anonymous
  • anonymous
Determine two pairs of polar coordinates for the point (5, -5) with 0° ≤ θ < 360°
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
https://i.gyazo.com/1a50524381eaebcd22762fa6292443fe.png
anonymous
  • anonymous
hey i actually got stuck @Astrophysics
Astrophysics
  • Astrophysics
Hey, no worries, where did you get stuck?

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Astrophysics
  • Astrophysics
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
anonymous
  • anonymous
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
anonymous
  • anonymous
r = 5sqrt(2)
Astrophysics
  • Astrophysics
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?
anonymous
  • anonymous
yea i got that
Astrophysics
  • Astrophysics
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
anonymous
  • anonymous
ok so tan(theta) = -5/5
anonymous
  • anonymous
???
Astrophysics
  • Astrophysics
Yes exactly!
anonymous
  • anonymous
tan(theta) = -1 from here to I inverse it?
Astrophysics
  • Astrophysics
Yup \[\theta = \tan^{-1}(-1)\]
anonymous
  • anonymous
ok give me a sec
anonymous
  • anonymous
7pi/4
Astrophysics
  • Astrophysics
Good so that is 315 degrees
anonymous
  • anonymous
ok so it would be 5sqrt(2),315
Astrophysics
  • Astrophysics
Yes, and the other you want for \[-5\sqrt{2}\]
Astrophysics
  • Astrophysics
Which is just 315-180
anonymous
  • anonymous
135
Astrophysics
  • Astrophysics
Bingo
anonymous
  • anonymous
so the answer is A?
Astrophysics
  • Astrophysics
Sounds good!
anonymous
  • anonymous
Thank you so much for the help! :)
Astrophysics
  • Astrophysics
Np :)
anonymous
  • anonymous
x is positive and y is negative so it is below x-axis x is positive , yis negative ,hence in fourth quadrant.
Astrophysics
  • Astrophysics
Yup I already had shown that in the previous post ;P
anonymous
  • anonymous
you mind helping me with one more? sorry. lol
Astrophysics
  • Astrophysics
Sure haha
anonymous
  • anonymous
Find all polar coordinates of point P = (2, 14°)
Astrophysics
  • Astrophysics
n is an integer
anonymous
  • anonymous
x = 1.94059 y = 0.48384
Astrophysics
  • Astrophysics
Ok maybe I shouldn't have wrote that just ignore it and look at what I said after
anonymous
  • anonymous
lol ok
anonymous
  • anonymous
so would it just be (2, 14 +/- 2npi) ???
Astrophysics
  • Astrophysics
Yes that works (2, 14 +360n) as we're using degrees
anonymous
  • anonymous
is that it?
Astrophysics
  • Astrophysics
yes or (-2, 14+90+360n)
anonymous
  • anonymous
wait im confused now where did the 90 come from?
Astrophysics
  • Astrophysics
Look at what I said twice, the polar representations
anonymous
  • anonymous
so all the polar coordinates are (2, 14 +360n) , (-2, 14+90+360n)
Astrophysics
  • Astrophysics
Yes that sounds good
anonymous
  • anonymous
ok thanks again. for all the help
Astrophysics
  • Astrophysics
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
anonymous
  • anonymous
oh ok. :)
Astrophysics
  • Astrophysics
wait wait that should be an OR not an AND haha ok
anonymous
  • anonymous
lol its alright
Astrophysics
  • Astrophysics
xD
Astrophysics
  • Astrophysics
Ok have fun and take care haha
anonymous
  • anonymous
you too. Thanks again
Astrophysics
  • Astrophysics
Np :)

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