5 s with a = -4 m/s² gives stopping time ts = (4)/4 = 1 s further. Displacement during braking: s₂ = v(5)·1 + ½(–4)·1² = 4 – 2 = 2 m. Total distance = s₁ + s₂. For the integral: s₁ = ∫₀⁵ (24 – 0.8t²) dt = [24t – (0.8/3)t³]₀⁵ = 120 – (0.8/3)×125 = 120 – ite integrals in piecewise motion and the physical interpretation of velocity-time graphs.
一辆汽车在 0 ≤ t ≤ 5 秒内沿直线行驶，速度为 v(t) = 24 – 0.8t²，v 的单位为 m/s。随后司机从 t = 5 开始以 4 m/s² 的恒定减速度刹车，直到汽车停下。我们需要计算从 t=0 到汽车停止点的总行驶距离。对于第一阶段，位移 s₁ 是 v(t) 从 0 到 5 的积分：s₁ = ∫₀⁵ (24 – 0.8t²) dt = [24t – (0.8/3)t³]₀⁵ = 120 – (0.8/3)×125 = 120 – 100/3 ≈ 86.67 m。在 t=5 时，速度 v(5) = 24 – 0.8×25 = 4 m/s。随后以 a = -4 m/s² 刹车，额外停止时间 ts = 4/4 = 1 秒。刹车期间的位移：s₂ = v(5)·1 + ½(–4)·1² = 4 – 2 = 2 m。总距离 = 86.67 + 2 = 88.67 m。此案例巩固了定积分在分段运动中的应用，以及对速度-时间图的物理解读。
4. Revenue and Elasticity of Demand | 收益与需求弹性
demand q for a product is related to the unit price p (in pounds) by the equation q = 800 – is given by R = p × q. To maximise revenue, we express R as a function of p: R(p) = p(800 – 20p) = 800p – 20p². Taking the derivative: R'(p) = 800 – 40p. Setting R' = 0 gives p = 20. At this price, q = 800 – 400 = 400, and maximum revenue is £ elasticity of demand E is defined as E = (p/q) × (dq/dp). Here dq/dp = -20, so E = (p/(800–20p)) × (-20) = -20p / (800– 400 = -1, which corresponds to unit elasticity, confirming that revenue is maximised. When |E| > 1, demand is elastic and raising price reduces revenue; when |E| illustrates the deep connection between calculus, economics, and optimisation strategies.
一家公司发现产品需求 q 与单价 p（英镑）的关系为 q = 800 – 20p。收益 R 由 R = p × q 给出。为使收益最大化，我们将 R 表示为 p 的函数：R(p) = p(800 – 20p) = 800p – 20p²。求导得：R'(p) = 800 – 40p。令 R'=0，得到 p=20。此时，q=400，最大收益为 £8000。需求弹性 E 的定义为 E = (p/q) × (dq/dp)。此处 dq/dp = -20，因此 E = (p/(800–20p)) × (-20) = -20p / (800–20p)。在 p=20 时，E = -400/400 = -1，对应单位弹性，这验证了收益达到最大化。当 |E|>1 时，需求富有弹性，提价会降低收益；当 |E|<1 时，需求缺乏弹性，提价会增加收益。本案例阐明了微积分、经济学与最优化策略之间的深层联系。
5. Trigonometric Height Measurement | 三角学高度测量
tres directly toward the mountain to point B, the angle of elevation becomes 38°. Let the mountain height be h, and let the distance from point B to the point directly below the summit be x. From triangle geometry: tan(22°) = h / (x + 400) and tan(38°) = h / x. Solving these two equations simultaneously: h = x tan(38°) and h = (x + 400) tan(22°). Equating gives x tan(38°) = (x + 400) tan(22°). Solving for x: x (tan(38°) – 0 tan(22°). Using approximate values tan(38°) ≈ 0.7813, tan(22°) ≈ 0.4040, we get x(0.7813 – ing skills and highlights the importance of bearings and angles of elevation in practical navigation and surveying.
一位测量员需要确定一座山的高度。从水平地面上的一点 A，测得山顶的仰角为 22°。朝着山体直线步行 400 米到达点 B 后，仰角变为 38°。设山高为 h，点 B 到山顶正下方点的距离为 x。根据三角形几何关系：tan(22°) = h / (x + 400) 且 tan(38°) = h / x。联立求解这两个方程：h = x tan(38°) 和 h = (x + 400) tan(22°)。令两者相等得 x tan(38°) = (x + 400) tan(22°)。求解 x：x (tan(38°) – tan(22°)) = 400 tan(22°)。利用近似值 tan(38°) ≈ 0.7813，tan(22°) ≈ 0.4040，可得 x(0.7813 – 0.4040) = 400×0.4040 → x(0.3773) = 161.6 → x ≈ 428.3 m。于是 h = 428.3 × 0.7813 ≈ 334.7 m。这一经典问题强化了三角方程求解技能，并突显了方位角与仰角在实际导航与测量中的重要性。
6. Exponential Decay in Medicine | 医药中的指数衰减
6 hours. The amount A(t) in mg after t hours satisfies A(t) = A₀ e⁻ᵏᵗ, where A₀ = 8 and k is the decay constant. Using the half-life, 4 = 8 e⁻ᵏ⁶, so e⁻ᵏ = (1/2)1/6 rs. The safe zone for another dose is when the level falls below 2 mg, so (ln 4)/k = (1.386)/0.1155 ≈ 12 hours. This case connects exponential functions, natural logarithms, and differentiable models to real-world pharmacokinetics, emphasising accurate algebraic manipulation.
一位患者注射了 8 mg 的药物。药物以与现存药量成正比的速度从体内消除，半衰期为 6 小时。t 小时后药物量 A(t) mg 满足 A(t) = A₀ e⁻ᵏᵗ，其中 A₀=8，k 是衰变常数。利用半衰期，4 = 8 e⁻ᵏ⁶，故 e⁻ᵏ = (1/2)1/6。因此 k = (ln 2)/6 ≈ 0.1155 h⁻¹。要计算药物水平降至 1 mg 所需的时间，解 1 = 8 e⁻⁰·¹¹⁵⁵ ᵗ，得 t = (ln 8)/k ≈ 2.079/0.1155 ≈ 18 小时。再次用药的安全区是当药量降至 2 mg 以下时，所以 (ln 4)/k = 1.386/0.1155 ≈ 12 小时。本案例将指数函数、自然对数与可微模型同现实中的药代动力学联系起来，着重强调了准确的代数处理能力。
7. Vectors in River Crossing | 向量在河流渡越中的应用
ht across the river, perpendicular to the banks. To achieve this, the boat must point upstream at an angle θ to counteract the current. The velocity vector of the boat relative to the ground is the sum of the boat's velocity relative to water (magnitude 5, direction θ upstream) and the water' ). For the resultant to have zero downstream component, the upstream component of the boat' k: 5 cos θ = 5 × (4/5) = 4 m/s. Thus the crossing time is 200 / 4 = 50 seconds. If the boat were to point directly across, it would be swept downstream, landing 200 m × (3/5) downstream. Vector decomposition elegantly solves navigation problems and appears frequently in SQA exam questions on forces and motion.
一艘船在静水中的速度为 5 m/s。它的目标是渡过一条 200 m 宽的河流，但河水的流速为 3 m/s，方向平行于河岸。船必须垂直于河岸径直渡河。为此，船头必须朝向上游某个角度 θ，以抵消水流的影响。船相对于地面的速度矢量等于船相对于水的速度（大小 5，方向与上游成 θ 角）与水流速度（3 m/s 向下游）之和。为使合速度的下游分量为零，船速的上游分量必须抵消水流：5 sin θ = 3，因此 sin θ = 3/5，得 θ ≈ 36.87°。最终的渡河速度是垂直于河岸的分量：5 cos θ = 5 × (4/5) = 4 m/s。所以渡河时间为 200 / 4 = 50 秒。如果船头直指对岸，它会被冲向下游，在下游方向偏离 200 m × (3/5) 处靠岸。向量分解巧妙地解决了航行问题，并频繁出现在 SQA 关于力与运动的考题中。
8. Probability and Expected Value in Game Design | 游戏设计中的概率与期望值
A board game uses a spinner with three sectors: red (40%), green (35%), and blue (25%). A player pays £1 to spin. Landing on red wins £2, green wins £ game for the player and determine whether the game is fair. Let X be the net profit: red: £+1, green: £0, blue: £– (X=0) = 0.35, P(X=-1) = 0.25. The expected value E(X) = 1×0.40 + 0×0.35 + (-1)×0.25 = 0.40 – 0.25 = £ e adjusted, or the prize for green could be reduced. This exercise builds fluency in discrete random variables and expectation, essential for statistical reasoning in SQA Advanced Mathematics.
某棋盘游戏使用一个分为三个扇区的转盘：红色（40%）、绿色（35%）和蓝色（25%）。玩家支付 £1 转动一次。停在红色赢得 £2，绿色赢得 £1，蓝色则一无所获。我们希望求玩家每局游戏的期望利润，并判断游戏是否公平。设 X 为净收益：红色：£+1，绿色：£0，蓝色：£–1。概率分布：P(X=1)=0.40，P(X=0)=0.35，P(X=-1)=0.25。期望值 E(X) = 1×0.40 + 0×0.35 + (-1)×0.25 = 0.40 – 0.25 = £0.15 每局。由于期望利润为正，长期来看游戏对玩家有利。要使游戏公平（E(X)=0），可以调整门票费或降低绿色奖赏。本练习提高了离散随机变量与期望值方面的熟练度，这对 SQA 进阶数学中的统计推理至关重要。
9. Sequences and Compound Interest | 数列与复利
An investor deposits £2000 into an account that pays 4.2% 2 = 0.0035. The balance after n months follows a geometric sequence: Bₙ = 2000 × (1 + r)ⁿ. After 10 years (120 months), B₁₂₀ = 2000 × (1.0035)120. Calculating (1.0035)120: let's use the approximation e120 ln 1.0035. ln 1.0035 ≈ 0.003493, product = 0.4192, so e0.4192 ≈ 1.5209. Thus balance ≈ 2000 × 1.5209 = £3041.80. Alternatively, the exact formula for compound interest is A = P (1 + r/n)nt. This directly links to the sum of a geometric series when considering regular deposits. Understanding sequences helps demystify long-term financial growth and illustrates the power of exponential models in personal finance.
一位投资者将 £2000 存入一个年利率为 4.2%、每月复利的账户。我们希望用等比数列求出 10 年后的余额。月利率为 r = 0.042 / 12 = 0.0035。n 个月后的余额遵循等比数列：Bₙ = 2000 × (1 + r)ⁿ。10 年后（120 个月），B₁₂₀ = 2000 × (1.0035)120。计算 (1.0035)120：利用近似式 e120 ln 1.0035，ln 1.0035 ≈ 0.003493，乘积为 0.4192，因此 e0.4192 ≈ 1.5209。所以余额 ≈ 2000 × 1.5209 = £3041.80。或者，复利的精确公式为 A = P (1 + r/n)nt。当考虑定期存款时，这直接与等比级数的求和联系起来。理解数列有助于揭开长期财务增长的奥秘，并展现指数模型在个人理财中的强大作用。
10. Integration by Substitution in Physics | 物理中的换元积分
) with respect to t: v(t) = ∫ (6t / (1 + t²)²) dt + C. Using the substitution u = 1 + t², then du = 2t dt, so 6t dt = 3 du. The integral becomes ∫ (3 / u²) du = –3 / u = –3 / (1 + t²). Thus v(t) = –3 / (1 + t²) + C. Applying v(0) = 2: 2 = –3/(1+0) + C → 2 = –3 + C → C = 5. Hence v(t) = 5 – when dealing with non-constant acceleration in mechanics, a key topic for SQA Advanced Mathematics.
一个粒子的加速度为 a(t) = 6t / (1 + t²)² m/s²。若初始速度 v(0) = 2 m/s，求速度函数 v(t)。我们对 a(t) 关于 t 积分：v(t) = ∫ (6t / (1 + t²)²) dt + C。使用换元法，令 u = 1 + t²，则 du = 2t dt，所以 6t dt = 3 du。积分变为 ∫ (3 / u²) du = –3 / u = –3 / (1 + t²)。因此 v(t) = –3 / (1 + t²) + C。应用 v(0)=2：2 = –3/(1+0) + C → 2 = –3 + C → C = 5。于是 v(t) = 5 – 3/(1 + t²)。当 t 很大时，v(t) 趋近于 5 m/s。这一案例展示了在处理力学中非恒定加速度时积分技巧（尤其是换元法）的必要性，这是 SQA 进阶数学的一个关键主题。
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