The number of year cells in a laboratory culture increases...

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?

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?

Transcript

00:02 Our goal in this problem is to find the values of a and b.

00:06 We have this function that relates the population of yeast to a and b.

00:12 And we have two pieces of information.

00:14 We know that at time zero, the population is 20, and that at time zero, the rate of change is 12.

00:21 And so in general, when we're solving for two unknowns, if we have two equations, that's great.

00:25 So we're going to be able to come up with two equations based on those two pieces of information.

00:30 So what we can do is we can find the derivative of this function with respect to time, and that will help us use the second piece of information.

00:38 So f prime of t would be, now first of all, i'm going to rewrite f of t as a times a quantity, 1 plus b, e to the negative 0 .7t to the negative first power.

00:51 That way i don't have to use the quotient rule.

00:53 So a is just a constant, and now we bring down the negative 1, and we raise the outside quantity, or we raise the end.

01:01 Inside quantity, i should say, to the negative 2 power, using the chain rule, and then we multiply by the derivative of the inside, and that's going to be negative 0 .7b.

01:12 Okay, so let's rewrite this derivative.

01:16 Notice that we have a negative in two places.

01:19 We have negative a, and we have negative 0 .7, so negative times negative is positive.

01:23 So we have 0 .7a .b over 1 plus b to the negative 0 .7t quantity squared.

01:34 So that is one relationship we have.

01:37 We have the derivative.

01:39 And what we know is that the derivative equals 12 when t is 0.

01:43 So f prime of 0 is 12.

01:46 So let's go ahead and put 12 in for t and see what happens.

01:50 12 equals 0 .7 ab over.

01:55 Now we're going to have e raised to the 0 power.

01:57 And what is e to the 0 power? it's 1.