Question

Rewrite the given equation using the substitutions x= r cos x and y= r sin x. Simplify your answer. x^2+y^2+3x=0

Answer 1

You must mean using the substitutions \(x=r\cos t\) and \(y=r\sin t\)...
\[x^2+y^2+3x=0~\Rightarrow~(r\cos t)^2+(r\sin t)^2+3(r\cos t)=0\]
Expanding the powers and whatnot:
\[r^2\cos^2t+r^2\sin^2t+3r\cos t=0\\
r^2\left(\cos^2t+\sin^2t\right)+3r\cos t=0\\
~~~~~~~~~~~~~~~~~~~~~~~~~~\vdots\]

Answer 2

So that simplifies to \[r(r+3\cos(t))=0\]

Answer 3

Is that correct?

Answer 4

Yes. And for \(r\not=0\), you can divide both sides by \(r\), giving you
\[r+3\cos t=0\]

Answer 5

Thanks for the help. I didn't see that solution until you helped. Can you explain why we're able to divide the outside r out?

Answer 6

You can do that for \(r\not=0\) only. If you don't have this restriction on \(r\), your answer would be the equation containing \(r^2\).