Differentiation

The derivative of $\boldsymbol{r}$ is defined the same way as for a real-valued function:

provided the limit exists.

For $\boldsymbol{r}(t) = \langle x(t), y(t), z(t) \rangle$, $\boldsymbol{r}$ is differentiable at $a$ iff each component function is differentiable at $a$, and

More generally, whenever the component derivatives exist,

For differentiable vector-valued functions $\boldsymbol{r}(t)$, $\boldsymbol{r}_1(t)$, $\boldsymbol{r}_2(t)$, differentiable real-valued function $f(t)$, scalar $k$, and constant vector $\vec{c}$:

• $\dfrac{\text{d}}{\text{d}t}[\vec{c}] = \vec{0}$
• $\dfrac{\text{d}}{\text{d}t}[\boldsymbol{r}_1(t) \pm \boldsymbol{r}_2(t)] = \boldsymbol{r}_1'(t) \pm \boldsymbol{r}_2'(t)$
• $\dfrac{\text{d}}{\text{d}t}[k\boldsymbol{r}(t)] = k\boldsymbol{r}'(t)$
• $\dfrac{\text{d}}{\text{d}t}[f(t)\boldsymbol{r}(t)] = f(t)\boldsymbol{r}'(t) + f'(t)\boldsymbol{r}(t)$
• $\dfrac{\text{d}}{\text{d}t}[\boldsymbol{r}_1(t) \cdot \boldsymbol{r}_2(t)] = \boldsymbol{r}_1(t) \cdot [\boldsymbol{r}_2'(t) + \boldsymbol{r}_1'(t) \cdot \boldsymbol{r}_2(t)$
• $\dfrac{\text{d}}{\text{d}t}[\boldsymbol{r}_1(t) \times \boldsymbol{r}_2(t)] = \boldsymbol{r}_1(t) \times [\boldsymbol{r}_2'(t) + \boldsymbol{r}_1'(t) \times \boldsymbol{r}_2(t)$
• $\dfrac{\text{d}}{\text{d}t}[\boldsymbol{r}(f(t))] = f'(t)\boldsymbol{r}'(f(t))$

Tangent Line

Let $P$ be a point on the graph of $\boldsymbol{r}(t)$, with $\boldsymbol{r}(t_0)$ the position vector of $P$. If $\boldsymbol{r}'(t_0)$ exists and is non-zero, $\boldsymbol{r}'(t_0)$ is a tangent vector to the graph at $P$. The tangent line through $P$ parallel to $\boldsymbol{r}'(t_0)$ has vector equation

If $\lVert \boldsymbol{r}(t) \rVert$ is constant for all $t$ in an interval, then $\boldsymbol{r}(t) \cdot \boldsymbol{r}'(t) = 0$. So $\boldsymbol{r}(t)$ and $\boldsymbol{r}'(t)$ are orthogonal for every $t$ in the interval.

Mean Value Theorem

Vector-valued functions do not satisfy a direct Mean Value Theorem of the form $\boldsymbol{r}(b) - \boldsymbol{r}(a) = \boldsymbol{r}'(c)(b-a)$.