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Can anyone help me with this problem?

Let the density per unit of volume in a cubical box of side length 2 vary directly as the distance from the center and inversely as $1+t^{2}$ where $t$ is the time. If the density at a corner of the box is 1 when $t=0$, find a formula for the density at any point at any time. What is the rate of change of the density at a point $\frac{1}{2}$ unit from the center of the box at time $t = 1$ ?

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  • $\begingroup$ ''Let the density per unit of volume in a cubical box of side length 2 directly as the distance from the center and inversely as 1+t2 where t is the time. ''Is this correctly stated? $\endgroup$ – Rafael Wagner Sep 21 '17 at 22:13
  • $\begingroup$ Excuse me, you're right, I wrote the statement wrong. $\endgroup$ – fer6268 Sep 21 '17 at 22:16
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If you define the density of your box as $u = u(r, \theta, \phi, t)$ then the description seems to suggest the differential equation; $$ \frac{\partial u}{\partial t} = \frac{r}{1+t^2}, \; u(\sqrt{3},.,.,0) = 1$$ So, $$ u = r (tan^{-1}t + \frac{1}{\sqrt{3}})$$ The rate of change of density at r=0.5 and t=1 is given by the differential equation and equals 0.25.

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