📚 Example 4.5.2: Tangent and Normal to a Curve | 例题 4.5.2：曲线的切线与法线
In AQA A Level Mathematics, Example 4.5.2 is a standard application of differentiation: finding the equations of the tangent and normal to a curve at a given point. This skill is assessed regularly in AS and A2 papers because it combines differentiation, coordinate geometry, and algebraic rearrangement in one compact problem.
在 AQA A Level 数学中，例题 4.5.2 是导数的典型应用：求曲线在给定点处的切线与法线方程。该技能经常在 AS 与 A2 试卷中出现，因为它将导数、坐标几何与代数整理综合到一个紧凑的问题中。
1. The Example Statement | 例题陈述
ions of the tangent and the normal to the curve y = x³ - 3x² + 2 at the point where x = 1.
我们使用一个具有代表性的例题 4.5.2：求曲线 y = x³ - 3x² + 2 在 x = 1 处的切线与法线方程。
This wording is typical of AQA questions. You are not asked to sketch the curve or to find the maximum or minimum; you only need the two straight lines that touch the curve in specific ways at the given point.
这一题干在 AQA 考题中很典型。题目不要求你画曲线，也不要求找最大值或最小值；你只需要求出在给定点以特定方式接触该曲线的两条直线。
2. Background: Differentiation and Gradients | 背景：导数与斜率
The derivative dy/dx gives the gradient of the curve at any point where the function is differentiable. For a tangent, the gradient m₁ is exactly the value of dy/dx at the contact point.
导数 dy/dx 给出了函数可导的任意点处曲线的斜率。对于切线而言，斜率 m₁ 正好是接触点处 dy/dx 的值。
, the normal gradient m₂ satisfies the relationship m₁ × m₂ = -1, provided m₁ is not zero.
法线在接触点处与切线垂直。如果切线斜率为 m₁，则法线斜率 m₂ 满足关系 m₁ × m₂ = -1，只要 m₁ 不等于零。
3. Step 1: Find dy/dx | 第一步：求 dy/dx
rm. The derivative of x³ is 3x², the derivative of -3x² is -6x, and the derivative of the constant 2 is 0.
对 y = x³ - 3x² + 2 逐项求导。x³ 的导数是 3x²，-3x² 的导数是 -6x，常数 2 的导数是 0。
dy/dx = 3x² - 6x
Always write this line clearly before substituting any values. AQA examiners award the first method mark for a correct derivative, even if later steps contain errors.
在代入任何数值之前，请清楚地写出这一行导数。即使后续步骤有错误，AQA 考官也会对正确的导数给予第一个方法分。
4. Step 2: Locate the Point | 第二步：定位切点
the original equation to find the y-coordinate. We get y = 1³ - 3(1)² + 2 = 1 - 3 + 2 = 0.
将 x = 1 代入原方程以求 y 坐标。我们得到 y = 1³ - 3(1)² + 2 = 1 - 3 + 2 = 0。
The point of contact is therefore (1, 0). This coordinate pair must be used in the linear equations for both the tangent and the normal.
因此接触点是 (1, 0)。这一坐标对必须同时用于切线和法线的直线方程中。
5. Step 3: Gradient of the Tangent | 第三步：切线斜率
ute x = 1 into the derivative dy/dx = 3x² - 6x. This gives m₁ = 3(1)² - 6(1) = 3 - 6 = -3.
将 x = 1 代入导数 dy/dx = 3x² - 6x。得到 m₁ = 3(1)² - 6(1) = 3 - 6 = -3。
that the curve is decreasing at x = 1, so the tangent slopes downwards from left to right.
负斜率说明曲线在 x = 1 处是递减的，因此切线从左向右向下倾斜。
6. Step 4: Tangent Equation | 第四步：切线方程
form y - y₁ = m₁(x - x₁). With the point (1, 0) and m₁ = -3, we obtain y - 0 = -3(x - 1).
使用点斜式 y - y₁ = m₁(x - x₁)。代入点 (1, 0) 与 m₁ = -3，我们得到 y - 0 = -3(x - 1)。
y = -3x + 3
rite it in the equivalent general form 3x + y - 3 = 0, which some AQA mark schemes prefer.
这是切线方程的斜截式。你也可以将它写成等价的一般式 3x + y - 3 = 0，AQA 的一些评分标准更偏好这种形式。
7. Step 5: Normal Equation | 第五步：法线方程
ent is the negative reciprocal. Using m₁ × m₂ = -1, we solve -3 × m₂ = -1 to get m₂ = 1/3.
由于切线斜率为 m₁ = -3，法线斜率是它的负倒数。利用 m₁ × m₂ = -1，我们解 -3 × m₂ = -1，得到 m₂ = 1/3。
Apply y - y₁ = m₂(x - x₁) with the same point (1, 0). This gives y - 0 = (1/3)(x - 1).
将同一点 (1, 0) 代入 y - y₁ = m₂(x - x₁)，得到 y - 0 = (1/3)(x - 1)。
y = (1/3)x - 1/3
3y - 1 = 0. Both forms are mathematically correct, but the integer form is often cleaner.
为避免最终答案中出现分数，可以将等式两边乘以 3，得到一般式 x - 3y - 1 = 0。两种形式在数学上都是正确的，但整数形式通常更简洁。
8. Verifying Your Answers | 验证答案
y = -3(1) + 3 = 0. For the normal x - 3y - 1 = 0, substituting (1, 0) gives 1 - 0 - 1 = 0.
检查两条直线是否都经过 (1, 0)。对于切线 y = -3x + 3，代入 x = 1 得 y = -3(1) + 3 = 0。对于法线 x - 3y - 1 = 0，代入 (1, 0) 得 1 - 0 - 1 = 0。
A quick numerical check like this can catch sign errors, especially when the derivative involves negative coefficients. It is worth ten seconds in the exam.
像这样的快速数值检验可以发现符号错误，尤其是当导数包含负系数时。在考试中花十秒钟这样做是值得的。
9. Common Mistakes | 常见错误
Some students use x = 1 as the point, but a line equation needs the full coordinate pair.
一个常见错误是在建立直线方程之前忘记求 y 坐标。有些学生把 x = 1 当作点，但直线方程需要完整的坐标对。
Another error is using the same gradient for the tangent and the normal. Remember that the normal is perpendicular, so its gradient must be the negative reciprocal unless the tangent is horizontal.
另一个错误是将切线与法线使用相同的斜率。请记住法线是垂直的，所以其斜率必须是负倒数，除非切线是水平的。
- Forgetting to differentiate before substituting x = 1
- 在代入 x = 1 之前忘记求导
- Calculating the derivative incorrectly, especially the middle term -6x
- 导数计算错误，尤其是中间项 -6x
- Writing the normal gradient as 3 instead of 1/3
- 将法线斜率写成 3 而不是 1/3
10. Exam Technique and Mark Allocation | 考试技巧与分值分配
tuting x = 1, for using the perpendicular gradient rule, and for a correct final equation.
在 AQA 评分标准中，这类题目通常值 4 到 6 分。你可以因求导、代入 x = 1、使用垂直线斜率规则以及正确的最终方程而获得方法分。
| Step 步骤 | Typical mark 典型分值 |
| Correct dy/dx 正确的 dy/dx | 1 |
| Point coordinates 点坐标 | 1 |
| Tangent gradient and equation 切线斜率与方程 | 2 |
| Normal gradient and equation 法线斜率与方程 | 2 |
Always show your working line by line. Even if the final answer is incorrect, clear intermediate steps can secure most of the marks.
务必逐行展示你的解答过程。即使最终答案有误，清晰的中间步骤也能拿到大部分分数。
11. Practice Variation | 练习变式
find the point (3, 4), then the tangent gradient m₁ = 2 and the normal gradient m₂ = -1/2.
尝试一道类似题目：求 y = x² - 4x + 7 在 x = 3 处的切线与法线。先计算 dy/dx = 2x - 4，再求出点 (3, 4)，然后得到切线斜率 m₁ = 2 与法线斜率 m₂ = -1/2。
, and the normal equation should be y = -1/2 x + 11/2, or x + 2y - 11 = 0 in integer form.
切线方程应为 y = 2x - 2，法线方程应为 y = -1/2 x + 11/2，或整数形式 x + 2y - 11 = 0。
12. Summary | 总结
To solve any tangent and normal question, differentiate first, substitute the given x-value to locate the point, evaluate the derivative to find the tangent gradient, then use the negative reciprocal for the normal gradient.
要求解任何切线与法线问题，首先求导，代入给定的 x 值以确定点，计算导数以求得切线斜率，然后使用负倒数作为法线斜率。
Finally, apply the point-slope formula to write both line equations. If time allows, substitute the original point into both answers to verify they are correct.
最后，应用点斜式写出两条直线方程。如果时间允许，将原点代入两个答案以验证其正确性。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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