Applications of Calculus and Mathematical Models | 微积分与数学模型的应用

📚 Applications of Calculus and Mathematical Models | 微积分与数学模型的应用

Mathematical modelling lies at the heart of the IB Mathematics curriculum, bridging theory and real-world scenarios. Whether you are analysing the spread of a virus, maximising profit for a business, or predicting the motion of a planet, calculus and function models provide the essential toolkit. This article explores ten key application areas, each presented with intuitive explanations and practical examples that align with the IB syllabus. By connecting abstract derivatives and integrals to tangible quantities like velocity, cost, and volume, you will strengthen your conceptual understanding and problem-solving confidence.

数学建模是 IB 数学课程的核心，它在理论和现实场景之间架起了一座桥梁。无论你是在分析病毒的传播、为企业最大化利润，还是预测行星的运动，微积分和函数模型都提供了不可或缺的工具。本文探讨了十个重要的应用领域，每个领域都配有符合 IB 大纲的直观解释和实际例子。通过将抽象的导数和积分与速度、成本、体积等具体量联系起来，你将加深概念理解并增强解题的信心。

1. Optimization Problems | 最优化问题

Optimization uses derivatives to find maximum or minimum values of a function. For a quantity f(x), setting the first derivative f'(x) = 0 locates critical points, and the second derivative test or a sign diagram confirms the nature. In IB problems, you often model a real-life situation — like minimising the surface area of a can for a fixed volume — then differentiate to find the optimal dimensions.

最优化通过导数来寻找函数的最大值或最小值。对于量 f(x)，令一阶导数 f'(x) = 0 可找到临界点，再利用二阶导数检验或符号表确认其性质。在 IB 题目中，你常常需要先对现实情景建模——比如固定体积下最小化罐子的表面积——然后通过求导找到最优尺寸。

A typical example: a box with a square base and an open top must hold a volume of 32 000 cm³. If the base side length is x cm and height h cm, then x²h = 32 000, so h = 32 000/x². The surface area S = x² + 4xh = x² + 128 000/x. Differentiating gives dS/dx = 2x − 128 000/x². Setting this to zero yields x³ = 64 000, so x = 40 cm, h = 20 cm, and S is minimised. The second derivative S”(x) = 2 + 256 000/x³ > 0 confirms a minimum.

一个典型例子：一个无盖方底盒子的容积必须为 32 000 cm³。设底面边长为 x cm，高为 h cm，则 x²h = 32 000，即 h = 32 000/x²。表面积 S = x² + 4xh = x² + 128 000/x。求导得 dS/dx = 2x − 128 000/x²。令其为零得到 x³ = 64 000，于是 x = 40 cm，h = 20 cm，此时表面积最小。二阶导数 S”(x) = 2 + 256 000/x³ > 0 确认了最小值。

---

2. Related Rates | 相关变化率

Related rates problems involve finding how fast one quantity changes by relating it to another whose rate of change is known. The chain rule is the key tool: if two variables x and y are linked by an equation, differentiating with respect to time t yields relationships among dx/dt and dy/dt. Common contexts include inflating a balloon, water draining from a tank, or a sliding ladder.

相关变化率问题要求通过将待求量和一个已知变化率的量关联起来，从而求出变化有多快。链式法则是关键工具：若两个变量 x 和 y 通过一个方程关联，那么对方程关于时间 t 求导就能得到 dx/dt 和 dy/dt 之间的关系。常见的场景包括气球充气、水从水箱排出或滑动的梯子。

For instance, a spherical balloon is filled with air at 100 cm³/s. Find the rate of increase of the radius when the radius is 5 cm. Volume V = (4/3)πr³, so dV/dt = 4πr² (dr/dt). Given dV/dt = 100, r = 5, we solve 100 = 4π(5)² (dr/dt) → dr/dt = 1/π cm/s ≈ 0.318 cm/s. The key steps are: write an equation linking the variables; differentiate implicitly with respect to t; substitute known values; solve for the unknown rate.

例如，一个球形气球以 100 cm³/s 的速度充气。当半径为 5 cm 时，求半径的增加速率。体积 V = (4/3)πr³，于是 dV/dt = 4πr² (dr/dt)。已知 dV/dt = 100，r = 5，求解 100 = 4π(5)² (dr/dt) → dr/dt = 1/π cm/s ≈ 0.318 cm/s。关键步骤为：写出关联变量的方程；关于 t 隐式求导；代入已知值；解出未知速率。

---

3. Kinematics: Position, Velocity, Acceleration | 运动学：位置、速度、加速度

Kinematics describes the motion of a particle along a straight line. Given the displacement function s(t), the velocity is v(t) = s'(t), and acceleration is a(t) = v'(t) = s”(t). Integration reverses the process: v(t) = ∫ a(t) dt and s(t) = ∫ v(t) dt. Problems often ask for the distance travelled (total area under the speed–time graph) versus displacement, and for times when the particle changes direction (v(t) = 0).

运动学描述质点沿直线的运动。给定位移函数 s(t)，速度是 v(t) = s'(t)，加速度是 a(t) = v'(t) = s”(t)。积分可以反过来：v(t) = ∫ a(t) dt，s(t) = ∫ v(t) dt。题目经常要求计算总路程（速度-时间图下的总面积）与位移的区别，以及质点改变方向的时刻（v(t) = 0）。

An IB-style question: s(t) = t³ − 6t² + 9t (m), t ≥ 0. Then v(t) = 3t² − 12t + 9 = 3(t−1)(t−3), so the particle is at rest at t = 1, 3 s. Displacement from t=0 to t=3 is s(3)−s(0)=0; however, distance travelled requires integrating |v(t)| from 0 to 1, 1 to 3: total distance = |s(1)−s(0)| + |s(3)−s(1)| = 4 + 4 = 8 m.

一道 IB 风格的题目：s(t) = t³ − 6t² + 9t (m)，t ≥ 0。那么 v(t) = 3t² − 12t + 9 = 3(t−1)(t−3)，因此质点在 t = 1, 3 s 时静止。从 t=0 到 t=3 的位移为 s(3)−s(0)=0；然而总路程需要对 |v(t)| 从 0 到 1，1 到 3 进行积分：总路程 = |s(1)−s(0)| + |s(3)−s(1)| = 4 + 4 = 8 m。

v = ds/dt, a = dv/dt, Δs = ∫ v dt

---

4. Area and Accumulation | 面积与累积

The definite integral ∫ₐᵇ f(x) dx represents the signed area between the curve y = f(x) and the x-axis from x = a to x = b. In applications, this can represent total accumulated change, such as total rainfall from a rate-of-rainfall function, or total charge from a current over time. If the graph crosses the x‑axis, the integral yields net area, while the total area is ∫ₐᵇ |f(x)| dx.

定积分 ∫ₐᵇ f(x) dx 表示曲线 y = f(x) 与 x 轴之间从 x = a 到 x = b 的有符号面积。在应用中，这可以代表累积变化总量，例如由降雨速率函数得到的总降雨量，或电流随时间累积的总电荷量。如果图像穿过 x 轴，积分给出的是净面积，而总面积则是 ∫ₐᵇ |f(x)| dx。

For a region bounded by two curves, area = ∫ₐᵇ (upper curve − lower curve) dx. For example, area between y = x² and y = x from x=0 to x=1 is ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6. Always sketch the region and find intersection points to set limits.

对于两条曲线围成的区域，面积 = ∫ₐᵇ (上曲线 − 下曲线) dx。例如，y = x² 和 y = x 从 x=0 到 x=1 之间的面积为 ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6。务必先画出草图并求出交点以确定积分限。

Total water = ∫ₜ₁ᵗ² r(t) dt, where r(t) is flow rate

---

5. Volumes of Revolution | 旋转体体积

A solid of revolution is created by rotating a curve about an axis. The disk method gives volume V = π ∫ₐᵇ [f(x)]² dx for rotation about the x‑axis, or V = π ∫꜀ᵈ [g(y)]² dy about the y‑axis. When the region is bounded by two curves, use the washer method: V = π ∫ₐᵇ (R² − r²) dx, where R and r are the outer and inner radii.

旋转体是由一条曲线绕某轴旋转而成的立体。圆盘法给出绕 x 轴旋转的体积 V = π ∫ₐᵇ [f(x)]² dx，绕 y 轴旋转则用 V = π ∫꜀ᵈ [g(y)]² dy。当区域由两条曲线围成时，使用圆环法：V = π ∫ₐᵇ (R² − r²) dx，其中 R 和 r 分别为外半径和内半径。

Example: Find the volume when the area between y = √x, x = 0, and y = 2 is revolved about the y‑axis. Rewrite x = y², limits y=0 to 2, so V = π ∫₀² (y²)² dy = π ∫₀² y⁴ dy = π [y⁵/5]₀² = 32π/5. Always consider which variable to integrate with respect to for simplicity.

例子：求 y = √x、x = 0 和 y = 2 围成的区域绕 y 轴旋转的体积。改写为 x = y²，y 从 0 到 2，则 V = π ∫₀² (y²)² dy = π ∫₀² y⁴ dy = π [y⁵/5]₀² = 32π/5。为了简便，始终要考虑对哪个变量积分更合适。

---

6. Exponential Growth and Decay Models | 指数增长与衰减模型

Many natural phenomena follow exponential patterns: population growth, radioactive decay, Newton’s law of cooling. The general form is y(t) = A eᵏᵗ, where A is the initial value and k is the relative growth rate (k > 0 for growth, k < 0 for decay). The differential equation dy/dt = ky says the rate of change is proportional to the amount present, and its solution is the exponential function.

许多自然现象遵循指数模式：种群增长、放射性衰变、牛顿冷却定律等。一般形式为 y(t) = A eᵏᵗ，其中 A 是初始值，k 是相对增长率（k > 0 表示增长，k < 0 表示衰减）。微分方程 dy/dt = ky 表明变化率与当前量成正比，它的解就是指函数。

Half-life questions are common: if a radioactive substance decays so that half remains after 5 years, find k. Using y = A eᵏᵗ, when t=5, A/2 = A e⁵ᵏ → e⁵ᵏ = ½ → 5k = ln(½) → k = −(ln 2)/5. To find the time for 90% decay, set 0.1A = A eᵏᵗ → ln(0.1) = kt → t = ln(0.1)/k. These models require careful algebraic manipulation of exponentials and logarithms.

半衰期问题很常见：如果一种放射性物质衰变，5 年后剩下原来的一半，求 k。利用 y = A eᵏᵗ，当 t=5 时，A/2 = A e⁵ᵏ → e⁵ᵏ = ½ → 5k = ln(½) → k = −(ln 2)/5。要找到衰变 90% 所需时间，设 0.1A = A eᵏᵗ → ln(0.1) = kt → t = ln(0.1)/k。这些模型需要仔细处理指数和对数的代数运算。

---

7. Differential Equations and Modeling | 微分方程与建模

Beyond exponential growth, differential equations model more complex behaviour like logistic growth (dP/dt = rP(1 − P/K)), mixing problems, and motion with resistance. IB students learn to verify solutions, solve separable equations dy/dx = g(x)h(y) by separating variables, and interpret slope fields to sketch solution curves.

除了指数增长，微分方程还可以建立更复杂行为的模型，例如逻辑斯蒂增长 (dP/dt = rP(1 − P/K))、混合问题以及有阻力的运动。IB 学生需要学会验证解、通过分离变量法求解可分离方程 dy/dx = g(x)h(y)，以及解释斜率场来勾画解曲线。

For a separable equation: dy/dx = xy, separate as (1/y) dy = x dx, integrate to get ln|y| = x²/2 + C, then y = k e^(x²/2). Initial conditions fix the constant. In modelling, always define variables clearly and state assumptions. For a pollutant concentration in a lake, we might set dC/dt = input rate − output rate − decay rate, leading to a first‑order linear differential equation.

对于可分离方程：dy/dx = xy，分离得 (1/y) dy = x dx，积分得到 ln|y| = x²/2 + C，然后 y = k e^(x²/2)。利用初始条件确定常数。在建模时，务必清晰定义变量并陈述假设。对于湖泊中的污染物浓度，我们可能会设 dC/dt = 输入速率 − 输出速率 − 降解速率，从而得到一个一阶线性微分方程。

---

8. Cost, Revenue, and Profit Optimization | 成本、收入与利润优化

In economics, calculus helps determine the production level that maximises profit. Given cost function C(x) and revenue R(x) = p(x)·x, profit Π(x) = R(x) − C(x). The marginal cost C'(x) and marginal revenue R'(x) express the cost/revenue of producing one more unit. Profit is maximised when R'(x) = C'(x), i.e., where marginal revenue equals marginal cost, provided Π”(x) < 0.

在经济学中，微积分有助于确定使利润最大化的生产水平。给定成本函数 C(x) 和收入 R(x) = p(x)·x，利润 Π(x) = R(x) − C(x)。边际成本 C'(x) 和边际收入 R'(x) 表示多生产一个单位所带来的成本/收入变化。当 R'(x) = C'(x)，即边际收入等于边际成本时，利润最大化，前提是 Π”(x) < 0。

Example: A company’s cost is C(x) = 0.5x² + 10x + 500 and demand is p = 50 − 0.2x (price per unit). Revenue R(x) = x(50 − 0.2x) = 50x − 0.2x². Profit Π = 50x − 0.2x² − (0.5x² + 10x + 500) = 40x − 0.7x² − 500. Then Π'(x) = 40 − 1.4x = 0 → x ≈ 28.57, and Π”(x) = −1.4 < 0 confirms a maximum. Round to a sensible integer and check endpoint yields.

例子：某公司的成本为 C(x) = 0.5x² + 10x + 500，需求为 p = 50 − 0.2x（单价）。收入 R(x) = x(50 − 0.2x) = 50x − 0.2x²。利润 Π = 50x − 0.2x² − (0.5x² + 10x + 500) = 40x − 0.7x² − 500。则 Π'(x) = 40 − 1.4x = 0 → x ≈ 28.57，Π”(x) = −1.4 < 0 确认为最大值。再取合理的整数并验证端点值。

---

9. Trigonometric Models and Waves | 三角模型与波

Periodic phenomena like tides, sound waves, and seasonal temperatures are modelled by functions of the form f(t) = A sin(B(t−C)) + D (or cosine). Here A is amplitude, period = 2π/|B|, C is horizontal shift, and D is the vertical shift (midline). Questions often involve finding these parameters from real data, evaluating the function at given times, or solving for when the function exceeds a threshold.

周期现象如潮汐、声波和季节温度可以用形如 f(t) = A sin(B(t−C)) + D（或余弦）的函数来建模。这里 A 是振幅，周期 = 2π/|B|，C 是水平位移，D 是垂直位移（中线）。常见问题包括从实际数据求这些参数、在给定时间求函数值，或求解函数何时超过某个阈值。

Given a tide height h(t) = 2.5 sin(π/6 (t − 3)) + 5, amplitude is 2.5 m, period = 2π/(π/6) = 12 hours, midline is 5 m. The maximum height is 5+2.5=7.5 m and minimum is 2.5 m. To find when the tide first reaches 6 m, solve 2.5 sin(π/6 (t−3)) + 5 = 6 → sin(π/6 (t−3)) = 0.4, then use inverse sine and periodicity.

给定潮汐高度 h(t) = 2.5 sin(π/6 (t − 3)) + 5，振幅为 2.5 m，周期 = 2π/(π/6) = 12 小时，中线为 5 m。最高高度为 5+2.5=7.5 m，最低为 2.5 m。要找到潮汐首次到达 6 m 的时刻，解方程 2.5 sin(π/6 (t−3)) + 5 = 6 → sin(π/6 (t−3)) = 0.4，然后利用反正弦和周期性。

---

10. Logarithmic Models (pH, Richter scale) | 对数模型（pH、里氏震级）

Logarithmic functions appear when quantities span several orders of magnitude. The pH of a solution is pH = −log₁₀[H⁺]; the Richter magnitude M is M = log₁₀(I) or a related formula using amplitudes. In IB, you might need to compare intensities: if M = log(I/I₀), an increase of 1 in magnitude means the intensity is multiplied by 10. Similarly, sound pressure level in decibels uses a logarithmic scale.

当量值跨越多个数量级时，就会用到对数函数。溶液的 pH 为 pH = −log₁₀[H⁺]；里氏震级 M 为 M = log₁₀(I) 或使用振幅的相关公式。在 IB 中，你可能需要比较强度：如果 M = log(I/I₀)，那么震级每增加 1，强度就乘以 10。类似地，声压级（分贝）也采用对数刻度。

Sample problem: An earthquake measured 6.2 on the Richter scale, and its aftershock measured 4.2. How many times more intense was the main quake? Intensity ratio I₁/I₂ = 10^(6.2−4.2) = 10² = 100 times. Logarithm properties (log a − log b = log(a/b)) are essential. Always check the definition; sometimes the formula uses (2/3)log₁₀(E) for energy, requiring careful substitution.

示例问题：一次地震的里氏震级为 6.2，余震为 4.2。主震的强度是余震的多少倍？强度比 I₁/I₂ = 10^(6.2−4.2) = 10² = 100 倍。对数性质（log a − log b = log(a/b)）至关重要。务必检查定义；有时公式采用 (2/3)log₁₀(E) 表示能量，需要仔细代入。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）