📚 Example 5.5.1: Tangent and Normal to a Cubic Curve | 例 5.5.1：三次曲线的切线与法线
In this worked example, we tackle a classic AQA A-Level Pure Mathematics problem: finding the equations of the tangent and normal to a cubic curve at a given point. The method combines coordinate geometry with differentiation, and it appears frequently in Paper 1.
在这个例题中，我们处理一道典型的 AQA A-Level 纯数学题：求三次曲线在给定点处的切线与法线方程。该方法将坐标几何与微分结合起来，经常出现在 Paper 1 中。
1. The Problem Statement | 题目陈述
2. Find the equation of the tangent and the normal to the curve at the point where x = 2.
一条曲线的方程为 y = x³ − 3x + 2。求该曲线在 x = 2 处的切线方程和法线方程。
y = x³ − 3x + 2, x = 2
2. Why Differentiation Gives the Gradient | 为什么微分能给出斜率
dy/dx gives the gradient of the tangent at a general point (x, y). This is because the derivative is the limit of the chord gradient as two points on the curve become infinitely close.
对于任意曲线 y = f(x)，导数 dy/dx 给出一般点 (x, y) 处切线的斜率。这是因为导数是曲线上两点无限接近时割线斜率的极限。
On AQA papers, you are expected to recognise that 'gradient of the curve' and 'gradient of the tangent' mean exactly the same thing as dy/dx evaluated at the given x-coordinate.
在 AQA 试题中，你应该认识到“曲线的斜率”和“切线的斜率”与在给定 x 坐标处求得的 dy/dx 是完全相同的意思。
3. Step 1 - Find the y-coordinate | 第一步：求 y 坐标
e original curve equation. This gives the point of contact between the curve and the tangent.
题目只给出 x 坐标，因此我们必须先把 x = 2 代入原曲线方程。这样就得到曲线与切线的切点。
y = (2)³ − 3(2) + 2 = 8 − 6 + 2 = 4
Therefore the point of contact is (2, 4).
因此切点坐标为 (2, 4)。
4. Step 2 - Differentiate the Function | 第二步：对函数求导
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 的导数是 3x²，−3x 的导数是 −3，常数 2 的导数是 0。
dy/dx = 3x² − 3
Keep this derivative in its simplified form before substituting any value. It is the gradient function for the whole curve.
在代入任何数值之前，保持导数为最简形式。它是整条曲线的斜率函数。
5. Step 3 - Evaluate the Gradient at x = 2 | 第三步：在 x = 2 处计算斜率
2 into the gradient function to obtain the gradient of the tangent at the required point.
将 x = 2 代入斜率函数，得到该点处切线的斜率。
mt = 3(2)² − 3 = 12 − 3 = 9
The tangent gradient is positive, so the curve is increasing at x = 2.
切线斜率为正，说明曲线在 x = 2 处是上升的。
6. Step 4 - Equation of the Tangent | 第四步：求切线方程
Use the point-gradient form y − y₁ = m(x − x₁) with (x₁, y₁) = (2, 4) and m = 9.
使用点斜式 y − y₁ = m(x − x₁)，其中 (x₁, y₁) = (2, 4)，m = 9。
y − 4 = 9(x − 2)
panding and simplifying gives the tangent equation in the slope-intercept form y = mx + c.
展开并化简，得到斜截式方程 y = mx + c。
y = 9x − 14
Equivalently, the tangent can be written in the general form 9x − y = 14.
等价地，切线可以写成一般式 9x − y = 14。
7. Step 5 - Equation of the Normal | 第五步：求法线方程
The normal is perpendicular to the tangent at the point of contact. If the tangent has gradient 9, then the normal gradient is the negative reciprocal: mn = −1/9.
法线在切点处与切线垂直。如果切线斜率为 9，那么法线斜率是其负倒数：mn = −1/9。
mn = −1/9
Using the same point (2, 4) and the point-gradient form gives the normal equation.
使用同一个点 (2, 4) 和点斜式，得到法线方程。
y − 4 = −(1/9)(x − 2)
Multiply through by 9 to avoid fractions: 9y − 36 = −x + 2, so x + 9y = 38.
两边同时乘以 9 以避免分数：9y − 36 = −x + 2，因此 x + 9y = 38。
y = −x/9 + 38/9
cceptable in AQA mark schemes, but the integer form x + 9y = 38 reduces arithmetic errors.
这两种形式在 AQA 评分标准中都是可以接受的，但整系数形式 x + 9y = 38 能减少算术错误。
8. Interpreting the Results | 解释结果
to the tangent. The two lines are perpendicular because the product of their gradients is −1.
切线 y = 9x − 14 在 (2, 4) 处与三次曲线刚好相切，而法线 x + 9y = 38 与切线成直角穿过三次曲线。这两条线互相垂直，因为它们的斜率乘积为 −1。
Sketching the tangent and normal together with the curve can help you check whether the signs and positions make sense.
将切线、法线与曲线画在一起，可以帮助你检查斜率的正负和位置是否合理。
9. Common Mistakes to Avoid | 应避免的常见错误
- Forgetting to find the y-coordinate before writing the line equation. — 在写直线方程之前忘记求 y 坐标。
- Using the normal gradient as 9 instead of −1/9. — 把法线斜率写成 9，而不是 −1/9。
- Expanding brackets incorrectly in the point-gradient form. — 在点斜式中括号展开出错。
- Confusing the tangent and normal equations in the final answer. — 在最终答案中混淆切线方程和法线方程。
- Writing dy/dx = 3x² − 3 + 2 instead of dropping the constant. — 将导数写成 dy/dx = 3x² − 3 + 2，而忘记常数的导数为 0。
Each of these errors loses method or accuracy marks, even if the final answer looks close.
即使最终答案看起来接近，以上每一个错误都会失去方法分或准确分。
10. AQA Exam Tips | AQA 应试建议
In AQA mark schemes, method marks are awarded for differentiating correctly, evaluating the gradient, and applying the point-gradient form. Show all steps clearly, even if the arithmetic is simple.
在 AQA 评分标准中，正确求导、计算斜率以及使用点斜式都会获得方法分。即使计算简单，也要清晰展示所有步骤。
and x + 9y = 38. If you leave the answer as y − 4 = 9(x − 2), you may lose accuracy marks.
考官通常会要求写出具体形式 y = 9x − 14 和 x + 9y = 38。如果只写成 y − 4 = 9(x − 2)，可能会失去准确分。
11. Practice Variation | 练习变式
− 6 at x = 1. Substitute, differentiate, evaluate the gradient, and write both equations.
用同样的方法处理曲线 y = x³ − 6x² + 11x − 6 在 x = 1 处的问题。先代入、再求导、计算斜率，然后写出两条直线方程。
You should obtain y = 0, dy/dx = 3x² − 12x + 11, mt = 2, tangent y = 2x − 2, and normal y = −½x + ½.
你应该得到 y = 0，dy/dx = 3x² − 12x + 11，mt = 2，切线 y = 2x − 2，法线 y = −½x + ½。
12. Summary | 方法总结
To find a tangent or normal: find the point, differentiate, evaluate the gradient, then apply the point-gradient formula. For the normal, use the negative reciprocal of the tangent gradient.
求切线或法线的方法：求点、求导、计算斜率，然后应用点斜式。对于法线，使用切线斜率的负倒数。
Mastering this five-step process will help you tackle many AQA differentiation questions involving tangents, normals, and stationary points.
掌握这五步法将帮助你解决许多涉及切线、法线和驻点的 AQA 微分问题。
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