## LordVictorius 2 years ago Explain polar coords and give an example.

1. Ajk

With rectangular coords we typically represent locations as the intersection of lines on a horizontal/vertical axes. With polar coords, we take the magnitude of the x and y parts that we are used to on a cartesian coord system. so if we had a point (3,4), to find the magnitude we would just that the square root of (3^2 + 4^2). Just think of it like finding the hypotenuse if you were doing pythagorean theoram. What you now have, in polar, is rhe "r", which is just the length of the ray (line) from the origin (0,0). But, we also need something else called the argument, which is the angle (or theta) from the x-axis. If we had some point on the cartesian plane, like (1,1) then it would have an "r" of sqrt(2), and an argument of pi/4. I would write it like sqrt(2)[cos(pi/4) + sin(pi/4)]. You might also be able to write it like sqrt(2)*e^i*(pi/4).

2. LordVictorius

What is the Cartesian Plane?

3. Ajk

just the normal x,y coords that we are used to.

4. LordVictorius

Okay.

5. LordVictorius

Where did you get the r's from (1,1)?

6. Ajk

if I take the sqrt of 1^2 + 1^2, I have sqrt(2). Just think of it like finding the hypotenuse of a right triangle.

7. LordVictorius

So if this (1,1) was (x,y), then it is x^2+y^2?

8. LordVictorius

Or the sqrt(x^2+y^2)?

9. Ajk

with x,y coords we look for whee they intersect and that is the point. with polar, we have a ray that extends from the origen. the length of that ray is found just by the same way that you would find the hyp of a right triangle.

10. Ajk

sqrt(x^2 + y^2)

11. LordVictorius

Oh! Okay, thanks!

12. Ajk

remember, that just gives you the length of the ray from the origin. You still need to include an angle, or else we would have no idea where the endpoint of the ray is.

13. LordVictorius

Wait, so how do I get the angle?

14. Ajk

You will likely have to use your trig and the length of the x (cosine) and y (sine) parts to determine angle. If you have an angle that can be represented at y/x, then you also have something that can be represented as tangent theta, right? (because sin/cos). So once you have that, then you take the arctan (y/x) and that will give you the angle.

15. Ajk

so for the example I used above, if I have x = 1 and y = 1, then what I really have is y/x = 1/1 = 1. So I take tangent theta = 1, ----> theta = arctan (or tan inverse) (1), and that is pi/4 (or 45 degrees if you are in degree mode)

16. LordVictorius

Can you help me with this problem? I think it might be simpler to explain the steps. Use your grapher to determine which of the graphs matches the polar equation r = 3 sin 2θ.

17. Ajk

what is the theta?

18. LordVictorius

IT says 3 sin 2(theta)

19. Ajk

Without rectangular coords to convert to polar, or a polar coord to convert back to rectangular, I'm not really sure what to do with that. Where are the graphs the question is asking you to compare it to?

20. LordVictorius

21. Ajk

yeah

22. LordVictorius

23. Ajk

ah, ok. hold on, let me plot it.

24. Ajk

it is the third option

25. LordVictorius

How did you get that?

26. Ajk

do you have a graphing calculator? On mine, I switched the graph-type from rectangular to polar, then I just entered what you gave me: 3sin(2theta). It plotted the 3rd option.

27. LordVictorius

Oh, okay.

28. LordVictorius

Can you do another one? Find the rectangular coordinates of the point with the polar coordinates (8,3/2*pi ).

29. Ajk

My calculator isn't cooperating, but lets see if we can reason the answer. We know r=8, and we have a theta of 3/2*pi. So I think then that the point is on the y-axis below zero. We have a problem trying to use tangent right away, because cosine of (3/2*pi) = 0, that means tangent give a divide by zero error. Even so, we know the ray is length 8, so I think the rectangular coords are (0,-8)

30. LordVictorius

Okay,

31. LordVictorius

32. LordVictorius

Wait, can I try one, and tell me if it's correct/

33. Ajk

Sure

34. LordVictorius

Find the rectangular coordinates of the point with the polar coordinates . This is the question.

35. LordVictorius

(-3, 5pi/8)

36. LordVictorius

Those are the coords.

37. LordVictorius

So r=-3, right?

38. LordVictorius

Agh, I'm confused.

39. Ajk

r = -3, but remember that r is sqrt(x^2 + y^2). We get the negative sign because 5/8*pi is in quadrant II, which means the cosine (x) part is a negative number.

40. LordVictorius

So I will need a unit circle, right?

41. Ajk

That is probably what I would do, look at tan(5/8pi) first

42. LordVictorius

It's not on the unit circle.

43. Ajk

if you divide pi (the top half of the circle) into 8, you can see that 5/8pi is just a little past pi/2

44. LordVictorius

So pi/2?

45. LordVictorius

0,1

46. Ajk

what I did to find x was 3cos(5/8*pi), and to find y was 3sin(5/8*pi). I had to round off a bit, but I got x = 1.15 and y = -2.78. Just to doublecheck, I did sqrt((1.15)^2 + (-.2.78)^2) and it did come back as 3.

47. Ajk

Since 5/8pi is in quadrant 2, I know the x value must be negative, and the y value will be positive.

48. LordVictorius

Can you tell me the equation you useed>

49. Ajk

I only used what you gave me. So, my reasoning went like this: first, locate what quadrant 5/8*pi is located. It is in Q2, so we are dealing with a negative x and a positive y. Then, to find x, I just entered 3cos(5/8*pi) into my calculator and got -1.15 (rounded). To find y, I entered 3sin(5/8*pi) into my calculator, and it gave me 2.78 (rounded). So with that I now have my x and y values to plot.

50. LordVictorius

So for r, you used x(sin or cos) of y, right?

51. Ajk

Yes, but I just thought of something. If that r is -3, then I should have used it in my computations for x and y, which would have switched the signs of both x and y. I think we need to change my answer to 1.15 and -2.78, sorry about that.

52. LordVictorius

So x =-2.78?

53. Ajk

nope, x = 1.15, and y = -2.78. Cosine gives us the x part, and Sine gives us the y part.

54. LordVictorius

Wait, I thought x was negative.

55. Ajk

it was, but look at that r. r = -3, so we have to include that in our calculation. If we have some negative value for x and multiply by -3, the answer will turn positive, right?

56. LordVictorius

Oh, okay.

57. LordVictorius

But I don't have that answer choice.

58. Ajk

same for y. It would be positive, but times -3 it turns negative.

59. Ajk

what are the options?

60. LordVictorius

I only have (-3/2,3/2) (3sqrt(3)/2),-3/2) (-3/2, 3sqrt(3)/2) and 3/2, -3/2

61. LordVictorius

Hello?

62. LordVictorius

I need to leave in half an hour, with the assignment done and Istill dont understand polar coords. I'm osrry, but I need a brief descriptiion that wil help get my assignment done.

63. LordVictorius

64. Ajk

It looks like there is probably a much easier/better way to do this problem than how I've been working on it. You might be better off spending that 30 minutes reviewing the textbook or getting help from someone better at this than I am.

65. LordVictorius

Soeey Ajk, maybe someome else can explani it better.

66. LordVictorius

Thanks anyway.

67. LordVictorius

:(

68. LordVictorius

@cwrw238

69. LordVictorius

@dpaInc

70. Ajk

I'd be interested to see an explanation, if someone else helps. I'll check back and hopefully learn something too.

71. LordVictorius

@satellite73

72. LordVictorius

@dpaInc