Question

Find the rectangular coordinates of the point with the polar coordinates (7, 2pi/3)

Answer 1

Awesome...
Polar coordinates
\[\huge (r , \theta)\]
Then your rectangular coordinates are given by
\[\huge (x,y) = (r \cos \ \theta, r \sin \ \theta)\]

Answer 2

@PeterPan so i would put
\[(7\cos \theta) , (\frac{ 2\Pi }{ 3 }\sin \theta)\]
into my calculator?

Answer 3

Here, \(\theta = \dfrac{2\pi}{3}\). And don't do the commas.

Answer 4

Nope... Remember,
\[\huge (r, \theta) = (7, \frac{2\pi}{3})\]

Answer 5

So your r is only ever 7
And your theta is only ever two-thirds of pi :)

Answer 6

Figure out what \(7\cos \left(\frac{2\pi}3{}\right)\) is, then figure out what \(7\sin\left(\frac{2\pi}{3}\right)\) is. Don't forget the comma and parentheses.

Answer 7

I didn't know Peter Pan was so good at mathematics. :-)

Answer 8

When you have all the time in the world, sometimes, it pays to learn stuff :D

Answer 9

@PeterPan and @ParthKohli oh wow, that makes so much more sense!!
so then \[x=\frac{ -7 }{2} y=\frac{ 7\sqrt{3} }{2}\]

Answer 10

Awesome work :)

Answer 11

wow thank you both so much!! you both made it much easier than i was!! @PeterPan @ParthKohli

Answer 12

Polar to Rectangular is such...
Now the other way around, not as simple :P

Answer 13

@PeterPan :( that's what i have to do now! can you help me with that also?

Answer 14

I'm at your call :)

Answer 15

@peterpan Find all polar coordinates of point P where P = \[( 1, \frac{ \Pi }{ 3 })\]

Answer 16

Are you sure it's polar coordinates we want and not rectangular? The presence of a 'pi' in your coordinates is worth considering...

Answer 17

@PeterPan

Answer 18

I think P is already in polar, and we're just trying to find which other polar coordinates are equivalent to it.

Answer 19

@PeterPan i don't know how to do that.. this section that i'm doing is the toughest for me..

Answer 20

Well, think of it this way... your theta, or your angle, if you rotate it one full rotation, or 2pi, it's basically the same angle, right?

Answer 21

@PeterPan yes

Answer 22

So you can add 2pi to it, and it stays the same, right?

Answer 23

@PeterPan yes.. ?

Answer 24

You can add it as many times as you want, right?

Answer 25

@PeterPan yes

Answer 26

So you can add 2npi, where n is any integer. Got that so far? :)

Answer 27

@PeterPan i was just about to ask what n was, okay i understand that now!

Answer 28

However, that's not the only way to rewrite your polar coordinates.
You can also negate the radius (make it negative) and then, rotate the angle by a half-turn, or just pi (instead of 2pi)

Answer 29

@PeterPan i understand that too

Answer 30

and once you've added pi to it, you can once again add 2pi as many times as you want.
So it's
2npi + pi, right?

Answer 31

@PeterPan yes

Answer 32

so it's (2n+1)pi, if you factor out the pi :P

Answer 33

@PeterPan and that right there will allow me to find any coordinate on the plane?

Answer 34

Well, to summarise, given our polar coordinates
\[\large (r, \theta)\]It is always the same when we add any multiple of 2pi to the angle:
\[\huge (r, \ \theta + 2n \pi)\]
It is also the same if we negate the radius, and then rotate by a half turn by adding pi...
\[\large (-r, \ \theta + \pi)\]
But much like the first case, we can also keep on adding 2pi to this new angle and have it stay the same...
\[\large (-r , \ \theta + \pi + 2n \pi)\]
factoring out pi
\[\huge (-r , \ \theta +(2n+1) \pi)\]

Answer 35

@PeterPan omg okay, that makes so much sense! thank youu!!

Answer 36

:)

Answer 37

@PeterPan do you mind helping me with one more? :)

Answer 38

Sure... let me have it :)

Answer 39

@PeterPan
Determine two pairs of polar coordinates for the point (3, -3) with 0° ≤ θ < 360°.

Answer 40

Okay, here's the thing about rectangular to polar...
It's simple getting r
You know how?

Answer 41

@PeterPan so is this the opposite of what we doing earlier?

Answer 42

Yep...

Answer 43

@PeterPan you would use \[r=\sqrt{x ^{2}+y ^{2}} , \theta=\tan^{-1} (\frac{ y }{ x })\]

Answer 44

Yeah, that's good. Seems you already know what you're doing :)
go ahead and plug in these formulas with
x = 3
y = -3

Answer 45

Not yet done, though, you're asked for two pairs of coordinates. Work out one pair, first, though

Answer 46

@PeterPan well r would equal \[\sqrt{18}\]
and \[\theta= -.785\]
that's what i got..

Answer 47

Remember, we're working with degrees this time. Try to redo the theta part.

Answer 48

@PeterPan i would plug \[\tan^{-1}( \frac{ -3 }{ 3 })\]
into my calculator correct?

Answer 49

Yeah, but make sure it's set to degrees, and not radians.

Answer 50

@peterpan i got -45
i forgot to change it into degrees! thanks

Answer 51

Okay, so you got -45, but that doesn't really fall within the 0° ≤ θ < 360°
restriction you wanted, so what to do?

Answer 52

@PeterPan change it to positive! or make r negative

Answer 53

change it to positive... how?
By the way, there's no need to keep tagging me, it's like you're calling out my name every time you answer me :D

Answer 54

oh, i wasn't sure if you would still get the notification! but okay

but i'm honestly not too sure what to do since -45 isn't in the range...

Answer 55

Well, remember, when we have θ, we can keep adding 2pi to it, and it basically stays the same?
Well, when we have degrees, we can keep adding 360 to it, without it changing !
:)

Answer 56

so we just add 2pi and it'll be postive right?

Answer 57

This is in degrees, so you'll want to add 360 degrees...

Answer 58

oh so it will be 315

Answer 59

That's right :)

Answer 60

But that's not the only pair of coordinates...

Answer 61

there's another one..

Answer 62

mhmm
but before that, what was your first pair?

Answer 63

\[(3\sqrt{2},315)\]

Answer 64

lol okay, but remember what we said?
It would be the same if we negate the radius (r)
and then add (or subtract, it don't matter, really) pi to the angle.
So, instead of pi, we add (or subtract) 180 degrees.
Go ahead now :)

Answer 65

it'll equal 135

Answer 66

will there be a negative in front of \[(3\sqrt{2})\]

Answer 67

Yes :)

Answer 68

So what are your two pairs?

Answer 69

(3sprt2, 315) and (-3sqrt2, 135) :)

Answer 70

thank you so much!! i appreciate your help so much!!!

Answer 71

No problem :)