Derivative of vector-valued function

The derivative of a vector-valued function $\mathbf{r}(t)$ is the componentwise derivative:

${\mathbf{r}}^{\prime }(t)={\lim }_{\Delta t\to 0}\frac{\mathbf{r}(t+\Delta t)-\mathbf{r}(t)}{\Delta t}=⟨x^{\prime }(t),y^{\prime }(t),z^{\prime }(t)⟩.$

So to differentiate a vector-valued function, differentiate each component.

Geometric meaning

The difference quotient $(\mathbf{r}(t+\Delta t)-\mathbf{r}(t))/\Delta t$ is a chord of the curve, scaled. As $\Delta t\to 0$, the chord aligns with the tangent line at the point $\mathbf{r}(t)$. So ${\mathbf{r}}^{\prime }(t)$ is a tangent vector to the curve at $\mathbf{r}(t)$, pointing in the direction of increasing $t$.

The magnitude $|{\mathbf{r}}^{\prime }(t)|$ is the speed, how fast the parameter point moves along the curve. The Unit tangent vector is $\mathbf{T}(t)={\mathbf{r}}^{\prime }(t)/|{\mathbf{r}}^{\prime }(t)|$, the direction of motion separated from the speed.

If $\mathbf{r}(t)$ represents the position of a particle at time $t$, then ${\mathbf{r}}^{\prime }(t)$ is velocity and ${\mathbf{r}}^{\prime \prime }(t)$ is acceleration.

Smoothness

A curve is smooth if ${\mathbf{r}}^{\prime }(t)$ is continuous and nonzero everywhere. The nonzero condition rules out “stalls” (places where the curve halts and reverses) and corners (where the tangent direction jumps). A curve made of finitely many smooth pieces glued at corners is piecewise smooth, the standard assumption for line integrals.

Differentiation rules

These mirror single-variable calculus, with one wrinkle for the cross product.

Rule	Statement
Sum	$(\mathbf{r}+\mathbf{s}{)}^{\prime }={\mathbf{r}}^{\prime }+{\mathbf{s}}^{\prime }$
Scalar multiple	$(c\mathbf{r}{)}^{\prime }=c{\mathbf{r}}^{\prime }$
Scalar × vector	$(f\mathbf{r}{)}^{\prime }=f^{\prime }\mathbf{r}+f{\mathbf{r}}^{\prime }$
Dot product	$(\mathbf{r}\cdot \mathbf{s}{)}^{\prime }={\mathbf{r}}^{\prime }\cdot \mathbf{s}+\mathbf{r}\cdot {\mathbf{s}}^{\prime }$
Cross product	$(\mathbf{r}\times \mathbf{s}{)}^{\prime }={\mathbf{r}}^{\prime }\times \mathbf{s}+\mathbf{r}\times {\mathbf{s}}^{\prime }$
Chain	$(\mathbf{r}(f(t)){)}^{\prime }=f^{\prime }(t){\mathbf{r}}^{\prime }(f(t))$

The cross-product rule looks like the ordinary product rule but order matters, since the Cross product isn’t commutative. Keep ${\mathbf{r}}^{\prime }$ on the left of $\mathbf{s}$ in the first term and $\mathbf{r}$ on the left of ${\mathbf{s}}^{\prime }$ in the second.

A useful identity

If $\mathbf{r}(t)$ has constant magnitude, then $\mathbf{r}(t)$ and ${\mathbf{r}}^{\prime }(t)$ are perpendicular.

Proof: $|\mathbf{r}{|}^2=\mathbf{r}\cdot \mathbf{r}$ constant implies $(\mathbf{r}\cdot \mathbf{r}{)}^{\prime }=2\mathbf{r}\cdot {\mathbf{r}}^{\prime }=0$. So $\mathbf{r}\cdot {\mathbf{r}}^{\prime }=0$.

Geometric content: a particle constrained to a sphere of fixed radius has velocity always tangent to the sphere — never radial.

In context

${\mathbf{r}}^{\prime }(t)$ shows up directly in:

• Unit tangent vector $\mathbf{T}={\mathbf{r}}^{\prime }/|{\mathbf{r}}^{\prime }|$.
• Arc length $L={\int }_a^b|{\mathbf{r}}^{\prime }(t)|dt$.
• Line integral integrand $\mathbf{F}\cdot d\mathbf{r}=\mathbf{F}(\mathbf{r}(t))\cdot {\mathbf{r}}^{\prime }(t)dt$.

It’s the conversion factor turning a “differential parameter step” into a “differential curve step,” the same role $du$ plays in $u$-substitution.