Respuesta :
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
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 [tex]k[/tex] is [tex]\fbox{\begin\\\ \math k=3ln2\\\end{minispace}}[/tex].
Part(b): The population of the bacteria after [tex]8[/tex] hours is [tex]\fbox{\begin\\\ 671088640\\\end{minispce}}[/tex] cells.
Part(d): The rate of growth after [tex]8[/tex] hours is [tex]\fbox{\begin\\\ 16777216\\\end{minispace}}[/tex] cells per hour.
Part(e): The time required for the population of the bacteria to reach a count of [tex]1[/tex] million is [tex]\fbox{\begin\\\ 4.87\\\end{minispace}}[/tex] hours.
Further explanation:
In the question it is given that a cell of the bacteria Bacterium Escherichia coli divides into two cells in every [tex]20[/tex] minutes.
According to the data given in the question the initial population of the bacteria is [tex]40[/tex] cells.
Consider the function for increase in the population of the bacteria as follows:
[tex]\fbox{\begin\\\ \math P(t)=P_{0}e^{(kt)}\\\end{minispace}}[/tex]
In the above equation [tex]P_{0}[/tex] represents the initial population, [tex]t[/tex] represents the time, [tex]P(t)[/tex] is the population after [tex]t[/tex] hours and [tex]k[/tex] is the relative growth.
It is given that the initial population is [tex]40[/tex] cells so, the value of [tex]P_{0}[/tex] is [tex]40[/tex].
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:
[tex]P(t)=P_{0}e^{(kt)}[/tex] (1)
Since, each cell of the bacteria divides into two cells in every [tex]20[/tex] minutes or [tex]\dfrac{1}{3}[/tex] hours.
Since, the initial population is [tex]40[/tex] cells so the population after [tex]\dfrac{1}{3}[/tex] hours is [tex]80[/tex] cells.
To obtain the value of [tex]k[/tex] substitute [tex]\dfrac{1}{3}[/tex] for [tex]t[/tex], [tex]40[/tex] for [tex]P_{0}[/tex] and [tex]80[/tex] for [tex]P(t)[/tex] in equation (1).
[tex]\begin{aligned}80&=40\times e^{(k/3)}e^{(k/3)}\\ \dfrac{80}{40}e^{(k/3)}&=2\end{aligned}[/tex]
Take antilog in the above equation.
[tex]\begin{aligned}\dfrac{k}{3}&=ln2\\k&=3ln2\end{aligned}[/tex]
Therefore, the value of [tex]k[/tex] is [tex]3ln2[/tex].
Thus, the relative growth of the bacteria is [tex]\fbox{\begin\\\ \math k=3ln2\\\end{minispace}}[/tex].
Part(b):Determine the population of the bacteria after [tex]\bf8[/tex] hours.
The equation to determine the population after [tex]t[/tex] hours is as follows:
[tex]\fbox{\begin\\\ \math P(t)=P_{0}e^{(kt)}\\\end{minispace}}[/tex]
Substitute [tex]40[/tex] for [tex]P_{0}[/tex], [tex]3ln2[/tex] for [tex]k[/tex], [tex]8[/tex] for [tex]t[/tex] in the above equation.
[tex]\begin{aligned}P(8)&=40e^{(8\eimes 3ln2)}\\&=40e^{(24ln2)}\\&=40\times 2^{24}\\&=671088640\end{alighned}[/tex]
Therefore, the population of the bacteria after [tex]8[/tex] hours is [tex]671088640[/tex] cells.
Part(d): Determine the rate of growth after [tex]\bf8[/tex] hours.
The rate of growth is defined as the ratio of the population of the bacteria after [tex]t[/tex] hours to the initial population of the bacteria.
Substitute [tex]8[/tex] for [tex]t[/tex] in the equation [tex]P(t)=P_{0}e^{(kt)}[/tex].
[tex]\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}[/tex]
Therefore, the value of [tex]\dfrac{P(8)}{P(0)}[/tex] is [tex]\fbox{\begin\\\ 16777216\\\end{minispace}}[/tex].
This implies that the rate of growth of bacteria after [tex]8[/tex] hours is [tex]16777216[/tex] cells per hour.
Part(e):Determine the time in which the population of the bacteria becomes [tex]1[/tex] million cells.
Consider the time in which the population of the bacteria reaches a count of [tex]1[/tex] million cells as [tex]t[/tex] hours.
Substitute [tex]1000000[/tex] for [tex]P(t)[/tex], [tex]40[/tex] for [tex]P_{0}[/tex] and [tex]3ln2[/tex] for [tex]k[/tex] in the equation [tex]P(t)=P_{0}e^{(kt)}[/tex].
[tex]\begin{aligned}1000000&=40e^{(3t\times ln2)}\\e^{(ln2^{(3t)})}&=25000\\2^{(3t)}&=25000\\8^t&=25000\end{aligned}[/tex]
Take antilog in the above equation.
[tex]\begin{aligned}t&=\dfrac{ln25000}{ln2}\\t&=4.87\end{aligned}[/tex]
Therefore, the time required for the population of the bacteria to reach a count of [tex]1[/tex] million is [tex]4.87[/tex] 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,
