## anonymous 5 years ago Convert the equation to polar form. 5x=5y AND x^2+y^2=16

1. anonymous

x=r cosƟ, y=r sinƟ, x^2+y^2=r^2.

2. anonymous

I don't get the second one. x^2+y^2=r^2

3. anonymous

No, not really. Trig and I don't get along at all

4. myininaya

ok loco do you have that picture i drawed you where I graph (x,y) and called the angle between the initial ray and I guess we can call it the vector with endpoints at the origin and at the point (x,y)

5. anonymous

yeah, it's open in my pdf

6. myininaya

k and I put inside a circle i called the radius of that circle r

7. myininaya

that trangle is a right triangle so we can use Pythagorean them to find what r is in terms of x and y

8. myininaya

r=sqrt(x^2+y^2)

9. myininaya

or like anwar said before x^2+y^2=r^2

10. anonymous

yeah, i don't this equation? what do i use it for?

11. myininaya

we want to convert x^2+y^2=16 to a polar equation so that means we want it in terms of r and pheta but what is x^2+y^2=.....?

12. myininaya

loco, do you know what to fill that blank with?

13. anonymous

5x^2+5y^2=16

14. myininaya

if we are doing 5x^2+5y^2=16 then we can factor out 5 so we have 5(x^2+y^2)=16 but what does x^2+y^2=----?

15. anonymous

does it equal r?

16. myininaya

right!

17. anonymous

sorry, if i take too long, trying to figure this out

18. myininaya

5r=16 is the polar equation of the cartesian equation 5x^2+5y^2=16

19. myininaya

these equations give the same graph using a little different way of graphing

20. myininaya

these are equations are equivalent

21. myininaya

polar is in the form (r,pheta) Cartesian is in the form (x,y)

22. anonymous

what do u mean equivalent? = to what? 16

23. myininaya

no to each other

24. anonymous

so 5x=5y, is that what u mean?

25. myininaya

5r=16 is equivalent to 5x^2+5y^2=16 because they are the same graph

26. anonymous

oh ok...

27. myininaya

so you want to convert 5x=5y to polar equation?

28. anonymous

yes

29. myininaya

so remember polar has (r,pheta) form so x=rcos(pheta) and y=rsin(pheta) 5rcos(pheta)=5rsin(pheta) we could divide both sides by 5 so we have rcos(pheta)=rsin(pheta) rcos(pheta)-rsin(pheta)=0 r(cos(pheta)-sin(pheta))=0 so we have r=0 and cos(pheta)-sin(pheta)=0

30. anonymous

wow... um.. let me try to figure all this out

31. anonymous

ok

32. anonymous

There?

33. myininaya

hey i was talking to my hubby. hes gonna be home soon so I'm going to leave soon but we were done with writing 5x=5y to a polar equation

34. anonymous

yes

35. myininaya

so did you have a question or something?

36. anonymous

this might sound arrogant but, how do i set it up and how do i get the answer

37. myininaya

how you have to do is place x^2+y^2 with r and replace x with rcos(pheta) and y with rsin(pheta) and usually I tried to simplified like I did above

38. myininaya

all not how

39. myininaya

write with no x and no y

40. myininaya

your goal is to write with r and pheta

41. anonymous

ok

42. myininaya

x^2+y^2 with r^2*

43. myininaya

do you see my mistake way up there? i forgot r^2 when i wrote 5r^2=16 is equivalent to 5x^2+5y^2=16

44. anonymous

so it would be 5^2+5^2=

45. myininaya

?

46. myininaya

where did you get x=5 and y=5?

47. anonymous

oh.. crap. i got them from the original equation

48. myininaya

hey anwar is going to come back and help you i have to leave k?

49. anonymous

ok thank you very much for all ur help though :)

50. myininaya

i think you are over thinking this stuff

51. anonymous

i tend to do that a lot :P

52. myininaya

i hope you get it

53. anonymous

thank you

54. anonymous

i got the whole night, don't worry i'll get it

55. anonymous

Just gimme a minute. I'll help you when I finish with another question if you still need help.

56. anonymous

ok

57. anonymous

I don't know where you have reached. But I'll start by these formula in polar coordinates: $x=r \cos \theta .... (1)$ $y=r \sin \theta .... (2)$ $x^2+y^2=r^2 .... (3)$

58. anonymous

We're trying to convert the given equations that are in terms of x and y to an a form in terms of r and theta, which is called polar coordinates. Is that clear so far?

59. anonymous

Good, Now consider the first equation: 5x=5y. Divide both sides by 5 you get: $x=y$ But we can substitute fot x with equation (1) and for y with equation (2). That's: $x=y \implies r \cos \theta=r \sin \theta$

60. anonymous

ok

61. anonymous

Is is clear so far?

62. anonymous

Is it*

63. anonymous

For the 2nd equation: $x^2+y^2=16$ we will just apply directly the third formula, since x^2+y^2 is just r^2. Then: $x^2+y^2=16 \implies r^2=16$

64. anonymous

is this the answer

65. anonymous

Yep. You can simplify it more, but this is OK.