Question

Express r=6(costheta+sintheta) in rectangular coordinates

Answer 1

without the 6 it is a circle with center (0,0) and radius 1

Answer 2

Yeah I just have no idea how to do it

Answer 3

i should know this, let me think
one first step would be to square both sides because \(r^2=x^2+y^2\)

Answer 4

we can do this, just about there

Answer 5

ok got it
ready?

Answer 6

yeah hold on im going to do it on a separate piece of paper

Answer 7

and it is always true that \(x=r\cos(\theta), y=r\sin(\theta)\)
or \(\frac{x}{r}=\cos(\theta), \frac{y}{r}=\sin(\theta)\)

Answer 8

so would x=6cos^2theta and y=6sin^2theta

Answer 9

you have \[r=6(\cos(\theta)+\sin(\theta))\]
replace \(\cos(\theta)\) by \(\frac{x}{r}\) and \(\sin(\theta)\) by \(\frac{y}{r}\) to get
\[r=\frac{6x}{r}+\frac{6y}{r}\]
so far so good?

Answer 10

to answer your question, no, it is not that

Answer 11

yeah I think I got it so far

Answer 12

ok to recap \[\frac{x}{r}=\cos(\theta), \frac{y}{r}=\sin(\theta)\] is always true, so we are just going to substitute in \[r=6(\cos(\theta)+\sin(\theta))\] to get \[r=\frac{6x}{r}+\frac{6y}{r}\] only recapping because i remember being confused the first time i saw it
ok now multiply both sides by \(r\)

Answer 13

we get
\[r^2=6x+6y\] and since \(r^2=x^2+y^2\) always we have
\[x^2+y^2=6x+6y\] and we are either done, or we can write this in the standard form of a circle

Answer 14

Okay I think i get it but I dont know what happened to it being \[r ^{2}=x ^{2}+y ^{2}\] ?

Answer 15

that is always true

Answer 16

ohhhh never mind I see what you did

Answer 17

k good

Answer 18

So would that be in rectangular coordinates?

Answer 19

yes or you can write
\[x^2-6x+y^2-6y=0\]and complete the square to get the standard form of the circle

Answer 20

gotta run good luck

Answer 21

wait so either one of those answers is in rectangular coordinates?

Answer 22

Okay thank you for your help (: