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Polar coordinates: e_r and e_theta

  1. Mar 13, 2009 #1
    1. The problem statement, all variables and given/known data[/b]

    Let e_r=(cos[tex]\theta[/tex],sin[tex]\theta[/tex]) and e_theta=(-sin[tex]\theta[/tex],cos[tex]\theta[/tex]).
    Let P(r,[tex]\theta[/tex]) be a point with e_r and e_theta at that point.
    What can you say about the three quantities (e_r, e_theta and the point P) as r and [tex]\theta[/tex] vary?

    2. Relevant equations

    r: distance from origin
    [tex]\theta[/tex]: angle

    3. The attempt at a solution

    As r moves around, the e_r and e_theta change place. As r increases or decreases e_r and e_theta don't change place but as theta changes, they do change place since e_theta is always orthogonal to point P.

    I feel like I'm not putting enough information. Is there something i didn't mention?
    Thank you.
     
  2. jcsd
  3. Mar 13, 2009 #2

    djeitnstine

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    Yes there is some information missing. You have to note that as r changes, [tex]\theta[/tex] changes with the rate that r changes, because [tex]e_{\theta}[/tex] is the derivative of [tex]e_{r}[/tex]
     
  4. Mar 13, 2009 #3
    What do you mean, I dont understand why.
     
  5. Mar 14, 2009 #4

    djeitnstine

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    Ok, starting with a (position) vector function, (I will use the cartesian coordinate system to make things appear obvious) lets say [tex]v=t^{2}i+2tj[/tex] The derivative is [tex]v'=2ti+2j[/tex]. Then how does [tex]v[/tex] vary with [tex]v'[/tex]? Since [tex]v'[/tex] is the derivative of [tex]v[/tex], then [tex]v'[/tex] must vary with [tex]v[/tex]'s rate of change. Looking at your position functions, [tex]-\sin{(\theta)}[/tex] is the derivative of [tex]\cos{(\theta)}[/tex]. Likewise for [tex]sin{(\theta)}[/tex] and [tex]cos{(\theta)}[/tex]
     
  6. Mar 14, 2009 #5
    Oh right, i see.
    Thank you
     
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