Question

A bacteria culture starts with 400 bacteria and grows at a rate proportional to its size. After 4 hours, there are 9000 bacteria. A. Find an expression for the number of bacteria after t hours. 400e^((.7783)t) B. Find the number of bacteria after 5 hours. 19592 C. After how many hours will the population reach 30000

Answer 1

Answer:

A) The expression for the number of bacteria is $P(t) = 400e^{0.7783t}$.

B) After 5 hours there will be 19593 bacteria.

C) After 5.55 hours the population of bacteria will reach 30000.

Step-by-step explanation:

A) Here we have a problem with differential equations. Recall that we can interpret the rate of change of a magnitude as its derivative. So, as the rate change proportionally to the size of the population, we have

$P' = kP$

where $P$ stands for the population of bacteria.

Writing $P'$ as $\frac{dP}{dt}$, we get

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

Notice that this is a separable equation, so

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

Then, integrating in both sides of the equality:

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

We have,

$\ln P = kt+C$.

Now, taking exponential

$P(t) = Ce^{kt}$.

The next step is to find the value for the constant $C$. We do this using the initial condition $P(0)=400$. Recall that this is the initial population of bacteria. So,

$400 = P(0) = Ce^{k0}=C$.

Hence, the expression becomes

$P(t) = 400e^{kt}$.

Now, we find the value for $k$. We are going to use that $P(4)=9000$. Notice that

$9000 = 400e^{k4}$.

Then,

$\frac{90}{4} = e^{4k}$.

Taking logarithm

$\ln\frac{90}{4} = 4k$, so $\frac{1}{4}\ln\frac{90}{4} = k$.

So, $k=0.7783788273$, and approximating to the fourth decimal place we can take $k=0.7783$. Hence,

$P(t) = 400e^{0.7783t}$.

B) To find the number of bacteria after 5 hours, we only need to evaluate the expression we have obtained in the previous exercise:

$P(5) =400e^{0.7783*5} = 19593.723 \approx 19593$.  

C) In this case we want to do the reverse operation: we want to find the value of t such that

$30000 = 400e^{0.7783t}$.

This expression is equivalent to

$75 = e^{0.7783t}$.

Now, taking logarithm we have

$\ln 75 = 0.7783t$.

Finally,

$t = \frac{\ln 75}{0.7783} \approx 5.55$.

So, after 5.55 hours the population of bacteria will reach 30000.