Question

the rate at which a spherical pill(table) dissolves in the body after being consumed is described by the rate of change of its volume with time dv/dt. given that the rate of change of volume is directly proportional to its surface area, compute the time in which 80% of a pill of size 4cm would dissolve, given that it dissolve completely in 2hrs.

Answer 1

             V = 4π R³ / 3  of a spherical pill
           dV/ dt = 4 π R²  = surface area of sphere
             let Z be a constant of proportionality.

Given  the size = diameter at t = 0 is,  4 cm.  So radius = 2 cm.
Volume  at t = 0, is  V₀ = 4 π (2)³ / 3 =  32π/3  cm³
Rate of change of Volume = d V / dt  = - Z * 4π R²  = - Z * S

[tex]R^3=\frac{3V}{4 \pi},\ \ S=4\pi (\frac{9V^2}{16 \pi^2})^{\frac{1}{3}}=k\ V^{\frac{2}{3}},\ k=4.836..\\\\\frac{dV}{dt}=-Zk\ V^{\frac{2}{3}},\ \ \ \frac{dV}{V^{\frac{2}{3}}}=-Z*k\ dt\\\\Integrating,\ \ 3V^{\frac{1}{3}}=C-Z*k\ t + C\\\\So,\ \ C=3*(\frac{32 \pi}{3})^{\frac{1}{3}}=9.672\ cm\\\\80\ percent\ V_0\ dissolves\ at\ t. \ V=0.20\ V_0\\\\(\frac{V}{V_0})^{\frac{1}{3}}=\frac{C-kt}{C}\\\\=t=\frac{C}{Z*k}*[ 1-\frac{V^3}{V_0^3} ]\\\\t=2\ hrs\ for\ V=0\\\\Z=1\ unit,\ \frac{C}{k}=2\\\\t=2[1-\sqrt[3]{0.20}]\ hours\\\\=0.83\ hrs[/tex]

Time to dissolve 80% of the pill = 0.83 * 60 = 49.8 minutes.