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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°.

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  1. Astrophysics
    • one year ago
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    \[x = rcos \theta\] \[y = rsin \theta\] For polar to cartesian coordinates. For cartesian to polar: \[r^2 = x^2+y^2 ~~~~ \theta = \tan ^{-1} \left( \frac{ y }{ x } \right)\]

  2. Astrophysics
    • one year ago
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    Note you want r so \[r = \sqrt{x^2+y^2}\]

  3. anonymous
    • one year ago
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    r = sqrt(8) or 2.828

  4. Astrophysics
    • one year ago
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    Sure, now find theta

  5. anonymous
    • one year ago
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    theta = 45

  6. anonymous
    • one year ago
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    now how do we find which quadrant it is in?

  7. Astrophysics
    • one year ago
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    Yeah but notice |dw:1435521133327:dw| theta gives us the value for the first and we need the fourth quadrant.

  8. Astrophysics
    • one year ago
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    Nice, you figured that out

  9. anonymous
    • one year ago
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    oh that makes sense

  10. anonymous
    • one year ago
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    so 315

  11. Astrophysics
    • one year ago
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    Yeah that seems reasonable so 7pi/4

  12. anonymous
    • one year ago
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    which is \[\large(\frac{ \sqrt{2} }{ 2 }, \frac{ \sqrt{-2} }{ 2 })\]

  13. anonymous
    • one year ago
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    on the unit circel

  14. anonymous
    • one year ago
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    circle*

  15. Astrophysics
    • one year ago
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    When you're looking for theta, you need unit circle :)

  16. Astrophysics
    • one year ago
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    Seems you got a good grasp on it though, nice work.

  17. Astrophysics
    • one year ago
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    \[\sqrt{8} \implies 2\sqrt{2}\]

  18. anonymous
    • one year ago
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    so what is the answer exactly? I need 2 points in Polar Coordinates of the original thing

  19. anonymous
    • one year ago
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    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°)

  20. Astrophysics
    • one year ago
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    Oh it wants two pairs

  21. Astrophysics
    • one year ago
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    So now what do you think the other will be?

  22. anonymous
    • one year ago
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    Im blanking @Astrophysics

  23. anonymous
    • one year ago
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    I think it's the third option. Am I right @zepdrix ?

  24. Astrophysics
    • one year ago
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    Oh I see a mistake

  25. Astrophysics
    • one year ago
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    What does \[\theta = \tan^{-1}\left( \frac{ -2 }{ 2 } \right)\] give?

  26. zepdrix
    • one year ago
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    |dw:1435522015498:dw|So you found your first polar pair by rotating 315, and then extending outward 2sqrt2? Looks good.

  27. anonymous
    • one year ago
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    -45 @Astrophysics

  28. zepdrix
    • one year ago
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    |dw:1435522105721:dw|For your second option, you're rotating 180 degrees less around, and moving radially backwards.

  29. zepdrix
    • one year ago
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    ya third option looks good bro!

  30. zepdrix
    • one year ago
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    for the second coordinate pair i meant*

  31. anonymous
    • one year ago
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    Thanks both of you!

  32. anonymous
    • one year ago
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    I have another question but ill post a new form.

  33. Astrophysics
    • one year ago
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    If we could only give more than one medal :P

  34. anonymous
    • one year ago
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    haha true

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