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?
Best Answer
So you know $P=ce^{kt}$. The population doubles in $24$ hours, or $2P=ce^{k(t+24)}$. Now can you find $k$ by dividing the two equations?
As a side note, you might want to know how that formula was obtained.$$\frac{dP}{dt}=kP\implies\frac{dP}P=k~dt$$Now integrate both sides,$$\int_{P(t=0)}^{P(t=t)}\frac{dP}P=k\int_{t=0}^{t=t}dt$$giving you $P(t)=P(0)e^{kt}$.