Question

Can someone please explain polar coordinates to me? Huge test tomorrow. Please Help!

Answer 1

I need to know how to go from rectangular to polar, polar to rectangular, the 4 different types of polar graphs, the equations for each, converting parametric to rectangular, and rectangular to parametric given two equations (i.e. y=6(x^2) + 1 and x - 3 = t

Answer 2

That is quite a laundry list of things to study there, @sjiwani
Best get started now, right? ^.^

Answer 3

Make your presence felt, @sjiwani ...

Answer 4

I'm trying to study right now, but I don't know how to go from rectangular to polar (i.e. (4,5) as rectangular, I don't know how to get to the polar version.

Answer 5

Well, that's why I'm here :)
Rectangular to Polar transformation is decidedly trickier than vice versa, but it's not THAT tricky.
\[\huge (x,y)\rightarrow (r,\theta)\]

Answer 6

^Yeah, that. Though if you're using a calculator, things may not turn out quite the way you want...

Answer 7

Can you do something simple as an example for me? Like, (4,5) to polar

Answer 8

Of course.
Using the formulas stated above...
\[\huge (x,y) \rightarrow (r,\theta)\]
\[\huge r=\sqrt{x^2+y^2}\]

Having said that, what's the value for r?

Answer 9

\[\sqrt{20}\] using that formula

Answer 10

Hardly. :)
You're to add the squares of 4 and 5, and then get the square root of the sum.

Answer 11

So,
\[\huge r=\sqrt{4^2+5^2}\]

Answer 12

16+25=41... I have \[\sqrt{20}\] written down for some reason.... idk why...

Answer 13

It'd be \[\sqrt{41}\]

Answer 14

Much better.
It's finding theta that's tricky, but that doesn't show here...
The nominal way to find theta is through this formula
\[\huge \theta=\tan^{-1}\frac{y}x\]

Answer 15

okay, so 51.3

Answer 16

WHat's an example of a tricky one?

Answer 17

Yeah. That was easy wasn't it? Unfortunately that formula is dodgy, when, for instance
\[\huge (x,y) = (-4,-5)\]
Try finding r and theta now..

Answer 18

Finding r is still straightforward, by the way.

Answer 19

And so yes, the polar coordinate of (4,5) would be \[(\sqrt{41}, 51.3)\]

Answer 20

so, one sec

Answer 21

Yes. But what are the polar coordinates of (-4,-5)?

Answer 22

well, they're both negative, so they would be in quadrant III, and just a guess here, 51.3-180?

Answer 23

letting r remain constant

Answer 24

Almost. They're both negative, so they're in Q3, that's a very good observation :)
So what you do is add 180 degrees, not subtract, even though they have the same effect ^.^

Answer 25

Is there a reason we add and not subtract? or would both be acceptable? I'm guessing both wouldn't be.

Answer 26

Both, but... we usually take a positive theta. For standardization purposes ^.^

Answer 27

Gotcha'
okay, also, can you tell me me we would convert \[r = 3\sin \theta \]
to rectangular?

Answer 28

Well, I'd say just replace everything by its equivalent in rectangular. That's the most direct approach, so say,
replace all
\[\large r \rightarrow \sqrt{x^2+y^2}\]

Answer 29

That's not the end of things yet, mind you ^

Answer 30

so what do we do with the 3 sin thetea?

Answer 31

I was getting to that :)
You can certainly replace theta itself with
\[\huge \theta =\tan^{-1}\frac{y}x\]But that complicates things. We can derive the following if you want, but feel free to use them :)
\[\huge \sin\theta=\frac{y}{\sqrt{x^2+y^2}}\]
\[\huge \cos\theta=\frac{x}{\sqrt{x^2+y^2}}\]

Answer 32

I have down that \[3\sin \theta \] for some reason is squared, and the equation ends up as \[x^2 + y^2 = y^2\]

Answer 33

We don't square it. We have
\[\huge \sqrt{x^2+y^2}=\frac{3y}{\sqrt{x^2+y^2}}\]

Answer 34

So we multiply everything by \(\large \sqrt{x^2+y^2}\) And we get
\[\huge x^2+y^2=3y\]
Clearly an equation of a circle, no? :)

Answer 35

Or not...

Answer 36

Something else...

Answer 37

Just had a thought, I think I got it by this logic:
\[x^2 + y^2 = r^2\] thus far, then since \[r = 3\sin \theta\] I squared it. Making a final equation to look like \[x^2 + y^2 = y^2\] because of something with the sin being equal to y somehow

Answer 38

No "somehow"s in Maths, though. I think we better derive this...
\[\huge \tan \theta=\frac{y}r\]right?

Answer 39

I have no idea..... *sigh*

Answer 40

I'm so confused...

Answer 41

Yeah, don't worry, we can do this :)
\[\huge \tan\theta = \frac{y}r\]you do agree to this, right?

Answer 42

yes

Answer 43

Well, let's visualise this...

Answer 44

So as you can see, tan θ = y/x

Answer 45

okay

Answer 46

I've got that, now how does that help us?

Answer 47

Using the pythagorean theorem, clearly (I hope)
the length of the hypotenuse is