Question

Convert the polar equation to rectangular form.
r=2/1+sin(theta)

Answer 1

Well, how do you write r in terms of Cartesian coordinates x and y?
And then how do you write theta?

Answer 2

...and sin(theta)?

Answer 3

I'm not really sure.

Answer 4

You know that
x = r . cos theta
y = r . sin theta

What are the equations the other way

r = ....
theta = ....
sin(theta) = ...
cos(theta) = ...

Answer 5

r = sqrt(x^2 + y^2), yes?

What about
theta = .... ?

Answer 6

Look, we have that
x = r . cos theta
y = r . sin theta

Hence
sin theta = ... ?

Answer 7

y?

Answer 8

y = r.sin(theta)
hence
sin(theta) = y/r
Yes?

Answer 9

Yep, got that :)

Answer 10

So now, substitute these expressions into the original equation
\[ r = \frac{2}{1 + \sin\theta} \]

Answer 11

Alright, I did. I just don't see how I can possibly get to the answer: "x^2 +4y -4"

Answer 12

Let me do a couple of steps for you
r = 2/(1 + sin theta)
r = 2/(1 + y/r)
r(1 + y/r) = 2
r + y = 2

sqrt(x^2 + y^2) = 2 - y

Now, you take it from there

Answer 13

ugh. Cat just stepped all over keyboard.

Writing it out again

r = 2/(1 + sin theta)
r = 2/(1 + y/r)
r ( 1 + y/r) = 2
r + y = 2
r = 2 - y

Now you take it from there

Answer 14

That's the part I always get stuck on though. :( sorry.

Answer 15

What is r in terms of x and y?

Answer 16

I end up with 2y + x -2 = 0 and that's not the answer.

Answer 17

x+y

Answer 18

Wrong. r is the radial distance of a point (x,y) from the origin. What is the distance of (x,y) from the origin? Use Pythagorus's theorem

Answer 19

sqrt(x^2+y^2)

Answer 20

Yes. Hence
r = 2 - y
is
sqrt(x^2 + y^2) = 2 - y

Simplify that.

Answer 21

That's the part I'm stuck on.

Answer 22

Square roots signs are always pesky. Square both sides to get rid of it.

Answer 23

Finished the problem. :) Thanks so much!!

Answer 24

good. Draw a diagram of the figure so you know what this looks like and make sure it jives with your understanding of r and theta.