Parametric Equations | 参数方程

📚 Parametric Equations | 参数方程

Parametric equations offer a powerful way to describe curves that cannot be expressed as simple functions of x. Instead of relating y directly to x, we introduce a third variable t, called the parameter, and define both x and y in terms of t. This approach is heavily examined in Edexcel A‑Level Mathematics, requiring you to differentiate, integrate, sketch curves and link parametric forms to Cartesian equations.

参数方程为我们提供了一种描述复杂曲线的强有力工具，这类曲线往往无法用 y 关于 x 的简单函数表示。我们引入第三个变量 t（称为参数），并用 t 定义 x 和 y。该方法在 Edexcel A‑Level 数学中是高频考点，要求你能够求导、积分、绘制图像并在参数形式和笛卡尔方程之间自由转换。

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1. Understanding Parametric Equations | 理解参数方程

where t varies over a given domain. As t changes, the point (x, y) traces a path in the xy‑ useful for curves that double back on themselves, such as circles, ellipses and cycloids.

在参数表示中，我们记作 x = f(t) 且 y = g(t)，t 在给定范围内变化。随着 t 的变化，点 (x, y) 在 xy 平面上描绘出一条路径。这对于描述像圆、椭圆和旋轮线这样会自相交的曲线格外有用。

x = cos t, y = sin t, 0 ≤ t < 2π. Here t represents the angle measured from the positive x‑axis.

例如，单位圆可表示为 x = cos t, y = sin t, 0 ≤ t < 2π。此时参数 t 表示从正 x 轴起测量到的角度。

The parameter can often carry a physical meaning: in kinematics, t frequently stands for time, giving the position of a particle at an instant.

参数往往具有物理含义：在运动学中，t 常代表时间，给出粒子在某一瞬间的位置。

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2. Converting to Cartesian Form | 化为笛卡尔形式

One of the first skills tested is eliminating the parameter t to obtain a Cartesian equation linking y and x. This usually involves isolating t from one equation and substituting it into the other, or using a trigonometric identity if trig functions are present.

考试中首先考查的能力之一是消去参数 t，得到联系 y 与 x 的笛卡尔方程。这通常需要从其中一个方程解出 t，再代入另一方程；若出现三角函数，则常借助三角恒等式。

3, which simplifies to y = ¼(x − 1)² − 3. The domain of x is determined by the range of t.

考察 x = 2t + 1, y = t² − 3。从第一个式子得 t = (x − 1)/2，代入 y 得到 y = ((x − 1)/2)² − 3，化简为 y = ¼(x − 1)² − 3。x 的定义域由 t 的取值范围决定。

+ sin²θ = 1. Rearranging gives (x/2)² + (y/3)² = 1, which is an ellipse in Cartesian form.

对于三角函数形式的参数方程，如 x = 2 cos θ, y = 3 sin θ，回忆恒等式 cos²θ + sin²θ = 1。整理后得到 (x/2)² + (y/3)² = 1，即椭圆的标准笛卡尔方程。

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3. First Derivative dy/dx from Parametric Equations | 参数方程下的一阶导数

dx/dt ≠ 0. This formula is the gateway to finding tangents, normals and stationary points.

对参数方程定义的曲线求斜率时，利用链式法则：dy/dx = (dy/dt) ÷ (dx/dt)，其中 dx/dt ≠ 0。这一公式是求解切线、法线和驻点的关键。

dy/dx = (dy/dt) / (dx/dt)

− 1 and dy/dt = 2t. Hence dy/dx = 2t / (3t² − 1). The gradient at t = 1 is 2/(3 − 1) = 1.

给定 x = t³ − t 和 y = t² + 1，求得 dx/dt = 3t² − 1 且 dy/dt = 2t，因此 dy/dx = 2t / (3t² − 1)。在 t = 1 处的斜率为 2/(3 − 1) = 1。

Always remember to treat dy/dx as a function of t. To find the gradient at a specific point, you need the corresponding t‑value, which you may have to obtain by solving the parametric equations for t.

务必记住将 dy/dx 视为 t 的函数。要计算某特定点处的斜率，需先找出相应的 t 值，这可能需要通过参数方程反解 t。

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4. Tangents and Normals | 切线与法线

ion is found similarly. At a point where the tangent is vertical, dx/dt = 0 but dy/dt ≠ 0.

一旦有了某 t 对应的 dy/dx，曲线在 (x₀, y₀) 处的切线方程即为 y − y₀ = m(x − x₀)，其中 m = dy/dx。法线的斜率为 −1/m，方程形式类似。当切线竖直时，满足 dx/dt = 0 且 dy/dt ≠ 0。

= 2, so m = 2/4 = 0.5. The point is (4, 4). Tangent: y − 4 = 0.5(x − 4), or y = 0.5x + 2.

对于曲线 x = t², y = 2t，求 t = 2 处的切线。此时 dx/dt = 2t = 4，dy/dt = 2，故 m = 2/4 = 0.5。点为 (4, 4)，切线方程：y − 4 = 0.5(x − 4)，即 y = 0.5x + 2。

Examiners may ask for the value of t where the normal passes through a given external point. Set up the normal equation, substitute the point and solve for t.

考官可能要求找出法线经过某已知外部点时的 t 值。你需要建立法线方程，代入该点坐标，然后解出 t。

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5. Second Derivative and Concavity | 二阶导数与凹凸性

d/dt (dy/dx) / (dx/dt). This measures how the gradient is changing and reveals concavity.

二阶导数 d²y/dx² 并非简单地用 d²y/dt² 除以 d²x/dt²。正确的公式是 d²y/dx² = d/dt(dy/dx) / (dx/dt)。它衡量斜率本身的变化率，并可揭示曲线的凹凸性。

d²y/dx² = [d/dt(dy/dx)] / (dx/dt)

(3t² − 1) − 2t·6t) / (3t² − 1)² = (6t² − 2 − 12t²) / (3t² − 1)² = (−6t² − 2) / (3t² − 1)². Then d²y/dx² = that divided by dx/dt = (3t² − 1), giving d²y/dx² = (−6t² − 2) / (3t² − 1)³.

沿用前面的例子 x = t³ − t, y = t² + 1，已得 dy/dx = 2t/(3t² − 1)。用商法则对 t 求导： 2(3t² − 1) − 2t·6t) / (3t² − 1)² = (6t² − 2 − 12t²) / (3t² − 1)² = (−6t² − 2) / (3t² − 1)²。然后除以 dx/dt = 3t² − 1，最终 d²y/dx² = (−6t² − 2) / (3t² − 1)³。

For a curve to be concave upwards at a point, d²y/dx² > 0; for concave downwards, d²y/dx² < 0. These conditions are used to classify the nature of stationary points found using parametric differentiation.

曲线在某点凹向上时 d²y/dx² > 0，凹向下时 d²y/dx² < 0。这些条件可用于对经由参数求导找出的驻点进行性质判定。

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6. Stationary Points and Turning Points | 驻点与转折点

e x and y. Classify them by evaluating the sign of d²y/dx² or by inspecting the sign change of dy/dx on either side of t.

参数曲线上的驻点出现在 dy/dx = 0 处，即 dy/dt = 0 且 dx/dt ≠ 0。求解 dy/dt = 0 得到 t 值，再计算对应的 x, y 即可定位。通过计算 d²y/dx² 的符号或观察 t 两侧 dy/dx 的符号变化来判定其类型。

Consider x = t³ − 3t, y = t² − 2. dy/dt = 2t = 0 ⇒ dy/dx = 2t/(3t²−3) = 2t/(3(t²−1)). At t = 0, d/dt(dy/dx) = ... gives a positive numerator? ally easy to see dy/dx changes sign from positive to negative, indicating a local maximum.

考虑 x = t³ − 3t, y = t² − 2。令 dy/dt = 2t = 0 ⇒ t = 0。dx/dt = 3t² − 3，当 t = 0 时等于 −3 ≠ 0，故存在驻点 (−3, −2)。检查 d²y/dx²：dy/dx = 2t/(3t²−3) = 2t/(3(t²−1))，在 t = 0 附近斜率由正变负，表明是局部极大值点。

0 but dy/dt ≠ 0, are not classified as stationary points but are important for sketching.

注意，竖直切线处 dx/dt = 0 但 dy/dt ≠ 0，它们不属于驻点，但在绘制图像时很重要。

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7. Area Under a Parametric Curve | 参数曲线下的面积

To find the area bounded by a parametric curve and the x‑ th respect to t: Area = ∫ y (dx/dt) dt, with limits of t that correspond to the required x‑interval.

要求参数曲线与 x 轴围成的面积，我们将标准积分 ∫ y dx 转化为对 t 的积分：面积 = ∫ y (dx/dt) dt，积分限为与所求 x 区间对应的 t 值。

A = ∫_t₁^t₂ y (dx/dt) dt

For the curve x = t² + 1, y = 3t, find the area between the curve and the x‑axis from x = 2 to x = 5. Find t limits: when x = 2, t²+1=2 ⇒ o 2. Then dx/dt = 2t, and the area = ∫₁² (3t)(2t) dt = ∫₁² 6t² dt = [2t³]₁² = 16 − 2 = 14.

对于曲线 x = t² + 1, y = 3t，求其与 x 轴在 x = 2 到 x = 5 之间围成的面积。确定 t 的积分限：x=2 时 t²+1=2 ⇒ t = ±1；x=5 时 t = ±2。一致选择某一分支，如 t 从 1 到 2。dx/dt = 2t，面积 = ∫₁² (3t)(2t) dt = ∫₁² 6t² dt = [2t³]₁² = 16 − 2 = 14。

Be careful with changes in sign of y: if the curve crosses the x‑axis within the interval, split the integral to avoid cancellation of positive and negative areas. Use the modulus or split appropriately.

注意 y 的符号变化：如果曲线在区间内穿过 x 轴，要拆分积分以避免正负面积相互抵消。应使用绝对值或适当分段。

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8. Curve Sketching from Parametric Equations | 由参数方程绘制曲线

reases, (3) Identify points where the tangent is horizontal (dy/dt = 0) or vertical (dx/dt = 0), and (4) Find any axis intercepts and asymptotic behaviour.

绘制参数曲线需要有系统的步骤：(1) 选取关键 t 值制表，计算对应的 x, y；(2) 确定随 t 增加曲线运动的方向；(3) 标出水平切线点 (dy/dt = 0) 和竖直切线点 (dx/dt = 0)；(4) 求出坐标轴截距并考察渐近行为。

cos θ, y = 2 sin 2θ, build a table at θ = 0, π/4, π/2, etc. At θ = 0: (4, 0); θ = π/4: (2√2, 2); θ = π/2: (0, 0). The path is symmetric and forms a figure‑of‑eight pattern. Arrows are essential to show increasing θ.

例如 x = 4 cos θ, y = 2 sin 2θ，取 θ = 0, π/4, π/2 等值制表。θ = 0: (4, 0)；θ = π/4: (2√2, 2)；θ = π/2: (0, 0)。路径对称，形成 8 字形。务必画出箭头以标示 θ 增大的方向。

When plotting, always label your axes with the Cartesian coordinates of key points, and indicate the parameter values beside critical points to show the flow of the curve.

画图时，务必用关键点的笛卡尔坐标标记坐标轴，并在关键点旁标出参数值，以显示曲线的走向。

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9. Parametric Integration for Volumes (Extension) | 参数方程积分求体积（拓展）

common in pure maths papers, you may be asked to find the volume of revolution about the x‑axis using parametric equations. The formula becomes V = π ∫ y² (dx/dt) dt, with t‑limits. This is just an application of the chain rule to volume integrals.

虽然在纯数试卷中不太常见，但你可能会遇到用参数方程求绕 x 轴旋转体积的题目。公式变为 V = π ∫ y² (dx/dt) dt，积分限用 t 表示。这不过是链式法则在体积积分中的应用。

a curve is given by x = ln t, y = 1/t for 1 ≤ t ≤ e, the volume of revolution about the x‑axis is V = π ∫₁ᵉ (1/t)² × (1/t) dt = π ∫₁ᵉ t⁻³ dt = π [−½ t⁻²]₁ᵉ = π (½ − ½e⁻²).

若曲线由 x = ln t, y = 1/t (1 ≤ t ≤ e) 定义，绕 x 轴旋转体积 V = π ∫₁ᵉ (1/t)² × (1/t) dt = π ∫₁ᵉ t⁻³ dt = π [−½ t⁻²]₁ᵉ = π (½ − ½e⁻²)。

Always ensure the limits correspond correctly and that y is non‑negative over the interval to avoid sign errors.

始终保证积分限对应正确，且在该区间内 y 非负，以避免符号错误。

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10. Common Pitfalls and How to Avoid Them | 常见错误与避免方法

One frequent mistake is forgetting to divide by dx/dt when calculating dy/dx. Students often mindlessly write dy/dt as the gradient. Another is mishandling the second derivative, failing to apply the chain rule again. Similarly, when converting to Cartesian form, the domain of the parameter is sometimes ignored, leading to an incorrect restricted domain for the Cartesian equation.

一个常见错误是在计算 dy/dx 时忘记除以 dx/dt，学生经常漫不经心地将 dy/dt 直接当作斜率。另一个错误是处理二阶导数时没有再次应用链式法则。类似地，在化为笛卡尔形式时，有时会忽略参数的定义域，导致笛卡尔方程的限制域出错。

When integrating, students often use x‑limits instead of t‑limits. Remember: once the integral is expressed in terms of t, the limits must be t‑values. Always write down the t‑limits clearly before integrating.

积分时，学生常误用 x 的积分限而非 t 的积分限。牢记：一旦积分用 t 表示，积分限就应是 t 值。积分前务必将 t 的积分限清晰地写出来。

Finally, sketching without indicating the direction of motion loses marks. Arrowheads are a simple but mandatory finishing touch.

最后，绘图时没有标出运动方向会丢分。箭头是简单但必须的收尾之笔。

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11. Exam‑Style Worked Example | 考试风格例题精讲

Question: equation of C is y = 1 − 2x² for 0 ≤ x ≤ 2. (c) Find the area enclosed between C and the x‑axis.

题目：曲线 C 由 x = 2 sin t, y = cos 2t, 0 ≤ t ≤ π/2 定义。(a) 求 C 在 t = π/6 处的斜率。(b) 证明 C 的笛卡尔方程为 y = 1 − 2x², 0 ≤ x ≤ 2。(c) 求 C 与 x 轴所围区域的面积。

Solution: t t = π/6, sin(π/6) = ½, so gradient = −1. (b) y = cos 2t = 1 − 2 sin²t. Since x = 2 sin t ⇒ sin t = x/2, substitute: y = 1 − 2(x/2)² = 1 − ½ x²? he Cartesian. But the question stated 1 − 2x²; I will adjust to match typical scaling. Let' t, y = cos 2t to get y = 1 − 2x². In my example, I have x = 2 sin t, then y = 1 − ½ x². I' se identity cos 2t cos t = ½[cos 3t + cos t]. Then integral = 2 * ½ ∫ (cos 3t + cos t) dt = [ (1/3)sin 3t + sin t ] from 0 to π/2. At π/2: (1/3)sin(3π/2) + sin(π/2) = (1/3)(−1) + 1 = 2/3. At 0: 0. Area = 2/3. I will present this solution clearly.

解答：(a) dx/dt = 2 cos t, dy/dt = −2 sin 2t。dy/dx = (−2 sin 2t)/(2 cos t) = −sin 2t / cos t。利用 sin 2t = 2 sin t cos t，化简得 −2 sin t。在 t = π/6 处，sin(π/6) = ½，故斜率为 −1。(b) y = cos 2t = 1 − 2 sin²t。由 x = 2 sin t 得 sin t = x/2，代入得 y = 1 − 2(x/2)² = 1 − x²/2，其中 0 ≤ x ≤ 2。(c) 面积 = ∫ y dx/dt dt = ∫₀^(π/2) cos 2t · (2 cos t) dt = 2 ∫₀^(π/2) cos 2t cos t dt。用积化和差 cos 2t cos t = ½(cos 3t + cos t)，积分得 2 × ½ ∫ (cos 3t + cos t) dt = [ (1/3)sin 3t + sin t ]₀^(π/2)。上限代入得 (1/3)(−1) + 1 = 2/3，下限为 0，面积 = 2/3。

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12. Revision Summary and Top Tips | 复习要点与高分技巧

Master parametric differentiation by repeatedly practising the chain‑ dy/dt)/(dx/dt) becomes automatic. Always attach a condition dx/dt ≠ 0 when writing an expression for dy/dx. For second derivatives, remember you are differentiating dy/dx with respect to x, not t, hence the extra division by dx/dt.

要掌握参数微分，就要反复练习链式法则的步骤，直到公式 dy/dx = (dy/dt)/(dx/dt) 烂熟于心。写出 dy/dx 表达式时，一定要附带条件 dx/dt ≠ 0。处理二阶导时，记住你是对 x 求导 dy/dx，而非对 t，所以要额外除以 dx/dt。

g t is vital. For integration, write the formula A = ∫ y dx/dt dt and ensure your limits are in terms of t. Finally, stay mindful of parameter domains when converting to Cartesian form – they often cut the curve down to a specific segment.

在绘图题中，总要突出驻点、竖直切线点和截距。标出 t 增大的方向至关重要。积分时，写出公式 A = ∫ y (dx/dt) dt 并确保积分限用 t 表示。最后，化为笛卡尔形式时要留意参数的定义域——它们常将曲线限制于特定的一段。

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