Solving Motion Problems Using Parametric and Vector-Valued Functions · 使用参数方程和向量值函数求解运动问题
Full planar motion, start to finish
- Now combine everything: a particle moving in the plane, described by $\mathbf{r}(t)=\langle x(t),y(t)\rangle$.
- Differentiate for velocity and acceleration; integrate to go the other way.
- Compute speed and distance traveled from the velocity components.
- This lesson is the toolbox for any BC plane-motion problem.
完整的平面运动,从头到尾
- 现在把一切结合:一个在平面中运动的质点,由 $\mathbf{r}(t)=\langle x(t),y(t)\rangle$ 描述。
- 求导得速度与加速度;积分则反向。
- 由速度分量计算速率与行进距离。
- 这一课是任何 BC 平面运动问题的工具箱。
Position, velocity, acceleration
- Position: $\mathbf{r}(t)=\langle x(t),\,y(t)\rangle$.
- Velocity: $\mathbf{v}(t)=\langle x'(t),\,y'(t)\rangle$ — direction of motion, tangent to the path.
- Acceleration: $\mathbf{a}(t)=\langle x''(t),\,y''(t)\rangle$.
- Move up the chain by differentiating, down by integrating with initial conditions.
位置、速度、加速度
- 位置: $\mathbf{r}(t)=\langle x(t),\,y(t)\rangle$。
- 速度: $\mathbf{v}(t)=\langle x'(t),\,y'(t)\rangle$——运动方向,与路径相切。
- 加速度: $\mathbf{a}(t)=\langle x''(t),\,y''(t)\rangle$。
- 通过求导沿链向上,通过带初始条件积分向下。
The velocity vector is tangent to the path; its magnitude is the ____. · 速度矢量与路径相切；其模是 ____。
Speed = magnitude of velocity. · 速率 = 速度的模。
Speed and distance traveled
- Speed at time $t$: $|\mathbf{v}(t)|=\sqrt{\big(x'(t)\big)^2+\big(y'(t)\big)^2}$ — a scalar.
- Distance traveled over $[a,b]$: integrate the speed, $\displaystyle\int_a^b |\mathbf{v}(t)|\,dt$.
- That's the same arc-length integral — total path length, never negative.
- Speed answers "how fast right now"; the integral answers "how far overall."
速率与行进距离
- 时刻 $t$ 的速率:$|\mathbf{v}(t)|=\sqrt{\big(x'(t)\big)^2+\big(y'(t)\big)^2}$——一个标量。
- $[a,b]$ 上的行进距离:对速率积分,$\displaystyle\int_a^b |\mathbf{v}(t)|\,dt$。
- 那就是同一个弧长积分——总路径长度,永不为负。
- 速率回答"此刻多快";积分回答"总共多远"。
A velocity component over time · 速度分量随时间变化
y = bx
Speed combines the velocity components; integrating speed over time gives the total distance the particle travels. · 速率结合速度分量；对速率关于时间积分得到粒子行进的总距离。
For · 支持 $\mathbf{v}(t)=\langle 2t,3\rangle$, find the speed at $t=2$. · 对于 $\mathbf{v}(t)=\langle 2t,3\rangle$，求在 $t=2$ 处的速率。
$\sqrt{4^2+3^2}=5$.
Distance traveled over $[a,b]$ is... · $[a,b]$ 期间行进的距离为...
Integrate the speed (magnitude of velocity). · 积分速率（速度的模）。
Position at a later time
- To find where the particle is at time $b$: start from a known position and add the accumulated change.
- $x(b)=x(a)+\displaystyle\int_a^b x'(t)\,dt$, and likewise for $y$ — each component separately.
- This is "final = initial + accumulated change," done per coordinate.
- Combine the two results into the position vector at time $b$.
之后某时刻的位置
- 要求质点在时刻 $b$ 在哪里:从已知位置出发,加上累积变化。
- $x(b)=x(a)+\displaystyle\int_a^b x'(t)\,dt$,$y$ 也类似——每个分量分别。
- 这是"最终 = 初始 + 累积变化",逐坐标做。
- 把两个结果合成时刻 $b$ 的位置向量。
Which of these are vectors (not scalars)? · 以下哪项是矢量（非标量）？
Velocity and acceleration are vectors; speed and distance are scalars. · 速度和加速度是矢量；速率和距离是标量。
The $x$-position at time $b$ is $x(a)$ plus... · $x$时刻的$b$位置是$x(a)$加上...
Final = initial + accumulated change of $x$. · 最终 = 初始 + $x$ 的累积变化量。
Distance traveled equals the magnitude of the displacement vector. · 行进距离等于位移矢量的模。
Distance is $\int|\mathbf{v}|\,dt$; displacement magnitude can be smaller. · 距离是 $\int|\mathbf{v}|\,dt$；位移模可能更小。
Keep the vectors and scalars straight: velocity and acceleration are vectors (direction + size); speed and distance are scalars. Distance traveled is $\int|\mathbf{v}|\,dt$ (the speed integral), not the displacement $\int \mathbf{v}\,dt$ or $|\int\mathbf{v}\,dt|$. And find a later position by adding the component integrals to the initial position.
把向量和标量分清:速度和加速度是向量(方向 + 大小);速率和距离是标量。行进距离是 $\int|\mathbf{v}|\,dt$(速率积分),而非位移 $\int \mathbf{v}\,dt$ 或 $|\int\mathbf{v}\,dt|$。而求之后的位置,把分量积分加到初始位置上。
A particle has $\mathbf{v}(t)=\langle 2t,\ 3\rangle$. Find its speed at $t=2$ and set up the distance traveled on $[0,2]$.
- Speed at $t=2$: $|\mathbf{v}(2)|=\sqrt{(4)^2+3^2}=\sqrt{25}=5$.
- Distance on $[0,2]$: $\displaystyle\int_0^2\sqrt{(2t)^2+3^2}\,dt=\int_0^2\sqrt{4t^2+9}\,dt$ (evaluate numerically).
一个质点有 $\mathbf{v}(t)=\langle 2t,\ 3\rangle$。求 $t=2$ 处的速率,并建立 $[0,2]$ 上的行进距离。
- $t=2$ 处速率:$|\mathbf{v}(2)|=\sqrt{(4)^2+3^2}=\sqrt{25}=5$。
- $[0,2]$ 上距离:$\displaystyle\int_0^2\sqrt{(2t)^2+3^2}\,dt=\int_0^2\sqrt{4t^2+9}\,dt$(数值求值)。
For plane motion $\mathbf{r}(t)=\langle x,y\rangle$: velocity $\mathbf{v}=\mathbf{r}'$, acceleration $\mathbf{a}=\mathbf{r}''$ (vectors); speed $|\mathbf{v}|=\sqrt{(x')^2+(y')^2}$ and distance traveled $\int_a^b|\mathbf{v}|\,dt$ (scalars). Find a later position by adding the component integrals of $x',y'$ to the initial position.
对平面运动 $\mathbf{r}(t)=\langle x,y\rangle$:速度 $\mathbf{v}=\mathbf{r}'$、加速度 $\mathbf{a}=\mathbf{r}''$(向量);速率 $|\mathbf{v}|=\sqrt{(x')^2+(y')^2}$ 与行进距离 $\int_a^b|\mathbf{v}|\,dt$(标量)。求之后的位置,把 $x',y'$ 的分量积分加到初始位置。