Answer:
a) [tex]\mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}[/tex]
b) [tex]\mathbf{x = 2000 - 2000e^{-0.015t}}[/tex]
c) the steady state mass of the drug is 2000 mg
d) t ≅ 153.51 minutes
Step-by-step 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 :
[tex]Rate _{(in)}= 500 \ mg/L \times 0.06 \ L/min = 30 mg/min[/tex]
[tex]Rate _{(out)}=\dfrac{x}{4} \ mg/L \times 0.06 \ L/min = 0.015x \ mg/min[/tex]
Therefore;
[tex]\dfrac{dx}{dt} = Rate_{(in)} - Rate_{(out)}[/tex]
with respect to x(0) = 0
[tex]\mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}[/tex]
b) Solve the initial value problem and graph both the mass of the drug and the concentration of the drug.
[tex]\dfrac{dx}{dt} = -0.015(x - 2000)[/tex]
[tex]\dfrac{dx}{(x - 2000)} = -0.015 \times dt[/tex]
By Using Integration Method:
[tex]ln(x - 2000) = -0.015t + C[/tex]
[tex]x -2000 = Ce^{(-0.015t)[/tex]
[tex]x = 2000 + Ce^{(-0.015t)}[/tex]
However; if x(0) = 0 ;
Then
C = -2000
Therefore
[tex]\mathbf{x = 2000 - 2000e^{-0.015t}}[/tex]
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
[tex]\mathbf{x = 2000 - 2000e^{-0.015 \times \infty }}[/tex]
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;
[tex]\mathbf{1800 = 2000 - 2000e^{(-0.015t)}}[/tex]
[tex]0.1 = e^{(-0.015t)[/tex]
[tex]ln(0.1) = -0.015t[/tex]
[tex]t = -\dfrac{In(0.1)}{0.015}[/tex]
t = 153.5056729
t ≅ 153.51 minutes
