Case Study Practice: Advanced Mathematics | 进阶数学案例分析实战演练

📚 Case Study Practice: Advanced Mathematics | 进阶数学案例分析实战演练

Case studies in advanced mathematics bridge the gap between theoretical concepts and real-world applications. In this article, we will work through a series of carefully chosen problems that illustrate how mathematical models are built, analysed, and solved. Each case will challenge you to apply your knowledge of algebra, functions, geometry, trigonometry, and calculus to practical scenarios, improving both your problem-solving skills and your intuitive understanding of mathematical structures.

进阶数学中的案例分析能够弥合理论概念与现实应用之间的鸿沟。在本文中，我们将演练一系列精心挑选的问题，展示如何建立、分析和求解数学模型。每个案例都会要求你综合运用代数、函数、几何、三角和微积分知识解决实际问题，从而提升你的解题能力与对数学结构的直觉理解。

---

1. Case Study Approach: From Words to Equations | 案例分析方法：从文字到方程

The first step in any case study is to read the problem carefully and identify the quantities involved. Decide what variables to use, and translate the given information into mathematical expressions or equations. This usually means defining x for an unknown, writing relationships as functions, and noting any constraints.

任何案例分析的第一步都是仔细阅读问题，确定所涉及的各个量。决定使用哪些变量，并将已知信息转化为数学表达式或方程。这通常意味着用 x 表示未知数，用函数写出关系，并记录所有约束条件。

Next, build a model—whether it is a linear equation, a quadratic function, or a trigonometric expression. Solve the model using appropriate techniques, and always interpret your result back in the context of the original problem. Ask yourself: does the solution make sense? Are the units correct? Is it within feasible limits?

接下来，建立模型——无论是线性方程、二次函数还是三角表达式。使用合适的技巧求解模型，并始终将结果放回原问题的情境中解释。问问自己：答案合理吗？单位正确吗？是否在可行范围内？

A final reflection step helps reinforce learning: think about how the model might change if the assumptions were different, or how the method could be applied to similar problems. This structured approach will be used throughout the following examples.

最后的反思环节有助于巩固学习：思考如果假设不同，模型会如何变化，或者这种方法如何应用于类似问题。这个结构化方法将贯穿接下来的所有范例。

---

2. Modelling with Quadratic Functions | 二次函数建模

Profit function: P(x) = −2x² + 160x − 2000

A small business selling handmade crafts finds that its monthly profit P (in dollars) is given by the quadratic function above, where x is the number of items produced and sold. The owner wants to know the production level that maximises profit and the maximum profit achievable.

一家销售手工制品的小企业发现，其月利润 P（美元）由上述二次函数给出，其中 x 是生产和销售的件数。店主希望知道使利润最大化的产量以及可达到的最大利润。

using x = −b/(2a). With a = −2 and b = 160, we get x = −160 / (2 × −2) = −160 / (−4) = 40.

由于 x² 的系数为 −2，抛物线开口向下，因此顶点对应最大值。顶点的 x 坐标由 x = −b/(2a) 求出。这里 a = −2，b = 160，得到 x = −160 / (2 × −2) = −160 / (−4) = 40。

3200 + 6400 − 2000 = 1200. Thus, producing and selling 40 items yields a maximum profit of $1200.

将 x = 40 代回函数：P(40) = −2(40)² + 160(40) − 2000 = −3200 + 6400 − 2000 = 1200。因此，生产并销售 40 件产品可获得最大利润 1200 美元。

This case illustrates how optimisation problems in business can be solved using elementary quadratic analysis and how the vertex formula gives a direct route to the optimal decision.

本案例展示了商业中的优化问题如何借助基本的二次分析解决，以及顶点公式如何直接给出最优决策。

---

3. Exponential Growth in Population Studies | 人口研究中的指数增长

Population: P(t) = 500e^0.03t (t in years)

t a rate proportional to its size. The population after t years is modelled by P(t) = 500e^0.03t. The biologist needs to determine how long it will take for the colony to double in size.

一位生物学家正在监测一个细菌菌落，其数量以连续比例增长。t 年后的种群数量满足 P(t) = 500e^0.03t。生物学家需要确定菌落数量翻倍所需的时间。

Setting P(t) equal to 1000 gives the equation 500e^0.03t = 1000. Dividing both sides by 500 yields e^0.03t = 2. Take the natural logarithm of each side: ln(e^0.03t) = ln 2, so 0.03t = ln 2.

令 P(t) = 1000，得到方程 500e^0.03t = 1000。两边同除以 500 得 e^0.03t = 2。对两边取自然对数：ln(e^0.03t) = ln 2，因此 0.03t = ln 2。

3 ≈ 23.1 years. The colony will double in roughly 23.1 years under the given growth model.

解出 t = (ln 2) / 0.03。使用计算器，ln 2 ≈ 0.6931，所以 t ≈ 0.6931 / 0.03 ≈ 23.1 年。在该增长模型下，菌落大约在 23.1 年后翻倍。

Exponential models like this appear in finance, biology, and physics. Recognising how to manipulate logarithms to solve for an exponent is a key skill in advanced mathematics.

此类指数模型在金融、生物学和物理学中均有出现。掌握如何运用对数求解指数是高阶数学的一项关键技能。

---

广告 / Sponsored

4. Trigonometric Analysis of Wave Motion | 波的三角分析

Height: h(t) = 3 sin(πt/6), t in seconds

s and t is in seconds. Analysts want to find the maximum height, the period of the motion, and the times when the buoy is at its mean position within the first 12 seconds.

海面浮标的垂直位移可以用 h(t) = 3 sin(πt/6) 描述，其中 h 单位为米，t 单位为秒。分析人员希望求出最大高度、运动周期以及在前 12 秒内浮标处于平衡位置的时刻。

metres and the minimum is −3 metres. The period T for a sine function b sin(ct) is T = 2π/|c|. Here c = π/6, so T = 2π / (π/6) = 12 seconds.

振幅是系数 3，因此相对平衡位置的最大高度为 3 米，最小为 −3 米。对于正弦函数 b sin(ct)，周期 T = 2π/|c|。这里 c = π/6，因此 T = 2π / (π/6) = 12 秒。

The buoy is at its mean position when h(t) = 0. Solve 3 sin(πt/6) = 0 ⇒ 6 = nπ, i.e. t = 6n for integer n. Within 0 ≤ t ≤ 12, the values are t = 0, 6, 12 seconds.

浮标处于平衡位置时 h(t)=0。解 3 sin(πt/6) = 0 ⇒ sin(πt/6) = 0。当 πt/6 = nπ，即 t = 6n（n 为整数）时成立。在 0 ≤ t ≤ 12 范围内，时刻为 t = 0, 6, 12 秒。

Trigonometric functions are essential for modelling periodic phenomena such as tides, sound waves, and alternating current. Understanding amplitude, period, and zero-crossings allows us to interpret real cyclic behaviour.

三角函数对于潮汐、声波和交流电等周期现象的建模至关重要。理解振幅、周期和过零点能让我们解释真实的循环行为。

---

5. Vector Applications in Navigation | 向量在导航中的应用

Resultant velocity: v_R = v_b + v_c

A motorboat heads due east at a speed of 8 km/h in still water. A current flows south-east at 3 km/h. The captain needs to find the actual velocity of the boat relative to the ground, both its speed and direction.

一艘摩托艇在静水中以 8 km/h 的速度向正东行驶。水流以 3 km/h 的速度向东南方向流动。船长需要求出艇相对于地面的实际速度，包括速率和方向。

Represent the boat's velocity as vector v_b = (8, 0) and the current as v_c = (3 cos 315°, 3 sin 315°) = (3 × √2/2, 3 × (−√2/2)) ≈ (2.12, −2.12). The resultant velocity is v_R = (8+2.12, 0−2.12) = (10.12, −2.12).

将艇速表示为向量 v_b = (8, 0)，水流表示为 v_c = (3 cos 315°, 3 sin 315°) = (3 × √2/2, 3 × (−√2/2)) ≈ (2.12, −2.12)。合速度为 v_R = (8+2.12, 0−2.12) = (10.12, −2.12)。

The speed is the magnitude: |v_R| = √(10.12² + (−2.12)²) ≈ √102.4 + 4.49 ≈ √106.9 ≈ 10.34 km/h. The direction angle θ is given by tan⁻¹(|−2.12|/10.12) ≈ tan⁻¹(0.209) ≈ 11.8° south of east.

速率即模长：|v_R| = √(10.12² + (−2.12)²) ≈ √102.4 + 4.49 ≈ √106.9 ≈ 10.34 km/h。方向角 θ 满足 tan⁻¹(|−2.12|/10.12) ≈ tan⁻¹(0.209) ≈ 11.8°，即东偏南 11.8°。

Vector addition is the foundation for solving navigation problems involving wind or currents. Breaking vectors into components and reassembling them gives a clear picture of the resultant motion.

向量加法是解决涉及风或水流导航问题的基础。将向量分解为分量再合成，可以清晰地描述合运动。

---

6. Optimising Area with Calculus | 使用微积分优化面积

Area: A = x(100 − 2x)

A farmer has 100 metres of fencing and wants to enclose a rectangular area against a straight river bank, using the river as one side of the rectangle. Let the side perpendicular to the river be x metres. Find the dimensions that give the maximum enclosed area.

一位农民有 100 米围栏，想利用笔直的河岸作为矩形一边来围出一个矩形区域。设垂直于河岸的边长为 x 米。试求使围成面积最大的尺寸。

2x. The area A = x(100 − 2x) = 100x − 2x². To maximise A, differentiate: dA/dx = 100 − 4x.

若垂直于河岸的边长为 x，则有两条这样的边，平行于河岸的边用剩余围栏：长度 = 100 − 2x。面积 A = x(100 − 2x) = 100x − 2x²。为最大化 A，求导：dA/dx = 100 − 4x。

Set dA/dx = 0 ⇒ 100 − 4x = 0 ⇒ e parallel side length is 100 − 2 × 25 = 50 metres. The maximum area is 25 × 50 = 1250 m². Published by TutorHao | Year 9 进阶数学 Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）