Question

polar to cartesian
\[r=2\sin(\theta)+2\cos(\theta)\]
Describe the resulting curve.

Answer 1

Would we want to multiply by y?

Answer 2

I see :)
If nothing else comes to mind, just replace all instances of...
\[\Large r = \sqrt{x^2+y^2}\]\[\Large \sin(\theta) = \frac{y}r\]\[\Large \cos (\theta) = \frac{x}r\]

Answer 3

r, not y, excuse me.
And that helps.

Answer 4

But most of these problems involve a little creativity...
Here's a tip, if you see any sines or cosines on the other side of the r, multiply everything by r, yeah :D

Answer 5

So you'd get.
\[\Large r^2 = 2r\sin(\theta) + 2r\cos(\theta)\]

Answer 6

Which may now be expressed in cartesian, using the very useful fact that
\[\Large r^2= x^2+y^2\]

Answer 7

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

Answer 8

And of course, these *familiar* things.
\[\Large r^2 = 2\color{blue}{r\sin(\theta)} + 2\color{red}{r\cos(\theta)}\]

Answer 9

Oh, there you go :)
Which is obviously an equation for a/an....?

Answer 10

Well, that isnt standard form for anything I dont believe....
gimme a minute :P

Answer 11

No rush :)

Answer 12

Is it a form of a circle?

Answer 13

Yup :)

Answer 14

\[\large x^2 - 2x +y^2 - 2y =0\\\large x^2-2x+1+y^2-2y+1= 2\\\large (x-1)^2+(y-1)^2=2\]

Answer 15

o.O
What in the blazes is that?

Answer 16

That is a graph good sir!

Answer 17

TJ.
And I can see that, but what does it have to do with this question?

Answer 18

I may have mis-typed something, but I believe that is the graph of this problem?

Answer 19

oh, there you go.

Answer 20

Fixed it.

Answer 21

They are going to want to know where this thing starts and in what direction it moves?

Answer 22

Well, that's simple enough, hang on

Answer 23

How do you know?

Answer 24

Sorry about that, my internet is not at peak condition at present.

Answer 25

Let's sketch it.
we know it should end up a circle centred around (1,1) with radius about 1.414 but let's see its direction...

Answer 26

Radius is about that yeah, sqrt2 to be exact. right?

Answer 27

sqrt(2) to be exact, yes.

So, we always have theta start at 0, and make it bigger...
so, at theta = 0, we have
r = 2sin(0) + 2cos(0)
which is 1, right?

Answer 28

Whoops, I meant r = 2
Sorry, I forgot the coefficient

Answer 29

Ok, i was gonna mention that.

Answer 30

Let's have another theta, say, pi/6

Answer 31

We're increasing theta starting from zero, in case you haven't gotten along with the gig :D

Answer 32

I noticed :P

Answer 33

at pi/6, we have
r = 2sin (pi/6) + 2cos(pi/6)
\[\Large r = 1 + \sqrt 3\]
\[\Large r \approx 2.732\]

Answer 34

So it's here... approximately...
So maybe you can get the feel that the direction of the circle is counter clockwise? :)