Derivative of a vector expression?

quarky2001
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I'm trying to find the time derivative of the following function, where E is a constant, spatially uniform vector field, and B is as well, but B varies with time.

<br /> \frac{d}{dt}\left(\frac{\vec{E}\times\vec{B}}{B^2}\right)<br />

Remembering that B is time dependent and E is not, I've calculated the derivative of this as (where a dot above the letter indicates a time derivative):

<br /> \frac{1}{B^4}\left[B^2(\vec{E}\times\dot{\vec{B}})-2B\dot{B}(\vec{E}\times\vec{B})\right]<br />

Can anyone tell me if this is correct?
 
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That looks fine to me, assuming you mean a dot product between B and its derivative.
 
Whoops, forgot that! Thanks for checking it though. It's a shame it doesn't simplify more, but so long as it's correct I'm fine with it.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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