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anonymous

  • one year ago

Determine two pairs of polar coordinates for the point (5, -5) with 0° ≤ θ < 360°

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  1. anonymous
    • one year ago
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    https://i.gyazo.com/1a50524381eaebcd22762fa6292443fe.png

  2. anonymous
    • one year ago
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    hey i actually got stuck @Astrophysics

  3. Astrophysics
    • one year ago
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    Hey, no worries, where did you get stuck?

  4. Astrophysics
    • one year ago
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    I will just put this for a reference \[x = r \cos \theta\]\[y= r \sin \theta\]\[r^2 = x^2+y^2 \implies r = \sqrt{x^2+y^2}\], and lets go over it

  5. anonymous
    • one year ago
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    ok so i plugged in everything... r=sqrt(5^2 + (-5)^2) but when i solved for theta my answer wasnt one of the options

  6. anonymous
    • one year ago
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    r = 5sqrt(2)

  7. Astrophysics
    • one year ago
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    Ok lets see, so you did \[r = \sqrt{x^2+y^2} \implies r = \sqrt{(5)^2+(-5)^2} = \sqrt{50} = 5\sqrt{2}\] so far so good?

  8. anonymous
    • one year ago
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    yea i got that

  9. Astrophysics
    • one year ago
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    Ok cool, so the angles can be quite tricky that's why I had drawn the triangle for you earlier, so here we have to use the tan ratio to find the angle

  10. anonymous
    • one year ago
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    ok so tan(theta) = -5/5

  11. anonymous
    • one year ago
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    ???

  12. Astrophysics
    • one year ago
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    Yes exactly!

  13. anonymous
    • one year ago
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    tan(theta) = -1 from here to I inverse it?

  14. Astrophysics
    • one year ago
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    Yup \[\theta = \tan^{-1}(-1)\]

  15. anonymous
    • one year ago
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    ok give me a sec

  16. anonymous
    • one year ago
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    7pi/4

  17. Astrophysics
    • one year ago
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    Good so that is 315 degrees

  18. anonymous
    • one year ago
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    ok so it would be 5sqrt(2),315

  19. Astrophysics
    • one year ago
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    Yes, and the other you want for \[-5\sqrt{2}\]

  20. Astrophysics
    • one year ago
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    Which is just 315-180

  21. anonymous
    • one year ago
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    135

  22. Astrophysics
    • one year ago
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    Bingo

  23. anonymous
    • one year ago
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    so the answer is A?

  24. Astrophysics
    • one year ago
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    Sounds good!

  25. anonymous
    • one year ago
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    Thank you so much for the help! :)

  26. Astrophysics
    • one year ago
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    Np :)

  27. anonymous
    • one year ago
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    x is positive and y is negative so it is below x-axis x is positive , yis negative ,hence in fourth quadrant.

  28. Astrophysics
    • one year ago
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    Yup I already had shown that in the previous post ;P

  29. anonymous
    • one year ago
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    you mind helping me with one more? sorry. lol

  30. Astrophysics
    • one year ago
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    Sure haha

  31. anonymous
    • one year ago
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    Find all polar coordinates of point P = (2, 14°)

  32. Astrophysics
    • one year ago
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    n is an integer

  33. anonymous
    • one year ago
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    x = 1.94059 y = 0.48384

  34. Astrophysics
    • one year ago
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    Ok maybe I shouldn't have wrote that just ignore it and look at what I said after

  35. anonymous
    • one year ago
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    lol ok

  36. anonymous
    • one year ago
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    so would it just be (2, 14 +/- 2npi) ???

  37. Astrophysics
    • one year ago
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    Yes that works (2, 14 +360n) as we're using degrees

  38. anonymous
    • one year ago
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    is that it?

  39. Astrophysics
    • one year ago
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    yes or (-2, 14+90+360n)

  40. anonymous
    • one year ago
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    wait im confused now where did the 90 come from?

  41. Astrophysics
    • one year ago
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    Look at what I said twice, the polar representations

  42. anonymous
    • one year ago
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    so all the polar coordinates are (2, 14 +360n) , (-2, 14+90+360n)

  43. Astrophysics
    • one year ago
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    Yes that sounds good

  44. anonymous
    • one year ago
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    ok thanks again. for all the help

  45. Astrophysics
    • one year ago
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    The polar representation should be \[(r, \theta) = (r, \theta+2 \pi n)~~~\text{and}~~~(-r, \theta+(2n+1) \pi)\] I think I made a mistake, but your answers are right in any case

  46. anonymous
    • one year ago
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    oh ok. :)

  47. Astrophysics
    • one year ago
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    wait wait that should be an OR not an AND haha ok

  48. anonymous
    • one year ago
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    lol its alright

  49. Astrophysics
    • one year ago
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    xD

  50. Astrophysics
    • one year ago
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    Ok have fun and take care haha

  51. anonymous
    • one year ago
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    you too. Thanks again

  52. Astrophysics
    • one year ago
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    Np :)

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