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

1. anonymous
2. anonymous

hey i actually got stuck @Astrophysics

3. Astrophysics

Hey, no worries, where did you get stuck?

4. 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

5. 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

6. anonymous

r = 5sqrt(2)

7. 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?

8. anonymous

yea i got that

9. 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

10. anonymous

ok so tan(theta) = -5/5

11. anonymous

???

12. Astrophysics

Yes exactly!

13. anonymous

tan(theta) = -1 from here to I inverse it?

14. Astrophysics

Yup $\theta = \tan^{-1}(-1)$

15. anonymous

ok give me a sec

16. anonymous

7pi/4

17. Astrophysics

Good so that is 315 degrees

18. anonymous

ok so it would be 5sqrt(2),315

19. Astrophysics

Yes, and the other you want for $-5\sqrt{2}$

20. Astrophysics

Which is just 315-180

21. anonymous

135

22. Astrophysics

Bingo

23. anonymous

24. Astrophysics

Sounds good!

25. anonymous

Thank you so much for the help! :)

26. Astrophysics

Np :)

27. anonymous

x is positive and y is negative so it is below x-axis x is positive , yis negative ,hence in fourth quadrant.

28. Astrophysics

29. anonymous

you mind helping me with one more? sorry. lol

30. Astrophysics

Sure haha

31. anonymous

Find all polar coordinates of point P = (2, 14°)

32. Astrophysics

n is an integer

33. anonymous

x = 1.94059 y = 0.48384

34. Astrophysics

Ok maybe I shouldn't have wrote that just ignore it and look at what I said after

35. anonymous

lol ok

36. anonymous

so would it just be (2, 14 +/- 2npi) ???

37. Astrophysics

Yes that works (2, 14 +360n) as we're using degrees

38. anonymous

is that it?

39. Astrophysics

yes or (-2, 14+90+360n)

40. anonymous

wait im confused now where did the 90 come from?

41. Astrophysics

Look at what I said twice, the polar representations

42. anonymous

so all the polar coordinates are (2, 14 +360n) , (-2, 14+90+360n)

43. Astrophysics

Yes that sounds good

44. anonymous

ok thanks again. for all the help

45. 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

46. anonymous

oh ok. :)

47. Astrophysics

wait wait that should be an OR not an AND haha ok

48. anonymous

lol its alright

49. Astrophysics

xD

50. Astrophysics

Ok have fun and take care haha

51. anonymous

you too. Thanks again

52. Astrophysics

Np :)