Question

good evening. I'm Jastine, a mathematics student i need your help for my requirements. we all know that the polar coordinate system is a two-dimensional coordinate system in which each point on a plane is determined by a distance from a fixed point and an angle from a fixed direction. will you please show me other explanation of polar coordinates?

Answer 1

Can you be more specific about the explanation you so desire?

Answer 2

@terenzreignz what i mean is please explain polar coordinates further more

Answer 3

Maybe learn how to convert from rectangular coordinates to polar?

Answer 4

yes, do you mind explain it to me?

Answer 5

Of course. So, the coordinates in rectangular are usually noted as (x , y) right?
Well, the coordinates in polar are noted as (r , θ)
Here's the illustration...

Say this is your point.

Answer 6

In rectangular, the coordinates are found like this:

Answer 7

In polar, the coordinates are found this way...

Does that give you a clearer picture?

Answer 8

yes thank you. the polar coordinates r and \[\theta \] are defined in terms of cartesian coordinates by x=rcos\[\theta \] and y=rsin\[\theta \] right?

Answer 9

Correct. So actually, I find getting x and y, given r and θ is easier than the other way around.
Nonetheless, you have to know it.

I trust you can see from the picture how to get r?
(Hint: Pythaogorean Theorem)

Answer 10

yes of course, r is the radial distance from the origin, and is the counterclockwise angle from the x-axis right? In terms of x and y is it \[r = \sqrt{x^2 + y^2}\] \[\theta = \tan^{-1} \frac{ x }{ y} \] right?

Answer 11

Whoops, a bit off there...
\[\large \theta = \tan^{-1}\frac{y}{x}\]

Answer 12

oh sorry typo error :)

Answer 13

of course, it isn't always that simple to get θ...

Answer 14

So, for instance, what if both x and y are negative?

Answer 15

ahm i don't have an idea. please explain it to me

Answer 16

The way I do it is take the angle I'll call

\[\large \alpha=\tan^{-1}\left|\frac{y}{x}\right|\]

Answer 17

Do you express your angle in degrees or in radians?

Answer 18

i express it in radian. i don't get it please elaborate :)

Answer 19

Okay, because x and y could either be positive or negative, get that angle alpha, where you just take the inverse tangent of the absolute value of y/x

Catch me so far?
I'll explain further in a bit.

Answer 20

oh i see i got it, is it only that way? just take the inverse tangent of the value of x/y?

Answer 21

not x/y..... y/x
:P
And that's not done yet.
So you have your angle alpha, right?

Answer 22

sorry typo error again. yes i have the angle alpha then?

Answer 23

So, now that you have alpha, consider these scenarios...
If both x and y are positive, then\[\huge \theta = \alpha\]
If x is negative and y is positive, then \[\huge \theta = \pi-\alpha\]
If both x and y are negative, then \[\huge \theta=\pi+\alpha\]
If x is positive and y is negative, then\[\huge \theta=2\pi-\alpha\]

Answer 24

thanks now i know..

Answer 25

Okay... great :)
These will be useful once you get to complex numbers, where you do something eerily similar :)

Answer 26

in finding the area of a region for polar graphs i use this formula \[A = \int\limits_{\alpha}^{\beta} \frac{ 1 }{ 2 } r^2 d \theta \] is there other formula i can use to find its area?

Answer 27

I'm afraid not... Integrals are usually the way to do them anyway, just practice real well :)

Answer 28

oh ok thanks, but how about the length of a region for polar graphs how was it anyway?

Answer 29

Assuming r = f(θ)
or r is a function of θ, then the arclength from a to b is given by
\[\huge \int_a^b\sqrt{\left[f(\theta)\right]^2+\left[f'(\theta)\right]^2}d\theta\]

Answer 30

is it just substituting the given or i will find first the derivative of the given equation?

Answer 31

seeing as there's a f'(θ), you have to get the derivative of the function anyway :P

Answer 32

sorry i mean integral not derivative

Answer 33

Yep, just substitute.

Answer 34

ah ok thanks a lot

Answer 35

No problem. :)

Answer 36

there are different kinds of polar graphs right?

Answer 37

Yeap

Answer 38

if the equation is \[r = 2(1 + \cos \theta ) \] how was it determine easily?

Answer 39

Hang on... afk
Meanwhile, ask some of the other users, I'm sure they'll be only too happy to help you.

Answer 40

ah ok thank you so much for your time i learned a lot :)