# Math Insight

### Applet: Parametrized ellipse

The vector-valued function $\dllp(t)=(3\cos t)\vc{i} + (2 \sin t) \vc{j}$ parametrizes an ellipse, shown in green. This ellipse is the image of the interval $[0,2\pi]$ (shown in red) under the mapping of $\dllp$. For each value of $t$, the blue vector is $\dllp(t)$. As you change $t$ by moving the blue point along the interval $[0,2\pi]$, the head of the arrow traces out the ellipse.

Can you show mathematically that $\dllp(t)$ traces out an ellipse? To do so, let $(x,y)$ be the point defined by $(x,y)=\dllp(t)$ for some value of $t$. Since $\dllp(t) = (3\cos t, 2 \sin t)$, we conclude that $x = 3\cos t$ and $y=2\sin t$. Can you see that $x$ and $y$ satisfy \begin{align*} \frac{x^2}{3^2} + \frac{y^2}{2^2} = 1, \end{align*} which is the equation for the above ellipse? (You remember that $\cos^2t + \sin^2t = 1$ for any $t$. You can calculate expressions for $x^2/3^2$ and $y^2/2^2$, and add them together.)

Applet file: parametrized_ellipse.ggb

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