Question

During a research​ experiment, it was found that the number of bacteria in a culture grew at a rate proportional to its size. At 1010​:00 AM there were 4 comma 0004,000 bacteria present in the culture. At​ noon, the number of bacteria grew to 4 comma 6004,600. How many bacteria will there be at​ midnight?

Answer 1

Answer:

68600 will there be at​ midnight ( approx )

Step-by-step explanation:

Let P shows the population of the bacteria,

Since, the number of bacteria in a culture grew at a rate proportional to its size,

$\implies \frac{dP}{dt}\propto P$

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

Where, k is the constant of proportionality,

$\frac{dP}{P}=kdt$

$\int \frac{dP}{P}=\int kdt$

$ln P=kt + C_1$

$P=e^{kt+C_1}$

$P=e^{kt}.e^{C_1}=C e^{kt}$

Now, let the population of bacteria is estimated from 10:00 AM,

So, at t = 0, P = 4,000 ( given )

$4000 = Ce^{0}$

$\implies C=4000$

Now, at noon there are 4,600 bacterias,

That is, at t = 2, P = 4600

$4600=Ce^{2k}$

$4600 = 4000 e^{2k}$

$\implies e^{2k}=\frac{4600}{4000}=1.15$

$2k=ln(1.5)\implies k=\frac{ln(1.5)}{2}=0.202732554054\approx 0.203$

Hence, the equation that represents the population of bacteria after t hours,

$P=4000 e^{0.203t}$

Therefore, the population of the bacteria at midnight ( after 14 hours ),

$P=4000 e^{0.203\times 14}=4000 e^{2.842}= 68600.1252903\approx 68600$