Question

A bacteria culture starts with 660 bacteria and grows at a rate proportional to its size. After 6 hours there will be 3960 bacteria. (a) Express the population after t hours as a function of t. population: (function of t) (b) What will be the population after 3 hours?

Answer 1

Answer:

a) $P(t) = 660 e^{\frac{ln(6)}{6} t}$

b) $P(t=3) = 660 e^{\frac{ln(6)}{6}*3} = 1616.663 \approx 1617$

Step-by-step explanation:

For this case we have the following info:

$P(0) = 660$ represent the initial amount of bacteria

$P(6) = 3960$ represent the population of bacteria after 6 hours

Part a

For this case we use the proportional model given by:

$\frac{dP}{dt}= kP$

Where P represent the population at time t and k is a constant of proportionality.

We can rewrite the model like this:

$\frac{dP}{P} = k dt$

And if we integrate both sides we got:

$ln |P|= kt + C$

Now if we apply exponential we got:

$P = e^{kt +c}= e^{kt} e^C$

$P(t)= P_o e^{kt}$

For this case using the initial condition:

$660 = P_o e^{0k}, P_o = 660$

We have the following model:

$P(t) = 660 e^{kt}$

Using the second condition $P(6) = 3960$ we have this:

$3960 = 660e^{6k}$

We can divide both sides by 660 and we got:

$6= e^{6k}$

Now we can apply natural log on both sides and we got:

$ln (6)= 6k$

$k = \frac{ln(6)}{6}=0.2986265782$

So then the model would be given by:

$P(t) = 660 e^{\frac{ln(6)}{6} t}$

Part b

For this case we just need to replace t=3 into the model and we got:

$P(t=3) = 660 e^{\frac{ln(6)}{6}*3} = 1616.663 \approx 1617$