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Problem 84 Hard Difficulty

The rate of growth of a fish population was modeled by the equation
$$ G(t) = \frac{60,000e^{-0.6t}}{(1 + 5e^{-0.6t})^2} $$
where $ t $ is measured in years and $ G $ in kilograms per year. If the biomass was 25,000 kg in the year 2000, what is the predicted biomass for the year 2020?

Answer

41666 $\mathrm{kg}$

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Video Transcript

So if we want to try to predict the biomass for the year 2020 Um, since this is giving the biomass from the year 2000, So essentially, this is a like time is equal to zero at 2000. So maybe I should say that up here, time is a good zero. And then down here, after 20 years past, we want to figure out Well, what is it going to be then? Now to get this, What we can do first is find our net change, and then once we get our net change, we can add that to our initial amount. So the net change, remember, is gonna be equal to just from a to B. Whatever time interval were interesting of our rate we integrate that were perspective time. So we just pick this up here, then go ahead and scoot this down, and then we're going to evaluate from 0 to 20. Okay, so now, in order for us to integrate this, um, notice how what we have in the denominator. If we take the derivative that it kind of looks like what we would get in the numerous Well off on the side I'm going to say this u is equal to one plus five e to the negative 0.16. And then if we take the derivative of this we get do you by d. T is equal to so that one goes away, we use ah, chain rolling all that. So it would be negative three e to the negative 0.60 And what we could do at this point, it's solve for DT some of the multiply DT over and divide by all that. So that's going to give us negative three e to negative 0.16 d You is eager to DT So now we can take this, replace it or DT and I'll leave these bounds off for just one moment. Um, so, yeah, that would be 60,000 times e to the negative 0.60 And then this would be times negative one third e to the negative 0.16 that we have d u. And then outside here. That would just be you squared now Uh huh. And then notice that these counselor with each other and then the three and that 60,000 which has become 20,000 per very negative. 20,000. Let's get rid of that. And now our new bounds of integration. What we're going to need to do is take zero, plug it into here and then take 20 and plug it into their. So, um, if I plug zero into here, um, that would just be one plus five. So it would just be six and then our new upper bound if we were to plug in 20. That doesn't look like it would be really nice, but I was kind of right it out. It would be one plus five e to the so we would do 0.6 times. 20 would be negative, tops the E to the negative 12. All right, so now let's actually simplify. All this down s would be negative. 20,000, integral from six toe one plus five e to the negative 12. And then this one over you squared. We could rewrite is you to the negative squared. Do you Now we can use power rule to integrate this. And so remember, power rule says we add one to the powers will be you to the negative first. And then we divide by the new powers. Would be negative one, and then we evaluate from six to the one plus five e to the negative 12. Uh, Now, these negatives actually cancel with each other and then you to the negative ones. Just one over you. So this is going to be equal to 1/1 plus five e to the negative 12. Um, and then minus 1/6 s. So this is going to be our exact net change. Um oh, actually, I almost forgot the 20,000 on the outside here. I was wondering why this was so small compared to what I got before. Yeah, so we have the 20,000 out here, So this is our net change. So net change. And now what was our initial uh, 25,000. So now we're going to come over here and add our initial to this. So plus our initial, which was 25,000, and so that would give the mass in 2020 to be approximately so I'll just write this out exactly. First, and then we can kind of estimate it. So 20,001 over one plus five Eat Negative. 12 minus 1/6. Okay, so let's go ahead and calculate on. I almost forgot the plus 25,000 there. Right? So first five times e raised to the negative 12 and then one minus that one over. So this first thing is going to be really close toe one and then minus 1/6 and then we multiply that by 20,000. So all of this is about 16,007 plus 25,000. And then if we add these together, this is going to say that our mass is we're going to expect 41,667 kilograms. Um, yeah. So this is what we would expect our biomass to end up being.

Bobby B.

University of North Texas

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Problem

Dialysis treatment removes urea and other waste p…

Problem 84 Hard Difficulty

The rate of growth of a fish population was modeled by the equation
$$ G(t) = \frac{60,000e^{-0.6t}}{(1 + 5e^{-0.6t})^2} $$
where $ t $ is measured in years and $ G $ in kilograms per year. If the biomass was 25,000 kg in the year 2000, what is the predicted biomass for the year 2020?

Answer

41666 $\mathrm{kg}$

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Calculus: Early Transcendentals

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