📚 Case Study in Motion: Parametric Curves and Kinematics | 运动学案例研究：参数曲线与运动分析
In this article we will work through a full worked example of a kinematics problem involving parametric equations – a topic that frequently appears in CIE Additional Mathematics. By following the step-by-step case study, you will learn how to extract velocity, acceleration, rest times, path tangents and arc length from a pair of parametric equations describing planar motion. The methods shown here are directly applicable to exam questions and help build a strong, intuitive understanding of vector calculus in a practical context.
本文将带您完整演练一个涉及参数方程的运动学问题——这是 CIE 进阶数学中频繁出现的主题。通过逐步分析这个案例，您将学会如何从描述平面运动的一对参数方程中提取速度、加速度、静止时刻、路径切线以及弧长。这里展示的方法直接适用于考试题目，并且有助于在实践情境中建立起对向量微积分的扎实、直观的理解。
1. Problem Statement | 问题陈述
measured in metres. We wish to analyse the motion of the object: determine its velocity and acceleration, find when it is at rest, compute its speed and direction at specific instants, obtain the equation of the tangent to its path, evaluate the overall displacement and total distance travelled, and finally identify the moments when it strikes the x-axis.
考虑一个在 xy 平面内运动的物体。其在 t 秒时的位置向量为 r(t) = (t³ − 3t) i + (t² − 2t) j，其中 0 ≤ t ≤ 3，距离以米为单位。我们希望分析该物体的运动：确定其速度和加速度，求出何时静止，计算特定时刻的速率和方向，获得路径的切线方程，评估总位移和总路程，并最终确定物体何时撞击 x 轴。
2. Parametric Equations and Initial Conditions | 参数方程与初始条件
ts time, so the path is traversed as t increases from 0 to 3 seconds. We will refer to these parametric functions throughout the analysis.
位置分量为 x(t) = t³ − 3t 和 y(t) = t² − 2t。在 t = 0 时，物体位于原点，因为 x(0) = 0 且 y(0) = 0。参数 t 代表时间，因此随着 t 从 0 秒增加到 3 秒，物体遍历该路径。我们将在整个分析中引用这些参数函数。
3. Velocity Vector | 速度向量
. This vector describes both the instantaneous speed and the direction of motion at any time t.
将位置向量对 t 求导，得到速度向量：v(t) = dr/dt。因此，v(t) = (dx/dt) i + (dy/dt) j = (3t² − 3) i + (2t − 2) j。该向量描述了任意时刻 t 的瞬时速率和运动方向。
= 9 i + 2 j (m/s). The horizontal component is 9 m/s and the vertical component is 2 m/s.
对于 t = 2 秒，速度为 v(2) = (3·4 − 3) i + (4 − 2) j = 9 i + 2 j（米/秒）。水平分量为 9 m/s，垂直分量为 2 m/s。
4. Speed of the Object | 物体的速率
Speed is the magnitude of the velocity vector: |v(t)| = √[(3t² − 3)² + (2t − 2)²]. At t = 2, we substitute: |v(2)| = √9² + 2² = √81 + 4 = √85 alue gives the rate at which the object is moving along its path, regardless of direction.
速率是速度向量的大小：|v(t)| = √[(3t² − 3)² + (2t − 2)²]。在 t = 2 时，代入得：|v(2)| = √9² + 2² = √81 + 4 = √85 ≈ 9.22 m/s。该值表示物体沿路径运动的快慢，与方向无关。
5. Direction of Motion | 运动方向
vector is usually given by the angle θ it makes with the positive x-axis, where tan θ = vy / vx (provided vx ≠ 0). For t ≠ 1, vx = 3t² − 3 and vy t = 2 s the object is moving in a direction approximately 12.5° above the positive x-axis.
速度向量的方向通常用其与 x 轴正方向的夹角 θ 表示，其中 tan θ = vy / vx（假设 vx ≠ 0）。当 t ≠ 1 时，vx = 3t² − 3，vy = 2t − 2。因此，tan θ = (2t − 2) / (3t² − 3)。对于 t = 2，tan θ = 2/9，所以 θ = arctan(2/9) ≈ 12.5°。因此，在 t = 2 秒时，物体运动方向大约在 x 轴正方向上方 12.5° 处。
6. Acceleration Vector | 加速度向量
upward) and grows linearly in the i direction. At t = 2, a(2) = 12 i + 2 j, with magnitude √144 + 4 = √148 ≈ 12.17 m/s².
加速度是速度的导数：a(t) = dv/dt = (6t) i + 2 j。该向量在 j 方向上是恒定的（向上 2 m/s²），而在 i 方向上线性增长。在 t = 2 时，a(2) = 12 i + 2 j，其大小为 √144 + 4 = √148 ≈ 12.17 m/s²。
zero (as shown next). This is typical of a turning point where the object momentarily stops before changing direction.
注意，当 t = 1 时，速度虽然为零（见下文），但加速度并不为零。这是转折点的典型特征：物体在改变方向前瞬间静止。
7. Rest and Turning Points | 静止与转折点
(t > 0). The second equation gives t = 1. Hence, both components vanish exactly at t = 1 s – 1 − 2) = (−2, −1). This is a turning point; the path changes direction after this moment.
当速度向量为零向量时，物体静止。令 v(t) = 0：3t² − 3 = 0 且 2t − 2 = 0。第一个方程给出 t² = 1，所以 t = 1（t > 0）。第二个方程给出 t = 1。因此，两个分量恰好在 t = 1 秒时同时为零——物体瞬间静止。该时刻的位置为 r(1) = (1 − 3, 1 − 2) = (−2, −1)。这是一个转折点，此后路径改变方向。
Before t = 1, for example at t = 0.5, vx = 3·0.25 − 3 = −2.25 (motion to the left), vy = 1 − 2 = −1 (downward). After t = 1, say at t = 1.5, vx = 3·2.25 − 3 = 3.75 (rightward), vy 1 (upward). The object reverses both horizontal and vertical motion at the turning point.
在 t = 1 之前，例如 t = 0.5 时，vx = 3·0.25 − 3 = −2.25（向左运动），vy = 1 − 2 = −1（向下）。在 t = 1 之后，例如 t = 1.5 时，vx = 3·2.25 − 3 = 3.75（向右），vy = 3 − 2 = 1（向上）。物体在转折点处同时逆转了水平和垂直运动。
8. Tangent Line to the Path | 路径的切线
y) = (2,0) and the gradient is dy/dx = 2/9. The equation of the tangent line can be written as y − 0 = (2/9)(x − 2) or y = (2/9)x − 4/9.
路径上任一点的梯度由 dy/dx = (dy/dt) / (dx/dt) 给出，前提是 dx/dt ≠ 0。当 t ≠ 1 时，dy/dx = (2t − 2) / (3t² − 3)。在 t = 2 时，点为 (x,y) = (2,0)，梯度为 dy/dx = 2/9。切线方程可写为 y − 0 = (2/9)(x − 2) 或 y = (2/9)x − 4/9。
This tangent line represents the instantaneous direction of the path and can be used to approximate the object's trajectory near the point of tangency.
该切线表示路径的瞬时方向，可用于近似物体在切点附近的轨迹。
9. Displacement and Distance Travelled | 位移与路程
6) = (18, 3). Therefore, the displacement vector is Δr = 18 i + 3 j, and its magnitude is √18² + 3² ement is a vector, while distance is a scalar that accounts for the entire path travelled.
位移仅仅是运动始末位置的变化。r(0) = (0,0)，r(3) = (27 − 9, 9 − 6) = (18, 3)。因此，位移向量为 Δr = 18 i + 3 j，其大小为 √18² + 3² = √333 ≈ 18.25 m。位移是一个向量，而路程是一个标量，考虑物体走过的整个路径。
distance travelled (path length) is the integral of speed over the time interval: L = ∫₀³ |v(t)| dt = ∫₀³ √9t⁴ − 14t² − 8t + 13 ded with a table of values or asked to use the trapezium rule. We will demonstrate Simpson's rule with n = 6 (h = 0.5) for an approximation.
总路程（路径长度）是速率对时间区间的积分：L = ∫₀³ |v(t)| dt = ∫₀³ √9t⁴ − 14t² − 8t + 13 dt。该积分没有初等原函数，因此需要数值方法。在考试中，可能会提供数值表格或要求使用梯形法则。这里我们使用辛普森法则，取 n = 6（h = 0.5）进行近似。
| t | 0 | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 |
| |v(t)| | √13 ≈ 3.606 | √9·0.0625 −14·0.25 −4+13 ≈ √0.5625−3.5−4+13=√6.0625≈2.462 | √9−14−8+13=√0=0 | √9·5.0625 −14·2.25 −12+13=√45.5625−31.5−12+13=√15.0625≈3.881 | √85 ≈ 9.220 | √9·39.0625 −14·6.25 −20+13=√351.5625−87.5−20+13=√257.0625≈16.034 | √9·81 −14·9 −24+13=√729−126−24+13=√592≈24.331 |
.937 + 4·22.377 + 18.44] = (0.1667)(27.937 + 89.508 + 18.44) = 0.1667·135.885 ≈ 22.65 m. Thus the total distance travelled is about 22.7 m, notably larger than the magnitude of displacement (18.25 m) because the path is curved and the object went back on itself.
应用辛普森法则： [27.937 + 4·22.377 + 18.44] = (0.1667)(27.937 + 89.508 + 18.44) = 0.1667·135.885 ≈ 22.65 m。因此，总路程约为 22.7 m，明显大于位移的大小（18.25 m），因为路径是弯曲的，且物体出现折返。
10. Impact with the x-axis | 与 x 轴的撞击
. At t = 2, the position is (2³ − 3·2, 2² − 2·2) = (8 − 6, 4 − 4) = (2,0). So the second impact occurs at (2,0), moving with velocity v(2) = 9 i + 2 j. The speed at impact is √85 above the horizontal, indicating the object leaves the x-axis immediately after touching it.
当 y 坐标为零时，物体撞击 x 轴。解 y(t) = t² − 2t = 0，得 t = 0 和 t = 2。t = 0 时物体位于起点 (0,0)。t = 2 时，位置为 (2³ − 3·2, 2² − 2·2) = (8 − 6, 4 − 4) = (2,0)。因此第二次撞击发生在 (2,0) 点，此时速度为 v(2) = 9 i + 2 j。撞击时速率为 √85 ≈ 9.22 m/s，方向在水平上方 12.5°，表明物体接触 x 轴后立即离开。
This illustrates that even though y returns to zero, the vertical velocity is not zero (2 m/s upward), so the object does not stop there – it merely crosses the axis.
这表明尽管 y 回到了零，但垂直速度并非为零（向上 2 m/s），所以物体并未在那里停下，而只是穿过 x 轴。
11. Summary and Key Takeaways | 总结与关键要点
Through this case study we have applied differentiation and integration to parametric equations in a real motion context. The object’s journey included a full stop at (−2,−1), a reversal of direction, a curved path, and two crossings of the x-axis. Velocity and acceleration vectors gave insights into instantaneous behavior, while numerical integration provided the path length when exact integration was impractical. Mastering these techniques will prepare you for a wide range of kinematics and calculus problems in CIE Additional Mathematics.
通过这个案例研究，我们在真实运动情境中应用了微分和积分于参数方程。物体的旅程包括在 (−2,−1) 处短暂停顿、方向逆转、弯曲路径以及两次穿过 x 轴。速度与加速度向量揭示了瞬时的运动行为，而当精确积分不可行时，数值积分给出了路径长度。掌握这些技巧，将使您为应对 CIE 进阶数学中广泛的运动学和微积分问题做好充分准备。
Published by TutorHao | Additional Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图