IB Mathematics: Exponential Growth and Decay Models | IB数学：指数增长与衰减模型

📚 IB Mathematics: Exponential Growth and Decay Models | IB数学：指数增长与衰减模型

Exponential growth and decay models describe quantities that change at a rate proportional to their current size. These models appear throughout IB Mathematics, from the natural sciences to finance, and mastering them is essential for Paper 1 and Paper 2 questions.

指数增长与衰减模型描述的是数量以与其当前大小成正比的速率进行变化的过程。这些模型贯穿IB数学，从自然科学到金融领域，掌握它们对Paper 1和Paper 2的解题至关重要。

1. The Exponential Function | 指数函数

The general exponential model is written as \(y(t) = y_0 e^{kt}\), where \(y_0\) is the initial value at \(t=0\), \(e\) is Euler’s number (\(e \approx 2.718\)), and \(k\) is a constant. If \(k > 0\), the model represents growth; if \(k < 0\), it represents decay.

一般指数模型写作 \(y(t) = y_0 e^{kt}\)，其中 \(y_0\) 是 \(t=0\) 时的初始值，\(e\) 是欧拉数（\(e \approx 2.718\)），\(k\) 是常数。若 \(k > 0\)，该模型表示增长；若 \(k < 0\)，则表示衰减。

The base \(e\) is used because it simplifies the differential calculus of the model. However, the same model can be expressed with any positive base, such as \(y(t) = y_0 a^{bt}\), which is equivalent to \(e^{bt \ln a}\).

使用底数 \(e\) 是因为它简化了模型的微积分运算。然而，同一模型也可以用任何正数为底表示，如 \(y(t) = y_0 a^{bt}\)，这等价于 \(e^{bt \ln a}\)。

2. The Differential Equation Model | 微分方程模型

A quantity \(y\) that grows or decays exponentially satisfies the first-order differential equation \(\frac{dy}{dt} = ky\). This equation states that the instantaneous rate of change of \(y\) is proportional to \(y\) itself, with proportionality constant \(k\).

一个呈指数增长或衰减的量 \(y\) 满足一阶微分方程 \(\frac{dy}{dt} = ky\)。该方程表明 \(y\) 的瞬时变化率与 \(y\) 自身成正比，比例常数为 \(k\)。

Solving this equation using separation of variables gives:

通过分离变量法解此方程，可得：

y = y₀ e^kt

where \(y_0 = y(0)\) is the initial condition. This solution can be verified by differentiating: \(\frac{dy}{dt} = y_0 k e^{kt} = k y\).

其中 \(y_0 = y(0)\) 是初始条件。此解可通过求导验证：\(\frac{dy}{dt} = y_0 k e^{kt} = k y\)。

3. Doubling Time and Half-Life | 倍增时间与半衰期

In exponential growth, the doubling time \(T_d\) is the time required for a quantity to double. Setting \(y = 2y_0\) gives \(2 = e^{kT_d}\), so \(T_d = \frac{\ln 2}{k}\) for \(k > 0\).

在指数增长中，倍增时间 \(T_d\) 是数量翻倍所需的时间。令 \(y = 2y_0\)，得 \(2 = e^{kT_d}\)，因此 \(T_d = \frac{\ln 2}{k}\)（\(k > 0\)）。

In exponential decay, the half-life \(T_{1/2}\) is the time required for half of the original quantity to remain. Setting \(y = \frac{1}{2}y_0\) gives \(\frac{1}{2} = e^{kT_{1/2}}\), so \(T_{1/2} = \frac{\ln 2}{-k}\) for \(k < 0\).

在指数衰减中，半衰期 \(T_{1/2}\) 是原始数量剩下一半所需的时间。令 \(y = \frac{1}{2}y_0\)，得 \(\frac{1}{2} = e^{kT_{1/2}}\)，因此 \(T_{1/2} = \frac{\ln 2}{-k}\)（\(k < 0\)）。

Quantity	Formula	Condition
Doubling time	T_d = ln 2 / k	k > 0
Half-life	T_1/2 = ln 2 / (-k)	k < 0

4. Applications of Exponential Growth | 指数增长的应用

Exponential growth models are widely used in population biology. When a population has unlimited resources, its size can be modelled by \(N(t) = N_0 e^{rt}\), where \(r\) is the intrinsic growth rate.

指数增长模型在种群生物学中被广泛使用。当种群资源无限时，其数量可用 \(N(t) = N_0 e^{rt}\) 建模，其中 \(r\) 是内禀增长率。

In finance, compound interest follows \(A = P(1 + \frac{r}{n})^{nt}\), which approaches \(A = Pe^{rt}\) as the compounding frequency \(n\) tends to infinity (continuous compounding).

在金融学中，复利公式为 \(A = P(1 + \frac{r}{n})^{nt}\)，当计息频率 \(n\) 趋于无穷大（连续复利）时，它趋近于 \(A = Pe^{rt}\)。

Other examples include the spread of viral infections in the early growth phase and the growth of certain investments. In IB problems, always identify which variable is “growing” and whether the growth rate is given as a percentage per unit time.

其他例子包括病毒感染在早期增长阶段的传播以及某些投资组合的增长。在IB题目中，务必确定哪个变量在“增长”，以及增长率是否以每单位时间的百分比形式给出。

5. Applications of Exponential Decay | 指数衰减的应用

Radioactive decay is the classic exponential decay example. The mass of a radioactive isotope follows \(m(t) = m_0 e^{-\lambda t}\), where \(\lambda\) is the decay constant, and the half-life is \(T_{1/2} = \frac{\ln 2}{\lambda}\).

放射性衰变是指数衰减的经典例子。放射性同位素的质量遵循 \(m(t) = m_0 e^{-\lambda t}\)，其中 \(\lambda\) 是衰变常数，半衰期为 \(T_{1/2} = \frac{\ln 2}{\lambda}\)。

In pharmacology, drug concentration in the bloodstream often decays exponentially, \(C(t) = C_0 e^{-kt}\). Doctors use the half-life to determine dosing intervals.

在药理学中，血液中的药物浓度通常呈指数衰减，\(C(t) = C_0 e^{-kt}\)。医生利用半衰期来确定给药间隔。

Newton’s law of cooling states that the temperature difference between an object and its surroundings decays exponentially. If \(T(t) – T_s = (T_0 – T_s)e^{-kt}\), the object cools at a rate proportional to the difference.

牛顿冷却定律指出，物体与环境之间的温差呈指数衰减。若 \(T(t) – T_s = (T_0 – T_s)e^{-kt}\)，物体以与该温差成正比的速率冷却。

6. Graphs and Key Properties | 图像与关键性质

The graph of \(y = y_0 e^{kt}\) has the following properties:

函数 \(y = y_0 e^{kt}\) 的图像具有以下性质：

It passes through the point \((0, y_0)\) on the y-axis.
It passes through the point \((0, y_0)\)，即与y轴交于该点。
The x-axis (\(y = 0\)) is a horizontal asymptote, but the function never actually reaches zero.
x轴（\(y = 0\)）是水平渐近线，但函数永远无法真正达到零。
If \(k > 0\), the graph increases at an increasing rate (convex upward).
若 \(k > 0\)，图像以越来越快的速率上升（向上凸）。
If \(k < 0\), the graph decreases at a decreasing rate, approaching the asymptote from above.
若 \(k < 0\)，图像以越来越慢的速率下降，从上方接近渐近线。

For initial value \(y_0 > 0\), the function is always positive. For \(y_0 < 0\), the graph is reflected across the x-axis.

当初始值 \(y_0 > 0\) 时，函数值恒为正。当 \(y_0 < 0\) 时，图像关于x轴对称。

7. Transformations and Parameters | 变换与参数

Changing the parameter \(y_0\) scales the graph vertically. If \(y_0\) is larger, the graph starts higher but the growth or decay rate \(k\) remains unchanged.

改变参数 \(y_0\) 会对图像进行垂直缩放。若 \(y_0\) 更大，图像起点更高，但增长或衰减速率 \(k\) 保持不变。

Changing \(k\) alters the steepness of the curve. A larger positive \(k\) gives faster growth; a more negative \(k\) gives faster decay. In the equation \(y = y_0 a^{bx}\), the continuous rate \(k\) equals \(b \ln a\), allowing conversion between different bases.

改变 \(k\) 会改变曲线的陡峭程度。正 \(k\) 越大增长越快；负 \(k\) 绝对值越大衰减越快。在方程 \(y = y_0 a^{bx}\) 中，连续变化率 \(k\) 等于 \(b \ln a\)，这允许在不同底数之间进行转换。

Horizontal shifts correspond to changing the initial time. The model \(y(t) = y_0 e^{k(t-t_0)}\) represents the same shape but shifted so that \(y = y_0\) at \(t = t_0\).

水平位移对应改变初始时间。模型 \(y(t) = y_0 e^{k(t-t_0)}\) 表示相同的形状，但平移使得在 \(t = t_0\) 时 \(y = y_0\)。

8. Linearization and Parameter Estimation | 线性化与参数估计

Given experimental data, we can estimate \(y_0\) and \(k\) by taking the natural logarithm of both sides in the model \(y = y_0 e^{kt}\):

给定实验数据，我们可以通过对模型 \(y = y_0 e^{kt}\) 两边取自然对数来估计 \(y_0\) 和 \(k\)：

ln y = ln y₀ + kt

This is the equation of a straight line with slope \(k\) and intercept \(\ln y_0\). Plotting \(\ln y\) against \(t\) should produce a scatter plot that is approximately linear if the exponential model is appropriate.

这是一条直线的方程，斜率为 \(k\)，截距为 \(\ln y_0\)。如果指数模型适用，绘制 \(\ln y\) 对 \(t\) 的散点图应当近似为一条直线。

In IB questions, you may be given a set of points and asked to find the line of best fit for \(\ln y\) vs \(t\). Use your GDC to compute the linear regression, then convert back to the exponential form by taking \(y_0 = e^{\text{intercept}}\).

在IB问题中，你可能会得到一组数据点，并要求你求出 \(\ln y\) 对 \(t\) 的拟合直线。使用图形计算器（GDC）进行线性回归，然后通过 \(y_0 = e^{\text{截距}}\) 将其转换回指数形式。

9. Common Pitfalls and IB Exam Tips | 常见陷阱与IB考试提示

Confusing exponential with polynomial: Exponential functions grow much faster than any polynomial. Do not apply linear or quadratic regression when the data shows a constant percentage rate of change.

混淆指数与多项式： 指数函数比任何多项式增长得快得多。当数据显示恒定百分比变化率时，不要应用线性或二次回归。

Misinterpreting the growth rate: If the growth rate is given as \(r\%\) per unit time, then \(k = r/100\) only for continuous exponential models. For discrete models, the factor is \(1 + r/100\) per step.

误解增长率： 若增长率以每单位时间的 \(r\%\) 给出，只有在连续指数模型中才有 \(k = r/100\)。对于离散模型，每一步的因子为 \(1 + r/100\)。

Forgetting units: Always state units for half-life, doubling time, and rates. In IB exams, marks are often awarded for correct units in final answers.

忽略单位： 始终写明半衰期、倍增时间和速率的单位。在IB考试中，最终答案的正确单位通常会得分。

Not checking the definition of \(t\): Some questions define \(t\) as the number of hours, days, or years. Adjust the equation accordingly, and use the exact time specified in the problem.

未检查 \(t\) 的定义： 有些题目将 \(t\) 定义为小时、天或年。请相应调整方程，并使用问题中指定的确切时间。

10. Worked Example: Population Growth | 例题：种群增长

A bacteria population doubles every 3 hours. If the initial population is 500, find the population after 8 hours.

某种细菌每3小时翻倍。若初始种群为500，求8小时后的种群数量。

First, find the continuous growth rate \(k\). Since doubling time \(T_d = \frac{\ln 2}{k} = 3\), we have \(k = \frac{\ln 2}{3} \approx 0.231\) per hour.

首先求连续增长速率 \(k\)。由倍增时间 \(T_d = \frac{\ln 2}{k} = 3\)，得 \(k = \frac{\ln 2}{3} \approx 0.231\) 每小时。

Using \(P(t) = 500 e^{kt}\), at \(t = 8\):

利用 \(P(t) = 500 e^{kt}\)，在 \(t = 8\) 时：

P(8) = 500 e^{0.231 × 8} ≈ 500 e^{1.848} ≈ 3174

Thus the population after 8 hours is approximately 3174 bacteria.

因此8小时后种群数量约为3174个细菌。

11. Worked Example: Radioactive Decay | 例题：放射性衰变

The half-life of carbon-14 is 5730 years. A sample initially contains 100 g of carbon-14. Determine the mass remaining after 1000 years and the time for it to reduce to 10 g.

碳-14的半衰期为5730年。某样品最初含有100 g碳-14。求1000年后剩余的质量，以及减少到10 g所需的时间。

First find the decay constant \(\lambda\) from \(\frac{\ln 2}{\lambda} = 5730\), so \(\lambda = \frac{\ln 2}{5730} \approx 1.209 \times 10^{-4}\) per year.

首先由 \(\frac{\ln 2}{\lambda} = 5730\) 求衰变常数 \(\lambda\)，得 \(\lambda = \frac{\ln 2}{5730} \approx 1.209 \times 10^{-4}\) 每年。

After 1000 years, \(m(1000) = 100 e^{-0.0001209 \times 1000} = 100 e^{-0.1209} \approx 88.6\) g.

1000年后，\(m(1000) = 100 e^{-0.0001209 \times 1000} = 100 e^{-0.1209} \approx 88.6\) g。

For \(m = 10\): \(10 = 100 e^{-\lambda t}\), so \(0.1 = e^{-\lambda t}\), giving \(t = \frac{\ln 10}{\lambda} \approx \frac{2.3026}{0.0001209} \approx 19030\) years.

对于 \(m = 10\)：\(10 = 100 e^{-\lambda t}\)，即 \(0.1 = e^{-\lambda t}\)，得 \(t = \frac{\ln 10}{\lambda} \approx \frac{2.3026}{0.0001209} \approx 19030\) 年。

12. Summary and Final Advice | 总结与最终建议

Exponential growth and decay models are powerful tools for describing real-world phenomena where the rate of change is proportional to the quantity. Master the differential equation, the standard solution \(y = y_0 e^{kt}\), and the relationships involving doubling time and half-life.

指数增长与衰减模型是描述“变化率与数量成正比”的现实世界现象的强有力工具。掌握微分方程、标准解 \(y = y_0 e^{kt}\) 以及倍增时间和半衰期的关系。

Be comfortable with converting between exponential forms, using logarithms to solve for exponents, and interpreting graphs. Practice using your GDC to calculate \(e^{x}\) and natural logarithms quickly and accurately.

要熟练地进行指数形式之间的转换，使用对数求解指数，以及解读图像。练习使用图形计算器快速准确地计算 \(e^{x}\) 和自然对数。

Finally, pay attention to the wording of IB questions: identify whether the given rate is continuous or discrete, choose the correct variable names, and always show your working to earn method marks.

最后，注意IB题目的表述：判断给定的速率是连续还是离散的，选择正确的变量名，并始终展示你的解题过程以获得方法分。

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IB Mathematics: Exponential Growth and Decay Models | IB数学：指数增长与衰减模型