- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
Logistic growth and phase lines practice question 7
A chemical is pumped into a tank at a constant grams per minute, and an enzyme converts it at a rate that levels off at 6 grams per minute when there is plenty of chemical. The amount of chemical in the tank, in grams, satisfies the equation below, and the tank starts with none. What happens in the long run when , and when ?
A long-run level needs a steady state with that an empty tank moves towards. For each , is there one?
For each , look for steady states by solving , keeping only solutions with .
The conversion rate is always less than 6. What does that say about the sign of when ?
Theory UBC Math 100 §13.2Key results
Worked solution
Strategy
For each inflow, find the steady states with S ≥ 0 and the sign of dS/dt on either side. A steady state needs the enzyme to be able to match the inflow.
Working
Clear the fraction: this is the only steady state.
Conversion grows with S and overtakes the inflow above 4, so 4 is stable and the empty tank fills up to it.
The enzyme can never keep up, so the amount rises by more than 1 g every minute, without bound.
Answer
The answer is with R = 4 the amount levels off at 4 grams; with R = 7 it grows without bound.
Check
At S = 4 the enzyme converts 24/6 = 4 g a minute, matching R = 4, while even S = 1000 g gets only 6000/1002 ≈ 5.99, short of 7.
Where this goes wrong
Expecting a steady state for every inflow. For R = 7, solving 7(2 + S) = 6S gives S = −14, which is not a possible amount: the enzyme can never convert 7 g a minute, so the chemical keeps piling up.
Leans on
Limits of rational functions at infinity, from Limits
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