Question

r=sin(theta)+2cos(theta)

Answer 1

i need it to go into rectangular for i missed this class and have no idea what to do

Answer 2

ok... well, first off, let us recall something for conversion between rectangular and polars
\(\bf rcos(\theta)=x\qquad rsin(\theta)=y\qquad x^2+y^2=r^2\)

so \(\bf r=sin(\theta)+2cos(\theta)\qquad \textit{let's multiply both sides by "r"}\qquad thus\\ \quad \\
r=sin(\theta)+2cos(\theta)\implies r^2=rsin(\theta)+2rcos(\theta)\)

so... what do you think... can we use any or all of out 3 identities above?

Answer 3

so then x ^{2}+y ^{2}=y+2rcos(theta)

Answer 4

yes, well \(2rcos(\theta) \implies 2x\)

Answer 5

so thats the final answer or do you combine the like terms?

Answer 6

so \(\bf r^2=rsin(\theta)+2rcos(\theta)\\ \quad \\\implies x^2+y^2=y+2x\implies x^2+y^2-y-2x=0\

so ... now all you have to do is "completing the square", you know what a perfect square trinomial is?

Answer 7

ack, woops, anyhow lemme redo that
so \(\bf r^2=rsin(\theta)+2rcos(\theta)\\ \quad \\\implies x^2+y^2=y+2x\implies x^2+y^2-y-2x=0\)

so ... now all you have to do is "completing the square", you know what a perfect square trinomial is?

Answer 8

yes

Answer 9

ok... let us do the squares for both sides.... gimme a sec

Answer 10

\(\bf x^2+y^2-y-2x=0\implies (x^2-2x+\square^2)+(y^2-y+\square^2)=0\\ \quad \\
\textit{so.... let's see what our values are}\\ \quad \\
-------------------------\\
2x\square =2x\implies \square =\cfrac{2x}{2x}=1\qquad \implies (x^2-2x+1^2)\\ \quad \\
2y\square =y\implies \square =\cfrac{y}{2y}=\cfrac{1}{2}\qquad \implies \left[y^2-y+\left(\cfrac{1}{2}\right)^2\right]\\ \quad \\-------------------------\\
\textit{so we added those values, so we also have to subtract them}\\
\textit{because all we're really doing is borrowing from "0" zero}\\ \quad \\
\textit{like for example }25\implies 25+0\implies 25+1000-1000\implies 25\qquad so\\ \quad \\

(x^2-2x+1^2)+\left[y^2-y+\left(\cfrac{1}{2}\right)^2\right]-1^2-\left(\cfrac{1}{2}\right)^2=0\\ \quad \\
(x-1)^2+\left[y-\cfrac{1}{2}\right]-1-\cfrac{1}{4}=0\implies
(x-1)^2+\left[y-\cfrac{1}{2}\right]=\cfrac{5}{4}\)

Answer 11

hmm forgot the exponent... anyhow \(\bf (x-1)^2+\left[y-\cfrac{1}{2}\right]^2-1-\cfrac{1}{4}=0\implies
(x-1)^2+\left[y-\cfrac{1}{2}\right]^2=\cfrac{5}{4}\)

Answer 12

hmmm.... I was going to divide... some but I think. there's no need

Answer 13

that's the equation, a circle :)

Answer 14

\(\bf (x-1)^2+\left(y-\frac{1}{2}\right)^2=\cfrac{5}{4}\)

so there, that's the rectangular form