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The number of year cells in a laboratory culture increases rapidly initially but levels off eventually. The population is modeled by the function $ n = f(t) = \frac {a}{1 + be^{-0.7t}} $where $ t $ is measured in hours, At time $ t = 0 $ the population is 20 cells and increasing at a rate of 12 cells/ hour. Find the values of $ a $ and $ b. $ According to this model, what happens to the yeast population in the long run?

$a=140, b=6$ The yeast will reach a population of 140

01:09

Amrita B.

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 7

Rates of Change in the Natural and Social Sciences

Derivatives

Differentiation

Oregon State University

Harvey Mudd College

Idaho State University

Boston College

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Our goal in this problem is to find the values of A and B. We have this function that relates the ah population of yeast to A and B, and we have two pieces of information. We know that a time zero, the population is 20 and that at time zero the rate of change is 12. And so in general, when we're solving for two unknowns, if we have two equations, that's great. So we're going to be able to come up with two equations based on those two pieces of information. So what we can do is we confined the derivative of this function with respect to time and that will help us use the second piece of information so f prime of tea would be. No. First of all, I'm going to rewrite FFT as a times a quantity one plus B e to the negative 0.72 the negative first power. That way I don't have to use the quotient rule, so a it's just a constant and now we bring down the negative one and we raised the outside quantity or we raised the inside quantity. I should say to the negative to power using the chain rule, and then we multiply by the derivative of the inside. And that's going to be negative. 0.7 b. Okay, so let's rewrite this derivative notice that we have a negative in two places. We have negative and we have negatives. Your 20.7 so negative times negative is positive. So we have 0.7 a b over one plus B E to the negative 0.70 quantity squared. So that is one relationship we have. We have the derivative on. What we know is that the derivative equals 12 when t is Europe, so f prime of zero is 12. So let's go ahead and put 12 in for tea and see what happens. 12 equals 0.7 a b over. Now we're going to have e raise to the zero power and what is E to the zero power? It's one. So we're going to have be times one, so we're going to have one plus B quantity squared. Okay, so we're going to take this as our first relationship between A and B, and now we need to get the second relationship between A and B. So we're going to go back to F zero equals 20. Going back to the original function f of zero would be a over one plus B times each of the zero. And again, each of the zero is one. So this is just f of zero equals a over one plus B, and that is going to equal 20. So we have a second relationship between A and B. Okay, remember that our goal is to solve for A and B and they're several directions. We could go from here. We could take this Ah, equation that we just found We could isolate a and do a substitution or we could isolate be into a substitution. Or I think this might work out nicely. We could take the entire, um, a over one plus B that we see in the other equation and remove it and put a 20 in its place. So let's try that and see what happens. So with the second equation, we have 12 equals 0.7 times 20 times be over the remaining one plus b. See, we took out the A over one plus B and we replaced it with 20. Now let's saw that for B. So we're going to multiply both sides by one plus B and we get 12 plus 12 b equals 14 b and subtract 12 b from both sides. We get 12 equals to be and then divide by two. We get be equal six. Okay, so now if we know that be equal. Six We confined A because a over one plus six would be 20. So a over seven is 20. So multiply both sides by seven and A is 1 40 Now if we substitute those numbers into the model we have, the population is 1 40 over one plus six times E to the negatives. You're a 60.70 now. What happens? The final question is what happens in the long run, meaning what happens as t goes toward infinity. So let's take the limit and find out. So if t is going toward infinity, then this is like e to the negative infinity, which is very, very close to zero, so six times very, very close to zero is also very, very close to zero. So we basically have 1 40 divided by one. So it's heading toward 1 40 as time goes on

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