Calc
posted by Joe .
A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrientbroth medium divides into two cells every 20 minutes. The initial population of a culture is 59 cells. (The tolerance is 1% unless your instructor states otherwise.)
(a) Find the relative growth rate.
(b) Find an expression for the number of cells after t hours.
(c) Find the number of cells after 5 hours.
(d) Find the rate of growth after 5 hours.

Number = 59(2)^(t/20) where t is the number of minutes
so after 5 hours
Number = 59(2)^15
= 1933312
rate of growth
= d(Number)/dt
= (59ln2(2)^(t/20))/20
= (59/20)ln(2)^(t/20)