Velocity and Acceleration in 3D
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 3.4: “Motion in Space” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/3-4-motion-in-space |
| Textbook used in class | Stewart, Calculus, Section 13.4: “Motion in Space: Velocity and Acceleration” (Examples 1, 2) |
Quick Reference
If $\mathbf{r}(t)$ is the position vector of a particle at time $t$:
| Quantity | Formula |
|---|---|
| Velocity | $\mathbf{v}(t) = \mathbf{r}'(t)$ |
| Speed | $v(t) = |\mathbf{v}(t)| = |\mathbf{r}'(t)|$ |
| Acceleration | $\mathbf{a}(t) = \mathbf{v}'(t) = \mathbf{r}''(t)$ |
| Position from velocity | $\mathbf{r}(t) = \int \mathbf{v}(t)\,dt + \mathbf{C}$ |
Motivation
Every physical motion -- a satellite in orbit, a thrown baseball, a particle in a magnetic field -- is described by a position vector $\mathbf{r}(t)$ that depends on time. The derivative of position is velocity (how fast and in what direction the object is moving), and the derivative of velocity is acceleration (how velocity is changing). These are the same relationships from single-variable calculus, now applied componentwise to vectors.
The power of the vector approach is that a single equation like $\mathbf{a}(t) = \mathbf{r}''(t)$ simultaneously encodes the $x$-, $y$-, and $z$-components of Newton’s second law.
Key Concepts
1. Velocity and Speed
The velocity vector $\mathbf{v}(t) = \mathbf{r}'(t)$ is tangent to the path and points in the direction of motion. Its components give the rates of change in each coordinate direction.
The speed is the scalar magnitude of velocity: $|\mathbf{v}(t)|$. Speed is always nonneg and equals the rate at which arc length is accumulating: $ds/dt = |\mathbf{v}(t)|$.
2. Acceleration
The acceleration vector $\mathbf{a}(t) = \mathbf{r}''(t)$ is the derivative of velocity. Acceleration has two sources: changing speed (tangential) and changing direction (normal). These are separated formally in the next skill.
3. Recovering Position from Acceleration
Given $\mathbf{a}(t)$ and initial conditions $\mathbf{r}(t_0)$ and $\mathbf{v}(t_0)$, integrate twice: $$\mathbf{v}(t) = \int \mathbf{a}(t)\,dt, \quad \mathbf{r}(t) = \int \mathbf{v}(t)\,dt.$$ Each integration introduces a vector constant of integration, determined by the initial conditions.
Worked Example
A particle has acceleration $\mathbf{a}(t) = \langle 2, 6t, -\cos t\rangle$. Given $\mathbf{v}(0) = \langle 1, 0, 2\rangle$ and $\mathbf{r}(0) = \langle 0, 1, 0\rangle$, find $\mathbf{r}(t)$. (Adapted from Stewart 13.4, Example 2.)
Step 1. Integrate $\mathbf{a}(t)$ to get $\mathbf{v}(t)$. $$\mathbf{v}(t) = \int \langle 2, 6t, -\cos t\rangle\,dt = \langle 2t, 3t^2, -\sin t\rangle + \mathbf{C}_1.$$
Apply $\mathbf{v}(0) = \langle 1, 0, 2\rangle$: $$\langle 0, 0, 0\rangle + \mathbf{C}_1 = \langle 1, 0, 2\rangle \implies \mathbf{C}_1 = \langle 1, 0, 2\rangle.$$ $$\mathbf{v}(t) = \langle 2t + 1,\; 3t^2,\; -\sin t + 2\rangle.$$
Step 2. Integrate $\mathbf{v}(t)$ to get $\mathbf{r}(t)$. $$\mathbf{r}(t) = \int \langle 2t+1, 3t^2, -\sin t + 2\rangle\,dt = \langle t^2 + t,\; t^3,\; \cos t + 2t\rangle + \mathbf{C}_2.$$
Apply $\mathbf{r}(0) = \langle 0, 1, 0\rangle$: $$\langle 0, 0, 1\rangle + \mathbf{C}_2 = \langle 0, 1, 0\rangle \implies \mathbf{C}_2 = \langle 0, 1, -1\rangle.$$ $$\boxed{\mathbf{r}(t) = \langle t^2 + t,\; t^3 + 1,\; \cos t + 2t - 1\rangle.}$$
Check: $\mathbf{r}(0) = \langle 0, 1, 1-1\rangle = \langle 0,1,0\rangle$. Correct.
speed is the third component of velocity, not a separate quantity. Speed $|\mathbf{v}(t)|$ is a scalar computed from ALL components of $\mathbf{v}$. It is NOT one of the components. A particle with velocity $\langle 3, 4, 0\rangle$ has speed $\sqrt{9+16} = 5$, not speed 3 or 4.
Common Misconceptions
the speed of a particle is one of the component values of its velocity vector.
This is the input-output-confusion error. Speed $|\mathbf{v}(t)|$ is the scalar magnitude of the velocity vector, computed from all components by $|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}$. It is not equal to any single component. A particle with velocity $\langle 3, 4, 0\rangle$ has speed 5, not 3 or 4. The Pythagorean theorem must be applied across all three components to obtain the speed.
Leveled Practice
Problem 1. A particle moves with $\mathbf{r}(t) = \langle t^2, e^t, \sin t\rangle$. Find $\mathbf{v}(t)$, $\mathbf{a}(t)$, and the speed at $t = 0$.
Show answer
$\mathbf{v}(t) = \langle 2t, e^t, \cos t\rangle$.
$\mathbf{a}(t) = \langle 2, e^t, -\sin t\rangle$.
Speed at $t=0$: $|\mathbf{v}(0)| = |\langle 0, 1, 1\rangle| = \sqrt{0+1+1} = \sqrt{2}$.