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OpenStudy (anonymous):
\[F(x, y, z) = {GmM \over {\sqrt{x^2+y^2+z^2}}}\]
calculate
\[\frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial y^2} + \frac{\partial^2 F}{\partial z^2}\]
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OpenStudy (dumbcow):
do you mean partial derivatives with respect to x^2, y^2, z^2 ?
OpenStudy (dumbcow):
or is it just 2nd derivatives with respect to x,y,z
OpenStudy (anonymous):
yes, I think that notation means take the second derivatives
OpenStudy (anonymous):
= 2 ( x² + y² + z²) / √(x² + y² + z²)
OpenStudy (anonymous):
I found it 0
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OpenStudy (anonymous):
it is odd
OpenStudy (anonymous):
what do you mean by odd?
OpenStudy (anonymous):
strange..
OpenStudy (anonymous):
but it coul be thoug..
OpenStudy (anonymous):
\[\frac{\partial F}{\partial x} = -\frac{xGmM}{(x^2+y^2+z^2)^{3/2}}\\\frac{\partial^2F}{\partial x^2} = \frac{GmM(2x^2-y^2-z^2)}{(x^2+y^2+z^2)^{5/2}}\]
does this look right?
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OpenStudy (anonymous):
you are right..
OpenStudy (anonymous):
and total is zero..
OpenStudy (anonymous):
It says to simplify the answer as much as possible, I guess that 0 will work :)
OpenStudy (anonymous):
I found
\[\frac{\partial^2 F}{\partial x^2}=3x^2(x^2+y^+z^2)^{-5/2}-(x^2+y^2+z^2){-3/2}\]
it is the same thing.
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OpenStudy (anonymous):
your answer is big probably zero..
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