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Solving cylindrical coordinates system, just want to check my answer

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data
    .. Here is the question;

    In cylindrical coordinate system ,
    (a) If r = 2 meters , [itex]\varphi[/itex] = 35° , z = 1 meter , what are x,y,z?
    (b) if (x,y,z) = (3,2,4) meters, what are (r, [itex]\varphi[/itex], z)

    2. Relevant equations
    x = r cos [itex]\varphi[/itex]
    y = r sin [itex]\varphi[/itex]
    z = z

    r = [itex]\sqrt{(x)^{2}+(y)^{2}}[/itex]
    [itex]\varphi[/itex]= [itex]tan^{-1}[/itex] [itex]\frac{y}{x}[/itex]
    z=z


    3. The attempt at a solution

    here is my answer, i just want to know if i'm correct :))

    for a.

    x = r cos [itex]\varphi[/itex]
    = 2 cos 35
    = 1.64

    y = r sin [itex]\varphi[/itex]
    = 2 sin 35°
    = 1.15

    z = z
    z = 1


    for b.


    r = [itex]\sqrt{(x)^{2}+(y)^{2}}[/itex]
    = [itex]\sqrt{(3)^{2}+(2)^{2}}[/itex]
    = 3.6 ≈ 4


    [itex]\varphi[/itex]= [itex]tan^{-1}[/itex] [itex]\frac{y}{x}[/itex]
    = [itex]tan^{-1}[/itex] [itex]\frac{2}{3}[/itex]
    = 33.69° ≈ 34°

    z = z
    z = 4
     
    Last edited: Jan 30, 2012
  2. jcsd
  3. Jan 30, 2012 #2
    Yep, looks good. If you want to check yourself, you can always draw it out in both coordinate systems (they should be close to the same place if the graphs are drawn hastily).
     
  4. Jan 30, 2012 #3
    thanks.. I was just confused because the other books I saw use [itex]\rho[/itex] instead of r.. i thought i still need to do something to r to make it [itex]\rho[/itex] .. hehe.. Thanks. :)))
     
  5. Jan 30, 2012 #4
    I always used ρ for spherical coordinates and r for the xy-plane. But as long as your equations are consistent with what you are trying to do, it doesn't matter if you draw a little duckie as a variable.
     
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