Year 10 Eduqas Further Mathematics: Case Study Masterclass | 10年级Eduqas进阶数学：案例分析实战演练

📚 Year 10 Eduqas Further Mathematics: Case Study Masterclass | 10年级Eduqas进阶数学：案例分析实战演练

Welcome to a practical masterclass designed to sharpen your skills in Eduqas further mathematics. Through ten carefully selected case studies, you will see how topics such as exponential growth, quadratic optimisation, calculus, matrix transformations, coordinate geometry, iterative methods, trigonometry, and probability come alive in real-world contexts. Each case study breaks down the problem, models it with mathematics, and leads you to a clear solution.

欢迎参加这个专为提升Eduqas进阶数学能力而设计的实战演练。通过十个精心选取的案例，你将看到指数增长、二次优化、微积分、矩阵变换、坐标几何、迭代法、三角学和概率等主题如何在真实情境中发挥作用。每个案例都会拆解问题，用数学建模，并引导你得出清晰的解答。

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1. Compound Interest: Tracking Investment Growth | 复利：追踪投资增长

Case: You invest £1000 in a savings account that pays 5% compound interest per year. After how many years will your investment double to £2000?

案例：你将1000英镑存入一个年复利5%的储蓄账户。多少年后投资会翻倍至2000英镑？

The amount after n years is given by A = P(1 + r/100)ⁿ, where P = 1000, r = 5. Setting A = 2000 gives 1000 × 1.05ⁿ = 2000, so 1.05ⁿ = 2.

n年后的金额为 A = P(1 + r/100)ⁿ，其中 P=1000, r=5。令 A=2000 得 1000 × 1.05ⁿ = 2000，即 1.05ⁿ = 2。

≈ 0.0212, therefore n ≈ 0.3010 / 0.0212 ≈ 14.2. So after 15 years the balance will exceed £2000.

取对数（以10为底或自然对数）得 n = log(2) / log(1.05)。代入对数近似值：log 2 ≈ 0.3010，log 1.05 ≈ 0.0212，因此 n ≈ 0.3010 / 0.0212 ≈ 14.2。因此大约15年后余额将超过2000英镑。

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2. Maximising Revenue: Ticket Pricing Model | 最大化收入：门票定价模型

A concert venue finds that when tickets cost £p, the number of tickets sold is q = 120 – 2p. What ticket price maximises revenue?

某音乐厅发现当票价为£p时，售出门票数量为 q = 120 – 2p。什么样的票价能使收入最大化？

Revenue R = price × quantity = p(120 – 2p) = 120p – 2p². This is a quadratic function with a maximum turning point.

收入 R = 价格 × 数量 = p(120 – 2p) = 120p – 2p²。这是一个二次函数，具有最大值的转折点。

Using calculus: dR/dp = 120 – 4p. Set dR/dp = 0: 120 – 4p = 0 → p = 30.

使用微积分：dR/dp = 120 – 4p。令 dR/dp = 0：120 – 4p = 0 → p = 30。

Alternatively, complete the square: R = –2(p² – 60p) = –2[(p – 30)² – 900] = –2(p – 30)² + 1800. Maximum revenue is £1800 when p = 30.

或者配方：R = –2(p² – 60p) = –2[(p – 30)² – 900] = –2(p – 30)² + 1800。最大收入为£1800，此时 p=30。

The ticket price should be set at £30 to maximise revenue.

票价应定为30英镑以实现收入最大化。

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3. Projectile Motion: Finding the Peak of a Throw | 抛射体运动：求抛掷最高点

A ball is thrown straight up. Its height h (metres) after t seconds is modelled by h(t) = –5t² + 20t + 1. Find the maximum height reached.

案例：一个球被垂直上抛。其高度h（米）与时间t（秒）的关系为 h(t) = –5t² + 20t + 1。求达到的最大高度。

The function is a downward-opening parabola. Using differentiation: h'(t) = –10t + 20. Setting h'(t) = 0 gives t = 2 seconds.

这是一个开口向下的抛物线。使用微分：h'(t) = –10t + 20。令导数等于零得 t = 2 秒。

Substitute t = 2 into h: h(2) = –5(2)² + 20(2) + 1 = –20 + 40 + 1 = 21 metres.

将 t=2 代入 h：h(2) = –5(2)² + 20(2) + 1 = –20 + 40 + 1 = 21 米。

21 metres at t = 2 s. The symmetry of the parabola also confirms the middle of the flight.

因此球在 t=2 秒时达到最大高度21米。抛物线的对称性同样证实了飞行中点的位置。

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4. Matrix Transformations: Rotating a Logo | 矩阵变换：旋转一个标识

A design company wants to rotate the point A(1, 0) by 45° anticlockwise about the origin. Use a rotation matrix to find the coordinates of the rotated point A'.

一家设计公司想要将点A(1, 0)绕原点逆时针旋转45°。用旋转矩阵求旋转后点 A' 的坐标。

The rotation matrix for an angle θ is:

旋转θ角的旋转矩阵为：

cos θ	–sin θ
sin θ	cos θ

For θ = 45°, cos 45° = sin 45° = √2/2. So the matrix becomes:

对于 θ=45°，cos 45° = sin 45° = √2/2。因此矩阵为：

√2/2	–√2/2
√2/2	√2/2

Multiply this matrix by the column vector (1, 0):

将此矩阵乘以列向量 (1, 0)：

x' = (√2/2)×1 + (–√2/2)×0 = √2/2 ≈ 0.707

y' = (√2/2)×1 + (√2/2)×0 = √2/2 ≈ 0.707

So A' int has moved from the x-axis onto the line y = x, exactly as expected for a 45° rotation.

因此 A' ≈ (0.707, 0.707)。该点从x轴转动到了直线 y=x 上，完全符合45°旋转的预期。

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5. Intersection of a Line and a Circle: Detector Range | 直线与圆的交点：探测器范围

Case: A sensor is located at the origin and can detect objects within a radius of √10 ath y = 2x + 1. Determine the points where the object enters and exits the detection zone.

案例：一个传感器位于原点，可探测半径为√10单位内的物体（圆方程 x² + y² = 10）。一移动物体沿直线 y = 2x + 1 前进。求物体进入和离开探测区域的点。

= 2x + 1 into the circle equation: x² + (2x + 1)² = 10 → x² + 4x² + 4x + 1 = 10 → 5x² + 4x – 9 = 0.

将 y = 2x + 1 代入圆的方程：x² + (2x + 1)² = 10 → x² + 4x² + 4x + 1 = 10 → 5x² + 4x – 9 = 0。

Factorise: (5x + 9)(x – 1) = 0, so x = –9/5 = –1.8 or x = 1.

因式分解：(5x + 9)(x – 1) = 0，得 x = –9/5 = –1.8 或 x = 1。

Find corresponding y: for x = –1.8, y = 2(–1.8) + 1 = –2.6; for x = 1, y = 3.

求对应y值：x = –1.8 时 y = 2(–1.8) + 1 = –2.6；x = 1 时 y = 3。

The object enters at (–1.8, –2.6) and exits at (1, 3). The line is a secant to the circle.

物体在 (–1.8, –2.6) 进入，在 (1, 3) 离开。这条直线是圆的一条割线。

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6. Iterative Methods: Solving a Cubic Equation | 迭代法：解三次方程

Case: Use the Newton–Raphson method to approximate a root of f(x) = x³ – 2x – 5 = 0, starting with x₀ = 2.

案例：用牛顿-拉弗森法求方程 f(x) = x³ – 2x – 5 = 0 的一个近似根，起始值 x₀ = 2。

The derivative is f'(x) = 3x² – 2. The Newton–Raphson iteration formula is x_n+1 = x_n – f(x_n)/f'(x_n).

导数为 f'(x) = 3x² – 2。牛顿-拉弗森迭代公式为 x_n+1 = x_n – f(x_n)/f'(x_n)。

Step 1: n = 0, x₀ = 2. f(2) = 8 – 4 – 5 = –1, f'(2) = 12 – 2 = 10. x₁ = 2 – (–1)/10 = 2.1.

第一步：n=0, x₀=2。f(2)=8–4–5=–1, f'(2)=12–2=10。x₁ = 2 – (–1)/10 = 2.1。

Step 2: x₁ = 2.1. f(2.1) = (2.1)³ – 2(2.1) – 5 = 9.261 – 4.2 – 5 = 0.061, f'(2.1) = 3(4.41) – 2 = 13.23 – 2 = 11.23. x₂ = 2.1 – 0.061/11.23 ≈ 2.094568.

第二步：x₁=2.1。f(2.1)=9.261–4.2–5=0.061, f'(2.1)=11.23。x₂ = 2.1 – 0.061/11.23 ≈ 2.094568。

After a couple of iterations the root is approximately 2.0946, which matches the exact root to four decimal places.

经过几次迭代，根约为2.0946，与精确解在小数点后四位一致。

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7. Trigonometry in Surveying: Measuring a Mountain Height | 测量学中的三角学：测量山高

Case: Surveyors measure the angle of elevation to the top of a mountain from two points 500 m apart on the same side of the mountain. From the nearer point the angle is 42°, from the farther point it is 28°. Find the height of the mountain.

案例：测量员从山同侧的两个点测量山顶仰角，两点相距500米。从较近的点测得仰角42°，较远的点测得28°。求山的高度。

e nearer point to the base. Then from the nearer point: tan 42° = h / d → d = h / tan 42°.

设山高为h，较近点到山脚的水平距离为d。则从较近点：tan 42° = h / d → d = h / tan 42°。

rom the farther point (distance d + 500): tan 28° = h / (d + 500) → d + 500 = h / tan 28°.

从较远点（距离 d+500）：tan 28° = h / (d+500) → d+500 = h / tan 28°。

Subtract the two equations: (d + 500) – d = h / tan 28° – h / tan 42° → 500 = h(1/tan 28° – 1/tan 42°).

两式相减：(d+500)–d = h / tan 28° – h / tan 42° → 500 = h(1/tan 28° – 1/tan 42°)。

0 = h × 0.7701, thus h ≈ 500 / 0.7701 ≈ 649.4 m. The mountain is approximately 649 m high.

计算三角值：tan 28°≈0.5317，tan 42°≈0.9004。因此 1/tan 28°≈1.8807，1/tan 42°≈1.1106。差值为0.7701。于是 500 = h × 0.7701，得 h ≈ 500 / 0.7701 ≈ 649.4 米。山高约为649米。

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8. Calculus Optimisation: Open Box Volume | 微积分优化：开口盒体积

Case: A rectangular sheet of cardboard measures 30 cm by 30 cm. Equal squares of side x cm are cut from each corner, and the sides are folded up to make an open box. Find the value of x that gives the maximum volume.

案例：一张30cm × 30cm的正方形纸板，从每个角剪去边长为x cm的正方形，然后将四边折起制成一个无盖盒。求使体积最大的x值。

After cutting, the base dimensions are (30 – 2x) by (30 – 2x), and the height is x. Volume V = x(30 – 2x)², for 0 < x < 15.

剪去后底部尺寸为 (30–2x)×(30–2x)，高度为x。体积 V = x(30–2x)²，定义域 0

Expand: V = x(900 – 120x + 4x²) = 4x³ – 120x² + 900x.

展开：V = x(900 – 120x + 4x²) = 4x³ – 120x² + 900x。

Differentiate: dV/dx = 12x² – 240x + 900. Set dV/dx = 0: 12x² – 240x + 900 = 0 → divide by 12: x² – 20x + 75 = 0.

求导：dV/dx = 12x² – 240x + 900。令导数为零：12x²–240x+900=0 → 除以12：x²–20x+75=0。

Factorise: (x – 15)(x – x = 15. x = 15 is at the boundary (volume zero), so the maximum volume occurs at x = 5 cm.

因式分解：(x–15)(x–5)=0，得 x=5 或 x=15。x=15 是边界（体积为零），因此最大体积出现在 x=5 cm。

Second derivative test: d²V/dx² = 24x – 240. At x = 5, d²V/dx² = 120 – 240 = –120 < 0, confirming a maximum. Maximum volume = 5 × (20)² = 2000 cm³.

二阶导数检验：d²V/dx² = 24x–240。x=5时，d²V/dx² = 120–240 = –120 < 0，确认是极大值。最大体积 = 5×(20)² = 2000 cm³。

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9. Conditional Probability: Medical Test Accuracy | 条件概率：医学检测准确性

Case: A disease affects 1% of the population. A test gives a positive result for 95% of people who have the disease (true positive) and a false positive for 2% of healthy people. If a person tests positive, what is the probability they actually have the disease?

案例：某疾病影响1%的人口。一项检测对患病者有95%的阳性检出率（真阳性），对健康人有2%的假阳性。如果一个人检测呈阳性，他实际患病的概率是多少？

Let D = having the disease, T⁺ = positive test. P(D) = 0.01, P(T⁺|D) = 0.95, P(T⁺|not D) = 0.02.

设 D=患病，T⁺=检测阳性。P(D)=0.01，P(T⁺|D)=0.95，P(T⁺|not D)=0.02。

We want P(D|T⁺). Using Bayes' theorem: P(D|T⁺) = [P(T⁺|D) × P(D)] / [P(T⁺|D)P(D) + P(T⁺|not D)P(not D)].

我们要求 P(D|T⁺)。使用贝叶斯定理：P(D|T⁺) = [P(T⁺|D)×P(D)] / [P(T⁺|D)P(D) + P(T⁺|not D)P(not D)]。

.95 × 0.01 = 0.0095. Denominator = 0.0095 + 0.02 × 0.99 = 0.0095 + 0.0198 = 0.0293. So P(D|T⁺) ≈ 0.0095 / 0.0293 ≈ 0.324 (about 32.4%).

计算：分子 = 0.95×0.01 = 0.0095。分母 = 0.0095 + 0.02×0.99 = 0.0095+0.0198 = 0.0293。因此 P(D|T⁺) ≈ 0.0095/0.0293 ≈ 0.324（约32.4%）。

Even with a positive test, the probability of actually having the disease is only about 32% – a surprising but important insight into the influence of low base rates.

即使检测呈阳性，实际患病的概率也仅约32%——这令人惊讶，却揭示了低基础率的重要影响。

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10. Arithmetic Series: Saving Plan with Regular Increments | 等差数列：定期递增的储蓄计划

Case: You start a savings plan where you deposit £20 in the first week and increase the deposit by £5 each week (so £20, £25, £30, ...). How much will you have saved in total after 20 weeks?

案例：你开始一个储蓄计划，第一周存入20英镑，每周增加5英镑（即20, 25, 30, ...）。20周后总共存了多少钱？

ithmetic sequence: first term a = 20, common difference d = 5. The number of terms n = 20.

存款构成等差数列：首

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