📚 Worked Example 7.3.1: Tangent and Normal to a Curve | 例题 7.3.1：曲线的切线与法线
In this AQA A Level Mathematics worked example, we focus on a classic application of differentiation: finding the equations of the tangent and normal to a curve at a given point. The example follows the style of Example 7.3.1 and tests the chain of skills required in Paper 1 Pure Mathematics.
这篇 AQA A Level 数学例题解析聚焦于微分的经典应用：求曲线在某一点的切线和法线方程。该例题采用 Example 7.3.1 的风格，考查 Pure Mathematics Paper 1 所需的一连串技能。
1. Restating the Problem | 题目重述
- 3x + 2. Find the equations of the tangent and the normal to C at the point where x = 2.
曲线 C 的方程为 y = x³ - 3x + 2。求曲线 C 在 x = 2 处的切线和法线方程。
urve at that point with the same gradient, while the normal is perpendicular to the tangent at the same point.
该题采用典型的 AQA 出题风格。已知曲线 C 的方程为 y = x³ - 3x + 2，我们需要求出该曲线在 x = 2 处的两条相关直线。切线在该点与曲线具有相同的斜率，而法线在同一点与切线垂直。
2. Core Idea: Tangents and Normals | 核心概念：切线与法线
For any differentiable curve, the derivative dy/dx gives the gradient of the tangent at a point. The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of the tangent gradient. If the tangent gradient is mt, then the normal gradient is mn = -1 / mt, provided mt is not zero.
对于任何可微曲线，导数 dy/dx 给出曲线上某一点切线的斜率。法线与切线垂直，因此法线的斜率是切线斜率的负倒数。如果切线斜率为 mt，则法线斜率 mn = -1 / mt，前提是 mt 不为零。
mn = -1 / mt
This relationship is central to the question. Once the tangent gradient is known, the normal gradient follows immediately, and both straight lines can be written using the same point of contact.
这一关系是本题的核心。一旦求出切线斜率，法线斜率即可直接得到，而两条直线都可以利用同一个切点写出方程。
3. Step 1: Differentiate the Curve | 步骤 1：求曲线导数
Differentiate term by term with respect to x. The derivative of x³ is 3x², the derivative of -3x is -3, and the derivative of the constant 2 is 0.
对 x 逐项求导。x³ 的导数为 3x²，-3x 的导数为 -3，常数 2 的导数为 0。
dy/dx = 3x² - 3
It is important to show this step clearly in an AQA solution because the derivative is the key to obtaining both gradients. A small slip in differentiation will usually affect all remaining marks.
在 AQA 的解答中，清晰写出这一步非常重要，因为导数是求两个斜率的关键。求导时的小错误通常会影响后续所有得分。
4. Step 2: Find the Point Coordinates | 步骤 2：求切点坐标
ation to find the y-coordinate. We use the original curve equation, not the derivative, because we need the point on the curve.
将 x = 2 代入原方程求 y 坐标。这里使用原曲线方程，而不是导数，因为我们需要的是曲线上的点。
y = (2)³ - 3(2) + 2 = 8 - 6 + 2 = 4
So the point of contact is P(2, 4). This point belongs to both the tangent and the normal, so it will be used in both equations later.
因此切点坐标为 P(2, 4)。该点既在切线上，也在法线上，所以稍后将用于两条直线的方程。
5. Step 3: Gradient of the Tangent | 步骤 3：求切线斜率
Substitute x = 2 into dy/dx to find the gradient of the tangent at P.
将 x = 2 代入 dy/dx，求出曲线在 P 点切线的斜率。
dy/dx = 3(2)² - 3 = 12 - 3 = 9
Therefore the tangent gradient is mt = 9. This means that for every 1 unit moved to the right, the tangent rises by 9 units.
因此切线斜率为 mt = 9。这意味着每向右移动 1 个单位，切线上升 9 个单位。
6. Step 4: Gradient of the Normal | 步骤 4：求法线斜率
Use the negative reciprocal of the tangent gradient. Since mt = 9, the normal gradient is computed as follows.
利用切线斜率的负倒数。因为 mt = 9，法线斜率计算如下。
mn = -1 / 9
The negative sign is essential. A common error is to write 1/9 instead of -1/9, which would give a line parallel to the tangent rather than perpendicular to it.
负号至关重要。一个常见错误是写成 1/9 而不是 -1/9，这样得到的直线将与切线平行，而不是与切线垂直。
7. Step 5: Equation of the Tangent | 步骤 5：求切线方程
gradient form y - y₁ = m(x - x₁). With P(2, 4) and m = 9, this gives the tangent equation.
使用点斜式 y - y₁ = m(x - x₁)。代入 P(2, 4) 和 m = 9，得到切线方程。
y - 4 = 9(x - 2) ⇒ y = 9x - 14
n an exam, either the expanded form or the original point-gradient form is accepted unless a specific form is requested.
展开括号得 y - 4 = 9x - 18，两边加 4 得到最终切线方程 y = 9x - 14。在考试中，除非题目要求特定形式，否则展开式或原始点斜式都可以接受。
8. Step 6: Equation of the Normal | 步骤 6：求法线方程
Use the same point P(2, 4) with the normal gradient -1/9. Substitute into the point-gradient form.
使用同一点 P(2, 4) 和法线斜率 -1/9，代入点斜式。
y - 4 = -1/9 (x - 2)
+ 2. Rearranging all terms to one side produces the normal equation in a tidy linear form.
两边乘以 9 消去分数，得 9y - 36 = -x + 2。将所有项移到一边，得到简洁的线性法线方程。
x + 9y - 38 = 0
rmal equation can be quoted in either equivalent form, but writing it with integer coefficients is often clearer and easier to check.
这等价于 x + 9y = 38。法线方程可以写成任一等价形式，但使用整数系数通常更清晰，也更容易检验。
9. Quick Check and Common Mistakes | 快速检验与常见错误
1, confirming perpendicularity. Substituting P(2, 4) into both line equations also confirms that the point lies on both lines.
检验切线和法线的斜率乘积是否为 -1。这里 9 × -1/9 = -1，确认两线垂直。将 P(2, 4) 代入两条直线方程，也可以确认该点同时在两条线上。
dx instead of the original equation for y, and confusing the tangent and normal equations.
常见错误包括求导错误、法线斜率的符号写错、将 x = 2 代入导数而不是原方程求 y 坐标，以及混淆切线和法线方程。
- Differentiate every term carefully, including constants.
- Write the normal gradient as the negative reciprocal, not the reciprocal.
- Use the original curve equation for the y-coordinate, not the derivative.
- Always quote the final line in the form requested, often ax + by + c = 0.
- 仔细对每一项求导，包括常数项。
- 法线斜率要写成负倒数，而不是倒数。
- 求 y 坐标时使用原曲线方程，而不是导数。
- 始终按照题目要求的形式写出最终直线方程，通常为 ax + by + c = 0。
10. Exam Tips for AQA | AQA 考试技巧
In AQA Pure Mathematics, this type of question is often worth 6 to 8 marks. The marks are allocated for correct differentiation, finding the coordinates, obtaining the tangent gradient, forming the normal gradient, and writing both equations.
在 AQA Pure Mathematics 中，这类题目通常占 6 到 8 分。分值分配给正确求导、求坐标、求切线斜率、求法线斜率以及写出两条直线方程。
Show all steps clearly because method marks can be awarded even if a numerical slip occurs later. If you obtain an unusual fraction, do not force it into a decimal; AQA accepts exact fractions such as -1/9.
清晰展示所有步骤，因为即使后面出现数值计算错误，方法分仍然可以获得。如果得到不常见的分数，不要强行化成小数；AQA 接受像 -1/9 这样的精确分数。
Finally, always read the question carefully to see whether it asks for the tangent, the normal, or both. In this example both are required, and both must be clearly labelled in your answer.
最后，务必仔细审题，看题目要求的是切线、法线还是两者都要。本题两者都要求，因此在答案中必须清楚标出。
11. Summary | 小结
e, find the point, compute the tangent gradient, take the negative reciprocal for the normal, then use the point-gradient form for both lines.
对于曲线 y = x³ - 3x + 2 在 x = 2 处，切线为 y = 9x - 14，法线为 x + 9y - 38 = 0。解题方法为：求导、求点坐标、计算切线斜率、取负倒数求法线斜率，然后用点斜式写出两条直线方程。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图