Answer:
a) ![\mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7Bdx%7D%7Bdt%7D%20%3D%2030%20-%200.015%20x%7D)
b) ![\mathbf{x = 2000 - 2000e^{-0.015t}}](https://tex.z-dn.net/?f=%5Cmathbf%7Bx%20%3D%202000%20-%202000e%5E%7B-0.015t%7D%7D)
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 :
![Rate _{(in)}= 500 \ mg/L \times 0.06 \ L/min = 30 mg/min](https://tex.z-dn.net/?f=Rate%20_%7B%28in%29%7D%3D%20500%20%5C%20mg%2FL%20%20%5Ctimes%200.06%20%5C%20%20L%2Fmin%20%3D%2030%20mg%2Fmin)
![Rate _{(out)}=\dfrac{x}{4} \ mg/L \times 0.06 \ L/min = 0.015x \ mg/min](https://tex.z-dn.net/?f=Rate%20_%7B%28out%29%7D%3D%5Cdfrac%7Bx%7D%7B4%7D%20%5C%20mg%2FL%20%20%5Ctimes%200.06%20%5C%20%20L%2Fmin%20%3D%200.015x%20%5C%20%20mg%2Fmin)
Therefore;
![\dfrac{dx}{dt} = Rate_{(in)} - Rate_{(out)}](https://tex.z-dn.net/?f=%5Cdfrac%7Bdx%7D%7Bdt%7D%20%3D%20Rate_%7B%28in%29%7D%20-%20Rate_%7B%28out%29%7D)
with respect to x(0) = 0
![\mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7Bdx%7D%7Bdt%7D%20%3D%2030%20-%200.015%20x%7D)
b) Solve the initial value problem and graph both the mass of the drug and the concentration of the drug.
![\dfrac{dx}{dt} = -0.015(x - 2000)](https://tex.z-dn.net/?f=%5Cdfrac%7Bdx%7D%7Bdt%7D%20%3D%20-0.015%28x%20-%202000%29)
![\dfrac{dx}{(x - 2000)} = -0.015 \times dt](https://tex.z-dn.net/?f=%5Cdfrac%7Bdx%7D%7B%28x%20-%202000%29%7D%20%3D%20-0.015%20%5Ctimes%20dt)
By Using Integration Method:
![ln(x - 2000) = -0.015t + C](https://tex.z-dn.net/?f=ln%28x%20-%202000%29%20%3D%20-0.015t%20%2B%20C)
![x -2000 = Ce^{(-0.015t)](https://tex.z-dn.net/?f=x%20-2000%20%3D%20Ce%5E%7B%28-0.015t%29)
![x = 2000 + Ce^{(-0.015t)}](https://tex.z-dn.net/?f=x%20%3D%202000%20%2B%20Ce%5E%7B%28-0.015t%29%7D)
However; if x(0) = 0 ;
Then
C = -2000
Therefore
![\mathbf{x = 2000 - 2000e^{-0.015t}}](https://tex.z-dn.net/?f=%5Cmathbf%7Bx%20%3D%202000%20-%202000e%5E%7B-0.015t%7D%7D)
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 \times \infty }}](https://tex.z-dn.net/?f=%5Cmathbf%7Bx%20%3D%202000%20-%202000e%5E%7B-0.015%20%5Ctimes%20%5Cinfty%20%7D%7D)
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)}}](https://tex.z-dn.net/?f=%5Cmathbf%7B1800%20%3D%202000%20-%202000e%5E%7B%28-0.015t%29%7D%7D)
![0.1 = e^{(-0.015t)](https://tex.z-dn.net/?f=0.1%20%3D%20e%5E%7B%28-0.015t%29)
![ln(0.1) = -0.015t](https://tex.z-dn.net/?f=ln%280.1%29%20%3D%20-0.015t)
![t = -\dfrac{In(0.1)}{0.015}](https://tex.z-dn.net/?f=t%20%3D%20-%5Cdfrac%7BIn%280.1%29%7D%7B0.015%7D)
t = 153.5056729
t ≅ 153.51 minutes