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Vector calc derivation

  • Thread starter srvs
  • Start date
  • #1
31
0
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". :(
 

Answers and Replies

  • #2
Cyosis
Homework Helper
1,495
0
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]
 
  • #3
dx
Homework Helper
Gold Member
2,011
18
[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:
  • #4
31
0
Shame on me. Thank you.
 

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