Derivatives of Vector Functions
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 3.2: “Calculus of Vector-Valued Functions” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/3-2-calculus-of-vector-valued-functions |
| Textbook used in class | Stewart, Calculus, Section 13.2: “Derivatives and Integrals of Vector Functions” (Examples 1, 2, 3) |
Opening Scenario
If $\mathbf{r}(t)$ gives the position of a particle at time $t$, then $\mathbf{r}'(t)$ gives the velocity: the instantaneous rate of change of position. The derivative of a vector function is defined exactly like the scalar derivative -- via a limit of difference quotients -- and the result is a vector pointing in the direction the curve is moving.
Quick Reference
Derivative: $\mathbf{r}'(t) = \langle f'(t), g'(t), h'(t)\rangle$ (differentiate each component).
The derivative as limit: $\mathbf{r}'(t) = \lim_{h\to 0}\dfrac{\mathbf{r}(t+h) - \mathbf{r}(t)}{h}$.
Tangent vector: $\mathbf{r}'(t)$ is tangent to the curve at the point $\mathbf{r}(t)$, pointing in the direction of motion.
Unit tangent vector: $\mathbf{T}(t) = \dfrac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}$.
Differentiation rules:
- $(c\mathbf{r})' = c\,\mathbf{r}'$
- $(\mathbf{r}_1 + \mathbf{r}_2)' = \mathbf{r}_1' + \mathbf{r}_2'$
- $(f\mathbf{r})' = f'\mathbf{r} + f\mathbf{r}'$
- $(\mathbf{r}_1\cdot\mathbf{r}_2)' = \mathbf{r}_1'\cdot\mathbf{r}_2 + \mathbf{r}_1\cdot\mathbf{r}_2'$
- $(\mathbf{r}_1\times\mathbf{r}_2)' = \mathbf{r}_1'\times\mathbf{r}_2 + \mathbf{r}_1\times\mathbf{r}_2'$
- $(\mathbf{r}(u(t)))' = u'(t)\,\mathbf{r}'(u(t))$ (chain rule)
Key Concepts
1. Computing the Derivative
Example 1. Differentiate $\mathbf{r}(t) = \langle t^3, \sin t, e^{2t}\rangle$ and find the unit tangent at $t = 0$. (Stewart 13.2, Example 1.)
$\mathbf{r}'(t) = \langle 3t^2, \cos t, 2e^{2t}\rangle$.
At $t = 0$: $\mathbf{r}'(0) = \langle 0, 1, 2\rangle$. $|\mathbf{r}'(0)| = \sqrt{0 + 1 + 4} = \sqrt{5}$.
$\mathbf{T}(0) = \dfrac{1}{\sqrt{5}}\langle 0, 1, 2\rangle$.
2. The Constant-Magnitude Rule
If $|\mathbf{r}(t)|$ is constant, then $\mathbf{r}(t)\cdot\mathbf{r}'(t) = 0$ (the position and velocity are always perpendicular). Proof: $d/dt\,|\mathbf{r}|^2 = 2\mathbf{r}\cdot\mathbf{r}' = 0$.
This is why a planet moves perpendicularly to the radius vector at each instant (Kepler’s second law has a geometric shadow here).
differentiating the magnitude instead of the vector. $\dfrac{d}{dt}|\mathbf{r}(t)|$ is a scalar -- the rate of change of the length of $\mathbf{r}$. $\mathbf{r}'(t)$ is a vector -- the rate of change of $\mathbf{r}$ itself. These are different objects. For a particle moving on a circle, $|\mathbf{r}(t)|$ is constant but $\mathbf{r}'(t) \neq \mathbf{0}$.
Common Errors Summary
| Error | Correction |
|---|---|
| Differentiating $|\mathbf{r}|$ instead of $\mathbf{r}$ | Differentiate each component; the result is a vector, not a scalar |
| Applying the cross-product rule in the wrong order | $(\mathbf{r}_1\times\mathbf{r}_2)' \neq \mathbf{r}_1'\times\mathbf{r}_2' $; use the product rule keeping order intact |
Common Misconceptions
the derivative of the magnitude of a vector function equals the magnitude of the derivative.
This is the height-vs-slope error in the vector setting. The scalar $d|\mathbf{r}(t)|/dt$ measures how the length of $\mathbf{r}$ changes, while the vector $\mathbf{r}'(t)$ measures how $\mathbf{r}$ itself changes. For a particle moving on a circle, $|\mathbf{r}(t)|$ is constant so $d|\mathbf{r}|/dt = 0$, yet $\mathbf{r}'(t) \neq \mathbf{0}$ because the direction is changing. The two quantities are related by $\mathbf{r} \cdot \mathbf{r}' = |\mathbf{r}|\,(d|\mathbf{r}|/dt)$, which shows they are generally different.
Leveled Practice
Problem 1. Find $\mathbf{r}'(t)$ and $\mathbf{T}(\pi)$ for $\mathbf{r}(t) = \langle \cos t, \sin t, t\rangle$.
Show answer
$\mathbf{r}'(t) = \langle -\sin t, \cos t, 1\rangle$. At $t = \pi$: $\mathbf{r}'(\pi) = \langle 0, -1, 1\rangle$. $|\mathbf{r}'(\pi)| = \sqrt{2}$. $\mathbf{T}(\pi) = \langle 0, -1/\sqrt{2}, 1/\sqrt{2}\rangle$.