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# In a fish farm, a population of fish is introduced into a pond and harvested regularly. A model for the rate of change of the fish population is given by the equation$\frac {dP}{dt} = r_o (1 - \frac {P(t)}{Pc}) P(t) - \beta P(t)$where $r_o$ is the birth rate of the fish, $Pc$ is the maximum population that the pond can sustain (called the carrying capacity ), and $\beta$ is the percentage of the population that is harvested.(a) What value of $dP/dt$ corresponds to a stable population? (b) If the pond can sustain 10.000 fish, the birth rate is $5\%,$ and the harvesting rate is $4\%,$ find the stable population level.(c) What happens if $\beta$ is raised to $5\%?$

## a) 0b) 2,000c) no stable population

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here we have the equation for the rate of change of the fish population and for the first part of the problem, we want to know. Ah, what would the rate of change be if the population was stable? Well, a stable population would not be growing or shrinking, so the rate of change would be zero for Part B. We want to know if we use the numbers, given what the stable population level would be. So using the numbers given and using zero for the rate of change, let's find P of t the population level so we can substitute zero in our equation for DP DT, we were told, are not as 5%. So we're going to use 50.5 We have one minus p of tea divided by 10,000. And then we have the unknown p of tea again minus 0.4 That's the beta value times pft. So our goal is to solve this equation for pft. So let's start by adding 0.4 p of tea to both sides and we have 0.4 p of tea equals 0.5 times one minus pft over 10,000 times Pft. Now, I don't know about you, but I would rather not have all these decimal numbers in here. So I think at this point, it would be nice to stop and multiply both sides of the equation by 100. So now we have four p of tea equals five times p of tea, times one minus p of tea over 10,000. All right, now notice that we have a factor of p of tea on both sides. Let's go ahead and divide by it. And now we have four equals five times one minus p of tea over 10,000. All right, so let's go ahead and distribute the five. Get rid of those parentheses. And now we have four equals five minus five pft. Over 10,000. Well, that fraction could be reduced. Five over 10,000 is just one over 2001 over 2000. Yeah, that's right. So we'll do that on the next step. Let's subtract five from both sides. We get negative. One equals negative pft over 2000. Now we can multiply both sides of the equation by negative 2000 and we get p of T equals 2000. So what is that telling us? That's telling us for those particular values that beta value that rate at everything the stable population would be 2000 fish. And now, for part C, we're going to look at what would happen if we changed beta 2.5 instead of 0.4 Now I think we can safely go back through without redoing all the steps and just look at what would happen if our 0.4 became a 0.5 This would be a 0.5 This would be a four. We're sorry. This would be a five. This would be a five, and at that point we would subtract five from both sides, and this would be a zero. Then we would multiply both sides by negative 2000 and we would get the population is zero, and that would tell us that basically, there's no way to have a stable population with that beta value.

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