I have a cube of size $h$, which is 'accreting' at a fixed rate $R$ - i.e. volume ($V$) is added in proportion to the surface area ($A$) of the cube, where $R$ is in units of $\frac{V}{A t}$.
I know the initial size of the cube ($h_0$) and the final size of the cube ($h_1$), and the elapsed time ($t_1$) - how do I calculate $R$?
The system is described by:
$$ \frac{dV}{dt} = R A $$
Because the structure is a cube, I know that:
$$ V = h^3 \\ A = 6 h^2 = 6 V ^{2/3} $$
Thus: $$ \frac{dV}{dt} = R 6 V^{2/3} $$
And that's where I get stuck... any suggestions very welcome!
In case context makes a difference, I'm trying to estimate the growth rate of a cubic crystal normalised to its surface area (i.e. $mol\ m^{-2}\ s^{-1}$).
Never mind, I found a solution:
$$ \frac{dV}{dT} = R\ 6 V^{2/3} \\ \frac{1}{6 V^{2/3}} dV = R\ dt \\ \int{\frac{1}{6 V^{2/3}} dV} = \int{R\ dt} \\ \frac{V^{1/3}}{2} + C = R t + C \\ V^{1/3} = 2 R t + C \\ V = (2 R t + C)^3 \\ R = \frac{V^{1/3} - C}{2t} $$
What is C? We know that when $t = 0$, $V = V_0$, so: $$ V = (2 R t + C)^3 \\ V_0 = C^3 \\ C = V_0^{1/3} $$ so: $$ R = \frac{V^{1/3} - V_0^{1/3}}{2t} $$