Infusion: An IV line provides a continuous flow of a drug directly into the bloodstream. Assuming A no initial drug presence, the amount of the drug aftert hours is given by m(t) = ÷ (1 – e-kt) for t > 0, where A is the rate the drug flows in milligrams per hour. We say the steady state level is the amount of drug remaining after a long time has passed. k 1. Suppose an antibiotic with a half-life of 12 hours is given intravenously with A = 50 . Find the rate constant k and graph m(t) for 0

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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4. m(t) = 400e^-0.462t

Infusion: An IV line provides a continuous flow of a drug directly into the bloodstream. Assuming
A
no initial drug presence, the amount of the drug aftert hours is given by m(t) = ÷ (1 – e-kt) for
t > 0, where A is the rate the drug flows in milligrams per hour. We say the steady state level is
the amount of drug remaining after a long time has passed.
k
1. Suppose an antibiotic with a half-life of 12 hours is given intravenously with A = 50 . Find
the rate constant k and graph m(t) for 0 <t< 48. Hint: you can find k as above, so maybe
use the formula that you derived earlier in question 4.
2. Using the plot from question 1, explain why the graph for infusion looks so different from the
graph for injection.
Transcribed Image Text:Infusion: An IV line provides a continuous flow of a drug directly into the bloodstream. Assuming A no initial drug presence, the amount of the drug aftert hours is given by m(t) = ÷ (1 – e-kt) for t > 0, where A is the rate the drug flows in milligrams per hour. We say the steady state level is the amount of drug remaining after a long time has passed. k 1. Suppose an antibiotic with a half-life of 12 hours is given intravenously with A = 50 . Find the rate constant k and graph m(t) for 0 <t< 48. Hint: you can find k as above, so maybe use the formula that you derived earlier in question 4. 2. Using the plot from question 1, explain why the graph for infusion looks so different from the graph for injection.
Expert Solution
Solution:

From question 4, we have m(t)=400e-0.462t.

Here, the amount of drug after t hours is m(t)=Ak1-e-kt.

The half life is 12 hours and A=50 mghr

So, m(12)=400e-0.462121.564

Use this amount and obtain the value of k:

1.564=50k1-e-12k

The value of k is 31.969.

Thus, we have

The amount of drug after t hours is m(t)=5031.9691-e-31.969t.

The graph of  is obtained as shown below:

Advanced Math homework question answer, step 2, image 1

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