Normal Equations of Curves | IB数学：曲线的法线方程

📚 Normal Equations of Curves | IB数学：曲线的法线方程

In coordinate geometry, the normal to a curve at a given point is the straight line perpendicular to the tangent at that point. Understanding how to derive and use normal equations is a fundamental skill in IB Mathematics, particularly in the Analysis and Approaches (AA) and Applications and Interpretation (AI) courses.

在坐标几何中，曲线在某一点的法线，是指在该点与切线垂直的直线。掌握如何求法线方程并加以运用，是IB数学中的基础技能，尤其适用于 Analysis and Approaches (AA) 与 Applications and Interpretation (AI) 课程。

1. Tangent and Normal | 切线与法线

For a differentiable function \(y = f(x)\), the tangent line at a point \((x₁, y₁)\) is the line that “touches” the curve at that point and has the same instantaneous rate of change (derivative). The normal line is perpendicular to the tangent at the point of contact.

对于可导函数 \(y = f(x)\)，曲线在点 \((x₁, y₁)\) 处的切线是“接触”该点且具有相同瞬时变化率（导数）的直线。法线则是切点处与切线垂直的直线。

If the tangent has slope \(mₜ\), then the normal has slope \(mₙ\), provided the tangent is not vertical/horizontal in a degenerate way. The relationship is:

若切线斜率为 \(mₜ\)，则法线斜率为 \(mₙ\)，前提是切线没有退化为竖直或水平等特殊情况。其关系为：

mₜ × mₙ = −1 ⇔ mₙ = −1 / mₜ (mₜ ≠ 0)

When the tangent is horizontal (slope 0), the normal is vertical; when the tangent is vertical, the normal is horizontal.

当切线水平（斜率为0）时，法线竖直；当切线竖直时，法线水平。

2. General Procedure | 一般求解步骤

To find the equation of the normal to a curve at a given point, follow these steps:

要求曲线在某一点的法线方程，可按以下步骤进行：

Find the derivative \(\frac{dy}{dx}\) of the curve.
求曲线的导数 \(\frac{dy}{dx}\)。
Evaluate \(\frac{dy}{dx}\) at the given x-coordinate \(x₁\) to obtain the tangent slope \(mₜ\).
将给定点的横坐标 \(x₁\) 代入导数，得到切线斜率 \(mₜ\)。
Compute the normal slope \(mₙ = -\frac{1}{mₜ}\) (if \(mₜ ≠ 0\)).
计算法线斜率 \(mₙ = -\frac{1}{mₜ}\)（若 \(mₜ ≠ 0\)）。
Use the point-slope form: \(y – y₁ = mₙ (x – x₁)\).
使用点斜式：\(y – y₁ = mₙ (x – x₁)\)。
Simplify to the required form, e.g. \(y = mx + c\) or \(ax + by + c = 0\).
化简为所需形式，如 \(y = mx + c\) 或 \(ax + by + c = 0\)。

3. Worked Example: Parabola | 例题：抛物线

Let us find the normal to the parabola \(y = x²\) at the point where \(x = 1\).

我们求抛物线 \(y = x²\) 在 \(x = 1\) 处的法线。

When \(x = 1\), \(y = 1² = 1\), so the point is \((1, 1)\). The derivative is \(\frac{dy}{dx} = 2x\). At \(x = 1\), the tangent slope is \(mₜ = 2\). Therefore the normal slope is

当 \(x = 1\) 时，\(y = 1² = 1\)，所以点为 \((1, 1)\)。导数为 \(\frac{dy}{dx} = 2x\)。在 \(x = 1\) 处，切线斜率 \(mₜ = 2\)。因此法线斜率为

mₙ = −1/2

Using point-slope form:

使用点斜式：

y − 1 = −1/2 (x − 1) ⇒ y = −1/2 x + 3/2

Thus the normal equation is \(y = -\frac{1}{2}x + \frac{3}{2}\). In standard form: \(x + 2y – 3 = 0\).

因此法线方程为 \(y = -\frac{1}{2}x + \frac{3}{2}\)。标准形式为 \(x + 2y – 3 = 0\)。

4. Implicit Differentiation | 隐函数求导

When the curve is given implicitly, e.g. \(x² + y² = 25\), we differentiate both sides with respect to \(x\).

当曲线以隐式给出，例如 \(x² + y² = 25\)，我们对两边关于 \(x\) 求导。

For \(x² + y² = 25\), differentiating gives \(2x + 2y \frac{dy}{dx} = 0\), so

对于 \(x² + y² = 25\)，求导得 \(2x + 2y \frac{dy}{dx} = 0\)，因此

\frac{dy}{dx} = -\frac{x}{y}

At a point \((x₁, y₁)\) on the circle, the tangent slope is \(-\frac{x₁}{y₁}\), and the normal slope is the negative reciprocal:

在圆上一点 \((x₁, y₁)\) 处，切线斜率为 \(-\frac{x₁}{y₁}\)，法线斜率为其负倒数：

mₙ = \frac{y₁}{x₁} (x₁ ≠ 0)

Example: At \((3, 4)\), the normal slope is \(4/3\), so the normal is \(y – 4 = \frac{4}{3}(x – 3)\). Simplified: \(4x – 3y = 0\), which passes through the origin. Indeed, for a circle centred at the origin, the normal at any point passes through the centre.

例如：在点 \((3, 4)\) 处，法线斜率为 \(4/3\)，所以法线为 \(y – 4 = \frac{4}{3}(x – 3)\)。化简得 \(4x – 3y = 0\)，它经过原点。事实上，对于圆心在原点的圆，任意点处的法线都经过圆心。

5. Parametric Equations | 参数方程

For a parametric curve \(x = f(t), y = g(t)\), the slope of the tangent is \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\), provided \(\frac{dx}{dt} ≠ 0\).

对于参数曲线 \(x = f(t), y = g(t)\)，切线斜率为 \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\)，前提是 \(\frac{dx}{dt} ≠ 0\)。

Then the normal slope is the negative reciprocal. For example, consider the parametric curve \(x = t²\), \(y = t³\) at \(t = 1\).

然后法线斜率为负倒数。例如，考虑参数曲线 \(x = t²\)，\(y = t³\) 在 \(t = 1\) 处。

At \(t = 1\): \(x = 1, y = 1\). Now \(\frac{dx}{dt} = 2t = 2\), \(\frac{dy}{dt} = 3t² = 3\). The tangent slope is \(\frac{3}{2}\), so the normal slope is \(-\frac{2}{3}\). The normal equation is

当 \(t = 1\) 时：\(x = 1, y = 1\)。 \(\frac{dx}{dt} = 2t = 2\)，\(\frac{dy}{dt} = 3t² = 3\)。切线斜率为 \(\frac{3}{2}\)，所以法线斜率为 \(-\frac{2}{3}\)。法线方程为

y − 1 = −2/3 (x − 1) ⇒ 2x + 3y − 5 = 0

6. Normal Passing Through a Given Point | 过定点的法线

A common IB question gives a fixed point \(P(a, b)\) and asks for the point(s) on the curve where the normal passes through \(P\). This means the normal line at \(Q(x₁, y₁)\) must satisfy the condition that \(P\) lies on it.

一类常见的IB问题给出定点 \(P(a, b)\)，要求曲线上法线经过 \(P\) 的点。这意味着在 \(Q(x₁, y₁)\) 处的法线必须满足 \(P\) 在其上。

If the normal slope at \(Q\) is \(mₙ\), then the line is \(y – y₁ = mₙ(x – x₁)\). Substituting \(P\) gives:

若 \(Q\) 处法线斜率为 \(mₙ\)，则直线为 \(y – y₁ = mₙ(x – x₁)\)。代入 \(P\) 得：

b − y₁ = mₙ (a − x₁)

This yields an equation in \(x₁\) (or \(t\)), which can be solved to find the required points.

这给出关于 \(x₁\)（或参数 \(t\)）的方程，求解即可得到所求点。

Example: For \(y = x²\), find the point where the normal passes through \((0, 1)\). Let the point be \((a, a²)\). The derivative is \(2a\), so the normal slope is \(-1/(2a)\). The normal line is \(y – a² = -\frac{1}{2a}(x – a)\). Substituting \((0, 1)\):

例：对于 \(y = x²\)，求法线经过 \((0, 1)\) 的点。设点为 \((a, a²)\)。导数为 \(2a\)，所以法线斜率为 \(-1/(2a)\)。法线为 \(y – a² = -\frac{1}{2a}(x – a)\)。代入 \((0, 1)\)：

1 − a² = −1/(2a)(0 − a) = 1/2

Thus \(a² = 1/2\), so \(a = ±1/\sqrt{2}\). Hence the points are \(( ±1/\sqrt{2}, 1/2)\).

于是 \(a² = 1/2\)，所以 \(a = ±1/\sqrt{2}\)。因此点为 \(( ±1/\sqrt{2}, 1/2)\)。

7. Intersection with Axes | 与坐标轴的交点

Often the normal is required in order to find its intercepts with the \(x\)- or \(y\)-axis. These intercepts are useful in solving geometry or optimization problems.

通常需要求出法线与 \(x\) 轴或 \(y\) 轴的交点，这些截距可用于解决几何或最优化问题。

For example, the normal to \(y = x²\) at \((1, 1)\) is \(y = -\frac{1}{2}x + \frac{3}{2}\). Its \(x\)-intercept is obtained by setting \(y = 0\):

例如，\(y = x²\) 在 \((1, 1)\) 处的法线为 \(y = -\frac{1}{2}x + \frac{3}{2}\)。令 \(y = 0\) 求其 \(x\) 截距：

0 = −1/2 x + 3/2 ⇒ x = 3

So the normal cuts the \(x\)-axis at \((3, 0)\).

因此法线与 \(x\) 轴交于 \((3, 0)\)。

8. Normal and Tangent Perpendicularity | 法线与切线的垂直关系

By definition, the normal and tangent at the same point are always perpendicular. This fact is sometimes used to verify results or to construct normals without calculus.

根据定义，同一点处的法线与切线始终垂直。这一事实有时可用于验证结果，或在不用微积分的情况下构造法线。

For a circle, each normal is a radius line; for other curves, the normal plays an important role in physics (e.g. acceleration in circular motion, reflection laws).

对于圆，每条法线都是半径所在直线；对于其他曲线，法线在物理学中（如圆周运动中的加速度、反射定律）发挥重要作用。

9. Common Pitfalls | 常见易错点

Forgetting to take the negative reciprocal of the tangent slope; using the tangent slope instead of the normal slope.
忘记对切线斜率取负倒数，而误用切线斜率作为法线斜率。
When \(\frac{dy}{dx} = 0\), students may incorrectly write \(mₙ = -∞\). The normal is simply the vertical line \(x = x₁\).
当 \(\frac{dy}{dx} = 0\) 时，学生可能错误地写为 \(mₙ = -∞\)。此时法线就是竖直线 \(x = x₁\)。
For implicit curves, forgetting to include the \(\frac{dy}{dx}\) factor when differentiating \(y²\) terms.
对于隐式曲线，对含 \(y²\) 的项求导时忘记乘以 \(\frac{dy}{dx}\)。
Substituting the given \(x\)-value into \(f'(x)\) correctly, but then using the original \(x\)-value in point-slope form instead of the point’s \(y\)-coordinate.
用给定的 \(x\) 值正确计算 \(f'(x)\)，但在点斜式中代入原 \(x\) 值而不是点的纵坐标。

10. Normal Equation from First Principles | 由第一性原理求法线

If the curve is a straight line \(y = mx + c\), then its normal at any point is the line with slope \(-1/m\) passing through that point. For a circle, the normal at \((x₁, y₁)\) has slope \(\frac{y₁}{x₁}\) (if \(x₁ ≠ 0\)). These can be derived directly from geometric perpendicularity without calculus.

如果曲线本身就是直线 \(y = mx + c\)，则其上任意一点处的法线是经过该点且斜率为 \(-1/m\) 的直线。对于圆，在 \((x₁, y₁)\) 处法线的斜率为 \(\frac{y₁}{x₁}\)（若 \(x₁ ≠ 0\)）。这些可以直接由几何垂直关系得出，无需微积分。

However, for general curves, calculus is the most reliable method.

但对于一般曲线，微积分是最可靠的方法。

11. IB Exam-Style Question | IB风格考题

Let us work through a typical IB problem step by step.

我们逐步完成一道典型的IB问题。

Question: Consider the curve \(y = \frac{1}{x}\). Find the equation of the normal at the point where \(x = 2\).

问题：考虑曲线 \(y = \frac{1}{x}\)。求 \(x = 2\) 处的法线方程。

Step 1: When \(x = 2\), \(y = \frac{1}{2}\), so the point is \((2, 0.5)\).

第一步：当 \(x = 2\) 时，\(y = \frac{1}{2}\)，所以点为 \((2, 0.5)\)。

Step 2: \(\frac{dy}{dx} = -\frac{1}{x²}\). At \(x = 2\), \(mₜ = -\frac{1}{4}\).

第二步：\(\frac{dy}{dx} = -\frac{1}{x²}\)。在 \(x = 2\) 处，\(mₜ = -\frac{1}{4}\)。

Step 3: Normal slope \(mₙ = -1 / (-\frac{1}{4}) = 4\).

第三步：法线斜率 \(mₙ = -1 / (-\frac{1}{4}) = 4\)。

Step 4: \(y – \frac{1}{2} = 4(x – 2)\), so \(y = 4x – \frac{15}{2}\).

第四步：\(y – \frac{1}{2} = 4(x – 2)\)，即 \(y = 4x – \frac{15}{2}\)。

The final answer is \(y = 4x – 7.5\).

最终答案为 \(y = 4x – 7.5\)。

12. Summary | 总结

The normal line at a point on a curve is the perpendicular to the tangent. Its slope is the negative reciprocal of the derivative at that point, except in the special cases of horizontal or vertical tangents. The standard method is: differentiate, evaluate \(mₜ\), compute \(mₙ\), and write the equation using the point-slope form.

曲线上一点处的法线是与切线垂直的直线。其斜率为该点导数的负倒数，除切线水平或竖直的特殊情况外。标准方法是：求导，计算 \(mₜ\)，求出 \(mₙ\)，再用点斜式写出方程。

Mastering normal equations is essential for success in both the calculator and non-calculator papers of IB Mathematics, as it links differentiation with coordinate geometry and has many applications.

掌握法线方程是IB数学中计算器与非计算器试卷取得成功的关键，它将微分与坐标几何联系起来，并具有广泛应用。

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Normal Equations of Curves | IB数学：曲线的法线方程