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.

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    $\begingroup$ You don't mean line in a pure geometric sense here -- you mean a 1-D curve, right? The shape you get is not linear in the geometric sense at all... $\endgroup$ – gt6989b Nov 28 '16 at 18:33
  • $\begingroup$ Yes sorry, I meant a curve - thanks for the edit. $\endgroup$ – Resquiens Nov 28 '16 at 18:35

Maybe I misunderstood your question, but the curve that connects the points is the one you gave

$$ \mathbf{r}(t) = \hat{x}\cos t + \hat{y}\sin t + \hat{y} t $$


$$ \mathbf{r}(0) = \hat{x} = (1, 0, 0) ~~~\mbox{and}~~~ \mathbf{r}(2\pi) = \hat{x} + 2\pi\hat{z}= (1, 0, 2\pi) $$

enter image description here

The way you create this in MATLAB is

ezplot3('cos(t)', 'sin(t)', 't', [0, 2 * pi])

  • $\begingroup$ Ah I see, thank you very much. My misunderstanding came from thinking that $r$ was a vector field, instead of a simple vector, and therefore I thought that the line that I was trying to plot was a level curve of some sort. Now I see that this isn't the case; moreover because 3 dimensional vector fields have level surfaces as opposed to the level curves of 2 dimensional vector fields since the contour would be an equation of 3 variables. Just adding this for anybody who happens to come across this answer with the same confusion that I had. $\endgroup$ – Resquiens Nov 28 '16 at 20:10

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