1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question about converting x = pi/6 to polar form

  1. Feb 4, 2010 #1
    1. The problem statement, all variables and given/known data
    Convert t = pi/6 into polar form.


    2. Relevant equations
    x = r*cos(t)
    y = r*sin(t)



    3. The attempt at a solution
    t = pi/6
    cos(t) = cos(pi/6)
    cos(t) = sqrt(3)/2
    x/r = sqrt(3)/2
    sqrt(3)r = 2x
    sqrt(3)sqrt(x^2 + y^2) = 2x
    sqrt(3x^2 + 3y^2) = 2x
    3x^2 + 3y^2 = 4x^2
    0 = x^2 - 3y^2

    However, the book says to use tangent instead, as shown below:

    t = pi/6
    tan t = sqrt(3)/3
    y/x = sqrt(3)/3
    sqrt(3)x - 3y = 0

    This is the correct answer. However, I do not understand why the answer I obtained is incorrect. Is it legal to cosine both sides as I did?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 4, 2010 #2

    Mentallic

    User Avatar
    Homework Helper

    Actually the answer you got and the correct answer are nearly the same.

    [tex]\sqrt{3}x-3y=0[/tex]

    [tex]x-\sqrt{3}y=0[/tex]

    [tex]x=\sqrt{3}y[/tex]

    [tex]x^2=3y^2[/tex]

    [tex]x^2-3y^2=0[/tex]

    which is what you had. The only problem is that when you squared both sides you have introduced more solutions which are not correct. If we go backwards such as taking the square root of both sides we get:

    [tex]x^2=3y^2[/tex]

    [tex]x=\pm \sqrt{3} y[/tex]

    and the solutions [itex]x=-\sqrt{3}y[/itex] are not correct.

    It is difficult to avoid squaring in this line you had: [tex]\sqrt{3(x^2 + y^2)} = 2x[/tex] which is why using the tangent is the best approach.
     
  4. Feb 4, 2010 #3
    Ahh, that makes a lot of sense. Thanks a lot!
     
  5. Feb 4, 2010 #4

    Mentallic

    User Avatar
    Homework Helper

    You're welcome :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook