Modelling With Differential Equations

Turning a described rate into an equation.

12 problems · about 21 minutes · FUN-7.A

NameDate

Figure — use these to answer the problems

A modelled equation produces a field like this. The initial condition picks which curve the situation follows.

k, the constant-1

Starting x0

Starting y2

Figure 1

Straightforward practice. Get the method working first.

5 problems

In at 15 and out at 6 litres per minute. What is the net rate?
Answer
2
What does an initial condition do?
- It selects one solution from the family the equation describes.
- It makes the equation solvable.
In at 12 and out at 5 litres per minute. Net rate?
Answer
In at 3 and out at 8 litres per minute. Net rate?
Answer
A cup cools in a 25°C room. What temperature does it approach?
Answer

The same ideas with more to keep track of.

3 problems

For a cooling object, is k positive or negative? 1 positive, 2 negative.
Answer
Does an initial condition select one solution from a family? 1 yes, 0 no.
Answer
"The rate is proportional to the amount present" gives dy/dt = k times what? 1 y, 2 t.
Answer

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

4 problems

9
Match each description with its equation form.
Draw a line from each item on the left to its match on the right.
- Rate proportional to the amount
- Rate proportional to a difference from a constant
- Rate with a ceiling on growth
- dy by dt equals k y
- dT by dt equals k times (T minus Ts)
- The logistic equation
In at 9 and out at 9 litres per minute. Net rate?
Answer
The Room
A cup cools in a 20°C room. What temperature does it approach, in degrees?
Answer
The Tank
Water flows in at 10 litres per minute and out at 4. What is the net rate, in litres per minute?
AnswerL/min