📚 Integrating Vectors | 向量积分
In A-level Mathematics, particularly within the mechanics strand of the Edexcel specification, integrating vector functions is an essential technique. It enables us to move from acceleration to velocity, and from velocity to displacement, all while working in two or three dimensions. Mastering vector integration unlocks a deeper understanding of kinematics and provides a powerful tool for solving real-world motion problems with variable forces.
在A-level数学中，尤其是在爱德思考试局的力学部分，向量函数的积分是一项核心技术。它使我们能够从加速度求速度，再从速度求位移，一切都是在二维或三维空间中进行的。掌握向量积分可以加深对运动学的理解，并为求解具有变力的真实运动问题提供强有力的工具。
1. Introduction to Integrating Vectors | 向量积分简介
Integrating a vector function simply means integrating each of its scalar components independently with respect to the same variable, typically time t. If a velocity vector is given as v(t) = f(t) i + g(t) j, then its indefinite integral is ∫ v(t) dt = (∫ f(t) dt) i + (∫ g(t) dt) j + C, where C is a constant vector. This component-wise approach is the key to all vector integration.
积分一个向量函数，就是对其每一个标量分量关于同一个变量（通常是时间 t）分别积分。若速度向量为 v(t) = f(t) i + g(t) j，则其不定积分为 ∫ v(t) dt = (∫ f(t) dt) i + (∫ g(t) dt) j + C，其中 C 是一个常向量。这种逐分量处理的方法是所有向量积分的关键。
2. Definite and Indefinite Integrals of Vector Functions | 向量函数的定积分与不定积分
An indefinite integral of a vector produces a family of vector functions differing by an arbitrary constant vector C = c₁i + c₂j. A definite integral between limits t = a and t = b, written as ∫ab v(t) dt, gives a specific vector representing the net change. For instance, ∫02 (2t i + 3t² j) dt = [t² i + t³ j]02 = 4i + 8j.
向量函数的不定积分会得到一族相差一个任意常向量 C = c₁i + c₂j 的函数。而在上下限 t = a 和 t = b 之间的定积分 ∫ab v(t) dt 则给出一个表示净变化的确定向量。例如，∫02 (2t i + 3t² j) dt = [t² i + t³ j]02 = 4i + 8j。
Note that the definite integral of a velocity vector over a time interval directly yields the displacement vector of the particle. The constant vector C cancels out in the evaluation, which is why we use indefinite integration with initial conditions when finding position functions.
请注意，速度向量在时间区间上的定积分直接给出粒子的位移向量。在计算过程中，常向量 C 会被抵消，这就是为什么在求位置函数时，我们常常采用带初始条件的不定积分方法。
3. Integrating Velocity to Find Displacement | 积分速度求位移
In kinematics, if a particle moves with velocity v(t), then its displacement vector Δr from time t₁ to t₂ is the definite integral Δr = ∫t₁t₂ v(t) dt. This vector represents the straight-line difference between the final and initial positions, independent of the path taken.
在运动学中，若粒子以速度 v(t) 运动，则从时间 t₁ 到 t₂ 的位移向量 Δr 就是定积分 Δr = ∫t₁t₂ v(t) dt。该向量表示最终位置与初始位置之间的直线差，与所经过的路径无关。
To obtain the position vector r(t) at any time, we use an indefinite integral: r(t) = ∫ v(t) dt + r0, where r0 is the initial position. This is fundamental for problems where we track motion from a known starting point.
为得到任意时刻的位置向量 r(t)，我们采用不定积分：r(t) = ∫ v(t) dt + r0，其中 r0 为初始位置。这对于从已知起点追踪运动的问题至关重要。
4. Integrating Acceleration to Find Velocity | 积分加速度求速度
When acceleration a(t) is known, the velocity vector is the integral of acceleration plus an initial velocity vector: v(t) = ∫ a(t) dt + v0. This is the vector analogue of the scalar equation v = u + ∫ a dt from constant-acceleration formulas.
当已知加速度 a(t) 时，速度向量就是加速度的积分加上初速度向量：v(t) = ∫ a(t) dt + v0。这正是一维匀加速公式 v = u + ∫ a dt 在向量空间中的推广。
v(t) = ∫ a(t) dt + v0
For example, if a = (6t i − 4 j) ms⁻² and the particle starts from rest at the origin, then v(t) = (3t² i − 4t j) ms⁻¹. Always remember to include the vector constant of integration determined by initial conditions.
例如，若 a = (6t i − 4 j) ms⁻²，且粒子从原点静止出发，则 v(t) = (3t² i − 4t j) ms⁻¹。请始终记住要包含由初始条件确定的向量积分常数。
5. Initial Conditions and Constants of Integration | 初始条件与积分常数
One of the most common pitfalls is forgetting that the constant of integration is a vector. When you integrate a(t) to get v, you must add C = c₁i + c₂j, not just a scalar C. Use a given velocity at a specific time (often t = 0) to solve for both components of C simultaneously.
最常见的陷阱之一，是忘记积分常数是一个向量。当你对 a(t) 积分求 v 时，必须加上 C = c₁i + c₂j，而不只是一个标量 C。利用某个特定时刻（通常为 t = 0）的速度，可以同时解出 C 的两个分量。
Similarly, when integrating velocity to get position, the constant vector D = d₁i + d₂j is determined by the initial position r(0). Write out the component equations and substitute the given vectors to find the unknown constants.
同样地，当对速度积分求位置时，常向量 D = d₁i + d₂j 由初始位置 r(0) 决定。请写出分量方程并代入已知向量，以求出未知常数。
6. Working with Parametric Vectors | 参数向量处理
Velocity is frequently expressed as a function of time in the form v(t) = p(t)i + q(t)j, where p and q are known scalar functions. Integrating component by component to find displacement is straightforward, but you must treat each component separately and then combine the results into a single vector expression.
速度常常以时间函数的形式表示为 v(t) = p(t)i + q(t)j，其中 p 和 q 是已知的标量函数。逐分量积分以求出位移是直观的，但你必须分别处理每个分量，然后将结果合并为一个向量表达式。
If the motion is in three dimensions, simply extend the method to include a k component. For instance, v = 3t² i + et j + sin t k integrates to r = t³i + etj − cos t k + C.
若运动发生在三维空间，只需延伸该方法以纳入 k 分量。例如，v = 3t² i + et j + sin t k 积分后得到 r = t³i + etj − cos t k + C。
7. Integrating to Find Position Vectors from Velocity | 由速度积分求位置向量
Once the velocity function is determined, finding the position vector requires a second integration. If v(t) = (4t+1)i + 2tj (using a power-like example carefully), the position is r(t) = ∫ [(4t+1)i + 2tj] dt = (2t²+
Published by TutorHao | A-Level 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