Question

In a certain culture of bacteria the rate of increase is proportional to the no.of bacteria present at that instant it is found that there are 10000 bacteria present in 3 hours and 40000 bacteria at the 5 hours the number of bacteria present in the beginning is?

Answer 1

We can use the general formula for exponential growth to solve this problem:

The formula is:

\[ N(t) = N_0 \cdot e^{kt} \] where:

• \(N(t)\) is the number of bacteria at time \(t\),
• \(N_0\) is the initial number of bacteria,
• k is the constant of proportionality (the growth rate),
• e is the base of the natural logarithm, approximately equal to 2.71828...

To find the value of \(N_0\), we use the information given in the problem. We are told that the rate of increase is proportional to the number of bacteria present, which means that:

\[ \frac{dN}{dt} = kN \] where \(\frac{dN}{dt}\) is the rate of change of the number of bacteria with respect to time. We can solve this differential equation by separating the variables and integrating both sides:

 

\[ \frac{dN}{N} = k \, dt \] Integrating both sides: \[ \ln(N) = kt + C \] Exponentiating both sides gives: \[ N = C e^{kt} \] Here, \(C\) is the constant of integration, which we can determine using the initial condition that there are 10000 bacteria present at 3 hours. Plugging in these values: \[ 10000 = C e^{3k} \] Similarly, we can use the condition that there are 40000 bacteria present at 5 hours: \[ 40000 = C e^{5k} \] Now we can solve these two equations simultaneously to find the values of \(C\) and \(k\): \[ 10000 = C e^{3k} \] \[ 40000 = C e^{5k} \] Dividing the second equation by the first equation: \[ \frac{40000}{10000} = \frac{C e^{5k}}{C e^{3k}} \quad \Rightarrow \quad 4 = e^{2k} \] Taking the natural logarithm of both sides:

Answer 2

Given:
\(N(t) = N0 \cdot e^{kt}\) (exponential growth model)
\(N(3) = 10000\)
\(N(5) = 40000\)

Using \(N(3)\):
Substitute t=3 into the exponential growth model:
\(N(3) = N0 \cdot e^{3k} = 10000\)

Using \(N(5)\):
Substitute t=5 into the exponential growth model:
\(N(5) = N0 \cdot e^{5k} = 40000\)

Divide the equations:
Divide the equation for N(5) by the equation for N(3):
\(\frac{N(5)}{N(3)} = \frac{40000}{10000} = 4\)
\(e^{5k} / e^{3k} = e^{2k} = 4\)

Find k:
Take the natural logarithm on both sides:
\(2k = \ln(4)\)
\(k = \frac{\ln(4)}{2}\)
\(k = \frac{\ln(2^2)}{2}\)
\(k = \ln(2)\)

Find N0:
Substitute \(k = \ln(2)\) back into \(N(3)\):
\(N0 \cdot e^{3 \ln(2)} = 10000\)
\(N0 \cdot (e^{\ln(2)})^3 = 10000\)
\(N0 \cdot 2^3 = 10000\)
\(N0 \cdot 8 = 10000\)
\(N0 = \frac{10000}{8}\)
\(N0 = 1250\)

So, the answer is 1250.