# Vector calc derivation

1. May 12, 2009

### srvs

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. May 12, 2009

### Cyosis

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

$$\frac{1}{2} \frac{\mu}{r^3}\frac{d (\vec{r} \cdot \vec{r})}{dt}=\frac{1}{2} \frac{\mu}{r^3}\frac{d r^2}{dt}=\frac{1}{2}\frac{\mu}{r^3} 2 r \frac{dr}{dt}=\frac{\mu}{r^2}\frac{dr}{dt}$$

Now the second term of the third equation.

$$\frac{d}{dt}\frac{-\mu}{r}=\frac{-1}{r^2}\frac{d-\mu r}{dt}=\frac{\mu}{r^2}\frac{dr}{dt}$$

3. May 12, 2009

### dx

$$\frac{1}{2} \frac{\mu}{r^3} \frac{d}{dt}r\cdot r = \frac{1}{2} \frac{\mu}{r^3} \frac{d}{dt}r^2 = \frac{1}{2} \frac{\mu}{r^3} 2r \frac{dr}{dt} = \frac{\mu}{r^2}\frac{dr}{dt} = -\mu \frac{dr^{-1}}{dt} = \frac{d}{dt} \frac{-\mu}{r}$$

EDIT: Cyosis beat me to it!

Last edited: May 12, 2009
4. May 12, 2009

### srvs

Shame on me. Thank you.

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