Cogito
Differential Equations · Chapter 3 · Lesson 2
Mixing and Circuit Problems
Rate in minus rate out, written as an equation.
12 problems · about 24 minutes · A-CED.A.1, A-CED.A.2
What this lesson teaches
The student sets up and interprets linear models for mixing tanks and simple circuits.
- For anything that accumulates, the rate of change is the rate in minus the rate out.
- The outflow depends on the current concentration, which makes the equation linear.
- Setting the rate to zero gives the steady state the system approaches.
Warm Up
Straightforward practice. Get the method working first.
5 problemsdS/dt = 8 − S/30. What is the steady-state amount?
Answer 240
Why 240.
How is a mixing problem set up?
Answer The rate of change equals the rate in minus the rate out.
Why In minus out.
2 grams per litre entering at 10 litres per minute. What is the rate in?
Answer 20
Why Multiply.
dS/dt = 20 − S/50. What is the steady-state amount?
Answer 1000
Why 20 times 50.
A 300 litre tank holding 900 grams of salt. What is the concentration in grams per litre?
Answer 3
Why 900 divided by 300.
Build It Up
The same ideas with more to keep track of.
3 problemsInflow 4 litres per minute and outflow 4 litres per minute. Does the volume change? 1 yes, 0 no.
Answer 0
Why They balance exactly.
dS/dt = 12 − S/25. What is the steady-state amount?
Answer 300
Why 12 times 25.
At the steady state, what is dS/dt?
Answer 0
Why Nothing is accumulating.
Stretch Yourself
Mixed problems. Work out what kind of question it is before you start.
4 problemsMatch each part of the model to what it represents.
Answer Inflow rate times incoming concentration → Rate in; Outflow rate times current concentration → Rate out; The difference between them → The rate of change of the amount
Why Accumulation is in minus out.
A 100 litre tank of pure water, so no salt at all. What is the initial concentration?
Answer 0
Why No salt present.
The Inflow: Brine at 4 grams per litre entering at 6 litres per minute. What is the rate in, in grams per minute?
Answer 24
Why 24.
The Steady State: dS/dt = 15 − S/40. What is the steady-state amount of salt, in grams?
Answer 600
Why 600 grams.