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egbeach

  • one year ago

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

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  1. Mertsj
    • one year ago
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    Jim will take over.

  2. jim_thompson5910
    • one year ago
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    similar to what Mertsj drew |dw:1437356798358:dw|

  3. jim_thompson5910
    • one year ago
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    |dw:1437356851407:dw|

  4. jim_thompson5910
    • one year ago
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    |dw:1437356864345:dw|

  5. jim_thompson5910
    • one year ago
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    and that 45 is actually -45 degrees because we're going clockwise instead of counterclockwise

  6. egbeach
    • one year ago
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    is it (2 square root of 2, 225°), (-2 square root of 2, 45°)

  7. jim_thompson5910
    • one year ago
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    |dw:1437357081640:dw|

  8. jim_thompson5910
    • one year ago
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    one possible representation is (2*sqrt(2), 315)

  9. jim_thompson5910
    • one year ago
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    315 - 180 = 135 degrees |dw:1437357159428:dw| so another possible polar representation is (-2*sqrt(2), 135) the negative r value means you walk backwards 2*sqrt(2) units while facing the 135 degree direction

  10. anonymous
    • one year ago
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    You can determine by this way: (2, -2) that is x =2, y =-2 hence \(r=\sqrt{2^2+(-2)^2}=2\sqrt2\) \(\theta =arctan(\dfrac{y}{x}) = arctan(-1)\) Hence on the unit circle, \(\theta = 3\pi/4\) or \(\theta =-\pi/4\) |dw:1437357509404:dw|

  11. anonymous
    • one year ago
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    Therefore, the first pair is \((2\sqrt2, -\pi/4) \)

  12. anonymous
    • one year ago
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    Moreover, the opposite side of the terminal point will stop at 3pi/4. But to get the point (2,-2) we need negative value of r, hence, the second point will be \((-2\sqrt2, 3\pi/4)\) |dw:1437357640591:dw|

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