Solving Motion Problems Using Parametric and Vector-Valued Functions
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.
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.
The velocity vector is tangent to the path; its magnitude is the ____.
Speed = magnitude of velocity.
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."
A velocity component over time
y = bx
Speed combines the velocity components; integrating speed over time gives the total distance the particle travels.
For $\mathbf{v}(t)=\langle 2t,3\rangle$, find the speed at $t=2$.
$\sqrt{4^2+3^2}=5$.
Distance traveled over $[a,b]$ is...
Integrate the speed (magnitude of velocity).
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 $b$.
Which of these are vectors (not scalars)?
Velocity and acceleration are vectors; speed and distance are scalars.
The $x$-position at time $b$ is $x(a)$ plus...
Final = initial + accumulated change of $x$.
Distance traveled equals the magnitude of the displacement vector.
Distance is $\int|\mathbf{v}|\,dt$; displacement magnitude can be smaller.
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.
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).
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.