1. ## Questions part 1

1) The number of yeast cells in a laboratory culture increases rapidly at first but levels off eventually. The population is modeled by the function below, where t is measured in hours. At time t = 0 the population is 10 cells and is increasing at a rate of 2 cells/hour.

a = ?
b = ?

According to this model, at what number of cell does the yeast population stabilize in the long run?

2) Suppose that a population of bacteria triples every hour and starts with 800 bacteria.

(a) Find an expression for the number n of bacteria after t hours.

f(t) = (800)3^t <- I did that already

f'(2.5) = ?

2. Originally Posted by CFem
1) The number of yeast cells in a laboratory culture increases rapidly at first but levels off eventually. The population is modeled by the function below, where t is measured in hours. At time t = 0 the population is 10 cells and is increasing at a rate of 2 cells/hour.

a = ?
b = ?

According to this model, at what number of cell does the yeast population stabilize in the long run?

t = 0 the population is 10 cells and is increasing at a rate of 2 cells/hour.

Therefore you have the data points $(0,10)$ and $(1,12)$

Find a and b by using the data given.

after you have a and b, take the limit of the function for $t \rightarrow \infty$

3. When you say solve for a and be using the information given, does that mean like a system of equations?

Because I did that and solved for a and got the wrong answer. Unless there's an easier way to do it? .-.

4. Yes, that's exactly what that means. You have $f(t)= \frac{a}{1+be^{-0.3t}}$ and you know that f(0)= 10 and f'(0)= 2. f(0)= 10 gives you the equation $\frac{a}{1+ b}= 10$. To use f'(0)= 2, you will need to differentiate f(x). You might find it simplest to write f(t) as $a(1+ be^{-0.3t})^{-1}$ and using the chain rule. Or else use the quotient rule.