Answer:
Explanation:
Given:
[tex]\begin{gathered} x(\theta)=3\cos\theta \\ y(\theta)=2\sin\theta \\ where\text{ }\theta\text{ lies between }\frac{\pi}{2}\text{ and }\frac{3\pi}{2} \end{gathered}[/tex]
To find:
The graph of the parametric equations
When theta = pi/2, we'll have;
[tex]\begin{gathered} x(\frac{\pi}{2})=3\cos\frac{\pi}{2}=3*0=0 \\ y(\frac{\pi}{2})=2\sin\frac{\pi}{2}=2*1=2 \end{gathered}[/tex]
When theta = 3pi/4;
[tex]\begin{gathered} x(\frac{3\pi}{4})=3\cos\frac{3\pi}{4}=\frac{-3\sqrt{2}}{2} \\ y(\frac{3\pi}{4})=2\sin\frac{3\pi}{4}=\sqrt{2} \end{gathered}[/tex]
When theta = pi;
[tex]\begin{gathered} x(\pi)=3\cos\pi=3*(-1)=-3 \\ y(\pi)=2\sin\pi=2*(0)=0 \end{gathered}[/tex]
When theta = 5pi/4;
[tex]\begin{gathered} x(\frac{5\pi}{4})=3\cos\frac{5\pi}{4}=-\frac{3\sqrt{2}}{2} \\ y(\frac{5\pi}{4})=2\sin\frac{5\pi}{4}=-\sqrt{2} \end{gathered}[/tex]
When theta = 3pi/2;
[tex]\begin{gathered} x(\frac{3\pi}{2})=3\cos\frac{3\pi}{2}=0 \\ y(\frac{3\pi}{2})=2\sin\frac{3\pi}{2}=-2 \end{gathered}[/tex]
We can now go ahead and sketch the graph as seen below;
