# Showing that $(5 \cos t, 5 \sin t, 10t)$ parameterizes an ellipse

How can I understand this parametrization belongs to an ellipse?

$$\vec{r}(t) = (5 \cos t, 5 \sin t, 10t); \; 0 < t < 2\pi$$

My attempt:

$$x(t) = 5 \cos t$$ $$y(t) = 5 \sin t$$ $$z(t) = 10t$$

I know the O-centered ellipse parametrization is

$$\vec{r}(t) = (a \cos t, b \sin t); \; t \in [0,2\pi]$$

While I'm not familiar yet with curves in $$\mathbb{R}^{3}$$, what about the z-component? How can I plot this curve?

Thanks in advance

• It parametrizes a helix. Notice that the projection on the $xy$-plane is a circle. – Kal S. Aug 31 '18 at 11:48

## 2 Answers

It is not an ellipse, it is a vertical helix.

More in detail the projection onto x-y plane is a circle centered at the origin with radius $R=5$ indeed

$$x^2+y^2=(5\cos t)^2+(5\sin t)^2=25$$

The ellipse will "seem" to be moving along the z axis. Imagine drawing an ellipse but at the same time, your pen moves normal to the paper.

If you want to plot it, try MATLAB (edit: or Wolfram Alpha is easier (type plot x=5cos(t), y=5sin(t), z=10t )).