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

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.

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
`(dP)/(dt)`, we get

`(dP)/(dt) = kP`.

Notice that this is a separable equation, so

`(dP)/(P) = kdt`.

Then, integrating in both sides of the equality:

`\int(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,

`(90)/(4) = e^(4k)`.

Taking logarithm

`\ln(90)/(4) = 4k`, so
`(1)/(4)\ln(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 = (\ln 75)/(0.7783) \approx 5.55`.

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