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anonymous

  • 5 years ago

Convert r=3sin(theta) into a cartesian equation.

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  1. anonymous
    • 5 years ago
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    \[r=\sqrt (x^2+y^2)\] \[\tan^{-1} (y/x)=\theta\]

  2. anonymous
    • 5 years ago
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    Put them in place, and try simplifying, the resulting equation will be in terms of x and y

  3. nowhereman
    • 5 years ago
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    No need to make it so complicated for θ. You should know that polar coordinates mean \[\sin θ = y\]

  4. anonymous
    • 5 years ago
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    rsin(theta) = y. I just need to know how to deal with the r being on the left side of the equation basically.

  5. anonymous
    • 5 years ago
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    Good point @nowhereman

  6. nowhereman
    • 5 years ago
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    a, right forgot that r

  7. nowhereman
    • 5 years ago
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    you can divide it away then

  8. anonymous
    • 5 years ago
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    In r you can put r=(√x^2+y^2)

  9. anonymous
    • 5 years ago
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    Ok so you use the equation of a circle? Not sure exactly where that's coming from...

  10. anonymous
    • 5 years ago
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    \[x^2+y^2=3y\]

  11. anonymous
    • 5 years ago
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    Sorry, you got to square the right side

  12. nowhereman
    • 5 years ago
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    That would just be the definition of polar coordinates. \[r = 3\sinθ ⇔ r = 3ry ⇔ 1 = 3y ∨ r = 0 ⇔ y = \frac 1 3 ∨ x = y = 0\]

  13. anonymous
    • 5 years ago
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    Kinda abstract...

  14. nowhereman
    • 5 years ago
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    yeah, that was math is about.

  15. anonymous
    • 5 years ago
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    Yeah I understand that but the pathetic American Education System doesn't go into the abstract much. So call me dumb but that's how it is. lol

  16. amistre64
    • 5 years ago
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    the graph of this thing is a circle centered on the yaxis stretching from the origin to the y=3 mark...

  17. amistre64
    • 5 years ago
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    it should end up as: x^2 + (y -1/2)^2 = 3/2 if i see it right :)

  18. nowhereman
    • 5 years ago
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    mmh, yeah I put the r on the wrong side... guess I'm not awake after all

  19. amistre64
    • 5 years ago
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    well (3/2)^2 :)

  20. anonymous
    • 5 years ago
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    Amistre, what was your process for getting that answer?

  21. amistre64
    • 5 years ago
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    seeing the graph in my head really :) the algebra is this: r = 3sin(t) ; * r r^2 = 3rsin(t) ; r^2 = x^2 + y^2 ; and rsin(t) = y x^2 + y^2 = 3y ; -3y x^2 +y^2 -3y = 0 ; complete the square x^2 +y^2 -3y + (3/2)^2 = (3/2)^2 ; clean up the square.. x^2 +(y-3/2)^2 = (3/2)^2

  22. amistre64
    • 5 years ago
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    i was off by a y -1/2 the first time lol....but then again i am not right in the head ;)

  23. anonymous
    • 5 years ago
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    Ok I think I'm beginning to see it. Thanks.

  24. amistre64
    • 5 years ago
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    any ?s speak up ;)

  25. anonymous
    • 5 years ago
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    I may come up with another one in a minute. Just digesting it right now.

  26. amistre64
    • 5 years ago
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    at sin(0) we are at the origin; at sin(90) we are at 3; then we swing back to 0 on our way thru and that makes the circle that sits on the origin stretched to y=3 in my head

  27. amistre64
    • 5 years ago
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    center at (0,3/2)

  28. amistre64
    • 5 years ago
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    but yet i cant seem to get how to turn parametric equations into cartesian equation lol

  29. anonymous
    • 5 years ago
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    I think I typed the right answer when I typed \[x^2+y^2=3y\] didn't I?

  30. amistre64
    • 5 years ago
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    iam; that was a good start yes :), just needed to turn it into the circle equation

  31. anonymous
    • 5 years ago
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    I am not getting you, if we just write it as \[x^2+y^2-3 y = 0 \], then it represents a circle doesn't it?

  32. amistre64
    • 5 years ago
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    been wondering how to make money doing this.... got $60 left in the bank.... :/

  33. amistre64
    • 5 years ago
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    that form is not the standard for a circle equation.... just needs dressed up a bit ;)

  34. anonymous
    • 5 years ago
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    Huh? What do you mean?

  35. anonymous
    • 5 years ago
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    "been wondering how to make money doing this.... got $60 left in the bank.... :/"??

  36. amistre64
    • 5 years ago
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    just broke, thats all :)

  37. anonymous
    • 5 years ago
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    I know some methods, but those are not similar to these

  38. amistre64
    • 5 years ago
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    :) I saw something today about a cat.... but it looked used up...

  39. amistre64
    • 5 years ago
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    id be great if I could get a job tutoring at the college..

  40. anonymous
    • 5 years ago
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    I can help you if you want to tutor people online by getting paid at the same time

  41. amistre64
    • 5 years ago
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    thatd be good if it works out :)

  42. amistre64
    • 5 years ago
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    i got aloooottt of time, and aloooott of math in me lol

  43. anonymous
    • 5 years ago
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    Are you serious about it?

  44. anonymous
    • 5 years ago
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    I mean do you really want to do it?

  45. amistre64
    • 5 years ago
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    yeah, this experience here has taught me that I am actually good at it :)

  46. anonymous
    • 5 years ago
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    You are really good

  47. anonymous
    • 5 years ago
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    So if you are interested, I can help you start, but you got to email me when you wish to begin

  48. amistre64
    • 5 years ago
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    tony031172@gmail.com is my junkmmail catcher, I check it still, but it keeps my good email from the clutter...

  49. anonymous
    • 5 years ago
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    I have your email address, did you forget?

  50. amistre64
    • 5 years ago
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    lol.... just wasnt sure ;)

  51. anonymous
    • 5 years ago
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    I have your real email address a*******@*****.com

  52. amistre64
    • 5 years ago
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    :) ill hit you up tomorrow, my borrowed time i up here at the library :/

  53. anonymous
    • 5 years ago
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    Sure, I hope I am disturbing all of them who are sitting here

  54. anonymous
    • 5 years ago
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    Bye

  55. amistre64
    • 5 years ago
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    Ciao :)

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