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I'm particularly interested in how to plot a parametric vector-valued function with an open interval in the following ellipse:

$\vec{r}(t) = 5\cos(3t)\vec{i} + 7\sin(3t)\vec{j} \, , \quad t \in \left( \dfrac{\pi}{6}, \dfrac{5\pi}{6} \right)$

If the trigonometric functions argument was only $ t $ instead of $ 3t $, I figure that the curve would only be the arc formed between the angles of $\dfrac{\pi}{6}$ and $\dfrac{5\pi}{6}$ with open circles at each end. Is that correct?

However, making use of parametric plotters available on the internet, the argument multiplied by three allows a complete ellipse to be drawn, which makes sense, after all I'm increasing the angles faster.

Thus, since if it's an open interval, how would the open intervals be displayed on the ellipse curve if the ellipse is drawn entirely?

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  • $\begingroup$ Simply add hollow circles at the ends of your curve. $\endgroup$ Commented Mar 19, 2017 at 22:21
  • $\begingroup$ But since the ellipse is drawn entirely, where the hollow circles would reside? The ends of the curve are pretty obvious in the case I mentioned of an interval ]pi/6, 5pi/6[ if the arguments were "t" instead of "3t". Although, in this case would be just a regular full ellipse with hollow circles at the aforementioned interval? That's my doubt. $\endgroup$
    – MCarsten
    Commented Mar 20, 2017 at 3:51

1 Answer 1

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Simply denote an exclusive point with an empty/hollow circle.

enter image description here

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