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 1.What happens if there are no predators? No prey? 2.Explain the coefficients of the RF terms in both equations. 3.What happens when both R = 0 and F = 0? 4.Are there other situations in which both populations are constant? 5.Modify the system so that the prey grows logistically if there are no predators.

Exercises Page 164, 1-6. I will assign either system (i) or (ii).

Graphing solutions Here are some solutions to 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) = 4 F(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.

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).

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 = d 2 y/dt 2. Now our DE becomes a system: 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 , #7, 9, 11, 15, 17, 19 Core: p , #10, 16, 20, 21 Some of the problems in this section are really wordy. You don’t have to copy them into your HW.