Question

This is a warmup for your first project. The drug valium is eliminated from the bloodstream through a decay process with a half-life of 36 hours. This means that no matter how much drug is in your body, 36 hours later, half will be left. An initial dose of 20 milligrams of valium is taken at midnight. This is actually a version of the mixing-tank problem with the tank representing the circulatory system. If we don’t know something, give it a name (e.g. flow in = rin). Assume that the volume of blood in a person does not fluctuate much, so we can assume it is a constant.
(a) Define your dependent and independent variable with units. Typical volume units for blood is in terms of Liters.
(b) What is the flow rate in? flow rate out? concentration going in? concentration going out? initial condition?
(c) Write the initial value problem (IVP).
(d) Solve your initial value problem.
(e) Use information given to you to determine all constants in your solution.
(f) What is the amount of valium in your body at noon the next day? How long does it take for the drug to reach 10% of its initial level?

Answer 1

Step-by-step explanation:

a. We have v as the amount of drug(mg) in the blood at time t (hrs)

b. Flow rate in = k litre/he

For a circulatory system we have the flow rate in and flow rate out to be the same

Therefore,

Flow rate out = k litre/hr

concentration of drug going in = 0mg/litre

Concentration of drug going out = v mg/litre

V(0) = 20mg since 20mg was taken at midnight

V(0) = 20

Half life t1/2 = 36 hours

V(36) = v(0)/2

= 20/2

= 10mg

C. $ivp = \frac{dv}{dt}$$= -kv$

$v(0) = 20mg\\v(36) = 10mg$

d. solution

$\frac{dv}{dt} = -kt\\ln(v) = ln(c) - kt$

$\frac{v}{c} = e^{-kt} = v=ce^{-kt}$

v(0) = 20

$ce^{-k(0)} =20$

c = 20

so

$v(t) = 20e^{-kt}$

e.

$t\frac{1}{2}=36hours\\ v(36)= 10$

$10 = 20e^{-36k}$

$\frac{1}{2} =e^{-36k}$

we take log

$k=\frac{ln(2)}{36}$

please check attachment for answer f