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lilsis76

  • 3 years ago

graph each point in a polar coordinate system then convert the given polar coordinates to rectangluar coordinates. can someone help me do this step by step so i understand please. 1) a) (3, 2pi/3)

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  1. ByteMe
    • 3 years ago
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    |dw:1351734596209:dw| notice that the point will have a negative x, positive y coordinates...

  2. lilsis76
    • 3 years ago
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    why would it be negative?

  3. ByteMe
    • 3 years ago
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    use... \(\large x=rcos\theta \) \(\large y=rsin\theta \)

  4. ByteMe
    • 3 years ago
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    because the point is in the second quadrant....

  5. ByteMe
    • 3 years ago
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    |dw:1351734804583:dw|

  6. lilsis76
    • 3 years ago
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    okay so r is the radius, and the radius is 3.

  7. lilsis76
    • 3 years ago
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    |dw:1351734929718:dw|2pi/3....i dont see how that can be at that angle.

  8. ByteMe
    • 3 years ago
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    yes... r=3; \(\theta=\frac{2\pi}{3} \)

  9. lilsis76
    • 3 years ago
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    okay i see that .....

  10. ByteMe
    • 3 years ago
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    |dw:1351735004666:dw|

  11. lilsis76
    • 3 years ago
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    but shouldnt the 2pi/3 go on the bottom like in the 270 degree? ugh...or do i use calcutore to solve?

  12. lilsis76
    • 3 years ago
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    oh lol sorry, i got them mixed up

  13. lilsis76
    • 3 years ago
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    and why is it to the left of the graph? arent they positive?

  14. ByteMe
    • 3 years ago
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    |dw:1351735188538:dw|

  15. ByteMe
    • 3 years ago
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    what are you asking in your last post?

  16. lilsis76
    • 3 years ago
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    okay, u see how u found the point in the left of the graph chart, why is it to the left. isnt it ( - , +) we have a (+,+)

  17. lilsis76
    • 3 years ago
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    do u get what i mean? cuz i see a positive point

  18. ByteMe
    • 3 years ago
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    oh... you're referring to the point \(\large (3, \frac{2\pi}{3}) \)..... that point is represented in POLAR form, \(\large (r, \theta) \) and not cartesian form (x, y)

  19. lilsis76
    • 3 years ago
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    |dw:1351735526253:dw|

  20. lilsis76
    • 3 years ago
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    okay but why doesnt the 3 go to the right?

  21. ByteMe
    • 3 years ago
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    3 is your radius... NOT your x coordinate...

  22. ByteMe
    • 3 years ago
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    here... click on this link... http://www.mathwords.com/p/polar_rectangular_conversion_formulas.htm

  23. lilsis76
    • 3 years ago
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    okay then, so looking at the unit circle its the point then, and like u said the 3 is the radius. so thats the reason why its to the left. It says now to convert the given polar coordinates to rectangular coordinates

  24. lilsis76
    • 3 years ago
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    how would i start this one?

  25. ByteMe
    • 3 years ago
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    no... the reason why it's on the left of the y-axis is because the angle theta, 2pi/3 resides in the second quadrant.

  26. ByteMe
    • 3 years ago
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    here... this is a better explanation of polar coordinates: http://www.mathsisfun.com/polar-cartesian-coordinates.html

  27. lilsis76
    • 3 years ago
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    okay ill look at it

  28. ByteMe
    • 3 years ago
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    so those formulas i gave you converts the given point in POLAR form to RECTANGULAR form...

  29. lilsis76
    • 3 years ago
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    okay. let me try on here and u let me know if i do it wrong. please.

  30. ByteMe
    • 3 years ago
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    ok...

  31. lilsis76
    • 3 years ago
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    x= r cos theta --> 3 cos 2pi/6 --> 3(1/2) --> 3/2 y= r sin theta --> 3 sin 2pi/6 --> 3(sqrt.3 /2) --> 3/2 sqrt3

  32. ByteMe
    • 3 years ago
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    why is the angle 2pi/6 ??? i thought it was 2pi/3 ???

  33. lilsis76
    • 3 years ago
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    AH! sorry, haha i was looking at a 6. let me try

  34. lilsis76
    • 3 years ago
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    x= r cos theta --> 3 cos 2pi/3 --> 3(1/2) --> 3/2 y= r sin theta --> 3 sin 2pi/3 --> 3(sqrt.3 /2) --> 3/2 sqrt3

  35. ByteMe
    • 3 years ago
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    careful.... \(\large cos(\frac{2\pi}{3})=-\frac{1}{2} \)

  36. ByteMe
    • 3 years ago
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    your y coordinate is correct...

  37. lilsis76
    • 3 years ago
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    oops, thanks, okay so then x= r cos theta --> 3 cos 2pi/6 --> 3(- 1/2) --> - 3/2 y= r sin theta --> 3 sin 2pi/6 --> 3(sqrt.3 /2) --> 3/2 sqrt3

  38. ByteMe
    • 3 years ago
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    yes... so the x y coordinate for the point is \(\large (-\frac{3}{2},\frac{3\sqrt3}{2}) \)

  39. lilsis76
    • 3 years ago
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    |dw:1351736659036:dw| then the coordinate - 3/2, 3 sqrt 3 /2 would be in the same area right?

  40. ByteMe
    • 3 years ago
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    it is the SAME point.... only expressed in cartesian form

  41. lilsis76
    • 3 years ago
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    oh...okay, let me try the other problems and ill be back online if I need help. THANK YOU!!!

  42. ByteMe
    • 3 years ago
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    yw... glad i could help...

  43. ByteMe
    • 3 years ago
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    good luck... :)

  44. lilsis76
    • 3 years ago
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    thanks

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