Question

Determine two pairs of polar coordinates for the point (2, -2) with 0° ≤ θ < 360°
(2 square root of 2, 225°), (-2 square root of 2, 45°)

(2 square root of 2, 135°), (-2 square root of 2, 315°)

(2 square root of 2, 315°), (-2 square root of 2, 135°)

(2 square root of 2, 45°), (-2 square root of 2, 225°)

Answer 1

@Hero @KingGeorge @Psymon ?

Answer 2

Yo, I have been summoned

Answer 3

haha yes, again.

Answer 4

So first we need to translate this into polar coordinates. There are several conversions we need to be able to remember for these conversions back and forth
\[x = rcos \theta \]
\[y = rsin \theta \]
\[x ^{2}+y ^{2} = r ^{2}\]
\[\tan \theta = \frac{ y }{ x }\]

These arethemain ones we need to be aware of. Now, polar coordinates are like:
\[(r, \theta)\] So using these conversions, we'll turn our rectangular coordinates into polar coordinates.

Answer 5

So given our point (2, -2), we can start off by finding r. Given our conversion above to get r, I'll use the pythagorean theorem one and do:
\[\sqrt{x ^{2}+y ^{2}} = r\]
\[\sqrt{(2)^{2}+(-2)^{2}} = \sqrt{8} = 2\sqrt{2}\]
So we've determined that r, the first coordinate for our polar coordinate point is 2sqrt2. We good with that? :P

Answer 6

Si

Answer 7

いいよ！
Okay, on to theta!

Answer 8

So the best way is usually to use the conversion:
\[\tan \theta = \frac{ y }{ x }\]

Answer 9

So doing that we can say:
\[\tan \theta = \frac{ -2 }{ 2 }\]
\[\tan \theta = -1\]
Now there are two points where tan = -1. So to determine which one we want, we just look at our rectangular coordinates. an x value of 2 and a y value of -2 tells us we are in the 4th quadrant. So in the 4th quadrant, I can say tan = -1 at the angle 7pi/4, which is also 315 degrees.

Now apparently, that gives us the answer right away, but let's look at how we would get the 2nd point if you don't mind :3

Answer 10

So I don't use theta -1 for both?

Answer 11

Well, we'll get to that :P Do you understand how I got the first point, though?

Answer 12

yeshh

Answer 13

Alright, awesome. So now what the point:
\[(2\sqrt{2}, 315)\] means that to get to my point, I need to face 315 degrees and then walk forward 2sqrt2 units. Now there are two ways to get to a point in polar coordinates. We can face the point and walk forward OR we can face 180 degrees in the opposite direction and walk backwards. So if we face the opposite direction, 180 degrees in the other direction, I am then facing 315 - 180 = 135 degrees.
SO facing 135 degrees, I now need to walk backwards. Now the radius doesn't change, that's the same, but when you walk backwards, your radius becomes negative. So that being said, your second point is
\[(-2\sqrt{2}, 135)\]
Kinda make sense? :P

Answer 14

sorry bout that, yes it does!

Answer 15

Alright, awesome xD

Answer 16

Find all polar coordinates of point P = (9, 75°).
{9,75+2pin}, {-9,75(2n+1)pi}
Is this correct?

Answer 17

Seems fine to me, lol.

Answer 18

okay sweet! what about this one:Find the rectangular coordinates of the point with the polar coordinates (8, 3 divided by 2 pi).

(0, -8)

Answer 19

sorry, I'm trying to get an A in my class and i have a 89 -_-

Answer 20

\[\frac{ 3 }{ 2\pi }?\]

Answer 21

3/2 (pi)

Answer 22

Lol, that makes more sense then xD

Answer 23

Well, now we just use these conversions:
\[x = rcos \theta \]
\[y = \sin \theta \]
So if theta is 3pi/2, we just need cos and sin at those points and we multiply by 8. SO cos at 3pi/2 is 0, meaning x = 0. sin at 3pi/2 = -1, times 8 means y = -8. So our point is
(0, -8)
Which itlooks like you got, lol.

Answer 24

Missed the r on the sin conversion, lol. Oh well :P

Answer 25

hallelujah, THANK YOU

Answer 26

Mhm. These arent bad once you understand whats going on :P

Answer 27

Yeah, I have my final soon and this is helping a lot since I tend to get distracted easily lol

Answer 28

Lol, I have to multitask. Im playing chess and playing a Japanese game while I do this xD So I get distracted, but I don't get TOO distracted.

Answer 29

If you have your final soon, have you made it to conic sections yet?

Answer 30

Yes, I wasn't have too much trouble with that. It's just the sequences and stuff

Answer 31

Sequences? You have sequences in your trig work? xD

Answer 32

It's precalc ): I purposely failed in school so I can retake it online and get an A lol

Answer 33

Odd idea to do, lol. I never had sequences like a lot of people have. So the basic sequences I'm not actually up to speed on. When I did sequences it was in calc 2, so we didn't even do the basic ones, lol.

Answer 34

Well see if you got some other questions ya need. I'll be back in 10-15mins.

Answer 35

okay thanks

Answer 36

mkay o.o

Answer 37

Okay found one: lim x^2-2x/(x^4) as x -->0

Answer 38

Interesting. Limits, eh O.o Well, let's see.

Answer 39

haha do you know them?

Answer 40

Infinity, lol. You'd either say infinity or does not exist, depending on what level we're dealing with. And yeah, I know of them. First topic you see in calculus 1.

Answer 41

how'd you get that?

Answer 42

Well, again, I don't know if you're expected to put does not exist or infinity, but I'll show you both. And I'm sorry, negative infinity, not infinity xD

Answer 43

they use DNE

Answer 44

Here is the visual explanation. This is the graph of the function you gave me:



And DNE stands for does not exist.