Definition: Vector valued function
A vector valued function if a function of the form
$$\begin{array}{rcl} \vec{r}(t) &=& \left(\begin{array}{r} x(t)\\y(t) \end{array}\right) =x(t)\vec{i} + y(t)\vec{j} \\\\ \vec{r}(t) &=& \left(\begin{array}{r} x(t)\\y(t)\\z(t) \end{array}\right) =x(t)\vec{i} + y(t)\vec{j} + z(t)\vec{k} \end{array}$$where $x(t)$, $y(t)$ and $z(t)$ are called the component functions. Each component functions is a real valued function.
Equivalently
$$\begin{array}{rcl}\displaystyle \lim_{t\to a}\vec{r}(t) &=& \left(\displaystyle\lim_{t\to a} x(t)\right)\vec{i} +\left(\displaystyle \lim_{t\to a} y(t)\right)\vec{j} +\left(\displaystyle\lim_{t\to a} z(t)\right)\vec{k} \end{array}$$Equivalently
$$\begin{array}{rcl}\displaystyle \lim_{t\to a}\vec{r}(t) &=& \left(\displaystyle\lim_{t\to a} x(t)\right)\vec{i} +\left(\displaystyle \lim_{t\to a} y(t)\right)\vec{j} +\left(\displaystyle\lim_{t\to a} z(t)\right)\vec{k} \end{array}$$The derivative of a vector valued function \(\)is
\[ \frac{\mathrm{d}\,{\vec{r}(t)}}{\mathrm{d}\,{t}} =\left[\vec{r}(t)\right]' =\vec{r}'(t) = \lim_{\Delta t \to 0} \frac{\vec{r}\left(t+\Delta t\right) - \vec{r}\left(t\right)}{\Delta t} \]If \(\vec{r}'(t)\) exists then \(\vec{r}(t)\) is differentiable at \(t\). If \(\vec{r}\) is differentiable for all \(t\in(a,b)\) then \(\vec{r}\) is differentiable on \((a,b)\). If in addition
\begin{eqnarray*} \left[\vec{r}(a)\right]' =\vec{r}'(a) &= & \lim_{\Delta t \to 0^+} \frac{\vec{r}\left(a+\Delta t\right) - \vec{r}\left(a\right)}{\Delta t} \\\\ \left[\vec{r}(b)\right]' =\vec{r}'(b)& =& \lim_{\Delta t \to 0^-} \frac{\vec{r}\left(b+\Delta t\right) - \vec{r}\left(b\right)}{\Delta t} \end{eqnarray*}then \(\vec{r}\) is differentiable on \([a,b]\).
Let \(C\) be a curve defined by a vector valued function \(\vec{r}(t)\). Suppose \(\displaystyle \left[\vec{r}(t_0)\right]'\) exists, then it is called a tangent vector to \(C\) at \(\vec{r}(t_0)\). If in addition \(\displaystyle \left|\left[\vec{r}(t_0)\right]'\right|\ne0\), then the principle unit tangent vector is
$$ \vec{T}(t) = \frac{\vec{r}'(t)}{\|\vec{r}'(t)\|} $$Definition: Indefinite integral
Let \(\)\(\vec{r}(t)\) be a vector valued function
$$\begin{array}{rcl} \vec{r}(t) & =& \left[x(t)\right]\vec{i} + \left[y(t)\right]\vec{j} + \left[z(t)\right]\vec{k} \end{array}$$The indefinite integral of \(\vec{r}(t)\) is
$$\begin{array}{rcl} \int\vec{r}(t)\,\mathrm{d}\,{t} & =& \int\left[\left[x(t)\right]\vec{i} + \left[y(t)\right]\vec{j} + \left[z(t)\right]\vec{k}\right]\,\mathrm{d}\,{t} \\\\ &=& \left[\int x(t)\,\mathrm{d}\,{t}\right]\vec{i} + \left[\int y(t)\,\mathrm{d}\,{t}\right]\vec{j} + \left[\int z(t)\,\mathrm{d}\,{t}\right]\vec{k} \end{array}$$Let \(\)\(\vec{r}(t)\) be a vector valued function
$$\begin{array}{rcl} \vec{r}(t) & =& \left[x(t)\right]\vec{i} + \left[y(t)\right]\vec{j} + \left[z(t)\right]\vec{k} \end{array}$$The definite integral of \(\vec{r}(t)\) is
$$\begin{array}{rcl} \int_{a}^{b}\vec{r}(t) \,\mathrm{d}\,{t} & =& \int_{a}^{b}\left[\left[x(t)\right]\vec{i} + \left[y(t)\right]\vec{j} + \left[z(t)\right]\vec{k}\right]\,\mathrm{d}\,{t} \\\\ &=& \left[\int_{a}^{b} x(t)\,\mathrm{d}\,{t}\right]\vec{i} + \left[\int_{a}^{b} y(t)\,\mathrm{d}\,{t}\right]\vec{j} + \left[\int_{a}^{b} z(t)\,\mathrm{d}\,{t}\right]\vec{k} \end{array}$$For a smooth in \([a,b]\) vector valued function
$$ \vec{r}(t) = x(t) \vec{i} + y(t) \vec{j} + z(t) \vec{k} $$the arc length in the interval \([a,b]\) is
$$ \begin{array}{rcl} L & =&\int_{a}^{b} \sqrt{ {\left(x'(t)\right)}^2 +{\left(y'(t)\right)}^2 +{\left(z'(t)\right)}^2 }\mathrm{d}{t} =\int_{a}^{b} \left\|\left[\vec{r}(t)\right]'\right\|\mathrm{d}{t} \end{array} $$Definition: arc length function
For a smooth in \([a,b]\) vector valued function
$$ \vec{r}(t) = x(t) \vec{i} + y(t) \vec{j} + z(t) \vec{k} $$the arc length function in the interval \([a,b]\) is
$$ \begin{array}{rcl} s(t) & =&\int_{a}^{t} \sqrt{ {\left(x'(u)\right)}^2 +{\left(y'(u)\right)}^2 +{\left(z'(u)\right)}^2 }\,\mathrm{d}{u} =\int_{a}^{t} \left\|\left[\vec{r}(u)\right]'\right\|\,\mathrm{d}{u} \end{array} $$if \(\displaystyle \left\|\left[\vec{r}(u)\right]'\right\| = 1\) then the curve is said to be parametrized with respect to arc length and
$\vec{r}(t) = x(t) \vec{i} + y(t) \vec{j} + z(t) \vec{k} $
$$ \begin{array}{rcl} L(t) & =& \int_{a}^{t} \left\|\left[\vec{r}(u)\right]'\right\|\mathrm{d}{t} \end{array} $$from
$$ \begin{array}{rcl} s = L(t) & \Rightarrow & t = L^{-1}(s) \end{array} $$Arc length parametrization is
$$\vec{g}(t) = \vec{r}\left(L^{-1}(s)\right)$$For a smooth in \(\) \([a,b]\) vector valued function parametrized by arc length
$$ \begin{array}{rcl} \vec{r}(s) &=& x(s) \vec{i} + y(s) \vec{j} + z(s) \vec{k} \end{array} $$the curvature \(\displaystyle \kappa = \kappa(s)\) is
$$ \begin{array}{rcl} \kappa(s) & =& \left\|\frac{\mathrm{d}\,\vec{T}(s)}{\mathrm{d}\,{s}}\right\| =\left\|{\left[\vec{T}(s)\right]'}\right\| \end{array} $$For a position vector valued function
$$ \begin{array}{rcl} \vec{r}(t) &=& x(t) \vec{i} + y(t) \vec{j} + z(t) \vec{k} \end{array} $$ we have $$ \begin{array}{rcl} \vec{v}(t) &=& \left[\vec{r}(t)\right]' \qquad \textrm{velocity}\quad \int_{a}^{b}\vec{v}(t)\mathrm{d}t \quad\textrm{ displacement} \\[2ex] {s}(t) &=& \left|\vec{v}(t)\right| = \left|\left[\vec{r}(t)\right]'\right| \qquad \textrm{speed} \\\\ \vec{a}(t) &=& \left[\vec{v}(t)\right]' = \left[\vec{r}(t)\right]'' \qquad \textrm{accelaration} \end{array} $$