Question

Parametric and Polar Curves

My question requires that I convert polar equations to Cartesian Coordinates. And then Describe the resulting curve.
I am given \[r=3csc(\theta)\]

Answer 1

Another amusement?

Answer 2

:D

Answer 3

Unfortunately no.... D:

Answer 4

Well, this is a horizontal line :D

Answer 5

I know that I need to multiply each side by r.
\[r^2=3r\csc(\theta)\]

Answer 6

\[r^2=x^2+y^2\]

Answer 7

Forget that,
\[\Large r = \frac3{\sin(\theta)}\]\[\Large r\sin(\theta) = 3\]

Answer 8

hehe
\[\Large \sin(\theta) = \frac{y}r\]

Answer 9

Everything okay? :)

Answer 10

I'm trying do decipher lol.

Answer 11

Sorry...
\[\Large \sin(\theta) = \frac{y}r\]\[\Large y = r\sin(\theta)\]

That said...
\[\Large r = 3\csc(\theta)\]\[\Large r\sin(\theta) = 3\]\[\huge y = 3\]

Answer 12

As well, I am trying to figure out why the example from the book is ZERO HELP!!!!!!

Answer 13

Zero?

Answer 14

The book example is
\[r=6\sin(\theta)\]
\[r^2=6r\sin(\theta)\]
\[x^2+y^2=6r\sin(\theta)\]
AND EVIDENTLY
\[y=r\sin(\theta)\]
which gives
\[x^2+y^2-6y=0\]
\[x^2+(y-3)^2=9\]
Circle radius 3, centered at (0,3)

Answer 15

I mean, where do they get this arbitrary equation
\[y=r\sin(\theta)\]

Answer 16

That's not arbitrary :)
It's Polar coordinates :D
Say that is your point (x,y)

Answer 17

So clearly,
\[\large \sin \theta = \frac{y}r\]
r represents the distance of said point from the origin .

Answer 18

AAAAHAAAAA!
y u no explain this to me book!!!??!?!?!?!?//1/!?

Answer 19

Book is arrogant :3

Answer 20

It really is a very poorly written book.

Answer 21

LOL
Issue a protest :D

Answer 22

Tried that....

Answer 23

OH well, that's what OS is for :3

Answer 24

err, dealing with the arrogance of said books :D

Answer 25

Thank you so much. It gives one example and expects us to apply it to tons of other problems.

Answer 26

Would you care to explain another one? I think I have an idea but the concept is still rough.

Answer 27

Sure :)

Answer 28

I shall post another question so that you may be justly rewarded.