📚 Example 4.5.1: Tangent and Normal to a Cubic Curve | 示例4.5.1：三次曲线的切线与法线
In this revision article we examine Example 4.5.1 from the AQA A-Level Mathematics pure mathematics specification. The example asks us to find the equations of the tangent and the normal to the curve y = x³ − 3x + 1 at the point where x = 2. This worked solution demonstrates the key techniques of differentiation, the equation of a straight line, and the perpendicularity condition for a normal.
在本复习文章中，我们分析 AQA A-Level 数学纯数学部分中的示例4.5.1。该示例要求我们求曲线 y = x³ − 3x + 1 在 x = 2 处的切线与法线方程。该解答演示了微分、直线方程以及法线垂直条件等关键技巧。
1. Understanding the problem | 理解问题
Before carrying out any calculations, it is important to restate what is required. A tangent is a straight line that just touches the curve at the point of interest, and it has the same gradient as the curve at that exact point. A normal is a straight line perpendicular to the tangent at the point of contact. To find both, we first need the coordinates of the point on the curve and the gradient of the curve there.
在进行任何计算之前，我们先重新说明题目要求。切线是一条与曲线在给定点仅接触且在该点与曲线斜率相同的直线。法线是在接触点垂直于切线的直线。为了找到两者，我们首先需要曲线上该点的坐标以及曲线在该点的斜率。
2. The derivative: power rule | 导数：幂法则
The gradient of a curve at any point is given by the derivative dy/dx. Our function is y = x³ − 3x + 1. Using the power rule, we differentiate term by term: the derivative of x³ is 3x²; the derivative of −3x is −3; the derivative of the constant 1 is 0. Therefore the derivative is:
曲线上任意一点的斜率由导数 dy/dx 给出。我们的函数是 y = x³ − 3x + 1。运用幂法则逐项求导：x³ 的导数为 3x²；−3x 的导数为 −3；常数 1 的导数为 0。因此导数为：
dy/dx = 3x² − 3
3. Gradients at the point | 求该点处的斜率
Substitute x = 2 into the derivative to obtain the gradient of the curve at the given point:
将 x = 2 代入导数，得到曲线在该给定点处的斜率：
dy/dx = 3(2²) − 3 = 3(4) − 3 = 12 − 3 = 9
So the gradient of the tangent at x = 2 is 9. This value is positive, which tells us that the curve is rising steeply at that point. Since the gradient is not zero, the point is not a stationary point and the tangent is not horizontal.
因此切线在 x = 2 处的斜率为 9。该值为正，说明曲线在该点急剧上升。由于该斜率不为零，该点不是驻点，切线也不水平。
4. Coordinates of the point | 点的坐标
We must also know the y-coordinate of the point on the curve. Substituting x = 2 into the original equation gives:
我们还必须知道曲线上该点的 y 坐标。将 x = 2 代入原方程，得到：
y = 2³ − 3(2) + 1 = 8 − 6 + 1 = 3
Hence the point of contact is (2, 3). This coordinate pair will be used in both the tangent equation and the normal equation.
因此切点为 (2, 3)。该坐标对将同时用于切线方程和法线方程。
5. Equation of the tangent | 切线方程
Using the point-slope form of a straight line, y − y₁ = m(x − x₁), with m = 9 and (x₁, y₁) = (2, 3), we get:
运用直线的点斜式 y − y₁ = m(x − x₁)，其中 m = 9，(x₁, y₁) = (2, 3)，得到：
y − 3 = 9(x − 2)
Expanding and simplifying: y − 3 = 9x − 18, so y = 9x − 15. Therefore the equation of the tangent is y = 9x − 15. This is the standard linear form that is easiest to interpret and use.
展开并化简：y − 3 = 9x − 18，所以 y = 9x − 15。因此切线方程为 y = 9x − 15。这是最便于解释和使用的标准线性形式。
6. Equation of the normal | 法线方程
The normal is perpendicular to the tangent. If two lines are perpendicular, the product of their gradients is −1. Thus the gradient of the normal, m_normal, satisfies 9 × m_normal = −1, giving m_normal = −1/9.
法线垂直于切线。若两条直线垂直，则它们斜率的乘积为 −1。因此法线的斜率 m_normal 满足 9 × m_normal = −1，得 m_normal = −1/9。
m_normal = −1/9
Using the point-slope form again with x₁ = 2 and y₁ = 3:
再次使用点斜式，其中 x₁ = 2，y₁ = 3：
y − 3 = −1/9 (x − 2)
Multiply through by 9: 9(y − 3) = −(x − 2). So 9y − 27 = −x + 2, which simplifies to x + 9y − 29 = 0. In gradient-intercept form, we can write y = −1/9 x + 29/9.
两边乘以 9：9(y − 3) = −(x − 2)。所以 9y − 27 = −x + 2，化简得 x + 9y − 29 = 0。用斜截式表示，y = −1/9 x + 29/9。
7. Interpreting the results | 结果的含义
The tangent y = 9x − 15 has a steep positive slope, consistent with the cubic rising quickly at that point. The normal has a slight negative slope, −1/9, which is the negative reciprocal of 9. Both lines pass through (2, 3), as required. The tangent touches the curve at that single point, while the normal intersects it there at right angles.
切线 y = 9x − 15 的斜率较大为正，这与三次曲线在该点快速上升一致。法线的斜率较小为负，−1/9，是 9 的负倒数。两条直线都经过 (2, 3)，满足要求。切线在该点仅与曲线相切，而法线在该点与曲线成直角相交。
8. Checking the answers | 检查答案
We can verify the tangent equation by substituting x = 2 into y = 9x − 15: y = 18 − 15 = 3, which matches the y-coordinate of the point. Similarly, for the normal in the form x + 9y − 29 = 0, substituting (2, 3) gives 2 + 27 − 29 = 0, so the point lies on the normal. Furthermore, the product of the two gradients is 9 × (−1/9) = −1, confirming that the lines are perpendicular.
我们可以通过代入验证切线方程：将 x = 2 代入 y = 9x − 15，得到 y = 18 − 15 = 3，与点的 y 坐标一致。同理，对于法线 x + 9y − 29 = 0，代入 (2, 3)：2 + 27 − 29 = 0，所以该点在法线上。此外，两斜率乘积为 9 × (−1/9) = −1，确认两直线垂直。
9. Common mistakes | 常见错误
Many students forget to find the y-coordinate before writing the line equation. Another frequent error is using the tangent gradient directly for the normal without taking the negative reciprocal. Some students also fail to simplify the equation to a standard form, which can make later work more difficult. Always check whether the point satisfies the line equation and whether the perpendicularity condition holds.
许多学生忘记先求 y 坐标就直接写直线方程。另一个常见错误是求法线时未取切线斜率的负倒数而直接使用。一些学生也未将方程化简为标准形式，这会使后续工作更加困难。始终检查该点是否满足直线方程，以及垂直条件是否成立。
10. Exam tips | 考试提示
In an exam, write down the derivative first, then the gradient, then the point, and then the line equations. Use clear notation such as m_T for the tangent gradient and m_N for the normal gradient. When showing working, the equation m_T × m_N = −1 is an important line that earns method marks. If the derivative evaluates to zero, the tangent is horizontal and the normal is vertical; handle that case separately.
在考试中，先写出导数，再写斜率，再写点坐标，然后写直线方程。使用清晰的记号，如 m_T 表示切线斜率，m_N 表示法线斜率。在书写过程中，m_T × m_N = −1 这一行至关重要，可获得方法分。如果导数值为零，切线水平，法线竖直；这种情况需单独处理。
11. Further exploration | 拓展思考
Suppose the point had been chosen where x = −1. Then y = (−1)³ − 3(−1) + 1 = −1 + 3 + 1 = 3, and the derivative at x = −1 is 3(−1)² − 3 = 0. The tangent would be horizontal, y = 3, and the normal would be vertical, x = −1. This special case demonstrates why the negative reciprocal rule cannot be used directly when the tangent gradient is zero; the normal is a vertical line with undefined gradient.
假设选择的点为 x = −1。那么 y = (−1)³ − 3(−1) + 1 = −1 + 3 + 1 = 3，且 x = −1 处导数为 3(−1)² − 3 = 0。切线将是水平线 y = 3，而法线将是竖直线 x = −1。这个特殊情况说明为什么当切线斜率为零时不能直接使用负倒数规则；法线是斜率未定义的竖直线。
12. Conclusion | 总结
Example 4.5.1 illustrates a complete process for finding tangents and normals to curves. The key steps are differentiating, evaluating the derivative, locating the point, and applying the point-slope formula. Remember that the normal’s gradient is the negative reciprocal of the tangent’s gradient. With practice, this type of question will become straightforward in the AQA A-Level Mathematics exam.
示例4.5.1展示了求曲线切线与法线的完整过程。关键步骤是求导、计算导数值、确定点坐标并应用点斜式公式。记住法线的斜率为切线斜率的负倒数。通过练习，这类题目在 AQA A-Level 数学考试中将变得简单。
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