Asked by luna
Warfarin is a drug used as an anticoagulant. After administration of the drug is stopped, the quantity remaining in a patient's body decreases at a rate proportional to the quantity remaining. Suppose that the half-life of warfarin in the body is 35 hours.
Write a differential equation satisfied by Q:
How many days does it take for the drug level in the body to be reduced to 30 percent of its original level?
Write a differential equation satisfied by Q:
How many days does it take for the drug level in the body to be reduced to 30 percent of its original level?
Answers
Answered by
Steve
dQ/dt = kQ
dQ/Q = k dt
lnQ = kt + c
Q = c*e^(kt)
since 2 = e^(ln2), that can of course be written by massaging the constants as
Q(t) = c*2^(-t/k)
where k=35
Q(t) = c*2^(-t/35)
Since we are given no actual amounts, we can set c=1, so that as a fraction of the initial amount,
Q(t) = 2^(-t/35)
So, just solve that when Q=.30
dQ/Q = k dt
lnQ = kt + c
Q = c*e^(kt)
since 2 = e^(ln2), that can of course be written by massaging the constants as
Q(t) = c*2^(-t/k)
where k=35
Q(t) = c*2^(-t/35)
Since we are given no actual amounts, we can set c=1, so that as a fraction of the initial amount,
Q(t) = 2^(-t/35)
So, just solve that when Q=.30
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