A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth 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.
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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)
To solve this problem, we can use the formula for exponential growth:
N(t) = Nā * e^(rt),
where N(t) is the number of cells at time t, Nā is the initial number of cells, r is the relative growth rate, and e is the base of the natural logarithm.
(a) To find the relative growth rate, we can use the given information that the bacterium divides into two cells every 20 minutes. We need to convert minutes to hours since the units of time in the formula are in hours. There are 60 minutes in an hour, so the relative growth rate can be calculated as:
r = ln(2) / (20/60) = ln(2) / (1/3) = ln(2) * 3 ā 2.079.
Therefore, the relative growth rate is approximately 2.079.
(b) We can derive an expression for the number of cells after t hours by substituting Nā = 59 and r = 2.079 into the exponential growth formula:
N(t) = 59 * e^(2.079t).
(c) To find the number of cells after 5 hours, we substitute t = 5 into the expression found in part (b):
N(5) = 59 * e^(2.079*5).
Using a calculator or computer software, we can evaluate this expression to find the number of cells after 5 hours.
(d) The rate of growth after 5 hours can be found by calculating the derivative of the number of cells with respect to time t:
dN(t) / dt = 59 * 2.079 * e^(2.079t).
By substituting t = 5 into this expression, we can find the rate of growth after 5 hours.