Question

Fan + Medal
Determine two pairs of polar coordinates for the point (2, -2) with 0° ≤ θ < 360°.

Answer 1

\[x = rcos \theta\]
\[y = rsin \theta\]
For polar to cartesian coordinates.

For cartesian to polar: \[r^2 = x^2+y^2 ~~~~ \theta = \tan ^{-1} \left( \frac{ y }{ x } \right)\]

Answer 2

Note you want r so \[r = \sqrt{x^2+y^2}\]

Answer 3

r = sqrt(8) or 2.828

Answer 4

Sure, now find theta

Answer 5

theta = 45

Answer 6

now how do we find which quadrant it is in?

Answer 7

Yeah but notice theta gives us the value for the first and we need the fourth quadrant.

Answer 8

Nice, you figured that out

Answer 9

oh that makes sense

Answer 10

so 315

Answer 11

Yeah that seems reasonable so 7pi/4

Answer 12

which is \[\large(\frac{ \sqrt{2} }{ 2 }, \frac{ \sqrt{-2} }{ 2 })\]

Answer 13

on the unit circel

Answer 14

circle*

Answer 15

When you're looking for theta, you need unit circle :)

Answer 16

Seems you got a good grasp on it though, nice work.

Answer 17

\[\sqrt{8} \implies 2\sqrt{2}\]

Answer 18

so what is the answer exactly? I need 2 points in Polar Coordinates of the original thing

Answer 19

these are my options:
(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 20

Oh it wants two pairs

Answer 21

So now what do you think the other will be?

Answer 22

Im blanking @Astrophysics

Answer 23

I think it's the third option. Am I right @zepdrix ?

Answer 24

Oh I see a mistake

Answer 25

What does \[\theta = \tan^{-1}\left( \frac{ -2 }{ 2 } \right)\] give?

Answer 26

So you found your first polar pair by rotating 315, and then extending outward 2sqrt2?
Looks good.