Section 8.1 Modelling with the "Rate In" - "Rate Out" Modelling Paradigm
- the size of a population (taking into account birth and death rates, or rates of immigration and emigration, for example)
- the amount of drug in the body (taking into account administration or drug intake and elimination or breakdown of the drug),
- the amount of a polluting chemical in a lake (taking into account inflow and outflow or removal),
- the amount of carbon dioxide in the lungs (taking into account inhalation/ventilation, absorption, and excretion), etc.
Subsection 8.1.1 Intravenous Drug Dosing
Suppose that a drug is administered to a patient through an intravenous line at a rate of 8 mg/hr, and that the drug has a half-life of 24 hours. Here, the quantity of interest is the amount of drug in the blood as a function of time. We let y(t) be the amount of the drug in the blood, measured in mg, at time t, measured in hours. If we let Rin be the rate at which the amount of the drug increases, then we can write Rin=8. We are given the half-life of the drug, T1/2=24. This implies that, in the absence of drug administration, the amount of drug decays exponentially with a rate constant of λ=ln(2)T1/2=ln(2)24. If we let Rout be the rate at which the amount of the drug decreases, then we can write Rout=−λy. where we are using the World's Most Important Differential Equation to model exponential decay. Putting these two rates together, the net rate of change is
dydt=Rin−Rout=8−λy.
Checkpoint 8.1.1.
Solution
- The units of \(dy/dt\) are the units of \(y\) divided by the units of \(t\text{,}\) that is, mg/hr.
- The units of \(R_{in}\) are mg/hr (given).
- The units of \(\lambda\) are 1 divided by the units of \(T_{1/2}\text{,}\) that is, 1/hr.
- The units of \(R_{in}\) are the units of \(\lambda\) times the units of are \(y\text{,}\) that is 1/hr times mg, or mg/hr.
In conclusion, the units of each term are the same, namely mg/hr.
Checkpoint 8.1.2.
Subsection 8.1.2 Hormone Production and Removal
We consider H(t), the amount of a certain hormone in the blood as a function of time t. Suppose that the rate of hormone production as a function of time is Rin=p(t), and the rate of deactivation of the hormone as a function of time is Rout=r(t), then the net rate of change of H(t) is
dHdt=Rin−Rout=p(t)−r(t).
In class, we considered the situation where p(t) and q(t) are periodic functions, with a period of 24 hours, and with different phases, as illustrated below.

Checkpoint 8.1.4.
Answer
The area of the shaded region is $$ \int_{3}^{15} \left( p(t) - r(t) \right) \ dt = \int_{3}^{15} H'(t) \ dt, $$ which is the net increase of the amount of hormone between 3 AM and 3 PM.
Subsection 8.1.3 Summary
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Differential equations often are set up with the "Rate In - Rate Out" modelling paradigm: dydt=Rin−Rout, where
- Rin is the rate at which y is produced, or the rate at which y enters a compartment;
- Rout is the rate at which y is destroyed/broken down, or the rate at which y leaves a compartment.
Subsection 8.1.4 Don't Forget
Don't forget to return to eClass to complete the pre-class quiz.Subsection 8.1.5 Further Study
Remember that the notes presented above only serve as an introduction to the topic. Further study of the topic will be required. This includes working through the pre-class quizzes, reviewing the lecture notes, and diligently working through the homework problems. As you study, you should reflect on the following learning outcomes, and critically assess where you are on the path to achieving these learning outcomes:Learning Outcomes
- Based on a narrative, interpret or set up a differential equation using the "Rate In - Rate Out" modelling paradigm.
Learning Outcome | Video | Textbook Section |
---|---|---|
1 | N/A | 7.4 (Exercises 43, 45-48) |