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Populations of aphids and ladybugs are modeled by the equations $ \frac {dA}{dt} = 2A - 0.01AL $$ \frac {dL}{dt} =0.5L + 0.0001AL $(a) Find the equilibrium solutions and explain their significance.(b) Find an expression for $ dL/dA. $(c) The direction field for the differential equation in part (b) is shown. Use it to sketch a phase portrait. What do the phase trajectories have in common?(d) Suppose that at time $ t = 0 $ there are 1000 aphids and 200 ladybugs. Draw the corresponding phase trajectory and use it to describe how both populations change.(e) Use part (d) to make rough sketches of the aphid and ladybug populations as function of $ t. $ How are the graphs related to each other?

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Calculus 2 / BC

Chapter 9

Differential Equations

Section 6

Predator-Prey Systems

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

Lectures

13:37

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation (ODE) is a differential equation containing one or more derivatives of a function and their rates of change with respect to the function itself; it can be used to model a wide variety of phenomena. Differential equations can be used to describe many phenomena in physics, including sound, heat, electrostatics, electrodynamics, fluid dynamics, elasticity, quantum mechanics, and general relativity.

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Populations of aphids and …

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03:05

Aphid-ladybug dynamics Pop…

07:48

In Exercise 10 we modeled …

02:35

Modified aphid-ladybug dyn…

So if a and l are constant than a prime and l prime equals 0, so we have 0 equals 2, a minus 0.01, a l, and then we have 0 equals negative 0.5 l plus 0.0001, a l and that goes to 0 equals a times. 2. Minus 0.01 l and 0 equals l times negative 0.5 plus 0.001, a for part b. We have d l over d, a is equal to d l d, t over d, a d t and that is negative: 0.5 l plus 0.0001, a l divided by 2, a minus 0.01, a l for part c. The solution curves are all closed. Curves that have the equilibrium .5200 inside them. For part d, we have at p not 1 thousamdand, 200 d. A d t is equal to 0 and d. L over dt is equal to negative 80, which is less than 0, so the number of lady bugs is decreasing and is going and the counter clockwise direction and at p 1 equals 5100. The lady bug population increases dramatically with a maximum of p 21425200 point. So for part e, both graphs have the same period and the graph of l peaks about a quarter of a cycle after the graph of a to kind of show you that here's the graph over here is a over here is l. We have 30000 point, and then here over here we have 900 and then 100 okay. So we have a and l and if we start with l, l starts about below 10000 decreases. First goes up to about 15000 decreases again, and then it reaches this axis. At 200 and then for a a increases first below 5000 peaks at about 15000, just like l goes down and drops very low, then goes all the way back up peaks just right before l does and it ends way below 100 point.

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