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Dot Product of a Vector with its Derivative |
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| Apr14-11, 06:07 PM | #1 |
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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. |
| Apr14-11, 06:17 PM | #2 |
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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 |
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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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