lilsis76
  • lilsis76
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
|dw:1351734596209:dw| notice that the point will have a negative x, positive y coordinates...
lilsis76
  • lilsis76
why would it be negative?
anonymous
  • anonymous
use... \(\large x=rcos\theta \) \(\large y=rsin\theta \)

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anonymous
  • anonymous
because the point is in the second quadrant....
anonymous
  • anonymous
|dw:1351734804583:dw|
lilsis76
  • lilsis76
okay so r is the radius, and the radius is 3.
lilsis76
  • lilsis76
|dw:1351734929718:dw|2pi/3....i dont see how that can be at that angle.
anonymous
  • anonymous
yes... r=3; \(\theta=\frac{2\pi}{3} \)
lilsis76
  • lilsis76
okay i see that .....
anonymous
  • anonymous
|dw:1351735004666:dw|
lilsis76
  • lilsis76
but shouldnt the 2pi/3 go on the bottom like in the 270 degree? ugh...or do i use calcutore to solve?
lilsis76
  • lilsis76
oh lol sorry, i got them mixed up
lilsis76
  • lilsis76
and why is it to the left of the graph? arent they positive?
anonymous
  • anonymous
|dw:1351735188538:dw|
anonymous
  • anonymous
what are you asking in your last post?
lilsis76
  • lilsis76
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 (+,+)
lilsis76
  • lilsis76
do u get what i mean? cuz i see a positive point
anonymous
  • anonymous
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)
lilsis76
  • lilsis76
|dw:1351735526253:dw|
lilsis76
  • lilsis76
okay but why doesnt the 3 go to the right?
anonymous
  • anonymous
3 is your radius... NOT your x coordinate...
anonymous
  • anonymous
here... click on this link... http://www.mathwords.com/p/polar_rectangular_conversion_formulas.htm
lilsis76
  • lilsis76
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
lilsis76
  • lilsis76
how would i start this one?
anonymous
  • anonymous
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.
anonymous
  • anonymous
here... this is a better explanation of polar coordinates: http://www.mathsisfun.com/polar-cartesian-coordinates.html
lilsis76
  • lilsis76
okay ill look at it
anonymous
  • anonymous
so those formulas i gave you converts the given point in POLAR form to RECTANGULAR form...
lilsis76
  • lilsis76
okay. let me try on here and u let me know if i do it wrong. please.
anonymous
  • anonymous
ok...
lilsis76
  • lilsis76
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
anonymous
  • anonymous
why is the angle 2pi/6 ??? i thought it was 2pi/3 ???
lilsis76
  • lilsis76
AH! sorry, haha i was looking at a 6. let me try
lilsis76
  • lilsis76
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
anonymous
  • anonymous
careful.... \(\large cos(\frac{2\pi}{3})=-\frac{1}{2} \)
anonymous
  • anonymous
your y coordinate is correct...
lilsis76
  • lilsis76
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
anonymous
  • anonymous
yes... so the x y coordinate for the point is \(\large (-\frac{3}{2},\frac{3\sqrt3}{2}) \)
lilsis76
  • lilsis76
|dw:1351736659036:dw| then the coordinate - 3/2, 3 sqrt 3 /2 would be in the same area right?
anonymous
  • anonymous
it is the SAME point.... only expressed in cartesian form
lilsis76
  • lilsis76
oh...okay, let me try the other problems and ill be back online if I need help. THANK YOU!!!
anonymous
  • anonymous
yw... glad i could help...
anonymous
  • anonymous
good luck... :)
lilsis76
  • lilsis76
thanks

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