A drug is fused into a patients bloodstream at a constant ra

A drug is fused into a patient\'s bloodstream at a constant rate of 2mg oer hr and us eliminated from the bloodstream at a rate proportional to the amount of the drug present at time t. Initially(when t=0), the patient\'s blood contains no drug.

(a) Set up and solve a differential euqation for the amount of drug A(T) in the patient\'s bloodstream at any time t, in hr.(Note:((dA)/(dt))=(infusion)-(elimination), r=infusion, KA=elimination and K is constant)

(b) Suppose that a simple is taken indicating there are 1.69mg of the drug in the patient\'s bloodstream after 1hr. Show that k?0.35

(c) How much mediation is present after 6 hrs? how muchg mediation is present t???

Please show the step by step solution, thanks

Solution

(A) dA/dt = r - kA

since r = 2 mg

=> dA/(2-kA) = dt

=> (-1/k)*ln(2-kA) = t + C

At t=0 , A = 0

=> (-1/k)ln(2) = C

=> (-1/k)*ln((2-kA)/2) = t

=> (2-kA)/2 = e^-kt

=> A = (2/k)*(1-e^-kt)

(b) 1.69 = (2/k)*(1-e^-k)

Solving the eqn u get

k = 0.3468 0.35

(C) After 6 hrs

A₆ = (2/0.35)*(1-e^-0.35*6) = 5.01 mg

At t =

A = 2/k = 5.714 mg