Question Statement

In a culture, the rate of growth of bacteria is proportional to the population present. If the population of bacteria becomes four times its initial size in two days, what will the population be after ten days if the initial population was 20?

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Background and Explanation

This problem is a classic example of exponential growth. When the rate of change of a variable is proportional to the variable itself, it is modeled by a first-order differential equation, leading to a solution where the population grows exponentially over time.

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Solution

Let $P$ be the population of bacteria at any time $t$. According to the problem, the rate of growth $\frac{dP}{dt}$ is proportional to $P$.

$\begin{array}{rl}\frac{dP}{dt}& \propto P\\ \frac{dP}{dt}& =kP\end{array}$

To solve this, we use the method of separation of variables:

$\frac{dP}{P}=kdt$

Integrating both sides:

$\begin{array}{rl}\int \frac{dP}{P}& =k\int dt\\ \ln P& =kt+c\\ P& =e^{kt+c}\end{array}$

Finding the constant $c$

We are given that the initial population ($t=0$) is 20. Substituting $t=0$ and $P=20$ into equation (1):

$\begin{array}{rl}20& =e^{k(0)+c}\\ 20& =e^{0+c}\\ 20& =e^c\\ \ln 20& =c\end{array}$

Substituting $e^c=20$ back into equation (1):

$\begin{array}{rl}P& =e^{kt}\cdot e^c\\ P& =e^{kt}\cdot 20\\ P& =20e^{kt}\end{array}$

Finding the growth constant $k$

The problem states that the population becomes four times its initial size in two days. When $t=2$, $P=4\times 20=80$. Substituting these values into equation (2):

$\begin{array}{rl}80& =20e^{k(2)}\\ \frac{80}{20}& =e^{2k}\\ 4& =e^{2k}\\ \ln 4& =2k\\ k& =\frac{1}{2}\ln 4\\ k& =\ln (4^{1/2})\\ k& =\ln 2\end{array}$

Now, substitute $k=\ln 2$ back into equation (2) to get the general growth equation:

$\begin{array}{rl}P& =20e^{(\ln 2)t}\\ P& =20e^{t\ln 2}\end{array}$

Calculating the population after 10 days

To find the population when $t=10$, we substitute $t=10$ into equation (3):

$\begin{array}{rl}P& =20e^{10\ln 2}\\ & =20e^{\ln 2^{10}}\\ & =20(2^{10})\\ & =20(1024)\\ P& =20480\end{array}$

The population of bacteria after ten days will be 20,480.

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Key Formulas or Methods Used

• Differential Equation for Growth: $\frac{dP}{dt}=kP$
• Separation of Variables: Moving all $P$ terms to one side and $t$ terms to the other before integrating.
• Exponential Form: $P=P_0{e}^{kt}$
• Logarithmic Properties:
  • $\ln (a^b)=b\ln a$
  • $e^{\ln x}=x$

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Summary of Steps

1. Set up the differential equation based on the proportionality statement.
2. Solve the differential equation using integration to find the general expression for $P$.
3. Apply the initial condition ($t=0,P=20$) to find the integration constant.
4. Use the second condition ($t=2,P=80$) to solve for the growth rate constant $k$.
5. Substitute the target time ($t=10$) into the specific growth equation to find the final population.