## Callisto 2 years ago How do the plane polar coordinates work?

1. experimentX

you know polar coordinate system right!!|dw:1349364467374:dw|

2. Callisto

Yes.. a little... But NOT in vectors!!

3. experimentX

no they are just same.|dw:1349364550790:dw|

4. Callisto

|dw:1349364634178:dw|

5. experimentX

we know that we write (x, y) as our point. In polar coordinate we write (r, theta) <--- this means this is a vector space. you have been using it unknowingly.

6. Callisto

Since it's related to sine, and cosine, it's not linear translation, right?

7. experimentX

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8. experimentX

yep. now let's go back to unit vector for a while. in cartesian how do you represent a vector?

9. Callisto

xi + yj? i and j are unit vectors

10. experimentX

yes ... and in cartesian coordinate?

11. Callisto

Huh?! Are they different??

12. experimentX

Woops sorry ... polar coordinate??

13. Callisto

Well... Isn't it what I got stuck?

14. experimentX

$r \hat r + \theta \hat \theta$

15. experimentX

now we need to find the relation between $$\hat r , \hat \theta$$ and $$\hat i, \hat j$$

16. experimentX

for that we know

17. experimentX

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18. experimentX

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19. experimentX

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20. experimentX

how do we define unit vectors?

21. Callisto

vector/length of vector?

22. experimentX

yep ...

23. experimentX

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24. experimentX

can you write the relation between $$\hat r$$ and i,j

25. Callisto

unit vector of r = (xi + yj) / (x^2 + y^2) ? I know it's strange...

26. experimentX

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27. Callisto

Oh, should I express it in terms of theta?

28. experimentX

yep

29. Callisto

unit vector of r = (xi + yj) / (x^2 + y^2) $$= \frac{(rcos) \theta i +(rsin\theta)j}{r^2cos^2\theta+r^2 sin^2 \theta}$$ $$= \frac{(rcos) \theta i +(rsin\theta)j}{r^2 }$$ Hmm.. doesn't look good, sorry!!

30. experimentX

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31. experimentX

can you check page no 14 again?

32. Callisto

Yes....

33. experimentX

you got your first relation right?

34. Callisto

Eh.... I was about to say no, but then, yes :|

35. experimentX

for second relation there are many tricks. this is my favourite. polar coordinate is orthogonal coordinate system. $\hat r \cdot \hat \theta = 0$ so change $$\theta = \theta + 90$$ you will get second relation.

36. Callisto

Hmm, sorry!! OS is not loading properly here... And... please give me sometime to work it out first.. :(

37. experimentX

sure till then i'll work out on the original method|dw:1349366641329:dw|

38. Callisto

Sorry... May I know what that e is?? I meant he wrote $$\hat e_\theta$$ or $$\hat e_r$$ there.

39. experimentX

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40. experimentX

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41. Callisto

So, they are the unit vectors?

42. experimentX

yep!!

43. experimentX

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44. Callisto

45. Callisto

PQ as a vector? or...?

46. experimentX

yep. PQ

47. experimentX

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48. experimentX

i hope you are understanding.

49. Callisto

PQ (vector) =$$r (cos(\theta +d\theta) -cos\theta )i + r (sin(\theta+d\theta) -sin\theta)j$$ Sum-to-product formula?

50. experimentX

yep .. .cos C - cos D .. and ...

51. Callisto

$PQ = r[cos(\theta+d\theta)-cos\theta] i + r[sin(\theta+d\theta)-sin\theta]j$ $=-2rsin(\theta+\frac{d\theta}{2})sin\frac{d\theta}{2}i +2rcos(\theta+\frac{d\theta}{2})sin\frac{d\theta}{2}j$

52. experimentX

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53. Callisto

Isn't that |PQ| something horrible??

54. experimentX

lol ... no. Looks like i forgot cos C - cos D

55. experimentX

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56. Callisto

57. experimentX

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58. experimentX

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59. experimentX

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60. experimentX

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61. experimentX

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62. experimentX

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63. experimentX

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64. experimentX

similarly for theta |dw:1349370057135:dw|