📚 Example 6.6.1: Equation of a Tangent | 例6.6.1：切线的方程
In this example, we explore how to find the equation of the tangent line to a given curve at a specific point. This is a core skill in A-Level mathematics, combining differentiation with coordinate geometry.
在本例中，我们将探索如何求曲线在某一点处的切线方程。这是 A-Level 数学的核心技能，将微分与坐标几何结合起来。
1. The Problem | 问题陈述
− 2x + 1. We are asked to find the equation of the tangent line at the point where x = 2.
考虑由方程 y = x³ − 2x + 1 定义的曲线。我们被要求在 x = 2 处求该曲线的切线方程。
Step 1: the point of tangency. Substitute x = 2 into the curve equation to find the y-coordinate.
步骤1：确定切点。将 x = 2 代入曲线方程，求出 y 坐标。
y = (2)³ − 2(2) + 1 = 8 − 4 + 1 = 5
So the point of tangency is (2, 5).
因此切点为 (2, 5)。
2. Differentiation | 微分
To find the gradient of the tangent, we need the derivative of the curve with respect to x. Differentiate each term using the power rule.
为了求切线的斜率，我们需要对曲线关于 x 求导。使用幂法则对每一项求导。
dy/dx = 3x² − 2
The derivative dy/dx gives the gradient of the tangent at any point x.
导数 dy/dx 给出任意点 x 处切线的斜率。
3. Finding the Gradient | 求斜率
the derivative at x = 2 to obtain the slope of the tangent line at the point of tangency.
在 x = 2 处计算导数值，以获得切点处切线的斜率。
dy/dx = 3(2)² − 2 = 3 × 4 − 2 = 12 − 2 = 10
Thus, the gradient m = 10.
因此，斜率 m = 10。
4. Equation of the Tangent | 切线方程
tangent. The general form is y − y₁ = m(x − x₁), where (x₁, y₁) is the point of tangency.
利用直线的点斜式，我们可以写出切线方程。一般形式为 y − y₁ = m(x − x₁)，其中 (x₁, y₁) 是切点。
Substitute m = 10, x₁ = 2, and y₁ = 5:
代入 m = 10, x₁ = 2, y₁ = 5：
y − 5 = 10(x − 2)
y − 5 = 10x − 20
y = 10x − 15
Therefore, the equation of the tangent line is y = 10x − 15.
因此，切线方程为 y = 10x − 15。
5. Visual Interpretation | 图像解释
l. The slope of the tangent is positive, indicating that the curve is increasing at x = 2.
切线恰好与曲线在点 (2, 5) 相切。在该点附近，切线很好地近似于曲线。切线的斜率为正，表明曲线在 x = 2 处递增。
- The gradient of the tangent equals the derivative of the curve at the point of contact.
- 切线的斜率等于曲线在接触点处的导数值。
- The tangent line has exactly one common point with the curve at the point of tangency (for a curve that is smooth).
- 对于光滑曲线，切线与曲线在切点处恰好有一个公共点。
6. Checking with a Table | 表格验证
| x | Curve y = x³ − 2x + 1 | Tangent y = 10x − 15 |
| 1.9 | 1.9³ − 3.8 + 1 ≈ 4.059 | 10(1.9) − 15 = 4.0 |
| 2.0 | 5 | 5 |
| 2.1 | 2.1³ − 4.2 + 1 ≈ 6.061 | 10(2.1) − 15 = 6.0 |
very close near x = 2, confirming that the tangent is indeed a good linear approximation.
在 x = 2 附近二者的值非常接近，证实切线确实是一个良好的线性近似。
7. Common Mistakes | 常见错误
Students often make the following errors when solving this type of problem.
学生在解决这类问题时经常犯以下错误。
- Forgetting to substitute the x-value into the original equation to find the y-coordinate. Instead, they may only compute the derivative.
- 忘记将 x 值代入原方程求 y 坐标，而只计算导数。
- Making an arithmetic slip when evaluating the derivative. For instance, 3(2)² may be mistakenly computed as 12 instead of 12 (which is correct), or sometimes as 36.
- 在计算导数值时出现算术错误。例如，3(2)² 可能被误算为 12（这是正确的），但有时会被误算为 36。
- Writing the final equation incorrectly, such as forgetting to expand the bracket or misplacing a sign.
- 写出最终方程时出错，例如忘记展开括号或弄错符号。
8. Extending the Idea | 拓展延伸
Once the tangent is found, we often also find the equation of the normal line, which is perpendicular to the tangent. The gradient of the normal is the negative reciprocal of the tangent's gradient.
求出切线后，我们通常还会求法线方程，法线与切线垂直。法线的斜率是切线斜率的负倒数。
For this example, the gradient of the normal is −1/10. Using point (2, 5):
对于本例，法线的斜率为 −1/10。使用点 (2, 5)：
y − 5 = −(1/10)(x − 2)
y = −x/10 + 1/5 + 5 = −x/10 + 26/5
This is a good exercise to reinforce the relationship between perpendicular gradients.
这是一个很好的练习，可以加深对垂直斜率之间关系的理解。
9. Practice Problem | 练习题目
n: Find the equation of the tangent to the curve y = x² + 3x − 4 at the point where x = 1.
请自行尝试类似题目：求曲线 y = x² + 3x − 4 在 x = 1 处的切线方程。
- First find the point: y = 1² + 3(1) − 4 = 0, so point (1, 0).
- 首先求点：y = 1² + 3(1) − 4 = 0，因此点为 (1, 0)。
- Then differentiate: dy/dx = 2x + 3. At x = 1, gradient = 5.
- 然后求导：dy/dx = 2x + 3。在 x = 1 处，斜率为 5。
- Equation: y − 0 = 5(x − 1) = 5x − 5.
- 方程：y − 0 = 5(x − 1) = 5x − 5。
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