Question

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. Find the values of a and b. According to this model, what happens to the yeast population in the long run?

Answer 1

Answer:

Step-by-step explanation:

Given the population of yeast modelled by the function f(t)=a/1+be^-0.7t where t is measured in hours.

If at time t = 0 the population is 20 cells, then the equation becomes:

20 = a/1+be^-0.7t

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

20 = a/1+be^-0

20 = a/1+b(1)

a/1+b = 20

a = 20(1+b) ........ equation 1

Also if the population is increasing at a rate of 12 cells/hour, then d(f(t)/dt = 12

Differentiate the expression with respect to time

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

d(f(t)/dt =

a{-(1+be^-0.7t)^-2×(-0.7be^-0.7t)

a{-(1+be^-0.7t)^-2×(-0.7be^-0.7t) = 12

0.7abe^-0.7t/(1+be^-0.7t)² = 12 ..... equation 2

at t = 0, the equation becomes

0.7abe^-0/(1+be^-0)² = 12

0.7ab/(1+b)²= 12

0.7ab = 12(1+b)²

Substitute 1 into 2

0.7×20(1+b)b = 12(1+b)²

14(1+b)b = 12(1+b)²

Divide both sides by 1+b

14b = 12(1+b)

14b = 12+12b

14b-12b = 12

2b = 12

b = 6

Substitute b= 6 into the equation a = 20(1+b) to get a.

a = 20(1+6)

a = 20×7

a = 140

For us to be able to determine what happens to the yeast population in the long run. we will take the limit of f(t) as t approaches infinity.

Lim t -->/infty a/1+be^-0.7t

= a/1+be^-(infty)

= a/1+b(0)

= a

Since a = 140, hence the population of the yeast tends to 140 on the long run