Graph of r(t) = costi + sintj + costk

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The graph of r(t) = costi + sintj + costk is confirmed to be elliptical rather than a helix. The z-coordinate returns to its original position at t = 2pi, indicating it cannot be a helix. The x and y coordinates describe a unit circle, while the y and z coordinates also form a unit circle. The x and z coordinates create a line in the xz plane, leading to the conclusion that the trajectory is elliptical. The introduction of a new unit vector h clarifies the elliptical nature of the graph.
Jared596
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Can someone tell me the general shape of the graph:

r(t) = costi + sintj + costk

I've been told it's an ellipse, but I thought it was a helix...
 
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You can tell it can't be a helix because the z-coordinate will return to its original postion when t = 2pi. Looking at just the x and y coordinates, you see that they're just describing the unit circle. If you look at just the y and z coordinates, they also describe the unit circle. The x and z coordinates, on the other hand, just describe a line in the xz plane that goes between x = z = -1 and x = z = 1. So, why not define a new unit vector \mathsf{h} = \frac{1}{\sqrt{2}}(\mathsf{i} + \mathsf{k})? This gives you

\mathsf{r}(t) = \sqrt{2} \cos t \mathsf{h} + \sin t \mathsf{j}

Now do you see why the trajectory is elliptical?
 
Last edited:
Yes i do. Thanks for your help.
 

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