section 2.1 modeling via systems. a tale of rabbits and foxes suppose you have two populations:...
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Section 2.1
MODELING VIA SYSTEMS
A tale of rabbits and foxes
Suppose you have two populations: rabbits and foxes.
R(t) represents the population of rabbits at time t.
F(t) represents the population of foxes at time t.
• What happens to the rabbits if there are no foxes?Try to write a DE.
• What happens to the foxes if there are no rabbits?Try to write a DE.
• What happens when a rabbit meets a fox?• If R is the number of rabbits and F is the number of foxes, the
number of “rabbit-fox interactions” should be proportional to what quantity?
The predator-prey system
A system of DEs that might describe the behavior of the populations of predators and prey is
• What happens if there are no predators? No prey?• Explain the coefficients of the RF terms in both equations.• What happens when both R = 0 and F = 0?• Are there other situations in which both populations are
constant?• Modify the system so that the prey grows logistically if there are
no predators.
€
dR
dt= 2R−1.2RF
dF
dt= −F + 0.9RF
Exercises
Page 164, 1-6. I will assign either system (i) or (ii).
Graphing solutions
Here are some solutions to
€
dR
dt= 2R−1.2RF
dF
dt= −F + 0.9RF
prey
predators
P(0) = 0
predators
prey
R(0) = 0
A startling picture!
Here’s what happens if we start with R(0) = 4 and F(0) = 1.
prey
predators
The phase plane
Look at PredatorPrey demo.
R(0) = 4F(0) = 1
This is the graph of the parametric equation (x,y) = (R(t), F(t)) for the IVP.
Exercises
• p. 165 #7a, 8ab• Look at GraphingSolutionsQuiz in the Differential
Equations software (hard!)
Spring break!
Now for something completely different…
Suppose a mass is suspended on a spring.• Assume the only force acting on the mass is the force
of the spring.• Suppose you stretch the spring and release it. How
does the mass move?
Quantities:y(t) = the position of the mass at time t. – y(0) = resting– y(t) > 0 when the spring is stretched – y(t) < 0 when the spring is compressed
Newton’s Second Law: force = mass acceleration
Hooke’s law of springs: the force exerted by a spring is proportional to the spring’s displacement from rest.
k is called the spring constant and depends on how powerful the spring is.
€
F = md2y
dt2
€
Fs = −ky
DE for a simple harmonic oscillator
Combine Newton and Hooke:
Sooo….
which is the equation for a simple (or undamped) harmonic oscillator. It is a second-order DE because it contains a second derivative (duh).
€
Fs = −ky = md2y
dt2
€
d2y
dt2+k
my = 0
How to solve it!
Now we do something really clever. We don’t have any methods to solve second-order DEs.
Let v(t) = velocity of the mass at time t.
Then v(t) = dy/dt and dv/dt = d2y/dt2. Now our DE becomes a system:
€
dy
dt= v
dv
dt= −
k
my
Comes from our assumption
Comes from the original DE
Exercises
p. 167 #19• Rewrite the DE as a system of first-order DEs.• Do (a) and (b).• Check (b) using the MassSpring tool.• Do (c) and (d).
Homework (due 5pm Thursday)
• Read 2.1• Practice: p. 164-7, #7, 9, 11, 15, 17, 19• Core: p. 164-7, #10, 16, 20, 21
Some of the problems in this section are really wordy. You don’t have to copy them into your HW.
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