Question

Convert the equation x2+y2-8y=0 from rectangular to polar

Answer 1

the 2's are squaring the x and y, by the way

Answer 2

Complete the square in 'y' first to proceed, can you? @rjeffries96

Answer 3

yeah, let me do that real quick

Answer 4

Sure, tyt in fact!

Answer 5

hmm, im kinda stuck on the 8y, how can you bring that over as the loose term if the y is with the 8? (i havent done this in a couple of years)

Answer 6

No probs, let's work this out ;)
SO I am looking at \(y^2 - 8y\), right?
If I compare this with the standard form \(a^2 - 2ab + b^2\), I think we have the \(a^2\) term, and... as well as the \(2ab\) term - what do you think?

Answer 7

yeh, you just dont have the b^2, is that right?

Answer 8

?

Answer 9

yes exactly! but for that we must first find what 'b' is, and that we can comparing '8y' with '2ab'. 'a' is already 'y', so what would 'b' be?

Answer 10

would it have to be x?

Answer 11

Nope. where does 'x' come into the picture here? let's do it this way:

2ab = 8y = 2.4.y

so, a = y, and b = 4 ---> understandable?

Answer 12

ohhhhh i get what you mean. Sorry about that haha

Answer 13

No probs :)
So, what's the equation that you get now? How would you complete the square?

Answer 14

so you would have y^2-8y+16=0?

Answer 15

yeah! but since am adding that '16', I need to subtract a '16' as well!

So the whole equations transforms to something like:
\(x^2 + y^2 -8y + 16 -16=0\)
\(= x^2 +(y-4)^2 = 16\)
\( = x^2 +(y-4)^2 = 4^2\)

Okay with this?

Answer 16

yeah that makes sense I think

Answer 17

so, you have your equation in standard form.
Now can you use what Mukushla wrote right on top to get your Polar form?

Answer 18

Yeh, i see it, that x=rcos(theta), y=rsin(theta), and x^2+y^2=r^2

Answer 19

Yeah, but one moment, this isn't over yet.
This circle's not centered at the origin, but has been translated along the Y-axis, if you look carefully at the equation we just derived. So, 'y-8' indicates an translation of '8' units towards the positive y-axis, so, in the polar coordinates, I add '8' to the ordinate.
Hence,
x = r cos(theta)
and
y = 8 + sin(theta)

Now you're done!


(don't need that "x^2+y^2=r^2" that was just posted by him as a hint that you must bring it to the standard form)

Answer 20

Oh ok, that (actually) makes sense haha. So the question I'm trying to answer tells me to then solve for r, so i would just solve for r in the x equation and not worry about the y equation?

Answer 21

'r' is the radius lol!

x^2 + y^2 = r^2 ---> compare your equation with this, and you 'll find your 'r'!
(lol how did I forget finding out 'r')

Answer 22

so would that be r^2cos^2(theta)+64+sin^2(theta)=r^2?

Answer 23

but then substitue r for 4, is that right?

Answer 24

No no, you don't add that up lol!
Polar form is just written in the form " x = rcos(theta), y = rsin(theta) "
and here, comparing with the standard equation, (our equation is \(x^2+(y−4)^2=4^2\)),

r^2 = 4^2
so, r=4


So, our Polar form is:

\(x=4 \cos\theta , y = 8 + 4 \sin\theta\)

That's it^

Answer 25

oh, so you dont use the x^2+y^2=r^2?

Answer 26

Nope. The polar is just represented the other way. Not in one single equation.

Answer 27

Oh ok. I need to review this stuff haha. Thank you sooo much for all your help, I really appreciate it

Answer 28

(because if you substitute and simply the equation, you end up with something like r^2 = r^2, which is useless for us)

Yeah, anytime! No worries :]