Math 107-250/350 Answers to Homework #5 9 Feb ’07
. 5 1. Warfarin is a drug used as a an anticoagulant. (It is also used in rat poison.) After
administration of the drug is stopped, the quantity in a patient’s body decreases at a
rate proportional to the quanitity remaining, i.e., the differential equation is
dW
dt =kW.
A patient took an unknown dose of Warfarin at 8am on Monday. Testing indicates
that at 10am on Tuesday, the patient contains 4.00mg of Warfarin and at 10am on
Wednesday, it contains 2.55mg. What was the unknown dose of Warfarin?
Solution. We are looking for a function W(t), the amount of Warfarin
in the patient’s body, in milligrams, where tis hours since 8am Monday.
The answer to the question will be W(0). We know that W(26) = 4.00,
W(50) = 2.55 and W(t) satisfies the differential equation
dW
dt =kW,
for some value of k.
Separating variables in the DE,
1
WdW =k dt
Z1
WdW =Zk dt
ln W=kt +C
W=ekt+C
W=Kekt where K=eC
To find Kand k, we need two equations, namely
4.00 = Ke26k,2.55 = K e50k
Dividing the first equation by the second, we have
4.00
2.55 =Ke26k
Ke50k=e−24k.
Thus, −24k= ln(1.569) = 0.450 and so k=−0.0188. Substituting this
value of kinto the first equation, we have
4.00 = Ke(−0.0188)26 =K e−0.0489 =K·.613
Thus, K= 4/.613 = 6.52.
So W(t)=6.52e−0.0188kand W(0) = 6.52. The patient took a dose of 6.52
mg of Warfarin at 8am on Monday.
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