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Math · Differential Equations

Chapter 3: Linear First-Order Equations

Mixing and Circuit Problems

Rate in minus rate out, written as an equation.

Lesson
2
Time
About 24 minutes
0 of 12 done

Step 1: Let's Learn

Read it, or press Listen and follow the words.

For anything that accumulates, the rate of change is what comes in minus what goes out. That single sentence builds most of these models.

Rate in

Multiply the inflow rate by the incoming concentration. This is usually a constant.

Rate out

Multiply the outflow rate by the current concentration, which is the amount present divided by the volume.

Why it is linear

The outflow term is proportional to the amount present, so the equation lands in exactly the form of the previous lesson.

Steady state

Setting the rate of change to zero gives the long-run amount. The tank eventually matches the incoming concentration.

Circuits use the same equation

A resistor and capacitor charging toward a supply voltage obey the identical mathematics, with charge in place of salt.

Rate in minus rate out

The change in the amount of substance equals what enters minus what leaves. Writing that sentence as an equation is the whole modelling step, and the rest is technique.

The outflow depends on the current amount

Concentration is amount over volume, so the rate out involves the unknown function. That dependence is what makes the problem a differential equation rather than arithmetic.

Watch whether the volume changes

If inflow and outflow rates differ, the volume is a function of time and appears in the concentration. Assuming constant volume when it is not is the standard modelling error here.

Circuits have the same structure

An RL or RC circuit produces a linear first-order equation identical in form to a mixing problem. The same mathematics describing brine tanks describes current in a coil, which is a genuine unification.

Step 2: Try It Yourself

Tap and try it out.

A tank approaching its steady state charges the same way a capacitor does: fast at first, then slowing as the gap closes.
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y = 3 · -0.75^x + 0
  • Point(2, 0.03)

Step 3: Watch an Example

One step at a time.

Watch Dmitri Build a Tank Model

Dmitri has a 200 litre tank with brine at 3 grams per litre entering at 5 litres per minute, and the well-mixed solution leaving at the same rate.

  1. Step 1

    He computes the rate in: 5 litres per minute times 3 grams per litre, which is 15 grams per minute.

Step 4: Your Turn

Practice makes it stick.

The Inflow

Problem 1 of 2

Brine at 4 grams per litre entering at 6 litres per minute. What is the rate in, in grams per minute?

The Steady State

Problem 2 of 2

dS/dt = 15 − S/40. What is the steady-state amount of salt, in grams?

In Minus Out

1 of 8

2 grams per litre entering at 10 litres per minute. What is the rate in?

2 of 8

dS/dt = 20 − S/50. What is the steady-state amount?

3 of 8

A 300 litre tank holding 900 grams of salt. What is the concentration in grams per litre?

4 of 8

Inflow 4 litres per minute and outflow 4 litres per minute. Does the volume change? 1 yes, 0 no.

5 of 8

dS/dt = 12 − S/25. What is the steady-state amount?

6 of 8

At the steady state, what is dS/dt?

7 of 8

Match each part of the model to what it represents.

Tap a card on the left to start.

8 of 8

A 100 litre tank of pure water, so no salt at all. What is the initial concentration?

Step 5: Quick Check

Show what you know.

Question 1 of 2

dS/dt = 8 − S/30. What is the steady-state amount?

Question 2 of 2

How is a mixing problem set up?

What You Learned

  • 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.