e e c f c f x f d d d a f a f y f b b b \u00b5 \u00b5 4 y 15 x Shark y and sardine x

E e c f c f x f d d d a f a f y f b b b µ µ 4 y 15

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e e c f c f x f d d d a f a f y f b b b + = = + = + = = = µ µ 0 4 y 15 0 x Shark ( y ) and sardine ( x ) trajectories The trajectories are closed curves representing periodic motion of both sharks and sardines. The trajectories look like the trajectories of the unfished case in Example 3 except the equilibrium point has moved to the right (more prey) and down (fewer predators). (b) With the parameters in part (a) and 0.5 f = the equilibrium point is (7, 1.5). This compares with the equilibrium point (6, 2) in the unfished case. As the fishing rate f increases from 0 to 2, the equilibrium point moves along the line from the unfished equilib- rium at (6, 2) to (10, 0). Hence, the fish- ing of each population at the same rate benefits the sardines ( x ) and drives the sharks ( y ) to extinction. This is illustrated in the figure. 8 4 y x 12 6 2 10 2 (6, 2) (10, 0) (sharks) (sardines) 1 (7, 1.5) (c) You should fish for sardines when the sardine population is increasing and sharks when the shark population is increasing. In both cases, more fishing tends to move the populations closer to equilibrium while maintaining higher populations in the low parts of the cycle. (d) If we look at the insecticide model and assume both the good guys (predators) and bad guys (prey) are harvested at the same rate, the good guys will also be diminished and the bad guys peak again. As 1 f (try 0.8 f = ) the predators get decimated first, then the prey can peak again. If you look at part (a), you see that the predator/prey model does not allow either population to go below zero, as the x - and y -axes are solutions and the solutions move along the axes, thus it is impossible for other solutions to cross either of these axes. You might continue this exploration with the IDE tool, Lotka-Volterra with Harvest, as in Problem 24. Full file at
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188 CHAPTER 2 Linearity and Nonlinearity Analyzing Competition Models 11. (1200 2 3 ) dR R R S dt = , (500 ) dS S R S dt = Rabbits are reproducing at the astonishing rate of 1200 per rabbit per unit time, in the absence of competition. However, crowding of rabbits decreases the population at a rate double the population. Furthermore, competition by sheep for the same resources diminishes the rabbit population by three times the number of sheep! Sheep on the other hand reproduce at a far slower (but still astonishing) rate of 500 per sheep per unit time. Competition among themselves and with rabbits diminishes merely one to one with the number of rabbits and sheep. Equilibria occur at 0 0 600 300 , , , and 0 500 0 200 ⎡ ⎤ ⎡ ⎤ ⎡ ⎢ ⎥ ⎢ ⎥ ⎢ ⎣ ⎦ ⎣ ⎦ ⎣ . The equilibria on the axes that are not the origin are the points toward which the populations head. Which species dies out depends on where they start. See Figure, where x and y are measured in hundreds.
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