📚 Year 13 WJEC Maths: Case Study in Action | WJEC 13年级数学：案例分析实战演练
In WJEC Year 13 Mathematics, exam questions often present a multi‑step scenario that tests your ability to connect pure maths with real‑world modelling. This article walks you through a full case study – an inverted conical water tank – which draws on differentiation, related rates, differential equations, and integration. By working through every stage, you will see how to structure clear, logical solutions that earn high marks under timed conditions.
在 WJEC 13 年级数学考试中，题目常以多步骤的实际情景出现，考察你将纯数学知识应用于现实建模的能力。本文将通过一个完整的案例——倒置圆锥形水箱——带你演练微分、相关变化率、微分方程和积分。逐步攻克每一问，你将学会如何在限时条件下构建清晰、逻辑严谨的高分解答。
1. Problem Statement | 问题陈述
A water tank is shaped as an inverted right circular cone with vertical height 10 m and base radius 4 m. Water is poured into the tank at a constant rate of 2 m³ per minute. Initially, the tank is empty. Let h be the depth of water (in metres) and V the volume of water (in m³) at time t minutes. The surface of the water is a circle of radius r.
一个水箱呈倒置的正圆锥形状，垂直高度为 10 m，底面半径为 4 m。水以恒定速率 2 m³每分钟注入水箱。初始水箱是空的。设 h 为水深（米），V 为时刻 t（分钟）时水的体积（m³）。水面是一个半径为 r 的圆。
You are required to find expressions for rates of change, form a differential equation for h, solve it, determine the time to fill the tank, and investigate how the surface area changes.
你需要求出变化率的表达式，建立关于 h 的微分方程并求解，确定注满水箱的时间，并探究水面面积如何变化。
2. Setting Up Variables and Relations | 设定变量与关系式
From the cone’s geometry, similar triangles give the ratio between the water radius r and depth h: r / h = 4 / 10, so r = (2/5)h. The volume of water in the cone is V = (1/3)πr²h. Substituting r yields an expression purely in h:
根据圆锥的几何形状，由相似三角形可得水面半径 r 与水深 h 的比值：r / h = 4 / 10，即 r = (2/5)h。水的体积为 V = (1/3)πr²h，代入 r 得到仅含 h 的表达式：
V = (1/3)π (4/25)h³ = (4π/75)h³
Write down the constant rate of inflow: dV/dt = 2 m³ min⁻¹. This is the key information linking volume and time.
记下恒定的进水速率：dV/dt = 2 m³ min⁻¹。这是联系体积与时间的关键信息。
3. Part (a) – Related Rates | 第(a)问 – 相关变化率
Differentiate V = (4π/75) h³ with respect to t using the chain rule:
利用链式法则对 V = (4π/75) h³ 关于 t 求导：
dV/dt = (4π/75) × 3h² × dh/dt = (4π/25) h² dh/dt
Substitute the known dV/dt = 2 to find an expression for dh/dt:
代入已知 dV/dt = 2，求出 dh/dt 的表达式：
2 = (4π/25) h² dh/dt ⇒ dh/dt = 25/(2πh²)
Hence, when the depth is 5 m, dh/dt m min⁻¹). This shows the water level rises more slowly as depth increases, because the tank widens.
因此，当水深为 5 m 时，dh/dt = 25/(2π×25) = 1/(2π) m min⁻¹（约 0.159 m min⁻¹）。这表明随着水深增加，水面上升变慢——因为水箱在变宽。
4. Part (b) – Forming a Differential Equation | 第(b)问 – 建立微分方程
Since dV/dt is constant, the relationship between h and t can be expressed as a differential equation. From the previous derivative we already have dh/dt = 25/(2πh²). Multiplying both sides by h² gives a separable form:
由于 dV/dt 是常数，h 与 t 的关系可表示为一个微分方程。由上一步导数我们已有 dh/dt = 25/(2πh²)。两边同乘 h² 得到可分离变量的形式：
h² dh/dt = 25/(2π)
This is the differential equation governing the depth. It is first order and separable, which makes it straightforward to integrate.
这就是控制水深的微分方程。它是一阶可分离的，因此容易积分求解。
5. Part (c) – Solving the DE | 第(c)问 – 解微分方程
Integrate both sides with respect to t:
对两边关于 t 积分：
∫ h² dh = ∫ (25/(2π)) dt
The left side becomes (1/3)h³; the right side is (25/(2π))t + C. Applying the initial condition h=0 when t=0 gives C=0, so the solution is:
左边积分得 (1/3)h³，右边为 (25/(2π))t + C。利用初始条件 t=0 时 h=0，得 C=0，于是解为：
(1/3)h³ = (25/(2π))t ⇒ h³ = (75/(2π))t
Thus h is proportional to the cube root of time: h = ³√( (75/(2π))t ). This model allows us to predict the depth at any instant.
因此 h 与时间的立方根成正比：h = ³√( (75/(2π))t )。这个模型使我们能预测任意时刻的水深。
6. Part (d) – Time to Fill the Tank | 第(d)问 – 注满时间
The tank is full when h = 10 m. Substitute into the cubic relation:
当 h = 10 m 时水箱注满。代入三次方程：
10³ = 1000 = (75/(2π))T
Solving for T gives:
解出 T：
T = 1000 × (2π/75) = (2000π)/75 = (80π)/3 minutes
6). The result is left in exact form unless the question requests a decimal approximation.
其结果约为 83.8 分钟（π ≈ 3.1416）。除非题目特别要求，答案应保留精确形式。
7. Part (e) – Rate of Change of Surface Area | 第(e)问 – 水面面积变化率
The water surface is a circle with area A = πr². Using r = (2/5)h, we write A = π(4/25)h² = (4π/25)h². Now differentiate with respect to t:
水面为圆形，面积 A = πr²。由 r = (2/5)h 得 A = π(4/25)h² = (4π/25)h²。对 t 求导：
dA/dt = (4π/25) × 2h × dh/dt = (8π/25) h dh/dt
Substitute the earlier expression for dh/dt = 25/(2πh²):
代入前面求得的 dh/dt = 25/(2πh²)：
dA/dt = (8π/25) h × 25/(2πh²) = 4 / h
This is a remarkably simple result. For example, when h = 5 m, dA/dt = 4/5 = 0.8 m² min⁻¹. As the tank fills, the surface area grows at a decreasing rate – initially extremely fast, but slowing as h gets large.
这个结果异常简洁。例如 h = 5 m 时，dA/dt = 0.8 m² min⁻¹。随着水箱注满，水面面积的增长速率递减——开始时极快，但随 h 增大而变慢。
8. Examiner’s Tips for WJEC Case Studies | 考官的案例题实战技巧
Always start by drawing a large, labelled diagram. This helps you identify similar triangles, constant rates, and the variables that need linking.
始终从绘制一幅大而清晰的标注图开始。这能帮你识别相似三角形、恒定速率以及需要联系的变量。
Write down every piece of given data in symbol form before any calculation. Define your variables explicitly – e.g., “let V be the volume of water in m³ at time t minutes”. This gives structure to your solution and impresses examiners.
在一切计算之前，用符号形式写下所有已知数据。明确定义变量——例如“设 V 为 t 分钟时水的体积（m³）”。这使你的解答有条理，并能给考官留下良好印象。
Use the chain rule carefully when dealing with related rates. It is safer to write dV/dt = dV/dh × dh/dt rather than jumping to a final expression. Marks are awarded for method, not just the answer.
处理相关变化率时，要谨慎使用链式法则。写出 dV/dt = dV/dh × dh/dt 比直接跳到最终表达式更稳妥。评分侧重于方法，而不仅仅是答案。
When a differential equation emerges, identify its type – here separable – and show the integration steps clearly. Don’t forget the constant of integration and use the initial conditions to find it. Leaving the final answer in exact form (with π, surds) will often match the mark scheme.
当推导出微分方程时，判断其类型——此处为可分离型——并清晰展示积分步骤。不要忘记积分常数，并利用初始条件确定它。最终答案保留精确形式（含 π、根号）通常与评分标准一致。
Finally, re‑read the question and check if interpretations are asked for. In this case study, commenting on why dh/dt decreases as h increases turned a purely algebraic answer into a full explanation, which can secure the final mark on a longer question.
最后，重读一遍题目，确认是否有解释要求。在此案例中，说明为什么 dh/dt 随 h 增大而减小，把纯代数答案变为完整的解释，这在较长的题目中有助于拿到最后一分。
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