Question

Convert polar to rectangular coordinate
r= 2/(1-cos theta)

Answer 1

\[r^2=(\frac{ 2 }{ 1-\cos \theta })r\]

Answer 2

\[r^2=(\frac{ 4r }{1r-rcos \theta) }\]

Answer 3

@freckles :)

Answer 4

@jim_thompson5910 sorry..me again :P

Answer 5

Eyyy

Answer 6

I'm not quite sure what you did with the r there. hmmm

Answer 7

We starting with this?\[\Large\rm r=\frac{2}{1-\cos \theta}\]

Answer 8

Yeah, and then I multiplied both sides by r to get r^2
but I think I distributed it wrong on the right side xP

Answer 9

Yah darn ;c

I guess I would recommend a different first step.
Multiply both sides by (1-cos theta)

Answer 10

(oops idk where that four came from either..hah)
ok

r(1-cos theta) = 2

Answer 11

distribute the r :)

Answer 12

r-rcos(theta)=2

Answer 13

Mmm k, what can we do with the rcos(theta)?

Answer 14

r-x=2

Answer 15

Ooo! ok that works out nicely.
Let's add the x to the other side.
r=x+2

Answer 16

how bout that lonely r?
Any ideas? :)

Answer 17

square it! ..? :P

Answer 18

So you want to square it because you're probably used to using this identity,\[\Large\rm r^2=x^2+y^2\]But we actually do have a way of dealing with an r,
if we square root this equation we get,\[\Large\rm r=\sqrt{x^2+y^2}\]Ya?
And then we can avoid introducing another silly r.

Answer 19

ou

Answer 20

\[\Large\rm \sqrt{x^2+y^2}=x+2\]That step make sense? :o

Answer 21

Yep yep

Answer 22

Hmm what next? :)
Any ideas?

Answer 23

umm can we simplify that into just
x+y = x+2? or is that illegal?

Answer 24

that is very illegal! :)

Answer 25

To deal with the square root, we'll have to `square` both sides.

Answer 26

haha k I thought that sounded bad xD

Answer 27

x^2+y^2=x^2+4x+4

Answer 28

Cooool, looks good.

Answer 29

...is that the end?

Answer 30

No. Cancel some stuff out if you're able.
If this is not a circle, then we'd like to get it in the form y=(x stuff).

Answer 31

y^2=4x+4?

Answer 32

Ya that's a good step :)
Hmm your y is squared...

Answer 33

how can we deal with the squareeeee? :D

Answer 34

sq root?

Answer 35

yeaaaaaa

Answer 36

square rooting a square, something special happens... member?

Answer 37

oh gosh..err

Answer 38

wouldn't it just become y?

Answer 39

Mmm close,\[\Large\rm \sqrt{a^2}=\pm a\]

Answer 40

Here is what it looks like without the plus/minus.
https://www.desmos.com/calculator/q6rbowsuio
The red is the entire bowl opening to the right.
You can see that the `positive root` only accounts for half of that bowl, ya? :)

Answer 41

positive root is the blue

Answer 42

oo yes I see.

Answer 43

\[\pm (y)=\sqrt{4x+4}\]

Answer 44

so we have to figure out what quadrant it is in

Answer 45

\[\Large\rm y^2=4x+4\]\[\Large\rm \pm y=\sqrt{4x+4}\]\[\Large\rm y=\pm\sqrt{4x+4}\]

Answer 46

Nahhh we don't need to do any fancy business like that.
Do we? Hmm.

I mean, we need both the plus and the minus.
But do you have some method for checking usually?

Answer 47

haha ok. Er usually my prof just decides on his own lol

Answer 48

XD

Answer 49

Yayyy you did it! \c:/
Good Job Ms Miriam!

Answer 50

thank you! good job to you too!
ey there's no place for putting plus or minus in the answer box :o

Answer 51

is there place for a comma? because you could input:
\(\Large\rm \sqrt{4x+4},\qquad-\sqrt{4x+4}\)

Answer 52

oh good idea. Well sometimes the answer is not so far into the equation. For example:
I put the answer
r= sq(7sec2(theta))
and they wanted (r^2(cos2theta))=7 ..a much less simplified version o.0

Answer 53

yikes :O

Answer 54

What is your profile picture?
Are those ... light bulbs ... being used as planters or something? 0_o

Answer 55

Yeah D: it's sometimes a guessing game -.-
so maybe they want y^2=4x+4 actually ?!

Answer 56

lolxD yes!!

Answer 57

oo interesting :3

Answer 58

Yah maybe that's what they want.

Answer 59

haha I think it's a good idea x)

Answer 60

Yep that's what it wanted :) thank you!

Answer 61

oh cool c: