Calculating r(t) with line integrals I have $F(x,y)$ equalling some $a \mathrm{i}+b\mathrm{j}+c\mathrm{k}$ is that all $r(t)$ is?
What if all of $a,b,c$ are not in terms of $t$?

Note: My $F(x,y)$ is a vector field.

Or does it come from my $C$ where $C$ is a rectangle?
Example: Vector field: $F(x,y) = x^2 y^2 \;\mathrm{i} + x \;\mathrm{j}$. Let $C$ be a rectangle with vertices $(0,0),(4,0),(4,4),(0,4)$ , let $T$ denote the unit tangent vector $C$ directed anti clockwise around $C$
I want $T$, which I know requires $r(t)$
 A: You can parameterize the four legs of the rectangle as follows:
\begin{align}
\vec{r}_1(t) =& (t, 0) & 0 \leq t \leq 4\\
\vec{r}_2(t) = & (4, t) & 0 \leq t \leq 4 \\
\vec{r}_3(t) = & (4 - t, 4) & 0 \leq t \leq 4 \\
\vec{r}_4(t) = & (0, 4 - t) & 0 \leq t \leq 4
\end{align}
But don't forget that you need to plug in each of those values into $\vec{F}$:
$$
\int\limits_0^4 \vec{F}\left(\vec{r}_1\right)\circ d\vec{r}_1 + \int\limits_0^4 \vec{F}\left(\vec{r}_2\right)\circ d\vec{r}_2 + \int\limits_0^4 \vec{F}\left(\vec{r}_3\right)\circ d\vec{r}_3 + \int\limits_0^4 \vec{F}\left(\vec{r}_4\right)\circ d\vec{r}_4
$$
You can combine this into a single integral, but you still need to evaluate each $\vec{F}\left(r_i\right)\circ d\vec{r}_i$.  For example, here is $\vec{r}_1$:
$$
x = t, y = 0\\
d\vec{r}_1 = \left(1, 0\right)dt \\
\vec{F}\left(\vec{r}_1\right) = \left(t^2\cdot 0^2, t\right) = \left(0, t\right) \\
\vec{F}\left(\vec{r}_1\right) \circ d\vec{r}_1 = 0 dt
$$
Now you need to do the same for the other three.  You can probably guess that it will vanish again for $\vec{r}_4$, but it probably won't for the other two.
