📚 AQA A-Level Mathematics Exercise 6C: Parametric Equations and Differentiation | AQA A-Level 数学 Exercise 6C：参数方程与微分
Exercise 6C in the AQA A-Level Mathematics course focuses on differentiating parametric equations. In this exercise, you will learn how to find the gradient of a curve when x and y are both expressed in terms of a third variable, usually t or θ. The techniques you master here are essential for later topics such as rates of change, curve sketching, and applied mechanics.
AQA A-Level 数学课程中的 Exercise 6C 聚焦于参数方程的微分。在本练习中，你将学会当 x 和 y 都表示为第三个变量（通常是 t 或 θ）的函数时，如何求曲线的斜率。你在这里掌握的技巧对后续主题（如变化率、曲线绘制和应用力学）至关重要。
1. Understanding Parametric Equations | 理解参数方程
A parametric equation defines a curve using two separate equations. Instead of writing y in terms of x directly, both x and y are written in terms of a parameter. For example, x = t² and y = t³ together describe a curve, and every value of t gives one point (x, y) on the curve.
参数方程使用两个独立的方程来定义一条曲线。与其直接将 y 表示为 x 的函数，不如将 x 和 y 都写成某个参数的函数。例如，x = t² 和 y = t³ 共同描述一条曲线，每一个 t 值都对应曲线上的一点 (x, y)。
The parameter often represents time in mechanics, or an angle in geometry. The same curve can have many different parameterisations, so it is important to focus on the relationship between x and y rather than the exact choice of parameter.
在力学中，参数通常表示时间；在几何中，参数通常表示角度。同一条曲线可以有许多不同的参数化方式，因此重点应放在 x 与 y 之间的关系上，而不是参数本身的具体选择。
2. The Chain Rule Connection | 链式法则的联系
To find the gradient dy/dx for a parametric curve, you use the chain rule. Since y is given as a function of t and x is also given as a function of t, the derivative of y with respect to x can be written as the ratio of two separate derivatives.
要求参数曲线的斜率 dy/dx，需要使用链式法则。由于 y 是 t 的函数，x 也是 t 的函数，因此 y 对 x 的导数可以写成两个独立导数的比值。
dy/dx = (dy/dt) ÷ (dx/dt), provided dx/dt ≠ 0
This relationship is often memorised as “the derivative of y with respect to t, divided by the derivative of x with respect to t.” It works because dy/dt = (dy/dx) × (dx/dt), and then dividing both sides by dx/dt gives the formula above.
这个关系通常被记作“y 对 t 的导数除以 x 对 t 的导数”。它之所以成立，是因为 dy/dt = (dy/dx) × (dx/dt)，两边同时除以 dx/dt 即可得到上述公式。
3. Step-by-Step Differentiation Process | 微分分步过程
Follow this systematic process for any parametric differentiation problem:
在解决任何参数微分问题时，请按以下系统化步骤进行：
- Step 1: Differentiate x with respect to t to find dx/dt.
- 第一步：对 x 关于 t 求导，得到 dx/dt。
- Step 2: Differentiate y with respect to t to find dy/dt.
- 第二步：对 y 关于 t 求导，得到 dy/dt。
- Step 3: Divide dy/dt by dx/dt to obtain dy/dx.
- 第三步：用 dy/dt 除以 dx/dt，得到 dy/dx。
- Step 4: Simplify the expression, cancelling common factors where possible.
- 第四步：化简表达式，尽可能约去公因子。
- Step 5: If a particular t value is given, substitute it to find the gradient at that point.
- 第五步：若给出了具体的 t 值，代入以求出该点处的斜率。
4. Worked Example 6C.1 | 例题 6C.1
Consider the curve defined by x = t² and y = 2t³. Find dy/dx in terms of t.
考虑由 x = t² 和 y = 2t³ 定义的曲线，求用 t 表示的 dy/dx。
First differentiate x with respect to t: dx/dt = 2t. Then differentiate y with respect to t: dy/dt = 6t². Therefore:
首先对 x 关于 t 求导：dx/dt = 2t。然后对 y 关于 t 求导：dy/dt = 6t²。因此：
dy/dx = 6t² ÷ 2t = 3t
Notice that the factor t cancels, leaving the simple result dy/dx = 3t. This gradient is valid for t ≠ 0, because when t = 0 the original expression dx/dt equals 0, meaning the tangent is vertical.
注意 t 因子被约去，得到简单结果 dy/dx = 3t。该斜率在 t ≠ 0 时成立，因为当 t = 0 时，原始表达式 dx/dt = 0，这意味着切线是竖直的。
5. Worked Example 6C.2 | 例题 6C.2
Now consider a curve where both x and y are trigonometric functions of θ: x = a cos θ, y = b sin θ. This represents an ellipse. Find dy/dx.
现在考虑一条曲线，其中 x 和 y 都是 θ 的三角函数：x = a cos θ, y = b sin θ。这表示一个椭圆。求 dy/dx。
Differentiate: dx/dθ = -a sin θ, and dy/dθ = b cos θ. Hence:
求导：dx/dθ = -a sin θ，dy/dθ = b cos θ。因此：
dy/dx = (b cos θ) ÷ (-a sin θ) = -(b cos θ)/(a sin θ) = -(b/a) cot θ
This result is very useful for finding the gradients of tangent lines to an ellipse at a given angle θ. It also shows that the gradient is undefined when θ = 0, π, etc., because sin θ = 0 and hence dx/dθ = 0.
这个结果对于求椭圆在给定角度 θ 处的切线斜率非常有用。它也表明当 θ = 0, π 等值时斜率无定义，因为此时 sin θ = 0，从而 dx/dθ = 0。
6. Finding Tangents and Normals | 求切线与法线
Once you have dy/dx for a parametric curve, you can find the equation of the tangent at a specific point. Substitute the given parameter value into x and y to find the coordinates, and into dy/dx to find the gradient. Then use the straight-line equation:
一旦求出了参数曲线的 dy/dx，就可以找到特定点处的切线方程。将给定参数值代入 x 和 y 以得到坐标，代入 dy/dx 以得到斜率，然后使用直线方程：
y – y₁ = m(x – x₁)
For the normal, the gradient is the negative reciprocal of the tangent gradient: m_normal = -1/m_tangent, provided m_tangent ≠ 0.
对于法线，其斜率是切线斜率的负倒数：m_normal = -1/m_tangent，前提是 m_tangent ≠ 0。
For example, using the earlier curve x = t², y = 2t³, at t = 1 we have x = 1, y = 2, and dy/dx = 3. The tangent equation is y – 2 = 3(x – 1), which simplifies to y = 3x – 1.
例如，对于之前的曲线 x = t², y = 2t³，在 t = 1 处，x = 1, y = 2，dy/dx = 3。切线方程为 y – 2 = 3(x – 1)，化简得 y = 3x – 1。
7. Second Derivatives of Parametric Curves | 参数曲线的二阶导数
Many students make the mistake of thinking that d²y/dx² is obtained by differentiating dy/dx with respect to t and dividing by d²x/dt². This is incorrect. The correct method treats dy/dx as a new function of t, and then applies the same parametric differentiation rule again.
许多学生错误地认为 d²y/dx² 可以通过对 dy/dx 关于 t 求导，再除以 d²x/dt² 来得到。这是不正确的。正确的方法是将 dy/dx 视为一个关于 t 的新函数，然后再次应用相同的参数微分规则。
d²y/dx² = [d/dt(dy/dx)] ÷ (dx/dt)
Let us verify with x = t² and y = 2t³. We already found dy/dx = 3t. Then d/dt(dy/dx) = 3, and dx/dt = 2t. Therefore d²y/dx² = 3/(2t).
我们用 x = t², y = 2t³ 来验证。我们已经求出 dy/dx = 3t。那么 d/dt(dy/dx) = 3，并且 dx/dt = 2t。因此 d²y/dx² = 3/(2t)。
8. Common Mistakes and Pitfalls | 常见错误与陷阱
| Mistake | Correction |
| Writing dy/dx = dx/dt ÷ dy/dt | Always use dy/dx = (dy/dt) ÷ (dx/dt) |
| Forgetting dx/dt = 0 condition | If dx/dt = 0, the tangent is vertical and dy/dx is undefined |
| Second derivative: d²y/dx² = (d²y/dt²) ÷ (d²x/dt²) | Use d²y/dx² = [d/dt(dy/dx)] ÷ (dx/dt) |
| Substituting values before differentiating | Differentiate first, then substitute the parameter value |
The table above summarises the most common errors seen in Exercise 6C. Always check that dx/dt is not zero before dividing, and always simplify your final gradient expression.
上表总结了 Exercise 6C 中最常见的错误。在除法之前始终检查 dx/dt 是否为零，并始终化简最终的斜率表达式。
9. Exam-Style Questions | 考试风格题
Here is a typical AQA exam-style question: A curve is given by x = eᵗ and y = t². Find dy/dx in terms of t, and then find the equation of the tangent at t = 0.
下面是一道典型的 AQA 考试风格题：一条曲线由 x = eᵗ 和 y = t² 给出。用 t 表示 dy/dx，并求 t = 0 处的切线方程。
We have dx/dt = eᵗ and dy/dt = 2t. Therefore dy/dx = 2t / eᵗ = 2t e⁻ᵗ. At t = 0, we get x = e⁰ = 1, y = 0, and dy/dx = 0. So the tangent is horizontal: y = 0.
我们有 dx/dt = eᵗ，dy/dt = 2t。因此 dy/dx = 2t / eᵗ = 2t e⁻ᵗ。在 t = 0 时，x = e⁰ = 1，y = 0，dy/dx = 0。所以切线是水平的：y = 0。
Another question: The curve x = t² + 1, y = t³ – 3t. Find the values of t where the gradient of the tangent is zero.
另一个问题：曲线 x = t² + 1, y = t³ – 3t。求切线斜率为零时的 t 值。
Compute dx/dt = 2t, dy/dt = 3t² – 3. Thus dy/dx = (3t² – 3)/(2t). Setting dy/dx = 0 requires the numerator to be zero: 3t² – 3 = 0, so t² = 1, giving t = ±1. Check that dx/dt ≠ 0 at these values: dx/dt = 2t = ±2, which is fine. Therefore the tangent is horizontal at t = 1 and t = -1.
计算 dx/dt = 2t, dy/dt = 3t² – 3。因此 dy/dx = (3t² – 3)/(2t)。令 dy/dx = 0 需要分子为零：3t² – 3 = 0，所以 t² = 1，得到 t = ±1。检查这些值处 dx/dt ≠ 0：dx/dt = 2t = ±2，没有问题。因此切线在 t = 1 和 t = -1 处是水平的。
10. Summary and Revision Checklist | 总结与复习清单
Parametric differentiation is a core skill for AQA A-Level Mathematics. The key formula dy/dx = (dy/dt) ÷ (dx/dt) appears frequently in both pure mathematics and applications. You must also be able to find tangents and normals, and to compute second derivatives correctly.
参数微分是 AQA A-Level 数学的核心技能。关键公式 dy/dx = (dy/dt) ÷ (dx/dt) 在纯数学和应用中频繁出现。你还必须能够求切线与法线，并正确计算二阶导数。
- Always write down dx/dt and dy/dt separately before forming the ratio.
- 始终先分别写出 dx/dt 和 dy/dt，再构成比值。
- State when dy/dx is undefined, i.e. when dx/dt = 0.
- 说明 dy/dx 何时无定义，即 dx/dt = 0 时。
- Remember the second derivative rule: differentiate dy/dx with respect to t, then divide by dx/dt.
- 记住二阶导数规则：先对 dy/dx 关于 t 求导，再除以 dx/dt。
- Practice with polynomial, exponential, trigonometric, and logarithmic parameters.
- 练习多项式、指数、三角函数和对数形式的参数。
By completing Exercise 6C carefully, you will build the fluency needed to tackle parametric problems in the AQA examinations with confidence.
通过认真完成 Exercise 6C，你将建立起解决 AQA 考试中参数问题所需的熟练度，从而充满信心地应对考试。
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