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Differential equation word problems

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Most drugs are eliminated from the body according to a strict exponential decay law. Here are two problems that illustrate the process.
1. The drug Valium has a half-life in the blood of 36 hours. Assume that a 50-milligram dose of Valium is taken at time t=0. Let m(t) be the amount of drug in the blood in milligrams t hours after the dose. Plot the function m(t) as it varies with time. After how many hours will the drug reach 10% of its initial value? After how many hours will it reach 1% of its initial value?

2. Now imagine that a drug (such as aspirin or an antibiotic) with a half-life of 12 hours is taken regularly every eight hours. Assume that the first dose is taken at time t=0.
(a) What is the amount of drug in the blood at t=8 hours just prior to the second dose?
(b) What is the amount of drug in the blood at t=8 hours just after the second dose?
(c) What is the amount of drug in the blood at t=16 hours just prior to the third dose?
(d) What is the amount of drug in the blood at t=16 hours just after the third dose?
(e) Now generalize: what is the amount of the drug in the blood at t=8(n-1) hours just prior to the nth dose where n=1,2,3,...?
(f) What is teh amount of drug in the blood at t=8(n-1) just after the nth dose?
(g) What can you say about the long-term amount of the drug in the blood? Does it continue to increase without bound or does it approach a steady state level? If you argue for the latter choice, find the steady state value of the amount of drug. Justify you conclusion.
(h) Quickly apply the periodic doses problem to teh following problem: A fish hatchery harvests 1/3 of its current fish population at the end of each year, and then immediately replenishes the population with 500 new fish. Assuming no deaths an an initial fish population of 1000 fish, what is the steady state population in the hatchery?

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Solution Summary

This uses differential equations to solve problems relating to the amount of drugs in blood at a given time.

Solution Preview

half life = th = 36 hours
m(t) = mo*exp(-k*t)
where, k = ln(2)/th = 0.693/36 = 0.019254 per hour
mo = initial mass = 50 milligram

For plot see attachment.

when, m(t) = 10% of mo = 0.1*50 = 5 mg
=> 5 = 50*exp(-0.019254*t)
=> 0.019254*t = ln(10) = 2.302585
=> t = 2.302585/0.019254
=> t =119.589955 == 120 hours --Answer

m(t) = 1% of mo = 0.5 mg
=> 0.5 =50*exp(-0.019254*t)
=> 0.019254*t = 2*ln(10) = 2*2.302585
=> t = 4.60517/0.019254 = 239.1799 == 239.2 hours --Answer

t(1/2) = 12 hours
let initial mass of each dose = mo
m(t) = mo*exp(-k*t)
k = 0.693/12 = 0.05776
m(t) = mo*exp(-0.05776*t) ...(1)

at t = 8 hours, just before the second drug:
m(8, before second drug) = mo*exp(-0.05776*8) = 0.62997*mo
i.e, 62.997 == 63% ...

Solution provided by:
  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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