📚 Example 3.3.2: Finding the Equation of a Tangent Using Differentiation | 例题 3.3.2：用微分求切线方程
applies to any differentiable function and is regularly tested in AQA Paper 1 and Paper 2.
在这篇 AQA A-level 纯数学复习笔记中，我们将完整讲解例题 3.3.2：求曲线 y = x³ - 4x² + 5 在 x = 2 处的切线方程。同样的方法适用于任何可导函数，也是 AQA 试卷 1 和试卷 2 中的常见考点。
1. The Problem Statement | 问题陈述
need to find both the gradient of the tangent and the coordinates of the point of contact.
例题 3.3.2 要求我们求出曲线在 x = 2 处的切线方程。曲线方程为 y = x³ - 4x² + 5。我们需要同时求出切线斜率以及切点坐标。
y = x³ - 4x² + 5
This is an application of differentiation to coordinate geometry: the derivative gives the gradient of the tangent at any point on the curve.
这是微分在坐标几何中的应用：导数给出曲线上任意一点处切线的斜率。
2. The Derivative as a Gradient Function | 导数作为斜率函数
is a function of x, so its value changes from point to point. At a specific value x = a, f'(a) is the gradient of the curve at that point.
对于任意光滑曲线 y = f(x)，导数 dy/dx 是一般点处切线的斜率。它是关于 x 的函数，因此其值随点而变化。在特定值 x = a 处，f'(a) 就是曲线在该点的斜率。
dy/dx = f'(x)
n function. We will then substitute x = 2 to obtain the numerical gradient of the tangent.
为了求切线方程，我们首先对给定函数求导。然后将 x = 2 代入，得到切线的具体斜率值。
3. Step 1: Find the Point on the Curve | 步骤 1：求曲线上点的坐标
x = 2, but we also need the y-coordinate. We substitute x = 2 into the original equation.
切线只在唯一一点处与曲线接触，因此我们必须找出该点。已知 x = 2，但我们还需要 y 坐标。将 x = 2 代入原方程即可。
y = (2)³ - 4(2)² + 5 = 8 - 16 + 5 = -3
Therefore, the point of contact is (2, -3). This coordinate pair will be used in the straight-line equation later.
因此，切点坐标为 (2, -3)。这个坐标将在后面的直线方程中使用。
4. Step 2: Differentiate the Function | 步骤 2：对函数求导
We differentiate term by term using the standard power rule. For a term axⁿ, the derivative is naxⁿ⁻¹. The derivative of a constant is zero.
我们使用标准幂函数求导法则逐项求导。对于项 axⁿ，其导数为 naxⁿ⁻¹。常数的导数为零。
dy/dx = 3x² - 8x
- The derivative of x³ is 3x².
- The derivative of -4x² is -8x.
- The derivative of 5 is 0.
This derivative function gives the gradient of the tangent at any chosen x-coordinate.
这个导数函数给出了任意选定 x 坐标处切线的斜率。
5. Step 3: Evaluate the Gradient at x = 2 | 步骤 3：计算 x = 2 处的斜率
tive to find the gradient of the tangent at the point of contact. This value is often denoted by m.
现在我们将 x = 2 代入导数，求出切点处切线的斜率。该值通常用 m 表示。
m = 3(2)² - 8(2) = 12 - 16 = -4
eans the curve is decreasing at x = 2, so the tangent slopes downwards from left to right.
因此切线的斜率为 -4。这个负值说明曲线在 x = 2 处是下降的，所以切线从左到右向下倾斜。
6. Step 4: Use the Straight-Line Equation | 步骤 4：使用直线方程
A tangent is a straight line, so we can apply the point-gradient form of the equation of a line. If a line has gradient m and passes through the point (x₁, y₁), its equation is:
切线是一条直线，因此我们可以使用直线的点斜式方程。如果一条直线的斜率为 m，且经过点 (x₁, y₁)，则其方程为：
y - y₁ = m(x - x₁)
int is (2, -3). Substituting these values gives the tangent equation in unsimplified form.
我们已知 m = -4，点为 (2, -3)。代入这些值，得到未化简的切线方程。
y - (-3) = -4(x - 2)
This step is straightforward but must be written clearly to avoid sign errors.
这一步很直接，但必须书写清晰，以避免符号错误。
7. Step 5: Simplify and State the Final Equation | 步骤 5：化简并写出最终方程
fy the equation so that it is in the standard form y = mx + c, where c is the y-intercept.
现在我们将方程化简为 y = mx + c 的标准形式，其中 c 为 y 轴截距。
y + 3 = -4x + 8
y = -4x + 5
nt to y = x³ - 4x² + 5 at x = 2 is y = -4x + 5. This is the final answer to Example 3.3.2.
因此，曲线 y = x³ - 4x² + 5 在 x = 2 处的切线方程为 y = -4x + 5。这就是例题 3.3.2 的最终答案。
8. Normals: The Perpendicular Line | 法线：垂直线
A normal is the line perpendicular to the tangent at the point of contact. If the tangent gradient is m₁, the normal gradient m₂ is the negative reciprocal of m₁.
法线是在切点处与切线垂直的直线。如果切线斜率为 m₁，则法线斜率 m₂ 是 m₁ 的负倒数。
m₂ = -1 / m₁
This relationship is extremely common in AQA questions. Students are often asked to find either the normal or the tangent, so both processes must be familiar.
这一关系在 AQA 试题中非常常见。学生经常被要求求法线或切线，因此两种过程都必须熟悉。
9. Worked Example: Normal to a Trigonometric Curve | 例题：三角曲线的法线
quation of the normal to y = sin x at x = π/4. First, we find the y-coordinate: sin(π/4) = √2/2. Next, differentiate: dy/dx = cos x. The tangent gradient at x = π/4 is cos(π/4) = √2/2.
求 y = sin x 在 x = π/4 处的法线方程。首先，求 y 坐标：sin(π/4) = √2/2。接着求导：dy/dx = cos x。在 x = π/4 处的切线斜率为 cos(π/4) = √2/2。
m_tangent = √2/2
The normal gradient is therefore the negative reciprocal: -1 ÷ (√2/2) = -√2. Using the point (π/4, √2/2) and the point-gradient form, we obtain the normal equation.
因此法线斜率为负倒数：-1 ÷ (√2/2) = -√2。利用点 (π/4, √2/2) 和点斜式方程，我们得到法线方程。
y - √2/2 = -√2(x - π/4)
This can be left unsimplified unless the question asks for a particular form. The key idea is to use the correct perpendicular gradient.
除非题目要求特定形式，否则可以保留未化简的形式。关键是要使用正确的垂直斜率。
10. AQA Exam-Style Practice | AQA 考试风格练习
quation y = x² - 4x + 7. Find the equation of the normal to the curve at the point (3, 4).
一条曲线方程为 y = x² - 4x + 7。求曲线在点 (3, 4) 处的法线方程。
t gradient is 2(3) - 4 = 2. The normal gradient is therefore -1/2. Using the point (3, 4):
解法：求导得 dy/dx = 2x - 4。在 x = 3 处，切线斜率为 2(3) - 4 = 2。因此法线斜率为 -1/2。利用点 (3, 4)：
y - 4 = -1/2(x - 3)
y = -x/2 + 11/2
This practice question mirrors the structure of many AQA coordinate geometry and differentiation problems.
这道练习题反映了许多 AQA 坐标几何与微分题目的结构。
11. Common Pitfalls | 常见错误
- Forgetting to find the y-coordinate before forming the line equation. The original function must be used, not the derivative.
- Using the tangent gradient when the question asks for the normal. Always check which line is required.
- Sign errors when simplifying y - y₁ = m(x - x₁). Work slowly with brackets.
- Differentiating incorrectly by applying the power rule in the wrong order or forgetting that the derivative of a constant is zero.
These errors can cost marks even when the main method is correct. Careful layout and checking the question wording are essential.
这些错误即使在主要方法正确的情况下也会失分。仔细书写步骤并检查题目要求至关重要。
12. Summary and Key Takeaways | 小结与关键要点
e given x-value to find m, and use y - y₁ = m(x - x₁). Simplify to y = mx + c if required.
要求切线方程，首先利用原方程求出切点。然后求导得到斜率函数，代入给定的 x 值求出 m，并使用 y - y₁ = m(x - x₁)。如有需要，化简为 y = mx + c。
For a normal, take the negative reciprocal of the tangent gradient. These steps apply to polynomial, trigonometric, exponential and logarithmic functions, making this one of the most transferable skills in AQA A-level Mathematics.
对于法线，取切线斜率的负倒数。这些步骤适用于多项式、三角函数、指数函数和对数函数，是 AQA A-level 数学中最具迁移性的技能之一。
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