Problem 10 Easy Difficulty

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?

Answer

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Video Transcript

we have that The A over D T equals to a minus 0.1 A, l and D l over. DT equals minus 0.5 l plus 0.0 one a Oh, we want to find d l over D A. Well, that equals d l over d t divided by d a o r a d t. So that just gets us minus 0.5 l plus 0.18 l divided by to a minus 0.1 a o. So we're giving the direction field. We want to sketch a face portrait. Well, if we look at the direction field, we see that everything sort of circles around 5000. Uh, Abel's 5000 l equals 200. So all the face poor treats look like these shorter Ko Val's. Now, if at time Teeples here there are 1000 efforts and 200 ladybugs then, uh, the corresponding phase trajectory is going to look like one of these ovals. And let's see if we're at 1000 a and 200 right? So let's say in particular, it's this one, the outer one so that it would look like that and, uh, so if we want to, then graph aphids as a function of time. We see that same time, people zero, they start at some point, then they grow and decrease, grow and decrease growing decrease. In fact, the same thing will be true for the ladybugs grow and decrease growing decrease growing.