# The number of yeast cells in a laboratory culture increases rapidly initially but levels off eventuallyThe number of yeast cells in a laboratory culture increases rapidly initially but levels off...

The number of yeast cells in a laboratory culture increases rapidly initially but levels off eventually

The number of yeast cells in a laboratory culture increases rapidly initially but levels off eventually. The population is modeled by the function

n = f(t) = a/(1 +be^(-0.7t))

where t is measured in hours. At time t = 0 the population is 20 cells and is increasing at a rate of 12 cells/hour.

a) Find the values of a and b.

b) According to this model, what happens to the yeast population in the long run?

(Please Show all work).

(a) To determine a and b, we need to set-up two equations. So, substitute t=0 and f(t)=20 to the given function.

`f(t)=a/(1+be^(-0.7t))`

`20=a/(1+be^(-0.7*0))`

`20=a/(1+b) ` (Let this be our EQ1.)

Next, take the derivative of f(t) with respect to t. Note that in taking the derivative, a and b are constant.

`f'(t)= d/(dt) (a/(1+be^(-0.7t)))`

`f'(t) = d/(dt) a(1+be^(-0.7t))^(-1)`

`f'(t)=a*-1*(1+be^(-0.7t))^(-2)*(1+be^(-0.7t))'`

`f'(t)=-a(1+be^(-0.7t))^(-2)*be^(-0.7t)*(-0.7)`

`f'(t)=(0.7abte^(-0.7t))/(1+be^(-0.7t))^2`

The value of f'(t) is the population rate of change of the cells which is 12cells/hr, when t=0. So,

`12=(0.7abe^(-0.7*0))/(1+be^(-0.7*0))`

`12=(0.7ab)/(1+b) ` (Let this be our EQ2.)

Then, let's apply substitution method of system of equations. Since a/(1+b)=20, substitute EQ1 to EQ2.

`12=(0.7b)*a/(1+b)`

`12=0.7b*20`

`12=14b`

`6/7=b`

Then, substitute value of b to EQ1. And solve for a.

`20=a/(1+6/7)`

`20=a/(13/7)`

`20*13/7=a`

`260/7=a`

**Hence, `a=260/7` and `b=6/7` .**

(b) Next, substitute the values of a and b to f(t).

`f(t)=(260/7)/(1+6/7e^(-0.7t))`

Then, multiply the top and bottom by the denominator 7, to simplify.

`f(t)=(260/7)/(1+6/7e^(-0.7t))*7/7`

So, the function of f(t) is:

`f(t)=260/(7+6e^(-0.7t))`

To determine the population in the long run, take the limit of f(t) as t approaches infinity.

`lim_(t->oo) f(t) = lim_(t->oo)` `260/(7+6e^(-0.7t)) = 260/(7+6e^(-0.7*oo))`

Note that if we multiply -0.7 with a very large number, the product is a very large number with a negative sign.

`=260/(7+6e^(-oo)) `

And note that `e^(-oo)=0` .

`=260/(7+6*0)=260/7=34.14~~35`

**Hence, in the long run, the population of the cell increases up to 35 cells only. **