Verified Answer :
a.) P(t) = P0exp(kt)
P(20/60) = 40 exp(20k/60)
80 = 40 exp(k/3)
exp(k/3) = 80/40 = 2
k/3 = ln(2)
k = 3ln(2)
b.) P(8) = 40(2)^24 = 40(16777216) = 671088640 cells
d.) Rate of change = exp(8k) = exp(8(3ln(2)) = exp(24ln(2)) = exp(16.6355) = 16777216 cells / hour
e.) P(t) = 40(2)^3t; t in hours
1,000,000 = 40(8)^t
25,000 = 8^t
ln(25,000) = t ln(8)
t = ln(25,000)/ln(8) = 4.87 hours
Thus, the final answer for the different parts is as follows:
Part(a): The relative growth or the value of \(k\) is \(\fbox{\begin\\\ \math k=3ln2\\\end{minispace}}\).
Part(b): The population of the bacteria after \(8\) hours is \(\fbox{\begin\\\ 671088640\\\end{minispce}}\) cells.
Part(d): The rate of growth after \(8\) hours is \(\fbox{\begin\\\ 16777216\\\end{minispace}}\) cells per hour.
Part(e): The time required for the population of the bacteria to reach a count of \(1\) million is \(\fbox{\begin\\\ 4.87\\\end{minispace}}\) hours.
Further explanation:
In the question it is given that a cell of the bacteria Bacterium Escherichia coli divides into two cells in every \(20\) minutes.
According to the data given in the question the initial population of the bacteria is \(40\) cells.
Consider the function for increase in the population of the bacteria as follows:
\(\fbox{\begin\\\ \math P(t)=P_{0}e^{(kt)}\\\end{minispace}}\)
In the above equation \(P_{0}\) represents the initial population, \(t\) represents the time, \(P(t)\) is the population after \(t\) hours and \(k\) is the relative growth.
It is given that the initial population is \(40\) cells so, the value of \(P_{0}\) is \(40\).
Part(a): Determine the relative growth or the value of k.
The function which represents the growth in the population of the bacteria is as follows:
\(P(t)=P_{0}e^{(kt)}\) (1)
Since, each cell of the bacteria divides into two cells in every \(20\) minutes or \(\dfrac{1}{3}\) hours.
Since, the initial population is \(40\) cells so the population after \(\dfrac{1}{3}\) hours is \(80\) cells.
To obtain the value of \(k\) substitute \(\dfrac{1}{3}\) for \(t\), \(40\) for \(P_{0}\) and \(80\) for \(P(t)\) in equation (1).
\(\begin{aligned}80&=40\times e^{(k/3)}e^{(k/3)}\\ \dfrac{80}{40}e^{(k/3)}&=2\end{aligned}\)
Take antilog in the above equation.
\(\begin{aligned}\dfrac{k}{3}&=ln2\\k&=3ln2\end{aligned}\)
Therefore, the value of \(k\) is \(3ln2\).
Thus, the relative growth of the bacteria is \(\fbox{\begin\\\ \math k=3ln2\\\end{minispace}}\).
Part(b):Determine the population of the bacteria after \(\bf8\) hours.
The equation to determine the population after \(t\) hours is as follows:
\(\fbox{\begin\\\ \math P(t)=P_{0}e^{(kt)}\\\end{minispace}}\)
Substitute \(40\) for \(P_{0}\), \(3ln2\) for \(k\), \(8\) for \(t\) in the above equation.
\(\begin{aligned}P(8)&=40e^{(8\eimes 3ln2)}\\&=40e^{(24ln2)}\\&=40\times 2^{24}\\&=671088640\end{alighned}\)
Therefore, the population of the bacteria after \(8\) hours is \(671088640\) cells.
Part(d): Determine the rate of growth after \(\bf8\) hours.
The rate of growth is defined as the ratio of the population of the bacteria after \(t\) hours to the initial population of the bacteria.
Substitute \(8\) for \(t\) in the equation \(P(t)=P_{0}e^{(kt)}\).
\(\begin{aligned}P(8)&=P_{0}e^{(8\times 3ln2)}\\\dfrac{P(8)}{P(0)}&= e^{(24ln2)}\\\dfrac{P(8)}{P(0)}&=16777216\end{aligned}\)
Therefore, the value of \(\dfrac{P(8)}{P(0)}\) is \(\fbox{\begin\\\ 16777216\\\end{minispace}}\).
This implies that the rate of growth of bacteria after \(8\) hours is \(16777216\) cells per hour.
Part(e):Determine the time in which the population of the bacteria becomes \(1\) million cells.
Consider the time in which the population of the bacteria reaches a count of \(1\) million cells as \(t\) hours.
Substitute \(1000000\) for \(P(t)\), \(40\) for \(P_{0}\) and \(3ln2\) for \(k\) in the equation \(P(t)=P_{0}e^{(kt)}\).
\(\begin{aligned}1000000&=40e^{(3t\times ln2)}\\e^{(ln2^{(3t)})}&=25000\\2^{(3t)}&=25000\\8^t&=25000\end{aligned}\)
Take antilog in the above equation.
\(\begin{aligned}t&=\dfrac{ln25000}{ln2}\\t&=4.87\end{aligned}\)
Therefore, the time required for the population of the bacteria to reach a count of \(1\) million is \(4.87\) hours.
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Answer details:
Grade: High school
Subject: Mathematics
Chapter: Exponential function
Keywords: Functions, exponential function, rate of growth, Bacterium Escherichia coli, relative growth, population, cells, relative growth, cell divides in two, growth function, decay function,