Equilibria and Stability

The values a system settles into, and the ones it runs from.

12 problems · about 23 minutes · F-IF.B.4, F-IF.C.7

What this lesson teaches

The student finds equilibrium solutions and classifies them as stable or unstable.

An equilibrium is a constant solution, found by setting the rate to zero.
It is stable when nearby solutions return, and unstable when they run away.
Carrying capacities, terminal velocities and steady temperatures are all stable equilibria.

Straightforward practice. Get the method working first.

5 problems

dy/dt = y(9 − y). What is the stable equilibrium?
Answer 9
Why 9.
What makes an equilibrium stable?
Answer Nearby solutions move back toward it.
Why Disturbances die away.
dy/dt = y − 7. What is the equilibrium?
Answer 7
Why Set the rate to zero.
dy/dt = y − 7. Is that equilibrium stable? 1 yes, 0 no.
Answer 0
Why Above it the rate is positive, so solutions run away.
dy/dt = 7 − y. Is the equilibrium stable? 1 yes, 0 no.
Answer 1
Why Above it the rate is negative, so solutions come back.

The same ideas with more to keep track of.

3 problems

dy/dt = y(6 − y). How many equilibria are there?
Answer 2
Why Two factors, two roots.
dy/dt = y(6 − y). What is the stable equilibrium?
Answer 6
Why The carrying capacity.
dT/dt = −0.5(T − 15). What is the long-run temperature?
Answer 15
Why The stable equilibrium.

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

4 problems

Sort each behaviour by the kind of equilibrium it describes.
Answer Stable: Nearby solutions return to it, A small disturbance dies away · Unstable: Nearby solutions run away from it, A small disturbance grows
Why Ask what a small nudge does.
A solution starting exactly at an unstable equilibrium. Does it move? 1 yes, 0 no.
Answer 0
Why Exactly at it, the rate is still zero.
The Capacity: dP/dt = P(500 − P). What is the non-zero equilibrium population?
Answer 500
Why 500.
The Cooling Cup: dT/dt = −0.2(T − 20). What is the equilibrium temperature?
Answer 20
Why 20.