📚 Case Study Practical Exercises for OCR Year 11 Further Maths | Year 11 OCR 进阶数学：案例分析实战演练
In OCR Level 2 Further Mathematics, applying abstract techniques to real-world scenarios is essential for mastering problem-solving skills. This article presents a series of case studies that connect topics such as binomial expansions, introductory calculus, matrices, vectors, and trigonometric functions to practical contexts. Each case study is followed by step-by-step analysis and bilingual commentary, allowing you to see how theory meets application.
在 OCR Level 2 进阶数学中，将抽象技巧应用到实际场景是掌握问题解决能力的关键。本文通过一系列案例分析，将二项式展开、微积分入门、矩阵、向量和三角函数等主题与具体情境联系起来。每个案例都配有逐步分析和中英双语解说，帮助你理解理论与应用的结合。
1. Optimising a Box Design Using Differentiation | 利用微分优化盒子设计
A packaging company wishes to produce an open-top box with a square base of side length x cm and height h cm. The volume must be 500 cm³. The cost of materials is proportional to the surface area. We need to find the dimensions that minimise the surface area, thereby minimising cost.
一家包装公司希望生产一个无盖方底盒子，底面边长为 x cm，高度为 h cm。容积必须为 500 cm³。材料成本与表面积成正比。我们需要找出使表面积最小的尺寸，从而将成本降到最低。
tiate: dS/dx = 2x − 2000/x². Set dS/dx = 0 → 2x = 2000/x² → x³ = 1000 → x = 10. Then h = 500/100 = 5. The second derivative d²S/dx² = 2 + 4000/x³ > 0 for all positive x, confirming a minimum.
容积约束条件给出 h = 500 / x²。无盖盒子的表面积 S = x² + 4xh。代入 h 得到 S(x) = x² + 4x × (500/x²) = x² + 2000/x。求导：dS/dx = 2x − 2000/x²。令 dS/dx = 0 → 2x = 2000/x² → x³ = 1000 → x = 10。于是 h = 500/100 = 5。二阶导数 d²S/dx² = 2 + 4000/x³ 对所有正数 x 均大于 0，确认为极小值。
ght of 5 cm. This case illustrates how calculus can solve practical optimisation problems.
因此，最优尺寸为底面 10 cm × 10 cm，高 5 cm。本案例展示了微积分如何解决实际优化问题。
2. Modelling Rocket Height with Kinematics and Functions | 用运动学与函数模拟火箭高度
seconds is given by h(t) = 40t − 5t². This is a quadratic function that models the rocket' n it reaches the ground, and the instantaneous velocity at any time using differentiation.
一枚模型火箭垂直向上发射。t 秒后的高度（米）由 h(t) = 40t − 5t² 给出。这是一个二次函数，模拟火箭的飞行直至落地。我们可以求出最大高度、回到地面的时间以及任意时刻的瞬时速度，利用微分计算。
The velocity v(t) = h' ² = 0 → 5t(8 − t) = 0, so t = 0 (launch) or t = 8 s. Therefore, the total flight time is 8 seconds.
速度 v(t) = h'(t) = 40 − 10t。在最高点，v(t) = 0 → 40 − 10t = 0 → t = 4 秒。最大高度 h(4) = 40×4 − 5×4² = 160 − 80 = 80 米。火箭落地时 h(t) = 0 → 40t − 5t² = 0 → 5t(8 − t) = 0，所以 t = 0（发射时刻）或 t = 8 秒。因此，总飞行时间为 8 秒。
ndicates downward motion. This case study reinforces interpreting quadratic models and their derivatives.
求落地时的速度：v(8) = 40 − 10×8 = −40 m/s，负号表示向下运动。本案例分析强化了对二次模型及其导数的理解。
3. Matrix Transformations in Logo Design | 标志设计中的矩阵变换
A graphic designer uses matrix transformations to manipulate a simple triangular logo. The initial vertices are A(1,0), B(0,2), and C(−1,0). The designer applies a rotation of 90° anticlockwise about the origin, followed by an enlargement with scale factor 2 centred at the origin. We can represent these transformations as matrices and find the new coordinates.
一位平面设计师运用矩阵变换处理一个简单的三角形标志。初始顶点为 A(1,0)、B(0,2) 和 C(−1,0)。设计师先绕原点逆时针旋转 90°，然后以原点为中心进行缩放因子为 2 的放大。我们可以将这些变换用矩阵表示，并求出新顶点的坐标。
M = E × R = [[2,0],[0,2]] × [[0,−1],[1,0]] = [[0,−2],[2,0]]. Apply M to each vertex (as column vectors): A' = M × [1;0] = [0;2]; B' = M × [0;2] = [−4;0]; C' = M × [−1;0] = [0;−2]. So the new triangle has vertices (0,2), (−4,0), and (0,−2).
90° 旋转矩阵为 R = [[0,−1],[1,0]]。缩放因子为 2 的放大矩阵为 E = [[2,0],[0,2]]。组合变换矩阵为 M = E × R = [[2,0],[0,2]] × [[0,−1],[1,0]] = [[0,−2],[2,0]]。将 M 作用于每个顶点（写成列向量）：A' = M × [1;0] = [0;2]；B' = M × [0;2] = [−4;0]；C' = M × [−1;0] = [0;−2]。因此，新三角形的顶点为 (0,2)、(−4,0) 和 (0,−2)。
This demonstrates how matrices provide a systematic method for transforming shapes in computer graphics.
这展示了矩阵如何为计算机图形中的形状变换提供一种系统化方法。
4. Binomial Expansion in Probability of Defective Items | 二项式展开在次品概率中的应用
A factory produces components where the probability of a defective item is 0.02. A quality control inspector randomly selects 10 items. The probability that exactly k items are defective follows a binomial distribution. The expansion of (0.98 + 0.02)¹⁰ can be used to find the probabilities of different numbers of defectives.
某工厂生产部件，次品率为 0.02。一名质检员随机抽取 10 件产品。恰好有 k 件次品的概率服从二项分布。(0.98 + 0.02)¹⁰ 的展开式可用于计算不同次品数的概率。
The binomial expansion gives: (0.98 + 0.02)¹⁰ = Σ from k=0 to 10 of [¹⁰Cₖ × (0.98)10−k × (0.02)k × (0.98)⁸ × (0.02)² ≈ 0.0153, and so on. The sum of all these probabilities is exactly 1.
二项式展开给出：(0.98 + 0.02)¹⁰ = Σ 从 k=0 到 10 [¹⁰Cₖ × (0.98)10−k × (0.02)k]。每一项对应 P(X = k)。例如，P(X = 0) = (0.98)¹⁰ ≈ 0.817，P(X = 1) = 10 × (0.98)⁹ × 0.02 ≈ 0.1667，P(X = 2) = 45 × (0.98)⁸ × (0.02)² ≈ 0.0153，以此类推。所有这些概率之和恰好为 1。
Using the expansion avoids calculating each term from scratch and links algebraic expansions to real-world probability distributions.
利用展开式可以避免每次从零开始计算，并将代数展开与实际概率分布联系起来。
5. Vector Navigation: Resolving Current and Wind Drift | 向量导航：水流与风偏的分解
A boat needs to travel directly across a river of width 300 m. The boat’s speed in still water is 5 m/s, and the river current flows at 2 m/s perpendicular to the bank. We determine the heading the boat must take and the resultant velocity using vector addition.
一艘船需要径直穿过一条宽 300 米的河流。船在静水中的速度为 5 m/s，而河水流速为垂直于河岸 2 m/s。我们需要利用向量加法确定船应采取的航向和合速度。
et the desired resultant velocity be perpendicular to the bank (straight across). The boat' velocity components are: 5sinθ against the current and 5cosθ across. To cancel the current, we require 5sinθ = 2 → sinθ = 0.4 → θ ≈ 23.6°. The resultant speed across is 5cosθ = 5 × √1−0.16 = 5 × 0.9165 = 4.58 m/s.
令所求的合速度垂直于河岸（径直穿过）。船相对于地面的速度是船相对于水的速度与水流速度的向量和。若船向上游偏航角度 θ，其速度分量为：逆流方向 5sinθ，横渡方向 5cosθ。为抵消水流，需要满足 5sinθ = 2 → sinθ = 0.4 → θ ≈ 23.6°。横渡的合速度为 5cosθ = 5 × √1−0.16 = 5 × 0.9165 = 4.58 m/s。
8 ≈ 65.5 s. This application shows vectors resolving motion into perpendicular components.
渡河时间等于距离除以横渡合速度：t = 300 / 4.58 ≈ 65.5 秒。这一应用展示了向量将运动分解为垂直分量的方法。
6. Trigonometric Modelling of Daylight Hours | 日照时间的三角函数建模
for January 1st). This case study explores maximum and minimum daylight hours and when they occur using trigonometric analysis.
某城市的日照时数可以用正弦函数建模：H(t) = 12.2 + 3.8sin(2π(t−80)/365)，其中 t 为一年中的天数（1月1日 t=1）。本案例分析将利用三角函数分析探索最大和最小日照时数及其发生时间。
The function has amplitude 3.8 and vertical shift 12.2. Maximum daylight occurs when sin(…) = 1 → 12.2 + 3.8 = 16.0 hours. Minimum when sin(… tice). Similarly, for minimum set the argument to 3π/2 → t−80 = 3×365/4 = 273.75 → t = 353.75 (around December 20th, winter solstice).
该函数的振幅为 3.8，垂直位移为 12.2。当日照函数中 sin(…) = 1 时取最高日照时数 → 12.2 + 3.8 = 16.0 小时。当 sin(…) = −1 时取最低日照时数 → 12.2 − 3.8 = 8.4 小时。求解 sin(2π(t−80)/365) = 1 时的 t：2π(t−80)/365 = π/2 → t−80 = 365/4 = 91.25 → t = 171.25（大约 6 月 20 日，夏至）。类似地，求最小值时令自变量为 3π/2 → t−80 = 3×365/4 = 273.75 → t = 353.75（大约 12 月 20 日，冬至）。
This model illustrates how trigonometric graphs can describe seasonal patterns, and how to interpret parameters of a sine function in context.
此模型说明了三角函数图像如何描述季节性规律，以及如何在具体情境中诠释正弦函数参数的含义。
7. Instantaneous Rate of Change: Medicine Concentration Decay | 瞬时变化率：药物浓度的衰减
ne, the concentration C (mg/L) in the bloodstream over time t hours is given by C(t) = 20e−0.4t. We can find the rate at which the concentration is decreasing at any moment by differentiation, and determine after how long the rate of decrease is larger than a certain threshold.
服药后，血液中的药物浓度 C (mg/L) 随时间 t 小时的变化为 C(t) = 20e−0.4t。我们可以通过求导得出浓度在任意时刻的下降速率，并判断经过多久后下降速率超过某个阈值。
Differentiate: dC/dt = 20 × (−0.4)e−0.4t = −8e−0.4t e initial decrease rate is −8 mg/L/h. The negative sign indicates the concentration is falling. As time increases, the magnitude of the rate decreases (|−8e−0.4t| gets smaller). To find when the rate is −0.5 mg/L/h, solve −8e−0.4t = −0.5 → e−0.4t = 0.0625 → −0.4t = ln(0.0625) = −2.7726 → t = 6.93 hours.
求导：dC/dt = 20 × (−0.4)e−0.4t = −8e−0.4t。该式给出了浓度的瞬时变化率。在 t=0 时，初始下降速率为 −8 mg/L/h。负号表示浓度正在下降。随着时间推移，速率绝对值逐渐减小（|−8e−0.4t| 越来越小）。要求下降速率为 −0.5 mg/L/h 的时刻，解 −8e−0.4t = −0.5 → e−0.4t = 0.0625 → −0.4t = ln(0.0625) = −2.7726 → t = 6.93 小时。
This case shows how differentiation provides the rate of change of an exponential decay model, common in pharmacokinetics.
本案例体现了微分如何为指数衰减模型（常见于药代动力学）提供变化速率。
8. Using Identities to Solve Geometric Problems: Ladder Angle Analysis | 利用恒等式解决几何问题：梯子角度分析
A ladder of length L metres leans against a vertical wall. The angle it makes with the ground is θ. The ladder's base is at distance d from the wall and the top touches the wall at height h. Suppose the ladder experiences a small shift so that the angle increases by δθ. Using trigonometric identities, we can approximate the resulting change in height without recalculating fully.
一把长 L 米的梯子斜靠在竖直墙壁上，与地面的夹角为 θ。梯子的底部距墙 d 米，顶部触墙高度为 h 米。假设梯子发生轻微滑动，使角度增加 δθ。利用三角恒等式，我们可以近似计算高度变化，而无需完全重新计算。
1, sin(δθ) ≈ δθ (in radians). Then Δh ≈ L( sinθ×1 + cosθ×δθ − sinθ ) = L cosθ δθ = d δθ. So the change in height is approximately the base distance multiplied by the small change in angle (in radians). If L=5 m, θ=60°, then d=5 cos60°=2.5 m. A 1° increase (≈0.01745 rad) results in Δh ≈ 2.5×0.01745 = 0.0436 m ≈ 4.36 cm.
已知 h = L sinθ, d = L cosθ。高度的变化量 Δh = L(sin(θ+δθ) − sinθ)。利用恒等式 sin(A+B) = sin A cos B + cos A sin B，并且当 δθ 很小时，cos(δθ) ≈ 1，sin(δθ) ≈ δθ（弧度制）。于是 Δh ≈ L( sinθ×1 + cosθ×δθ − sinθ ) = L cosθ δθ = d δθ。所以高度的变化约等于底部距离乘以角度的微小变化量（弧度）。若 L=5 m，θ=60°，则 d=5 cos60°=2.5 m。增加 1°（≈0.01745 rad）导致 Δh ≈ 2.5×0.01745 = 0.0436 m ≈ 4.36 厘米。
This application demonstrates the power of trigonometric identities and small-angle approximations in solving practical geometric variations.
这一应用展示了三角恒等式和小角度近似在解决实际几何变化时的强大功能。
9. Area Under a Velocity-Time Graph: Distance Travelled by a Car | 速度-时间图下的面积：汽车行驶距离
s. The distance travelled is the area under the velocity-time curve. Using definite integration, we can find the exact distance covered during this interval.
一辆汽车从静止开始加速，前 10 秒内速度 v(t) = 3t²。行驶距离等于速度-时间曲线下的面积。我们可以通过定积分求出这段时间内行驶的准确距离。
ows a basic use of integration: accumulating a quantity (distance) from a rate (velocity).
距离 s = ∫ 从 0 到 10 v(t) dt = ∫ 从 0 到 10 3t² dt = [t³] 从 0 到 10 = 1000 − 0 = 1000 米。这显示了积分的基本用途：从速率（速度）累积出数量（距离）。
(200+10) − 0 = 210 m. Definite integrals thus model a wide range of real motion scenarios.
进一步巩固，假设速度函数为 v(t) = 4t + 1。则距离 = ∫ 从 0 到 10 (4t+1) dt = [2t² + t] 从 0 到 10 = (200+10) − 0 = 210 米。定积分因此可以模拟各种实际运动场景。
10. Solving Simultaneous Equations Using Matrices: Market Equilibrium
.5Q, where P is price and Q is quantity. The equilibrium is found by solving these two equations simultaneously. This can be expressed in matrix form and solved by finding the inverse matrix.
在经济学中，产品的供给与需求可以用线性模型来表示。假设需求函数为 P = 50 − 2Q，供给函数为 P = 10 + 1.5Q，其中 P 为价格，Q 为数量。均衡点通过联立求解这两个方程得到。这可以写成矩阵形式，并通过求逆矩阵来解答。
epresent the system in standard form: 2Q + P = 50, −1.5Q + P = 10. In matrix form: [[2,1],[−1.5,1]] × [Q;P] = [50;10]. The coefficient matrix has determinant = 2×1 − 1×(−1.5) = 2 + 1.5 = 3.5. Inverse matrix = 1/3.5 × [[1,−1],[1.5,2]] = [[0.2857,−0.2857],[0.4286,0.5714]] (approx). Multiply by constants vector gives [Q;P] = [11.43;27.14].
首先，令方程相等：50 − 2Q = 10 + 1.5Q → 40 = 3.5Q → Q = 40/3.5 = 80/7 ≈ 11.43，则 P = 50 − 2×(80/7) = 50 − 160/7 = (350−160)/7 = 190/7 ≈ 27.14。现在用标准形式表示方程组：2Q + P = 50，−1.5Q + P = 10。矩阵形式：[[2,1],[−1.5,1]] × [Q;P] = [50;10]。系数矩阵的行列式 = 2×1 − 1×(−1.5) = 2 + 1.5 = 3.5。逆矩阵 = 1/3.5 × [[1,−1],[1.5,2]] = [[0.2857,−0.2857],[0.4286,0.5714]] (近似值)。乘以常数列向量得到 [Q;P] = [11.43;27.14]。
Using matrices provides an alternative method that is particularly efficient for larger systems and shows the deep connection between algebra and economic modelling.
使用矩阵提供了一种替代方法，特别适用于大型方程组，并揭示了代数与经济建模之间的深层联系。
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