SOLUTION: 0 The population P of bacteria in an experiment grows according to the equation \frac{dP}{dt}=kP, where k is a constant and t is measured in hours. If the population of bacter

Algebra ->  Test -> SOLUTION: 0 The population P of bacteria in an experiment grows according to the equation \frac{dP}{dt}=kP, where k is a constant and t is measured in hours. If the population of bacter      Log On


   

Question 1143638: 0

The population P of bacteria in an experiment grows according to the equation \frac{dP}{dt}=kP, where k is a constant and t is measured in hours. If the population of bacteria doubles every 24 hours, what is the value of k?
I was given this problem and I'm not sure what to do with it. I know the formula for this kind of equation is ce^{kx}. But,how do you plug in the values given?

Answer by greenestamps(13216) About Me  (Show Source):
You can put this solution on YOUR website!


The exponential growth function for the population is

P%28t%29+=+P%280%29e%5E%28kt%29

The rate of growth is the derivative of the function:

dP%2Fdt+=+k%2Ap%280%29e%5E%28kt%29

which is equal to kP.

We are told that the population doubles in 24 hours -- i.e., P(24) = 2P(0).

2P%280%29+=+P%280%29e%5E%2824k%29
2+=+e%5E%2824k%29
ln%282%29+=+24k
k+=+ln%282%29%2F24

The constant k is ln(2)/24.

The growth function is

P%28t%29+=+P%280%29e%5E%28%28ln%282%29%2F24%29t%29

Given P(0) = 1000, this function gives us

P%280%29+=+P%280%29e%5E%28%28ln%282%29%2F24%290%29+=+P%280%29e%5E0+=+P%280%29+=+1000





... showing that the population doubles every 24 hours.

Note there is nothing special about the 24 hours doubling time given in this problem.

In general, if the doubling time is d (with time measured in hours, or days, or centuries, or milliseconds, or whatever), then the constant k is ln(2)/d.