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

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2\[x = rcos \theta\] \[y = rsin \theta\] For polar to cartesian coordinates. For cartesian to polar: \[r^2 = x^2+y^2 ~~~~ \theta = \tan ^{1} \left( \frac{ y }{ x } \right)\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Note you want r so \[r = \sqrt{x^2+y^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0r = sqrt(8) or 2.828

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Sure, now find theta

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now how do we find which quadrant it is in?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Yeah but notice dw:1435521133327:dw theta gives us the value for the first and we need the fourth quadrant.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Nice, you figured that out

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Yeah that seems reasonable so 7pi/4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which is \[\large(\frac{ \sqrt{2} }{ 2 }, \frac{ \sqrt{2} }{ 2 })\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2When you're looking for theta, you need unit circle :)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Seems you got a good grasp on it though, nice work.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2\[\sqrt{8} \implies 2\sqrt{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so what is the answer exactly? I need 2 points in Polar Coordinates of the original thing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0these 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°)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Oh it wants two pairs

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2So now what do you think the other will be?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im blanking @Astrophysics

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think it's the third option. Am I right @zepdrix ?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Oh I see a mistake

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2What does \[\theta = \tan^{1}\left( \frac{ 2 }{ 2 } \right)\] give?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0dw:1435522015498:dwSo you found your first polar pair by rotating 315, and then extending outward 2sqrt2? Looks good.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0dw:1435522105721:dwFor your second option, you're rotating 180 degrees less around, and moving radially backwards.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0ya third option looks good bro!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0for the second coordinate pair i meant*

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have another question but ill post a new form.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2If we could only give more than one medal :P
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