What is the Derivation of Vector Calculus Equations?

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Homework Help Overview

The discussion revolves around the derivation of equations in vector calculus, specifically transitioning from one equation to another as illustrated in a provided image link. Participants are examining the terms involved and their transformations.

Discussion Character

  • Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster questions the transition between specific terms in the equations, particularly the sign and coefficient of a term involving mu and r. Some participants attempt to clarify the derivation steps for the second term of the equations, while others provide their interpretations of the transformations involved.

Discussion Status

Participants are actively engaging with the derivation process, with some providing detailed breakdowns of the terms and their derivatives. There is acknowledgment of contributions, indicating a collaborative effort to clarify the reasoning behind the equations.

Contextual Notes

The original poster mentions a lack of clarity from their professor, who suggested focusing on the derivations. This context highlights the challenge of understanding the mathematical transformations without direct instruction.

srvs
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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". :(
 
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Your professor was right. Let's 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]
 
[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:
Shame on me. Thank you.
 

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