Differentiation of Parametric Equations | 参数方程形式的函数求导

📚 Differentiation of Parametric Equations | 参数方程形式的函数求导

When a curve is expressed in parametric form, both x and y are given as functions of a third variable, usually t. We can still find the gradient dy/dx without eliminating t, using a simple quotient of derivatives. This is a core technique in A-Level and IB calculus, and it often appears in differentiation questions involving curves, tangents, normals and stationary points.

当曲线以参数方程形式给出时,x 与 y 都表示为第三个变量(通常为 t)的函数。我们不必消去参数 t,也能直接求出斜率 dy/dx;只需将两个导函数相除。这是 A-Level 和 IB 数学微积分中的核心技巧,也经常出现在涉及曲线、切线、法线与驻点的求导题中。


1. What Are Parametric Equations? | 什么是参数方程?

A curve can be written in Cartesian form as y = f(x), or implicitly as f(x, y) = 0. In parametric form, we use a third variable t to describe both coordinates at the same time.

曲线通常可以写成直角坐标形式 y = f(x),或隐式形式 f(x, y) = 0。在参数形式中,我们引入第三个变量 t,同时描述两个坐标。

  • Example: x = t², y = 2t. As t varies, the point (t², 2t) traces out a parabola.

    例如:x = t²,y = 2t。当 t 变化时,点 (t², 2t) 描出一条抛物线。

  • Example: x = r cos θ, y = r sin θ describes a circle when θ varies from 0 to 2π.

    例如:x = r cos θ,y = r sin θ,当 θ 在 0 到 2π 间变化时,表示一个圆。

Parametric equations are especially useful in mechanics, because t can represent time. The position of a moving particle can be described by its x and y coordinates at each time t.

参数方程在力学中尤其有用,因为 t 可以表示时间。一个运动质点的位置,可以由它在每个时刻 t 的 x 坐标和 y 坐标来描述。


2. The Key Formula | 核心公式

Suppose a curve is given by x = f(t) and y = g(t). To find the gradient of the curve, we use the following formula.

设曲线由 x = f(t),y = g(t) 给出。要求曲线的斜率,我们使用下面这个公式。

dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0

In words: differentiate y with respect to t, then divide by the derivative of x with respect to t.

换句话说:先对 t 求 y 的导数,再除以 x 对 t 的导数。

The condition dx/dt ≠ 0 is important. If dx/dt = 0, then a small change in t gives no change in x, so the tangent to the curve is vertical and dy/dx is not defined.

条件 dx/dt ≠ 0 很重要。如果 dx/dt = 0,那么 t 的微小变化不会引起 x 的变化,此时切线是竖直的,dy/dx 不存在。


3. Derivation: Why Does It Work? | 推导:为什么可行?

We can derive the formula using the chain rule. If y is considered as a function of x, and x is a function of t, then y is a composite function of t.

我们可以用链式法则来推导。如果把 y 看作 x 的函数,而 x 又是 t 的函数,那么 y 就是关于 t 的复合函数。

dy/dt = (dy/dx) × (dx/dt)

Dividing both sides by dx/dt gives the required formula.

两边同时除以 dx/dt,就得到所需公式。

Another way to see it is through small changes. Let Δt be a small change in t, producing small changes Δx and Δy.

我们也可以通过微小增量来理解。设 Δt 是 t 的微小改变量,它分别引起 Δx 和 Δy 的微小变化。

Δy/Δx = (Δy/Δt) / (Δx/Δt)

Taking the limit as Δt → 0 gives dy/dx = (dy/dt) / (dx/dt).

令 Δt → 0 取极限,即可得到 dy/dx = (dy/dt) / (dx/dt)。


4. Worked Example 1: Polynomial Parameter | 例 1:多项式参数

Find dy/dx for the curve x = t² + 1 and y = t³ + t. Evaluate the gradient at t = 2.

求曲线 x = t² + 1,y = t³ + t 的 dy/dx,并计算 t = 2 处的斜率。

Differentiate x and y with respect to t.

分别对 t 求 x 和 y 的导数。

dx/dt = 2t, dy/dt = 3t² + 1

Therefore:

因此:

dy/dx = (3t² + 1) / (2t)

At t = 2:

当 t = 2 时:

dy/dx = (12 + 1) / 4 = 13/4

The corresponding point on the curve is x = 5, y = 10. So the gradient at (5, 10) is 13/4.

对应的曲线点为 x = 5,y = 10。因此点 (5, 10) 处的斜率为 13/4。

Notice that if t = 0, then dx/dt = 0 and the gradient is undefined. This suggests a vertical tangent on the curve.

注意,如果 t = 0,则 dx/dt = 0,斜率不存在。这说明曲线上可能有一条竖直切线。


5. Tangents and Normals | 切线与法线

Once dy/dx is known, the tangent and normal lines can be found using the point corresponding to a given parameter value.

求出 dy/dx 之后,就可以利用给定参数值对应的点来求切线方程和法线方程。

For a point (x₀, y₀) with gradient m:

对于点 (x₀, y₀),斜率为 m:

Tangent: y − y₀ = m(x − x₀)

Normal: y − y₀ = −(1/m)(x − x₀), m ≠ 0

Using the previous example at t = 2, we have x₀ = 5, y₀ = 10 and m = 13/4.

继续使用上一个例子,在 t = 2 处,x₀ = 5,y₀ = 10,m = 13/4。

The tangent equation is:

切线方程为:

y − 10 = (13/4)(x − 5)

This can be simplified to 4y = 13x − 25.

化简可得 4y = 13x − 25。

The normal has gradient −4/13, so its equation is:

法线的斜率为 −4/13,因此法线方程为:

y − 10 = −(4/13)(x − 5)

which simplifies to 13y + 4x = 150.

化简可得 13y + 4x = 150。


6. The Second Derivative | 二阶导数

The second derivative d²y/dx² measures how the gradient changes along the curve. It is used to determine concavity and points of inflection.

二阶导数 d²y/dx² 衡量的是曲线上斜率的变化率,常用于判断凹凸性和拐点。

To find d²y/dx², first write p = dy/dx as a function of t. Then:

要求 d²y/dx²,先把 p = dy/dx 写成 t 的函数,然后:

d²y/dx² = (dp/dt) / (dx/dt)

Equivalently:

等价地:

d²y/dx² = [d/dt(dy/dx)] / (dx/dt)

A common mistake is to compute (d²y/dt²) / (d²x/dt²). This is wrong. The correct approach is to differentiate dy/dx with respect to t first, and then divide by dx/dt once more.

一个常见错误是直接去算 (d²y/dt²) / (d²x/dt²)。这是错误的。正确做法是先用 t 对 dy/dx 求导,然后再除以一个 dx/dt。

Continuing the earlier example, dy/dx = (3t² + 1)/(2t). We can rewrite this as:

继续前面的例子,dy/dx = (3t² + 1)/(2t)。我们可以将其改写为:

dy/dx = 3t/2 + 1/(2t)

Differentiate this with respect to t:

对 t 求导:

d/dt(dy/dx) = 3/2 − 1/(2t²)

Now divide by dx/dt = 2t:

再除以 dx/dt = 2t:

d²y/dx² = [3/2 − 1/(2t²)] / (2t) = 3/(4t) − 1/(4t³)

This gives the exact second derivative in terms of the parameter t.

这样就得到了以参数 t 表示的二阶导数。


7. Worked Example 2: Trigonometric Parameter | 例 2

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