📚 Worked Example 2.5.2: Tangents and Normals to a Curve | 例题 2.5.2：曲线的切线与法线
In this AQA A-level Maths worked example, we show how to find the equations of the tangent and the normal to a curve at a given point. The method combines differentiation, coordinate substitution, and the perpendicular gradient rule.
在这道 AQA A-level 数学例题中，我们将演示如何求曲线在给定点处的切线方程和法线方程。该方法综合运用了求导、坐标代入以及垂直斜率关系。
1. The Problem Statement | 问题陈述
= x³ − 3x + 2, find the equations of the tangent and the normal at the point where x = 2.
对于曲线 y = x³ − 3x + 2，求 x = 2 处的切线方程和法线方程。
This is a standard AQA Pure topic that often appears in AS and A-level exam papers. You must show clear working for both gradients and both equations.
这是 AQA 纯数学中的一个标准考点，经常出现在 AS 和 A-level 试卷中。你必须清晰展示两个斜率和两个方程的求解过程。
2. Understanding Tangents and Normals | 理解切线与法线
A tangent is a straight line that touches a curve at exactly one point without crossing it at that point. Its gradient is equal to the derivative dy/dx at that point.
切线是与曲线在一点处相切且在该点处不穿过曲线的直线。它的斜率等于函数在该点处的导数 dy/dx。
A normal is the line perpendicular to the tangent at the point of contact. If the tangent gradient is m, the normal gradient is −1/m, provided m ≠ 0.
法线是在切点处垂直于切线的直线。如果切线斜率为 m，则法线斜率为 −1/m，前提是 m ≠ 0。
ar gradient rule is essential here: m₁ × m₂ = −1 for two perpendicular non-vertical lines.
这里必须使用垂直斜率关系：对于两条垂直的非竖直直线，m₁ × m₂ = −1。
3. Step 1: Differentiate the Function | 第一步：对函数求导
We differentiate y = x³ − 3x + 2 term by term using the power rule: d(xⁿ)/dx = n xⁿ⁻¹.
我们使用幂法则 d(xⁿ)/dx = n xⁿ⁻¹ 对 y = x³ − 3x + 2 逐项求导。
dy/dx = 3x² − 3
The derivative gives the gradient of the curve at any point x. At this stage, do not substitute the x-value until the derivative is fully simplified.
导数给出了曲线在任意 x 值处的斜率。在这个阶段，先不要代入 x 值，要将导数完全化简。
4. Step 2: Find the Gradient at x = 2 | 第二步：求 x = 2 处的斜率
Substitute x = 2 into dy/dx = 3x² − 3 to obtain the tangent gradient at that point.
将 x = 2 代入 dy/dx = 3x² − 3，得到该点处的切线斜率。
mt = 3(2)² − 3 = 12 − 3 = 9
fore the tangent has gradient 9. This also means the curve is increasing steeply at x = 2.
因此切线的斜率为 9。这也表明曲线在 x = 2 处上升得很快。
5. Step 3: Find the Coordinates of the Point | 第三步：求切点坐标
on the line. The tangent and normal both pass through the point on the curve where x = 2.
要写出直线方程，我们需要直线上的一点。切线和法线都经过曲线上 x = 2 的点。
Substitute x = 2 into the original equation y = x³ − 3x + 2.
将 x = 2 代入原方程 y = x³ − 3x + 2。
y = (2)³ − 3(2) + 2 = 8 − 6 + 2 = 4
So the point of contact is (2, 4). This coordinate pair is used for both the tangent and normal equations.
所以切点坐标是 (2, 4)。这组坐标将同时用于切线和法线方程。
6. Step 4: Equation of the Tangent | 第四步：切线方程
t-gradient form of a straight line: y − y₁ = m(x − x₁), where (x₁, y₁) = (2, 4) and m = 9.
使用直线的点斜式：y − y₁ = m(x − x₁)，其中 (x₁, y₁) = (2, 4)，m = 9。
y − 4 = 9(x − 2)
Expand and simplify to give the tangent in the form y = mx + c.
展开并化简，将切线写成 y = mx + c 的形式。
y = 9x − 14
tion. You can verify that when x = 2, y = 4, so the line passes through the contact point.
这就是所求的切线方程。你可以验证当 x = 2 时 y = 4，说明该直线经过切点。
7. Step 5: Gradient of the Normal | 第五步：法线的斜率
Since the normal is perpendicular to the tangent, its gradient is the negative reciprocal of the tangent gradient.
由于法线垂直于切线，它的斜率是切线斜率的负倒数。
mn = −1 / 9
Do not forget the negative sign. A common error is to use 1/9 or −9. Check that mt × mn = −1.
不要漏掉负号。常见错误是写成 1/9 或 −9。请检查 mt × mn = −1。
Here 9 × (−1/9) = −1, so the two gradients are indeed perpendicular.
这里 9 × (−1/9) = −1，因此两条直线的斜率确实互相垂直。
8. Step 6: Equation of the Normal | 第六步：法线方程
Again use the point-gradient form with (x₁, y₁) = (2, 4) and mn = −1/9.
再次使用点斜式，代入 (x₁, y₁) = (2, 4) 和 mn = −1/9。
y − 4 = −1/9 (x − 2)
Multiply through by 9 to avoid fraction complications: 9(y − 4) = −(x − 2).
两边乘以 9 以消除分数：9(y − 4) = −(x − 2)。
9y − 36 = −x + 2
Rearrange to a tidy standard form: x + 9y = 38. You can also write y = −x/9 + 38/9.
整理为标准形式：x + 9y = 38。你也可以写成 y = −x/9 + 38/9。
Both forms are acceptable in AQA exams unless a specific form is requested.
除非题目要求特定的形式，否则这两种写法在 AQA 考试中都可以接受。
9. Visual Check and Alternative Forms | 图形验证与等价形式
sitive gradient, so the tangent y = 9x − 14 should lie close to the curve near that point.
快速画图有助于验证结果。在 x = 2 附近，曲线 y = x³ − 3x + 2 经过 (2, 4) 且斜率为较大的正值，因此切线 y = 9x − 14 在该点附近应紧贴曲线。
small negative gradient (−1/9), so it should cross the tangent at (2, 4) at a right angle.
法线 x + 9y = 38 的斜率为较小的负值 (−1/9)，因此它应在 (2, 4) 处与切线成直角相交。
= 0. Here the tangent can be written as 9x − y − 14 = 0 and the normal as x + 9y − 38 = 0.
在考试中，你可能会被要求将答案写成 ax + by + c = 0 的形式。这里切线可以写成 9x − y − 14 = 0，法线可以写成 x + 9y − 38 = 0。
10. Common Mistakes | 常见错误
One common mistake is to forget to find the y-coordinate before using the point-gradient formula. Without the full point, the equation will be incorrect even if the gradient is right.
一个常见错误是在使用点斜式之前忘记求 y 坐标。如果缺少完整的点坐标，即使斜率正确，方程也会出错。
Another error is confusing the derivative dy/dx with the tangent equation. The derivative is a gradient function, not the equation of the tangent line.
另一个错误是将导数 dy/dx 与切线方程混淆。导数是一个斜率函数，而不是切线方程。
Also, when finding the normal gradient, many students write 1/9 without the negative sign, or use −9 instead of −1/9. Always check the perpendicular condition mt × mn = −1.
此外，在求法线斜率时，许多学生漏写负号，或者把 −9 当成 −1/9。一定要检验垂直条件 mt × mn = −1。
11. Exam Tips for AQA | AQA 考试提示
Always show the derivative step clearly before substituting the x-value. In AQA mark schemes, method marks are awarded for correct differentiation and correct use of the perpendicular gradient rule.
在代入 x 值之前，一定要清晰展示求导步骤。在 AQA 评分标准中，正确求导和正确使用垂直斜率关系都能得到方法分。
Leave fractions in their exact form unless the question asks for decimals. For example, −1/9 is preferred to −0.111...
除非题目要求用小数，否则请保留分数的精确形式。例如，−1/9 比 −0.111... 更合适。
9y = 38. Underline or box your final answers if that helps the examiner follow your work.
将最终方程写成清晰的形式，例如 y = 9x − 14 和 x + 9y = 38。如果有助于考官阅读，你可以在最终答案下划线或画框。
12. Practice Variant and Summary | 变式练习与总结
question: for the curve y = 2x³ − x + 1, find the tangent and normal equations at x = −1.
尝试一道类似的题目：对于曲线 y = 2x³ − x + 1，求 x = −1 处的切线方程和法线方程。
= 5. The y-coordinate is y = 2(−1)³ − (−1) + 1 = −2 + 1 + 1 = 0, so the point is (−1, 0).
首先求导得到 dy/dx = 6x² − 1。当 x = −1 时，斜率为 6 − 1 = 5。y 坐标为 y = 2(−1)³ − (−1) + 1 = −2 + 1 + 1 = 0，因此点为 (−1, 0)。
normal gradient is −1/5, giving y − 0 = −1/5 (x + 1), which simplifies to 5y + x + 1 = 0.
切线为 y − 0 = 5(x + 1)，即 y = 5x + 5。法线斜率为 −1/5，得到 y − 0 = −1/5 (x + 1)，化简后为 5y + x + 1 = 0。
In summary, to solve a tangent and normal problem, differentiate first, substitute the x-coordinate into the gradient function, find the y-coordinate from the original curve, then use the point-gradient form for both lines.
总之，求解切线和法线问题时，先求导，将 x 坐标代入斜率函数，从原曲线求出 y 坐标，然后对两条直线使用点斜式。
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