Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Set up and symbolically solve (by hand) the differential equation for each probl

ID: 2886870 • Letter: S

Question

Set up and symbolically solve (by hand) the differential equation for each problem. State your solution and answer all parts.

3. 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?

Explanation / Answer

Rate(in) = 500 * 0.06 = 30 mg/min

Rate(out) = x/4 * 0.06 = 0.015x mg/min

So, we have
dx/dt = Rate(in) - Rate(out)

dx/dt = 30 - 0.015x -----> ANS
with x(0) = 0
------------------------------------------------------------------------

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

dx/(x - 2000) = -0.015dt

Integrating :
ln(x - 2000) = -0.015t + C

x -2000 = Ce^(-0.015t)

x = 2000 + Ce^(-0.015t)

Now, using x(0) = 0 , we get C = -2000

So,
x = 2000 - 2000e^(-0.015t) -----> ANS

-----------------------------------------------------------------------

c)
When t ---> inf,
we get x = 2000 - 0

x = 2000

So, the steady state mass of the drug is :
2000 mg ---> ANS

----------------------------------------------------------------------

d)
90% of 2000 is 1800

So, we have
1800 = 2000 - 2000e^(-0.015t)

0.1 = e^(-0.015t)

ln(0.1) = -0.015t

So,
t = -ln(0.1) / 0.015

t = 153.5056728662697123

So,
t = 153.5 minutes approx -----> ANS

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote