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

NameDate

Figure — use these to answer the problems

A tank approaching its steady state charges the same way a capacitor does: fast at first, then slowing as the gap closes.

y = 3 · -0.75^x + 0

the starting gap3

how fast the gap closes-0.75

the time shown2

Figure 1

Straightforward practice. Get the method working first.

5 problems

dS/dt = 8 − S/30. What is the steady-state amount?
Answer
2
How is a mixing problem set up?
- The rate of change equals the rate in minus the rate out.
- The rate of change equals the rate in plus the rate out.
2 grams per litre entering at 10 litres per minute. What is the rate in?
Answer
dS/dt = 20 − S/50. What is the steady-state amount?
Answer
A 300 litre tank holding 900 grams of salt. What is the concentration in grams per litre?
Answer

The same ideas with more to keep track of.

3 problems

Inflow 4 litres per minute and outflow 4 litres per minute. Does the volume change? 1 yes, 0 no.
Answer
dS/dt = 12 − S/25. What is the steady-state amount?
Answer
At the steady state, what is dS/dt?
Answer

Mixed problems. Work out what kind of question it is before you start.

4 problems

9
Match each part of the model to what it represents.
Draw a line from each item on the left to its match on the right.
- Inflow rate times incoming concentration
- Outflow rate times current concentration
- The difference between them
- Rate in
- Rate out
- The rate of change of the amount
A 100 litre tank of pure water, so no salt at all. What is the initial concentration?
Answer
The Inflow
Brine at 4 grams per litre entering at 6 litres per minute. What is the rate in, in grams per minute?
Answer
The Steady State
dS/dt = 15 − S/40. What is the steady-state amount of salt, in grams?
Answer