Question

A common inhabitant of human intestines is the bacterium Escherichia coli, named after the German pediatrician Theodor Escherich, who identified it in 1885. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 50 cells.
(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 6 hours.
(d) Find the rate of growth after 6 hours. (Round your answer to the nearest integer.)
(e) When will the population reach a million cells?

Answer 1

Answer:

a). k = 2.0794

b). $A_{t}=A_{0}(2^{3t} )$

c). 13107200 cells

d). rate of growth after 6 hours = 27255112

e). 4.76 hours

Step-by-step explanation:

From the formula of bacterial population,

$A_{t}=A_{0}e^{kt}$

Where $A_{t}$ = Bacterial population after time t

$A_{0}$ = Initial population

k = relative growth factor

t = duration or time

a). For $A_{t}=2\times 50=100$ cells

and $A_{0}=50$ cells

Time t = 20 minutes = $\frac{1}{3}$ hours

Now we plug in these values in the formula,

100 = $50(e^{\frac{k}{3}})$

$e^{\frac{k}{3}}=2$

By taking natural log on both the sides of the equation,

$\frac{k}{3}(lne)=ln(2)$

k = 3ln(2) = 2.0794

b). To get the expression we will plug in the value of k in the formula.

$A_{t}=A_{0}e^{kt}$

Since k = 3ln(2)

$A_{t}=A_{0}e^{3t(ln2)}$

Let y = $e^{3t(ln2)}$

By taking natural log on both the sides of the equation,

lny = $ln(e^{3t(ln2)} )$

lny = 3t(ln2)ln(e)

ln(y) = 3t(ln2)

ln(y) = $ln(2)^{3t}$

y = $2^{3t}$

Now our expression will be $A_{t}=A_{0}(2^{3t} )$

c). Number of cells after t = 6 hours

$A_{6}=50(2^{3\times 6} )$

$A_{6}=50(2^{18})$

$A_{6}=50\times (262144)=13107200$

d). We can get the rate of growth by finding derivative of the expression

$\frac{d}{dt}(A_{t})=\frac{d}{dt}[A_{0}(e^{kt}})]$

          = $A_{0}[ke^{kt}]$

Now $\frac{d}{dt}(A_{6})=k.A_{6}$

                   = 2.0794×13107200

                   = 27255112

e) From the given formula,

$A_{t}=A_{0}(2^{3t} )$

$1000000=50(2^{3t} )$

$2^{3t}=20000$

3t(ln2) = ln(20000)

t = $\frac{ln(20000)}{3(ln2)}$

t = $\frac{9.90349}{2.07944}=4.76$ hours