# A common assumption in modeling drug assimilation is that the blood volume in a person is a single compartment that behaves like a stirred tank. Suppo

Modeling
A common assumption in modeling drug assimilation is that the blood volume in a person is a single compartment that behaves like a stirred tank. Suppose the blood volume is a four-liter tank that initially has a zero concentration of a particular drug. At time $$\displaystyle{t}={0}$$, an intravenous line is inserted into a vein (into the tank) that carries a drug solution with a concentration of 500 mg/L. The inflow rate is 0.06 L/min. Assume the drug is quickly mixed thoroughly in the blood and that the volume of blood remains constant. a. Write an initial value problem that models the mass of the drug in the blood, for $$\displaystyle{t}\ \geq\ {0}$$. Solve the initial value problem, and graph both the mass of the drug and the concentration of the drug. c. What is the steady-state mass of the drug in the blood? d. After how many minutes does the drug mass reach 90% of its steady-state level?

2021-01-03

Step 1 a) Use stirred tank model with constants $$\displaystyle{V}={4},{R}={0.06},{C}_{{{i}}}={500},{m}_{{{0}}}={0}$$
$$\displaystyle{m}'{\left({t}\right)}=-{\frac{{{0.06}}}{{{4}}}}{m}{\left({t}\right)}+{500}\cdot{0.06}$$
$$\displaystyle{m}{\left({0}\right)}={0}$$

Step 2

b) Solution is given with $$\displaystyle{m}{\left({t}\right)}={\left({m}_{{{0}}}-{C}_{{{i}}}{V}\right)}{e}-{\frac{{{R}}}{{{V}}}}{t}+{C}_{{{i}}}{V}$$ Solution is $$\displaystyle{m}{\left({t}\right)}={\left({0}-{500}\cdot{4}\right)}{e}-{\frac{{{0.06}}}{{{4}}}}{t}+{500}\cdot{4}$$
$$\displaystyle{m}{\left({t}\right)}=-{2000}{e}^{{-{0.015}{t}}}+{2000}$$ Graph of mass of the drug m(t) in mg over time 19610300291.jpg

Step 3

Graph of concentration of the drug $$\displaystyle{\frac{{{m}{\left({t}\right)}}}{{{V}}}}$$ in mg/L over time 19610300292.jpg

Step 4

c) From the first graph it is visible that steady state mass of the drug is 2000 mg.

Step 5 c) 90% of steady state level drug mass is 1800. Solve $$\displaystyle{m}{\left({t}\right)}={1800}$$ to obtain t $$\displaystyle{2000}{e}^{{-{0.015}{t}}}+{2000}={1800}$$
$$\displaystyle{e}^{{-{0.015}{t}}}={0.1}$$
$$\displaystyle-{0.015}{t}={\ln{{\left({0.1}\right)}}}t \approx 153.5$$