1. Not finding help here? Sign up for a free 30min 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!

Vector Coordinate Conversions

  1. Feb 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Transform the following vector into cylindrical coordinates and then evaluate them at the indicated points:

    [tex]\vec A = (x + y)\hat x [/tex]


    [tex]P_1 (1, 2, 3)[/tex]

    2. Relevant equations
    [tex]r = \sqrt{x^2 + y^2}[/tex]

    [tex]\phi = \tan^{-1}(\frac{y}{x})[/tex]

    [tex]z = z[/tex]

    3. The attempt at a solution
    [tex]r = \sqrt{x^2 + 0^2} = x[/tex]

    [tex]\phi = \tan^{-1}(\frac{0}{x}) = 0[/tex]

    [tex]z = z = 0[/tex]

    [tex]\vec A = x\hat r [/tex] at point [tex]P_1 (1, 2, 3) \Longrightarrow \hat r[/tex]

    Could someone please check if this is correct? There are a few more of these, but if I can do this one, then the rest are no problem. Thanks.
    Last edited: Feb 9, 2008
  2. jcsd
  3. Feb 9, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    Can I assume that [itex]\hat{x}[/itex] is the unit vector in the x direction? If so then [itex](x+ y)\hat{x}[/itex] is not a "vector", it is a "vector field"- a vector at each point in the xy-plane. At (1, 2, 3) (surprising how often that point shows up!), that is the vector [itex]3 \hat{x}[/itex], of length 3 pointing in the x-direction. That vector has no z-component. The projection of the vector <2, 0> in the direction of the <1, 2> vector will be the [itex]\hat{r}[/itex] component. <2, 0> minus that projection will be the component in the [itex]\theta[/itex] direction.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?