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,

which is: N(t) = N0 * e^(kt) where: N(t) is the number of bacteria at time t N0 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 N0, we need to 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: dN/dt = kN where 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:

(dN/N) = k*dt ln(N) = kt + C N = e^(kt+C) N = Ce^(kt)

Here, C is the constant of integration, which we can determine using the initial condition that there are 10000 bacteria present in 3 hours. Plugging in these values, we get: 10000 = Ce^(3k) Similarly,

we can use the condition that there are 40000 bacteria present at 5 hours: 40000 = Ce^(5k)

Now we can solve these two equations simultaneously to find the values of C and k: 10000 = Ce^(3k) 40000 = Ce^(5k)

Dividing the second equation by the first equation, we get: 4 = e^(2k) Taking the natural logarithm of both sides, we get: ln(4) = 2k

Solving for k, we get: k = ln(4) / 2 = 0.6931

Substituting this value of k into either of the two equations, we can solve for

C: 10000 = Ce^(3k) 10000 = Ce^(2.079) C = 10000 / e^(2.079) = 2593.39 (rounded to two decimal places)

Therefore, the initial number of bacteria is: N0 = Ce^(kt) = 2593.39 * e^(0.6931*0) = 2593.39 So there were approximately 2593 bacteria present in the beginning.