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Convertion from and to spherical - cartesian

  1. Jun 9, 2012 #1
    I googled it, and it says:
    [itex]\dot{x}[/itex]=[itex]\dot{r}[/itex]sinθcos∅ + (rcosθcos∅)[itex]\dot{θ}[/itex] - (rsinθsin∅)[itex]\dot{∅}[/itex]
    .
    .
    and so on for [itex]\dot{y}[/itex] & [itex]\dot{z}[/itex]

    And then they wrote "We will also need the inverse transformation obtained by solving the equations above w.r.t [itex]\dot{r}[/itex], [itex]\dot{θ}[/itex], and [itex]\dot{∅}[/itex]

    for example they got:
    [itex]\dot{r}[/itex]=sinθcos∅[itex]\dot{x}[/itex]+sinθsin∅[itex]\dot{y}[/itex] + cosθ[itex]\dot{z}[/itex]
    .
    .
    and so on, my question how did they deduced the latter from the former equations?

    Here's the link if you want to see them clearer:
    http://www.physics.sc.edu/~yar/Phys701_2009/homework/hw9_solutions.pdf [Broken]

    Just what's the procedure?
    Thanks
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 9, 2012 #2

    tiny-tim

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    Hi M. next! :smile:
    It's just algebra (plus standard trigonometric identities) …

    for example, the r' component of the RHSs is r'(sin2θcos2∅ + sin2θsin2∅ + cos2θ) = r' :wink:
     
  4. Jun 9, 2012 #3
    Hmm.. I mean how do we get r' from (x', y', z')?
     
  5. Jun 9, 2012 #4
    I solved three equations with three unknowns and it worked but after so much effort! Thanks anyways Tim.
     
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