Work, Velocity & Force vectors Help

AI Thread Summary
The discussion centers on understanding a physics equation related to work and energy, specifically the derivation of F . dr = (MV2^2)/2 - (MV1^2)/2. Key questions include the origin of the number 2 in the denominator and the incorporation of the velocity vector into the derivative. The response suggests applying the product rule to clarify the derivation, noting that it reflects the vector equivalent of differentiating a square. The conversation also includes a request for assistance with using LaTeX for mathematical expressions. Overall, the thread emphasizes the importance of understanding derivatives in the context of classical mechanics.
Jonnyb42
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This is probably more of a math question, but it pertains to physics.

Today in Classical Mechanics the professor was showing the following,

F . dr = (MV2^2)/2 - (MV1^2)/2

but in great length, one of the steps along the way I did not understand how he did it... it was this:

screw latex, if someone could please post a link to somewhere on the forums that shows how to use latex properly, I can't use it to save my life...
here is the step with good old microsoft:

[PLAIN]http://mynqa.com/Cargo/unk.bmp

My questions are, where did the number 2 come from in the denominator? Also, how did he bring in the velocity vector into the derivative like that?

Thanks for any help.
 
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Jonnyb42 said:
screw latex, if someone could please post a link to somewhere on the forums that shows how to use latex properly, I can't use it to save my life...

See my sig :wink:

here is the step with good old microsoft:

[PLAIN]http://mynqa.com/Cargo/unk.bmp

My questions are, where did the number 2 come from in the denominator? Also, how did he bring in the velocity vector into the derivative like that?

Thanks for any help.

Just apply the product rule to the right-hand side of the equation...
 
Last edited by a moderator:
It is essentially the vector equivalent of the derivative of a square.

d/dt(x2)=2xdx/dt
 
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