📚 Tangents and Normals | 切线与法线
Differentiation opens the door to a geometric understanding of curves. By computing the derivative of a function, we can determine the precise slope of a curve at any point. Two fundamental lines associated with every smooth point on a curve are the tangent and the normal. The tangent touches the curve gently, sharing its instantaneous direction, while the normal stands perpendicular to the tangent, giving a complementary view of the curve's geometry. This article explores how to find and interpret tangents and normals using the tools of IB Mathematics, covering explicit, implicit, and parametric functions.
微分打开了对曲线进行几何理解的大门。通过计算函数的导数，我们可以确定曲线上任意一点的精确斜率。与曲线上每个光滑点相关联的两条基本直线是切线和法线。切线轻轻触碰曲线，与其瞬时方向重合；法线则与切线垂直，提供了曲线几何的补充视角。本文将探索如何使用 IB 数学工具求取并理解切线与法线，涵盖显函数、隐函数与参数函数。
1. What Is a Tangent? | 什么是切线？
A tangent to a curve at a point P is a straight line that just 'touches' the curve at P without crossing it (locally). It captures the direction in which the curve is heading at that instant. In calculus, the slope of the tangent is given by the derivative of the function at that point.
曲线在点 P 处的切线是一条刚好“触碰”曲线于 P 点而不（局部地）穿过的直线。它抓住了曲线在该瞬间的走向。在微积分中，切线的斜率就是函数在该点的导数。
For a function y = f(x), the gradient mtan of the tangent at x = a is f ' on the curve coalesce, a process that leads directly to the definition of the derivative.
对于函数 y = f(x)，在 x = a 处切线的斜率 mtan 为 f '(a)。这个斜率是当曲线上两点无限接近时割线斜率的极限，这一过程直接引出了导数的定义。
2. The Equation of a Tangent | 切线方程
Once the slope m = f ' nd the point of contact is (a, f(a)), the tangent line can be written in point-slope form:
一旦知道了斜率 m = f '(a) 以及切点 (a, f(a))，切线就可以写成点斜式：
y − f(a) = f '(a)(x − a)
For example, consider f(x) = x² at a = 3. Here f(3) = 9 and f '(x) = 2x, so f '(3) = 6. The tangent is y − 9 = 6(x − 3), which simplifies to y = 6x − 9.
例如，考虑 f(x) = x² 在 a = 3 处。此时 f(3) = 9，f '(x) = 2x，故 f '(3) = 6。切线为 y − 9 = 6(x − 3)，化简得 y = 6x − 9。
ues. The principle remains: find the gradient at the point, then apply y − y₁ = m(x − x₁).
如果曲线以 y = f(x) 的形式给出，这个过程很直接。然而，曲线可能由隐式或参数方程定义，需要不同的求导技术。原则不变：找到该点的斜率，然后应用 y − y₁ = m(x − x₁)。
3. What Is a Normal? | 什么是法线？
ₙ = −1/mₜ. The normal is as steeply sloped as needed to be at right angles to the tangent.
曲线在某点的法线是垂直于该点切线的直线。如果切线的斜率为 mₜ（≠ 0），法线的斜率为 mₙ = −1/mₜ。法线的倾斜程度恰好使其与切线成直角。
versely, if the tangent is vertical (undefined slope), the normal is horizontal: y = f(a).
当 mₜ = 0（水平切线）时，法线为铅直线，其方程就是 x = a。反之，如果切线为铅直线（斜率无定义），法线为水平线：y = f(a)。
, mₜ = 6, so mₙ = −1/6. The normal equation is y − 9 = (−1/6)(x − 3) or y = −(1/6)x + 9.5.
对于同一抛物线 f(x) = x² 在 a = 3 处，mₜ = 6，所以 mₙ = −1/6。法线方程为 y − 9 = (−1/6)(x − 3)，即 y = −(1/6)x + 9.5。
4. Tangents and Normals for Polynomial Functions | 多项式函数的切线与法线
s are the simplest functions to differentiate because of the power rule. For f(x) = axⁿ, f '(x) = naxⁿ⁻¹. This allows quick calculations of tangents and normals at any point.
多项式由于幂法则而成为最容易求导的函数。对于 f(x) = axⁿ，f '(x) = naxⁿ⁻¹。这使得在任何点求切线和法线的计算都非常快捷。
xample: Find the tangent and normal to f(x) = x³ − 3x at x = 2. First, f(2) = 8 − 6 = 2. f '(x) = 3x² − 3, so f ' 2) → y = 9x − 16. Normal: gradient = −1/9, so y − 2 = (−1/9)(x − 2) → y = −(1/9)x + 20/9.
示例：求 f(x) = x³ − 3x 在 x = 2 处的切线与法线。首先，f(2) = 8 − 6 = 2。f '(x) = 3x² − 3，故 f '(2) = 9。切线：y − 2 = 9(x − 2) → y = 9x − 16。法线：斜率 = −1/9，故 y − 2 = (−1/9)(x − 2) → y = −(1/9)x + 20/9。
Even with higher-degree polynomials, the procedure is identical. Always check that the derivative is correct and that the point of contact lies on the curve - a common mistake is to use a point not satisfying f(x).
即使是更高次的多项式，步骤也是一样的。务必确保导数正确，且切点确实在曲线上——一个常见错误是使用了不满足 f(x) 的点。
5. Tangents and Normals Using Implicit Differentiation | 利用隐函数求导的切线与法线
easily isolate y as a function of x. Instead, we differentiate both sides with respect to x, treating y as an implicit function of x. This yields an expression for dy/dx involving both x and y.
当曲线由像 x² + y² = r² 这样的方程隐式定义时，我们不太容易将 y 表示为 x 的函数。此时，我们对方程两边关于 x 求导，将 y 视为 x 的隐函数。这就得到了一个同时含有 x 和 y 的 dy/dx 表达式。
e tangent slope is −3/4, so the tangent is y − 4 = (−3/4)(x − 3). The normal slope is 4/3.
对于圆 x² + y² = 25，求导：2x + 2y(dy/dx) = 0 → dy/dx = −x/y。在点 (3,4) 处，切线斜率为 −3/4，因此切线为 y − 4 = (−3/4)(x − 3)。法线斜率为 4/3。
y²) = −3x² − y → dy/dx = −(3x² + y)/(x + 3y²). Substitute coordinates to get the gradient.
对于像 x³ + xy + y³ = 3 这样更复杂的关系，逐项求导：3x² + (1·y + x·dy/dx) + 3y²(dy/dx) = 0。提取 dy/dx：dy/dx(x + 3y²) = −3x² − y → dy/dx = −(3x² + y)/(x + 3y²)。代入坐标即可得到斜率。
6. Parametric Curves: Tangents and Normals | 参数曲线：切线与法线
y(t). The chain rule gives the derivative dy/dx as (dy/dt) / (dx/dt), provided dx/dt ≠ 0.
参数方程通过将 x 和 y 表示为第三个参数 t 的函数来描述曲线：x = x(t)，y = y(t)。根据链式法则，导数 dy/dx = (dy/dt) / (dx/dt)，条件是 dx/dt ≠ 0。
nt is (3−1)/(2) = 1. Tangent: y − 0 = 1(x − 1) → y = x − 1. Normal: y = −(x − 1) = −x + 1.
示例：x = t²，y = t³ − t。导数为 dx/dt = 2t，dy/dt = 3t² − 1。因此 dy/dx = (3t² − 1)/(2t)。在 t = 1 处，点为 (1,0)，斜率为 (3−1)/(2) = 1。切线：y − 0 = 1(x − 1) → y = x − 1。法线：y = −(x − 1) = −x + 1。
dy/dt = 0 gives an indeterminate form; further limit analysis is needed to find the slope.
当 dx/dt = 0 且 dy/dt ≠ 0 时，切线为垂直。例如，对于 x = t²，y = t³ 在 t = 0 处，dx/dt = 0，dy/dt = 0 形成一个不定式；需要通过极限进一步分析来求斜率。
7. Horizontal and Vertical Tangents | 水平切线与垂直切线
nts occur where dy/dx = 0, meaning the curve momentarily levels out. For y = f(x), solve f ' dy/dt = 0 while dx/dt ≠ 0. The normal at such a point is vertical, given by x = constant.
水平切线出现在 dy/dx = 0 处，意味着曲线瞬间趋于平坦。对 y = f(x)，解 f '(x) = 0。对于参数曲线，令 dy/dt = 0 且 dx/dt ≠ 0。在该点处的法线为铅直线，即 x = 常数。
(x), this can happen at cusps or where the function has an undefined derivative (e.g., y = ∛ ns are dx/dt = 0 and dy/dt ≠ 0. The tangent is x = x₀ and the normal is horizontal y = y₀.
垂直切线出现在导数趋于无穷大处。对于 y = f(x)，这可能发生在尖点或函数导数不存在之处（如 y = ∛x² 在 x=0 处有无穷斜率）。对于参数曲线，条件为 dx/dt = 0 且 dy/dt ≠ 0。切线为 x = x₀，法线为水平线 y = y₀。
ends to infinity, indicating a vertical tangent x = 1. The normal is y = 0 (since f(1)=0).
示例：曲线 y = (x − 1)²/³。导数为 (2/3)(x − 1)⁻¹/³。在 x = 1 处，导数趋于无穷大，表明有垂直切线 x = 1。法线为 y = 0（因 f(1)=0）。
8. Intersection of Tangents and Normals | 切线与法线的交点
Sometimes we need to find where the tangent at one point meets the tangent or normal at another point, or where a normal crosses the curve again. These intersection problems involve solving simultaneous equations of the line with another line or with the original curve.
有时我们需要求出某点的切线与另一点的切线或法线的交点，或者法线与曲线再次相交于何处。这类交点问题需要联立直线方程与另一条直线方程或与原曲线方程进行求解。
olving 2x − 1 = −2x − 1 gives 4x = 0 → x = 0, y = −1. So the tangents intersect at (0,−1).
例如，求抛物线 y = x² 在 x = 1 和 x = −1 处的切线的交点。在 x=1 处，点 (1,1)，切线 y − 1 = 2(x − 1) → y = 2x − 1。在 x=−1 处，点 (−1,1)，切线 y − 1 = −2(x + 1) → y = −2x − 1。解 2x − 1 = −2x − 1 得 4x = 0 → x = 0，y = −1。因此两切线交于点 (0,−1)。
/x = 1, so normal slope = −1. Normal: y = −(x − 1) → y = −x + 1. The x-intercept is at y=0 ⇒ s at the point of tangency; to find a second intersection we would need a different point.
另一个经典问题：求 y = ln x 在 x = 1 处的法线与 x 轴的再次交点。在 x=1 处，y=0，导数 1/x = 1，法线斜率 = −1。法线：y = −(x − 1) → y = −x + 1。x 轴截距在 y=0 ⇒ x=1。因此它只在切点处相交；要寻求第二个交点需要另一个点。
9. Angle between Two Curves and Their Tangents | 两曲线及其切线间的夹角
When two curves intersect, the angle between them is defined as the angle between their tangents at the point of intersection. If the tangents have gradients m₁ and m₂, the acute angle θ satisfies:
当两条曲线相交时，它们之间的夹角定义为在交点处两条切线的夹角。如果切线斜率分别为 m₁ 和 m₂，则锐角 θ 满足：
tan θ = |(m₁ − m₂) / (1 + m₁ m₂)|
e orthogonal (perpendicular). This concept appears when proving that circles with different radii intersecting at a common point might have orthogonal tangents if certain conditions hold.
如果乘积 m₁ m₂ = −1，则曲线正交（垂直）。当证明不同半径的圆在公共点相交时，如果满足一定条件，其切线可能会正交，这时就会用到这一概念。
Example: The curves y = x² and y = √x intersect at (1,1). For y = x², m₁ = 2x = 2 at x=1. For y = √x, differentiate: y = x^½ so dy/dx = (1/2)x−½ = 1/(2√x), m₂ = 1/2. Then tan θ = |(2 − 1/2) / (1 + 2·½)| = |(3/2) / (2)| = 3/4. Hence θ = arctan(3/4) ≈ 36.9°.
示例：曲线 y = x² 与 y = √x 交于 (1,1)。对 y = x²，m₁ = 2x 在 x=1 时为 2。对 y = √x，求导：y = x^½，故 dy/dx = (1/2)x−½ = 1/(2√x)，m₂ = 1/2。然后 tan θ = |(2 − 1/2) / (1 + 2·½)| = |(3/2) / (2)| = 3/4。因此 θ = arctan(3/4) ≈ 36.9°。
10. Practical Tips and Common Pitfalls | 实用技巧与常见误区
Always confirm the given point lies on the curve before computing a tangent or normal. Plug the x-coordinate into f(x) to verify the y-value matches.
在计算切线或法线之前，务必确认给定点确实在曲线上。将 x 坐标代入 f(x)，核对 y 值是否匹配。
For implicit differentiation, remember to apply the chain rule to terms involving y. A missing dy/dx factor is a frequent error.
对于隐函数求导，要记住对含有 y 的项使用链式法则。遗漏 dy/dx 因子是一个常见错误。
When finding normals, it is easy to forget the negative reciprocal. Double-check the sign: mnormal = −1/mtangent.
求法线时，很容易忘记负倒数。仔细检查符号：m法线 = −1/m切线。
ifferentiate y with respect to x directly; many candidates mistakenly write dy/dx = dy/dt.
对于参数曲线，斜率为 (dy/dt)/(dx/dt)。不要直接对 x 求导 y；许多考生错误地写成 dy/dx = dy/dt。
al and vice versa. State the equation carefully: x = a for vertical, y = b for horizontal.
如果切线是垂直的，法线就是水平的，反之亦然。仔细写出方程：垂直为 x = a，水平为 y = b。
When finding the angle between two curves, use the absolute value to obtain the acute angle.
求两曲线之间的夹角时，要使用绝对值以得到锐角。
11. Worked Problem Set | 精选题集
Problem 1: For the curve y = e²ˣ + 3x, find the tangent at x = 0 and the normal at the same point. f '(x) = 2e²ˣ = 0, point is (0, e⁰ + 0) = (0,1). Tangent slope = 2·1 + 3 = 5. Tangent: y − 1 = 5(x − 0) ⇒ y = 5x + 1. Normal slope = −1/5, so normal: y − 1 = −⅕ x.
问题1：对于曲线 y = e²ˣ + 3x，求在 x = 0 处的切线与法线。f '(x) = 2e²ˣ + 3。在 x = 0 处，点为 (0, e⁰ + 0) = (0,1)。切线斜率 = 2·1 + 3 = 5。切线：y − 1 = 5(x − 0) ⇒ y = 5x + 1。法线斜率 = −1/5，法线：y − 1 = −⅕ x。
oint where t = π/4. First, dx/dt = −2sin t, dy/dt = 3cos t. At t = π/4, sin(π/4)=cos(π/4)=√2/2. Then dx/dt = −2·√2/2 = −√2, dy/dt = 3·√2/2. Gradient dy/dx = (3√2/2)/(−√2) = −3/2. Point is (2·√2/2, 3·√2/2) = (√2, (3√2)/2). Tangent: y − (3√2)/2 = −3/2 (x − √2).
问题2：参数方程为 x = 2cos t，y = 3sin t。求在 t = π/4 处的切线方程。首先，dx/dt = −2sin t，dy/dt = 3cos t。在 t = π/4 时，sin(π/4)=cos(π/4)=√2/2。则 dx/dt = −2·√2/2 = −√2，dy/dt = 3·√2/2。斜率 dy/dx = (3√2/2)/(−√2) = −3/2。点为 (2·√2/2, 3·√2/2) = (√2, (3√2)/2)。切线：y − (3√2)/2 = −3/2 (x − √2)。
y/dx = −2/(1+2) = −2/3. Tangent gradient = −2/3, normal gradient = 3/2. Normal equation: y
Published by TutorHao | IB Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图