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# In the study of ecosystems, predator-prey models are often used to study the interaction between species. Consider populations of tundra wolves, given by $W(t),$ and caribou, given by $C(t),$ in northern Canada. The interaction has been modeled by the equations$\frac {dC}{dt} = aC - bCW$ $\frac {dW}{dt} = -cW + dCW$(a) What values of $dC/dt$ and $dW/dt$ correspond to stable populations?(b) How would the statement "The caribou go extinct" be represented mathematically?(c) Suppose that $a = 0.05, b = 0.001, c = 0.05,$ and $d = 0.0001.$ Find all population pairs (C, W) that lead to stable populations. According to this model, is it possible for the two species to live in balance or will one or both species become extinct?

## a) $\frac{d C}{d t}=0$ and $\frac{d W}{d t}=0$b) $C=0$c) $(0,0),(500,50)$yes

Derivatives

Differentiation

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

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all right in this problem, DC DT represents the rate of change of the caribou population and E W D. T represents the rate of change of the wolf population. And so if the population's air stable, those rates of change would be zero. The populations would not be increasing or decreasing for Part B. If the caribou were to go extinct. That would mean that the equation see of tea was zero. Because CFT represents the caribou population, never part c were given a bunch of numbers and weaken. Substitute those into the equations that we have, and our goal is to find values of C and W that would be solutions to those equations. So we're going to substitute the numbers into D C D T. And we get G C D T equals 0.5 c minus 0.1 C. W. And we wanted to be stable, so that's going to equal zero and 40 Wgt. We substitute the given numbers in and we get negative 0.5 w plus 0.1 C w. And we want that to equal zero for another stable population so what we have here are two equations with two unknowns. We can call this equation one, and we can call this Equation two. And we're going to solve the system for C and W. And what I'm going to do is I'm going to take care of all these decimals. I would like to multiply the first equation by 1000. So Equation one, when multiplied by 1000 is going to be 50 c minus C W equals zero. And then I would like to multiply equation to buy 10,000 and that's going to be negative. 500 w plus c times w equals zero. Okay, now, what I would like to do is add these equations together using the elimination method for solving a system, and we end up with 50 c minus 500 w equals zero. That gives us a more simplified relationship between C and W. And what we can do is add 500 w to both sides and divide both sides by 50. And now we know that C equals 10 w. The population of caribou is 10 times the population of wolves. We still don't know the values of C and W But what we can dio is we can take that and substitute it into one of our equations. I'm gonna go back to Equation one. I'm going to substitute 10 w in for C, so I'll have 50 times 10 w minus 10 w times w equals zero. So that would be 500 w minus 10 w squared equals zero. Let's factor 10 w out of both of those which we probably could have done earlier and not, uh and we could have saved a step. So 10 w times 50 minus w equals zero. So then set each of those equal to zero when we get 10. W equals zero when we get 50 minus w equals zero. Let's solve each of those and see what we get. For the 1st 1 we get w equals zero. And for the 2nd 1 we get w equals 50. So that's telling us there are two ways to have stable populations. One way is if we have no wolves. And the other way is if we have 50 wolves and if we have no wolves, we have no caribou since caribou equals 10 times the wolves. And if we have 50 wolves. We have 500 caribou since caribou equals 10 times the wolves. So those are the two ways to get a stable population, so it is possible to live in balance.

Oregon State University

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Derivatives

Differentiation

##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp