Question

Convert the Cartesian equation x^2 + y^2 - 6x = 0 to a polar equation.

Answer 1

so, what are the polar forms of "x" and "y"?

Answer 2

\[x=r\cos\theta\\y=r\sin\theta\]

Answer 3

now, plug these for "x" and "y" in the given equation.

Answer 4

Okay, hold on. & Thank you!

Answer 5

Well obviously you get (rcostheta)^2 + (rsintheta)^2 - 6(rcostheta)=0 but I don't really know where to go from there. But I have to get r alone, right?

Answer 6

correct. so you get
\[r^2\cos^2\theta+r^2\sin^2\theta-6r\cos\theta=0\]
factor out \(r^2\) from the first two terms!

Answer 7

To factor it out it would give me r^2(cos^2theta + sin^2theta) - 6rcostheta = 0
Right?

Answer 8

keep going..
see any Identities you could use there?

Answer 9

\[\sin^2\theta+\cos^2\theta=?\]

Answer 10

Doesn't that equal 1?

Answer 11

exacly.

Answer 12

use the equation button below to enter the next step that you have

Answer 13

\[r^2 - 6xcos^2\theta = 0\]
& I didn't even know that button existed... Sorry about that

Answer 14

then it would be \[r^2=6rcos^2\theta\]

Answer 15

Then you would take the square roots, right?

Answer 16

well, we try not to do that second step.

Answer 17

Oh so would the answer be \[r^2= 6\cos \theta\]

Answer 18

\[r^2-6r\cos\theta=0\\
r(r-6\cos\theta)=0\\
r=0\qquad{\rm OR}\qquad r-6\cos\theta=0\implies\Large\boxed{r=6\cos\theta}\]

Answer 19

the first one is rejecected since it is the "origin" and not a curve.

Answer 20

Ohhh. I see. Thank you for your help :)