I have the equation $$r=\hat{i}\cos(t)+\hat{j}\sin(t)+\hat{k}t.$$
What I would like to know is how I would obtain an equation for the curve connecting the two points $(1,0,0)$ and $(1,0,2\pi)$ via the individual vectors in the vector field that are connected together (in two dimensions this is called a flow line, although I'm not sure whether the term is the same for 3 dimensions).
I do know how to do this in two dimensions by evaluating the gradient of a function as the $\hat{k}$ component over the $\hat{i}$ component and then integrating to find the general equation for all the flow lines then finding the specific flow line by finding the constant of integration by plugging in a point on your desired flow line $(x_1,y_1)$. However, I cannot really see how to do this for the three dimensional equivalent?
P.S. My motivation for doing this is because I wish to plot the aforementioned path in Matlab, as I am learning about line integrals and this is the path that I am integrating along. I have tried using quiver3, but this obviously gives me all of the possible vectors and not just the path I want.
Maybe I misunderstood your question, but the curve that connects the points is the one you gave
$$ \mathbf{r}(t) = \hat{x}\cos t + \hat{y}\sin t + \hat{y} t $$
where
$$ \mathbf{r}(0) = \hat{x} = (1, 0, 0) ~~~\mbox{and}~~~ \mathbf{r}(2\pi) = \hat{x} + 2\pi\hat{z}= (1, 0, 2\pi) $$
The way you create this in MATLAB is
ezplot3('cos(t)', 'sin(t)', 't', [0, 2 * pi])