Question

What is the equation of this limacon curve?

Answer 1

it's [-5, 5] by [-5, 5] btw

Answer 2

looks like a circle doesn't it?
(x-h)^2+(y-k)^2=r^2
the center looks like it can be on the x-axis on the negative side

Answer 3

yes i'm just not sure how to get the equation

Answer 4

(x-h)^2+(y-k)^2=r^2 is equation for a circle with center (h,k) and radius r

Answer 5

you already have the equation just need to find the center (h,k) and r

Answer 6

i'm doing practice problems and my options are:
r = 2 - 2 cos θ
r = 3 - 2 cos θ
r = 3 - cos θ
r = 3 - sin θ

Answer 7

\[(x-h)^2+(y-k)^2=r^2 \\ x^2-2xh+h^2+y^2-2ky+k^2=r^2 \\ \text{ recall } x^2+y^2=r^2 \\ \text{ so we have } r^2-2xh+h^2+y^2-2ky+k^2=r^2 \\ \text{ subtract } r^2 \text{ on both sides gives } -2xh+h^2+k^2-2ky=0 \\ \text{ so also recall } x=r \cos(\theta) \text{ and } y =r \sin(\theta) \\ -2 r \cos(\theta) h +h^2+k^2-2k r \sin(\theta) =0\]

\[\text{ factor } r \text{ from the two terms that have it } \\ r( -2 \cos(\theta) h -2 k \sin(\theta))+h^2+k^2=0\]

Now solve for r
\[r=\frac{-(h^2+k^2)}{-2 \cos(\theta) h-2 k \sin(\theta)} \\ \text{ multiply by } \frac{-1}{-1} \\ r=\frac{h^2+k^2}{2( h \cos(\theta)+k \sin(\theta))}\]

but anyways...
if I did that right I think we can also write it in this form
\[r=2a \cos(\theta)+2 b \sin(\theta) \text{ this is suppose to be a circle with radius } \sqrt{a^2+b^2} \\ \text{ and center} (a,b)\]

But none of your equations seem to fit this.

None of the equations seem to fit that http://www.wolframalpha.com/input/?i=t%3D3-2cos%28theta%29+polar+
You can look at the graphs here for each one of the equations.
None of them come out to be a circle with a center away from the y-axis

Answer 8

http://assets.openstudy.com/updates/attachments/552d7b31e4b07e661d0f60c2-katiexo-1429044028465-0806_g2_q3.jpg
are you sure this right?

Answer 9

Anyways maybe I'm wrong or whatever
I must go now sorry

Answer 10

where is the -2 coming from? did you see my coordinates of [-5, 5] by [-5, 5]

Answer 11

so the options that you said were options are not options for the graph?

Answer 12

you can use the equation I gave above

Answer 13

\[r=2a \cos(\theta)+2 bcos(\theta)\]
this has center (a,b)
and it looks like your b is 0
so you have
\[r=2a \cos(\theta)\]

Answer 14

a should be negative since it is to the center it to the left of the origin

Answer 15

but none your options are given by this

Answer 16

do you remember this is what you posted
"i'm doing practice problems and my options are:
r = 2 - 2 cos θ
r = 3 - 2 cos θ
r = 3 - cos θ
r = 3 - sin θ"

Answer 17

r=3-2cos(theta) isn't a circle but it probably most similar to your graph