Transcript section2.1
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 dR 2R 1.2RF dt dF F 0.9RF dt 1. 2. 3. 4. 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? 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 P(0) = 0 dR 2R 1.2RF dt dF F 0.9RF dt prey predators R(0) = 0 predators prey A startling picture! Here’s what happens if we start with R(0) = 4 and F(0) = 1. predators prey The phase plane Look at PredatorPrey demo. This is the graph of the parametric equation (x,y) = (R(t), F(t)) for the IVP. R(0) = 4 F(0) = 1 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 d2y Fm 2 dt Hooke’s law of springs: the force exerted by a spring is proportional to the spring’s displacement from rest. Fs ky k is called the spring constant and depends on how powerful the spring is. DE for a simple harmonic oscillator Combine Newton and Hooke: d2y Fs ky m 2 dt Sooo…. d2y k y0 2 m dt which is the equation for a simple (or undamped) oscillator. It is a second-order DE harmonic 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 = d2y/dt2. Now our DE becomes a system: dy v dt dv k y dt m 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.