📚 Case Study in Action: Practical Problem-Solving with Year 11 Eduqas Further Mathematics | 案例分析实战演练：运用Year 11 Eduqas进阶数学解决问题
This article takes a hands-on approach to Year 11 Eduqas Further Mathematics by presenting real-world case studies where core techniques – from differentiation to matrices – are used to model and solve genuine problems. Each section introduces a scenario, builds the mathematical model, and walks through the solution, reinforcing the relevance of advanced mathematics in engineering, economics, biology and beyond.
本文通过一系列真实世界案例，展示Year 11 Eduqas进阶数学的核心技巧——从微分到矩阵——如何用于建模和解决实际问题。每个部分都将引入一个情境，建立数学模型并逐步求解，以此强化进阶数学在工程、经济学、生物学等领域的实际意义。
1. Modelling Population Growth with Differential Equations | 用微分方程建模人口增长
fter one hour the count reaches 800. We can model this with the differential equation dP/dt = kP, where P is the population and k is a constant. Solving by separation of variables yields P = P₀ eᵏᵗ. Substituting t=0 gives P₀=500. Using P(1)=800: 800 = 500 eᵏ, so eᵏ = 1.6 and k = ln 1.6 ≈ 0.470. The model predicts P(3) = 500 × e3×0.470 urs. This exponential model helps the biologist schedule interventions when the population reaches a critical threshold.
一位生物学家观察到细菌菌落的生长速率与当前数量成正比。初始时有500个细菌，一小时后数量达到800。我们可以用微分方程 dP/dt = kP 建模，其中 P 为种群数量，k 为常数。分离变量求解得 P = P₀ eᵏᵗ。代入 t=0 得 P₀=500。利用 P(1)=800：800 = 500 eᵏ，故 eᵏ = 1.6 且 k = ln 1.6 ≈ 0.470。模型预测 t=3 时 P(3) = 500 × e3×0.470 ≈ 500 × 4.096 = 2048 个细菌。该指数模型可帮助生物学家在种群达到临界阈值时安排干预措施。
2. Optimising Revenue with Quadratic Functions | 用二次函数优化收益
A coffee shop owner finds that when a cup of coffee is priced at £ n as R(x) = (250 + x)(300 - 0.5x) in pence, where x is the increase from 250p. Expanding: R(x) = 75000 + 300x - 125x - 0.5x² = 75000 + 175x - 0.5x². This is a concave quadratic; the maximum occurs at x = -b/(2a) = -175/(2×-0.5) = 175 pence, i.e. an increase of £1.75 giving a price of £ um daily revenue is R(175) = (250+175)(300 - 0.5×175) = 425 × (300 - 87.5) = 425 × 212.5 = £903.125. Thus the optimal price is £4.25.
一家咖啡店老板发现，每杯咖啡定价2.50英镑时每天售出300杯。价格每上涨10便士，销量减少5杯。设价格上涨 x 便士（相对250便士）后的收益 R（便士）为 R(x) = (250 + x)(300 - 0.5x)。展开得 R(x) = 75000 + 300x - 125x - 0.5x² = 75000 + 175x - 0.5x²。这是一个上凸二次函数，最大值出现在 x = -b/(2a) = -175/(2×-0.5) = 175 便士，即上涨1.75英镑，售价为4.25英镑。最大日收益为 R(175) = (250+175)(300 - 0.5×175) = 425 × 212.5 = 90312.5便士，即903.13英镑。因此最优价格为4.25英镑。
3. Analysing Structural Forces with Vectors | 用向量分析结构受力
tor force. The vector AH = (1.5-0, 2-0, 5-0) = (1.5, 2, 5) and BH = (1.5-3, 2-4, 5-0) = (-1.5, -2, 5). The magnitudes are |AH| = √1.5²+2²+5² = √2.25+4+25 = √31.25 ≈ 5.59 m, and |BH| = √((-1.5)²+(-2)²+5²) = √2.25+4+25 = √31.25 as well. If the load is 1000 N vertically downwards, equilibrium requires T₁·(AH/|AH|) + T₂·(BH/|BH|) + (0,0,-1000) = 0. By symmetry, T₁ = T₂ = T. The z-components give 2T(5/√31.25) = 1000, so T = 500√31/5 ≈ 559 N. Vectors provide a clear method for calculating tensions in spatial structures.
一台起重机用两根缆绳吊起重物，缆绳固定在点 A(0,0,0) 和 B(3,4,0)，挂钩位于 H(1.5,2,5)。每根缆绳的张力为一向量力。向量 AH = (1.5, 2, 5)，BH = (-1.5, -2, 5)。模长分别为 |AH| = √1.5²+2²+5² = √31.25 ≈ 5.59 m，|BH| 同理。如果负载为竖直向下的 1000 N，平衡条件为 T₁·(AH/|AH|) + T₂·(BH/|BH|) + (0,0,-1000) = 0。由对称性 T₁ = T₂ = T。z 方向合力：2T(5/√31.25) = 1000，解得 T = 500√31.25/5 ≈ 559 N。向量方法为空间结构受力计算提供了清晰的途径。
4. Predicting Outcomes with Probability Matrices | 用概率矩阵预测结果
e the rows represent current brand choice and columns next month’s choice. If initially 60% buy A and 40% uy B, the state vector after one month is v₁ = M × v₀ = [[0.7,0.3],[0.4,0.6]] × [0.6, 0.4]ᵀ s = [s₁, s₂] satisfying (M - I)s = 0 and s₁+s₂=1. Solving: -0.3s₁+0.3s₂=0 gives s₁=s₂=0.5. So long-term each brand holds 50% of the market.
一项市场调查表明，顾客每月在两个品牌 A 和 B 之间转换，转移矩阵为 M = [[0.7, 0.3], [0.4, 0.6]]，行表示当前选择的品牌，列表示下个月的选择。若最初 60% 购买 A、40% 购买 B，一个月后的状态向量为 v₁ = M × v₀ = [0.54, 0.48]。经过多个月后系统趋于稳态向量 s = [s₁, s₂]，满足 (M - I)s = 0 且 s₁+s₂=1。求解 -0.3s₁+0.3s₂=0 得 s₁=s₂=0.5。长远来看每个品牌各占 50% 的市场份额。
5. Curve Fitting and Least Squares Regression | 曲线拟合与最小二乘回归
s proposed. Using least squares, we minimise Σ(yᵢ - a - bxᵢ)². Normal equations: Σy = na + bΣx, Σxy = aΣx + bΣx². Here n=4, Σx=10, Σy=20.1, Σx²=30, Σxy=1·2.1+2·4.0+3·5.9+4·8.1 = 2.1+8.0+17.7+32.4=60.2. Substituting: 20.1 = 4a + 10b; 60.2 = 10a + 30b. Solving gives b = (4×60.2 - 10×20.1)/(4×30 - 10²) = (240.8 - 201)/(120 - 100) = 39.8/20 = 1.99, and a = (20.1 - 10×1.99)/4 = (20.1 - 19.9)/4 = 0.05. The best-fit line is y = 0.05 + 1.99x, which implies a spring constant of about 1.99 mm/kg.
工程师获得弹簧在载荷 x (kg) 下的伸长量 y (mm) 数据：(1, 2.1)、(2, 4.0)、(3, 5.9)、(4, 8.1)。拟用直线模型 y = a + bx。最小二乘法要求最小化 Σ(yᵢ - a - bxᵢ)²。正规方程为 Σy = na + bΣx，Σxy = aΣx + bΣx²。此处 n=4，Σx=10，Σy=20.1，Σx²=30，Σxy=60.2。代入：20.1 = 4a + 10b；60.2 = 10a + 30b。解得 b = 1.99，a = 0.05。最佳拟合直线为 y = 0.05 + 1.99x，表明弹簧刚度约为 1.99 mm/kg。
6. Solving Systems of Equations in Electrical Circuits | 解决电路中的方程组问题
= 10; 4I₂ - 3I₃ = 2. Represent this as an augmented matrix and use row reduction. Matrix: ⌈1 -1 -1 | 0⌉; ⌈2 4 0 | 10⌉; ⌈0 4 -3 | 2⌉. R₂ → R₂ - 2R₁ gives ⌈1 -1 -1 | 0⌉; ⌈0 6 2 | 10⌉; ⌈0 4 -3 | 2⌉. Then R₃ → 3R₃ - 2R₂: 3×(0,4,-3|2) = (0,12,-9|6); 2×(0,6,2|10) = (0,12,4|20); subtract: (0,0,-13| = 17/13 A, I₁ = I₂ + I₃ = 31/13 A. Matrix methods systematically solve circuit equations.
在直流电路中，基尔霍夫定律给出三个关于电流 I₁、I₂、I₃ 的线性方程：I₁ - I₂ - I₃ = 0；2I₁ + 4I₂ = 10；4I₂ - 3I₃ = 2。将其表示为增广矩阵并进行行化简。矩阵：⌈1 -1 -1 | 0⌉；⌈2 4 0 | 10⌉；⌈0 4 -3 | 2⌉。R₂ → R₂ - 2R₁ 得 ⌈1 -1 -1 | 0⌉；⌈0 6 2 | 10⌉；⌈0 4 -3 | 2⌉。然后 R₃ → 3R₃ - 2R₂ 得 (0,0,-13|-14)，故 I₃ = 14/13 A。回代得 I₂ = 17/13 A，I₁ = 31/13 A。矩阵方法可系统解决电路方程。
7. Maximising Area with Integration | 用积分最大化面积
e enclosed area. Let length parallel to river be L, width be W. Constraint: L + 2W = 200, so L = 200 - 2W. Area A = L×W = (200-2W)W = 200W - 2W². This quadratic is maximised at W = 50 m (by -b/2a), giving L = 100 m, Area = 5000 m². Alternatively, using calculus: dA/dW = 200 - 4W = 0 ⇒ urve is ∫₀⁸ (8x - x²) dx = [4x² - x³/3]₀⁸ = 256 - 512/3 = 256/3 ≈ 85.33 m². Integration allows precise area calculations for irregular boundaries.
一位农民用200米围栏围出一个一边靠河（河流为直线，该边无需围栏）的矩形田地，求最大化面积的尺寸。设平行于河岸的长度为 L，宽度为 W。约束条件：L + 2W = 200，则 L = 200 - 2W。面积 A = L×W = (200-2W)W = 200W - 2W²。此二次函数在 W = 50 m 处取得最大值（利用 -b/2a），得 L = 100 m，面积为5000 m²。也可用微积分：dA/dW = 200 - 4W = 0 ⇒ W=50。进一步，假设河岸为曲线 y = 8x - x²（x=0到x=8），并用直线连接端点形成围栏。曲线下的面积 = ∫₀⁸ (8x - x²) dx = [4x² - x³/3]₀⁸ = 256 - 512/3 = 256/3 ≈ 85.33 m²。积分可以精确计算不规则边界的面积。
8. Cryptography with Matrices | 矩阵密码学
ng matrix E = [[3,2],[1,1]]. To decrypt, the inverse E⁻¹ = [[1,-2],[-1,3]] is used. Encode "HI": H=8, I=9 → vector [8,9]ᵀ ×9, 1×8+1×9] = [24+18, 8+9] = [42, 17]. The sent cipher is 42, 17. Decoding: E⁻¹ × [42,17]ᵀ he original. Matrix ciphers demonstrate how linear algebra underpins secure communication.
一种简单的加密方案将消息编码为数字序列（A=1, B=2, ...），分组为列向量并乘以 2×2 的编码矩阵 E = [[3,2],[1,1]]。解密时使用逆矩阵 E⁻¹ = [[1,-2],[-1,3]]。加密 "HI"：H=8, I=9 → 向量 [8,9]ᵀ。乘以 E：[3×8+2×9, 1×8+1×9] = [42, 17]。发送的密文为 42, 17。解码：E⁻¹ × [42,17]ᵀ = [1×42 + (-2)×17, -1×42 + 3×17] = [8, 9]，恢复出原始向量。矩阵密码展示了线性代数如何支撑安全通信。
9. Kinematics and Parametric Equations | 运动学与参数方程
n speed is minimised. Velocity components: vₓ = dx/dt = 6t - 2, vᵧ = dy/dt = 3t² - 8t. Speed s = √((6t-2)² + (3t²-8t)²). To minimise s, minimise f(t) = (6t-2)² + (3t²-8t)². Expand: f = 36t² - 24t + 4 + 9t⁴ - 48t³ + 64t² = 9t⁴ - 48t³ + 100t² - 24t + 4. Differentiate: f' d ≈ √((6×0.13-2)² + (3×0.0169 - 8×0.13)²) ≈ 1.95 m/s. Parametric equations elegantly describe curved motion in vector form.
一质点运动的位置由参数方程 x = 3t² - 2t、y = t³ - 4t² + 2 (t ≥ 0) 给出。求其速度向量及速率最小时刻。速度分量：vₓ = dx/dt = 6t - 2，vᵧ = dy/dt = 3t² - 8t。速率 s = √((6t-2)² + (3t²-8t)²)。为最小化 s，最小化 f(t) = (6t-2)² + (3t²-8t)²。展开：f = 36t² - 24t + 4 + 9t⁴ - 48t³ + 64t² = 9t⁴ - 48t³ + 100t² - 24t + 4。求导：f'(t) = 36t³ - 144t² + 200t - 24。令为零，数值求解得 t≈0.13 s 时速率很低。验证端点，实际最小值在 t=0.13 s，速率约为 1.95 m/s。参数方程以向量形式优雅地描述了曲线运动。
10. Financial Modelling with Geometric Series | 用几何级数进行金融建模
A savings account pays 4% annual interest compounded monthly. If £200 is deposited at the start of each month, how much is accumulated after 5 years? , second for 59 months to 200r⁵⁹, ..., last for 1 month to 200r. Total A = 200(r + r² + ... + r⁶⁰). This equals 200r(1 - r⁶⁰)/(1 - r). With r = 301/300 ≈ 1.003333, r⁶⁰ ≈ (1.003333)⁶⁰ ≈ 1.2214. Then A ≈ 200 × 1.003333 × (1 - 1.2214)/(1 - 1.003333) = 200 × 1.003333 × (-0.2214)/(-0.003333) ≈ 200 × 1.003333 × 66.44 ≈ 13330. This demonstrates the power of regular savings and geometric series.
一个储蓄账户年利率4%按月复利。若每月初存入200英镑，5年后总共能积累多少？月增长因子 r = 1 + 0.04/12 = 1 + 1/300 = 301/300。60次存款构成等比数列：第一次存款增长60个月至 200r⁶⁰，第二次增长59个月至 200r⁵⁹……最后一次增长1个月至 200r。总额 A = 200(r + r² + ... + r⁶⁰) = 200r(1 - r⁶⁰)/(1 - r)。代入 r = 301/300 ≈ 1.003333，r⁶⁰ ≈ 1.2214，则 )/(1 - 1.003333) = 200 × 1.003333 × (-0.2214)/(-0.003333) ≈ 200 × 1.003333 × 66.44 ≈ 13330 英镑。这展示了定期储蓄和几何级数的强大之处。
Published by TutorHao | Further Mathematics Revision Series | aleveler.com
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