Convert Rectangular points into Polar points:
(4,0)
When converting I got (4,0) using the methods of.
sqrt(x^2+y^2)
sqrt(4^2+0^2)
sqrt(4^2)=4
x=4
y=tan^-1(y/x)
tan^-1(0/4)=0
Where do I go from here?

- anonymous

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- johnweldon1993

And we know that the tangent of 0 is 0 so that is is
Your polar coordinate will be (4,0^\circ)\]

- johnweldon1993

Ahh wrote that bad...
\[\large (4,0^\circ)\]

- anonymous

Ah I recognize you p:
Well thank you again!

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## More answers

- johnweldon1993

lol of course! :P

- anonymous

Really quick question
I also have to transform (0,3)
and I got (3,0deg)
is that right?
@johnweldon1993

- johnweldon1993

If it's weird to think of a 0 degree...just remember
A "polar coordinate" is a point, broken into a radius and an angle from the positive 'x-axis'...for example let's take a random point (5,4)
|dw:1433133598811:dw|
So if you have a point (4,0)
|dw:1433133646240:dw|
As you can see, we are already ON the x-axis so there is no angle to make!

- johnweldon1993

Hmm...
(0,3)
Tell me, what is \(\large \tan^{-1}(\frac{3}{0})\)

- johnweldon1993

Dividing by 0
Tricky huh?
Now obviously this has "imaginary numbers" written all over it
BUT!!!! We can look at it graphically as I have posted above
|dw:1433133926883:dw|

- anonymous

My calculator says
(tan^-1(0/3)=0

- johnweldon1993

Right (0/3) is indeed 0
But the point you have provided is (0,3) which would lead to \(\large \tan^{-1}(\frac{3}{0})\)

- johnweldon1993

Because remember it is
\[\large \theta = \tan^{-1}(\frac{y}{x})\]

- anonymous

Opp. :( silly mistakes will be the end of me!

- johnweldon1993

Lol well remember the original slope "rise over run"
rise = vertical = y
run = horizontal = x
:)

- anonymous

Now I'm still confused because I'm getting 0 still.

- johnweldon1993

Or of course, graph it out to make sure you can at least see why
|dw:1433134326239:dw|
Now you are SOLVING for \(\large \theta\) using tangent which you know is opposite/adjacent so y/x !

- johnweldon1993

And your calculator might just be "erroring" out
We know very well that anything divided by 0 is "indeterminant" right?

- johnweldon1993

The calculator might just be saying "oh that's 0"
But now, here it is MUCH easier to look at it from the graphical standpoint to visualize it!

- anonymous

AH now I"m thoroughly confused.
So now I'm finding the angle.
By using 3/0?

- johnweldon1993

Lol you should be! :P
Right you are using the fact that your point is (0,3)
and the formula you are using is indeed \(\large \tan^{-1}(\frac{y}{x})\) meaning \(\large \tan^{-1}(\frac{3}{0})\)
which yeah at first...makes NO sense...you CAN'T divide by 0 right?

- anonymous

Yeah, or if you do it's always 0.
So that's why I assumed that it would become 0 degree's.
But since it's not what do I do? Haha.

- johnweldon1993

Oh god take THAT idea out of your brain! lol "or if you do it's always 0" Nope, in algebra we are usually taught this INCORRECT statement for no reason...
Now that we have more advanced math behind us...we know better that dividing by 0 will leave us with an error or with an indeterminant decision (meaning we need to look at another way to approach this)....or usually (more often than not) it will actually give us \(\large \infty\)
So...here...it will not be 0 degrees because we cannot say 3/0 = 0 ....we say 3/0 does not exist using what we know so far...we need to turn to other methods
Like looking at a graph of whats happening :P

- johnweldon1993

So to your question "what do I do now" I refer you back to...
|dw:1433135202301:dw|

- anonymous

How do I find that? Haha, I feel like the more we dive into this the more confused I get p:

- anonymous

OH WAIT

- johnweldon1993

Oh....I think you have a breakthrough!... lol

- anonymous

Isn't it 90?

- anonymous

because each quadrant has 90degrees right?

- johnweldon1993

YES!! :D

- anonymous

OH wow that was like.

- anonymous

Okay so the final answer is
(3,90deg)
?

- johnweldon1993

Lol that was literally it, everything else was just from questions that arose from that dividing by 0 thing :P
And they might want it in \(\large \pi\) form so probably stick with
\[\large (3,(\pi/2))\]

- anonymous

Gonna go have aneurysm.

- johnweldon1993

Lol nahhh you're good! Have a popsicle! :D lol

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