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

  1. Nov 13, 2007 #1
    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...
  2. jcsd
  3. Nov 13, 2007 #2


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    Homework Helper

    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 [itex]\mathsf{h} = \frac{1}{\sqrt{2}}(\mathsf{i} + \mathsf{k})[/itex]? This gives you

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

    Now do you see why the trajectory is elliptical?
    Last edited: Nov 13, 2007
  4. Nov 14, 2007 #3
    Yes i do. Thanks for your help.
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