Derivative of a Vector Function

A vector-valued function $\mathbf{r}(t) = \langle f(t),\, g(t),\, h(t)\rangle$ traces a curve in space as $t$ varies. Its derivative comes from the same limit as in single-variable calculus,

$\mathbf{r}'(t) = \lim_{\Delta t \to 0}\frac{\mathbf{r}(t + \Delta t) - \mathbf{r}(t)}{\Delta t}$

and because vector subtraction and scalar division act componentwise, this collapses to the rule you actually use:

$\mathbf{r}'(t) = \langle f'(t),\, g'(t),\, h'(t)\rangle$

It exists only where every component is differentiable — one non-smooth component kills the whole derivative.

What it means geometrically. $\mathbf{r}'(t)$ is a vector, not a slope: it points along the curve in the direction of increasing $t$ and is tangent to it. If $\mathbf{r}$ is position, then $\mathbf{r}'$ is velocity, $\|\mathbf{r}'\|$ is speed (a scalar), and $\mathbf{r}''$ is acceleration.

With $\mathbf{u}, \mathbf{v}$ differentiable and $c$ a scalar function:

$\frac{d}{dt}[\mathbf{u} + \mathbf{v}] = \mathbf{u}' + \mathbf{v}', \qquad \frac{d}{dt}[c\,\mathbf{u}] = c'\mathbf{u} + c\,\mathbf{u}'$

$\frac{d}{dt}[\mathbf{u} \cdot \mathbf{v}] = \mathbf{u}' \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{v}' \quad (\text{a scalar})$

$\frac{d}{dt}[\mathbf{u} \times \mathbf{v}] = \mathbf{u}' \times \mathbf{v} + \mathbf{u} \times \mathbf{v}' \quad (\text{order is fixed})$

$\frac{d}{dt}[\mathbf{r}(s(t))] = \mathbf{r}'(s(t))\,s'(t)$

Speed is the magnitude $\|\mathbf{r}'(t)\| = \sqrt{f'^2 + g'^2 + h'^2}$, and arc length on $[a,b]$ is $\int_a^b \|\mathbf{r}'(t)\|\,dt$.

The unit tangent requires $\mathbf{r}'(t) \neq \mathbf{0}$:

$\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}$

A curve with $\mathbf{r}' \neq \mathbf{0}$ everywhere is called smooth; where $\mathbf{r}' = \mathbf{0}$ the curve can have a cusp even though every component is differentiable.

• Reversing a cross product. $\mathbf{u}' \times \mathbf{v} + \mathbf{u} \times \mathbf{v}'$ — swapping either factor flips a sign, because $\mathbf{a} \times \mathbf{b} = -\,\mathbf{b} \times \mathbf{a}$.
• Confusing $\|\mathbf{r}\|'$ with $\|\mathbf{r}'\|$. They are different: for $\mathbf{r} = \langle\cos t, \sin t\rangle$ the first is $0$ and the second is $1$.
• Treating $\mathbf{r}'$ as a slope. It is a vector; the "slope" of a space curve is not defined.
• Forgetting the chain rule inside a component. $\frac{d}{dt}e^{3t} = 3e^{3t}$, not $e^{3t}$.
• Normalising before differentiating. $\mathbf{T}'(t)$ is not $\mathbf{r}''/\|\mathbf{r}''\|$; differentiate $\mathbf{r}$ first, then normalise.
• Dividing by zero at a cusp. Check $\mathbf{r}'(t) \neq \mathbf{0}$ before forming $\mathbf{T}(t)$.