Growth and Decay | 增长与衰减

📚 Growth and Decay | 增长与衰减

cs curriculum, understanding the differential equation dy/dt = ky and its solution y = A e^kt is essential for solving real-world problems involving growth and decay.

指数增长与衰减是数学中的基本概念，描述了量随时间与其当前值成正比的变化方式。这些模型出现在种群生物学、放射性衰变、金融及许多其他领域。在IB数学课程中，理解微分方程 dy/dt = ky 及其解 y = A e^kt 对于解决涉及增长与衰减的实际问题至关重要。

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1. Introduction to Exponential Models | 指数模型简介

y captures this relationship. The constant k is called the growth rate (positive) or decay rate (negative). The model is memoryless in the sense that the future state depends only on the present value, not the history.

指数模型假设一个量的变化率与其自身成正比。若 y 表示时刻 t 的量，则微分方程 dy/dt = ky 描述了这种关系。常数 k 称为增长率（正）或衰减率（负）。该模型具有无记忆性，即未来状态只依赖于当前值，与历史无关。

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2. Solving the Differential Equation | 解微分方程

Separating variables gives dy/y = k dt. Integrating both sides yields ln|y| = kt + C, where C is the constant of integration. Exponentiating gives |y| = e^{kt + C} = e^C e^kt. Letting A = ±e^C (or simply A = e^C if the initial value is positive) we obtain the general solution.

分离变量得 dy/y = k dt。两边积分得到 ln|y| = kt + C，其中 C 为积分常数。两边指数化得 |y| = e^{kt + C} = e^C e^kt。令 A = ±e^C（若初值为正，可直接取 A = e^C）即得通解。

y = A e^kt

If t = 0, then y(0) = A e⁰ ity. The behaviour depends entirely on the sign of k: growth when k > 0, decay when k < 0. For IB examinations, students must be able to derive this solution and interpret the parameters A and k in context.

若 t = 0，则 y(0) = A e⁰ = A，因此 A 代表初始量。行为完全取决于 k 的符号：k > 0 为增长，k < 0 为衰减。在 IB 考试中，学生必须能够推导此解，并结合实际情境解释参数 A 与 k。

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3. Growth Models: k > 0 | 增长模型：k > 0

When k > 0, the exponential function y = A e^kt increases without bound as t increases. The graph passes through (0, A) and is concave up. A classic example is a population of bacteria that doubles at regular intervals under unlimited resources. The larger the value of k, the steeper the curve.

当 k > 0 时，指数函数 y = A e^kt 随 t 增加而无界增长。图像经过 (0, A) 并呈凹向上形态。一个经典例子是在资源无限条件下，细菌数量每隔固定时间翻倍。k 值越大，曲线越陡峭。

The doubling time T_d is the period required for the quantity to double. By setting y(T_d) = 2A, we have 2A = A e^{k T_d}, which simplifies to e^{k T_d} = 2. Taking natural logs gives k T_d = ln 2, hence T_d = (ln 2) / k. This formula is widely used in biology and finance.

倍增时间 T_d 是量翻倍所需的时间。令 y(T_d) = 2A，则有 2A = A e^{k T_d}，化简得 e^{k T_d} = 2。取自然对数得 k T_d = ln 2，故 T_d = (ln 2) / k。该公式在生物学和金融中应用广泛。

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4. Decay Models: k < 0 | 衰减模型：k < 0

For k < 0, it is convenient to write k = -λ with λ > 0, so y = A e^-λt. The quantity decreases by a fixed proportion per unit time, approaching zero asymptotically. Radioactive substances lose mass according to this law, as does the discharge of a capacitor in physics.

对于 k < 0，通常记 k = -λ，λ > 0，因此 y = A e^-λt。量每单位时间以固定比例减少，逐渐趋近于零，形成水平渐近线。放射性物质的质量依此规律减少，物理学中电容器的放电过程亦然。

The half-life T_1/2 is the time taken for the quantity to fall to half its original value. Setting y(T_1/2) = A/2 gives A/2 = A e^{-λ T_1/2}, leading to e^{-λ T_1/2} = 1/2. Taking logarithms yields -λ T_1/2 = -ln 2, thus T_1/2 = (ln 2) / λ = (ln 2) / |k|.

半衰期 T_1/2 是量减少到初始值一半所需的时间。令 y(T_1/2) = A/2 得 A/2 = A e^{-λ T_1/2}，则 e^{-λ T_1/2} = 1/2。取对数得 -λ T_1/2 = -ln 2，故 T_1/2 = (ln 2) / λ = (ln 2) / |k|。

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5. Half-life and Doubling Time – Key Relationships | 半衰期与倍增时间 – 关键关系

The two key time intervals are summarised below. Note the structural symmetry: both formulas involve ln 2 divided by the magnitude of the rate constant.

两个关键时间间隔总结如下。注意结构对称性：两个公式均涉及 ln 2 除以速率常数的绝对值。

Doubling time T_d = ln 2 / k ,   Half-life T_1/2 = ln 2 / |k|

vice versa. For example, if a population doubles in 5 hours, then k = ln 2 / 5 ≈ 0.1386 hr^-1.

在处理真实数据问题时，学生常常由给定的半衰期或倍增时间求 k，或反向计算。例如，若某族群 5 小时翻倍，则 k = ln 2 / 5 ≈ 0.1386 hr^-1。

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6. Continuous Compound Interest | 连续复利

s, when the principal P earns an annual nominal rate r compounded continuously, is A = P e^rt. Here r corresponds to the growth constant k.

指数增长最实际的应用之一是金融中的连续复利。当本金 P 以名义年利率 r 连续复利时，t 年后账户总额 A(t) 为 A = P e^rt。此处 r 对应于增长常数 k。

limit of discrete compounding. If interest is compounded n times per year, A = P (1 + r/n)^nt. As n → ∞, this expression approaches P e^rt. The effective annual rate is e^r - 1.

该公式来自离散复利的极限。若每年复利 n 次，A = P (1 + r/n)^nt。当 n → ∞ 时，该表达式趋近于 P e^rt。有效年利率为 e^r - 1。

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7. Newton’s Law of Cooling | 牛顿冷却定律

Newton’s law of cooling describes how a hot object cools in a constant-temperature environment. The rate of change of temperature T is proportional to the difference between T and the ambient temperature T_env: dT/dt = -k (T - T_env), with k > 0.

牛顿冷却定律描述了热物体在恒温环境中的冷却过程。温度 T 的变化率与 T 和环境温度 T_env 之差成正比：dT/dt = -k (T - T_env)，其中 k > 0。

ial equation can be solved by substitution or integrating factor. The solution is T(t) = T_env + (T₀ - T_env) e^-kt, where T₀ is the initial temperature. This shows exponential decay of the temperature difference towards zero.

该微分方程可通过换元或积分因子法求解。解为 T(t) = T_env + (T₀ - T_env) e^-kt，其中 T₀ 是初始温度。这表明温度差按指数衰减至零。

A typical IB problem gives temperatures at two different times and asks to find k or predict future values. The process always involves forming two equations to eliminate T_env or using given ambient temperature directly.

典型的 IB 问题会给出两个时刻的温度，要求计算 k 或预测未来值。解题过程往往包括建立两个方程以消去 T_env，或者直接利用已知环境温度。

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8. Logistic Growth – Bounded Populations | 逻辑斯谛增长 – 有限制的种群模型

When P is small, growth is roughly exponential; as P approaches K, growth slows and halts.

现实中资源有限。逻辑斯谛模型通过引入环境承载力 K 修正了指数增长。微分方程为 dP/dt = r P (1 - P/K)。当 P 很小时，增长近似于指数型；当 P 趋近 K 时，增长减缓并停滞。

The solution is an S-shaped curve given by P(t) = K / (1 + ((K - P₀) / P₀) e^-rt /2, where the growth rate is maximum. This model is particularly relevant in biology and ecology, and may appear in HL AA or AI examinations.

其解为 S 形曲线 P(t) = K / (1 + ((K - P₀) / P₀) e^-rt)。拐点出现在 P = K/2 处，此时增长率最大。该模型在生物学和生态学中尤为重要，可能出现在 HL AA 或 AI 考试中。

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9. Graphical Interpretation and Transformations | 图形解释与变换

Understanding the graphs of exponential functions is crucial. For y = A e^kt Changing A scales the graph vertically, while changing k stretches or compresses it horizontally.

理解指数函数的图像至关重要。对于 y = A e^kt，y 轴截距为 A。若 k > 0，曲线急剧上升；若 k < 0，曲线向水平渐近线 y = 0 衰减。改变 A 会纵向拉伸图形，改变 k 则水平伸缩图形。

IB students often encounter questions on transforming y = e^x: for example, y = 2 e^-3x + 5 involves a vertical stretch by factor 2, reflection across the y-axis? No, it is horizontal compression and reflection across y-axis? Actually e^-3x sion. The translation +5 shifts the asymptote to y = 5. Such interpretations are standard.

IB 学生常遇到关于变换 y = e^x 的问题：例如 y = 2 e^-3x + 5 包含了纵向拉伸因子 2、关于 y 轴反射及水平压缩，再加平移 +5 使渐近线变为 y = 5。此类解释属于常规考查内容。

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10. Solving Real-World Problems – Strategies | 求解实际问题 – 策略

For example, given y(0) = 100 and y(3) = 250, find A and k. Substitute to get 250 = 100 e^3k, then solve for k = (1/3) ln(2.5).

典型的 IB 问题会提供两个数据点或某一速率，要求求出模型参数 A 和 k。可采用联立方程或对数变换。例如，已知 y(0) = 100 和 y(3) = 250，求 A 和 k。代入得 250 = 100 e^3k，解得 k = (1/3) ln(2.5)。

Another common task is to predict future values or past values using the established model. Students must be adept at rearranging exponential expressions, applying natural logs, and correctly interpreting the physical domain (e.g., time cannot be negative in some contexts).

另一常见任务是运用已建立的模型预测未来或回溯过去的值。学生必须熟练重排指数式、

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