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Dot Product of a Vector with its Derivative

 
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Apr14-11, 06:07 PM   #1
 

Dot Product of a Vector with its Derivative


Hi Guys,

I am doing an Electromagnetism course at uni and we just derived Poynting's Theorem in class. However, he left steps for us to fill in and that is why I have a question.

In the derivation we get the dot product:

B dot dB/dt

and

E dot dE/dt

where both B and E are vectors. The answer that was given was

B dot dB/dt = 1/2 d/dt(B^2)

I dont quite understand where this comes from. I can kind of justify the d/dt (B^2) with my knowledge of dot product but I dont see where the half comes in.

Thanks for any help.
 
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Apr14-11, 06:17 PM   #2
 
Consider

[tex]
\frac{\partial}{\partial t} \left( B^2\right) = \frac{\partial}{\partial t} \left( \mathbf{B \cdot B} \right)
[/tex]
If you don't know what to do with the dot product, you can expand it out. You should find that you are really just taking the derivative of a sum of three scalar functions. Use the standard differentiation rules for sums and products, then figure out how to re-writeyour answer in terms of
[tex] \mathbf{B \cdot} \frac{ \partial \mathbf{B}}{\partial t}} [/tex].


jason
 
Apr14-11, 06:49 PM   #3
 
Ah I see now. I forgot I had two functions that I had to use the product rule on. Silly me!

Thanks for the help Jason.
 
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