📚 A-Level Mathematics: Exponential Models and Their Applications | A-Level 数学：指数模型及其应用
Exponential models describe how quantities grow or decay at a rate proportional to their current size. They appear throughout A-Level Mathematics, from pure algebra to real-world applications such as radioactive decay, population growth and finance. This article explains the key ideas, gives worked examples, and highlights common pitfalls.
指数模型描述数量以与其当前大小成正比的速率增长或衰减。它们贯穿 A-Level 数学，从纯代数到放射性衰变、人口增长和金融等实际应用。本文将解释关键概念，给出例题，并指出常见错误。
1. The Exponential Function | 指数函数
An exponential function has the form f(x) = aˣ tive constant and a ≠ 1. When a > 1, the function increases; when 0 < a < 1, it decreases.
指数函数的形式为 f(x) = aˣ，其中底数 a 是正数且 a ≠ 1。当 a > 1 时，函数递增；当 0 < a < 1 时，函数递减。
limit e = limₙ→∞ (1 + 1/n)ⁿ and has the remarkable property that its derivative is itself.
最重要的底数是自然数 e ≈ 2.71828。它来自极限 e = limₙ→∞ (1 + 1/n)ⁿ，并具有导数为自身的特殊性质。
d/dx(eˣ) = eˣ
Also e⁰ = 1, and the exponent laws hold: eˣ⁺ʸ = eˣeʸ, e⁻ˣ = 1/eˣ, and (eˣ)ⁿ = eⁿˣ.
此外 e⁰ = 1，指数法则成立：eˣ⁺ʸ = eˣeʸ、e⁻ˣ = 1/eˣ、(eˣ)ⁿ = eⁿˣ。
2. The General Exponential Model | 一般指数模型
ity at t = 0, t is time, and k is the growth constant if k > 0 or decay constant if k < 0.
一般指数模型常写作 y = A₀aᵏᵗ 或等价地 y = A₀eᵏᵗ，其中 A₀ 是 t = 0 时的初始量，t 是时间，k 是增长常数（k > 0）或衰减常数（k < 0）。
by the identity aᵏᵗ = eᵏᵗ·ln a. For example, y = 100 × 2ᵗ can be written as y = 100eᵗ·ln2.
这两种形式通过恒等式 aᵏᵗ = eᵏᵗ·ln a 联系。例如，y = 100 × 2ᵗ 可写成 y = 100eᵗ·ln2。
Here the constant k = ln a. If a > 1, then k > 0; if 0 < a < 1, then k < 0.
这里 k = ln a。若 a > 1，则 k > 0；若 0 < a < 1，则 k < 0。
3. Exponential Growth | 指数增长
eases by the same percentage over equal time intervals. The model is y = A₀eᵏᵗ with k > 0.
当数量在相等的时间间隔内增加相同百分比时，发生指数增长。模型为 y = A₀eᵏᵗ，其中 k > 0。
r. After t hours, the population is y = 100 × 2ᵗ = 100eᵗ·ln2, so k = ln2 ≈ 0.693 per hour.
例如，假设细菌种群从 100 个开始，每小时翻倍。t 小时后，种群为 y = 100 × 2ᵗ = 100eᵗ·ln2，所以 k = ln2 ≈ 0.693 每小时。
The doubling time T is the time needed to double the initial quantity. Setting A₀eᵏᵀ = 2A₀ gives eᵏᵀ = 2, hence T = ln2/k. For the bacteria, T = 1 hour.
倍增时间 T 是使初始量翻倍所需的时间。令 A₀eᵏᵀ = 2A₀，得 eᵏᵀ = 2，故 T = ln2/k。对细菌而言，T = 1 小时。
4. Exponential Decay | 指数衰减
= A₀e⁻ᵏᵗ, where k > 0 is the decay constant and the negative exponent causes the decrease.
当数量每单位时间减少固定百分比时，发生指数衰减。其模型为 y = A₀e⁻ᵏᵗ，其中 k > 0 是衰减常数，负指数导致减少。
isotope with half-life T½, then after time t the mass is m(t) = m₀e⁻ᵏᵗ, where k = ln2/T½.
放射性衰变是经典例子。若样品最初含有 m₀ 克放射性同位素，半衰期为 T½，则经过时间 t 后质量为 m(t) = m₀e⁻ᵏᵗ，其中 k = ln2/T½。
The half-life is defined by m(T½) = m₀/2. Substituting gives m₀e⁻ᵏᵀ½ = m₀/2, so e⁻ᵏᵀ½ = 1/2. Thus T½ = ln2/k. This formula is frequently tested in exams.
半衰期定义为 m(T½) = m₀/2。代入得 m₀e⁻ᵏᵀ½ = m₀/2，所以 e⁻ᵏᵀ½ = 1/2。因此 T½ = ln2/k。这个公式在考试中经常出现。
5. Linearising Exponential Data | 指数数据的线性化
. This shows that ln y is a linear function of t, with slope k and vertical intercept ln A₀.
指数关系可以通过取对数转化为线性关系。若 y = A₀eᵏᵗ，则 ln y = ln A₀ + kt。这表明 ln y 是 t 的线性函数，斜率为 k，纵截距为 ln A₀。
In practice, if you are given data that appears exponential, plot ln y against t. If the plot is approximately a straight line, then an exponential model is appropriate.
实践中，如果给定数据看起来是指数形式，可以绘制 ln y 对 t 的图像。如果图像近似为直线，则指数模型适用。
Similarly, for a model of the form y = abˣ, taking natural logs gives ln y = ln a + x ln b, which is linear in x.
类似地，对于形如 y = abˣ 的模型，取自然对数得 ln y = ln a + x ln b，这是关于 x 的线性式。
6. Modelling with Exponential Functions | 用指数函数建模
Exponential models are used in many fields. Common examples include population growth, radioactive decay, Newton's law of cooling, medical drug dosage, and compound interest.
指数模型广泛应用于许多领域。常见例子包括人口增长、放射性衰变、牛顿冷却定律、药物剂量和复利。
When building an exponential model, follow these steps:
建立指数模型时，可遵循以下步骤：
Identify the initial value A₀ at t = 0.
确定 t = 0 时的初始值 A₀。
Determine whether the situation involves growth or decay.
判断情况涉及增长还是衰减。
nown conditions to find k. For example, if the quantity doubles in 3 hours, solve 2 = e³ᵏ.
利用已知条件求 k。例如，如果数量在 3 小时内翻倍，则解 2 = e³ᵏ。
Write the model in the form y = A₀eᵏᵗ or y = A₀e⁻ᵏᵗ and use it to predict other values.
将模型写成 y = A₀eᵏᵗ 或 y = A₀e⁻ᵏᵗ 形式，并用于预测其他值。
7. Continuous Compound Interest | 连续复利
nnual rate r compounded n times per year, then after t years the amount is A = P₀(1 + r/n)nt.
金融数学提供了一个直接应用。若本金为 P₀，年利率为 r，每年复利 n 次，则 t 年后金额为 A = P₀(1 + r/n)nt。
As n becomes very large, this expression approaches A = P₀eʳᵗ. This is known as continuous compounding.
当 n 变得非常大时，该表达式趋近于 A = P₀eʳᵗ。这称为连续复利。
For example, £1000 invested at 5% compounded continuously for 3 years gives A = 1000e0.05 × 3 = 1000e⁰·¹⁵ ≈ £1161.83.
例如，1000 英镑以 5% 连续复利投资 3 年，得到 A = 1000e0.05 × 3 = 1000e⁰·¹
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