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Vector calc derivation

  1. May 12, 2009 #1
    Hi,

    This is probably really simple but I can't figure out how they go from eq. 2 to eq. 3 here:
    http://img179.imageshack.us/my.php?image=vcalc.jpg

    First term I see, however the - mu / r I don't. Shouldn't this be + (1/2) mu / r?

    Asked my professor but he said "just do the derivations". :(
     
  2. jcsd
  3. May 12, 2009 #2

    Cyosis

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    Homework Helper

    Your professor was right. Lets take the second term of the second equation first.

    [tex]
    \frac{1}{2} \frac{\mu}{r^3}\frac{d (\vec{r} \cdot \vec{r})}{dt}=\frac{1}{2} \frac{\mu}{r^3}\frac{d r^2}{dt}=\frac{1}{2}\frac{\mu}{r^3} 2 r \frac{dr}{dt}=\frac{\mu}{r^2}\frac{dr}{dt}
    [/tex]

    Now the second term of the third equation.

    [tex]
    \frac{d}{dt}\frac{-\mu}{r}=\frac{-1}{r^2}\frac{d-\mu r}{dt}=\frac{\mu}{r^2}\frac{dr}{dt}
    [/tex]
     
  4. May 12, 2009 #3

    dx

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    [tex] \frac{1}{2} \frac{\mu}{r^3} \frac{d}{dt}r\cdot r = \frac{1}{2} \frac{\mu}{r^3} \frac{d}{dt}r^2 = \frac{1}{2} \frac{\mu}{r^3} 2r \frac{dr}{dt} = \frac{\mu}{r^2}\frac{dr}{dt} = -\mu \frac{dr^{-1}}{dt} = \frac{d}{dt} \frac{-\mu}{r} [/tex]

    EDIT: Cyosis beat me to it!
     
    Last edited: May 12, 2009
  5. May 12, 2009 #4
    Shame on me. Thank you.
     
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