3.8 Q-22

Question Statement

In a culture, bacteria increase at a rate proportional to the number of bacteria present. If the initial number of bacteria is 200 and they double in 2 hours, find the number of bacteria present four hours later.

Background and Explanation

This is a growth problem involving exponential growth, where the rate of change of the population is proportional to the current population. The differential equation that describes this type of growth is:

\frac{dP}{dt} = K P

where $P$ is the number of bacteria, $t$ is time, and $K$ is the growth constant. The solution involves solving this differential equation and using the given conditions to find the constant $K$ , and ultimately the population at any time $t$ .

Solution

Set up the differential equation:

The rate of growth of the bacteria is proportional to the current population. The equation is:

\frac{dP}{dt} = K P \quad (K > 0)

Separate the variables:

Rearranging the equation to separate variables gives:

\frac{1}{P} , dP = K , dt

Integrate both sides:

Now, integrate both sides:

\int \frac{1}{P} , dP = \int K , dt

The integrals give us:

\ln P = Kt + C_1

Simplifying:

P = e^{Kt + C_1}

This can be rewritten as:

P = e^{Kt} \cdot e^{C_1} = C e^{Kt} \quad \text{(where $C = e^{C_1}$)}

Use the initial condition:

We know that when $t = 0$ , $P = 200$ . Substituting this into the equation:

200 = C e^{K(0)}

Since $e^{0} = 1$ , we get:

200 = C

So, $C = 200$ . Therefore, the equation for $P$ becomes:

P = 200 e^{Kt} \tag{1}

Use the condition at $t = 2$ :

We are told that the population doubles in 2 hours. So, when $t = 2$ , $P = 400$ . Substituting into the equation:

400 = 200 e^{K(2)}

Divide both sides by 200:

2 = e^{2K}

Take the natural logarithm of both sides:

\ln 2 = 2K

Solve for $K$ :

K = \frac{1}{2} \ln 2

Find the number of bacteria at $t = 4$ :

Now that we know $K$ , substitute it into the equation for $P$ :

P = 200 e^{Kt}

Substituting $K = \frac{1}{2} \ln 2$ and $t = 4$ :

P = 200 e^{\left(\frac{1}{2} \ln 2\right) 4}

Simplifying:

P = 200 e^{2 \ln 2}

Using the property of logarithms $e^{\ln a^b} = a^b$ :

P = 200 \cdot 2^2 = 200 \cdot 4 = 800

Therefore, the number of bacteria after 4 hours is 800.

Key Formulas or Methods Used

Exponential Growth Equation: $\frac{dP}{dt} = K P$
Separation of Variables: To separate $P$ and $t$ and integrate both sides.
Exponential Function Properties: $e^{\ln a^b} = a^b$ for simplifying the result.

Summary of Steps

Start with the differential equation $\frac{dP}{dt} = K P$ .
Separate the variables and integrate to get $P = C e^{Kt}$ .
Use the initial condition $P = 200$ when $t = 0$ to find $C = 200$ .
Use the condition that the population doubles in 2 hours to find $K = \frac{1}{2} \ln 2$ .
Substitute $K$ into the equation for $P$ and calculate the population at $t = 4$ , which is 800.