Question

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 that the blood volume is a four-liter tank that initially has a zero concentration of a particular drug. At time 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 t20.
b) 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 stead-state level?

Answer 1

a)
`\mathbf{(dx)/(dt) = 30 - 0.015 x}`

b)
`\mathbf{x = 2000 - 2000e^(-0.015t)}`

c) the steady state mass of the drug is 2000 mg

d) t ≅ 153.51 minutes

Explanation:

From the given information;

At time t= 0

an intravenous line is inserted into a vein (into the tank) that carries a drug solution with a concentration of 500

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.

The objective of the question is to calculate the following :

a) Write an initial value problem that models the mass of the drug in the blood for t ≥ 0.

From above information given :

`Rate _((in))= 500 \ mg/L * 0.06 \ L/min = 30 mg/min`

`Rate _((out))=(x)/(4) \ mg/L * 0.06 \ L/min = 0.015x \ mg/min`

Therefore;

`(dx)/(dt) = Rate_((in)) - Rate_((out))`

with respect to x(0) = 0

`\mathbf{(dx)/(dt) = 30 - 0.015 x}`

b) Solve the initial value problem and graph both the mass of the drug and the concentration of the drug.

`(dx)/(dt) = -0.015(x - 2000)`

`(dx)/((x - 2000)) = -0.015 * dt`

By Using Integration Method:

`ln(x - 2000) = -0.015t + C`

`x -2000 = Ce^{(-0.015t)`

`x = 2000 + Ce^((-0.015t))`

However; if x(0) = 0 ;

Then

C = -2000

Therefore

`\mathbf{x = 2000 - 2000e^(-0.015t)}`

c) What is the steady-state mass of the drug in the blood?

the steady-state mass of the drug in the blood when t = infinity

`\mathbf{x = 2000 - 2000e^(-0.015 * \infty )}`

x = 2000 - 0

x = 2000

Thus; the steady state mass of the drug is 2000 mg

d) After how many minutes does the drug mass reach 90% of its stead-state level?

After 90% of its steady state level; the mas of the drug is 90% × 2000

= 0.9 × 2000

= 1800

Hence;

`\mathbf{1800 = 2000 - 2000e^((-0.015t))}`

`0.1 = e^{(-0.015t)`

`ln(0.1) = -0.015t`

`t = -(In(0.1))/(0.015)`

t = 153.5056729

t ≅ 153.51 minutes