📚 Gradients, Tangents and Normals | 梯度、切线与法线
Finding the gradient of a curve and using it to write down the equations of tangents and normals is one of the most reliable sources of marks in A-Level Mathematics. The topic sits at the meeting point of two big ideas: differentiation, which tells you how steep a curve is at an instant, and coordinate geometry, which lets you turn that steepness into a straight-line equation. Once you can move confidently between a curve, its derivative and the straight lines that touch or cross it at right angles, a whole family of exam questions becomes routine.
求曲线的梯度，并用它写出切线与法线的方程，是 A-Level 数学中最稳定的得分点之一。本专题处在两大思想的交汇处：微分告诉你曲线在某一瞬间有多陡，坐标几何则让你把这个陡峭程度转化为一条直线的方程。一旦你能在曲线、它的导数、以及与它相切或垂直的直线之间自如切换，一整类考题就会变得十分套路化。
1. Gradients and the Derivative | 梯度与导数
For a straight line the gradient is constant, so a single number describes the whole line. A curve is different: its steepness changes from point to point. The gradient of a curve at a particular point is defined as the gradient of the tangent to the curve at that point - the straight line that just touches the curve there and matches its direction exactly.
对于直线，梯度是常数，一个数字就能描述整条直线。曲线则不同：它的陡峭程度逐点变化。曲线在某一点的梯度，定义为该点处切线的梯度——这条直线恰好在该点与曲线接触，并且方向与曲线完全一致。
just a number, for this instantaneous gradient. If y = f(x), then the derivative dy/dx = f' function: substitute an x-value and it returns the gradient of the curve at that x-value.
微分给出的不是一个数字，而是一个求瞬时梯度的公式。若 y = f(x)，则导数 dy/dx = f'(x) 就是梯度函数：代入一个 x 值，它就返回曲线在该 x 处的梯度。
dy/dx = f'(x) = gradient of the tangent at the point (x, f(x))
t is 2 × 3 = 6; at x = -1 it is 2 × (-1) = -2; at x = 0 it is 0, which is why the curve is momentarily flat at its minimum point.
对于简单的二次函数 y = x²，微分得到 dy/dx = 2x。因此任意 x 坐标处的梯度都是该值的两倍。当 x = 3 时梯度为 2 × 3 = 6；当 x = -1 时为 2 × (-1) = -2；当 x = 0 时为 0，这正是曲线在最低点处瞬时水平的原因。
- Positive gradient - the curve is rising as x increases. | 梯度为正——随着 x 增大，曲线上升。
- Negative gradient - the curve is falling as x increases. | 梯度为负——随着 x 增大，曲线下降。
- Zero gradient - the tangent is horizontal; usually a stationary point. | 梯度为零——切线水平，通常是驻点。
2. The Tangent Line at a Point | 某点处的切线方程
Once you know the gradient m of the curve at a point, the tangent is simply the straight line through that point with gradient m. The standard tool is the point-gradient form of a line, which works whenever you know one point and the slope.
一旦你知道曲线在某点的梯度 m，切线就是过该点、梯度为 m 的直线。标准工具是直线的点斜式，只要知道一个点和斜率就能使用。
y - y₁ = m(x - x₁)
Here (x₁, y₁) is the point of contact on the curve and m is the value of dy/dx at that same x-value. The method is always the same three steps: find the y-coordinate by substituting into the original equation, differentiate to get the gradient function, then substitute to get the numerical gradient.
其中 (x₁, y₁) 是曲线上的切点，m 是 dy/dx 在同一 x 值处的取值。方法永远是同样的三步：代回原方程求出 y 坐标；求导得到梯度函数；再代入求出数值梯度。
| Step | 步骤 | Action | 操作 | Why | 原因 |
| 1 | Substitute x into y = f(x) | 将 x 代入 y = f(x) | You need both coordinates of the point | 需要点的两个坐标 |
| 2 | Differentiate to get dy/dx | 求导得 dy/dx | You need the gradient function | 需要梯度函数 |
| 3 | Substitute x into dy/dx | 将 x 代入 dy/dx | You need the numerical gradient m | 需要数值梯度 m |
3. Worked Example: Tangent to a Quadratic | 例题：二次曲线的切线
g the y-coordinate: y = 3² - 4(3) + 7 = 9 - 12 + 7 = 4, so the point of contact is (3, 4).
求曲线 y = x² - 4x + 7 在 x = 3 处的切线方程。先求 y 坐标：y = 3² - 4(3) + 7 = 9 - 12 + 7 = 4，所以切点为 (3, 4)。
3 gives m = 2(3) - 4 = 2. The tangent therefore has gradient 2 and passes through (3, 4).
逐项求导：dy/dx = 2x - 4。代入 x = 3 得 m = 2(3) - 4 = 2。因此切线的梯度为 2，且经过点 (3, 4)。
y - 4 = 2(x - 3) → y = 2x - 2
n: at x = 3, y = 2(3) - 2 = 4, which matches the point of contact, so the line is correct.
写成 ax + by + c = 0 的形式为 2x - y - 2 = 0。务必把点代回检验：当 x = 3 时 y = 2(3) - 2 = 4，与切点一致，说明直线正确。
the quadratic has already passed its minimum at x = 2, the curve is indeed rising at x = 3.
顺便检验梯度符号很有价值：梯度为正意味着切线从左向右上升。由于 2 > 0，且该二次函数在 x = 2 处已过最低点，曲线在 x = 3 处确实在上升。
4. The Normal Line and the Negative Reciprocal | 法线与负倒数
The normal to a curve at a point is the straight line through that point which is perpendicular to the tangent. Because it is perpendicular, its gradient is the negative reciprocal of the tangent gradient. This single rule converts almost every tangent question into a normal question.
曲线在某点的法线，是过该点且垂直于切线的直线。由于垂直，其梯度是切线梯度的负倒数。仅凭这一条规则，几乎所有切线题都能转化为法线题。
If the tangent gradient is m, the normal gradient is -1/m
verify with any pair of perpendicular lines you already know, such as y = 2x and y = -½x.
背后的垂直条件是两条直线梯度的乘积等于 -1，你可以用任何一对已知的垂直直线验证，例如 y = 2x 与 y = -½x。
m₁ × m₂ = -1 ⇌ m₂ = -1 ÷ m₁
e normal is horizontal with equation y = y₁. Vertical tangents occur where dy/dx is undefined or infinite, for example where a curve has a cusp or where a denominator in dy/dx becomes zero.
两种特殊情况需要留意。若切线水平（m = 0），法线垂直，方程直接写作 x = x₁；若切线垂直，法线水平，方程为 y = y₁。切线垂直出现在 dy/dx 无定义或趋于无穷的地方，例如曲线出现尖点，或 dy/dx 的分母为零时。
- Tangent gradient 3 → normal gradient -1/3. | 切线梯度 3 → 法线梯度 -1/3。
- Tangent gradient -2 → normal gradient ½. | 切线梯度 -2 → 法线梯度 ½。
- Tangent gradient ⅗ → normal gradient -5/3. | 切线梯度 ⅗ → 法线梯度 -5/3。
5. Worked Example: Tangent and Normal Together | 例题：切线与法线一起求
t where x = 2. First, the y-coordinate: y = 2³ - 3(2) = 8 - 6 = 2, so the point is (2, 2).
求曲线 y = x³ - 3x 在 x = 2 处的切线与法线方程。先求 y 坐标：y = 2³ - 3(2) = 8 - 6 = 2，所以点为 (2, 2)。
= 12 - 3 = 9. The tangent is therefore y - 2 = 9(x - 2), which simplifies to y = 9x - 16.
导数为 dy/dx = 3x² - 3。代入 x = 2 得 m = 3(4) - 3 = 12 - 3 = 9。因此切线为 y - 2 = 9(x - 2)，化简得 y = 9x - 16。
tiplying through by 9 to clear the fraction gives 9y - 18 = -x + 2, hence x + 9y - 20 = 0.
法线梯度为 -1/9，因此法线为 y - 2 = -1/9(x - 2)。两边乘以 9 消去分母得 9y - 18 = -x + 2，因此 x + 9y - 20 = 0。
Tangent: y = 9x - 16 Normal: x + 9y - 20 = 0
e perpendicular. Mark schemes often award a method mark for the negative reciprocal even if an arithmetic slip ruins the final equation, so always write the step down.
注意两个答案必须自洽：两条直线都过 (2, 2)，且 9 × (-1/9) = -1 说明它们互相垂直。评分标准通常会对负倒数这一步给出方法分，即使后续算错导致最终方程有误，所以这一步一定要写出来。
6. Finding Points Where the Gradient Is Known | 已知梯度求点
Examiners frequently reverse the question: instead of giving you the x-coordinate, they tell you something about the tangent and expect you to solve for the point. The key insight is that any statement about the direction of a tangent is really a statement about dy/dx.
考官常把问题反过来问：不给 x 坐标，而是告诉你关于切线的某个条件，让你解出点的位置。关键认识是：任何关于切线方向的描述，本质上都是对 dy/dx 的描述。
- Tangent parallel to the x-axis → set dy/dx = 0. | 切线与 x 轴平行 → 令 dy/dx = 0。
- Tangent parallel to the line y = 4x + 1 → set dy/dx = 4. | 切线与直线 y = 4x + 1 平行 → 令 dy/dx = 4。
- llel to the x-axis → set dy/dx equal to infinity, i.e. find where the tangent is vertical. | 法线与 x 轴平行 → 切线必须垂直，即找 dy/dx 无定义之处。
- Normal with gradient -1/2 → tangent gradient is 2, so set dy/dx = 2. | 法线梯度为 -1/2 → 切线梯度为 2，令 dy/dx = 2。
- 6 + 9 + 2 = 6, and at x = 3, y = 27 - 54 + 27 + 2 = 2. The tangents parallel to the x-axis touch the curve at (1, 6) and (3, 2).
考虑 y = x³ - 6x² + 9x + 2。求导得 dy/dx = 3x² - 12x + 9 = 3(x - 1)(x - 3)。令 dy/dx = 0 得 x = 1 或 x = 3。代回原式：x = 1 时 y = 1 - 6 + 9 + 2 = 6；x = 3 时 y = 27 - 54 + 27 + 2 = 2。与 x 轴平行的切线在 (1, 6) 与 (3, 2) 两处与曲线相切。
ing y-values are 2 and 4²... evaluate carefully: at x = 0, y = 2; at x = 4, y = 64 - 96 + 36 + 2 = 6. Both tangents then have gradient 9.
如果要求切线平行于 y = 9x 的点，则解 3x² - 12x + 9 = 9，得 3x² - 12x = 0，即 x = 0 或 x = 4。对应 y 值需仔细计算：x = 0 时 y = 2；x = 4 时 y = 64 - 96 + 36 + 2 = 6。这两处的切线梯度都是 9。
7. Why a Repeated Root Means Tangency | 为什么重根意味着相切
There is a second, algebraic way to test whether a line is a tangent. If a straight line meets a curve, substituting the line into the curve equation produces a polynomial equation whose roots are the x-coordinates of the intersections. A tangent touches at exactly one point, and the algebra reflects this by giving a repeated root.
还有第二种用代数判断相切的方法。若直线与曲线相交，把直线代入曲线方程会得到一个多项式方程，其根就是交点横坐标。切线只在一个点接触，代数上表现为出现重根。
ch factorises as (x - 3)² = 0. The single repeated root x = 3 confirms tangency at (3, 4).
取曲线 y = x² - 4x + 7 与前面求出的直线 y = 2x - 2。令二者相等得 x² - 4x + 7 = 2x - 2，即 x² - 6x + 9 = 0，因式分解为 (x - 3)² = 0。唯一的重根 x = 3 证实它们在 (3, 4) 处相切。
Equivalently, use the discriminant of the quadratic ax² + bx + c = 0, namely Δ = b² - 4ac.
等价地，可用二次方程 ax² + bx + c = 0 的判别式 Δ = b² - 4ac 来判断。
| Discriminant | 判别式 | Meaning | 含义 |
| Δ > 0 | Two distinct intersections; the line is a chord or secant | 两个不同交点；直线是割线 |
| Δ = 0 | Repeated root; the line is a tangent | 重根；直线是切线 |
| Δ < 0 | No real intersection; the line misses the curve | 无实交点；直线与曲线不相交 |
² - x - k = 0 must have Δ = 1 + 4k = 0, so k = -1/4 - a neat confirmation that the tangent to y = x² at x = ½ is indeed y = x - ¼.
当题目要求你求某个未知常数时，判别式法特别有用。例如若 y = x + k 是 y = x² 的切线，则 x² - x - k = 0 必须满足 Δ = 1 + 4k = 0，即 k = -1/4，这正好印证 y = x² 在 x = ½ 处的切线确实是 y = x - ¼。
8. Tangents and Normals to Circles | 圆及其他隐式曲线的切线法线
vative is found by differentiating both sides with respect to x. Remember that y is a function of x, so the derivative of y² is 2y dy/dx.
圆通常以隐式给出，例如 x² + y² = 25，其导数可通过对两边关于 x 求导得到。注意 y 是 x 的函数，所以 y² 的导数是 2y dy/dx。
2x + 2y dy/dx = 0 → dy/dx = -x/y
nt gradient is -3/4. The tangent is y - 4 = -3/4(x - 3), which simplifies to 3x + 4y = 25.
对于圆 x² + y² = 25 上点 (3, 4)，切线梯度为 -3/4。切线为 y - 4 = -3/4(x - 3)，化简得 3x + 4y = 25。
through the centre, and here the centre is the origin, so the line y = 4x/3 is exactly the radius extended.
法线梯度为 4/3，因此法线为 y - 4 = 4/3(x - 3)，化简得 y = 4x/3。这并非巧合：圆上任一点的法线都通过圆心，本题圆心为原点，所以直线 y = 4x/3 正是半径的延长线。
- Radius and tangent meet at 90° at the point of contact. | 半径与切线在切点处成 90°。
- Therefore the normal of a circle is along the radius. | 因此圆的法线沿半径方向。
- For a
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