differential equations

Question

The population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time t. After 3 hours it is observed that 600 bacteria are present. After 10 hours 4000 bacteria are present. What was the initial number of bacteria? (Round your answer to the nearest integer.)

_______________ bacteria

Answer 1

The phrase about the culture growing at a rate that is proportional to the number of bacteria present allows to to come up with a solution to a differential equation. Let P represent the population of the bacteria at time t.

dP/dt = kP (the rate of change of the population is proportional to the population (kP) ).

So,

dP/P = k dt (separate the variables)

ln|P| = kt +c

P = e^kt+c = e^ce^kt = A e^kt (where e^c - a constant - has been replaced by A).

Now use the facts that when t = 3 , P = 600 and when t = 10, P = 4000 to solve for k and A.

4000 = Ae^10k

600 = A e^3k

Divide top equation by 2nd:

(20/3) = e^7k

^{ln(20/3) = 7k}

^{k = ln(20/3) / 7}

And A = 600/e^3k

To determine the number of bacteria present initially, find P(0).

Check it out here:

desmos.com/calculator/sbp7hjg2ay

After you have done a few problems like this you will go straight to:

P(t) = A e^kt