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EN
University of Utah

# A population is modeled by the differential equation $\frac {dP}{dt} = 1.2 P \left(1 - \frac{P}{4200} \right)$ (a) For what values of $P$ is the population increasing?(b) For what values of $P$ is the population decreasing?(c) What are the equilibrium solutions?

## (a) $\frac{d P}{d t}=1.2 P\left(1-\frac{P}{4200}\right), \quad$ Now $\frac{d P}{d t}>0 \Rightarrow 1-\frac{P}{4200}>0 \quad[\text { assuming that } P>0] \Rightarrow \frac{P}{4200}<1 \Rightarrow$$P<4200 \Rightarrow the population is increasing for 0<P<4200$$$\text { (b) } \frac{d P}{d t}<0 \Rightarrow P>4200\text { (c) } \frac{d P}{d t}=0 \Rightarrow P=4200 \text { or } P=0

#### Topics

Differential Equations

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##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

The first thing to notice is that this is a quadratic equation. If we distribute one point two p, the highest power term we get its Peace Square to the square term is a quadratic, which means if we graph it, it's going to look like a parabola. Because the coefficients next to the P squared is negative, it's going to be parable A that opens downwards, so we conduct graft. This instead of the ax axis, will have the P and for the wine we have t P T T. And here, when the derivative is positive, is what it's going to be above the P access, which also corresponds to the increase in the population. Since the derivative represents the change in population and similarly, when it's going to be negative, it's going to correspond to the area below the P access and thie decrease in population. We already know that it's going to look like a quadratic, so we'LL have two points better on the P access and we want to figure out what those points are. We already know that at those two points, the derivative, or DPT, is Cyril, but we're not sure what P values those will correspond to. So the next step is to set our differential equations two zero we want to solve for the p values and we'LL keep isolating P and multiplying both sides by four thousand two hundred. Get that the P here's four thousand two hundred or zero. So going back to our previous work way see that those two points correspond to peeping four thousand two hundred and derivative beings Europe War, both P and the derivative if, um, beings here. So now we are prepared to answer for all the parts. So part a asks about the increase in the population, and we see that this a curves when it's going to be above the P access. So P is between zero and four thousand two hundred. So the decrease in population happens at those two parts of the graph. So one part is not relevant because to get there, we have to have negative population. But of course, that doesn't make sense to have a negative population, so we'LL ignore this and we'LL focus on this part of the graph. So when P is four thousand two hundred or larger, will have a decrease in population, and the equilibrium occurs when the derivative is zero. So DPD tea in zero and that occurs at those two points and we already solved. For those, um, that's when P S zero or P is four thousand two hundred.

EN
University of Utah

#### Topics

Differential Equations

##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

Join Bootcamp