Dot Product of a Vector with its Derivative

[Ajihood]

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 don't quite understand where this comes from. I can kind of justify the d/dt (B^2) with my knowledge of dot product but I don't see where the half comes in.

Thanks for any help.

 

[jasonRF]

Consider

<br /> \frac{\partial}{\partial t} \left( B^2\right) = \frac{\partial}{\partial t} \left( \mathbf{B \cdot B} \right)<br />
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
\mathbf{B \cdot} \frac{ \partial \mathbf{B}}{\partial t}}.jason

 

[Ajihood]

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.