📚 Tangent and Normal Line Equations | 切线与法线的方程求解
In IB Mathematics (Analysis and Approaches & Applications and Interpretation), finding the equations of tangent and normal lines is a core skill in differential calculus. It combines the concept of the derivative as a slope with the algebraic form of a straight line.
在 IB 数学（分析与方法 & 应用与解释）中，求切线与法线的方程是微积分中的核心技能。它将导数作为斜率的概念与直线方程的代数形式结合起来。
1. The Derivative as Slope | 导数就是斜率
For a function y = f(x), the derivative f'(x) at a point x = a gives the slope of the tangent line to the curve at that point. The slope is denoted by m_tangent.
对于函数 y = f(x)，在点 x = a 处的导数 f'(a) 给出了曲线在该点的切线斜率。该斜率记为 m_切线。
m_tangent = f'(a)
This slope represents the instantaneous rate of change of the function at x = a.
这个斜率表示函数在 x = a 处的瞬时变化率。
2. Equation of a Straight Line | 直线方程
To write the equation of any line, we need a point (x₁, y₁) and a slope m. The point-slope form is:
要写出任意直线方程，我们需要一个点 (x₁, y₁) 和斜率 m。点斜式方程为：
y − y₁ = m(x − x₁)
Alternatively, the equation can be rearranged into slope-intercept form y = mx + c.
也可以整理成斜截式 y = mx + c。
3. Tangent Line at a Point | 在某点处的切线
Given y = f(x), the tangent at x = a is the line that touches the curve at (a, f(a)) and has the same slope as the curve at that instant.
给定 y = f(x)，在 x = a 处的切线是指与曲线在点 (a, f(a)) 相切且在该处与曲线具有相同斜率的直线。
The equation of the tangent is:
切线方程为：
y − f(a) = f'(a)(x − a)
Example: For f(x) = x², find the tangent at x = 3.
例：对于 f(x) = x²，求 x = 3 处的切线。
f(3) = 9, f'(x) = 2x, so f'(3) = 6. Tangent: y − 9 = 6(x − 3), i.e. y = 6x − 9.
f(3) = 9，f'(x) = 2x，所以 f'(3) = 6。切线：y − 9 = 6(x − 3)，即 y = 6x − 9。
4. Normal Line at a Point | 在某点处的法线
The normal line is perpendicular to the tangent at the point of contact. If the tangent slope is m, the normal slope is −1/m (provided m ≠ 0).
法线是与切线在接触点处垂直的直线。若切线斜率为 m，则法线斜率为 −1/m（前提是 m ≠ 0）。
Equation of the normal:
法线方程为：
y − f(a) = −1/f'(a) (x − a)
For the example above, f'(3) = 6, so the normal slope is −1/6. Normal: y − 9 = −1/6 (x − 3).
对于上面的例子，f'(3) = 6，所以法线斜率为 −1/6。法线：y − 9 = −1/6 (x − 3)。
5. Horizontal and Vertical Cases | 水平与垂直的情况
If f'(a) = 0, the tangent is horizontal: y = f(a). The normal is then vertical, with equation x = a.
如果 f'(a) = 0，切线水平：y = f(a)。此时法线垂直，方程为 x = a。
If the tangent is vertical (which occurs when f'(a) is undefined but the curve has a vertical tangent), the normal is horizontal: y = f(a).
如果切线垂直（当 f'(a) 无定义但曲线存在垂直切线时），法线水平：y = f(a)。
Always check these special cases in exam questions.
在考试题目中务必检查这些特殊情况。
6. Finding the Point from a Given Slope | 由已知斜率求切点
Sometimes the question gives the slope of the tangent and asks for the point of tangency. Set f'(x) equal to the given slope and solve for x.
有时题目给出切线斜率，要求切点。令 f'(x) 等于给定斜率，解出 x。
Example: Find the point on y = x² − 4x + 3 where the tangent has slope 2.
例：求 y = x² − 4x + 3 上切线斜率为 2 的点。
f'(x) = 2x − 4. Set 2x − 4 = 2 → x = 3. Then y = 9 − 12 + 3 = 0. Point: (3, 0).
f'(x) = 2x − 4。令 2x − 4 = 2 → x = 3。则 y = 9 − 12 + 3 = 0。点为 (3, 0)。
7. Tangent Passing Through an External Point | 过曲线外一点的切线
If a tangent passes through a point (x₁, y₁) that is not necessarily on the curve, let the point of tangency be (a, f(a)). The slope of the line joining the two points must equal f'(a).
如果一条切线通过一个不一定在曲线上的点 (x₁, y₁)，设切点为 (a, f(a))。连接两点的直线斜率必须等于 f'(a)。
(y₁ − f(a)) / (x₁ − a) = f'(a)
Solve this equation for a. There may be more than one solution, giving multiple tangents.
解这个方程求 a。可能有多个解，从而得到多条切线。
8. Parametric Curves | 参数曲线
For a curve defined parametrically by x = x(t) and y = y(t), the slope of the tangent is dy/dx = (dy/dt)/(dx/dt), provided dx/dt ≠ 0.
对于由参数方程 x = x(t) 和 y = y(t) 定义的曲线，切线斜率为 dy/dx = (dy/dt)/(dx/dt)，前提是 dx/dt ≠ 0。
At the point corresponding to t₀, the tangent line is:
在对应 t₀ 的点处，切线为：
y − y(t₀) = (dy/dx)(t₀) · (x − x(t₀))
For example, x = t², y = t³. Then dy/dx = (3t²)/(2t) = 3t/2 (for t ≠ 0). At t = 2, point (4, 8), slope = 3, so tangent: y − 8 = 3(x − 4).
例如，x = t²，y = t³。则 dy/dx = (3t²)/(2t) = 3t/2（t ≠ 0）。在 t = 2 时，点为 (4, 8)，斜率 = 3，因此切线：y − 8 = 3(x − 4)。
9. Implicit Differentiation | 隐函数求导
For curves defined implicitly, such as x² + y² = 25, differentiate both sides with respect to x, treating y as a function of x.
对于隐式定义的曲线，如 x² + y² = 25，两边对 x 求导，将 y 视为 x 的函数。
Example: For x² + y² = 25, we get 2x + 2y·dy/dx = 0, so dy/dx = −x/y.
例：对于 x² + y² = 25，得到 2x + 2y·dy/dx = 0，所以 dy/dx = −x/y。
At (3, 4), slope = −3/4. Tangent: y − 4 = −3/4 (x − 3). Normal slope = 4/3.
在 (3, 4) 处，斜率 = −3/4。切线：y − 4 = −3/4 (x − 3)。法线斜率 = 4/3。
10. Using the GDC / Calculator | 使用图形计算器 / 计算器
In IB exams, a GDC can be used to find tangent lines. On most calculators, use the “dy/dx” or “tangent” feature after graphing the function.
在 IB 考试中，可以使用图形计算器求切线。在大多数计算器上，绘制函数后使用 “dy/dx” 或 “tangent” 功能。
However, you must show the analytic method in the working for full marks. Use the GDC to verify your answer.
然而，要获得满分，你必须在解题过程中展示解析方法。使用计算器来验证答案。
| Method | When to Use |
| Point-slope form | Point and slope known |
| Set derivative equal to slope | Slope given, find point |
| External point condition | Tangent passes through external point |
| Implicit differentiation | Curve not explicitly solved for y |
11. Common Exam Mistakes | 常见考试错误
- Forgetting to find the y-coordinate: always evaluate f(a) first.
- 忽略求 y 坐标：务必先计算 f(a)。
- Using the original function instead of its derivative for the slope.
- 用原函数而不是其导数求斜率。
- Taking the negative reciprocal incorrectly: the reciprocal of m is 1/m, then change the sign.
- 错误地取负倒数：m 的倒数是 1/m，然后变号。
- Confusing 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。
12. Worked Example: Full Process | 综合例题：完整过程
Find the equations of the tangent and normal to f(x) = x³ − 3x + 2 at x = 1.
求 f(x) = x³ − 3x + 2 在 x = 1 处的切线与法线方程。
Step 1: f(1) = 1 − 3 + 2 = 0. Point (1, 0).
步骤 1：f(1) = 1 − 3 + 2 = 0。点为 (1, 0)。
Step 2: f'(x) = 3x² − 3. f'(1) = 0.
步骤 2：f'(x) = 3x² − 3。f'(1) = 0。
Step 3: Tangent slope = 0, so tangent is horizontal: y = 0.
步骤 3：切线斜率 = 0，所以切线水平：y = 0。
Step 4: Normal is vertical: x = 1.
步骤 4：法线垂直：x = 1。
Always interpret zero and undefined slopes correctly.
务必正确理解零斜率和无定义斜率的情况。
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