parametrization of the ellipse

[shivers20]

x=4sint, y=2cost, 0<t<pi

x^2/16 + y^2/4 = 1

when t=0 x= 4sin(0) y=2cos(0)

x=4, y=2 (4,2)

I am having difficulty plotting the points.

Its an eclipse so my starting motion point is (4,2) and stops at halfway through the ellipse at pi.

Is this correct?

 

[stapel]

Are you required to graph from the parametrizations of x and y, rather than from the explicit formula?

Thank you.

Eliz.

 

[skeeter]

I am having difficulty plotting the points.

no wonder ... 4sin(0) = 0, not 4

you can't sketch the ellipse x^2/16 + y^2/4 = 1 ?

 

[soroban]

Hello, shivers201

You had the right idea . . .

\(\displaystyle \begin{array}{cc}x\:=\:4\sin t \\ y\:=\:2\cos t\end{array}\;\;0\,<\,t\,<\,\pi\)

\(\displaystyle \L\:\frac{x^2}{16}\,+\,\frac{y^2}{4}\:=\:1\) . . . correct!

When \(\displaystyle t\,=\,0:\;\;\begin{array}{cc}x\,=\,4\sin(0)\,=\,0 \\ y\,=\,2\cos(0)\,=\,2\end{array}\;\;\Rightarrow\;\;(0,2)\)

When \(\displaystyle t\,=\,\frac{\pi}{2}:\;\;\begin{array}{cc}x\,=\,4\sin\left(\frac{\pi}{2}\right) \,= \,4 \\ y\,=\,2\cos\left(\frac{\pi}{2}\right)\,=\,0\end{array}\;\;\Rightarrow\;\;(4,0)\)

When \(\displaystyle t\,=\,\pi:\;\;\begin{array}{cc}x\,=\,4\sin(\pi)\,=\,0 \\ y\,=\,2\cos(\pi)\,=\,-2\end{array}\;\;\Rightarrow\;\;(0,-2)\)


The ellipse is formed from the 12:00 position,
. . clockwise to the 6:00 position.