Parametric Equations | 参数方程考点精讲

📚 Parametric Equations | 参数方程考点精讲

Parametric equations allow us to describe curves using a third variable, often denoted t, which is called the parameter. Instead of a direct relationship like y = f(x), both x and y are expressed as functions of t: x = f(t), y = g(t). This approach is particularly powerful for modelling motion, generating complex curves, and simplifying calculus operations such as differentiation. In the OCR IGCSE Mathematics specification, parametric equations test your ability to switch between parametric and Cartesian forms, differentiate with respect to the parameter, find gradients, tangents, normals, and analyse stationary points. Mastering these skills not only secures marks on dedicated parametric questions but also enhances your overall graphical and algebraic fluency.

参数方程使用第三个变量（通常记作 t，称为参数）来描述曲线。与 y = f(x) 这样的直接关系不同，x 和 y 都表示为 t 的函数：x = f(t)，y = g(t)。这种方法特别擅长为运动建模、生成复杂曲线，以及简化求导等微积分运算。在 OCR IGCSE 数学大纲中，参数方程考察你是否能在参数形式与直角坐标形式之间转换、对参数求导、求斜率、切线和法线，以及分析驻点。掌握这些技巧不仅能在专门的参数题中得分，还能提升你整体的图像与代数能力。

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

In a Cartesian equation, y is expressed directly in terms of x, such as y = 2x + 3 or x² + y² = 25. A parametric equation introduces an independent parameter, typically t, and defines both x and y separately: x = x(t), y = y(t). As t varies, the point (x, y) traces out a curve. For instance, x = t², y = 2t represents a parabola. Think of t as time: at each moment, the particle’s position is given by the pair (x(t), y(t)). This temporal interpretation is why parametric equations are widely used in physics and engineering.

在直角坐标方程中，y 直接用 x 表示，例如 y = 2x + 3 或 x² + y² = 25。参数方程引入一个独立参数（通常是 t），并分别定义 x 和 y：x = x(t)，y = y(t)。当 t 变化时，点 (x, y) 随之描绘出曲线。比如，x = t²，y = 2t 表示一条抛物线。你可以把 t 看作时间：每一时刻，质点的位置由 (x(t), y(t)) 给出。这种时间性解释正是参数方程在物理和工程中广泛应用的原因。

2. Eliminating the Parameter to Obtain the Cartesian Equation | 消去参数得出直角坐标方程

To convert a parametric pair into a single Cartesian equation relating x and y, you must eliminate the parameter t. The simplest method is solving for t from one equation and substituting it into the other. For example, given x = 2t + 1, y = t² – 3, we write t = (x – 1)/2 and substitute into y to obtain y = ((x – 1)/2)² – 3, which simplifies to y = (x – 1)²/4 – 3. Always check the domain: if t has a restricted range, so do x and y.

要将参数对转换成只含 x 和 y 的直角坐标方程，必须消去参数 t。最简单的方法是从一个方程解出 t，代入另一个方程。例如，已知 x = 2t + 1，y = t² – 3，可写出 t = (x – 1)/2，代入 y 得到 y = ((x – 1)/2)² – 3，化简为 y = (x – 1)²/4 – 3。注意检查定义域：若 t 的范围受限，x 和 y 的范围也会受限。

When trigonometric functions are involved, use identities like sin² t + cos² t = 1, 1 + tan² t = sec² t or sinh² t + cosh² t =… (but hyperbolic functions are beyond IGCSE). For x = a cos t, y = b sin t, divide each by a and b, square and add: (x/a)² + (y/b)² = cos² t + sin² t = 1, giving an ellipse.

当出现三角函数时，利用恒等式如 sin² t + cos² t = 1。对于 x = a cos t，y = b sin t，可分别除以 a 和 b，平方相加：(x/a)² + (y/b)² = cos² t + sin² t = 1，得到椭圆方程。

3. Differentiation of Parametric Equations: The Chain Rule | 参数方程求导：链式法则

To find dy/dx from x = f(t), y = g(t), we use the parametric chain rule: dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0. This formula comes from the differentials cancelling as if they were fractions: dy/dx = (dy/dt) · (dt/dx) = (dy/dt) / (dx/dt). Remember to differentiate both x and y with respect to t separately, then form the quotient. For example, if x = t³ + t, y = 2t² + 1, then dx/dt = 3t² + 1, dy/dt = 4t, so dy/dx = 4t / (3t² + 1).

要从 x = f(t)，y = g(t) 求 dy/dx，我们使用参数链式法则：dy/dx = (dy/dt) / (dx/dt)，前提是 dx/dt ≠ 0。这个公式源自微分的约简：dy/dx = (dy/dt) · (dt/dx) = (dy/dt) / (dx/dt)。记得分别对 x 和 y 关于 t 求导，然后构成商。例如，若 x = t³ + t，y = 2t² + 1，则 dx/dt = 3t² + 1，dy/dt = 4t，所以 dy/dx = 4t / (3t² + 1)。

Once you have dy/dx in terms of t, you can evaluate the gradient at a specific t-value or at a point by first finding t. This is particularly useful in determining the nature of stationary points, where dy/dx = 0.

一旦得到用 t 表示的 dy/dx，就可以计算特定 t 值处的斜率，或先求出 t 再计算某点的斜率。这尤其适用于确定驻点（dy/dx = 0）的性质。

4. Finding Tangents and Normals | 求切线和法线

The equation of a tangent to a parametric curve at a point corresponding to t = t₀ uses the gradient m = dy/dx evaluated at t₀. First find the coordinates (x(t₀), y(t₀)). Then the tangent line is y – y(t₀) = m (x – x(t₀)). The normal is perpendicular, so its gradient is -1/m (provided m ≠ 0). Equation: y – y(t₀) = (-1/m)(x – x(t₀)). Both are written in their simplest linear form.

参数曲线上对应于 t = t₀ 的点处的切线方程，使用斜率 m = dy/dx 在 t₀ 处的值。首先求出坐标 (x(t₀), y(t₀))，则切线为 y – y(t₀) = m (x – x(t₀))。法线与之垂直，因此斜率为 -1/m（若 m ≠ 0）。方程为 y – y(t₀) = (-1/m)(x – x(t₀))。两者均需写成最简线性形式。

Example: Curve is x = 2cos t, y = 3sin t. At t = π/4, x = 2cos(π/4) = √2, y = 3sin(π/4) = (3√2)/2. Compute dx/dt = -2sin t, dy/dt = 3cos t. At t = π/4, dx/dt = -√2, dy/dt = (3√2)/2. Gradient m = (3√2/2) / (-√2) = -3/2. Tangent: y – (3√2)/2 = -3/2 (x – √2). Normal gradient = 2/3. Normal: y – (3√2)/2 = (2/3)(x – √2). Make sure to simplify.

例题：曲线为 x = 2cos t，y = 3sin t。在 t = π/4 处，x = 2cos(π/4) = √2，y = 3sin(π/4) = (3√2)/2。计算 dx/dt = -2sin t，dy/dt = 3cos t。t = π/4 时 dx/dt = -√2，dy/dt = (3√2)/2。斜率 m = (3√2/2) / (-√2) = -3/2。切线：y – (3√2)/2 = -3/2 (x – √2)。法线斜率为 2/3。法线：y – (3√2)/2 = (2/3)(x – √2)。务必化简。

5. Second Derivative d²y/dx² for Parametric Curves | 参数曲线的二阶导数 d²y/dx²

The second derivative measures the rate of change of the gradient and helps determine concavity and the nature of stationary points. To find d²y/dx² from parametric equations, we differentiate dy/dx (which is a function of t) with respect to x using the chain rule again: d²y/dx² = d/dx (dy/dx) = [d/dt (dy/dx)] / (dx/dt). Notice it is NOT simply the second derivative of y over the second derivative of x. A common mistake is writing (d²y/dt²) / (d²x/dt²) – this is incorrect.

二阶导数衡量斜率的变化率，有助于确定曲线的凹凸性和驻点的性质。要从参数方程求 d²y/dx²，需要再次运用链式法则令 dy/dx（它是 t 的函数）对 x 求导：d²y/dx² = d/dx (dy/dx) = [d/dt (dy/dx)] / (dx/dt)。注意，它并不是简单地用 y 对 t 的二阶导数除以 x 对 t 的二阶导数。常见错误是写成 (d²y/dt²) / (d²x/dt²)——这是错误的。

Procedure: first find dx/dt and dy/dt, then form dy/dx = (dy/dt)/(dx/dt). Differentiate this expression with respect to t (using quotient or product rules as necessary) to get d/dt (dy/dx). Finally divide by dx/dt. Simplify as much as possible. For example, if x = t², y = t³ – t, then dx/dt = 2t, dy/dt = 3t² – 1, dy/dx = (3t² – 1)/(2t). Differentiate dy/dx wrt t: d/dt[(3t² – 1)/(2t)] = ( (6t)(2t) – (3t² – 1)(2) ) / (4t²) = (12t² – 6t² + 2) / (4t²) = (6t² + 2)/(4t²) = (3t² + 1)/(2t²). Then d²y/dx² = [ (3t² + 1)/(2t²) ] / (2t) = (3t² + 1)/(4t³).

步骤：先求 dx/dt 和 dy/dt，构成 dy/dx = (dy/dt)/(dx/dt)。将此表达式对 t 求导（必要时使用商法则或积法则）得到 d/dt (dy/dx)。最后除以 dx/dt。尽可能化简。例如，若 x = t²，y = t³ – t，则 dx/dt = 2t，dy/dt = 3t² – 1，dy/dx = (3t² – 1)/(2t)。对 dy/dx 关于 t 求导：d/dt[(3t² – 1)/(2t)] = ( (6t)(2t) – (3t² – 1)(2) ) / (4t²) = (12t² – 6t² + 2) / (4t²) = (6t² + 2)/(4t²) = (3t² + 1)/(2t²)。然后 d²y/dx² = [ (3t² + 1)/(2t²) ] / (2t) = (3t² + 1)/(4t³)。

6. Sketching Parametric Curves | 绘制参数曲线

To sketch a curve given parametrically, a table of values for t, x, and y is extremely helpful. Choose a range of t values (often from a given domain), compute the corresponding (x, y) points, and plot them. Observe how the curve progresses as t increases – indicate direction with arrows. If the Cartesian equation is recognisable (e.g., a parabola, circle, or ellipse), you can use its standard shape to guide the sketch. Always label key points like intercepts with axes, maxima/minima, and any cusps or loops.

要绘制以参数形式给出的曲线，构建 t、x、y 的取值表格非常有用。选取一系列 t 值（通常来自给定范围），计算对应的 (x, y) 点，然后绘制。观察曲线随 t 增大的走向——用箭头标出方向。如果能识别出直角坐标方程（如抛物线、圆或椭圆），可借助其标准形状辅助绘图。务必标出关键点，如与坐标轴的交点、极大值/极小值，以及任何尖点或环。

Some curves form loops or self-intersections, which become apparent when you evaluate the parameter at different points giving the same (x, y). For instance, x = t² – 1, y = t³ – t shows a loop near the origin. Understanding the parametric behaviour allows you to interpret motion along the curve, which is tested in context‑based problems.

有些曲线会形成环或自交点，当你计算不同参数点得出相同 (x, y) 时就会显现。例如 x = t² – 1，y = t³ – t 在原点附近有一个环。理解参数行为有助于解释沿曲线的运动，这在实际应用题中会考查。

7. Common Parametric Forms: Circle, Ellipse, Parabola | 常见参数形式：圆、椭圆、抛物线

OCR IGCSE exams frequently present standard curves in parametric form. Recognising them saves time and reduces algebraic errors.
• Circle centred at (0,0) radius r: x = r cos t, y = r sin t, 0 ≤ t < 2π.
• Ellipse centred at (0,0) semi-axes a, b: x = a cos t, y = b sin t, 0 ≤ t < 2π.
• Parabola with vertex at origin, opening right: x = at², y = 2at (or y = at², x = 2at for upward). The parameter t often has geometric meaning, like slope of tangent.

OCR IGCSE 考试常以参数形式给出标准曲线。认出它们能节省时间并减少代数错误。
• 圆心 (0,0) 半径 r：x = r cos t，y = r sin t，0 ≤ t < 2π。
• 椭圆中心 (0,0) 半轴 a, b：x = a cos t，y = b sin t，0 ≤ t < 2π。
• 顶点在原点、开口向右的抛物线：x = at²，y = 2at（或向上开口 y = at²，x = 2at）。参数 t 通常具有几何意义，如切线斜率。

You may need to adapt them for translations: (x – h)/a = cos t, (y – k)/b = sin t gives an ellipse centred at (h, k). Similarly, a line can be parameterised as x = x₁ + tΔx, y = y₁ + tΔy, which is the vector form of a straight line.

你可能需要为平移做调整：(x – h)/a = cos t，(y – k)/b = sin t 表示中心在 (h, k) 的椭圆。类似地，直线可参数化为 x = x₁ + tΔx，y = y₁ + tΔy，即直线的向量形式。

8. Stationary Points and Their Nature | 驻点及其性质

Stationary points on a parametric curve occur where dy/dx = 0, i.e., dy/dt = 0 and dx/dt ≠ 0. Set dy/dt = 0 to find the corresponding t-values. Then determine the coordinates (x, y). To classify the stationary point as a local maximum, minimum or point of inflection, compute the second derivative d²y/dx² at that t-value. If d²y/dx² > 0, it’s a minimum; if < 0, a maximum; if = 0, further investigation (or the first derivative test) is needed – but for IGCSE, the sign of the second derivative is usually sufficient.

参数曲线上的驻点出现在 dy/dx = 0 处，即 dy/dt = 0 且 dx/dt ≠ 0。令 dy/dt = 0 求出对应 t 值，然后确定坐标 (x, y)。要将驻点分类为局部极大、极小或拐点，需计算该 t 值处的二阶导数 d²y/dx²。若 d²y/dx² > 0，为极小值；若 < 0，为极大值；若 = 0，则需进一步检验（或一阶导数检验）——但在 IGCSE 中，二阶导数的符号通常已足够。

Points where dx/dt = 0 and dy/dt ≠ 0 give vertical tangents; gradients become infinite. Although not always called stationary points, they are important for sketching. If both derivatives vanish simultaneously (singular point), the curve may have a cusp or a self-intersection – this is rarely required at IGCSE level but could be mentioned.

若 dx/dt = 0 且 dy/dt ≠ 0，则切线垂直，斜率趋于无穷。这些点虽不总称驻点，但对绘图很重要。若两个导数同时为零（奇点），曲线可能有尖点或自交——这在 IGCSE 层面较少要求，但可提及。

9. Practical Applications: Kinematics in One Dimension Extended | 实际应用：一维运动学的延展

Parametric equations naturally model the motion of a particle where x and y represent horizontal and vertical displacements as functions of time t. Velocity components are v_x = dx/dt, v_y = dy/dt; speed is √( (dx/dt)² + (dy/dt)² ). Acceleration components are a_x = d²x/dt², a_y = d²y/dt². The tangent to the path gives the direction of motion. Questions may ask for the position at a given time, the vector of velocity, or the Cartesian equation of the path. This bridges pure parametric calculus with applied mathematics.

参数方程自然地为质点运动建模，其中 x 和 y 表示水平与竖直位移关于时间 t 的函数。速度分量为 v_x = dx/dt，v_y = dy/dt；速率为 √( (dx/dt)² + (dy/dt)² )。加速度分量为 a_x = d²x/dt²，a_y = d²y/dt²。路径的切线给出运动方向。考题可能要求给出某时刻的位置、速度向量或路径的直角坐标方程。这搭建了纯参数微积分与应用数学的桥梁。

A typical problem: “A particle moves along a curve with position x = 2t + 1, y = t² – 4t. Find its velocity at t = 3, and determine the time when it crosses the x-axis.” Solve y = t² – 4t = 0, giving t = 0, 4. Then compute dx/dt = 2, dy/dt = 2t – 4; speed at t=3 = √(2² + (6-4)²) = √(4+4) = 2√2.

典型题目：”质点沿曲线运动，位置为 x = 2t + 1，y = t² – 4t。求 t = 3 时的速度，并确定它何时穿过 x 轴。” 解 y = t² – 4t = 0，得 t = 0, 4。然后计算 dx/dt = 2，dy/dt = 2t – 4；t=3 时速率为 √(2² + (6-4)²) = √(4+4) = 2√2。

10. Worked Example: From Parameter to Tangent, Normal and Stationary Points | 例题精讲：从参数到切线、法线与驻点

Consider the curve defined by x = t³ – 3t, y = t² – 2 for t ∈ ℝ. Find the equation of the tangent at t = 2, the normal at that point, and any stationary points.
1. Derivatives: dx/dt = 3t² – 3, dy/dt = 2t.
2. At t = 2: x = 8 – 6 = 2, y = 4 – 2 = 2. dx/dt = 3(4)-3 = 9, dy/dt = 4. Gradient m = 4/9. Tangent: y – 2 = (4/9)(x – 2) → 9y – 18 = 4x – 8 → 9y = 4x + 10 → y = (4/9)x + 10/9.
3. Normal gradient = -9/4. Normal: y – 2 = (-9/4)(x – 2) → 4y – 8 = -9x + 18 → 9x + 4y = 26.
4. Stationary points: set dy/dx = 0 → 2t/(3t² – 3) = 0 → t = 0. Check dx/dt at t=0: -3 ≠ 0. At t=0, x = 0, y = -2. Find d²y/dx²: first dy/dx = 2t/(3(t²-1)). d/dt of that = [2(3t²-3) – 2t(6t)] / (3(t²-1))² = [6t²-6 -12t²] /9(t²-1)² = (-6t² -6)/9(t²-1)² = – (t²+1)/(1.5(t²-1)²). Then d²y/dx² = [ d/dt(dy/dx) ] / (dx/dt) = [ -(t²+1)/(1.5(t²-1)²) ] / (3(t²-1)) = -(t²+1) / (4.5 (t²-1)³). At t=0, denominator negative? (t²-1)³ = (-1)³ = -1, so d²y/dx² = – (1) / (4.5 * -1) = 1/4.5 > 0, hence minimum at (0, -2).

考虑由 x = t³ – 3t，y = t² – 2，t ∈ ℝ 定义的曲线。求 t = 2 处的切线方程、该点法线方程以及任何驻点。
1. 导数：dx/dt = 3t² – 3，dy/dt = 2t。
2. t = 2 时：x = 8 – 6 = 2，y = 4 – 2 = 2。dx/dt = 3(4)-3 = 9，dy/dt = 4。斜率 m = 4/9。切线：y – 2 = (4/9)(x – 2) → 9y – 18 = 4x – 8 → 9y = 4x + 10 → y = (4/9)x + 10/9。
3. 法线斜率 = -9/4。法线：y – 2 = (-9/4)(x – 2) → 4y – 8 = -9x + 18 → 9x + 4y = 26。
4. 驻点：令 dy/dx = 0 → 2t/(3t² – 3) = 0 → t = 0。检验 dx/dt 在 t=0：-3 ≠ 0。t=0 时 x=0, y=-2。求 d²y/dx²：先 dy/dx = 2t/(3(t²-1))。关于 t 求导 = [2(3t²-3) – 2t(6t)] / (3(t²-1))² = [6t²-6 -12t²] /9(t²-1)² = (-6t² -6)/9(t²-1)² = -(t²+1)/(1.5(t²-1)²)。然后 d²y/dx² = [ d/dt(dy/dx) ] / (dx/dt) = [ -(t²+1)/(1.5(t²-1)²) ] / (3(t²-1)) = -(t²+1) / (4.5 (t²-1)³)。t=0 时分母 (t²-1)³ = (-1)³ = -1，所以 d²y/dx² = -(1) / (4.5 * -1) = 1/4.5 > 0，故极小值点为 (0, -2)。

11. Common Mistakes and Tips | 常见错误与提分技巧

• Forgetting the chain rule structure: Writing dy/dx as dy/dt × dx/dt instead of dividing. Remember dy/dx = (dy/dt) / (dx/dt).
• Mishandling trigonometric elimination: Always check that the identity matches the signs and coefficients; e.g., x = 2sin t, y = 3cos t requires (x/2)² + (y/3)² = sin²t + cos²t = 1, not swapping sine/cosine.
• Second derivative pitfall: Many students incorrectly compute d²y/dx² as (d²y/dt²)/(d²x/dt²). Always differentiate dy/dx with respect to t and then divide by dx/dt.
• Domain issues: When eliminating t, note any restrictions on t that restrict x and y. For example, t ≥ 0 may give only the right branch of a parabola.
• Arithmetic errors: Double-check substitution and simplification, especially with negative signs and fractions.

• 忘记链式法则结构：把 dy/dx 写成 dy/dt × dx/dt 而不是相除。记住 dy/dx = (dy/dt) / (dx/dt)。
• 三角函数消元失误：务必核对恒等式的符号和系数；例如 x = 2sin t，y = 3cos t 需用 (x/2)² + (y/3)² = sin²t + cos²t = 1，不要互换正弦与余弦。
• 二阶导数误区：许多学生错误地将 d²y/dx² 计算为 (d²y/dt²)/(d²x/dt²)。一定要先对 t 求导 dy/dx，再除以 dx/dt。
• 定义域问题：消去 t 时，注意 t 的任何限制会约束 x 和 y 的范围。例如 t ≥ 0 可能只给出抛物线的右支。
• 算术错误：仔细检查代入和化简，特别是有负号和分数时。

12. Summary and Exam Strategy | 总结与应试策略

Parametric equations consolidate your understanding of algebra, trigonometry, and differentiation. In the exam, read the question carefully: identify what is given (x(t), y(t), domain of t, a specific point or parameter value). Show each step methodically – finding derivatives, eliminating parameters, and substituting t-values. When a question asks for a tangent or normal, always state the gradient clearly before writing the line equation. If a second derivative is requested, apply the correct formula and simplify fully. For sketching, use a table and mark direction. Check if the answer makes sense relative to any given domain. With consistent practice, parametric questions become a reliable opportunity to secure high marks.

参数方程整合了你在代数、三角和求导方面的理解。在考试中，仔细读题：明确已知条件（x(t)、y(t)、t 的定义域、特定点或参数值）。有条有理地展示每个步骤——求导、消参、代入 t 值。当题目要求求切线或法线时，务必先清楚写出斜率再写直线方程。若要求二阶导数，应用正确公式并彻底化简。绘制曲线时使用表格并标出方向。检查答案是否符合给定定义域。通过持续练习，参数题将成为你稳拿高分的机会。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Parametric Equations | 参数方程 考点精讲