📚 Year 12 CCEA Mathematics: Case Study Practice | CCEA 12年级数学：案例分析实战演练
Developing the ability to apply mathematical concepts to real-world scenarios is key to success in CCEA Year 12 Mathematics. This collection of case studies bridges pure and applied topics, giving you the practice needed to tackle exam-style contextual problems with confidence.
培养将数学概念应用于实际情境的能力是CCEA 12年级数学成功的关键。这组案例研究连接了纯数学与应用数学主题，让你通过实战演练，自信地应对考试中的情境问题。
1. Projectile Motion & Quadratic Functions | 抛物运动与二次函数
A ball is thrown upward from a platform 2 m above the ground. Its height h (in metres) after t seconds is modelled by h(t) = -4.9t² + 14.7t + 2. Determine the maximum height reached and the time when the ball strikes the ground.
一个球从离地2米高的平台向上抛出。抛出t秒后的高度h（米）由函数h(t) = -4.9t² + 14.7t + 2 给出。求球达到的最大高度以及球落地的时间。
To find the maximum height, differentiate h(t): h'(t) = -9.8t + 14.7. Set h'(t) = 0 to locate the stationary point: t = 14.7 ÷ 9.8 = 1.5 s. Substituting back gives the maximum height.
为求最大高度，对h(t)求导：h'(t) = -9.8t + 14.7。令导数为零求驻点：t = 14.7 ÷ 9.8 = 1.5 秒。代回原函数得最大高度。
h(1.5) = -4.9(1.5)² + 14.7(1.5) + 2 = 13.025 m
The ball hits the ground when h(t) = 0. Solve -4.9t² + 14.7t + 2 = 0 using the quadratic formula.
球落地时h(t)=0。使用求根公式解方程 -4.9t² + 14.7t + 2 = 0。
t = [ -14.7 ± √(14.7² – 4(-4.9)(2)) ] / (2 × -4.9)
Negative time is discarded; the positive solution t ≈ 3.13 s gives the time of impact.
舍弃负的时间；正根 t ≈ 3.13 秒即为落地时间。
2. Exponential Decay & Half-Life | 指数衰变与半衰期
A radioactive substance decays according to m = m₀ e⁻ᵏᵗ, where m₀ is the initial mass and k is a positive constant. The half-life of the substance is 8 days. Find k and the time taken for the mass to reduce to 10% of its original value.
某种放射性物质按规律 m = m₀ e⁻ᵏᵗ 衰变，其中m₀为初始质量，k为正常数。该物质的半衰期为8天。求k的值，以及质量降至初始值10%所需的时间。
At half-life, m = ½ m₀, t = 8 days. Substitute: ½ m₀ = m₀ e⁻⁸ᵏ leads to e⁻⁸ᵏ = ½. Taking natural logs gives -8k = ln ½, so k = (ln 2) / 8 ≈ 0.0866 day⁻¹.
在半衰期，m = ½ m₀，t = 8天。代入得 ½ m₀ = m₀ e⁻⁸ᵏ，即 e⁻⁸ᵏ = ½。取自然对数得 -8k = ln ½，因此 k = (ln 2)/8 ≈ 0.0866 天⁻¹。
For 10% remaining, 0.10 m₀ = m₀ e⁻ᵏᵗ. Thus e⁻ᵏᵗ = 0.10, so -kt = ln 0.10. Using k ≈ 0.0866, t = -ln 0.10 / 0.0866 ≈ 26.6 days.
降至10%时，0.10 m₀ = m₀ e⁻ᵏᵗ，即 e⁻ᵏᵗ = 0.10。因此 -kt = ln 0.10。代入k ≈ 0.0866，得 t = -ln 0.10 / 0.0866 ≈ 26.6 天。
3. Optimisation: Maximum Area & Minimum Cost | 最优化问题：面积最大与成本最小
A farmer has 200 m of fencing to enclose a rectangular area against a straight riverbank (no fencing needed along the river). Find the dimensions that maximise the enclosed area.
一位农夫有200米长的围栏，想沿一道笔直的河岸围出一个矩形区域（河岸一侧无需围栏）。求能围出最大面积的矩形尺寸。
Let the side parallel to the river be y and each perpendicular side be x. Then 2x + y = 200, so y = 200 – 2x. The area A = x y = x(200 – 2x) = 200x – 2x².
设平行于河岸的边长为y，垂直于河岸的边长为x。则 2x + y = 200，故 y = 200 – 2x。面积 A = x y = x(200 – 2x) = 200x – 2x²。
Differentiate: dA/dx = 200 – 4x. Set equal to zero: x = 50 m. Then y = 200 – 2(50) = 100 m. The maximum area occurs at these dimensions.
求导：dA/dx = 200 – 4x。令导数为零得 x = 50 米。则 y = 200 – 2(50) = 100 米。此尺寸下面积最大。
Check using second derivative: d²A/dx² = -4 < 0, confirming a maximum.
用二阶导数检验：d²A/dx² = -4 < 0，确认为极大值。
4. Trigonometric Modelling: Tidal Heights | 三角函数模型：潮汐高度
The height of water at a dock, H metres, is modelled by H(t) = 5 + 3 sin(πt/6), where t is the time in hours after midnight. Find the maximum and minimum water heights and the time of the first high tide after 6:00 a.m.
码头水深H（米）由模型 H(t) = 5 + 3 sin(πt/6) 给出，其中t为午夜后的时间（小时）。求最高与最低水位，以及早上6点后第一次高潮的时间。
The sine function oscillates between -1 and 1, so H ranges from 5 – 3 = 2 m to 5 + 3 = 8 m. The minimum depth is 2 m and the maximum is 8 m.
正弦函数在-1与1之间振荡，因此H的范围为 5-3=2 米至 5+3=8 米。最低水深2米，最高水深8米。
High tide occurs when sin(πt/6) = 1, i.e. πt/6 = π/2 + 2πk. Solving for k = 0 gives t = 3 (3:00 a.m.). For the first high tide after 6:00 a.m., use k = 1: πt/6 = π/2 + 2π → t = 15, i.e. 3:00 p.m.
高潮发生在 sin(πt/6)=1 时，即 πt/6 = π/2 + 2πk。取k=0得t=3（凌晨3:00）。早上6点后的第一次高潮对应 k=1：πt/6 = π/2 + 2π → t=15，即下午3:00。
5. Normal Distribution & Quality Control | 正态分布与质量控制
Bolts produced by a machine have diameters normally distributed with mean μ = 10.00 mm and standard deviation σ = 0.15 mm. Bolts are rejected if their diameter is less than 9.80 mm or greater than 10.25 mm. Find the percentage rejected.
一台机器生产的螺栓直径服从正态分布，均值μ = 10.00 mm，标准差σ = 0.15 mm。直径小于9.80 mm或大于10.25 mm的螺栓视为不合格。求不合格的百分比。
Standardise the lower limit: z₁ = (9.80 – 10.00) / 0.15 = -1.333. Using normal tables, P(Z < -1.333) ≈ 0.0912.
将下限标准化：z₁ = (9.80 – 10.00)/0.15 = -1.333。查正态分布表得 P(Z < -1.333) ≈ 0.0912。
Upper limit: z₂ = (10.25 – 10.00) / 0.15 = 1.667. P(Z > 1.667) = 1 – P(Z < 1.667) ≈ 1 - 0.9522 = 0.0478.
上限：z₂ = (10.25 – 10.00)/0.15 = 1.667。P(Z > 1.667) = 1 – P(Z < 1.667) ≈ 1 - 0.9522 = 0.0478。
Total rejection rate = 0.0912 + 0.0478 = 0.1390, i.e. about 13.9%.
总不合格率 = 0.0912 + 0.0478 = 0.1390，即约13.9%。
6. Kinematics: Displacement, Velocity & Acceleration | 运动学：位移、速度与加速度
A particle moves along a straight line such that its displacement s metres from a fixed point after t seconds is s(t) = t³ – 6t² + 9t. Find the velocity and acceleration functions, the times when the particle is at rest, and the total distance travelled in the first 4 seconds.
一个质点沿直线运动，t秒时相对固定点的位移s（米）为 s(t) = t³ – 6t² + 9t。求速度和加速度函数、质点静止的时刻，以及最初4秒内经过的总路程。
Velocity v(t) = ds/dt = 3t² – 12t + 9. Acceleration a(t) = dv/dt = 6t – 12.
速度 v(t) = ds/dt = 3t² – 12t + 9。加速度 a(t) = dv/dt = 6t – 12。
Set v(t) = 0: 3t² – 12t + 9 = 0 → t² – 4t + 3 = 0 → (t – 1)(t – 3) = 0. The particle is at rest at t = 1 s and t = 3 s.
令 v(t)=0：3t² – 12t + 9 = 0 → t² – 4t + 3 = 0 → (t-1)(t-3)=0。质点在 t=1 秒和 t=3 秒静止。
To find total distance, evaluate s at t = 0, 1, 3, 4: s(0)=0, s(1)=4, s(3)=0, s(4)=4. Distance = |4-0| + |0-4| + |4-0| = 4 + 4 + 4 = 12 m.
为求总路程，计算t=0,1,3,4时的位移：s(0)=0, s(1)=4, s(3)=0, s(4)=4。路程 = |4-0|+|0-4|+|4-0| = 4+4+4 = 12 米。
7. Logarithmic Scales: pH and Sound Intensity | 对数尺度：pH值与声强
The pH of a solution is pH = -log₁₀[H⁺], where [H⁺] is the hydrogen ion concentration in mol/L. Lemon juice has a hydrogen ion concentration of 0.01 mol/L. Calculate its pH. If a cleaner has pH 11, determine the hydrogen ion concentration.
溶液的pH值计算公式为 pH = -log₁₀[H⁺]，其中[H⁺]为氢离子浓度（mol/L）。柠檬汁的氢离子浓度为0.01 mol/L。计算其pH。若某种清洁剂的pH为11，求其氢离子浓度。
For lemon juice: pH = -log₁₀(0.01) = -log₁₀(10⁻²) = 2.
柠檬汁：pH = -log₁₀(0.01) = -log₁₀(10⁻²) = 2。
For the cleaner: 11 = -log₁₀[H⁺] → log₁₀[H⁺] = -11 → [H⁺] = 10⁻¹¹ mol/L.
清洁剂：11 = -log₁₀[H⁺] → log₁₀[H⁺] = -11 → [H⁺] = 10⁻¹¹ mol/L。
8. Linear Regression & Correlation | 线性回归与相关性
A study records the number of hours spent revising (x) and test scores (y) for six students: (2,55), (3,60), (5,70), (6,75), (8,85), (9,90). Find the equation of the regression line y on x and use it to predict the score for a student who revises 7 hours.
一项研究记录了六名学生的复习小时数(x)与考试成绩(y)：（2,55),(3,60),(5,70),(6,75),(8,85),(9,90）。求y对x的回归直线方程，并用该方程预测复习7小时的成绩。
Compute means: x̄ = (2+3+5+6+8+9)/6 = 5.5, Ȳ = (55+60+70+75+85+90)/6 = 72.5. Calculate Sxy and Sxx.
计算均值：x̄ = 5.5, Ȳ = 72.5。计算Sxy和Sxx。
Sxy = Σ(xᵢ – x̄)(yᵢ – Ȳ) = 175, Sxx = Σ(xᵢ – x̄)² = 35. Thus gradient b = Sxy / Sxx = 175/35 = 5. Intercept a = Ȳ – b x̄ = 72.5 – 5(5.5) = 45.
Sxy = Σ(xᵢ – x̄)(yᵢ – Ȳ) = 175, Sxx = Σ(xᵢ – x̄)² = 35。因此斜率 b = Sxy/Sxx = 175/35 = 5。截距 a = Ȳ – b x̄ = 72.5 – 5(5.5) = 45。
Regression line: y = 45 + 5x. For x = 7, predicted score y = 45 + 5(7) = 80.
回归直线：y = 45 + 5x。复习7小时，预测成绩 y = 45 + 5(7) = 80。
9. Conditional Probability & Tree Diagrams | 条件概率与树形图
In a bag there are 5 red sweets and 3 blue sweets. A sweet is taken at random and eaten; a second sweet is then taken. Find the probability that (a) both are red, (b) the second is red given that the first was blue.
一个袋子里有5颗红色糖果和3颗蓝色糖果。随机取出一颗吃掉，然后再取出一颗。求：(a) 两颗都是红色的概率；(b) 已知第一颗是蓝色，第二颗是红色的概率。
Draw a tree: 1st draw P(R) = 5/8, P(B) = 3/8. If 1st is R, remainder 4R,3B; P(R|1st R) = 4/7. If 1st is B, remainder 5R,2B; P(R|1st B) = 5/7.
画树形图：第一次取 P(R)=5/8, P(B)=3/8。若第一次为红，剩下4红3蓝，P(第二次红|第一次红)=4/7。若第一次为蓝，剩下5红2蓝，P(第二次红|第一次蓝)=5/7。
(a) P(both red) = 5/8 × 4/7 = 20/56 = 5/14. (b) P(2nd red | 1st blue) = 5/7.
(a) 两颗皆红概率 = 5/8 × 4/7 = 20/56 = 5/14。(b) 已知第一颗蓝，第二颗红概率 = 5/7。
10. Geometric Sequences & Compound Interest | 等比数列与复利
An investment account pays 4% interest per annum, compounded annually. A lump sum of £2000 is deposited. How much will the account be worth after 6 years? How long will it take for the amount to double?
一个投资账户年利率4%，每年复利一次。存入一次性本金£2000。6年后账户价值多少？需要多少年才能翻倍？
The amount after n years is A = 2000(1.04)ⁿ. For n = 6, A = 2000(1.04)⁶. Calculate (1.04)⁶ ≈ 1.2653, so A ≈ £2530.60.
n年后的金额为 A = 2000(1.04)ⁿ。n=6时，A = 2000(1.04)⁶。算得 (1.04)⁶ ≈ 1.2653，故 A ≈ £2530.60。
For doubling
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