Since a vector-valued function is essentially three garden-variety scalar functions, the key to doing calculus with vector-valued functions is to apply everything you already know one variable at a time.

For instance, suppose that ${\bf r}(t) = \langle x(t), y(t), z(t) \rangle$ and that $\displaystyle{\lim_{t \to a} {\bf r}(t) = {\bf u}} = \langle u_1, u_2, u_3 \rangle$. This means that, whenever $t$ is close to $a$, ${\bf r(t)}$ is close to ${\bf u}$. But that's the same thing as $x(t)$ being close to $u_1$, $y(t)$ being close to $u_2$ and $z(t)$ being close to $u_3$. In other words, $$\lim_{t \to a} {\bf r}(t) = \langle \lim_{t \to a} x(t), \lim_{t \to a} y(t), \lim_{t \to a} z(t). \rangle$$

Differentiating vector functions: What's the derivative of a vector-valued function ${\bf r}(t)$? The definition of ${\bf r}'(t)$ is just the same as for a scalar-valued function, namely $${\bf r}'(t)\ = \ \lim_{h\,\to\, 0}\,\frac{{\bf r}(t+h) - {\bf r}(t)}{h}.$$

To get a feel for what this means with vectors, use the interactive figure to the right. This plots a graph $y=f(x)$ with the parametrization $x=t$, $y=f(t)$, or equivalently ${\bf r}(t) = t {\bf i} + f(t) {\bf j}$. Use the Triangle Law to interpret ${\bf r}(t+h) - {\bf r}(t)$ as the green vector.
  (i) How does this green vector correspond to the 'rise' in the usual definition of $f'(x)$??
  (ii) What properties of vectors are needed to identify the quotient $\Delta {\bf r}/\Delta t$ with the brown vector? How does the direction of the brown vector correspond to slope?
  (iii) The quotient $\Delta {\bf r}/\Delta t$ approaches a vector tangent to the graph of ${\bf r}(t)$ at $P$ as $h \to 0$.


The vector ${\bf r}'(t)$ can be identified with the velocity vector for the graph of ${\bf r}$ at $P$ shown in red; it gives the rate of change of ${\bf r}(t)$ at $P$. As with limits, we can compute derivatives one variable at a time (aka component-wise). To compute ${\bf r}'(t)$ in coordinates: $${\bf r}'(t) \ = \ \lim_{h\,\to\,0}\, \frac{{\bf r}(t+h) - {\bf r}(t)}{h} \ = \ \lim_{h\,\to\,0}\, \frac{\bigl(x(t+h){\bf i} + y(t+h){\bf j} \bigr) - \bigl(x(t){\bf i} + y(t) {\bf j}\bigr)} {h} $$ $$=\ \lim_{h\,\to\,0}\, \left(\frac{x(t+h) - x(t)} {h}\right){\bf i} + \lim_{h\,\to\,0}\, \left(\frac{y(t+h) - y(t)} {h}\right) {\bf j} \,.$$ For the specific parametrization ${\bf r}(t) = t{\bf i} + f(t) {\bf j}$, we have ${\bf r}'(t) = {\bf i} + f'(t) {\bf j}$. In general, for any ${\bf r} : U \subseteq {\mathbb R} \to {\mathbb R}^3$. $${\bf r}'(t) \ = \ x'(t)\, {\bf i} + y'(t)\, \ {\bf j} +z'(t)\, {\bf k} \ = \ \langle\,x'(t),\, y'(t),\, z'(t) \rangle\,.$$

Integrals work the same way. If ${\bf f}(t) = \langle f_1(t), f_2(t), f_3(t) \rangle$, then we break the interval $[a,b]$ into $N$ pieces and compute \begin{eqnarray*} \int_a^b {\bf f}(t) dt & = & \lim_{N \to \infty} \sum_{i=1}^N {\bf f}(t_i^*) \Delta t \cr & = & \lim_{N \to \infty} \langle \sum_{i=1}^N f_1(t_i^*) \Delta t, \sum_{i=1}^N f_2(t_i^*) \Delta t, \sum_{i=1}^N f_3(t_i^*) \Delta t \rangle \cr & = & \left \langle \int_a^b f_1(t) dt, \int_a^b f_2(t) dt, \int_a^b f_3(t) dt. \right \rangle \end{eqnarray*}

Both forms of the fundamental theorem of calculus are also unchanged.

To see this, just apply the regular FTC to each component.