Question

the letter x and y represent rectangular coordinates, write the equation in polar coordinates (r,Ɵ):
2x²+3y²=6

Answer 1

\[x=Rcos(\theta) \\ \\ y=Rsin(\theta) \\ \\ 2x^2+3y^2=6 \\ \\ 2R^2\cos^2\theta+3R^2\sin^2\theta=6 \\ \\ R^2(2\cos^2\theta+3\sin^2\theta)=6\]

Answer 2

Let \(x=r\cos\theta,y=r\sin\theta\).$$2(r\cos\theta)^2+3(r\sin\theta)^2=6\\2r^3\cos^2\theta+3r^2\sin^2\theta=6\\r^2\left(2\cos^2\theta+3\sin^2\theta\right)=6\\r^2=\frac6{2\cos^2\theta+3\sin^3\theta}$$Now let's play with this to make it look a little nicer. Recall that the Pythagorean identity gives us \(\cos^2\theta+\sin^2\theta=1\), so \(\cos^2\theta=1-\sin^2\theta\). Substitute that in:$$r^2=\frac6{2-2\sin^2\theta+3\sin^2\theta}=\frac6{2+\sin^2\theta}$$

Answer 3

I done it in
\[2R^2(1-\sin^2\theta)+3R^2\sin^2\theta=6 \\ \\ 2R^2+R^2\sin^2\theta=6 \\ \\ R^2(2+\sin^2\theta)-6=0 \\ \\ R^2=\frac{6}{2+\sin^2\theta}\]
but I don't think that's necessary

Answer 4

Now merely take the square root to yield:$$r=\frac{\sqrt6}{\sqrt{2+\sin^2\theta}}$$... and it appears we can't really do anything else. There is also a possibility your teacher does not want you to do the Pythagorean identity step, but I prefer it personally.

Answer 5

Its just express in terms of R and \(theta\) so, \[R^2(2\cos^2\theta+3\sin^2\theta)=6 \]
should be fine