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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


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    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.
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