- Difficulty
- Standard
- Suggested time
- 4 min
- Marks
- 2
Logistic growth and phase lines practice question 4
A patient receives a drug through an intravenous drip at 10 mg per hour. The liver removes 4 % of the drug in the body each hour, and the kidneys remove a further 1 % each hour. Write a differential equation for the amount of drug in the body, in mg, and find the amount it approaches in the long run.
Give the amount in milligrams.
Build the equation as rate in minus rate out. Which term is a fixed amount per hour, and which depend on how much drug is present?
Rate of change = rate in − rate out. The drip adds 10 mg each hour, and the two organs together remove .
In the long run nothing changes: set and solve for .
A whole number.
Theory UBC Math 100 §13.2Key results
Worked solution
Strategy
Input minus removal gives an equation of the form dA/dt = a − bA. Its single steady state, a/b, is stable, so every solution ends up there.
Working
Each organ removes a fixed fraction of what is present, so the two removal terms add.
At the steady state the drip exactly replaces what is removed.
Both arrows point at 200, so it is stable and A(t) approaches it from any starting amount.
Answer
The answer is dA/dt = 10 − 0.05A, and the amount approaches 200 mg.
Check
At 200 mg the liver removes 8 mg/h and the kidneys 2 mg/h, together exactly the 10 mg/h dripping in.
Where this goes wrong
Using one removal route only, which gives 10/0.04 = 250 mg (liver) or 10/0.01 = 1000 mg (kidneys). Both organs clear the drug at once, so the fractions add and the removal rate is 0.05A.
Leans on
Equations of the form y′ = a − by, from Exponential growth and decay
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