A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 420.
(a) Find an expression for the number of bacteria after t hours.
$y(t)=y(0)e^{kt}=100e^{kt}$
$y(0)=100$
$y(1)=420$
$y(1)=\frac{100e^{k(1)}}{100}=\frac{420}{100}$
$\ln\ e^{k\cdot1}=\ln\ 4.2$
$k=ln\ 4.2$
Thus, $y(t)=100e^{(\ln\ 4.2)\cdot t}$
(b) Find the rate of growth after 3 hours
I am having a hard time understanding how to utilize the formula given:
The instructions are the following:
What is the significance of the proportionality constant K? In the
context of population growth, where P(t) is the size of a population
at time $t$, we write$\frac{dP}{dt}= kP$
or
$\frac{1}{p}\ \frac{dP}{dt} = k$
I don't understand how to utilize this information. To be quite honest, I'm not sure what the fraction $\frac{dP}{dt}$ really means. Can someone explain at my level?
Best Answer
$\frac{dP}{dt}$ is the “instantaneous” rate of change of the population. Formally, it’s the derivative of the population size function with respect to time. So, the equation $\frac{dP}{dt}=kP$ just says that the growth rate at any instant is proportional to the population at that time, which you already knew.
Now that you’ve computed $k$, you just plug it into the growth rate formula above, setting $P$ to the population size at $3$ hours, i.e., $\frac{dP}{dt}(3 hrs) = kP(3 hrs) = \ln 4.2 \cdot 100e^{3 \ln{4.2}} = {4.2}^3\cdot100\cdot\ln{4.2}$.