Understanding the Dot Product Formula: An Explanation and Example

AI Thread Summary
The discussion centers on understanding the dot product in the context of the equation F = mv' = P'. The confusion arises from the relationship v ⋅ dv/dt and its equivalence to 1/2 d(v²)/dt, which is clarified through the product rule of differentiation. Participants explain that differentiating v², represented as v ⋅ v, yields the expression v' ⋅ v + v ⋅ v', helping to clarify the initial misunderstanding. The conversation emphasizes the importance of recognizing the product rule in this context for better comprehension. Overall, the thread provides insights into the mathematical principles underlying the dot product and its application in physics.
Seung Ju Yoo
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In a book I was reading, it says
F=mv'=P'

so dot producting on both sides with v

Fv = mv ⋅ dv/dt = 1/2 m d(v2)/dt = d(1/2 m v^2)/dtI really don't get how v ⋅ dv/dt = 1/2 d(v2)/dt.
I have seen few threads and they say it's about product rule, but they don't really explain in detail.

Could anyone help me with this?
 
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Seung Ju Yoo said:
In a book I was reading, it says
F=mv'=P'

so dot producting on both sides with v

Fv = mv ⋅ dv/dt = 1/2 m d(v2)/dt = d(1/2 m v^2)/dtI really don't get how v ⋅ dv/dt = 1/2 d(v2)/dt.
I have seen few threads and they say it's about product rule, but they don't really explain in detail.

Could anyone help me with this?

Welcome to PF!

##v^2 = \textbf{v.v}##

Can you now differentiate that equation?
 
V^2 = V . V

And the time derivative of it is V' . V + V . V'
 
PeroK said:
Welcome to PF!

##v^2 = \textbf{v.v}##

Can you now differentiate that equation?
jedishrfu said:
V^2 = V . V

And the time derivative of it is V' . V + V . V'
Oh.. I see. I did not now that d(x ⋅ y)/dt = x' ⋅ y + x ⋅ y'

Knowing this, going right from left is easy, but I guess going left to right needs some practice to spot!

Thank you both peroK and Jedishrfu!
 
In Cartesian component form, what is ##\vec{v}\centerdot d\vec{v}##?

Chet
 
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