Modeling Bacterial Growth with Differential Equations

[thereddevils]

Homework Statement

A cultured bacteria of a species multiplied at a rate that is directly proportional to the number of cultured bacteria in the culture. If x is the number of bacteria in the culture in time t seconds write down the differential equation that describes the growth of the bacteria. At the beginning of the experiment, there were 1000 bacteria of a certain species. It was known that the cultured bacteria multiply at a rate of 1.5 times per hour. Find the number of bacteria in the culture after 3 hours.

Homework Equations

The Attempt at a Solution

The differential equation is dx/dt=kx with solution x=1000e^(kt)

It was known that the cultured bacteria multiply at a rate of 1.5 times per hour. What did this line imply?

since in one hour it multiplied 1.5 times, then in 3 hours would be 4.5 times. Simply take 1000 multiplied by 4.5 to get the answer?

 

[Mentallic]

It was known that the cultured bacteria multiply at a rate of 1.5 times per hour. What did this line imply?

After 1 hour the bacteria has multiplied by 1.5 times. You need to use this information to find the constant of exponentiation k.

since in one hour it multiplied 1.5 times, then in 3 hours would be 4.5 times. Simply take 1000 multiplied by 4.5 to get the answer?

Definitely not. This is not a linear problem.
In the second hour, we already have 1.5x the original bacteria population and now we're going to take 1.5x that population, not the original.

 

[thereddevils]

After 1 hour the bacteria has multiplied by 1.5 times. You need to use this information to find the constant of exponentiation k.


Definitely not. This is not a linear problem.
In the second hour, we already have 1.5x the original bacteria population and now we're going to take 1.5x that population, not the original.

ok thanks,

dx/dt=1.5=kx

k=1.5/x

x=1000e^(1.5t/x)

Then when t=3, x=1000e^(4.5/x)

so i solve for x here?

 

[Mentallic]

You have x=1000e^{kt}

And we are given that at time t=0 there are 1000 bacteria, so x=1000 at t=0, which is true by the formula given. We are also given that at time t=1 (we are assuming t is in hours here to make things simple) that the bacteria is 1.5 times the original number, so x=1500 at t=1. Here if we plug this into the equation we can solve for k to find its value. Now we use this value of k to find the value of x at t=3.

 

[thereddevils]

You have x=1000e^{kt}

And we are given that at time t=0 there are 1000 bacteria, so x=1000 at t=0, which is true by the formula given. We are also given that at time t=1 (we are assuming t is in hours here to make things simple) that the bacteria is 1.5 times the original number, so x=1500 at t=1. Here if we plug this into the equation we can solve for k to find its value. Now we use this value of k to find the value of x at t=3.

Erm that got me confused a little again, if we can do like that why can't the problem be treated like a geometric progression?

1000 , 1000(1.5) , 1000(1.5)^2 ,... ??

 

[Mentallic]

It can, but it would be hard to figure out what happens in fractions of an hour that way.

 

[thereddevils]

It can, but it would be hard to figure out what happens in fractions of an hour that way.

true , thanks for your guidance.

 

[Mentallic]

No worries

Just as a note that has helped me in the past for these kinds of problems. Remember to use all the info given. It's usually needed to find some unknown constants which otherwise wouldn't give you the answer you're looking for.

This happens a lot in integration too for example when you're trying to find what the constant C must be in a physical question.

Say for this question you didn't use the extra info given, you'll try to find the population at time t=3, which gives x=1000e^{3k} but of course this isn't an answer we're looking for, which means you missed something.

Good luck!