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.
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V = 4π R³ / 3 of a spherical pill
dV/ dt = 4 π R² = surface area of sphere
let Z be a constant of proportionality.
Rate of change of Volume = d V / dt = - Z * 4π R² = - Z * S
![R^3=\frac{3V}{4 \pi},\ \ S=4\pi R^2=4\pi (\frac{9V^2}{16 \pi^2})^{\frac{1}{3}}=K_0\ V^{\frac{2}{3}}, \ where\ K_0=4.836..\\\\\frac{dV}{dt}=-Z*K_0\ V^{\frac{2}{3}},\ \ \ \frac{dV}{V^{\frac{2}{3}}}=-Z*K_0\ dt\\\\Integrating,\ \ 3V^{\frac{1}{3}}=C-Z*K_0\ t + C,\ \ \ V=\frac{1}{27}(C-Z*K_0t)^3\\\\at\ t=0,\ V=V_0=\frac{1}{3}4 \pi (2\ cm)^3=\frac{32 \pi}{3}\ cm^3\\\\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-K_0t}{C}\\\\=t=\frac{C}{Z*K_0}*[ 1-\frac{V^3}{V_0^3} ]\\\\t=2\ hrs\ for\ V=0\\\\Z=1\ unit,\ \frac{C}{K_0}=2\\\\t=2[1-\sqrt[3]{0.20}]\\\\=0.83\ hrs\ units](https://tex.z-dn.net/?f=R%5E3%3D%5Cfrac%7B3V%7D%7B4+%5Cpi%7D%2C%5C+%5C+S%3D4%5Cpi+R%5E2%3D4%5Cpi+%28%5Cfrac%7B9V%5E2%7D%7B16+%5Cpi%5E2%7D%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%3DK_0%5C+V%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%2C+%5C+where%5C+K_0%3D4.836..%5C%5C%5C%5C%5Cfrac%7BdV%7D%7Bdt%7D%3D-Z%2AK_0%5C+V%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%2C%5C+%5C+%5C+%5Cfrac%7BdV%7D%7BV%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%7D%3D-Z%2AK_0%5C+dt%5C%5C%5C%5CIntegrating%2C%5C+%5C+3V%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%3DC-Z%2AK_0%5C+t+%2B+C%2C%5C+%5C+%5C+V%3D%5Cfrac%7B1%7D%7B27%7D%28C-Z%2AK_0t%29%5E3%5C%5C%5C%5Cat%5C+t%3D0%2C%5C+V%3DV_0%3D%5Cfrac%7B1%7D%7B3%7D4+%5Cpi+%282%5C+cm%29%5E3%3D%5Cfrac%7B32+%5Cpi%7D%7B3%7D%5C+cm%5E3%5C%5C%5C%5CSo%2C%5C+%5C+C%3D3%2A%28%5Cfrac%7B32+%5Cpi%7D%7B3%7D%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%3D9.672%5C+cm%5C%5C%5C%5C80%5C+percent%5C+V_0%5C+dissolves%5C+at%5C+t.+%5C+V%3D0.20%5C+V_0%5C%5C%5C%5C%28%5Cfrac%7BV%7D%7BV_0%7D%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%3D%5Cfrac%7BC-K_0t%7D%7BC%7D%5C%5C%5C%5C%3Dt%3D%5Cfrac%7BC%7D%7BZ%2AK_0%7D%2A%5B+1-%5Cfrac%7BV%5E3%7D%7BV_0%5E3%7D+%5D%5C%5C%5C%5Ct%3D2%5C+hrs%5C+for%5C+V%3D0%5C%5C%5C%5CZ%3D1%5C+unit%2C%5C+%5Cfrac%7BC%7D%7BK_0%7D%3D2%5C%5C%5C%5Ct%3D2%5B1-%5Csqrt%5B3%5D%7B0.20%7D%5D%5C%5C%5C%5C%3D0.83%5C+hrs%5C+units "R^3=\frac{3V}{4 \pi},\ \ S=4\pi R^2=4\pi (\frac{9V^2}{16 \pi^2})^{\frac{1}{3}}=K_0\ V^{\frac{2}{3}}, \ where\ K_0=4.836..\\\\\frac{dV}{dt}=-Z*K_0\ V^{\frac{2}{3}},\ \ \ \frac{dV}{V^{\frac{2}{3}}}=-Z*K_0\ dt\\\\Integrating,\ \ 3V^{\frac{1}{3}}=C-Z*K_0\ t + C,\ \ \ V=\frac{1}{27}(C-Z*K_0t)^3\\\\at\ t=0,\ V=V_0=\frac{1}{3}4 \pi (2\ cm)^3=\frac{32 \pi}{3}\ cm^3\\\\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-K_0t}{C}\\\\=t=\frac{C}{Z*K_0}*[ 1-\frac{V^3}{V_0^3} ]\\\\t=2\ hrs\ for\ V=0\\\\Z=1\ unit,\ \frac{C}{K_0}=2\\\\t=2[1-\sqrt[3]{0.20}]\\\\=0.83\ hrs\ units")
time to dissolve 80% of the pill = 0.83 * 60 = 49.8 minutes
dV/ dt = 4 π R² = surface area of sphere
let Z be a constant of proportionality.
Rate of change of Volume = d V / dt = - Z * 4π R² = - Z * S
time to dissolve 80% of the pill = 0.83 * 60 = 49.8 minutes
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