anonymous
  • anonymous
Fan + Medal Determine two pairs of polar coordinates for the point (2, -2) with 0° ≤ θ < 360°.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Astrophysics
  • Astrophysics
\[x = rcos \theta\] \[y = rsin \theta\] For polar to cartesian coordinates. For cartesian to polar: \[r^2 = x^2+y^2 ~~~~ \theta = \tan ^{-1} \left( \frac{ y }{ x } \right)\]
Astrophysics
  • Astrophysics
Note you want r so \[r = \sqrt{x^2+y^2}\]
anonymous
  • anonymous
r = sqrt(8) or 2.828

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Astrophysics
  • Astrophysics
Sure, now find theta
anonymous
  • anonymous
theta = 45
anonymous
  • anonymous
now how do we find which quadrant it is in?
Astrophysics
  • Astrophysics
Yeah but notice |dw:1435521133327:dw| theta gives us the value for the first and we need the fourth quadrant.
Astrophysics
  • Astrophysics
Nice, you figured that out
anonymous
  • anonymous
oh that makes sense
anonymous
  • anonymous
so 315
Astrophysics
  • Astrophysics
Yeah that seems reasonable so 7pi/4
anonymous
  • anonymous
which is \[\large(\frac{ \sqrt{2} }{ 2 }, \frac{ \sqrt{-2} }{ 2 })\]
anonymous
  • anonymous
on the unit circel
anonymous
  • anonymous
circle*
Astrophysics
  • Astrophysics
When you're looking for theta, you need unit circle :)
Astrophysics
  • Astrophysics
Seems you got a good grasp on it though, nice work.
Astrophysics
  • Astrophysics
\[\sqrt{8} \implies 2\sqrt{2}\]
anonymous
  • anonymous
so what is the answer exactly? I need 2 points in Polar Coordinates of the original thing
anonymous
  • anonymous
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°)
Astrophysics
  • Astrophysics
Oh it wants two pairs
Astrophysics
  • Astrophysics
So now what do you think the other will be?
anonymous
  • anonymous
Im blanking @Astrophysics
anonymous
  • anonymous
I think it's the third option. Am I right @zepdrix ?
Astrophysics
  • Astrophysics
Oh I see a mistake
Astrophysics
  • Astrophysics
What does \[\theta = \tan^{-1}\left( \frac{ -2 }{ 2 } \right)\] give?
zepdrix
  • zepdrix
|dw:1435522015498:dw|So you found your first polar pair by rotating 315, and then extending outward 2sqrt2? Looks good.
anonymous
  • anonymous
-45 @Astrophysics
zepdrix
  • zepdrix
|dw:1435522105721:dw|For your second option, you're rotating 180 degrees less around, and moving radially backwards.
zepdrix
  • zepdrix
ya third option looks good bro!
zepdrix
  • zepdrix
for the second coordinate pair i meant*
anonymous
  • anonymous
Thanks both of you!
anonymous
  • anonymous
I have another question but ill post a new form.
Astrophysics
  • Astrophysics
If we could only give more than one medal :P
anonymous
  • anonymous
haha true

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